Molar Mass of Iron(III) oxide

Fe2O3 — 159.687 g/mol, worked out element by element.

Molar mass of Iron(III) oxide, Fe₂O₃

159.687 g/mol

Molecular weight 159.687 · 5 atoms per formula unit

Also known as: fe2o3, ferric oxide, rust, hematite.

Rust, and the ore blast furnaces reduce; the standard empirical-formula worked example.

How 159.687 g/mol is worked out

Mass contribution of each element
ElementAtomsAtomic massMass% of total
Iron255.845111.69069.94%
Oxygen315.99947.99730.06%
Total5159.687100%

M = 2 × 55.845 + 3 × 15.999 = 159.687 g/mol

Grams to moles for Iron(III) oxide

Mass to moles conversions
MassMolesFormula units
1 g0.0063 mol3.771 × 10^21
10 g0.0626 mol3.771 × 10^22
25 g0.1566 mol9.428 × 10^22
50 g0.3131 mol1.886 × 10^23
100 g0.6262 mol3.771 × 10^23
250 g1.5656 mol9.428 × 10^23
159.687 g1 mol6.022 × 10^23

For any other amount, use the molar mass calculator, or the molarity calculator to make up a solution of a given concentration.

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Frequently Asked Questions

What is the molar mass of Iron(III) oxide?

The molar mass of Iron(III) oxide (Fe2O3) is 159.687 g/mol. That is the sum of the standard atomic weights of all 5 atoms in one formula unit: 2 × 55.845 + 3 × 15.999 = 159.687.

What is the molecular weight of Iron(III) oxide?

159.687 — the same number as the molar mass, but as a dimensionless ratio rather than in g/mol. The two terms describe the same arithmetic, so a molecular weight of 159.69 and a molar mass of 159.69 g/mol are the same answer.

How many moles are in 100 g of Iron(III) oxide?

100 ÷ 159.687 = 0.6262 mol. To convert any mass to moles, divide by the molar mass; to go the other way, multiply.

What is the percent composition of Iron(III) oxide?

By mass, Iron(III) oxide is 69.94% iron, 30.06% oxygen. Each element's share is its total mass in the formula divided by 159.687.

How many atoms are in one formula unit of Iron(III) oxide?

5: 2 iron, 3 oxygen. One mole contains 6.02214076 × 10²³ formula units, so a mole of Iron(III) oxide contains 5 × 6.022 × 10²³ atoms in total.