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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?

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

Let the number of $M^{2+}$ ion = $x$
Then the number of $M^{3+}$ ion will be =$(0.96-x)$
Total no. of cations =$2x+3(0.96-x)$
$\qquad\qquad\quad\quad\;\;=2x+2.88-3x$
$\qquad\qquad\quad\quad\;\;=-x+2.88$
No. of anions $O^{2-}=-2\times 1=-2$
No. of cations =No. of anions
$-x+2.88=2$
$-x=-2.88+2$
$-x=-0.88$
% 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\%$