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

V_{2} O_{3}

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V O

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V_{2} O_{5}

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V_{2} O_{7}

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Solution

The correct option is C $V_{2} O_{5}$
Metal oxide $= 2.74 g$ ;
wt. of vanadium $= 1.53 g$
$%$ of $V = \frac{1.53}{2.74} \times 100 = 55.83$
Thus, $%$ of $O = 100 - 55.83 = 44.17$
No. of moles of $V = \frac{55.83}{52} = 1.1$
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 $= V_{2} O_{5}$

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2.74 g

1.53 g

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2.74 g

1.53 g

2.73

1.53

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