Question

Empirical formula of a compound is CH₂O. If its empirical formula mass is equal to its vapour density. Calculate the molecular formula of the compound.

Answer 1

We know that molecular mass and vapour density of a gas is given by the following relation:
Vapour density $\times$ 2 = Molecular mass
Vapour density = $\frac{\mathrm{Molecular}\mathrm{mass}}{2}$ ....(i)
In the question, it is stated that vapour density is equal to empirical formula mass of the compound CH₂O.
∴ Vapour density = Empirical formula mass ....(ii)
Equating equations (i) and (ii), we get:

$$\begin{gathered} \frac{\mathrm{Molecular}\mathrm{mass}}{2}=\mathrm{Empirical}\mathrm{formula}\mathrm{mass} \\ \mathrm{Molecular}\mathrm{mass}=2\times \mathrm{Empirical}\mathrm{formula}\mathrm{mass} \end{gathered}$$

Thus, molecular formula of the compound is double the empirical formula. Hence, the molecular formula of the compound is C₂H₄O₂ or CH₃COOH.