A-level chemistry · Chapter 3
Amount of substance
Moles, gases, titrations, yield
Avogadro constant NA = 6.022 × 10²³ mol⁻¹ — the number of particles in one mole.
n = m / M · n = c × V(dm³) · pV = nRT (R = 8.314 J K⁻¹ mol⁻¹)
- Ideal gas: p in Pa, V in m³ (cm³ × 10⁻⁶), T in K (°C + 273). Most exam errors are unit conversions.
- Titration: n(known) from c×V → use mole ratio from the equation → 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
Worked example. 25.0 cm³ of NaOH is neutralised by 21.40 cm³ of 0.0850 mol dm⁻³ HCl. Find [NaOH].
- n(HCl) = c × V = 0.0850 × 0.02140 = 1.819 × 10⁻³ mol
- Equation: HCl + NaOH → NaCl + H₂O — ratio 1 : 1
- n(NaOH) = 1.819 × 10⁻³ mol
- c = n ÷ V = 1.819 × 10⁻³ ÷ 0.0250 = 0.0728 mol dm⁻³
- 3 significant figures — match the data given.
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).
Worked example. What volume does 0.100 mol of gas occupy at 298 K and 100 kPa?
- V = nRT ÷ p = (0.100 × 8.31 × 298) ÷ 100 000 = 2.48 × 10⁻³ m³ = 2.48 dm³
3.1.2.5Percentage uncertainty
- % uncertainty = (uncertainty ÷ measured value) × 100. Doubles when a quantity is a difference of two readings (burette titre, ΔT, mass by difference).
- To reduce it: make the measured quantity bigger (larger titre, larger ΔT) — not more repeats (repeats reduce random error, not % uncertainty of the instrument).
Exam tip. Empirical formula questions: divide each element's mass (or %) by its Ar, then divide all by the smallest. Only round at the very end — 1.33 is a ratio of 4:3, not "1".
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/divided quantities, add the percentage uncertainties. A titration's total = balance % + pipette % + burette % — 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. Used whenever the substance won't titrate directly.
Extended worked example — back-titration. 1.00 g of impure CaCO₃ is treated with 50.0 cm³ of 1.00 mol dm⁻³ HCl; the excess needs 24.0 cm³ of 1.00 mol dm⁻³ NaOH.
- Acid added 0.0500 mol; acid left 0.0240 mol → acid used 0.0260 mol.
- CaCO₃ + 2HCl: n(CaCO₃) = 0.0130 mol → mass 1.30 g?? Greater than the sample — so recheck: 0.0130 × 100.1 = 1.30 g means the sample can't be pure… and indeed the question's numbers force purity = impossible → always sanity-check. With 40.0 cm³ NaOH instead: acid used 0.0100 mol → 0.500 g → 50.0% pure.
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.
Worked example. What volume does 0.250 mol of gas occupy at 100 kPa and 25 °C? V = nRT/p = (0.250 × 8.31 × 298) ÷ 100 000 = 6.19 × 10⁻³ m³ = 6.19 dm³. Cross-check: near room conditions one mole occupies ≈ 24 dm³, so a quarter mole ≈ 6 dm³. ✓
Formula-finding toolkit
- Empirical formula: divide each element's mass (or %) by its Aᵣ, divide through by the smallest, scale to whole numbers. 40.0% C, 6.7% H, 53.3% O → 3.33 : 6.7 : 3.33 → CH₂O.
- Molecular formula: divide Mᵣ (from the mass spectrum) by the empirical mass: 180 ÷ 30 = 6 → C₆H₁₂O₆.
- Water of crystallisation: heat to constant mass, then n(salt) : n(H₂O). 2.46 g of hydrated MgSO₄ leaving 1.20 g anhydrous → n(MgSO₄) = 0.00997, n(H₂O) = 1.26 ÷ 18 = 0.0700 → ratio 7 → MgSO₄·7H₂O.
Titration calculations — one template for all of them
1. n = cV/1000 for the known solution. 2. Use the mole ratio from the balanced equation. 3. 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 — a statement about how well your experiment went (losses to side reactions, transfer, purification).
- % atom economy = Mᵣ of desired product ÷ Mᵣ of ALL products × 100 — 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.
Limiting reagent example. 5.00 g of Fe (0.0895 mol) with 5.00 g of S (0.156 mol) for Fe + S → FeS: the 1 : 1 ratio makes Fe limiting, so n(FeS) = 0.0895 mol → 7.87 g. Always divide moles by the stoichiometric coefficient to compare fairly.
Concentration conversions. mol dm⁻³ × Mᵣ = g dm⁻³. Parts per million: mass of solute ÷ mass of solution × 10⁶ — used for pollutants and hard-water ions. Dilution: c₁V₁ = c₂V₂ (moles are conserved when only water is added).
Mastery vault🏛 Every remaining spec point, banked
Avogadro's constant in anger
How many atoms are in 0.500 g of calcium? n = 0.500/40.1 = 0.0125 mol → atoms = 0.0125 × 6.022 × 10²³ = 7.53 × 10²¹. For molecules, multiply again by atoms per molecule: 0.10 mol of CO₂ contains 1.8 × 10²³ atoms. Ionic: 0.1 mol CaCl₂ releases 0.3 mol of ions.
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
Problem. 1.00 g of impure CaCO₃ is dissolved in 50.0 cm³ of 0.500 mol dm⁻³ HCl (an excess). The leftover acid needs 40.0 cm³ of 0.200 mol dm⁻³ NaOH. Find the purity.
n(HCl total) = 0.0500 × 0.500 = 0.0250 mol. n(NaOH) = 0.0400 × 0.200 = 0.00800 mol = n(HCl left over). n(HCl that reacted with the carbonate) = 0.0250 − 0.00800 = 0.0170 mol. CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂, so n(CaCO₃) = 0.0170 ÷ 2 = 0.00850 mol → m = 0.00850 × 100.1 = 0.851 g → purity = 85.1%. The template: total acid − titrated leftover = reacted acid → apply the ratio → mass → % of sample.
n(HCl total) = 0.0500 × 0.500 = 0.0250 mol. n(NaOH) = 0.0400 × 0.200 = 0.00800 mol = n(HCl left over). n(HCl that reacted with the carbonate) = 0.0250 − 0.00800 = 0.0170 mol. CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂, so n(CaCO₃) = 0.0170 ÷ 2 = 0.00850 mol → m = 0.00850 × 100.1 = 0.851 g → purity = 85.1%. The template: total acid − titrated leftover = reacted acid → apply the ratio → mass → % of sample.
Uncertainty arithmetic
- % uncertainty = (absolute uncertainty ÷ measured value) × 100. Burette titre: ±0.10 cm³ (two readings) — on 23.50 cm³ that is 0.43%.
- Adding/subtracting quantities: add ABSOLUTE uncertainties (a mass by difference doubles the balance's ±0.005 g).
- Multiplying/dividing: add PERCENTAGE uncertainties. Overall % uncertainty of a concentration = balance% + volumetric flask% + pipette% + burette%.
- To reduce: bigger titres, bigger masses, 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 limiting reagent by n ÷ coefficient; compute product from the smaller; state the excess left over (n(excess) − ratio-scaled n(limiting)).
- % purity = mass of pure substance (from titration/gas volume) ÷ mass of sample × 100 — the back-titration's usual finale.
- Significant figures: answer to the fewest s.f. in the data (usually 3); keep guard digits until the end; never round intermediate titres.
Molar-volume flexibility. 24.0 dm³ mol⁻¹ (RTP) is a convenience number; if the question gives T and p, they want pV = nRT instead. Seeing "at 298 K and 100 kPa" is the signal: molar volume = RT/p = 8.31 × 298 ÷ 100 000 = 0.0248 m³ = 24.8 dm³.
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