How much iron is present in 7.94 g of iron(III) oxide (Fe2O3)?

Method 1 :

Molar mass of Fe = 55.8 g/mol
Molar mass of Fe₂O₃ = (55.8×2 + 16.0×3) g/mol = 159.6 g/mol

Mass fraction of Fe in Fe₂O₃ = (55.8 × 2)/159.6 = 111.6/159.6
Mass of Fe = (7.94 g) × (111.6/159.6) = 5.55 g
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Method 2 :

Molar mass of Fe = 55.8 g/mol
Molar mass of Fe₂O₃ = (55.8×2 + 16.0×3) g/mol = 159.6 g/mol

No. of moles of Fe₂O₃ = (7.94 g) / (159.6 g/mol) = 0.0497 mol

1 mole of Fe₂O₃ contains 2 moles of Fe.
No. of moles of Fe = (0.0497 mol) × 2 = 0.0994 mol
Mass of Fe = (0.0994 mol) × (55.8 g/mol) = 5.55 g