# Question #c4c4e

Oct 3, 2016

The empirical formula given should be ${C}_{2} {H}_{4} O$.

Then let the molecular formula be $\left({C}_{2} {H}_{4} O\right) n$.

Considering atomic masses as

$C \to 12 \text{g/mol}$

$O \to 16 \text{g/mol}$

$H \to 1 \text{g/mol}$

We can get calculated molar mass of the compound is $\left(2 \cdot 12 + 4 \cdot 1 + 16 \cdot 1\right) n = 44 n \text{ g/mol}$

Equating it with given molar mass we get

$44 n = 88.12$

$\implies n = \frac{88.12}{44} \approx 2$

Hence the MF of the compound is

${C}_{4} {H}_{8} {O}_{2}$