The mole and the Avogadro constant

O Level / IGCSE Chemistry · Stoichiometry. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.

Test this topic freeAll O Level / IGCSE Chemistry topics

The idea

The mole is the unit of amount of substance. One mole of anything contains 6.02 × 1023 particles, a number called the Avogadro constant. The mass of one mole, the molar mass, is the Ar or Mr written in grams, so one mole of water has a mass of 18 g.

Almost every calculation follows the same three steps: change what you are given into moles, use the numbers in the balanced equation to find the moles of the substance you want, then change those moles into a mass, a gas volume or a concentration.

If amounts of two reactants are given, the one that runs out first is the limiting reactant. It decides how much product forms.

Rules to remember

Common mistake

Students forget to change cm3 into dm3 before using a concentration: 25.0 cm3 is 0.0250 dm3, not 25 dm3. Also use the ratio from the balanced equation instead of assuming it is 1 : 1.

Worked example

Magnesium reacts with excess dilute hydrochloric acid: Mg + 2HCl → MgCl2 + H2. What volume of hydrogen, measured at r.t.p., is produced from 4.8 g of magnesium? (Mg 24; molar gas volume 24 dm3 at r.t.p.)

  1. Amount of Mg = mass ÷ molar mass = 4.8 ÷ 24 = 0.20 mol.
  2. From the equation, 1 mol of Mg gives 1 mol of H2, so 0.20 mol of Mg gives 0.20 mol of H2.
  3. Volume of H2 = amount × 24 = 0.20 × 24 = 4.8 dm3.

Answer: 4.8 dm3 of hydrogen (4800 cm3).

Practice questions

Try each one, then open the answer.

1. 25.0 cm3 of 0.10 mol/dm3 sodium hydroxide is exactly neutralised by 20.0 cm3 of hydrochloric acid. What is the concentration of the acid? (NaOH + HCl → NaCl + H2O)

  1. A
    0.080 mol/dm3
  2. B
    0.10 mol/dm3
  3. C
    0.125 mol/dm3
  4. D
    0.20 mol/dm3
Show answer

Answer: C. moles NaOH = 0.10 × 0.025 = 0.0025 mol = moles HCl (1 : 1). Concentration = 0.0025 ÷ 0.020 = 0.125 mol/dm3.

2. Zinc sulfate is made by adding 6.5 g of zinc to 50 cm3 of 1.0 mol/dm3 dilute sulfuric acid. When the reaction has stopped, what mass of zinc is left over? (Zn 65)

  1. A
    0 g
  2. B
    6.5 g
  3. C
    1.6 g
  4. D
    3.25 g
Show answer

Answer: D. Zn + H2SO4 → ZnSO4 + H2. Acid = 0.050 × 1.0 = 0.050 mol; zinc = 6.5 ÷ 65 = 0.10 mol. Only 0.050 mol of zinc reacts, leaving 0.050 mol = 3.25 g. Using excess zinc makes sure that all the acid reacts, and the leftover metal is filtered off.

3. What mass of sodium hydroxide is needed to make 250 cm3 of a 0.40 mol/dm3 solution? (Na 23, O 16, H 1)

  1. A
    0.10 g
  2. B
    1.6 g
  3. C
    4.0 g
  4. D
    16 g
Show answer

Answer: C. moles = 0.40 × 0.250 = 0.10 mol. Mr of NaOH = 40, so mass = 0.10 × 40 = 4.0 g. Using 1 dm3 by mistake gives 16 g.

More questions on this topic

Read the full notes

Keep going

← Relative masses of atoms and moleculesElectrolysis →All rules on one pageStuck? Ask a question