3.3 The Mole and the Avogadro Constant

1. Overview

This topic introduces the "mole," the fundamental unit of measurement in chemistry that allows scientists to count atoms and molecules by weighing them. Understanding the mole is essential for performing quantitative analysis, predicting how much product will form in a reaction, and determining the concentrations of solutions used in laboratory work.

Key Definitions

• Mole (mol): The unit of amount of substance; one mole contains $6.02 \times 10^{23}$ particles.
• Avogadro Constant: The number of particles in one mole of a substance, equal to $6.02 \times 10^{23}$.
• Molar Mass ($M_r$ or $A_r$): The mass of one mole of a substance, expressed in $g/mol$.
• Molar Gas Volume: The volume occupied by one mole of any gas at room temperature and pressure (r.t.p.), which is $24\ dm^3$.
• Concentration: The amount of solute dissolved in a specific volume of solvent, measured in $g/dm^3$ or $mol/dm^3$.
• Empirical Formula: The simplest whole-number ratio of atoms of each element in a compound.
• Molecular Formula: The actual number of atoms of each element in one molecule of a compound.
• Limiting Reactant: The reactant that is completely used up first in a reaction, determining the maximum amount of product formed.

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Core Content

Concentration is a measure of how much "stuff" (solute) is dissolved in a liquid (solvent). In IGCSE Chemistry, you must recognize two ways to express this:

1. Mass Concentration ($g/dm^3$): How many grams of a substance are in $1\ dm^3$ of solution.
2. Molar Concentration ($mol/dm^3$): How many moles of a substance are in $1\ dm^3$ of solution.

Note: $1\ dm^3 = 1000\ cm^3 = 1\ Litre$

Worked Example: If $10\ g$ of Sodium Chloride is dissolved in $0.5\ dm^3$ of water: $$\text{Concentration} = \frac{10\ g}{0.5\ dm^3} = 20\ g/dm^3$$

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Extended Content (Extended Curriculum Only)

The Mole and Avogadro's Constant

One mole of any substance contains $6.02 \times 10^{23}$ particles (atoms, ions, or molecules).

• $1\ mol$ of $C$ atoms = $6.02 \times 10^{23}$ atoms.
• $1\ mol$ of $H_2O$ molecules = $6.02 \times 10^{23}$ molecules.

Calculations using Mass and Moles

Use the formula: $\text{moles (n)} = \frac{\text{mass (m)}}{\text{molar mass (M)}}$

Worked Example: Calculating Mass Find the mass of $0.2\ mol$ of Calcium Carbonate ($CaCO_3$).

1. Calculate $M_r$: $Ca(40) + C(12) + O \times 3(16 \times 3) = 100\ g/mol$.
2. $\text{Mass} = \text{moles} \times M_r = 0.2 \times 100 = 20\ g$.

Molar Gas Volume

At room temperature and pressure (r.t.p.), one mole of any gas occupies $24\ dm^3$ (or $24,000\ cm^3$).

• $\text{Volume of gas (dm}^3\text{)} = \text{moles} \times 24$

Reacting Masses and Stoichiometry

Stoichiometry uses the balanced equation to find unknown amounts. Example: $2Mg(s) + O_2(g) \rightarrow 2MgO(s)$ Word Equation: Magnesium (s) + Oxygen (g) $\rightarrow$ Magnesium oxide (s) If you have $2\ moles$ of $Mg$, you need $1\ mole$ of $O_2$ to produce $2\ moles$ of $MgO$.

Titration Calculations

To calculate an unknown concentration from titration data:

1. Calculate moles of the "known" solution: $n = \text{conc} \times \text{vol (dm}^3\text{)}$.
2. Use the balanced equation ratio to find moles of the "unknown" solution.
3. Calculate concentration: $\text{conc} = \frac{n}{\text{vol (dm}^3\text{)}}$.

Empirical and Molecular Formulae

Empirical Formula Steps:

1. List the mass (or %) of each element.
2. Divide by $A_r$ to find moles.
3. Divide all by the smallest mole value to find the ratio.

Molecular Formula: $\text{Multiplier} = \frac{\text{Relative Molecular Mass (given)}}{\text{Empirical Formula Mass}}$

Percentage Yield, Composition, and Purity

• % Yield = $\frac{\text{Actual Mass}}{\text{Theoretical Mass}} \times 100$
• % Purity = $\frac{\text{Mass of pure substance}}{\text{Total mass of sample}} \times 100$
• % Composition by mass = $\frac{A_r \times \text{number of atoms}}{M_r \text{ of compound}} \times 100$

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Key Equations

Concept	Equation	Units
Moles (Solids)	$n = \frac{m}{M}$	$m$ (g), $M$ (g/mol)
Moles (Gases)	$n = \frac{V}{24}$	$V$ ($dm^3$) at r.t.p.
Moles (Solutions)	$n = C \times V$	$C$ ($mol/dm^3$), $V$ ($dm^3$)
Concentration Conversion	$C (g/dm^3) = C (mol/dm^3) \times M_r$	-
Particles	$N = n \times (6.02 \times 10^{23})$	$n$ (mol)

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Common Mistakes to Avoid

• ❌ Wrong: Using $cm^3$ directly in concentration formulas.
  • ✓ Right: Always convert $cm^3$ to $dm^3$ by dividing by 1000.
• ❌ Wrong: Forgetting that diatomic gases like Oxygen ($O_2$) or Nitrogen ($N_2$) have a molar mass double their $A_r$.
  • ✓ Right: $M_r$ of $O_2 = 16 \times 2 = 32\ g/mol$.
• ❌ Wrong: Including the large coefficients from balanced equations when calculating $M_r$.
  • ✓ Right: The $M_r$ of $2H_2O$ is just the $M_r$ of $H_2O$ ($18\ g/mol$); the "2" is only used for the mole ratio.

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Exam Tips

• Command Words:
  • "Calculate": Show every step of your working. Marks are often awarded for the method even if the final answer is wrong.
  • "Determine": Usually requires you to use data from a previous part of the question.
• Significant Figures: Always give your final answer to 3 significant figures unless specified otherwise.
• Real-world contexts: Expect calculations based on the Haber Process ($N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$) or neutralization reactions in titrations.
• Standard Values: Always use the $A_r$ values provided on the Periodic Table included in your exam paper (e.g., $Cu = 64$ or $63.5$ depending on the syllabus year).