What was the mass of FeSO4⋅7H2O in the 2.955 g sample?

Method 1:

Molar mass of FeSO₄•7H₂O = (55.85 + 32.06 + 16.00×11 + 1.008×14) g/mol = 278.02 g/mol
Molar mass of Fe₂O₃ = (55.85×2 + 16.00×3) g/mol = 159.70 g/mol

The series of changes:
2 mol FeSO₄•7H₂O → 2 mol Fe³⁺ → 1 mol Fe₂O₃•xH₂O → 1 mol Fe₂O₃
Mole ratio  FeSO₄•7H₂O : Fe₂O₃ = 2 : 1

Moles of Fe₂O₃ = (0.624 g) / (159.70 g/mol) = 0.0039073 mol
Moles of FeSO₄•7H₂O = (0.003973 mol) × 2 = 0.007815 mol
Mass of FeSO₄•7H₂O = (0.007815 mol) × (278.02 g/mol) = 2.173 g

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Method 2:

(0.624 g Fe₂O₃) × (1 mol Fe₂O₃ / 159.7 g Fe₂O₃) × (2 mol FeSO₄•7H₂O / 1 mol Fe₂O₃) × (278.02 g FeSO₄•7H₂O / 1 mol FeSO₄•7H₂O)
= 2.173 g FeSO₄•7H₂O

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Method 3:

Mass of Fe in Fe₂O₃ = (0.624 g) × (55.85×2 / 159.70) = 0.43645 g
Mass of Fe in FeSO₄•7H₂O = 0.43645 g
Mass of FeSO₄•7H₂O = (0.43645 g) × (278.02/55.85) = 2.173 g