Molar Mass of C6H12N4 (Methenamine)

Molar Mass Calculator

What is the molar mass of C₆H₁₂N₄?

Answer

M(C₆H₁₂N₄) =

140.190 g/mol

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

Contribution of each element
Element	Atoms, n	Atomic weight, A	n × A, g/mol Mass fraction, w	Mass fraction, w
Ccarbon	6	12.011	72.066 w ≈ 51.41%	≈ 51.41%
Hhydrogen	12	1.0080	12.0960 w ≈ 8.63%	≈ 8.63%
Nnitrogen	4	14.007	56.028 w ≈ 39.97%	≈ 39.97%
Σ = 140.190 g/mol w = 100%				100%

Each fraction is rounded on its own, so the printed ones add up to 100.01%; unrounded, they make exactly 100%.

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

Mass fractions of the elements

C 51.41%N 39.97%

H 8.63%

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Molar masses of substances nearby

Formula	M, g/mol
K₂CO₃ (potassium carbonate)	138.204
La (lanthanum)	138.91
Ce (cerium)	140.12
C₆H₁₂N₄ (methenamine)	140.190
Pr (praseodymium)	140.91
NaMnO₄ (sodium permanganate)	141.924
AsH₃O₄ (arsenic acid)	141.942

FAQ

What is the molar mass of C6H12N4?

M(C₆H₁₂N₄) = 140.190 g/mol

How to calculate the molar mass of C₆H₁₂N₄?

Add up the atomic weights of all the atoms in the formula: 6 × 12.011 + 12 × 1.0080 + 4 × 14.007 = 140.190 g/mol.

How many grams are in 2 moles of C₆H₁₂N₄?

m = n × M = 2 mol × 140.190 g/mol = 280.380 g.

See Also