Question:
An oxide of iron contains 69.9% iron. Find its empirical formula.
(Given: Atomic masses Fe = 56, O = 16)
An oxide of iron contains 69.9% iron. Find its empirical formula.
(Given: Atomic masses Fe = 56, O = 16)
(Given: Atomic masses Fe = 56, O = 16)
Updated On: Apr 8, 2026
- \( \text{Fe}_3\text{O}_4 \)
- \( \text{Fe}_2\text{O}_3 \)
- \( \text{FeO}_3 \)
- \( \text{FeO} \)
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The Correct Option is B
Solution and Explanation
Given that the oxide contains 69.9% iron, we can calculate the moles of iron and oxygen in the compound. Assume 100 g of the oxide.
- Mass of iron \( \text{Fe} = 69.9 \, \text{g} \)
- Mass of oxygen \( \text{O} = 100 - 69.9 = 30.1 \, \text{g} \)
Now, calculate the moles of iron and oxygen:
- Moles of iron:
\[ \text{Moles of Fe} = \frac{69.9}{56} = 1.25 \, \text{mol} \] - Moles of oxygen:
\[ \text{Moles of O} = \frac{30.1}{16} = 1.88 \, \text{mol} \] Now, find the ratio of moles of iron to oxygen: \[ \frac{1.25}{1.88} \approx 0.67 \] This gives the simplest whole number ratio of Fe to O as 2:3. Hence, the empirical formula of the compound is \( \text{Fe}_2\text{O}_3 \).
Final Answer: \( \text{Fe}_2\text{O}_3 \)
\[ \text{Moles of Fe} = \frac{69.9}{56} = 1.25 \, \text{mol} \] - Moles of oxygen:
\[ \text{Moles of O} = \frac{30.1}{16} = 1.88 \, \text{mol} \] Now, find the ratio of moles of iron to oxygen: \[ \frac{1.25}{1.88} \approx 0.67 \] This gives the simplest whole number ratio of Fe to O as 2:3. Hence, the empirical formula of the compound is \( \text{Fe}_2\text{O}_3 \).
Final Answer: \( \text{Fe}_2\text{O}_3 \)
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