Molar Mass of AgBr (Silver Bromide)

Molar Mass Calculator

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	Mass fraction, w
Agsilver	1	107.87	107.87 w ≈ 57.45%	≈ 57.45%
Brbromine	1	79.904	79.904 w ≈ 42.55%	≈ 42.55%
Σ = 187.774 ≈ 187.77 g/mol w = 100%				100%

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

Mass fractions of the elements

Ag 57.45%Br 42.55%

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Molar masses of substances nearby

Formula	M, 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

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