What is the empirical formula of vanadium oxide if 2.74g of the proof vests is 55.83% of V, 44.17% of O?( at.wt=51 for V) 

Mass of oxide =mass of metal +mass of oxygen

$\begin{array}{l}2.74=1.53+{\mathrm{W}}_{\text{oxygen }}\Rightarrow {\mathrm{W}}_{\text{oxygen }}=1.21 \mathrm{g}\\ \text{ moles of }\mathrm{V}=\frac{1.53}{51}=0.03\\ \text{ moles of }\mathrm{O}=\frac{1.21}{16}=0.075\\ {\mathrm{V}}_{0.03}{\mathrm{O}}_{0.075}\\ ={\mathrm{V}}_2{\mathrm{O}}_5\end{array}$