With complete combustion of the hydrocarbon, 13.2 g of carbon dioxide and 2.7 g of water were obtained.

With complete combustion of the hydrocarbon, 13.2 g of carbon dioxide and 2.7 g of water were obtained. The vapor density of the substance is 3.482 g / l. Derive the molecular formula of the hydrocarbon.

Given:
m (CO2) = 13.2 g
m (H2O) = 2.7 g
ρ (HC) = 3.482 g / l

To find:
UV -?

Decision:
1) n (CO2) = m (CO2) / Mr (CO2) = 13.2 / 44 = 0.3 mol;
2) n (C in HC) = n (C in CO2) = n (CO2) = 0.3 mol;
3) n (H2O) = m (H2O) / Mr (H2O) = 2.7 / 18 = 0.15 mol;
4) n (H in HC) = n (H in H2O) = n (H2O) * 2 = 0.15 * 2 = 0.3 mol;
5) C (x) H (y)
x: y = m (C in HC): m (C in HC) = 0.3: 0.3 ≈ 1: 1;
CH is the simplest formula;
6) Mr (HC) = ρ (HC) * Vm = 3.482 * 22.4 = 78 g / mol;
7) Mr (CH) = Ar (C) + Ar (H) = 12 + 1 = 13 g / mol;
8) Mr (HC) = Mr (CH) * 6;
9) Unknown substance – C6H6 – benzene.

Answer: Unknown substance – C6H6 – benzene.