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    Molar Mass of FeSO4·7H2O (Ferrous Sulfate Heptahydrate)

    Molar Mass Calculator
    Chemical formula

    What is the molar mass of FeSO4·7H2O?

    Answer

    M(FeSO4·7H2O) ≈
    278.01 g/mol

    Atomic weights: IUPAC standard values (CIAAW, 2024); the sum is rounded to the decimal places of the least precise one — S: 32.06.

    7 H2O — water of crystallization, part of the molar mass: 7 × 18.015 = 126.105.

    FeSO4·7H2O = FeSO4 + 7 H2O → Fe 1, S 1, O 11, H 14

    Contribution of each element
    Element Atoms, n Atomic weight, A n × A, g/mol Mass fraction, w
    Feiron155.84555.845 w ≈ 20.09%
    Ssulfur132.0632.06 w ≈ 11.53%
    Ooxygen1115.999175.989 w ≈ 63.30%
    Hhydrogen141.008014.1120 w ≈ 5.08%
    Σ = 278.006 ≈ 278.01 g/mol w = 100%

    Atomic weights: CIAAW, Abridged Standard Atomic Weights (2024)

    Mass fractions of the elements

    Fe 20.09%S 11.53%H 5.08%

    See Also

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    Molar Mass of FeSO4·7H2O (Ferrous Sulfate Heptahydrate). M(FeSO4·7H2O) ≈ 278.01 g/mol: 55.845 (Fe) + 32.06 (S) + 11 × 15.999 (O) + 14 × 1.0080 (H). Mass fractions: O 63.30%, Fe 20.09%, S 11.53%.
    https://calculat.io/en/other/molar-mass/feso4.7h2o/generated.jpgCopy

    FAQ

    What is the molar mass of FeSO4·7H2O?

    M(FeSO₄·7H₂O) ≈ 278.01 g/mol

    Is the water counted in FeSO₄·7H₂O?

    Yes: the water of crystallization is part of the formula and of the molar mass. 7 H₂O add 7 × 18.015 = 126.105 g/mol; M(FeSO₄·7H₂O) ≈ 278.01 g/mol, while anhydrous M(FeSO₄) ≈ 151.90 g/mol.

    How many grams are in 2 moles of FeSO₄·7H₂O?

    m = n × M = 2 mol × 278.01 g/mol = 556.02 g.