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Stoichiometry Step-by-Step

By the Periodixy Editorial Team · Last reviewed July 10, 2026

Stoichiometry answers the question every reaction raises: “If I start with this much of one substance, how much of another do I get or need?” It sounds intimidating, but every stoichiometry problem — no matter how it is dressed up — follows the same four-step path.

Laboratory glassware set up for a quantitative chemistry experiment
Photo by olia danilevich on Pexels

The universal path

The route every problem takes

grams A → moles A → (mole ratio) → moles B → grams B

  1. Balance the equation — coefficients are the recipe.
  2. Convert the known amount to moles (divide grams by molar mass).
  3. Apply the mole ratio from the coefficients (multiply by target ÷ known).
  4. Convert to the requested unit (multiply moles by molar mass for grams).
💡 Tip: Moles are the only currency that reaction coefficients understand. Grams must always be exchanged into moles before the mole ratio, and back out afterwards if needed.

Fully worked example

Burning methane

How many grams of CO₂ form when 8.0 g of CH₄ burns completely? (CH₄ + 2O₂ → CO₂ + 2H₂O)

  1. The equation is already balanced ✓
  2. Moles of CH₄ = 8.0 g ÷ 16.04 g/mol = 0.499 mol
  3. Mole ratio CO₂ : CH₄ = 1 : 1 → 0.499 mol CO₂
  4. Grams of CO₂ = 0.499 mol × 44.01 g/mol

Answer: ≈ 22 g of CO₂

A ratio that isn't 1:1

How many grams of H₂O are produced from 4.04 g of H₂? (2H₂ + O₂ → 2H₂O)

  1. Moles of H₂ = 4.04 ÷ 2.016 = 2.00 mol
  2. Ratio H₂O : H₂ = 2 : 2 = 1 : 1 → 2.00 mol H₂O
  3. Mass = 2.00 × 18.02 g/mol

Answer: ≈ 36.0 g of water

Dimensional analysis: the one-line version

Once comfortable, chain the conversions so units cancel in one line:

One-line setup for the methane example

8.0 g CH₄ × (1 mol CH₄ / 16.04 g) × (1 mol CO₂ / 1 mol CH₄) × (44.01 g / 1 mol CO₂) ≈ 22 g

If the units cancel down to what the question asks for, the setup is almost certainly right. The Stoichiometry Calculator shows this chain for any equation.

Common pitfalls

  • Skipping the balance check — ratios from an unbalanced equation are meaningless.
  • Using the coefficient ratio on grams directly (2 g H₂ does NOT give 2 g H₂O).
  • Inverting the mole ratio — write it as target-over-known and double-check with units.
  • Using the molar mass of the wrong substance in step 4.

Summary

  • Every problem: balance → moles → mole ratio → requested unit.
  • Coefficients give mole ratios, never gram ratios.
  • Dimensional analysis chains the steps and self-checks through unit cancellation.

Frequently asked questions

What is a mole ratio?

The ratio of coefficients between two substances in a balanced equation. In 2H₂ + O₂ → 2H₂O, the ratio of H₂O to O₂ is 2:1.

Do I always need molar masses?

Only when the problem involves grams. A moles-to-moles question needs just the mole ratio.

What if I'm given amounts of two reactants?

That's a limiting reactant problem — one reactant runs out first. See [Limiting Reactants Explained](/study-guides/limiting-reactants-explained).

Practise with these tools

Stoichiometry CalculatorConvert between grams and moles of any two substances in a balanced equation using mole ratios.Molar Mass CalculatorCalculate the molar mass of any chemical formula with a full element-by-element breakdown and percent composition.Limiting Reactant CalculatorFind the limiting reactant, theoretical yield and leftover excess for any balanced reaction.Chemical Equation BalancerBalance any chemical equation instantly and check atom conservation on both sides.

Related study guides

How to Balance Chemical EquationsLearn to balance chemical equations step by step using the counting method, with combustion and metal-acid examples plus the most common mistakes.Molar Mass: Definition and How to Calculate ItWhat is molar mass, and how do you find it? Add up the atomic masses in a formula to get grams per mole. Worked examples for H2O, glucose, Ca(OH)2 and NaCl.Limiting Reactants ExplainedThe limiting reactant runs out first and caps the product. Learn the divide-by-coefficient method, theoretical yield and excess with worked examples.