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

Titrations tell you exactly how much of one solution reacts with another. The method always follows the same three steps:

1. Calculate moles of the solution you know: $n = \text{conc} \times \text{vol (in dm}^3\text{)}$.
2. Use the balanced equation to find how many moles of the other solution reacted.
3. Calculate the unknown concentration: $\text{conc} = n / \text{vol}$.

Worked Example: Finding acid concentration from a titration $25.0\text{ cm}^3$ of sodium hydroxide solution ($0.10\text{ mol/dm}^3$) exactly neutralises $20.0\text{ cm}^3$ of hydrochloric acid. $$NaOH + HCl \rightarrow NaCl + H_2O$$

1. Moles of NaOH = $0.10 \times 25.0/1000 = 0.0025\text{ mol}$
2. From the equation, the ratio is 1:1, so moles of HCl = $0.0025\text{ mol}$
3. Concentration of HCl = $0.0025 / (20.0/1000) = 0.0025 / 0.020 = 0.125\text{ mol/dm}^3$
4. To convert to g/dm\u00b3: $0.125 \times 36.5 = 4.56\text{ g/dm}^3$

The hardest part is remembering to convert cm\u00b3 to dm\u00b3 (divide by 1000) before using the formula.

Worked Example: Mole ratio to volume of gas Calcium oxide reacts with ammonium chloride: $CaO + 2NH_4Cl \rightarrow CaCl_2 + 2NH_3 + H_2O$. If $2.8\text{ g}$ of CaO reacts, what volume of ammonia is produced at r.t.p.?

1. Moles of CaO = $2.8 / 56 = 0.050\text{ mol}$ (Mr of CaO = 40+16 = 56)
2. From the equation: 1 mol CaO produces 2 mol NH\u2083, so moles of NH\u2083 = $0.050 \times 2 = 0.10\text{ mol}$
3. Volume at r.t.p. = $0.10 \times 24{,}000 = 2{,}400\text{ cm}^3$

This pattern \u2014 mass \u2192 moles \u2192 ratio \u2192 volume \u2014 is the core method for all stoichiometry questions involving gases.

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.

Worked Example: Finding the empirical formula A compound contains $2.4\text{ g}$ of carbon and $0.8\text{ g}$ of hydrogen. Find its empirical formula.

1. Moles of C = $2.4 / 12 = 0.20$
2. Moles of H = $0.8 / 1 = 0.80$
3. Divide by the smallest (0.20): C = $0.20/0.20 = 1$, H = $0.80/0.20 = 4$
4. Empirical formula: $CH_4$ (methane)

The trick is step 3: always divide every value by the smallest to get whole numbers. If you get 1:1.5, multiply both by 2 to get 2:3.

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 = $\frac{A_r \times \text{number of atoms}}{M_r \text{ of compound}} \times 100$

Worked Example: Percentage yield A student heats $6.0\text{ g}$ of magnesium in air. The equation is: $2Mg + O_2 \rightarrow 2MgO$. The theoretical yield of MgO is $10.0\text{ g}$, but the student only collects $8.5\text{ g}$. $$\% \text{ Yield} = \frac{8.5}{10.0} \times 100 = 85\%$$ Percentage yield is always less than 100% because some product is lost during transfer, or the reaction may not go to completion.

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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).

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0620 Theory papers.

Exam-Style Question 1 — Short Answer [5 marks]

Question:

Magnesium reacts with hydrochloric acid according to the following equation:

$Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)$

(a) Define the term 'mole'. [2]

(b) Calculate the number of moles of magnesium in 4.86 g of magnesium. The relative atomic mass of magnesium is 24.3. [2]

(c) State the volume, in dm³, occupied by 0.5 moles of hydrogen gas at room temperature and pressure (r.t.p.). [1]

Worked Solution:

(a)

1. A mole is the amount of substance [Definition of 'amount of substance'.]

2. containing the same number of particles (atoms, ions, molecules, etc.) as there are atoms in 12 grams of carbon-12. [Definition of 'Avogadro's number'.]

How to earn full marks:

• Correctly define 'amount of substance' (or equivalent) for 1 mark.
• Mention Avogadro's number or relate it to 12g of carbon-12 for 1 mark.

(b)

1. $n = \frac{m}{M_r}$ Formula relating moles, mass and molar mass.

2. $n = \frac{4.86}{24.3} = 0.2$ Correct substitution and calculation.

How to earn full marks:

• Write the correct formula or show mass/Mr for 1 mark.
• Correct answer with unit for 1 mark.

(c)

1. $0.5 \times 24 = 12$ Multiply moles by molar gas volume.

How to earn full marks:

• Correct answer with unit for 1 mark.

Answer: (a) See above (b) $\boxed{0.2 \text{ mol}}$ (c) $\boxed{12 \text{ dm}^3}$

Common Pitfall: Remember to include units in your final answer. Also, make sure you know the definition of a mole, including the link to Avogadro's constant and carbon-12.

Exam-Style Question 2 — Extended Response [8 marks]

Question:

A student investigates the reaction between zinc and hydrochloric acid:

$Zn(s) + 2HCl(aq) \rightarrow ZnCl_2(aq) + H_2(g)$

The student adds 6.54 g of zinc to 50 cm³ of 2.0 mol/dm³ hydrochloric acid.

(a) Calculate the number of moles of zinc used in the experiment. The relative atomic mass of zinc is 65.4. [2]

(b) Calculate the number of moles of hydrochloric acid used in the experiment. [2]

(c) Determine which reactant, zinc or hydrochloric acid, is the limiting reactant. Explain your reasoning. [3]

(d) Calculate the maximum volume of hydrogen gas, in dm³, that can be produced at r.t.p. [1]

Worked Solution:

(a)

1. $n = \frac{m}{M_r}$ Formula relating moles, mass and molar mass.

2. $n = \frac{6.54}{65.4} = 0.1$ Correct substitution and calculation.

How to earn full marks:

• Write the correct formula or show mass/Mr for 1 mark.
• Correct answer with unit for 1 mark.

(b)

1. $n = c \times v$ Formula relating moles, concentration and volume.

2. $v = 50 \text{ cm}^3 = 0.05 \text{ dm}^3$ Convert cm³ to dm³.

3. $n = 2.0 \times 0.05 = 0.1$ Correct substitution and calculation.

How to earn full marks:

• Write the correct formula or show concentration x volume for 1 mark.
• Correct answer with unit for 1 mark.

(c)

1. From the equation, 1 mole of Zn reacts with 2 moles of HCl. State the stoichiometric ratio.

2. 0.1 moles of Zn would require 0.2 moles of HCl. Calculate the required amount of HCl.

3. There are only 0.1 moles of HCl, so HCl is the limiting reactant. Conclude that HCl is limiting, based on the calculation.

How to earn full marks:

• State the mole ratio between Zn and HCl for 1 mark.
• Calculate the required amount of HCl or Zn for 1 mark.
• Correctly identify the limiting reactant and justify the answer for 1 mark.

(d)

1. From the equation, 2 moles of HCl produce 1 mole of $H_2$. State the stoichiometric relationship between HCl and hydrogen.

2. 0.1 moles of HCl will produce 0.05 moles of $H_2$. Calculate moles of hydrogen.

3. $0.05 \times 24 = 1.2$ Multiply moles of hydrogen by molar gas volume.

How to earn full marks:

• Correct answer with unit for 1 mark.

Answer: (a) $\boxed{0.1 \text{ mol}}$ (b) $\boxed{0.1 \text{ mol}}$ (c) HCl is the limiting reactant because there is not enough HCl to react completely with the Zn. (d) $\boxed{1.2 \text{ dm}^3}$

Common Pitfall: When determining the limiting reactant, make sure you compare the mole ratio from the balanced equation to the actual mole ratio in the reaction mixture. Don't forget to convert cm³ to dm³ when using the formula n = c x v.

Exam-Style Question 3 — Short Answer [6 marks]

Question:

Ethanol ($C_2H_5OH$) is a liquid at room temperature and pressure.

(a) Define the term 'relative molecular mass'. [2]

(b) Calculate the relative molecular mass of ethanol ($C_2H_5OH$). The relative atomic masses are: $A_r$ (C) = 12.0, $A_r$ (H) = 1.0, $A_r$ (O) = 16.0. [2]

(c) A student burns 4.6 g of ethanol in excess oxygen. Calculate the number of molecules of ethanol burned. Avogadro's constant is $6.02 \times 10^{23}$. [2]

Worked Solution:

(a)

1. The relative molecular mass is the sum of the relative atomic masses [Definition of 'relative'.]

2. of all the atoms in a molecule. [Definition of 'molecular'.]

How to earn full marks:

• Mention the sum of atomic masses for 1 mark.
• Mention atoms in a molecule for 1 mark.

(b)

1. $M_r = (2 \times 12.0) + (6 \times 1.0) + 16.0$ Correctly identify and multiply the atomic masses.

2. $M_r = 24.0 + 6.0 + 16.0 = 46.0$ Correct summation.

How to earn full marks:

• Correctly multiply atomic masses by the number of atoms for 1 mark.
• Correct final answer with no units for 1 mark.

(c)

1. $n = \frac{m}{M_r}$ Formula relating moles, mass and molar mass.

2. $n = \frac{4.6}{46.0} = 0.1$ Correct substitution and calculation.

3. Number of molecules = $0.1 \times 6.02 \times 10^{23} = 6.02 \times 10^{22}$ Multiply moles by Avogadro's constant.

How to earn full marks:

• Calculate moles of ethanol correctly for 1 mark.
• Multiply moles by Avogadro's constant for 1 mark.

Answer: (a) See above (b) $\boxed{46.0}$ (c) $\boxed{6.02 \times 10^{22}}$

Common Pitfall: Remember that relative molecular mass ($M_r$) has no units. When calculating the number of molecules, make sure you use Avogadro's constant ($6.02 \times 10^{23}$) and not just the number of moles.

Exam-Style Question 4 — Extended Response [9 marks]

Question:

A fertiliser contains ammonium sulfate, $(NH_4)_2SO_4$. A student wants to determine the percentage purity of the ammonium sulfate in the fertiliser. The student dissolves 5.00 g of the fertiliser in water and reacts it with excess sodium hydroxide solution. The ammonia gas produced is absorbed in 50.0 cm³ of 1.00 mol/dm³ hydrochloric acid. The excess hydrochloric acid is then titrated with 1.00 mol/dm³ sodium hydroxide solution, requiring 25.0 cm³ for complete neutralisation.

The equations for the reactions are:

$(NH_4)_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4(aq) + 2NH_3(g) + 2H_2O(l)$

$NH_3(g) + HCl(aq) \rightarrow NH_4Cl(aq)$

$NaOH(aq) + HCl(aq) \rightarrow NaCl(aq) + H_2O(l)$

(a) Calculate the number of moles of hydrochloric acid originally added. [1]

(b) Calculate the number of moles of sodium hydroxide used in the titration. [1]

(c) Calculate the number of moles of hydrochloric acid that reacted with the ammonia. [2]

(d) Calculate the number of moles of ammonia produced. [1]

(e) Calculate the mass of ammonium sulfate, $(NH_4)_2SO_4$, in the 5.00 g of fertiliser. The relative formula mass of ammonium sulfate is 132. [2]

(f) Calculate the percentage purity of the ammonium sulfate in the fertiliser. [2]

Worked Solution:

(a)

1. $n = c \times v$ Formula relating moles, concentration and volume.

2. $v = 50.0 \text{ cm}^3 = 0.050 \text{ dm}^3$ Convert cm³ to dm³.

3. $n = 1.00 \times 0.050 = 0.050$ Correct substitution and calculation.

How to earn full marks:

• Correct answer with unit for 1 mark.

(b)

1. $n = c \times v$ Formula relating moles, concentration and volume.

2. $v = 25.0 \text{ cm}^3 = 0.025 \text{ dm}^3$ Convert cm³ to dm³.

3. $n = 1.00 \times 0.025 = 0.025$ Correct substitution and calculation.

How to earn full marks:

• Correct answer with unit for 1 mark.

(c)

1. Moles of HCl reacted with $NH_3$ = Initial moles of HCl - Moles of NaOH used in titration. State the relationship.

2. $0.050 - 0.025 = 0.025$ Correct subtraction.

How to earn full marks:

• State that you need to subtract the moles of NaOH from the initial moles of HCl for 1 mark.
• Correctly calculate the moles of HCl reacted with ammonia for 1 mark.

(d)

1. 1 mole of $NH_3$ reacts with 1 mole of $HCl$. State the stoichiometric relationship.

2. Moles of $NH_3$ = Moles of HCl reacted with $NH_3 = 0.025$ Equate moles of ammonia and reacted HCl.

How to earn full marks:

• Correct answer with unit for 1 mark.

(e)

1. From the equation, 2 moles of $NH_3$ are produced from 1 mole of $(NH_4)_2SO_4$. State the stoichiometric ratio.

2. Moles of $(NH_4)_2SO_4 = \frac{0.025}{2} = 0.0125$ Divide the moles of ammonia by 2.

3. Mass of $(NH_4)_2SO_4 = n \times M_r = 0.0125 \times 132 = 1.65$ Multiply moles of ammonium sulfate by its molar mass.

How to earn full marks:

• Divide the moles of ammonia by 2 to find moles of ammonium sulfate for 1 mark.
• Multiply moles of ammonium sulfate by its molar mass for 1 mark.

(f)

1. Percentage purity = $\frac{\text{Mass of }(NH_4)_2SO_4}{\text{Mass of fertiliser}} \times 100$ Formula for percentage purity.

2. Percentage purity = $\frac{1.65}{5.00} \times 100 = 33.0$ Correct substitution and calculation.

How to earn full marks:

• Correct formula or show mass of ammonium sulfate / mass of fertiliser for 1 mark.
• Correct answer with unit for 1 mark.

Answer: (a) $\boxed{0.050 \text{ mol}}$ (b) $\boxed{0.025 \text{ mol}}$ (c) $\boxed{0.025 \text{ mol}}$ (d) $\boxed{0.025 \text{ mol}}$ (e) $\boxed{1.65 \text{ g}}$ (f) $\boxed{33.0 \%}$

Common Pitfall: This multi-step calculation requires careful attention to the stoichiometry of each reaction. Be sure to track which substance is in excess and how the moles of each substance relate to each other according to the balanced equations. Remember to use the correct units and significant figures throughout the calculation.