Molar mass of iron(II) sulfate (FeSO4)

The molar mass of iron(II) sulfate (FeSO4) is 151.901 g/mol.

ferrous sulfate

Step-by-step calculation

ElementAtomsAtomic massSubtotal
Iron × 1 55.845 = 55.845
Sulfur × 1 32.06 = 32.06
Oxygen × 4 15.999 = 63.996
Total151.901 g/mol

Percent composition by element

  • Iron 36.76%
  • Sulfur 21.11%
  • Oxygen 42.13%

Try another formula

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

Could not read that formula

Related compounds

Fe2(SO4)3 Iron(III) 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 FeSO4?

The molar mass of iron(II) sulfate (FeSO4) is 151.901 g/mol.

How do you calculate the molar mass of FeSO4?

Multiply the atomic mass of each element by the number of its atoms in FeSO4, then add the results: 1 × 55.845 (Fe) + 1 × 32.06 (S) + 4 × 15.999 (O) = 151.901 g/mol.

What percentage of FeSO4 is Oxygen?

Oxygen makes up 42.13% of the mass of iron(II) sulfate (FeSO4).