Grade 11 Physical Sciences
Term 3 · Weeks 2–5

Quantitative Aspects of Chemical Change

Paper 2 — Chemistry · Grade 11

Stoichiometry is the branch of chemistry that uses the mole concept to relate masses, volumes, and particle counts in chemical reactions. Mastering mole calculations unlocks every quantitative problem in chemistry.

Week 2

8.1 The Mole and Avogadro's Law

Define mole concept: n = m/M, n = N/NA, n = V/22.4 (at STP)Calculate concentration: c = n/V

Definition

Avogadro's Law

Equal volumes of gases, at the same temperature and pressure, contain the same number of molecules.

The mole is chemistry's counting unit. One mole of any substance contains 6,022 × 10²³ particles (Avogadro's number, NA). Because of Avogadro's Law, one mole of ANY gas occupies 22,4 dm³ at STP (Standard Temperature and Pressure: 0 °C and 101,325 kPa).

Mole Triangle — n = m/Mm(grams, g)n(moles, mol)M(g·mol⁻¹)n = m/Mm = n × MM = m/n
Figure 8.1 — The mole triangle. Cover the quantity you need: n = m/M, m = n × M, M = m/n. Always use mass in grams and molar mass in g·mol⁻¹.

Formula

Moles from mass

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

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

SI unit: mol

Formula

Moles from number of particles

n=NNAn = \frac{N}{N_A}

n = moles (mol), N = number of particles, NA = 6,022 × 10²³ mol⁻¹

SI unit: mol

Formula

Moles of gas at STP

n=V22,4n = \frac{V}{22{,}4}

n = moles (mol), V = volume (dm³) at STP

SI unit: mol

Formula

Concentration

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

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

SI unit: mol·dm⁻³

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Exam Tip

STP is 0 °C (273 K) and 101,325 kPa. If the problem says 'room temperature and pressure' (RTP), use 24 dm³·mol⁻¹ instead of 22,4 dm³·mol⁻¹.

Worked Example

How many moles are in 36 g of water? (M(H₂O) = 18 g·mol⁻¹)

Given

  • m = 36 g
  • M(H₂O) = 18 g·mol⁻¹

Find

n = ?

Solution

  1. 1n = m/M
  2. 2n = 36/18
  3. 3n = 2 mol
Answer: n = 2 mol

Worked Example

Calculate the concentration of a solution in which 5,3 g of Na₂CO₃ (M = 106 g·mol⁻¹) is dissolved in 500 cm³ of water.

Given

  • m = 5,3 g
  • M(Na₂CO₃) = 106 g·mol⁻¹
  • V = 500 cm³ = 0,5 dm³

Find

c = ?

Solution

  1. 1n = m/M = 5,3/106 = 0,05 mol
  2. 2c = n/V = 0,05/0,5
  3. 3c = 0,1 mol·dm⁻³
Answer: c = 0,1 mol·dm⁻³
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Practice Question

Calculate the number of moles in 4,48 dm³ of CO₂ gas measured at STP.

(3 marks)

Week 3

8.2 Stoichiometric Calculations and Mole Ratios

Use mole ratios from balanced equations in stoichiometric calculations

A balanced chemical equation gives the ratio in which reactants combine and products form. These ratios are called mole ratios. For example, in 2H₂ + O₂ → 2H₂O the mole ratio H₂ : O₂ : H₂O = 2 : 1 : 2. You MUST use a balanced equation before applying mole ratios.

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Watch Out

Always balance the equation FIRST. Using mole ratios from an unbalanced equation is one of the most common errors in stoichiometry.

Step-by-step method for stoichiometry

  1. Write and balance the chemical equation.
  2. Convert the given quantity to moles (n = m/M or n = V/22,4).
  3. Use the mole ratio from the balanced equation.
  4. Convert moles of the required substance to the desired unit (mass, volume, etc.).

Worked Example

How many grams of water are produced when 4 g of H₂ burns completely in oxygen? (M(H₂) = 2 g·mol⁻¹, M(H₂O) = 18 g·mol⁻¹)

Given

  • m(H₂) = 4 g
  • 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: m(H₂O) = 36 g

Worked Example

What volume of CO₂ (at STP) is produced by burning 12 g of carbon? (Ar: C = 12; C + O₂ → CO₂)

Given

  • m(C) = 12 g
  • M(C) = 12 g·mol⁻¹
  • Balanced: C + O₂ → CO₂

Find

V(CO₂) at STP

Solution

  1. 1n(C) = 12/12 = 1 mol
  2. 2Mole ratio C : CO₂ = 1 : 1, so n(CO₂) = 1 mol
  3. 3V = n × 22,4 = 1 × 22,4 = 22,4 dm³
Answer: V(CO₂) = 22,4 dm³
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Practice Question

In the reaction N₂ + 3H₂ → 2NH₃, what mass of NH₃ is produced from 14 g of N₂? (M(N₂) = 28 g·mol⁻¹, M(NH₃) = 17 g·mol⁻¹)

(5 marks)

Periodic Table TrendsLiFCsatomic radius ↑ionisation energy →electronegativity →Groups →
Figure 8.2 — Use the periodic table to find molar masses. The number at the bottom of each element tile is the relative atomic mass (Ar) in g·mol⁻¹.
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Real World

Industrial chemists use stoichiometry to scale laboratory reactions up to factory size — calculating exactly how many tonnes of raw material are needed to produce a target mass of product while minimising waste.

Week 4–5

8.3 Limiting Reagent, Theoretical Yield, and Percentage Purity

Identify limiting and excess reagentsCalculate theoretical yield and percentage yield: % yield = (actual/theoretical) × 100Calculate percentage purity

Definition

Limiting reagent

The limiting reagent is the reactant that is completely consumed in a chemical reaction, thereby determining how much product can form. All other reagents present in amounts greater than required are said to be in excess.

Definition

Excess reagent

The excess reagent is the reactant that remains when the reaction is complete because it was present in a greater quantity than was needed to react with the limiting reagent. The amount of excess reagent does not affect the amount of product formed.

When two reactants are mixed, one runs out first — that is the limiting reagent. It determines the maximum amount of product that can form (the theoretical yield). The other reactant is in excess and some of it remains unreacted.

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Exam Tip

To identify the limiting reagent: divide the moles of EACH reactant by its coefficient in the balanced equation. The reactant with the SMALLEST result is the limiting reagent.

Formula

Percentage yield

% yield=mactualmtheoretical×100\%\,\text{yield} = \dfrac{m_{\text{actual}}}{m_{\text{theoretical}}} \times 100

m_actual = mass of product obtained in experiment (g); m_theoretical = mass of product calculated from stoichiometry (g)

SI unit: %

Formula

Percentage purity

%=mpuremsample×100\% = \frac{m_{\text{pure}}}{m_{\text{sample}}} \times 100

mass of pure substance = mass of desired compound only; mass of sample = total mass including impurities

SI unit: %

Limiting vs Excess Reagent

PropertyLimiting ReagentExcess Reagent
Amount usedCompletely consumedPartially consumed
Controls productYes — determines yieldNo
Left over?NoYes
Found by…Smallest moles/coefficient ratioLarger moles/coefficient ratio

Worked Example

2 mol of H₂ and 2 mol of O₂ react: 2H₂ + O₂ → 2H₂O. Identify the limiting reagent and calculate the theoretical yield of water.

Given

  • n(H₂) = 2 mol
  • n(O₂) = 2 mol
  • Balanced: 2H₂ + O₂ → 2H₂O

Find

Limiting reagent and theoretical yield of H₂O

Solution

  1. 1Divide by coefficient: H₂ → 2/2 = 1; O₂ → 2/1 = 2
  2. 2Smallest ratio: H₂ (ratio = 1) → H₂ is the limiting reagent
  3. 3n(H₂O) = n(H₂) × (2/2) = 2 mol (mole ratio H₂:H₂O = 1:1)
  4. 4m(H₂O) = 2 × 18 = 36 g
Answer: H₂ is the limiting reagent; theoretical yield = 36 g of H₂O

Worked Example

A student obtains 28 g of H₂O in the reaction above. Calculate the percentage yield.

Given

  • Actual yield = 28 g
  • Theoretical yield = 36 g

Find

% yield

Solution

  1. 1% yield = (actual/theoretical) × 100
  2. 2% yield = (28/36) × 100
  3. 3% yield = 77,8%
Answer: % yield = 77,8%

Worked Example

A 5,0 g sample of CaCO₃ contains 4,6 g of pure CaCO₃. Calculate the percentage purity.

Given

  • Mass of sample = 5,0 g
  • Mass of pure CaCO₃ = 4,6 g

Find

% purity

Solution

  1. 1% purity = (mass of pure substance / mass of sample) × 100
  2. 2% purity = (4,6/5,0) × 100
  3. 3% purity = 92%
Answer: % purity = 92%
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Watch Out

Do not confuse % yield with % purity. Percentage yield compares what you got in an experiment to what stoichiometry predicts. Percentage purity measures how much of a sample is the desired compound.

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Practice Question

Fe₂O₃ + 3CO → 2Fe + 3CO₂. If 160 g of Fe₂O₃ (M = 160 g·mol⁻¹) and 42 g of CO (M = 28 g·mol⁻¹) are used, identify the limiting reagent and calculate the theoretical yield of Fe (M = 56 g·mol⁻¹).

(7 marks)

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Energy and Chemical Change

Quantitative Aspects of Chemical Change Grade 11 Physical Sciences CAPS Notes | MathSciBuddy