Establish the molecular formula of an alkene, the hydration of which produces alcohol

Establish the molecular formula of an alkene, the hydration of which produces alcohol, the vapor density of which is 2.07 times heavier than air.

Since the condition does not say otherwise, we assume that the alcohol obtained in this reaction is saturated and monohydric.

Let’s write the reaction equation:

CnH2n + H2O (H +, t) = CnH2n + 1OH, where CnH2n + 1 is an unknown alkyl radical.

Let’s find the molecular weight of the unknown alcohol:

M (CnH2n + 1OH) = M (air) * Dair (CnH2n + 1OH) = 29 * 2.07 = 60 (amu).

Let’s find the molecular weight of the alkyl radical:

M (CnH2n + 1) = M (CnH2n + 1OH) – M (OH) = 60 – 17 = 43 (amu).

Let’s compose and solve the equation:

12 * n + 1 * (2n + 1) = 43,

n = 3, whence the molecular formula of alkene: C3H6 (which corresponds to propene or propylene).

Answer: C3H6.