Set the molecular formula of gaseous carbon in which the mass fraction of carbon is 85.7℅

Set the molecular formula of gaseous carbon in which the mass fraction of carbon is 85.7℅, and the mass of its vapors with a volume of 10 dm3 is 18.75.

Given:
CxHy
ω (C in CxHy) = 85.7%
V (CxHy) = 10 dm3 = 10 l
m (CxHy) = 18.75 g

To find:
CxHy -?

Decision:
1) n (CxHy) = V (CxHy) / Vm = 10 / 22.4 = 0.45 mol;
2) M (CxHy) = m (CxHy) / n (CxHy) = 18.75 / 0.45 = 42 g / mol;
3) Mr (CxHy) = M (CxHy) = 42;
4) N (C in CxHy) = (ω (C in CxHy) * Mr (CxHy)) / (Ar (C) * 100%) = (85.7% * 42) / (12 * 100%) = 3 ;
5) ω (H in CxHy) = 100% – ω (C in CxHy) = 100% – 85.7% = 14.3%;
6) N (H in CxHy) = (ω (H in CxHy) * Mr (CxHy)) / (Ar (H) * 100%) = (14.3% * 42) / (1 * 100%) = 6 ;
Unknown substance – C3H6 – propene.

Answer: Unknown substance – C3H6 – propene.