Molar mass of iron(II) sulfate heptahydrate (FeSO4·7H2O)

The molar mass of iron(II) sulfate heptahydrate (FeSO4·7H2O) is 278.006 g/mol.

ferrous sulfate heptahydrate · green vitriol

Step-by-step calculation

Element	Atoms	Atomic mass	Subtotal
Iron	× 1	55.845	= 55.845
Sulfur	× 1	32.06	= 32.06
Oxygen	× 11	15.999	= 175.989
Hydrogen	× 14	1.008	= 14.112
Total			278.006 g/mol

Percent composition by element

Iron 20.09%
Sulfur 11.53%
Oxygen 63.3%
Hydrogen 5.08%

Try another formula

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

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Related compounds

(NH₄)₂Fe(SO₄)₂·6H₂O Ammonium iron(II) sulfate hexahydrate H₂SO₄ Sulfuric acid KAl(SO₄)₂·12H₂O Potassium alum CuSO₄·5H₂O Copper(II) sulfate pentahydrate H₂SO₃ Sulfurous acid FeSO₄ Iron(II) sulfate

FAQ

What is the molar mass of FeSO4·7H2O?

The molar mass of iron(II) sulfate heptahydrate (FeSO4·7H2O) is 278.006 g/mol.

How do you calculate the molar mass of FeSO4·7H2O?

Multiply the atomic mass of each element by the number of its atoms in FeSO4·7H2O, then add the results: 1 × 55.845 (Fe) + 1 × 32.06 (S) + 11 × 15.999 (O) + 14 × 1.008 (H) = 278.006 g/mol.

What percentage of FeSO4·7H2O is Oxygen?

Oxygen makes up 63.3% of the mass of iron(II) sulfate heptahydrate (FeSO4·7H2O).