What Stoichiometry Is
Stoichiometry is the bookkeeping of chemical reactions — it lets you predict *how much* product a reaction makes, or *how much* reactant you need, using the balanced equation as a recipe. The coefficients in a balanced equation give the ratio of particles (and therefore moles) that react and form. Master one idea and the rest follow: the coefficients are a mole ratio.
Before you start
Molar mass and the mole are covered in full in The Mole, Molar Mass & Particle Count; here we use them as ready-made tools. Every problem below also assumes your equation is already balanced — unbalanced coefficients give wrong ratios and wrong answers.
Step 1 — The Mole Ratio
A balanced equation's coefficients tell you the ratio in which substances react and form. For the synthesis of ammonia: N₂ + 3H₂ → 2NH₃ the coefficients are 1, 3, and 2. Any two of them make a mole ratio — a conversion factor you can flip either way (for example 3 mol H₂ / 2 mol NH₃ *or* 2 mol NH₃ / 3 mol H₂). Choose the version that cancels the unit you have and leaves the unit you want.
| Mole ratio | Reads as |
|---|---|
| 1 N₂ : 3 H₂ | 1 mol N₂ reacts with 3 mol H₂ |
| 1 N₂ : 2 NH₃ | 1 mol N₂ produces 2 mol NH₃ |
| 3 H₂ : 2 NH₃ | 3 mol H₂ produce 2 mol NH₃ |
Balance first, always
Mole ratios come from the coefficients of a balanced equation. If the equation isn't balanced, every ratio — and every answer built on it — is wrong. Balancing changes coefficients only; never change a subscript to balance, because that changes the substance (O₂ is oxygen gas, O₃ is ozone).
Mole-Mole Problems
When both the known and the unknown are already in moles, the problem is a single step: multiply by the mole ratio from the balanced equation. Set up the ratio so the known unit cancels.
Example 1 — moles → moles. Using N₂ + 3H₂ → 2NH₃, how many moles of NH₃ form from 5.0 mol H₂ (excess N₂)? 5.0 mol H₂ × (2 mol NH₃ / 3 mol H₂) = 3.3 mol NH₃ Example 2. How many moles of H₂ are needed to react completely with 4.0 mol N₂? 4.0 mol N₂ × (3 mol H₂ / 1 mol N₂) = 12 mol H₂
Let the units steer
Write every conversion factor with units and cancel them like numbers. If 'mol H₂' is on top of what you have, put 'mol H₂' on the bottom of the ratio so it cancels and leaves 'mol NH₃'. If your units don't cancel to the target, the setup is upside-down.
Mass-Mass Problems: The Conversion Road Map
Balances measure grams, but the equation only speaks moles. So a grams-to-grams problem is really three steps: convert to moles, cross the mole bridge, convert back to grams. This is the stoichiometry road map: grams of A → moles of A → moles of B → grams of B.
| To convert | Do this | Result |
|---|---|---|
| grams of A | ÷ molar mass of A | moles of A |
| moles of A | × (coefficient B ÷ coefficient A) | moles of B |
| moles of B | × molar mass of B | grams of B |
Example — grams → grams. Using N₂ + 3H₂ → 2NH₃, how many grams of NH₃ form from 50.0 g N₂ (excess H₂)? Molar masses: N₂ = 28.02 g/mol, NH₃ = 17.03 g/mol. 1. grams → moles: 50.0 g N₂ ÷ 28.02 g/mol = 1.784 mol N₂ 2. mole ratio: 1.784 mol N₂ × (2 mol NH₃ / 1 mol N₂) = 3.569 mol NH₃ 3. moles → grams: 3.569 mol NH₃ × 17.03 g/mol = 60.8 g NH₃
Grams can't ride the ratio
The mole ratio only works on moles, never grams. Skipping the grams→moles step and multiplying masses directly by the coefficient ratio is the single most common stoichiometry mistake. Always convert to moles first, use the ratio, then convert back to grams.
Limiting Reactant
When you're given amounts of two or more reactants, one usually runs out first — the limiting reactant. It determines how much product forms; whatever is left over is the excess reactant. Don't guess from mass or even from moles alone — compare each reactant's moles divided by its coefficient. The smallest quotient is limiting.
Example. For N₂ + 3H₂ → 2NH₃, you mix 28.0 g N₂ with 10.0 g H₂. Which is limiting, and what mass of NH₃ can form? Molar masses: N₂ = 28.02, H₂ = 2.016, NH₃ = 17.03 g/mol. Convert each to moles: 28.0 g ÷ 28.02 = 1.00 mol N₂; 10.0 g ÷ 2.016 = 4.96 mol H₂. Then divide by the coefficients:
| Reactant | Moles | ÷ coefficient | Quotient |
|---|---|---|---|
| N₂ | 1.00 mol | ÷ 1 | 1.00 |
| H₂ | 4.96 mol | ÷ 3 | 1.65 |
N₂ has the smaller quotient, so N₂ is limiting and H₂ is in excess. Base the product on the limiting reactant: 1.00 mol N₂ × (2 mol NH₃ / 1 mol N₂) × 17.03 g/mol = 34.1 g NH₃ (theoretical yield) H₂ actually used = 1.00 mol N₂ × (3 mol H₂ / 1 mol N₂) = 3.00 mol, so 4.96 − 3.00 = 1.96 mol H₂ (≈ 3.95 g) is left over.
Fewer grams ≠ limiting
The limiting reactant is not automatically the one with the smallest mass, or even the fewest moles — coefficients matter. Here H₂ has far more moles than N₂, yet you can't tell which limits until you divide moles by the coefficients (1 and 3). Always compare moles ÷ coefficient.
Theoretical Yield & Percent Yield
Theoretical yield is the maximum product the stoichiometry allows — calculated from the limiting reactant, assuming the reaction goes perfectly. Actual yield is what you really collect in the lab, always measured, and usually less. Percent yield compares them: percent yield = (actual yield ÷ theoretical yield) × 100% Both yields must be in the same unit (usually grams).
Example. From the limiting-reactant problem above, the theoretical yield is 34.1 g NH₃. Suppose a chemist actually isolates 28.0 g NH₃. percent yield = (28.0 g ÷ 34.1 g) × 100% = 82.1%
Why yield is rarely 100%
Real reactions lose product to side reactions, incomplete reactions, impurities, and losses during transfer, filtering, and drying. A percent yield above 100% almost always means the product is still wet or impure (extra mass), not that you beat the stoichiometry.
Try it yourself
stoichiometry-reaction-trials.csvDownload stoichiometry-reaction-trials.csv: https://restedteacher.com/practice-files/chemistry/stoichiometry-reaction-trials.csv
Five lab trials of the single-displacement reaction 2Al + 3CuSO4 -> Al2(SO4)3 + 3Cu, each with a starting mass of aluminum, a starting mass of copper(II) sulfate, and the actual mass of copper collected.
Open it as a spreadsheet
Open stoichiometry-reaction-trials.csv in Google Sheets or Excel. Add columns for moles of Al, moles of CuSO4, theoretical yield, and percent yield so you can work through all five trials side by side.
The balanced equation is 2Al + 3CuSO₄ → Al₂(SO₄)₃ + 3Cu. Molar masses: Al = 26.98 g/mol, CuSO₄ = 159.62 g/mol, Cu = 63.55 g/mol. 1. For trial 1, determine which reactant is limiting. 2. For trial 1, calculate the theoretical yield of Cu in grams (round to 3 significant figures), then use the actual yield in the file to calculate the percent yield (round to 3 significant figures). 3. Which trial has the highest percent yield?