Who this is for: Cambridge IGCSE Coordinated Science (0654) students who want stoichiometry — balancing equations and calculating reacting masses — to become a reliable source of marks instead of a trial-and-error balancing exercise.
What query it owns: how to understand and revise stoichiometry in Cambridge IGCSE Coordinated Science.
Why this is safe: this page owns the stoichiometry revision-guide angle, while Tutopiya’s Stoichiometry subtopic page owns the learning resource and the free Stoichiometry quiz owns the practice.

Stoichiometry is the maths of chemistry — using balanced equations to work out how much reactant you need or product you get. Cambridge IGCSE Coordinated Science (0654) expects you to balance equations, use mole ratios, calculate reacting masses, and find percentage yield. This guide links each skill to what examiners reward.

Key takeaways

• A balanced equation has equal numbers of each atom on both sides.
• Mole ratios come from the coefficients in a balanced equation.
• Reacting mass calculations: moles = mass ÷ Mᵣ, then use mole ratio, then mass = moles × Mᵣ.
• Percentage yield = (actual yield ÷ theoretical yield) × 100%.
• State symbols (s), (l), (g), (aq) show physical state in equations.

What is stoichiometry in Cambridge IGCSE Coordinated Science?

Stoichiometry uses the law of conservation of mass — atoms are neither created nor destroyed in a chemical reaction. A balanced chemical equation shows the mole ratio in which reactants combine and products form. From this, you can calculate masses, identify limiting reagents, and determine percentage yield when a reaction does not go to completion.

You can read the full explanation, worked examples and notes on Tutopiya’s Stoichiometry subtopic page before you attempt questions.

Steps for reacting mass calculations

Step	Action	Example (Mg + 2HCl → MgCl₂ + H₂)
1	Write balanced equation	Mg + 2HCl → MgCl₂ + H₂
2	Find moles of known substance	n(Mg) = 24/24 = 1 mol
3	Use mole ratio	1 mol Mg → 1 mol H₂
4	Convert moles to mass	m(H₂) = 1 × 2 = 2 g

Key formulas

Calculation	Formula	Notes
Moles from mass	n = m / Mᵣ	Mᵣ = relative formula mass
Mass from moles	m = n × Mᵣ	Use mole ratio between substances
Percentage yield	(actual ÷ theoretical) × 100%	Theoretical = maximum possible from equation
Mole ratio	From equation coefficients	2H₂ + O₂ → 2H₂O gives H₂:O₂ = 2:1

Stoichiometry in past-paper wording: command words that matter

Command word / phrase	What the question wants	Typical stoichiometry stem
Balance	Write balanced equation	”Balance the equation for the reaction of magnesium with oxygen.”
Calculate	Reacting mass or yield	”Calculate the mass of water formed from 4 g of hydrogen.”
Deduce	Work out from given data	”Deduce the mass of oxygen needed to react with 6 g of carbon.”
State the mole ratio	Coefficients from equation	”State the mole ratio of H₂ to O₂ in the reaction.”

Worked exam-style stems (how to answer the wording)

1. “Calculate the mass of magnesium oxide formed when 12 g of magnesium burns in oxygen. (Aᵣ: Mg = 24, O = 16)” Equation: 2Mg + O₂ → 2MgO. Mᵣ(MgO) = 40. n(Mg) = 12/24 = 0.5 mol. Ratio 2Mg:2MgO = 1:1, so n(MgO) = 0.5 mol. m(MgO) = 0.5 × 40 = 20 g. Mark-scheme reward: balanced equation, moles, ratio and final mass.
2. “In a reaction the theoretical yield is 10 g but the actual yield is 7 g. Calculate the percentage yield.” (7 ÷ 10) × 100 = 70%. Reward: correct formula and answer.
3. “Balance the equation: _Fe + _O₂ → _Fe₂O₃” 4Fe + 3O₂ → 2Fe₂O₃. Reward: correct coefficients with equal atoms on each side.

Test yourself with the Stoichiometry quiz once you can balance equations and calculate reacting masses without a formula sheet.

How stoichiometry connects to the rest of Coordinated Science

Stoichiometry builds on The Mole and Atomic Structure And The Periodic Table. It underpins titration calculations, electrolysis and industrial chemistry. The Cambridge IGCSE Coordinated Science resource hub links every Stoichiometry subtopic.

Common mistakes students make

• Using an unbalanced equation for mole ratios.
• Forgetting to convert mass → moles → mass (skipping the mole step).
• Using atomic mass instead of formula mass for compounds.
• Balancing by changing subscripts instead of coefficients.
• Calculating percentage yield as theoretical ÷ actual (should be actual ÷ theoretical).

When you need more support

If stoichiometry questions keep costing marks, work through the Stoichiometry quiz, then get focused help from a Cambridge IGCSE Coordinated Science tutor.

Frequently asked questions

Is stoichiometry hard in Coordinated Science? The method is consistent — balance, find moles, use ratio, convert back to mass. Marks are lost on unbalanced equations and skipped steps.

How do you balance a chemical equation? Adjust coefficients (numbers in front) so atoms of each element are equal on both sides; never change subscripts.

What is percentage yield? The percentage of the theoretical maximum product actually obtained: (actual yield ÷ theoretical yield) × 100%.

How do I revise stoichiometry effectively? Practise balancing five equations, work through reacting mass problems step by step, then take the Stoichiometry quiz.

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