2 Finding the empirical formula

Percentage composition means mass of each element in 100 g of the compound. Steps:

1. Write the mass of each element (percentage → grams in a 100 g sample). If one percentage is missing, subtract the others from 100.
2. Divide each mass by its atomic mass to get moles of atoms.
3. Divide all by the smallest number of moles.
4. If a ratio is not a whole number, multiply all by the smallest integer that makes them whole: 0.5 → ×2; 0.33 or 0.67 → ×3; 0.25 or 0.75 → ×4; 0.2 → ×5. Do not just round ratios such as 1.5 or 2.33.

Worked example 5.1 (pattern of compound X). Compound with 85.7 % C and 14.3 % H [C = 12, H = 1]. C: 85.7/12 = 7.14; H: 14.3/1 = 14.3. Divide by 7.14: C = 1, H = 2.0. Empirical formula CH₂.

Worked example 5.2 — three elements. A compound is 40 % C, 6.67 % H and 53.33 % O [C = 12, H = 1, O = 16]. C: 3.33; H: 6.67; O: 3.33. Ratio 1 : 2 : 1 → CH₂O. If M = 180 g/mol, n = 180/30 = 6, molecular formula C₆H₁₂O₆ (glucose).

Worked example 5.3 — multiplier needed. Aluminium oxide: 52.94 % Al, 47.06 % O [Al = 27, O = 16]. Al: 1.96; O: 2.94. Ratio O/Al = 1.5, so Al : O = 1 : 1.5 → ×2 → Al₂O₃.

Worked example 5.4 — phosphorus oxide. 43.7 % P, 56.3 % O [P = 31, O = 16]. P: 1.41; O: 3.52; ratio 1 : 2.5 → ×2 → P₂O₅. If M = 284 g/mol, the empirical mass = 142, n = 2, molecular formula P₄O₁₀ (this matches the P₄O₁₀ in the chapter exercises).

Worked example 5.5 — masses given. 1.2 g of carbon is combined with 0.3 g of hydrogen: C = 0.1 mol, H = 0.3 mol → CH₃. With M = 30 g/mol: empirical mass 15, n = 2 → C₂H₆.