Molar mass of iron(III) oxide (Fe2O3)

The molar mass of iron(III) oxide (Fe2O3) is 159.687 g/mol.

rust · ferric oxide

Step-by-step calculation

Element	Atoms	Atomic mass	Subtotal
Iron	× 2	55.845	= 111.69
Oxygen	× 3	15.999	= 47.997
Total			159.687 g/mol

Iron(III) oxide is a reddish-brown solid that occurs naturally as the mineral hematite and, in hydrated form, makes up most of the rust on iron and steel.
Hematite is a major iron ore, and the oxide is also used as a red pigment and as the polishing powder known as jeweler's rouge.
Its reaction with powdered aluminum, called the thermite reaction, releases enough heat to produce molten iron.

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

Could not read that formula

FeSO₄ Iron(II) sulfate FeO Iron(II) oxide Fe₃O₄ Iron(II,III) oxide Fe(OH)₃ Iron(III) hydroxide LiFePO₄ Lithium iron phosphate Fe(NO₃)₃ Iron(III) nitrate

What is the molar mass of Fe2O3?

The molar mass of iron(III) oxide (Fe2O3) is 159.687 g/mol.

How do you calculate the molar mass of Fe2O3?

Multiply the atomic mass of each element by the number of its atoms in Fe2O3, then add the results: 2 × 55.845 (Fe) + 3 × 15.999 (O) = 159.687 g/mol.

What percentage of Fe2O3 is Iron?

Iron makes up 69.94% of the mass of iron(III) oxide (Fe2O3).