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)

A
$V_{2} O_{3}$
B
VO
C
$V_{2} O_{5}$
D
$V_{2} O_{7}$

Solution:

Mass of oxide =mass of metal +mass of oxygen

$\begin{array}{l} 2.74 = 1.53 + W_{oxygen} \Rightarrow W_{oxygen} = 1.21 g \\ moles of V = \frac{1.53}{51} = 0.03 \\ moles of O = \frac{1.21}{16} = 0.075 \\ V_{0.03} O_{0.075} \\ = V_{2} O_{5} \end{array}$

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