An organic compound has an empirical formula (${\mathrm{CH}}_2\mathrm{O}$). Its vapour density is 45. The molecular formula of the compound is

$\text{ Mol. wt. }=2\times \text{ vapour density }=2\times 45=90$

Empirical formula weight = 12 + 2 +16 = 30

$\mathrm{n}=\frac{\text{ mol. wt. }}{\text{ empirical formula wt. }}=\frac{90}{30}=3$

$\therefore \text{ Molecular formula of the compound }={\left({\mathrm{CH}}_2\mathrm{O}\right)}_3={\mathrm{C}}_3{\mathrm{H}}_6{\mathrm{O}}_3$