The hydrocarbon contains 82.76% carbon. determine its molecular formula, given that the density in air is 2.

Given:
CxHy
ω (C in CxHy) = 82.76%
D air. (CxHy) = 2

To find:
CxHy -?

Decision:
1) M (CxHy) = D air. (CxHy) * M (air) = 2 * 29 = 58 g / mol;
2) Mr (CxHy) = M (CxHy) = 58;
3) N (C in CxHy) = (ω (C in CxHy) * Mr (CxHy)) / (Ar (C) * 100%) = (82.76% * 58) / (12 * 100%) = 4 ;
4) ω (H in CxHy) = 100% – ω (C in CxHy) = 100% – 82.76% = 17.24%;
5) N (H in CxHy) = (ω (H in CxHy) * Mr (CxHy)) / (Ar (H) * 100%) = (17.24% * 58) / (1 * 100%) = 10 ;
6) Unknown substance – C4H10 – butane.

Answer: Unknown substance – C4H10 – butane.