Molar mass of iron(III) nitrate nonahydrate (Fe(NO3)3·9H2O)

The molar mass of iron(III) nitrate nonahydrate (Fe(NO3)3·9H2O) is 403.992 g/mol.

ferric nitrate nonahydrate

Step-by-step calculation

Element	Atoms	Atomic mass	Subtotal
Iron	× 1	55.845	= 55.845
Nitrogen	× 3	14.007	= 42.021
Oxygen	× 18	15.999	= 287.982
Hydrogen	× 18	1.008	= 18.144
Total			403.992 g/mol

Percent composition by element

Iron 13.82%
Nitrogen 10.4%
Oxygen 71.28%
Hydrogen 4.49%

Try another formula

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

Could not read that formula

Related compounds

(NH₄)₂Fe(SO₄)₂·6H₂O Ammonium iron(II) sulfate hexahydrate NH₄NO₃ Ammonium nitrate C₈H₁₀N₄O₂ Caffeine HNO₃ Nitric acid NH₄OH Ammonium hydroxide HNO₂ Nitrous acid

FAQ

What is the molar mass of Fe(NO3)3·9H2O?

The molar mass of iron(III) nitrate nonahydrate (Fe(NO3)3·9H2O) is 403.992 g/mol.

How do you calculate the molar mass of Fe(NO3)3·9H2O?

Multiply the atomic mass of each element by the number of its atoms in Fe(NO3)3·9H2O, then add the results: 1 × 55.845 (Fe) + 3 × 14.007 (N) + 18 × 15.999 (O) + 18 × 1.008 (H) = 403.992 g/mol.

What percentage of Fe(NO3)3·9H2O is Oxygen?

Oxygen makes up 71.28% of the mass of iron(III) nitrate nonahydrate (Fe(NO3)3·9H2O).