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    Molar Mass of FeH8N2O8S2 (Ferrous Ammonium Sulfate)

    Molar Mass Calculator
    Chemical formula

    What is the molar mass of FeH8N2O8S2?

    Answer

    M(FeH8N2O8S2) ≈
    284.04 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.

    Contribution of each element
    Element Atoms, n Atomic weight, A n × A, g/mol Mass fraction, w
    Feiron155.84555.845 w ≈ 19.66%
    Hhydrogen81.00808.0640 w ≈ 2.84%
    Nnitrogen214.00728.014 w ≈ 9.86%
    Ooxygen815.999127.992 w ≈ 45.06%
    Ssulfur232.0664.12 w ≈ 22.57%
    Σ = 284.035 ≈ 284.04 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

    S 22.57%Fe 19.66%N 9.86%H 2.84%

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    Molar Mass of FeH8N2O8S2 (Ferrous Ammonium Sulfate). M(FeH8N2O8S2) ≈ 284.04 g/mol: 55.845 (Fe) + 8 × 1.0080 (H) + 2 × 14.007 (N) + 8 × 15.999 (O) + 2 × 32.06 (S). Mass fractions: O 45.06%, S 22.57%, Fe 19.66%.
    https://calculat.io/en/other/molar-mass/feh8n2o8s2/generated.jpgCopy

    FAQ

    What is the molar mass of FeH8N2O8S2?

    M(FeH₈N₂O₈S₂) ≈ 284.04 g/mol

    How to calculate the molar mass of FeH₈N₂O₈S₂?

    Add up the atomic weights of all the atoms in the formula: 55.845 + 8 × 1.0080 + 2 × 14.007 + 8 × 15.999 + 2 × 32.06 = 284.035 ≈ 284.04 g/mol.

    How many grams are in 2 moles of FeH₈N₂O₈S₂?

    m = n × M = 2 mol × 284.04 g/mol = 568.08 g.