The cyclic hydrocarbon having no branches in the cyclic chain has a vapor density in air of 1.931.

The cyclic hydrocarbon having no branches in the cyclic chain has a vapor density in air of 1.931. the mass fraction of carbon in this substance is 85.7%. Determine the formula of the hydrocarbon and write down its structural formula.

Given:
CxHy
D air. (CxHy) = 1.931
ω (C in CxHy) = 85.7%

To find:
CxHy -?

Decision:
1) M (CxHy) = D air. (CxHy) * M (air) = 1.931 * 29 = 56 g / mol;
2) Mr (CxHy) = M (CxHy) = 56;
3) N (C in CxHy) = (ω (C in CxHy) * Mr (CxHy)) / (Ar (C) * 100%) = (85.7% * 56) / (12 * 100%) = 4 ;
4) ω (H in CxHy) = 100% – ω (C in CxHy) = 100% – 85.7% = 14.3%;
5) N (H in CxHy) = (ω (H in CxHy) * Mr (CxHy)) / (Ar (H) * 100%) = (14.3% * 56) / (1 * 100%) = 8 ;
6) Unknown substance – C4H8 – cyclobutane.

Answer: Unknown substance – C4H8 – cyclobutane.