Mole Calculations

Every mole calculation follows the same path: convert what you know into moles, use the mole ratio from the balanced equation, then convert back into what you want. Gases: $n=\frac{V}{24}$ ( $\mathrm{dm^3}$ at room temperature and pressure). Solutions: $n=c\times V$ ( $V$ in $\mathrm{dm^3}$ ). Percentage yield $=\frac{\text{actual}}{\text{theoretical}}\times100$ .

Cambridge 5070 Paper 1Cambridge 5070 Paper 2Cambridge 0620 Papers 2, 4Cambridge 0620 Papers 1, 3Edexcel 4CH1 Paper 1CEdexcel 4CH1 Paper 2C≈ 50 minRevision 1 · updated 2026-10-30

Section 1 of 14

Learning objectives

Calculate reacting masses using mole ratios.
Use the molar gas volume.
Calculate concentrations and do titration calculations.
Calculate percentage yield and purity, and identify a limiting reactant.

Pause and think

A recipe needs 2 eggs per cake. You have 7 eggs and flour for 5 cakes. How many cakes can you make?

Section 2 of 14

Core idea

Start hereRecipes for factories

A fertiliser plant needs to know exactly how many tonnes of ammonia it will get from its hydrogen. Too little hydrogen wastes nitrogen; too much wastes money. Mole calculations are the recipe book.

The big idea

Turn every given quantity into moles, use the ratio from the balanced equation, then convert back. Gases: $n=V/24$ . Solutions: $n=cV$ .

1. Reacting masses

Worked exampleIron from iron(III) oxide

$\mathrm{Fe_2O_3+3CO\rightarrow2Fe+3CO_2}$ . What mass of iron forms from $320$ g of $\mathrm{Fe_2O_3}$ ? ( $M_r=160$ ; Fe 56)

Goal: Mass of Fe.
Strategy: Moles → ratio → mass.

$n(\mathrm{Fe_2O_3})=\frac{320}{160}=2.0$
Mass to moles.
$n(\mathrm{Fe})=2\times2.0=4.0$
Ratio $1:2$ from the equation.
$m=4.0\times56=224\text{ g}$
Moles to mass.

Answer

224

✓ Check

\mathrm{Fe_2O_3}

\frac{112}{160}=70\%

iron, and

70\%

320=224

✓.

2. Gases and solutions

$1$ mol of gas $=24\ \mathrm{dm^3}=24\,000\ \mathrm{cm^3}$ at r.t.p.
$c=\frac{n}{V}$ with $V$ in $\mathrm{dm^3}$ .
$\mathrm{g/dm^3}=\mathrm{mol/dm^3}\times M_r$ .

Worked exampleTitration

$25.0\ \mathrm{cm^3}$ of $\mathrm{Na_2CO_3}$ needs $20.0\ \mathrm{cm^3}$ of $0.150\ \mathrm{mol/dm^3}$ HCl. $\mathrm{Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2}$

Goal: $c(\mathrm{Na_2CO_3})$ .
Strategy: Moles of the known solution, ratio, divide by volume.

$n(\mathrm{HCl})=0.150\times0.0200=0.00300$
$n=cV$ with $V$ in dm³.
$n(\mathrm{Na_2CO_3})=0.00150$
Ratio $1:2$ .
$c=\frac{0.00150}{0.0250}=0.0600\ \mathrm{mol/dm^3}$
Divide by volume in dm³.

Answer

0.0600\ \mathrm{mol/dm^3}

✓ CheckSmaller than the acid concentration, as expected since it needs twice as much acid per mole ✓.

3. Yield and limiting reactants

Why yields fall short

Drawing diagram…

Actual yield is almost always less than theoretical.

Exam tipLimiting reactant Work out moles of each reactant, divide by its coefficient: the smaller result is the limiting reactant, and all product amounts are based on it.

Pause and think

A student gets 105% yield. What went wrong?

Summary

Moles → ratio → answer.
$V=24n$ ; $n=cV$ .
Yield = actual ÷ theoretical × 100.

Limiting reactant: The reactant that is used up first.
Concentration: Amount of solute per dm³ of solution.
Molar gas volume: 24 dm³ per mole at room temperature and pressure.

HistoryGay-Lussac. In 1808 Joseph Gay-Lussac noticed gases react in simple whole-number volume ratios, like 2 volumes of hydrogen with 1 of oxygen: the first clue to the mole ratio.

Fun factFactory yields. A drug made in 10 steps with 90% yield at each step ends up with only about 35% overall yield, which is one reason medicines can be expensive.

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