Overview
The transcript explains hydrates (ionic crystals containing water of crystallization) and shows how to calculate percentage water by mass and the number of moles of water of crystallization using stoichiometric methods.
Hydrates: Concepts
- Hydrates are ionic compounds with water molecules trapped in crystals.
- Water molecules are in a fixed ratio to the formula unit (e.g., CuSO4·5H2O is 1:5).
- The dot in formulas separates anhydrous compound from water; it does not mean multiply.
- Terms: hydrated (contains water), anhydrous (no water), water of crystallization.
Experimental Determination of Water Loss
- Weigh hydrated crystals, heat in a test tube over a Bunsen burner.
- Water evaporates; droplets appear on tube walls during heating.
- Continue heating until mass no longer decreases; reweigh to get anhydrous mass.
- Mass of water lost = hydrated mass − anhydrous mass.
Percentage of Water by Mass
- Goal: % by mass of H2O in a hydrate.
- Method: %H2O = (mass of H2O in formula unit / total molar mass of hydrate) × 100.
- Use atomic masses from the periodic table; include water in total mass.
Worked Example: CuCl2·2H2O
- Total molar mass: Cu (63.5) + 2×Cl (2×35.5) + 4×H (4×1) + 2×O (2×16) = 170.5 g/mol.
- Molar mass of water part: 4×H + 2×O = 36 g/mol.
- %H2O = (36 / 170.5) × 100 = 21.11%.
Finding Moles of Water of Crystallization (x or n)
- Strategy: Determine moles of anhydrous compound and moles of H2O from mass change.
- Form a ratio: (moles anhydrous) : (moles H2O) and simplify to 1 : x.
- Use n = mass / molar mass for each component.
Worked Example: AlCl3·xH2O
- Given: hydrated mass = 100.00 g; anhydrous mass = 55.28 g.
- Mass of H2O lost = 100.00 − 55.28 = 44.72 g.
- Moles AlCl3 = 55.28 / [Al (27) + 3×Cl (3×35.5) = 133.5] = 0.41408 mol.
- Moles H2O = 44.72 / 18 = 2.48444… mol.
- Ratio = 0.41408 : 2.48444 → divide both by 0.41408 → 1 : 6.00.
- Therefore, x = 6; formula is AlCl3·6H2O.
Past Paper Style Example: Na2CO3·xH2O
- Given: hydrated mass = 14.2 g; anhydrous (Na2CO3) mass = 5.3 g.
- Mass of H2O lost = 14.2 − 5.3 = 8.9 g.
- Moles Na2CO3 = 5.3 / [2×Na (2×23) + C (12) + 3×O (48) = 106] = 0.050 mol.
- Moles H2O = 8.9 / 18 = 0.49444… mol.
- Ratio = 0.050 : 0.49444 → divide by 0.050 → 1 : 9.89 ≈ 1 : 10.
- Therefore, x = 10; formula is Na2CO3·10H2O.
Summary Table: Methods and Examples
| Task | Key Steps | Formula(s) | Example Result |
|---|
| Identify hydrate composition | Recognize dot notation; fixed ratio water:formula unit | — | CuSO4·5H2O → 1:5 |
| Experimental water loss | Heat until constant mass; subtract masses | mass H2O = m(hydrated) − m(anhydrous) | 44.72 g H2O (from 100 g to 55.28 g) |
| % water by mass | Compute molar masses; percent composition | %H2O = (molar mass H2O part / total molar mass) × 100 | CuCl2·2H2O → 21.11% |
| Find x in hydrate | Convert masses to moles; ratio to 1:x | n = m/M; simplify ratio | AlCl3·xH2O → x = 6 |
| Find x (past paper) | Same as above with given data | n = m/M; simplify ratio | Na2CO3·xH2O → x = 10 |
Key Terms & Definitions
- Hydrate: Ionic crystal containing water molecules in fixed ratio.
- Water of crystallization: Water molecules incorporated in the crystal lattice.
- Hydrated: Compound that contains water of crystallization.
- Anhydrous: Compound without water of crystallization.
- Percentage composition (by mass): Fraction of a component’s mass relative to total mass, as a percentage.
- Stoichiometry: Quantitative relationships in chemical formulas and reactions.
Action Items / Next Steps
- Practice calculating molar masses, ensuring inclusion of water part in hydrates.
- Work through ratio-to-1:x conversions without premature rounding.
- Apply the heating-to-constant-mass method conceptually in problems.