Stoichiometry: IGCSE Chemistry Exam Guide

Stoichiometry carries more marks across 0620 than any other single topic. Mole calculations appear on Paper 2 (MCQ), Paper 4 (structured and free-response), and they feed into questions on acids, electrolysis, energetics and organic chemistry. A student who cannot do moles reliably loses marks in at least four different topic areas. The detailed calculation technique is in mole calculations technique; this guide covers the exam strategy for the whole stoichiometry section.

The four calculation types and how to spot them

Every stoichiometry calculation on 0620 falls into one of four types. Recognising which type you are in tells you which formula to start with.

1. Reacting masses

Signal words: “Calculate the mass of… produced from… g of…”

Method:

1. Write the balanced equation.
2. Calculate Mr of the substance you know the mass of.
3. Find moles of that substance: moles = mass / Mr.
4. Use the mole ratio from the equation to find moles of the target substance.
5. Calculate the mass: mass = moles x Mr of the target.

2. Gas volume calculations

Signal words: “Calculate the volume of gas produced at RTP.”

Method:

1. Find moles of the gas (from a reacting mass or mole ratio).
2. Volume = moles x 24 dm3 (at room temperature and pressure).
3. Convert to cm3 if the question asks for it (x 1000).

The 24 dm3 value is provided on the paper, but you must know to use it and when.

3. Concentration and titration calculations

Signal words: “concentration”, “titre”, “mol/dm3”, “g/dm3”.

Method:

1. moles = concentration x volume (volume in dm3, not cm3).
2. Use the mole ratio to find moles of the other substance.
3. Calculate the concentration of the unknown.

The most common unit error: the volume from a burette is in cm3, but the formula needs dm3. Divide by 1000 before substituting.

4. Empirical and molecular formula

Signal words: “Determine the empirical formula”, “the compound contains…% by mass”.

Method:

1. Convert percentages to masses (assume 100 g total).
2. Divide each mass by the Ar of that element to get moles.
3. Divide all mole values by the smallest to get the simplest ratio.
4. For molecular formula, divide the actual Mr by the empirical formula Mr and multiply through.

The non-negotiable rule: show every step

Examiners award method marks at each step. A wrong final answer with correct working can still earn 3 out of 4 marks. A correct final answer with no working earns 1 mark and, if you copied it wrong from the calculator, zero. The guide on showing working in calculations covers this in detail.

Label each line:

• “Mr of CaCO3 = 40 + 12 + (16 x 3) = 100”
• “moles of CaCO3 = 5.0 / 100 = 0.05 mol”
• “from the equation, 1 mol CaCO3 gives 1 mol CaO”
• “mass of CaO = 0.05 x 56 = 2.8 g”

Each labelled line is a potential mark. Each unlabelled jump is a potential zero.

Mole ratios: the step students skip

The balanced equation tells you the ratio in which substances react. In CaCO3 -> CaO + CO2, the ratio is 1:1:1. In 2Mg + O2 -> 2MgO, the ratio is 2:1:2.

The most common error is assuming a 1:1 ratio without checking. When the question gives you N2 + 3H2 -> 2NH3 and asks for the volume of hydrogen needed for a given volume of nitrogen, the answer is three times the nitrogen volume, not equal to it. Read the coefficients.

Unit traps the exam sets

Every series, the mark scheme shows examiners deducting marks for unit errors. The three most frequent:

1. cm3 vs dm3. Titration volumes come in cm3. The concentration formula uses dm3. Divide by 1000.
2. g/dm3 vs mol/dm3. The question specifies which unit. If it asks for g/dm3, multiply moles by Mr after finding concentration in mol/dm3.
3. Missing units on the final answer. A number without a unit is incomplete. Write “2.8 g”, not “2.8”.

Percentage yield and atom economy (Extended only)

These are definition-and-substitute questions. The formulas:

• Percentage yield = (actual yield / theoretical yield) x 100
• Atom economy = (Mr of desired product / total Mr of all products) x 100

The exam gives you the numbers; the marks are for correct substitution and the percentage sign. The conceptual follow-up: “Suggest why the percentage yield is less than 100%” expects answers like incomplete reaction, loss during transfer, side reactions — not “the experiment went wrong”.

Worked exam question

Q (Paper 4): 6.5 g of zinc reacts with excess hydrochloric acid. Zn + 2HCl -> ZnCl2 + H2 Calculate the volume of hydrogen gas produced at RTP. (Ar: Zn = 65, H = 1, Cl = 35.5) (3 marks)

Model answer:

1. Moles of Zn = 6.5 / 65 = 0.10 mol (1)
2. From the equation, 1 mol Zn produces 1 mol H2, so moles of H2 = 0.10 mol (1)
3. Volume of H2 = 0.10 x 24 = 2.4 dm3 (1)

Three steps, three marks. Omit step 2 and the examiner cannot see that you used the ratio, even if your answer is correct.

Exam strategy for stoichiometry

• Read the question twice before writing anything. Identify: what substance do I know? What substance do I need? What is the link (equation ratio)?
• Write the balanced equation even if it is given. Check the coefficients against your calculation.
• Circle the units the question asks for. If it says cm3, convert at the end. If it says g/dm3, do not stop at mol/dm3.
• Estimate before calculating. If 100 g of CaCO3 gives 56 g of CaO (1:1 ratio, Mr 100 and 56), then 5 g should give roughly 2.8 g. If your calculator says 28 g, you moved a decimal place.

Stoichiometry is the topic that most reliably separates A/A* students from B/C students on 0620. If mole calculations are a consistent weak point, targeted practice with feedback is the fastest fix. A trial lesson focused on your specific calculation errors can diagnose the exact step where marks are being lost.