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$1.25 \mathrm{~g}$ of a metal (M) reacts with oxygen completely to produce $1.68 \mathrm{~g}$ of metal oxide. The empirical formula of the metal oxide is
[molar mass of $\mathrm{M}$ and $\mathrm{O}$ are $69.7 \mathrm{~g} \mathrm{~mol}^{-1}$ and $16.0 \mathrm{~g} \mathrm{~mol}^{-1}$, respectively]
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[molar mass of $\mathrm{M}$ and $\mathrm{O}$ are $69.7 \mathrm{~g} \mathrm{~mol}^{-1}$ and $16.0 \mathrm{~g} \mathrm{~mol}^{-1}$, respectively]
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Verified Answer
The correct answer is:
$\mathrm{M}_{2} \mathrm{O}_{3}$
$\Rightarrow \quad \frac{1.25}{\mathrm{E}}=\frac{1.68}{\mathrm{E}+8}$ $\Rightarrow \quad \mathrm{E}=23.25$
$\quad \mathrm{n}-$ factor $=\frac{69.7}{23.25} \approx 3$
$\therefore \quad$ Emprical formula $=\mathrm{M}_{2} \mathrm{O}_{3}$
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