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 the crystal.

91.75% and 8.3%

The correct answer is option 1. 91.75% and 8.3%.

The formula \(M_{0.96}O_{1.00}\) shows that \(M : O=\, \  0.96:1.00 =\, \  96:100\). Thus, if there are 100 oxygen atoms, then \(M\), then \(M\) atoms \( = 96\)

Charge on \(100\) \(O^{2-}\) ions\( = 100 × (-2) = -200\)

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, \(x = 88\)

∴ \(\% \text{of M as }M^{2+} = \frac{88}{96} × 100 = 91.7\%\)

∴ \(\% \text{ of M as }M^{3+} = 100 - 91.7 = 8.3\%\)