Summary — What This Chapter Covers
- The unified atomic mass unit is one twelfth of the mass of a carbon-12 atom; all relative masses are measured against it.
- One mole contains 6.02 × 10²³ particles (the Avogadro constant, L).
- n = m/M, n = cV, and for gases at room conditions n = V/24.0 dm³.
- Ionic formulas come from ionic charges; equations must balance in atoms and charge, with state symbols.
- Empirical formula: simplest whole-number ratio. Molecular formula: actual number of each atom.
- Hydrated salts contain water of crystallisation; anhydrous salts do not.
- Reacting-mass, gas-volume, concentration, limiting-reagent and percentage-yield calculations all go through moles.
1. Relative Masses of Atoms and Molecules
The carbon-12 scale
Definitions in terms of the unified atomic mass unit
The unified atomic mass unit (u) is defined as one twelfth of the mass of a carbon-12 atom (1 u ≈ 1.66 × 10⁻²⁷ kg).
| Term | Definition |
|---|
| Relative isotopic mass | The mass of an atom of an isotope compared with one twelfth of the mass of an atom of carbon-12. |
| Relative atomic mass, Ar | The weighted mean mass of an atom of an element compared with one twelfth of the mass of an atom of carbon-12. |
| Relative molecular mass, Mr | The weighted mean mass of a molecule compared with one twelfth of the mass of an atom of carbon-12. |
| Relative formula mass | The weighted mean mass of one formula unit (used for ionic compounds) compared with one twelfth of the mass of an atom of carbon-12. |
Relative masses have no units. "Weighted mean" means each isotope counts in proportion to its abundance.
Weighted mean
Ar = Σ(isotopic mass × % abundance) ÷ 100
Worked Example
Chlorine is 75.8% ³⁵Cl and 24.2% ³⁷Cl. Calculate Ar(Cl).
Worked Example
Calculate Mr of CuSO₄·5H₂O. (Ar: Cu 63.5, S 32.1, O 16.0, H 1.0)
Practice Question
Boron has Ar = 10.8 and two isotopes, ¹⁰B and ¹¹B. Find the percentage of ¹⁰B.
2. The Mole and the Avogadro Constant
The mole
A counting unit
One mole of a substance contains the same number of particles as there are atoms in exactly 12 g of carbon-12. That number is the Avogadro constant, L = 6.02 × 10²³ mol⁻¹.
The mass of one mole (the molar mass, M, in g mol⁻¹) is numerically equal to Ar or Mr.
Mole relationships
n = m/M number of particles = n × L
Worked Example
How many oxygen atoms are in 8.8 g of CO₂? (Mr = 44.0)
Practice Question
Find the mass of 3.01 × 10²² molecules of water.
3. Formulas
Writing formulas
From charges and oxidation numbers
Roman numeral
| Ion | Formula | Ion | Formula |
|---|
| Nitrate | NO₃⁻ | Ammonium | NH₄⁺ |
| Carbonate | CO₃²⁻ | Zinc | Zn²⁺ |
| Sulfate | SO₄²⁻ | Silver | Ag⁺ |
| Hydroxide | OH⁻ | Hydrogencarbonate | HCO₃⁻ |
| Phosphate | PO₄³⁻ | | |
Balance the charges: aluminium sulfate is Al₂(SO₄)₃ (2 × 3+ = 3 × 2−); ammonium phosphate is (NH₄)₃PO₄.