Establish the molecular formula of a substance that contains hydrogen, carbon and oxygen .

Establish the molecular formula of a substance that contains hydrogen, carbon and oxygen in a mass ratio of 1: 6: 4. The hydrogen density of this substance is 22.

Given:
CxHyOz
m (H): m (C): m (O) = 1: 6: 4
D H2 (CxHyOz) = 22

Find:
CxHyOz -?

Solution:
1) M (CxHyOz) = D H2 (CxHyOz) * M (H2) = 22 * 2 = 44 g / mol;
2) Mr (CxHyOz) = M (CxHyOz) = 44;
3) Let m (CxHyOz) = 11 g;
4) ω (H in CxHyOz) = m (H in CxHyOz) * 100% / m (CxHyOz) = 1 * 100% / 11 = 9.09%;
5) ω (C in CxHyOz) = m (C in CxHyOz) * 100% / m (CxHyOz) = 6 * 100% / 11 = 54.55%;
6) ω (O in CxHyOz) = 100% – ω (H in CxHyOz) – ω (C in CxHyOz) = 100% – 9.09% – 54.55% = 36.36%;
7) N (H in CxHyOz) = (ω (H in CxHyOz) * Mr (CxHyOz)) / (Ar (H) * 100%) = (9.09% * 44) / (1 * 100%) = 4 ;
8) N (C in CxHyOz) = (ω (C in CxHyOz) * Mr (CxHyOz)) / (Ar (C) * 100%) = (54.55% * 44) / (12 * 100%) = 2 ;
9) N (O in CxHyOz) = (ω (O in CxHyOz) * Mr (CxHyOz)) / (Ar (O) * 100%) = (36.36% * 44) / (16 * 100%) = 1 ;
Unknown substance – C2H4O – acetaldehyde.

Answer: Unknown substance – C2H4O – acetaldehyde.