5.5 - Water of Crystallisation

Hydrated and anhydrous substances

• Some solids have water built into their crystals. In these solids, water molecules are fitted into the lattice alongside the ions.
• Each water molecule is attracted to the surrounding ions and held in place.
• The water isn't damp on the surface - it's chemically combined as a fixed part of the solid.

Hydrated vs anhydrous

• Hydrated - A substance whose crystals contain water of crystallisation.
• Anhydrous - A substance containing no water at all.
• Hydrated and anhydrous copper(II) sulfate are the same salt, with and without that water: the hydrated form is blue and the anhydrous form is white.

Water counts towards the mass

Because the water is part of the solid, it's included in the mass of the crystals when weighed on a balance.

Water of crystallisation and the dot notation

• Water held in a crystal lattice is called water of crystallisation.
• One mole of a hydrated substance always holds a fixed whole number of moles of water, recorded after a dot in the formula.
• Hydrated copper(II) sulfate is CuSO₄·5H₂O - five moles of water for every one mole of copper(II) sulfate.
• That number is the degree of hydration:
  • Cobalt(II) chloride, CoCl₂·6H₂O → degree of hydration 6
  • Barium chloride, BaCl₂·2H₂O → degree of hydration 2
  • Anhydrous barium chloride, with no water of crystallisation, is BaCl₂

Water in the relative formula mass

• Water of crystallisation counts towards the compound's relative formula mass ($M_r$).
• Add up the relative atomic masses from the periodic table, reading the dot as "plus that many lots of H₂O".
• For CuSO₄·5H₂O ($A_r$: Cu = 64, S = 32, O = 16, H = 1):

$$\begin{array}{cc}{M}_r& =64+32+(4\times 16)+5\times (2+16)\\ & =250\end{array}$$

The percentage by mass of water is found by dividing the mass of the water ($18n$) by the total $M_r$, then multiplying by 100:

$$\begin{array}{cc}\text{Percentage of water}& =\frac{5\times 18}{250}\times 100\\ & =36\%\end{array}$$

Heating a hydrate to constant mass

• Heating a hydrated salt drives off the water of crystallisation as steam and leaves the anhydrous salt behind, decreasing the solid's mass.
• If heating is stopped too early, some water remains locked in the crystals as a mixture of hydrated and anhydrous salt.
• To ensure all water has been removed, heat to constant mass:
  1. Measure the mass of the empty container (crucible or evaporating basin).
  2. Add the hydrated salt and measure the mass of the container and contents.
  3. Heat the solid, allow it to cool, and measure the mass again.
  4. Repeat heating, cooling and measuring until two consecutive readings give the same mass.

Example set of readings

• The last two readings agree, so constant mass was reached at 27.80 g.
• Mass of hydrated salt = 29.60 - 24.60 = 5.00 g
• Mass of anhydrous salt left = 27.80 - 24.60 = 3.20 g
• Mass of water lost = 5.00 - 3.20 = 1.80 g