Using Moles - Complete Theory & Formulas

Cambridge IGCSE Chemistry - Grade 9

1. The Mole Concept - Core Theory

Definition of a Mole

A mole is the amount of substance that contains as many elementary entities (atoms, molecules, ions) as there are atoms in exactly 12 grams of carbon-12.

Avogadro's Constant

N_A = 6.022 × 10²³ mol^-1

This means 1 mole of any substance contains 6.022 × 10²³ particles

Key Principle: One mole of any substance contains the same number of particles, but the mass depends on the type of particle.

2. Fundamental Formulas

2.1 Mole-Mass Relationship

Number of moles (n) = Mass (m) ÷ Molar mass (M)

n = m/M

Step 1: Find the molar mass using A_r or M_r values

Step 2: Apply the formula n = m/M

Step 3: Check units (moles should be dimensionless)

2.2 Molar Mass Calculation

Molar Mass = A_r or M_r expressed in g/mol

Examples:

Substance	Formula	Calculation	Molar Mass (g/mol)
Sodium	Na	23	23
Water	H₂O	(2×1) + 16	18
Carbon dioxide	CO₂	12 + (2×16)	44
Calcium carbonate	CaCO₃	40 + 12 + (3×16)	100

3. Gas Volume Calculations

Avogadro's Law: Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.

At Room Temperature and Pressure (RTP: 20°C, 1 atm):
1 mole of any gas occupies 24 dm³

3.1 Gas Volume Formulas

Volume (V) = Number of moles (n) × 24 dm³

V = n × 24

3.2 Combined Gas-Mass Formula

Volume = (Mass ÷ Molar Mass) × 24

V = (m/M) × 24 dm³

Example: Volume of 22g CO₂ at RTP

Given: Mass = 22g, M_r of CO₂ = 44

Step 1: n = 22 ÷ 44 = 0.5 moles

Step 2: V = 0.5 × 24 = 12 dm³

4. Solution Concentration

4.1 Concentration Formulas

Concentration (mol/dm³) = Amount of solute (mol) ÷ Volume of solution (dm³)

C = n/V

Amount of solute (mol) = Concentration (mol/dm³) × Volume (dm³)

n = C × V

4.2 Unit Conversions

Volume conversions:
1 dm³ = 1 liter = 1000 cm³ = 1000 mL

Concentration units:
mol/dm³ = M (Molar) = mol/L

Example: Preparing 0.1 M NaOH solution

To make 250 cm³ of 0.1 M NaOH:

Volume = 250 cm³ = 0.25 dm³

Moles needed = 0.1 × 0.25 = 0.025 mol

Mass needed = 0.025 × 40 = 1.0 g NaOH

5. Chemical Equations and Stoichiometry

5.1 Mole Ratios from Equations

Chemical equations show the mole ratio in which substances react.

Example: 2H₂ + O₂ → 2H₂O
This means: 2 mol H₂ : 1 mol O₂ : 2 mol H₂O

Mole ratio = Coefficient in balanced equation

Actual moles = (Given moles × Required coefficient) ÷ Given coefficient

5.2 Mass Calculations from Equations

Step 1: Write balanced equation

Step 2: Calculate molar masses

Step 3: Use mole ratios

Step 4: Convert to required units

Example: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Question: What mass of CO₂ is produced from 10g CaCO₃?

Solution:

1. Molar masses: CaCO₃ = 100 g/mol, CO₂ = 44 g/mol

2. Moles of CaCO₃ = 10 ÷ 100 = 0.1 mol

3. From equation: 1 mol CaCO₃ → 1 mol CO₂

4. Mass of CO₂ = 0.1 × 44 = 4.4 g

6. Finding Chemical Formulas

6.1 Empirical Formula

Empirical Formula: Shows the simplest whole number ratio of atoms in a compound.

Step 1: Find mass of each element

Step 2: Convert mass to moles (÷ A_r)

Step 3: Find simplest ratio

Step 4: Write empirical formula

6.2 Molecular Formula

n = M_r ÷ Empirical mass

Molecular formula = n × Empirical formula

Example: Finding molecular formula

Given: Empirical formula = CH₂, M_r = 42

Empirical mass = 12 + (2×1) = 14

n = 42 ÷ 14 = 3

Molecular formula = 3 × CH₂ = C₃H₆

7. Percentage Composition

7.1 Percentage by Mass

% of element = (Mass of element in compound ÷ Total mass of compound) × 100%

% of element = (Number of atoms × A_r ÷ M_r) × 100%

Example: % composition of H₂SO₄

M_r of H₂SO₄ = (2×1) + 32 + (4×16) = 98

% H = (2×1 ÷ 98) × 100% = 2.04%

% S = (32 ÷ 98) × 100% = 32.65%

% O = (4×16 ÷ 98) × 100% = 65.31%

8. Yield and Purity Calculations

8.1 Percentage Yield

% Yield = (Actual yield ÷ Theoretical yield) × 100%

Theoretical yield: Maximum amount calculated from equation
Actual yield: Amount actually obtained in experiment

8.2 Percentage Purity

% Purity = (Mass of pure substance ÷ Mass of impure sample) × 100%

Example: Purity calculation

A 5.0g sample of limestone produces 2.2g CO₂ when treated with acid.

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Moles of CO₂ = 2.2 ÷ 44 = 0.05 mol

Moles of CaCO₃ = 0.05 mol (1:1 ratio)

Mass of pure CaCO₃ = 0.05 × 100 = 5.0g

But this equals the total sample mass, so purity = 100%

9. Advanced Applications

9.1 Limiting Reagent

The limiting reagent is the reactant that is completely consumed first, limiting the amount of product formed.

Step 1: Calculate moles of each reactant

Step 2: Use mole ratios to find which runs out first

Step 3: Base calculations on limiting reagent

9.2 Back Titration

Moles reacted = Initial moles - Final moles

10. Quick Reference Formulas

Essential Formula Sheet

Calculation	Formula	Units
Number of moles	n = m/M	mol
Gas volume (RTP)	V = n × 24	dm³
Concentration	C = n/V	mol/dm³
Percentage yield	% = (actual/theoretical) × 100	%
Percentage purity	% = (pure/impure) × 100	%
Percentage composition	% = (part/whole) × 100	%

Important Constants

Avogadro's constant: N_A = 6.022 × 10²³ mol^-1

Molar volume at RTP: 24 dm³/mol

RTP conditions: 20°C (293K), 1 atm pressure

"Master these formulas and you master the language of chemistry!"

Cambridge IGCSE Chemistry - Using Moles Complete Theory & Formulas

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