Question

The empirical formula of a compound is $C{H_2}O$ and its vapour density is $30.$. The molecular formula of the compound is:
A. ${C_3}{H_6}{O_3}$
B. ${C_2}{H_4}{O_2}$
C. $C{H_2}O$
D. ${C_2}{H_4}O$

Answer 1

Molecular formulas give the exact number of atoms of each element present in the molecular compound whereas Empirical formula gives the simplest whole-number ratio of atoms in a compound.
If the empirical formula and molecular mass of the compound is known we can calculate the molecular formula of the compound.

Complete step by step solution:
The molecular formula is the formula derived from molecules and is representative of the total number of individual atoms present in a molecule of a compound.
The molecular mass (or molecular weight) of a compound is a multiple of the empirical formula mass and uses a subscript that reports the actual number of each type of atom in a molecule of the compound. The molecular formula is commonly used and is a multiple of the empirical formula.
The general statement relating molecular formula and the empirical formula expressed as, $Molar{\text{ }}mass{\text{ }} = {\text{ }}n{\text{ }} \times {\text{ }}empirical{\text{ }}formula$
where $n$ is a whole number then $n = \dfrac{{{\text{ }}Molar{\text{ }}mass{\text{ }}}}{{empirical{\text{ }}formula}}$
Given, empirical formula of a compound = $C{H_2}O$
vapour density of a compound = $30$
Relation between Molar mass and Vapour density expressed as - $Molar{\text{ }}mass{\text{ }}of{\text{ }}compound{\text{ }} = {\text{ }}2 \times Vapour{\text{ }}density$
$= {\text{ }}2{\text{ }} \times 30{\text{ }} = {\text{ }}60g$
Empirical formula of a compound, $C{H_2}O$= $(Mass{\text{ }}of{\text{ }}C{\text{ }} + {\text{ }}2 \times Mass{\text{ }}of{\text{ }}H + {\text{ }}Mass{\text{ }}of{\text{ }}O)$
Total mass = $(12 + 2 + 16) = 30$
Dividing molar mass by Empirical formula mass = $n{\text{ }} = \dfrac{{Molar{\text{ }}mass}}{{empirical{\text{ }}formula}}$
$n{\text{ }} = \dfrac{{60}}{{30}} = 2$
Putting value of $n = 2$ in the empirical formula we get molecular formula as
the molecular formula for the compound is ${C_2}{H_4}{O_2}$

**So the correct option is (B)

Note:**
Divide the molar mass of the compound by the empirical formula mass. The result should be a whole number or very close to a whole number. This is because different elements combine in the ratio of whole numbers to form a compound according to the law of definite proportions. Sometimes, the empirical formula and molecular formula both can be the same.