Grade 10 Physical Sciences
Term 3 · Weeks 2–3

Quantitative Aspects of Chemical Change

Paper 2 — Chemistry · Grade 10

The mole is chemistry's counting unit — just as a dozen means 12, a mole means 6,02 × 10²³ particles. All stoichiometric calculations flow from the single formula n = m/M.

Week 2

The Mole Concept

Define mole and Avogadro's numberCalculate n = m/MCalculate number of particles N = n × NₐCalculate molar mass Mr from formula
mol

Definition

Mole (mol)

The mole is the amount of substance that contains as many elementary entities as there are atoms in exactly 12 g of ¹²C; 1 mol = 6,02 × 10²³ particles.

Nₐ

Definition

Avogadro's number (Nₐ)

Avogadro's number is the number of particles in one mole of any substance: Nₐ = 6,02 × 10²³ mol⁻¹.

M

Definition

Molar mass (M)

The molar mass is the mass of one mole of a substance, numerically equal to its relative atomic or molecular mass; units g·mol⁻¹.

Ar

Definition

Relative atomic mass (Ar)

The relative atomic mass is the weighted average mass of the atoms of an element relative to ¹²C = 12.

Mole Triangle — n = m/Mm(grams, g)n(moles, mol)M(g·mol⁻¹)n = m/Mm = n × MM = m/n
Figure 16.1 — The mole triangle. Cover the quantity you want to find: n = m/M, m = n × M, M = m/n. Always use mass in grams (g) and molar mass in g·mol⁻¹.

Formula

Amount of substance

n=mMn = \frac{m}{M}

n = moles (mol), m = mass (g), M = molar mass (g·mol⁻¹)

SI unit: mol

Formula

Number of particles

N=n×NAN = n \times N_A

N = number of particles, n = moles (mol), Nₐ = 6,02 × 10²³ mol⁻¹

SI unit: dimensionless (count)

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Exam Tip

To find M for a compound: multiply each element's Ar by the number of that atom in the formula, then add. M(H₂O) = 2(1) + 16 = 18 g·mol⁻¹.

Worked Example

Calculate the number of moles in 44 g of CO₂. (M(CO₂) = 44 g·mol⁻¹)

Given

  • m = 44 g
  • M(CO₂) = 44 g·mol⁻¹

Find

n = ?

Solution

  1. 1n = m/M
  2. 2n = 44/44
  3. 3n = 1 mol
Answer: n = 1 mol

Worked Example

Calculate the molar mass of Ca(OH)₂ and the number of moles in 74 g of it. (Ar: Ca = 40, O = 16, H = 1)

Given

  • Formula: Ca(OH)₂ contains 1 Ca, 2 O, and 2 H
  • m = 74 g

Find

M(Ca(OH)₂) and n

Solution

  1. 1M(Ca(OH)₂) = 40 + 2(16 + 1) = 40 + 2(17) = 40 + 34 = 74 g·mol⁻¹
  2. 2n = m/M = 74/74 = 1 mol
Answer: M(Ca(OH)₂) = 74 g·mol⁻¹; n = 1 mol

Worked Example

How many molecules are in 3 mol of O₂? (Nₐ = 6,02 × 10²³ mol⁻¹)

Given

  • n = 3 mol
  • Nₐ = 6,02 × 10²³ mol⁻¹

Find

N = ?

Solution

  1. 1N = n × Nₐ
  2. 2N = 3 × 6,02 × 10²³
  3. 3N = 1,806 × 10²⁴ molecules
Answer: N = 1,806 × 10²⁴ molecules
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Practice Question

Calculate the number of moles in: (a) 36 g of H₂O, (b) 16 g of O₂, (c) 5,6 g of Fe. (Ar: H = 1, O = 16, Fe = 56)

(6 marks)

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Practice Question

A sample contains 3,01 × 10²³ atoms of sodium. Calculate: (a) the number of moles of Na atoms, (b) the mass of the sample. (M(Na) = 23 g·mol⁻¹; Nₐ = 6,02 × 10²³ mol⁻¹)

(4 marks)

Week 3

Concentration and Stoichiometry

Calculate concentration c = n/VConvert between cm³ and dm³Use mole ratios from balanced equationSolve stoichiometry problems
c

Definition

Concentration (c)

Concentration is the number of moles of solute per unit volume of solution; c = n/V, measured in mol·dm⁻³.

Definition

Stoichiometry

Stoichiometry is the calculation of the quantities (mass, moles, volume) of reactants and products in a chemical reaction using a balanced equation.

Definition

Molar volume

The molar volume is the volume occupied by one mole of gas at STP (standard temperature and pressure: 0°C and 101,3 kPa); molar volume = 22,4 dm³·mol⁻¹.

Formula

Concentration

c=nVc = \frac{n}{V}

c = concentration (mol·dm⁻³), n = moles (mol), V = volume (dm³)

SI unit: mol·dm⁻³

ℹ

Note

Volume conversion: 1 dm³ = 1 L = 1 000 cm³ = 1 000 mL. To convert cm³ to dm³: divide by 1 000. E.g., 250 cm³ = 0,250 dm³.

Worked Example

Calculate the concentration of a solution made by dissolving 8 g of NaOH in 500 cm³ of water. M(NaOH) = 40 g·mol⁻¹

Given

  • m = 8 g
  • M(NaOH) = 40 g·mol⁻¹
  • V = 500 cm³ = 0,5 dm³

Find

c = ?

Solution

  1. 1n = m/M = 8/40 = 0,2 mol
  2. 2V = 500 cm³ ÷ 1 000 = 0,5 dm³
  3. 3c = n/V = 0,2/0,5 = 0,4 mol·dm⁻³
Answer: c = 0,4 mol·dm⁻³

Worked Example

In the reaction 2H₂ + O₂ → 2H₂O, how many grams of H₂O are produced from 4 g of H₂? (M(H₂) = 2 g·mol⁻¹; M(H₂O) = 18 g·mol⁻¹)

Given

  • m(H₂) = 4 g
  • M(H₂) = 2 g·mol⁻¹
  • M(H₂O) = 18 g·mol⁻¹
  • Balanced equation: 2H₂ + O₂ → 2H₂O

Find

m(H₂O) = ?

Solution

  1. 1n(H₂) = m/M = 4/2 = 2 mol
  2. 2Mole ratio H₂ : H₂O = 2 : 2 = 1 : 1
  3. 3n(H₂O) = 2 mol
  4. 4m(H₂O) = n × M = 2 × 18 = 36 g
Answer: 36 g of H₂O produced
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Practice Question

How many grams of NaCl are needed to make 250 cm³ of a 2,0 mol·dm⁻³ solution? (M(NaCl) = 58,5 g·mol⁻¹)

(5 marks)

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Practice Question

In the reaction N₂ + 3H₂ → 2NH₃, how many moles of NH₃ are produced from 0,6 mol of H₂?

(3 marks)

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Practice Question

Calculate the volume of CO₂ produced at STP when 5 mol of CaCO₃ decomposes: CaCO₃ → CaO + CO₂. (Molar volume = 22,4 dm³·mol⁻¹)

(3 marks)

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Vectors and Scalars

Quantitative Aspects of Chemical Change Grade 10 Physical Sciences CAPS Notes | MathSciBuddy