Molar Mass of PbCrO4 (Lead(II) Chromate)

Molar Mass Calculator

What is the molar mass of PbCrO₄?

Answer

M(PbCrO₄) ≈

323.2 g/mol

Atomic weights: IUPAC standard values (CIAAW, 2024); the sum is rounded to the decimal places of the least precise one — Pb: 207.2.

Contribution of each element
Element	Atoms, n	Atomic weight, A	n × A, g/mol Mass fraction, w	Mass fraction, w
Pblead	1	207.2	207.2 w ≈ 64.11%	≈ 64.11%
Crchromium	1	51.996	51.996 w ≈ 16.09%	≈ 16.09%
Ooxygen	4	15.999	63.996 w ≈ 19.80%	≈ 19.80%
Σ = 323.192 ≈ 323.2 g/mol w = 100%				100%

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

Mass fractions of the elements

Pb 64.11%

O 19.80%Cr 16.09%

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Molar masses of substances nearby

Formula	M, g/mol
ZnI₂ (zinc iodide)	319.18
C₁₆H₁₈ClN₃S (methylene blue)	319.85
Na₂SO₄·10H₂O (sodium sulfate decahydrate)	322.19
PbCrO₄ (lead(II) chromate)	323.2
Hg(NO₃)₂ (mercuric nitrate)	324.60
(CH₃COO)₂Pb (lead(II) acetate)	325.3
CCs₂O₃ (cesium carbonate)	325.83

FAQ

What is the molar mass of PbCrO4?

M(PbCrO₄) ≈ 323.2 g/mol

How to calculate the molar mass of PbCrO₄?

Add up the atomic weights of all the atoms in the formula: 207.2 + 51.996 + 4 × 15.999 = 323.192 ≈ 323.2 g/mol.

How many grams are in 2 moles of PbCrO₄?

m = n × M = 2 mol × 323.2 g/mol = 646.4 g.

See Also