A metal oxide contains $70 %$ metal with a molar mass of metal is $56$ and the molar mass of the compound is $160$ . Its empirical formula will be:

M_{2} O_{5}

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M_{2} O_{3}

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M O

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M O_{3}

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Solution

The correct option is A $M_{2} O_{3}$
Let us consider the problem
Given that
% of metal $= 70 %$
% of oxygen $100 - 70 = 30 %$
Calculate the empirical formula formula :-
Metal (M) :- Atomic mass $= 56$
Atomic ratio $= \frac{70}{56} = 1.25$
Simplest ratio $= \frac{1.25}{1.25} = 1$
Simplest whole number ratio $= 2$
Similarly ;
Oxygen (o) $\to$ Atomic mass $16$
Atomic ratio $= \frac{30 - 0}{16} = 1.875$
Simplest ratio $= \frac{1.875}{1.35} = 1.5$
Simplest whole number ratio $= 3$
Hence the empirical formula of metal oxide is $M_{2} O_{3}$

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An oxide of metal M has 40% by mass of oxygen. Metal M has relative atomic mass of 24g/mol. The empirical formula of the oxide is:

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