Stoichiometry

Moles, gases, titrations, yield — AQA's “Amount of substance”

For the ideal gas equation, p must be in Pa, V in m³ (cm³ × 10⁻⁶) and T in K (°C + 273). Most exam errors here are unit conversions, not chemistry.

Every titration follows the same route: find n(known) from c×V, apply the mole ratio from the equation, then c(unknown) = n/V. Concordant titres agree within 0.10 cm³ — use their mean.

Atom economy is about the reaction type (addition = 100%); yield is about practical losses. Examiners ask you to distinguish them.

3.1.2.4Full titration calculation — the 5-step routine

3.1.2.3Using pV = nRT without unit disasters

Convert first, every time: kPa → Pa (×1000) · cm³ → m³ (×10⁻⁶) · dm³ → m³ (×10⁻³) · °C → K (+273). Do the conversions before touching the equation and the marks look after themselves.

3.1.2.5Percentage uncertainty

% uncertainty = (uncertainty ÷ measured value) × 100. It doubles whenever a quantity is the difference of two readings — a burette titre, a ΔT, a mass by difference — because each reading brings its own uncertainty.

To reduce it, make the measured quantity bigger: a larger titre or a larger ΔT. More repeats will not do it — repeats reduce random error, not the % uncertainty of the instrument.

Deep dive📚 The rest of the chapter, in full

The ideal gas equation, without unit disasters

pV = nRT with R = 8.31 J K⁻¹ mol⁻¹ demands SI: pressure in Pa (kPa × 10³), volume in m³ (cm³ × 10⁻⁶, dm³ × 10⁻³), temperature in K (°C + 273). Most lost marks are unit conversions, not chemistry.

Formula-finding toolkit

Empirical formula: divide each element's mass (or %) by its Aᵣ, divide through by the smallest, then scale to whole numbers. For 40.0% C, 6.7% H and 53.3% O the ratio is 3.33 : 6.7 : 3.33, which gives CH₂O.

Molecular formula: divide Mᵣ (from the mass spectrum) by the empirical mass. Here 180 ÷ 30 = 6, so the molecular formula is C₆H₁₂O₆.

Water of crystallisation: heat to constant mass, then compare n(salt) : n(H₂O). If 2.46 g of hydrated MgSO₄ leaves 1.20 g anhydrous, n(MgSO₄) = 0.00997 and n(H₂O) = 1.26 ÷ 18 = 0.0700 — a ratio of 7, so the salt is MgSO₄·7H₂O.

Titration calculations — one template for all of them

The template never changes. First find n = cV/1000 for the known solution, then apply the mole ratio from the balanced equation, then convert back to the wanted quantity (c = 1000n/V, or m = nM). Write the three steps explicitly — error-carried-forward marks depend on visible working. For back-titrations: total acid added − acid titrated by the base = acid that reacted with the sample.

Yield vs atom economy — different questions

% yield = actual moles ÷ theoretical moles × 100. It is a statement about how well your experiment went — losses to side reactions, transfer and purification all drag it down.

% atom economy = Mᵣ of desired product ÷ Mᵣ of ALL products × 100. This is a property of the reaction itself, calculated from the equation: addition reactions are 100% atom-economical; substitutions and eliminations are not.

Green chemistry favours high atom economy — less waste, better use of feedstocks. Industry may accept a lower-yield route if its atom economy, energy cost, or catalyst recyclability is better.

Extended🎓 Beyond the standard course

Where the ideal gas model fails: pV = nRT assumes point molecules with no attractions. At high pressure, molecular volume matters (real V > ideal); at low temperature, attractions bite (real p < ideal). Gases are most ideal hot and dilute.

Combining uncertainties: for multiplied or divided quantities, add the percentage uncertainties. A titration's total = balance % + pipette % + burette %, and the dominant term tells you which instrument to upgrade.

Back-titration: react an insoluble sample (e.g. CaCO₃) with a measured excess of acid, then titrate the leftover acid. Sample moles = acid added − acid left. It is the standard move whenever the substance won't titrate directly.

Mastery vault🏛 Every remaining spec point, banked

Avogadro's constant in anger

Gas volume reasoning (Avogadro's law)

Equal volumes of gases at the same T and p contain equal moles — so volume ratios ARE mole ratios. 20 cm³ of propane burns in 5 × 20 = 100 cm³ of oxygen (C₃H₈ + 5O₂ → 3CO₂ + 4H₂O). Eudiometry questions give you combustion volumes and ask for the formula: work the ratios backwards.

The back-titration, fully worked

Uncertainty arithmetic

% uncertainty = (absolute uncertainty ÷ measured value) × 100. A burette titre carries ±0.10 cm³ because it is two readings — on 23.50 cm³ that is 0.43%.

When adding or subtracting quantities, add the ABSOLUTE uncertainties: a mass by difference doubles the balance's ±0.005 g. When multiplying or dividing, add the PERCENTAGE uncertainties, so the overall % uncertainty of a concentration = balance% + volumetric flask% + pipette% + burette%.

To reduce it, use bigger titres, bigger masses and more precise glassware. The absolute uncertainty is fixed by the instrument, so grow the denominator.

Making a standard solution — the six steps with reasons

Weigh by difference (eliminates residue error) → dissolve completely in a beaker (all solute enters solution) → transfer with rinsings (no moles left behind) → make up to the mark, bottom of meniscus at eye level (calibrated volume) → stopper and invert repeatedly (uniform concentration). Then c = 1000 × (m/Mᵣ)/V. A suitable primary standard (e.g. anhydrous Na₂CO₃) is pure, stable, non-hygroscopic and has a high Mᵣ to shrink weighing error.

Excess and percentage-purity patterns

Identify the limiting reagent by n ÷ coefficient, compute the product from the smaller value, then state the excess left over (n(excess) − ratio-scaled n(limiting)).

% purity = mass of pure substance (from titration or gas volume) ÷ mass of sample × 100 — the back-titration's usual finale.

Give the answer to the fewest s.f. in the data (usually 3). Keep guard digits until the end and never round intermediate titres.

Hydrates, dilutions and limiting reagents — the next layer

These problems reuse n = m/M and n = cV, but bury the trick in an extra step: a mass loss on heating, a series of dilutions, or a reagent that runs out first.

Concentration units. Convert with M, and treat dilute aqueous solutions as 1.00 g cm⁻³:

Precipitation by ionic equation. For BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq), cancel the spectators Na⁺ and Cl⁻ to leave Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s). The 1:1 ratio means 0.0100 mol Ba²⁺ yields 0.0100 mol BaSO₄ = 0.0100 × 233.4 = 2.33 g of precipitate.