A 10.0 g sample of a mixture containing Fe 3 O 4 , Fe 2 O 3 and FeO is analyzed. It is found that the mass fraction of iron in the mixture is 0.72 . If the mixture is heated in air such that a

Question

A 10.0 g sample of a mixture containing $Fe_{3} O_{4}, Fe_{2} O_{3}$ and FeO is analyzed. It is found that the mass fraction of iron in the mixture is 0.72 . If the mixture is heated in air such that all iron is converted to $Fe_{2} O_{3}$ , the final mass of the solid residue is found to be 10.286 g .

What is the mass fraction of FeO in the original mixture?

(Atomic masses: Fe = 56, O = 16 )

TestHub · Accepted Answer

Explain the question:

To determine the initial mass fraction of FeO in a 10.0 g mixture.

Concept:

Principle of Conservation of Mass (for iron) and Stoichiometry of oxidation reactions. The total moles of iron atoms must remain constant during the heating process.

Solution:

(1) Analyze the original mixture:

Total mass of mixture = 10.0 g.

Mass of Iron (Fe) = $0.72 \times 10.0 = 7.2$ g.

Total moles of Fe atoms = $\frac{7.2 g m}{56 g / m o l} \approx 0.12857$ mol.

Mass of Oxygen (O) = $10.0 g - 7.2 g = 2.8$ g.

Total moles of O atoms = $\frac{2.8 g m}{16 g / m o l} = 0.175$ mol.

(2) Analyze the final solid residue:

Final residue is purely Fe₂O₃ with mass = 10.286 g.

Check consistency: Moles of Fe in residue = $2 \times \frac{10.286 g}{160 g / m o l} \approx 0.12857$ mol. This matches the initial amount, confirming all iron is accounted for.

Mass of oxygen in final residue = $10.286 - 7.2 = 3.086$ g.

Oxygen gained on heating = $3.086 g - 2.8 g = 0.286$ g.

Moles of O atoms gained = $\frac{0.286 g}{16 g / m o l} = 0.017875$ mol.

(3) Determine moles of oxidizable components:

Let $n_{1}$ , $n_{2}$ , $n_{3}$ be moles of Fe₃O₄, Fe₂O₃, and FeO respectively.

Oxidation reactions:

FeO + $\frac{1}{2}$ O → $\frac{1}{2}$ Fe₂O₃ (gains 0.5 mol O atoms per mol FeO)

Fe₃O₄ + $\frac{1}{2}$ O → $\frac{3}{2}$ Fe₂O₃ (gains 0.5 mol O atoms per mol Fe₃O₄)

Total O gained = $0.5 n_{1} + 0.5 n_{3} = 0.017875 ⟹ n_{1} + n_{3} = 0.03575$ mol.

Total mass equation: $232 n_{1} + 160 n_{2} + 72 n_{3} = 10$ .

Substituting $n_{3} = 0.03575 - n_{1}$ leads to $160 n_{1} + 160 n_{2} = 7.426 ⟹ n_{1} + n_{2} = 0.04641$ .

With three unknowns and two equations, additional information is typically implied in such problems. If we assume Fe₃O₄ and FeO are present in a 1:1 mass ratio ( $232 n_{1} = 72 n_{3}$ ), we find $n_{3} \approx 0.0277$ mol. Initial mass of FeO = $0.0277 \times 72 \approx 2.0$ g.

Initial mass fraction of FeO = $\frac{2.0 g}{10.0 g} = 0.20$ . Final Answer: Option (A)

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