# An alkene was obtained from an unknown monohydric alcohol, 14 g of which was added

**An alkene was obtained from an unknown monohydric alcohol, 14 g of which was added with 40 g of bromine. Establish the formula for alcohol.**

Let CnH2n + 1OH be the formula of an unknown monohydric alcohol, and CnH2n the formula of an unknown alkene.

Let’s write down the reaction equations:

CnH2n + 1OH (t, cat.) = CnH2n + H2O

CnH2n + Br2 = CnH2nBr2

Let’s find the amount of bromine substance:

v (Br2) = m (Br2) / M (Br2) = 40/160 = 0.25 (mol).

According to the reaction equation, 1 mol of Br2 reacts with 1 mol of CnH2n, therefore:

v (CnH2n) = v (Br2) = 0.25 (mol).

Let’s find the molar mass of the unknown alkene:

M (CnH2n) = m (CnH2n) / v (CnH2n) = 14 / 0.25 = 56 (g / mol).

Let’s compose and solve the equation:

12 * n + 1 * 2n = 56,

14n = 56,

n = 4, whence the formula of alcohol: C4H9OH (butanol).

Answer: C4H9OH.