Molar Mass of Mg3(PO4)2 (Magnesium Phosphate)

Molar Mass Calculator

What is the molar mass of Mg₃(PO₄)₂?

Answer

M(Mg₃(PO₄)₂) =

262.855 g/mol

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

Mg₃(PO₄)₂ = 3 Mg + 2 × (P + 4 O) → Mg 3, P 2, O 8

Contribution of each element
Element	Atoms, n	Atomic weight, A	n × A, g/mol Mass fraction, w	Mass fraction, w
Mgmagnesium	3	24.305	72.915 w ≈ 27.74%	≈ 27.74%
Pphosphorus	2	30.974	61.948 w ≈ 23.57%	≈ 23.57%
Ooxygen	8	15.999	127.992 w ≈ 48.69%	≈ 48.69%
Σ = 262.855 g/mol w = 100%				100%

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

Mass fractions of the elements

O 48.69%

Mg 27.74%P 23.57%

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Molar masses of substances nearby

Formula	M, g/mol
C₆H₅Na₃O₇ (trisodium citrate)	258.069
Ba(NO₃)₂ (barium nitrate)	261.34
Na₂Cr₂O₇ (sodium dichromate)	261.965
Mg₃(PO₄)₂ (magnesium phosphate)	262.855
AlBr₃ (aluminum bromide)	266.694
PbCO₃ (lead carbonate)	267.2
HgCl₂ (mercuric chloride)	271.49

FAQ

What is the molar mass of Mg3(PO4)2?

M(Mg₃(PO₄)₂) = 262.855 g/mol

How to calculate the molar mass of Mg₃(PO₄)₂?

Add up the atomic weights of all the atoms in the formula: 3 × 24.305 + 2 × 30.974 + 8 × 15.999 = 262.855 g/mol.

How many grams are in 2 moles of Mg₃(PO₄)₂?

m = n × M = 2 mol × 262.855 g/mol = 525.710 g.

See Also