Question

Analysis shows that a metal oxide has the empirical formula M0.96 O1.00. Calculate the percentage of M2+ and M3+
ions in this crystal?

Answer 1

Formula of metal oxide is given, M₀.₉₆O₁.₀₀
Let the number of M²⁺ ions in metal oxide = x
Then, number of M³⁺ ions in metal oxide = 0.96 - x
so, total positive charge = number of M²⁺ ions × oxidation of M²⁺ + number of M³⁺ ions × oxidation of M³⁺
= 2x + 3(0.96 - x) = 2x + 2.88 - 3x
= 2.88 - x

Because metal oxide is neutral so, total positive charge = total negative charge
2.88 - x = 2 [ ∵ oxidation of O is - 2]
2.88 - 2 = x ⇒x = 0.88
so, number of M³⁺ ions in metal oxide = 0.96 - 0.88 = 0.08
Now, percentage of M²⁺ ions in Metal oxide = 0.88/0.96 × 100 =91.67%
percentage of M³⁺ ions in metal oxide = 0.08/0.96 × 100 = 8.33%

Answer 2

Answer:

The ratio of M2+ and O2− ions in the pure sample of metal oxide = 1:1. Let x ions M2+ be replaced by M3+ ions in the given sample.

No. of M2+ 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 M3+ions in the metal oxide=No. if M3+ionsTotal /no. of M atoms×100=0.080.96×100=8.33%

%of M2+ions in the metal oxide=(100−8.33)=91.67%