AP Chemistry Stoichiometry

The convert-to-moles method, and three full worked examples.

Stoichiometry is the calculation that shows up more than any other on the AP Chemistry exam, buried inside limiting reactant problems, gas law problems, titrations, and electrochemistry alike. The method itself is short and doesn't change from problem to problem; what changes is which units you're converting into and out of moles.

The Method

Every stoichiometry problem follows the same three-step chain:

Convert your given amount to moles (divide a mass by molar mass, if that's what you're given).
Apply the mole ratio from the balanced equation to convert to moles of the substance you want.
Convert those moles to whatever final unit the question asks for (grams, particles, liters of gas, and so on).

This works because atoms are conserved in a chemical reaction — nothing is created or destroyed, only rearranged — so the coefficients of a balanced equation give you an exact proportionality between the substances involved.

Worked Example 1: Mole-to-Mole

N₂ + 3H₂ → 2NH₃. If 2.5 mol of N₂ reacts completely with excess H₂, how many moles of NH₃ form?

The mole ratio of N₂ to NH₃ is 1:2 (from the coefficients).
2.5 mol N₂ × (2 mol NH₃ / 1 mol N₂) = 5.0 mol NH₃

Worked Example 2: Mass-to-Mass

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. If 22.0 g of propane (C₃H₈) burns completely, what mass of CO₂ is produced?

Molar mass of C₃H₈ = 44.09 g/mol.
22.0 g ÷ 44.09 g/mol = 0.499 mol C₃H₈
Mole ratio C₃H₈ to CO₂ is 1:3:
0.499 mol × 3 = 1.50 mol CO₂
Molar mass of CO₂ = 44.01 g/mol:
1.50 mol × 44.01 g/mol = 65.9 g CO₂

Worked Example 3: Mole-to-Mass

2KClO₃ → 2KCl + 3O₂. If 0.75 mol of KClO₃ decomposes completely, what mass of O₂ gas is produced?

Mole ratio KClO₃ to O₂ is 2:3:
0.75 mol × (3/2) = 1.125 mol O₂
Molar mass of O₂ = 32.00 g/mol:
1.125 mol × 32.00 g/mol = 36.0 g O₂

When Two Reactant Amounts Are Given

Every example above starts from a single reactant, with the other assumed to be in excess. If a problem instead gives you starting amounts of two reactants, you can't skip straight to the mole-ratio step: you first have to figure out which reactant runs out first, since that's the one that actually limits how much product can form. See limiting reactant and percent yield for the full method and worked examples covering that case.

Common Mistakes

Skipping the mole conversion. Mole ratios from a balanced equation only apply to moles, not directly to grams or any other unit. Converting straight from grams of one substance to grams of another without passing through moles gives a wrong answer.
Using an unbalanced equation. The mole ratio comes from the coefficients, so balance the equation first, every time.
Flipping the mole ratio. The ratio is moles of what you want over moles of what you were given, not the other way around.
Forgetting excess reactant is assumed. Unless a problem gives you starting amounts of more than one reactant, assume everything except the substance named is present in excess and isn't the limiting factor.

Frequently Asked Questions

What is stoichiometry?

Stoichiometry is calculating the amount of one substance in a chemical reaction from a known amount of another substance, using the mole ratios given by a balanced equation's coefficients. Because atoms are conserved in a reaction, you can always calculate product amounts from reactant amounts, or reactant amounts from product amounts.

What's the general method for solving a stoichiometry problem?

Convert your given amount to moles, apply the mole ratio from the balanced equation to convert to moles of what you want, then convert those moles to whatever final unit the question asks for (grams, particles, liters of gas, and so on). This convert-to-moles, apply-the-ratio, convert-back three-step chain works for essentially every stoichiometry problem.

Do I need to balance the equation first?

Yes, always. The mole ratio you use comes directly from the coefficients of a balanced equation. An unbalanced equation gives you the wrong ratio and every answer built on it will be wrong, even if the rest of your setup is correct.

What if I'm given starting amounts of two reactants instead of one?

Then you have a limiting reactant problem, not a basic stoichiometry problem. You have to determine which reactant runs out first before you can calculate how much product forms. See the full method with worked examples on this site's limiting reactant and percent yield page.

Can stoichiometry combine with gas laws or molarity?

Yes. Once you have moles of a gas, the ideal gas law (PV = nRT) converts that to a volume, pressure, or temperature. Once you have moles of a solute, molarity (M = mol/L) converts that to a solution volume or concentration. The College Board's own AP Chemistry framework explicitly lists both as ways stoichiometric calculations combine with other topics.

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