Each 9.4 gm of compound contains 7.2 gm carbon, 0.6 gm hydrogen, and the rest oxygen. The empirical formula of the compound is :

Given,

Carbon has 7.2 gm

Hydrogen has 0.6 gm

$\left.\begin{array}{l}{n}_{\mathrm{C}}=\frac{7.2}{12}=0.6\\ n_{\mathrm{H}}=\frac{0.6}{1}=0.6\\ n_{\mathrm{O}}=\frac{1.6}{16}=0.1\end{array}\right\}{\mathrm{C}}_6{\mathrm{H}}_6\mathrm{O}$

Thus,  the empirical formula of the compound is ${\mathrm{C}}_6{\mathrm{H}}_6\mathrm{O}$.