Stoichiometry – IGCSE Chemistry Definition
IGCSE Chemistry definition of stoichiometry: the study of quantitative relationships in chemical reactions. Covers balancing equations and mole-ratio calculations.
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Chemistry teaching team: K. S. Tan (15+ years teaching IGCSE Chemistry) and Ms Yash (10+ years teaching IGCSE Chemistry) and Ms Kartini (15+ years teaching IGCSE Chemistry).
Mapped to Cambridge IGCSE Chemistry 0620 (2026–2028). Last updated 2026-08-19.
Stoichiometry is the area of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. On the 0620 syllabus, stoichiometric calculations are the backbone of Paper 4 and the Supplement content. Every reacting-mass, gas-volume, and titration calculation is a stoichiometry problem.
The 0620 definition
Stoichiometry is the calculation of the quantities of reactants and products in chemical reactions, using balanced equations and mole ratios.
The general method
All stoichiometric calculations follow the same steps:
- Write a balanced equation — the coefficients give the mole ratio
- Calculate moles of the substance you know (from mass, volume, or concentration)
- Use the mole ratio from the equation to find moles of the substance you want
- Convert to the required quantity (mass, volume, or concentration)
Balancing equations
A balanced equation has the same number of each type of atom on both sides. The coefficients (big numbers in front) adjust the ratio — you never change the formulae themselves.
Example: balancing the combustion of methane
CH₄ + O₂ → CO₂ + H₂O (unbalanced)
CH₄ + 2O₂ → CO₂ + 2H₂O (balanced: 1C, 4H, 4O each side)
Mole ratios in action
From CH₄ + 2O₂ → CO₂ + 2H₂O:
| Substance | CH₄ | O₂ | CO₂ | H₂O |
|---|---|---|---|---|
| Mole ratio | 1 | 2 | 1 | 2 |
| If 0.5 mol CH₄ reacts | 0.5 | 1.0 | 0.5 | 1.0 |
Types of stoichiometric calculation on the 0620 syllabus
- Reacting masses: mass → moles → ratio → moles → mass
- Gas volumes at RTP: using molar gas volume (24 dm³/mol)
- Solution volumes: using concentration (mol/dm³) and volume (dm³)
- Titration calculations: combining concentration, volume, and mole ratio
- Percentage yield and purity: comparing actual vs theoretical
Worked exam question
What mass of sodium hydroxide is needed to neutralise 50 cm³ of 0.2 mol/dm³ sulfuric acid? H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O (Ar: Na = 23, O = 16, H = 1) (3)
Mark scheme
Moles of H₂SO₄ = 0.2 x (50/1000) = 0.01 mol [1]; from equation, moles of NaOH = 2 x 0.01 = 0.02 mol [1]; mass of NaOH = 0.02 x 40 = 0.8 g [1]
Common exam mistakes
- Forgetting to balance the equation before reading off the mole ratio. An unbalanced equation gives the wrong ratio and the wrong answer.
- Ignoring the mole ratio and assuming 1:1. Always check the coefficients — H₂SO₄ reacts with 2NaOH, not 1.
- Converting units incorrectly. Make sure volumes are in dm³ for concentration calculations and that you use the correct molar gas volume.
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