Sec 3 Chemistry Mole & Stoichiometry Notes

Overview

What you’ll learn

Calculate relative molecular and formula masses from relative atomic masses, including brackets and water of crystallisation.
Use the mole: moles = mass ÷ molar mass, and state the Avogadro constant in particle conversions.
Calculate percentage composition by mass and use experimental masses to find empirical and molecular formulae.
Convert between moles, concentration in mol/dm³, concentration in g/dm³ and solution volume.
Use the molar gas volume of 24 dm³ at r.t.p. to convert between moles and gas volumes.
Interpret balanced equations as mole ratios and carry out reacting-mass and gas-volume calculations.
Identify the limiting reagent and calculate the amount of product and of excess reagent left.
Write and balance full equations and ionic equations, with state symbols.

In 30 seconds

1One mole is 6.02 × 10²³ particles (the Avogadro constant) — but in practice you count moles by weighing: moles = mass ÷ Mr.
2Mr is the sum of the Ar values in the formula: Ca(OH)₂ = 40 + 2(16 + 1) = 74 — multiply everything inside brackets.
3Equations read as mole ratios: from N₂ + 3H₂ → 2NH₃, 1 mol of N₂ makes 2 mol of NH₃.
4Percentage by mass of element X = (total Ar of X in the formula ÷ Mr) × 100 — oxygen in CaCO₃ is 48%.
5Empirical formula = simplest mole ratio from experimental masses; molecular formula = (empirical formula)ₙ.
6Concentration: mol/dm³ = moles ÷ volume in dm³, and g/dm³ = mol/dm³ × Mr; 1 dm³ = 1000 cm³.
7One mole of any gas occupies 24 dm³ at r.t.p.: volume (dm³) = moles × 24.
8The limiting reagent runs out first and caps the product; always calculate moles and compare with the equation ratio.

Notes · 1 of 5

The mole: counting particles by weighing

Chemists count with a balance: Mr grams of a substance is one mole of it.

Atoms are far too small to count individually, so chemistry uses a bundle: the mole, containing 6.02 × 10²³ particles (the Avogadro constant, L). The bundle is sized so that one mole of any substance weighs exactly its relative formula mass in grams — 18 g of water is 1 mol of water molecules; 44 g of carbon dioxide is 1 mol of CO₂ molecules.

Formula masses from the data booklet
Substance	Working	Mr / g per mol
H₂O	2(1) + 16	18
CaCO₃	40 + 12 + 3(16)	100
Ca(OH)₂	40 + 2(16 + 1)	74
Al₂(SO₄)₃	2(27) + 3(32 + 4(16))	342
CuSO₄·5H₂O	64 + 32 + 4(16) + 5(18)	250

Notes · 2 of 5

Percentage composition and formulae from experiment

Percentages come from the formula; the empirical formula comes from percentages.

The percentage by mass of an element is its share of the formula mass. Limestone is calcium carbonate, CaCO₃ (Mr = 100), and its calcium content is (40 ÷ 100) × 100 = 40%; oxygen is (48 ÷ 100) × 100 = 48%. The reverse journey — from experimental masses to a formula — uses the same numbers backwards.

1Masses to moles
Divide each element’s mass by its Ar: 7.0 g Fe → 7.0 ÷ 56 = 0.125 mol; 3.0 g O → 3.0 ÷ 16 = 0.1875 mol.
2Mole ratio
Divide by the smallest: 0.125 ÷ 0.125 = 1 (Fe); 0.1875 ÷ 0.125 = 1.5 (O).
3Clear fractions
× 2 gives Fe : O = 2 : 3 — the empirical formula is Fe₂O₃.
4Molecular formula
If Mr of the true molecule is given, compare: n = molecular Mr ÷ empirical Mr; multiply the empirical formula by n.

Notes · 3 of 5

Concentrations of solutions

Two currencies: mol/dm³ for equations, g/dm³ for the shelf bottle.

Dissolve 4 g of sodium hydroxide (Mr = 40) in water and make up to 500 cm³. Moles of NaOH = 4 ÷ 40 = 0.1 mol. Concentration = 0.1 mol ÷ 0.5 dm³ = 0.2 mol/dm³ — and in mass terms, 4 g in 0.5 dm³ is 8 g/dm³. The two currencies are linked by Mr: g/dm³ = mol/dm³ × Mr (0.2 × 40 = 8 ✓).

Conversion kit
To find	Use	Watch out for
moles in solution	concentration (mol/dm³) × volume (dm³)	cm³ must be divided by 1000
moles from mass	mass (g) ÷ Mr	Use the formula mass of the dissolved substance
g/dm³ from mol/dm³	multiply by Mr	Say which substance the concentration refers to
titration unknown	moles from the known solution, then ratio	Balance the equation before using the ratio

Notes · 4 of 5

Gas volumes at r.t.p.

One mole of any gas fills 24 dm³ at room temperature and pressure — volume counts moles, not mass.

At room temperature and pressure (r.t.p.), one mole of any gas occupies 24 dm³ (24 000 cm³). The beauty of this is that the volume of a gas measures its moles directly — 0.25 mol of carbon dioxide fills 0.25 × 24 = 6.0 dm³, whether it came from burning carbon or fizzing marble.

Two cautions earn marks. First, the 24 dm³ rule applies only to gases — liquids and solids have their own volumes unrelated to moles. Second, it is quoted at r.t.p.; different conditions change the molar volume, and the question must say so.

Notes · 5 of 5

Reacting masses and the limiting reagent

Convert to moles, apply the ratio, convert back — and check who runs out first.

1Balance the equation
The ratio is meaningless until the equation is balanced.
2Convert everything to moles
mass ÷ Mr for solids and solutions; volume ÷ 24 for gases at r.t.p.
3Apply the mole ratio
Scale from what you have to what you want — and identify the limiting reagent first.
4Convert back
× Mr for masses, × 24 for gas volumes, and round only at the end (3 s.f.).

Ionic equations follow the same logic with charges. Writing the limestone reaction ionically: CaCO₃(s) + 2H⁺(aq) → Ca²⁺(aq) + H₂O(l) + CO₂(g) — the spectator chloride ions are dropped, and the charge on each side sums to +2. Check atoms and charges every time.

Revise

Key terms

The words examiners expect you to use precisely.

Mole: The amount of substance containing 6.02 × 10²³ particles; the mass of one mole in grams equals the relative formula mass.
Avogadro constant (L): 6.02 × 10²³ particles per mole.
Relative molecular mass (Mr): The sum of the relative atomic masses in a molecular formula.
Relative formula mass: The summed mass of all atoms in an ionic formula, used exactly like Mr.
Percentage composition: The mass percentage of each element in a compound.
Empirical formula: The simplest whole-number ratio of atoms of each element in a compound.

Show 6 more key terms

Molecular formula: The actual number of atoms of each element in one molecule: (empirical formula)ₙ.
Concentration: Amount of solute per volume of solution, in mol/dm³ or g/dm³.
Molar gas volume: The volume of one mole of any gas: 24 dm³ at room temperature and pressure.
Limiting reagent: The reactant that is used up first, capping the amount of product formed.
Ionic equation: An equation showing only the species that change, with spectator ions removed and charges balanced.
Spectator ion: An ion present in the reaction mixture that takes no part in the reaction and is omitted from the ionic equation.

Revise

Common mistakes

Where students most often lose marks in this chapter — and what to do instead.

Mistake: Dividing by Mr when you should multiply, e.g. computing moles of 36 g of water as 36 × 18.
Moles = mass ÷ Mr; multiplying gives mol² nonsense.
Instead: moles = mass ÷ Mr; mass = moles × Mr — division goes one way, multiplication the other.
Mistake: Ignoring brackets in formula masses, e.g. Ca(OH)₂ = 40 + 16 + 1 = 57.
The 2 multiplies everything inside the bracket: 40 + 2(17) = 74.
Instead: Expand brackets fully before summing.

Show 10 more mistakes

Mistake: Using the equation’s subscript numbers instead of the balancing coefficients as the mole ratio.
Subscripts describe one particle; the ratio comes from the numbers in front of the formulae.
Instead: In 2H₂ + O₂ → 2H₂O the ratio is 2 : 1 : 2 — coefficients only.
Mistake: Quoting concentrations with cm³ in the denominator, e.g. "0.002 mol/cm³".
Standard concentration units are mol/dm³; 1 dm³ = 1000 cm³.
Instead: Convert the volume first: 25.0 cm³ = 0.0250 dm³.
Mistake: Using 24 dm³ for solids or liquids, e.g. computing the volume of 1 mol of water as 24 dm³.
The molar gas volume applies to gases only.
Instead: 24 dm³ per mole is a gas-law result at r.t.p.; water is 18 cm³ per mole as a liquid.
Mistake: Stopping the empirical-formula ratio at 1 : 1.5 and writing FeO₁.₅.
Empirical formulae need whole numbers.
Instead: Multiply through by 2 to clear the fraction: Fe₂O₃.
Mistake: Mixing up empirical and molecular formulae, e.g. calling C₄H₁₀ the empirical formula of butane.
C₄H₁₀ simplifies to C₂H₅ — the empirical formula is the simplest ratio.
Instead: Divide by n; the molecular formula is n × empirical formula.
Mistake: Computing product from the reagent you have more of, not the limiting one.
The excess reagent’s moles overstate the product.
Instead: Convert both to moles, divide each by its coefficient; the smaller result is limiting.
Mistake: Forgetting to balance charges in ionic equations, e.g. writing H⁺ + OH⁻ → H₂O with leftover ions.
Both atoms and charges must balance on each side.
Instead: Check sums of charges: H⁺ + OH⁻ (+1 + −1 = 0) → H₂O (0).
Mistake: Reporting 0.05 mol as "50 mol" after mis-handling cm³ → dm³ conversion.
0.05 dm³ is 50 cm³, not the other way round.
Instead: Divide cm³ by 1000 to get dm³; keep units in every line.
Mistake: Rounding mid-calculation, e.g. using 0.7 instead of 0.7142…, then quoting 3 s.f. at the end.
Early rounding compounds into a wrong final answer.
Instead: Carry extra digits; round only the final answer to 3 s.f.
Mistake: Writing "1 mol of O₂ = 16 g".
O₂ is diatomic: one mole of oxygen molecules has mass 2 × 16 = 32 g.
Instead: Use the molecular formula: O₂ → Mr = 32; atoms vs molecules must be distinguished.

Check yourself

Quick quiz

20 questions with instant explanations. About 5 minutes.

Question 1 of 3

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Recall the mole

The Avogadro constant is

Practise

Topical questions

Exam-style practice in three tiers, from recall to data questions.

4Foundation
5Standard
3Challenging

12 questions worth 52 marks, each with a model answer, mark scheme and examiner tips.

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Exam practice

Practice paper

A 30-mark paper set out like a school paper, with a separate mark scheme. Give yourself 40 min.

Question paper

Chapter 5 Practice Paper

30 marks · 40 min · 4 pages

5 multiple-choice questions and structured questions with answer lines. Chapter 5 (formula masses, moles, concentrations, gas volumes, reacting masses, limiting reagents).

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Mark scheme

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Chapter 5 Practice Paper — mark scheme

Answers · Marking points · 2 pages

Model answers with a mark for each creditworthy point — for marking at home or in class.

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Where this chapter is tested

Each school sets its own scope — check your assessment outline.

Questions

Frequently asked questions

What does Sec 3 Chemistry Chapter 5 cover?

It covers sub-topic 2.4 of O-Level Pure Chemistry 6092: formula masses, the mole and Avogadro constant, percentage composition, empirical and molecular formulae, concentrations in mol/dm³ and g/dm³, gas volumes at r.t.p., reacting-mass and limiting-reagent calculations, and writing balanced full and ionic equations.

What is the best order for a stoichiometry calculation?

Balance the equation, convert everything to moles (mass ÷ Mr, concentration × volume in dm³, or gas volume ÷ 24), identify the limiting reagent, apply the coefficient ratio, and only then convert back to mass, volume or concentration — rounding to 3 s.f. at the very end.

When do I use 24 dm³?

Only for gases, only at room temperature and pressure. Convert moles of gas to volume with × 24 dm³, or volume to moles with ÷ 24. Solids and liquids never use the molar gas volume.

How do I spot the limiting reagent quickly?

Convert each reactant to moles and divide each by its balancing coefficient. The smaller result is the limiting reagent — every product calculation then keys off that reactant, and the difference for the other one is what remains unreacted.

When is this chapter usually tested in school?

The mole concept sits mid-Sec 3 and appears in Weighted Assessment 2, the end-of-year examination, and then in every titration and gas-volume question through Sec 4 and the O-Level papers. Scope varies by school, so check your school’s assessment outline.

Sec 3 Chemistry Mole Concept & Stoichiometry notes