An organic compound in which the mass fraction of carbon is 64.9%, oxygen is 21.6%,

An organic compound in which the mass fraction of carbon is 64.9%, oxygen is 21.6%, hydrogen is 13.5%, and the relative density of vapor in air is 2.55. determine the molecular formula of the substance, make up the formulas of possible isomers and name them according to the systematic nomenclature.

Given:
CxOyHz
ω (C) = 64.9%
ω (O) = 21.6%
ω (H) = 13.5%
D air. (CxOyHz) = 2.55

Find:
CxOyHz -?

1) M (CxOyHz) = D air. (CxOyHz) * M (air) = 2.55 * 29 = 74 g / mol;
2) Mr (CxOyHz) = M (CxOyHz) = 74;
3) x = (ω (C) * Mr (CxOyHz)) / (Ar (C) * 100%) = (64.9% * 74) / (12 * 100%) = 4;
4) y = (ω (O) * Mr (CxOyHz)) / (Ar (O) * 100%) = (21.6% * 74) / (16 * 100%) = 1;
5) z = (ω (H) * Mr (CxOyHz)) / (Ar (H) * 100%) = (13.5% * 74) / (1 * 100%) = 10;
6) C4OH10;
Answer: Unknown substance – C4OH10.