Atoms, Molecules and Stoichiometry at AS

This guide covers Topic 2, Atoms, molecules and stoichiometry, subtopics 2.1 to 2.4, from Cambridge International AS & A Level Chemistry 9701, 2025–2027 series. This is AS Level content.

Before studying this — and how this differs from the IGCSE/O Level guide

If you’ve studied Cambridge IGCSE 0620 or O Level 5070, Formulae, Equations and the Mole already covers relative mass, the mole, balancing equations and reacting-mass calculations at that level. That resource does not cover 9701’s AS requirements — it exists for a different, earlier qualification, and this page does not assume you’ve read it, though the two are consistent with each other.

9701 goes further in three specific ways: writing ionic formulas directly from oxidation numbers (rather than a fixed list of named ions), constructing proper ionic equations that omit spectator ions, and a wider range of stoichiometric calculations — gas volumes, solution concentrations, limiting reagent, and percentage yield — all performed with the significant-figure discipline the syllabus explicitly requires.

Syllabus coverage

CAMBRIDGE INTERNATIONAL AS & A LEVEL CHEMISTRY 9701 — AS Level, Topic 2

2.1 Relative masses of atoms and molecules — the unified atomic mass unit, defined as one twelfth of the mass of a carbon-12 atom; relative atomic mass, relative isotopic mass, relative molecular mass and relative formula mass, defined in terms of it.

2.2 The mole and the Avogadro constant — defining and using the mole in terms of the Avogadro constant.

2.3 Formulas — writing formulas of ionic compounds from ionic charges and oxidation numbers (including predicting ionic charge from Periodic Table position, and recalling the formulas of NO₃⁻, CO₃²⁻, SO₄²⁻, OH⁻, NH₄⁺, Zn²⁺, Ag⁺, HCO₃⁻, PO₄³⁻); writing and constructing balanced equations, including ionic equations that omit spectator ions, with state symbols; the terms empirical and molecular formula; the terms anhydrous, hydrated and water of crystallisation; calculating empirical and molecular formulas from given data.

2.4 Reacting masses and volumes (of solutions and gases) — calculations involving reacting masses (including percentage yield), volumes of gases, volumes and concentrations of solutions, and limiting/excess reagent; deducing stoichiometric relationships from such calculations. Answers must reflect the number of significant figures given or asked for in the question.

There is no Core/Extended tiering at 9701 — every outcome above is required for every AS candidate.

Ionic formulas from oxidation numbers

Rather than memorising a formula for every compound, 9701 expects you to build one from the ionic charge (shown as a Roman numeral oxidation number where needed) and the Periodic Table position of each element:

Iron(III) oxide: Fe³⁺ and O²⁻. Balancing charge needs the lowest common multiple of 3 and 2, which is 6 — two Fe³⁺ and three O²⁻, giving Fe₂O₃.

You’re also expected to recall the formulas of seven specific polyatomic ions without being given them: NO₃⁻, CO₃²⁻, SO₄²⁻, OH⁻, NH₄⁺, HCO₃⁻, PO₄³⁻. Two more worth memorising alongside them, Zn²⁺ and Ag⁺, are monatomic metal ions whose charge isn’t obvious from the Periodic Table group number.

Ionic equations without spectator ions

A full symbol equation shows every species; an ionic equation shows only the species that actually change, omitting anything that appears unchanged on both sides (the spectator ions):

Full:   Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq)
Ionic:  Pb2+(aq) + 2I-(aq) → PbI2(s)

K⁺ and NO₃⁻ appear on both sides in identical form and are omitted. Only species that change state or combine are kept — which is why the solid product, PbI₂, stays fully written out rather than being split into ions.

Stoichiometric calculations

Percentage yield

percentage yield = (actual yield / theoretical yield) × 100

The theoretical yield is calculated from the limiting reagent using the balanced equation’s mole ratio; the actual yield is what was obtained experimentally.

Worked example. Reacting 5.00 g of magnesium (Ar = 24.3) with excess dilute hydrochloric acid produces 0.450 g of hydrogen gas. What is the percentage yield? (Mg + 2HCl → MgCl₂ + H₂; Ar(H) = 1.0)

moles of Mg = 5.00 / 24.3 = 0.2058 mol
mole ratio Mg : H2 is 1 : 1, so theoretical moles of H2 = 0.2058 mol
theoretical mass of H2 = 0.2058 × 2.0 = 0.4115 g
percentage yield = (0.450 / 0.4115) × 100 = 109%

A yield over 100% is chemically impossible and signals an error — most likely that the hydrogen collected wasn’t pure, or a measurement was imprecise. This is a deliberately awkward result: real calculations don’t always land on a tidy answer, and recognising an impossible result is itself part of the skill.

Limiting reagent

When a question gives quantities of two reactants, you must first identify which one runs out first (the limiting reagent) before calculating a product’s mass or volume — using the reagent in excess will overstate the answer.

Worked example. 4.00 g of hydrogen reacts with 4.00 g of oxygen. Which is limiting? (2H₂ + O₂ → 2H₂O; Ar(H) = 1.0, Ar(O) = 16.0)

moles of H2 = 4.00 / 2.0 = 2.00 mol
moles of O2 = 4.00 / 32.0 = 0.125 mol
ratio required (from equation) is 2 : 1, so 2.00 mol H2 would need 1.00 mol O2
only 0.125 mol O2 is available — oxygen is the limiting reagent

Every product quantity must be based on the 0.125 mol of oxygen, not the larger quantity of hydrogen: water formed = 2 × 0.125 = 0.250 mol = 4.50 g.

The excess reagent is different — it is found as initial amount minus amount consumed, which needs the initial hydrogen figure:

H2 consumed  = 2 × 0.125 = 0.250 mol  (0.500 g)
H2 unreacted = 2.00 − 0.250 = 1.75 mol = 1.75 × 2.0 = 3.50 g
check: 0.500 g H2 + 4.00 g O2 = 4.50 g H2O

Gas volumes and solution concentrations

Reacting-volume calculations for gases and solutions follow the same three-step logic as reacting masses — convert what you’re given to moles, apply the equation’s mole ratio, convert the answer to what’s asked for — with the AS syllabus expecting fluency across masses, gas volumes and solution concentrations within a single multi-step question, not each in isolation as separate question types.

Common mistakes

• Using a memorised list of ion formulas instead of building them from charge. 9701 expects you to predict charge from Periodic Table position and apply it — memorising is a poor substitute when an unfamiliar ion appears.
• Leaving spectator ions in an “ionic equation.” If every species from the full equation is still present, it isn’t an ionic equation.
• Basing a limiting-reagent calculation on whichever reactant’s mass is given first, rather than actually comparing the mole ratio required against the mole ratio available.
• Reporting more significant figures than the data supports, or rounding too early in a multi-step calculation and compounding the error.
• Assuming this resource replaces the IGCSE/O Level mole guide. It builds on it — the underlying mole concept doesn’t change, but the range of formula-writing and calculation types genuinely does.

Quick revision checklist

• The unified atomic mass unit, and Ar/Mr/relative formula mass defined from it
• Writing ionic formulas from charge, including the seven named polyatomic ions and Zn²⁺/Ag⁺
• Writing full and ionic equations, with correct state symbols
• Empirical vs molecular formula, and calculating each from data
• Percentage yield, limiting reagent, gas volume and solution concentration calculations, worked to the correct number of significant figures

Related resources

• Formulae, Equations and the Mole — the IGCSE/O Level foundation this resource builds on, not a substitute for it
• Atomic Structure: Orbitals and Ionisation Energy — the previous AS topic
• Chemical Bonding: Shapes and Intermolecular Forces — the next AS topic
• Amount of Substance (OCR A Level) — the equivalent AS-level content from a different awarding body
• Cambridge AS & A Level Chemistry hub

Written against Cambridge International AS & A Level Chemistry 9701, 2025–2027 series. Always check the current syllabus for your examination year.