An aromatic hydrocarbon weighing 2.65 g, which is a homologue of benzene, was burned to obtain 4.48

An aromatic hydrocarbon weighing 2.65 g, which is a homologue of benzene, was burned to obtain 4.48 liters of carbon monoxide (4) Determine the molecular formula of this hydrocarbon.

Given:
m (CxHy) = 2.65 g
V (CO2) = 4.48 l

Find:
CxHy -?

Solution:
1) n (CO2) = V (CO2) / Vm = 4.48 / 22.4 = 0.2 mol;
2) n (C) = n (CO2) = 0.2 mol;
3) m (C) = n (C) * M (C) = 0.2 * 12 = 2.4 g;
4) m (H) = m (CxHy) – m (C) = 2.65 – 2.4 = 0.25 g;
5) n (H) = m (H) / M (H) = 0.25 / 1 = 0.25 mol;
6) x: y = n (C): n (H) = 0.2: 0.25 = 1: 1.25 = 4: 5 = 8: 10;
7) C8H10.

Answer: Unknown substance – C8H10.