The mole concept

Unit 5 The mole concept

Key unit competence: Apply the mole concept in quantitative chemical calculations involving mass, moles and chemical formulae.

Key ideas

• A mole is the amount of substance containing 6.02 × 10²³ particles (the Avogadro constant, N_A). Number of particles = moles × 6.02 × 10²³.
• Relative atomic mass (Ar) compares the mass of an atom with 1/12 of the mass of a carbon-12 atom. Relative molecular/formula mass (Mr) = sum of the Ar values in the formula, e.g. Mr of H₂O = 2(1) + 16 = 18.
• Molar mass (M) is the mass of one mole in g/mol (numerically equal to Mr). moles = mass ÷ molar mass (n = m/M).
• Percentage of an element = (number of atoms × Ar ÷ Mr) × 100. The empirical formula is the simplest whole-number ratio of atoms; the molecular formula is a whole-number multiple of it (n = Mr ÷ empirical formula mass).
• Stoichiometry: use the mole ratio in the balanced equation to find amounts of reactants or products. The limiting reactant is used up first and decides the amount of product; the other is in excess.
• Molar gas volume: one mole of any gas occupies 22.4 dm³ at STP (0 °C, 1 atm) and about 24 dm³ at room temperature and pressure. moles of gas = volume ÷ molar volume.

Common misconceptions

• "One mole of every substance has the same mass." No: One mole always has the same NUMBER of particles; the mass depends on the molar mass.
• "The empirical formula is always the same as the molecular formula." No: Glucose C₆H₁₂O₆ has empirical formula CH₂O.
• "The reactant with the smaller mass is always the limiting reactant." No: Compare MOLES using the equation ratio, not masses.

Worked examples

Practice questions

1. Define one mole.

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   The amount of substance that contains 6.02 × 10²³ particles.

   Common mistake: Saying 'the mass of one atom'.

2. Calculate the Mr of Ca(OH)₂ (Ca 40, O 16, H 1).

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   40 + 2 × (16 + 1) = 74.

   Common mistake: Forgetting to multiply OH by 2.

3. How many moles are in 10 g of CaCO₃ (Mr 100)?

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   0.1 mol.

   Common mistake: Multiplying instead of dividing.

4. What mass of NaOH (Mr 40) is in 0.25 mol?

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   10 g.

   Common mistake: Dividing 0.25 by 40.

5. Find the volume of 0.3 mol of nitrogen at STP.

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   0.3 × 22.4 = 6.72 dm³.

   Common mistake: Using 24 dm³ when STP is stated.

6. Calculate the percentage of carbon in methane, CH₄.

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   12 ÷ 16 × 100 = 75%.

   Common mistake: Using 12 ÷ 4.

7. Find the empirical formula of a compound with 2.4 g carbon and 0.6 g hydrogen.

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   C: 0.2 mol; H: 0.6 mol; ratio 1 : 3; CH₃.

   Common mistake: Using mass ratio directly (4 : 1).

8. How many atoms are in 0.5 mol of NH₃?

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   0.5 × 4 × 6.02 × 10²³ = 1.204 × 10²⁴ atoms.

   Common mistake: Forgetting the 4 atoms per molecule.

9. What is meant by the limiting reactant?

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   The reactant that is completely used up first and limits the amount of product.

   Common mistake: Saying it is the one present in smaller mass.

10. What mass of oxygen is needed to burn 6 g of carbon to CO₂?

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    n(C) = 0.5 mol; O₂ = 0.5 mol; mass = 16 g.

    Common mistake: Using O = 16 instead of O₂ = 32.

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. How many molecules are in 0.25 mol of CO2?	a. 24 g
   2. Moles of a gas = volume ÷ ______	b. 1.505 × 1023
   3. What mass of magnesium (Ar = 24) contains the same number of atoms as 12 g of carbon (Ar = 12)?	c. 44.8 dm3
   4. 4 g of hydrogen gas (H2, Mr = 2) occupies what volume at STP?	d. molar volume

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   1 → b (1.505 × 10²³) · 2 → d (molar volume) · 3 → a (24 g) · 4 → c (44.8 dm³)

   • 0.25 × 6.02 × 10²³ = 1.505 × 10²³.
   • n = V ÷ Vm.
   • 12 g C is 1 mol; 1 mol Mg = 24 g.
   • n = 4 ÷ 2 = 2 mol; V = 2 × 22.4 = 44.8 dm³.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   CH₂O  ·  is completely used up first  ·  C₆H₁₂O₆  ·  1 : 1
   1. The limiting reactant is the reactant that .
   2. A compound has empirical formula CH₂O and Mr = 180. Its molecular formula is .
   3. In 2H₂ + O₂ → 2H₂O, the mole ratio of hydrogen to water is .
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   1: is completely used up first · 2: C₆H₁₂O₆ · 3: 1 : 1

   • It limits the amount of product formed.
   • Empirical mass 30; 180 ÷ 30 = 6.
   • 2 mol H₂ gives 2 mol H₂O.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. The Avogadro constant is:	a. 6.02 × 1023 per mole
   2. Percentage of nitrogen in NH4NO3 (N = 14, H = 1, O = 16):	b. 0.5 mol
   3. Percentage of oxygen in water (Mr = 18):	c. 88.9%
   4. How many moles are in 11.2 dm3 of oxygen at STP?	d. 35%

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   1 → a (6.02 × 10²³ per mole) · 2 → d (35%) · 3 → c (88.9%) · 4 → b (0.5 mol)

   • This is the number of particles in one mole.
   • Mr = 80; N = 28; 28 ÷ 80 × 100 = 35%.
   • 16 ÷ 18 × 100 = 88.9%.
   • 11.2 ÷ 22.4 = 0.5.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   100  ·  CH₂  ·  56  ·  98 g/mol
   1. The relative formula mass of CaCO₃ (Ca = 40, C = 12, O = 16) is .
   2. The empirical formula of C₂H₄ is .
   3. The molar mass of H₂SO₄ (H = 1, S = 32, O = 16) is .
   Show answer

   1: 100 · 2: CH₂ · 3: 98 g/mol

   • 40 + 12 + 48 = 100.
   • Divide by 2 to get the simplest ratio.
   • 2 + 32 + 64 = 98.

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