1.25 g of a metal (M) reacts with oxygen completely to produce 1.68 g of metal oxide. The empirical formula of the metal oxide is
[molar mass of M and O are 69.7 $g m o l^{- 1}$ and 16.0 $g m o l^{- 1}$ , respectively]

M_{2} O

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M_{2} O_{3}

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M O_{2}

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M_{3} O_{4}

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Solution

The correct option is B $M_{2} O_{3}$
$M + O_{2} \to M_{x} O_{y} P e r c e n t a g e o f M i n m e t a l o x i d e = \frac{1.25}{1.68} \times 100 = 74 P e r c e n t a g e o f O i n m e t a l o x i d e = 1 - 74 = 26$
Divide percentage of component present by molar mass,
$M : \frac{74}{69.7} = 1.06 O : \frac{26}{16} = 1.625$
Multiply by 2 to get whole number.
So the empirical formula of oxide is $M_{2} O_{3}$

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