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    Molar Mass of AgBr (Silver Bromide)

    Molar Mass Calculator
    Chemical formula

    What is the molar mass of AgBr?

    Answer

    M(AgBr) ≈
    187.77 g/mol

    Atomic weights: IUPAC standard values (CIAAW, 2024); the sum is rounded to the decimal places of the least precise one — Ag: 107.87.

    Contribution of each element
    Element Atoms, n Atomic weight, A n × A, g/mol Mass fraction, w
    Agsilver1107.87107.87 w ≈ 57.45%
    Brbromine179.90479.904 w ≈ 42.55%
    Σ = 187.774 ≈ 187.77 g/mol w = 100%

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

    Mass fractions of the elements

    Related Calculations

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    <a href="https://calculat.io/en/other/molar-mass/agbr">Molar Mass of AgBr (Silver Bromide): Step by Step</a>Copy
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    Molar Mass of AgBr (Silver Bromide): Step by Step. M(AgBr) ≈ 187.77 g/mol: 107.87 (Ag) + 79.904 (Br). Mass fractions: Ag 57.45%, Br 42.55%.
    https://calculat.io/en/other/molar-mass/agbr/generated.jpgCopy

    Molar masses of substances nearby

    FormulaM, g/mol
    MgBr₂ (magnesium bromide)184.113
    Re (rhenium)186.21
    Cu(NO₃)₂ (copper(II) nitrate)187.554
    AgBr (silver bromide)187.77
    KC₄H₅O₆ (potassium hydrogen tartrate)188.176
    Zn(NO₃)₂ (zinc nitrate)189.39
    SnCl₂ (stannous chloride)189.61
    Recalculate with different values ↑

    FAQ

    What is the molar mass of AgBr?

    M(AgBr) ≈ 187.77 g/mol

    How to calculate the molar mass of AgBr?

    Add up the atomic weights of all the atoms in the formula: 107.87 + 79.904 = 187.774 ≈ 187.77 g/mol.

    How many grams are in 2 moles of AgBr?

    m = n × M = 2 mol × 187.77 g/mol = 375.54 g.

    See Also