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Stoichiometry has a reputation for being the point where chemistry stops making sense. It doesn’t deserve it. Nearly every stoichiometry problem you’ll meet in an introductory course is the same three-step journey, and once you can see the map, the problems stop looking different from each other.
The word itself comes from Greek stoicheion (element) and metron (measure). It just means measuring the amounts of substances in a reaction.
The key idea: equations count in moles
A balanced equation like
2H₂ + O₂ → 2H₂O
is a recipe written in particles: two molecules of hydrogen react with one molecule of oxygen to make two molecules of water. Scale it up by 6.022 × 10²³ and it’s equally true in moles: 2 mol of H₂ + 1 mol of O₂ → 2 mol of H₂O.
What the equation does not say is that 2 grams of hydrogen react with 1 gram of oxygen. Coefficients are mole ratios, never mass ratios. That’s the single most important sentence in this whole topic.
The mole map
Every problem follows this route:
Given quantity → moles of given → moles of wanted → wanted quantity
- Step 1: Convert whatever you’re given into moles. Grams? Divide by molar mass. A solution? Multiply molarity by volume in litres. A gas? Use PV = nRT. A number of particles? Divide by Avogadro’s number.
- Step 2: Cross the bridge using the mole ratio from the balanced equation.
- Step 3: Convert the moles of the wanted substance into whatever the question asks for.
Step 2 is the only step that involves the reaction at all. Steps 1 and 3 are just unit conversions you already know.
Example 1: mass to mass
How many grams of carbon dioxide are produced when 100 g of propane, C₃H₈, burns completely?
Balance first: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Step 1 — grams of propane to moles. Molar mass of C₃H₈ = 3(12.011) + 8(1.008) = 44.10 g/mol. 100 g ÷ 44.10 g/mol = 2.268 mol C₃H₈
Step 2 — mole ratio. The equation says 1 C₃H₈ produces 3 CO₂. 2.268 mol × (3 mol CO₂ / 1 mol C₃H₈) = 6.803 mol CO₂
Step 3 — moles of CO₂ to grams. Molar mass of CO₂ = 44.01 g/mol. 6.803 mol × 44.01 g/mol = 299 g CO₂
Burning 100 g of fuel makes almost 300 g of carbon dioxide. That surprises people — where does the extra mass come from? From the oxygen in the air, which ends up in the CO₂ and the water.
Example 2: mass to moles
How many moles of oxygen gas are needed to burn that same 100 g of propane?
We already have 2.268 mol of propane from step 1. The ratio is 1 C₃H₈ : 5 O₂.
2.268 × 5 = 11.34 mol O₂
No step 3 needed, because the question asked for moles.
Example 3: with a solution
What volume of 0.500 M hydrochloric acid reacts completely with 5.00 g of calcium carbonate?
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
Step 1: CaCO₃ = 100.09 g/mol. 5.00 ÷ 100.09 = 0.04996 mol. Step 2: 1 CaCO₃ : 2 HCl → 0.09991 mol HCl. Step 3: volume = moles ÷ molarity = 0.09991 ÷ 0.500 = 0.1998 L = 200 mL.
Example 4: with a gas
What volume does the CO₂ from example 3 occupy at 25 °C and 1.00 atm?
Step 2: 1 CaCO₃ : 1 CO₂ → 0.04996 mol CO₂. Step 3: V = nRT ÷ P = 0.04996 × 0.08206 × 298.15 ÷ 1.00 = 1.22 L.
Four quite different-looking questions, one route.
A checklist that prevents most mistakes
- Is the equation balanced? If not, nothing else will be right.
- Did you convert to moles before using the ratio? Never put grams into a mole ratio.
- Is the ratio the right way up? Wanted on top, given on the bottom — the given unit should cancel.
- Are diatomic gases written as diatomic? O₂, not O.
- Is there more than one reactant amount given? Then you have a limiting reagent problem — work out which one runs out first before calculating any product.
- Does the answer make sense? If 5 g of reactant “produces” 5 kg of product, something went wrong.
Quick answers
Why do we have to use moles? Because atoms of different elements have different masses. Equal masses of two substances almost never contain equal numbers of particles, and reactions happen particle by particle.
What’s a mole ratio? The ratio of coefficients between two substances in a balanced equation, used as a conversion factor between their moles.
Does stoichiometry work for real reactions? It gives the theoretical amounts. Real reactions often fall short, which is what percent yield measures.
Tools that help
The equation balancer handles step zero, the molar mass calculator and grams to moles converter handle steps 1 and 3, and the limiting reagent calculator runs the whole map at once when you have amounts of every reactant.
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