Section 1 of 4

Section 1: Principles of Chemistry

Principles of chemistry notes covering states of matter, elements, compounds, atomic structure, periodic table, formulae, calculations, bonding, and electrolysis.

Edexcel IGCSE Hub Section 1

Key Definitions

Atom
The smallest part of an element that can exist.
Element
A substance made of only one type of atom.
Compound
A substance made of two or more elements chemically combined in fixed proportions.
Mixture
A combination of two or more substances that are not chemically joined together.
Isotopes
Atoms of the same element with the same number of protons but different numbers of neutrons.
Relative Atomic Mass (Ar)
The weighted average mass of an atom of an element compared with 1/12th of the mass of a carbon-12 atom.
Empirical Formula
The simplest whole-number ratio of atoms of each element in a compound.
Electrolysis
The decomposition of an electrolyte by an electric current.

States of Matter

States of Matter Particle Arrangements SOLID LIQUID GAS
Interactive Particle Simulator Drag particles or adjust temperature to see states of matter behavior:
Temperature: Medium
Solid: Vibrate in place Liquid: Flow & slide Gas: Fast & random
Particle models of the three states of matter. Solids are rigidly packed, liquids are loosely packed and free to flow, and gases are widely spaced and highly energetic.

Solids

Particles packed tightly in a fixed, regular pattern (lattice). Strong forces hold them in place. Particles vibrate on the spot. Fixed shape and volume; cannot be compressed.

Liquids

Particles close together but arranged randomly. Weaker forces than solids. Particles can move and slide past each other. Fixed volume but take the shape of their container.

Gases

Particles very far apart, arranged randomly. Very weak forces. Particles move quickly and randomly. No fixed shape or volume; easily compressed.

Changes of State

When a substance changes state, the particles gain or lose energy. The chemical bonds are not broken - only the forces between particles change.

  • Melting: Solid → Liquid (particles gain energy, overcome some forces, start to move)
  • Boiling/Evaporating: Liquid → Gas (particles gain enough energy to overcome all intermolecular forces)
  • Freezing: Liquid → Solid (particles lose energy, forces pull them into fixed positions)
  • Condensing: Gas → Liquid (particles lose energy, forces bring them closer)
  • Sublimation: Solid → Gas directly (e.g. Dry ice, iodine)
During a change of state, the mass is conserved - no atoms are created or destroyed, they simply rearrange. The temperature stays constant during a change of state as energy is used to break or form intermolecular forces.

State Symbols

In equations, state symbols show the physical state: (s) solid, (l) liquid, (g) gas, (aq) aqueous (dissolved in water).

AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)
Remember: during changes of state, the substance does not become a new substance - it is a physical change. The particles stay the same; only the arrangement and energy change.

Limitations of the Particle Model Extended

The simple particle model is useful for explaining states of matter and changes of state, but it has limitations:

  • It assumes particles are solid, inelastic spheres; in reality, atoms have complex internal structures with electron clouds.
  • It ignores the forces between particles; in reality, intermolecular forces (such as van der Waals forces and hydrogen bonds) play a crucial role in determining properties.
  • It ignores the spaces between particles; in reality, the distance between particles varies significantly and affects properties like compressibility.
  • It does not show the relative sizes of particles in different states: particles do not actually change size when a substance changes state.
In a Higher Tier exam, you may be asked to explain why the particle model is a simplification. Always refer to the assumptions it makes (solid spheres, no forces, no spaces) and explain how real particles differ.

Diffusion & Dilution Experiments

Diffusion is the net movement of particles from an area of higher concentration to an area of lower concentration, down a concentration gradient. It occurs because particles move randomly and collide with each other. This is a passive process that does not require external energy.

Experiment 1: Diffusion of Bromine Gas

A gas jar containing red-brown bromine gas is placed underneath an inverted gas jar containing air, separated by a glass cover. When the cover is removed, the red-brown colour slowly spreads upwards until both jars are a uniform light brown.

Explanation: Bromine and air particles move randomly and collide. Even though bromine is denser than air, the random motion of particles causes them to mix evenly over time.

Experiment 2: Ammonia and Hydrogen Chloride Diffusion

This classic experiment demonstrates that different gases diffuse at different rates:

  • A long glass tube has cotton wool soaked in concentrated ammonia solution at one end (releasing NH3 gas) and cotton wool soaked in concentrated hydrochloric acid at the other end (releasing HCl gas).
  • After a few minutes, a white ring of solid ammonium chloride (NH4Cl) forms inside the tube.
  • Key Observation: The white ring forms closer to the hydrochloric acid end of the tube.
  • Equation: NH3(g) + HCl(g) → NH4Cl(s)
  • Explanation: Ammonia particles have a lower relative molecular mass (Mr = 17) than hydrogen chloride particles (Mr = 36.5). Therefore, ammonia particles travel faster and diffuse more quickly through the air, meeting the slower HCl particles closer to the acid end.

Experiment 3: Dilution of Potassium Manganate(VII)

When a crystal of potassium manganate(VII) is placed in a beaker of water, the purple colour slowly spreads out from the crystal until the entire solution becomes purple. If more water is added (dilution), the purple colour becomes lighter.

Explanation: Water particles and potassium manganate(VII) particles move randomly, causing the dissolved particles to diffuse throughout the liquid. Dilution spreads the particles further apart, decreasing their concentration.

Solubility & Solubility Curves Extended

In chemistry, we use specific terms to describe how substances dissolve to form mixtures:

  • Solute: The substance that dissolves in a liquid (e.g., salt, sugar).
  • Solvent: The liquid in which the solute dissolves (e.g., water, ethanol).
  • Solution: The mixture formed when a solute has dissolved in a solvent.
  • Saturated Solution: A solution that contains the maximum amount of dissolved solute at a specific temperature. Any further solute added will not dissolve.

Definition of Solubility

Solubility is defined as the maximum mass of solute (in grams) that can dissolve in 100 g of solvent at a specific temperature.

For example, if the solubility of sodium chloride in water is 36 g/100 g H₂O at 20°C, it means that a maximum of 36 grams of NaCl will dissolve in 100 grams of water at 20°C.

Core Practical 1.7C: Investigate the Solubility of a Solid in Water

This experiment explains how to determine the solubility of a salt (e.g., ammonium chloride, NH₄Cl) at a specific temperature.

Procedure:
  1. Set up a water bath at a specific temperature (e.g., 40°C) and place a boiling tube containing water and excess ammonium chloride into the water bath. Stir the mixture until no more salt dissolves to ensure the solution is saturated.
  2. Weigh an empty evaporating basin and record its mass.
  3. Carefully decant (pour off) some of the clear saturated solution into the evaporating basin, ensuring that no undissolved solid is transferred.
  4. Weigh the evaporating basin containing the solution and record the mass.
  5. Heat the evaporating basin gently using a Bunsen burner to evaporate the water.
  6. Once the water has mostly evaporated, heat the basin gently to avoid spitting (where the solid jumps out of the basin).
  7. Allow the basin to cool, then weigh it. Re-heat, cool, and re-weigh the basin until a constant mass is obtained. This ensures that all the water has been completely evaporated.
Calculations:

From the recorded masses, we can determine the solubility:

  • Mass of saturated solution = (Mass of basin + solution) - (Mass of empty basin)
  • Mass of dry residue (solute) = (Mass of basin + dry residue) - (Mass of empty basin)
  • Mass of water evaporated (solvent) = (Mass of saturated solution) - (Mass of dry residue)
Solubility (g/100 g water) = (Mass of dry residue ÷ Mass of water evaporated) × 100
Analysis of Errors:
  • Spitting of solid: If the solution is heated too strongly, the solid can "spit" out of the basin. This decreases the measured mass of the dry residue, leading to an underestimation of the solubility.
  • Incomplete evaporation: If the solid is not heated to a constant mass, water remains in the residue. This increases the measured mass of the residue, leading to an overestimation of the solubility.

Solubility Curves

A solubility curve shows how the solubility of different solutes changes as the temperature of the solvent changes. For most solids, solubility increases with temperature because the water molecules have more kinetic energy to break solute bonds.

Solubility Curves Solubility (g per 100g water) Temperature (°C) 0 30 60 90 120 150 0 20 40 60 80 100 KNO₃ NaCl Ce₂(SO₄)₃

Solubility curves for potassium nitrate (solubility increases steeply), sodium chloride (solubility increases very slightly), and cerium(III) sulfate (solubility decreases as temperature rises).

Cooling & Precipitation Calculations

If a saturated solution is cooled, the solubility decreases, and some of the dissolved solute precipitates (crystallises) out of the solution. To calculate the mass of crystals formed, follow this method:

  1. Find the solubility of the salt at the higher temperature and calculate the mass of solute dissolved in the given mass of water.
  2. Find the solubility of the salt at the lower temperature and calculate the mass of solute that remains dissolved in the same mass of water.
  3. Subtract the mass of solute remaining dissolved from the mass originally dissolved.

Worked Example

A saturated solution of potassium nitrate (KNO₃) in 50 g of water is cooled from 60°C to 20°C. Using the solubility curve data below, calculate the mass of crystals that will precipitate out of the solution:

  • Solubility of KNO₃ at 60°C = 110 g / 100 g H₂O
  • Solubility of KNO₃ at 20°C = 32 g / 100 g H₂O

Step 1: Calculate mass dissolved at 60°C:

In 100 g water, 110 g KNO₃ dissolves.

In 50 g water: 110 × (50 ÷ 100) = 55 g KNO₃ dissolved.

Step 2: Calculate mass dissolved at 20°C:

In 100 g water, 32 g KNO₃ dissolves.

In 50 g water: 32 × (50 ÷ 100) = 16 g KNO₃ remains dissolved.

Step 3: Subtract to find the mass of crystals formed:

Mass of crystals = 55 g - 16 g = 39 g

The mass of potassium nitrate crystals precipitated is 39 g.

Elements, Compounds & Mixtures

What is an Atom?

An atom is the smallest part of an element that can exist. Think of it as the ultimate LEGO brick. Everything in the universe - the air you breathe, the water you drink, and even you - is made of atoms.

Atoms are incredibly empty! If an atom were the size of a football stadium, its nucleus (the central part) would only be the size of a marble.

Elements

An element is a substance made of only one type of atom. There are about 118 known elements, and each has a unique atomic number (the number of protons in its nucleus). All known elements are catalogued in the periodic table. Examples include Hydrogen (H), Oxygen (O), Carbon (C), and Iron (Fe).

Compounds

A compound is a substance formed when two or more elements are chemically bonded together. The atoms in a compound are held together by chemical bonds (ionic or covalent), and the compound has different properties from the individual elements. A compound can only be separated into its elements by chemical reactions, not by physical methods.

H₂ + Cl₂ → 2HCl

In this example, hydrogen and chlorine (both elements) react to form hydrogen chloride (a compound).

When describing the difference between a mixture and a compound, always state that the elements in a compound are chemically bonded, while in mixtures they are not. A compound has properties different from its constituent elements.

Chemical Equations

Chemical reactions can be represented using word equations and balanced symbol equations. A balanced equation has the same number of each type of atom on both sides - this reflects the law of conservation of mass.

Writing Word Equations

A word equation names the reactants (what goes in) and the products (what comes out), separated by an arrow.

magnesium + oxygen → magnesium oxide

Balancing Symbol Equations

In a balanced symbol equation, the number of atoms of each element must be the same on both sides. You balance by placing numbers in front of formulae - never change the small (subscript) numbers.

Balance: Mg + O₂ → MgO

Step 1: Count atoms: Left has 1 Mg and 2 O. Right has 1 Mg and 1 O - not balanced.

Step 2: Put a 2 in front of MgO: Mg + O₂ → 2MgO. Now right has 2 Mg and 2 O.

Step 3: Put a 2 in front of Mg: 2Mg + O₂ → 2MgO. Both sides: 2 Mg and 2 O ✓

Balanced equation: 2Mg + O₂ → 2MgO

Never change the small (subscript) numbers in a formula to balance an equation - only change the large numbers in front. Changing subscripts would change the substance itself!

Mixtures

A mixture consists of two or more substances that are not chemically bonded together. The substances retain their individual properties and can be separated by physical methods.

Key Properties of Mixtures

  • The components are not chemically combined.
  • They retain their original properties.
  • They can be separated by physical processes (filtration, distillation, etc.).
  • No chemical reaction is needed to separate them.

Examples include air (a mixture of gases), salt water, and alloys like steel.

Particle models comparing elements, compounds, and mixtures. Element Only one type of atom Compound Different atoms chemically bonded Mixture Multiple substances not chemically bonded
Particle models comparing elements, compounds, and mixtures.

Separation Techniques

Rf Values

Each substance in a chromatogram can be identified by its Rf value - a ratio that is unique to each substance under specific conditions (same solvent, same temperature).

Rf = distance moved by substance ÷ distance moved by solvent

Calculating an Rf value

Given: A spot has moved 4.2 cm and the solvent front has moved 6.0 cm.

Rf = 4.2 ÷ 6.0 = 0.70

This value can then be compared to known Rf values to identify the substance.

Testing for Purity

Chromatography can determine whether a substance is pure or a mixture:

  • A pure substance produces a single spot on the chromatogram.
  • A mixture produces two or more spots.
You should be able to describe when each technique should be used. For example, filtration for an insoluble solid from a liquid; distillation for a solvent from a solution; chromatography for dissolved substances like dyes. Remember: Rf values always fall between 0 and 1.

Atomic Structure

Subatomic Particles

The anatomy of an atom showing protons, neutrons, and electrons. ELECTRON Charge: -1 Mass: ~0 PROTON Charge: +1 Mass: 1 NEUTRON Charge: 0 Mass: 1 NUCLEUS = Protons + Neutrons
The anatomy of an atom showing protons, neutrons, and electrons.

Atoms are made up of three types of subatomic particles, each with specific properties:

  • Proton: Found in the nucleus. Relative mass = 1, Relative charge = +1.
  • Neutron: Found in the nucleus. Relative mass = 1, Relative charge = 0.
  • Electron: Found in shells orbiting the nucleus. Relative mass ≈ 1/2000 (negligible), Relative charge = −1.
The number of protons defines which element an atom is. This is called the atomic number (Z). The mass number (A) is the total number of protons + neutrons. In a neutral atom, the number of electrons equals the number of protons.
To find the number of neutrons: Neutrons = Mass Number − Atomic Number. Remember that in a neutral atom, protons = electrons.

How many neutrons does an atom of aluminium-27 have?

Step 1: Aluminium has atomic number 13 (from periodic table), so it has 13 protons.

Step 2: Neutrons = Mass number − Atomic number = 27 − 13 = 14 neutrons

Atomic Size & Mass Distribution

The Scale of an Atom

Atoms are unimaginably small, and the nucleus is even smaller in comparison.

  • The atomic radius is approximately 1 × 10⁻¹⁰ metres.
  • The radius of the nucleus is about 1 × 10⁻¹⁴ metres.

This means the nucleus is roughly 10,000 times smaller than the entire atom. If an atom were the size of a football stadium, the nucleus would be the size of a pea placed in the centre.

Distribution of Mass

Despite its tiny size, the nucleus contains nearly all of the atom's mass. This is because protons and neutrons (found in the nucleus) each have a relative mass of 1. Electrons have a negligible mass (approximately 1/2000th of a proton or neutron). Therefore, when calculating the mass of an atom, the mass of the electrons is considered to be zero.

Be prepared to use standard form to compare the sizes of the atom and its nucleus. For example, an atom (radius 1 × 10⁻¹⁰ m) is 10,000 (or 10⁴) times larger than its nucleus (radius 1 × 10⁻¹⁴ m).

Isotopes & Relative Atomic Mass

The three isotopes of Carbon. The proton number stays the same, only the neutrons change. Carbon-12 6 Protons, 6 Neutrons 98.9% Abundant Carbon-13 6 Protons, 7 Neutrons 1.1% Abundant Carbon-14 6 Protons, 8 Neutrons < 0.1% (Radioactive)
The three isotopes of Carbon. The proton number stays the same, only the neutrons change.

Isotopes are atoms of the same element that have the same number of protons but a different number of neutrons. Because isotopes have the same number of protons, they also have the same number of electrons - so they have identical chemical properties.

Understanding Isotopes

The key difference between isotopes is their mass number. Let's look at carbon as an example - all carbon atoms have 6 protons:

  • Carbon-12: 6 protons + 6 neutrons (Mass number = 12)
  • Carbon-13: 6 protons + 7 neutrons (Mass number = 13)
  • Carbon-14: 6 protons + 8 neutrons (Mass number = 14)

Relative Atomic Mass (Ar)

The relative atomic mass is the weighted mean mass of an atom of an element, taking into account its naturally occurring isotopes. This is why some elements on the periodic table have mass numbers that are not whole numbers.

Calculating the Ar of Chlorine

In nature, chlorine exists as two isotopes: 75% is Chlorine-35 and 25% is Chlorine-37.

Step 1: Multiply abundance by mass:
(75 × 35) = 2625
(25 × 37) = 925

Step 2: Add: 2625 + 925 = 3550

Step 3: Divide by 100: 3550 ÷ 100 = 35.5

Electron Shells & Configuration

The way electrons are arranged in an atom is fundamental to chemistry. It dictates how an element behaves, why it reacts, and where it sits in the periodic table.

Rules for Filling Shells

  • Electrons always occupy the lowest available energy level first (the innermost shell).
  • Shell 1: holds a maximum of 2 electrons.
  • Shell 2: holds a maximum of 8 electrons.
  • Shell 3: holds a maximum of 8 electrons.

How to Work Out Electron Configurations

Sodium (Na) - Atomic number 11

First shell fills with 2 electrons.

Second shell fills with 8 electrons. (2 + 8 = 10 used).

Remaining 1 electron goes into the third shell.

Electron configuration: 2.8.1

Chlorine (Cl) - Atomic number 17

First shell: 2 electrons.
Second shell: 8 electrons.
Third shell: 7 electrons.

Electron configuration: 2.8.7

Calcium (Ca) - Atomic number 20

First shell: 2 electrons.
Second shell: 8 electrons.
Third shell: 8 electrons.
Fourth shell: 2 electrons.

Electron configuration: 2.8.8.2

Why Electron Arrangement Matters

The electrons in the outermost shell are called valence electrons. The number of valence electrons determines the chemical properties of an element. Elements in the same group have the same number of valence electrons - which is why they have similar chemical properties.

An atom is most stable when it has a full outer shell. The noble gases (Group 0) have full outer shells, which is why they are so unreactive. Other atoms react by gaining, losing, or sharing electrons to achieve a stable, full outer shell - this is the basis for all chemical bonding.

Build your own atoms with protons, neutrons, and electrons using our Atom Builder tool.

The Periodic Table

The periodic table is a masterfully organised chart of all the known chemical elements. It's arranged to reveal patterns in the properties of elements - a concept known as periodicity.

1H
2He
3Li
4Be
5B
6C
7N
8O
9F
10Ne
11Na
12Mg
13Al
14Si
15P
16S
17Cl
18Ar
19K
20Ca
21Sc
22Ti
23V
24Cr
25Mn
26Fe
27Co
28Ni
29Cu
30Zn
31Ga
32Ge
33As
34Se
35Br
36Kr
37Rb
38Sr
39Y
40Zr
41Nb
42Mo
43Tc
44Ru
45Rh
46Pd
47Ag
48Cd
49In
50Sn
51Sb
52Te
53I
54Xe
55Cs
56Ba
*57-71
72Hf
73Ta
74W
75Re
76Os
77Ir
78Pt
79Au
80Hg
81Tl
82Pb
83Bi
84Po
85At
86Rn
87Fr
88Ra
**89-103
104Rf
105Db
106Sg
107Bh
108Hs
109Mt
110Ds
111Rg
112Cn
113Nh
114Fl
115Mc
116Lv
117Ts
118Og
57La
58Ce
59Pr
60Nd
61Pm
62Sm
63Eu
64Gd
65Tb
66Dy
67Ho
68Er
69Tm
70Yb
71Lu
89Ac
90Th
91Pa
92U
93Np
94Pu
95Am
96Cm
97Bk
98Cf
99Es
100Fm
101Md
102No
103Lr
Alkali Metals Alkaline Earth Transition Metals Post-Transition Metalloids Non-Metals Halogens Noble Gases Lanthanides Actinides
The modern periodic table of elements, colour-coded by element group.

How the Table is Arranged

The modern periodic table arranges elements in order of increasing atomic number (Z). The atomic number is the number of protons - the element's unique ID number.

A common mistake is to say the table is ordered by relative atomic mass. Always state that the modern periodic table is ordered by atomic number.

Periods (Horizontal Rows)

The rows across the periodic table are called periods. The period number tells you how many occupied electron shells an element has.

  • Period 1 elements (H, He) have 1 electron shell.
  • Period 2 elements (Li to Ne) have 2 electron shells.
  • Period 3 elements (Na to Ar) have 3 electron shells.

Groups (Vertical Columns)

The columns down the periodic table are called groups. For the main groups, the group number tells you the number of electrons in the element's outermost shell (valence electrons).

  • Group 1 (Alkali Metals): 1 outer electron - very reactive.
  • Group 7 (Halogens): 7 outer electrons - very reactive (need one more for a full shell).
  • Group 0 (Noble Gases): Full outer shell - stable and very unreactive.
🔬

Explore every element and its properties on our Interactive Periodic Table.

Metals & Non-Metals

The periodic table has a fundamental dividing line that separates elements into metals (left and centre) and non-metals (right).

Physical Properties

Property Metals Non-Metals
Appearance Shiny (lustrous) Dull
State at room temp Mostly solid (except mercury) Solid, liquid or gas
Conductivity Good conductors of heat & electricity Poor conductors (insulators)
Malleability Malleable & ductile Brittle (if solid)
Melting/Boiling points Generally high Generally low
Density Generally high Generally low
Comparing typical properties of metals and non-metals.

How Metals and Non-Metals Form Ions

When metals react, they lose their outer electrons to form positive ions (cations). For example, sodium (2.8.1) loses 1 electron to become Na⁺ (2.8).

When non-metals react, they gain electrons to form negative ions (anions). For example, chlorine (2.8.7) gains 1 electron to become Cl⁻ (2.8.8).

Reactions of Oxides

Metal Oxides are Basic

Metal oxides react with acids in neutralisation reactions to produce a salt and water.

MgO + 2HCl → MgCl₂ + H₂O

Non-Metal Oxides are Acidic

Non-metal oxides typically react with water to form acidic solutions.

SO₂ + H₂O → H₂SO₃
When asked to classify an element, first state its position on the periodic table. Then support this by citing two typical physical or chemical properties. Don't confuse groups (vertical columns) and periods (horizontal rows)!

Group 0: The Noble Gases

The elements in Group 0 (Helium, Neon, Argon, etc.) are known as the noble gases. They are chemically inert - extremely unreactive.

Why are Noble Gases So Unreactive?

They all have a full outer shell of electrons:

  • Helium: 2
  • Neon: 2.8
  • Argon: 2.8.8

A full outer shell is the most stable arrangement. They have no tendency to lose, gain, or share electrons, so they exist as individual monatomic atoms.

Trends in Group 0

Boiling points increase as you go down the group. As atoms become larger with more electrons, the weak intermolecular forces of attraction between them become stronger. More energy is needed to overcome these forces.

Uses of Noble Gases

Their inertness makes noble gases useful where unreactive atmospheres are needed:

  • Helium: Balloons and airships - low density and non-flammable (safer than hydrogen).
  • Neon: Advertising signs - glows bright red-orange when electricity passes through it.
  • Argon: Filament lamps and welding - provides an inert atmosphere that prevents the hot metal from reacting with oxygen.
Whenever you're asked why noble gases are unreactive, you must mention that they have a stable electron arrangement because they have a full outer shell.

Formulae, Equations & Calculations

Conservation of Mass

The law of conservation of mass states that no atoms are lost or made during a chemical reaction. The total mass of the products equals the total mass of reactants.

This is because the same atoms are present before and after the reaction: they have just been rearranged. This is why we must balance chemical equations.

Conservation of Mass Balance Scale C O₂ Reactants O C O Products =

The same atoms exist before and after: mass is always conserved

Atoms are neither created nor destroyed in a chemical reaction: only rearranged into different substances.

Apparent Changes in Mass

Sometimes, the mass of a reaction vessel appears to change:

  • If a gas escapes (e.g., CO₂ from a thermal decomposition), mass appears to decrease.
  • If a gas is gained from the air (e.g., oxygen during oxidation of metals), mass appears to increase.

In reality, if the system were sealed, the total mass would remain unchanged.

Chemical Measurements & Uncertainty

Whenever a measurement is made, there is always some uncertainty about the result obtained. No measurement is perfectly accurate; all instruments have limitations and all humans introduce small errors when reading scales.

Improving the Quality of Measurements

To increase confidence in results, scientists:

  • Repeat measurements and calculate a mean (average).
  • Identify and exclude anomalous results (outliers): values that don't fit the pattern. These are usually caused by human error or equipment malfunction.
  • Use the range of the measurements about the mean as a measure of uncertainty.
Uncertainty = range ÷ 2. The range is the difference between the highest and lowest values in a set of repeat measurements (after removing anomalies).

A student records three titre values: 24.50, 24.60, and 24.40 cm³. Calculate the uncertainty.

Step 1: Mean = (24.50 + 24.60 + 24.40) ÷ 3 = 24.50 cm³

Step 2: Range = 24.60 – 24.40 = 0.20 cm³

Step 3: Uncertainty = 0.20 ÷ 2 = ± 0.10 cm³

Result: 24.50 ± 0.10 cm³

If a question asks you to "estimate the uncertainty," calculate half the range. A smaller uncertainty means the results are more precise (close together). Don't confuse precision with accuracy: accurate means close to the true value.

Relative Formula Mass (Mr)

The relative formula mass of a compound is the sum of the relative atomic masses of all the atoms in its formula. Ar values are found on the periodic table.

Calculate the Mr of calcium carbonate (CaCO₃)

Ca = 40, C = 12, O = 16

Mr = 40 + 12 + (3 × 16) = 40 + 12 + 48 = 100

Calculate the Mr of magnesium hydroxide (Mg(OH)₂)

Mg = 24, O = 16, H = 1

Mr = 24 + 2 × (16 + 1) = 24 + 34 = 58

Watch out for brackets! In Mg(OH)₂, the subscript 2 applies to both the O and the H inside the brackets, so there are 2 oxygen atoms and 2 hydrogen atoms.

Calculate the Mr of aluminium sulfate, Al₂(SO₄)₃

Al = 27, S = 32, O = 16

There are: 2 Al, 3 S, 12 O (3 × 4 = 12)

Mr = (2 × 27) + (3 × 32) + (12 × 16) = 54 + 96 + 192 = 342

For formulae like Al₂(SO₄)₃, multiply everything inside the brackets by the subscript outside. Here the 3 applies to one S and four O atoms, giving 3 S and 12 O.

Moles & Avogadro's Constant

A mole is simply a number: 6.02 × 10²³ (Avogadro's constant). One mole of any substance contains exactly this number of particles (atoms, molecules, or ions).

moles = mass ÷ Mr

The Moles Triangle

Moles Formula Triangle mass mol × Mr
mass = mol × Mr
mol = mass ÷ Mr
Mr = mass ÷ mol

How many moles in 11 g of CO₂?

Mr of CO₂ = 12 + (2 × 16) = 44

Moles = 11 ÷ 44 = 0.25 mol

What mass is 3 moles of water (H₂O)?

Mr of H₂O = (2 × 1) + 16 = 18

Mass = mol × Mr = 3 × 18 = 54 g

This triangle is your best friend for calculations. Cover what you want to find, and the remaining two values show you what to do. If they are side by side, multiply. If one is on top, divide.
🧮

Check your mole calculations instantly with our Moles Calculator - converts between mass, moles, and molar mass.

Reacting Masses

Use a balanced equation plus the moles formula to predict masses used or produced in reactions. Follow these three steps every time:

Reacting Masses Calculation Process Step 1 Find moles of known Step 2 Use mole ratio Step 3 Convert to mass/vol

What mass of magnesium oxide is produced from 6 g of magnesium?

Equation: 2Mg + O₂ → 2MgO

Step 1: Moles of Mg = 6 ÷ 24 = 0.25 mol

Step 2: Ratio: 2Mg : 2MgO = 1:1, so moles of MgO = 0.25 mol

Step 3: Mass of MgO = 0.25 × 40 = 10 g

What mass of iron is produced when 80 g of iron(III) oxide is reduced?

Equation: Fe₂O₃ + 3CO → 2Fe + 3CO₂

Step 1: Mr of Fe₂O₃ = (2 × 56) + (3 × 16) = 160.
Moles = 80 ÷ 160 = 0.5 mol

Step 2: Ratio: 1 Fe₂O₃ : 2 Fe, so moles of Fe = 0.5 × 2 = 1.0 mol

Step 3: Mass of Fe = 1.0 × 56 = 56 g

Always check the mole ratio carefully - it is NOT always 1:1! Read the balanced equation coefficients to find the correct ratio.

Limiting Reactants

In many reactions, one reactant is used up before the others. This reactant is called the limiting reactant: it limits the amount of product that can be formed.

The other reactant(s) are said to be in excess.

Limiting Reactants Represented as Particles

Limiting

Used up first

Excess Reactants Represented as Particles

Excess

Left over

The amount of product formed is always determined by the limiting reactant, not the one in excess.

4.8 g of Mg reacts with 14.6 g of HCl. Which is the limiting reactant?

Equation: Mg + 2HCl → MgCl₂ + H₂

Step 1: Moles of Mg = 4.8 ÷ 24 = 0.2 mol

Step 2: Moles of HCl = 14.6 ÷ 36.5 = 0.4 mol

Step 3: The ratio is 1 Mg : 2 HCl. So 0.2 mol Mg needs 0.2 × 2 = 0.4 mol HCl.

Result: We have exactly 0.4 mol HCl - both reactants run out at the same time. Neither is in excess.

If we only had 0.3 mol HCl, then HCl would be limiting (not enough to react with all the Mg).

To identify the limiting reactant: (1) calculate moles of each reactant, (2) use the balanced equation ratios to compare, (3) the one that runs out first is the limiting reactant.

Concentration

Concentration tells you how much solute is dissolved in a given volume of solution.

Comparison of dilute and concentrated solutions Dilute Low concentration (Few solute particles) Concentrated High concentration (Many solute particles)
Dilute vs concentrated solutions at the same total volume.
concentration (g/dm³) = mass of solute (g) ÷ volume (dm³)

The Concentration Triangle

Concentration Formula Triangle mass conc × vol
mass = conc × vol
conc = mass ÷ vol
vol = mass ÷ conc
Remember: 1 dm³ = 1000 cm³. To convert cm³ to dm³, divide by 1000.

2.5 g of NaOH dissolved in 500 cm³. Find concentration in g/dm³.

Volume = 500 ÷ 1000 = 0.5 dm³

Concentration = 2.5 ÷ 0.5 = 5 g/dm³

Concentration in mol/dm³

concentration (mol/dm³) = moles of solute ÷ volume (dm³)

4 g of NaOH is dissolved in 250 cm³ of water. Find the concentration in mol/dm³.

Step 1: Mr of NaOH = 23 + 16 + 1 = 40

Step 2: Moles = 4 ÷ 40 = 0.1 mol

Step 3: Volume = 250 ÷ 1000 = 0.25 dm³

Step 4: Concentration = 0.1 ÷ 0.25 = 0.4 mol/dm³

If a question gives you mass in grams and asks for mol/dm³, you must first convert mass to moles using mol = mass ÷ Mr, then divide by the volume in dm³. Always convert cm³ to dm³ first!
🧪

Need to work through a titration calculation? Use our Titration Calculator for step-by-step working with automatic unit conversion.

Percentage Yield Extended

The percentage yield compares the actual amount of product obtained to the theoretical maximum (calculated from stoichiometry).

% yield = (actual yield ÷ theoretical yield) × 100

Three main reasons why yields are always less than 100%:

  • The reaction is reversible and does not go to completion.
  • Some product is lost during transfer (e.g., filtration, evaporation).
  • Side reactions produce unwanted by-products.

A student calculates they should make 10 g of copper sulfate. They actually collect 7.5 g. What is the percentage yield?

% yield = (7.5 ÷ 10) × 100 = 75%

Atom Economy Extended

Atom economy measures the proportion of reactant atoms that become useful product. It evaluates the efficiency of the reaction pathway itself.

atom economy = (Mr of desired product ÷ total Mr of all products) × 100
Addition vs Substitution Reactions Addition Reaction 100% All atoms form ONE product + Desired Substitution Reaction <100% Waste by-products form + Desired + Waste
High atom economy means fewer waste products and maximum efficiency.

High atom economy is desirable. It means less waste, lower costs, and a more sustainable process.

Yield vs Atom Economy

Feature % Yield Atom Economy
Measures Lab performance Pathway efficiency
Needs experiment? Yes No (equation only)
Type of waste Spills, losses By-products
Ideal value 100% 100%

Calculate the atom economy for producing hydrogen from the reaction:

Zn + H₂SO₄ → ZnSO₄ + H₂

Step 1: Mr of desired product (H₂) = 2

Step 2: Mr of all products = ZnSO₄ (161) + H₂ (2) = 163

Step 3: Atom economy = (2 ÷ 163) × 100 = 1.2%

This is very low - most of the atoms end up in the by-product (ZnSO₄), not in the desired product (H₂).

You need both high percentage yield and high atom economy for a truly efficient, sustainable reaction. AQA loves comparing these two metrics.

Titrations Extended

A titration is a technique used to find the concentration of an unknown acid or alkali by reacting it with one of known concentration.

Titration Apparatus Burette Contains acid of known concentration Tap / Stopcock Controls flow Conical Flask Contains alkali + indicator solution Clamp Stand
The burette delivers a measured volume of acid into the alkali below.

The Method

  1. Use a pipette to measure a fixed volume of alkali into a conical flask.
  2. Add a few drops of indicator (e.g., phenolphthalein or methyl orange).
  3. Fill a burette with acid of known concentration.
  4. Add acid gradually, swirling, until the indicator permanently changes colour: the end point.
  5. Record the titre (volume of acid added). Repeat until concordant results are achieved (within 0.10 cm³).
When describing the titration method, always mention: using a pipette for the fixed volume, a burette for the variable volume, and an indicator for the end point. Repeat to get concordant results.

Titration Calculations

25 cm³ of NaOH (unknown concentration) is neutralised by 20 cm³ of 0.5 mol/dm³ HCl. Find the concentration of NaOH.

Equation: NaOH + HCl → NaCl + H₂O

Step 1: Moles of HCl = conc × vol = 0.5 × (20 ÷ 1000)
= 0.5 × 0.02 = 0.01 mol

Step 2: Ratio NaOH : HCl = 1:1, so moles of NaOH = 0.01 mol

Step 3: Volume of NaOH = 25 ÷ 1000 = 0.025 dm³

Step 4: Concentration of NaOH = 0.01 ÷ 0.025 = 0.4 mol/dm³

Titration calculations always follow the same pattern: find moles of the known substance → use the mole ratio → calculate the unknown concentration. Be careful to convert cm³ to dm³!

Molar Volume of Gases Extended

At room temperature and pressure (RTP: 20°C, 1 atm), one mole of any gas occupies a volume of 24 dm³ (or 24,000 cm³).

volume (dm³) = moles × 24
moles = volume (dm³) ÷ 24

Gas Volume Triangle

Gas Volume Formula Triangle vol mol × 24
vol = mol × 24
mol = vol ÷ 24
(at RTP)

What volume does 0.5 mol of oxygen gas occupy at RTP?

Volume = 0.5 × 24 = 12 dm³

What volume of CO₂ is produced when 5 g of CaCO₃ is decomposed? (at RTP)

Equation: CaCO₃ → CaO + CO₂

Step 1: Mr of CaCO₃ = 40 + 12 + (3 × 16) = 100

Step 2: Moles of CaCO₃ = 5 ÷ 100 = 0.05 mol

Step 3: Ratio 1:1, so moles of CO₂ = 0.05 mol

Step 4: Volume = 0.05 × 24 = 1.2 dm³ (or 1200 cm³)

The molar volume (24 dm³) only applies at RTP. If a question states different conditions, you may need the ideal gas equation instead (not required at GCSE). Questions often combine mass→moles→volume in multi-step problems.
🧮

Explore gas behaviour with our Gas Law Calculator - see how pressure, volume, and temperature are connected.

Empirical & Molecular Formulae

Empirical Formula: The simplest whole-number ratio of the atoms of each element present in a compound.

Molecular Formula: The actual number of atoms of each element present in one molecule of a compound.

Calculating Empirical Formula from Mass or Percentage Composition

To find the empirical formula, follow these steps:

  1. Write down the mass or percentage of each element.
  2. Divide each mass/percentage by the relative atomic mass (Ar) of that element to find the number of moles.
  3. Divide each number of moles by the smallest mole value obtained in step 2.
  4. If necessary, multiply all numbers to get whole numbers (e.g., if you get 1.5, multiply all by 2).

Worked Example

A compound contains 82.7% carbon and 17.3% hydrogen by mass. Calculate its empirical formula.

Element Carbon (C) Hydrogen (H)
Mass / % 82.7 g 17.3 g
Ar 12 1
Moles (Mass ÷ Ar) 82.7 ÷ 12 = 6.89 mol 17.3 ÷ 1 = 17.3 mol
Divide by smallest (6.89) 6.89 ÷ 6.89 = 1.0 17.3 ÷ 6.89 = 2.51
Multiply by 2 for integers 1.0 × 2 = 2 2.5 × 2 = 5

The empirical formula is C2H5.

Calculating Molecular Formula from Empirical Formula and Mr

  1. Calculate the relative formula mass (Mr) of the empirical formula.
  2. Divide the given actual Mr of the compound by the empirical formula mass.
  3. Multiply the empirical formula subscripts by this factor.

Worked Example

The empirical formula of a compound is C2H5 and its relative molecular mass (Mr) is 58. Find its molecular formula.

Step 1: Empirical formula mass of C2H5 = (2 × 12) + (5 × 1) = 29

Step 2: Factor = Actual Mr ÷ Empirical mass = 58 ÷ 29 = 2

Step 3: Molecular formula = (C2H5) × 2 = C4H10

Molar Gas Volume at RTP Extended

Avogadro's Law states that equal volumes of gases at the same temperature and pressure contain the same number of molecules.

This means that one mole of any gas (whether carbon dioxide, oxygen, hydrogen, or any other gas) always occupies the same volume under the same conditions:

  • At Room Temperature and Pressure (RTP) (defined as 20°C / 293 K and 1 atmosphere pressure), this volume is 24 dm³ (or 24,000 cm³).
  • This volume is called the molar volume of a gas.

Formulae for Gas Volumes

To convert between gas volume and moles, use the following relationships:

Volume (dm³) = Moles × 24
Volume (cm³) = Moles × 24,000
Moles = Volume (dm³) ÷ 24
Moles = Volume (cm³) ÷ 24,000

Worked Examples

Example 1: Volume from Moles

Calculate the volume, in dm³, occupied by 0.25 moles of carbon dioxide gas at RTP.

Working:

Volume = Moles × 24

Volume = 0.25 × 24 = 6.0 dm³

Answer: 6.0 dm³

Example 2: Volume from Mass

Calculate the volume, in cm³, occupied by 8.8 g of propane gas (C₃H₈) at RTP. (A_r: C = 12, H = 1)

Working:

  1. Find M_r of C₃H₈: (3 × 12) + (8 × 1) = 44
  2. Calculate moles: Moles = Mass ÷ M_r = 8.8 ÷ 44 = 0.20 mol
  3. Calculate volume in cm³: Volume = Moles × 24,000 = 0.20 × 24,000 = 4,800 cm³

Answer: 4,800 cm³

Example 3: Stoichiometry & Reacting Gas Volumes

On heating, calcium carbonate decomposes to form calcium oxide and carbon dioxide gas:
CaCO₃(s) → CaO(s) + CO₂(g)
Calculate the volume of carbon dioxide gas, in dm³, produced at RTP when 15 g of calcium carbonate is fully decomposed. (M_r of CaCO₃ = 100)

Working:

  1. Calculate moles of CaCO₃: Moles = Mass ÷ M_r = 15 ÷ 100 = 0.15 mol
  2. Determine mole ratio from balanced equation: 1 mole of CaCO₃ decomposes to produce 1 mole of CO₂. Therefore, Moles of CO₂ = 0.15 mol.
  3. Calculate volume of CO₂: Volume = Moles × 24 = 0.15 × 24 = 3.6 dm³

Answer: 3.6 dm³

Ionic Bonding

Ionic bonding happens between a metal and a non-metal. It involves the transfer of electrons from the metal atom to the non-metal atom.

Animated Ionic Bonding: NaCl Formation ANIMATION: FORMATION OF NaCl Na + Sodium Atom (2.8.1) Sodium Ion [2.8]⁺ Cl - Chlorine Atom (2.8.7) Chloride Ion [2.8.8]⁻

How Ionic Bonds Form

  1. Electron Transfer: The metal atom loses outer electrons → positive ion (cation). The non-metal atom gains electrons → negative ion (anion).
  2. Electrostatic Attraction: The oppositely charged ions are strongly attracted to each other. This powerful force is the ionic bond.

Formation of Sodium Chloride (NaCl)

Sodium (2.8.1) loses 1 electron → Na⁺ (2.8)

Chlorine (2.8.7) gains 1 electron → Cl⁻ (2.8.8)

The Na⁺ and Cl⁻ ions are held together by strong electrostatic attraction.

Formation of Magnesium Oxide (MgO)

Magnesium (2.8.2) loses 2 electrons → Mg²⁺ (2.8)

Oxygen (2.6) gains 2 electrons → O²⁻ (2.8)

Two electrons are transferred. The Mg²⁺ and O²⁻ ions have a strong electrostatic attraction.

Formation of Magnesium Chloride (MgCl₂)

Magnesium (2.8.2) loses 2 electrons - one to each of two chlorine atoms.

Each Chlorine (2.8.7) gains 1 electron → Cl⁻ (2.8.8)

This gives the formula MgCl₂ - one Mg²⁺ ion for every two Cl⁻ ions.

Dot-and-Cross Diagrams for Ionic Bonds

In dot-and-cross diagrams, dots (•) represent electrons from one atom and crosses (×) represent electrons from the other. For ionic bonds, the transferred electrons are shown in the outer shell of the ion that gained them. Square brackets and charges are used around each ion.

When drawing dot-and-cross diagrams for ionic compounds, remember to show: the electron transfer, square brackets around each ion, and the charge on each ion (e.g. [Na]⁺ and [Cl]⁻).

The Giant Ionic Lattice

Ionic compounds form a giant ionic lattice - a regular, repeating 3D arrangement of cations and anions. Each ion is strongly attracted to all surrounding ions of the opposite charge.

Ionic Bonding and Giant Ionic Lattice ELECTRON TRANSFER GIANT IONIC LATTICE Na Sodium (2.8.1) Cl Chlorine (2.8.7) 1 electron Forms Ions Na + Sodium Ion [2.8]⁺ Cl - Chloride Ion [2.8.8]⁻ Na⁺ Ion Cl⁻ Ion
The transfer of an electron from a sodium atom to a chlorine atom forms oppositely charged ions, which arrange into a massive 3D giant ionic lattice.
A complete answer on ionic bonding must describe both the transfer of electrons AND the resulting electrostatic attraction between oppositely charged ions. For MgO, state that two electrons are transferred.

Properties of Ionic Compounds

High Melting & Boiling Points

Ionic compounds have very high melting and boiling points (e.g., NaCl melts at 801°C). A large amount of energy is needed to overcome the strong electrostatic forces between the ions.

Electrical Conductivity

Whether an ionic compound conducts depends on its state:

  • Solid: Does not conduct - ions are held in fixed positions and cannot move.
  • Molten/Dissolved: Conducts electricity - ions are free to move and carry charge.

Solubility

Many ionic compounds dissolve in water. Water molecules surround the individual ions, breaking down the lattice and allowing the ions to move freely.

When answering about ionic compound properties, always link to the structure. Use "strong electrostatic forces" for melting points, and "ions are free to move" for conductivity.

Covalent Bonding

Covalent bonding occurs when two non-metal atoms share pairs of electrons to achieve a full outer shell. Each atom contributes one or more electrons to form a shared pair.

Animated Covalent Bonding: H₂ Formation ANIMATION: COVALENT H₂ FORMATION H Hydrogen Atom (1) H Hydrogen Atom (1) Hydrogen Molecule (H₂) Shared pair of electrons forms a single covalent bond

Common Covalent Molecules

Single Bonds (one shared pair)

  • H₂ (hydrogen) - each H shares 1 electron
  • HCl (hydrogen chloride) - H and Cl each share 1 electron
  • H₂O (water) - oxygen shares 1 electron with each of 2 hydrogens
  • CH₄ (methane) - carbon shares 1 electron with each of 4 hydrogens

Double Bonds (two shared pairs)

  • O₂ (oxygen) - each oxygen shares 2 electrons (O=O)

Triple Bonds (three shared pairs)

  • N₂ (nitrogen) - each nitrogen shares 3 electrons (N≡N)

Dot-and-Cross Diagrams for Covalent Bonds

In covalent dot-and-cross diagrams, show the outer shell electrons only. Dots represent electrons from one atom and crosses from the other. The shared pair sits in the overlapping region between the two atoms.

When drawing covalent dot-and-cross diagrams, you only need to show the outer shell electrons. For double bonds, show two shared pairs, and for triple bonds, three shared pairs. Do NOT use square brackets (those are only for ionic diagrams).

Simple Molecular Substances

Made of individual, discrete molecules. Within each molecule, atoms are held by very strong covalent bonds. However, forces between molecules (intermolecular forces) are very weak.

Covalent Bonding in Water and Methane WATER (H₂O) METHANE (CH₄) H H O V-shaped (Bent) molecule H H H H C Tetrahedral molecule
Simple molecules like water (H₂O) and methane (CH₄) are formed by atoms sharing electron pairs. Strong covalent bonds hold the atoms together within each molecule.
When discussing simple molecules, always distinguish between the strong covalent bonds within molecules and the weak intermolecular forces between them. It's the weak forces that are broken during melting or boiling - the covalent bonds are NOT broken.

Properties of Simple Molecules

  • Low melting/boiling points: Only a small amount of energy is needed to overcome the weak intermolecular forces. Many are gases or liquids at room temperature.
  • Poor electrical conductors: The molecules are neutral with no free-moving electrons or ions to carry charge.

Examples: Hydrogen (H₂), Chlorine (Cl₂), Water (H₂O), Methane (CH₄).

Giant Covalent Structures

Huge numbers of non-metal atoms joined by a continuous network of strong covalent bonds. No separate molecules and no weak intermolecular forces.

Giant Covalent Structures: Diamond and Graphite DIAMOND GRAPHITE Rigid tetrahedral lattice Hexagonal layers that can slide
Left: Diamond's 3D tetrahedral structure. Right: Graphite's layered structure with weak intermolecular forces between layers.

Diamond

Each carbon atom forms four strong covalent bonds in a rigid tetrahedral arrangement. Extremely hard, very high melting point, does not conduct electricity (no delocalised electrons).

Graphite

Each carbon forms three covalent bonds, creating flat layers of hexagonal rings. Layers held by weak forces - can slide (soft and slippery). One delocalised electron per carbon allows electrical conductivity along the layers.

Silicon Dioxide (SiO₂)

Similar structure to diamond. Each silicon bonded to four oxygens, each oxygen bonded to two silicons. Very hard, high melting point, does not conduct electricity.

When explaining properties of giant covalent structures, always link to the specific structure. E.g., "Diamond is hard because each carbon is held in a rigid lattice by four strong covalent bonds."

Metallic Bonding

In metals, atoms lose outer electrons to form a regular lattice of positive ions surrounded by a "sea" of delocalised electrons. The metallic bond is the strong electrostatic attraction between the positive ions and the delocalised electrons.

Metallic Bonding in Pure Metals and Alloys PURE METAL ALLOY + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + Neat layers can slide easily over each other. + + + + + + + + + + + + + + + + + + + + + + + + + + Different sized atoms distort layers, preventing sliding.
Left: Pure metals form layers of identical positive ions that can slide easily. Right: Alloys contain mixed atoms of different sizes, distorting the regular layers and making them harder.

How Metallic Bonding Explains Metal Properties

  • Good electrical conductors: Delocalised electrons can move throughout the structure and carry charge.
  • Good thermal conductors: Delocalised electrons carry kinetic energy from hotter to cooler regions.
  • Malleable & ductile: Layers of ions can slide over each other without breaking the metallic bond.
  • High melting points: Strong metallic bonds require a lot of energy to break.

Alloys

Alloys are mixtures of a metal with other elements. Different-sized atoms distort the layers, so they cannot slide as easily. This makes alloys harder and often stronger than pure metals.

Common Alloys

Alloy Composition Use
Steel Iron + carbon Construction, tools, vehicles
Stainless steel Iron + chromium + nickel Cutlery, surgical instruments
Bronze Copper + tin Statues, medals, ship propellers
Brass Copper + zinc Musical instruments, door handles
Gold alloys Gold + copper/silver Jewellery (harder than pure gold)
Common alloys, their compositions and uses.
Most everyday metals are actually alloys. Pure metals are often too soft because their identical atoms form neat layers that slide easily.
When explaining why alloys are harder than pure metals, you must mention that the different-sized atoms distort the regular layers, preventing them from sliding.

Electrolysis

4.4.3 Electrolysis

Electrolysis uses electricity to decompose an ionic compound. For electrolysis to work, the ions must be free to move. This means the compound must be either molten or dissolved in water (aqueous).

In the solid state, the strong electrostatic forces in the ionic lattice hold the ions in fixed positions, preventing them from carrying charge. When melted or dissolved, the lattice breaks down and the ions become mobile.

Key terminology

  • Electrolyte: the ionic compound, either molten or in aqueous solution.
  • Electrode: a solid conductor through which the current enters and leaves the electrolyte. Usually made of inert materials like graphite or platinum.
  • Cathode (negative electrode): attracts positive ions (cations). Reduction occurs here (cations gain electrons).
  • Anode (positive electrode): attracts negative ions (anions). Oxidation occurs here (anions lose electrons).
Use the mnemonic PANIC: Positive Anode, Negative Is Cathode. Or remember AN OX, RED CAT: Anode = Oxidation, Reduction = Cathode.
Simple Electrolysis Cell Diagram D.C. Power Supply + - e⁻ e⁻ Anode (+) Cathode (-) Oxidation Reduction Electrolyte (molten or aqueous ions) + + - -

A simple electrolysis setup. Positive cations migrate to the negative cathode, while negative anions migrate to the positive anode.

Electrolysis of Molten Compounds

For a simple molten binary ionic compound (one containing only two elements), the products are straightforward:

  • The metal is always produced at the cathode.
  • The non-metal is always produced at the anode.

Example: lead(II) bromide

PbBr₂(l) → Pb(l) + Br₂(g)

At the cathode: Pb²⁺ ions gain electrons and form molten lead.

Pb²⁺(l) + 2e⁻ → Pb(l)

At the anode: Br⁻ ions lose electrons and form bromine gas (brown vapour).

2Br⁻(l) → Br₂(g) + 2e⁻
When writing half-equations for electrolysis, check that the electrons are on the correct side. Cathode = electrons on the left (gained). Anode = electrons on the right (lost).

Extraction of Aluminium

Aluminium is too reactive to be extracted by reduction with carbon. It must be extracted by electrolysis of aluminium oxide (Al₂O₃), purified from bauxite ore.

Why is cryolite used?

Pure aluminium oxide has an extremely high melting point (over 2000°C). Melting it directly would be hugely expensive. Instead, the aluminium oxide is dissolved in molten cryolite (Na₃AlF₆), which lowers the operating temperature to about 950°C. This significantly reduces energy costs.

At the electrodes

Both electrodes are made from carbon (graphite).

  • Cathode: Al³⁺ + 3e⁻ → Al (molten aluminium sinks to the bottom and is tapped off).
  • Anode: 2O²⁻ → O₂ + 4e⁻ (oxygen gas is produced).
The carbon anodes must be replaced regularly. At the high operating temperatures, the oxygen produced reacts with the carbon anodes to form CO₂, causing them to burn away. This is a major ongoing cost of the process.
A common exam question: "Why must the carbon anodes be replaced?" Answer: because the oxygen produced at the anode reacts with the hot carbon, forming carbon dioxide, which gradually wears the anodes away.

Electrolysis of Aqueous Solutions Extended

When an ionic compound is dissolved in water, there is an added complication: water itself partially ionises, introducing extra H⁺ and OH⁻ ions into the solution. This means there are competing ions at each electrode.

Rules for the cathode (negative electrode)

Both the metal cations and H⁺ ions from water migrate to the cathode. Which one is discharged depends on reactivity:

  • If the metal is more reactive than hydrogen (e.g. Sodium, calcium, aluminium), hydrogen gas is produced at the cathode.
  • If the metal is less reactive than hydrogen (e.g. Copper, silver), the metal is deposited at the cathode.

Rules for the anode (positive electrode)

Both the non-metal anions and OH⁻ ions from water migrate to the anode:

  • If halide ions (Cl⁻, Br⁻, I⁻) are present, the halogen is produced (e.g. Chlorine gas, bromine).
  • If no halide ions are present, oxygen gas is produced (from the discharge of OH⁻ ions).
This is Required Practical 3. You need to be able to predict the products at each electrode for any given aqueous solution, and describe the observations (gas bubbles, metal plating, colour changes).

Predicting products: electrolysis of copper sulfate solution

Cathode: Cu²⁺ and H⁺ are present. Copper is less reactive than hydrogen, so copper metal is deposited. Observation: pink/brown solid coats the cathode.

Anode: SO₄²⁻ and OH⁻ are present. No halide ions are present, so oxygen gas is produced. Observation: bubbles at the anode.

Predicting products: electrolysis of sodium chloride solution

Cathode: Na⁺ and H⁺ are present. Sodium is more reactive than hydrogen, so hydrogen gas is produced. Observation: bubbles; squeaky pop with a lighted splint.

Anode: Cl⁻ and OH⁻ are present. Halide ions are present, so chlorine gas is produced. Observation: bubbles; bleaches damp litmus paper.

Predicting products: electrolysis of copper(II) bromide solution

Cathode: Cu²⁺ and H⁺ are present. Copper is less reactive than hydrogen, so copper metal is deposited. Observation: brown/pink solid coats the cathode.

Anode: Br⁻ and OH⁻ are present. Bromide is a halide ion, so bromine is produced. Observation: orange/brown colour near the anode.

Practice: write the half-equations for the electrolysis of copper(II) bromide solution

Cathode (reduction): Cu²⁺(aq) + 2e⁻ → Cu(s)

Anode (oxidation): 2Br⁻(aq) → Br₂(aq) + 2e⁻

Check: 2 electrons lost at the anode = 2 electrons gained at the cathode. ✔

Electrolysis of Brine

Brine is a concentrated solution of sodium chloride (NaCl). Its electrolysis is an important industrial process because it produces three useful products:

  • Chlorine gas (Cl₂) at the anode - used in bleach, PVC plastics, and water purification/disinfection.
  • Hydrogen gas (H₂) at the cathode - used as a fuel and in the manufacture of margarine (hardening vegetable oils).
  • Sodium hydroxide solution (NaOH) left in the solution - used in soap, paper and ceramics manufacturing, and oven cleaners.

Half-equations

2Cl⁻(aq) → Cl₂(g) + 2e⁻  (anode)
2H⁺(aq) + 2e⁻ → H₂(g)  (cathode)

Testing the products

  • Chlorine: bleaches damp litmus paper (turns it white).
  • Hydrogen: squeaky pop with a lighted splint.
  • Sodium hydroxide: turns universal indicator blue/purple (alkaline).
The sodium ions (Na⁺) and hydroxide ions (OH⁻) remain in solution because sodium is too reactive to be discharged. They combine to form NaOH, which is why the remaining solution is alkaline.

Ionic Half-Equations in Electrolysis

At the electrodes, ions either gain or lose electrons. These reactions are represented by ionic half-equations. By looking at these, we can identify which species are oxidized and which are reduced.

OIL RIG: Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons).

  • At the Cathode (negative electrode), positive ions (cations) gain electrons. This is reduction.
  • At the Anode (positive electrode), negative ions (anions) lose electrons. This is oxidation.

Worked Examples of Electrode Reactions

1. Electrolysis of Molten Lead(II) Bromide (PbBr2)

  • At the Cathode: Lead ions gain electrons to form lead metal (reduction).
    Pb2+ + 2e- → Pb(l)
  • At the Anode: Bromide ions lose electrons to form bromine gas (oxidation).
    2Br- → Br2(g) + 2e-

2. Electrolysis of Aqueous Sodium Chloride (NaCl)

  • At the Cathode: Hydrogen ions from water are reduced in preference to sodium ions (reduction).
    2H+ + 2e- → H2(g)
  • At the Anode: Chloride ions are oxidized in preference to hydroxide ions (oxidation).
    2Cl- → Cl2(g) + 2e-

3. Electrolysis of Dilute Sulfuric Acid (H2SO4)

  • At the Cathode: Hydrogen ions are reduced to hydrogen gas.
    2H+ + 2e- → H2(g)
  • At the Anode: Hydroxide ions from water are oxidized to oxygen gas and water.
    4OH- → O2(g) + 2H2O(l) + 4e-

4. Electrolysis of Aqueous Copper(II) Sulfate (CuSO4)

  • Using Carbon (Inert) Electrodes:
    • Cathode: Copper ions are reduced, forming a brown copper coating.
      Cu2+ + 2e- → Cu(s)
    • Anode: Hydroxide ions are oxidized, forming oxygen bubbles.
      4OH- → O2(g) + 2H2O(l) + 4e-
  • Using Active Copper Electrodes:
    • Cathode: Copper ions are reduced to copper metal (the cathode increases in mass).
      Cu2+ + 2e- → Cu(s)
    • Anode: The copper anode itself oxidizes and dissolves (the anode decreases in mass).
      Cu(s) → Cu2+ + 2e-