Lesson overview
The mole: How many particles are in a spoonful?
Weigh a spoonful of blue copper(II) sulfate crystals and make it into exactly 250.0 cm³ of solution in a volumetric flask, filling the last drops to the line by hand. Work out its concentration, count the copper ions in a single drop, then plan and make a paler solution.
Suggested sequence
- Lesson 1 · How many molecules are in a spoonful of water? The mole, the Avogadro constant and n = m/M. Read together and discuss (15–20 min), practise with self-checking (20–25 min).
- Lesson 2 · How much solute is in the solution? Concentration in g/dm³ and mol/dm³. Read together (10–15 min), show or explore the experiment (10 min), practise with self-checking (15–20 min).
- End of module · final check. 6 new questions, 20–30 min, without hints or notes; pupils check their answers once they have answered them all.
- One to two weeks later · return questions. 4 questions, 15 min, lesson closed.
Learning goals
Part 1: How many molecules are in a spoonful of water?
- State that the mole, mol, is the unit of amount of substance and that one mole contains 6.02 × 10²³ particles, e.g. atoms, ions, molecules; this number is the Avogadro constant
- Use the relationship amount of substance (mol) = mass (g) / molar mass (g / mol) to calculate: (a) amount of substance (b) mass (c) molar mass (d) relative atomic mass or relative molecular / formula mass (e) number of particles, using the value of the Avogadro constant
Part 2: How much solute is in the solution?
- Use the relationship amount of substance (mol) = mass (g) / molar mass (g / mol) to calculate: (a) amount of substance (b) mass (c) molar mass (d) relative atomic mass or relative molecular / formula mass (e) number of particles, using the value of the Avogadro constant
- State that concentration can be measured in g / dm³ or mol / dm³
- Calculate volumes of solutions and concentrations of solutions expressed in g / dm³ and mol / dm³, including conversion between cm³ and dm³ (the solution part of 3.3.5)
Misconceptions to listen for
- “They have the same mass, because each is 1 mol.” — One mole always holds the same number of particles, but a zinc atom is about twice as heavy as a sulfur atom, so 1 mol of zinc (65 g) is heavier than 1 mol of sulfur (32 g).
- “The zinc holds more atoms, because zinc atoms are heavier.” — A mole is a number of particles, whatever they weigh. Heavier atoms make the mole heavier, not more numerous.
- “Each sample has a mass of 6.02 × 10²³ grams.” — 6.02 × 10²³ is a number of particles, not a number of grams. The mass of a mole is its molar mass.
- “Molar mass has no unit, like Mr.” — Mr compares masses and has no unit, but molar mass is the mass of one mole, so it is in g/mol.
- “Amount = molar mass ÷ mass.” — Divide the mass by the molar mass, n = m/M. The units show it: g ÷ g/mol leaves mol.
- “NH3 has M = 15 g/mol.” — The 3 belongs to the hydrogen: three H atoms, so M = 14 + 3 × 1.
- “Mass = amount ÷ molar mass.” — Rearrange n = m/M by multiplying both sides by M: m = n × M.
- “Molar mass = amount ÷ mass.” — That gives mol/g, the moles in each gram. Molar mass is grams in each mole: mass ÷ amount.
- “A mole of molecules holds 6.02 × 10²³ atoms.” — It holds 6.02 × 10²³ molecules. Multiply by the number of atoms in each molecule to count the atoms.
- “MgCl2 gives two ions: Mg2+ and Cl2−.” — Chloride ions are single, Cl−. Each MgCl2 gives one Mg2+ and two separate Cl− ions.
Scope
The mole (3.3 Core 1, Supplement 2, 3 and part of 5): concentration in g/dm³ and mol/dm³, the mole and the Avogadro constant, n = m/M for amount, mass, molar mass, relative mass and number of particles, and the volumes and concentrations of solutions with cm³ and dm³. Molar gas volume, reacting masses and gas volumes from equations, limiting reactants, empirical and molecular formulae, percentage yield and purity, and titration are left to other modules.