A sample of a metal oxide has formula M_(0.83) O_(1.00). The metal M can exist in two oxidation states +2 and +3.
In the sample of M_(0.83) O_(1.00), the percentage of metal ions existing in +2 oxidation state is __________ %. (nearest integer)

Solution

Chemistry ma’am · 4 steps

Two unknowns, two facts: the metal ions add up to 0.83 and their charges must balance the oxide.

1. Set up unknowns. Let a be the number of M²⁺ ions and b the number of M³⁺ ions per formula unit. Together they make up all the metal, so a + b = 0.83.
2. Balance the charge. One O²⁻ ion carries a charge of -2, and the compound is neutral. So the metal ions must supply a total charge of +2.

   2a + 3b = 2

3. Solve. Put a = 0.83 - b into the charge equation: 1.66 + b = 2, so b = 0.34 and a = 0.49.
4. So. Percentage of M²⁺ = 0.49/0.83 × 100 = 59.04%, which rounds to 59.

The common slip: 41. Reporting the share of M³⁺, 0.34/0.83 × 100, gives 41, not 59.

The idea to keep. For non-stoichiometric oxides, use two equations: the total metal count and electrical neutrality. Take the percentage out of the total metal, not out of 1.

Answer from the NTA final answer key. Solution by PrepWiser.