Part 1: Moles, particles and formulae

Use a measurable mass to count chemical entities, then infer a formula from experimental evidence. Keep the identity of the particles visible at every step.

A mole counts a specified entity

The mole, symbol mol, is the unit of amount of substance. It counts a very large fixed number of specified entities: atoms, molecules, ions or formula units. Edexcel calculations use the Avogadro constant L = 6.02 × 10²³ mol⁻¹ unless a different value is supplied. The modern SI definition fixes the exact number at 6.02214076 × 10²³; a mole is not itself a mass in grams.

One mole of O₂ molecules contains two moles of oxygen atoms. One mole of solid CaCl₂ contains one mole of calcium ions and two moles of chloride ions; it is better described using formula units than molecules. State what N counts before multiplying by L. Equal amounts of two substances contain equal numbers of the chosen entities, although their masses can be very different.

Molar mass connects counting to weighing

Molar mass M is mass per mole, in g mol⁻¹. Add the relative atomic masses from the periodic table with the formula subscripts: M[Mg(NO₃)₂] = 24.3 + 2(14.0 + 3 × 16.0) = 148.3 g mol⁻¹. Brackets multiply every atom inside them. Relative formula mass is a ratio with no unit; molar mass has units even though the numerical values match when expressed in g mol⁻¹.

For 0.740 g of Ca(OH)₂, M = 40.1 + 2(16.0 + 1.0) = 74.1 g mol⁻¹. The amount of formula units is 0.740/74.1 = 0.0099865 mol. There are twice as many hydroxide ions: n(OH⁻) = 0.019973 mol, so N(OH⁻) = 0.019973 × 6.02 × 10²³ = 1.20 × 10²² ions, to three significant figures. Multiplying the initial amount by L alone would count formula units, not hydroxide ions.

Mass must match M's units. Convert 35.0 mg to 0.0350 g before using a molar mass in g mol⁻¹. For a pure liquid, density can supply the mass: 2.00 cm³ of a liquid of density 0.800 g cm⁻³ has mass 1.60 g. This is a density calculation, not a solution concentration calculation.

Empirical formula: divide masses by atomic masses

An empirical formula gives the simplest whole-number ratio of atoms of each element in a compound. A molecular formula gives the actual numbers in one molecule. The mass ratio is not the atom ratio because different atoms have different masses. Dividing each mass by its relative atomic mass converts all elements onto the common scale of amount in moles.

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Imagine a 100 g sample. Amounts are C: 40.0/12.0 = 3.333; H: 6.7/1.0 = 6.7; O: 53.3/16.0 = 3.331 mol. Divide all three by 3.331: approximately 1 : 2 : 1, giving CH₂O. The small deviation reflects rounded experimental data.

Do not round a ratio of 1 : 1.50 directly to 1 : 2. Multiply all ratios by two to obtain 2 : 3. Ratios near 1.33 or 1.67 often need multiplication by three. An implausible ratio can also indicate poor data or a wrong assumption, so check the experiment before forcing integers.

Molecular formula needs an additional molar mass

Different molecules can share an empirical formula. The empirical formula mass of CH₂O is 30.0. If an independent measurement gives a molar mass of 90.0 g mol⁻¹, the multiplier is 90.0/30.0 = 3. The molecular formula is C₃H₆O₃. Multiply every subscript; C₃H₂O would no longer have the same atom ratio.

Part 3 obtains a molar mass from a gas or completely vaporised liquid using pV = nRT. That measurement and an elemental composition answer different questions: composition gives relative atom numbers, while molar mass fixes the number of empirical units in each molecule.

Water of crystallisation from mass loss

A hydrate contains a fixed number of water molecules per formula unit in its crystal. A 2.460 g sample of MgSO₄·xH₂O leaves 1.200 g of anhydrous MgSO₄ after suitable heating. Water lost = 1.260 g. Using M(MgSO₄) = 120.4 and M(H₂O) = 18.0, n(salt) = 1.200/120.4 = 0.0099668 mol and n(water) = 1.260/18.0 = 0.07000 mol. Their ratio gives x = 7.02, consistent with MgSO₄·7H₂O within the experimental precision.

Weigh a clean dry crucible and lid, then the hydrate. Heat gently at first to reduce spitting, cool and reweigh; repeat heating, cooling and weighing until constant mass. A partly open lid allows water vapour to escape while limiting loss of solid. Cool before weighing because a hot object causes unstable balance readings and can damage the balance. Use tongs, a heatproof mat and eye protection.

Incomplete dehydration leaves the final mass too high: inferred water loss is too small and inferred anhydrous salt amount too large, so x is too low. Loss of solid by spitting makes the apparent water loss too large and residual salt mass too small, so x is too high. Constant mass supports completion but does not prove that only water was lost; decomposition requires a different method or temperature.

Formula from a mass gain: magnesium oxide

For magnesium heated in air, measure the mass of magnesium and the final oxide; oxygen mass is the gain, not the whole product mass. For 0.486 g Mg and 0.806 g product, oxygen mass = 0.320 g. Amounts are 0.486/24.3 = 0.0200 mol Mg and 0.320/16.0 = 0.0200 mol O, giving MgO.

A lid limits escape of the white powder, but it must be lifted briefly to admit oxygen. Heat, cool and reweigh to constant mass. Powder loss decreases the apparent oxygen gain and produces an artificially Mg-rich ratio. Magnesium can also react with nitrogen in air; follow the supplied procedure for treating this contamination rather than assuming every mass gain is oxygen. Do not stare at intensely burning magnesium.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.

Sources