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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 sample. |
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Answer» The ratio of `M^(2+)` and `O^(2-)` ions in the pure sample of metal oxide = `1:1`. Let x ions `M^(2+)` be replaced by `M^(3+)` ions in the given sample. No. of `M^(2+)` ions present = `(0.96-x)` Since the oxide is neutral in nature, Total charge on M atoms = Charge on oxygen atoms `2(0.96-x)+3x=2` `1.92-2x+3x=2` `x=2-1.92=0.08` `%"of M"^(3+)"ions in the metal oxide"=("No. if M"^(3+)"ions")/("Total no. of M atoms")xx100=(0.08)/(0.96)xx100=8.33%` `%"of M"^(2+)"ions in the metal oxide"=(100-8.33)=91.67%` |
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