Lesson 2: Amount of Substance — The Mole

The mole is the chemist's counting unit — the bridge between the atomic world and the laboratory bench. Every titration, every yield calculation and every gas measurement in A-level chemistry runs through moles. This lesson covers the Avogadro constant, molar mass, empirical and molecular formulae, reacting-mass and percentage-yield calculations, atom economy, gas volumes including the ideal gas equation pV = nRT, solution concentrations, and the titration calculations that dominate the required practicals. Fluency here is non-negotiable: examiners assume you can convert between mass, moles, gas volume and concentration without hesitation.

The mole and the Avogadro constant

One mole is the amount of substance containing exactly 6.02 × 10²³ particles — the Avogadro constant, L. The particles may be atoms, molecules, ions or electrons; always state which. The molar mass of a substance is the mass of one mole, numerically equal to its relative atomic or molecular mass in grams: Mᵣ of H₂SO₄ is 98.1, so one mole weighs 98.1 g. The central equation is n = m / Mᵣ.

Rearranging gives m = n × Mᵣ (weighing out a required number of moles) and Mᵣ = m / n (identifying unknowns). Give answers to the same precision as the data, usually 3 significant figures. A common error is using atomic numbers instead of relative atomic masses, or forgetting subscripts — in Ca(OH)₂ the Mᵣ is 40.1 + 2 × (16.0 + 1.0) = 74.1, not 57.1.

The mole concept extends to volumes. At room temperature and pressure (RTP: 20 °C, 1 atm), one mole of any gas occupies 24.0 dm³ — the molar gas volume — so n = V / 24.0 for gas volumes in dm³ at RTP. For solutions, concentration in mol dm⁻³ gives n = c × V, with V in dm³ (divide cm³ by 1000). These three conversions — mass, gas volume, concentration — are the entire toolkit of quantitative chemistry.

Empirical and molecular formulae

The empirical formula is the simplest whole-number ratio of atoms in a compound; the molecular formula is the actual number of atoms in a molecule. From percentage composition: divide each percentage by the Aᵣ to get moles, then divide all mole values by the smallest. A compound with 40.0% C, 6.7% H, 53.3% O gives moles 3.33 : 6.7 : 3.33, ratio 1 : 2 : 1, empirical formula CH₂O.

If the relative molecular mass is known, divide it by the empirical formula mass: CH₂O (mass 30) with Mᵣ = 180 gives a multiplier of 6 and molecular formula C₆H₁₂O₆. Mass spectrometry (Lesson 1) supplies Mᵣ via the molecular ion peak. Combustion analysis also yields empirical formulae: burning a hydrocarbon produces CO₂ and H₂O, and moles of C and H come from the product masses; oxygen is found by difference. Always show the divide-by-smallest step — examiners award method marks even when the final ratio needs rounding (1 : 1.5 doubles to 2 : 3).

Reacting masses, percentage yield and atom economy

Reacting-mass calculations follow a fixed method: write the balanced equation, convert the given mass to moles, use the molar ratio to find moles of the desired substance, convert back to mass. Example: what mass of MgO forms from 6.0 g of Mg? 2Mg + O₂ → 2MgO. Moles Mg = 6.0 / 24.3 = 0.247; the ratio Mg : MgO is 1 : 1, so moles MgO = 0.247; mass = 0.247 × 40.3 = 9.95 g (3 s.f.).

Percentage yield = (actual yield / theoretical yield) × 100. Yields are never 100% in practice: reactions may be reversible, side reactions occur, product is lost in transfer and purification, and reactants may be impure. Atom economy = (Mᵣ of desired product / ΣMᵣ of all products) × 100 — how efficiently reactant atoms end up in the useful product. Addition reactions have 100% atom economy; substitution and elimination reactions produce waste. When evaluating a synthetic route, weigh percentage yield (conversion efficiency) against atom economy (waste): a high-yield route with poor atom economy may still be uneconomic, which matters for green chemistry and sustainability.

Gases — molar volume and the ideal gas equation

At RTP one mole of gas occupies 24.0 dm³, so 0.50 mol of CO₂ occupies 12.0 dm³. At other temperatures and pressures the molar volume changes, which is why the ideal gas equation is needed: pV = nRT, where p is pressure in pascals, V is volume in m³, R = 8.31 J K⁻¹ mol⁻¹ and T is temperature in kelvin.

Unit conversions are where marks are lost. Convert °C to K by adding 273; cm³ to m³ by dividing by 10⁶; kPa to Pa by multiplying by 1000. Worked example: find the Mᵣ of a gas where 0.40 g occupies 150 cm³ at 100 kPa and 25 °C.

V = 150 × 10⁻⁶ = 1.50 × 10⁻⁴ m³ ; p = 100 000 Pa ; T = 298 K
n = pV / RT = (100 000 × 1.50 × 10⁻⁴) / (8.31 × 298) = 0.00606 mol
Mᵣ = m / n = 0.40 / 0.00606 = 66.0

The ideal gas equation assumes perfectly elastic collisions, negligible molecular volume and no intermolecular forces. Real gases deviate at high pressure (molecules close together, so their volume is significant) and at low temperature (forces cause condensation). Always convert to SI units before substituting into pV = nRT.

Concentrations and titration calculations

Concentration in mol dm⁻³ is moles per cubic decimetre: c = n / V. To prepare a standard solution, weigh the solid accurately, dissolve it in distilled water, transfer to a volumetric flask with washings, make up to the graduation mark, then invert to mix (Required Practical 1). Dilution calculations use c₁V₁ = c₂V₂, with consistent volume units.

Titration calculations are the most examined quantitative skill. Method: from the titre (mean of concordant results, within 0.10 cm³ of each other), calculate moles of the known solution (n = cV), use the equation's molar ratio to find moles of the unknown, then divide by its volume. Example: 25.0 cm³ of NaOH is neutralised by 20.0 cm³ of 0.100 mol dm⁻³ H₂SO₄. Moles acid = 0.100 × 0.0200 = 0.00200 mol; ratio H₂SO₄ : NaOH is 1 : 2, so moles NaOH = 0.00400 mol; concentration = 0.00400 / 0.0250 = 0.160 mol dm⁻³.

Technique earns the practical marks: rinse the burette with the solution it will contain, read the meniscus at eye level to 0.05 cm³, add dropwise near the end-point, and use a white tile to see colour changes. Record burette readings to 2 decimal places. Percentage uncertainty in a titre = (2 × 0.05 / titre) × 100; examiners ask you to calculate it and suggest improvements, such as using a larger titre.

Worked examples — quantitative chemistry in action

Worked example 1: water of crystallisation. A 5.00 g sample of hydrated copper(II) sulphate is heated to constant mass, leaving 3.20 g of anhydrous CuSO₄ (Mᵣ 159.6). Moles of CuSO₄ = 3.20 / 159.6 = 0.0201 mol. Mass of water lost = 5.00 − 3.20 = 1.80 g, so moles of H₂O = 1.80 / 18.0 = 0.100 mol. Ratio H₂O : CuSO₄ = 0.100 / 0.0201 ≈ 5, so the formula is CuSO₄·5H₂O. "Heated to constant mass" guarantees all the water has been driven off — examiners test that phrase.

Worked example 2: percentage yield. Heating 6.50 g of CaCO₃ (Mᵣ 100.1) gives 3.10 g of CaO (Mᵣ 56.1). Moles of CaCO₃ = 6.50 / 100.1 = 0.0649 mol, so the theoretical mass of CaO = 0.0649 × 56.1 = 3.64 g. Percentage yield = 3.10 / 3.64 × 100 = 85.1%. The shortfall is explained by product lost in transfer and material left unreacted in the cooler parts of the apparatus — not by "the reaction not finishing", unless the reaction is reversible.

Going deeper — limiting reagents and back titrations

Real analyses rarely mix exact stoichiometric amounts. The limiting reagent is the reactant that is completely consumed; all others are in excess, and the product yield is calculated from the limiting reagent alone. Method: calculate moles of each reactant, divide each by its stoichiometric coefficient, and the smallest quotient identifies the limiting reagent. Example: 4.0 g of Mg (0.165 mol) with 100 cm³ of 1.00 mol dm⁻³ HCl (0.100 mol): the equation Mg + 2HCl → MgCl₂ + H₂ needs two moles of acid per mole of metal, so 0.100 mol of HCl can consume only 0.0500 mol of Mg — the acid limits, the H₂ produced is 0.0500 mol (1.20 dm³ at RTP), and 0.115 mol of Mg remains unreacted.

Back titration handles analytes that react slowly or are insoluble: add a known excess of reagent, let the reaction complete, then titrate what is left. The standard example is the purity of limestone: a weighed CaCO₃ sample is dissolved in a measured excess of HCl; the leftover acid is titrated with NaOH; the difference gives the acid consumed by the carbonate. Whenever a direct titration would be sluggish, heterogeneous or lacking a sharp end-point, expect a back titration — and set the calculation out as three steps: total reagent added, reagent left over, reagent used by the analyte.

Common mistakes

• Leaving volumes in cm³ inside n = cV. The formula needs dm³: divide cm³ by 1000 before multiplying, or every titration answer is 1000× too big.
• Using 24.0 dm³ mol⁻¹ at non-RTP conditions. The molar gas volume applies only at room temperature and pressure; any other conditions demand pV = nRT with SI units throughout.
• Atom economy with the wrong denominator. The denominator is the total Mᵣ of all products, not all reactants — though by mass conservation the two totals are equal, so either route works only if you include every product.
• Rounding mid-calculation. Keep full calculator precision through chained steps; rounding the moles to 2 s.f. early can drag the final answer outside the mark scheme's tolerance.

Exam Experience

Q1. A gas cylinder contains 0.88 g of a gas occupying 440 cm³ at 100 kPa and 27 °C. Calculate the relative molecular mass of the gas. [3] Q2. An impure sample of calcium carbonate of mass 1.20 g reacts with excess hydrochloric acid. The carbon dioxide produced occupies 216 cm³ at RTP. (a) Calculate the percentage purity of the sample. [3] (b) State two reasons why the volume of CO₂ collected might be less than the value expected for a pure sample, other than impurity. [2]

Mark scheme. Q1: T = 300 K [1]; n = pV/RT = (100 000 × 4.40 × 10⁻⁴)/(8.31 × 300) = 0.0176 mol [1]; Mᵣ = 0.88/0.0176 = 50.0 [1]. Q2(a): n(CO₂) = 216/24 000 = 9.00 × 10⁻³ mol [1]; ratio CaCO₃ : CO₂ = 1 : 1, so mass of pure CaCO₃ = 9.00 × 10⁻³ × 100.1 = 0.901 g [1]; purity = 0.901/1.20 × 100 = 75.1% [1]. (b) Any two: some CO₂ dissolves in the acid/water; gas leaks from the apparatus; the reaction had not finished when the volume was read; the sample was not fully reacted because the acid became locally depleted around lumps [2].