The hydrocarbon has an elemental composition of 82.76% carbon and 17.24% hydrogen (by weight).

The hydrocarbon has an elemental composition of 82.76% carbon and 17.24% hydrogen (by weight). When chlorination (radical) hydrocarbon forms two isomeric monochlorides – primary and tertiary. Determine the structure of the starting hydrocarbon.

Given:
CxHy
ω (C in CxHy) = 82.76%
ω (H in CxHy) = 17.24%

To find:
CxHy -?

Decision:
1) Let m (CxHy) = 100 g;
2) m (C in CxHy) = ω (C in CxHy) * m (CxHy) / 100% = 82.76% * 100/100% = 82.76 g;
3) n (C in CxHy) = m (C in CxHy) / M (C) = 82.76 / 12 = 6.897 mol;
4) m (H in CxHy) = ω (H in CxHy) * m (CxHy) / 100% = 17.24% * 100/100% = 17.24 g;
5) n (H in CxHy) = m (H in CxHy) / M (H) = 17.24 / 1 = 17.24 mol;
6) x: y = n (C in CxHy): n (H in CxHy) = 6.897: 17.24 = 1: 2.5 = 4: 10;
7) Unknown substance – C4H10 – methylpropane.

Answer: Unknown substance – C4H10 – methylpropane.