Molar mass of iron(III) sulfate (Fe2(SO4)3)

The molar mass of iron(III) sulfate (Fe2(SO4)3) is 399.858 g/mol.

ferric sulfate

Step-by-step calculation

Element	Atoms	Atomic mass	Subtotal
Iron	× 2	55.845	= 111.69
Sulfur	× 3	32.06	= 96.18
Oxygen	× 12	15.999	= 191.988
Total			399.858 g/mol

Percent composition by element

Iron 27.93%
Sulfur 24.05%
Oxygen 48.01%

Try another formula

H₂O NaCl CO₂ C₆H₁₂O₆ H₂SO₄ Ca(OH)₂ CuSO₄·5H₂O C₈H₁₀N₄O₂

Could not read that formula

Related compounds

FeSO₄ Iron(II) sulfate FeSO₄·7H₂O Iron(II) sulfate heptahydrate (NH₄)₂Fe(SO₄)₂·6H₂O Ammonium iron(II) sulfate hexahydrate H₂SO₄ Sulfuric acid Fe₂O₃ Iron(III) oxide CuSO₄ Copper(II) sulfate

FAQ

What is the molar mass of Fe2(SO4)3?

The molar mass of iron(III) sulfate (Fe2(SO4)3) is 399.858 g/mol.

How do you calculate the molar mass of Fe2(SO4)3?

Multiply the atomic mass of each element by the number of its atoms in Fe2(SO4)3, then add the results: 2 × 55.845 (Fe) + 3 × 32.06 (S) + 12 × 15.999 (O) = 399.858 g/mol.

What percentage of Fe2(SO4)3 is Oxygen?

Oxygen makes up 48.01% of the mass of iron(III) sulfate (Fe2(SO4)3).