Mass percentage of Fe=69.9%
Mass percentage of O=100%−69.9%=30.1%
Number of moles of Fe=5669.9=1.248
Number of moles of O=1630.1=1.881
Molar ratio of Fe:O=1.248:1.881
Dividing by the smallest value, the ratio is 1.2481.248:1.2481.881=1:1.5
Multiplying by 2 to get the simplest whole number ratio, we get 2:3.
The empirical formula is Fe2O3.
Answer: Fe2O3
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.)
JEE Main 2026 — Chemistry Physical 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.)
Held on 6 Apr 2026 · Verified 1 Jul 2026.
Options
- A
FeO
- B
Fe2O3
- C
Fe3O4
- D
FeO3
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