30 g of saturated monohydric alcohol, taken in excess, was passed over red-hot copper (II) oxide.

30 g of saturated monohydric alcohol, taken in excess, was passed over red-hot copper (II) oxide. After the end of the reaction and the separation of water, the mass of organic reaction products was 29.2 g. When an excess of metallic potassium was added to this mixture, 1.12 liters of gas (NU) were released. Determine the molecular formula of the alcohol.

Given:
m (СnH2n + 1OH) = 30 g
m (org.product) = 29.2 g
V (gas) = ​​1.12 l

Find:
СnH2n + 1OH -?

Solution:
1) СnH2n + 1OH + CuO => СnH2nO + Cu + H2O;
2СnH2n + 1OH + 2K => 2СnH2n + 1OK + H2 ↑;
2) m (H gone from alcohol) = m (alcohol) – m (org.prod.) = 30 – 29.2 = 0.8 g;
3) n (H gone from alcohol) = m (H gone from alcohol) / M (H) = 0.8 / 1 = 0.8 mol;
4) n1 (alcohol) = n (H escaped from alcohol) / 2 = 0.8 / 2 = 0.4 mol;
5) n (H2) = V (H2) / Vm = 1.12 / 22.4 = 0.05 mol;
6) n2 (alcohol) = n (H2) * 2 = 0.05 * 2 = 0.1 mol;
7) n total (alcohol) = n1 (alcohol) + n2 (alcohol) = 0.4 + 0.1 = 0.5 mol;
8) M (alcohol) = m (alcohol) / n (alcohol) = 30 / 0.5 = 60 g / mol;
9) M (СnH2n + 1OH) = Mr (СnH2n + 1OH) = Ar (C) * N (C) + Ar (H) * N (H) + Ar (O) * N (O);
60 = 12 * n + 1 * (2n + 1) + 16 + 1;
14n = 42;
n = 3;
Unknown alcohol – C3H7OH – propanol.

Answer: Unknown alcohol – C3H7OH – propanol.