Molar mass of zinc sulphate heptahydrate (ZnSO₄·7H₂O)

The molar mass of zinc sulphate heptahydrate, ZnSO₄·7H₂O, is 287.54 g/mol. One mole of it, which is 6.022 × 10²³ formula units, has a mass of 287.54 grams.

About zinc sulphate heptahydrate

White vitriol is the common crystalline form of zinc sulphate, with seven water molecules.

Other names: white vitriol.

The working

The formula ZnSO₄·7H₂O has 1 zinc atom, 1 sulphur atom, 11 oxygen atoms and 14 hydrogen atoms, 27 atoms in all. Multiply each count by the atomic weight of the element and add:

65.38 + 32.06 + 11 × 15.999 + 14 × 1.0080 = 287.541 g/mol

With the rounded atomic masses used in many school books (Zn = 65, S = 32, O = 16, H = 1) the answer is 287 g/mol. The difference from 287.54 is too small to matter in most school sums.

The water in the crystal

The part after the dot is water of crystallisation: 7 molecules of water for each ZnSO₄. That water has a mass of 7 × 18.015 = 126.11 g/mol, which is 43.9% of the crystal. Without it, the anhydrous part ZnSO₄ has a molar mass of 161.44 g/mol. Heating drives the water off, so a sample loses up to 43.9% of its mass. How this is used to find the number of waters is shown on the hydrates page.

Percentage composition

By mass, ZnSO₄·7H₂O is 22.74% zinc, 11.15% sulphur, 61.20% oxygen and 4.91% hydrogen. The largest share comes from oxygen. In a 100 g sample there are 22.7 g of zinc, 11.1 g of sulphur, 61.2 g of oxygen and 4.9 g of hydrogen.

Notice that hydrogen has more atoms in the formula (14) than oxygen (11), 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 25 g of ZnSO₄·7H₂O? Divide the mass by the molar mass: 25 ÷ 287.54 = 0.08694 mol.

Moles to grams. What is the mass of 0.1 mol of ZnSO₄·7H₂O? Multiply: 0.1 × 287.54 = 28.75 g.

Counting particles. 25 g contains 0.08694 mol × 6.022 × 10²³ = 5.236 × 10²² formula units.

Making a solution. A 0.1 mol/L solution contains 0.1 × 287.54 = 28.75 g of ZnSO₄·7H₂O in each litre. For 250 mL, weigh 7.189 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

• Ammonium iron(II) sulphate hexahydrate FeSO₄·(NH₄)₂SO₄·6H₂O · 392.12
• 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

All hydrates · All compounds