A compound having empirical formula (C₃H₄O)_n has vapour density 84. The molecular formula of this compound is

$\text{ Molecular mass }=2\times \text{ V.D }=2\times 84=168$

$\begin{array}{r}n=\frac{\text{ Molecular mass }}{\text{ Empirical formula mass }}\\ =\frac{168}{12\times 3+4+16}=\frac{168}{56}=3\end{array}$

$\text{ Thus, }\text{ molecular formula }=\left({\mathrm{C}}_3{\mathrm{H}}_4\mathrm{O}\right)\times 3$

$={\mathrm{C}}_9{\mathrm{H}}_{12}{\mathrm{O}}_3$