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The Mole And The Avogadro Constant

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Notes de leçon

The Mole and Avogadro Constant

  • The mole (mol) is the SI unit for amount of substance.
  • One mole of any substance contains 6.02 × 10²³ particles (Avogadro constant).
  • Particles can be atoms, molecules, ions, or formula units.
  • Example: 1 mol Na = 6.02 × 10²³ Na atoms; 1 mol H₂ = 6.02 × 10²³ H₂ molecules.
  • Molar mass is the mass of one mole of a substance (g/mol).
  • For elements, molar mass = relative atomic mass in grams.
  • For compounds, molar mass = relative molecular/formula mass in grams.

Mole–Mass–Mr Relationship

  • Use the formula triangle: moles = mass / Mr.
  • Mass = moles × Mr; Mr = mass / moles.
  • Always show working to avoid errors and gain credit.
  • Example: 0.250 mol Zn has mass = 0.250 × 65 = 16.25 g.
  • Example: 2.64 g sucrose (Mr = 342) gives moles = 2.64 / 342 = 7.72 × 10⁻³ mol.

Formula triangle for moles, mass and molar mass

Formula triangle for moles, mass and molar mass

Mole and Volume of Gas

  • Avogadro's Law: equal moles of any gas occupy equal volume at same T and P.
  • At RTP (20 °C, 1 atm), molar gas volume = 24 dm³ (24 000 cm³).
  • Volume = moles × 24 dm³ (or × 24 000 cm³).
  • Moles = volume (dm³) / 24 (or volume (cm³) / 24 000).
  • Example: 3 mol H₂ occupies 3 × 24 = 72 dm³.
  • Example: 1200 cm³ O₂ gives moles = 1200 / 24 000 = 0.05 mol.

Molar gas volume (dm³) formula triangle

Molar gas volume (dm³) formula triangle

Calculating Number of Particles

  • Number of particles = moles × Avogadro constant (6.02 × 10²³).
  • For compounds, count atoms per formula unit.
  • Example: 1 mol MgCl₂ contains 6.02 × 10²³ molecules, 3 × 6.02 × 10²³ atoms.
  • Example: 15.7 g H₂O (Mr = 18) → moles = 15.7/18 = 0.872 mol → molecules = 0.872 × 6.02 × 10²³ = 5.25 × 10²³.

Reacting Masses

  • Use balanced equation to find mole ratio between reactants/products.
  • Convert given mass to moles using moles = mass / Mr.
  • Use ratio to find moles of required substance.
  • Convert back to mass using mass = moles × Mr.
  • Example: 6.0 g Mg (Mr = 24) → 0.25 mol Mg → 1:1 ratio → 0.25 mol MgO → mass = 0.25 × 40 = 10 g.

Limiting Reactant

  • Limiting reactant is used up first and determines amount of product.
  • Excess reactant remains after reaction stops.
  • To find limiting reactant: calculate moles of each, compare with balanced equation ratio.
  • Example: 9.2 g Na (0.40 mol) + 8.0 g S (0.25 mol); equation 2Na + S → Na₂S requires 0.40 mol Na to react with 0.20 mol S; S (0.25 mol) is excess, Na is limiting.

Diapos

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  1. 1.What is the Avogadro constant?

    Easy
    • A6.02 × 1023
    • B6.02 × 1022
    • C6.02 × 1024
    • D6.02 × 1021

  2. 2.How many moles are there in 2.64 g of sucrose (C12H22O11, Mr = 342)? Give your answer in standard form to 3 significant figures.

    Medium
    • A7.72 × 10⁻³
    • B7.72 × 10⁻²
    • C1.30 × 10²
    • D2.72 × 10⁻³

  3. 3.State the volume, in dm³, occupied by one mole of any gas at room temperature and pressure.

    Easy

  4. 4.One mole of any substance contains 6.02 × 1023 particles.

    Easy

    True or false?

  5. 5.Complete the sentence.

    Easy

    The mass of one mole of a substance is called its ____ mass.

  6. 6.How many moles of carbon dioxide gas are present in 960 dm³ at RTP? (Molar volume = 24 dm³/mol)

    Easy
    • A40
    • B36
    • C48
    • D44

  7. 7.Match each quantity to its correct value.

    Medium
    • Number of atoms in 1 mole of MgCl2
    • Number of chloride ions in 1 mole of MgCl2
    • Number of molecules in 1 mole of MgCl2
    • 6.02 × 1023
    • 1.204 × 1024
    • 1.806 × 1024

  8. 8.Arrange the following steps in the correct order to calculate the mass of product from a given mass of reactant in a chemical reaction.

    Medium
    • Calculate moles of reactant using mass ÷ Mr
    • Use the balanced equation to find the mole ratio
    • Calculate mass of product using moles × Mr

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