Hydrocarbon chemical analysis shows that it contains 85.71% C and 14.29% H.

Hydrocarbon chemical analysis shows that it contains 85.71% C and 14.29% H. The relative density of the hydrocarbon in air is 1.45. Determine the molecular formula of the hydrocarbon.

According to the condition, in the unknown hydrocarbon the following mass ratio:

w (C): w (H) = 85.71: 14.29

which, taking into account atomic weights, turns into the following quantitative ratio:

n (C): n (H) = 85.71 / 12: 14.29 / 1 = 7.14: 14.29 = 1: 2

Thus, the general formula for the unknown hydrocarbon is CnH2n.

Molecular weight of unknown hydrocarbon:

M (CnH2n) = Dair (CnH2n) * M (air) = 1.45 * 29 = 42 (amu)

Let’s compose and solve the equation:

12n + 2n = 42, n = 3,

whence the hydrocarbon formula: C3H6, which corresponds to propene and cyclopropane