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A particular non-stoichiometric metal oxide has the determined formula . Assuming the metallic elemeRedox Reactions and Volumetric Analysis Chemistry Question

Question

A particular non-stoichiometric metal oxide has the determined formula $M_{0.98}O$ . Assuming the metallic element $M$ structurally exists within the crystal lattice strictly as a mixture of $M^{2+}$ and $M^{3+}$ ions, what is the exact numerical percentage of metal $M$ present specifically in the $+3$ oxidation state?

Answer: A

💡 Solution & Explanation

Let the number of trivalent $M^{3+}$ ions in the formula be denoted as $x$ . The number of divalent $M^{2+}$ ions must be $0.98 - x$ . To maintain total electrical neutrality against the single oxygen atom's $-2$ charge: $3x + 2(0.98 - x) = 2 \Rightarrow 3x + 1.96 - 2x = 2 \Rightarrow x = 0.04$ . The total combined number of metal ions is $0.98$ . The precise fractional abundance of $M^{3+}$ is $0.04 / 0.98 \approx 0.04081$ . Converting to a percentage gives approximately $4.08\%$ .

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