Part 1: Moles, formulae and hydrated salts

Convert particles and masses into moles, use composition to find formulae, and interpret mass changes in practical work.

The mole connects counting with weighing

A mole is an amount of substance containing Avogadro’s number of specified entities. Use NA = 6.02 × 10²³ mol⁻¹ for OCR calculations unless given another value. State the entities: atoms, molecules, ions or formula units. Molar mass M is mass per mole, in g mol⁻¹.

Worked example: 4.40 g CO₂ contains n = 4.40/44.0 = 0.100 mol molecules, or 6.02 × 10²² molecules. Each contains two oxygen atoms, giving 1.20 × 10²³ oxygen atoms to three significant figures.

Convert 1000 mg to 1 g before using g mol⁻¹. Relative formula mass has no unit; molar mass does. For a pure liquid with density 0.800 g cm⁻³, 10.0 cm³ has mass 8.00 g; then divide by M to find moles.

Count the entity actually requested

A mole is a counting unit for an enormous specified collection, just as a dozen counts twelve specified objects. One mole of water molecules contains one mole of oxygen atoms and two moles of hydrogen atoms. It does not contain only one mole of atoms altogether. The numerical mass of a mole depends on what is counted; equal moles of H₂ and O₂ have equal molecule counts but different masses.

Worked chain: 5.00 cm³ of a pure liquid has density 0.800 g cm⁻³ and molar mass 40.0 g mol⁻¹. Its mass is 5.00 × 0.800 = 4.00 g; its amount is 4.00/40.0 = 0.100 mol; its molecule count is 0.100 × 6.02 × 10²³ = 6.02 × 10²². Multiplying volume directly by Avogadro’s constant has no chemical meaning.

For 0.0250 mol Al₂(SO₄)₃, count 0.0500 mol aluminium ions and 0.0750 mol sulfate ions. There are 0.300 mol oxygen atoms within the sulfates. Dissolving this salt does not release free O²⁻ ions: the oxygen remains part of SO₄²⁻. Distinguish ions in solution from atoms inside a polyatomic ion.

Empirical and molecular formulae

An empirical formula is the simplest whole-number ratio of atoms of each element. A molecular formula gives the actual numbers in one molecule. Divide each mass or mass percentage by Ar, divide by the smallest molar amount and scale fractional ratios to whole numbers. Do not round 1.5 to 2: multiply every ratio by 2.

Worked example: 40.0% C, 6.67% H, 53.3% O gives 40.0/12.0 : 6.67/1.0 : 53.3/16.0 = 3.333 : 6.67 : 3.331 ≈ 1 : 2 : 1. Empirical formula CH₂O has mass 30.0. If Mr = 180, the multiplier is 6 and molecular formula C₆H₁₂O₆.

In combustion analysis, each mole of CO₂ accounts for one mole of C atoms and each mole of water for two moles of H atoms. For a compound containing only C, H and O, find oxygen mass by subtracting calculated C and H masses from the original sample mass.

Why a ratio of 1 : 1.5 becomes 2 : 3

An oxide contains 2.70 g Al and 2.40 g O. Using Ar 27.0 and 16.0 gives 0.100 and 0.150 mol. Dividing by 0.100 gives 1 : 1.5. Multiplying both terms by two gives Al₂O₃. Rounding 1.5 to 2 would produce AlO₂, which changes the measured ratio.

Ratios close to 1.33, 1.50, 1.67 and 1.25 often suggest multiplication by 3, 2, 3 and 4 respectively. These are clues, not permission to force poor data into any desired formula. Retain sufficient decimal places until the simplest plausible integer ratio is clear. Mass percentages can be treated as masses in a hypothetical 100 g sample because only their ratio matters.

Worked combustion analysis: recover the original atoms

A 1.50 g compound containing only C, H and O burns completely to give 2.20 g CO₂ and 0.900 g H₂O. The oxygen in those products comes partly from the oxygen supply, so do not use all product oxygen as the oxygen originally in the compound.

From CO₂: n = 2.20/44.0 = 0.0500 mol, so n(C) = 0.0500 mol and m(C) = 0.600 g. From water: n = 0.900/18.0 = 0.0500 mol; each water contains two H atoms, so n(H) = 0.100 mol and m(H) = 0.100 g.

The original oxygen mass is 1.50 − 0.600 − 0.100 = 0.800 g, giving n(O) = 0.0500 mol. The ratio C:H:O is 1:2:1, so empirical formula CH₂O. If independent evidence gives Mr = 60.0, double every subscript to obtain C₂H₄O₂. This formula does not yet identify one unique structure.

Water of crystallisation and constant mass

A hydrated salt contains water in its crystalline structure; an anhydrous salt has no water of crystallisation. In salt·xH₂O, x is the mole ratio water : anhydrous salt, not a mass ratio.

Illustrative data: 2.46 g MgSO₄·xH₂O leaves 1.20 g MgSO₄ (M = 120 g mol⁻¹). Water lost = 1.26 g; n(water) = 1.26/18.0 = 0.0700 mol; n(salt) = 1.20/120 = 0.0100 mol, so x = 7.

Weigh a dry crucible and lid, then the sample; heat, cool and reweigh until constant mass. Avoid losing solid and avoid temperatures that decompose the salt. Cool before weighing; use tongs. Incomplete dehydration makes calculated x too low. Loss of salt misidentified as water can make x too high. This is a PAG 1 measurement context.

A complementary mass-gain experiment burns cleaned Mg ribbon in a weighed crucible with a lid. Lift the lid briefly to admit oxygen while limiting loss of MgO smoke, then cool and reweigh to constant mass. Mass gain gives oxygen mass; divide Mg and O masses by their Ar values to find the empirical ratio. Use 2Mg + O₂ → 2MgO, not an unbalanced one-Mg equation. Incomplete oxidation, MgO loss or nitride side-products can bias the result.

Work in either direction with hydrated salts

When predicting mass remaining, use the molar mass of the material initially weighed. For 4.92 g MgSO₄·7H₂O, take M(MgSO₄) = 120 and M(H₂O) = 18. The hydrate has M = 246, so n(hydrate) = 4.92/246 = 0.0200 mol. Each mole contains one mole of MgSO₄, leaving 0.0200 × 120 = 2.40 g anhydrous salt after complete dehydration.

The seven multiplies water, not sulfate: n(water) = 7 × 0.0200 = 0.140 mol and water mass = 2.52 g. Check 2.40 + 2.52 = 4.92 g. Dividing the original mass by 120 would incorrectly pretend that the original water had no mass.

Quick checks

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

Sources