Molar mass of iron(III) nitrate (Fe(NO3)3)

The molar mass of iron(III) nitrate (Fe(NO3)3) is 241.857 g/mol.

ferric nitrate

Step-by-step calculation

Element	Atoms	Atomic mass	Subtotal
Iron	× 1	55.845	= 55.845
Nitrogen	× 3	14.007	= 42.021
Oxygen	× 9	15.999	= 143.991
Total			241.857 g/mol

Percent composition by element

Iron 23.09%
Nitrogen 17.37%
Oxygen 59.54%

Try another formula

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

Could not read that formula

Related compounds

Fe(NO₃)₃·9H₂O Iron(III) nitrate nonahydrate (NH₄)₂Fe(SO₄)₂·6H₂O Ammonium iron(II) sulfate hexahydrate Fe₂O₃ Iron(III) oxide AgNO₃ Silver nitrate NaNO₃ Sodium nitrate NH₄NO₃ Ammonium nitrate

FAQ

What is the molar mass of Fe(NO3)3?

The molar mass of iron(III) nitrate (Fe(NO3)3) is 241.857 g/mol.

How do you calculate the molar mass of Fe(NO3)3?

Multiply the atomic mass of each element by the number of its atoms in Fe(NO3)3, then add the results: 1 × 55.845 (Fe) + 3 × 14.007 (N) + 9 × 15.999 (O) = 241.857 g/mol.

What percentage of Fe(NO3)3 is Oxygen?

Oxygen makes up 59.54% of the mass of iron(III) nitrate (Fe(NO3)3).