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Analysis shows that a metal oxide has the empirical formula $M_{0.96}O_{1.00}$.Calculate the percentage of $M^{2+}$ and $M^{3+}$ ions in the crystal?

$\begin{array}{1 1}81.67\%,7.33\%\\91.67\%,8.33\%\\71.67\%,6.33\%\\61.67\%,5.33\%\end{array} $

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Let the number of $M^{2+}$ ion = $x$
Then the number of $M^{3+}$ ion will be =$(0.96-x)$
Total no. of cations =$2x+3(0.96-x)$
No. of anions $O^{2-}=-2\times 1=-2$
No. of cations =No. of anions
% of $M^{2+}$ ion =$\large\frac{0.88}{0.96}$$\times 100=91.67\%$
% of $M^{3+}$ ion =$\large\frac{0.08}{0.96}$$\times 100=8.33\%$
answered Aug 4, 2014 by sreemathi.v

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