IGCSE Chemistry moles and stoichiometry: diagnose errors step by step

Where does the first stoichiometry error begin?

An incorrect final answer is often only a symptom. The actual error may be that:

• the chemical equation is not balanced;
• masses are treated directly in the ratio of the coefficients;
• the starting quantity is not converted to moles;
• a volume remains in cm³ while concentration is in mol/dm³;
• a relationship is applied to the wrong substance;
• rounding happens too early;
• a chemically impossible result is not checked.

The diagnostic question is: which was the first step after which later, mathematically correct operations could no longer produce the correct answer?

Five decisions before every calculation

1. 1What is required?
2. 2What is given?
3. 3Is an equation ratio needed?
4. 4Which quantity relationship is needed?
5. 5Is the result reasonable?

1. What is required? Mass, amount of substance, concentration, volume, yield or formula?
2. What is given? Write it with the quantity name and unit.
3. Is an equation ratio needed? Usually yes when moving between two different substances.
4. Which relationship is needed? For example n = m / Mr, n = cV, or the gas-quantity relationship specified by the syllabus.
5. Is it reasonable? Check the unit, order of magnitude and equation ratio.

If these five decisions cannot be named, do not start with the calculator. In Paper 1/2 MCQs, distractors exploit precisely these wrong decisions.

A relationship map of chemical quantities

Mass
m ÷ M_rAmount
n / molSolution
c × VGas volume
V ÷ molar volumeNumber of particles
N ÷ N_A

Amount of substance, n, is the central bridge. Convert mass with m / Mr and a solution with c × V, where V is in dm³. The coefficients in a balanced equation connect the moles of two substances. From the moles of the required substance, move back to mass, concentration or volume.

Do not draw the entire map automatically for every question. Mark the shortest valid route—for example: mass A → moles A → equation ratio → moles B → mass B.

A balanced equation gives a mole ratio, not a mass ratio

2H₂+O₂→2H₂O

2 mol:1 mol:2 mol

2H2 + O2 → 2H2O means that 2 mol of hydrogen reacts with 1 mol of oxygen to produce 2 mol of water under ideal stoichiometric conditions. It does not mean that 2 g of hydrogen requires 1 g of oxygen. Relative molecular masses are needed before working with mass.

A useful visible line is:

n(starting) × required coefficient / starting coefficient = n(required)

This is not a new formula but a visible version of the ratio. Always label the substances by name or formula so they are not reversed.

Complete worked example: calcium carbonate and hydrochloric acid

Worked example

What mass of carbon dioxide forms when 5.00 g of pure CaCO₃ reacts completely?

1. Balanced equation

CaCO3 + 2HCl → CaCl2 + H2O + CO2

2. Starting amount of substance

Mr(CaCO3) = 40 + 12 + 3 × 16 = 100

n(CaCO3) = m / Mr = 5.00 g / 100 g mol−1 = 0.0500 mol

3. Equation ratio

CaCO₃ : CO₂ = 1 : 1, so n(CO2) = 0.0500 mol.

4. Required mass

Mr(CO2) = 12 + 2 × 16 = 44

m(CO2) = n × Mr = 0.0500 mol × 44 g mol−1 = 2.20 g

5. Check

The product mass is smaller than the 5.00 g of starting solid, which is possible because only part of the CaCO₃ mass enters CO₂; the remaining atoms enter other products. The unit is grams and the 1:1 ratio was used.

If the question also specified an 80% percentage yield, calculate actual yield after the theoretical 2.20 g: 2.20 × 0.80 = 1.76 g. Do not apply yield randomly before the mole ratio; name which quantity is theoretical and which is actual.

Solution calculations: volume units are the common breaking point

If c is in mol/dm³ and V in cm³, then V(dm3) = V(cm3) / 1000. For example, the amount of NaOH in 25.0 cm³ of 0.200 mol/dm³ solution is:

V = 25.0 / 1000 = 0.0250 dm3

n = cV = 0.200 mol dm−3 × 0.0250 dm3 = 0.00500 mol

If the question now asks about another substance, use the equation ratio next. If it instead asks for the mass concentration or number of particles in this solution, a different final conversion is needed.

In a titration-style question, write the two solutions separately. c1V1 = c2V2 is not a universal reaction formula: it applies directly only to a 1:1 stoichiometric relationship. Other coefficients require the mole ratio as well.

Gas quantities, limiting reactants and yield at Extended level

An Extended question may combine several bridges:

• gas volume ↔ amount of substance: use the molar gas volume and conditions specified in the syllabus or question;
• limiting reactant: calculate possible product from both reactants; the smaller quantity limits;
• percentage yield: actual / theoretical × 100%;
• percentage purity: only the mass of the pure component enters the mole calculation;
• empirical formula: mass or percentage → moles → divide by the smallest → whole-number ratio.

Do not confuse yield and purity. Purity identifies the reacting portion of the starting sample; yield compares product actually obtained with the theoretical amount.

The first wrong step: targeted repair from the symptom

Cannot start↓

Name the starting and required quantities.

Wrong mole ratio↓

Balance the equation and label its coefficients.

Factor-of-1000 error↓

Check the cm³ ↔ dm³ conversion.

Correct number, wrong unit↓

Write the unit during substitution.

Do not keep only the final result in the error log. A useful entry is: “The equation was balanced, but I reversed the 2:1 ratio; next task: three ratio-transfer questions without a calculator.” A weak entry is: “I cannot do moles.”

If the calculation is correct but a Paper 3/4 mark is lost, the issue may be communication: a missing unit, premature rounding, an invisible ratio or failure to name the required substance. The structured-answer guide treats this separately.

A five-part reasonableness check

✓ Sign✓ Order of magnitude✓ Unit✓ Ratio✓ Physical possibility

• Sign: a negative mass or amount of substance is impossible in this context.
• Order of magnitude: hundreds of moles from 25 cm³ of dilute solution usually indicate a unit error.
• Unit: the question asks for grams but the result remains in moles? A conversion back is missing.
• Ratio: with a 1:2 equation ratio, did the mole value change in the correct direction?
• Physical possibility: percentage yield is 140%? A real product might be impure or wet, but in an examination calculation recheck the data and definition.

Independent mini-task

2Al + 3Cl2 → 2AlCl3

What mass of aluminium chloride can theoretically form from the complete reaction of 0.120 mol chlorine when aluminium is in excess? Use Ar(Al) = 27.0, Ar(Cl) = 35.5.

Answer and diagnostic points

Cl₂ : AlCl₃ = 3 : 2, so n(AlCl3) = 0.120 × 2/3 = 0.0800 mol. Mr(AlCl3) = 27.0 + 3 × 35.5 = 133.5. Therefore m = 0.0800 × 133.5 = 10.68 g, appropriately reported as 10.7 g, for example. A result of 0.180 mol reverses the ratio; a mass near 16.0 g probably omits the 3:2 conversion.

Practice system: mix decision errors, not just quantity types

A useful 30-minute block contains:

1. two short mass–mole questions;
2. one solution and unit conversion;
3. one two-substance equation ratio;
4. one Extended addition if it belongs to the student’s route;
5. correction back to the first wrong step;
6. a new similar question 48–72 hours later.

Do not complete ten identical substitutions in a row. The examination decision is selecting the necessary relationship. Stoichiometry appears early in the 8-week plan, then returns in every paper format.