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Stoichiometry: A Step-by-Step Guide

Use balanced equations to convert between grams, moles and product amounts with a repeatable problem-solving process.

◎ Clear explanations◉ Worked examples▤ Practice problems⬇ Free resources

What is stoichiometry?

Stoichiometry relates reactant and product amounts using the coefficients of a balanced chemical equation. Coefficients are ratios of particles or moles—not grams.

Illustrated Stoichiometry: A Step-by-Step Guide study overview
IntoChemistry visual guide. Follow the worked examples below for the complete reasoning.

The reliable five-step method

  1. Balance the chemical equation.
  2. Convert the given quantity into moles if needed.
  3. Apply the mole ratio from the balanced coefficients.
  4. Convert the result to grams, liters, or particles as requested.
  5. Check cancellation of units and significant figures.

Follow the calculation roadmap

Roadmap of the unit conversions in a typical stoichiometry problem.
Roadmap of the unit conversions in a typical stoichiometry problem.

Worked stoichiometry examples

Worked example

How much water forms from 3.00 mol O₂ and excess H₂?

  1. Balance the equation: 2 H₂ + O₂ → 2 H₂O.
  2. Use the balanced mole ratio: 3.00 mol O₂ × (2 mol H₂O / 1 mol O₂).
  3. Verify O₂ cancels and the remaining unit is mol H₂O.

Answer: 6.00 mol H₂O.

Worked example

How many grams of CO₂ form from 8.00 g CH₄?

  1. Balance: CH₄ + 2 O₂ → CO₂ + 2 H₂O (oxygen in excess).
  2. Convert CH₄ to mol: 8.00 g ÷ 16.05 g/mol = 0.4984 mol.
  3. The CH₄:CO₂ coefficient ratio is 1:1.
  4. Multiply by the CO₂ molar mass, 44.01 g/mol.

Answer: 21.9 g CO₂ (three significant figures).

Worked example

Find the limiting reactant: 3.00 mol H₂ + 2.00 mol O₂

  1. Use 2 H₂ + O₂ → 2 H₂O.
  2. Reacting all 3.00 mol H₂ requires 1.50 mol O₂, which is available.
  3. Therefore, H₂ runs out first and controls product amount.

Answer: H₂ is limiting; 3.00 mol H₂O can form.

Common mistakes

  • Using an equation before balancing it.
  • Using molar masses in place of coefficient ratios.
  • Skipping conversion from grams into moles.
  • Choosing the excess reagent as the limiting reactant.

Quick self-check

Why is the mole ratio 2:1 for H₂O:O₂?

Because the balanced equation 2 H₂ + O₂ → 2 H₂O has coefficients 2 and 1 for those species.

Interactive Stoichiometry: A Step-by-Step Guide Practice

Apply what you have learned using immediate feedback.

Practice tool

Stoichiometry: Mole Ratios

All equations are balanced. Assume that any other reactant is available in excess. Give an answer in moles; three significant figures are sufficient.

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