What is the empirical formula of vanadium oxide, if $2.74\text{g}$ of the metal oxide contains $1.53\text{g}$ of metal?
(Molar Mass of Vanadium = $52\text{g}$)

The correct option is C $V_2{O}_5$
Metal oxide $=2.74\text{g}$;
wt. of vanadium $=1.53\text{g}$
$\%$ of $V=\frac{1.53}{2.74}\times 100=55.83$
Thus, $\%$ of $O=100-55.83=44.17$
No. of moles of $V=\frac{55.83}{52}\approx 1.1$
No. of moles of $O=\frac{44.17}{16}=2.76$
Simplest ratio of $V$ and $O=1:2.5$ or $2:5$
Hence, the empirical formula $=V_2{O}_5$