Derive the molecular formula of a hydrocarbon containing 88.89% carbon, by mass

Derive the molecular formula of a hydrocarbon containing 88.89% carbon, by mass, if you know that its density by air is 1.862. Give all possible formulas for the structural isomers of this substance.

Given:
CxHy
ω (C) = 88.89%
D air. (CxHy) = 1.862

To find:
CxHy -?

1) Calculate the mass fraction of H in CxHy:
ω (H) = 100% – ω (C) = 100% – 88.89% = 11.11%;
2) Calculate the molar mass of CxHy:
M (CxHy) = D air. (CxHy) * M (air) = 1.862 * 29 = 54 g / mol;
3) Determine the relative molecular weight of CxHy:
Mr (CxHy) = M (CxHy) = 54;
4) Calculate the number of C atoms in CxHy:
x = (ω (C in CxHy) * Mr (CxHy)) / (Ar (C) * 100%) = (88.89% * 54) / (12 * 100%) = 4;
5) Calculate the number of H atoms in CxHy:
y = (ω (H in CxHy) * Mr (CxHy)) / (Ar (H) * 100%) = (11.11% * 54) / (1 * 100%) = 6;
6) Unknown compound – C4H6.

Answer: Unknown compound – C4H6.