Q1 / 75 · JEE Main 2026 (06 April Shift 1) · Chemistry
An oxide of iron contains $69.9\%$ iron, its empirical formula, is: (Given: Molar mass of Fe and O are $56$ and $16$ g mol$^{-1}$ respectively.)
- A$\text{FeO}$
- Correct answerB$\text{Fe}_2\text{O}_3$
- C$\text{Fe}_3\text{O}_4$
- D$\text{FeO}_3$
Solution
Mass percentage of $\text{Fe} = 69.9\%$
Mass percentage of $\text{O} = 100\% - 69.9\% = 30.1\%$
Number of moles of $\text{Fe} = \dfrac{69.9}{56} = 1.248$
Number of moles of $\text{O} = \dfrac{30.1}{16} = 1.881$
Molar ratio of $\text{Fe} : \text{O} = 1.248 : 1.881$
Dividing by the smallest value, the ratio is $\dfrac{1.248}{1.248} : \dfrac{1.881}{1.248} = 1 : 1.5$
Multiplying by $2$ to get the simplest whole number ratio, we get $2 : 3$.
The empirical formula is $\text{Fe}_2\text{O}_3$.
Answer: $\text{Fe}_2\text{O}_3$
Mass percentage of $\text{O} = 100\% - 69.9\% = 30.1\%$
Number of moles of $\text{Fe} = \dfrac{69.9}{56} = 1.248$
Number of moles of $\text{O} = \dfrac{30.1}{16} = 1.881$
Molar ratio of $\text{Fe} : \text{O} = 1.248 : 1.881$
Dividing by the smallest value, the ratio is $\dfrac{1.248}{1.248} : \dfrac{1.881}{1.248} = 1 : 1.5$
Multiplying by $2$ to get the simplest whole number ratio, we get $2 : 3$.
The empirical formula is $\text{Fe}_2\text{O}_3$.
Answer: $\text{Fe}_2\text{O}_3$
Hence, the correct answer is option: 'B'.