Molar Mass of C12H25NaO4S (Sodium Lauryl Sulfate)

Molar Mass Calculator

What is the molar mass of C₁₂H₂₅NaO₄S?

Answer

M(C₁₂H₂₅NaO₄S) ≈

288.38 g/mol

Atomic weights: IUPAC standard values (CIAAW, 2024); the sum is rounded to the decimal places of the least precise one — S: 32.06.

Contribution of each element
Element	Atoms, n	Atomic weight, A	n × A, g/mol Mass fraction, w	Mass fraction, w
Ccarbon	12	12.011	144.132 w ≈ 49.98%	≈ 49.98%
Hhydrogen	25	1.0080	25.2000 w ≈ 8.74%	≈ 8.74%
Nasodium	1	22.990	22.990 w ≈ 7.97%	≈ 7.97%
Ooxygen	4	15.999	63.996 w ≈ 22.19%	≈ 22.19%
Ssulfur	1	32.06	32.06 w ≈ 11.12%	≈ 11.12%
Σ = 288.378 ≈ 288.38 g/mol w = 100%				100%

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

Mass fractions of the elements

C 49.98%

O 22.19%S 11.12%H 8.74%Na 7.97%

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Molar masses of substances nearby

Formula	M, g/mol
Na₂CO₃·10H₂O (sodium carbonate decahydrate)	286.138
ZnSO₄·7H₂O (zinc sulfate heptahydrate)	287.54
CaWO₄ (calcium tungstate)	287.91
C₁₂H₂₅NaO₄S (sodium lauryl sulfate)	288.38
CaI₂ (calcium iodide)	293.88
K₂Cr₂O₇ (potassium dichromate)	294.181
HgSO₄ (mercuric sulfate)	296.65

FAQ

What is the molar mass of C12H25NaO4S?

M(C₁₂H₂₅NaO₄S) ≈ 288.38 g/mol

How to calculate the molar mass of C₁₂H₂₅NaO₄S?

Add up the atomic weights of all the atoms in the formula: 12 × 12.011 + 25 × 1.0080 + 22.990 + 4 × 15.999 + 32.06 = 288.378 ≈ 288.38 g/mol.

How many grams are in 2 moles of C₁₂H₂₅NaO₄S?

m = n × M = 2 mol × 288.38 g/mol = 576.76 g.

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