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.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Define Variables
Let:
- \( x \) = number of \( M^{2+} \) ions
- \( 0.96 - x \) = number of \( M^{3+} \) ions (since the total number of metal ions is 0.96)

### Step 2: Set Up the Charge Balance Equation
The total positive charge from the metal ions must balance the negative charge from the oxide ions. The charge balance equation can be set up as follows:
\[ 2x + 3(0.96 - x) - 2 = 0 \]
Here, \( 2x \) is the contribution from \( M^{2+} \) ions, \( 3(0.96 - x) \) is the contribution from \( M^{3+} \) ions, and \( -2 \) accounts for the two negative charges from the two oxide ions.

### Step 3: Simplify the Equation
Expanding the charge balance equation:
\[ 2x + 2.88 - 3x - 2 = 0 \]
Combining like terms gives:
\[ - x + 0.88 = 0 \]

### Step 4: Solve for \( x \)
Rearranging the equation:
\[ x = 0.88 \]

### Step 5: Calculate the Number of \( M^{3+} \) Ions
Now, substituting \( x \) back into the expression for \( M^{3+} \):
\[ M^{3+} = 0.96 - x = 0.96 - 0.88 = 0.08 \]

### Step 6: Calculate the Percentages
Now, we can calculate the percentages of \( M^{2+} \) and \( M^{3+} \):
- Percentage of \( M^{2+} \):
\[ \text{Percentage of } M^{2+} = \left( \frac{x}{0.96} \right) \times 100 = \left( \frac{0.88}{0.96} \right) \times 100 \approx 91.67\% \]

- Percentage of \( M^{3+} \):
\[ \text{Percentage of } M^{3+} = 100 - \text{Percentage of } M^{2+} = 100 - 91.67 \approx 8.33\% \]

### Final Results
- Percentage of \( M^{2+} \) ions: **91.67%**
- Percentage of \( M^{3+} \) ions: **8.33%**

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