Establish the molecular formula of a substance with a mass fraction of carbon 85.71%, hydrogen 14.29%.

Establish the molecular formula of a substance with a mass fraction of carbon 85.71%, hydrogen 14.29%. The relative vapor density of this substance for nitrogen is 1.5

Given:
CxHy
ω (C in CxHy) = 85.71%
ω (H in CxHy) = 14.29%
D N2 (CxHy) = 1.5

To find:
CxHy -?

Decision:
1) M (CxHy) = D N2 (CxHy) * M (N2) = 1.5 * 28 = 42 g / mol;
2) Mr (CxHy) = M (CxHy) = 42;
3) N (C in CxHy) = (ω (C in CxHy) * Mr (CxHy)) / (Ar (C) * 100%) = (85.71% * 42) / (12 * 100%) = 3 ;
4) N (H in CxHy) = (ω (H in CxHy) * Mr (CxHy)) / (Ar (H) * 100%) = (14.29% * 42) / (1 * 100%) = 6 ;
5) The molecular formula of the substance is C3H6 – propene.

Answer: The molecular formula of the substance is C3H6 – propene.