Question

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)

Hint:

In stoichiometry problems, carefully balance the charges and mole ratios to determine the proportions of each oxidation state.

Answer 1

Correct Answer: 59

To determine the percentage of metal ions in the +2 oxidation state in the compound $M_{0.83}O_{1.00}$, we follow these steps:

The metal $M$ exists in two oxidation states: +2 and +3. The overall charge of the compound must be zero. Given the formula $M_{0.83}O_{1.00}$, this implies 0.83 moles of $M$ combines with 1.00 mole of oxygen (Oxygen having a -2 charge).

Let's assume $x$ moles of $M$ are in the +2 oxidation state and $y$ moles of $M$ are in the +3 oxidation state. Then:

• The total moles of $M$: $x + y = 0.83$
• Charge balance equation: $2x + 3y = 2 \times 1.00 = 2$

We have these two equations:
$x + y = 0.83$
$2x + 3y = 2$

Solving for $x$ and $y$:

1. From $x + y = 0.83$, we get $y = 0.83 - x$
2. Substitute $y$ in the second equation: $2x + 3(0.83 - x) = 2$
3. Simplifying: $2x + 2.49 - 3x = 2$, which gives $-x + 2.49 = 2$, thus $x = 0.49$
4. As $x + y = 0.83$, $y = 0.34$

Therefore, the moles of $M^{2+}$ ions are 0.49. The percentage of $M$ ions in the +2 oxidation state is calculated by:

$\text{Percentage of } M^{2+} = \left(\frac{x}{x+y}\right) \times 100 = \left(\frac{0.49}{0.83}\right) \times 100 ≈ 59$

The computed percentage ≈ 59, which lies within the expected range.