Molar MassMolar mass of iron(III) sulphate (Fe2(SO4)3)
The molar mass of iron(III) sulphate, Fe2(SO4)3, is 399.86 g/mol. One mole of it, which is 6.022 × 1023 formula units, has a mass of 399.86 grams.
399.86 g/mol
2 × 55.845 + 3 × 32.06 + 12 × 15.999 = 399.858 · 17 atoms in the formula · also called ferric sulfate
About iron(III) sulphate
Iron(III) sulphate is used in water treatment and as a mordant in dyeing. It has two iron atoms and three sulphate groups in each formula unit.
Other names: ferric sulfate.
The working
The formula Fe2(SO4)3 has 2 iron atoms, 3 sulphur atoms and 12 oxygen atoms, 17 atoms in all. Multiply each count by the atomic weight of the element and add:
| Element | Atoms | Atomic weight | Total | Share by mass |
|---|---|---|---|---|
| Iron (Fe) | 2 | 55.845 | 111.690 | 27.93% |
| Sulphur (S) | 3 | 32.06 | 96.180 | 24.05% |
| Oxygen (O) | 12 | 15.999 | 191.988 | 48.01% |
2 × 55.845 + 3 × 32.06 + 12 × 15.999 = 399.858 g/mol
With the rounded atomic masses used in many school books (Fe = 56, S = 32, O = 16) the answer is 400 g/mol. The difference from 399.86 is too small to matter in most school sums.
Percentage composition
By mass, Fe2(SO4)3 is 27.93% iron, 24.05% sulphur and 48.01% oxygen. The largest share comes from oxygen. In a 100 g sample there are 27.9 g of iron, 24.1 g of sulphur and 48.0 g of oxygen.
Worked examples
Grams to moles. How many moles are in 5 g of Fe2(SO4)3? Divide the mass by the molar mass: 5 ÷ 399.86 = 0.0125 mol.
Moles to grams. What is the mass of 0.25 mol of Fe2(SO4)3? Multiply: 0.25 × 399.86 = 99.96 g.
Counting particles. 5 g contains 0.0125 mol × 6.022 × 1023 = 7.530 × 1021 formula units.
Making a solution. A 0.1 mol/L solution contains 0.1 × 399.86 = 39.99 g of Fe2(SO4)3 in each litre. For 250 mL, weigh 9.996 g and make it up to the mark with water.
Quick conversion tables
| Amount | Mass |
|---|---|
| 0.1 mol | 39.99 g |
| 0.5 mol | 199.9 g |
| 1 mol | 399.9 g |
| 2 mol | 799.7 g |
| 5 mol | 1999 g |
| Mass | Moles |
|---|---|
| 1 g | 0.002501 mol |
| 10 g | 0.02501 mol |
| 100 g | 0.2501 mol |
| 1 kg | 2.501 mol |
For other amounts use the grams and moles converter.
Questions people ask
What is the molar mass of Fe2(SO4)3?
The molar mass of iron(III) sulphate (Fe2(SO4)3) is 399.86 g/mol, or 399.858 g/mol before rounding, using IUPAC abridged atomic weights.
How many moles are in 100 g of Fe2(SO4)3?
100 ÷ 399.86 = 0.2501 mol.
What is the mass of one formula unit of Fe2(SO4)3?
Divide the molar mass by the Avogadro constant: 399.86 ÷ 6.022 × 1023 = 6.640 × 10-22 g.
What is the percentage of oxygen in Fe2(SO4)3?
Oxygen makes up 48.01% of the mass of Fe2(SO4)3: 12 × 15.999 ÷ 399.858 × 100.
Which values? This site uses the abridged standard atomic weights published by IUPAC (CIAAW), read on 12 November 2026. Your textbook or question paper may round them differently, for example H = 1 and O = 16, so its answer can differ slightly. In an exam, use the values the paper gives.