Find the molecular formula of a hydrocarbon with a mass fraction of hydrogen of 15.79%.

Find the molecular formula of a hydrocarbon with a mass fraction of hydrogen of 15.79%. The relative density of the vapor of this substance with respect to nitrogen

Given:
CxHy
ω (H) = 15.79%
D N2 (CxHy) = 4.07

To find:
CxHy -?

1) ω (C) = 100% – ω (H) = 100% – 15.79% = 84.21%;
2) M (N2) = Mr (N2) = Ar (N) * N (N) = 14 * 2 = 28 g / mol;
3) M (CxHy) = D N2 (CxHy) * M (N2) = 4.07 * 28 = 114 g / mol;
4) Mr (CxHy) = M (CxHy) = 114;
5) x = (ω (C in CxHy) * Mr (CxHy)) / (Ar (C) * 100%) = (84.21% * 114) / (12 * 100%) = 8;
6) y = (ω (H in CxHy) * Mr (CxHy)) / (Ar (H) * 100%) = (15.79% * 114) / (1 * 100%) = 18;
7) Unknown compound – C8H18.

Answer: Unknown compound – C8H18.