Analysis tells you what a compound is made of, by mass. Divide by the atomic masses and you have the ratio of atoms; one more number, the relative molecular mass, turns that ratio into the real formula. The same method finds the water in a hydrated salt.
The empirical formula is the simplest whole-number ratio of the atoms of each element in a compound; the molecular formula is the actual number of atoms of each element in one molecule. To find the empirical formula, divide each mass or percentage by Ar and then divide by the smallest answer. The molecular formula is n × the empirical formula, where n = Mr ÷ empirical formula mass.
Fig. 1: from percentages to glucose in four steps
Section 01
Four steps to an empirical formula.
Enter a composition by mass in the simulator and follow its working: divide by Ar, divide by the smallest, then scale to whole numbers. Add an Mr and it finds the molecular formula too.
1. Mass or %: write down each element's mass (or percentage, which is the mass in 100 g).
2. Divide by Ar: this gives the moles of atoms of each element.
3. Divide by the smallest: gives the ratio with the smallest number set to 1.
4. Whole numbers: if a value ends near .5, double everything; near .33 or .67, multiply by 3. Round only when a value is within about 0.1 of a whole number.
Stage 1 · Learn
Check what the sim just showed you
Four quick checks. Each correct answer earns XP and records this skill on your profile.
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The empirical formula of a compound is:
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What is the empirical formula of glucose, C6H12O6?
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A hydrocarbon contains carbon and hydrogen atoms in the ratio 1 : 2.5. Its empirical formula is:
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A compound has the empirical formula CH2 and Mr = 70.0. Its molecular formula is:
Section 02
From empirical to molecular.
Several different compounds can share one empirical formula. The relative molecular mass tells them apart.
Empirical formula CH2 has mass 14.0. For an alkene with Mr = 56.0, n = 56.0 ÷ 14.0 = 4, so the molecular formula is C4H8.
Glucose (C6H12O6), ethanoic acid (C2H4O2) and methanal (CH2O) all have the empirical formula CH2O.
Ionic compounds only have an empirical formula: NaCl describes the ratio of ions in a lattice, not a molecule.
Section 03
Combustion data and water of crystallisation.
Two classic variants: find carbon and hydrogen from the carbon dioxide and water formed on burning, and find x in a hydrated salt from the mass lost on heating.
Combustion: 0.290 g of a compound of C, H and O burns to give 0.660 g CO2 and 0.270 g H2O. Mass of C = 0.660 × 12.0/44.0 = 0.180 g; mass of H = 0.270 × 2.0/18.0 = 0.030 g; mass of O = 0.290 − 0.180 − 0.030 = 0.080 g. Moles: C 0.0150, H 0.030, O 0.0050, a ratio of 3 : 6 : 1, so C3H6O.
Hydrated salt: 2.50 g of CuSO4·xH2O is heated to constant mass, leaving 1.60 g of anhydrous CuSO4 (Mr 159.6). Water lost = 0.90 g = 0.0500 mol; CuSO4 = 0.0100 mol; x = 0.0500 ÷ 0.0100 = 5.
Examiner trap
Do not round a ratio like 1 : 1.5 to 1 : 2 or 1 : 1.33 to 1 : 1. Those are real ratios of 2 : 3 and 3 : 4, and rounding them early gives the wrong formula (Fe2O3 is 1 : 1.5). In the hydrate experiment, heat to constant mass: if some water remains, the water lost is too small and x comes out too low.
Stage 2 · Exam
Exam-style questions
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A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. What is its empirical formula? (Ar: H 1.0, C 12.0)
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0.72 g of magnesium burns to form 1.20 g of oxide. What is the empirical formula of the oxide? (Ar: O 16.0, Mg 24.3)
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4.92 g of hydrated magnesium sulfate, MgSO4·xH2O, is heated to constant mass, leaving 2.40 g of MgSO4 (Mr 120.4). What is x? (Mr of H2O = 18.0)
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Complete combustion of 0.100 mol of a hydrocarbon produces 0.400 mol of CO2 and 0.500 mol of H2O. What is its molecular formula?
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Empirical and molecular formulae, mastered.
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