When 1.45 g of organic matter is burned, 1.12 l of carbon dioxide and 0.9 g of water are obtained.

When 1.45 g of organic matter is burned, 1.12 l of carbon dioxide and 0.9 g of water are obtained. The hydrogen vapor density of the substance is 45. Determine the formula of the organic substance.

Given:
m (RH) = 1.45 g
V (CO2) = 1.12 l
m (H2O) = 0.9 g
D H2 (ОВ) = 45

To find:
ОВ -?

Decision:
1) n (CO2) = V (CO2) / Vm = 1.12 / 22.4 = 0.05 mol;
2) n (C) = n (CO2) = 0.05 mol;
3) m (C) = n (C) * M (C) = 0.05 * 12 = 0.6 g;
4) n (H2O) = m (H2O) / M (H2O) = 0.9 / 18 = 0.05 mol;
5) n (H) = n (H2O) * 2 = 0.05 * 2 = 0.1 mol;
6) m (H) = n (H) * M (H) = 0.1 * 1 = 0.1 g;
7) m (O) = m (ОВ) – m (C) – m (H) = 1.45 – 0.6 – 0.1 = 0.75 g;
8) n (O) = m (O) / M (O) = 0.75 / 16 = 0.05 mol;
9) x: y: z = n (C): n (H): n (O) = 0.05: 0.1: 0.05 = 1: 2: 1;
10) M (CH2O) = Mr (CH2O) = Ar (C) * N (C) + Ar (H) * N (H) + Ar (O) * N (O) = 12 * 1 + 1 * 2 + 16 * 1 = 30 g / mol;
11) M (ОВ) = D H2 (ОВ) * M (H2) = 45 * 2 = 90 g / mol;
12) M (OB) / M (CH2O) = 90/30 = 3;
13) C3H6O3.

Answer: Unknown substance – C3H6O3.