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Hydrates and Water of Crystallization

Sep 17, 2025

Overview

This lecture covers the concept of hydrates (hydrated ionic compounds), water of crystallization, and methods to calculate both the percentage by mass of water and the number of moles of water of crystallization in a given sample.

Hydrates and Water of Crystallization

  • Hydrates are ionic compounds with water molecules trapped in their crystal structure.
  • Water molecules attach to formula units in a specific ratio, called water of crystallization.
  • The formula of a hydrate is written as the salt’s formula followed by a dot and the number of Hâ‚‚O molecules (e.g., CuSO₄·5Hâ‚‚O).
  • An anhydrous compound is a salt that no longer contains water of crystallization.

Experimental Determination of Hydrates

  • Hydrated compounds can be heated to remove water, leaving the anhydrous salt behind.
  • The mass difference before and after heating gives the mass of water lost.
  • Heating continues until the mass no longer changes, indicating all water has been removed.

Calculating Percentage of Water by Mass

  • Percentage of water by mass = (mass of Hâ‚‚O in compound / total mass of compound) × 100.
  • To find total mass, sum the atomic masses of all elements in the hydrate formula, including water.
  • Example: For CuCl₂·2Hâ‚‚O, total mass = mass of Cu + 2×Cl + 4×H + 2×O = 170.5 g/mol.
  • Mass of water = 4×H + 2×O = 36 g/mol; percentage = (36/170.5) × 100 ≈ 21.11%.

Calculating Moles of Water of Crystallization

  • Subtract mass of anhydrous salt from hydrated mass to get mass of water lost.
  • Convert both mass of water and mass of anhydrous salt to moles: moles = mass / molar mass.
  • Find the simplest ratio (1:n) between moles of anhydrous salt and moles of water.
  • Example: If 100 g of hydrated AlCl₃ becomes 55.28 g after heating, mass of water lost = 44.72 g.
  • Calculate moles of AlCl₃ and Hâ‚‚O; ratio gives n, the number of moles of water of crystallization.

Worked Example: Sodium Carbonate Hydrate

  • Hydrated mass: 14.2 g; anhydrous mass: 5.3 g; water lost: 8.9 g.
  • Find moles: Naâ‚‚CO₃ (5.3 g ÷ 106 g/mol = 0.05 mol), Hâ‚‚O (8.9 g ÷ 18 g/mol ≈ 0.494 mol).
  • Ratio: 0.05 : 0.494 → 1 : 10, so the formula is Naâ‚‚CO₃·10Hâ‚‚O.

Key Terms & Definitions

  • Hydrate — Ionic compound with water molecules incorporated into its crystal structure.
  • Water of crystallization — The fixed amount of water in a hydrate.
  • Anhydrous — The form of a compound without water of crystallization.
  • Percentage composition by mass — The mass percentage of each component in a compound.

Action Items / Next Steps

  • Practice calculating percentage of water by mass and moles of water of crystallization using sample problems.
  • Review stoichiometry and use periodic table for molar mass calculations.