By burning carbon with a mass of 11.2 g, 35.2 g were obtained. carbon monoxide and 14.4 g of water.

By burning carbon with a mass of 11.2 g, 35.2 g were obtained. carbon monoxide and 14.4 g of water. The relative density of substances in the air is 1.93. Find the molecular formula of carbon.

Given:
m (in-va) = 11.2 g
m (CO2) = 35.2 g
m (H2O) = 14.4 g
D air. (in-va) = 1.93

Find: substance -?

Decision:
1) n (CO2) = m (CO2) / M (CO2) = 35.2 / 44 = 0.8 mol;
2) n (C in water) = n (C in CO2) = n (CO2) = 0.8 mol;
3) m (C in in-ve) = n (C in in-ve) * Ar (C) = 0.8 * 12 = 9.6 g;
4) n (H2O) = m (H2O) / M (H2O) = 14.4 / 18 = 0.8 mol;
5) n (H in in-ve) = n (H in H2O) = n (H2O) * 2 = 0.8 * 2 = 1.6 mol;
6) m (H in in-ve) = n (H in in-ve) * Ar (H) = 1.6 * 1 = 1.6 g;
7) m (O in in-ve) = m (in-v) – m (C in in-ve) – m (H in in-ve) = 11.2 – 9.6 – 1.6 = 0 g ;
8) C (x) H (y)
x: y = n (C in in-ve): n (H in in-ve) = 0.8: 1.6 = 1: 2;
CH2 is the simplest formula;
9) M (in-va) = D air. (in-va) * M (air) = 1.93 * 29 = 55.97 ≈ 56 g / mol;
10) M (CH2) = Mr (CH2) = Ar (C) + Ar (H) = 12 + 1 * 2 = 14 g / mol;
11) M (in-va) = M (CH2) * 4;
12) Unknown substance – C4H8 – butene.

Answer: Unknown substance – C4H8 – butene.