Molar mass of iron(II,III) oxide (Fe3O4)

The molar mass of iron(II,III) oxide (Fe3O4) is 231.531 g/mol.

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Step-by-step calculation

ElementAtomsAtomic massSubtotal
Iron × 3 55.845 = 167.535
Oxygen × 4 15.999 = 63.996
Total231.531 g/mol

Percent composition by element

  • Iron 72.36%
  • Oxygen 27.64%

Try another formula

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

Could not read that formula

Related compounds

Fe2O3 Iron(III) oxide FeSO4 Iron(II) sulfate FeO Iron(II) oxide Fe(OH)3 Iron(III) hydroxide LiFePO4 Lithium iron phosphate Fe(NO3)3 Iron(III) nitrate

FAQ

What is the molar mass of Fe3O4?

The molar mass of iron(II,III) oxide (Fe3O4) is 231.531 g/mol.

How do you calculate the molar mass of Fe3O4?

Multiply the atomic mass of each element by the number of its atoms in Fe3O4, then add the results: 3 × 55.845 (Fe) + 4 × 15.999 (O) = 231.531 g/mol.

What percentage of Fe3O4 is Iron?

Iron makes up 72.36% of the mass of iron(II,III) oxide (Fe3O4).