Hydrates Knowledge-Based Study

Study notes

1. What is a Hydrate?

   A hydrate is an ionic compound that contains a specific number of water molecules chemically trapped within its crystal lattice. These incorporated water molecules are known as waters of crystallization or waters of hydration.

   When writing the chemical formula of a hydrate, a centered dot ($\cdot$) separates the formula of the ionic salt from the water molecules, indicating that the water is bound within the structure in a fixed stoichiometric ratio rather than forming a distinct chemical solution.

   • Example: $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$ (copper(II) sulfate pentahydrate) indicates that 5 moles of water molecules are bound to every 1 mole of copper(II) sulfate formula units.
   • Anhydrous Salt: When all water of crystallization is removed, the remaining salt is termed anhydrous (e.g., $\text{CuSO}_4$).

   Molar Mass Determination

   To determine the molar mass of a hydrate, the mass of the anhydrous salt and the total mass of the bound water molecules must be combined.

   Consider cobalt(II) chloride hexahydrate, $\text{CoCl}_2 \cdot 6\text{H}_2\text{O}$:

   • Molar mass of $\text{CoCl}_2$ = $58.93 + (2 \times 35.45) = 129.83\text{ g/mol}$
   • Molar mass of $6\text{H}_2\text{O}$ = $6 \times (2 \times 1.008 + 16.00) = 6 \times 18.016 = 108.10\text{ g/mol}$
   • Molar mass of $\text{CoCl}_2 \cdot 6\text{H}_2\text{O}$ = $129.83 + 108.10 = 237.93\text{ g/mol}$

   The mass percentage of water in a hydrate can be calculated using the following relationship:

   $$\text{Mass \% of } \text{H}_2\text{O} = \frac{\text{mass of water in 1 mole of hydrate}}{\text{molar mass of hydrate}} \times 100\%$$

   Thermal Decomposition of Hydrates

   Because the water molecules in a hydrate are held by relatively weak ion-dipole interactions compared to the strong ionic bonds holding the salt itself, heating a hydrate drives off the water as vapor, leaving the dry anhydrous solid behind:

   $$\text{Hydrate}(s) \xrightarrow{\Delta} \text{Anhydrous Salt}(s) + x\text{H}_2\text{O}(g)$$

   By measuring masses before and after heating, the mass of water driven off is determined experimentally by applying the law of conservation of mass:

   • Mass of water lost = $\text{mass of initial hydrate} - \text{mass of anhydrous residue}$
   • Heating is typically continued until a constant mass is obtained (known as heating to constant mass) to ensure that all water has completely evaporated.

   Calculating the Value of $x$

   The mole ratio between the anhydrous salt and water reveals the empirical formula of the hydrate, $\text{Salt} \cdot x\text{H}_2\text{O}$. Follow these systematic steps:

   1. Determine masses: Measure the mass of the dry anhydrous salt residue and calculate the mass of evaporated water.
   2. Convert masses to moles: Divide each mass by its corresponding molar mass:
      $n_{\text{salt}} = \frac{m_{\text{salt}}}{M_{\text{salt}}}$ and $n_{\text{H}_2\text{O}} = \frac{m_{\text{H}_2\text{O}}}{M_{\text{H}_2\text{O}}}$
   3. Determine the mole ratio: Divide the moles of water by the moles of anhydrous salt to find the integer $x$:
      $$x = \frac{n_{\text{H}_2\text{O}}}{n_{\text{salt}}}$$

   Round $x$ to the nearest whole integer to write the complete empirical formula of the hydrate.