A hydrated (or crystalline) salt is an ionic compound that contains a definite number of water molecules incorporated into its crystal lattice. The water present in the crystal is called the water of crystallization. A salt with its water removed (by heating) is called an anhydrous salt. Hydrated salts are usually written as:
$$ \text{Salt} \cdot x\text{H}_2\text{O} $$ where \(x\) is the number of moles of water of crystallization per mole of anhydrous salt.
General laboratory approach (typical experiment):
Example 1: A 5.00 g sample of a hydrated copper(II) sulphate was heated and 3.20 g of anhydrous CuSO4 remained. Determine the formula of the hydrated salt (i.e. find \(x\) in CuSO4·xH2O).
Step 1 — mass of water lost:
$$ m_{\text{water}} = 5.00\ \text{g} - 3.20\ \text{g} = 1.80\ \text{g} $$Step 2 — molar masses (to 5 s.f where useful):
$$ CuSO_4 = 63.55 + 32.06 + 4\times16.00 $$
$$= 159.61\ \text{g mol}^{-1}$$.
\(M(\text{H}_2\text{O}) = 18.015\ \text{g mol}^{-1}\).
Step 3 — moles of anhydrous CuSO4: $$ n_{\text{CuSO}_4} = \frac{3.20}{159.61} $$ So $$ n_{\text{CuSO}_4} \approx 0.02004\ \text{mol}. $$
Step 4 — moles of water: $$ n_{\text{H}_2\text{O}} = \frac{1.80}{18.015} $$ So $$ n_{\text{H}_2\text{O}} \approx 0.09994 \approx 0.1000\ \text{mol}. $$
Step 5 — ratio:
$$ \frac{n_{\text{H}_2\text{O}}}{n_{\text{CuSO}_4}} = \frac{0.1000}{0.02004} \approx 4.99 \approx 5 $$Answer: The hydrated salt is CuSO4·5H2O (copper(II) sulphate pentahydrate).
Example 2: A 28.51 g sample of hydrated sodium carbonate was heated and 10.60 g of Na2CO3 (anhydrous) remained. Determine the formula of the hydrated salt (Na2CO3·xH2O).
Step 1 — mass of water lost:
$$ m_{\text{water}} = 28.51\ \text{g} - 10.60\ \text{g} = 17.91\ \text{g} $$Step 2 — molar masses:
$$\small{({Na}_2\text{CO}_3) = 2(22.99) + 12.01 + 3(16.00)}$$
$$= 105.99\ \text{g mol}^{-1} $$.
\(M(\text{H}_2\text{O}) = 18.015\ \text{g mol}^{-1}\).
Step 3 — moles of anhydrous Na2CO3: $$ n_{\text{Na}_2\text{CO}_3} = \frac{10.60}{105.99} $$ So $$ n_{\text{Na}_2\text{CO}_3} \approx 0.1000\ \text{mol}. $$
Step 4 — moles of water: $$ n_{\text{H}_2\text{O}} = \frac{17.91}{18.015} $$ So $$ n_{\text{H}_2\text{O}} \approx 0.994\ \text{mol}. $$
Step 5 — ratio:
$$ \frac{n_{\text{H}_2\text{O}}}{n_{\text{Na}_2\text{CO}_3}} = \frac{0.994}{0.1000} \approx 9.94 \approx 10 $$Answer: The hydrated salt is Na2CO3·10H2O (sodium carbonate decahydrate, washing soda).
Question 3: A 4.929 g sample of hydrated magnesium sulphate was heated and 2.407 g of anhydrous MgSO4 remained. Find the value of \(x\) in MgSO4·xH2O.
Step 1 — mass of water:
$$ m_{\text{water}} = 4.929\ \text{g} - 2.407\ \text{g} = 2.522\ \text{g} $$Step 2 — molar masses:
$$\small{({MgSO}_4) = 24.31 + 32.06 + 4(16.00)} $$
$$ = 120.37\ \text{g mol}^{-1} $$.
\(M(\text{H}_2\text{O}) = 18.015\ \text{g mol}^{-1}\).
Step 3 — moles of anhydrous MgSO4: $$ n_{\text{MgSO}_4} = \frac{2.407}{120.37} $$ So $$ n_{\text{MgSO}_4} \approx 0.02000\ \text{mol}. $$
Step 4 — moles of water: $$ n_{\text{H}_2\text{O}} = \frac{2.522}{18.015} $$ So $$ n_{\text{H}_2\text{O}} \approx 0.1400\ \text{mol}. $$
Step 5 — ratio:
$$ \frac{n_{\text{H}_2\text{O}}}{n_{\text{MgSO}_4}} = \frac{0.1400}{0.02000} = 7.00 $$Answer: MgSO4·7H2O (magnesium sulphate heptahydrate; Epsom salt).
Question 4: A 4.885 g sample of hydrated barium chloride was heated and 4.165 g of anhydrous BaCl2 remained. Determine the formula BaCl2·xH2O.
Step 1 — mass of water lost:
$$ m_{\text{water}} = 4.885\ \text{g} - 4.165\ \text{g} = 0.720\ \text{g} $$Step 2 — molar masses:
$$({BaCl}_2) = 137.33 + 2(35.45) $$
$$= 208.23\ \text{g mol}^{-1}$$.
\(M(\text{H}_2\text{O}) = 18.015\ \text{g mol}^{-1}\).
Step 3 — moles of anhydrous BaCl2: $$ n_{\text{BaCl}_2} = \frac{4.165}{208.23} $$ So $$ n_{\text{BaCl}_2} \approx 0.02000\ \text{mol}. $$
Step 4 — moles of water: $$ n_{\text{H}_2\text{O}} = \frac{0.720}{18.015} $$ So $$ n_{\text{H}_2\text{O}} \approx 0.04000\ \text{mol}. $$
Step 5 — ratio:
$$ \frac{n_{\text{H}_2\text{O}}}{n_{\text{BaCl}_2}} = \frac{0.04000}{0.02000} = 2.00 $$Answer: The hydrated salt is BaCl2·2H2O (barium chloride dihydrate).
Question 5: A hydrated salt of copper(II) sulphate was analysed and found to contain 36.0% water by mass. Determine the value of \(x\) in CuSO4·xH2O.
Step 1 — assume 100.0 g of the hydrated salt so percentages translate to grams:
$$ m_{\text{water}} = 36.0\ \text{g}, \qquad $$ $$ m_{\text{CuSO}_4} = 100.0 - 36.0 = 64.0\ \text{g}. $$Step 2 — molar masses (as before):
\(M(\text{CuSO}_4) = 159.61\ \text{g mol}^{-1}\).
\(M(\text{H}_2\text{O}) = 18.015\ \text{g mol}^{-1}\).
Step 3 — moles present in 100 g sample: $$ n_{\text{CuSO}_4} = \frac{64.0}{159.61} $$ So $$ n_{\text{CuSO}_4} \approx 0.4009\ \text{mol}. $$ $$ n_{\text{H}_2\text{O}} = \frac{36.0}{18.015} $$ So $$ n_{\text{H}_2\text{O}} \approx 1.998 \approx 2.000\ \text{mol}. $$
Step 4 — ratio:
$$ \frac{n_{\text{H}_2\text{O}}}{n_{\text{CuSO}_4}} = \frac{2.000}{0.4009} \approx 4.99 \approx 5 $$Answer: CuSO4·5H2O (copper(II) sulphate pentahydrate). The percentage value 36.0% water corresponds closely to the pentahydrate.