Question

The empirical formula of a compound is $C{H_2}O$ . If its vapour density is $90$; find out the molecular formula of the compound.
(A) ${C_5}{H_{10}}{O_5}$
(B) ${C_4}{H_8}{O_4}$
(C) ${C_6}{H_{12}}{O_6}$
(D) ${C_3}{H_6}{O_3}$

Answer 1

Hint: The simplest whole number ratio of different atoms present in a compound is known as empirical formula. Whereas, molecular formula gives the exact number of different atoms present in a compound. If the empirical formula and molar mass of a compound is known we can calculate the molecular formula of the compound.

Complete step by step answer:
$\dfrac{{\text{molar mass}}}{{\text{empirical formula mass}}} = n $
For calculation of molecular mass in this question:
Given, empirical formula of compound = $C{H_2}O$.
And vapour density of compound = \[90\].
Molar mass of the compound = $ \text{ vapour density} \times 2 \\
   = 90 \times 2 \\
   = 180g \\
$
Empirical formula mass = $
  12 + 2(1) + 16 \\
   = 12 + 2 + 16 \\
   = 30g \\
 $
Dividing molar mass by empirical formula mass
$ \dfrac{{\text{molar mass}}}{{\text{empirical formula mass}}} = n \\
   = \dfrac{{180g}}{{30g}} = 6 \\
 $
Therefore, the molecular formula of the compound is ${C_6}{H_{12}}{O_6}$.

Hence, the correct option is (C) ${C_6}{H_{12}}{O_6}$ .

Note:
When molar mass is divided by empirical formula mass the result should always be in whole numbers. This is because different elements combine in the ratio of whole numbers to form a compound according to the law of definite proportions.