9701 Chemistry · Topic 2 · AS Level
Stoichiometry and Mole Calculations Cheat Sheet — A Level Chemistry 9701
Almost every 9701 paper carries marks that come down to a mole calculation done cleanly. These two sheets cover the mole triangle in all its forms, empirical and molecular formulae, limiting reagents, percentage yield and atom economy, then move on to titration arithmetic, back titrations and the ideal gas equation — with worked examples set out the way an examiner wants to see them.
✅ 29 worked sections
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What’s on this cheat sheet
Atoms, Molecules and Stoichiometry
01 · Relative masses
Ar — weighted mean mass of an atom relative to ¹²C taken as exactly 12. Mr — the same for a molecule or formula unit. Both are ratios, so no units.
Ar = Σ(isotopic mass × %abundance) ÷ 100. Read abundances off a mass spectrum (x‑axis m/z, y‑axis relative abundance).
02 · The mole
N = n × L (L = 6.02 × 10²³ mol⁻¹)
One mole is the amount containing as many particles as there are atoms in 12 g of ¹²C. Always state what the particles are — atoms, molecules or ions.
03 · Solutions and gases
pV = nRT · Vm = 24.0 dm³ at r.t.p.
Ideal‑gas units: p in Pa, V in m³ (1 dm³ = 1 × 10⁻³ m³), T in K, R = 8.31 J K⁻¹ mol⁻¹. cm³ ÷ 1000 = dm³.
04 · Empirical and molecular formulae
Empirical — simplest whole‑number ratio of atoms. Molecular — the actual number in one molecule; it is a whole‑number multiple of the empirical formula.
C 40.0 %, H 6.7 %, O 53.3 %:
3.33 : 6.7 : 3.33 → 1 : 2 : 1 → CH₂O
If Mr = 180, 180 ÷ 30 = 6 → C₆H₁₂O₆
05 · Balancing and state symbols
Balance atoms first, then charge. Every equation needs (s) (l) (g) (aq). Ionic equations show only the species that change — cancel the spectator ions.
Half‑equations: balance atoms, then O with H₂O, then H with H⁺, then charge with e⁻.
06 · Reacting‑mass calculations
mass or volume → moles → mole ratio from the equation → moles of the wanted species → mass or volume.
Never compare masses directly across an equation — only moles carry the ratio.
07 · Worked example — limiting reactant
Mg + 2HCl → MgCl₂ + H₂
n(Mg) = 4.00 ÷ 24.3 = 0.165 mol
n(HCl) = 2.00 × 0.100 = 0.200 mol
HCl needed for all the Mg = 2 × 0.165 = 0.329 mol > 0.200
→ HCl is limiting
n(H₂) = 0.200 ÷ 2 = 0.100 mol
V(H₂) = 0.100 × 24.0 = 2.40 dm³ at r.t.p.
08 · Worked example — reacting masses
CaCO₃ → CaO + CO₂
n(CaCO₃) = 25.0 ÷ 100.1 = 0.250 mol
ratio 1 : 1 → n(CaO) = 0.250 mol
m(CaO) = 0.250 × 56.1 = 14.0 g
Check by difference: the 11.0 g lost is the CO₂.
09 · Types of formula
| Type | Ethanoic acid |
|---|---|
| empirical | CH₂O |
| molecular | C₂H₄O₂ |
| structural | CH₃COOH |
| displayed | every bond drawn |
An ionic compound has only an empirical formula — the lattice has no molecules.
10 · Worked example — combustion data
n(C) = 0.440 ÷ 44.0 = 0.0100 mol
n(H) = 2 × (0.270 ÷ 18.0) = 0.0300 mol
ratio C : H = 1 : 3 → CH₃
If Mr = 30, molecular formula = C₂H₆
11 · Avogadro’s law
Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. So for gases the volume ratio equals the mole ratio — no need to convert.
N₂ + 3H₂ → 2NH₃: 10 cm³ of N₂ needs 30 cm³ of H₂ and gives 20 cm³ of NH₃, all measured under the same conditions.
12 · Percentage composition
(2 × 14.0) ÷ 80.0 × 100 = 35.0 %
Compare fertilisers this way — urea CO(NH₂)₂ is 46.7 % N.
13 · The four routes to moles
From mass — n = m ÷ M
From solution — n = c × V(dm³)
From gas volume — n = V ÷ 24.0 dm³ at r.t.p., or n = pV ÷ RT
From particles — n = N ÷ 6.02 × 10²³
14 · Worked example — gas volumes
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O(l)
O₂ needed = 5 × 50 = 250 cm³ → 50 cm³ O₂ left over
CO₂ formed = 3 × 50 = 150 cm³
Water is a liquid at r.t.p., so it adds no volume.
Final gas volume = 150 + 50 = 200 cm³
15 · Yield and atom economy
atom economy = Mr(desired) ÷ ΣMr(products) × 100
Theoretical yield comes from the limiting reactant. Losses come from side reactions, reversible reactions and transfer losses. An addition reaction has 100 % atom economy.
Titrations and Gas Calculations
16 · Standard solutions
Dissolve a weighed mass, transfer with washings to a volumetric flask, make up to the mark, invert to mix. Concentration in mol dm⁻³ or g dm⁻³; g dm⁻³ = mol dm⁻³ × M.
Dilution: c₁V₁ = c₂V₂. Moles are unchanged by dilution.
17 · Titration technique
Pipette the aliquot, rinse the burette with the titrant, record initial and final readings to 0.05 cm³, discard the rough titre and average only concordant titres (within 0.10 cm³).
Rinse the conical flask with distilled water only. Adding more water does not change the moles present, so it does not affect the titre.
18 · Worked example — titration
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
n(H₂SO₄) = 0.100 × 0.02240 = 2.24 × 10⁻³ mol
n(NaOH) = 2 × 2.24 × 10⁻³ = 4.48 × 10⁻³ mol
c(NaOH) = 4.48 × 10⁻³ ÷ 0.0250 = 0.179 mol dm⁻³
Give 3 s.f. — the data limits it.
19 · Back titration
Used when the sample is insoluble or reacts slowly (an impure carbonate, an antacid). React with a known excess of acid, then titrate the leftover acid with standard alkali.
20 · Worked example — gas law
V = 240 × 10⁻⁶ m³, p = 1.00 × 10⁵ Pa
n = pV ÷ RT = (1.00 × 10⁵ × 2.40 × 10⁻⁴) ÷ (8.31 × 300)
n = 24.0 ÷ 2493 = 9.63 × 10⁻³ mol
M = 0.240 ÷ 9.63 × 10⁻³ = 24.9 g mol⁻¹
21 · Ideal gases
Assumptions: negligible molecular volume, no intermolecular forces, elastic collisions, random rapid motion.
Real gases deviate most at high pressure and low temperature, where molecules are close together and attractions matter. Behaviour is most ideal at high T and low p.
22 · Water of crystallisation
For MX·xH₂O, find moles of the anhydrous salt and moles of water lost on heating, then take the ratio.
n(CuSO₄) = 1.60 ÷ 159.6 = 0.01003 mol
n(H₂O) = 0.90 ÷ 18.0 = 0.0500 mol
ratio 1 : 4.99 → x = 5
23 · Apparatus and uncertainty
| Apparatus | Uncertainty |
|---|---|
| burette (per reading) | ± 0.05 cm³ |
| pipette 25.0 cm³ | ± 0.06 cm³ |
| balance (per reading) | ± 0.001 g |
A titre needs two burette readings, so the total uncertainty is ± 0.10 cm³. % uncertainty = uncertainty ÷ measurement × 100; reduce it by using a larger titre or a bigger sample mass.
24 · Worked example — back titration
n(HCl) added = 0.0300 mol
n(HCl) left = 0.0220 × 0.500 = 0.0110 mol
n(HCl) reacted = 0.0190 mol
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
n(CaCO₃) = 0.00950 mol
m = 0.00950 × 100.1 = 0.951 g → purity 95.1 %
25 · Ionic equations to know
H⁺ + OH⁻ → H₂O
Carbonate + acid
CO₃²⁻ + 2H⁺ → H₂O + CO₂
Precipitation
Ag⁺ + Cl⁻ → AgCl(s)
Ba²⁺ + SO₄²⁻ → BaSO₄(s)
Metal + acid
Mg + 2H⁺ → Mg²⁺ + H₂
26 · Worked example — dilution
c₁V₁ = c₂V₂
V₁ = (0.150 × 250) ÷ 2.00
V₁ = 18.8 cm³
Pipette this into a 250 cm³ volumetric flask and make up to the mark.
27 · Sources of error in a titration
Rinsing the burette with water rather than the titrant dilutes it and the titre reads high. An air bubble in the jet leaves the same false high reading when it clears.
Overshooting the end point, misreading the meniscus, and swirling too little all shift the titre. Read the bottom of the meniscus at eye level against a white tile.
Repeat until two titres agree within 0.10 cm³, and average only those.
28 · Worked example — purity
Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂
n(HCl) = 2.00 × 10⁻³ mol
n(Na₂CO₃) in 25 cm³ = 1.00 × 10⁻³ mol
in 250 cm³ = 0.0100 mol
m = 0.0100 × 106.0 = 1.06 g
purity = 1.06 ÷ 2.50 × 100 = 42.4 %
Marks lost here
— Leaving volumes in cm³ inside n = cV, or in dm³ inside pV = nRT.
— Ignoring the mole ratio and assuming 1 : 1.
— Forgetting to test which reactant is limiting.
— Rounding mid‑calculation, or quoting more significant figures than the data allows.
— Omitting state symbols, or leaving spectator ions in an ionic equation.
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Stoichiometry and Mole Calculations — Frequently Asked Questions
How do you decide which reactant is limiting?
Divide the moles of each reactant by its coefficient in the balanced equation. The smallest answer is the limiting reagent, and every subsequent calculation — yield, enthalpy, gas volume — must be based on that reactant.
What is the difference between percentage yield and atom economy?
Percentage yield compares the mass you actually made with the maximum the equation allows. Atom economy compares the mass of the desired product with the total mass of all products, so it is a property of the equation itself and does not change with how well the experiment was run.
Which units must be used in pV = nRT?
Pressure in pascals, volume in cubic metres and temperature in kelvin, with R = 8.31 J K⁻¹ mol⁻¹. Converting cm³ to m³ means dividing by 1 000 000 — the single most common slip in this calculation.
