← Topic 3
⚡ Challenge Paper Preparation

Challenge Prep: Stoichiometry

IGCSE Chemistry 0620 — Topic 3

Stoichiometry is where the maths meets the chemistry. Challenge papers don't just ask you to balance an equation or calculate a mass — they combine mole calculations with limiting reagents, percentage yield, gas volumes, and empirical formulae, often in a single multi-step question. The chemistry knowledge itself isn't harder, but the chain of reasoning is longer, and one small slip (forgetting to convert cm³ to dm³, or using the wrong mole ratio) can send the whole calculation off track. This guide shows you every trap examiners set, walks you through the reasoning step by step, and gives you practice questions that mirror exactly what you'll face on the day.

⚠️ Common Traps & Misconceptions

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Ten mistakes that cost students marks every single exam session. Learn the trap, understand the truth, and see how examiners exploit each one.

⚠️ TRAP
Trap 1: Forgetting to multiply atom counts by subscripts in formulae
The TrapWhen balancing equations or calculating Mr, students look at the elements present but forget to multiply by the subscript number. For example, in Ca(OH)2, they count 1 O and 1 H instead of 2 O and 2 H.
The TruthThe subscript outside a bracket multiplies everything inside the bracket. Ca(OH)2 means Ca + 2 × (O + H) = 1 Ca, 2 O, 2 H. Similarly, Mg(NO3)2 means 1 Mg, 2 N, and 6 O (because 2 × 3 = 6).
Why It MattersIf you miscount atoms, your Mr is wrong, which makes your moles wrong, which makes your reacting mass wrong — every calculation in the chain collapses from one small counting error at the start.
Example Question"Calculate the relative formula mass of aluminium sulfate, Al2(SO4)3." [Answer: 2(27) + 3(32) + 12(16) = 342 — not 2(27) + 1(32) + 4(16)]
⚠️ TRAP
Trap 2: Confusing empirical formula with molecular formula
The TrapStudents give the empirical formula when the question asks for the molecular formula, or vice versa. They calculate CH2O and write that as their final answer when the question says Mr = 180.
The TruthThe empirical formula is the simplest whole-number ratio of atoms (e.g. CH2O). The molecular formula is the actual number of atoms in one molecule. To get from empirical to molecular: divide the given Mr by the empirical formula mass, then multiply all subscripts by that factor. For CH2O: empirical mass = 12 + 2 + 16 = 30. Factor = 180 ÷ 30 = 6. Molecular formula = C6H12O6.
Why It MattersChallenge papers almost always give you Mr alongside the composition data and ask for the molecular formula. Stopping at the empirical formula loses you all the marks for the second step.
Example Question"A compound contains 40.0% C, 6.7% H, and 53.3% O by mass. Its Mr is 60. Determine the molecular formula."
⚠️ TRAP
Trap 3: Not converting cm³ to dm³ for concentration calculations
The TrapWhen using the formula concentration = moles ÷ volume, students plug in the volume in cm³ instead of dm³, giving an answer 1000 times too large (or too small).
The TruthConcentration in mol/dm³ requires volume in dm³. To convert: divide cm³ by 1000. So 250 cm³ = 0.250 dm³, and 25.0 cm³ = 0.0250 dm³. Alternatively, you can use the formula: moles = concentration × volume(cm³) ÷ 1000.
Why It MattersChallenge papers deliberately give volumes in cm³ — typically 25.0 cm³ titration volumes or 250 cm³ volumetric flask volumes — to test whether you convert properly. An answer that is 1000× too big is an instant zero.
Example Question"25.0 cm³ of sodium hydroxide solution of concentration 0.100 mol/dm³ is used. Calculate the number of moles of NaOH."
⚠️ TRAP
Trap 4: Using the wrong mole ratio from the balanced equation
The TrapStudents correctly calculate the moles of one substance but then assume a 1:1 ratio with the other substance, ignoring the actual coefficients in the balanced equation.
The TruthThe mole ratio comes from the balanced equation coefficients, not from the formula subscripts. In 2Mg + O2 → 2MgO, the ratio of Mg to O2 is 2:1 (not 1:1). So if you have 0.4 mol Mg, you need 0.2 mol O2.
Why It MattersChallenge papers sometimes give equations with coefficients like 2:5:4 or other non-obvious ratios. Writing the mole ratio explicitly on your working paper is the best way to avoid this trap.
Example Question"2Al + 3H2SO4 → Al2(SO4)3 + 3H2. Calculate the mass of aluminium needed to react with 14.7 g of sulfuric acid."
⚠️ TRAP
Trap 5: Confusing percentage yield with percentage purity
The TrapStudents mix up these two related-but-different concepts. They apply the yield formula when the question asks about purity, or vice versa.
The TruthPercentage yield compares actual product obtained to the theoretical (calculated) maximum: (actual mass ÷ theoretical mass) × 100%. Percentage purity tells you what fraction of a sample is the desired substance: (mass of pure substance ÷ total mass of sample) × 100%. They answer completely different questions.
Why It MattersChallenge papers often combine both in one question: "A 95% pure sample of limestone weighing 10 g is heated. If the yield is 85%, what mass of CaO is obtained?" You must apply purity first (to get the actual mass of CaCO3), then calculate the theoretical yield, then apply percentage yield.
Example Question"8.0 g of impure iron is reacted with excess hydrochloric acid. The iron is 70% pure. Calculate the mass of iron(II) chloride produced."
⚠️ TRAP
Trap 6: Using the wrong molar volume for gas calculations
The TrapStudents either forget the molar volume value altogether or use 22,400 cm³ (STP) when the question specifies room temperature and pressure (RTP), where the value is 24,000 cm³ (or 24 dm³).
The TruthAt room temperature and pressure (RTP), 1 mole of any gas occupies 24 dm³ (= 24,000 cm³). The formula is: volume = moles × 24 dm³ (or moles × 24,000 cm³). If the question says "at STP" instead, it will tell you the value to use.
Why It MattersChallenge papers will say "at room temperature and pressure" in the question stem. The molar volume of 24 dm³ is given on the data sheet, but under pressure many students forget to look it up or use the wrong one.
Example Question"Calculate the volume of carbon dioxide gas produced at RTP when 5.0 g of calcium carbonate reacts with excess hydrochloric acid."
⚠️ TRAP
Trap 7: Not identifying the limiting reagent
The TrapWhen both reactant masses are given, students pick one at random to base their calculation on, instead of checking which one runs out first (the limiting reagent).
The TruthThe limiting reagent is the reactant that is completely used up first — it determines the maximum amount of product. To find it: calculate the moles of each reactant, then compare using the mole ratio from the equation. The reactant that gives fewer moles of product is the limiting reagent.
Why It MattersChallenge papers deliberately give you both reactant masses and make one of them the excess reagent. If you base your calculation on the excess reagent instead of the limiting one, your answer will be too large.
Example Question"3.0 g of magnesium is added to 50.0 cm³ of 1.0 mol/dm³ hydrochloric acid. Calculate the volume of hydrogen gas produced at RTP."
⚠️ TRAP
Trap 8: Including spectator ions in ionic equations
The TrapStudents write the full formula equation instead of the net ionic equation, or they remove the wrong ions. They include sodium ions and chloride ions that don't actually participate in the reaction.
The TruthSpectator ions appear on both sides of the equation unchanged — they don't take part in the reaction. To write a net ionic equation: (1) write the full ionic equation showing all ions, (2) cancel any ion that appears identically on both sides, (3) what remains is the net ionic equation.
Why It MattersChallenge papers ask for ionic equations to test whether you understand what's actually reacting. For example, any acid + alkali neutralisation is simply H+(aq) + OH−(aq) → H2O(l), regardless of which acid or alkali is used.
Example Question"Write the ionic equation for the reaction between hydrochloric acid and sodium hydroxide solution."
⚠️ TRAP
Trap 9: Rounding mole calculations too early
The TrapStudents round their moles to 1 significant figure halfway through a multi-step calculation, and the rounding error compounds, giving a final answer that is noticeably wrong.
The TruthKeep at least 3 significant figures throughout every intermediate step. Only round your final answer to the number of significant figures appropriate for the data given (usually 3 s.f.). Premature rounding is a silent killer of marks.
Why It MattersChallenge papers with multi-step calculations (e.g. titration → moles → mole ratio → concentration → mass) accumulate rounding errors across 4 or 5 steps. If your final answer is outside the acceptable range because of rounding, you lose the final mark.
Example Question"A student calculated 0.004167 mol but rounded it to 0.004 mol before the next step. By the final answer, their result was 4% too low and fell outside the mark scheme range."
⚠️ TRAP
Trap 10: Confusing the formula of a diatomic element with its atom
The TrapWhen calculating gas volumes or writing equations, students write O instead of O2, or N instead of N2, or H instead of H2. This throws off the entire mole calculation because the Mr is halved.
The TruthSeven elements exist as diatomic molecules: H2, N2, O2, F2, Cl2, Br2, I2. Whenever these appear in an equation as elements (not in compounds), they must be written as X2. The Mr of oxygen gas is 32 (not 16), the Mr of hydrogen gas is 2 (not 1), etc.
Why It MattersChallenge papers might ask "Calculate the mass of 0.5 mol of chlorine gas." If you use 35.5 (atomic mass of Cl) instead of 71 (Mr of Cl2), your answer is exactly half the correct value.
Example Question"Calculate the volume of oxygen gas at RTP needed to completely burn 0.1 mol of methane. [CH4 + 2O2 → CO2 + 2H2O]"

🧩 Multi-Step Reasoning Walkthroughs

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Five challenging questions broken down step by step. Try each step yourself before revealing the next.

Walkthrough 1 — Reacting Masses with Limiting Reagent4.8 g of magnesium is added to 100 cm³ of 1.0 mol/dm³ sulfuric acid. Mg + H2SO4 → MgSO4 + H2. Calculate the maximum mass of magnesium sulfate produced.
1

Decode

Both reactant amounts are given — that's the signal for a limiting reagent question. You must find which reactant runs out first, then calculate the product from that reactant only.

2

Find moles

Moles of Mg = mass ÷ Mr = 4.8 ÷ 24 = 0.20 mol.

Moles of H2SO4 = concentration × volume(dm³) = 1.0 × (100 ÷ 1000) = 1.0 × 0.100 = 0.10 mol.

3

Compare using mole ratio

From the equation, the mole ratio is 1 Mg : 1 H2SO4. We have 0.20 mol Mg but only 0.10 mol acid. The acid runs out first — H2SO4 is the limiting reagent. There is excess magnesium left over.

4

Build the answer

From the equation: 1 mol H2SO4 → 1 mol MgSO4. So 0.10 mol acid produces 0.10 mol MgSO4.

Mass of MgSO4 = moles × Mr = 0.10 × (24 + 32 + 64) = 0.10 × 120 = 12.0 g.

Final AnswerMaximum mass of MgSO4 = 12.0 g.
Examiner's NoteIf you had used the moles of Mg (0.20) instead of the acid (0.10), you'd have got 24.0 g — exactly double the correct answer. Whenever both reactant amounts are given, always check for the limiting reagent.
Walkthrough 2 — Empirical to Molecular FormulaA compound contains 52.2% C, 13.0% H, and 34.8% O by mass. Its relative molecular mass is 46. Determine the molecular formula.
1

Decode

You're given percentage composition AND Mr. That means the answer requires TWO steps: first find the empirical formula, then scale it up to the molecular formula. Stopping at the empirical formula will lose marks.

2

Divide % by Ar

C: 52.2 ÷ 12 = 4.35
H: 13.0 ÷ 1 = 13.0
O: 34.8 ÷ 16 = 2.175

Divide by the smallest (2.175):
C: 4.35 ÷ 2.175 = 2
H: 13.0 ÷ 2.175 = 5.98 ≈ 6
O: 2.175 ÷ 2.175 = 1

Empirical formula: C2H6O

3

Compare masses

Empirical formula mass = 2(12) + 6(1) + 16 = 46.
Given Mr = 46. Factor = 46 ÷ 46 = 1.

So the molecular formula is the same as the empirical formula: C2H6O (ethanol).

4

Sanity check

Check: C2H6O has Mr = 46 ✔. Contains C, H, O ✔. Percentages: C = 24/46 × 100 = 52.2%, H = 6/46 × 100 = 13.0%, O = 16/46 × 100 = 34.8% ✔.

Final AnswerMolecular formula = C2H6O.
Examiner's NoteThe factor won't always be 1. If the Mr had been 92, the factor would be 2 and the molecular formula would be C4H12O2. Always complete this second step — the question is testing whether you can go beyond the empirical formula.
Walkthrough 3 — Titration Calculation25.0 cm³ of sodium hydroxide solution is neutralised by exactly 20.0 cm³ of 0.50 mol/dm³ hydrochloric acid. NaOH + HCl → NaCl + H2O. Calculate the concentration of the NaOH solution in mol/dm³.
1

Decode

This is a titration calculation. You know the acid's volume and concentration (so you can find its moles). You know the NaOH volume but not its concentration. The chain is: moles of acid → mole ratio → moles of NaOH → concentration of NaOH.

2

Moles of HCl

moles = concentration × volume(dm³) = 0.50 × (20.0 ÷ 1000) = 0.50 × 0.0200 = 0.0100 mol.

3

Mole ratio

From the equation: NaOH : HCl = 1 : 1. So moles of NaOH = 0.0100 mol.

4

Calculate concentration

concentration = moles ÷ volume(dm³) = 0.0100 ÷ (25.0 ÷ 1000) = 0.0100 ÷ 0.0250 = 0.40 mol/dm³.

Final AnswerConcentration of NaOH = 0.40 mol/dm³.
Examiner's NoteThe classic error is forgetting to convert cm³ to dm³. If you had left the volume as 25.0 instead of 0.0250, you'd get 0.0004 mol/dm³ — clearly absurdly dilute. Use the mental check: "Is my answer a reasonable concentration?" Concentrations in IGCSE problems are usually between 0.05 and 2.0 mol/dm³.
Walkthrough 4 — Percentage Yield with ImpurityA sample of limestone contains 80% calcium carbonate by mass. 25.0 g of this limestone is heated strongly. CaCO3 → CaO + CO2. The percentage yield of the reaction is 90%. Calculate the mass of calcium oxide obtained.
1

Decode

This is a three-layer question: (1) purity adjustment, (2) stoichiometry calculation, (3) yield adjustment. You must do them in order. Many students skip the purity step and use 25.0 g as pure CaCO3.

2

Actual mass of CaCO3

Mass of pure CaCO3 = 80% × 25.0 = 20.0 g.

3

Theoretical mass of CaO

Moles of CaCO3 = 20.0 ÷ 100 = 0.200 mol.
From equation: 1 mol CaCO3 → 1 mol CaO.
Theoretical moles of CaO = 0.200 mol.
Theoretical mass of CaO = 0.200 × 56 = 11.2 g.

4

Actual yield

Actual mass = 90% × 11.2 = 0.90 × 11.2 = 10.08 g ≈ 10.1 g (3 s.f.).

Final AnswerMass of CaO obtained = 10.1 g.
Examiner's NoteThe order matters: purity first, then stoichiometry, then yield. Confusing the order or skipping purity is worth losing 2-3 marks. Label each step clearly in your working — examiners award method marks even if the final number is slightly off.
Walkthrough 5 — Gas Volume from EquationWhat volume of hydrogen gas, measured at RTP, is produced when 0.54 g of aluminium reacts with excess hydrochloric acid? 2Al + 6HCl → 2AlCl3 + 3H2
1

Decode

Mass of solid → moles → mole ratio → moles of gas → volume. The trap is the 2:3 mole ratio (not 1:1) between Al and H2.

2

Moles

Moles of Al = 0.54 ÷ 27 = 0.020 mol.

3

Mole ratio

From the equation: 2 mol Al produces 3 mol H2.
So 0.020 mol Al produces (3/2) × 0.020 = 0.030 mol H2.

4

Gas volume at RTP

Volume = moles × 24,000 cm³ = 0.030 × 24,000 = 720 cm³.

(Or in dm³: 0.030 × 24 = 0.72 dm³.)

Final AnswerVolume of H2 at RTP = 720 cm³.
Examiner's NoteThe common error is using a 1:1 ratio (giving 480 cm³) or a 2:6 ratio by accident. Write the ratio explicitly from the equation: "2 mol Al : 3 mol H2." Also check your answer makes sense — 720 cm³ is about the volume of a large water bottle, which is physically reasonable for this amount of metal.

🔍 Spot the Difference

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Pairs of questions that look nearly identical but require different methods. Spot the crucial difference before revealing the answer.

Question A
Calculate the mass of CO2 produced when 10 g of CaCO3 reacts with excess HCl.
Moles CaCO3 = 10/100 = 0.1. Ratio 1:1. Mass CO2 = 0.1 × 44 = 4.4 g.
Question B
Calculate the volume of CO2 at RTP produced when 10 g of CaCO3 reacts with excess HCl.
Moles CaCO3 = 0.1. Ratio 1:1. Volume = 0.1 × 24,000 = 2400 cm³.
Key DifferenceSame reaction, same starting point. Question A asks for mass (multiply moles by Mr). Question B asks for volume (multiply moles by 24,000 cm³). Read the question word carefully: mass vs volume requires a completely different final step.
Question A
Find the empirical formula of a compound containing 75% carbon and 25% hydrogen by mass.
C: 75/12 = 6.25. H: 25/1 = 25. Ratio = 1:4. Empirical formula: CH4.
Question B
Find the molecular formula of a compound containing 75% carbon and 25% hydrogen. Mr = 16.
Empirical formula: CH4 (mass = 16). Factor = 16/16 = 1. Molecular formula: CH4.
Key DifferenceQuestion A stops at the empirical formula. Question B requires you to go further and find the molecular formula by dividing Mr by the empirical formula mass. In this case they happen to be the same, but that won't always be true — don't assume it.
Question A
Calculate the concentration in mol/dm³ of a solution containing 4.0 g of NaOH in 500 cm³.
Moles = 4.0/40 = 0.10. Volume = 0.500 dm³. Concentration = 0.10/0.500 = 0.20 mol/dm³.
Question B
Calculate the concentration in g/dm³ of a solution containing 4.0 g of NaOH in 500 cm³.
No need for moles! Concentration = 4.0/0.500 = 8.0 g/dm³.
Key DifferenceQuestion A asks for mol/dm³ (you must convert mass to moles first). Question B asks for g/dm³ (just divide mass by volume directly). Always check the units the question asks for — they determine your method.
Question A
5.0 g of zinc is added to excess dilute HCl. What mass of ZnCl2 is formed?
Excess acid = zinc is the limiting reagent. Use moles of Zn to calculate product.
Question B
5.0 g of zinc is added to 50 cm³ of 0.50 mol/dm³ HCl. What mass of ZnCl2 is formed?
Both amounts given = must find the limiting reagent by comparing moles before calculating.
Key DifferenceQuestion A says "excess" acid, so zinc is automatically the limiting reagent. Question B gives both amounts, forcing you to calculate moles of each and determine which runs out first. The word "excess" vs two given amounts changes the entire approach.
Question A
What is the percentage yield if 8.0 g of product was obtained when 10.0 g was expected?
Yield = (8.0/10.0) × 100 = 80%.
Question B
A sample weighs 10.0 g. Chemical analysis shows it contains 8.0 g of pure iron. What is the percentage purity?
Purity = (8.0/10.0) × 100 = 80%.
Key DifferenceBoth calculations give 80%, but they mean different things. Percentage yield compares actual product to theoretical maximum. Percentage purity compares pure substance to total sample mass. In multi-step problems, you apply them at different stages, and mixing them up changes the answer.

🔗 Stoichiometry Concept Map

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Click each node to see how the subtopics connect. Every challenge question is a chain of 2-4 of these concepts linked together.

⭐ CENTRAL HUB
The Mole — the connecting concept for ALL stoichiometry
Mass ↔ Moles ▶
Formula: moles = mass ÷ Mr
Connects to: Reacting masses, empirical formula, percentage composition
Challenge link: Used in every stoichiometry question as the first or last step. Always the starting point when given a mass.
Moles ↔ Gas Volume ▶
Formula: volume = moles × 24 dm³ (at RTP)
Connects to: Reacting volumes, gas collection experiments
Challenge link: Questions switch between cm³ and dm³ to test unit conversion. Gas volume replaces mass as the final step when the product is a gas.
Moles ↔ Concentration & Volume ▶
Formula: moles = concentration × volume (in dm³)
Connects to: Titrations, solution preparation, dilution
Challenge link: Titration questions are the most common multi-step type. They chain: known substance moles → mole ratio → unknown substance moles → concentration.
Mole Ratios (from balanced equations) ▶
Rule: Coefficients in a balanced equation give the mole ratio of reactants and products.
Connects to: Every reacting mass/volume calculation. Also connects to limiting reagent identification.
Challenge link: The ratio is the bridge between "substance A" and "substance B." Getting it wrong (e.g. using 1:1 when it should be 1:2) makes everything after it wrong.
Empirical & Molecular Formulae ▶
Method: % ÷ Ar → simplest ratio → empirical formula. Then scale using Mr.
Connects to: Percentage composition (reverse direction), combustion analysis
Challenge link: Challenge papers give unfamiliar compounds (vitamin C, aspirin) and expect you to apply the same method regardless of context.
Percentage Yield & Percentage Purity ▶
Yield: (actual ÷ theoretical) × 100%. Purity: (pure ÷ total) × 100%.
Connects to: Applied after (or before) a stoichiometry calculation to adjust for real-world conditions.
Challenge link: Combined questions: use purity to adjust the starting mass, then do stoichiometry, then apply yield to the final answer. Three layers of calculation in one question.
Limiting Reagent ▶
Method: Calculate moles of each reactant. Use mole ratio to find which one runs out first.
Connects to: Reacting masses, excess calculations
Challenge link: Signalled by giving BOTH reactant amounts (instead of one + "excess"). The product must be calculated from the limiting reagent only.

❌ "Why Is This Wrong?" Exercises

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Real-style student answers with hidden mistakes. Find the flaw yourself, then reveal the explanation.

Exercise 1: "Calculate the mass of water produced when 4.0 g of hydrogen reacts with excess oxygen. 2H2 + O2 → 2H2O"
Student's Answer"Moles of H = 4.0 ÷ 1 = 4.0 mol. Ratio 1:1 so moles of water = 4.0. Mass = 4.0 × 18 = 72 g."
The FlawThe student used the Ar of a single hydrogen atom (1) instead of the Mr of hydrogen gas (H2 = 2). Also, the mole ratio from the equation is 2:2 (i.e. 1:1 for H2:H2O), but because moles of H2 were calculated wrong, the final answer is double what it should be.
Correct AnswerMoles of H2 = 4.0 ÷ 2 = 2.0 mol. Ratio H2:H2O = 2:2 = 1:1. Moles of H2O = 2.0. Mass = 2.0 × 18 = 36 g.
Key RuleHydrogen exists as H2, not H. Always use Mr = 2 for hydrogen gas, Mr = 32 for oxygen gas, Mr = 28 for nitrogen gas.
Exercise 2: "25.0 cm³ of 0.200 mol/dm³ NaOH is used. Calculate the number of moles."
Student's Answer"Moles = 0.200 × 25.0 = 5.0 mol."
The FlawThe student multiplied concentration by the volume in cm³ instead of dm³. 25.0 cm³ = 0.0250 dm³. The answer is 1000 times too large.
Correct AnswerMoles = 0.200 × (25.0 ÷ 1000) = 0.200 × 0.0250 = 0.00500 mol.
Key RuleIf concentration is in mol/dm³, volume must be in dm³. Divide cm³ by 1000. An answer of 5.0 mol from 25 cm³ of dilute solution is physically absurd — always sense-check.
Exercise 3: "Calculate the Mr of calcium hydroxide, Ca(OH)2."
Student's Answer"Ca = 40, O = 16, H = 1. Mr = 40 + 16 + 1 = 57."
The FlawThe student ignored the subscript 2 outside the brackets. (OH)2 means there are two OH groups, so 2 oxygen atoms and 2 hydrogen atoms, not 1 of each.
Correct AnswerMr = 40 + 2(16) + 2(1) = 40 + 32 + 2 = 74.
Key RuleSubscripts outside brackets multiply everything inside. Write out the full atom count before adding: 1 Ca, 2 O, 2 H.
Exercise 4: "Determine the empirical formula of a compound containing 2.4 g of carbon and 0.8 g of hydrogen."
Student's Answer"C: 2.4 g, H: 0.8 g. Ratio = 2.4 : 0.8 = 3 : 1. Empirical formula: C3H."
The FlawThe student divided the masses directly instead of first converting to moles. You must divide each mass by the element's Ar before finding the ratio.
Correct AnswerC: 2.4 ÷ 12 = 0.2 mol. H: 0.8 ÷ 1 = 0.8 mol. Ratio: 0.2 : 0.8 = 1 : 4. Empirical formula: CH4.
Key RuleEmpirical formula uses the ratio of moles, not masses. Convert mass → moles first, then simplify.
Exercise 5: "2Mg + O2 → 2MgO. 2.4 g of Mg reacts with excess O2. Calculate the mass of MgO."
Student's Answer"Moles Mg = 2.4/24 = 0.1. Ratio Mg:MgO = 2:1 so moles MgO = 0.05. Mass = 0.05 × 40 = 2.0 g."
The FlawThe student read the ratio as 2 Mg : 1 MgO, but the equation shows 2 Mg : 2 MgO, which is 1:1. The "2" in front of MgO was overlooked.
Correct AnswerMoles Mg = 0.1 mol. Ratio Mg:MgO = 2:2 = 1:1. Moles MgO = 0.1. Mass = 0.1 × 40 = 4.0 g.
Key RuleRead ALL coefficients from the balanced equation carefully. Write out the ratio explicitly: "2 mol Mg : 2 mol MgO = 1:1" to avoid misreading.
Exercise 6: "A reaction has a theoretical yield of 20 g. The actual yield is 15 g. Calculate the percentage yield."
Student's Answer"Percentage yield = (20/15) × 100 = 133%."
The FlawThe student put the numbers in the wrong order — theoretical on top, actual on bottom. Percentage yield can never exceed 100%. If your answer is over 100%, the fraction is upside down.
Correct AnswerPercentage yield = (actual ÷ theoretical) × 100 = (15 ÷ 20) × 100 = 75%.
Key RuleActual goes on top, theoretical on the bottom. Think: "What fraction of the possible product did I actually get?"
Exercise 7: "Calculate the volume of CO2 at RTP produced from 0.25 mol CaCO3."
Student's Answer"Volume = 0.25 × 24 = 6.0 cm³."
The FlawThe student multiplied by 24 and wrote the unit as cm³. But 24 is the molar volume in dm³. The answer should be 6.0 dm³ OR they should multiply by 24,000 to get 6000 cm³.
Correct AnswerVolume = 0.25 × 24 = 6.0 dm³, or 0.25 × 24,000 = 6000 cm³.
Key RuleMolar volume = 24 dm³ = 24,000 cm³. Choose the right multiplier for the unit you want. Always state the unit clearly.
Exercise 8: "Write the ionic equation for zinc reacting with copper sulfate solution."
Student's Answer"Zn + CuSO4 → ZnSO4 + Cu"
The FlawThis is the full equation, not the ionic equation. The sulfate ion (SO42−) is a spectator ion — it appears on both sides unchanged and should be removed.
Correct AnswerZn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s). The SO42− ions are spectators.
Key RuleTo write an ionic equation: split soluble ionic compounds into ions, keep metals/gases/water/precipitates as whole formulae, then cancel ions that appear identically on both sides.

✍️ Ultra-Detailed Practice Questions

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Ten Cambridge-style challenge questions. Write your answer in the box, then reveal the model answer with full mark scheme and examiner's notes.

Question 1
[3 marks]
Calculate the relative formula mass (Mr) of ammonium sulfate, (NH4)2SO4. Show your working clearly. [Ar: N = 14, H = 1, S = 32, O = 16]
Model AnswerAtom count: 2 N, 8 H, 1 S, 4 O [1 mark for correct atom count]
Mr = 2(14) + 8(1) + 32 + 4(16) [1 mark for method]
= 28 + 8 + 32 + 64 = 132 [1 mark for correct answer]
Examiner's NotesThe (NH4)2 means everything in the bracket is multiplied by 2: 2 nitrogen atoms and 2 × 4 = 8 hydrogen atoms. The most common error is writing 2 H instead of 8 H, giving Mr = 128.
Question 2
[4 marks]
Riya's grandmother in Jaipur gives her a sample of baking soda (sodium hydrogen carbonate, NaHCO3). She heats 8.4 g of it: 2NaHCO3 → Na2CO3 + H2O + CO2. Calculate the mass of sodium carbonate (Na2CO3) produced. [Ar: Na = 23, H = 1, C = 12, O = 16]
Model AnswerMr of NaHCO3 = 23 + 1 + 12 + 48 = 84 [1]
Moles of NaHCO3 = 8.4 ÷ 84 = 0.10 mol [1]
Mole ratio: 2 mol NaHCO3 : 1 mol Na2CO3, so moles Na2CO3 = 0.10 ÷ 2 = 0.050 mol [1]
Mr of Na2CO3 = 2(23) + 12 + 3(16) = 106. Mass = 0.050 × 106 = 5.3 g [1]
Examiner's NotesThe critical step is the 2:1 ratio. Many students assume 1:1 and get 10.6 g. Write the ratio from the equation explicitly in your working. Also show your Mr calculation — method marks are awarded even if the arithmetic slips.
Question 3
[4 marks]
A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Its Mr is 60. Determine the molecular formula. [Ar: C = 12, H = 1, O = 16]
Model AnswerC: 40.0/12 = 3.33. H: 6.7/1 = 6.7. O: 53.3/16 = 3.33 [1]
Divide by smallest (3.33): C = 1, H = 2.01 ≈ 2, O = 1 [1]
Empirical formula: CH2O. Empirical mass = 12 + 2 + 16 = 30 [1]
Factor = 60 ÷ 30 = 2. Molecular formula: C2H4O2 (ethanoic acid) [1]
Examiner's NotesThe question asks for the molecular formula, not the empirical formula. Writing CH2O loses 2 marks. Always check whether Mr is given — if it is, you must go beyond the empirical formula.
Question 4
[4 marks]
25.0 cm³ of potassium hydroxide solution is exactly neutralised by 30.0 cm³ of 0.10 mol/dm³ hydrochloric acid. KOH + HCl → KCl + H2O. Calculate the concentration of KOH in (a) mol/dm³ and (b) g/dm³. [Ar: K = 39, O = 16, H = 1]
Model AnswerMoles HCl = 0.10 × 30.0/1000 = 0.00300 mol [1]
Ratio KOH:HCl = 1:1, so moles KOH = 0.00300 mol [1]
(a) Concentration = 0.00300 ÷ (25.0/1000) = 0.00300 ÷ 0.0250 = 0.12 mol/dm³ [1]
(b) Mr of KOH = 39 + 16 + 1 = 56. Concentration in g/dm³ = 0.12 × 56 = 6.72 g/dm³ [1]
Examiner's NotesPart (b) tests the conversion between mol/dm³ and g/dm³. Many students forget this link: concentration(g/dm³) = concentration(mol/dm³) × Mr. Both volume conversions (cm³ to dm³) are tested here.
Question 5
[4 marks]
3.0 g of magnesium ribbon is added to 100 cm³ of 0.50 mol/dm³ hydrochloric acid. Mg + 2HCl → MgCl2 + H2. Calculate the volume of hydrogen gas produced at RTP. [Ar: Mg = 24]
Model AnswerMoles Mg = 3.0/24 = 0.125 mol. Moles HCl = 0.50 × 0.100 = 0.050 mol [1]
From equation: 1 mol Mg needs 2 mol HCl. For 0.125 mol Mg, we'd need 0.250 mol HCl — but we only have 0.050. So HCl is the limiting reagent. [1]
From equation: 2 mol HCl → 1 mol H2. So 0.050 mol HCl → 0.025 mol H2. [1]
Volume = 0.025 × 24,000 = 600 cm³ [1]
Examiner's NotesThe trap is using Mg to calculate — that gives 3000 cm³, which is 5 times too large. Both reactant amounts are given, so you MUST find the limiting reagent. The HCl runs out long before the Mg is used up.
Question 6
[3 marks]
A sample of impure iron ore contains 70% Fe2O3 by mass. Calculate the mass of iron that can be extracted from 500 g of the ore. Fe2O3 + 3CO → 2Fe + 3CO2. [Ar: Fe = 56, O = 16]
Model AnswerMass of Fe2O3 = 70% × 500 = 350 g [1]
Mr Fe2O3 = 2(56) + 3(16) = 160. Moles = 350/160 = 2.1875 mol [1]
Ratio: 1 mol Fe2O3 : 2 mol Fe. Moles Fe = 2 × 2.1875 = 4.375. Mass Fe = 4.375 × 56 = 245 g [1]
Examiner's NotesTwo layers here: (1) apply purity first, (2) then do stoichiometry with the 1:2 ratio. If you skip purity and use 500 g directly, you get 350 g — too high. If you use a 1:1 ratio instead of 1:2, you get half the correct answer.
Question 7
[3 marks]
Write the ionic equation for the reaction between dilute sulfuric acid and sodium hydroxide solution. Include state symbols.
Model AnswerFull equation: H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l) [1]
Full ionic: 2H+(aq) + SO42−(aq) + 2Na+(aq) + 2OH−(aq) → 2Na+(aq) + SO42−(aq) + 2H2O(l)
Cancel spectators (Na+ and SO42−):
H+(aq) + OH−(aq) → H2O(l) [2]
Examiner's NotesALL acid-alkali neutralisation ionic equations simplify to H+ + OH− → H2O. State symbols are required for full marks. Many students lose a mark by omitting (aq) and (l).
Question 8
[4 marks]
In a school experiment in Mumbai, Arun reacted 6.5 g of zinc with excess dilute sulfuric acid. He collected 1800 cm³ of hydrogen gas at RTP. Zn + H2SO4 → ZnSO4 + H2. Calculate the percentage yield. [Ar: Zn = 65]
Model AnswerMoles Zn = 6.5/65 = 0.10 mol [1]
Ratio Zn:H2 = 1:1, so theoretical moles H2 = 0.10 mol [1]
Theoretical volume = 0.10 × 24,000 = 2400 cm³. Actual volume = 1800 cm³ [1]
Percentage yield = (1800/2400) × 100 = 75% [1]
Examiner's NotesYou can calculate percentage yield using volumes just as easily as masses — the formula works the same way as long as both values use the same unit. The yield is less than 100% because some gas dissolves in the water or escapes before being collected.
Question 9
[3 marks]
A student burns 0.92 g of ethanol (C2H5OH) completely in oxygen. C2H5OH + 3O2 → 2CO2 + 3H2O. Calculate the volume of carbon dioxide gas produced at RTP. [Ar: C = 12, H = 1, O = 16]
Model AnswerMr of C2H5OH = 2(12) + 6(1) + 16 = 46. Moles = 0.92/46 = 0.020 mol [1]
Ratio C2H5OH : CO2 = 1 : 2, so moles CO2 = 0.040 mol [1]
Volume = 0.040 × 24,000 = 960 cm³ [1]
Examiner's NotesCommon errors: (1) counting H atoms wrong in C2H5OH (it's 6 H, not 5 — the OH group contains an extra H). (2) Using a 1:1 ratio instead of 1:2. (3) Forgetting to convert moles to volume at the end.
Question 10
[5 marks]
Priya has a 250 cm³ volumetric flask. She wants to prepare a solution of sodium carbonate (Na2CO3) with a concentration of 0.20 mol/dm³. (a) Calculate the mass of Na2CO3 she needs. (b) She dissolves this mass in water. 25.0 cm³ of this solution exactly neutralises 50.0 cm³ of hydrochloric acid. Na2CO3 + 2HCl → 2NaCl + H2O + CO2. Calculate the concentration of the HCl. [Ar: Na = 23, C = 12, O = 16]
Model Answer(a) Moles needed = 0.20 × (250/1000) = 0.20 × 0.250 = 0.050 mol [1]
Mr = 2(23) + 12 + 3(16) = 106. Mass = 0.050 × 106 = 5.30 g [1]

(b) Moles Na2CO3 in 25.0 cm³ = 0.20 × 0.0250 = 0.0050 mol [1]
Ratio Na2CO3 : HCl = 1 : 2, so moles HCl = 0.010 mol [1]
Concentration HCl = 0.010 ÷ 0.050 = 0.20 mol/dm³ [1]
Examiner's NotesThis is a classic 5-mark structured question. Part (a) tests solution preparation. Part (b) chains titration calculation with the 1:2 mole ratio. The trap is using 250 cm³ (the whole flask) instead of 25.0 cm³ (the portion actually titrated) in part (b).
IGCSE Chemistry 0620 — Topic 3 Challenge Prep Guide. Work through every section, revisit the traps the night before your exam.