Question

a compound had a molecular weight of 112.124 atomic mass units and the empirical formula C3H4O what is the molecular formula of the compound?

Answer 1

Answer : The molecular of the compound is, $C_6H_8O_2$

Solution : Given,

Molecular weight of compound = 112.124 amu

The Empirical formula = $C_3H_4O_1$

First we have to calculate the empirical formula weight.

The empirical formula weight = 12(3) + 4(1) + 1(16) = 56 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}\\\\n=\frac{112.124}{56}=2$

Molecular formula = $(C_3H_4O_1)_n=(C_3H_4O_1)_2=C_6H_8O_2$

Therefore, the molecular of the compound is, $C_6H_8O_2$