4 Combustion analysis

When a compound is burned completely, all its carbon appears as CO₂ and all its hydrogen as H₂O. Therefore

mass of C = mass of CO₂ × (12/44), mass of H = mass of H₂O × (2/18).

• If the compound contains only C and H, these two masses add up to the sample mass.
• If they add up to less, the difference is oxygen (found by subtraction).

Worked example 5.8 — hydrocarbon (pattern of the book). A hydrocarbon of molar mass 78 g/mol, mass 3.9 g, burns to give 13.2 g CO₂ and 2.7 g H₂O [CO₂ = 44, H₂O = 18]. Mass of C = 13.2 × 12/44 = 3.6 g; mass of H = 2.7 × 2/18 = 0.3 g (total 3.9 g, so no oxygen). Moles: C = 0.3; H = 0.3 → CH (mass 13); n = 78/13 = 6 → C₆H₆.

Worked example 5.9 — compound with oxygen. 2.3 g of an organic compound C, H, O gives 4.4 g CO₂ and 2.7 g H₂O. C = 4.4 × 12/44 = 1.2 g; H = 2.7 × 2/18 = 0.3 g; O = 2.3 − 1.5 = 0.8 g. Moles: C 0.1, H 0.3, O 0.05. Divide by 0.05: C₂H₆O (empirical). With molar mass 46 it is also the molecular formula.

Worked example 5.10 — finding x and y. A hydrocarbon CₓH_y (M = 44) yields 13.2 g CO₂ and 7.2 g H₂O from 4.4 g. C = 13.2 × 12/44 = 3.6 g, H = 7.2 × 2/18 = 0.8 g. Moles C 0.3, H 0.8 → ratio 1 : 2.67 → ×3 → C₃H₈ (44). So x = 3, y = 8.