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Atoms, Molecules and Stoichiometry at AS

Ionic formulae from Roman-numeral oxidation numbers, ionic equations, and stoichiometric calculations including limiting reagent and percentage yield, for Cambridge International AS & A Level Chemistry 9701.

Subject
Chemistry
Level
AS LEVEL
Topic
Atoms, molecules and stoichiometry
Updated

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 nine specific polyatomic ions without being given them: NO₃⁻, CO₃²⁻, SO₄²⁻, OH⁻, NH₄⁺, Zn²⁺, Ag⁺, HCO₃⁻, PO₄³⁻.

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 subsequent calculation in this question — mass of water formed, mass of unreacted hydrogen — must be based on the 0.125 mol of oxygen, not the larger quantity of hydrogen present.

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 nine named polyatomic ions
  • 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

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

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