Question

An oxide of iron contains 69.9% iron. Find its empirical formula.
(Given: Atomic masses Fe = 56, O = 16)

• A. \( \text{Fe}_3\text{O}_4 \)
• B. \( \text{Fe}_2\text{O}_3 \)
• C. \( \text{FeO}_3 \)
• D. \( \text{FeO} \)

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to find the empirical formula of the iron oxide using the mass percentages of the constituent elements.
Step 2: Key Formula or Approach:
Convert the mass percentage to moles and find the simplest whole-number ratio.
Step 3: Detailed Explanation:
1. Percentage of Fe = 69.9%.
2. Percentage of O = 100% - 69.9% = 30.1%.
Calculate moles in 100g of substance:
- Moles of Fe = \( \frac{69.9}{56} \approx 1.25 \) mol.
- Moles of O = \( \frac{30.1}{16} \approx 1.88 \) mol.
Divide by the smallest value (1.25) to find the ratio:
- Fe : \( \frac{1.25}{1.25} = 1 \)
- O : \( \frac{1.88}{1.25} \approx 1.5 \)
To get whole numbers, multiply the ratio by 2:
- Fe = \(1 \times 2 = 2\)
- O = \(1.5 \times 2 = 3\)
The empirical formula is \(Fe_2O_3\).
Step 4: Final Answer:
The empirical formula of the oxide is \(Fe_2O_3\).