Every lesson, checkpoint, FRQ practice set and test of this unit, in course order. Answer keys are marked ✓ KEY. Feedback, model responses and rubrics are shown in full.
7 lessons
Outcome: State that a physical change alters a substance's properties but not its composition.
A student leaves an open dish holding 50 g of water on a windowsill.
A week later, the dish is empty.
The liquid water turned into water vapor in the room's air.
The water looks gone for good.
But every molecule of that vapor is still H₂O.
You've already seen that a formula names one substance: H₂O names water, in the dish or in the air.
So the vapor in the air and the water in the dish are the same substance.
Did the water stop being water? No.
Evaporating changed the water's state, its shape, and its place.
Evaporating did not change what the water is made of.
What a substance is made of is called its composition.
A change that alters a substance's properties but not its composition is called a physical change.
A physical change alters a substance's properties but not its composition — the evaporating water stayed H₂O throughout.
State that a physical change alters a substance's properties but not its composition.
Check — multiple choice
A bar of candle wax melts into a clear liquid. The liquid is still wax. Which statement correctly describes the melting?
Feedback if chosen: Correct. The wax's state changed — a property. What the wax is made of did not change, so the melting is a physical change.
Feedback if chosen: This option is wrong — you turned a state change into a new substance. The liquid is still wax; its composition did not change.
Feedback if chosen: This option is wrong — you swapped the two words. The melting changed a property, the state; the composition stayed the same.
Feedback if chosen: This option is wrong — you erased the change. The state is a property, and the state did change; the composition did not.
Check — multiple choice
A film of spilled gasoline dries off a driveway in minutes. A student says the gasoline stopped being gasoline when it disappeared. Is the student correct?
Feedback if chosen: This option is wrong — you treated evaporating as making a new substance. Every molecule of the vapor is still gasoline.
Feedback if chosen: This option is wrong — you tied composition to the liquid state. A substance keeps its composition in any state.
Feedback if chosen: Correct. The gasoline evaporated — a change of state. Its composition did not change.
Feedback if chosen: This option is wrong — you gave the right verdict for a wrong reason. Liquids do change state; the drying film is doing exactly that.
Check — multiple choice
Which of these changes is a physical change?
Feedback if chosen: This option is wrong — the green layer is a new substance on the copper. The roof's composition changed.
Feedback if chosen: This option is wrong — ash and smoke are new substances. The log's composition changed.
Feedback if chosen: This option is wrong — the sponge is a new substance made from the batter. The batter's composition changed.
Feedback if chosen: Correct. The butter's state changed and nothing else. Melted butter is still butter — same composition.
Outcome: Classify phase changes and the formation or separation of mixtures as physical changes.
A student stirs 10 g of sodium chloride into 200 mL of water.
The sodium chloride vanishes from view.
Then the student boils the water away.
At the bottom of the dish sits 10 g of white sodium chloride, unchanged.
The mixing seemed to destroy the sodium chloride.
The boiling gave the sodium chloride back, grain for grain.
Did the sodium chloride or the water ever stop being what they are? No.
Two whole families of change work this way.
You've already seen that a physical change alters a substance's properties but not its composition.
You've already seen mixtures: two or more substances together, each keeping its own composition.
The first family: every phase change — melting, freezing, boiling, condensing — changes a substance's state and nothing else.
So every phase change is a physical change.
The second family: making a mixture moves substances together without altering any of them.
Separating a mixture moves substances apart without altering any of them.
So making or separating a mixture is a physical change.
| Family | One example | What changes | What stays the same |
|---|---|---|---|
| Phase change | Ice melts into liquid water | The state | The composition of the water |
| Making or separating a mixture | Sodium chloride stirred into water | Where the substances are | The composition of each substance |
Classify every phase change, and every making or separating of a mixture, as a physical change — each substance's composition stays the same.
Classify phase changes and the formation or separation of mixtures as physical changes.
Check — multiple choice
A square of chocolate melts in a warm pocket. Which family of physical change is the melting?
Feedback if chosen: Correct. Melting changes the chocolate's state and nothing else — a phase change.
Feedback if chosen: This option is wrong — nothing mixed. The chocolate's state changed, so the melting is a phase change.
Feedback if chosen: This option is wrong — nothing separated. The chocolate's state changed, so the melting is a phase change.
Feedback if chosen: This option is wrong — the melted chocolate is still chocolate. A change that leaves the composition alone is a physical change.
Check — multiple choice
A camper stirs instant coffee powder into hot water until every granule of the powder vanishes from view. Which family of physical change is the stirring?
Feedback if chosen: This option is wrong — the powder did not change state; it spread through the water. The stirring formed a mixture.
Feedback if chosen: Correct. The stirring moved powder and water together. Each substance keeps its own composition — a mixture formed.
Feedback if chosen: This option is wrong — the substances moved together, not apart. The stirring formed a mixture.
Feedback if chosen: This option is wrong — the powder's substances are still there in the water. Making a mixture leaves every composition alone, so the stirring is physical.
Check — multiple choice
A filter catches sand out of muddy water. Which family of physical change is the filtering?
Feedback if chosen: This option is wrong — no substance changed state. The filter moved sand and water apart: a separation.
Feedback if chosen: This option is wrong — the mud was already a mixture. The filter moved its substances apart: a separation.
Feedback if chosen: Correct. The filter moved sand and water apart. Each substance keeps its own composition — a mixture separated.
Feedback if chosen: This option is wrong — the sand is still sand and the water is still water. Separating a mixture is a physical change.
Check — multiple choice
A cook pours pulpy orange juice through a fine sieve, then leaves the strained juice in a warm oven until all the water in it has evaporated. Which family or families of physical change happen here?
Feedback if chosen: This option is wrong — you counted the evaporating but not the parting. The sieve pulls the pulp out of the juice: a separation.
Feedback if chosen: This option is wrong — you counted the parting but not the evaporating. The water changes state: liquid to gas.
Feedback if chosen: Correct. The water changes state (a phase change), and the sieving and the evaporating each pull the juice mixture apart (separations).
Feedback if chosen: This option is wrong — both families are here: the water changes state, and the juice mixture is pulled apart.
Outcome: State that a chemical change transforms substances into new substances, typically with different compositions.
A teacher passes an electric current through a dish of water.
Gas bubbles stream up both wires.
After ten minutes, 20 mL of gas sits above one wire and 10 mL sits above the other.
Near a flame, the larger sample pops with a squeak — that gas is hydrogen.
The smaller sample relights a glowing splint — that gas is oxygen.
Neither gas behaves like water.
Where did the water go?
The current turned the water into two new substances.
The larger sample is hydrogen gas, H₂.
The smaller sample is oxygen gas, O₂.
Neither formula is H₂O, so neither gas is water.
A change that transforms substances into new substances is called a chemical change.
A physical change leaves every substance's composition alone.
A chemical change replaces the substances themselves.
A chemical change transforms substances into new substances, typically with different compositions — the current turned H₂O into the new substances H₂ and O₂.
State that a chemical change transforms substances into new substances, typically with different compositions.
Check — multiple choice
An egg white turns from a clear liquid to a white solid as it cooks, and no amount of cooling brings the clear liquid back. What kind of change is the cooking?
Feedback if chosen: Correct. The white solid is a new substance — cooling cannot undo the change, because the clear liquid no longer exists.
Feedback if chosen: This option is wrong — you read liquid-to-solid as freezing. A frozen substance melts back; the cooked egg white never turns back, because it is a new substance.
Feedback if chosen: This option is wrong — nothing mixed into the egg white. The egg white itself became a new substance.
Feedback if chosen: This option is wrong — nothing was pulled apart. The egg white itself became a new substance.
Check — multiple choice
A camping fuel tablet burns until nothing visible is left. A student says the tablet's substance simply disappeared. Is the student correct?
Feedback if chosen: This option is wrong — a chemical change replaces substances with new substances; nothing simply disappears.
Feedback if chosen: This option is wrong — heat is not a substance. The burning made new substances: gases the eye cannot see.
Feedback if chosen: Correct. Burning transformed the tablet into new gaseous substances. The eye loses them, but they exist in the air.
Feedback if chosen: This option is wrong — you made the change physical. The tablet burned: its substances became new substances.
Check — multiple choice
Which of these changes to a spoonful of sugar is a chemical change?
Feedback if chosen: This option is wrong — stirred-in sugar forms a mixture. The sugar is still sugar in the water.
Feedback if chosen: Correct. Black, bitter carbon is a new substance with a different composition — the sugar is gone.
Feedback if chosen: This option is wrong — ground sugar is still sugar. Only the shape of the pieces changed.
Feedback if chosen: This option is wrong — melted sugar is still sugar. Only the state changed: a phase change.
Outcome: Identify heat or light, gas formation, precipitate formation, and color change as the four types of observation that give possible evidence of a chemical change.
A student pours 20 mL of one clear, colorless solution into 20 mL of another.
The instant the liquids touch, a bright yellow solid appears and sinks through the mixture.
Neither starting solution contained any solid.
Neither starting solution was yellow.
The student cannot see molecules.
So the student cannot see directly whether a new substance formed.
Which observations should count as evidence that one did?
Four kinds of observation carry the evidence.
**1.** The mixture can release heat or light.
**2.** A gas can form.
**3.** A solid that does not dissolve can appear when two solutions mix — chemists call that solid a precipitate.
**4.** A color can change.
| Type | One example |
|---|---|
| Heat or light released | A hand warmer's pouch grows hot |
| Gas formation | Bubbles stream from a tablet in water |
| Precipitate formation | A yellow solid sinks when two clear solutions mix |
| Color change | A clear mixture turns deep orange |
The yellow solid in the beaker is a precipitate — the third type.
Each of the four is possible evidence that a new substance formed.
When you watch a change, your first move is to name the type of observation you saw.
Heat or light, a gas forming, a precipitate forming, and a color change are the four types of observation that give possible evidence of a chemical change.
Identify heat or light, gas formation, precipitate formation (a solid that does not dissolve, appearing when solutions mix), and color change as the four types of observation giving possible evidence of a chemical change.
Check — multiple choice
A student drips vinegar onto baking soda. The mixture froths as bubbles stream out. Which type of evidence is the frothing?
Feedback if chosen: Correct. The bubbles are a gas forming in the mixture — possible evidence of a chemical change.
Feedback if chosen: This option is wrong — a precipitate is a solid that sinks, not bubbles that rise.
Feedback if chosen: This option is wrong — no color changed. The observation is bubbles: a gas forming.
Feedback if chosen: This option is wrong — nothing glowed or grew hot. The observation is bubbles: a gas forming.
Check — multiple choice
Two clear solutions are mixed, and a fine white solid settles to the bottom of the flask. Which type of evidence is the solid?
Feedback if chosen: This option is wrong — the solid settles down through the liquid; a gas would bubble up and out.
Feedback if chosen: This option is wrong — white solid appearing is a new solid, not a change in the liquid's color.
Feedback if chosen: This option is wrong — nothing glowed or grew hot. A solid appearing from two solutions is a precipitate.
Feedback if chosen: Correct. A solid that appears when two solutions mix, and does not dissolve, is a precipitate.
Check — multiple choice
A slice of apple left in the air slowly turns brown. Which type of evidence is the browning?
Feedback if chosen: This option is wrong — no bubbles formed. The apple's surface changed color.
Feedback if chosen: This option is wrong — no solid appeared from mixed solutions. The apple's surface changed color.
Feedback if chosen: Correct. The surface went from pale to brown — a color change, possible evidence of a chemical change.
Feedback if chosen: This option is wrong — the apple released no heat or light you could observe. The color changed.
Check — multiple choice
A camper snaps a glow stick, and the liquid inside shines green for hours. Which type of evidence is the glow?
Feedback if chosen: This option is wrong — no bubbles formed. The stick releases light.
Feedback if chosen: Correct. The stick releases light — one of the four observation types, possible evidence of a chemical change.
Feedback if chosen: This option is wrong — no solid appeared. The stick releases light.
Feedback if chosen: This option is wrong — the glow is released light, not a substance changing color. Light released is its own type.
Outcome: Explain why heat or light, a new gas, a new solid, or a new color can signal a chemical change: new substances bring different properties.
The yellow solid is still sitting in its beaker — and it is still a puzzle.
Both starting solutions were colorless.
The student added nothing else.
Yet a bright yellow solid now rests at the bottom.
A color cannot come from nowhere.
Which substance in the beaker carries that yellow?
You've already seen that a chemical change transforms substances into new substances.
A new substance brings its own properties.
The yellow belongs to a substance that did not exist before the solutions mixed.
The same logic covers all four observation types.
A new solid can appear because a newly formed substance does not dissolve in the mixture.
A new gas can appear because a newly formed substance is a gas at that temperature.
A new color can appear because a newly formed substance absorbs different light than the starting substances.
Released heat or light can appear because forming a new substance can release energy.
Read each observation as a property. Then ask which substance owns that property.
Those observations can signal a chemical change because a chemical change produces new substances, and new substances bring different properties — released energy, a new gas, a new solid, or a new color.
Explain why heat or light, gas formation, precipitate formation, and color change can signal a chemical change: new substances bring different properties.
Check — multiple choice
Two colorless solutions are mixed, and the mixture turns deep orange. Which explanation of the orange color would mean a chemical change happened?
Feedback if chosen: Correct. A new substance brings its own properties. A substance that absorbs different light than its colorless parents is a new substance — a chemical change.
Feedback if chosen: This option is wrong — both starting solutions are colorless, and blending two colorless liquids gives a colorless mixture. Blending is physical anyway.
Feedback if chosen: This option is wrong — mixing spreads substances out; it cannot concentrate them. Concentrating a substance is not a chemical change anyway.
Feedback if chosen: This option is wrong — warming changes a property, not a composition. A warm liquid is still the same substance.
Check — multiple choice
A tablet dropped into water sends up a steady stream of bubbles. Which new-substance property would explain the bubbles?
Feedback if chosen: This option is wrong — absorbing different light explains a color change, not bubbles.
Feedback if chosen: This option is wrong — released energy explains heat or light, not bubbles.
Feedback if chosen: Correct. Bubbles form when a newly made substance is a gas at that temperature — the gas escapes upward.
Feedback if chosen: This option is wrong — a substance that does not dissolve explains a sinking solid, not rising bubbles.
Check — multiple choice
The powder in a hand warmer changes inside its pouch, and the pouch grows hot for hours. Which explanation of the heat would mean a chemical change happened?
Feedback if chosen: This option is wrong — pressing grains together is physical. No new substance forms, so no chemical change follows.
Feedback if chosen: This option is wrong — releasing stored warmth is physical, and the pouch grows hotter than the room around it.
Feedback if chosen: This option is wrong — dissolving is physical. The powder's substances would all still be there, just spread through the moisture.
Feedback if chosen: Correct. A chemical change makes new substances, and forming a new substance can release energy — the heat you feel.
Outcome: Explain why each type of observation is possible rather than conclusive evidence of a chemical change.
A pan of water heats on a stove.
At 100 °C the water bubbles hard, and gas streams up through the liquid.
You've already seen that a forming gas counts as evidence of a chemical change.
But every bubble here holds plain water vapor — still H₂O, only hotter.
The rule just called a physical change chemical.
Can the four observations be trusted at all?
Here is the catch: each of the four observations can also accompany a purely physical change.
Boiling makes bubbles, and no new substance forms.
Condensing releases heat, and no new substance forms.
Mixing yellow paint into blue paint changes the color, and no new substance forms.
A warming glass of fizzy water releases a stream of gas, and no new substance forms — the dissolved gas leaves the mixture.
| Observation | A chemical case | A physical look-alike |
|---|---|---|
| Gas forms | A tablet fizzes in water | Water boils into vapor |
| Heat released | A hand warmer's powder changes | Steam condenses on a window |
| Color change | An apple slice browns in air | Yellow paint mixes into blue paint |
| Solid appears | A yellow solid forms from two solutions | Ice forms in freezing water |
So each observation is possible evidence, never proof.
The honest verdict weighs the observation against both explanations.
Ask: did a new substance form, or does a physical change explain the same sight?
Each observation is possible rather than conclusive evidence of a chemical change, because the same signs can accompany purely physical changes.
Explain why each type of observation is possible rather than conclusive evidence of a chemical change.
Check — multiple choice
A student sees bubbles rising through a liquid and says bubbles always mean a chemical change. Is the student correct?
Feedback if chosen: This option is wrong — a boiling liquid bubbles with its own vapor. No new substance is needed.
Feedback if chosen: This option is wrong — any liquid can boil. Bubbles from any boiling liquid are that same substance as a gas.
Feedback if chosen: This option is wrong — you over-corrected. Bubbles are possible evidence; they are just not proof.
Feedback if chosen: Correct. Boiling is a physical change that makes bubbles, so a gas is possible evidence, never proof.
Check — multiple choice
Water vapor from the air turns to liquid droplets on the outside of a cold drink can, and the can warms slightly. What does the released heat show?
Feedback if chosen: This option is wrong — you treated heat as proof. Condensing is a phase change, and it releases heat with no new substance.
Feedback if chosen: This option is wrong — the can only warmed. Warming changes a property, not the metal's composition.
Feedback if chosen: Correct. The vapor condensed — a phase change — and released heat. Every molecule stayed H₂O.
Feedback if chosen: This option is wrong — this very case is a physical change releasing heat. That is why heat is possible evidence, never proof.
Check — multiple choice
Why is a color change possible evidence rather than proof of a chemical change?
Feedback if chosen: This option is wrong — color changes are easy to observe. The problem is that a physical change can produce one too.
Feedback if chosen: Correct. Mixing paints changes the color, and mixing is physical — so a color change alone cannot prove a chemical change.
Feedback if chosen: This option is wrong — a substance's color is its own property. The real reason color is not proof: a physical mixing can change color too.
Feedback if chosen: This option is wrong — no count of observation types adds up to proof. Each type, alone or combined, can come from a physical change.
Outcome: Judge whether a described change is chemical or physical from the observed evidence, naming the type of evidence relied on.
A shiny iron nail lies outdoors for a month.
The nail turns brown and flaky.
The nail's mass rises from 5.0 g to 5.3 g, though nobody added anything to it.
In the same month, a puddle beside the nail shrinks and vanishes.
One of these two changes made a new substance. The other change did not.
Which is which — and which observation settles each case?
Work every verdict the same way.
**1.** Name the type of evidence you observed: heat or light, a gas, a precipitate, or a color change.
**2.** Ask which explanation covers the observation better — a new substance, or the same substance in a new form.
The nail first: the brown, flaky surface is a color change.
The extra 0.3 g means something from the air joined the iron.
A new substance covers both observations: the flaky brown layer is built from the iron and the air's oxygen. The nail's change is chemical.
The puddle next: no color change, no new solid, no released heat.
The same water in a new state covers everything observed: the puddle evaporated. The puddle's change is physical.
Judge whether a described change is chemical or physical from the observed evidence, and name the type of evidence you relied on.
Judge whether a described change is chemical or physical from the observed evidence, naming the type of evidence relied on.
Check — multiple choice
A pale strip of copper is heated in a flame. The strip turns black, and the black layer stays after the strip cools. Is the strip's change chemical or physical?
Feedback if chosen: Correct. The black layer stays after cooling — a new substance formed on the copper.
Feedback if chosen: This option is wrong — a heated-only strip would return to pale copper on cooling. The black layer stayed: a new substance.
Feedback if chosen: This option is wrong — nothing here changed state or mixed. The one change observed made a new substance.
Feedback if chosen: This option is wrong — the strip's surface visibly changed from pale to black, and the layer stayed.
Check — multiple choice
In the copper strip case, which type of evidence does the verdict rely on?
Feedback if chosen: This option is wrong — no bubbles or escaping gas were described.
Feedback if chosen: This option is wrong — no solid appeared from mixed solutions; the change happened on the strip's own surface.
Feedback if chosen: Correct. The observation was pale copper turning black — a color change that outlasted the heating.
Feedback if chosen: This option is wrong — the flame supplied the heat; the strip was not observed releasing heat or light.
Check — multiple choice
A bottle of perfume is left open. In a week, half the liquid is gone, and the room smells of perfume. Is the perfume's change chemical or physical?
Feedback if chosen: This option is wrong — no observation needs a new substance. The perfume evaporated and mixed with the air.
Feedback if chosen: Correct. The liquid evaporated and mixed into the room's air — the same substance in a new form.
Feedback if chosen: This option is wrong — everything observed is covered by evaporation and mixing alone; no new substance is needed.
Feedback if chosen: This option is wrong — half the liquid left the bottle. A change happened: a physical one.
Check — multiple choice
In the perfume case, which reasoning settles the verdict?
Feedback if chosen: Correct. When the same substance in a new form covers every observation, the verdict is physical.
Feedback if chosen: This option is wrong — a falling level shows the liquid left, not that a new substance formed. Evaporation covers it.
Feedback if chosen: This option is wrong — smell shows perfume vapor reached your nose. The vapor is still perfume: a mixture with air.
Feedback if chosen: This option is wrong — a liquid turning to gas is a phase change; the substance stays itself.
5 lessons
Outcome: Classify a process that breaks and/or forms chemical bonds as a chemical change.
A camping stove burns methane gas.
In ten minutes, the flame heats 500 g of water from 20 °C to boiling.
The flame releases heat and light.
You've already seen that heat and light are possible evidence of a chemical change.
But that evidence describes only the outside of the flame.
Inside the flame sit methane molecules, CH₄, and oxygen molecules, O₂.
What happens to those molecules while the methane burns?
You've already seen that a covalent bond is one or more pairs of electrons shared between two atoms.
In the flame, the C–H bonds inside each methane molecule break.
The O=O bonds inside the oxygen molecules break.
New C=O bonds form, and new O–H bonds form.
The same atoms, joined by new bonds, now make carbon dioxide, CO₂, and water, H₂O.
Seen from the inside, that is what makes the burning chemical: bonds broke, and new bonds formed.
You've already seen that the four observations are possible evidence, never proof.
The bonds settle the verdict: any process that breaks or forms chemical bonds is a chemical change — burning methane breaks C–H and O=O bonds and forms C=O and O–H bonds.
Classify a process that breaks and/or forms chemical bonds as a chemical change.
Check — multiple choice
Inside a lightning spark, N≡N bonds in N₂ molecules and O=O bonds in O₂ molecules break, and new N–O bonds form. Classify the change.
Feedback if chosen: Correct. Bonds broke and new bonds formed — that is what makes a process a chemical change.
Feedback if chosen: This option is wrong — a physical change leaves every bond intact. Here N≡N and O=O bonds broke.
Feedback if chosen: This option is wrong — one process happened, and it broke and formed bonds: chemical.
Feedback if chosen: This option is wrong — breaking and forming bonds is exactly what a chemical change does.
Check — multiple choice
Which fact about a process guarantees that the process is a chemical change?
Feedback if chosen: This option is wrong — condensing steam releases heat with every bond intact. Heat is possible evidence, never proof.
Feedback if chosen: Correct. The four visible signs are possible evidence; broken or formed bonds settle the verdict.
Feedback if chosen: This option is wrong — boiling water bubbles with every bond intact. A gas is possible evidence, never proof.
Feedback if chosen: This option is wrong — mixed paints change color with every bond intact. A color change is possible evidence, never proof.
Check — multiple choice
A student stretches a rubber band and lets it snap back. Can the stretch be classified as a chemical change?
Feedback if chosen: This option is wrong — shape is a property. A change of shape with every bond intact is physical.
Feedback if chosen: This option is wrong — stored energy alone makes nothing chemical. With every bond intact, the stretch is physical.
Feedback if chosen: Correct. Stretching only straightens the rubber's long molecules; no bonds break or form. A chemical change needs bonds to break or form, so the stretch is not chemical.
Feedback if chosen: This option is wrong — you reached the right verdict for a wrong reason. Rubber is full of covalent bonds; they simply stayed intact.
Outcome: Classify a process that changes only the attractions between intact molecules, such as a phase change, as a physical change.
A kettle boils 1 L of water at 100 °C.
Steam bursts from the spout.
You've already seen that a chemical change breaks bonds.
Boiling looks violent enough to break anything.
Does the boiling break the O–H bonds inside each water molecule?
Two kinds of attraction act in liquid water.
Strong covalent bonds join the O atom and the H atoms inside each molecule.
You've already seen the weaker kind: intermolecular forces, the attractions that pull neighboring molecules toward each other.
Here are the two, drawn side by side: solid lines for bonds, dashed lines for attractions between molecules.
Boiling gives each molecule enough energy to escape its neighbors.
Every O–H bond stays intact.
The steam is made of the same whole H₂O molecules, just far apart.
So boiling changes only the attractions between intact molecules — and that is why a phase change is physical.
Any process that changes only the attractions between intact molecules, such as a phase change, is a physical change — boiling separates whole H₂O molecules while every O–H bond stays intact.
Classify a process that changes only the attractions between intact molecules, such as a phase change, as a physical change.
Check — multiple choice
A spoonful of solid coconut oil melts in a student's warm hand. The student says the melting shows the bonds inside the oil's molecules breaking. Is the student correct?
Feedback if chosen: This option is wrong — the melted oil is the same substance. The molecules escaped their fixed positions with every bond intact.
Feedback if chosen: This option is wrong — the hand's warmth overcomes the attractions between molecules; the bonds inside them need far more energy.
Feedback if chosen: Correct. Melting overcomes the attractions between molecules. The bonds inside each molecule survive.
Feedback if chosen: This option is wrong — the molecules are full of bonds. Those bonds simply stay intact while the molecules separate from their fixed positions.
Check — multiple choice
Solid iodine warms gently and turns straight into a purple vapor of I₂ molecules. Each I–I bond stays intact. Classify the change.
Feedback if chosen: This option is wrong — no bond broke or formed. Only the attractions between the I₂ molecules changed.
Feedback if chosen: This option is wrong — nothing chemical happened: every I–I bond stayed intact.
Feedback if chosen: This option is wrong — a change did happen: the solid became a gas.
Feedback if chosen: Correct. The I₂ molecules separated from one another with every bond intact — a physical change.
Check — multiple choice
Which of these processes is a physical change?
Feedback if chosen: This option is wrong — breaking bonds makes a process chemical.
Feedback if chosen: Correct. The molecules stay intact, and only the attractions between them change — physical.
Feedback if chosen: This option is wrong — forming new bonds makes a process chemical.
Feedback if chosen: This option is wrong — breaking bonds makes a process chemical.
Outcome: Explain why dissolving an ionic solid in water can be argued to be either physical or chemical.
A student stirs 5 g of sodium chloride, NaCl, into 250 mL of water.
The sodium chloride vanishes from view.
You've already seen the ionic lattice: a repeating grid of positive and negative ions, held together by ionic bonds.
Dissolving pulls that lattice of ions apart.
Ionic bonds break.
So the bond rule answers: chemical change.
But you've already seen dissolving classified as a physical change — the making of a mixture.
Which answer is right?
The honest answer: chemists argue the case both ways.
The chemical argument first: dissolving breaks the ionic bonds of the lattice.
You've already seen the ion-dipole attraction: an ion pulls the charged ends of nearby water molecules toward itself.
New attractions then form: each Na⁺ ion and each Cl⁻ ion pulls water molecules around itself — ion-dipole attractions.
Bonds broke and new attractions formed — that reads like a chemical change.
The physical argument next: no new substance forms.
Boil the water away, and the same 5 g of sodium chloride returns.
No new substance — that reads like a physical change.
AP asks you to explain both arguments, not to pick a winner.
Dissolving an ionic solid in water can be argued either way: ionic bonds in the lattice break, while new ion-dipole attractions form between the ions and the water molecules — and yet no new substance forms.
Explain why dissolving an ionic solid in water can be argued to be either a physical or a chemical change, giving both arguments.
Check — drawing task (pen and paper, self-marked)
A student stirs potassium chloride, KCl, into water, and the KCl dissolves. Take a pen and paper and write the two arguments — one for calling the dissolving chemical, one for calling it physical. Then sketch the before and after: a fragment of the KCl lattice, then a K⁺ ion and a Cl⁻ ion each surrounded by water molecules. Draw bonds as solid lines and attractions as dashed lines. When you are done, select Check to compare your answer with the model answer.
Model answer shown to the student
Self-mark checklist — the drawing should show:
Check — multiple choice
When magnesium chloride, MgCl₂, dissolves in water, which bonds or attractions break, and which form?
Feedback if chosen: This option is wrong — the water molecules stay intact. The lattice's ionic bonds break.
Feedback if chosen: This option is wrong — you reversed the order of events. The lattice breaks apart first; the ion-dipole attractions then form.
Feedback if chosen: Correct. The lattice's ionic bonds break, and each freed ion pulls water molecules around itself — ion-dipole attractions form.
Feedback if chosen: This option is wrong — dissolving an ionic solid pulls its lattice apart, so ionic bonds do break.
Check — multiple choice
Why can chemists argue that dissolving an ionic solid is physical, even though ionic bonds break?
Feedback if chosen: This option is wrong — ionic bonds are strong. The physical argument rests on no new substance forming.
Feedback if chosen: This option is wrong — the water molecules stay intact, and nothing cancels. The argument rests on no new substance forming.
Feedback if chosen: This option is wrong — an ion-dipole attraction acts between an ion and a whole molecule; no electrons are shared.
Feedback if chosen: Correct. Nothing new exists in the beaker: the same ions, the same water. Boiling the water away returns the same solid.
Outcome: Explain the visible characteristics of a chemical change in terms of the bonds broken and formed.
A teacher lights a 5 cm strip of magnesium ribbon.
The ribbon burns with a white light too bright to watch.
Heat pours off the flame.
The ribbon contains only magnesium atoms.
The air supplies only oxygen molecules.
Where does all that energy come from?
The visible signs come from invisible events: bonds break, and bonds form.
Burning magnesium breaks the O=O bonds in the oxygen molecules.
New, stronger bonds then form between magnesium and oxygen in the new substance, magnesium oxide, MgO.
Forming stronger bonds releases energy.
That released energy leaves the flame as the heat and light you observe.
Read every visible sign this way: the bonds broken and the bonds formed explain what the eye sees.
What you observe on the outside of a chemical change is explained by the bonds broken and formed inside it — burning magnesium releases heat and light as new, stronger bonds form in MgO.
Explain the visible characteristics of a chemical change in terms of the chemical bonds broken and formed.
Check — multiple choice
When hydrogen burns, the flame releases heat. Where does the released energy come from?
Feedback if chosen: This option is wrong — breaking a bond costs energy. The release comes from the new bonds forming.
Feedback if chosen: Correct. Forming the new, stronger O–H bonds in water releases energy, and that energy leaves as heat.
Feedback if chosen: This option is wrong — rubbing is not a chemical event. The energy comes from new bonds forming.
Feedback if chosen: This option is wrong — the oxygen is not cooling; the oxygen is bonding. Forming the new bonds releases the energy.
Check — multiple choice
A student says a flame's heat comes from bonds breaking apart. Is the student correct?
Feedback if chosen: This option is wrong — pulling bonded atoms apart costs energy, the way pulling magnets apart does.
Feedback if chosen: This option is wrong — breaking any bond costs energy, double bonds included.
Feedback if chosen: This option is wrong — a flame plainly releases heat and light. The source is the new bonds forming.
Feedback if chosen: Correct. Breaking bonds costs energy; forming the new, stronger bonds releases more than the breaking cost.
Check — multiple choice
A pad of damp steel wool rusts in a jar: its iron slowly joins with oxygen from the air, and the jar stays warm for hours. Why does the rusting release heat?
Feedback if chosen: Correct. Iron and oxygen atoms form new, stronger bonds, and forming bonds releases energy — felt as hours of warmth.
Feedback if chosen: This option is wrong — breaking bonds costs energy. The release comes from the new iron-oxygen bonds forming.
Feedback if chosen: This option is wrong — stored factory warmth would fade in minutes. New bonds forming release the heat for hours.
Feedback if chosen: This option is wrong — no squeezing happens. The heat comes from new bonds forming between iron and oxygen.
Outcome: Explain the visible characteristics of a physical change in terms of the attractions between particles that are overcome or formed.
A pan of water boils on full power.
The stove adds heat to the water every second.
Yet the temperature stays stuck at 100 °C, minute after minute.
You've already seen that released energy comes from new, stronger bonds forming.
Boiling forms no new bonds.
So where does the stove's energy go?
Attractions still cost energy to overcome.
Every escaping molecule must pull away from the neighbors that attract it.
The stove's heat supplies that energy, molecule by molecule.
The mirror case releases energy: when steam condenses, the attractions between the molecules form again.
Forming attractions releases energy.
So a physical change shows visible energy signs too.
| A chemical change | A physical change | |
|---|---|---|
| What breaks | Bonds inside the particles | Attractions between the particles |
| What forms | New bonds inside new substances | Attractions between the same particles |
| Energy released when | New, stronger bonds form | Attractions between particles form |
| Energy needed when | Bonds are broken | Attractions are overcome |
| One example | Magnesium burns: heat and light | Steam condenses: heat released |
Read those signs as attractions overcome or formed — never as bonds broken.
The visible characteristics of a physical change are explained by the attractions between particles that are overcome or formed — boiling water needs a continuous energy input to overcome the attractions between H₂O molecules.
Explain the visible characteristics of a physical change in terms of the attractions between particles that are overcome or formed.
Check — multiple choice
Sweat evaporates from your skin, and your skin cools. Why does the evaporating sweat take energy from the skin?
Feedback if chosen: This option is wrong — evaporation leaves every O–H bond intact. The energy overcomes attractions between whole molecules.
Feedback if chosen: Correct. Escaping molecules must overcome the attractions to their neighbors, and the energy comes from your skin — so the skin cools.
Feedback if chosen: This option is wrong — the molecules survive whole; they only separate from their neighbors.
Feedback if chosen: This option is wrong — the vapor is still water. Evaporation is a physical change: attractions overcome, nothing new formed.
Check — multiple choice
On a cold evening, water vapor in the air turns into tiny fog droplets. Does the condensing absorb energy or release energy?
Feedback if chosen: This option is wrong — that is evaporating's account, reversed. Attractions form here, and forming attractions releases energy.
Feedback if chosen: This option is wrong — the molecules lock back into contact with their neighbors, and forming those attractions releases energy.
Feedback if chosen: Correct. Condensing lets the attractions between molecules form again, and forming attractions releases energy.
Feedback if chosen: This option is wrong — physical changes move energy too. Forming attractions between intact molecules releases energy.
Check — multiple choice
A pool of liquid wax cools and hardens, releasing heat as it does. No bonds break or form. Which account explains the released heat?
Feedback if chosen: This option is wrong — the stem says no bonds form. The molecules stay intact and only lock together.
Feedback if chosen: This option is wrong — the stem says no bonds break. The molecules stay whole.
Feedback if chosen: This option is wrong — heat does not sit stored inside the liquid wax. The melting's energy went into pulling molecules apart; forming those attractions again releases it.
Feedback if chosen: Correct. As the molecules lock into the solid, attractions between them form — and forming attractions releases energy.
Check — multiple choice
Four changes are observed. Which one is a physical change?
Feedback if chosen: This option is wrong — you classified browning as physical. The browned crust is a new substance with a new color, so the dough's composition has changed.
Feedback if chosen: Correct. Frost is water vapor becoming solid water — a phase change. The water's composition never changes, so the change is physical.
Feedback if chosen: This option is wrong — you treated the color change as only a look. Sunlight turns the paper's compounds into new, yellow substances.
Feedback if chosen: This option is wrong — you missed the heat and light. A firework's contents transform into new substances, releasing that energy.
Check — multiple choice
A strip of metal is dropped into a clear liquid. The mixture fizzes, and the test tube becomes noticeably warmer. Which two types of evidence of a possible chemical change are observed?
Feedback if chosen: Correct. The fizzing is a gas forming, and the warming is heat being produced — two of the four types of evidence.
Feedback if chosen: This option is wrong — you counted a color change, but nothing in the observation changed color. The warming is production of heat.
Feedback if chosen: This option is wrong — you called the fizzing a precipitate. A precipitate is a new solid; fizzing is a gas forming.
Feedback if chosen: This option is wrong — neither observation fits. The fizzing is a gas forming, and the warming is production of heat.
Check — multiple choice
A student rapidly bends a paper clip back and forth, and the bent spot becomes hot. The student claims the heat proves a chemical change happened in the metal. Why is the heat not conclusive evidence?
Feedback if chosen: This option is wrong — you ruled heat out entirely. Production of heat is one of the four types of evidence; it just cannot prove a chemical change on its own.
Feedback if chosen: This option is wrong — you required a second sign. One type of evidence can stand alone; the problem is that physical changes can produce the same sign.
Feedback if chosen: Correct. The flexing warms the metal with no new substance forming — a physical change makes the same observation, so heat alone cannot prove a chemical change.
Feedback if chosen: This option is wrong — metals do undergo chemical changes. The student's error is about what the heat can prove, not about metals.
Check — multiple choice
A drop of one clear liquid is added to another clear liquid. A white solid immediately appears and settles at the bottom of the tube. Which judgment of this change is correct?
Feedback if chosen: This option is wrong — you stopped at the mixing. A solid appearing from two liquids is a precipitate, one of the four types of evidence of a chemical change.
Feedback if chosen: This option is wrong — you blamed freezing, but nothing cooled the liquids. A solid appearing where the two liquids mix is evidence that a new substance has formed.
Feedback if chosen: This option is wrong — the verdict is right but the evidence named is not. No heat was observed; the evidence here is the precipitate.
Feedback if chosen: Correct. A new solid appearing when two liquids mix is a precipitate — evidence that a new substance, with new properties, has formed.
Check — multiple choice
Liquid acetone evaporates from an open dish. Which description of what happens to the particles is correct?
Feedback if chosen: Correct. Evaporation only overcomes the attractions between intact molecules — no bonds break — so it is a physical process.
Feedback if chosen: This option is wrong — you broke the bonds inside the molecules. Evaporation overcomes only the attractions between molecules; every bond within each molecule survives.
Feedback if chosen: This option is wrong — you made new substances. The vapor is still acetone; only the spacing between the molecules has changed.
Feedback if chosen: This option is wrong — you let the molecules change size. Each molecule stays the same; it is the distance between molecules that grows.
Check — multiple choice
Solid calcium bromide, CaBr₂, dissolves in water, and the mixture stays clear. Why can this dissolving be argued to be either a physical or a chemical process?
Feedback if chosen: This option is wrong — you left the particles unchanged. The lattice's ionic bonds do break, which is exactly what gives the chemical-process argument its footing.
Feedback if chosen: This option is wrong — you invented new elements. The Ca²⁺ and Br⁻ ions themselves are unchanged; what changes is what holds them.
Feedback if chosen: Correct. Breaking ionic bonds looks chemical; forming attractions between intact ions and water molecules looks physical. Both happen at once, so the process can be argued either way.
Feedback if chosen: This option is wrong — you made heat the deciding test. The deciding evidence is at the particle level: bonds broken on one side, new attractions formed on the other.
Check — multiple choice
An oil lamp burns lamp oil, and the flame releases heat and light. Which statement explains where the released energy comes from?
Feedback if chosen: This option is wrong — you had bond-breaking release energy. Breaking a bond always takes energy in; it is bond-forming that releases energy.
Feedback if chosen: Correct. The new substances' bonds are stronger than the old ones, so forming them releases more energy than the breaking took in — the difference leaves as heat and light.
Feedback if chosen: This option is wrong — you reached for the attractions between molecules. Those belong to physical changes; a flame's energy comes from bonds breaking and forming.
Feedback if chosen: This option is wrong — you destroyed matter. Every atom survives the burning; the energy comes from the difference between bonds broken and bonds formed.
Check — multiple choice
An ice cube in a glass of lemonade melts, steadily drawing thermal energy from the drink. What is that absorbed energy used for?
Feedback if chosen: This option is wrong — you spent the energy on bonds. Melting leaves every bond inside each molecule intact; the energy works on the attractions between molecules.
Feedback if chosen: This option is wrong — you let the molecules change size. Each molecule stays the same; what changes is how freely the molecules move past one another.
Feedback if chosen: This option is wrong — you made a new substance. The liquid is still water; only the arrangement of the intact molecules has changed.
Feedback if chosen: Correct. A physical change like melting needs energy to loosen the attractions between intact molecules, so they can leave the fixed arrangement and move past one another.
8 lessons
Outcome: Identify the parts of a chemical equation: reactants, products, the arrow, coefficients, and subscripts.
So far, every chemical change took a whole sentence of prose to describe.
Chemists write the same event as one short line: 2H₂ + O₂ → 2H₂O.
The line says: hydrogen and oxygen turn into water.
But the line contains five 2s.
A 2 written before a formula does one job.
A 2 written inside a formula does a different job.
What does each part of the line mean?
The starting substances stand on the left — chemists call them the reactants.
The new substances stand on the right — chemists call them the products.
The arrow between them means "turn into".
The plus sign means "and".
The 2 written before H₂ states how many hydrogen molecules take part — a number written before a formula is called a coefficient.
The 2 written inside H₂ fixes which substance the formula names — a number written inside a formula is called a subscript.
A line built this way is called a chemical equation.
The chemical change an equation describes is called a reaction.
Read it as a sentence: two hydrogen molecules and one oxygen molecule turn into two water molecules.
A chemical equation names its parts: reactants on the left, products on the right, the arrow between them, coefficients for relative amounts, and subscripts fixing each substance's identity.
Identify the parts of a chemical equation: reactants on the left, products on the right, the arrow, coefficients (relative amounts), and subscripts (fixed identity).
Check — multiple choice
In the chemical equation 2CO + O₂ → 2CO₂, what is the 2 written before CO?
Feedback if chosen: Correct. A number written before a formula is a coefficient — here, two CO molecules take part.
Feedback if chosen: This option is wrong — a subscript sits inside a formula. This 2 stands before the formula: a coefficient.
Feedback if chosen: This option is wrong — the reactants are the substances CO and O₂ themselves, not the numbers before them.
Feedback if chosen: This option is wrong — the product is CO₂, on the right of the arrow. The 2 before CO is a coefficient.
Check — multiple choice
In the same equation, 2CO + O₂ → 2CO₂, what does the 2 written inside CO₂ tell you?
Feedback if chosen: This option is wrong — the amount is the coefficient's job, the 2 before CO₂. The 2 inside is a subscript.
Feedback if chosen: This option is wrong — CO₂ stands on the right of the arrow, so CO₂ is a product.
Feedback if chosen: This option is wrong — no part of a formula marks an equation as finished. The 2 inside CO₂ fixes the substance's identity.
Feedback if chosen: Correct. The subscript fixes the identity: two O atoms per carbon makes the substance CO₂, carbon dioxide.
Check — multiple choice
In 2Na + Cl₂ → 2NaCl, which substances are the reactants?
Feedback if chosen: This option is wrong — NaCl stands on the right of the arrow: the product.
Feedback if chosen: This option is wrong — you took one from each side. The reactants all stand on the left of the arrow.
Feedback if chosen: Correct. The starting substances, Na and Cl₂, stand on the left of the arrow — the reactants.
Feedback if chosen: This option is wrong — Cl₂ also stands on the left of the arrow, so Cl₂ is a reactant too.
Check — multiple choice
Which sentence reads the equation S + O₂ → SO₂ correctly?
Feedback if chosen: This option is wrong — you read the arrow backward. The reactants, S and O₂, stand on the left; they turn into SO₂.
Feedback if chosen: Correct. The plus sign reads 'and'; the arrow reads 'turn into': sulfur and oxygen turn into sulfur dioxide.
Feedback if chosen: This option is wrong — the plus sign joins the two reactants on the left. O₂ is a starting substance, not a product.
Feedback if chosen: This option is wrong — SO₂ stands on the right of the arrow, so SO₂ is the product, not a reactant.
Outcome: Identify the state symbols (s), (l), (g), and (aq) as labels for solid, liquid, gas, and dissolved-in-water substances.
Sodium chloride appears in two different equations.
In the first equation, the NaCl is a dry white solid.
In the second equation, the NaCl is dissolved in water, invisible.
The formula NaCl looks identical in both.
A reader cannot tell the solid from the solution.
How does an equation show the state of each substance?
A short label in parentheses follows each formula.
(s) means solid.
(l) means liquid.
(g) means gas.
(aq) means dissolved in water — a dissolved substance is called aqueous.
| Symbol | Meaning | One example |
|---|---|---|
| (s) | solid | NaCl(s) — dry solid sodium chloride |
| (l) | liquid | H₂O(l) — liquid water |
| (g) | gas | O₂(g) — oxygen gas |
| (aq) | aqueous: dissolved in water | NaCl(aq) — sodium chloride dissolved in water |
These four labels are the state symbols.
Now the two versions read differently: NaCl(s) is the dry solid; NaCl(aq) is sodium chloride dissolved in water.
From now on, every equation carries a state symbol on every formula.
The state symbols (s), (l), (g) and (aq) label solid, liquid, gas and dissolved-in-water substances in an equation — NaCl(aq) is sodium chloride dissolved in water.
Identify the state symbols (s), (l), (g), and (aq) as labels for solid, liquid, gas, and dissolved-in-water (aqueous) substances in an equation.
Check — multiple choice
What does the (g) in CO₂(g) tell you?
Feedback if chosen: This option is wrong — dissolved in water is (aq). The label (g) means gas.
Feedback if chosen: Correct. (g) is the state symbol for a gas.
Feedback if chosen: This option is wrong — liquid is (l). The label (g) means gas.
Feedback if chosen: This option is wrong — solid is (s). The label (g) means gas.
Check — multiple choice
In the equation 2K(s) + Br₂(l) → 2KBr(s), what state is the bromine, Br₂?
Feedback if chosen: Correct. Br₂(l) — the (l) labels the bromine as a liquid.
Feedback if chosen: This option is wrong — you read the potassium's label. Br₂ carries (l).
Feedback if chosen: This option is wrong — a gas would carry (g). Br₂ carries (l).
Feedback if chosen: This option is wrong — dissolved in water would be (aq). Br₂ carries (l).
Check — multiple choice
Vitamin C powder is stirred into a glass of water until every grain dissolves. Which label follows the vitamin C's formula in an equation?
Feedback if chosen: This option is wrong — (l) is for a pure liquid, like a melted solid. Dissolved-in-water substances carry (aq).
Feedback if chosen: This option is wrong — (s) is for the solid grains before stirring. The dissolved vitamin C carries (aq).
Feedback if chosen: This option is wrong — no gas formed. The dissolved vitamin C carries (aq).
Feedback if chosen: Correct. The vitamin C is dissolved in water — aqueous — so its formula carries (aq).
Outcome: State that every physical and chemical change can be represented by an equation whose atom counts balance.
You've already seen 50 g of water evaporate from an open dish.
Evaporation is a physical change — no new substance forms.
So far, equations have described chemical changes.
Does a physical change get an equation too?
What would sit on the right of the arrow, if nothing new forms?
Every change, physical or chemical, can be written as an equation.
Evaporation reads: H₂O(l) → H₂O(g).
The formula stays the same, because the substance stays the same.
Only the state symbol changes.
One rule governs every equation in chemistry: the atom counts on the two sides must match.
Count this one: two H and one O on the left; two H and one O on the right.
An equation whose atom counts match on the two sides is called a balanced equation.
Every physical and chemical change can be represented by an equation whose atom counts balance — evaporation is written H₂O(l) → H₂O(g).
State that every physical and chemical change can be represented symbolically by an equation whose atom counts balance.
Check — multiple choice
Ice melts in a glass. Which equation represents the melting?
Feedback if chosen: This option is wrong — you wrote freezing. Melting starts from the solid: H₂O(s) on the left.
Feedback if chosen: This option is wrong — melting makes no new substances. H₂ and O₂ would mean a chemical change.
Feedback if chosen: This option is wrong — (g) is a gas. Melted ice is liquid water: H₂O(l).
Feedback if chosen: Correct. The substance stays H₂O; only the state symbol changes, solid to liquid.
Check — multiple choice
What makes an equation a balanced equation?
Feedback if chosen: This option is wrong — state symbols report states. Balance is about the atom counts matching.
Feedback if chosen: Correct. Balanced means every element's atom count is the same on both sides of the arrow.
Feedback if chosen: This option is wrong — the number of formulas can differ. What must match is each element's atom count.
Feedback if chosen: This option is wrong — a physical change carries the same substance on both sides, and its equation still balances.
Check — multiple choice
Solid gold melts in a furnace. Which equation represents the change?
Feedback if chosen: This option is wrong — you wrote the gold hardening. Melting starts from the solid: Au(s) on the left.
Feedback if chosen: This option is wrong — (aq) means dissolved in water. Melted gold is a pure liquid: Au(l).
Feedback if chosen: Correct. The gold stays gold; the state symbol changes from (s) to (l).
Feedback if chosen: This option is wrong — melting gives a liquid, not a gas, and no coefficients are needed here.
Outcome: Explain that a chemical change rearranges atoms, so an equation must show equal atom counts of every element on both sides.
The camping stove burns methane again.
The equation reads: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g).
Look at the carbon atom inside the CO₂.
That carbon atom did not exist before the flame — or did it?
New substances formed. Were new atoms made too?
A chemical change creates no atoms.
A chemical change destroys no atoms.
The change rearranges the atoms into new groupings.
The carbon atom inside the CO₂ molecule came from the methane molecule.
The two oxygen atoms beside it came from the O₂ molecules.
Count the atoms across the equation: one C, four H, and four O on each side.
So an equation must show equal atom counts for every element — the same atoms, regrouped.
A chemical change is a rearrangement of atoms into new combinations, so its equation must show equal numbers of atoms of every element on both sides — burning methane redistributes the same C, H, and O atoms into CO₂ and H₂O.
Explain that a chemical change is a rearrangement of atoms, so an equation representing it must show equal atom counts of every element before and after.
Check — multiple choice
A fuel gas burns: C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(g). Where did the carbon atoms inside the CO₂ come from?
Feedback if chosen: Correct. A chemical change creates no atoms. The two carbon atoms in each C₂H₄ molecule regrouped into CO₂ molecules.
Feedback if chosen: This option is wrong — heat is energy, not matter. No flame can create an atom.
Feedback if chosen: This option is wrong — O₂ molecules hold only oxygen atoms. The carbon came from the fuel.
Feedback if chosen: This option is wrong — the water is a product, made at the same time. The carbon came from the fuel.
Check — multiple choice
A sheet of paper burns in a metal dish, leaving a pinch of ash. Most of the paper's atoms are no longer in the dish. What happened to them?
Feedback if chosen: This option is wrong — a chemical change destroys no atoms. The missing atoms left as gases.
Feedback if chosen: This option is wrong — atoms do not shrink. They were regrouped into gases and left with the smoke.
Feedback if chosen: This option is wrong — heat and light are not substances, and atoms cannot become them. The atoms left inside gas molecules.
Feedback if chosen: Correct. The paper's atoms were regrouped into gaseous new substances that drifted away. Every atom survived.
Check — multiple choice
Why must a chemical equation show equal atom counts of every element on both sides?
Feedback if chosen: This option is wrong — readability is not the reason. The counts match because every atom survives the change.
Feedback if chosen: This option is wrong — the number of formulas can differ across the arrow. The atom counts are what must match.
Feedback if chosen: Correct. The same atoms exist before and after, only regrouped — so every element's count is identical on both sides.
Feedback if chosen: This option is wrong — atoms never multiply in a chemical change. The counts match because the same atoms regroup.
Outcome: State that equal atom counts on the two sides of a balanced equation demonstrate conservation of mass.
A student burns 4 g of hydrogen gas in open air.
The flame produces 36 g of water.
The water's mass is nine times the fuel's mass.
Nine grams out for every gram in looks like creation from nothing.
But no atom was created.
Where did the extra 32 g come from?
Every atom survives a chemical change.
Each atom carries a fixed mass.
Same atoms on each side means same mass on each side.
The missing 32 g came from oxygen atoms in the air: 2H₂(g) + O₂(g) → 2H₂O(g).
4 g of hydrogen and 32 g of oxygen make 36 g of water — the mass in equals the mass out.
The matched atom counts of a balanced equation demonstrate this rule — chemists call it the conservation of mass.
The atoms lead the bookkeeping, and the mass follows.
Equal atom counts of every element on the two sides of a balanced equation demonstrate conservation of mass.
State that equal atom counts of every element on the two sides of a balanced equation demonstrate conservation of mass.
Check — multiple choice
In the balanced equation 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s), the atom counts of iron and oxygen match on both sides. What do the matched counts demonstrate?
Feedback if chosen: Correct. The same atoms, each with its fixed mass, appear on both sides — so the total mass is conserved.
Feedback if chosen: This option is wrong — the sides hold different numbers of formulas (two and one). The atom counts are what match.
Feedback if chosen: This option is wrong — new substances form here, so the change is chemical. Its balanced equation still conserves mass.
Feedback if chosen: This option is wrong — atoms never change element. Each element's own count matches across the equation.
Check — multiple choice
A tuft of steel wool with a mass of 2.0 g burns in air, and the gray product's mass is 2.4 g. Where did the extra 0.4 g of mass come from?
Feedback if chosen: This option is wrong — heat is not a substance and has no mass to give. The extra mass is oxygen atoms from the air.
Feedback if chosen: This option is wrong — atoms carry fixed masses. The product gained whole new atoms: oxygen from the air.
Feedback if chosen: This option is wrong — mass is conserved, and it is: oxygen atoms joined the iron, so the product holds the iron's mass plus theirs.
Feedback if chosen: Correct. The burning bonded oxygen atoms from the air to the iron. The product holds the iron's atoms plus the oxygen atoms — and their mass.
Check — multiple choice
A sealed jar holds a lit candle and air, with a total mass of 305.00 g. The candle burns until the flame goes out. What is the jar's total mass now?
Feedback if chosen: This option is wrong — nothing left the sealed jar. Every atom, and so every gram, is still inside.
Feedback if chosen: This option is wrong — nothing entered the sealed jar. No atoms arrived, so no mass arrived.
Feedback if chosen: Correct. The jar is sealed: the same atoms are inside, regrouped into new substances, with the same total mass.
Feedback if chosen: This option is wrong — conservation of mass predicts it exactly: same atoms in the jar, same 305.00 g.
Outcome: State that equal total charge on the two sides of an equation demonstrates conservation of charge.
A strip of zinc metal stands in a blue solution of copper ions.
Each copper ion, Cu²⁺, carries a charge of 2+.
Copper metal grows on the zinc strip.
The copper metal carries no charge at all.
Charge never just vanishes.
So where did each ion's 2+ charge go?
Write the event as an equation: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s).
Sum the charges on the left: Zn is neutral, 0; Cu²⁺ carries 2+. The left side totals 2+.
Sum the charges on the right: Zn²⁺ carries 2+; Cu is neutral, 0. The right side totals 2+.
The 2+ now sits on the zinc ion instead of the copper ion — and the total never changed.
Equal total charge on the two sides demonstrates a second bookkeeping rule — chemists call it the conservation of charge.
Conservation of charge stands beside conservation of mass: an equation with ions must respect both.
Equal total charge on the two sides of an equation demonstrates conservation of charge — Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) carries 2+ on each side.
State that equal total charge on the two sides of an equation demonstrates conservation of charge.
Check — multiple choice
In the equation Mg(s) + Fe²⁺(aq) → Mg²⁺(aq) + Fe(s), what is the total charge on each side?
Feedback if chosen: This option is wrong — you missed Mg²⁺ on the right. Each side holds one 2+ ion and one neutral metal.
Feedback if chosen: This option is wrong — each side carries one ion. Fe²⁺ gives the left 2+, and Mg²⁺ gives the right 2+.
Feedback if chosen: Correct. Left: 0 + 2+ = 2+. Right: 2+ + 0 = 2+. The totals match.
Feedback if chosen: This option is wrong — you added the two sides together. Sum each side separately: 2+ on each.
Check — multiple choice
What does equal total charge on the two sides of an equation demonstrate?
Feedback if chosen: Correct. Charge moves between particles, but the total never changes — conservation of charge.
Feedback if chosen: This option is wrong — conservation of mass is demonstrated by the matched atom counts, not the charge totals.
Feedback if chosen: This option is wrong — the sides can be full of ions. Only the totals must match.
Feedback if chosen: This option is wrong — ions can appear on both sides. The totals match even so.
Check — multiple choice
A student checks that the atom counts match in an equation involving ions, and declares the equation finished. What has the student skipped?
Feedback if chosen: This option is wrong — state symbols report states; no rule requires their counts to match.
Feedback if chosen: This option is wrong — the number of formulas can differ across the arrow.
Feedback if chosen: This option is wrong — with ions in the equation, a second rule applies: the charge totals must match too.
Feedback if chosen: Correct. Ions carry charge, and charge is conserved — so the two sides' charge totals must also be equal.
Outcome: Identify whether a given equation is balanced by counting the atoms of each element on both sides.
A textbook margin shows the equation: H₂(g) + O₂(g) → H₂O(l).
The line looks tidy.
Count the oxygen atoms: two on the left, one on the right.
One oxygen atom vanished between the sides.
Atoms cannot vanish.
So the equation is wrong. How do you catch errors like this every time?
Count one element at a time.
A coefficient multiplies every atom in the formula that follows it: 2H₂O contains four H atoms and two O atoms.
Compare each element's count on the left with its count on the right.
| Equation | Element | Left | Right | Verdict |
|---|---|---|---|---|
| C(s) + O₂(g) → CO₂(g) | C | 1 | 1 | |
| O | 2 | 2 | balanced | |
| H₂(g) + O₂(g) → H₂O(l) | H | 2 | 2 | |
| O | 2 | 1 | unbalanced |
Equal counts for every element: the equation is balanced.
Any element off by even one atom: the equation is unbalanced.
The check is mechanical.
Identify whether an equation is balanced by counting the atoms of each element on both sides — H₂(g) + O₂(g) → H₂O(l) is unbalanced, because two oxygen atoms become one.
Identify whether a given equation is balanced by counting the atoms of each element on both sides.
Check — numeric entry
How many oxygen atoms does 3CO₂ contain? Enter a whole number.
Feedback when correct: Correct. The coefficient 3 multiplies the subscript 2: 3 × 2 = 6 oxygen atoms.
Anticipated wrong answers (feedback shown):
| 2 | You read one molecule's subscript only. The coefficient 3 multiplies it: 3 × 2 = 6 oxygen atoms. |
| 3 | You counted the molecules, not the oxygen atoms. Each of the 3 molecules holds 2 O atoms: 3 × 2 = 6 O atoms. |
| 5 | You added the coefficient 3 and the subscript 2. The coefficient multiplies: 3 × 2 = 6 oxygen atoms. |
Check — numeric entry
How many hydrogen atoms does 4NH₃ contain? Enter a whole number.
Feedback when correct: Correct. The coefficient 4 multiplies the subscript 3: 4 × 3 = 12 hydrogen atoms.
Anticipated wrong answers (feedback shown):
| 3 | You read one molecule's subscript only. The coefficient 4 multiplies it: 4 × 3 = 12 hydrogen atoms. |
| 4 | You counted the molecules, not the hydrogen atoms. Each of the 4 molecules holds 3 H atoms: 4 × 3 = 12 H atoms. |
| 7 | You added the coefficient 4 and the subscript 3. The coefficient multiplies: 4 × 3 = 12 hydrogen atoms. |
Check — multiple choice
Is the equation N₂(g) + 2O₂(g) → 2NO₂(g) balanced?
Feedback if chosen: Correct. N: 2 and 2. O: 4 and 4. Every element's counts match.
Feedback if chosen: This option is wrong — count again: N₂ holds 2 N, and 2NO₂ holds 2 × 1 = 2 N. They match.
Feedback if chosen: This option is wrong — count again: 2O₂ holds 4 O, and 2NO₂ holds 2 × 2 = 4 O. They match.
Feedback if chosen: This option is wrong — formula counts never decide balance. Only the atom counts of each element matter.
Check — multiple choice
Is the equation Fe(s) + O₂(g) → Fe₂O₃(s) balanced?
Feedback if chosen: This option is wrong — count each element: Fe is 1 and 2; O is 2 and 3. Neither matches.
Feedback if chosen: This option is wrong — iron does fail (1 and 2), but oxygen fails too: 2 on the left, 3 on the right.
Feedback if chosen: Correct. Fe: 1 on the left, 2 on the right. O: 2 on the left, 3 on the right. Both fail.
Feedback if chosen: This option is wrong — oxygen does fail (2 and 3), but iron fails too: 1 on the left, 2 on the right.
Outcome: Explain why an equation is balanced by adjusting coefficients only and never subscripts.
The broken equation returns: H₂(g) + O₂(g) → H₂O(l).
A student repairs the oxygen count by rewriting the product as H₂O₂.
Count the atoms now: two H and two O on each side.
The counts pass.
But H₂O₂ is hydrogen peroxide — a bleach, not water.
The repair balanced the atoms by changing the substance.
Why is that repair illegal?
A coefficient and a subscript do different jobs.
A subscript fixes the identity of a substance: change the subscript, and you name a new substance.
A coefficient states the amount: change the coefficient, and the substance stays itself.
| The move | What stays | What changes |
|---|---|---|
| Change a coefficient: H₂O → 2H₂O | The substance: still water | The amount: two molecules |
| Change a subscript: H₂O → H₂O₂ | The amount: one molecule | The substance: now hydrogen peroxide |
Water stays water at any coefficient: H₂O, 2H₂O, and 6H₂O are all water.
H₂O₂ is not water at any amount.
The legal repair touches coefficients only: 2H₂(g) + O₂(g) → 2H₂O(l).
An equation is balanced by adjusting coefficients only, never subscripts — changing a subscript changes the identity of a substance rather than its amount.
Explain why an equation is balanced by adjusting coefficients only and never subscripts.
Check — multiple choice
Why may the unbalanced equation H₂(g) + O₂(g) → H₂O(l) not be fixed by rewriting the product as H₂O₂?
Feedback if chosen: This option is wrong — the H₂O₂ version balances perfectly. The problem is that it describes a different substance forming.
Feedback if chosen: Correct. A subscript fixes identity. Rewriting H₂O as H₂O₂ swaps water for hydrogen peroxide — a different substance.
Feedback if chosen: This option is wrong — subscripts may not be changed on either side. Any subscript change renames a substance.
Feedback if chosen: This option is wrong — H₂O₂ is real: hydrogen peroxide, a bleach. That is exactly why the repair misdescribes the change.
Check — multiple choice
The equation H₂(g) + Cl₂(g) → HCl(g) is unbalanced. Which repair is legal?
Feedback if chosen: Correct. The coefficient 2 fixes the counts — H: 2 and 2, Cl: 2 and 2 — and the product stays HCl.
Feedback if chosen: This option is wrong — you changed a subscript. HCl₂ names a different substance from HCl.
Feedback if chosen: This option is wrong — you changed subscripts. H₂Cl₂ names a different substance from HCl.
Feedback if chosen: This option is wrong — you rewrote the reactants. Hydrogen and chlorine exist as H₂ and Cl₂ molecules; changing their formulas renames them.
Check — multiple choice
Carbon burns in limited air, forming carbon monoxide, CO. The equation C(s) + O₂(g) → CO(g) is unbalanced, and a student repairs it to C(s) + O₂(g) → CO₂(g). The atom counts now match. What is wrong with the repair?
Feedback if chosen: This option is wrong — you accepted a repair that renames the product. The equation must describe the change that happened: CO formed.
Feedback if chosen: This option is wrong — rewriting O₂ as O renames the reactant. Oxygen gas exists as O₂ molecules.
Feedback if chosen: This option is wrong — CO₂ is a fine product of other changes. The problem here is that this change made CO.
Feedback if chosen: Correct. Writing CO₂ balances the atoms by describing a different event. The legal repair keeps CO: 2C(s) + O₂(g) → 2CO(g).
4 lessons
Outcome: Balance a chemical equation by adjusting coefficients to the smallest whole numbers that equalize every element's atom count.
Ethane burns in oxygen: __C₂H₆(g) + __O₂(g) → __CO₂(g) + __H₂O(g), with the coefficients missing.
A student tries 1, 1, 1, 1: the carbon count fails — two C on the left, one on the right.
The student fixes carbon with a 2 before CO₂: now the oxygen count fails.
Each repair breaks a different element.
Guessing could circle forever.
Is there an order of moves that always ends?
To balance an equation without guessing, work element by element, in a chosen order.
**1.** Balance the elements that appear in only one formula on each side first — here carbon and hydrogen.
**2.** Save for last any element that appears in several formulas on one side — here oxygen, which sits in both CO₂ and H₂O.
**3.** If a half remains, double every coefficient to clear it.
**4.** Finish on the smallest whole numbers that balance every element.
Now watch the routine applied to the ethane equation.
Carbon: C₂H₆ holds two C atoms, so write 2 before CO₂: C₂H₆(g) + __O₂(g) → 2CO₂(g) + __H₂O(g).
Hydrogen: C₂H₆ holds six H atoms, so write 3 before H₂O: C₂H₆(g) + __O₂(g) → 2CO₂(g) + 3H₂O(g).
Oxygen: the right side now holds 2 × 2 + 3 × 1 = 7 O atoms, so the left needs three and a half O₂ molecules.
A half may not stand, so double every coefficient: 2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(g).
| Step | Equation so far | C | H | O |
|---|---|---|---|---|
| Start | C₂H₆ + O₂ → CO₂ + H₂O | 2 | 1 | 6 | 2 | 2 | 3 |
| Carbon | C₂H₆ + O₂ → 2CO₂ + H₂O | 2 | 2 | 6 | 2 | 2 | 5 |
| Hydrogen | C₂H₆ + O₂ → 2CO₂ + 3H₂O | 2 | 2 | 6 | 6 | 2 | 7 |
| Oxygen, then double | 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O | 4 | 4 | 12 | 12 | 14 | 14 |
Check every element: C is 4 and 4; H is 12 and 12; O is 14 and 14.
The coefficients 2, 7, 4, 6 share no common factor, so they are the smallest whole numbers.
Balance a chemical equation by adjusting coefficients to the smallest whole numbers that equalize every element's atom count.
Balance a chemical equation by adjusting coefficients to the smallest whole numbers that equalize every element's atom count.
Check — numeric entry
Balance __C₃H₈(g) + __O₂(g) → __CO₂(g) + __H₂O(g). Carbon and hydrogen are already settled: 3 before CO₂ and 4 before H₂O. What coefficient goes before O₂? Enter a whole number.
Feedback when correct: Correct. The right side holds 3 × 2 + 4 × 1 = 10 O atoms, so the left needs 10 ÷ 2 = 5 O₂ molecules.
Anticipated wrong answers (feedback shown):
| 10 | You counted the O atoms needed, not the O₂ molecules. Each O₂ molecule carries two O atoms: 10 ÷ 2 = 5 O₂ molecules. |
| 7 | You added the coefficients 3 and 4. Count atoms instead: the right side holds 10 O atoms, so the left needs 5 O₂ molecules. |
| 6 | You counted 3 × 2 = 6 O atoms in the CO₂ only. The 4H₂O adds 4 more: 10 O atoms in all, so 5 O₂ molecules. |
Check — numeric entry
Balance __Al(s) + __O₂(g) → __Al₂O₃(s) with the smallest whole numbers. What coefficient goes before Al? Enter a whole number.
Feedback when correct: Correct. 4Al(s) + 3O₂(g) → 2Al₂O₃(s): Al is 4 and 4, and O is 6 and 6, on the smallest whole numbers.
Anticipated wrong answers (feedback shown):
| 2 | You balanced Al against one Al₂O₃. Oxygen forces two Al₂O₃ (6 O atoms, from 3O₂), and two Al₂O₃ hold four Al atoms. |
| 8 | You doubled past the smallest set. 8Al + 6O₂ → 4Al₂O₃ balances, but every coefficient halves: 4Al + 3O₂ → 2Al₂O₃. |
| 1 | One Al cannot supply the two Al atoms inside even one Al₂O₃. Balance O first: 3O₂ → 2Al₂O₃, which needs 4 Al. |
Check — multiple choice
Nitrogen and hydrogen form ammonia. Which is the balanced equation, on the smallest whole-number coefficients?
Feedback if chosen: This option is wrong — every count matches, but each coefficient halves. 2, 6, 4 shrink to 1, 3, 2.
Feedback if chosen: Correct. N: 2 and 2. H: 6 and 6. And 1, 3, 2 share no common factor — the smallest whole numbers.
Feedback if chosen: This option is wrong — count hydrogen: 2 on the left, 6 on the right. The left needs 3H₂.
Feedback if chosen: This option is wrong — count nitrogen: 2 on the left, 1 on the right. The right needs 2NH₃.
Check — drawing task (pen and paper, self-marked)
Methanol burns in oxygen: __CH₃OH(l) + __O₂(g) → __CO₂(g) + __H₂O(g). Take a pen and paper and write the complete balanced equation, using the smallest whole-number coefficients. Work element by element — carbon, hydrogen, then oxygen — and clear any half by doubling every coefficient. When you are done, select Check to compare your answer with the model answer.
Model answer shown to the student
Self-mark checklist — the drawing should show:
Outcome: Identify whether an equation involving ions is charge-balanced by summing the charges on each side.
A zinc strip stands in a solution of silver ions, Ag⁺.
Silver crystals grow on the strip.
A student writes: Zn(s) + Ag⁺(aq) → Zn²⁺(aq) + Ag(s).
Count the atoms: one zinc and one silver on each side — the atom check passes.
Now look at the charges: 1+ on the left becomes 2+ on the right.
The equation creates charge from nothing — an impossible event.
Which check catches the errors an atom count cannot see?
You've already seen that equal total charge on the two sides demonstrates conservation of charge.
Turn that rule into a routine: sum the charges on each side, sign included.
A neutral atom or molecule adds zero.
The left side, Zn + Ag⁺, sums to 0 + 1+ = 1+.
The right side, Zn²⁺ + Ag, sums to 2+ + 0 = 2+.
1+ does not equal 2+: the equation is not charge-balanced.
The repair doubles the silver: Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s).
Sum again: the left is 0 + 2+ = 2+, and the right is 2+ + 0 = 2+.
An equation with ions must pass both checks — atoms and charge — before chemists trust it.
Identify whether an equation involving ions is charge-balanced by summing the charges on each side — Zn + Ag⁺ → Zn²⁺ + Ag fails, because the sides sum to 1+ and 2+.
Identify whether an equation involving ions is charge-balanced by summing the charges on each side.
Check — multiple choice
Is the equation Ni(s) + Pb²⁺(aq) → Ni²⁺(aq) + Pb(s) charge-balanced? The ions carry the charges shown.
Feedback if chosen: Correct. Left: 0 + 2+ = 2+. Right: 2+ + 0 = 2+. The totals match.
Feedback if chosen: This option is wrong — you missed Pb²⁺ on the left. The neutral nickel adds 0, and Pb²⁺ adds 2+.
Feedback if chosen: This option is wrong — you doubled the right side. Ni²⁺ adds 2+, and the neutral lead adds 0.
Feedback if chosen: This option is wrong — you skipped the sums. One ion per side balances only when the two charges are equal, as they are here.
Check — multiple choice
Is the equation Al(s) + Sn²⁺(aq) → Al³⁺(aq) + Sn(s) charge-balanced? The ions carry the charges shown.
Feedback if chosen: This option is wrong — the right side holds Al³⁺, which adds 3+, not 2+.
Feedback if chosen: Correct. Left: 0 + 2+ = 2+. Right: 3+ + 0 = 3+. The totals differ, so the equation fails the charge check.
Feedback if chosen: This option is wrong — being metals cancels nothing. Sum the charges: 2+ on the left, 3+ on the right.
Feedback if chosen: This option is wrong — the neutral solids add 0, but each side also holds one ion: 2+ on the left, 3+ on the right.
Check — multiple choice
A student checks that the atom counts match in an equation involving ions, and concludes the equation must be charge-balanced too. Is the student correct?
Feedback if chosen: This option is wrong — an atom can appear as an ion on one side and neutral on the other, changing the charge without changing the count.
Feedback if chosen: This option is wrong — a one-silver-per-side equation can pass the atom check and still read 1+ against 2+.
Feedback if chosen: Correct. The two checks are independent: one silver atom per side passes the atom count, while 1+ against 2+ fails the charge sum.
Feedback if chosen: This option is wrong — plenty of equations involving ions pass both checks. The point is that the two checks are separate — do both every time.
Outcome: Evaluate a proposed chemical equation for atom- or charge-balance errors and correct any error found.
A homework page proposes: Mg(s) + O₂(g) → MgO(s).
Magnesium appears once on each side.
The line looks clean.
But count the oxygen: two O atoms enter, and one O atom leaves.
The AP exam supplies equations like this one, and asks whether they are right.
If an equation is wrong, you must name the error and make the legal fix.
Apply one checklist to every proposed equation.
**1.** Count the atoms of each element on both sides.
**2.** Sum the charges on each side, if any ion appears.
**3.** Confirm any repair touches coefficients only, and leaves them at the smallest whole numbers.
Name the error found, then correct it with coefficients.
Here, oxygen fails the count, and coefficients repair it: 2Mg(s) + O₂(g) → 2MgO(s).
Check the repair: Mg is 2 and 2; O is 2 and 2; the coefficients 2, 1, 2 are the smallest whole numbers.
Evaluate a proposed chemical equation for atom- or charge-balance errors, and correct any error you find by adjusting coefficients.
Evaluate a proposed chemical equation for atom- or charge-balance errors and correct any error found.
Check — multiple choice
A student proposes: 4H₂(g) + 2O₂(g) → 4H₂O(g). Every element's atom counts match. What, if anything, is wrong?
Feedback if chosen: This option is wrong — oxygen matches: 4 on each side. The failure is in the checklist's last line.
Feedback if chosen: This option is wrong — every formula is intact. The failure is in the coefficients.
Feedback if chosen: This option is wrong — the counts pass, but the checklist also demands the smallest whole numbers.
Feedback if chosen: Correct. 4, 2, 4 all halve: 2H₂(g) + O₂(g) → 2H₂O(g) says the same thing on the smallest whole numbers.
Check — multiple choice
Sodium burns in chlorine gas, forming sodium chloride. A homework page proposes: Na(s) + Cl₂(g) → NaCl(s). Which is the balanced equation, on the smallest whole-number coefficients?
Feedback if chosen: This option is wrong — you doubled everything. Chlorine now counts 4 and 2: still unbalanced.
Feedback if chosen: Correct. Na: 2 and 2. Cl: 2 and 2. Coefficients 2, 1, 2 are the smallest whole numbers, and every formula is intact.
Feedback if chosen: NaCl₂ is a different substance from the product, sodium chloride.
Feedback if chosen: This option is wrong — you rewrote the reactant. Chlorine gas exists as Cl₂ molecules; changing the formula renames the substance.
Check — multiple choice
A proposed equation reads: Ca(s) + Al³⁺(aq) → Ca²⁺(aq) + Al(s). Which flaw, if any, does the equation contain?
Feedback if chosen: This option is wrong — calcium counts 1 and 1. The atom check passes; the charge check is the one that fails.
Feedback if chosen: This option is wrong — every formula is intact. The charge sums are what differ.
Feedback if chosen: Correct. Left: 0 + 3+ = 3+. Right: 2+ + 0 = 2+. Coefficients repair it: 3Ca(s) + 2Al³⁺(aq) → 3Ca²⁺(aq) + 2Al(s), 6+ on each side.
Feedback if chosen: This option is wrong — the charge check fails: 3+ on the left against 2+ on the right.
Outcome: Write a balanced equation, with state symbols, for a physical change.
A 10 g block of dry ice — solid carbon dioxide — sits on a plate.
An hour later, the plate is empty and dry: no puddle, no crumbs.
The solid carbon dioxide went straight to gas.
The change is physical: the substance stayed CO₂ throughout.
Earlier lessons wrote equations for chemical changes.
Write this change as an equation — what goes on each side of the arrow?
You've already seen that every physical and chemical change can be represented by a balanced equation.
You've already seen the state symbols: (s), (l), (g), and (aq).
To write the equation for a physical change, put the same substance on both sides.
Then change only the state symbol.
Dry ice reads: CO₂(s) → CO₂(g).
The atom counts match automatically — no atom regroups in a physical change.
The state symbols carry the whole story: without them, CO₂ → CO₂ would show no change at all.
One equation shape now covers every change, physical and chemical alike.
Write a balanced equation, with state symbols, for a physical change — dry ice passing straight from solid to gas is CO₂(s) → CO₂(g).
Write a balanced equation, with state symbols, for a physical change.
Check — multiple choice
A pond freezes over. Which equation represents the freezing?
Feedback if chosen: Correct. The water stays H₂O; the state symbol changes from (l) to (s).
Feedback if chosen: This option is wrong — you wrote melting. Freezing starts from the liquid: H₂O(l) on the left.
Feedback if chosen: This option is wrong — freezing makes no new substances. H₂ and O₂ forming would be a chemical change.
Feedback if chosen: This option is wrong — (aq) means dissolved in water. Water cannot be dissolved in itself; frozen water is H₂O(s).
Check — multiple choice
Rubbing alcohol, C₃H₈O, evaporates from a countertop. Using the smallest whole-number coefficients, which equation represents the change?
Feedback if chosen: This option is wrong — you wrote condensing. Evaporation starts from the liquid: (l) on the left.
Feedback if chosen: This option is wrong — the counts match, but 2 and 2 are not the smallest whole numbers. Halve them: 1 and 1.
Feedback if chosen: Correct. The alcohol stays C₃H₈O; only the state symbol changes, (l) to (g), on the smallest whole numbers.
Feedback if chosen: This option is wrong — (aq) means dissolved in water. The alcohol left the counter as a gas: (g).
Check — drawing task (pen and paper, self-marked)
Overnight, water vapor in the air turns to liquid dew on cold grass. Take a pen and paper and write the balanced equation for this change, with a state symbol on every formula. When you are done, select Check to compare your answer with the model answer.
Model answer shown to the student
Self-mark checklist — the drawing should show:
7 lessons
Outcome: Classify a chemical equation as the molecular, complete ionic, or net ionic form of a reaction.
A student pours silver nitrate solution into sodium chloride solution.
A white solid appears at once: silver chloride.
The student writes the balanced equation: AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq).
That equation shows NaCl as one intact formula.
You've already seen what water does to a dissolved ionic compound: the water pulls each unit apart into separate ions.
So the beaker holds no intact NaCl units at all.
The equation is balanced — and the equation still hides what actually floats in the liquid.
Chemists keep three written forms for one reaction — the table shows all three for the silver chloride reaction.
| Form | The silver chloride reaction in that form | What the form shows |
|---|---|---|
| Molecular | AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq) | every compound as an intact formula |
| Complete ionic | Ag⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + Cl⁻(aq) → AgCl(s) + Na⁺(aq) + NO₃⁻(aq) | every dissolved compound as its separate ions |
| Net ionic | Ag⁺(aq) + Cl⁻(aq) → AgCl(s) | only the species that change |
The molecular form writes every compound as an intact formula.
The molecular form is easy to balance, but the molecular form does not show the separate ions in the liquid.
The complete ionic form writes every dissolved compound as its separate ions — the honest picture of the liquid.
The net ionic form keeps only the species that change — the chemistry alone.
All three forms describe the same reaction.
The question in front of you decides which form you write.
Classify molecular, complete ionic, and net ionic equations as three forms of the same reaction, chosen to fit the context.
Check — multiple choice
Classify this equation: Mg(NO₃)₂(aq) + 2KOH(aq) → Mg(OH)₂(s) + 2KNO₃(aq).
Feedback if chosen: This option is wrong — you classified intact formulas as separate ions. Every compound here is written as one intact formula, so this is the molecular form.
Feedback if chosen: Correct. Every compound is written as one intact formula — the molecular form.
Feedback if chosen: This option is wrong — a net ionic equation keeps only the species that change, written as ions. This equation still shows every compound intact.
Feedback if chosen: This option is wrong — this equation uses chemical formulas, not substance names. Every formula is intact, so this is the molecular form.
Check — multiple choice
Classify this equation: Mg²⁺(aq) + 2NO₃⁻(aq) + 2K⁺(aq) + 2OH⁻(aq) → Mg(OH)₂(s) + 2K⁺(aq) + 2NO₃⁻(aq).
Feedback if chosen: This option is wrong — you accepted unchanged ions in a net form. K⁺ and NO₃⁻ appear identical on both sides, so this is the complete ionic form.
Feedback if chosen: This option is wrong — this equation uses formulas with charges, not substance names. The dissolved compounds are split into ions: the complete ionic form.
Feedback if chosen: Correct. Every dissolved compound is written as its separate ions, and the solid stays intact — the complete ionic form.
Feedback if chosen: This option is wrong — a molecular equation writes every compound as an intact formula. Here the dissolved compounds are split into ions.
Check — multiple choice
Classify this equation: Mg²⁺(aq) + 2OH⁻(aq) → Mg(OH)₂(s).
Feedback if chosen: Correct. Only the species that change appear — the net ionic form.
Feedback if chosen: This option is wrong — a molecular equation writes intact compound formulas such as Mg(NO₃)₂. Here the reactants are separate ions.
Feedback if chosen: This option is wrong — a complete ionic equation lists every ion in the liquid, unchanged ions included. Here only the species that change remain.
Feedback if chosen: This option is wrong — this equation uses ion formulas with charges, not substance names. Only the species that change appear: the net ionic form.
Check — multiple choice
One mixing of two solutions was written three ways: as a molecular, a complete ionic, and a net ionic equation. Which statement is correct?
Feedback if chosen: This option is wrong — you tied each form to a new reaction. One reaction happened in the beaker; the three forms are three ways of writing it.
Feedback if chosen: This option is wrong — every correctly written form balances. The forms differ in which species they show, not in whether they balance.
Feedback if chosen: This option is wrong — all three forms show the same product forming. They differ only in how much of the liquid's detail they show.
Feedback if chosen: Correct. One reaction, three written forms — the question in front of you decides which form you write.
Outcome: State which species an ionic equation writes as separate aqueous ions, and which keep their intact formulas.
The silver chloride reaction has finished in its beaker.
The sodium chloride entered the beaker already dissolved in the water.
You've already seen water molecules pull each dissolved NaCl unit apart into separate Na⁺ and Cl⁻ ions.
The new silver chloride sits at the bottom of the beaker as one solid lump.
The figure zooms into both: separate Na⁺ and Cl⁻ ions free in the liquid, and one intact lump of solid AgCl.
On paper, the complete ionic equation splits compounds into ions.
Should the equation split both compounds — or only one?
An ionic compound that can dissolve in water is called soluble.
When water pulls a dissolved soluble ionic compound apart into separate free ions, that event is called dissociation.
A dissolved, dissociated ionic compound is written as its separate ions, each with its own charge and its own (aq).
Everything the water never pulled apart keeps its intact formula.
NaCl(aq) is dissolved and dissociated — write it as separate ions: Na⁺(aq) + Cl⁻(aq).
AgCl(s) is a solid — write it as its intact formula: AgCl(s).
H₂O(l) is a liquid — write it as its intact formula: H₂O(l).
Dissolved sugar is a dissolved molecular substance, not ions — write it as its intact formula: C₁₂H₂₂O₁₁(aq).
Write what the water actually did — nothing more.
State how species are written in ionic equations: dissolved soluble ionic compounds as separate aqueous ions; solids, liquids, gases, and dissolved molecular substances as intact formulas.
Check — multiple choice
Which species is written as separate aqueous ions in a complete ionic equation?
Feedback if chosen: Correct. KNO₃ is a dissolved soluble ionic compound, so the water has dissociated it — write K⁺(aq) + NO₃⁻(aq).
Feedback if chosen: This option is wrong — you split a solid. The water never pulled MgO(s) apart, so MgO(s) keeps its intact formula.
Feedback if chosen: This option is wrong — you split a dissolved molecular substance. Ethanol dissolves as whole molecules, so C₂H₅OH(aq) stays intact.
Feedback if chosen: This option is wrong — you split a liquid. Liquid water keeps its intact formula, H₂O(l).
Check — multiple choice
Which species keeps its intact formula in a complete ionic equation?
Feedback if chosen: This option is wrong — MgBr₂ is a dissolved soluble ionic compound, so the water has dissociated it into Mg²⁺(aq) + 2Br⁻(aq).
Feedback if chosen: This option is wrong — K₂SO₄ is a dissolved soluble ionic compound, so the water has dissociated it into 2K⁺(aq) + SO₄²⁻(aq).
Feedback if chosen: This option is wrong — NH₄Cl is a dissolved soluble ionic compound, so the water has dissociated it into NH₄⁺(aq) + Cl⁻(aq).
Feedback if chosen: Correct. Methanol is a molecular substance — it dissolves as whole molecules, so CH₃OH(aq) stays intact.
Check — multiple choice
Potassium sulfate, K₂SO₄, is soluble in water. What does dissociation do to a K₂SO₄ unit that dissolves?
Feedback if chosen: This option is wrong — you kept a soluble ionic compound intact. Dissociation means the water pulls the dissolved unit apart into its free ions.
Feedback if chosen: Correct. Dissociation frees each ion the compound contains: 2K⁺(aq) + SO₄²⁻(aq).
Feedback if chosen: This option is wrong — you glued the two potassium ions together. Each K⁺ ion goes free on its own; K₂²⁺ is not a real ion.
Feedback if chosen: This option is wrong — you broke a polyatomic ion apart. SO₄²⁻ travels as one intact ion; dissociation only separates the ions from each other.
Check — multiple choice
Glucose is a molecular substance, and glucose dissolves in water. A student writes dissolved glucose as separate ions. Which statement is correct?
Feedback if chosen: This option is wrong — you treated dissolving and dissociating as the same event. Only soluble ionic compounds dissociate; molecular substances dissolve as whole molecules.
Feedback if chosen: This option is wrong — dissociation frees ions, and a glucose molecule contains no ions to free.
Feedback if chosen: Correct. Glucose has no ions for the water to free, so each dissolved molecule stays whole: C₆H₁₂O₆(aq).
Feedback if chosen: This option is wrong — glucose does dissolve. It just dissolves as whole molecules, so it is written intact.
Outcome: Name the spectator ions in a complete ionic equation — the ions that appear unchanged on both sides.
Here is the silver chloride reaction as a complete ionic equation: Ag⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + Cl⁻(aq) → AgCl(s) + Na⁺(aq) + NO₃⁻(aq).
Na⁺ enters the beaker as a free ion in the water.
Na⁺ leaves the beaker as a free ion in the water, unchanged.
NO₃⁻ repeats the same story.
The figure shows the beaker before and after: Na⁺ and NO₃⁻ are drawn identical in both panels.
Only Ag⁺ and Cl⁻ changed — those two ions joined into the solid.
Half of the ions in this equation took no part in the change.
An ion that appears unchanged on both sides of a complete ionic equation took no part in the change.
Such an ion is called a spectator ion: the ion watches the reaction the way a spectator watches a match.
In this equation, Na⁺ and NO₃⁻ are the spectator ions.
The ions that do change carry the whole chemistry.
One read does the whole job: look at a complete ionic equation and name its spectators.
State that spectator ions appear unchanged on both sides of a complete ionic equation and take no part in the chemical change, and name them in a supplied equation.
Check — multiple choice
Here is a complete ionic equation: Ca²⁺(aq) + 2NO₃⁻(aq) + 2K⁺(aq) + 2F⁻(aq) → CaF₂(s) + 2K⁺(aq) + 2NO₃⁻(aq). Which ions are the spectator ions?
Feedback if chosen: This option is wrong — you picked the two ions that changed. Ca²⁺ and F⁻ joined into the solid CaF₂; the spectators are the ions left unchanged.
Feedback if chosen: This option is wrong — you kept Ca²⁺. Ca²⁺ ends up inside the solid, so Ca²⁺ changed; only K⁺ and NO₃⁻ stayed free and unchanged.
Feedback if chosen: Correct. K⁺ and NO₃⁻ appear as identical free ions on both sides — they took no part in the change.
Feedback if chosen: This option is wrong — you kept F⁻. F⁻ ends up inside the solid, so F⁻ changed; only K⁺ and NO₃⁻ stayed free and unchanged.
Check — multiple choice
What makes an ion a spectator ion?
Feedback if chosen: Correct. An ion unchanged on both sides took no part in the change — that is what makes it a spectator.
Feedback if chosen: This option is wrong — an ion only on the reactant side was consumed, so it changed. A spectator appears identical on both sides.
Feedback if chosen: This option is wrong — charge size does not decide the label. What decides it is whether the ion changed.
Feedback if chosen: This option is wrong — the ion that joins into the solid is exactly the ion that changed. A spectator stays free and unchanged.
Check — multiple choice
In Ni²⁺(aq) + 2Cl⁻(aq) + 2Na⁺(aq) + CO₃²⁻(aq) → NiCO₃(s) + 2Na⁺(aq) + 2Cl⁻(aq), a student says: "Cl⁻ appears in the equation, so Cl⁻ must have reacted." Which statement is correct?
Feedback if chosen: This option is wrong — the complete ionic form lists everything in the liquid, reacting or not. Appearing in the equation is not evidence of reacting.
Feedback if chosen: Correct. Cl⁻ enters free and leaves free, identical on both sides — a spectator ion.
Feedback if chosen: This option is wrong — you invented a change and an undo. The equation shows Cl⁻ identical on both sides; nothing happened to it.
Feedback if chosen: This option is wrong — Cl⁻ does appear, on both sides. It appears unchanged, which is exactly why it is a spectator.
Outcome: Write the complete ionic equation for a reaction by splitting each aqueous ionic compound in the balanced molecular equation into its ions.
An exam line supplies one molecular equation: K₂CO₃(aq) + CaCl₂(aq) → CaCO₃(s) + 2KCl(aq).
The instruction says: write the complete ionic equation.
You've already seen which species dissociate: dissolved soluble ionic compounds split into ions, and everything else stays intact.
You've already seen how ion charges fix formulas — here the ions are K⁺, CO₃²⁻, Ca²⁺, and Cl⁻.
Writing the ions yourself is a new job: every charge, every count, every state symbol must be right.
Split every aqueous ionic compound into its separate ions, each with its own charge and its own (aq).
Copy every solid, liquid, gas, and dissolved molecular substance through unchanged.
A coefficient multiplies every ion its compound releases: 2KCl(aq) becomes 2K⁺(aq) + 2Cl⁻(aq).
Then check atoms: every element's atom count must match across the arrow.
Then check charge: each side's total charge must match.
Q: What is the complete ionic equation for K₂CO₃(aq) + CaCl₂(aq) → CaCO₃(s) + 2KCl(aq)?
Split every aqueous ionic compound into its ions
K₂CO₃(aq) → 2K⁺(aq) + CO₃²⁻(aq)
CaCl₂(aq) → Ca²⁺(aq) + 2Cl⁻(aq)
2KCl(aq) → 2K⁺(aq) + 2Cl⁻(aq)
Copy the solid through unchanged
CaCO₃(s) stays CaCO₃(s)
Write the full equation
2K⁺(aq) + CO₃²⁻(aq) + Ca²⁺(aq) + 2Cl⁻(aq) → CaCO₃(s) + 2K⁺(aq) + 2Cl⁻(aq)
Check atoms and total charge
Atoms: 2 K, 1 C, 3 O, 1 Ca, 2 Cl on each side
Charge: (+2) + (−2) + (+2) + (−2) = 0 on the left, and 0 + (+2) + (−2) = 0 on the right
The AP exam asks for this equation hand-written on paper — so practice writing it on paper now.
Write the complete ionic equation by dissociating each aqueous ionic compound in the balanced molecular equation, with correct formulas, charges, and state symbols.
Check — drawing task (pen and paper, self-marked)
Take a pen and paper and write the complete ionic equation for Na₂CO₃(aq) + MgSO₄(aq) → MgCO₃(s) + Na₂SO₄(aq). When you are done, select Check to compare your answer with the model answer.
Model answer shown to the student
Self-mark checklist — the drawing should show:
Check — multiple choice
Silver is present as Ag⁺ in silver nitrate. Which is the complete ionic equation for AgNO₃(aq) + KBr(aq) → AgBr(s) + KNO₃(aq)?
Feedback if chosen: This option is wrong — you split the solid. AgBr forms as a solid, so AgBr(s) stays one intact formula on the right.
Feedback if chosen: This option is wrong — you left a dissolved ionic compound intact. AgNO₃(aq) is dissolved and dissociated, so it splits into Ag⁺(aq) + NO₃⁻(aq).
Feedback if chosen: This option is wrong — you dropped the charges and states. Every free ion carries its charge and its (aq); the solid carries its (s).
Feedback if chosen: Correct. The dissolved compounds split into their ions, and the solid AgBr(s) stays intact.
Outcome: Write the net ionic equation by canceling the spectator ions, then check that atoms and charge still balance.
The complete ionic equation of the silver chloride reaction lists seven species.
Four of them are the two spectator ions, Na⁺ and NO₃⁻, copied on both sides.
The chemistry lives in just three species: Ag⁺, Cl⁻, and AgCl.
Cancel every spectator ion that appears identical on both sides — the figure shows the crossing-out.
Whatever remains is the net ionic equation: Ag⁺(aq) + Cl⁻(aq) → AgCl(s).
Check atoms: 1 Ag and 1 Cl on each side.
Check charge: the left side sums to (+1) + (−1) = 0, and the right side is 0.
A cancellation needs identical species: same formula, same charge, same state, free in the water on both sides.
An ion locked inside a solid is a different species from the same ion free in the water — never cancel across that difference.
Mix any soluble silver compound with any soluble chloride: the same net ionic equation comes out — one form for the shared chemistry.
Write the net ionic equation by canceling the spectator ions from the complete ionic equation and verifying that atoms and charge still balance.
Check — drawing task (pen and paper, self-marked)
Zinc is present as Zn²⁺ in zinc chloride. Take a pen and paper and write the complete ionic equation and then the net ionic equation for ZnCl₂(aq) + Na₂S(aq) → ZnS(s) + 2NaCl(aq). When you are done, select Check to compare your answer with the model answer.
Model answer shown to the student
Self-mark checklist — the drawing should show:
Check — multiple choice
In Ca²⁺(aq) + 2NO₃⁻(aq) + 2K⁺(aq) + 2F⁻(aq) → CaF₂(s) + 2K⁺(aq) + 2NO₃⁻(aq), a student says: "F appears on both sides, so F⁻(aq) cancels against the F in CaF₂(s)." Which statement is correct?
Feedback if chosen: This option is wrong — cancellation works on identical species, not on repeated elements. The state must match too.
Feedback if chosen: Correct. A cancellation needs the same formula, charge, and state on both sides; a free F⁻(aq) and an F locked in CaF₂(s) are different species.
Feedback if chosen: This option is wrong — that cancellation would leave Ca²⁺ with no partner: +2 on the left against 0 on the right, so the charges no longer balance.
Feedback if chosen: This option is wrong — F⁻(aq) belongs in this net ionic equation. It changed: it joined Ca²⁺ inside the solid.
Outcome: Evaluate a proposed ionic equation and name its error: a wrong split, an uncanceled spectator, or broken atom or charge balance.
A classmate splits every species in the silver chloride reaction into ions — the solid included.
On paper, every ion now appears unchanged on both sides.
Cancel them all, and nothing remains.
The equation now claims no reaction happened — yet the white solid sits right there in the beaker.
One wrong split erased a real reaction.
Three errors cover nearly every wrong ionic equation.
Error 1: a solid, liquid, gas, or dissolved molecular substance was split into ions.
Error 2: a spectator ion was left standing after the cancellation.
Error 3: the atom counts or the total charges stopped matching across the arrow.
Read every proposed equation for the three errors, in that order.
The same three-error read checks your own equations before an examiner does.
Evaluate a proposed complete ionic or net ionic equation and identify the error if present: a wrong split, an uncanceled spectator ion, or broken atom or charge balance.
Check — multiple choice
Solutions of ammonium carbonate and barium nitrate are mixed, and barium carbonate forms as a solid. A classmate proposes this complete ionic equation: 2NH₄⁺(aq) + CO₃²⁻(aq) + Ba²⁺(aq) + 2NO₃⁻(aq) → Ba²⁺(aq) + CO₃²⁻(aq) + 2NH₄⁺(aq) + 2NO₃⁻(aq). What is wrong with it?
Feedback if chosen: This option is wrong — no cancellation has happened yet; this is a complete ionic equation. The defect is the solid written as free ions.
Feedback if chosen: This option is wrong — every atom count matches. The defect is the solid written as free ions.
Feedback if chosen: Correct. Barium carbonate forms as a solid, so the right side must read BaCO₃(s) — writing free Ba²⁺ and CO₃²⁻ erases the reaction.
Feedback if chosen: This option is wrong — the equation shows the solid's ions still free in the water, so it claims no change happened. The solid was split.
Check — multiple choice
Silver iodide forms as a solid when solutions of silver nitrate and sodium iodide are mixed. A classmate proposes this net ionic equation: Ag⁺(aq) + I⁻(aq) + Na⁺(aq) → AgI(s) + Na⁺(aq). What is wrong with it?
Feedback if chosen: Correct. Na⁺(aq) appears identical on both sides, so Na⁺ is a spectator — a net ionic equation cancels it.
Feedback if chosen: This option is wrong — AgI(s) is written intact, as it should be. The defect is the unchanged ion still standing on both sides.
Feedback if chosen: This option is wrong — each side sums to +1, so the charges match. The defect is the spectator left standing.
Feedback if chosen: This option is wrong — a net ionic equation keeps only the species that change, and Na⁺ did not change.
Check — multiple choice
Aluminum hydroxide forms as a solid when solutions of aluminum chloride and sodium hydroxide are mixed. Aluminum is present as Al³⁺. A classmate proposes this net ionic equation: Al³⁺(aq) + 2OH⁻(aq) → Al(OH)₃(s). What is wrong with it?
Feedback if chosen: This option is wrong — no unchanged ion appears on both sides. The defect is the balance: count the atoms and sum the charges.
Feedback if chosen: This option is wrong — Al(OH)₃(s) is written intact, as it should be. The defect is the balance.
Feedback if chosen: Correct. The left has 2 O and 2 H against 3 O and 3 H on the right.
The left's charge sums to +1 against 0 on the right.
The coefficient must be 3OH⁻.
Feedback if chosen: This option is wrong — check the two balances: 2 OH⁻ carries −2 against Al³⁺'s +3, and only 2 O reach a solid that needs 3.
Check — multiple choice
Magnesium fluoride forms as a solid when solutions of magnesium bromide and sodium fluoride are mixed. A classmate proposes this net ionic equation: Mg²⁺(aq) + 2F⁻(aq) → MgF₂(s). What is wrong with it?
Feedback if chosen: This option is wrong — MgF₂(s) is written intact on the right, exactly as a solid should be.
Feedback if chosen: This option is wrong — no ion appears unchanged on both sides; the spectators Na⁺ and Br⁻ were already canceled.
Feedback if chosen: This option is wrong — the left sums to (+2) + 2 × (−1) = 0, and the right is 0. The charges match.
Feedback if chosen: Correct. The split is right, the spectators are gone, and atoms and charge both balance — the equation survives all three error checks.
Outcome: Write the balanced equation — molecular or net ionic, as the question demands — from the names of the substances mixed.
An AP item names two solutions to mix: barium chloride and sodium sulfate.
The item adds one fact: barium sulfate is insoluble in water.
No equation appears anywhere on the page.
The item asks for the net ionic equation of the mixing.
Answering takes every ionic-equation skill you have, at once.
The ion charges settle the formulas: Ba²⁺ with Cl⁻ gives BaCl₂, and Na⁺ with SO₄²⁻ gives Na₂SO₄.
The supplied facts settle the state symbols: both compounds are dissolved, and barium sulfate forms as a solid.
Q: What is the net ionic equation for mixing aqueous barium chloride and aqueous sodium sulfate, given that barium sulfate is insoluble in water?
Write the balanced molecular equation
BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq)
Split every aqueous ionic compound into its ions
Ba²⁺(aq) + 2Cl⁻(aq) + 2Na⁺(aq) + SO₄²⁻(aq) → BaSO₄(s) + 2Na⁺(aq) + 2Cl⁻(aq)
Cancel the spectator ions
Na⁺ and Cl⁻ appear unchanged on both sides — cancel them
Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)
Check atoms and total charge
Atoms: 1 Ba, 1 S, 4 O on each side
Charge: (+2) + (−2) = 0 on the left, and 0 on the right
Stop at whichever form the question names — molecular, complete ionic, or net ionic.
The AP exam asks you to write exactly this, from exactly this much information.
Write the balanced equation — molecular or net ionic as the context demands — for a reaction specified by the identities of its reactants and products.
Check — drawing task (pen and paper, self-marked)
Iron is present as Fe³⁺ in iron(III) chloride. Solutions of iron(III) chloride and sodium hydroxide are mixed, and iron(III) hydroxide forms as an insoluble solid. Take a pen and paper and write the balanced molecular equation and the net ionic equation for the mixing. When you are done, select Check to compare your answer with the model answer.
Model answer shown to the student
Self-mark checklist — the drawing should show:
Check — multiple choice
Solutions of potassium phosphate, K₃PO₄, and calcium nitrate are mixed. Calcium phosphate, Ca₃(PO₄)₂, is insoluble in water. Which is the net ionic equation?
Feedback if chosen: This option is wrong — you stopped at the molecular form. The question asks for the net ionic form: split the dissolved compounds and cancel the spectators.
Feedback if chosen: This option is wrong — the atoms do not balance. One Ca and one PO₄ cannot build Ca₃(PO₄)₂; the coefficients must read 3Ca²⁺ and 2PO₄³⁻.
Feedback if chosen: Correct. 3Ca²⁺ and 2PO₄³⁻ build the solid, atoms match, and the charge sums to 0 on each side.
Feedback if chosen: This option is wrong — K⁺ appears unchanged on both sides. A net ionic equation cancels every spectator ion.
Outcome: Write net ionic equations that earn the point: right species split, spectators gone, atoms and charge balanced.
On the free-response section, an equation-writing point is all-or-nothing: one wrongly split species, one surviving spectator, or one unbalanced charge earns zero.
The graders' first check is the split: dissolved ionic compounds are written as separate aqueous ions, while every solid, liquid, gas, and dissolved molecular substance stays intact.
The second check is the spectators: any ion that appears unchanged on both sides is dropped from the net ionic equation.
The third check is a double balance: the atoms of every element AND the total charge must match across the arrow — graders check both.
State symbols are not required for the point — the scoring guidelines say so every year — but every ion's charge is checked, so write each charge every time.
When a part says write the net ionic equation, make the three checks in order — split, spectators, double balance — before moving on.
Free-response question
Source note (not shown as part of the question): authored to the write-equation archetype (FRQ playbook T8; modeled on the net-ionic points of 2021 FRQ Q3(a) and 2023 FRQ Q3(a))
Aqueous solutions of nickel(II) bromide and potassium sulfide react by the molecular equation NiBr₂(aq) + K₂S(aq) → NiS(s) + 2KBr(aq). The NiS is insoluble; the other three compounds are soluble and fully dissociated in water.
(a) Write the complete ionic equation for the reaction. [1 point]
(b) Write the net ionic equation for the reaction. [1 point]
(c) Identify the spectator ions. Justify your answer. [1 point]
✓ Model response (the key)
(a)
Ni²⁺(aq) + 2Br⁻(aq) + 2K⁺(aq) + S²⁻(aq) → NiS(s) + 2K⁺(aq) + 2Br⁻(aq)
(b)
Ni²⁺(aq) + S²⁻(aq) → NiS(s)
(c)
K⁺ and Br⁻ are the spectator ions.
Each appears unchanged, still dissolved, on both sides of the complete ionic equation, so neither takes part in the change.
Scoring rubric — 3 points available
Check — multiple choice
Four students answered part (b) of the question you just did. Which response earns the point?
Feedback if chosen: This option is wrong — you accepted a sentence in place of an equation. The point pays for the symbolic equation itself, with species and charges.
Feedback if chosen: Correct. The species that change appear with their charges, the spectators are gone, and atoms and charge both balance (2+ and 2− sum to zero).
Feedback if chosen: This option is wrong — you accepted the complete ionic equation. For the net ionic point, the unchanged K⁺ and Br⁻ must be canceled.
Feedback if chosen: This option is wrong — you accepted ions written without their charges. Ni²⁺ and S²⁻ carry charges, and graders check every charge.
3 lessons
Outcome: Draw a particulate representation of a balanced equation, with correct ratios, intact units, and every atom conserved.
An AP free-response box sits empty under one instruction: draw a particulate representation of 2H₂ + O₂ → 2H₂O.
The equation states the recipe: two hydrogen molecules and one oxygen molecule become two water molecules.
The drawing must show that recipe as circles on paper.
You've already drawn particle pictures of solids, liquids, and gases.
Nobody has asked you to draw a chemical reaction before.
Here is a correct drawing for 2H₂ + O₂ → 2H₂O — a Before box and an After box.
The particle ratio copies the coefficients: two H₂ and one O₂ in, two H₂O out.
Each unit stays joined: two touching H circles make one H₂ molecule.
In a reaction drawing, touching circles mean one unit — one molecule, or one formula unit of an ionic compound. In the solid and liquid boxes you drew in Unit 3, touching showed packing — separate particles crowded together; here, touching means joined.
A lone circle means a free atom — this equation never names one.
Count each element's atoms across the arrow: four H and two O on each side.
A correct drawing states the same conservation as the balanced equation — in circles instead of symbols.
A drawing may scale the recipe up: four H₂ and two O₂ making four H₂O keeps the same ratio, so that drawing is also correct.
Draw a particulate representation of a balanced equation, with correct particle ratios, correctly bonded or intact species, and every element's atom count conserved.
Check — multiple choice
The reaction is H₂ + Cl₂ → 2HCl, and it goes to completion. The Before box is shown. Which drawing shows a correct After box?
Feedback if chosen: This option is wrong — you lost atoms. The Before box holds 6 H and 6 Cl; three HCl units hold only 3 of each.
Feedback if chosen: Correct. Six HCl units use every atom — 6 H and 6 Cl — and each unit is drawn as one touching H–Cl pair.
Feedback if chosen: This option is wrong — you drew separated atoms. Hydrogen chloride exists as joined HCl molecules, so lone circles tell a false story.
Feedback if chosen: This option is wrong — you drew no reaction. The reaction goes to completion, so the After box must show the HCl the equation names.
Check — multiple choice
The drawing shows a sample of fluorine gas. What does one pair of touching F circles show?
Feedback if chosen: Correct. Touching circles mean one unit — here one F₂ molecule.
Feedback if chosen: This option is wrong — you read touching circles as free atoms. Touching circles mean one joined unit; a free atom is drawn as a lone circle.
Feedback if chosen: This option is wrong — you counted each circle as a molecule. The two touching circles together make one F₂ molecule.
Feedback if chosen: This option is wrong — one F atom would be one lone circle. Two touching circles make one two-atom molecule.
Check — drawing task (pen and paper, self-marked)
The equation is N₂ + 2O₂ → 2NO₂, and every substance is a gas. Take a pen and paper and draw a particulate representation of this equation — a Before box and an After box. When you are done, select Check to compare your drawing with the model answer.
Model answer shown to the student
Self-mark checklist — the drawing should show:
Outcome: Write the balanced equation a particulate diagram represents, with smallest whole-number coefficients.
A diagram shows nitrogen and hydrogen molecules becoming ammonia — here it is.
A student writes exactly what the boxes count: 2N₂ + 6H₂ → 4NH₃.
Every atom balances.
The AP answer reads N₂ + 3H₂ → 2NH₃ — and the student's unreduced line earns no point.
Both equations describe the same boxes.
What did the student miss?
Move 1: read each particle's label and write its formula.
Move 2: count the particles of each type on each side of the arrow, and write the counts as first coefficients.
Move 3: divide every count by the largest whole number that divides them all.
Coefficients state a ratio between the species, not a tally of drawn circles — that is why move 3 exists.
Q: A diagram shows 2 N₂ and 6 H₂ becoming 4 NH₃, all gases. What balanced equation, with smallest whole-number coefficients, does the diagram represent?
Read the labels and write the formulas
N₂ and H₂ before; NH₃ after
Count the particles and write the counts as first coefficients
2N₂ + 6H₂ → 4NH₃
Divide every count by the largest whole number that divides them all — here 2
N₂ + 3H₂ → 2NH₃
Check atoms, and add the state symbols
2 N and 6 H on each side
N₂(g) + 3H₂(g) → 2NH₃(g)
A physical change obeys the same three moves.
This diagram shows the same five molecules packed close as a liquid, then far apart as a gas.
The molecules themselves are unchanged, so this is a physical change — and the change still gets an equation.
Count and divide: five H₂O become five H₂O, so the equation is H₂O(l) → H₂O(g).
The state symbols carry the whole change.
Write the balanced equation a particulate diagram represents — reaction or physical change — dividing the particle counts down to the smallest whole-number coefficients.
Check — drawing task (pen and paper, self-marked)
The diagram shows a container before and after hydrogen peroxide, H₂O₂, breaks down. Hydrogen peroxide and water are liquids here; oxygen is a gas. Take a pen and paper and write the balanced equation, with smallest whole-number coefficients and state symbols, that the diagram represents. When you are done, select Check to compare your answer with the model answer.
Model answer shown to the student
Self-mark checklist — the drawing should show:
Check — multiple choice
The diagram shows a container of gases before and after a reaction. Which balanced equation, with smallest whole-number coefficients, does the diagram represent?
Feedback if chosen: This option is wrong — you copied the drawn tally. Coefficients state the ratio, so divide every count by the largest whole number that divides them all: 2.
Feedback if chosen: This option is wrong — the atoms do not balance: this line has 4 O on the left and only 3 O on the right.
Feedback if chosen: Correct. The counts 4, 2, 4 divide by 2 to give 2SO₂ + O₂ → 2SO₃, and 2 S and 6 O sit on each side.
Feedback if chosen: This option is wrong — you reversed the direction. The Before box holds SO₂ and O₂; the After box holds the SO₃.
Check — multiple choice
The diagram shows the same sample of bromine before and after a change, liquid first and gas after. Which equation, with smallest whole-number coefficients, does the diagram represent?
Feedback if chosen: This option is wrong — you copied the drawn tally. Five and five divide by 5: the equation reads Br₂(l) → Br₂(g).
Feedback if chosen: This option is wrong — you split the molecules. The After box shows the same joined Br₂ units, only farther apart.
Feedback if chosen: This option is wrong — the labels read Br₂: bromine's particles are two-atom molecules, not lone Br atoms.
Feedback if chosen: Correct. The molecules are unchanged, the counts divide down to 1 and 1, and the state symbols carry the whole change.
Outcome: Evaluate a particulate drawing and name its flaw: atoms not conserved, separated atoms, or joined pairs for a dissolved ionic compound.
A drawing claims to show sodium chloride dissolved in water — every Na circle sits glued to a Cl circle.
Touching circles mean one unit, so this picture shows intact NaCl units floating in the water.
You've already seen what the water actually did: the water pulled every unit apart into free Na⁺ and Cl⁻ ions.
The tidy drawing tells a false story.
Three flaws cover nearly every wrong particle drawing.
Flaw 1: an element's atom count changed across the arrow.
Flaw 2: a substance was drawn as loose separate atoms where the real substance exists as joined molecules.
Flaw 3: a dissolved ionic compound was drawn as joined pairs where the water actually freed the ions.
Read for the three flaws in that order.
State the flaw you find — or state that the drawing survives all three.
Evaluate a particulate representation and identify the flaw if present: atoms not conserved, separated atoms for intact molecules, or joined pairs for a dissolved ionic compound.
Check — multiple choice
The drawing shows a container before and after hydrogen and bromine react to form hydrogen bromide, HBr. Which statement about the drawing is correct?
Feedback if chosen: Correct. The Before box holds 4 H and 4 Br, but the After box holds only 3 of each — one of each atom vanished.
Feedback if chosen: This option is wrong — every unit in both boxes is drawn joined. The defect is the count: 4 of each atom before, 3 after.
Feedback if chosen: This option is wrong — nothing here is dissolved. The defect is the count: 4 of each atom before, 3 after.
Feedback if chosen: This option is wrong — count the atoms across the arrow: 4 H and 4 Br become 3 H and 3 Br, so atoms were lost.
Check — multiple choice
Oxygen exists as O₂ molecules. The drawing claims to show a sample of oxygen gas. Which statement about the drawing is correct?
Feedback if chosen: This option is wrong — there is no arrow to count across. The defect is the lone circles standing for a substance made of joined molecules.
Feedback if chosen: Correct. Oxygen's particles are joined O₂ molecules, so each unit must be two touching circles — lone circles tell a false story.
Feedback if chosen: This option is wrong — nothing here is dissolved. The defect is the lone circles standing for joined molecules.
Feedback if chosen: This option is wrong — the drawing shows free atoms, and oxygen gas contains none. Each O₂ molecule must be drawn as two touching circles.
Check — multiple choice
Potassium bromide is a soluble ionic compound. The drawing claims to show potassium bromide dissolved in water. Which statement about the drawing is correct?
Feedback if chosen: This option is wrong — there is no arrow to count across. The defect is the joined K–Br pairs in the water.
Feedback if chosen: This option is wrong — nothing here is drawn as loose atoms. The defect is the joined K–Br pairs in the water.
Feedback if chosen: This option is wrong — the water dissociates a dissolved soluble ionic compound, so the drawing should show free K⁺ and Br⁻ ions, not joined pairs.
Feedback if chosen: Correct. The water pulls every KBr unit apart, so the drawing should show free K⁺ and Br⁻ ions among the water molecules.
Check — multiple choice
Sodium fluoride is a soluble ionic compound. The drawing claims to show sodium fluoride dissolved in water. Which statement about the drawing is correct?
Feedback if chosen: This option is wrong — there is no arrow to count across, and no atoms are missing.
Feedback if chosen: This option is wrong — no joined Na–F pairs appear. Every ion floats free, which is exactly what dissociation produces.
Feedback if chosen: Correct. The water freed the ions, and the drawing shows exactly that: separate Na⁺ and F⁻ among the water molecules — it survives all three flaw checks.
Feedback if chosen: This option is wrong — the lone circles here are free ions of a dissolved ionic compound, and free is how the water actually leaves them.
Check — multiple choice
A car airbag inflates by the balanced equation 2NaN₃(s) → 2Na(s) + 3N₂(g). Each side shows 2 Na atoms and 6 N atoms. What do these matched atom counts demonstrate?
Feedback if chosen: This option is wrong — you kept the substances the same. NaN₃ is gone and Na and N₂ are new; it is the atoms, not the substances, that survive.
Feedback if chosen: Correct. Equal atom counts of every element on the two sides show that every atom is kept — which is conservation of mass.
Feedback if chosen: This option is wrong — you reached for charge. Charge conservation is shown by summing charges; matched atom counts demonstrate conservation of mass.
Feedback if chosen: This option is wrong — you read matched atom counts as matched amounts. The coefficients 2 and 3 say the amounts of Na and N₂ differ.
Check — multiple choice
Silane burns in oxygen: __SiH₄ + __O₂ → __SiO₂ + __H₂O. Which set of coefficients balances the equation with the smallest whole numbers?
Feedback if chosen: This option is wrong — you left oxygen unbalanced: 2 O atoms on the left, 4 on the right. The O₂ needs a coefficient of 2.
Feedback if chosen: Correct. Si: 1 = 1. H: 4 = 4. O: 2 × 2 = 4 on the left, 2 + 2 = 4 on the right — balanced, in the smallest whole numbers.
Feedback if chosen: This option is wrong — you counted 4 O atoms on the right and then wrote 4 O₂, which supplies 8. Each O₂ brings two O atoms, so 2 O₂ is enough.
Feedback if chosen: This option is wrong — the atoms balance, but every coefficient can be halved. A balanced equation uses the smallest whole numbers.
Check — multiple choice
A student proposes Ba + O₂ → BaO for barium metal forming barium oxide, BaO. The oxygen atoms are not conserved. Which corrected equation is right, using the smallest whole-number coefficients?
Feedback if chosen: Correct. Doubling the Ba and the BaO gives 2 Ba and 2 O atoms on each side — balanced by coefficients alone, in the smallest whole numbers.
Feedback if chosen: This option is wrong — you changed a subscript. BaO₂ is a different substance from the barium oxide the question names; balancing may only adjust coefficients.
Feedback if chosen: This option is wrong — you rewrote oxygen gas as single O atoms. Oxygen's formula is O₂; changing it changes the substance's identity.
Feedback if chosen: This option is wrong — the atoms balance, but every coefficient can be halved. Balance with the smallest whole numbers: 2Ba + O₂ → 2BaO.
Check — multiple choice
Solutions of strontium chloride, SrCl₂(aq), and lithium sulfate, Li₂SO₄(aq), are mixed, and solid SrSO₄ forms. (Lithium compounds are soluble — supplied.) Which ions are the spectator ions?
Feedback if chosen: This option is wrong — you picked the ions that react. Sr²⁺ and SO₄²⁻ join to build the solid, so they take part in the change.
Feedback if chosen: This option is wrong — you grouped the two positive ions. The Sr²⁺ leaves the solution to build the solid, so it is no spectator.
Feedback if chosen: This option is wrong — you grouped the two negative ions. The SO₄²⁻ leaves the solution to build the solid, so it is no spectator.
Feedback if chosen: Correct. Li⁺ and Cl⁻ stay dissolved and unchanged on both sides of the complete ionic equation — they take no part in the change.
Check — multiple choice
When Cu(NO₃)₂(aq) and Na₂CO₃(aq) are mixed, solid copper(II) carbonate forms. (CuCO₃ is insoluble — supplied.) Which is the correct net ionic equation?
Feedback if chosen: Correct. Only the ions that build the solid remain, and both sides balance: atoms match, and the charges sum to zero on each side.
Feedback if chosen: This option is wrong — you wrote the molecular equation. A net ionic equation keeps only the species that change.
Feedback if chosen: This option is wrong — you left the NO₃⁻ uncanceled. An ion that appears unchanged on both sides is a spectator and is removed.
Feedback if chosen: This option is wrong — you wrote the spectators as the reaction. Na⁺ and NO₃⁻ stay dissolved and separate; nothing joins them together.
Check — multiple choice
Mixing Zn(NO₃)₂(aq) with NaOH(aq) produces solid zinc hydroxide. (Zn(OH)₂ is insoluble — supplied.) Which equation shows only the species that change?
Feedback if chosen: This option is wrong — you dropped the coefficient 2. With one OH⁻ the hydrogen and oxygen atoms don't balance, and the charges sum to 1+ on the left and 0 on the right.
Feedback if chosen: This option is wrong — you gave the molecular equation. The context demands the net ionic form, which keeps only the species that change.
Feedback if chosen: Correct. Zn²⁺ and 2OH⁻ build the solid; atoms balance, and the charge sums to zero on each side. The Na⁺ and NO₃⁻ spectators are gone.
Feedback if chosen: This option is wrong — you precipitated the wrong pair. Nitrates are always soluble, so Zn²⁺ and NO₃⁻ stay dissolved; it is the hydroxide that forms the solid.
Check — multiple choice
Iodine vapor and chlorine gas react by I₂(g) + Cl₂(g) → 2ICl(g). The shared box shows the mixture before the reaction. Which drawing shows the box after the reaction goes to completion?
Feedback if chosen: This option is wrong — you read the coefficient 2 as the final count. Three I₂ molecules bring six I atoms, so six ICl molecules must form.
Feedback if chosen: Correct. The three I₂ and three Cl₂ supply 6 I and 6 Cl atoms, and six ICl molecules hold exactly 6 I and 6 Cl — every atom is conserved and every molecule is intact.
Feedback if chosen: This option is wrong — you kept the Cl₂. Three I₂ need exactly three Cl₂ (a 1:1 relationship), so all of the Cl₂ is used up.
Feedback if chosen: This option is wrong — you drew the atoms separated. The atoms regroup into intact ICl molecules; they never float free.
Check — multiple choice
The boxes show a gas before and after it decomposes. Which balanced equation, in smallest whole numbers, does the drawing represent?
Feedback if chosen: This option is wrong — you copied one of each formula without counting atoms: one O on the left cannot make an O₂ on the right.
Feedback if chosen: This option is wrong — you copied the drawn counts (4, 4, 2) without reducing them. Coefficients state the smallest whole-number relationship: 2, 2, 1.
Feedback if chosen: This option is wrong — you wrote single O atoms. The drawing shows the oxygen leaving as O₂ molecules, and O₂ is oxygen's formula as a gas.
Feedback if chosen: Correct. The drawn counts 4 : 4 : 2 reduce to 2 : 2 : 1, and 2N₂O → 2N₂ + O₂ balances: 4 N and 2 O atoms on each side.
Check — multiple choice
A student draws a particle-level picture of sodium iodide, NaI, dissolved in water, shown in the figure. Which evaluation of the drawing is correct?
Feedback if chosen: This option is wrong — you kept the ions paired. Water pulls the lattice apart into separate Na⁺ and I⁻ ions; nothing stays joined.
Feedback if chosen: This option is wrong — you faulted the count. Any number of NaI units may be shown; the flaw is that each unit is drawn as a touching pair.
Feedback if chosen: Correct. A dissolved ionic compound is drawn as separate aqueous ions — Na⁺ and I⁻ apart from each other, each surrounded by water.
Feedback if chosen: This option is wrong — you fused the particles. Dissolving separates the ions; it never clumps them into one cluster.
6 lessons
Outcome: Explain why a known reactant amount fixes the product amounts — and a known product amount fixes the reactant amounts.
A gas boiler will burn 400 kg of methane over the coming winter.
A regulator asks the engineer for the mass of carbon dioxide the boiler will release.
The engineer cannot burn the winter's gas early to find out.
Yet the engineer calculates the answer now — and the winter proves the calculation right.
How can arithmetic reach the mass of a substance before the substance forms?
The balanced equation for the burning is CH₄ + 2O₂ → CO₂ + 2H₂O.
You've already seen that a chemical change only rearranges atoms: the change creates no atom and destroys no atom.
Every carbon atom that enters the boiler inside a methane molecule leaves it inside a carbon dioxide molecule.
So whoever counts the carbon going in has already counted the carbon coming out.
This means the mass of methane burned fixes the mass of carbon dioxide released.
The logic reverses too: a measured product mass fixes the reactant masses that were consumed.
Every calculation in the rest of this unit stands on this one fact.
Explain that known reactant amounts fix product amounts, and known product amounts fix reactant amounts, because every element's atom count is identical before and after the change.
Free-response question
Source note (not shown as part of the question): authored to archetype: conservation-of-atoms explanation (EK 4.5.A.1)
Barbecue charcoal is essentially carbon. A cook burns 3.0 kg of charcoal completely: C(s) + O₂(g) → CO₂(g). Before lighting it, an engineer states the exact mass of CO₂ the burn will release. Write your answer on paper, one short sentence per step of your reasoning, then select Check to compare it with the model response.
(a) Explain why the measured mass of charcoal fixes the mass of CO₂ released, before the burn happens. [2 points]
✓ Model response (the key)
(a)
A chemical change creates no atom and destroys no atom.
So every carbon atom in the 3.0 kg of charcoal ends up inside one CO₂ molecule.
Counting the carbon going in has already counted the carbon coming out, so the charcoal mass fixes the CO₂ mass.
Scoring rubric — 2 points available
Check — multiple choice
A kerosene heater's flame burns all evening, consuming the fuel completely, and the fuel level falls. A student says: "The carbon in the kerosene disappears as the fuel burns away." Which statement is correct?
Feedback if chosen: This option is wrong — a chemical change destroys no atom. The carbon atoms leave the heater inside CO₂ molecules.
Feedback if chosen: This option is wrong — atoms never turn into heat. The burning releases heat, and every carbon atom still leaves inside a CO₂ molecule.
Feedback if chosen: Correct. The atoms survive the change — they leave the heater inside CO₂ molecules, invisible in the air.
Feedback if chosen: This option is wrong — a complete burn leaves no soot. Every carbon atom leaves the heater inside a CO₂ molecule, invisible in the air.
Check — multiple choice
A sealed test rig collects and measures all the CO₂ a known fuel releases as it burns completely. Can the mass of fuel that burned be calculated from the measured CO₂ mass?
Feedback if chosen: This option is wrong — the atom counts match in both directions, so the calculation goes backwards just as well as forwards.
Feedback if chosen: Correct. Every carbon atom in the measured CO₂ came out of the fuel, so the product amount fixes the fuel amount.
Feedback if chosen: This option is wrong — a balance is one way to get the mass, but the atom counts fix it by calculation too.
Feedback if chosen: This option is wrong — the atom-counting logic holds for any known fuel, not one particular gas.
Outcome: State the mole ratio between any pair of species in a balanced equation.
The balanced equation for burning hydrogen reads 2H₂ + O₂ → 2H₂O.
The coefficients count particles: two hydrogen molecules consume one oxygen molecule.
No laboratory counts particles one by one.
A torch will burn 10.0 mol of hydrogen from a tank.
The fitter must order the oxygen that exactly matches — no shortfall, no waste.
Can coefficients written for molecules speak for moles by the tank-load?
You've already seen that a mole is a fixed count of particles: 6.022 × 10²³ of them.
Multiply two molecules and one molecule by that same fixed count: the 2-to-1 ratio survives untouched.
So in 2H₂ + O₂ → 2H₂O, hydrogen pairs with oxygen at 2 mol to 1 mol.
So the torch's 10.0 mol of hydrogen consumes exactly 5.0 mol of oxygen.
Each such pairing is called a mole ratio.
A mole ratio exists between every pair of species in the equation — the table lists all three pairings.
| Pair of species | Mole ratio |
|---|---|
| H₂ and O₂ | 2 mol of H₂ to 1 mol of O₂ |
| H₂ and H₂O | 2 mol of H₂ to 2 mol of H₂O |
| O₂ and H₂O | 1 mol of O₂ to 2 mol of H₂O |
A mole ratio works like an exchange rate between two species: every calculation in this sequence passes through one.
State that the coefficients of a balanced equation give the mole ratio between every pair of species in the reaction.
Check — multiple choice
In a balanced equation, what does a mole ratio state?
Feedback if chosen: Correct. The coefficients pair every two species in moles — that pairing is the mole ratio.
Feedback if chosen: This option is wrong — coefficients count moles of particles, not grams. Masses enter later, through the mole equations.
Feedback if chosen: This option is wrong — that number is Avogadro's number, the count inside one mole. A mole ratio pairs two species.
Feedback if chosen: This option is wrong — a balanced equation carries no temperature. The coefficients state amounts.
Check — multiple choice
Hexane burns: 2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O. What is the mole ratio of C₆H₁₄ to O₂?
Feedback if chosen: This option is wrong — you flipped the pairing. The coefficient 2 belongs to C₆H₁₄ and the 19 belongs to O₂.
Feedback if chosen: This option is wrong — you paired C₆H₁₄ with CO₂'s coefficient. The question asks for the O₂ pairing: 2 mol to 19 mol.
Feedback if chosen: This option is wrong — you paired the two products' coefficients. The question asks about C₆H₁₄ and O₂: 2 mol to 19 mol.
Feedback if chosen: Correct. The coefficients read 2 for C₆H₁₄ and 19 for O₂, so the pair is 2 mol to 19 mol.
Check — multiple choice
Sodium burns: 4Na + O₂ → 2Na₂O. What is the mole ratio of Na to Na₂O?
Feedback if chosen: This option is wrong — you flipped the pairing. The coefficient 4 belongs to Na and the 2 belongs to Na₂O.
Feedback if chosen: Correct. The coefficients read 4 for Na and 2 for Na₂O — a mole ratio exists between every pair, products included.
Feedback if chosen: This option is wrong — you paired Na with O₂'s coefficient. Na₂O's own coefficient is 2.
Feedback if chosen: This option is wrong — you ignored the coefficients. They read 4 for Na and 2 for Na₂O.
Check — multiple choice
A student reads 2H₂ + O₂ → 2H₂O and says: "2 g of hydrogen react with 1 g of oxygen." Which statement is correct?
Feedback if chosen: This option is wrong — a coefficient counts particles, scaled up to moles. Grams enter only through each substance's own molar mass.
Feedback if chosen: This option is wrong — the two units cannot both fit: an H₂ molecule and an O₂ molecule have different masses.
Feedback if chosen: Correct. Coefficients speak in moles; turning moles into grams needs each substance's own molar mass.
Feedback if chosen: This option is wrong — you swapped the numbers but kept the wrong unit. Coefficients pair moles, not grams.
Outcome: Predict how changing one species' amount changes every other amount — direction and factor, before any arithmetic.
One camping stove burns methane all evening: CH₄ + 2O₂ → CO₂ + 2H₂O.
A second, identical stove burns twice as much methane in the same evening.
The second stove clearly consumes more oxygen and releases more carbon dioxide.
But does "more" mean a little more, or exactly twice as much?
Every reaction in this sequence finishes completely, with more than enough of every other reactant present.
Under that condition, all the amounts move together in the coefficients' ratio.
The table shows the same burning at three scales.
| Methane burned | Oxygen consumed | Carbon dioxide formed |
|---|---|---|
| 1 mol | 2 mol | 1 mol |
| 2 mol | 4 mol | 2 mol |
| ½ mol | 1 mol | ½ mol |
Twice the methane consumes twice the oxygen and forms twice the carbon dioxide.
Half the methane consumes half the oxygen and forms half the carbon dioxide.
The ratio never bends: 1 to 2 stays 1 to 2 at any scale.
Say the direction and the size of the change before any calculation — that habit catches wrong answers early.
Predict qualitatively how changing the amount of one species changes the amounts of the others, in the ratio fixed by the coefficients.
Check — multiple choice
A welding torch burns acetylene: 2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O. The burning finishes completely, with more than enough oxygen available. The torch burns three times as much acetylene. How much oxygen does it consume?
Feedback if chosen: Correct. All the amounts move together: three times the acetylene consumes three times the oxygen, still in the 2-to-5 ratio.
Feedback if chosen: This option is wrong — you scaled by the coefficient instead of the change. The coefficients set the ratio; the factor of three scales every amount.
Feedback if chosen: This option is wrong — more fuel burning must consume more oxygen. Every amount scales by the same factor of three.
Feedback if chosen: This option is wrong — you multiplied the factor by the coefficient. The factor of three alone scales every amount.
Check — multiple choice
A spirit burner burns ethanol: C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O. A student says: "Doubling the ethanol changes the mole ratio in which ethanol and oxygen react." Which statement is correct?
Feedback if chosen: This option is wrong — both amounts double at once, and doubling both sides of a ratio leaves the ratio unchanged.
Feedback if chosen: This option is wrong — twice the ethanol burning consumes twice the oxygen. What never changes is the ratio between them.
Feedback if chosen: This option is wrong — nothing flips the pairing. The coefficients still read 1 for ethanol and 3 for oxygen at any scale.
Feedback if chosen: Correct. The amounts scale together, so 1 mol to 3 mol holds at every scale — the ratio never bends.
Check — multiple choice
A camping stove burns butane: 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O, finishing completely with more than enough oxygen. The camper wants the stove to release half as much CO₂. What must change?
Feedback if chosen: This option is wrong — you scaled the wrong way. Twice the butane forms twice the CO₂; the target is half.
Feedback if chosen: Correct. The scaling applies in both directions: half the CO₂ means half of every amount, so burn half the butane.
Feedback if chosen: This option is wrong — the oxygen here is already more than enough, and the butane burned is what fixes the CO₂ formed.
Feedback if chosen: This option is wrong — the CO₂ formed follows the butane burned, in the ratio the coefficients fix.
Outcome: Calculate the moles of one species consumed or produced from the moles of another, using the mole ratio.
A camping supplier fills cylinders with propane.
One burner consumes 0.30 mol of propane per hour.
Air must supply oxygen at the matching rate.
The balanced equation reads C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.
The mole ratio pairs 1 mol of propane with 5 mol of oxygen.
So exactly how many moles of oxygen does 0.30 mol of propane consume?
Multiply the known moles by the mole ratio, written as a fraction: the wanted species on top, the known species underneath.
The ratio's own units cancel the known species and leave the wanted one.
Q: The equation is C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. How many moles of O₂ does 0.30 mol of C₃H₈ consume?
Write down the values in the question
moles of C₃H₈ = 0.30 mol
mole ratio: 5 mol of O₂ to 1 mol of C₃H₈
Write down the equation
moles of O₂ = moles of C₃H₈ × (5 mol O₂ / 1 mol C₃H₈)
Substitute in the values, and calculate
moles of O₂ = 0.30 × 5
moles of O₂ = 1.5 mol
The same one-step routine works in both directions — flip the fraction when the known and the wanted swap places.
Q: The equation is C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. How many moles of C₃H₈ must burn to produce 1.2 mol of CO₂?
Write down the values in the question
moles of CO₂ = 1.2 mol
mole ratio: 1 mol of C₃H₈ to 3 mol of CO₂
Write down the equation
moles of C₃H₈ = moles of CO₂ × (1 mol C₃H₈ / 3 mol CO₂)
Substitute in the values, and calculate
moles of C₃H₈ = 1.2 × 1/3
moles of C₃H₈ = 0.40 mol
The answer's unit names the wanted species, in moles.
Calculate the moles of one species consumed or produced from the moles of another using the mole ratio given by the coefficients.
Check — numeric entry
Phosphorus burns: P₄ + 5O₂ → P₄O₁₀. The burning finishes completely, with more than enough oxygen present. How many moles of O₂ does 0.20 mol of P₄ consume? Give your answer to 2 significant figures.
Tolerance: 0.01
Feedback when correct: Correct. moles of O₂ = 0.20 × (5/1) = 1.0 mol.
Anticipated wrong answers (feedback shown):
| 0.040 | You divided by the ratio instead of multiplying. The wanted species, O₂, goes on top: 0.20 × (5/1) = 1.0 mol. |
| 0.04 | You divided by the ratio instead of multiplying. The wanted species, O₂, goes on top: 0.20 × (5/1) = 1.0 mol. |
| 0.20 | You skipped the mole ratio. Each mole of P₄ consumes 5 mol of O₂: 0.20 × 5 = 1.0 mol. |
Worked solution revealed after a second miss
Write down the values in the question
moles of P₄ = 0.20 mol
mole ratio: 5 mol of O₂ to 1 mol of P₄
Write down the equation
moles of O₂ = moles of P₄ × (5 mol O₂ / 1 mol P₄)
Substitute in the values, and calculate
moles of O₂ = 0.20 × 5
moles of O₂ = 1.0 mol
Check — numeric entry
Heating potassium chlorate releases oxygen: 2KClO₃ → 2KCl + 3O₂. How many moles of KClO₃ must break down to release 0.75 mol of O₂? Give your answer to 2 significant figures.
Tolerance: 0.005
Feedback when correct: Correct. moles of KClO₃ = 0.75 × (2/3) = 0.50 mol.
Anticipated wrong answers (feedback shown):
| 1.125 | You flipped the mole ratio. The wanted species, KClO₃, goes on top: 0.75 × (2/3) = 0.50 mol. |
| 1.13 | You flipped the mole ratio. The wanted species, KClO₃, goes on top: 0.75 × (2/3) = 0.50 mol. |
| 0.75 | You skipped the mole ratio. 2 mol of KClO₃ releases 3 mol of O₂: 0.75 × (2/3) = 0.50 mol. |
Worked solution revealed after a second miss
Write down the values in the question
moles of O₂ = 0.75 mol
mole ratio: 2 mol of KClO₃ to 3 mol of O₂
Write down the equation
moles of KClO₃ = moles of O₂ × (2 mol KClO₃ / 3 mol O₂)
Substitute in the values, and calculate
moles of KClO₃ = 0.75 × 2/3
moles of KClO₃ = 0.50 mol
Outcome: Calculate the mass of product a reactant mass forms, by chaining mass → moles → mole ratio → moles → mass.
A cylinder delivers 10.0 g of propane to a burner: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.
A researcher must state the mass of carbon dioxide the burning releases.
The mole ratio speaks only in moles: 1 mol of propane to 3 mol of carbon dioxide.
No balance on Earth displays moles.
You've already seen the two mole equations: n = m/M turns a mass into moles, and m = nM turns moles back into a mass.
Can those two skills chain with the mole ratio into one route from grams to grams?
First, n = m/M turns the known mass into moles.
Then the mole ratio carries the amount across to the wanted substance.
Last, m = nM turns the wanted moles back into a mass.
Masses never cross between substances directly — the mole ratio is the only bridge.
Write the unit on every line: the units expose a wrong turn before it costs points.
Q: The equation is C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. What mass of CO₂ forms when 10.0 g of C₃H₈ burns completely? (molar masses: C₃H₈ 44.09 g/mol, CO₂ 44.01 g/mol)
Write down the values in the question
m of C₃H₈ = 10.0 g
M of C₃H₈ = 44.09 g/mol
M of CO₂ = 44.01 g/mol
mole ratio: 3 mol of CO₂ to 1 mol of C₃H₈
Turn the known mass into moles
n = m/M
n of C₃H₈ = 10.0 / 44.09
n of C₃H₈ = 0.2268 mol
Carry the moles across the mole ratio
n of CO₂ = 0.2268 × (3/1)
n of CO₂ = 0.6804 mol
Turn the wanted moles into a mass
m = nM
m of CO₂ = 0.6804 × 44.01
m of CO₂ = 29.9 g
Calculate the mass of product formed from the mass of a reactant by chaining mass → moles → mole ratio → moles → mass.
Check — multiple choice
Which equation turns a mass into moles?
Feedback if chosen: This option is wrong — m = nM is the reverse conversion: it turns moles into a mass.
Feedback if chosen: This option is wrong — multiplying the mass by the molar mass moves away from moles. Divide instead: n = m/M.
Feedback if chosen: This option is wrong — you flipped the fraction. The mass goes on top: n = m/M.
Feedback if chosen: Correct. Dividing the mass by the molar mass gives the moles: n = m/M.
Check — numeric entry
A kiln heats limestone, which breaks down completely: CaCO₃(s) → CaO(s) + CO₂(g). What mass of CO₂ does 25.0 g of CaCO₃ release? Give your answer to 3 significant figures. (molar masses: CaCO₃ 100.09 g/mol, CO₂ 44.01 g/mol)
Tolerance: 0.05
Feedback when correct: Correct. n of CaCO₃ = 25.0/100.09 = 0.2498 mol; the ratio is 1 to 1, so n of CO₂ = 0.2498 mol; m = 0.2498 × 44.01 = 11.0 g.
Anticipated wrong answers (feedback shown):
| 25.0 | You carried the mass straight across. Masses never cross between substances directly — turn the mass into moles, cross the mole ratio, then turn back. |
| 0.250 | You stopped at the moles of CaCO₃. Finish the chain: cross the 1-to-1 ratio, then m = nM = 0.2498 × 44.01 = 11.0 g. |
Worked solution revealed after a second miss
Write down the values in the question
m of CaCO₃ = 25.0 g
M of CaCO₃ = 100.09 g/mol
M of CO₂ = 44.01 g/mol
mole ratio: 1 mol of CO₂ to 1 mol of CaCO₃
Turn the known mass into moles
n = m/M
n of CaCO₃ = 25.0 / 100.09
n of CaCO₃ = 0.2498 mol
Carry the moles across the mole ratio
n of CO₂ = 0.2498 × (1/1)
n of CO₂ = 0.2498 mol
Turn the wanted moles into a mass
m = nM
m of CO₂ = 0.2498 × 44.01
m of CO₂ = 11.0 g
Check — numeric entry
A camping burner holds 6.41 g of methanol, which burns completely: 2CH₃OH + 3O₂ → 2CO₂ + 4H₂O. What mass of water forms? Give your answer to 3 significant figures. (molar masses: CH₃OH 32.04 g/mol, H₂O 18.02 g/mol)
Tolerance: 0.02
Feedback when correct: Correct. n of CH₃OH = 6.41/32.04 = 0.2001 mol; n of H₂O = 0.2001 × (4/2) = 0.4001 mol; m = 0.4001 × 18.02 = 7.21 g.
Anticipated wrong answers (feedback shown):
| 3.61 | You skipped the mole ratio. 2 mol of CH₃OH forms 4 mol of H₂O, so multiply the moles by 4/2 before m = nM. |
| 1.80 | You flipped the mole ratio. The wanted species, H₂O, goes on top: multiply by 4/2, not 2/4. |
| 12.8 | You multiplied the mass by the ratio directly. Masses never cross the ratio — turn the mass into moles first. |
Worked solution revealed after a second miss
Write down the values in the question
m of CH₃OH = 6.41 g
M of CH₃OH = 32.04 g/mol
M of H₂O = 18.02 g/mol
mole ratio: 4 mol of H₂O to 2 mol of CH₃OH
Turn the known mass into moles
n = m/M
n of CH₃OH = 6.41 / 32.04
n of CH₃OH = 0.2001 mol
Carry the moles across the mole ratio
n of H₂O = 0.2001 × (4/2)
n of H₂O = 0.4001 mol
Turn the wanted moles into a mass
m = nM
m of H₂O = 0.4001 × 18.02
m of H₂O = 7.21 g
Outcome: Calculate the reactant mass a target product mass demands, by walking the same chain backwards.
A demonstration will burn iron wool in pure oxygen: 4Fe + 3O₂ → 2Fe₂O₃.
The plan needs 25.0 g of iron(III) oxide — the product — at the end.
This time the unknown sits at the start: what mass of oxygen must the burning consume?
The chain is the same three conversions.
Which way do you walk it?
First, n = m/M turns the target product mass into moles.
Then the mole ratio carries the moles across — this time the product plays the known species.
Last, m = nM turns the reactant moles into the mass to weigh out.
Forward or backwards, the route never changes shape: mass → moles → mole ratio → moles → mass.
Only the choice of known species sets the direction.
Q: The equation is 4Fe + 3O₂ → 2Fe₂O₃. What mass of O₂ must be consumed to form 25.0 g of Fe₂O₃? (molar masses: Fe₂O₃ 159.70 g/mol, O₂ 32.00 g/mol)
Write down the values in the question
m of Fe₂O₃ = 25.0 g
M of Fe₂O₃ = 159.70 g/mol
M of O₂ = 32.00 g/mol
mole ratio: 3 mol of O₂ to 2 mol of Fe₂O₃
Turn the target mass into moles
n = m/M
n of Fe₂O₃ = 25.0 / 159.70
n of Fe₂O₃ = 0.1565 mol
Carry the moles across the mole ratio
n of O₂ = 0.1565 × (3/2)
n of O₂ = 0.2348 mol
Turn the reactant moles into a mass
m = nM
m of O₂ = 0.2348 × 32.00
m of O₂ = 7.51 g
Calculate the mass of reactant required to form a target mass of product by working the mass → moles → mole ratio → moles → mass chain in reverse.
Check — numeric entry
An aluminum smelter breaks down aluminum oxide: 2Al₂O₃ → 4Al + 3O₂. What mass of Al₂O₃ must break down to produce 10.8 g of aluminum? Give your answer to 3 significant figures. (molar masses: Al 26.98 g/mol, Al₂O₃ 101.96 g/mol)
Tolerance: 0.05
Feedback when correct: Correct. n of Al = 10.8/26.98 = 0.4003 mol; n of Al₂O₃ = 0.4003 × (2/4) = 0.2001 mol; m = 0.2001 × 101.96 = 20.4 g.
Anticipated wrong answers (feedback shown):
| 81.6 | You flipped the mole ratio. The wanted species, Al₂O₃, goes on top: multiply by 2/4, not 4/2. |
| 40.8 | You skipped the mole ratio. 4 mol of Al comes from 2 mol of Al₂O₃, so halve the moles before m = nM. |
| 5.40 | You multiplied the mass by the ratio directly. Masses never cross the ratio — turn the mass into moles first. |
Worked solution revealed after a second miss
Write down the values in the question
m of Al = 10.8 g
M of Al = 26.98 g/mol
M of Al₂O₃ = 101.96 g/mol
mole ratio: 2 mol of Al₂O₃ to 4 mol of Al
Turn the target mass into moles
n = m/M
n of Al = 10.8 / 26.98
n of Al = 0.4003 mol
Carry the moles across the mole ratio
n of Al₂O₃ = 0.4003 × (2/4)
n of Al₂O₃ = 0.2001 mol
Turn the wanted moles into a mass
m = nM
m of Al₂O₃ = 0.2001 × 101.96
m of Al₂O₃ = 20.4 g
Check — multiple choice
The equation is 2Ca + O₂ → 2CaO. To find the mass of calcium needed to make 14.0 g of calcium oxide, a student multiplies 14.0 g by the mole ratio 2/2 and answers 14.0 g. What did the student skip?
Feedback if chosen: Correct. The mole ratio pairs moles, never grams — the chain is mass → moles → mole ratio → moles → mass.
Feedback if chosen: This option is wrong — the equation is already balanced. The skipped step is the mass-to-moles conversion.
Feedback if chosen: This option is wrong — the 2/2 ratio reads the same either way. The skipped step is the mass-to-moles conversion.
Feedback if chosen: This option is wrong — the question asks for the calcium mass alone. The skipped step is the mass-to-moles conversion.
The next two questions do not name the chain's direction — read each question and choose the direction yourself.
Check — numeric entry
Potassium burns: 4K + O₂ → 2K₂O. What mass of K₂O forms when 7.82 g of potassium burns completely? Give your answer to 3 significant figures. (molar masses: K 39.10 g/mol, K₂O 94.20 g/mol)
Tolerance: 0.03
Feedback when correct: Correct. n of K = 7.82/39.10 = 0.200 mol; n of K₂O = 0.200 × (2/4) = 0.100 mol; m = 0.100 × 94.20 = 9.42 g.
Anticipated wrong answers (feedback shown):
| 18.8 | You skipped the mole ratio. 4 mol of K forms 2 mol of K₂O, so halve the moles before m = nM. |
| 37.7 | You flipped the mole ratio. The wanted species, K₂O, goes on top: multiply by 2/4, not 4/2. |
| 3.91 | You multiplied the mass by the ratio directly. Masses never cross the ratio — turn the mass into moles first. |
Worked solution revealed after a second miss
Write down the values in the question
m of K = 7.82 g
M of K = 39.10 g/mol
M of K₂O = 94.20 g/mol
mole ratio: 2 mol of K₂O to 4 mol of K
Turn the known mass into moles
n = m/M
n of K = 7.82 / 39.10
n of K = 0.200 mol
Carry the moles across the mole ratio
n of K₂O = 0.200 × (2/4)
n of K₂O = 0.100 mol
Turn the wanted moles into a mass
m = nM
m of K₂O = 0.100 × 94.20
m of K₂O = 9.42 g
Check — numeric entry
Carbon disulfide burns: CS₂ + 3O₂ → CO₂ + 2SO₂. What mass of CS₂ must burn to form 12.8 g of SO₂? Give your answer to 3 significant figures. (molar masses: SO₂ 64.06 g/mol, CS₂ 76.13 g/mol)
Tolerance: 0.03
Feedback when correct: Correct. n of SO₂ = 12.8/64.06 = 0.1998 mol; n of CS₂ = 0.1998 × (1/2) = 0.0999 mol; m = 0.0999 × 76.13 = 7.61 g.
Anticipated wrong answers (feedback shown):
| 30.4 | You flipped the mole ratio. The wanted species, CS₂, goes on top: multiply by 1/2, not 2/1. |
| 15.2 | You skipped the mole ratio. 1 mol of CS₂ forms 2 mol of SO₂, so halve the moles before m = nM. |
| 6.40 | You multiplied the mass by the ratio directly. Masses never cross the ratio — turn the mass into moles first. |
Worked solution revealed after a second miss
Write down the values in the question
m of SO₂ = 12.8 g
M of SO₂ = 64.06 g/mol
M of CS₂ = 76.13 g/mol
mole ratio: 1 mol of CS₂ to 2 mol of SO₂
Turn the target mass into moles
n = m/M
n of SO₂ = 12.8 / 64.06
n of SO₂ = 0.1998 mol
Carry the moles across the mole ratio
n of CS₂ = 0.1998 × (1/2)
n of CS₂ = 0.0999 mol
Turn the wanted moles into a mass
m = nM
m of CS₂ = 0.0999 × 76.13
m of CS₂ = 7.61 g
9 lessons
Outcome: State that the limiting reactant is the reactant completely consumed first, and that reactants left over are in excess.
A student lowers a 1 L glass jar over a burning candle.
The flame burns steadily for a few seconds.
Then the flame gutters out.
Plenty of wax remains.
The sealed jar held only a small amount of oxygen.
The flame used up that oxygen.
Why does the leftover wax not keep the flame burning?
The burning consumed two reactants: the wax and the oxygen.
The oxygen was completely consumed first.
At that moment the burning stopped, whatever amount of wax remained.
The reactant completely consumed first is called the limiting reactant.
The reactants left over are said to be in excess.
In the jar, the oxygen is the limiting reactant, and the wax is in excess.
The limiting reactant fixes how much product forms.
Every yield calculation in the rest of this unit starts by naming it.
State that the limiting reactant is the reactant completely consumed first, and that reactants left over are in excess.
Check — multiple choice
A welding torch burns acetylene gas with oxygen from a tank. Mid-job the oxygen tank empties, and the flame goes out while acetylene still flows. Which reactant is the limiting reactant?
Feedback if chosen: This option is wrong — you picked the reactant that remains. Acetylene still flows after the flame dies, so the acetylene is in excess.
Feedback if chosen: Correct. The oxygen was completely consumed first. The reaction stopped at that moment, so the oxygen is the limiting reactant.
Feedback if chosen: This option is wrong — the acetylene keeps flowing after the flame goes out, so only the oxygen was completely consumed.
Feedback if chosen: This option is wrong — one reactant was completely consumed first: the oxygen. The oxygen is the limiting reactant.
Check — multiple choice
A student watches the candle-jar experiment and claims the flame went out because the candle used up all its wax. Which statement is correct?
Feedback if chosen: This option is wrong — you trusted the claim over the jar. Plenty of wax remains after the flame dies, so the wax was not consumed first.
Feedback if chosen: This option is wrong — you assumed both reactants were used up at the same moment. The leftover wax shows only the oxygen was completely consumed.
Feedback if chosen: This option is wrong — you kept the reaction going past the point where one reactant was gone. A reaction stops the moment one reactant is completely consumed.
Feedback if chosen: Correct. The wax is still there. The oxygen in the jar was completely consumed first, so the oxygen — the limiting reactant — ended the burning.
Check — multiple choice
Zinc powder reacts with sulfur by Zn + S → ZnS. The reaction stops when all the sulfur is consumed, and zinc remains. Which reactant is limiting?
Feedback if chosen: This option is wrong — zinc remains when the reaction stops, so the zinc is in excess.
Feedback if chosen: Correct. The sulfur is completely consumed first, so the sulfur is the limiting reactant.
Feedback if chosen: This option is wrong — you named the product. ZnS is formed by the reaction; the limiting reactant is one of the starting substances.
Feedback if chosen: This option is wrong — the reaction stopped because one reactant was completely consumed: the sulfur.
Check — multiple choice
Charcoal burns in a sealed container of air by C + O₂ → CO₂. The burning stops when all the O₂ is consumed, and charcoal remains. Which reactant is in excess?
Feedback if chosen: This option is wrong — the O₂ was completely consumed, so the O₂ is the limiting reactant, not the excess.
Feedback if chosen: This option is wrong — you named the product. CO₂ is made by the burning; the reactants are the carbon and the O₂.
Feedback if chosen: Correct. Charcoal is left over when the burning stops, so the carbon is in excess.
Feedback if chosen: This option is wrong — one reactant is left over: the charcoal. A leftover reactant is in excess.
Check — multiple choice
Silver reacts with sulfur by 2Ag + S → Ag₂S. When the reaction stops, unreacted silver remains and no sulfur is left. Which reactant was the limiting reactant?
Feedback if chosen: Correct. No sulfur is left — the sulfur was completely consumed first, so the sulfur was the limiting reactant.
Feedback if chosen: This option is wrong — silver remains when the reaction stops, so the silver is in excess.
Feedback if chosen: This option is wrong — you named the product. Ag₂S is formed by the reaction; the limiting reactant is a starting substance.
Feedback if chosen: This option is wrong — only the sulfur was completely consumed; the silver is left over.
Outcome: Explain why the limiting reactant caps the amounts of products formed: the reaction stops when it is used up, whatever amounts of the other reactants remain.
A sealed box holds 4 molecules of carbon monoxide, CO, and 3 molecules of oxygen, O₂.
The two gases react by the equation 2CO + O₂ → 2CO₂.
The reaction makes 4 molecules of CO₂, then stops.
One molecule of O₂ sits in the box untouched.
A student pumps in 5 more molecules of O₂ to push the reaction further.
Not one extra molecule of CO₂ forms.
Why does extra O₂ make no extra product?
Every molecule of CO₂ is built from one molecule of CO.
When the last molecule of CO reacts, the building material for CO₂ is gone.
The reaction stops at that moment.
Adding more O₂ supplies more of the wrong ingredient.
So the limiting reactant caps the amounts of every product.
This cap is why a yield calculation starts from the limiting reactant, never from the reactant in excess.
Explain why the limiting reactant caps the amounts of products: the reaction stops when the limiting reactant is used up, whatever amounts of the other reactants remain.
Check — multiple choice
Phosphorus reacts with chlorine by P₄ + 6Cl₂ → 4PCl₃. In a flask, the reaction stops when all the P₄ is consumed, with Cl₂ left over. Why does the reaction stop?
Feedback if chosen: This option is wrong — you blamed the leftover Cl₂. The Cl₂ can still collide with other molecules; what is missing is the P₄ it must react with.
Feedback if chosen: This option is wrong — you gave the product a blocking role. PCl₃ does not stop the reaction; the consumed P₄ does.
Feedback if chosen: Correct. Each PCl₃ molecule is built from the P₄. When the last of the P₄ reacts, the building material is gone, so the reaction stops.
Feedback if chosen: This option is wrong — the flask itself disproves this. Cl₂ is left over, and the reaction has already stopped.
Check — multiple choice
Carbon disulfide burns by CS₂ + 3O₂ → CO₂ + 2SO₂. A reaction consumes all of the CS₂, and O₂ remains. A student pumps in more O₂ to make more CO₂. How much more CO₂ forms?
Feedback if chosen: Correct. The CS₂ is completely consumed, so the added O₂ has nothing left to react with — no extra CO₂ forms.
Feedback if chosen: This option is wrong — even a little extra CO₂ would need CS₂ to build it, and the CS₂ is completely consumed.
Feedback if chosen: This option is wrong — product scales with the limiting reactant, and the amount of CS₂ has not changed. It is gone.
Feedback if chosen: This option is wrong — temperature plays no part here. The CS₂ is completely consumed, so no amount of O₂ makes more CO₂.
Check — multiple choice
In the same carbon-disulfide vessel, why does the added O₂ make no extra CO₂?
Feedback if chosen: This option is wrong — all O₂ molecules are identical. The missing ingredient is the CS₂, not a livelier oxygen.
Feedback if chosen: This option is wrong — the SO₂ blocks nothing. The added O₂ simply has no CS₂ left to react with.
Feedback if chosen: This option is wrong — the student did pump the O₂ in, so the vessel held it. The missing ingredient is the CS₂.
Feedback if chosen: Correct. The carbon in each CO₂ molecule comes from a CS₂ molecule. With the CS₂ completely consumed, no amount of O₂ can build more CO₂.
Check — multiple choice
Sulfur dioxide reacts with oxygen by 2SO₂ + O₂ → 2SO₃, and the SO₂ is the limiting reactant. A chemist doubles the amount of O₂ supplied. What happens to the amount of SO₃ formed?
Feedback if chosen: This option is wrong — you scaled the product with the excess reactant. The SO₂ caps the SO₃, and the amount of SO₂ has not changed.
Feedback if chosen: This option is wrong — you let the extra O₂ push the reaction further. The reaction already stops when the SO₂ is consumed, so extra O₂ adds nothing.
Feedback if chosen: This option is wrong — adding a reactant never lowers the product. The limiting SO₂ fixes the SO₃ at the same amount as before.
Feedback if chosen: Correct. The limiting SO₂ caps the SO₃. Doubling the O₂ supplies more of the reactant that was already left over, so the SO₃ formed stays the same.
Outcome: Identify the limiting reactant by comparing the available mole ratio with the required mole ratio.
A tank supplies 3.0 mol of H₂ for the reaction 2H₂ + O₂ → 2H₂O.
A second tank supplies 2.0 mol of O₂.
A student names O₂ as the limiting reactant.
The student's reason: 2.0 mol is less than 3.0 mol.
The reaction actually uses up the H₂ first.
The rule 'fewer moles means limiting' picked the wrong reactant.
What comparison finds the true limiting reactant?
The balanced equation decides, not the raw amounts.
The equation demands 2 mol of H₂ for every 1 mol of O₂.
So the supplied 3.0 mol of H₂ needs only 1.5 mol of O₂.
The supply of 2.0 mol of O₂ is more than enough.
| O₂ the reaction requires | O₂ available |
|---|---|
| 1.5 mol (from 3.0 mol H₂ × the 1:2 mole ratio) | 2.0 mol |
| Available > required, so the H₂ is used up first | |
H₂ is therefore the limiting reactant.
Comparing the required amount with the available amount names the limiting reactant every time.
Identify the limiting reactant by comparing the available mole ratio of the two reactants with the required mole ratio from the equation.
Q: 2.0 mol of N₂ and 4.5 mol of H₂ are supplied for N₂ + 3H₂ → 2NH₃. Which reactant is limiting?
Write down the values from the question
N₂ supplied = 2.0 mol
H₂ supplied = 4.5 mol
Work out the amount the mole ratio requires
H₂ required = 3 × 2.0 = 6.0 mol(the equation demands 3 mol of H₂ per 1 mol of N₂)
Compare required with available
required 6.0 mol > available 4.5 mol
The H₂ is used up first — H₂ is the limiting reactant
Check — numeric entry
Silicon reacts with chlorine by Si + 2Cl₂ → SiCl₄. A flask is supplied with 0.30 mol of Si. How many moles of Cl₂ does complete reaction of the Si require? Give your answer in mol, to 2 significant figures.
Feedback when correct: Correct. The equation demands 2 mol of Cl₂ for every 1 mol of Si, so 2 × 0.30 = 0.60 mol of Cl₂.
Anticipated wrong answers (feedback shown):
| 0.15 | This value halves the silicon instead of doubling it. The equation demands 2 mol of Cl₂ per 1 mol of Si: 2 × 0.30 = 0.60 mol. |
| 0.30 | This value uses a 1:1 ratio. The equation demands 2 mol of Cl₂ per 1 mol of Si: 2 × 0.30 = 0.60 mol. |
Worked solution revealed after a second miss
Write down the values from the question
Si supplied = 0.30 mol
ratio: 2 mol Cl₂ per 1 mol Si
Substitute in the values, and calculate
Cl₂ required = 2 × 0.30
Cl₂ required = 0.60 mol
Check — multiple choice
A flask holds 0.30 mol of Si and 0.50 mol of Cl₂ for Si + 2Cl₂ → SiCl₄. Which reactant is limiting?
Feedback if chosen: This option is wrong — you compared the wrong way. The Si needs 0.60 mol of Cl₂ and gets only 0.50 mol, so the Cl₂ is used up first.
Feedback if chosen: Correct. Required Cl₂ (0.60 mol) is more than the available Cl₂ (0.50 mol), so the Cl₂ is completely consumed first — Cl₂ is limiting.
Feedback if chosen: This option is wrong — the amounts do not match. The Si demands 0.60 mol of Cl₂, and only 0.50 mol is available.
Feedback if chosen: This option is wrong — moles are exactly what decides. Comparing required Cl₂ (0.60 mol) with available Cl₂ (0.50 mol) names the Cl₂ as limiting.
Check — multiple choice
Sodium reacts with oxygen by 4Na + O₂ → 2Na₂O. A vessel is supplied with 2.0 mol of Na and 0.40 mol of O₂. Which reactant is limiting?
Feedback if chosen: This option is wrong — check what the Na requires. 2.0 mol of Na needs only 0.50 mol of O₂, but the O₂ available (0.40 mol) falls short first.
Feedback if chosen: This option is wrong — the amounts do not match. The 2.0 mol of Na demands 0.50 mol of O₂, and only 0.40 mol is available.
Feedback if chosen: Correct. 2.0 mol of Na requires only 0.50 mol of O₂ (1 mol per 4 mol of Na). Only 0.40 mol is available — the O₂ is limiting.
Feedback if chosen: This option is wrong — you named the product. Na₂O is formed by the reaction; the limiting reactant is one of the starting substances.
Check — multiple choice
For 4Li + O₂ → 2Li₂O, a vessel is supplied with 1.2 mol of Li and 1.0 mol of O₂. A student names O₂ as limiting because 1.0 mol is less than 1.2 mol. Which statement is correct?
Feedback if chosen: This option is wrong — you kept the fewer-moles rule. The equation demands only 1 mol of O₂ per 4 mol of Li, so the plentiful-looking Li is used up first.
Feedback if chosen: This option is wrong — being a gas plays no part. The mole ratio decides: 1.2 mol of Li needs just 0.30 mol of O₂, so the Li is limiting.
Feedback if chosen: This option is wrong — every reaction with fixed supplies has a limiting reactant. Here the Li is consumed first.
Feedback if chosen: Correct. Required O₂ = 1.2 ÷ 4 = 0.30 mol, far less than the 1.0 mol available. The Li is used up first, even though the O₂ has fewer moles.
Outcome: Calculate the amount of excess reactant left over after the limiting reactant is completely consumed.
We've seen 3.0 mol of H₂ react with 2.0 mol of O₂ by 2H₂ + O₂ → 2H₂O.
The reaction used up every molecule of H₂.
Some O₂ stayed behind in the flask, unreacted.
Guessing 2.0 mol ignores the O₂ that the reaction consumed.
Guessing 0 mol ignores the leftover gas.
Exactly how many moles of O₂ remain?
The amount remaining = the amount supplied − the amount consumed.
The mole ratio gives the consumed amount: 3.0 mol of H₂ consumes 3.0 ÷ 2 = 1.5 mol of O₂.
The flask supplied 2.0 mol of O₂.
So 2.0 mol − 1.5 mol = 0.5 mol of O₂ remains.
The AP exam asks for this leftover amount directly.
Calculate the amount of excess reactant left over after the limiting reactant is completely consumed, as amount supplied minus amount consumed.
Q: 3.0 mol of SO₂ and 2.5 mol of O₂ react by 2SO₂ + O₂ → 2SO₃ until all the SO₂ is consumed. How many moles of O₂ remain?
Write down the values from the question
SO₂ consumed = 3.0 mol
O₂ supplied = 2.5 mol
Work out the amount consumed
O₂ consumed = 3.0 ÷ 2 = 1.5 mol(the equation demands 1 mol of O₂ per 2 mol of SO₂)
Subtract consumed from supplied
O₂ remaining = 2.5 − 1.5
O₂ remaining = 1.0 mol
Check — numeric entry
Aluminum reacts with bromine by 2Al + 3Br₂ → 2AlBr₃. A vessel holds 1.0 mol of Al and 2.0 mol of Br₂, and the reaction continues until all the Al is consumed. How many moles of Br₂ does the reaction consume? Give your answer in mol, to 2 significant figures.
Feedback when correct: Correct. The equation demands 3 mol of Br₂ per 2 mol of Al, so 1.0 mol of Al consumes 1.0 × 3 ÷ 2 = 1.5 mol of Br₂.
Anticipated wrong answers (feedback shown):
| 2.0 | This value assumes the whole Br₂ supply reacts. The mole ratio caps consumption: 1.0 mol of Al consumes only 1.0 × 3 ÷ 2 = 1.5 mol of Br₂. |
| 3.0 | This value multiplies by 3 without dividing by 2. The ratio is 3 mol of Br₂ per 2 mol of Al: 1.0 × 3 ÷ 2 = 1.5 mol. |
| 0.67 | This value inverts the ratio. The equation demands 3 mol of Br₂ per 2 mol of Al: 1.0 × 3 ÷ 2 = 1.5 mol. |
Worked solution revealed after a second miss
Write down the values from the question
Al consumed = 1.0 mol
ratio: 3 mol Br₂ per 2 mol Al
Substitute in the values, and calculate
Br₂ consumed = 1.0 × 3 ÷ 2
Br₂ consumed = 1.5 mol
Check — numeric entry
In the same reaction, 2.0 mol of Br₂ was supplied and 1.5 mol of Br₂ was consumed. How many moles of Br₂ remain? Give your answer in mol, to 2 significant figures.
Feedback when correct: Correct. Remaining = supplied − consumed = 2.0 − 1.5 = 0.50 mol of Br₂.
Anticipated wrong answers (feedback shown):
| 2.0 | This value ignores the Br₂ the reaction consumed. Subtract: 2.0 − 1.5 = 0.50 mol remains. |
| 1.5 | This value is the consumed amount. The question asks what remains: 2.0 − 1.5 = 0.50 mol. |
Worked solution revealed after a second miss
Write down the values from the question
Br₂ supplied = 2.0 mol
Br₂ consumed = 1.5 mol
Write down the equation
remaining = supplied − consumed
Substitute in the values, and calculate
remaining = 2.0 − 1.5
remaining = 0.50 mol
Check — numeric entry
Nitrogen reacts with oxygen by N₂ + O₂ → 2NO. A vessel holds 0.80 mol of N₂ and 0.50 mol of O₂, and the reaction continues until all the O₂ is consumed. How many moles of N₂ remain? Give your answer in mol, to 2 significant figures.
Feedback when correct: Correct. The 1:1 ratio means 0.50 mol of O₂ consumes 0.50 mol of N₂. Remaining N₂ = 0.80 − 0.50 = 0.30 mol.
Anticipated wrong answers (feedback shown):
| 0.80 | This value ignores the N₂ the reaction consumed. The 0.50 mol of O₂ consumed 0.50 mol of N₂, so 0.80 − 0.50 = 0.30 mol remains. |
| 0.50 | This value is the amount consumed, not the amount remaining. Remaining N₂ = 0.80 − 0.50 = 0.30 mol. |
| 1.0 | This value doubles the O₂ using the NO coefficient. The N₂:O₂ ratio is 1:1, so 0.50 mol of N₂ is consumed and 0.30 mol remains. |
Worked solution revealed after a second miss
Write down the values from the question
N₂ supplied = 0.80 mol
O₂ consumed = 0.50 mol
ratio: 1 mol N₂ per 1 mol O₂
Write down the equation
remaining = supplied − consumed
Substitute in the values, and calculate
remaining = 0.80 − 0.50
remaining = 0.30 mol
Outcome: State that the theoretical yield is the maximum amount of product, fixed by complete consumption of the limiting reactant.
A student burns 8.0 g of methane, CH₄, in more than enough oxygen.
The reaction CH₄ + 2O₂ → CO₂ + 2H₂O turns the methane into carbon dioxide and water.
Before the burn, the student wants a target: the greatest mass of CO₂ the 8.0 g could possibly deliver.
The oxygen does not set that target.
You've already seen why: extra amounts of the excess reactant make no extra product.
The limiting methane caps the CO₂.
Complete consumption of the limiting reactant fixes the maximum amount of product.
The mole ratio converts the whole limiting amount into product.
That calculated maximum is called the theoretical yield.
The theoretical yield is a ceiling, not a promise.
A real burn can collect less than the theoretical yield, never more.
The rest of this sequence measures every real collection against that ceiling.
State that the theoretical yield is the maximum amount of product, fixed by complete consumption of the limiting reactant.
Check — multiple choice
Aluminum reacts with iron(III) oxide, Fe₂O₃, by 2Al + Fe₂O₃ → Al₂O₃ + 2Fe. The Al is the limiting reactant and the Fe₂O₃ is in excess. Which starting amount fixes the theoretical yield of Fe?
Feedback if chosen: This option is wrong — you started from the excess reactant. Some Fe₂O₃ is left over at the end, so it cannot set the ceiling.
Feedback if chosen: Correct. Complete consumption of the limiting Al fixes the maximum Fe the reaction can make — that maximum is the theoretical yield.
Feedback if chosen: This option is wrong — you combined the two supplies. Only the limiting reactant is completely consumed, so only the Al sets the ceiling.
Feedback if chosen: This option is wrong — mass alone decides nothing. The limiting reactant, found from the mole ratio, fixes the theoretical yield.
Check — multiple choice
A synthesis has a theoretical yield of 10.0 g. A student claims that a careful chemist could collect 12.0 g of product. What is the greatest mass any collection of this synthesis can reach?
Feedback if chosen: This option is wrong — collections below the ceiling are common, but the ceiling itself is the 10.0 g theoretical yield.
Feedback if chosen: Correct. The theoretical yield is the maximum. A real collection can match 10.0 g or fall below it, never pass it.
Feedback if chosen: This option is wrong — you accepted the claim. No collection can pass the 10.0 g theoretical yield.
Feedback if chosen: This option is wrong — the limiting reactant sets a hard ceiling: the 10.0 g theoretical yield.
Check — multiple choice
In the same synthesis, why can no collection pass 10.0 g?
Feedback if chosen: This option is wrong — the balance reads any mass placed on it. The reaction itself stops at 10.0 g.
Feedback if chosen: This option is wrong — you named the wrong reactant. The excess reactant is left over; the limiting reactant is the one used up.
Feedback if chosen: This option is wrong — the product does not destroy itself. The reaction simply stops when the limiting reactant is gone.
Feedback if chosen: Correct. The theoretical yield assumes every particle of the limiting reactant becomes product. With the limiting reactant gone, nothing more can form.
Check — multiple choice
What is the theoretical yield of a reaction?
Feedback if chosen: This option is wrong — the collected mass is a measurement made after the reaction. The theoretical yield is calculated before it, as the maximum.
Feedback if chosen: This option is wrong — the leftover excess is a reactant, not a product. The theoretical yield is the ceiling on the product.
Feedback if chosen: Correct. Complete consumption of the limiting reactant fixes the maximum product — the theoretical yield.
Feedback if chosen: This option is wrong — reactant mass is not product mass. The mole ratio converts only the limiting amount into product.
Check — multiple choice
Hydrogen reacts with iodine by H₂ + I₂ → 2HI. The H₂ is limiting and the I₂ is in excess. The theoretical yield of HI is calculated from which amount?
Feedback if chosen: Correct. The limiting H₂ is completely consumed, and the mole ratio converts that whole amount into the maximum HI.
Feedback if chosen: This option is wrong — the I₂ is in excess, so some of it never reacts. The leftover part cannot contribute to the ceiling.
Feedback if chosen: This option is wrong — averaging mixes the excess into the calculation. Only the limiting H₂ sets the maximum.
Feedback if chosen: This option is wrong — the collected mass is measured after the reaction. The theoretical yield is calculated from the limiting reactant before it.
Check — multiple choice
A reaction's theoretical yield is 6.0 g. Which of these collected masses is possible?
Feedback if chosen: This option is wrong — 6.6 g sits above the 6.0 g ceiling, and no collection can pass the theoretical yield.
Feedback if chosen: This option is wrong — 7.0 g sits above the 6.0 g ceiling, and no collection can pass the theoretical yield.
Feedback if chosen: Correct. A real collection can only match the 6.0 g ceiling or fall below it — 5.4 g is below the ceiling, so it is possible.
Feedback if chosen: This option is wrong — two of these masses sit above the theoretical yield, and no collection can pass the ceiling.
Outcome: Calculate the theoretical yield of a product from starting amounts of two or more reactants, by identifying the limiting reactant and chaining to the product amount.
A fuel cell combines 4.0 g of H₂ with 16.0 g of O₂ by 2H₂ + O₂ → 2H₂O.
A student must state the theoretical yield of H₂O in grams.
The 16.0 g of O₂ looks like the bigger supply — four times the mass of the H₂.
But moles decide, not grams.
The 4.0 g of H₂ is 2.0 mol.
The 16.0 g of O₂ is only 0.50 mol.
The calculation converts each mass into moles first.
The mole comparison names the limiting reactant.
The mole ratio then carries the limiting amount to moles of product.
The molar mass turns the product moles into grams.
In short: mass → moles → limiting reactant → mole ratio → moles → mass.
Calculate the theoretical yield of a product from the starting amounts of two or more reactants: convert to moles, identify the limiting reactant, then chain to the product mass.
Q: What is the theoretical yield of H₂O, in grams, from 4.0 g of H₂ and 16.0 g of O₂? (2H₂ + O₂ → 2H₂O; molar masses: H₂ 2.0 g/mol, O₂ 32.0 g/mol, H₂O 18.0 g/mol)
Write down the values from the question
m(H₂) = 4.0 g, M(H₂) = 2.0 g/mol
m(O₂) = 16.0 g, M(O₂) = 32.0 g/mol
Convert each mass to moles
n(H₂) = 4.0 ÷ 2.0 = 2.0 mol
n(O₂) = 16.0 ÷ 32.0 = 0.50 mol
Identify the limiting reactant
O₂ required = 2.0 ÷ 2 = 1.0 mol(the equation demands 1 mol of O₂ per 2 mol of H₂)
available 0.50 mol < required 1.0 mol — O₂ is limiting
Convert the limiting amount to moles of product
n(H₂O) = 2 × 0.50 = 1.0 mol(2 mol of H₂O per 1 mol of O₂)
Convert to mass
m(H₂O) = 1.0 × 18.0
m(H₂O) = 18 g
Check — numeric entry
Sodium reacts with chlorine by 2Na + Cl₂ → 2NaCl. A vessel is supplied with 4.6 g of Na (molar mass 23.0 g/mol) and 14.2 g of Cl₂ (molar mass 70.9 g/mol). How many moles of Na is 4.6 g? Give your answer in mol, to 2 significant figures.
Feedback when correct: Correct. n = m ÷ M = 4.6 ÷ 23.0 = 0.20 mol of Na.
Anticipated wrong answers (feedback shown):
| 0.10 | This value halves the sodium using the equation's coefficient. Converting a mass to moles uses only the molar mass: 4.6 ÷ 23.0 = 0.20 mol. |
| 105.8 | This value multiplies by the molar mass. Divide instead: n = m ÷ M = 4.6 ÷ 23.0 = 0.20 mol. |
Worked solution revealed after a second miss
Write down the values from the question
m(Na) = 4.6 g
M(Na) = 23.0 g/mol
Write down the equation
n = m ÷ M
Substitute in the values, and calculate
n = 4.6 ÷ 23.0
n = 0.20 mol
Check — multiple choice
In the same vessel, the 4.6 g of Na is 0.20 mol and the 14.2 g of Cl₂ is 0.20 mol. For 2Na + Cl₂ → 2NaCl, which reactant is limiting?
Feedback if chosen: This option is wrong — the 0.20 mol of Na needs only 0.10 mol of Cl₂, and 0.20 mol is available — the Cl₂ is in excess.
Feedback if chosen: This option is wrong — you named the product. NaCl is formed by the reaction; the limiting reactant is a starting substance.
Feedback if chosen: This option is wrong — equal moles are not a perfect match. The equation demands 2 mol of Na per 1 mol of Cl₂, so the Na is used up first.
Feedback if chosen: Correct. 0.20 mol of Na requires only 0.10 mol of Cl₂, so the Cl₂ outlasts the Na. The Na is completely consumed first — Na is limiting.
Check — numeric entry
Still with 2Na + Cl₂ → 2NaCl: the limiting reactant is 0.20 mol of Na, and the molar mass of NaCl is 58.4 g/mol. What is the theoretical yield of NaCl, in grams, to 2 significant figures?
Tolerance: 0.5
Feedback when correct: Correct. The ratio is 2 NaCl per 2 Na — one for one — so 0.20 mol of Na gives 0.20 mol of NaCl, and 0.20 × 58.4 = 12 g.
Anticipated wrong answers (feedback shown):
| 23 | This value converts the excess Cl₂ to product. The ceiling comes from the limiting Na: 0.20 mol of Na gives 0.20 mol of NaCl = 12 g. |
| 5.8 | This value halves the sodium. The ratio is 2 NaCl per 2 Na — one for one — so 0.20 mol of NaCl forms: 12 g. |
Worked solution revealed after a second miss
Write down the values from the question
n(Na) = 0.20 mol (limiting)
ratio: 2 mol NaCl per 2 mol Na
M(NaCl) = 58.4 g/mol
Convert the limiting amount to moles of product
n(NaCl) = 0.20 mol
Substitute in the values, and calculate
m = 0.20 × 58.4
m = 12 g
Check — numeric entry
Calcium oxide reacts with water by CaO + H₂O → Ca(OH)₂. A flask combines 11.2 g of CaO (molar mass 56.1 g/mol) with 5.4 g of H₂O (molar mass 18.0 g/mol). What is the theoretical yield of Ca(OH)₂ (molar mass 74.1 g/mol), in grams, to 2 significant figures?
Tolerance: 0.5
Feedback when correct: Correct. n(CaO) = 0.20 mol, n(H₂O) = 0.30 mol; the 1:1 ratio makes the CaO limiting. So 0.20 mol of Ca(OH)₂ forms: 0.20 × 74.1 = 15 g.
Anticipated wrong answers (feedback shown):
| 22 | This value converts the H₂O — the excess reactant. The limiting CaO (0.20 mol) caps the product: 0.20 × 74.1 = 15 g. |
| 0.20 | This value stops at the moles of product. Convert to mass with the molar mass: 0.20 × 74.1 = 15 g. |
Worked solution revealed after a second miss
Write down the values from the question
m(CaO) = 11.2 g, M = 56.1 g/mol
m(H₂O) = 5.4 g, M = 18.0 g/mol
M(Ca(OH)₂) = 74.1 g/mol
Convert each mass to moles
n(CaO) = 11.2 ÷ 56.1 = 0.20 mol
n(H₂O) = 5.4 ÷ 18.0 = 0.30 mol
Identify the limiting reactant
ratio 1:1 — the CaO (0.20 mol) is limiting
Convert the limiting amount to moles of product
n(Ca(OH)₂) = 0.20 mol
Convert to mass
m = 0.20 × 74.1
m = 15 g
Check — numeric entry
Aluminum reacts with chlorine by 2Al + 3Cl₂ → 2AlCl₃. A vessel combines 5.4 g of Al (molar mass 27.0 g/mol) with 28.4 g of Cl₂ (molar mass 70.9 g/mol). What is the theoretical yield of AlCl₃ (molar mass 133.3 g/mol), in grams, to 2 significant figures?
Tolerance: 1
Feedback when correct: Correct. n(Al) = 0.20 mol, n(Cl₂) = 0.40 mol; the Al requires 0.30 mol of Cl₂, so the Al is limiting. 0.20 mol of AlCl₃ forms: 0.20 × 133.3 = 27 g.
Anticipated wrong answers (feedback shown):
| 36 | This value converts the excess Cl₂ to product. The limiting Al (0.20 mol) caps the AlCl₃ at 0.20 mol: 0.20 × 133.3 = 27 g. |
| 18 | This value applies the 2-to-3 Al-to-Cl₂ ratio to the AlCl₃. The Al-to-AlCl₃ ratio is 2:2 — one for one — so 0.20 mol of AlCl₃ forms: 27 g. |
Worked solution revealed after a second miss
Write down the values from the question
m(Al) = 5.4 g, M = 27.0 g/mol
m(Cl₂) = 28.4 g, M = 70.9 g/mol
M(AlCl₃) = 133.3 g/mol
Convert each mass to moles
n(Al) = 5.4 ÷ 27.0 = 0.20 mol
n(Cl₂) = 28.4 ÷ 70.9 = 0.40 mol
Identify the limiting reactant
Cl₂ required = 0.20 × 3 ÷ 2 = 0.30 mol
available 0.40 mol > required 0.30 mol — Al is limiting
Convert the limiting amount to moles of product
n(AlCl₃) = 0.20 mol
Convert to mass
m = 0.20 × 133.3
m = 27 g
Outcome: State that percent yield is the actual yield divided by the theoretical yield, times 100%.
A student burns 8.0 g of CH₄ in more than enough oxygen, aiming at the 22 g theoretical yield of CO₂.
The collected gas has a mass of only 11 g.
A second student burns the same 8.0 g and collects 20 g.
Both collections fall short of the 22 g ceiling.
The first collection lost half of the promised CO₂.
The second collection lost only 2 g.
How should a chemist report how close a collection came to the ceiling?
The mass actually collected is called the actual yield.
The actual yield divided by the theoretical yield, times 100%, is called the percent yield.
percent yield = actual yield ÷ theoretical yield × 100%
The first collection: 11 g ÷ 22 g × 100% = a 50% yield.
The second collection: 20 g ÷ 22 g × 100% = a 91% yield.
One percentage compares any two experiments, whatever their scale.
State that percent yield is the actual yield divided by the theoretical yield, times 100%.
Check — multiple choice
A synthesis has a theoretical yield of 5.0 g, and the student collects 4.0 g of product. Which mass is the actual yield?
Feedback if chosen: Correct. The actual yield is the mass actually collected — here, 4.0 g.
Feedback if chosen: This option is wrong — 5.0 g is the theoretical yield, the calculated ceiling. The actual yield is the mass collected: 4.0 g.
Feedback if chosen: This option is wrong — 1.0 g is the shortfall between the two yields, not a yield itself.
Feedback if chosen: This option is wrong — you added the two masses. The actual yield is simply what was collected: 4.0 g.
Check — multiple choice
Two students attempt the same synthesis with a 15 g theoretical yield. One collects 6.0 g; the other collects 12 g. Which collection has the higher percent yield?
Feedback if chosen: This option is wrong — 6.0 g ÷ 15 g gives 40%, the lower fraction of the ceiling.
Feedback if chosen: Correct. 12 g ÷ 15 g × 100% = 80%, against 40% for the 6.0 g collection — the larger collection of the same ceiling has the higher percent yield.
Feedback if chosen: This option is wrong — a shared ceiling does not equalize the yields. The collected masses differ, so the fractions differ: 40% against 80%.
Feedback if chosen: This option is wrong — the reactant masses already did their work by fixing the 15 g ceiling. The comparison needs only the two collected masses.
Check — numeric entry
A synthesis has a theoretical yield of 8.0 g, and the student collects 6.0 g. What is the percent yield, in %?
Feedback when correct: Correct. percent yield = 6.0 ÷ 8.0 × 100% = 75%.
Anticipated wrong answers (feedback shown):
| 133 | This value divides the wrong way. Percent yield is actual ÷ theoretical: 6.0 ÷ 8.0 × 100% = 75%. |
| 0.75 | This value stops at the fraction. Multiply by 100% to report a percentage: 75%. |
Worked solution revealed after a second miss
Write down the values from the question
actual yield = 6.0 g
theoretical yield = 8.0 g
Write down the equation
percent yield = actual ÷ theoretical × 100%
Substitute in the values, and calculate
percent yield = 6.0 ÷ 8.0 × 100%
percent yield = 75%
Check — multiple choice
A student collects 9.0 g of product against a theoretical yield of 12.0 g. Which calculation gives the percent yield?
Feedback if chosen: This option is wrong — you divided the ceiling by the collection. Percent yield is actual ÷ theoretical: 9.0 ÷ 12.0 × 100%.
Feedback if chosen: This option is wrong — you subtracted the masses. Percent yield is a fraction of the ceiling, not a difference: 9.0 ÷ 12.0 × 100%.
Feedback if chosen: This option is wrong — you multiplied the two yields. Percent yield divides the actual by the theoretical: 9.0 ÷ 12.0 × 100%.
Feedback if chosen: Correct. percent yield = actual yield ÷ theoretical yield × 100% = 9.0 ÷ 12.0 × 100% = 75%.
Check — numeric entry
A synthesis has a theoretical yield of 6.0 g, and the actual yield is 4.8 g. What is the percent yield, in %?
Feedback when correct: Correct. percent yield = 4.8 ÷ 6.0 × 100% = 80%.
Anticipated wrong answers (feedback shown):
| 125 | This value divides the wrong way. Percent yield is actual ÷ theoretical: 4.8 ÷ 6.0 × 100% = 80%. |
| 0.80 | This value stops at the fraction. Multiply by 100% to report a percentage: 80%. |
Worked solution revealed after a second miss
Write down the values from the question
actual yield = 4.8 g
theoretical yield = 6.0 g
Write down the equation
percent yield = actual ÷ theoretical × 100%
Substitute in the values, and calculate
percent yield = 4.8 ÷ 6.0 × 100%
percent yield = 80%
Outcome: Calculate percent yield from actual and theoretical yields, rearranging when the unknown is the actual or theoretical yield.
A student's experiment produced a theoretical yield of 13.2 g of CO₂.
The collection flask ended the day with 11.2 g of CO₂ inside.
The percent yield follows in one substitution.
The next exercise flips the question: a synthesis with a 22.0 g theoretical yield reports an 80.0% yield, and no collected mass appears anywhere.
The same equation answers both — the working just makes a different quantity the subject.
Calculate percent yield from the actual and theoretical yields using percent yield = actual yield ÷ theoretical yield × 100%, rearranging when the unknown is the actual or the theoretical yield.
Q: The theoretical yield is 13.2 g and the collected mass is 11.2 g. What is the percent yield?
Write down the values from the question
actual yield = 11.2 g
theoretical yield = 13.2 g
Write down the equation
percent yield = actual ÷ theoretical × 100%
Substitute in the values, and calculate
percent yield = 11.2 ÷ 13.2 × 100%
percent yield = 84.8%
Q: A synthesis with a 22.0 g theoretical yield reports an 80.0% yield. What mass was actually collected?
Write down the values from the question
percent yield = 80.0%
theoretical yield = 22.0 g
Write down the equation
percent yield = actual ÷ theoretical × 100%
Make the actual yield the subject
actual = percent yield ÷ 100% × theoretical
Substitute in the values, and calculate
actual = 0.800 × 22.0
actual = 17.6 g
Check — numeric entry
A synthesis has a theoretical yield of 20.0 g, and the student collects 13.0 g. What is the percent yield, in %, to 3 significant figures?
Feedback when correct: Correct. percent yield = 13.0 ÷ 20.0 × 100% = 65.0%.
Anticipated wrong answers (feedback shown):
| 154 | This value divides the ceiling by the collection. Percent yield is actual ÷ theoretical: 13.0 ÷ 20.0 × 100% = 65.0%. |
| 0.65 | This value stops at the fraction. Multiply by 100% to report a percentage: 65.0%. |
Worked solution revealed after a second miss
Write down the values from the question
actual yield = 13.0 g
theoretical yield = 20.0 g
Write down the equation
percent yield = actual ÷ theoretical × 100%
Substitute in the values, and calculate
percent yield = 13.0 ÷ 20.0 × 100%
percent yield = 65.0%
Check — multiple choice
A synthesis reports a 75.0% yield with a theoretical yield of 16.0 g; the collected mass is unknown. Which line makes the actual yield the subject?
Feedback if chosen: This option is wrong — you divided the ceiling by the percentage. Rearranging percent = actual ÷ theoretical × 100% gives actual = percent ÷ 100% × theoretical.
Feedback if chosen: Correct. Rearranging percent yield = actual ÷ theoretical × 100% gives actual = 75.0 ÷ 100% × 16.0 = 12.0 g.
Feedback if chosen: This option is wrong — you dropped the ÷ 100%. A 75.0% yield is the fraction 0.750, so actual = 0.750 × 16.0.
Feedback if chosen: This option is wrong — you inverted the whole right-hand side. The rearrangement is actual = percent ÷ 100% × theoretical.
Check — numeric entry
A synthesis has a theoretical yield of 24.0 g, and the student collects 18.0 g. What is the percent yield, in %, to 3 significant figures?
Feedback when correct: Correct. percent yield = 18.0 ÷ 24.0 × 100% = 75.0%.
Anticipated wrong answers (feedback shown):
| 133 | This value divides the ceiling by the collection. Percent yield is actual ÷ theoretical: 18.0 ÷ 24.0 × 100% = 75.0%. |
| 6 | This value subtracts the masses. Percent yield is a fraction of the ceiling: 18.0 ÷ 24.0 × 100% = 75.0%. |
Worked solution revealed after a second miss
Write down the values from the question
actual yield = 18.0 g
theoretical yield = 24.0 g
Write down the equation
percent yield = actual ÷ theoretical × 100%
Substitute in the values, and calculate
percent yield = 18.0 ÷ 24.0 × 100%
percent yield = 75.0%
Check — numeric entry
A synthesis reports a 40.0% yield with a theoretical yield of 35.0 g. What mass was actually collected, in grams, to 3 significant figures?
Feedback when correct: Correct. actual = percent ÷ 100% × theoretical = 0.400 × 35.0 = 14.0 g.
Anticipated wrong answers (feedback shown):
| 87.5 | This value divides the ceiling by the fraction, which computes a theoretical yield. Make the actual yield the subject: 0.400 × 35.0 = 14.0 g. |
| 1400 | This value skips the ÷ 100%. A 40.0% yield is the fraction 0.400, so actual = 0.400 × 35.0 = 14.0 g. |
Worked solution revealed after a second miss
Write down the values from the question
percent yield = 40.0%
theoretical yield = 35.0 g
Write down the equation
percent yield = actual ÷ theoretical × 100%
Make the actual yield the subject
actual = percent ÷ 100% × theoretical
Substitute in the values, and calculate
actual = 0.400 × 35.0
actual = 14.0 g
Outcome: Explain why an actual yield can fall below the theoretical yield, and justify whether a stated percent yield is consistent with a described laboratory step.
A synthesis has a theoretical yield of 5.0 g of solid product.
The procedure ends with a rinse: 50 mL of water washes the solid on its filter paper.
One supplied fact matters: this solid is slightly soluble in water.
So some of the solid dissolves into the rinse water.
The dissolved portion drains away with the water.
The dried solid has a mass of 4.2 g — an 84% yield.
An AP question asks: does the rinse step explain a yield below 100%?
The theoretical yield assumes a perfect procedure.
Assumption one: every particle of the limiting reactant reacts.
Assumption two: every particle of product reaches the balance.
Real lab steps break those assumptions.
A reaction can stop before completion.
A rinse or a transfer can carry product away.
A justification names the specific lab step and states the product that step loses.
Here: the rinse dissolved some product, so the collected 4.2 g sits below the 5.0 g ceiling — 84% is consistent with the rinse.
The assumptions can also fail upward.
A solid still wet with rinse water carries that water onto the balance.
The balance then reads product plus water.
So a measured mass above the theoretical yield does not mean extra product formed.
It means the measured mass includes something that is not product.
Explain why an actual yield can fall below the theoretical yield, and justify whether a stated percent yield is consistent with a described laboratory step.
Check — multiple choice
A synthesis has a theoretical yield of 8.0 g. During filtering, some solid product passes through the filter paper into the waste flask. The dried collected solid is 6.8 g. Why is the actual yield below the theoretical yield?
Feedback if chosen: Correct. The lost solid went through the paper into the waste flask. Product that never reaches the balance never has its mass measured, so the collection falls below the ceiling.
Feedback if chosen: This option is wrong — you blamed the calculation. The ceiling is fixed by the limiting reactant; the description names a real loss: solid passing the filter.
Feedback if chosen: This option is wrong — drying removes water, not product. The described loss is the solid that passed through the filter paper.
Feedback if chosen: This option is wrong — the excess reactant does not soak up product. The description names the loss: solid passing through the filter into the waste.
Check — multiple choice
A student sees a percent yield of 84% and claims it proves the chemist who performed the synthesis made a mistake. Which statement is correct?
Feedback if chosen: This option is wrong — you tied every shortfall to error. The reaction itself can stop before completion, leaving reactant unconverted with no mistake made.
Feedback if chosen: This option is wrong — you invented a spill threshold. A reaction stopping short, a rinse, or a transfer can each lower the yield without any error.
Feedback if chosen: Correct. The ceiling assumes every particle of limiting reactant reacts. A reaction that stops before completion breaks that assumption by itself — no mistake is needed.
Feedback if chosen: This option is wrong — the right verdict needs the right reason. The size of the percentage does not judge care; the reaction stopping short can lower any yield.
Free-response question
Source note (not shown as part of the question): authored to archetype 2022 AP Chemistry FRQ Q1(b) (percent-yield justification)
A student mixes solutions of CaCl₂ and Na₂SO₄, and solid CaSO₄ forms by CaCl₂(aq) + Na₂SO₄(aq) → CaSO₄(s) + 2NaCl(aq). CaSO₄ is slightly soluble in water: a small amount of it dissolves in any rinse water. The theoretical yield of CaSO₄ is 6.0 g. Take a pen and paper and write your justifications; then reveal the model response to self-check.
(a) The student filters the solid, rinses it with 50 mL of water, dries it, and measures 5.1 g — an 85% yield. Justify whether the 85% yield is consistent with the described rinse step, in one or two sentences. [1 point]
(b) A second student performs the same synthesis, filters and rinses the solid, but skips the drying step. The still-wet solid has a measured mass of 6.5 g — a 108% yield. Justify whether a yield above 100% is consistent with the described procedure, in one or two sentences. [1 point]
✓ Model response (the key)
(a)
The rinse water dissolved a small amount of the CaSO₄, and that dissolved portion drained away with the water.
So the collected mass sits below the 6.0 g ceiling, and an 85% yield is consistent with the rinse step.
(b)
The solid was not dried, so the measured 6.5 g is the CaSO₄ plus the rinse water still held in it.
A measured mass above the theoretical yield is consistent with the skipped drying — the extra mass is water, not extra product.
Scoring rubric — 2 points available
Check — numeric entry
A synthesis has a theoretical yield of 12.5 g, and the student collects 10.0 g. What is the percent yield, in %, to 3 significant figures?
Feedback when correct: Correct. percent yield = 10.0 ÷ 12.5 × 100% = 80.0%.
Anticipated wrong answers (feedback shown):
| 125 | This value divides the ceiling by the collection. Percent yield is actual ÷ theoretical: 10.0 ÷ 12.5 × 100% = 80.0%. |
| 0.8 | This value stops at the fraction. Multiply by 100% to report a percentage: 80.0%. |
Worked solution revealed after a second miss
Write down the values from the question
actual yield = 10.0 g
theoretical yield = 12.5 g
Write down the equation
percent yield = actual ÷ theoretical × 100%
Substitute in the values, and calculate
percent yield = 10.0 ÷ 12.5 × 100%
percent yield = 80.0%
Check — multiple choice
Aluminum reacts with oxygen by 4Al + 3O₂ → 2Al₂O₃. A vessel is supplied with 3.0 mol of Al and 2.0 mol of O₂. Which reactant is limiting?
Feedback if chosen: This option is wrong — the Al needs 3.0 × 3 ÷ 4 = 2.25 mol of O₂. Only 2.0 mol is available, so the O₂ is used up first.
Feedback if chosen: Correct. The 3.0 mol of Al requires 2.25 mol of O₂, more than the 2.0 mol available — the O₂ is completely consumed first.
Feedback if chosen: This option is wrong — the amounts do not match. The Al demands 2.25 mol of O₂ and only 2.0 mol is available.
Feedback if chosen: This option is wrong — you named the product. Al₂O₃ is formed by the reaction; the limiting reactant is a starting substance.
Check — numeric entry
Iron reacts with sulfur by Fe + S → FeS. A crucible holds 0.60 mol of Fe and 0.40 mol of S, and the reaction continues until all the S is consumed. How many moles of Fe remain? Give your answer in mol, to 2 significant figures.
Feedback when correct: Correct. The 1:1 ratio means 0.40 mol of S consumes 0.40 mol of Fe. Remaining Fe = 0.60 − 0.40 = 0.20 mol.
Anticipated wrong answers (feedback shown):
| 0.40 | This value is the amount of Fe consumed. The question asks what remains: 0.60 − 0.40 = 0.20 mol. |
| 0.60 | This value ignores the Fe the reaction consumed. Subtract the consumed 0.40 mol: 0.20 mol remains. |
Worked solution revealed after a second miss
Write down the values from the question
Fe supplied = 0.60 mol
S consumed = 0.40 mol
ratio: 1 mol Fe per 1 mol S
Write down the equation
remaining = supplied − consumed
Substitute in the values, and calculate
remaining = 0.60 − 0.40
remaining = 0.20 mol
Check — numeric entry
Potassium reacts with bromine by 2K + Br₂ → 2KBr. A vessel combines 7.8 g of K (molar mass 39.1 g/mol) with 24.0 g of Br₂ (molar mass 159.8 g/mol). What is the theoretical yield of KBr (molar mass 119.0 g/mol), in grams, to 2 significant figures?
Tolerance: 0.5
Feedback when correct: Correct. n(K) = 0.20 mol, n(Br₂) = 0.15 mol; the K requires only 0.10 mol of Br₂, so the K is limiting. 0.20 mol of KBr forms: 0.20 × 119.0 = 24 g.
Anticipated wrong answers (feedback shown):
| 36 | This value converts the excess Br₂ to product. The limiting K (0.20 mol) caps the KBr at 0.20 mol: 0.20 × 119.0 = 24 g. |
| 12 | This value halves the potassium. The ratio is 2 KBr per 2 K — one for one — so 0.20 mol of KBr forms: 24 g. |
Worked solution revealed after a second miss
Write down the values from the question
m(K) = 7.8 g, M = 39.1 g/mol
m(Br₂) = 24.0 g, M = 159.8 g/mol
M(KBr) = 119.0 g/mol
Convert each mass to moles
n(K) = 7.8 ÷ 39.1 = 0.20 mol
n(Br₂) = 24.0 ÷ 159.8 = 0.15 mol
Identify the limiting reactant
Br₂ required = 0.20 ÷ 2 = 0.10 mol
available 0.15 mol > required 0.10 mol — K is limiting
Convert the limiting amount to moles of product
n(KBr) = 0.20 mol
Convert to mass
m = 0.20 × 119.0
m = 24 g
5 lessons
Outcome: Calculate the amount of a species produced or consumed when solutions react, using molarity × volume then the mole ratio.
A student pours 50.0 mL of 0.20 M silver nitrate, AgNO₃, into a beaker.
The student adds more than enough sodium chloride solution.
White silver chloride, AgCl, appears at once.
AgCl is insoluble in water — the question supplies this fact.
The equation reads AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq).
The question asks for the moles of AgCl formed.
Until now, every chain of conversions started from a mass in grams.
This time the only given quantities are 0.20 M on the label and 50.0 mL in the cylinder.
You've already seen the tool in Unit 3: moles of solute = molarity × volume in liters.
50.0 mL is 0.0500 L.
So 0.20 M × 0.0500 L = 0.010 mol of AgNO₃.
The mole ratio then converts 0.010 mol of AgNO₃ into 0.010 mol of AgCl.
The chain gained one new first step and changed nowhere else.
Most AP reactions happen in solution.
This one conversion opens every solution reaction to the mole ratio.
Calculate the amount of a species produced or consumed when solutions react, using molarity × volume to get moles, then the mole ratio.
Q: How many moles of AgCl form when 50.0 mL of 0.20 M AgNO₃ reacts with more than enough NaCl? (AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq))
Write down the values from the question
M(AgNO₃ solution) = 0.20 M
V = 50.0 mL = 0.0500 L
Write down the equation
moles of solute = molarity × volume (L)
Substitute in the values, and calculate
n(AgNO₃) = 0.20 × 0.0500
n(AgNO₃) = 0.010 mol
Apply the mole ratio
n(AgCl) = 0.010 mol(1 mol of AgCl per 1 mol of AgNO₃)
Check — multiple choice
You've already seen this tool in Unit 3. Which equation gives the moles of solute in a solution?
Feedback if chosen: This option is wrong — dividing molarity by volume gives neither moles nor a useful quantity. Moles of solute = molarity × volume in liters.
Feedback if chosen: This option is wrong — volume ÷ molarity has the wrong shape. Moles of solute = molarity × volume in liters.
Feedback if chosen: Correct. Molarity is moles per liter, so multiplying by the volume in liters returns the moles of solute.
Feedback if chosen: This option is wrong — you kept the volume in milliliters. Molarity is moles per liter, so the volume must be converted to liters first.
Check — multiple choice
A cylinder holds 25.0 mL of solution. What is that volume in liters?
Feedback if chosen: This option is wrong — you divided by 100. There are 1000 mL in a liter, so 25.0 mL = 25.0 ÷ 1000 = 0.0250 L.
Feedback if chosen: Correct. There are 1000 mL in a liter, so 25.0 mL = 25.0 ÷ 1000 = 0.0250 L.
Feedback if chosen: This option is wrong — you divided by 10 000. There are 1000 mL in a liter: 25.0 ÷ 1000 = 0.0250 L.
Feedback if chosen: This option is wrong — you divided by 10. There are 1000 mL in a liter: 25.0 ÷ 1000 = 0.0250 L.
Check — numeric entry
Sodium hydroxide reacts with magnesium chloride by 2NaOH(aq) + MgCl₂(aq) → Mg(OH)₂(s) + 2NaCl(aq); Mg(OH)₂ is insoluble in water (supplied). A student pours 40.0 mL of 0.25 M NaOH into more than enough MgCl₂ solution. How many moles of NaOH does the 40.0 mL deliver? Give your answer in mol, to 2 significant figures.
Feedback when correct: Correct. n = M × V = 0.25 × 0.0400 L = 0.010 mol of NaOH.
Anticipated wrong answers (feedback shown):
| 10 | This value keeps the volume in milliliters. Convert first: 40.0 mL = 0.0400 L, so n = 0.25 × 0.0400 = 0.010 mol. |
| 0.25 | This value is the molarity alone. Multiply by the volume in liters: 0.25 × 0.0400 = 0.010 mol. |
Worked solution revealed after a second miss
Write down the values from the question
M = 0.25 M
V = 40.0 mL = 0.0400 L
Write down the equation
moles of solute = molarity × volume (L)
Substitute in the values, and calculate
n = 0.25 × 0.0400
n = 0.010 mol
Check — numeric entry
Sodium hydroxide reacts with magnesium chloride by 2NaOH(aq) + MgCl₂(aq) → Mg(OH)₂(s) + 2NaCl(aq). A poured volume delivers 0.010 mol of NaOH into more than enough MgCl₂ solution. How many moles of Mg(OH)₂ form? Give your answer in mol, to 2 significant figures.
Feedback when correct: Correct. The equation makes 1 mol of Mg(OH)₂ from 2 mol of NaOH, so 0.010 ÷ 2 = 0.0050 mol.
Anticipated wrong answers (feedback shown):
| 0.010 | This value assumes a 1:1 ratio. The equation makes 1 mol of Mg(OH)₂ from 2 mol of NaOH: 0.010 ÷ 2 = 0.0050 mol. |
| 0.020 | This value doubles instead of halving. The ratio is 1 Mg(OH)₂ per 2 NaOH: 0.010 ÷ 2 = 0.0050 mol. |
Worked solution revealed after a second miss
Write down the values from the question
n(NaOH) = 0.010 mol
ratio: 1 mol Mg(OH)₂ per 2 mol NaOH
Substitute in the values, and calculate
n(Mg(OH)₂) = 0.010 ÷ 2
n(Mg(OH)₂) = 0.0050 mol
Check — numeric entry
Calcium chloride reacts with sodium carbonate by CaCl₂(aq) + Na₂CO₃(aq) → CaCO₃(s) + 2NaCl(aq); CaCO₃ is insoluble in water (supplied). 50.0 mL of 0.30 M CaCl₂ reacts with more than enough Na₂CO₃ solution. How many moles of CaCO₃ form? Give your answer in mol, to 2 significant figures.
Feedback when correct: Correct. n(CaCl₂) = 0.30 × 0.0500 L = 0.015 mol, and the 1:1 ratio gives 0.015 mol of CaCO₃.
Anticipated wrong answers (feedback shown):
| 15 | This value keeps the volume in milliliters. Convert first: 50.0 mL = 0.0500 L, so n = 0.30 × 0.0500 = 0.015 mol. |
| 0.0075 | This value halves the moles. The ratio here is 1 CaCO₃ per 1 CaCl₂, so no halving: 0.015 mol. |
Worked solution revealed after a second miss
Write down the values from the question
M = 0.30 M
V = 50.0 mL = 0.0500 L
ratio: 1 mol CaCO₃ per 1 mol CaCl₂
Write down the equation
moles of solute = molarity × volume (L)
Substitute in the values, and calculate
n(CaCl₂) = 0.30 × 0.0500 = 0.015 mol
n(CaCO₃) = 0.015 mol
Outcome: Calculate the solution volume or concentration required to deliver the needed amount of a dissolved reactant, using the mole ratio.
A marble chip of calcium carbonate, CaCO₃, has a mass of 0.50 g.
Hydrochloric acid dissolves the chip by CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g).
The acid bottle is labeled 0.10 M.
Too little acid leaves solid behind.
We've seen a calculation that starts from a known volume and asks for an amount.
This task supplies the amount and asks for the volume.
How many milliliters of the 0.10 M acid consume the whole chip?
The mass becomes moles first: 0.50 g ÷ 100.1 g/mol = 0.0050 mol of CaCO₃.
The mole ratio demands 2 mol of HCl per 1 mol of CaCO₃ — here, 0.010 mol.
You've already seen the definition: molarity = moles ÷ volume in liters.
Making the volume the subject gives: volume = moles ÷ molarity.
So 0.010 mol ÷ 0.10 M = 0.10 L, which is 100 mL.
Every dispensing question in AP labs asks this direction: which volume carries the needed moles?
Calculate the solution volume or concentration required to deliver the needed amount of a dissolved reactant, using the mole ratio.
Q: How many milliliters of 0.10 M HCl consume 0.50 g of CaCO₃? (CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g); molar mass of CaCO₃ = 100.1 g/mol)
Write down the values from the question
m(CaCO₃) = 0.50 g, M = 100.1 g/mol
HCl molarity = 0.10 M
Convert the mass to moles
n(CaCO₃) = 0.50 ÷ 100.1 = 0.0050 mol
Apply the mole ratio
n(HCl) = 2 × 0.0050 = 0.010 mol(2 mol of HCl per 1 mol of CaCO₃)
Make the volume the subject
volume = moles ÷ molarity(from molarity = moles ÷ volume)
Substitute in the values, and calculate
V = 0.010 ÷ 0.10 = 0.10 L
V = 100 mL
Check — multiple choice
You've already seen molarity defined in Unit 3. Which equation defines molarity?
Feedback if chosen: This option is wrong — multiplying moles by volume grows with both quantities. Molarity divides: moles of solute ÷ liters of solution.
Feedback if chosen: This option is wrong — you inverted the fraction. Molarity is moles of solute ÷ liters of solution.
Feedback if chosen: This option is wrong — you used grams. Molarity counts moles of solute per liter of solution.
Feedback if chosen: Correct. Molarity = moles of solute ÷ liters of solution.
Check — multiple choice
Starting from molarity = moles ÷ volume, which line makes the volume the subject?
Feedback if chosen: Correct. Multiplying both sides by volume and dividing both sides by molarity gives volume = moles ÷ molarity.
Feedback if chosen: This option is wrong — you inverted the fraction. Rearranging molarity = moles ÷ volume gives volume = moles ÷ molarity.
Feedback if chosen: This option is wrong — you multiplied where the rearrangement divides. Volume = moles ÷ molarity.
Feedback if chosen: This option is wrong — you inverted the whole expression. Volume = moles ÷ molarity.
Check — numeric entry
Magnesium reacts with hydrochloric acid by Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g). A student must dissolve 0.243 g of Mg (molar mass 24.3 g/mol) completely, using 0.40 M HCl. How many milliliters of the acid does the reaction require? Give your answer in mL, to 2 significant figures.
Feedback when correct: Correct. n(Mg) = 0.243 ÷ 24.3 = 0.0100 mol; the ratio demands 0.0200 mol of HCl; V = 0.0200 ÷ 0.40 = 0.050 L = 50 mL.
Anticipated wrong answers (feedback shown):
| 25 | This value skips the mole ratio. The equation demands 2 mol of HCl per 1 mol of Mg: 0.0200 mol of HCl, so V = 50 mL. |
| 0.050 | This value is in liters. The question asks for milliliters: 0.050 L = 50 mL. |
| 8.0 | This value multiplies moles by molarity. Volume = moles ÷ molarity = 0.0200 ÷ 0.40 = 0.050 L = 50 mL. |
Worked solution revealed after a second miss
Write down the values from the question
m(Mg) = 0.243 g, M = 24.3 g/mol
HCl molarity = 0.40 M
ratio: 2 mol HCl per 1 mol Mg
Convert the mass to moles
n(Mg) = 0.243 ÷ 24.3 = 0.0100 mol
Apply the mole ratio
n(HCl) = 2 × 0.0100 = 0.0200 mol
Make the volume the subject
volume = moles ÷ molarity
Substitute in the values, and calculate
V = 0.0200 ÷ 0.40 = 0.050 L
V = 50 mL
Check — numeric entry
Iron reacts with copper(II) sulfate by Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s). How many milliliters of 0.20 M CuSO₄ consume 0.112 g of Fe (molar mass 55.8 g/mol)? Give your answer in mL, to 2 significant figures.
Feedback when correct: Correct. n(Fe) = 0.112 ÷ 55.8 = 0.0020 mol; the 1:1 ratio needs 0.0020 mol of CuSO₄; V = 0.0020 ÷ 0.20 = 0.010 L = 10 mL.
Anticipated wrong answers (feedback shown):
| 0.010 | This value is in liters. The question asks for milliliters: 0.010 L = 10 mL. |
| 20 | This value doubles the moles. The Fe:CuSO₄ ratio is 1:1, so 0.0020 mol of CuSO₄ is needed: 10 mL. |
Worked solution revealed after a second miss
Write down the values from the question
m(Fe) = 0.112 g, M = 55.8 g/mol
CuSO₄ molarity = 0.20 M
ratio: 1 mol CuSO₄ per 1 mol Fe
Convert the mass to moles
n(Fe) = 0.112 ÷ 55.8 = 0.0020 mol
Apply the mole ratio
n(CuSO₄) = 0.0020 mol
Make the volume the subject
volume = moles ÷ molarity
Substitute in the values, and calculate
V = 0.0020 ÷ 0.20 = 0.010 L
V = 10 mL
Check — numeric entry
A reaction needs 0.030 mol of NaOH, and a bottle's whole 60.0 mL delivers exactly that amount. What is the molarity of the NaOH solution? Give your answer in M, to 2 significant figures.
Feedback when correct: Correct. molarity = moles ÷ volume = 0.030 ÷ 0.0600 L = 0.50 M.
Anticipated wrong answers (feedback shown):
| 0.0005 | This value divides by 60 as if the milliliters were liters. Convert first: 60.0 mL = 0.0600 L, so 0.030 ÷ 0.0600 = 0.50 M. |
| 2.0 | This value divides volume by moles. Molarity = moles ÷ volume in liters: 0.030 ÷ 0.0600 = 0.50 M. |
Worked solution revealed after a second miss
Write down the values from the question
n = 0.030 mol
V = 60.0 mL = 0.0600 L
Write down the equation
molarity = moles ÷ volume (L)
Substitute in the values, and calculate
molarity = 0.030 ÷ 0.0600
molarity = 0.50 M
Outcome: Calculate the pressure, volume, or temperature of a gaseous species via PV = nRT after the mole ratio supplies its moles.
A student heats 5.0 g of calcium carbonate, CaCO₃, in a test tube.
The heat decomposes the solid by CaCO₃(s) → CaO(s) + CO₂(g).
The carbon dioxide escapes into a collection vessel at 298 K and 1.0 atm.
The question asks for the volume of the collected CO₂, in liters.
The familiar chain converts 5.0 g of CaCO₃ into 0.050 mol of CO₂.
Moles are not yet a volume.
What converts 0.050 mol of gas into liters at 298 K and 1.0 atm?
You've already seen the tool in Unit 3: the ideal gas law, PV = nRT.
The mole ratio supplies n, the moles of gas.
The question states the temperature and the pressure — here, 298 K and 1.0 atm.
R is printed on the AP equation sheet: R = 0.08206 L·atm·mol⁻¹·K⁻¹.
With V as the subject, the substitution delivers the volume.
Nothing else in the chain changes.
The ideal gas law brings every gas-making reaction into the mole-ratio chain.
Calculate the pressure, volume, or temperature of a gaseous species via PV = nRT, after the mole ratio supplies its moles.
Q: What volume of CO₂, in liters, forms when 5.0 g of CaCO₃ decomposes completely at 298 K and 1.0 atm? (CaCO₃(s) → CaO(s) + CO₂(g); molar mass of CaCO₃ = 100.1 g/mol; R = 0.08206 L·atm·mol⁻¹·K⁻¹)
Write down the values from the question
m(CaCO₃) = 5.0 g, M = 100.1 g/mol
T = 298 K, P = 1.0 atm
R = 0.08206 L·atm·mol⁻¹·K⁻¹
Convert the mass to moles
n(CaCO₃) = 5.0 ÷ 100.1 = 0.050 mol
n(CO₂) = 0.050 mol(1 mol of CO₂ per 1 mol of CaCO₃)
Write down the equation
PV = nRT
Make V the subject
V = nRT ÷ P
Substitute in the values, and calculate
V = 0.050 × 0.08206 × 298 ÷ 1.0
V = 1.2 L
Check — multiple choice
You've already seen the ideal gas law in Unit 3. Which equation is the ideal gas law?
Feedback if chosen: This option is wrong — you divided by the temperature. The ideal gas law multiplies: PV = nRT.
Feedback if chosen: This option is wrong — you swapped T and V. Pressure × volume sits on the left: PV = nRT.
Feedback if chosen: This option is wrong — you multiplied the volume onto the right side. Pressure × volume equals nRT: PV = nRT.
Feedback if chosen: Correct. PV = nRT, with R = 0.08206 L·atm·mol⁻¹·K⁻¹ on the AP equation sheet.
Check — numeric entry
Heating decomposes magnesium carbonate by MgCO₃(s) → MgO(s) + CO₂(g). 8.4 g of MgCO₃ (molar mass 84.3 g/mol) decomposes completely. How many moles of CO₂ form? Give your answer in mol, to 2 significant figures.
Feedback when correct: Correct. n(MgCO₃) = 8.4 ÷ 84.3 = 0.10 mol, and the 1:1 ratio gives 0.10 mol of CO₂.
Anticipated wrong answers (feedback shown):
| 0.050 | This value halves the moles. The ratio here is 1 CO₂ per 1 MgCO₃, so no halving: 0.10 mol. |
| 708 | This value multiplies by the molar mass. Divide instead: n = 8.4 ÷ 84.3 = 0.10 mol. |
Worked solution revealed after a second miss
Write down the values from the question
m(MgCO₃) = 8.4 g
M = 84.3 g/mol
ratio: 1 mol CO₂ per 1 mol MgCO₃
Write down the equation
n = m ÷ M
Substitute in the values, and calculate
n = 8.4 ÷ 84.3 = 0.10 mol
n(CO₂) = 0.10 mol
Check — multiple choice
Starting from PV = nRT, which line makes V the subject?
Feedback if chosen: This option is wrong — you inverted the fraction. Dividing both sides of PV = nRT by P gives V = nRT ÷ P.
Feedback if chosen: This option is wrong — you swapped T and P. Dividing both sides by P gives V = nRT ÷ P.
Feedback if chosen: Correct. Dividing both sides of PV = nRT by P gives V = nRT ÷ P.
Feedback if chosen: This option is wrong — you moved n to the bottom. n stays on top: V = nRT ÷ P.
Check — numeric entry
A reaction produces 0.10 mol of CO₂, and the gas is collected at 298 K and 1.0 atm (R = 0.08206 L·atm·mol⁻¹·K⁻¹). What volume does the CO₂ fill? Give your answer in liters, to 2 significant figures.
Tolerance: 0.1
Feedback when correct: Correct. V = nRT ÷ P = 0.10 × 0.08206 × 298 ÷ 1.0 = 2.4 L.
Anticipated wrong answers (feedback shown):
| 0.41 | This value divides P by nRT. Make V the subject the other way: V = nRT ÷ P = 2.4 L. |
| 250 | This value uses R = 8.314 J·mol⁻¹·K⁻¹, the joule form. With pressure in atm and volume in liters, R = 0.08206 L·atm·mol⁻¹·K⁻¹: V = 2.4 L. |
Worked solution revealed after a second miss
Write down the values from the question
n = 0.10 mol
T = 298 K, P = 1.0 atm
R = 0.08206 L·atm·mol⁻¹·K⁻¹
Write down the equation
PV = nRT
Make V the subject
V = nRT ÷ P
Substitute in the values, and calculate
V = 0.10 × 0.08206 × 298 ÷ 1.0
V = 2.4 L
Check — numeric entry
Baking soda decomposes on heating by 2NaHCO₃(s) → Na₂CO₃(s) + H₂O(g) + CO₂(g). 8.4 g of NaHCO₃ (molar mass 84.0 g/mol) decomposes completely, and the CO₂ is collected at 298 K and 1.0 atm (R = 0.08206 L·atm·mol⁻¹·K⁻¹). What volume of CO₂ forms? Give your answer in liters, to 2 significant figures.
Tolerance: 0.05
Feedback when correct: Correct. n(NaHCO₃) = 8.4 ÷ 84.0 = 0.10 mol; the ratio gives 0.050 mol of CO₂; V = 0.050 × 0.08206 × 298 ÷ 1.0 = 1.2 L.
Anticipated wrong answers (feedback shown):
| 2.4 | This value skips the mole ratio. The equation makes 1 CO₂ per 2 NaHCO₃, so n(CO₂) = 0.050 mol and V = 1.2 L. |
| 0.61 | This value halves twice. Halve the 0.10 mol of NaHCO₃ once: n(CO₂) = 0.050 mol, so V = 1.2 L. |
Worked solution revealed after a second miss
Write down the values from the question
m(NaHCO₃) = 8.4 g, M = 84.0 g/mol
T = 298 K, P = 1.0 atm
R = 0.08206 L·atm·mol⁻¹·K⁻¹
Convert the mass to moles
n(NaHCO₃) = 8.4 ÷ 84.0 = 0.10 mol
n(CO₂) = 0.10 ÷ 2 = 0.050 mol
Write down the equation
PV = nRT
Make V the subject
V = nRT ÷ P
Substitute in the values, and calculate
V = 0.050 × 0.08206 × 298 ÷ 1.0
V = 1.2 L
Check — numeric entry
Heating decomposes potassium chlorate by 2KClO₃(s) → 2KCl(s) + 3O₂(g). 1.23 g of KClO₃ (molar mass 122.6 g/mol) decomposes completely, and the O₂ collects in a sealed 0.50 L vessel at 298 K (R = 0.08206 L·atm·mol⁻¹·K⁻¹). What pressure does the O₂ exert? Give your answer in atm, to 2 significant figures.
Tolerance: 0.02
Feedback when correct: Correct. n(KClO₃) = 1.23 ÷ 122.6 = 0.01003 mol; the ratio gives n(O₂) = 0.01505 mol; P = nRT ÷ V = 0.01505 × 0.08206 × 298 ÷ 0.50 = 0.74 atm.
Anticipated wrong answers (feedback shown):
| 0.49 | This value uses the KClO₃ moles for the gas. The ratio is 3 O₂ per 2 KClO₃: n(O₂) = 0.01505 mol, so P = 0.74 atm. |
| 0.37 | This value halves the O₂. The ratio multiplies by 3 ÷ 2: n(O₂) = 0.01505 mol, so P = 0.74 atm. |
Worked solution revealed after a second miss
Write down the values from the question
m(KClO₃) = 1.23 g, M = 122.6 g/mol
V = 0.50 L, T = 298 K
R = 0.08206 L·atm·mol⁻¹·K⁻¹
Convert the mass to moles
n(KClO₃) = 1.23 ÷ 122.6 = 0.01003 mol
n(O₂) = 0.01003 × 3 ÷ 2 = 0.01505 mol
keep the unrounded moles for the next step
Write down the equation
PV = nRT
Make P the subject
P = nRT ÷ V
Substitute in the values, and calculate
P = 0.01505 × 0.08206 × 298 ÷ 0.50
P = 0.74 atm
Outcome: Calculate the amount of a solid or dissolved species from the measured pressure, volume, and temperature of a gas consumed or produced.
A strip of magnesium ribbon fizzes away in hydrochloric acid.
The reaction reads Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g).
A gas syringe traps the hydrogen: 0.25 L at 298 K and 2.0 atm.
The acid dissolved every visible scrap of the magnesium.
The only surviving evidence is the trapped gas.
How does the trapped H₂ reveal the grams of Mg that reacted?
PV = nRT, with n as the subject, converts the gas data into moles: n = PV ÷ RT.
The mole ratio matches 1 mol of H₂ to 1 mol of Mg.
The molar mass then turns the moles of Mg into grams.
The gas syringe stood in for the balance.
AP questions use this direction often: the collected gas is the evidence for the reacting mass.
Calculate the amount of a solid or dissolved species from the measured pressure, volume, and temperature of a gas consumed or produced.
Q: What mass of Mg reacted, if the H₂ produced measures 0.25 L at 298 K and 2.0 atm? (Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g); molar mass of Mg = 24.3 g/mol; R = 0.08206 L·atm·mol⁻¹·K⁻¹)
Write down the values from the question
P = 2.0 atm, V = 0.25 L, T = 298 K
R = 0.08206 L·atm·mol⁻¹·K⁻¹
M(Mg) = 24.3 g/mol
Write down the equation
PV = nRT
Make n the subject
n = PV ÷ RT
Substitute in the values, and calculate
n(H₂) = 2.0 × 0.25 ÷ (0.08206 × 298)
n(H₂) = 0.0204 mol
Apply the mole ratio, and convert to mass
n(Mg) = 0.0204 mol(1 mol of Mg per 1 mol of H₂)
m(Mg) = 0.0204 × 24.3 = 0.50 g
Check — multiple choice
A gas is collected at 25 °C. Which temperature goes into PV = nRT?
Feedback if chosen: Correct. K = °C + 273 = 25 + 273 = 298 K.
Feedback if chosen: This option is wrong — 273 K is 0 °C. Add the 25: K = 25 + 273 = 298 K.
Feedback if chosen: This option is wrong — you kept the Celsius number. Convert: K = °C + 273, so 25 °C = 298 K.
Feedback if chosen: This option is wrong — you subtracted the wrong way. K = °C + 273 = 298 K, and kelvin temperatures are never negative.
Check — numeric entry
Zinc reacts with hydrochloric acid by Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g). The H₂ collected measures 0.50 L at 25 °C (298 K) and 1.0 atm (R = 0.08206 L·atm·mol⁻¹·K⁻¹). How many moles of H₂ is that? Give your answer in mol, to 2 significant figures.
Tolerance: 0.001
Feedback when correct: Correct. n = PV ÷ RT = 1.0 × 0.50 ÷ (0.08206 × 298) = 0.020 mol of H₂.
Anticipated wrong answers (feedback shown):
| 0.24 | This value uses 25 in place of the temperature. PV = nRT takes kelvin: T = 298 K, so n = 0.020 mol. |
Worked solution revealed after a second miss
Write down the values from the question
P = 1.0 atm, V = 0.50 L, T = 298 K
R = 0.08206 L·atm·mol⁻¹·K⁻¹
Write down the equation
PV = nRT
Make n the subject
n = PV ÷ RT
Substitute in the values, and calculate
n = 1.0 × 0.50 ÷ (0.08206 × 298)
n = 0.020 mol
Check — numeric entry
Zinc reacts with hydrochloric acid by Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g). A reaction produces 0.020 mol of H₂. How many grams of Zn reacted? (molar mass of Zn = 65.4 g/mol) Give your answer in g, to 2 significant figures.
Tolerance: 0.1
Feedback when correct: Correct. The Zn:H₂ ratio is 1:1, so n(Zn) = 0.020 mol, and m = 0.020 × 65.4 = 1.3 g.
Anticipated wrong answers (feedback shown):
| 0.65 | This value halves the zinc using the HCl coefficient. The Zn:H₂ ratio is 1:1: m = 0.020 × 65.4 = 1.3 g. |
| 2.6 | This value doubles the zinc using the HCl coefficient. The Zn:H₂ ratio is 1:1: m = 0.020 × 65.4 = 1.3 g. |
Worked solution revealed after a second miss
Write down the values from the question
n(H₂) = 0.020 mol
ratio: 1 mol Zn per 1 mol H₂
M(Zn) = 65.4 g/mol
Apply the mole ratio
n(Zn) = 0.020 mol
Substitute in the values, and calculate
m = 0.020 × 65.4
m = 1.3 g
Check — numeric entry
Iron reacts with hydrochloric acid by Fe(s) + 2HCl(aq) → FeCl₂(aq) + H₂(g). The H₂ collected measures 0.49 L at 298 K and 1.0 atm (R = 0.08206 L·atm·mol⁻¹·K⁻¹). What mass of Fe reacted? (molar mass of Fe = 55.8 g/mol) Give your answer in g, to 2 significant figures.
Tolerance: 0.05
Feedback when correct: Correct. n(H₂) = 1.0 × 0.49 ÷ (0.08206 × 298) = 0.020 mol; the 1:1 ratio gives 0.020 mol of Fe; m = 0.020 × 55.8 = 1.1 g.
Anticipated wrong answers (feedback shown):
| 0.56 | This value halves the iron using the HCl coefficient. The Fe:H₂ ratio is 1:1, so m = 0.020 × 55.8 = 1.1 g. |
| 2.2 | This value doubles the iron using the HCl coefficient. The Fe:H₂ ratio is 1:1, so m = 0.020 × 55.8 = 1.1 g. |
Worked solution revealed after a second miss
Write down the values from the question
P = 1.0 atm, V = 0.49 L, T = 298 K
R = 0.08206 L·atm·mol⁻¹·K⁻¹
M(Fe) = 55.8 g/mol
Make n the subject
n = PV ÷ RT
Substitute in the values, and calculate
n(H₂) = 1.0 × 0.49 ÷ (0.08206 × 298) = 0.020 mol
Apply the mole ratio, and convert to mass
n(Fe) = 0.020 mol
m = 0.020 × 55.8 = 1.1 g
Outcome: Calculate the answer to a multi-step reaction-amount problem chaining limiting-reactant analysis with mass, molarity, and ideal-gas conversions.
A student mixes 50.0 mL of 0.10 M barium chloride with 75.0 mL of 0.10 M sodium sulfate.
White barium sulfate, BaSO₄, appears as a solid.
The equation reads BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq).
BaSO₄ is insoluble in water — the question supplies this fact.
The question asks only for the mass of BaSO₄ formed, in grams.
Both reactant amounts hide behind molarities and volumes.
Either reactant could limit.
Which chain of steps reaches the grams of BaSO₄?
The plan chains stages this unit already taught.
Step 1: molarity × volume converts each solution into moles.
Step 2: the mole comparison names the limiting reactant.
Step 3: the mole ratio carries the limiting amount to moles of product.
Step 4: the molar mass converts the product moles to grams.
The AP exam's long questions chain these same single steps in fresh orders.
Calculate the answer to a multi-step reaction-amount problem by chaining limiting-reactant analysis with mass, molarity, and ideal-gas conversions.
Q: What mass of BaSO₄ forms when 50.0 mL of 0.10 M BaCl₂ is mixed with 75.0 mL of 0.10 M Na₂SO₄? (BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq); molar mass of BaSO₄ = 233.4 g/mol)
Write down the values from the question
BaCl₂: 0.10 M, 50.0 mL = 0.0500 L
Na₂SO₄: 0.10 M, 75.0 mL = 0.0750 L
M(BaSO₄) = 233.4 g/mol
Convert each solution to moles
n(BaCl₂) = 0.10 × 0.0500 = 0.0050 mol
n(Na₂SO₄) = 0.10 × 0.0750 = 0.0075 mol
Identify the limiting reactant
the ratio is 1:1 — the smaller amount is used up first
0.0050 mol < 0.0075 mol — BaCl₂ is limiting
Convert the limiting amount to moles of product
n(BaSO₄) = 0.0050 mol(1 mol of BaSO₄ per 1 mol of BaCl₂)
Convert to mass
m = 0.0050 × 233.4
m = 1.2 g
Check — multiple choice
Silver nitrate reacts with calcium chloride by 2AgNO₃(aq) + CaCl₂(aq) → 2AgCl(s) + Ca(NO₃)₂(aq); AgCl is insoluble in water (supplied). 40.0 mL of 0.25 M AgNO₃ is mixed with 30.0 mL of 0.20 M CaCl₂, and the question asks for the mass of AgCl. Which plan solves it, in order?
Feedback if chosen: This option is wrong — the mole ratio has nothing to convert until the moles exist. The chain starts from M × V for each solution.
Feedback if chosen: Correct. Convert each solution to moles, name the limiting reactant, carry its amount through the mole ratio, then convert to grams.
Feedback if chosen: This option is wrong — no reactant masses are given. The starting quantities are molarities and volumes, so the chain starts from M × V.
Feedback if chosen: This option is wrong — converting reactants to mass adds a dead-end step. After the moles, compare to find the limiting reactant, then go to the product.
Check — numeric entry
A chemist mixes 40.0 mL of 0.25 M AgNO₃ with 30.0 mL of 0.20 M CaCl₂; the equation is 2AgNO₃(aq) + CaCl₂(aq) → 2AgCl(s) + Ca(NO₃)₂(aq), and AgCl is insoluble in water. What mass of AgCl forms? (molar mass of AgCl = 143.3 g/mol) Give your answer in g, to 2 significant figures.
Tolerance: 0.05
Feedback when correct: Correct. n(AgNO₃) = 0.010 mol, n(CaCl₂) = 0.0060 mol; the CaCl₂ demands 0.012 mol of AgNO₃, so AgNO₃ is limiting; n(AgCl) = 0.010 mol; m = 0.010 × 143.3 = 1.4 g.
Anticipated wrong answers (feedback shown):
| 1.7 | This value converts the CaCl₂ — the excess reactant — to product. The limiting AgNO₃ (0.010 mol) caps the AgCl: 1.4 g. |
| 0.72 | This value halves the AgCl moles. The ratio is 2 AgCl per 2 AgNO₃ — one for one — so n(AgCl) = 0.010 mol: 1.4 g. |
Worked solution revealed after a second miss
Write down the values from the question
AgNO₃: 0.25 M, 40.0 mL = 0.0400 L
CaCl₂: 0.20 M, 30.0 mL = 0.0300 L
M(AgCl) = 143.3 g/mol
Convert each solution to moles
n(AgNO₃) = 0.25 × 0.0400 = 0.010 mol
n(CaCl₂) = 0.20 × 0.0300 = 0.0060 mol
Identify the limiting reactant
AgNO₃ required = 2 × 0.0060 = 0.012 mol
available 0.010 mol < required 0.012 mol — AgNO₃ is limiting
Convert the limiting amount to moles of product
n(AgCl) = 0.010 mol
Convert to mass
m = 0.010 × 143.3
m = 1.4 g
Check — numeric entry
Sodium carbonate reacts with hydrochloric acid by Na₂CO₃(aq) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g). 50.0 mL of 0.50 M Na₂CO₃ is mixed with 100.0 mL of 0.40 M HCl, and the CO₂ escapes at 298 K and 1.0 atm (R = 0.08206 L·atm·mol⁻¹·K⁻¹). What volume of CO₂ forms? Give your answer in liters, to 2 significant figures.
Tolerance: 0.02
Feedback when correct: Correct. n(Na₂CO₃) = 0.025 mol, n(HCl) = 0.040 mol; the Na₂CO₃ demands 0.050 mol of HCl, so HCl is limiting; n(CO₂) = 0.040 ÷ 2 = 0.020 mol; V = 0.020 × 0.08206 × 298 ÷ 1.0 = 0.49 L.
Anticipated wrong answers (feedback shown):
| 0.61 | This value converts the Na₂CO₃ — the excess reactant. The limiting HCl caps the CO₂ at 0.040 ÷ 2 = 0.020 mol: V = 0.49 L. |
| 0.98 | This value skips the mole ratio. The equation makes 1 CO₂ per 2 HCl, so n(CO₂) = 0.020 mol and V = 0.49 L. |
Worked solution revealed after a second miss
Write down the values from the question
Na₂CO₃: 0.50 M, 50.0 mL = 0.0500 L
HCl: 0.40 M, 100.0 mL = 0.1000 L
T = 298 K, P = 1.0 atm, R = 0.08206 L·atm·mol⁻¹·K⁻¹
Convert each solution to moles
n(Na₂CO₃) = 0.50 × 0.0500 = 0.025 mol
n(HCl) = 0.40 × 0.1000 = 0.040 mol
Identify the limiting reactant
HCl required = 2 × 0.025 = 0.050 mol
available 0.040 mol < required 0.050 mol — HCl is limiting
Convert the limiting amount to moles of product
n(CO₂) = 0.040 ÷ 2 = 0.020 mol
Make V the subject, and calculate
V = nRT ÷ P
V = 0.020 × 0.08206 × 298 ÷ 1.0 = 0.49 L
Outcome: Earn every point of a chained calculation: setup shown, every mole relationship written, sig figs kept, errors carried consistently.
Calculation points pay only with visible work: the exam booklet says you must show the method and the steps to receive credit — a bare answer earns zero, even when it is right.
Writing the bare formula is not work; substituted values are. Write every conversion with its numbers and units, including every mole relationship from the equation's coefficients.
In a two-point calculation, the first point usually pays for the setup and first conversion — so a chain with an arithmetic slip at the end still collects the setup point.
A wrong answer carried into a later part does not cost you twice: graders award the later point when your work uses the earlier value consistently.
Watch the flagged points: where a part demands the correct number of significant figures, a right value at the wrong precision loses the point.
Answer the question below on paper, writing each conversion as its own line with units. Then reveal the model and score yourself one rubric line at a time.
Free-response question
Source note (not shown as part of the question): authored to the multi-step calculation archetype (FRQ playbook T3; modeled on 2022 FRQ Q1(a–b) and 2024 FRQ Q2(c))
Lithium metal reacts with nitrogen gas to form solid lithium nitride: 6Li(s) + N₂(g) → 2Li₃N(s). A student reacts 2.08 g of lithium (molar mass 6.94 g/mol) with more than enough N₂, and the reaction goes to completion. (Molar mass of Li₃N: 34.83 g/mol.)
(a) Calculate the number of moles of Li in the 2.08 g sample. [1 point]
(b) Calculate the theoretical yield of Li₃N, in grams, reported to the correct number of significant figures. [2 points]
(c) The student collects 2.61 g of Li₃N. Calculate the percent yield. [1 point]
(d) The student claims that the percent yield below 100% proves some of the lithium never reacted. Do you agree with the claim? Justify your answer. [1 point]
✓ Model response (the key)
(a)
n(Li) = 2.08 g ÷ 6.94 g/mol = 0.300 mol
(b)
n(Li₃N) = 0.300 mol Li × (2 mol Li₃N / 6 mol Li) = 0.100 mol
m(Li₃N) = 0.100 mol × 34.83 g/mol = 3.48 g
(c)
percent yield = (2.61 g ÷ 3.48 g) × 100% = 75.0%
(d)
No — the claim is not justified.
The N₂ was in excess and the reaction went to completion, so all of the lithium could be consumed.
Product lost during collection or transfer lowers the collected mass, so a percent yield below 100% can occur with no lithium left unreacted.
Scoring rubric — 5 points available
5 lessons
Outcome: State that a titration determines the amount of an analyte by adding measured volumes of a titrant whose reacting species consumes the analyte in a specific, complete reaction.
A stockroom bottle of hydrochloric acid has lost its concentration label.
A graduated cylinder can measure the acid's volume to a tenth of a milliliter.
No instrument on the bench measures concentration directly.
The lab needs the molarity of that acid for tomorrow's experiment.
A bottle of sodium hydroxide beside it is labeled exactly: 0.100 M.
NaOH consumes HCl completely, by HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l).
How can the known solution measure the unknown one?
Chemists react the unknown sample with a solution of exactly known concentration.
The measuring procedure is called a titration.
The solution of exactly known concentration is called the titrant.
The titrant is added bit by bit from a buret — a tall graduated tube with a valve at the bottom.
The dissolved substance being measured is called the analyte.
The titrant's reacting species consumes the analyte in a specific, complete reaction.
The measured volume of titrant then reveals the amount of analyte.
Here the 0.100 M NaOH is the titrant, and the HCl in the flask is the analyte.
State that a titration determines the amount of an analyte by adding measured volumes of a titrant, a solution of known concentration whose reacting species consumes the analyte in a specific, complete reaction.
Check — multiple choice
A chemist fills a buret with 0.200 M HNO₃ and drains it into a 25.0 mL sample of KOH solution of unknown concentration (HNO₃(aq) + KOH(aq) → KNO₃(aq) + H₂O(l)). Name the titrant.
Feedback if chosen: This option is wrong — the KOH is the substance being measured. The titrant is the solution of exactly known concentration in the buret.
Feedback if chosen: Correct. The titrant is the solution of exactly known concentration, added from the buret — here, the 0.200 M HNO₃. An acid can be the titrant.
Feedback if chosen: This option is wrong — KNO₃ is a product of the reaction. The titrant is one of the two starting solutions: the known 0.200 M HNO₃.
Feedback if chosen: This option is wrong — the water is a product of the reaction. The titrant is the known solution added from the buret: the HNO₃.
Check — multiple choice
A student drains 0.150 M NaOH into a vinegar sample to measure its dissolved acid, HC₂H₃O₂ (HC₂H₃O₂(aq) + NaOH(aq) → NaC₂H₃O₂(aq) + H₂O(l)). Which species is the analyte?
Feedback if chosen: Correct. The analyte is the dissolved substance being measured — the HC₂H₃O₂ in the vinegar sample.
Feedback if chosen: This option is wrong — the NaOH is the known solution doing the measuring: the titrant.
Feedback if chosen: This option is wrong — NaC₂H₃O₂ forms during the reaction. The analyte is the dissolved substance the titration sets out to measure.
Feedback if chosen: This option is wrong — the water is only the sample's liquid. The analyte is the dissolved substance being measured: the HC₂H₃O₂.
Check — multiple choice
What does a titration determine?
Feedback if chosen: This option is wrong — color is not the target. A titration measures the amount of the analyte through a reaction with the titrant.
Feedback if chosen: This option is wrong — a cylinder measures volume directly. A titration measures the amount of the dissolved analyte.
Feedback if chosen: Correct. A titration determines the amount of an analyte from the measured volume of titrant that consumes it.
Feedback if chosen: This option is wrong — temperature plays no measuring role here. A titration measures the amount of the analyte.
Check — multiple choice
A buret delivers 0.100 M Ba(OH)₂ into a sample of HBr solution of unknown concentration (2HBr(aq) + Ba(OH)₂(aq) → BaBr₂(aq) + 2H₂O(l)). Name the titrant.
Feedback if chosen: This option is wrong — the HBr is the analyte, the dissolved substance being measured.
Feedback if chosen: This option is wrong — BaBr₂ is a product. The titrant is the known solution in the buret.
Feedback if chosen: This option is wrong — the water is a product. The titrant is the known solution in the buret.
Feedback if chosen: Correct. The Ba(OH)₂ solution has the exactly known concentration and sits in the buret — it is the titrant. A base can be the titrant.
Check — multiple choice
A student titrates a sample of dissolved Na₂CO₃ with 0.250 M HCl (Na₂CO₃(aq) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g)). Name the analyte.
Feedback if chosen: Correct. The Na₂CO₃ is the dissolved substance being measured — the analyte.
Feedback if chosen: This option is wrong — the HCl is the known solution doing the measuring: the titrant.
Feedback if chosen: This option is wrong — the CO₂ is a product of the reaction, not the substance being measured.
Feedback if chosen: This option is wrong — the NaCl is a product of the reaction, not the substance being measured.
Outcome: State that the equivalence point is reached when the moles of titrant added exactly match the moles needed to consume all the analyte, in the reaction's mole ratio.
A student drains 0.100 M NaOH from the buret into 20.0 mL of the unlabeled HCl (HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)).
Every drop of NaOH consumes some of the HCl.
Early in the titration, plenty of HCl survives in the flask.
Too far in, extra NaOH piles up with no HCl left to consume.
Somewhere between sits one exact moment.
At that moment, the added NaOH matches the HCl completely: nothing of either is left over.
What defines that moment — the volumes, or the moles?
The moles decide, in the reaction's own mole ratio.
HCl reacts with NaOH 1 mol to 1 mol.
So the exact moment arrives when the moles of NaOH added equal the moles of HCl in the flask.
That moment is called the equivalence point.
A 2-to-1 reaction shifts the target: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O demands 2 mol of NaOH per 1 mol of H₂SO₄.
Every titration calculation anchors itself at the equivalence point.
State that the equivalence point is reached when the moles of titrant added exactly match the moles needed to consume all the analyte, in the reaction's mole ratio.
Check — multiple choice
A student titrates an HNO₃ sample with KOH solution (HNO₃(aq) + KOH(aq) → KNO₃(aq) + H₂O(l)). When is the equivalence point reached?
Feedback if chosen: This option is wrong — you matched volumes. The two solutions can have different molarities, so equal volumes rarely mean matched moles.
Feedback if chosen: This option is wrong — warmth is not the definition. The equivalence point is the mole-matching moment, set by the reaction's ratio.
Feedback if chosen: This option is wrong — the first drop leaves nearly all the HNO₃ unconsumed. Equivalence needs the full matching amount of KOH.
Feedback if chosen: Correct. The reaction is 1:1, so equivalence arrives when the moles of KOH added exactly match the moles of HNO₃ in the sample.
Check — multiple choice
A 20.0 mL acid sample is titrated with a base solution, reacting 1 mol to 1 mol. A student predicts the equivalence point at exactly 20.0 mL of base, because the volumes must match. Which statement is correct?
Feedback if chosen: This option is wrong — you carried the mole ratio onto the volumes. A more concentrated base delivers the matching moles in less volume.
Feedback if chosen: Correct. Equivalence is about moles. The base volume that carries the matching moles depends on the two concentrations, so 20.0 mL is a coincidence, not a rule.
Feedback if chosen: This option is wrong — which solution sits in the buret changes nothing. Moles, not volumes, define the equivalence point.
Feedback if chosen: This option is wrong — 'always more' is another volume rule. A concentrated base reaches the matching moles in less volume than the sample's.
Check — multiple choice
Hydrochloric acid titrates a barium hydroxide sample by 2HCl(aq) + Ba(OH)₂(aq) → BaCl₂(aq) + 2H₂O(l). A student claims equivalence arrives when the moles of HCl added equal the moles of Ba(OH)₂ in the sample. How many moles of HCl does the equivalence point actually demand?
Feedback if chosen: This option is wrong — you halved in the wrong direction. Each mol of Ba(OH)₂ consumes 2 mol of HCl, so more HCl is needed, not less.
Feedback if chosen: This option is wrong — you accepted the claim. Equal moles is the 1:1 special case, and this reaction is 2:1.
Feedback if chosen: Correct. The mole ratio sets the target: 2 mol of HCl per 1 mol of Ba(OH)₂. Equal moles would stop halfway to equivalence.
Feedback if chosen: This option is wrong — you doubled twice. The ratio demands 2 mol of HCl per 1 mol of Ba(OH)₂, not 4.
Check — multiple choice
In the same titration, why does the equivalence point demand twice the moles of HCl?
Feedback if chosen: Correct. The coefficients set the target: 2 mol of HCl react with every 1 mol of Ba(OH)₂, so equivalence needs twice the moles.
Feedback if chosen: This option is wrong — the concentrations were never stated. The mole ratio, from the coefficients, sets the target.
Feedback if chosen: This option is wrong — HCl and NaOH react 1:1. The ratio comes from each equation's own coefficients, not from a general rule.
Feedback if chosen: This option is wrong — mass plays no part in equivalence. The coefficients count particles, and each Ba(OH)₂ consumes two HCl.
Check — multiple choice
A flask holds 0.0020 mol of HBr, titrated with NaOH (HBr(aq) + NaOH(aq) → NaBr(aq) + H₂O(l)). At which added amount of NaOH is the equivalence point reached?
Feedback if chosen: This option is wrong — you halved the target. The 1:1 ratio matches the analyte exactly: 0.0020 mol of NaOH.
Feedback if chosen: Correct. The 1:1 ratio means the moles of NaOH added must equal the 0.0020 mol of HBr.
Feedback if chosen: This option is wrong — you doubled the target. The 1:1 ratio matches the analyte exactly: 0.0020 mol of NaOH.
Feedback if chosen: This option is wrong — you multiplied the analyte's moles by four. The 1:1 ratio matches them exactly: 0.0020 mol of NaOH.
Check — multiple choice
A sample contains 0.0010 mol of dissolved Na₂CO₃, titrated with HCl (Na₂CO₃(aq) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g)). At which added amount of HCl is the equivalence point reached?
Feedback if chosen: This option is wrong — you halved instead of doubling. Each Na₂CO₃ consumes 2 HCl: 2 × 0.0010 = 0.0020 mol.
Feedback if chosen: This option is wrong — equal moles is the 1:1 case. This ratio demands 2 mol of HCl per 1 mol of Na₂CO₃: 0.0020 mol.
Feedback if chosen: Correct. The 2:1 ratio demands 2 × 0.0010 = 0.0020 mol of HCl to consume all the Na₂CO₃.
Feedback if chosen: This option is wrong — you multiplied by 4. The ratio is 2:1, so 2 × 0.0010 = 0.0020 mol of HCl.
Outcome: Classify the endpoint — the observable signal from an added indicator dye — as distinct from the equivalence point it is chosen to approximate.
The flask at the equivalence point looks exactly like the flask one drop earlier.
Both mixtures are colorless liquids.
No flash, no sound, no visible signal marks the matching moles.
So a chemist adds a few drops of phenolphthalein before starting.
Phenolphthalein is a dye added to supply a visible signal — a dye used this way is called an indicator.
At one particular drop of NaOH, the whole flask turns faint pink.
The student closes the valve and notes the delivered volume.
Is the pink moment the equivalence point itself, or only a stand-in?
The color change is the endpoint: the observable signal from the added indicator.
| Endpoint | Equivalence point | |
|---|---|---|
| What defines it | The indicator's observable color change | The moles of titrant matching the analyte in the mole ratio |
| What a chemist observes | A color change in the flask | Nothing — the moment is invisible |
| Role in the calculation | Its measured volume stands in | The moment the calculation assumes |
The equivalence point is the invisible mole-matching moment.
The endpoint is chosen to approximate the equivalence point.
The two are different events.
An AP answer names the observed color change the endpoint, never the equivalence point.
Classify the endpoint — the observable signal from an added indicator dye — as distinct from the equivalence point it is chosen to approximate.
Check — multiple choice
During a titration with phenolphthalein added, the flask turns faint pink and stays pink. In a titration, the color change marks the —
Feedback if chosen: Correct. The observable signal from the added indicator is the endpoint — chosen to approximate the equivalence point, but a different event.
Feedback if chosen: This option is wrong — the equivalence point is the invisible mole-matching moment. The visible color change is the endpoint.
Feedback if chosen: This option is wrong — you merged the two events. The indicator's signal lands close to the mole match, but the endpoint and the equivalence point remain different events.
Feedback if chosen: This option is wrong — the color change is exactly what the endpoint names: the indicator's observable signal.
Check — multiple choice
At one instant of a titration, the moles of NaOH added exactly match the moles of HCl in the flask, in the 1:1 ratio. In a titration, that mole match marks the —
Feedback if chosen: This option is wrong — the endpoint is the indicator's visible signal. The mole match itself, visible or not, is the equivalence point.
Feedback if chosen: Correct. The moles of titrant matching the analyte, in the reaction's ratio, is the definition of the equivalence point.
Feedback if chosen: This option is wrong — the mole match and the indicator's signal are two different events. The indicator is chosen so its signal lands close.
Feedback if chosen: This option is wrong — the mole match is exactly what the equivalence point names.
Check — multiple choice
A titration calculation assumes the moles of titrant matched the analyte. Which point does the calculation rest on?
Feedback if chosen: This option is wrong — the endpoint is the indicator's signal, not the mole match. The calculation rests on the mole-matching moment.
Feedback if chosen: Correct. The calculation assumes matched moles, and the equivalence point is exactly that mole-matching moment.
Feedback if chosen: This option is wrong — you made the two names synonyms. The endpoint is the signal; the equivalence point is the mole match it approximates.
Feedback if chosen: This option is wrong — a volume matters only because it delivers the matching moles. The mole match is the equivalence point.
Check — multiple choice
The equivalence point is invisible. Which measured volume stands in for it in the calculation?
Feedback if chosen: Correct. The indicator's signal is chosen to land close to the mole match, so the volume delivered at the endpoint stands in for the equivalence point's volume.
Feedback if chosen: This option is wrong — the sample's volume converts the analyte's moles to a molarity. The stand-in for the invisible moment is the endpoint's delivered volume.
Feedback if chosen: This option is wrong — the buret's capacity is fixed before any titrant flows, so it cannot measure this titration. The calculation uses the volume delivered at the endpoint.
Feedback if chosen: This option is wrong — the flask's total mixes both liquids together. The calculation needs the titrant's own delivered volume, read at the endpoint.
Check — multiple choice
A student sees the first permanent color change in the flask. In a titration, that observation marks the —
Feedback if chosen: This option is wrong — the equivalence point is the invisible mole match. The observed color change is the endpoint.
Feedback if chosen: This option is wrong — the color change and the mole match are two different events. The signal is chosen to land close to the match, nothing more.
Feedback if chosen: Correct. The observable signal from the added indicator is the endpoint.
Feedback if chosen: This option is wrong — the first permanent color change is exactly what the endpoint names.
Check — multiple choice
In one titration, no indicator was added — yet at some instant the moles of titrant matched the analyte. Which point still occurred?
Feedback if chosen: This option is wrong — with no indicator there is no signal, so no endpoint. The mole match still happened.
Feedback if chosen: This option is wrong — the endpoint needs the indicator's signal, and no indicator was added.
Feedback if chosen: Correct. The equivalence point is the mole match itself — it occurs whether or not anything shows it.
Feedback if chosen: This option is wrong — the moles still matched at some instant; that mole match is the equivalence point.
Outcome: Calculate the titrant volume required to reach the equivalence point from the analyte amount, the titrant molarity, and the mole ratio.
A sample flask contains 0.0025 mol of HCl.
The buret above the flask is filled with 0.100 M NaOH (HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)).
A buret carries only 50.0 mL in one filling.
Halfway through a titration is a bad time to refill.
The student wants the target volume in advance.
How many milliliters of the 0.100 M NaOH reach the equivalence point?
At the equivalence point, the moles of NaOH added equal the moles of HCl present.
The flask's 0.0025 mol of HCl therefore demands 0.0025 mol of NaOH.
You've already seen the volume step: volume = moles ÷ molarity.
So 0.0025 ÷ 0.100 = 0.0250 L — the target is 25.0 mL, well within one filling.
A 2-to-1 reaction doubles the demanded moles before the volume step.
AP items use this prediction to check one idea: the equivalence point is about moles, never about equal volumes.
Calculate the titrant volume required to reach the equivalence point from the analyte amount, the titrant molarity, and the mole ratio.
Q: How many milliliters of 0.100 M NaOH reach the equivalence point with 0.0025 mol of HCl? (HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l))
Write down the values from the question
n(HCl) = 0.0025 mol
NaOH molarity = 0.100 M
ratio: 1 mol NaOH per 1 mol HCl
Work out the moles the equivalence point demands
n(NaOH) = 0.0025 mol
Make the volume the subject
volume = moles ÷ molarity
Substitute in the values, and calculate
V = 0.0025 ÷ 0.100 = 0.0250 L
V = 25.0 mL
Check — multiple choice
You've already seen this tool earlier in this unit. Which line gives the volume of solution that delivers a needed amount of moles?
Feedback if chosen: This option is wrong — multiplying grows the volume with the concentration. Rearranging molarity = moles ÷ volume gives volume = moles ÷ molarity.
Feedback if chosen: This option is wrong — you inverted the fraction. Volume = moles ÷ molarity.
Feedback if chosen: This option is wrong — you inverted the whole expression. Volume = moles ÷ molarity.
Feedback if chosen: Correct. Rearranging molarity = moles ÷ volume gives volume = moles ÷ molarity.
Check — numeric entry
A flask contains 0.0040 mol of HNO₃, titrated with 0.200 M KOH (HNO₃(aq) + KOH(aq) → KNO₃(aq) + H₂O(l)). How many milliliters of the KOH reach the equivalence point? Give your answer in mL, to 3 significant figures.
Feedback when correct: Correct. The 1:1 ratio demands 0.0040 mol of KOH; V = 0.0040 ÷ 0.200 = 0.0200 L = 20.0 mL.
Anticipated wrong answers (feedback shown):
| 0.0200 | This value is in liters. The question asks for milliliters: 0.0200 L = 20.0 mL. |
| 10.0 | This value halves the moles. The 1:1 ratio demands the full 0.0040 mol of KOH: 20.0 mL. |
| 40.0 | This value doubles the moles. The 1:1 ratio demands exactly 0.0040 mol of KOH: 20.0 mL. |
Worked solution revealed after a second miss
Write down the values from the question
n(HNO₃) = 0.0040 mol
KOH molarity = 0.200 M
ratio: 1 mol KOH per 1 mol HNO₃
Work out the moles the equivalence point demands
n(KOH) = 0.0040 mol
Make the volume the subject
volume = moles ÷ molarity
Substitute in the values, and calculate
V = 0.0040 ÷ 0.200 = 0.0200 L
V = 20.0 mL
Check — numeric entry
A flask contains 0.0030 mol of H₂SO₄, titrated with NaOH by H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l). How many moles of NaOH does the equivalence point demand? Give your answer in mol, to 2 significant figures.
Feedback when correct: Correct. The ratio demands 2 mol of NaOH per 1 mol of H₂SO₄: 2 × 0.0030 = 0.0060 mol.
Anticipated wrong answers (feedback shown):
| 0.0030 | This value assumes a 1:1 ratio. The equation demands 2 mol of NaOH per 1 mol of H₂SO₄: 0.0060 mol. |
| 0.0015 | This value halves instead of doubling. The ratio is 2 NaOH per 1 H₂SO₄: 2 × 0.0030 = 0.0060 mol. |
Worked solution revealed after a second miss
Write down the values from the question
n(H₂SO₄) = 0.0030 mol
ratio: 2 mol NaOH per 1 mol H₂SO₄
Substitute in the values, and calculate
n(NaOH) = 2 × 0.0030
n(NaOH) = 0.0060 mol
Check — numeric entry
The NaOH in that titration is 0.150 M, and the equivalence point demands 0.0060 mol of NaOH. How many milliliters of the NaOH reach the equivalence point? Give your answer in mL, to 3 significant figures.
Feedback when correct: Correct. V = moles ÷ molarity = 0.0060 ÷ 0.150 = 0.0400 L = 40.0 mL.
Anticipated wrong answers (feedback shown):
| 0.0400 | This value is in liters. The question asks for milliliters: 0.0400 L = 40.0 mL. |
| 20.0 | This value halves the demanded moles. The full 0.0060 mol of NaOH is needed: 40.0 mL. |
Worked solution revealed after a second miss
Write down the values from the question
n(NaOH) = 0.0060 mol
NaOH molarity = 0.150 M
Make the volume the subject
volume = moles ÷ molarity
Substitute in the values, and calculate
V = 0.0060 ÷ 0.150 = 0.0400 L
V = 40.0 mL
Check — numeric entry
A flask contains 0.0075 mol of HBr, titrated with 0.250 M LiOH (HBr(aq) + LiOH(aq) → LiBr(aq) + H₂O(l)). How many milliliters of the LiOH reach the equivalence point? Give your answer in mL, to 3 significant figures.
Feedback when correct: Correct. The 1:1 ratio demands 0.0075 mol of LiOH; V = 0.0075 ÷ 0.250 = 0.0300 L = 30.0 mL.
Anticipated wrong answers (feedback shown):
| 0.0300 | This value is in liters. The question asks for milliliters: 0.0300 L = 30.0 mL. |
| 60.0 | This value doubles the moles. The 1:1 ratio demands exactly 0.0075 mol of LiOH: 30.0 mL. |
| 15.0 | This value halves the moles. The 1:1 ratio demands the full 0.0075 mol of LiOH: 30.0 mL. |
Worked solution revealed after a second miss
Write down the values from the question
n(HBr) = 0.0075 mol
LiOH molarity = 0.250 M
ratio: 1 mol LiOH per 1 mol HBr
Work out the moles the equivalence point demands
n(LiOH) = 0.0075 mol
Make the volume the subject
volume = moles ÷ molarity
Substitute in the values, and calculate
V = 0.0075 ÷ 0.250 = 0.0300 L
V = 30.0 mL
Outcome: Calculate the amount or concentration of an analyte from titration data at the equivalence point, assuming the titration reaction goes to completion.
The label-less acid bottle from the start of this sequence still waits.
A student titrates a 20.0 mL sample of the acid with 0.100 M NaOH (HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)).
The indicator turns faint pink after exactly 25.0 mL of NaOH.
The endpoint stands in for the equivalence point.
So the moles of NaOH delivered match the moles of HCl in the sample.
Every needed quantity now exists: 20.0 mL of acid, 25.0 mL of NaOH, 0.100 M.
What molarity belongs on the bottle's new label?
Step 1: molarity × volume converts the delivered NaOH into moles.
Step 2: the mole ratio places the matching moles of HCl in the sample.
Step 3: moles ÷ sample volume converts back to molarity.
The whole routine assumes the titration reaction goes to completion.
Unit 8 builds its further titration work on this exact chain.
Calculate the amount or concentration of an analyte from titration data at the equivalence point, assuming the titration reaction goes to completion.
Q: A 20.0 mL HCl sample reaches the equivalence point after 25.0 mL of 0.100 M NaOH. What is the molarity of the HCl? (HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l))
Write down the values from the question
NaOH: 0.100 M, 25.0 mL = 0.0250 L
HCl sample volume = 20.0 mL = 0.0200 L
ratio: 1 mol HCl per 1 mol NaOH
Convert the delivered titrant to moles
n(NaOH) = 0.100 × 0.0250 = 0.0025 mol
Apply the mole ratio
n(HCl) = 0.0025 mol
Convert back to molarity
molarity = moles ÷ volume
M(HCl) = 0.0025 ÷ 0.0200 = 0.125 M
Check — multiple choice
You've already seen this tool earlier in this unit. Which equation converts a solution's molarity and volume into moles of solute?
Feedback if chosen: This option is wrong — dividing by the volume shrinks the moles as the solution grows. Moles = molarity × volume in liters.
Feedback if chosen: This option is wrong — volume ÷ molarity has the wrong shape. Moles = molarity × volume in liters.
Feedback if chosen: This option is wrong — you kept the volume in milliliters. Molarity is moles per liter, so the volume must be in liters.
Feedback if chosen: Correct. Molarity is moles per liter, so multiplying by the volume in liters returns the moles of solute.
Check — numeric entry
A 25.0 mL sample of HBr solution reaches the equivalence point after 30.0 mL of 0.150 M KOH (HBr(aq) + KOH(aq) → KBr(aq) + H₂O(l)). How many moles of KOH were delivered? Give your answer in mol, to 2 significant figures.
Feedback when correct: Correct. n = M × V = 0.150 × 0.0300 L = 0.0045 mol of KOH.
Anticipated wrong answers (feedback shown):
| 4.5 | This value keeps the volume in milliliters. Convert first: 30.0 mL = 0.0300 L, so n = 0.150 × 0.0300 = 0.0045 mol. |
| 0.0038 | This value uses the sample's 25.0 mL. The delivered titrant is the KOH: 0.150 × 0.0300 = 0.0045 mol. |
Worked solution revealed after a second miss
Write down the values from the question
KOH: 0.150 M, 30.0 mL = 0.0300 L
Write down the equation
moles = molarity × volume (L)
Substitute in the values, and calculate
n = 0.150 × 0.0300
n = 0.0045 mol
Check — numeric entry
In the same titration, 0.0045 mol of KOH was delivered at the equivalence point (HBr + KOH react 1 mol to 1 mol). How many moles of HBr were in the sample? Give your answer in mol, to 2 significant figures.
Feedback when correct: Correct. At the equivalence point the 1:1 ratio matches the moles exactly: 0.0045 mol of HBr.
Anticipated wrong answers (feedback shown):
| 0.0090 | This value doubles the moles. The ratio is 1:1, so the HBr matches the KOH exactly: 0.0045 mol. |
| 0.0023 | This value halves the moles. The ratio is 1:1, so the HBr matches the KOH exactly: 0.0045 mol. |
Worked solution revealed after a second miss
Write down the values from the question
n(KOH) = 0.0045 mol
ratio: 1 mol HBr per 1 mol KOH
Apply the mole ratio
n(HBr) = 0.0045 mol
Check — numeric entry
The 0.0045 mol of HBr sat in the 25.0 mL sample. What is the molarity of the HBr solution? Give your answer in M, to 2 significant figures.
Feedback when correct: Correct. molarity = moles ÷ volume = 0.0045 ÷ 0.0250 L = 0.18 M.
Anticipated wrong answers (feedback shown):
| 0.00018 | This value divides by 25 as if the milliliters were liters. Convert first: 25.0 mL = 0.0250 L, so 0.0045 ÷ 0.0250 = 0.18 M. |
| 0.15 | This value copies the titrant's molarity. The sample's molarity comes from its own moles and volume: 0.0045 ÷ 0.0250 = 0.18 M. |
Worked solution revealed after a second miss
Write down the values from the question
n(HBr) = 0.0045 mol
sample volume = 25.0 mL = 0.0250 L
Write down the equation
molarity = moles ÷ volume (L)
Substitute in the values, and calculate
M = 0.0045 ÷ 0.0250
M = 0.18 M
Check — numeric entry
A 10.0 mL sample of H₂SO₄ solution reaches the equivalence point after 24.0 mL of 0.100 M NaOH (H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)). What is the molarity of the H₂SO₄? Give your answer in M, to 3 significant figures.
Feedback when correct: Correct. n(NaOH) = 0.100 × 0.0240 = 0.00240 mol; the 2:1 ratio gives n(H₂SO₄) = 0.00120 mol; M = 0.00120 ÷ 0.0100 = 0.120 M.
Anticipated wrong answers (feedback shown):
| 0.240 | This value skips the mole ratio. Each H₂SO₄ consumed 2 NaOH, so n(H₂SO₄) = 0.00240 ÷ 2 = 0.00120 mol: 0.120 M. |
| 0.480 | This value doubles instead of halving. The ratio gives n(H₂SO₄) = 0.00240 ÷ 2 = 0.00120 mol: 0.120 M. |
Worked solution revealed after a second miss
Write down the values from the question
NaOH: 0.100 M, 24.0 mL = 0.0240 L
sample volume = 10.0 mL = 0.0100 L
ratio: 1 mol H₂SO₄ per 2 mol NaOH
Convert the delivered titrant to moles
n(NaOH) = 0.100 × 0.0240 = 0.00240 mol
Apply the mole ratio
n(H₂SO₄) = 0.00240 ÷ 2 = 0.00120 mol
Convert back to molarity
M = 0.00120 ÷ 0.0100
M = 0.120 M
Check — numeric entry
A 50.0 mL sample contains dissolved KOH, the analyte. Titrated with 0.200 M HCl (HCl(aq) + KOH(aq) → KCl(aq) + H₂O(l)), it reaches the equivalence point after 15.0 mL of the HCl. What mass of KOH did the sample contain? (molar mass of KOH = 56.1 g/mol) Give your answer in g, to 2 significant figures.
Tolerance: 0.01
Feedback when correct: Correct. n(HCl) = 0.200 × 0.0150 = 0.0030 mol; the 1:1 ratio gives 0.0030 mol of KOH; m = 0.0030 × 56.1 = 0.17 g.
Anticipated wrong answers (feedback shown):
| 0.0030 | This value stops at the moles of KOH. The question asks for the mass: 0.0030 × 56.1 = 0.17 g. |
| 0.56 | This value uses the sample's 50.0 mL to compute moles. The delivered titrant fixes the moles: 0.200 × 0.0150 = 0.0030 mol, so 0.17 g. |
Worked solution revealed after a second miss
Write down the values from the question
HCl: 0.200 M, 15.0 mL = 0.0150 L
ratio: 1 mol KOH per 1 mol HCl
M(KOH) = 56.1 g/mol
Convert the delivered titrant to moles
n(HCl) = 0.200 × 0.0150 = 0.0030 mol
Apply the mole ratio
n(KOH) = 0.0030 mol
Convert to mass
m = 0.0030 × 56.1
m = 0.17 g
Check — multiple choice
Hydrogen sulfide burns by 2H₂S + 3O₂ → 2SO₂ + 2H₂O, going to completion with more than enough O₂. The amount of H₂S supplied is tripled, and the O₂ remains more than enough. What happens to the amount of SO₂ formed?
Feedback if chosen: This option is wrong — you treated the amounts as fixed by the equation. The equation fixes only the proportions; more H₂S means proportionally more SO₂.
Feedback if chosen: This option is wrong — you used the coefficient 2 as the change factor. Coefficients set the proportion between species, not how a change scales.
Feedback if chosen: Correct. Amounts scale in proportion: with the other reactant plentiful, three times the H₂S makes three times the SO₂.
Feedback if chosen: This option is wrong — you multiplied by the O₂ coefficient 3 as well. The supplied change is a factor of 3, and the amounts follow that factor alone.
Check — multiple choice
Hydrazine, N₂H₄ (molar mass 32.05 g/mol), burns by N₂H₄ + O₂ → N₂ + 2H₂O with more than enough O₂. What mass of H₂O (molar mass 18.02 g/mol) forms when 8.01 g of N₂H₄ burns completely?
Feedback if chosen: This option is wrong — you applied the 1:2 relationship upside down, halving the moles. Each N₂H₄ makes two H₂O: 0.250 × 2 = 0.500 mol.
Feedback if chosen: This option is wrong — you skipped the mole step in the middle: 0.250 mol of N₂H₄ makes 0.500 mol of H₂O, not 0.250 mol.
Feedback if chosen: Correct. 8.01 ÷ 32.05 = 0.250 mol N₂H₄; the coefficients give 0.250 × 2 = 0.500 mol H₂O; 0.500 × 18.02 = 9.01 g.
Feedback if chosen: This option is wrong — you treated the 8.01 g as 8.01 mol. Convert the mass to moles first: 8.01 ÷ 32.05 = 0.250 mol.
Check — multiple choice
Iron reacts with oxygen by 3Fe + 2O₂ → Fe₃O₄. A mixture of 0.60 mol of Fe and 0.50 mol of O₂ reacts to completion. Which reactant is the limiting reactant?
Feedback if chosen: This option is wrong — you compared the raw amounts. The comparison must use the 3:2 requirement: 0.60 mol of Fe needs 0.40 mol of O₂, and 0.50 mol is available.
Feedback if chosen: Correct. The required amount of O₂ (0.60 × 2/3 = 0.40 mol) is less than the 0.50 mol supplied, so the Fe is used up first.
Feedback if chosen: This option is wrong — an exact match would need 0.40 mol of O₂ for 0.60 mol of Fe. With 0.50 mol of O₂ supplied, the O₂ is in excess.
Feedback if chosen: This option is wrong — moles are exactly what the comparison needs. Coefficients relate moles directly; masses would first be converted to moles anyway.
Check — multiple choice
Magnesium reacts with nitrogen gas by 3Mg + N₂ → Mg₃N₂. A mixture of 6.0 mol of Mg and 3.0 mol of N₂ reacts to completion. What is the theoretical yield of Mg₃N₂?
Feedback if chosen: Correct. 6.0 mol of Mg needs 2.0 mol of N₂, and 3.0 mol is available — so the Mg is limiting, and 6.0 ÷ 3 = 2.0 mol of Mg₃N₂ forms.
Feedback if chosen: This option is wrong — you built the yield from the N₂, but the N₂ is in excess. The limiting Mg fixes the yield: 6.0 ÷ 3 = 2.0 mol.
Feedback if chosen: This option is wrong — you carried the Mg moles straight across. Three Mg are consumed for each Mg₃N₂ formed: 6.0 ÷ 3 = 2.0 mol.
Feedback if chosen: This option is wrong — you added the two starting amounts. Reactant amounts never add; the limiting reactant alone fixes the yield.
Check — multiple choice
A solution holds 0.0300 mol of Ni²⁺(aq). Solid nickel(II) hydroxide forms by Ni²⁺(aq) + 2OH⁻(aq) → Ni(OH)₂(s) — supplied. What volume of 0.150 M NaOH(aq) supplies exactly the OH⁻ needed to consume all of the Ni²⁺?
Feedback if chosen: This option is wrong — you slipped a factor of ten converting liters to milliliters. 0.0600 ÷ 0.150 = 0.400 L, which is 400 mL.
Feedback if chosen: This option is wrong — you applied the 1:2 relationship upside down, halving the Ni²⁺ moles. Each Ni²⁺ needs two OH⁻: 0.0600 mol in all.
Feedback if chosen: This option is wrong — you skipped the 1:2 requirement. The 0.0300 mol of Ni²⁺ needs 0.0600 mol of OH⁻, not 0.0300 mol.
Feedback if chosen: Correct. OH⁻ needed = 0.0300 × 2 = 0.0600 mol; V = 0.0600 ÷ 0.150 = 0.400 L = 400 mL.
Check — multiple choice
Solid ammonium nitrite decomposes by NH₄NO₂(s) → N₂(g) + 2H₂O(l). The N₂ collected at 25 °C and 1.00 atm occupies 4.89 L. What mass of NH₄NO₂ (molar mass 64.06 g/mol) decomposed? (R = 0.08206 L·atm·mol⁻¹·K⁻¹.)
Feedback if chosen: This option is wrong — you calculated the mass of the N₂ itself (0.200 × 28.02). The question asks for the NH₄NO₂ that decomposed.
Feedback if chosen: Correct. n(N₂) = PV/RT = (1.00 × 4.89) ÷ (0.08206 × 298) = 0.200 mol; the 1:1 relationship gives 0.200 mol of NH₄NO₂; 0.200 × 64.06 = 12.8 g.
Feedback if chosen: This option is wrong — you doubled the moles, but NH₄NO₂ and N₂ sit in a 1:1 relationship. The 2 belongs to the H₂O.
Feedback if chosen: This option is wrong — you substituted 25 into PV = nRT without converting to kelvin. Convert first: T = 25 + 273 = 298 K.
Check — multiple choice
50.0 mL of 0.200 M Pb(NO₃)₂(aq) is mixed with 50.0 mL of 0.150 M Na₂SO₄(aq), and PbSO₄(s) forms (insoluble — supplied; molar mass 303.3 g/mol; the ions react 1:1). What mass of PbSO₄ forms?
Feedback if chosen: This option is wrong — you halved the smaller amount, but the ions react 1:1. The 0.00750 mol of SO₄²⁻ makes 0.00750 mol of PbSO₄.
Feedback if chosen: Correct. Pb²⁺: 0.0500 × 0.200 = 0.0100 mol; SO₄²⁻: 0.0500 × 0.150 = 0.00750 mol — the smaller amount limits, and 0.00750 × 303.3 = 2.27 g.
Feedback if chosen: This option is wrong — you built the answer from the Pb²⁺, the reactant in excess. The SO₄²⁻ is used up first and sets the yield.
Feedback if chosen: This option is wrong — you added the two mole amounts. The ions pair 1:1, so the smaller amount alone fixes how much solid forms.
Check — multiple choice
During a titration, the dye added to the flask changes color, and the color persists. Which statement about this endpoint is correct?
Feedback if chosen: Correct. The color change is what the eye can see; the dye is chosen so that this signal lands close to the equivalence point it stands in for.
Feedback if chosen: This option is wrong — you merged the two ideas. The exact mole match is the equivalence point; the endpoint is the visible signal chosen to approximate it.
Feedback if chosen: This option is wrong — the titrant reacts from its first drop. The endpoint arrives near the end, when the analyte is essentially consumed.
Feedback if chosen: This option is wrong — you compared concentrations. The color change signals that the analyte is used up, not that one solution is stronger.
Check — multiple choice
A flask holds 0.00240 mol of oxalic acid, H₂C₂O₄. It is titrated with 0.160 M NaOH, reacting by H₂C₂O₄ + 2NaOH → Na₂C₂O₄ + 2H₂O — supplied. What volume of the NaOH reaches the equivalence point?
Feedback if chosen: This option is wrong — you halved the acid's moles instead of doubling them. Each H₂C₂O₄ consumes two NaOH: 0.00480 mol are needed.
Feedback if chosen: This option is wrong — you skipped the 1:2 requirement. The 0.00240 mol of acid needs 0.00480 mol of NaOH, not 0.00240 mol.
Feedback if chosen: Correct. NaOH needed = 0.00240 × 2 = 0.00480 mol; V = 0.00480 ÷ 0.160 = 0.0300 L = 30.0 mL.
Feedback if chosen: This option is wrong — you applied the factor 2 twice, once to the moles and again to the volume. Doubling happens once: 0.00480 mol, giving 30.0 mL.
Check — multiple choice
A 20.0 mL sample of HI(aq) reaches the equivalence point when 24.0 mL of 0.250 M KOH has been added, reacting by HI + KOH → KI + H₂O — supplied. What is the molarity of the HI sample?
Feedback if chosen: This option is wrong — you divided by the combined volume (44.0 mL). Molarity of the sample uses the sample's own volume, 20.0 mL.
Feedback if chosen: This option is wrong — you swapped the volumes in M₁V₁ = M₂V₂: 0.208 = 0.250 × 20.0/24.0. The titrant's molarity multiplies the titrant's own volume: 0.250 × 24.0/20.0 = 0.300 M.
Feedback if chosen: This option is wrong — you copied the titrant's molarity. The two solutions need different concentrations to balance different volumes.
Feedback if chosen: Correct. KOH added: 0.0240 × 0.250 = 0.00600 mol; the 1:1 relationship gives 0.00600 mol of HI; 0.00600 ÷ 0.0200 = 0.300 M.
9 lessons
Outcome: Identify a reaction as acid-base because it transfers one or more protons (H⁺) from a donor species to an acceptor species.
A drop of concentrated hydrochloric acid sits on one glass rod.
A drop of concentrated ammonia solution sits on a second rod, held 2 cm away.
A white ring of smoke forms in the air between the two rods.
The smoke is a cloud of tiny crystals of solid ammonium chloride, NH₄Cl.
The reaction made no water at all.
Yet chemists call this an acid-base reaction. What moved from the acid to the ammonia?
You've already seen that a hydrogen atom is one proton with one electron around it.
Take that electron away, and the H⁺ ion that remains is a bare proton.
In the smoke reaction, each HCl molecule hands one H⁺ ion — one proton — to an NH₃ molecule.
The NH₃ becomes NH₄⁺, and the Cl⁻ left behind pairs with it in the solid NH₄Cl.
A reaction that transfers one or more protons (H⁺) from one species to another is called an acid-base reaction.
The species that gives the proton away is the donor.
The species that takes the proton is the acceptor.
Here are three mixes, judged in the same frame: does a proton move?
| HCl + NH₃ | HCl + NaOH | AgNO₃ + NaCl | |
|---|---|---|---|
| Does a proton move? | yes | yes | no |
| Acid-base? |
In HCl + NH₃, a proton moves from the HCl to the NH₃ — this is an acid-base reaction.
In HCl + NaOH, a proton moves from the HCl to the OH⁻, making H₂O — this is an acid-base reaction too.
In AgNO₃ + NaCl, no proton moves anywhere — this is not an acid-base reaction.
Water forms in many acid-base reactions.
But the proton transfer, not the water, decides the label.
Identify a reaction as acid-base by checking whether it transfers one or more protons (H⁺) from a donor species to an acceptor species.
Check — multiple choice
What must a reaction do to earn the name acid-base?
Feedback if chosen: This option is wrong — you made water the test. Water forms in many acid-base reactions, but the proton transfer decides the label; the smoke reaction made no water.
Feedback if chosen: Correct. An acid-base reaction transfers one or more protons — H⁺ ions — from a donor species to an acceptor species.
Feedback if chosen: This option is wrong — a new solid signals a different kind of event. The acid-base test asks one question only: does a proton move?
Feedback if chosen: This option is wrong — you moved the wrong particle. The acid-base test tracks protons (H⁺ ions), not electrons.
Check — multiple choice
In which reaction does a proton move from one species to another?
Feedback if chosen: This option is wrong — no species here gives away an H⁺. The magnesium and the copper trade places, but no proton moves.
Feedback if chosen: This option is wrong — the silver and potassium compounds trade partners, but neither hands an H⁺ to the other.
Feedback if chosen: Correct. The HBr hands one H⁺ to the OH⁻, making H₂O — a proton moves, so the reaction is acid-base.
Feedback if chosen: This option is wrong — one compound breaks into two here, and no H⁺ passes from one species to another.
Check — multiple choice
HF(aq) + NH₃(aq) → NH₄F(aq) is an acid-base reaction. Which species donates the proton?
Feedback if chosen: This option is wrong — you swapped donor and acceptor. The NH₃ gains an H⁺ and becomes NH₄⁺, so NH₃ is the acceptor.
Feedback if chosen: This option is wrong — F⁻ is what remains of the HF after the proton has already left; it donates nothing.
Feedback if chosen: Correct. The HF gives up its H⁺ to the NH₃ — HF is the donor, and NH₃ is the acceptor.
Feedback if chosen: This option is wrong — NH₄⁺ is the product of the transfer: the NH₃ after it has accepted the proton.
Check — multiple choice
HNO₃(aq) + NH₃(aq) → NH₄NO₃(aq). A student claims the reaction cannot be acid-base, because no water forms. Which statement is correct?
Feedback if chosen: This option is wrong — you made water the test. Many acid-base reactions produce water, but the label is decided by the proton transfer alone.
Feedback if chosen: Correct. The HNO₃ hands one H⁺ to the NH₃, making NH₄⁺. A proton moves, so the reaction is acid-base — with or without water.
Feedback if chosen: This option is wrong — you counted products instead of tracking the proton. The number of products plays no part in the acid-base test.
Feedback if chosen: This option is wrong — you made the solvent the test. The label rests on the proton transfer: an H⁺ moves from the HNO₃ to the NH₃.
Outcome: State that oxidation is the loss of one or more electrons by a species, and recognize a loss of electrons in a shown change.
A student stands a coil of copper wire in 50 mL of colorless silver nitrate solution.
Within minutes, glittering crystals of silver grow on the wire.
The solution slowly turns pale blue.
Pale blue is the color of dissolved Cu²⁺ ions in water.
The wire began as neutral copper atoms.
You've already seen that a neutral atom carries equal numbers of protons and electrons.
Each copper atom that left the wire ended up as a Cu²⁺ ion. What did the atom lose?
Each copper atom released two electrons.
With two electrons gone, the atom carries two more protons than electrons.
So the atom became a Cu²⁺ ion and dissolved into the water.
A loss of one or more electrons by a species is called oxidation.
The copper atoms were oxidized.
The lost electrons did not vanish.
The silver ions in the solution caught them.
State that oxidation is the loss of one or more electrons by a species, and recognize a loss of electrons in a shown change.
Check — multiple choice
What is oxidation?
Feedback if chosen: This option is wrong — you flipped the direction. A species that gains electrons is doing the opposite of oxidation.
Feedback if chosen: This option is wrong — you moved the wrong particle. Oxidation tracks electrons, not protons.
Feedback if chosen: This option is wrong — dissolving alone changes nothing about a particle's electrons. Oxidation is defined by electrons leaving a species.
Feedback if chosen: Correct. Oxidation is the loss of one or more electrons by a species.
Check — multiple choice
A magnesium atom becomes a Mg²⁺ ion. What happened to the atom?
Feedback if chosen: Correct. Two electrons left the neutral atom, so the atom now carries two more protons than electrons — a 2+ charge. The atom was oxidized.
Feedback if chosen: This option is wrong — gaining electrons would make the atom negative. A 2+ charge means two electrons are missing.
Feedback if chosen: This option is wrong — losing protons would change which element the atom is. The nucleus stays intact; two electrons left instead.
Feedback if chosen: This option is wrong — atoms do not gain protons in a reaction. The 2+ charge comes from two electrons leaving.
Check — multiple choice
Which change is an oxidation?
Feedback if chosen: This option is wrong — the chlorine atom gains an electron. Oxidation is a loss of electrons, not a gain.
Feedback if chosen: Correct. The zinc atom loses two electrons — a loss of electrons is exactly what oxidation means.
Feedback if chosen: This option is wrong — melting rearranges whole molecules and moves no electrons between species.
Feedback if chosen: This option is wrong — the K⁺ ion gains an electron. Oxidation is the opposite change: electrons leave the species.
Check — multiple choice
A student watches the copper wire thin as the solution turns blue, and says the copper is simply dissolving, the way sugar dissolves in tea — each atom leaving unchanged. What actually happens to each copper atom that leaves the wire?
Feedback if chosen: This option is wrong — you accepted the student's picture. A neutral Cu atom would leave the water colorless; the blue color shows Cu²⁺ ions forming.
Feedback if chosen: This option is wrong — the atom stays in one piece. It gives up exactly two electrons and keeps everything else.
Feedback if chosen: Correct. The blue color is the fingerprint of Cu²⁺. Each departing atom lost two electrons — the copper was oxidized, not simply dissolved.
Feedback if chosen: This option is wrong — you reversed the change. Gaining two electrons would give the atom a 2− charge, not the 2+ charge the blue color shows.
Outcome: State that reduction is the gain of one or more electrons by a species, and tell a gain of electrons apart from a loss.
The silver crystals carry the other half of the copper-wire story.
The solution began full of silver ions, Ag⁺.
An Ag⁺ ion stays dissolved in the water.
But the crystals growing on the wire are built from neutral silver atoms.
Each Ag⁺ ion carried one positive charge too many to be neutral.
What did each silver ion gain?
Each Ag⁺ ion picked up one electron from the copper wire.
The extra electron canceled the ion's single positive charge.
The neutral silver atom then settled onto the wire as part of a solid crystal.
A gain of one or more electrons by a species is called reduction.
The Ag⁺ ions were reduced.
Here are the two halves of the beaker's story, side by side in identical frames.
| The copper atoms | The silver ions | |
|---|---|---|
| Electrons lost or gained? | Lose two | Gain one |
| What each becomes | Cu²⁺ in solution | Ag atom on the wire |
| The name | Oxidation | Reduction |
The copper atoms lost electrons — the copper was oxidized.
The silver ions gained electrons — the silver ions were reduced.
Oxidation supplies the electrons that reduction uses.
State that reduction is the gain of one or more electrons by a species, and tell a gain of electrons apart from a loss.
Check — multiple choice
An Fe³⁺ ion gains three electrons and becomes a neutral Fe atom. Which word names the change?
Feedback if chosen: This option is wrong — oxidation is a loss of electrons. The Fe³⁺ ion gained three.
Feedback if chosen: Correct. The ion gained electrons, and a gain of one or more electrons by a species is reduction.
Feedback if chosen: This option is wrong — dissociation is a dissolved compound separating into its ions; no electrons move between species there.
Feedback if chosen: This option is wrong — dissolving changes where a particle sits, not how many electrons it carries.
Check — multiple choice
Which change is a reduction?
Feedback if chosen: This option is wrong — the calcium atom loses electrons, and a loss of electrons is oxidation.
Feedback if chosen: This option is wrong — an unchanged ion has neither gained nor lost anything; no reduction happened.
Feedback if chosen: This option is wrong — the aluminum atom loses electrons; that is oxidation, the opposite change.
Feedback if chosen: Correct. The bromine atom gains one electron — a gain of electrons is reduction.
Check — multiple choice
A sulfur atom gains two electrons and becomes S²⁻. A lithium atom loses one electron and becomes Li⁺. Which labels fit the two changes?
Feedback if chosen: Correct. The sulfur gains electrons (reduction), and the lithium loses one (oxidation).
Feedback if chosen: This option is wrong — you swapped the two labels. Gaining electrons is reduction, so the sulfur is the reduced one.
Feedback if chosen: This option is wrong — only the sulfur gains electrons. The lithium loses one, so the lithium is oxidized.
Feedback if chosen: This option is wrong — only the lithium loses an electron. The sulfur gains two, so the sulfur is reduced.
Check — multiple choice
A species is reduced during a reaction. Where do the electrons it gains come from?
Feedback if chosen: This option is wrong — the water is a bystander here. The electrons come from the partner species that loses them.
Feedback if chosen: Correct. Electrons are never created — a reduced species gains exactly the electrons that an oxidized species loses.
Feedback if chosen: This option is wrong — charge cannot turn into particles. The charge is canceled by real electrons arriving from another species.
Feedback if chosen: This option is wrong — protons stay protons. The gained electrons travel from the species being oxidized.
Outcome: State the five rules for assigning oxidation numbers, including the peroxide and metal-hydride exceptions.
A kitchen stove burns about 1 g of methane each minute.
The equation reads CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g).
Every reactant and every product is a neutral molecule.
No ion appears anywhere in the equation.
The copper-wire story showed its electron transfer through visible charges: Cu²⁺ formed, and Ag⁺ vanished.
Here, nothing carries a charge, before or after.
Did any electrons move at all?
Chemists assign every atom in a formula its own bookkeeping charge, called an oxidation number.
Five short rules assign the oxidation number of any atom in any formula.
| The rule | One example | The named exception |
|---|---|---|
| An atom in a free element counts 0 | each O in O₂ is 0 | none |
| A monatomic ion counts its own charge | Cu²⁺ is +2 | none |
| Oxygen counts −2 | the O in H₂O is −2 | peroxides such as H₂O₂: each O is −1 |
| Hydrogen counts +1 | each H in H₂O is +1 | metal hydrides such as NaH: the H is −1 |
| The numbers in a species sum to its overall charge | H₂O sums to 0 | none |
Rule 1: an atom in a free element counts 0 — each O in O₂ counts 0, and Cu metal counts 0.
Rule 2: a monatomic ion counts its own charge — Cu²⁺ counts +2, and Cl⁻ counts −1.
Rule 3: oxygen counts −2 — except in peroxides such as H₂O₂, where each O counts −1.
Rule 4: hydrogen counts +1 — except in metal hydrides such as NaH, where the H counts −1.
Rule 5: the oxidation numbers of all the atoms in a species add up to the species' overall charge.
In H₂O, each hydrogen counts +1.
In H₂O, the oxygen counts −2.
The sum is (+1) + (+1) + (−2) = 0 — exactly the overall charge of the neutral molecule, as rule 5 requires.
With a number on every atom, moving electrons leave a visible trace — you will soon read that trace across a whole equation.
State the rules for assigning oxidation numbers: free element 0; monatomic ion = its charge; O −2 except peroxides; H +1 except metal hydrides; and a sum equal to the species' overall charge.
Check — multiple choice
What is the oxidation number of each atom in N₂ gas?
Feedback if chosen: Correct. N₂ is a free element, and every atom in a free element counts 0.
Feedback if chosen: This option is wrong — you gave nitrogen hydrogen's number. N₂ is a free element, so each atom counts 0.
Feedback if chosen: This option is wrong — −3 is what nitrogen takes in some compounds. In the free element, each atom counts 0.
Feedback if chosen: This option is wrong — −2 is oxygen's usual number in compounds. N₂ is a free element, so each atom counts 0.
Check — multiple choice
What is the oxidation number of the Ba in a Ba²⁺ ion?
Feedback if chosen: This option is wrong — 0 belongs to atoms in a free element. Ba²⁺ is a monatomic ion, so it counts its own charge.
Feedback if chosen: This option is wrong — you flipped the sign. The ion's charge is 2+, so its oxidation number is +2.
Feedback if chosen: Correct. A monatomic ion counts its own charge, and Ba²⁺ carries a 2+ charge.
Feedback if chosen: This option is wrong — the ion carries a 2+ charge, not 1+. A monatomic ion counts exactly its own charge.
Check — multiple choice
What is the oxidation number of each oxygen atom in hydrogen peroxide, H₂O₂?
Feedback if chosen: This option is wrong — you applied the usual oxygen rule to its named exception. H₂O₂ is a peroxide, where each O counts −1.
Feedback if chosen: This option is wrong — 0 belongs to free elements, and the oxygen here is bound inside a compound.
Feedback if chosen: This option is wrong — +1 is hydrogen's number in H₂O₂. The oxygen takes the peroxide value, −1.
Feedback if chosen: Correct. H₂O₂ is a peroxide — the named exception where each oxygen counts −1 instead of −2.
Check — multiple choice
What is the oxidation number of the hydrogen in the metal hydride NaH?
Feedback if chosen: Correct. NaH is a metal hydride — the named exception where hydrogen counts −1 instead of +1.
Feedback if chosen: This option is wrong — you applied the usual hydrogen rule to its named exception. In a metal hydride, the H counts −1.
Feedback if chosen: This option is wrong — 0 belongs to free elements, and this hydrogen is bound to sodium in a compound.
Feedback if chosen: This option is wrong — no rule gives hydrogen −2. In a metal hydride, the H counts −1.
Outcome: Identify the oxidation number of every atom in the reactants and products of a given equation.
The stove's equation reads CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g).
One carbon atom crosses the arrow, from CH₄ into CO₂.
The rules put that carbon at −4 in CH₄.
The rules put the same carbon at +4 in CO₂.
The carbon's oxidation number climbed from −4 to +4.
In these neutral molecules no electron is lost outright — the carbon lost part of its share of the bonding electrons.
Which atoms took the electron share the carbon gave up? Numbering every atom answers that.
You've already seen the five oxidation-number rules.
Work one formula at a time, never the whole equation at once.
Write the ruled atoms first: hydrogen +1, oxygen −2, atoms of a free element 0.
Rule 5 — the sum rule — then fixes the one atom the other rules left unknown.
Q: What is the oxidation number of every atom in CH₄ + 2O₂ → CO₂ + 2H₂O?
Write the ruled atoms in each formula
In CH₄: each H is +1
In O₂: each O is 0 (free element)
In CO₂: each O is −2; in H₂O: each H is +1, the O is −2
Use the sum rule on CH₄
C + 4(+1) = 0
C = −4
Use the sum rule on CO₂
C + 2(−2) = 0
C = +4
Here is the finished equation with every atom numbered.
A changed oxidation number is the visible trace of moved electrons.
Identify the oxidation number of every atom in the reactants and products of a given equation, using the five rules.
Check — multiple choice
In NH₃, each hydrogen counts +1, and the molecule's overall charge is 0. What is the oxidation number of the N?
Feedback if chosen: This option is wrong — you flipped the sign. The sum rule reads N + 3(+1) = 0, so N = −3.
Feedback if chosen: Correct. The sum rule fixes the unknown atom: N + 3(+1) = 0, so N = −3.
Feedback if chosen: This option is wrong — 0 would leave the three hydrogens' +3 uncanceled, and the sum must equal the molecule's overall charge of 0.
Feedback if chosen: This option is wrong — −1 gives a sum of +2, not 0. The three hydrogens contribute +3, so the N must be −3.
Check — numeric entry
2SO₂(g) + O₂(g) → 2SO₃(g). In SO₃, each oxygen counts −2. What is the oxidation number of the S in SO₃? Enter a whole number.
Feedback when correct: Correct. The three oxygens contribute 3(−2) = −6, and the molecule's overall charge is 0, so the S must be +6.
Anticipated wrong answers (feedback shown):
| -6 | You gave the sulfur the oxygens' total instead of what cancels it. The three O atoms sum to −6, so the S must be +6 for the molecule to sum to 0. |
| 4 | You numbered the S in SO₂, the reactant. In SO₃ there are three oxygens: S + 3(−2) = 0, so S = +6. |
| 2 | You used one oxygen instead of three. SO₃ carries three O atoms at −2 each: S + 3(−2) = 0, so S = +6. |
Worked solution revealed after a second miss
Write down what the rules give
Each O in SO₃ is −2
SO₃ is neutral: its numbers sum to 0
Write the sum rule for SO₃
S + 3(−2) = 0
Solve for S
S − 6 = 0
S = +6
Check — multiple choice
Fe(s) + 2HCl(aq) → FeCl₂(aq) + H₂(g). What is the oxidation number of each H atom in the product H₂?
Feedback if chosen: This option is wrong — +1 is hydrogen's number inside compounds like HCl. H₂ is a free element, so each atom counts 0.
Feedback if chosen: Correct. H₂ is a free element, and every atom in a free element counts 0.
Feedback if chosen: This option is wrong — −1 belongs to hydrogen in metal hydrides such as NaH. In the free element H₂, each atom counts 0.
Feedback if chosen: This option is wrong — no rule gives a hydrogen atom +2. In the free element H₂, each atom counts 0.
Check — multiple choice
Cl₂(g) + 2KBr(aq) → 2KCl(aq) + Br₂(l). In KCl, the potassium counts +1. What is the oxidation number of the Cl in KCl?
Feedback if chosen: This option is wrong — 0 belongs to the Cl in Cl₂, the free element. Inside KCl, the sum rule fixes the Cl at −1.
Feedback if chosen: This option is wrong — +1 is the potassium's number. The two numbers must sum to 0, so the Cl is −1.
Feedback if chosen: This option is wrong — −2 would make KCl sum to −1, not 0. With K at +1, the Cl must be −1.
Feedback if chosen: Correct. KCl is neutral, so K + Cl = 0. With the potassium at +1, the Cl counts −1.
Outcome: Identify a reaction as redox — or not — from oxidation-number changes between reactants and products.
You've already seen the stove's equation with every atom numbered: CH₄ + 2O₂ → CO₂ + 2H₂O.
The carbon's oxidation number climbed from −4 to +4.
Each oxygen's oxidation number dropped from 0 to −2.
A climbing oxidation number marks lost electron share — the carbon was oxidized.
A dropping oxidation number marks gained electron share — the oxygen was reduced.
So the burning methane moved electrons after all, with no ion anywhere in sight.
A reaction that transfers one or more electrons from one species to another is called an oxidation-reduction reaction — redox for short.
Oxidation numbers turn the verdict into a routine.
First, number every atom on both sides of the arrow.
Then look for change: a rising number marks oxidation, and a falling number marks reduction.
If no atom's number changes anywhere, the reaction is not redox.
Now apply the three steps to NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l).
Every number stays the same from left to right: Na +1, O −2, H +1, Cl −1.
No number changed, so no electrons moved — NaOH + HCl is not a redox reaction.
It is still an acid-base reaction: the HCl passed a proton to the OH⁻.
One reaction, two different tests — each test answers its own question.
Identify a reaction as oxidation-reduction (redox) when oxidation numbers change between reactants and products, and as not redox when none change.
Check — multiple choice
Which of these reactions is a redox reaction?
Feedback if chosen: This option is wrong — number the atoms: every one keeps its value. A proton moves here, but no electrons do.
Feedback if chosen: This option is wrong — the ions trade partners, but Ag stays +1, Cl stays −1, K stays +1. No number changes.
Feedback if chosen: Correct. The Zn climbs from 0 to +2 and the Cu²⁺ falls from +2 to 0 — oxidation numbers change, so electrons moved.
Feedback if chosen: This option is wrong — every atom keeps its oxidation number. The HF hands over a proton, not electrons.
Check — multiple choice
2Na(s) + Cl₂(g) → 2NaCl(s). Which verdict fits the reaction?
Feedback if chosen: Correct. The sodium's oxidation number climbs and the chlorine's falls — changed numbers are the trace of moved electrons.
Feedback if chosen: This option is wrong — number the atoms on both sides: the Na's number rises and the Cl's falls. Both change, so the reaction is redox.
Feedback if chosen: This option is wrong — no species here donates a proton; there is no H anywhere in the equation.
Feedback if chosen: This option is wrong — the equation is all the test needs. Number every atom on both sides and look for change.
Check — multiple choice
In 2Na(s) + Cl₂(g) → 2NaCl(s), which change is the evidence that the reaction is redox?
Feedback if chosen: This option is wrong — a solid product does form, but forming a solid moves no electrons. The redox evidence is a changed oxidation number.
Feedback if chosen: Correct. Changed oxidation numbers are the trace of moved electrons — that is the evidence the redox verdict rests on.
Feedback if chosen: This option is wrong — the shape of the equation is not evidence. The test reads oxidation numbers, and here they change.
Feedback if chosen: This option is wrong — every proper equation is balanced, redox or not. The evidence is the changed oxidation numbers.
Check — multiple choice
A reaction's oxidation numbers all stay the same from reactants to products. What is the verdict?
Feedback if chosen: This option is wrong — redox needs at least one number to change. Unchanged numbers mean no electrons moved.
Feedback if chosen: This option is wrong — not-redox does not force acid-base. The acid-base test is separate: does a proton move?
Feedback if chosen: Correct. Unchanged oxidation numbers mean no electrons transferred — the reaction is not redox.
Feedback if chosen: This option is wrong — temperature plays no part in the test. The verdict comes from the oxidation numbers alone.
Outcome: Identify which species is oxidized and which is reduced from the oxidation-number changes across an equation.
A 2.0 g zinc granule drops into hydrochloric acid.
Bubbles of hydrogen gas stream off the metal.
The equation reads Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g).
The zinc's oxidation number climbs from 0 to +2.
Each hydrogen's oxidation number falls from +1 to 0.
Here is the exam's question about this equation: which species is oxidized, and which is reduced?
Track each atom's oxidation number across the arrow.
The zinc's number climbed, so the zinc lost electrons — Zn is the species that is oxidized.
The hydrogen's number fell, so the H⁺ gained electrons — H⁺ is the species that is reduced.
AP answers name the species as written in the equation: H⁺ itself, not just 'hydrogen'.
Identify which species is oxidized and which is reduced from the oxidation-number changes across an equation.
Check — multiple choice
Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s). Which species is oxidized?
Feedback if chosen: This option is wrong — the Cu²⁺'s number falls from +2 to 0. A falling number marks the reduced species, not the oxidized one.
Feedback if chosen: Correct. The Fe climbs from 0 to +2 — a rising oxidation number marks the species that is oxidized.
Feedback if chosen: This option is wrong — Fe²⁺ is the product of the change. The species that is oxidized is the Fe as written on the reactant side.
Feedback if chosen: This option is wrong — the Cu is the product of the reduction half of the story; nothing about it rises.
Check — multiple choice
Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s). Which species is reduced?
Feedback if chosen: Correct. The Cu²⁺ falls from +2 to 0 — a falling oxidation number marks the species that is reduced.
Feedback if chosen: This option is wrong — the Fe's number climbs from 0 to +2, so the Fe is oxidized, not reduced.
Feedback if chosen: This option is wrong — Cu is the product of the reduction. The species that is reduced is the Cu²⁺ as written on the reactant side.
Feedback if chosen: This option is wrong — Fe²⁺ is a product, and its number was reached by climbing. Nothing about the iron falls.
Check — multiple choice
2Al(s) + 3Br₂(l) → 2AlBr₃(s). In AlBr₃, the Al counts +3 and each Br counts −1. Which species is oxidized?
Feedback if chosen: This option is wrong — the bromine falls from 0 to −1. A falling number marks reduction, not oxidation.
Feedback if chosen: This option is wrong — AlBr₃ is the product the changes end in, not the species whose number rose from the reactant side.
Feedback if chosen: Correct. The Al climbs from 0 to +3 — the rising oxidation number marks the aluminum as the oxidized species.
Feedback if chosen: This option is wrong — both elements change their numbers here: Al rises and Br falls. The riser, Al, is the oxidized one.
Check — multiple choice
2Al(s) + 3Br₂(l) → 2AlBr₃(s). In AlBr₃, the Al counts +3 and each Br counts −1. Which species is reduced?
Feedback if chosen: This option is wrong — the aluminum's number rises from 0 to +3, so the Al is oxidized, not reduced.
Feedback if chosen: This option is wrong — AlBr₃ is where the changes end. The reduced species is the Br₂ as written on the reactant side.
Feedback if chosen: This option is wrong — both numbers change: Al rises and Br falls. The faller, Br₂, is the reduced species.
Feedback if chosen: Correct. Each Br falls from 0 to −1 — the falling oxidation number marks the Br₂ as the reduced species.
Outcome: Classify combustion — a species reacting rapidly with oxygen gas — as a subclass of redox.
A teacher lights a 5 cm strip of magnesium ribbon.
The metal burns with a blinding white light.
The equation reads 2Mg(s) + O₂(g) → 2MgO(s).
The strip contains no carbon.
No carbon dioxide forms, and no water forms.
Is burning magnesium the same kind of event as a burning kitchen stove?
When a species reacts rapidly with oxygen gas, chemists call the reaction combustion.
Methane reacts rapidly with O₂ — this is combustion.
Magnesium reacts rapidly with O₂ — this is combustion too.
Sulfur reacts rapidly with O₂ — this is combustion too.
NaOH reacts with HCl, and no O₂ takes part — this is not combustion.
The fuel can be a metal, a nonmetal, or a compound — the O₂ is what makes a burn a burn.
Here are the stove's burn and the ribbon's burn, judged in identical frames.
| CH₄ + 2O₂ → CO₂ + 2H₂O | 2Mg + O₂ → 2MgO | |
|---|---|---|
| Reacts rapidly with O₂? | yes | yes |
| Oxidation numbers change? | C −4 to +4, O 0 to −2 | Mg 0 to +2, O 0 to −2 |
| Combustion AND redox |
In CH₄ + 2O₂ → CO₂ + 2H₂O, the carbon climbs from −4 to +4, and the oxygen falls from 0 to −2.
In 2Mg + O₂ → 2MgO, the magnesium climbs from 0 to +2, and the oxygen falls from 0 to −2.
In every burn, the oxygen's oxidation number falls from 0 to −2.
So every combustion transfers electrons — every combustion is a redox reaction.
Combustion is one subclass inside the redox class, not a separate class of its own.
Classify combustion — a species reacting rapidly with oxygen gas — as a subclass of oxidation-reduction reactions.
Check — multiple choice
A strip of zinc burns in oxygen: 2Zn(s) + O₂(g) → 2ZnO(s). The zinc reacts rapidly with the oxygen. Which labels fit the reaction?
Feedback if chosen: Correct. A species reacts rapidly with O₂, so the reaction is combustion — and the oxygen falls from 0 to −2, so it is redox as well.
Feedback if chosen: This option is wrong — check the numbers: Zn climbs from 0 to +2 and O falls from 0 to −2. Every combustion is also redox.
Feedback if chosen: This option is wrong — a species reacting rapidly with O₂ is exactly what combustion means, whatever the fuel is.
Feedback if chosen: This option is wrong — the reaction passes both tests: rapid reaction with O₂, and changed oxidation numbers.
Check — multiple choice
Which of these reactions is a combustion?
Feedback if chosen: This option is wrong — no O₂ takes part. The nitrogen reacts with hydrogen, and combustion needs oxygen gas.
Feedback if chosen: This option is wrong — this reaction is redox, but no O₂ takes part, so it is not a combustion.
Feedback if chosen: Correct. The carbon reacts with oxygen gas — a species reacting rapidly with O₂ is a combustion.
Feedback if chosen: This option is wrong — this is a proton transfer with no O₂ anywhere; it is acid-base, not combustion.
Check — multiple choice
A 3 cm ribbon of calcium burns brightly in oxygen: 2Ca(s) + O₂(g) → 2CaO(s). A student says the reaction is not combustion, because calcium contains no carbon. Which statement is correct?
Feedback if chosen: This option is wrong — you accepted a narrower definition than the real one. Combustion is any species reacting rapidly with O₂; carbon is not required.
Feedback if chosen: This option is wrong — CO₂ forms only when the fuel contains carbon. The test is the rapid reaction with O₂, which this reaction passes.
Feedback if chosen: This option is wrong — you reached the right verdict from the wrong evidence. Light is a byproduct; the O₂ reacting rapidly is what makes this combustion.
Feedback if chosen: Correct. Combustion is a species reacting rapidly with oxygen gas. The calcium does exactly that, so the burn is combustion — no carbon needed.
Check — multiple choice
Why is every combustion also a redox reaction?
Feedback if chosen: This option is wrong — heat is not the redox test. The test is a changed oxidation number, and combustion always changes oxygen's.
Feedback if chosen: Correct. The O₂ always ends up in a product with oxygen at −2. A changed oxidation number means transferred electrons — redox.
Feedback if chosen: This option is wrong — burning magnesium makes only solid MgO. The redox evidence is the changed oxidation numbers, not the products' state.
Feedback if chosen: This option is wrong — where the burn happens is not evidence. The evidence is the oxygen falling from 0 to −2 in every burn.
Outcome: State that complete combustion of a hydrocarbon produces carbon dioxide and water, and write the balanced equation for a given fuel.
A camping stove burns 5 g of propane, C₃H₈, each minute.
A cold pan held above the flame collects droplets of water.
Limewater held in the exhaust stream turns cloudy.
Cloudy limewater is the standard test signal for carbon dioxide.
Propane contains only carbon and hydrogen atoms.
Where did the water and the carbon dioxide come from?
A compound built from only carbon and hydrogen is called a hydrocarbon.
Propane is a hydrocarbon.
So is methane, CH₄.
Complete combustion sends every carbon atom in the fuel into carbon dioxide, CO₂.
Complete combustion sends every hydrogen atom in the fuel into water, H₂O.
So complete combustion of any hydrocarbon produces the same two products: carbon dioxide and water.
Methane's burn shows the pattern: CH₄ + 2O₂ → CO₂ + 2H₂O.
Knowing the two fixed products lets you write the equation for a fuel you have never seen.
You've already seen how to balance an equation by inspection — that balancing step finishes the job here.
State that complete combustion of a hydrocarbon (a compound of only carbon and hydrogen) produces carbon dioxide and water.
Check — multiple choice
Butane, C₄H₁₀, is a hydrocarbon. It burns completely in oxygen. Which products form?
Feedback if chosen: This option is wrong — the fuel's hydrogen ends up bonded to oxygen. Complete combustion sends every H atom into H₂O, not H₂.
Feedback if chosen: This option is wrong — you lost the hydrogen. The fuel's ten H atoms must end up somewhere, and complete combustion sends them into water.
Feedback if chosen: Correct. Complete combustion of any hydrocarbon makes the same two products: every C atom goes into CO₂, and every H atom goes into H₂O.
Feedback if chosen: This option is wrong — the fuel's carbon does not survive as the element. Complete combustion sends every C atom into CO₂.
Check — multiple choice
Which of these compounds is a hydrocarbon?
Feedback if chosen: This option is wrong — CH₃OH contains an oxygen atom. A hydrocarbon is built from carbon and hydrogen only.
Feedback if chosen: Correct. C₂H₆ contains only carbon and hydrogen atoms — a hydrocarbon.
Feedback if chosen: This option is wrong — CO₂ contains oxygen and no hydrogen at all. A hydrocarbon is carbon and hydrogen only.
Feedback if chosen: This option is wrong — H₂O contains no carbon. A hydrocarbon needs both carbon and hydrogen, and nothing else.
Check — drawing task (pen and paper, self-marked)
Take a pen and paper and write the balanced equation for the complete combustion of propane, C₃H₈(g) — its rapid reaction with O₂(g), making the two fixed products. At the flame's temperature every species is a gas. Include the coefficients and a state symbol on every species. When you are done, select Check to compare your answer with the model answer.
Model answer shown to the student
Self-mark checklist — the drawing should show:
4 lessons
Outcome: State that a precipitation reaction forms an insoluble or sparingly soluble ionic compound when ions in aqueous solutions mix.
A student pours 20 mL of potassium iodide solution into 20 mL of lead(II) nitrate solution.
Both starting solutions are colorless.
A brilliant yellow solid appears the instant the liquids touch.
The solid is lead(II) iodide, PbI₂.
You've already seen this yellow solid — it opened this unit as the unnamed solid of the very first lesson.
Now it has a name.
You've already seen that a solid appearing from mixed solutions is called a precipitate.
Which ions built the solid?
You've already seen that a dissolved ionic compound dissociates into separate aqueous ions.
After the pour, four kinds of ions drift freely in one beaker: Pb²⁺, NO₃⁻, K⁺, and I⁻.
Lead(II) iodide cannot stay dissolved in water.
So the Pb²⁺ ions and the I⁻ ions lock together into solid PbI₂.
When ions from mixed aqueous solutions join into an insoluble — or sparingly soluble — ionic compound, chemists call the event a precipitation reaction.
The net ionic equation records exactly that event.
You've already seen spectator ions: the K⁺ and NO₃⁻ stay dissolved, unchanged, on both sides.
The molecular equation for the same event reads Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq).
That form hides the spectators — the net ionic form shows the actual event: two kinds of ions joining into a solid.
State that a precipitation reaction forms an insoluble or sparingly soluble ionic compound — a precipitate — when ions in aqueous solutions are mixed.
Check — multiple choice
Solutions of silver nitrate and potassium bromide are mixed, and pale-yellow solid AgBr forms at once. What happened when the solutions mixed?
Feedback if chosen: This option is wrong — no species here donates a proton. Dissolved ions joined into a solid, and that event has its own name.
Feedback if chosen: Correct. Ions from the two mixed solutions — Ag⁺ and Br⁻ — joined into an insoluble ionic compound. That is a precipitation reaction.
Feedback if chosen: This option is wrong — number the atoms: every one keeps its oxidation number. The ions only paired up; no electrons moved.
Feedback if chosen: This option is wrong — a new solid appeared where there was none. Something happened: dissolved ions joined into insoluble AgBr.
Check — multiple choice
A student says the yellow lead(II) iodide was already floating in one of the two bottles before the pour. What is wrong with the claim?
Feedback if chosen: This option is wrong — you accepted the claim. A stirred-up solid would have colored its own bottle; both bottles were colorless.
Feedback if chosen: This option is wrong — neither bottle held any solid at all. Every compound in each bottle was dissolved as separate ions.
Feedback if chosen: Correct. Each bottle held only dissolved ions, so each was colorless. PbI₂ could form only after the pour put Pb²⁺ and I⁻ in the same water.
Feedback if chosen: This option is wrong — the yellow solid did form from two colorless solutions. The color belongs to the new compound the ions built, not to either bottle.
Check — multiple choice
Solutions of calcium chloride and sodium carbonate mix, and solid calcium carbonate forms. Which two ions joined into the solid?
Feedback if chosen: Correct. The solid is calcium carbonate, so the Ca²⁺ ions and the CO₃²⁻ ions locked together; the Na⁺ and Cl⁻ stayed dissolved as spectators.
Feedback if chosen: This option is wrong — Na⁺ and Cl⁻ are the spectators here. They stay dissolved, unchanged, on both sides of the event.
Feedback if chosen: This option is wrong — two positive ions repel each other. A precipitate pairs a positive ion with a negative ion.
Feedback if chosen: This option is wrong — two negative ions repel each other. The solid pairs the positive Ca²⁺ with the negative CO₃²⁻.
Check — multiple choice
What happens in every precipitation reaction?
Feedback if chosen: This option is wrong — a moving proton is the acid-base test. Precipitation is about ions joining into a solid.
Feedback if chosen: This option is wrong — a changed oxidation number is the redox test. In a precipitation, the ions keep their charges and simply pair up.
Feedback if chosen: This option is wrong — a gas leaving is a different event. The defining event here is a solid forming from dissolved ions.
Feedback if chosen: Correct. In every precipitation reaction, ions from mixed aqueous solutions join into an insoluble or sparingly soluble ionic compound.
Check — multiple choice
Solutions of iron(II) nitrate and sodium hydroxide are mixed. Iron(II) hydroxide is insoluble in water (supplied). Solid Fe(OH)₂ appears. Which name fits the event?
Feedback if chosen: Correct. Dissolved Fe²⁺ and OH⁻ ions joined into an insoluble ionic compound — a precipitation reaction.
Feedback if chosen: This option is wrong — dissociation is the reverse direction: a dissolved compound separating into its ions. Here ions joined into a solid.
Feedback if chosen: This option is wrong — no proton moves. The Fe²⁺ and OH⁻ ions simply lock together into a solid.
Feedback if chosen: This option is wrong — every ion keeps its charge, so no electrons moved. The event is ions joining into a solid.
Outcome: State that all sodium, potassium, ammonium, and nitrate salts are soluble in water, and that any other compound's solubility must be supplied.
A student mixes 25 mL of silver nitrate solution with 25 mL of sodium chloride solution.
White solid silver chloride precipitates at once.
The student then mixes sodium nitrate solution with potassium chloride solution.
No solid appears at all.
On paper, all four dissolved compounds look alike.
To predict the solid, the student must know which compounds refuse to stay dissolved.
Must an AP student memorize the solubility of every compound?
An ionic compound is called a salt.
Sodium chloride, silver nitrate, and potassium iodide are all salts.
The AP exam expects exactly four solubility facts from memory.
| Family | One example |
|---|---|
| sodium salts | NaCl |
| potassium salts | K₂SO₄ |
| ammonium salts | NH₄Br |
| nitrate salts | Ca(NO₃)₂ |
Every sodium salt is soluble in water.
Every potassium salt is soluble in water.
Every ammonium salt is soluble in water.
Every nitrate salt is soluble in water.
Those four families are the entire memory list.
Any other compound's solubility must arrive in the question, stated as a given fact.
AgCl precipitated — and a question about it would supply the fact that silver chloride is insoluble.
NaNO₃ and KCl stayed dissolved — the four-family list covers both, from memory.
State that all sodium, potassium, ammonium, and nitrate salts are soluble in water, and that any other compound's solubility must be supplied in the question.
Check — multiple choice
From the memory list alone, what do you know about potassium carbonate, K₂CO₃, in water?
Feedback if chosen: Correct. K₂CO₃ is a potassium salt, and every potassium salt is soluble — one of the four memorized families.
Feedback if chosen: This option is wrong — you ruled against a compound on the memory list. Every potassium salt is soluble in water.
Feedback if chosen: This option is wrong — no potassium salt is only sparingly soluble. The memory list makes every potassium salt fully soluble.
Feedback if chosen: This option is wrong — the list does decide this one. K₂CO₃ is a potassium salt, one of the four always-soluble families.
Check — multiple choice
From memory alone, which of these compounds is certainly soluble in water?
Feedback if chosen: This option is wrong — silver iodide sits outside the four families, so its solubility would have to be supplied in the question.
Feedback if chosen: Correct. NH₄Br is an ammonium salt, and every ammonium salt is soluble — one of the four memorized families.
Feedback if chosen: This option is wrong — barium sulfate sits outside the four families. Nothing about it can be claimed from memory.
Feedback if chosen: This option is wrong — lead(II) chloride sits outside the four families, so its solubility must arrive as a supplied fact.
Check — multiple choice
From the memory list alone, what do you know about copper(II) sulfide, CuS, in water?
Feedback if chosen: This option is wrong — CuS is not a sodium, potassium, ammonium, or nitrate salt, so memory licenses no verdict at all.
Feedback if chosen: This option is wrong — even a compound that happens to be insoluble cannot be called insoluble from memory. Outside the four families, the fact must be supplied.
Feedback if chosen: Correct. CuS sits outside the four always-soluble families, so its solubility must be supplied in the question.
Feedback if chosen: This option is wrong — the memory list carries no sparingly-soluble entries. Outside the four families, every verdict waits for supplied information.
Check — multiple choice
What does the word salt name?
Feedback if chosen: This option is wrong — sodium chloride is one salt among thousands. The word covers a whole class of compounds.
Feedback if chosen: This option is wrong — you defined salt by solubility. Plenty of salts, such as AgCl, barely dissolve at all.
Feedback if chosen: This option is wrong — a compound's color does not decide whether it is a salt. Lead(II) iodide is a bright yellow salt.
Feedback if chosen: Correct. A salt is any ionic compound — sodium chloride, silver nitrate, and potassium iodide are all salts.
Check — multiple choice
Which of these compounds is a salt?
Feedback if chosen: This option is wrong — CO₂ is a molecular compound of two nonmetals. A salt is an ionic compound.
Feedback if chosen: This option is wrong — H₂O is a molecular compound. A salt is an ionic compound.
Feedback if chosen: This option is wrong — NH₃ is a molecular compound; it dissolves in water, but dissolving does not make a compound a salt.
Feedback if chosen: Correct. LiCl is built from Li⁺ and Cl⁻ ions. An ionic compound is a salt.
Outcome: Predict whether a precipitate forms when two given solutions mix — and identify it — using the four-family memory list plus supplied solubility information.
A lab plan calls for mixing 25 mL of barium chloride solution with 25 mL of sodium sulfate solution.
An information line on the plan reads: barium sulfate is insoluble in water.
Four ions will share the water: Ba²⁺, Cl⁻, Na⁺, and SO₄²⁻.
Two new cation-anion pairings are possible: barium with sulfate, and sodium with chloride.
The student wants the verdict before pouring anything.
Which pairing, if either, drops out of the water as a solid?
First, list the ions the mixed solution will contain.
Then write the two new cation-anion pairings.
Then judge each pairing: check the four-family memory list first, and the supplied solubility line second.
A pairing judged insoluble is the precipitate.
A mixture with no insoluble pairing forms no precipitate at all.
You've already seen how to write a net ionic equation — the verdict lands as one.
Q: Does a precipitate form when barium chloride solution and sodium sulfate solution mix — and if so, which compound is it?
List the ions in the mixed solution
Ba²⁺, Cl⁻, Na⁺, SO₄²⁻
Write the two new pairings
barium with sulfate: BaSO₄
sodium with chloride: NaCl
Judge each pairing
NaCl: a sodium salt → soluble, from the memory list
BaSO₄: the plan supplies the fact → insoluble
Name the precipitate and write the net ionic equation
BaSO₄ precipitates
Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)
Predict whether a precipitate forms when two given solutions are mixed — and identify it — using the four-family solubility facts plus any supplied solubility information.
Check — multiple choice
Solutions of strontium nitrate, Sr(NO₃)₂, and sodium carbonate, Na₂CO₃, will be mixed. Which two new cation-anion pairings could form?
Feedback if chosen: This option is wrong — those are the two compounds you started with, not new pairings. Each cation pairs with the other solution's anion.
Feedback if chosen: This option is wrong — you paired the two cations together and the two anions together. Like charges repel; a pairing is one cation with one anion.
Feedback if chosen: Correct. Each cation takes the other solution's anion: Sr²⁺ with CO₃²⁻ makes SrCO₃, and Na⁺ with NO₃⁻ makes NaNO₃.
Feedback if chosen: This option is wrong — you dropped one pairing. Two cations and two anions always offer two new pairings to judge.
Check — multiple choice
Solutions of zinc chloride, ZnCl₂, and sodium carbonate, Na₂CO₃, are mixed. Supplied fact: zinc carbonate is insoluble in water. What precipitate, if any, forms?
Feedback if chosen: This option is wrong — NaCl is a sodium salt, on the memory list, so it stays dissolved.
Feedback if chosen: Correct. NaCl stays dissolved (memory list), and the supplied fact judges ZnCO₃ insoluble — ZnCO₃ is the precipitate.
Feedback if chosen: This option is wrong — one pairing fails the solubility test. The supplied fact says zinc carbonate is insoluble, so it drops out as a solid.
Feedback if chosen: This option is wrong — NaCl sits on the memory list: a sodium salt. Only the ZnCO₃ leaves the solution.
Check — multiple choice
Solutions of ammonium chloride, NH₄Cl, and potassium nitrate, KNO₃, are mixed. What forms?
Feedback if chosen: This option is wrong — KCl is a potassium salt, on the memory list, so it stays dissolved.
Feedback if chosen: This option is wrong — NH₄NO₃ is an ammonium salt and a nitrate salt, so the memory list keeps it dissolved.
Feedback if chosen: This option is wrong — NH₄Cl is an ammonium salt; it began dissolved and stays dissolved.
Feedback if chosen: Correct. Every possible pairing here is an ammonium, potassium, or nitrate salt — all soluble from memory — so no precipitate forms.
Check — drawing task (pen and paper, self-marked)
Solutions of copper(II) sulfate, CuSO₄, and sodium hydroxide, NaOH, are mixed. Supplied fact: copper(II) hydroxide is insoluble in water. Take a pen and paper and write the net ionic equation for the precipitation, with a state symbol on every species and the charge on every ion. When you are done, select Check to compare your answer with the model answer.
Model answer shown to the student
Self-mark checklist — the drawing should show:
Outcome: Classify a given reaction as acid-base, redox, or precipitation from its equation or description, citing the evidence for the class.
A student mixes 50 mL of silver nitrate solution with 50 mL of sodium chloride solution.
White silver chloride precipitates.
On the exam, the same event arrives as a bare equation: AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq).
The question asks for the class: acid-base, redox, or precipitation.
Unit 4 taught one test per class.
Which test settles this equation, and what evidence would you cite?
A transferred proton places a reaction in the acid-base class.
A changed oxidation number places a reaction in the redox class.
A new insoluble ionic solid places a reaction in the precipitation class.
| The class | The test | The evidence to cite | One example equation |
|---|---|---|---|
| acid-base | Does a proton move? | the H⁺ moving from donor to acceptor | HCl(g) + NH₃(g) → NH₄Cl(s) |
| redox | Does an oxidation number change? | the numbers that rise and fall | Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g) |
| precipitation | Do dissolved ions join into an insoluble solid? | the insoluble product | Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s) |
Apply all three tests to the silver equation.
No proton moves — there is no H anywhere in the equation.
Number the atoms: Ag +1, N +5, O −2, Na +1, Cl −1 — every number stays the same on both sides.
The solid AgCl is a new insoluble ionic compound, built from the dissolved Ag⁺ and Cl⁻.
The evidence settles the verdict: the reaction is a precipitation reaction.
Two tests can fire at once. Mix barium hydroxide solution with sulfuric acid: protons move to the OH⁻ ions AND insoluble BaSO₄ forms — that one reaction is both acid-base and precipitation.
On the exam, cite the evidence with the class: the insoluble product, the moved proton, or the changed oxidation numbers. Where two tests fire, cite both.
Classify a given reaction as acid-base, oxidation-reduction, or precipitation from its equation or description, citing the proton-transfer, oxidation-number, or insoluble-product evidence.
Check — multiple choice
HClO₄(aq) + KOH(aq) → KClO₄(aq) + H₂O(l). Classify the reaction.
Feedback if chosen: Correct. The HClO₄ passes a proton to the OH⁻, making H₂O — a transferred proton places the reaction in the acid-base class.
Feedback if chosen: This option is wrong — number the atoms: every one keeps its oxidation number. A proton moves here, not electrons.
Feedback if chosen: This option is wrong — every product stays dissolved except the water; no insoluble ionic solid forms.
Feedback if chosen: This option is wrong — one test does fire. The HClO₄ hands an H⁺ to the OH⁻, so the reaction is acid-base.
Check — multiple choice
HClO₄(aq) + KOH(aq) → KClO₄(aq) + H₂O(l). Which observation about the equation is the evidence for its class?
Feedback if chosen: This option is wrong — no solid forms here; every species except the water stays dissolved.
Feedback if chosen: This option is wrong — no gas appears in the equation. The evidence is the H⁺ moving to the OH⁻.
Feedback if chosen: Correct. The H⁺ moving from the HClO₄ to the OH⁻ is exactly the evidence the acid-base classification rests on.
Feedback if chosen: This option is wrong — check the numbers: H stays +1, O stays −2, K stays +1, and the Cl keeps its value. Nothing changes.
Check — multiple choice
AgNO₃(aq) + KI(aq) → AgI(s) + KNO₃(aq). Supplied fact: silver iodide is insoluble in water. A student classifies the reaction by its shape — the two compounds swap partners — and stops there. Which statement is correct?
Feedback if chosen: This option is wrong — the acid-base test asks whether a proton moves, and no species here carries an H to give.
Feedback if chosen: Correct. The class comes from evidence, not shape: dissolved Ag⁺ and I⁻ joined into insoluble AgI, so the reaction is a precipitation.
Feedback if chosen: This option is wrong — the classes rest on evidence: an insoluble product, a moved proton, or changed oxidation numbers. The swap shape names none of them.
Feedback if chosen: This option is wrong — the redox test reads oxidation numbers, and every atom here keeps its number. Shape decides nothing.
Free-response question
Source note (not shown as part of the question): authored to archetype classify-and-cite (LO 4.7.A)
A strip of magnesium is placed in a colorless solution of silver nitrate. The equation is Mg(s) + 2AgNO₃(aq) → Mg(NO₃)₂(aq) + 2Ag(s). Answer on paper, one short sentence per step of your reasoning, then select Check to score your answer against the model.
(a) Classify the reaction as acid-base, oxidation-reduction, or precipitation. [1 point]
(b) Cite the evidence for your classification in part (a), naming the species that change. [2 points]
✓ Model response (the key)
(a)
The reaction is an oxidation-reduction (redox) reaction.
(b)
The Mg's oxidation number climbs from 0 to +2, so the Mg is oxidized.
The Ag's oxidation number falls from +1 to 0, so the Ag⁺ is reduced.
Electrons move from the Mg to the Ag⁺ — that transfer is the redox evidence.
Scoring rubric — 3 points available
10 lessons
Outcome: State that a Brønsted-Lowry acid is a proton (H⁺) donor, and spot the donor in a shown proton transfer.
In water, hydrochloric acid reacts with sodium hydroxide.
Hydrogen chloride gas also reacts with ammonia gas in perfectly dry air.
Where the two gases mix, a white smoke of tiny solid crystals appears.
You've already seen that an acid-base reaction is a transfer of a proton — an H⁺ ion — from one species to another.
The two settings could hardly look more different.
In both settings, the HCl does the same single job.
The HCl hands a proton, H⁺, to the other species.
What exactly earns HCl the title "acid"?
An acid is a species that donates a proton, an H⁺ ion, to another species.
The chemists Brønsted and Lowry pinned the definition to that single act.
A proton donor is therefore called a Brønsted-Lowry acid.
Toward water, HCl donates a proton.
Toward ammonia, HCl donates a proton just the same: HCl + NH₃ → NH₄⁺ + Cl⁻.
The NH₄⁺ and Cl⁻ ions lock together as a white solid — the smoke you saw.
No water formed, yet the reaction is still acid-base.
The definition travels with the proton, never with the water.
State that a Brønsted-Lowry acid is a proton (H⁺) donor, and recognize the donor in a shown proton transfer.
Check — multiple choice
What makes a species a Brønsted-Lowry acid?
Feedback if chosen: Correct. Donating a proton is the single act the Brønsted-Lowry definition names — nothing else is required.
Feedback if chosen: This option is wrong — you described the receiving species. The Brønsted-Lowry acid is the species that gives the proton away.
Feedback if chosen: This option is wrong — you tied the definition to one product. HCl + NH₃ makes no water at all, yet HCl still acts as the acid.
Feedback if chosen: This option is wrong — containing hydrogen is not enough. NH₃ contains hydrogen, yet toward HCl it receives a proton rather than donating one.
Check — multiple choice
Hydrogen iodide reacts with ammonia: HI + NH₃ → NH₄⁺ + I⁻. The proton moves from HI to NH₃. Which species is the Brønsted-Lowry acid?
Feedback if chosen: This option is wrong — NH₃ received the proton. The acid is the species that donated it.
Feedback if chosen: Correct. HI let its proton go — the proton donor is the Brønsted-Lowry acid.
Feedback if chosen: This option is wrong — NH₄⁺ is a product. The acid label belongs to the reactant that donated the proton: HI.
Feedback if chosen: This option is wrong — I⁻ is what HI became after donating. The acid is HI itself, the species that gave the proton away.
Check — multiple choice
HCl gas reacts with NH₃ gas in dry air, with no water anywhere. A student claims the reaction cannot be acid-base, because no water forms. Which statement is correct?
Feedback if chosen: This option is wrong — water is one common outcome, not the definition. The proton transfer defines the reaction, and HCl + NH₃ has one.
Feedback if chosen: This option is wrong — the definition travels with the proton, not with the water. The dry-air transfer is acid-base without any water present.
Feedback if chosen: Correct. Proton transfer defines an acid-base reaction. Water plus a dissolved solid is just one common outcome, and this reaction skips it.
Feedback if chosen: This option is wrong — the smoke is solid ammonium chloride, built from NH₄⁺ and Cl⁻ ions. The right defense of the reaction is the proton transfer itself.
Check — multiple choice
During a reaction in water, an HNO₃ molecule turns into an NO₃⁻ ion. What did HNO₃ do?
Feedback if chosen: This option is wrong — accepting a proton would make the formula one H heavier, not one H lighter.
Feedback if chosen: This option is wrong — you reached for electron transfer. Compare the formulas: HNO₃ lost an H⁺, not an electron.
Feedback if chosen: This option is wrong — both formulas hold three oxygen atoms. The difference is one proton, H⁺.
Feedback if chosen: Correct. HNO₃ is one proton heavier than NO₃⁻, so HNO₃ donated a proton — it acted as a Brønsted-Lowry acid.
Check — multiple choice
During a reaction, one species hands an H⁺ ion to another. Which label fits the species that hands the H⁺ over?
Feedback if chosen: This option is wrong — an oxidized species gives away electrons. This species gave away a proton, H⁺.
Feedback if chosen: This option is wrong — a reduced species gains electrons. This species gave away a proton, H⁺.
Feedback if chosen: This option is wrong — a spectator ion takes no part in the reaction. This species handed over a proton.
Feedback if chosen: Correct. A species that donates a proton, H⁺, is a Brønsted-Lowry acid.
Outcome: State that a Brønsted-Lowry base is a proton (H⁺) acceptor, and tell the acid from the base in a shown transfer.
The white-smoke reaction has two actors.
The HCl donated the proton.
The ammonia played the opposite part.
NH₃ carries a lone pair of electrons on its nitrogen atom.
That lone pair grabbed the incoming H⁺.
The catch built the NH₄⁺ ion.
What do chemists call the species on the receiving end?
A species that accepts a proton during a reaction is called a Brønsted-Lowry base.
| Brønsted-Lowry acid | Brønsted-Lowry base | |
|---|---|---|
| What it does with the proton | donates H⁺ | accepts H⁺ |
| In HCl + NH₃ → NH₄⁺ + Cl⁻ | HCl | NH₃ |
The Brønsted-Lowry acid donates the proton.
The Brønsted-Lowry base accepts the proton.
NH₃ accepted the H⁺ from HCl.
So NH₃ acted as the base.
Every proton transfer therefore carries one acid and one base.
State that a Brønsted-Lowry base is a proton (H⁺) acceptor, and recognize the acceptor in a shown proton transfer.
Check — multiple choice
In HCl(aq) + OH⁻(aq) → H₂O(l) + Cl⁻(aq), the proton moves from HCl to OH⁻. Which species is the Brønsted-Lowry base?
Feedback if chosen: This option is wrong — HCl donated the proton, so HCl is the acid, not the base.
Feedback if chosen: Correct. OH⁻ accepted the proton — the proton acceptor is the Brønsted-Lowry base.
Feedback if chosen: This option is wrong — H₂O is a product. The base label belongs to the reactant that accepted the proton: OH⁻.
Feedback if chosen: This option is wrong — Cl⁻ is what HCl became after donating. It accepted nothing in this reaction.
Check — multiple choice
What makes a species a Brønsted-Lowry base?
Feedback if chosen: This option is wrong — donating a proton is the acid's job. The base does the opposite.
Feedback if chosen: This option is wrong — some bases do release OH⁻, but the Brønsted-Lowry definition asks only one thing: did the species accept a proton?
Feedback if chosen: This option is wrong — you reached for electron transfer. The Brønsted-Lowry definition is about the proton, H⁺.
Feedback if chosen: Correct. Accepting a proton is the single act that makes a species a Brønsted-Lowry base.
Check — multiple choice
Chlorous acid reacts with the carbonate ion: HClO₂(aq) + CO₃²⁻(aq) → ClO₂⁻(aq) + HCO₃⁻(aq). The proton moves from HClO₂ to CO₃²⁻. Which species acts as the Brønsted-Lowry base?
Feedback if chosen: This option is wrong — HClO₂ donated the proton, so it is the acid.
Feedback if chosen: This option is wrong — ClO₂⁻ is what the acid became after donating. It is a product, not the acceptor.
Feedback if chosen: Correct. CO₃²⁻ accepted the proton and became HCO₃⁻ — the acceptor is the Brønsted-Lowry base.
Feedback if chosen: This option is wrong — HCO₃⁻ is the product the acceptance built. The base label belongs to CO₃²⁻, the reactant that accepted the proton.
Check — multiple choice
H₂S donates a proton to NH₃. Which labels are right?
Feedback if chosen: Correct. H₂S donated the proton (acid), and NH₃ accepted it (base) — every proton transfer carries one of each.
Feedback if chosen: This option is wrong — you swapped the two jobs. The donor is the acid, and the acceptor is the base.
Feedback if chosen: This option is wrong — a proton cannot be donated by both species at once. One species donates, and the other accepts.
Feedback if chosen: This option is wrong — a proton cannot be accepted by both species at once. One species accepts, and the other donates.
Check — multiple choice
In a proton transfer, methylamine, CH₃NH₂, accepts an H⁺ ion and becomes CH₃NH₃⁺. Which label fits the methylamine?
Feedback if chosen: Correct. The methylamine accepted a proton, H⁺ — a proton acceptor is a Brønsted-Lowry base.
Feedback if chosen: This option is wrong — the acid is the species that donates the proton. The methylamine accepted one.
Feedback if chosen: This option is wrong — a reduced species gains electrons. The methylamine gained a proton, H⁺.
Feedback if chosen: This option is wrong — a spectator takes no part. The methylamine caught the proton and became CH₃NH₃⁺.
Outcome: Identify the Brønsted-Lowry acid and base in a printed proton-transfer equation by comparing each reactant's formula with its product's.
An AP item prints one bare equation: HBr + NH₃ → NH₄⁺ + Br⁻.
The item wants two labels: the acid and the base.
No lab scene backs the equation up.
Somewhere in those formulas, one proton changed hands.
Which species let the proton go, and which species caught it?
Compare each reactant's formula with its product's formula.
HBr turned into Br⁻ — one proton lighter.
The species that lost the proton is the acid.
NH₃ turned into NH₄⁺ — one proton heavier.
The species that gained the proton is the base.
The charge helps you check: losing H⁺ drops a formula's charge by one, and gaining H⁺ raises it by one.
Formula comparison alone settles both labels.
Identify which species acts as the Brønsted-Lowry acid and which as the base in a given proton-transfer equation.
Check — multiple choice
HF(aq) + OH⁻(aq) → F⁻(aq) + H₂O(l). Which species is the Brønsted-Lowry acid?
Feedback if chosen: Correct. HF turned into F⁻, one proton lighter — the species that lost the proton is the acid.
Feedback if chosen: This option is wrong — OH⁻ turned into H₂O, one proton heavier. It did not lose a proton.
Feedback if chosen: This option is wrong — F⁻ is a product. Compare reactants with their products: the reactant that lost the proton is HF.
Feedback if chosen: This option is wrong — H₂O is a product. The acid is the reactant that lost the proton: HF.
Check — multiple choice
HF(aq) + OH⁻(aq) → F⁻(aq) + H₂O(l). Which species is the Brønsted-Lowry base?
Feedback if chosen: This option is wrong — F⁻ is a product. The base is the reactant that gained the proton.
Feedback if chosen: Correct. OH⁻ turned into H₂O, one proton heavier — the species that gained the proton is the base.
Feedback if chosen: This option is wrong — H₂O is what the base became. The base label belongs to the reactant OH⁻.
Feedback if chosen: This option is wrong — HF lost a proton, so HF is the acid, not the base.
Check — multiple choice
HSO₄⁻(aq) + NH₃(aq) → SO₄²⁻(aq) + NH₄⁺(aq). Which species is the Brønsted-Lowry acid?
Feedback if chosen: This option is wrong — NH₃ turned into NH₄⁺, one proton heavier. It did not lose a proton.
Feedback if chosen: This option is wrong — SO₄²⁻ is a product. Compare each reactant with its product: HSO₄⁻ became SO₄²⁻.
Feedback if chosen: Correct. HSO₄⁻ turned into SO₄²⁻, one proton lighter — an ion can be the acid, as long as it lost the proton.
Feedback if chosen: This option is wrong — NH₄⁺ is a product. The acid is the reactant that lost the proton: HSO₄⁻.
Check — multiple choice
HSO₄⁻(aq) + NH₃(aq) → SO₄²⁻(aq) + NH₄⁺(aq). Which species is the Brønsted-Lowry base?
Feedback if chosen: This option is wrong — HSO₄⁻ lost a proton, so it is the acid.
Feedback if chosen: This option is wrong — SO₄²⁻ is what the acid became. Products carry no acid or base label here.
Feedback if chosen: This option is wrong — NH₄⁺ is the product the acceptance built. The base is the reactant NH₃.
Feedback if chosen: Correct. NH₃ turned into NH₄⁺, one proton heavier — the species that gained the proton is the base.
Outcome: Explain why a water molecule can donate a proton (forming OH⁻) or accept one (forming H₃O⁺, the hydronium ion).
Dissolve hydrogen chloride in water.
Each water molecule catches the proton that an HCl molecule releases.
The catch turns the H₂O into a new ion, H₃O⁺: HCl(aq) + H₂O(l) → H₃O⁺(aq) + Cl⁻(aq).
Water accepted a proton — water acted as the base.
Now dissolve ammonia in water instead.
Some water molecules hand one of their own protons to NH₃ molecules.
Each water molecule that donates is left as OH⁻, and each NH₃ that accepts becomes NH₄⁺.
Water donated a proton — water acted as the acid.
The same molecule played both parts.
What equips water to play both?
Check — multiple choice
First, a quick recall. Oxygen attracts shared electrons much more strongly than hydrogen does. What does that make each O–H bond in water?
Feedback if chosen: Correct. Clearly unequal pull on the shared pair makes the O–H bond polar — oxygen holds the electron density more tightly.
Feedback if chosen: This option is wrong — nearly equal pull would make it nonpolar, and O and H pull very unequally.
Feedback if chosen: This option is wrong — the pair is still shared between two nonmetals, so the bond is covalent, just polar.
Feedback if chosen: This option is wrong — each O–H bond shares one pair. The point here is the unequal pull on that pair.
Check — multiple choice
One more recall. How many lone pairs of electrons sit on the oxygen atom of a water molecule?
Feedback if chosen: This option is wrong — oxygen bonds twice and keeps the rest of its valence electrons as lone pairs.
Feedback if chosen: This option is wrong — count oxygen's six valence electrons: two bonds use two of them, leaving four electrons — two pairs.
Feedback if chosen: Correct. Oxygen's six valence electrons give two O–H bonds and two lone pairs.
Feedback if chosen: This option is wrong — four lone pairs would need eight nonbonding electrons, and oxygen has only four left after bonding.
Those two features are water's whole toolkit.
A polar O–H bond can release its hydrogen as H⁺.
The molecule that released it has acted as an acid.
The leftover species is OH⁻.
Oxygen's two lone pairs can also bind an extra H⁺.
The molecule that binds one has acted as a base.
The product of that catch, H₃O⁺, is called the hydronium ion.
Which part water plays depends on its partner: beside a proton donor, water accepts; beside a proton acceptor, water donates.
Explain why a water molecule can either donate a proton (forming OH⁻) or accept one (forming H₃O⁺, the hydronium ion): its polar O–H bonds can release H⁺, and the lone pairs on its oxygen can bind H⁺.
Check — multiple choice
Perchloric acid donates protons to water: HClO₄(aq) + H₂O(l) → H₃O⁺(aq) + ClO₄⁻(aq). Which part did water play?
Feedback if chosen: This option is wrong — water gained a proton here, becoming H₃O⁺. Donating would have left it as OH⁻.
Feedback if chosen: Correct. H₂O turned into H₃O⁺, one proton heavier — water accepted the proton, so it acted as the base.
Feedback if chosen: This option is wrong — water changed into H₃O⁺, so it took part in the transfer.
Feedback if chosen: This option is wrong — each water molecule played one part here: it accepted the proton HClO₄ donated.
Check — multiple choice
What equips one water molecule to act as an acid or as a base?
Feedback if chosen: Correct. Releasing an H⁺ from a polar O–H bond is the acid's act; binding an H⁺ with a lone pair is the base's act.
Feedback if chosen: This option is wrong — no O–H bond breaks into gases here. One proton, H⁺, moves; the rest of the molecule stays together.
Feedback if chosen: This option is wrong — flowing moves whole molecules, not protons. Water's O–H bonds can release H⁺, and its lone pairs can bind H⁺.
Feedback if chosen: This option is wrong — both hydrogen ends of water carry the same slight positive charge. The two jobs come from the bonds and the lone pairs.
Check — multiple choice
A student says: "Water is always the base in a proton transfer." Which statement is correct?
Feedback if chosen: This option is wrong — water also donates. Beside ammonia, water hands over a proton and is left as OH⁻.
Feedback if chosen: This option is wrong — water changes in both cases: to H₃O⁺ when it accepts, and to OH⁻ when it donates.
Feedback if chosen: This option is wrong — you flipped too far. Water does act as a base beside acids like HCl; it just is not locked into that part.
Feedback if chosen: Correct. Water's part depends on its partner: it accepts beside a donor, and it donates beside an acceptor like NH₃.
Check — multiple choice
The sulfide ion, S²⁻, takes a proton from a water molecule. Which species is the water left as?
Feedback if chosen: This option is wrong — H₃O⁺ forms when water gains a proton. Here water lost one.
Feedback if chosen: Correct. Water donated a proton to S²⁻, so the water is left one proton lighter: OH⁻.
Feedback if chosen: This option is wrong — HS⁻ is what the sulfide ion becomes after catching the proton, not what the water becomes.
Feedback if chosen: This option is wrong — the water gave a proton away, so its formula changed from H₂O to OH⁻.
Check — multiple choice
Which formula belongs to the hydronium ion?
Feedback if chosen: This option is wrong — OH⁻ is the species left behind when water donates a proton.
Feedback if chosen: This option is wrong — count the hydrogens: water plus one extra H⁺ carries three, not two.
Feedback if chosen: Correct. Hydronium is a water molecule that has bound one extra proton: H₃O⁺.
Feedback if chosen: This option is wrong — H⁺ is the bare proton itself. Hydronium is the ion built when water binds that proton.
Check — multiple choice
Which change builds a hydronium ion, H₃O⁺?
Feedback if chosen: This option is wrong — losing an H⁺ leaves the water as OH⁻, one proton lighter.
Feedback if chosen: This option is wrong — an electron changes the charge the wrong way, and no proton arrives. Hydronium is water plus one H⁺.
Feedback if chosen: This option is wrong — two joined waters would carry four hydrogens. Hydronium carries three: H₂O plus one H⁺.
Feedback if chosen: Correct. A water molecule that binds one extra proton becomes the hydronium ion, H₃O⁺.
Outcome: Write the ionization equation for a named acid or base in water, choosing → for a complete ionization and ⇌ for a partial one.
Pure hydrogen chloride is a molecular gas.
A tank of the pure gas conducts no electricity at all.
A student dissolves 1 g of the gas in 100 mL of water.
The solution now conducts strongly.
A solution conducts only when mobile ions move through it.
The gas brought no ions with it.
Where did the ions come from?
The water reacts with the dissolved HCl.
Each water molecule pulls the proton off an HCl molecule.
The transfer creates two ions: H₃O⁺ and Cl⁻.
A reaction in which a dissolved acid or base forms ions with water is called ionization.
The written record is an ionization equation: HCl(aq) + H₂O(l) → H₃O⁺(aq) + Cl⁻(aq).
HCl transfers its protons completely — virtually every dissolved HCl molecule lets its proton go.
A complete ionization keeps the forward arrow, →.
Some acids and bases transfer protons only partially — most of their dissolved molecules stay unchanged.
A partial ionization is written with the double arrow, ⇌.
The double arrow means the reaction does not go to completion — you'll study this in Unit 7.
Ammonia ionizes partially, so its equation carries ⇌: NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq).
Every question states whether the named acid or base ionizes completely or partially — you never have to remember which.
To write any ionization equation, follow three steps.
**1.** Decide which way the proton moves: a dissolved acid gives its proton to water, and a dissolved base takes a proton from water.
**2.** Write the products the transfer creates: the changed acid or base, plus H₃O⁺ (from an acid) or OH⁻ (from a base).
**3.** Pick the arrow the stated case demands: → for complete, ⇌ for partial.
Write the proton-transfer equation for an acid or base ionizing in water, using → for a complete ionization and ⇌ for a partial one.
Check — multiple choice
Phosphoric acid, H₃PO₄, ionizes partially in water, transferring one proton per molecule. Which species donates the proton?
Feedback if chosen: Correct. H₃PO₄ is the dissolved acid, so it gives its proton to water.
Feedback if chosen: This option is wrong — beside a dissolved acid, water accepts the proton rather than donating one.
Feedback if chosen: This option is wrong — H₃O⁺ is a product of the transfer, not a starting species.
Feedback if chosen: This option is wrong — OH⁻ appears when a dissolved base takes a proton from water. An acid's ionization makes H₃O⁺ instead.
Check — multiple choice
When H₃PO₄ transfers one proton to water, which two new species form?
Feedback if chosen: This option is wrong — OH⁻ forms when water donates a proton. Here water accepts one, becoming H₃O⁺.
Feedback if chosen: Correct. Water plus the proton is H₃O⁺, and H₃PO₄ minus the proton is H₂PO₄⁻.
Feedback if chosen: This option is wrong — H₄PO₄⁺ would mean the acid gained a proton. The acid loses one, becoming H₂PO₄⁻.
Feedback if chosen: This option is wrong — it reverses the transfer for both species: the acid loses a proton, and the water gains it.
Check — multiple choice
The ionization of H₃PO₄ is stated to be partial. Which arrow belongs in its ionization equation?
Feedback if chosen: This option is wrong — the forward arrow claims the reaction goes to completion, and the stem says the ionization is partial.
Feedback if chosen: This option is wrong — the arrow carries information. The stated case decides it: partial demands ⇌.
Feedback if chosen: Correct. A partial ionization carries the double arrow — the reaction does not go to completion.
Feedback if chosen: This option is wrong — every chemical equation carries an arrow between reactants and products.
Check — multiple choice
Which of these equations shows an ionization?
Feedback if chosen: This option is wrong — this is dissociation: the Na⁺ and Cl⁻ ions already existed in the solid and simply separated. No proton moved.
Feedback if chosen: Correct. A dissolved acid forms ions with water by proton transfer — that is ionization, and the partial case carries ⇌.
Feedback if chosen: This option is wrong — this is a precipitation reaction. Ions swap partners, but no new ions form by proton transfer with water.
Feedback if chosen: This option is wrong — hydrogen burns to water here. No dissolved acid or base forms ions with water.
Check — drawing task (pen and paper, self-marked)
Take a pen and paper and write the ionization equation for nitrous acid, HNO₂, which ionizes partially in water. Include state symbols. When you are done, select Check to compare your answer with the model answer.
Model answer shown to the student
Self-mark checklist — the drawing should show:
Check — drawing task (pen and paper, self-marked)
Take a pen and paper and write the ionization equation for hydrobromic acid, HBr, which ionizes completely in water. Include state symbols. When you are done, select Check to compare your answer with the model answer.
Model answer shown to the student
Self-mark checklist — the drawing should show:
Check — drawing task (pen and paper, self-marked)
Take a pen and paper and write the ionization equation for the carbonate ion, CO₃²⁻, which acts as a base and ionizes partially in water. Include state symbols. When you are done, select Check to compare your answer with the model answer.
Model answer shown to the student
Self-mark checklist — the drawing should show:
Check — multiple choice
Solid cesium chloride, CsCl, dissolves in water, and its Cs⁺ and Cl⁻ ions separate into the water. No proton moves. Which name fits the event?
Feedback if chosen: Correct. The Cs⁺ and Cl⁻ ions already existed in the solid and simply separated — that is dissociation.
Feedback if chosen: This option is wrong — ionization creates new ions by proton transfer with water. These ions already existed in the solid.
Feedback if chosen: This option is wrong — precipitation is the opposite change: dissolved ions joining into a solid.
Feedback if chosen: This option is wrong — an acid-base reaction needs a moving proton, and no proton moves here.
Outcome: State that a conjugate acid-base pair is two species whose formulas differ by exactly one proton, and judge whether two formulas form a pair.
Line up the four species of the ammonia equation: NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq).
You've already seen the double arrow: it means the reaction does not go to completion — you'll study this in Unit 7.
NH₃ and NH₄⁺ differ by exactly one proton.
H₂O and OH⁻ differ by exactly one proton too.
The four species pair off like before-and-after photographs.
The same pairings show up in every proton-transfer equation.
What should chemists call two species separated by a single proton?
Two species whose formulas differ by exactly one proton form a conjugate acid-base pair.
| The two formulas | The difference | Verdict |
|---|---|---|
| NH₄⁺ and NH₃ | one H⁺ | a conjugate pair |
| H₂CO₃ and CO₃²⁻ | two H⁺ | not a pair |
| Fe³⁺ and Fe²⁺ | no H⁺, only charge | not a pair |
For example, NH₄⁺ and NH₃ differ by exactly one proton. NH₄⁺ and NH₃ are a conjugate acid-base pair.
HF and F⁻ differ by exactly one proton. HF and F⁻ are a conjugate acid-base pair.
But H₂CO₃ and CO₃²⁻ differ by two protons. H₂CO₃ and CO₃²⁻ are not a conjugate acid-base pair.
And Fe³⁺ and Fe²⁺ differ by no proton at all — only by charge. Fe³⁺ and Fe²⁺ are not a conjugate acid-base pair.
Within a pair, the member holding the extra proton is the pair's acid.
The member missing the proton is the pair's base.
In the NH₄⁺ and NH₃ pair, NH₄⁺ is the acid and NH₃ is the base.
State that a conjugate acid-base pair is two species whose formulas differ by exactly one proton (H⁺), and name which member is the pair's acid.
Check — multiple choice
Which of these is a conjugate acid-base pair?
Feedback if chosen: This option is wrong — H₂S and S²⁻ differ by two protons, one too many for a conjugate pair.
Feedback if chosen: This option is wrong — Na⁺ and OH⁻ are unrelated formulas. Removing a proton from one never gives the other.
Feedback if chosen: Correct. HS⁻ minus one proton is S²⁻ — exactly one H⁺ apart, so the two form a conjugate pair.
Feedback if chosen: This option is wrong — Cl₂ and Cl⁻ differ by an atom and an electron, not by a proton.
Check — multiple choice
Which of these is NOT a conjugate acid-base pair?
Feedback if chosen: This option is wrong — HSO₄⁻ minus one proton is SO₄²⁻, so these two do form a pair.
Feedback if chosen: This option is wrong — H₂O minus one proton is OH⁻, so these two do form a pair.
Feedback if chosen: Correct. H₂SO₄ and SO₄²⁻ differ by two protons — a conjugate pair sits exactly one proton apart.
Feedback if chosen: This option is wrong — HCN minus one proton is CN⁻, so these two do form a pair.
Check — multiple choice
Two species form a conjugate acid-base pair when their formulas differ by exactly what?
Feedback if chosen: This option is wrong — electron transfer is redox territory. Conjugate pairs are built by the transfer of a proton.
Feedback if chosen: Correct. One proton, H⁺, is the exact difference — no more, no less.
Feedback if chosen: This option is wrong — OH⁻ is a whole ion. The pair's two formulas differ by the bare proton, H⁺.
Feedback if chosen: This option is wrong — the difference is a hydrogen ion, not an oxygen atom.
Check — multiple choice
HSO₃⁻ and SO₃²⁻ form a conjugate acid-base pair. Which member is the pair's acid?
Feedback if chosen: This option is wrong — SO₃²⁻ is the member missing the proton, so it is the pair's base.
Feedback if chosen: This option is wrong — within one pair, exactly one member holds the extra proton.
Feedback if chosen: This option is wrong — every conjugate pair has an acid member: the one holding the extra proton.
Feedback if chosen: Correct. HSO₃⁻ holds the extra proton, and the member holding the extra proton is the pair's acid.
Check — multiple choice
Which of these pairings is a conjugate acid-base pair?
Feedback if chosen: This option is wrong — H₂C₂O₄ and C₂O₄²⁻ differ by two protons, one too many for a conjugate pair.
Feedback if chosen: Correct. HC₂O₄⁻ minus one proton is C₂O₄²⁻ — exactly one H⁺ apart, so the two form a conjugate pair.
Feedback if chosen: This option is wrong — H₃O⁺ minus one proton is H₂O, not OH⁻. The two differ by two protons.
Feedback if chosen: This option is wrong — Mn³⁺ and Mn²⁺ differ by charge alone, with no proton anywhere in the difference.
Outcome: Identify both conjugate acid-base pairs in a proton-transfer equation.
The AP exam prints this equation: HF(aq) + H₂O(l) ⇌ F⁻(aq) + H₃O⁺(aq).
You've already seen the double arrow: it means the reaction does not go to completion — you'll study this in Unit 7.
The item asks for both conjugate acid-base pairs.
Four species sit in the equation.
A tempting move pairs the two reactants with each other.
That move fails every time.
Which species truly belongs with which?
Every conjugate pair has one member on each side of the arrow.
Start with the acid on the left.
HF loses its proton and becomes F⁻, on the right.
HF and F⁻ form the first pair.
H₂O gains that proton and becomes H₃O⁺, on the right.
H₂O and H₃O⁺ form the second pair.
Two reactants never form a conjugate pair with each other — the pair is a before-and-after, not a partnership.
Identify both conjugate acid-base pairs in a proton-transfer equation.
Check — multiple choice
HCOOH(aq) + H₂O(l) ⇌ HCOO⁻(aq) + H₃O⁺(aq). Which pairing is one of this equation's conjugate pairs?
Feedback if chosen: This option is wrong — you paired the two reactants. A conjugate pair has one member on each side of the arrow.
Feedback if chosen: Correct. HCOOH minus one proton is HCOO⁻ — one member on each side, exactly one H⁺ apart.
Feedback if chosen: This option is wrong — HCOOH and H₃O⁺ do sit on opposite sides, but they differ by far more than one proton.
Feedback if chosen: This option is wrong — you paired the two products. A conjugate pair spans the arrow: one member on each side.
Check — multiple choice
HCOOH(aq) + H₂O(l) ⇌ HCOO⁻(aq) + H₃O⁺(aq). Which of the following is one of this equation's conjugate pairs?
Feedback if chosen: This option is wrong — you paired the two reactants. The pair spans the arrow.
Feedback if chosen: This option is wrong — H₂O and HCOO⁻ sit on opposite sides, but removing a proton from H₂O gives OH⁻, not HCOO⁻.
Feedback if chosen: Correct. H₂O plus the caught proton is H₃O⁺ — one member on each side, one H⁺ apart.
Feedback if chosen: This option is wrong — you paired the two products. Each pair has one member on each side of the arrow.
Check — multiple choice
ClO⁻(aq) + H₂O(l) ⇌ HClO(aq) + OH⁻(aq). Which pairing is one of this equation's conjugate pairs?
Feedback if chosen: Correct. ClO⁻ plus one proton is HClO — one member on each side, exactly one H⁺ apart.
Feedback if chosen: This option is wrong — you paired the two reactants. A conjugate pair always spans the arrow.
Feedback if chosen: This option is wrong — ClO⁻ and OH⁻ sit on opposite sides, but no single proton turns one into the other.
Feedback if chosen: This option is wrong — you paired the two products. Each pair has one member on each side of the arrow.
Check — multiple choice
ClO⁻(aq) + H₂O(l) ⇌ HClO(aq) + OH⁻(aq). Which of the following is one of this equation's conjugate pairs?
Feedback if chosen: This option is wrong — H₂O and HClO sit on opposite sides, but removing a proton from HClO gives ClO⁻, not H₂O.
Feedback if chosen: This option is wrong — you paired the two reactants. The pair spans the arrow.
Feedback if chosen: This option is wrong — OH⁻ pairs with H₂O, its one-proton partner, not with ClO⁻.
Feedback if chosen: Correct. H₂O minus the donated proton is OH⁻ — one member on each side, one H⁺ apart.
Outcome: State that the stronger an acid, the weaker its conjugate base — and the reverse.
Two acids ionize in water: hydrochloric acid, HCl, and hydrofluoric acid, HF.
In an HCl solution, almost every HCl molecule has donated its proton.
In an HF solution, most HF molecules still keep their protons.
Chemists call the near-complete donor a strong acid — so HCl is a strong acid.
Chemists call the reluctant donor a weak acid — so HF is a weak acid.
Now look at the two leftover ions, Cl⁻ and F⁻.
Which of the two ions grabs protons more eagerly?
Within a conjugate pair, the two strengths are opposites: the stronger the acid, the weaker its conjugate base.
| Conjugate pair | How completely the acid donates protons | How readily the base accepts protons |
|---|---|---|
| HCl and Cl⁻ | almost completely | hardly at all |
| HF and F⁻ | only partially | noticeably more than Cl⁻ does |
A stronger acid donates its protons more completely.
Its conjugate base accepts protons more weakly.
HCl donates protons almost completely.
So Cl⁻ accepts protons hardly at all — a negligible base.
HF donates only some of its protons.
So F⁻ accepts protons noticeably — a stronger base than Cl⁻.
The rule: the stronger the acid, the weaker its conjugate base.
And the reverse: the weaker the acid, the stronger its conjugate base.
State the inverse strength relationship within a conjugate pair: the stronger the acid, the weaker its conjugate base, and the reverse — qualitatively, with no numbers.
Check — multiple choice
Hydrobromic acid, HBr, donates its protons almost completely in water — a strong acid. What does that make its conjugate base, Br⁻?
Feedback if chosen: Correct. The stronger the acid, the weaker its conjugate base — so Br⁻, partnered with a strong acid, accepts protons hardly at all.
Feedback if chosen: This option is wrong — you carried the strength across the pair. Within the pair, the stronger one partner, the weaker the other.
Feedback if chosen: This option is wrong — Br⁻ is the pair's base, the member missing the proton. The question is only how strong a base it is.
Feedback if chosen: This option is wrong — the two strengths are not equal; they are inverse. A near-complete donor leaves a near-useless acceptor.
Check — multiple choice
A student says: "HClO₄ is a strong acid, so its conjugate base, ClO₄⁻, must be a strong base." Which statement is correct?
Feedback if chosen: This option is wrong — strength flips across a pair; it never carries across. The relationship is inverse.
Feedback if chosen: Correct. HClO₄ donates protons almost completely, so its conjugate base has almost no pull on protons — a negligible base.
Feedback if chosen: This option is wrong — an ion that accepted protons almost completely would be a strong base, and a strong acid's partner is the opposite.
Feedback if chosen: This option is wrong — a base needs no hydrogen; it only needs to accept a proton. ClO₄⁻ is a base — just a negligible one.
Check — multiple choice
H₂S keeps most of its protons in water — a weak acid. Compared with the conjugate base of a strong acid, how readily does HS⁻ accept protons?
Feedback if chosen: This option is wrong — you applied the rule in the same direction. The weaker the acid, the stronger its conjugate base.
Feedback if chosen: This option is wrong — the two acids differ in strength, so their conjugate bases must differ the opposite way.
Feedback if chosen: Correct. H₂S is the weaker acid, so HS⁻ is the stronger base — it accepts protons more readily than a strong acid's conjugate base does.
Feedback if chosen: This option is wrong — a weak acid's conjugate base is the eager one. It is the strong acid's partner that barely accepts.
Outcome: Predict the relative strengths of two conjugate partners from supplied acid-strength information.
A test item supplies one fact: HF is a weaker acid than HCl.
The item then asks which conjugate base is stronger, F⁻ or Cl⁻.
The item supplies no other chemistry at all.
We've already compared the two members of one pair.
Can one supplied sentence rank two different pairs?
Use the rule once per pair.
HCl is the stronger acid. So its conjugate base, Cl⁻, is the weaker base.
HF is the weaker acid. So its conjugate base, F⁻, is the stronger base.
F⁻ therefore accepts protons more readily than Cl⁻.
The whole routine is one flip: the base order reverses the supplied acid order.
Never rank from memory — the acid-strength facts always arrive in the question.
Predict the relative strengths of two conjugate partners from supplied acid-strength information.
Check — multiple choice
Supplied fact: CH₃COOH is a stronger acid than HCN. Which is the stronger base, CH₃COO⁻ or CN⁻?
Feedback if chosen: This option is wrong — you kept the supplied order. The base order reverses it: the stronger acid owns the weaker base.
Feedback if chosen: This option is wrong — the acids differ in strength, so their conjugate bases must differ the opposite way.
Feedback if chosen: This option is wrong — one supplied acid ranking is exactly enough. Flip it once and the base ranking appears.
Feedback if chosen: Correct. HCN is the weaker acid, so CN⁻ is the stronger base — the base order reverses the supplied acid order.
Check — multiple choice
Supplied facts, weakest acid first: HBrO, then HNO₂, then HClO₃. Rank their conjugate bases from strongest to weakest.
Feedback if chosen: Correct. The weakest acid, HBrO, owns the strongest base, BrO⁻ — the base order reverses the supplied acid order.
Feedback if chosen: This option is wrong — you kept the acid order. Reverse it: the strongest acid, HClO₃, owns the weakest base.
Feedback if chosen: This option is wrong — you swapped the top two. HBrO is the weaker acid of the two, so BrO⁻ outranks NO₂⁻ as a base.
Feedback if chosen: This option is wrong — you swapped the bottom two. HNO₂ is a weaker acid than HClO₃, so NO₂⁻ outranks ClO₃⁻ as a base.
Check — multiple choice
Supplied fact: the nitrite ion, NO₂⁻, accepts protons more readily than the iodide ion, I⁻. Which is the stronger acid, HNO₂ or HI?
Feedback if chosen: This option is wrong — you kept the supplied order. HNO₂ partners the stronger base, so HNO₂ is the weaker acid.
Feedback if chosen: Correct. I⁻ is the weaker base, so its partner HI is the stronger acid — the flip works in either direction.
Feedback if chosen: This option is wrong — the bases differ in strength, so the acids must differ the opposite way.
Feedback if chosen: This option is wrong — a supplied base ranking decides an acid ranking the same way: flip it once.
Outcome: Label every species in a proton-transfer equation as an acid or a base, and pair each acid with its conjugate base.
One free-response line prints NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq).
You've already seen the double arrow: it means the reaction does not go to completion — you'll study this in Unit 7.
The task asks for a label on every species.
Two species must carry the label acid.
Two species must carry the label base.
Each acid must then be matched with its conjugate base.
One misplaced label breaks the whole map.
Can you place all four labels and match both conjugate pairs, unaided?
First find the proton that moves.
On the left, H₂O gives the proton away — H₂O is the acid.
On the left, NH₃ catches the proton — NH₃ is the base.
On the right, NH₄⁺ is NH₃ plus the proton — NH₄⁺ is NH₃'s conjugate acid.
On the right, OH⁻ is H₂O minus the proton — OH⁻ is H₂O's conjugate base.
Each side of the equation carries one acid and one base.
Four labels and two pairs all follow from that one moved proton.
Identify every species in a given proton-transfer reaction as a Brønsted-Lowry acid or base, and pair each acid with its conjugate base.
Check — multiple choice
CH₃COOH(aq) + H₂O(l) ⇌ CH₃COO⁻(aq) + H₃O⁺(aq). Which species is the Brønsted-Lowry acid on the left?
Feedback if chosen: Correct. CH₃COOH becomes CH₃COO⁻, one proton lighter — the donor on the left is the acid.
Feedback if chosen: This option is wrong — H₂O becomes H₃O⁺, one proton heavier. Water accepts here, so it is the base.
Feedback if chosen: This option is wrong — CH₃COO⁻ sits on the right. The question asks for the acid among the reactants.
Feedback if chosen: This option is wrong — H₃O⁺ sits on the right. The question asks for the acid among the reactants.
Check — multiple choice
CH₃COOH(aq) + H₂O(l) ⇌ CH₃COO⁻(aq) + H₃O⁺(aq). Which species is CH₃COOH's conjugate base?
Feedback if chosen: This option is wrong — you paired the two reactants. A conjugate pair has one member on each side of the arrow.
Feedback if chosen: This option is wrong — H₃O⁺ is water's conjugate acid, built when H₂O caught the proton.
Feedback if chosen: Correct. CH₃COOH minus its donated proton is CH₃COO⁻ — its conjugate base, across the arrow.
Feedback if chosen: This option is wrong — OH⁻ does not appear in this equation at all.
Check — multiple choice
In CH₃COOH(aq) + H₂O(l) ⇌ CH₃COO⁻(aq) + H₃O⁺(aq), why does H₂O carry the label base?
Feedback if chosen: This option is wrong — no OH⁻ appears in this equation. Here water gained the proton instead of losing one.
Feedback if chosen: Correct. H₂O turned into H₃O⁺, one proton heavier — accepting the proton is what earns the base label.
Feedback if chosen: This option is wrong — being a liquid moves no proton. The formula change — H₂O to H₃O⁺, a gained proton — is the evidence.
Feedback if chosen: This option is wrong — position in the equation carries no meaning. Compare each reactant's formula with its product's.
Check — multiple choice
HCO₃⁻(aq) + H₂O(l) ⇌ H₂CO₃(aq) + OH⁻(aq). Which species is the Brønsted-Lowry base on the left?
Feedback if chosen: This option is wrong — H₂O donates its proton in this equation, so H₂O is the acid, not the base.
Feedback if chosen: This option is wrong — H₂CO₃ sits on the right. The question asks for the base among the reactants.
Feedback if chosen: This option is wrong — OH⁻ sits on the right. The question asks for the base among the reactants.
Feedback if chosen: Correct. HCO₃⁻ becomes H₂CO₃, one proton heavier — the acceptor on the left is the base.
Check — multiple choice
HCO₃⁻(aq) + H₂O(l) ⇌ H₂CO₃(aq) + OH⁻(aq). Which species is H₂O's conjugate base?
Feedback if chosen: This option is wrong — removing one proton from H₂CO₃ does not give H₂O, so H₂CO₃ cannot be water's partner.
Feedback if chosen: Correct. H₂O minus its donated proton is OH⁻ — its conjugate base, across the arrow.
Feedback if chosen: This option is wrong — you paired the two reactants. The pair spans the arrow.
Feedback if chosen: This option is wrong — H₃O⁺ does not appear in this equation at all.
Check — multiple choice
HCO₃⁻(aq) + H₂O(l) ⇌ H₂CO₃(aq) + OH⁻(aq). Which pairing is one of the equation's conjugate pairs?
Feedback if chosen: This option is wrong — you paired the two reactants. A conjugate pair has one member on each side of the arrow.
Feedback if chosen: This option is wrong — you paired the two products. Each pair spans the arrow.
Feedback if chosen: Correct. HCO₃⁻ plus the caught proton is H₂CO₃ — one member on each side, one H⁺ apart.
Feedback if chosen: This option is wrong — HCO₃⁻ and OH⁻ sit on opposite sides, but no single proton turns one into the other.
Free-response question
Source note (not shown as part of the question): authored to archetype full-species map (LO 4.8.A)
The hydrogen phosphate ion reacts with water: HPO₄²⁻(aq) + H₂O(l) ⇌ H₂PO₄⁻(aq) + OH⁻(aq). Answer on paper, then reveal the model response and score yourself.
(a) Identify the Brønsted-Lowry acid and the Brønsted-Lowry base on the reactant side of the equation. Justify each label with the change in the species' formula. [2 points]
(b) Write the two conjugate acid-base pairs in the equation. [2 points]
✓ Model response (the key)
(a)
H₂O is the acid.
H₂O donates a proton, becoming OH⁻ — one proton lighter.
HPO₄²⁻ is the base.
HPO₄²⁻ accepts the proton, becoming H₂PO₄⁻ — one proton heavier.
(b)
H₂PO₄⁻ and HPO₄²⁻ form one conjugate pair.
H₂O and OH⁻ form the other conjugate pair.
Scoring rubric — 4 points available
Check — multiple choice
Silver phosphate forms by 3Ag⁺(aq) + PO₄³⁻(aq) → Ag₃PO₄(s). Using the oxidation-number rules, what is the oxidation number of P in PO₄³⁻?
Feedback if chosen: This option is wrong — you gave the P atom the whole ion's charge. The −3 belongs to the ion; the atoms' oxidation numbers must sum to it.
Feedback if chosen: This option is wrong — you flipped a sign along the way. Four O at −2 give −8, and P + (−8) = −3 requires P = +5.
Feedback if chosen: Correct. Each O is −2, so the four O contribute −8; the sum must equal the ion's charge: P + (−8) = −3, giving P = +5.
Feedback if chosen: This option is wrong — you set the sum to zero as if PO₄³⁻ were neutral. The oxidation numbers must sum to the ion's charge, −3.
Check — multiple choice
Powdered sulfur burns by S(s) + O₂(g) → SO₂(g). Which species is oxidized?
Feedback if chosen: Correct. S goes from 0 in the free element to +4 in SO₂ — a loss of electron share, which is oxidation. The O drops from 0 to −2 and is reduced.
Feedback if chosen: This option is wrong — the O₂'s oxidation number does change (0 to −2), but downward. Gaining electron share is reduction, not oxidation.
Feedback if chosen: This option is wrong — you named a product. Oxidized and reduced label the reacting species; SO₂ is where the atoms end up.
Feedback if chosen: This option is wrong — you judged that no electrons are transferred. Oxidation numbers reveal the electron shift even in molecular compounds: S rises to +4 and O falls to −2.
Check — multiple choice
Pentane, C₅H₁₂, is a hydrocarbon. When it burns completely in oxygen, which products form?
Feedback if chosen: This option is wrong — you left the carbon unburned as soot. Complete burning carries every C atom into CO₂.
Feedback if chosen: This option is wrong — you sent the hydrogen out unreacted. The H atoms combine with oxygen, leaving as H₂O.
Feedback if chosen: This option is wrong — you lost the hydrogen entirely. Pentane's twelve H atoms must leave in a product: H₂O.
Feedback if chosen: Correct. Complete combustion of a hydrocarbon always gives the same pair: every C ends in CO₂ and every H ends in H₂O.
Check — multiple choice
Solutions of ammonium sulfate, (NH₄)₂SO₄(aq), and potassium nitrate, KNO₃(aq), are mixed. What forms?
Feedback if chosen: Correct. Swapping partners could only give K₂SO₄ and NH₄NO₃, and potassium and ammonium compounds are always soluble — so every ion stays dissolved.
Feedback if chosen: This option is wrong — you precipitated a potassium compound. Potassium compounds are on the always-soluble list, so K₂SO₄ stays dissolved.
Feedback if chosen: This option is wrong — you precipitated an ammonium nitrate pairing. Both ammonium and nitrate compounds are always soluble.
Feedback if chosen: This option is wrong — you removed everything from solution. All four ions belong to the always-soluble families, so nothing leaves.
Check — multiple choice
Milk of magnesia neutralizes stomach acid by Mg(OH)₂(s) + 2HCl(aq) → MgCl₂(aq) + 2H₂O(l). How is this reaction classified?
Feedback if chosen: This option is wrong — you saw a solid and stopped. The solid is a reactant being consumed; a precipitation reaction forms a new solid from mixed solutions.
Feedback if chosen: Correct. Each HCl donates a proton (H⁺) to a hydroxide ion, forming water — proton transfer is the defining evidence of an acid-base reaction.
Feedback if chosen: This option is wrong — changing partners moves no electrons. Every oxidation number is unchanged: Mg stays +2, Cl stays −1, H stays +1, O stays −2.
Feedback if chosen: This option is wrong — the solid here dissolves away rather than forming. Only the proton-transfer evidence fits this equation.
Check — multiple choice
Ethylamine reacts completely with hydrobromic acid: C₂H₅NH₂(aq) + HBr(aq) → C₂H₅NH₃⁺(aq) + Br⁻(aq). Which species donates the proton?
Feedback if chosen: This option is wrong — you picked the acceptor. The ethylamine gains a proton, growing into C₂H₅NH₃⁺; the Brønsted-Lowry acid is the donor.
Feedback if chosen: Correct. The HBr hands its H⁺ to the ethylamine and leaves as Br⁻ — HBr is the proton donor, the Brønsted-Lowry acid.
Feedback if chosen: This option is wrong — you picked a product. C₂H₅NH₃⁺ is what the acceptor becomes after gaining the proton.
Feedback if chosen: This option is wrong — you picked what remains of the donor. Br⁻ is HBr after its proton has already left.
Check — multiple choice
Benzoic acid partially ionizes in water: C₆H₅COOH(aq) + H₂O(l) ⇌ C₆H₅COO⁻(aq) + H₃O⁺(aq). Which two pairs are the conjugate acid-base pairs?
Feedback if chosen: Correct. Each pair differs by exactly one H⁺: C₆H₅COOH loses a proton to become C₆H₅COO⁻, and H₂O gains one to become H₃O⁺.
Feedback if chosen: This option is wrong — you paired the two reactants together and the two products together. A conjugate pair spans the arrow, differing by one H⁺.
Feedback if chosen: This option is wrong — you paired the two acids together and the two bases together. Each conjugate pair contains one acid and its own base.
Feedback if chosen: This option is wrong — you left the water out. H₂O accepts the proton, so H₃O⁺/H₂O is the second conjugate pair.
Check — multiple choice
Supplied fact: HOBr is a weaker acid than HOCl. Which is the stronger base, OBr⁻ or OCl⁻?
Feedback if chosen: This option is wrong — you matched strong with strong. The relationship within a conjugate pair is inverse: the stronger acid keeps the weaker base.
Feedback if chosen: This option is wrong — differing by one proton makes them conjugates, not equals. Within each pair, the stronger the acid, the weaker its conjugate base.
Feedback if chosen: This option is wrong — acid strengths are exactly what settle it. The inverse relationship converts the acid comparison into the base comparison.
Feedback if chosen: Correct. HOBr holds its proton more tightly, so its conjugate base OBr⁻ pulls protons back more strongly — the weaker acid pairs with the stronger base.
Check — multiple choice
In NO₂⁻(aq) + H₂O(l) ⇌ HNO₂(aq) + OH⁻(aq), which labeling of the four species is correct?
Feedback if chosen: This option is wrong — you swapped donor and acceptor. The NO₂⁻ gains the proton (becoming HNO₂), so NO₂⁻ is the base and H₂O is the acid.
Feedback if chosen: This option is wrong — the acid and base labels are right, but the conjugates are mismatched. Each conjugate belongs to the species it differs from by one H⁺.
Feedback if chosen: Correct. H₂O donates the proton (acid) and becomes OH⁻, its conjugate base; NO₂⁻ accepts it (base) and becomes HNO₂, its conjugate acid.
Feedback if chosen: This option is wrong — you benched the water. H₂O donates the proton in this equation, so it acts as the Brønsted-Lowry acid.
5 lessons
Outcome: State that a half-reaction shows the oxidation or the reduction separately, with the electrons written explicitly as e⁻.
You've already seen the copper wire standing in silver nitrate solution: silver crystals grow on the wire, and the solution slowly turns blue.
The overall equation reads Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s).
Every atom balances.
The total charge balances too.
You've already seen that this reaction is redox: copper loses electrons, and the silver ions gain them.
Yet the equation never shows a single electron.
Where in the written equation did the electrons go?
Chemists write one electron as e⁻.
Split the reaction into its two halves.
| The half-reaction | The process | Where the electrons sit |
|---|---|---|
| Cu(s) → Cu²⁺(aq) + 2e⁻ | copper loses two electrons (oxidation) | product side |
| Ag⁺(aq) + e⁻ → Ag(s) | a silver ion gains one electron (reduction) | reactant side |
One half shows only the loss: Cu(s) → Cu²⁺(aq) + 2e⁻.
The copper atom loses two electrons, so the two e⁻ sit on the product side.
The other half shows only the gain: Ag⁺(aq) + e⁻ → Ag(s).
The silver ion gains one electron, so the e⁻ sits on the reactant side.
Each written half, electrons included, is called a half-reaction.
Half-reactions make the electron bookkeeping visible.
Unit 9's electrochemistry is built entirely from half-reactions.
State that a half-reaction shows the oxidation or the reduction process separately, with the electrons lost or gained written explicitly as e⁻.
Check — multiple choice
What does a half-reaction show?
Feedback if chosen: This option is wrong — you described the overall equation. A half-reaction shows one of the two processes, electrons included.
Feedback if chosen: Correct. A half-reaction isolates the oxidation or the reduction, with the electrons lost or gained written explicitly as e⁻.
Feedback if chosen: This option is wrong — spectator ions belong to ionic-equation work. A half-reaction shows an oxidation or a reduction.
Feedback if chosen: This option is wrong — mole ratios come from a balanced overall equation. A half-reaction's job is to show the electrons.
Check — multiple choice
Ni(s) → Ni²⁺(aq) + 2e⁻. Which process does this half-reaction show?
Feedback if chosen: This option is wrong — reduction gains electrons, so its e⁻ sit on the reactant side. These sit with the products.
Feedback if chosen: This option is wrong — one half-reaction shows exactly one of the two processes.
Feedback if chosen: Correct. The two e⁻ sit on the product side — the nickel atom lost them, and losing electrons is oxidation.
Feedback if chosen: This option is wrong — electrons appear explicitly, so one of the two redox processes is on show: here, oxidation.
Check — multiple choice
Pb²⁺(aq) + 2e⁻ → Pb(s). Which process does this half-reaction show?
Feedback if chosen: This option is wrong — oxidation loses electrons, so its e⁻ sit on the product side. These sit with the reactants.
Feedback if chosen: Correct. The two e⁻ sit on the reactant side — the lead(II) ion gains them, and gaining electrons is reduction.
Feedback if chosen: This option is wrong — one half-reaction shows exactly one of the two processes.
Feedback if chosen: This option is wrong — electrons appear explicitly, so one of the two redox processes is on show: here, reduction.
Check — multiple choice
A metal atom loses electrons during oxidation. On which side of its half-reaction do the lost electrons appear?
Feedback if chosen: This option is wrong — electrons on the reactant side would mean the species consumed them. A loser releases them.
Feedback if chosen: This option is wrong — electrons written on both sides would cancel and show nothing.
Feedback if chosen: This option is wrong — writing the electrons explicitly is the whole point of a half-reaction.
Feedback if chosen: Correct. The atom releases the electrons, so the released e⁻ appear with the products.
Outcome: Write the two balanced half-reactions for a given redox reaction, each balanced in atoms and charge.
A 3.0 g strip of zinc stands in blue copper(II) sulfate solution.
The overall equation reads Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s).
One free-response part asks for the oxidation half-reaction alone.
Copying half of the overall equation is not enough.
A half-reaction must balance its atoms and its charge, electrons included.
Which pieces belong in the zinc half, and how many electrons?
Assign oxidation numbers to every species.
Zn rises from 0 to +2 — the zinc atom is oxidized.
Cu falls from +2 to 0 — the copper(II) ion is reduced.
**1.** Write the oxidized species alone, before and after: Zn(s) → Zn²⁺(aq).
**2.** Add electrons to the side that needs them, until both sides carry equal charge.
Zn(s) → Zn²⁺(aq) + 2e⁻ balances at zero charge on each side.
**3.** Write the reduced species' half the same way, with its electrons on the reactant side: Cu²⁺(aq) + 2e⁻ → Cu(s).
The two balanced halves retell the whole equation with the electrons in plain sight.
Write the two balanced half-reactions — oxidation and reduction, each balanced in atoms and charge — for a given redox reaction.
Check — multiple choice
Nickel metal reacts with lead(II) ions: Ni(s) + Pb²⁺(aq) → Ni²⁺(aq) + Pb(s). Which species is oxidized?
Feedback if chosen: This option is wrong — Pb falls from +2 to 0. A falling oxidation number marks reduction.
Feedback if chosen: Correct. Ni rises from 0 to +2 — a rising oxidation number marks the oxidized species.
Feedback if chosen: This option is wrong — Ni²⁺ is the product of the oxidation, not the species that was oxidized.
Feedback if chosen: This option is wrong — Pb(s) is the product of the reduction. Compare each species' oxidation number before and after.
Check — multiple choice
For Ni(s) + Pb²⁺(aq) → Ni²⁺(aq) + Pb(s), which is the correct oxidation half-reaction?
Feedback if chosen: This option is wrong — you put the lost electrons on the reactant side. The nickel atom releases them, so they sit with the products.
Feedback if chosen: This option is wrong — you reversed the change. The equation starts from Ni(s), the metal.
Feedback if chosen: Correct. Ni loses two electrons, and with 2e⁻ on the product side each side carries zero total charge.
Feedback if chosen: This option is wrong — one electron leaves the charges unbalanced: 0 on the left against +2 − 1 = +1 on the right.
Check — multiple choice
For Ni(s) + Pb²⁺(aq) → Ni²⁺(aq) + Pb(s), which is the correct reduction half-reaction?
Feedback if chosen: Correct. The lead(II) ion gains two electrons, and with 2e⁻ on the reactant side each side carries zero total charge.
Feedback if chosen: This option is wrong — you put the gained electrons on the product side. The ion consumes them, so they sit with the reactants.
Feedback if chosen: This option is wrong — you reversed the change. The equation starts from Pb²⁺, the ion.
Feedback if chosen: This option is wrong — one electron leaves the charges unbalanced: +2 − 1 = +1 on the left against 0 on the right.
Check — drawing task (pen and paper, self-marked)
Magnesium metal reacts with tin(II) ions: Mg(s) + Sn²⁺(aq) → Mg²⁺(aq) + Sn(s). Take a pen and paper and write the two balanced half-reactions — the oxidation half and the reduction half. Include state symbols on the metals and ions. When you are done, select Check to compare your answer with the model answer.
Model answer shown to the student
Self-mark checklist — the drawing should show:
Outcome: Write the overall balanced redox equation from two half-reactions by scaling until the electrons cancel.
Aluminum foil dissolves in 100 mL of blue copper(II) chloride solution.
The oxidation half reads Al(s) → Al³⁺(aq) + 3e⁻.
The reduction half reads Cu²⁺(aq) + 2e⁻ → Cu(s).
The aluminum half releases three electrons.
The copper half consumes only two.
Electrons cannot be left over in a finished equation.
How do you match the release to the consumption?
Scale each half until the electrons released equal the electrons consumed.
The smallest number that both 3 and 2 divide into is 6.
Double the aluminum half: 2Al(s) → 2Al³⁺(aq) + 6e⁻.
Triple the copper half: 3Cu²⁺(aq) + 6e⁻ → 3Cu(s).
Add the two scaled halves.
The six electrons on each side cancel.
The overall equation appears: 2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s).
Check the result: the atoms balance, and each side carries a total charge of 6+.
Write the overall balanced redox equation from two half-reactions by scaling each until the electrons lost equal the electrons gained, then canceling the electrons.
Check — multiple choice
The halves Cr(s) → Cr³⁺(aq) + 3e⁻ and Ag⁺(aq) + e⁻ → Ag(s) are to be combined. By what must each half be multiplied so the electrons match, using the smallest whole numbers?
Feedback if chosen: This option is wrong — that releases nine electrons against one consumed. Scale the half that consumes too few.
Feedback if chosen: This option is wrong — that gives nine electrons released against three consumed. Match the totals, not the factors.
Feedback if chosen: Correct. One chromium half releases three electrons, and three silver halves consume three — the totals match at 3.
Feedback if chosen: This option is wrong — 6 and 6 do match, but the numbers are not the smallest whole numbers. Divide both factors by 2.
Check — multiple choice
After Cr(s) → Cr³⁺(aq) + 3e⁻ and Ag⁺(aq) + e⁻ → Ag(s) are scaled, added, and the electrons canceled, what is the overall equation?
Feedback if chosen: Correct. Tripling the silver half matches three electrons each way; adding and canceling leaves charge 3+ on each side.
Feedback if chosen: This option is wrong — you skipped the scaling. Three electrons released against one consumed leaves two unaccounted for.
Feedback if chosen: This option is wrong — the electrons must cancel. A finished overall equation never shows e⁻.
Feedback if chosen: This option is wrong — you scaled the wrong half. The silver half consumes too few electrons, so the silver half is tripled.
Check — multiple choice
Why must two half-reactions be scaled before they are added?
Feedback if chosen: This option is wrong — coefficients record amounts, not speed. The scaling is bookkeeping for the electrons.
Feedback if chosen: This option is wrong — the two halves hold different species, so their atom counts need not match. The electron counts must.
Feedback if chosen: This option is wrong — the matching number depends on the two halves. For 2e⁻ and 2e⁻ it is simply 2.
Feedback if chosen: Correct. Electrons cannot be left over in a finished equation, so the release must equal the consumption before the halves are added.
Check — drawing task (pen and paper, self-marked)
Take a pen and paper and combine Ni(s) → Ni²⁺(aq) + 2e⁻ with Ag⁺(aq) + e⁻ → Ag(s) into the overall balanced equation, using the smallest whole numbers. When you are done, select Check to compare your answer with the model answer.
Model answer shown to the student
Self-mark checklist — the drawing should show:
Outcome: Evaluate a proposed redox construction and name its error — or pass it — using three checks in order.
A classmate combines two half-reactions: Zn(s) → Zn²⁺(aq) + 2e⁻ and Ag⁺(aq) + e⁻ → Ag(s).
The submitted overall equation reads Zn(s) + Ag⁺(aq) → Zn²⁺(aq) + Ag(s).
Every atom appears balanced at a glance.
Now count the charge.
The left side carries a total charge of 1+.
The right side carries a total charge of 2+.
Somewhere an electron vanished from the bookkeeping.
What exactly went wrong?
Check any redox construction with three questions, asked in order.
**1.** Did the electrons released equal the electrons consumed?
**2.** Did every electron cancel from the final equation?
**3.** Do the atoms and the total charge balance on both sides?
The first question that fails names the error.
The classmate's halves traded two electrons for one — question 1 fails: an electron mismatch.
Doubling the silver half repairs the construction: Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s).
A construction that passes all three questions carries no error.
Evaluate a proposed half-reaction pair or combined redox equation, and identify the error if one is present: an electron mismatch, electrons not canceled, or an atom or charge imbalance.
Check — multiple choice
The halves Fe(s) → Fe²⁺(aq) + 2e⁻ and Ag⁺(aq) + e⁻ → Ag(s) are combined into Fe(s) + Ag⁺(aq) → Fe²⁺(aq) + Ag(s). Which error does the construction contain?
Feedback if chosen: Correct. The iron half releases two electrons and the silver half consumes one — question 1 fails before anything else.
Feedback if chosen: This option is wrong — no e⁻ appears in the submitted equation. The failure comes earlier: the release never matched the consumption.
Feedback if chosen: This option is wrong — the charge imbalance you can spot is a symptom. The first failing check, the electron mismatch, names the error.
Feedback if chosen: This option is wrong — check question 1: two electrons out against one in. The construction fails at the first step.
Check — multiple choice
The halves Mg(s) → Mg²⁺(aq) + 2e⁻ and Cu²⁺(aq) + 2e⁻ → Cu(s) are added to give Mg(s) + Cu²⁺(aq) + 2e⁻ → Mg²⁺(aq) + Cu(s) + 2e⁻. Which error does the construction contain?
Feedback if chosen: This option is wrong — two electrons out and two in: question 1 passes.
Feedback if chosen: Correct. The 2e⁻ on each side match, so they must be struck out — a finished equation never shows electrons.
Feedback if chosen: This option is wrong — with the e⁻ counted, both sides balance. The failure is that the electrons are still written.
Feedback if chosen: This option is wrong — question 2 fails: the matching electrons were left in the final equation.
Check — multiple choice
From 2Al(s) → 2Al³⁺(aq) + 6e⁻ and 3Pb²⁺(aq) + 6e⁻ → 3Pb(s), a student writes the overall equation 2Al(s) + 3Pb²⁺(aq) → 2Al³⁺(aq) + 2Pb(s). Which error does the construction contain?
Feedback if chosen: This option is wrong — six electrons out and six in: question 1 passes.
Feedback if chosen: This option is wrong — no e⁻ appears in the final equation: question 2 passes.
Feedback if chosen: Correct. Question 3 fails: three Pb enter on the left, but only two Pb appear on the right — an atom imbalance.
Feedback if chosen: This option is wrong — count the lead atoms: three on the left, two on the right.
Check — multiple choice
The halves Sn(s) → Sn²⁺(aq) + 2e⁻ and Pb²⁺(aq) + 2e⁻ → Pb(s) are combined into Sn(s) + Pb²⁺(aq) → Sn²⁺(aq) + Pb(s). Which error does the construction contain?
Feedback if chosen: This option is wrong — two electrons out and two in: question 1 passes.
Feedback if chosen: This option is wrong — no e⁻ appears in the final equation: question 2 passes.
Feedback if chosen: This option is wrong — one Sn and one Pb on each side, and a 2+ total charge on each side: question 3 passes.
Feedback if chosen: Correct. All three checks pass: matched electrons, all canceled, and atoms and charge balanced — the construction is sound.
Outcome: Write the balanced redox equation for a described process end-to-end, from oxidation numbers to the final atom and charge check.
A free-response prompt describes a lab in words alone.
A coil of aluminum wire rests in 200 mL of blue copper(II) sulfate solution.
Reddish solid copper collects on the wire.
The prompt asks for the balanced equation for the reaction.
No equation appears anywhere in the prompt.
You have already practiced every step the answer needs.
Can you walk from the description to the balanced equation, unaided?
To build the equation from a description, follow five steps.
**1.** Translate the description into species: the aluminum wire is Al(s), and the copper(II) sulfate solution supplies Cu²⁺(aq).
The sulfate ion takes no part — it is a spectator ion.
**2.** Assign oxidation numbers: Al rises from 0 to +3 (oxidized), and Cu falls from +2 to 0 (reduced).
**3.** Write both balanced half-reactions: Al(s) → Al³⁺(aq) + 3e⁻ and Cu²⁺(aq) + 2e⁻ → Cu(s).
**4.** Scale until the electrons match: double the aluminum half and triple the copper half, for six electrons each way.
**5.** Add the halves, cancel the electrons, and check the result: 2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s), atoms and charge balanced.
Write the balanced redox equation for a described process end-to-end: assign oxidation numbers, identify the oxidized and reduced species, write both half-reactions, equalize the electrons, and verify atom and charge balance.
Check — multiple choice
A strip of tin stands in silver nitrate solution. Gray silver crystals grow on the strip, and the tin forms tin(II) ions. Which species is oxidized?
Feedback if chosen: This option is wrong — Ag falls from +1 to 0 as the crystals grow. A falling oxidation number marks reduction.
Feedback if chosen: Correct. Sn rises from 0 to +2 — the rising oxidation number marks the oxidized species.
Feedback if chosen: This option is wrong — Sn²⁺ is the product of the oxidation, not the species that was oxidized.
Feedback if chosen: This option is wrong — Ag(s) is the product of the reduction: the grown crystals.
Check — multiple choice
For the tin strip in silver nitrate solution (tin forms tin(II) ions), which pair of half-reactions is correct?
Feedback if chosen: Correct. Tin releases two electrons on its product side, and each silver ion consumes one on its reactant side.
Feedback if chosen: This option is wrong — both halves carry their electrons on the wrong side: the loser releases, and the gainer consumes.
Feedback if chosen: This option is wrong — you reversed both changes. The description starts from tin metal and silver ions.
Feedback if chosen: This option is wrong — the electron counts are off: Sn²⁺ needs 2e⁻ released, and Ag⁺ consumes just 1e⁻.
Check — multiple choice
For Sn(s) → Sn²⁺(aq) + 2e⁻ and Ag⁺(aq) + e⁻ → Ag(s), what scaling matches the electrons with the smallest whole numbers?
Feedback if chosen: This option is wrong — that releases four electrons against one consumed.
Feedback if chosen: This option is wrong — that gives four released against two consumed. Match the totals.
Feedback if chosen: Correct. One tin half releases two electrons, and two silver halves consume two — matched at 2.
Feedback if chosen: This option is wrong — 4 and 4 do match, but they are not the smallest whole numbers. Halve both factors.
Check — drawing task (pen and paper, self-marked)
A magnesium ribbon stands in silver nitrate solution, and silver crystals grow on the ribbon. Magnesium forms Mg²⁺ ions. Silver forms Ag⁺ ions. Take a pen and paper and write the two half-reactions and the overall balanced equation. When you are done, select Check to compare your answer with the model answer.
Model answer shown to the student
Self-mark checklist — the drawing should show:
Check — drawing task (pen and paper, self-marked)
Iron filings are stirred into tin(II) chloride solution. Gray tin metal forms, and the iron forms iron(II) ions. Take a pen and paper and write the two half-reactions and the overall balanced equation. When you are done, select Check to compare your answer with the model answer.
Model answer shown to the student
Self-mark checklist — the drawing should show:
Check — multiple choice
Zn(s) → Zn²⁺(aq) + 2e⁻ shows one process with its electrons written out. What is such an equation called?
Feedback if chosen: This option is wrong — a net ionic equation drops spectators but never shows electrons.
Feedback if chosen: This option is wrong — a molecular equation writes whole compounds and never shows electrons.
Feedback if chosen: Correct. One process — here the oxidation — with its electrons written explicitly is a half-reaction.
Feedback if chosen: This option is wrong — an overall equation has had its electrons canceled. These 2e⁻ are still on show.
Check — multiple choice
Ni²⁺(aq) + 2e⁻ → Ni(s). Which process does this half-reaction show?
Feedback if chosen: This option is wrong — oxidation releases electrons on the product side. These 2e⁻ sit with the reactants.
Feedback if chosen: Correct. The nickel(II) ion gains two electrons — the e⁻ on the reactant side mark a reduction.
Feedback if chosen: This option is wrong — no solid forms from two mixed solutions here. An ion is gaining electrons.
Feedback if chosen: This option is wrong — dissociation frees ions from a dissolving solid. Here an ion gains electrons.
Check — multiple choice
Chromium is oxidized to Cr³⁺. How many electrons appear in the half-reaction Cr(s) → Cr³⁺(aq) + ?e⁻?
Feedback if chosen: This option is wrong — one electron leaves the right side at +3 − 1 = +2, not 0.
Feedback if chosen: This option is wrong — two electrons leave the right side at +3 − 2 = +1, not 0.
Feedback if chosen: Correct. Three electrons bring the right side to +3 − 3 = 0, matching the left side's 0.
Feedback if chosen: This option is wrong — six electrons would swing the right side to +3 − 6 = −3. Six belongs to scaled halves, not this one.
Check — multiple choice
The halves Cr(s) → Cr³⁺(aq) + 3e⁻ and Pb²⁺(aq) + 2e⁻ → Pb(s) are to be combined. What scaling matches the electrons with the smallest whole numbers?
Feedback if chosen: Correct. Two chromium halves release six electrons, and three lead halves consume six — matched at the smallest shared total.
Feedback if chosen: This option is wrong — that gives nine released against four consumed. Scale each half by the other's electron count.
Feedback if chosen: This option is wrong — 18 and 12 electrons do not match, and the factors are far from smallest.
Feedback if chosen: This option is wrong — unscaled, three electrons leave and only two arrive. One electron would be left over.
Check — multiple choice
The halves Zn(s) → Zn²⁺(aq) + 2e⁻ and Pb²⁺(aq) + 2e⁻ → Pb(s) are added to give Zn(s) + Pb²⁺(aq) + 2e⁻ → Zn²⁺(aq) + Pb(s) + 2e⁻. Which error does the construction contain?
Feedback if chosen: This option is wrong — two electrons out and two in: the first check passes.
Feedback if chosen: Correct. The matching 2e⁻ must be struck from both sides — a finished equation never shows electrons.
Feedback if chosen: This option is wrong — with the electrons counted, both sides balance. The fault is the uncanceled electrons.
Feedback if chosen: This option is wrong — the second check fails: the electrons are still written in the final equation.
Check — multiple choice
Cobalt powder is stirred into lead(II) nitrate solution; gray lead metal forms, and the cobalt forms cobalt(II) ions. Which is the balanced overall equation?
Feedback if chosen: This option is wrong — you reversed the description. The cobalt metal dissolves, and the lead metal forms.
Feedback if chosen: This option is wrong — count the charge: 2+ on the left against 4+ + 0 on the right. Both halves trade two electrons, so the ratio is 1:1.
Feedback if chosen: This option is wrong — the two electrons released and the two consumed cancel. A finished equation never shows e⁻.
Feedback if chosen: Correct. Each cobalt atom releases two electrons and each lead(II) ion consumes two, so the species pair 1:1 — atoms and charge (2+ each side) balance.
Free-response question
Source note (not shown as part of the question): authored to archetype end-to-end redox construction (LO 4.9.A)
A strip of chromium metal stands in a solution of tin(II) chloride, SnCl₂. Gray tin metal collects on the strip, and the chromium forms chromium(III) ions, Cr³⁺. Answer on paper, then reveal the model response and score yourself.
(a) Write the balanced oxidation half-reaction and the balanced reduction half-reaction for this process. [2 points]
(b) Write the balanced overall equation, showing with a sentence or a line of working that the electrons released equal the electrons consumed. [2 points]
✓ Model response (the key)
(a)
Oxidation: Cr(s) → Cr³⁺(aq) + 3e⁻.
Reduction: Sn²⁺(aq) + 2e⁻ → Sn(s).
(b)
Double the chromium half and triple the tin half: six electrons released, six consumed.
2Cr(s) → 2Cr³⁺(aq) + 6e⁻ and 3Sn²⁺(aq) + 6e⁻ → 3Sn(s).
Adding and canceling the electrons: 2Cr(s) + 3Sn²⁺(aq) → 2Cr³⁺(aq) + 3Sn(s).
Check: two Cr and three Sn on each side, and a total charge of 6+ on each side.
Scoring rubric — 4 points available
Outcome: Finish a real ten-point long question in 23 minutes and score it against the real rubric.
The free-response section gives you 7 questions in 105 minutes. Questions 1–3 are 10-point long questions, so each has a budget of about 23 minutes — a little over 2 minutes per point.
This question reaches back into Unit 3, as the real exam does: its molarity, dilution, and absorbance parts are Unit 3 skills. Its part (a) — the net ionic equation — is the Unit 4 skill you have just built.
Calculation parts pay only with visible work: write the setup, substitute the values with their units, then report the answer. A justification part pays fastest when the claim comes first, followed by the chain through the measured quantity.
Set a 23-minute timer, answer all seven parts (ten points) of the question below on paper under time, and only then reveal the model response and score yourself.
Free-response question
Source note (not shown as part of the question): 2021 FRQ Q3, complete; rubric from the 2021 scoring guidelines
A student is given the task of determining the molar concentration of a CuSO₄(aq) solution using two different procedures, precipitation and spectrophotometry. For the precipitation experiment, the student adds 20.0 mL of 0.200 M Ba(NO₃)₂(aq) to 50.0 mL of the CuSO₄(aq). The reaction goes to completion, and a white precipitate forms. The student filters the precipitate and dries it overnight. The data: volume of CuSO₄(aq) = 50.0 mL; volume of 0.200 M Ba(NO₃)₂(aq) = 20.0 mL; mass of dry filter paper = 0.764 g; mass of filter paper and dried precipitate = 1.136 g. (Molar mass of BaSO₄: 233.39 g/mol.)
(a) Write a balanced net ionic equation for the precipitation reaction. [1 point]
(b) Calculate the number of moles of precipitate formed. [2 points]
(c) Calculate the molarity of the original CuSO₄ solution. [1 point]
(d) For the spectrophotometry experiment, the student first makes a standard curve. The student uses a 0.1000 M solution of CuSO₄(aq) to make three more solutions of known concentration (0.0500 M, 0.0300 M, and 0.0100 M) in 50.00 mL volumetric flasks. Calculate the volume of 0.1000 M CuSO₄(aq) needed to make 50.00 mL of 0.0500 M CuSO₄(aq). [1 point]
(e) Briefly describe the procedure the student should follow to make 50.00 mL of 0.0500 M CuSO₄(aq) using 0.1000 M CuSO₄(aq), a 50.00 mL volumetric flask, and other standard laboratory equipment. Assume that all appropriate safety precautions will be taken. [2 points]
(f) The standard curve is given in the figure above. The absorbance of the CuSO₄ solution of unknown concentration is 0.219. Determine the molarity of the solution. [1 point]
(g) A second student performs the same experiment. There are a few drops of water in the cuvette before the second student adds the CuSO₄(aq) solution of unknown concentration. Will this result in a CuSO₄(aq) concentration for the unknown that is greater than, less than, or equal to the concentration determined in part (f)? Justify your answer. [2 points]
✓ Model response (the key)
(a)
Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)
(b)
mass of BaSO₄ = 1.136 g − 0.764 g = 0.372 g
moles of BaSO₄ = 0.372 g × (1 mol/233.39 g) = 0.00159 mol
(c)
0.00159 mol BaSO₄ × (1 mol CuSO₄/1 mol BaSO₄) = 0.00159 mol CuSO₄
molarity = 0.00159 mol ÷ 0.0500 L = 0.0318 M (0.0319 M if decimals are carried)
(d)
moles of CuSO₄ needed = 0.05000 L × 0.0500 M = 0.00250 mol
volume of stock = 0.00250 mol ÷ 0.1000 M = 0.0250 L = 25.0 mL
(e)
Measure 25.0 mL of the 0.1000 M CuSO₄(aq) with a 25.0 mL volumetric pipet (a graduated cylinder or buret is also acceptable).
Transfer the 25.0 mL of solution to the 50.00 mL volumetric flask, then add water up to the 50.00 mL mark.
(f)
slope of the standard curve = 0.630 ÷ 0.1000 M = 6.3 M⁻¹
concentration = 0.219 ÷ 6.3 M⁻¹ = 0.035 M (any value between 0.032 M and 0.038 M is accepted)
(g)
The concentration will be less than the concentration determined in part (f).
The additional water decreases the concentration of CuSO₄ in the cuvette, so the absorbance decreases (Beer-Lambert law), and the standard curve then gives a lower estimated concentration.
Scoring rubric — 10 points available
Check — multiple choice
When Co(NO₃)₂(aq) and (NH₄)₂CO₃(aq) are mixed, solid cobalt(II) carbonate forms. (CoCO₃ is insoluble — supplied.) Which is the correct net ionic equation?
Feedback if chosen: Correct. Only the two ions that build the solid remain; the atoms match, and each side's charges sum to zero.
Feedback if chosen: This option is wrong — you left the NH₄⁺ spectators uncanceled. An ion unchanged on both sides is removed from the net ionic equation.
Feedback if chosen: This option is wrong — you gave the molecular equation. The net ionic form keeps only the species that change.
Feedback if chosen: This option is wrong — you wrote the spectators as the reaction. NH₄⁺ and NO₃⁻ stay dissolved and separate throughout.
Check — multiple choice
Mixing K₂SO₃(aq) with excess HCl(aq) releases SO₂ gas and leaves water. (K₂SO₃ and KCl are soluble; the HCl is present as H⁺ and Cl⁻ ions — supplied.) Which net ionic equation represents the change?
Feedback if chosen: This option is wrong — you gave the molecular equation. The context demands the net ionic form: only the species that change.
Feedback if chosen: This option is wrong — you dropped the coefficient 2. With one H⁺ the hydrogens don't balance, and the left side's charge sums to 1− against 0 on the right.
Feedback if chosen: This option is wrong — you wrote the spectators as the reaction. K⁺ and Cl⁻ stay dissolved and separate; the change happens to the sulfite and the protons.
Feedback if chosen: Correct. The sulfite ion and two protons make the gas and the water; K⁺ and Cl⁻ are spectators, and atoms and charge (2− + 2+ = 0) both balance.
Check — multiple choice
Titanium reacts with chlorine gas by Ti + 2Cl₂ → TiCl₄. A mixture of 0.30 mol of Ti and 0.50 mol of Cl₂ reacts to completion. What is the theoretical yield of TiCl₄?
Feedback if chosen: This option is wrong — you halved the Ti amount, applying the Cl₂'s coefficient to the wrong reactant. The Cl₂ limits: 0.50 ÷ 2 = 0.25 mol.
Feedback if chosen: Correct. 0.30 mol of Ti would need 0.60 mol of Cl₂, but only 0.50 mol is available — the Cl₂ limits, and 0.50 ÷ 2 = 0.25 mol of TiCl₄ forms.
Feedback if chosen: This option is wrong — you built the yield from the Ti, the reactant in excess. The Cl₂ is used up first and caps the product.
Feedback if chosen: This option is wrong — you carried the Cl₂ moles straight across. Two Cl₂ are consumed for each TiCl₄ formed: 0.50 ÷ 2 = 0.25 mol.
Check — multiple choice
A 15.0 mL sample of K₂CO₃(aq) reaches the equivalence point when 22.5 mL of 0.200 M HBr has been added, reacting by K₂CO₃ + 2HBr → 2KBr + H₂O + CO₂ — supplied. What is the molarity of the K₂CO₃ sample?
Feedback if chosen: This option is wrong — you divided by the combined volume (37.5 mL). The sample's molarity uses the sample's own 15.0 mL.
Feedback if chosen: Correct. HBr added: 0.0225 × 0.200 = 0.00450 mol; the 1:2 relationship gives 0.00450 ÷ 2 = 0.00225 mol of K₂CO₃; 0.00225 ÷ 0.0150 = 0.150 M.
Feedback if chosen: This option is wrong — you skipped the 2:1 requirement. Two HBr are consumed per K₂CO₃, so the sample holds half the titrant's moles.
Feedback if chosen: This option is wrong — you applied the mole relationship upside down, doubling instead of halving: 0.00450 ÷ 2 = 0.00225 mol of K₂CO₃.
Check — multiple choice
Chlorine displaces iodide ions: Cl₂(aq) + 2I⁻(aq) → 2Cl⁻(aq) + I₂(aq). Which species is oxidized?
Feedback if chosen: This option is wrong — starting as the element says nothing by itself. Cl goes from 0 to −1, a gain of electrons: the Cl₂ is reduced.
Feedback if chosen: This option is wrong — you named a product. Cl⁻ is what the Cl₂ becomes after gaining electrons; oxidized and reduced label the reacting species.
Feedback if chosen: Correct. Each I⁻ loses an electron, rising from −1 to 0 — losing electrons is oxidation. The Cl₂ gains those electrons and is reduced.
Feedback if chosen: This option is wrong — you named a product. I₂ is what the oxidized I⁻ becomes, not the species that lost the electrons.
Check — multiple choice
Solutions of lead(II) acetate, Pb(C₂H₃O₂)₂(aq), and potassium bromide, KBr(aq), are mixed. (PbBr₂ is insoluble; lead and acetate compounds here are otherwise soluble — supplied.) What forms?
Feedback if chosen: This option is wrong — you ruled precipitates out entirely. When a new pairing of ions is insoluble, the solid forms the moment the solutions mix.
Feedback if chosen: This option is wrong — you precipitated a potassium compound. Potassium compounds are always soluble, so K⁺ and acetate stay dissolved.
Feedback if chosen: Correct. Swapping partners pairs Pb²⁺ with Br⁻, and the supplied fact says PbBr₂ is insoluble — so it leaves the solution as a solid.
Feedback if chosen: This option is wrong — you precipitated a starting compound. KBr was already dissolved; a precipitate is a new pairing of the mixed ions.
Check — multiple choice
In OBr⁻(aq) + H₂O(l) ⇌ HOBr(aq) + OH⁻(aq), which labeling of the four species is correct?
Feedback if chosen: Correct. H₂O donates the proton (acid) and becomes OH⁻, its conjugate base; OBr⁻ accepts it (base) and becomes HOBr, its conjugate acid.
Feedback if chosen: This option is wrong — you swapped donor and acceptor. The OBr⁻ gains the proton, becoming HOBr, so OBr⁻ is the base and H₂O is the acid.
Feedback if chosen: This option is wrong — the acid and base labels are right, but the conjugates are crossed. Each conjugate belongs to the species it differs from by one H⁺.
Feedback if chosen: This option is wrong — you benched the water. H₂O donates the proton in this equation, so it acts as the Brønsted-Lowry acid.
Check — multiple choice
A student combines the half-reactions Mn(s) → Mn²⁺(aq) + 2e⁻ and Fe³⁺(aq) + 3e⁻ → Fe(s) directly, writing Mn + Fe³⁺ → Mn²⁺ + Fe. Which evaluation is correct?
Feedback if chosen: This option is wrong — you skipped the electron check. The charge proves it: 3+ on the left against 2+ on the right.
Feedback if chosen: This option is wrong — you kept the electrons. In a correctly combined equation the electrons lost and gained cancel completely.
Feedback if chosen: This option is wrong — you changed a species to force the balance. Each half-reaction fixes its ion's real charge; only the scaling may change.
Feedback if chosen: Correct. Electrons lost must equal electrons gained. Scaling gives 3Mn → 3Mn²⁺ + 6e⁻ and 2Fe³⁺ + 6e⁻ → 2Fe, so 3Mn + 2Fe³⁺ → 3Mn²⁺ + 2Fe.
Check — multiple choice
Aluminum metal reacts with nickel(II) ions in solution. Which balanced equation represents the reaction, with electrons lost equal to electrons gained?
Feedback if chosen: This option is wrong — you added the halves without scaling: 3 electrons lost against 2 gained, and the charge reads 2+ on the left, 3+ on the right.
Feedback if chosen: This option is wrong — you bent aluminum's charge to force a 1:1 match. Aluminum forms Al³⁺; the fix is scaling the halves, not changing the ion.
Feedback if chosen: Correct. Two Al lose 6 electrons (2 × 3) and three Ni²⁺ gain 6 (3 × 2); atoms and charge (6+ each side) both balance.
Feedback if chosen: This option is wrong — you swapped the multipliers: this version loses 9 electrons and gains only 4, leaving the charge unbalanced.
Check — multiple choice
Four changes are observed. Which one is most likely a chemical change?
Feedback if chosen: Correct. The steady fizzing is a gas forming from the tablet and the water — one of the four types of evidence of a chemical change.
Feedback if chosen: This option is wrong — you classified evaporation as chemical. The water leaves as water vapor, the same substance in a new phase.
Feedback if chosen: This option is wrong — the olive oil turns solid as it cools, a phase change. The solid oil is the same substance, so no new substance appears.
Feedback if chosen: This option is wrong — melting is a phase change. The liquid is the same substance as the solid.
Check — multiple choice
Opening a bottle of sparkling water releases a rush of bubbles, and a student cites the bubbles as proof of a chemical change. Why is the student wrong?
Feedback if chosen: This option is wrong — you overcorrected. A new gas is one of the four types of evidence; it just cannot prove a chemical change on its own.
Feedback if chosen: This option is wrong — you required a second sign. One type of evidence can stand alone; the flaw is that physical changes can produce the same sign.
Feedback if chosen: Correct. The CO₂ was already there, dissolved; dropping the pressure lets it escape. A physical change makes the same observation, so the bubbles cannot prove a chemical change.
Feedback if chosen: This option is wrong — mixtures react all the time. The student's error is about the evidence, not about what a drink is.
Check — multiple choice
Mixing Cr(NO₃)₃(aq) and NaOH(aq) produces a solid. (Cr(OH)₃ is insoluble, and the chromium is present as Cr³⁺ — supplied.) Which is the correct net ionic equation?
Feedback if chosen: This option is wrong — you wrote the molecular equation. The net ionic form keeps only the species that change.
Feedback if chosen: This option is wrong — you dropped the coefficient 3. With one OH⁻ the atoms don't balance, and the charge reads 2+ on the left against 0 on the right.
Feedback if chosen: This option is wrong — you wrote the change backward. The solid forms from the mixed ions; it does not break apart into them.
Feedback if chosen: Correct. Cr³⁺ and 3OH⁻ build the solid; atoms balance, and each side's charge sums to zero. The Na⁺ and NO₃⁻ spectators are gone.
Check — multiple choice
Mixing Na₃PO₄(aq) and LiNO₃(aq) forms solid lithium phosphate. (Li₃PO₄ is insoluble — supplied.) A student writes the complete ionic equation as 3Na⁺(aq) + PO₄³⁻(aq) + 3Li⁺(aq) + 3NO₃⁻(aq) → 3Li⁺(aq) + PO₄³⁻(aq) + 3Na⁺(aq) + 3NO₃⁻(aq). What is the error?
Feedback if chosen: Correct. Only dissolved ionic compounds are split into ions. The insoluble Li₃PO₄ leaves the solution as a solid, so it is written as Li₃PO₄(s).
Feedback if chosen: This option is wrong — canceling spectators is the next step, the net ionic form. In a complete ionic equation the spectators legitimately appear on both sides.
Feedback if chosen: This option is wrong — the charges do sum equally on the two sides. The defect is a solid written as if it were still dissolved.
Feedback if chosen: This option is wrong — you accepted a right-hand side with no solid in it. The reaction's whole point is the Li₃PO₄ leaving the solution as Li₃PO₄(s).
Check — multiple choice
Tin reacts with chlorine gas to form liquid tin(IV) chloride. The boxes show a sample before and after the reaction. Which balanced equation, in smallest whole numbers, does the drawing represent?
Feedback if chosen: This option is wrong — you copied one of each formula without counting atoms: two Cl atoms on the left cannot make four on the right.
Feedback if chosen: Correct. The drawn counts 2 : 4 : 2 reduce to 1 : 2 : 1, and Sn + 2Cl₂ → SnCl₄ balances — 1 Sn and 4 Cl atoms on each side.
Feedback if chosen: This option is wrong — you copied the drawn counts without reducing them. Coefficients state the smallest whole-number relationship: 1, 2, 1.
Feedback if chosen: This option is wrong — you wrote single Cl atoms. The drawing shows the chlorine arriving as Cl₂ molecules, and Cl₂ is chlorine's formula as a gas.
Check — multiple choice
Nitrogen monoxide reacts with oxygen by 2NO(g) + O₂(g) → 2NO₂(g). A student draws the before and after boxes shown. Which evaluation of the drawing is correct?
Feedback if chosen: This option is wrong — you counted only the NO₂ units. Count atoms: the extra O₂ pushes the After box to 10 O atoms against 8 before.
Feedback if chosen: Correct. Before: 4 N and 8 O atoms. After: 4 NO₂ holds 4 N and 8 O — so the leftover O₂ adds two O atoms from nowhere. Four NO uses exactly two O₂, with none left.
Feedback if chosen: This option is wrong — molecules are drawn intact. The flaw is in the atom count, not in showing bonded molecules.
Feedback if chosen: This option is wrong — the NO is fully consumed: four NO needs exactly two O₂, and two O₂ were supplied.
Check — multiple choice
Four processes are described at the particle level. Which one is a chemical process?
Feedback if chosen: This option is wrong — you classified dissolving a gas as chemical. Every O₂ molecule stays intact; only the attractions around it change.
Feedback if chosen: This option is wrong — mixing rearranges where the molecules are, not what they are. No bond breaks or forms.
Feedback if chosen: Correct. Bonds breaking and new bonds forming is the defining mark of a chemical process — the atoms regroup into new substances.
Feedback if chosen: This option is wrong — condensing only brings intact molecules closer. Every bond inside each molecule survives, so the process is physical.
Check — multiple choice
A block of solid air freshener slowly shrinks as it turns straight into vapor, absorbing thermal energy from the room as it goes. What is that absorbed energy used for?
Feedback if chosen: This option is wrong — you spent the energy on bonds. The vapor is made of the same intact molecules; no bond inside any molecule breaks.
Feedback if chosen: This option is wrong — you made new substances. The scent molecules in the vapor are the same molecules that sat in the solid.
Feedback if chosen: This option is wrong — you spent the energy warming the vapor. The vapor leaves at the room's temperature; the energy is absorbed as molecules pull away from their neighbors.
Feedback if chosen: Correct. A physical change like this one needs energy only to overcome the attractions between intact molecules — that is what the room's thermal energy supplies.
Check — multiple choice
Tungsten metal is produced by WO₃ + 3H₂ → W + 3H₂O. What mass of H₂ (molar mass 2.016 g/mol) is required to convert 46.4 g of WO₃ (molar mass 231.84 g/mol) completely?
Feedback if chosen: This option is wrong — you applied the 1:3 relationship upside down, dividing by 3. Each WO₃ needs three H₂: 0.200 × 3 = 0.600 mol.
Feedback if chosen: This option is wrong — you skipped the mole relationship: 0.200 mol of WO₃ needs 0.600 mol of H₂, not 0.200 mol.
Feedback if chosen: Correct. 46.4 ÷ 231.84 = 0.200 mol WO₃; the coefficients give 0.200 × 3 = 0.600 mol H₂; 0.600 × 2.016 = 1.21 g.
Feedback if chosen: This option is wrong — you treated the 46.4 g as 46.4 mol. Convert the mass to moles first: 46.4 ÷ 231.84 = 0.200 mol.
Check — multiple choice
A synthesis has a theoretical yield of 6.40 g of product. The student collects 5.60 g. What is the percent yield?
Feedback if chosen: This option is wrong — you calculated the shortfall, (6.40 − 5.60) ÷ 6.40. Percent yield is the collected amount over the theoretical amount.
Feedback if chosen: Correct. Percent yield = actual ÷ theoretical × 100% = 5.60 ÷ 6.40 × 100% = 87.5%.
Feedback if chosen: This option is wrong — you inverted the ratio. A yield above 100% would mean more product appeared than the reactants could make.
Feedback if chosen: This option is wrong — the reactant masses have already done their work: they fixed the 6.40 g theoretical yield. Only the two yields enter the percentage.
Check — multiple choice
A 20.0 mL sample of NaOH(aq) reaches the equivalence point when 24.0 mL of 0.175 M HCl has been added, reacting by HCl + NaOH → NaCl + H₂O — supplied. What is the molarity of the NaOH sample?
Feedback if chosen: This option is wrong — you reported the moles of HCl as a molarity. Divide those moles by the sample's volume: 0.00420 ÷ 0.0200 = 0.210 M.
Feedback if chosen: This option is wrong — you swapped the volumes in M₁V₁ = M₂V₂: 0.146 = 0.175 × 20.0/24.0. The titrant's molarity multiplies the titrant's own volume: 0.175 × 24.0/20.0 = 0.210 M.
Feedback if chosen: This option is wrong — you copied the titrant's molarity. The smaller sample volume must hold the same moles, so its concentration is higher.
Feedback if chosen: Correct. HCl added: 0.0240 × 0.175 = 0.00420 mol; the 1:1 relationship gives 0.00420 mol of NaOH; 0.00420 ÷ 0.0200 = 0.210 M.
Check — multiple choice
Which statement correctly describes the equivalence point of a titration?
Feedback if chosen: Correct. The equivalence point is defined by the mole match — titrant moles exactly sufficient, in the reaction's ratio, to consume the analyte.
Feedback if chosen: This option is wrong — you described the endpoint. The color change is the visible signal chosen to approximate the equivalence point, not its definition.
Feedback if chosen: This option is wrong — you matched volumes. The defining match is between moles, and equal moles rarely arrive at equal volumes.
Feedback if chosen: This option is wrong — you matched concentrations. Two solutions of different concentrations reach equivalence through their volumes; the mole match is what defines it.
Check — multiple choice
Ammonia gas and hydrogen bromide gas react directly: NH₃(g) + HBr(g) → NH₄Br(s). How is this reaction classified?
Feedback if chosen: This option is wrong — no oxidation number changes: H stays +1, Br stays −1, N stays −3. No electrons move.
Feedback if chosen: This option is wrong — a precipitation reaction forms an insoluble solid from ions mixed in solution. Here two gases combine directly; the evidence to read is the proton transfer.
Feedback if chosen: This option is wrong — you made water the requirement. Proton transfer defines an acid-base reaction; water is just a common product, not a necessary one.
Feedback if chosen: Correct. The HBr donates H⁺ and the NH₃ accepts it, forming NH₄⁺ and Br⁻ — proton transfer is the defining evidence of an acid-base reaction.
Check — multiple choice
For which compound can the solubility in water be stated from the memorized solubility facts alone, with no supplied information?
Feedback if chosen: Correct. Ammonium compounds sit on the always-soluble list — sodium, potassium, ammonium, and nitrate compounds — so (NH₄)₃PO₄ is soluble, no lookup needed.
Feedback if chosen: This option is wrong — MnS contains none of the four always-soluble families, so its solubility would have to be supplied.
Feedback if chosen: This option is wrong — CuI contains none of the four always-soluble families, so its solubility would have to be supplied.
Feedback if chosen: This option is wrong — Fe(OH)₂ contains none of the four always-soluble families, so its solubility would have to be supplied.
Check — multiple choice
Propanoic acid reacts with ammonia: C₂H₅COOH(aq) + NH₃(aq) → C₂H₅COO⁻(aq) + NH₄⁺(aq). Which pair is one of the conjugate acid-base pairs?
Feedback if chosen: This option is wrong — you paired the two reactants. A conjugate pair spans the arrow, its two members differing by exactly one H⁺.
Feedback if chosen: Correct. C₂H₅COOH and C₂H₅COO⁻ differ by exactly one proton — the acid and the base it becomes after donating. (NH₄⁺/NH₃ is the other pair.)
Feedback if chosen: This option is wrong — you paired the two acids. Each conjugate pair holds one acid and its own base, one proton apart.
Feedback if chosen: This option is wrong — you paired the two products. The partners of a conjugate pair sit on opposite sides of the equation.
Check — multiple choice
Within a conjugate acid-base pair, how are the strengths of the two partners related?
Feedback if chosen: This option is wrong — you matched strong with strong. An acid that gives its proton up completely leaves a partner with almost no pull for protons.
Feedback if chosen: This option is wrong — you leveled the pair. The more completely the acid donates, the less its partner accepts; the two strengths are opposite.
Feedback if chosen: Correct. The more completely an acid donates its protons, the weaker its conjugate base's pull for protons — the relationship within a pair is inverse.
Feedback if chosen: This option is wrong — the two are locked together. They are the same species with and without one proton, so one partner's grip fixes the other's.
Check — multiple choice
A student combines the half-reactions Al(s) → Al³⁺(aq) + 3e⁻ and Ag⁺(aq) + e⁻ → Ag(s) directly, writing Al + Ag⁺ → Al³⁺ + Ag. Which evaluation is correct?
Feedback if chosen: Correct. Electrons lost must equal electrons gained. Scaling the silver half by 3 gives 3Ag⁺ + 3e⁻ → 3Ag, so Al + 3Ag⁺ → Al³⁺ + 3Ag.
Feedback if chosen: This option is wrong — you skipped the electron check. The charge proves it: 1+ on the left against 3+ on the right.
Feedback if chosen: This option is wrong — you kept the electrons. In a correctly combined equation the electrons lost and gained cancel completely.
Feedback if chosen: This option is wrong — you changed a species to force the balance. Each half-reaction fixes its ion's real charge; only the scaling may change.
Check — multiple choice
The half-reactions Cd(s) → Cd²⁺(aq) + 2e⁻ and Au³⁺(aq) + 3e⁻ → Au(s) are combined into one balanced equation. Which is correct?
Feedback if chosen: This option is wrong — you added the halves without scaling: 2 electrons lost against 3 gained, and the charge reads 3+ on the left, 2+ on the right.
Feedback if chosen: This option is wrong — you swapped the multipliers: this version loses 4 electrons and gains 9. Scale the Cd half by 3 and the Au half by 2.
Feedback if chosen: This option is wrong — you left the electrons in the overall equation. Once lost and gained electrons match, they cancel completely.
Feedback if chosen: Correct. Three Cd lose 6 electrons (3 × 2) and two Au³⁺ gain 6 (2 × 3); the electrons cancel, and atoms and charge (6+ each side) balance.
Free-response question
Source note (not shown as part of the question): authored to the redox + stoichiometry long archetype (FRQ playbook T3/T8; modeled on the shapes of 2021 FRQ Q1(e), 2024 FRQ Q3(a–c), and 2024 FRQ Q2(c)); no unit-4-pure real long FRQ exists in 2021–2026
A student adds 0.540 g of aluminum foil (molar mass of Al: 26.98 g/mol) to a flask of hydrochloric acid. The metal dissolves, and a gas forms by the equation 2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g). In trial 1 the acid is in excess, and the reaction goes to completion.
(a) Identify the oxidation number of Al in Al(s) and in AlCl₃(aq). [1 point]
(b) Identify the species that is oxidized and the species that is reduced. Justify your answer in terms of oxidation numbers. [1 point]
(c) Write the two half-reactions — oxidation and reduction — each balanced in atoms and charge. (In solution, the HCl is present as H⁺ and Cl⁻ ions.) [2 points]
(d) Calculate the number of moles of H₂ produced in trial 1. [1 point]
(e) Calculate the volume of the H₂, in liters, at 298 K and 1.00 atm, reported to the correct number of significant figures. (R = 0.08206 L·atm·mol⁻¹·K⁻¹.) [1 point]
(f) In trial 2 the student adds another 0.540 g of Al, but this time the flask holds only 0.0200 mol of HCl. Identify the limiting reactant in trial 2. Justify your answer by comparing the moles available with the moles required. [2 points]
(g) Calculate the number of moles of H₂ produced in trial 2. [1 point]
(h) A student claims that this reaction is an acid-base reaction rather than an oxidation-reduction reaction, because HCl is an acid. Do you agree with the claim? Justify your answer using oxidation numbers. [1 point]
✓ Model response (the key)
(a)
In Al(s), the oxidation number of Al is 0 (an atom in a free element).
In AlCl₃, each Cl is −1, so Al is +3.
(b)
Al is oxidized: its oxidation number rises from 0 to +3, a loss of electrons.
The H⁺ of the HCl is reduced: hydrogen's oxidation number falls from +1 to 0 in H₂, a gain of electrons.
(c)
Oxidation: Al → Al³⁺ + 3e⁻
Reduction: 2H⁺ + 2e⁻ → H₂
(d)
n(Al) = 0.540 g ÷ 26.98 g/mol = 0.0200 mol
n(H₂) = 0.0200 mol Al × (3 mol H₂ / 2 mol Al) = 0.0300 mol
(e)
V = nRT/P = (0.0300 × 0.08206 × 298) ÷ 1.00 = 0.734 L
(f)
The HCl is the limiting reactant.
The 0.0200 mol of Al would require 0.0200 × (6 mol HCl / 2 mol Al) = 0.0600 mol of HCl, and only 0.0200 mol is available — so the HCl is used up first.
(g)
n(H₂) = 0.0200 mol HCl × (3 mol H₂ / 6 mol HCl) = 0.0100 mol
(h)
No — the claim is not correct.
Oxidation numbers change — Al rises from 0 to +3 and H falls from +1 to 0 — so electrons transfer, and the reaction is oxidation-reduction.
No proton transfers to a base here: the H⁺ ions gain electrons instead.
Scoring rubric — 10 points available
Free-response question
Source note (not shown as part of the question): authored to the titration short archetype (FRQ playbook T3/T6/T9; modeled on the shapes of 2023 FRQ Q4(b–c) and 2024 FRQ Q7); no unit-4-pure real short FRQ exists in 2021–2026
A student determines the concentration of a strontium hydroxide solution by titration. A 25.0 mL sample of Sr(OH)₂(aq) is titrated with 0.200 M HNO₃(aq), reacting by Sr(OH)₂(aq) + 2HNO₃(aq) → Sr(NO₃)₂(aq) + 2H₂O(l). The indicator's color change persists when 30.0 mL of the HNO₃ has been added.
(a) Calculate the molarity of the Sr(OH)₂ sample. [2 points]
(b) The color change the student watched for is the endpoint. Explain why the endpoint is not, by definition, the same thing as the equivalence point. [1 point]
(c) Before filling the buret, the student rinsed it with distilled water but not with the 0.200 M HNO₃(aq). Rinsing with water leaves the buret's walls wet, so the titrant that follows is slightly diluted. Predict whether the calculated molarity of the Sr(OH)₂ will be greater than, less than, or equal to the true molarity. Justify your answer. [1 point]
✓ Model response (the key)
(a)
n(HNO₃) = 0.0300 L × 0.200 mol/L = 0.00600 mol
n(Sr(OH)₂) = 0.00600 mol HNO₃ × (1 mol Sr(OH)₂ / 2 mol HNO₃) = 0.00300 mol
molarity = 0.00300 mol ÷ 0.0250 L = 0.120 M
(b)
The equivalence point is the exact mole match: the moles of HNO₃ added equal what the 2:1 mole ratio requires to consume all of the Sr(OH)₂.
The endpoint is the observable signal — the persisting color change — chosen to approximate that point.
The indicator changes color close to, but not exactly at, the mole match, so the two can differ slightly.
(c)
The calculated molarity will be greater than the true molarity.
Water left in the buret dilutes the HNO₃, so each milliliter delivers fewer moles than the label says, and a larger volume is needed to reach the endpoint.
The student calculates with the labeled 0.200 M and the larger volume, so the calculated moles of HNO₃ — and the calculated Sr(OH)₂ molarity — come out too high.
Scoring rubric — 4 points available
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