Molar Mass of Mn(NO3)2 (Manganese Nitrate)

Molar Mass Calculator

What is the molar mass of Mn(NO₃)₂?

Answer

M(Mn(NO₃)₂) =

178.946 g/mol

Atomic weights: IUPAC standard values (CIAAW, 2024).

Mn(NO₃)₂ = Mn + 2 × (N + 3 O) → Mn 1, N 2, O 6

Contribution of each element
Element	Atoms, n	Atomic weight, A	n × A, g/mol Mass fraction, w	Mass fraction, w
Mnmanganese	1	54.938	54.938 w ≈ 30.70%	≈ 30.70%
Nnitrogen	2	14.007	28.014 w ≈ 15.66%	≈ 15.66%
Ooxygen	6	15.999	95.994 w ≈ 53.64%	≈ 53.64%
Σ = 178.946 g/mol w = 100%				100%

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

Mass fractions of the elements

O 53.64%

Mn 30.70%N 15.66%

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Molar masses of substances nearby

Formula	M, g/mol
BaF₂ (barium fluoride)	175.33
HC₆H₇O₆ (vitamin C)	176.124
Hf (hafnium)	178.49
Mn(NO₃)₂ (manganese nitrate)	178.946
C₆H₁₂O₆ (glucose)	180.156
C₉H₈O₄ (aspirin)	180.159
Ta (tantalum)	180.95

FAQ

What is the molar mass of Mn(NO3)2?

M(Mn(NO₃)₂) = 178.946 g/mol

How to calculate the molar mass of Mn(NO₃)₂?

Add up the atomic weights of all the atoms in the formula: 54.938 + 2 × 14.007 + 6 × 15.999 = 178.946 g/mol.

How many grams are in 2 moles of Mn(NO₃)₂?

m = n × M = 2 mol × 178.946 g/mol = 357.892 g.

See Also