Molar mass of ammonium iron(II) sulphate hexahydrate (FeSO₄·(NH₄)₂SO₄·6H₂O)

The molar mass of ammonium iron(II) sulphate hexahydrate, FeSO₄·(NH₄)₂SO₄·6H₂O, is 392.12 g/mol. One mole of it, which is 6.022 × 10²³ formula units, has a mass of 392.12 grams.

About ammonium iron(II) sulphate hexahydrate

Mohr's salt is a double salt of iron(II) sulphate and ammonium sulphate. It keeps better in air than green vitriol, which makes it the preferred primary standard for permanganate titrations in Indian school and college practicals.

Other names: Mohr's salt.

The working

The formula FeSO₄·(NH₄)₂SO₄·6H₂O has 1 iron atom, 2 sulphur atoms, 14 oxygen atoms, 2 nitrogen atoms and 20 hydrogen atoms, 39 atoms in all. Multiply each count by the atomic weight of the element and add:

55.845 + 2 × 32.06 + 14 × 15.999 + 2 × 14.007 + 20 × 1.0080 = 392.125 g/mol

With the rounded atomic masses used in many school books (Fe = 56, S = 32, O = 16, N = 14, H = 1) the answer is 392 g/mol. The difference from 392.12 is too small to matter in most school sums.

The water in the crystal

The part after the dot is water of crystallisation: 6 molecules of water for each FeSO₄·(NH₄)₂SO₄. That water has a mass of 6 × 18.015 = 108.09 g/mol, which is 27.6% of the crystal. Without it, the anhydrous part FeSO₄·(NH₄)₂SO₄ has a molar mass of 284.03 g/mol. Heating drives the water off, so a sample loses up to 27.6% of its mass. How this is used to find the number of waters is shown on the hydrates page.

Percentage composition

By mass, FeSO₄·(NH₄)₂SO₄·6H₂O is 14.24% iron, 16.35% sulphur, 57.12% oxygen, 7.14% nitrogen and 5.14% hydrogen. The largest share comes from oxygen. In a 100 g sample there are 14.2 g of iron, 16.4 g of sulphur, 57.1 g of oxygen, 7.1 g of nitrogen and 5.1 g of hydrogen.

Notice that hydrogen has more atoms in the formula (20) than oxygen (14), yet gives a smaller share of the mass. Percentage by mass depends on how heavy each atom is, not only on how many there are.

Worked examples

Grams to moles. How many moles are in 50 g of FeSO₄·(NH₄)₂SO₄·6H₂O? Divide the mass by the molar mass: 50 ÷ 392.12 = 0.1275 mol.

Moles to grams. What is the mass of 0.4 mol of FeSO₄·(NH₄)₂SO₄·6H₂O? Multiply: 0.4 × 392.12 = 156.9 g.

Counting particles. 50 g contains 0.1275 mol × 6.022 × 10²³ = 7.679 × 10²² formula units.

Making a solution. A 0.1 mol/L solution contains 0.1 × 392.12 = 39.21 g of FeSO₄·(NH₄)₂SO₄·6H₂O in each litre. For 250 mL, weigh 9.803 g and make it up to the mark with water.

Quick conversion tables

For other amounts use the grams and moles converter.

Questions people ask

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.

More hydrates

• Potassium aluminium sulphate dodecahydrate KAl(SO₄)₂·12H₂O · 474.37
• Barium chloride dihydrate BaCl₂·2H₂O · 244.26
• Calcium chloride dihydrate CaCl₂·2H₂O · 147.01
• Copper(II) sulphate pentahydrate CuSO₄·5H₂O · 249.68
• Sodium carbonate decahydrate Na₂CO₃·10H₂O · 286.14
• Magnesium sulphate heptahydrate MgSO₄·7H₂O · 246.47

All hydrates · All compounds