Cambridge AS Level Chemistry 9701

Moles and stoichiometry

One mole is the amount of substance containing $6.02 \times 10^{23}$ particles — Avogadro’s constant — and every stoichiometry calculation is the same three steps: convert what you are given into moles, use the balanced equation’s ratio to get moles of what you want, convert back into the unit the question asks for.

In Cambridge AS Chemistry 9701 this is syllabus topic 2, and it carries 76 of the 920 Paper 1 questions Quanta has mapped — 8.3%, the second-largest topic on the paper. Its true weight is higher still, because the calculations in energetics, equilibria and electrochemistry all run through it.

Updated 22 September 2026

The four conversions

Everything on this topic is one of four routes into moles. Learn them as a set, because a question tells you which one it wants by the unit it gives you:

Given	Moles	Notes
Mass of a solid	n = m ÷ M	M is the molar mass in g mol⁻¹, from the Periodic Table you are given.
Volume and concentration of a solution	n = c × V	V in dm³. If the question gives cm³, divide by 1000 first.
Volume of a gas	n = V ÷ 24.0 dm³	At room temperature and pressure. Use the ideal gas equation at any other condition.
Number of particles	n = N ÷ 6.02 × 10²³	Watch whether the question counts molecules, atoms or ions.

For gases outside room conditions, the ideal gas equation replaces the 24.0 dm³:

p V = n R T R = 8.31 J K^{- 1} m o l^{- 1}

Its units are the whole difficulty: $p$ in pascals, $V$ in cubic metres (so cm³ ÷ 10⁶, dm³ ÷ 10³) and $T$ in kelvin (°C + 273). A question that gives kPa and dm³ is testing the conversion, not the rearrangement.

The three-step method

Into moles. Whatever you are given — grams, cm³ of solution, dm³ of gas — convert it with the matching formula above.
Across the equation. Multiply by the ratio of coefficients from the balanced equation. This is the only step where the chemistry enters, and the only step where an unbalanced equation quietly ruins the answer.
Out of moles. Convert into the unit asked for, using the same formulae in reverse.

Write the three lines separately even in a multiple-choice paper. Every distractor on a Paper 1 stoichiometry question is generated by getting exactly one of them wrong — the ratio inverted, a factor of 1000 missing, the molar mass of the wrong species.

Worked example

25.0 cm³ of 0.100 mol dm⁻³ sulfuric acid reacts completely with magnesium carbonate. Calculate the mass of MgCO₃ that reacts and the volume of CO₂ produced at room temperature and pressure.

Equation, balanced first:

H_{2} S O_{4} + M g C O_{3} \to M g S O_{4} + H_{2} O + C O_{2}

Step 1 — into moles. $V = 25.0 \div 1000 = 0.0250 d m^{3}$ , so

n (H_{2} S O_{4}) = 0.100 \times 0.0250 = 2.50 \times 10^{- 3} m o l

Step 2 — across the equation. The ratio is 1 : 1 : 1, so $n (M g C O_{3}) = n (C O_{2}) = 2.50 \times 10^{- 3}$ mol.

Step 3 — out of moles. $M (M g C O_{3}) = 24.3 + 12.0 + 48.0 = 84.3 g m o l^{- 1}$ :

m = n M = 2.50 \times 10^{- 3} \times 84.3 = 0.211 g

V (C O_{2}) = 2.50 \times 10^{- 3} \times 24.0 = 0.0600 d m^{3} = 60.0 c m^{3}

Both answers to three significant figures, matching the data given. The trap here is the 1 : 1 ratio looking too easy — swap the carbonate for a Group 1 carbonate such as $N a_{2} C O_{3}$ and the acid ratio becomes 1 : 1 but $H C l$ would be 2 : 1. Balance before you calculate, every time.

Limiting reagent and percentage yield

When a question gives you quantities of two reactants, it is a limiting-reagent question. Find the moles of each, divide each by its coefficient in the balanced equation, and the smaller result is the limiting reagent — the one that runs out first and therefore fixes the yield. The other is in excess, and the amount left over is a common follow-up.

percentage yield = \frac{actual yield}{theoretical yield} \times 100

Both yields must be in the same unit — moles or grams, not one of each. A yield below 100% is normal: side reactions, incomplete reactions and losses on transfer and purification all account for it, and questions ask you to name one.

Atom economy is a different measure and is often confused with yield. It compares the mass of the desired product with the total mass of all products, so it is fixed by the equation itself and would be the same even for a perfect reaction:

atom economy = \frac{M_{r} of desired product}{total M_{r} of products} \times 100

Empirical and molecular formulae

The empirical formula is the simplest whole-number ratio of atoms; the molecular formula is the actual number in one molecule, always a whole-number multiple of it.

From percentage composition or masses

Take the percentage (or mass) of each element.
Divide each by that element’s relative atomic mass.
Divide every result by the smallest of them.
Scale to whole numbers if you are left with a .5 or a .33 — multiply through by 2 or 3 rather than rounding.

Then, for the molecular formula, divide the compound’s $M_{r}$ by the empirical formula’s mass and multiply the subscripts by that factor. A compound with empirical formula $C H_{2} O$ (mass 30.0) and $M_{r} = 180$ is $180 \div 30 = 6$ , so $C_{6} H_{12} O_{6}$ .

Titration calculations

A titration gives you the concentration of one solution from a known concentration of another. The calculation is the three-step method with $n = c V$ at both ends:

c_{1} V_{1} \times \frac{{ratio}_{2}}{{ratio}_{1}} = c_{2} V_{2}

Use the mean titre of the concordant results — the ones within 0.10 cm³ of each other — and never include a rough titre in the mean. Questions frequently hide the ratio: a diprotic acid such as $H_{2} S O_{4}$ neutralises twice its moles of $N a O H$ , so forgetting the 2 halves or doubles the answer, and both of those values will be among the options.

Common mistakes

1.Using an unbalanced equation
Step 2 is the ratio of coefficients. Balance first, and check it by counting atoms on both sides — every distractor in the option list is a plausible wrong ratio.
2.cm³ used where dm³ is needed
$n = c V$ needs $V$ in dm³: divide cm³ by 1000. This single factor generates the “×1000 out” distractor on nearly every solution question.
3.Molar mass of the wrong species
Calculate $M$ for the substance you are converting, not the one in the question stem. Water of crystallisation counts: $C u S O_{4} \cdot 5 H_{2} O$ is 249.6, not 159.6.
4.Rounding the empirical ratio
A ratio of 1 : 1.5 is 2 : 3, not 1 : 2. Multiply up to whole numbers instead of rounding down.
5.Ideal gas equation in the wrong units
Pa, m³ and K. kPa needs ×1000, dm³ needs ÷1000 and °C needs +273 — miss any one and the answer is out by a power of ten.
6.Assuming the reactant given is the limiting one
When two quantities are given, test both: moles ÷ coefficient, smaller wins. The excess reactant never controls the yield.

Common questions

What is a mole in chemistry?

The amount of substance that contains $6.02 \times 10^{23}$ particles — Avogadro’s constant. One mole of any substance has a mass in grams equal to its relative formula mass, which is what makes $n = m / M$ work.

How do you find the limiting reagent?

Convert each reactant to moles, divide each by its coefficient in the balanced equation, and the smallest value is limiting. That reactant is used up first, so it decides the maximum yield; the other is in excess.

What is the difference between empirical and molecular formula?

The empirical formula is the simplest whole-number ratio of atoms; the molecular formula is the actual number of each atom in a molecule. Divide the molecular mass by the empirical formula mass to get the multiplier between them — $C H_{2} O$ with $M_{r} = 180$ is $C_{6} H_{12} O_{6}$ .

What is the molar gas volume?

24.0 dm³ per mole at room temperature and pressure, for any gas. Outside those conditions use $p V = n R T$ with pressure in pascals, volume in cubic metres and temperature in kelvin.

How much of 9701 Paper 1 is stoichiometry?

76 of the 920 questions Quanta has mapped — 8.3% — are filed under topic 2 directly. Many more depend on it, since the calculations in energetics, equilibria and electrochemistry all begin by converting something into moles. See the full Paper 1 topic weighting →

Practise moles and stoichiometry against real mark schemes

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