The Mole and The Avogadro Constant

3.3 The Mole and the Avogadro Constant

Definition of the Mole

• A mole (mol) is the unit of amount of substance in chemistry.
• One mole contains exactly 6.02 × 10²³ particles (atoms, ions, molecules, or formula units).
• The number 6.02 × 10²³ is called the Avogadro constant (Nₐ).
• Examples:
  • 1 mol of carbon atoms = 6.02 × 10²³ carbon atoms
  • 1 mol of NaCl formula units = 6.02 × 10²³ NaCl units

Relationship Between Mass, Moles, and Molar Mass

• The formula connecting these quantities is:
  n = m ÷ M
  where:
  • n = amount of substance (mol)
  • m = mass (g)
  • M = molar mass (g/mol)
• Rearranged formulas:
  • m = n × M
  • M = m ÷ n
• Example:
  Mass of 0.5 mol of sodium chloride (NaCl):
  M(NaCl) = 23 + 35.5 = 58.5 g/mol
  m = 0.5 × 58.5 = 29.25 g

Relative Atomic Mass and Relative Molecular Mass in Moles

• Relative atomic mass (Aᵣ) is the average mass of the atoms of an element compared to 1/12th of the mass of a carbon-12 atom.
• Relative molecular mass or relative formula mass (Mᵣ) is the sum of the relative atomic masses of the atoms in the molecule or ionic compound.
• Molar mass (M) has the same numerical value as Mᵣ, but with units g/mol.

Number of Particles and the Avogadro Constant

• Relationship:
  N = n × Nₐ
  where:
  • N = number of particles (atoms, molecules, or ions)
  • n = amount of substance (mol)
  • Nₐ = 6.02 × 10²³ mol⁻¹
• Example:
  Number of molecules in 0.25 mol of H₂O:
  N = 0.25 × 6.02 × 10²³ = 1.505 × 10²³ molecules

The Molar Gas Volume at Room Temperature and Pressure (r.t.p.)

• At r.t.p. (25°C, 1 atm), 1 mole of any gas occupies 24 dm³.
• Formula:
  n = V ÷ 24
  where V = volume in dm³.
• Conversion:
  V(dm³) = V(cm³) ÷ 1000
• Example:
  Volume of 0.2 mol of oxygen gas at r.t.p.:
  V = 0.2 × 24 = 4.8 dm³

Concentration of Solutions

• Concentration can be measured in mol/dm³ or g/dm³.
• Relationship for molar concentration:
  c = n ÷ V
  where:
  • c = concentration (mol/dm³)
  • n = amount of substance (mol)
  • V = volume of solution (dm³)
• Relationship for mass concentration:
  c = m ÷ V
  where:
  • c = concentration (g/dm³)
  • m = mass of solute (g)
  • V = volume of solution (dm³)
• Conversion between g/dm³ and mol/dm³:
  c(mol/dm³) = c(g/dm³) ÷ M
• Example:
  58.5 g of NaCl dissolved in 1 dm³:
  c(mol/dm³) = 58.5 ÷ 58.5 = 1 mol/dm³

Stoichiometric Calculations

• Steps:
  1. Write the balanced chemical equation.
  2. Find moles of the known substance using n = m ÷ M or n = V ÷ 24 (for gases).
  3. Use the mole ratio from the equation to find moles of the unknown substance.
  4. Convert moles to mass, volume, or number of particles.

Limiting Reactants

• In a reaction, the limiting reactant is the one that is completely used up first.
• Steps to find limiting reactant:
  1. Calculate moles of each reactant.
  2. Compare the mole ratio with the balanced equation.
  3. The reactant with fewer moles than required is the limiting reactant.

Empirical and Molecular Formula Calculations

• Empirical formula: simplest whole number ratio of atoms in a compound.
  Steps:
  1. Find masses of each element.
  2. Convert to moles (m ÷ Aᵣ).
  3. Divide each by the smallest mole value.
  4. Multiply to get whole numbers if needed.
• Molecular formula: multiple of the empirical formula.
  Molecular formula mass ÷ Empirical formula mass = multiple

Percentage Yield

• % yield = (actual yield ÷ theoretical yield) × 100
• Theoretical yield: maximum possible mass of product, calculated from moles in the balanced equation.
• Actual yield: measured mass of product obtained.

Percentage Composition by Mass

• % by mass of element = (total mass of element in formula ÷ molar mass of compound) × 100

Percentage Purity

• % purity = (mass of pure substance ÷ mass of sample) × 100