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.