what is the empirical formula of vanadium oxide, if 2.74 g of the metal oxide contains 1.53 g of metal ?

Metal oxide = 2.74 g;
wt. of vanadium = 1.53 g 
$\mathrm{\%}\text{ of }\mathrm{V}=\frac{1.53}{2.74}\times 100=55.83$
Thus, % of O = 100  -55.83=44.17
No. of moles of $\mathrm{V}=\frac{55.83}{52}=1.1$
No. of mores of $\mathrm{O}=\frac{44.17}{16}=2.76$
Simplest ratio of V and O = 1:2.5 or 2 :5
Hence, the empirical formula : V₂O₅