Molar Mass of Fe(NO3)3 (Ferric Nitrate)

Molar Mass Calculator

What is the molar mass of Fe(NO₃)₃?

Answer

M(Fe(NO₃)₃) =

241.857 g/mol

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

Fe(NO₃)₃ = Fe + 3 × (N + 3 O) → Fe 1, N 3, O 9

Contribution of each element
Element	Atoms, n	Atomic weight, A	n × A, g/mol Mass fraction, w	Mass fraction, w
Feiron	1	55.845	55.845 w ≈ 23.09%	≈ 23.09%
Nnitrogen	3	14.007	42.021 w ≈ 17.37%	≈ 17.37%
Ooxygen	9	15.999	143.991 w ≈ 59.54%	≈ 59.54%
Σ = 241.857 g/mol w = 100%				100%

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

Mass fractions of the elements

O 59.54%

Fe 23.09%N 17.37%

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Molar masses of substances nearby

Formula	M, g/mol
PbO₂ (lead dioxide)	239.2
PbS (lead(II) sulfide)	239.3
Pb(OH)₂ (lead(II) hydroxide)	241.2
Fe(NO₃)₃ (ferric nitrate)	241.857
BaCl₂·2H₂O (barium chloride dihydrate)	244.26
C₁₀H₁₆N₂O₃S (biotin)	244.31
MgSO₄·7H₂O (magnesium sulfate heptahydrate)	246.47

FAQ

What is the molar mass of Fe(NO3)3?

M(Fe(NO₃)₃) = 241.857 g/mol

How to calculate the molar mass of Fe(NO₃)₃?

Add up the atomic weights of all the atoms in the formula: 55.845 + 3 × 14.007 + 9 × 15.999 = 241.857 g/mol.

How many grams are in 2 moles of Fe(NO₃)₃?

m = n × M = 2 mol × 241.857 g/mol = 483.714 g.

See Also