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    Molar Mass of Fe2(SO4)3 (Ferric Sulfate)

    Molar Mass Calculator
    Chemical formula

    What is the molar mass of Fe2(SO4)3?

    Answer

    M(Fe2(SO4)3) ≈
    399.86 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.

    Fe2(SO4)3 = 2 Fe + 3 × (S + 4 O) → Fe 2, S 3, O 12

    Contribution of each element
    Element Atoms, n Atomic weight, A n × A, g/mol Mass fraction, w
    Feiron255.845111.690 w ≈ 27.93%
    Ssulfur332.0696.18 w ≈ 24.05%
    Ooxygen1215.999191.988 w ≈ 48.01%
    Σ = 399.858 ≈ 399.86 g/mol w = 100%

    Each fraction is rounded on its own, so the printed ones add up to 99.99%; unrounded, they make exactly 100%.

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

    Mass fractions of the elements

    Fe 27.93%S 24.05%

    See Also

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    <a href="https://calculat.io/en/other/molar-mass/fe2%28so4%293">Molar Mass of Fe2(SO4)3 (Ferric Sulfate): Step by Step</a>Copy
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    Molar Mass of Fe2(SO4)3 (Ferric Sulfate): Step by Step. M(Fe2(SO4)3) ≈ 399.86 g/mol: 2 × 55.845 (Fe) + 3 × 32.06 (S) + 12 × 15.999 (O). Mass fractions: O 48.01%, Fe 27.93%, S 24.05%.
    https://calculat.io/en/other/molar-mass/fe2%28so4%293/generated.jpgCopy

    FAQ

    What is the molar mass of Fe2(SO4)3?

    M(Fe₂(SO₄)₃) ≈ 399.86 g/mol

    How to calculate the molar mass of Fe₂(SO₄)₃?

    Add up the atomic weights of all the atoms in the formula: 2 × 55.845 + 3 × 32.06 + 12 × 15.999 = 399.858 ≈ 399.86 g/mol.

    How many grams are in 2 moles of Fe₂(SO₄)₃?

    m = n × M = 2 mol × 399.86 g/mol = 799.72 g.