Composition of sample wurtzite is `Fe_(0.93)O_(1.0)`. What percentage of iron is present in the form of Fe(III) ?

Explanation

To determine the percentage of iron present in the form of Fe(III) in the compound Fe₀.93O₁.0, we can follow these steps:

### Step 1: Identify the composition of the compound
The given compound is Fe₀.93O₁.0, which indicates that there are 0.93 moles of iron (Fe) and 1.0 mole of oxygen (O).

### Step 2: Define the variables for iron oxidation states
Let:
- \( x \) = moles of Fe²⁺ (ferrous ion)
- \( 0.93 - x \) = moles of Fe³⁺ (ferric ion)

### Step 3: Write the charge balance equation
For electrical neutrality, the total positive charge must equal the total negative charge. The charges contributed by the iron ions are:
- From Fe²⁺: \( 2x \)
- From Fe³⁺: \( 3(0.93 - x) \)

The total negative charge from oxygen (O²⁻) is:
- From O²⁻: \( 2 \) (since there is 1 mole of O, and each O contributes a charge of -2)

Setting up the charge balance equation:
\[ 2x + 3(0.93 - x) = 2 \]

### Step 4: Solve the equation
Expanding the equation:
\[ 2x + 2.79 - 3x = 2 \]
Combining like terms:
\[ -1x + 2.79 = 2 \]
Rearranging gives:
\[ -x = 2 - 2.79 \]
\[ -x = -0.79 \]
Thus,
\[ x = 0.79 \]

### Step 5: Calculate the moles of Fe³⁺
Now, substituting \( x \) back to find the moles of Fe³⁺:
\[ \text{Moles of Fe³⁺} = 0.93 - x = 0.93 - 0.79 = 0.14 \]

### Step 6: Calculate the total mass of the sample
The molar mass of Fe is approximately 56 g/mol and the molar mass of O is approximately 16 g/mol. The total mass of the sample can be calculated as follows:
\[ \text{Total mass} = (0.93 \times 56) + (1.0 \times 16) = 51.68 + 16 = 67.68 \text{ g} \]

### Step 7: Calculate the mass of Fe³⁺
The mass of Fe³⁺ is:
\[ \text{Mass of Fe³⁺} = \text{Moles of Fe³⁺} \times \text{Molar mass of Fe} = 0.14 \times 56 = 7.84 \text{ g} \]

### Step 8: Calculate the percentage of Fe³⁺ in the sample
Finally, the percentage of iron present as Fe³⁺ is calculated as:
\[ \text{Percentage of Fe³⁺} = \left( \frac{\text{Mass of Fe³⁺}}{\text{Total mass}} \right) \times 100 = \left( \frac{7.84}{67.68} \right) \times 100 \approx 11.57\% \]

### Final Answer
The percentage of iron present in the form of Fe(III) is approximately **11.57%**.

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