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

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Solution

The formula $M_{0.96} O_{1.00}$ show that $M : O = 0.96 : 1.00 = 96 : 100$ . Thus, if there are $100$ O-atms, then M atoms $= 96$
Charge on $100 O^{2 -}$ ions $= 100 \times (- 2) = - 200$
Suppose M atoms as $M^{2 +} = x$ .
Then M atoms as $M^{3 +} = 96 - x$ .
Total charge on $M^{2 +}$ and $M^{3 +} = (+ 2) x + (+ 3) (96 - x) = (288 - x)$
As metal oxide is neutral, total charge on cations $=$ total charge on anions
$∴ (288 - x) = 200$ or $x = 88$

$∴ %$ of $M$ as $M^{2 +} = \frac{88}{96} \times 100 = 91.7 %$

$%$ of $M$ as $M^{3 +} = 100 - 91.7 = 9.3 %$

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