Pearson Edexcel International GCSE Chemistry 4CH1: States of matter; elements, compounds and mixtures – Study Guide

This guide teaches sub-topics 1(a) States of matter and 1(b) Elements, compounds and mixtures of the Pearson Edexcel International GCSE Chemistry (4CH1) specification, Issue 3 (September 2024). It covers specification points 1.1 to 1.13. The qualification is untiered. Points 1.5C, 1.6C and 1.7C are printed in bold with a “C” reference, so they are assessed on Paper 2 only; everything else can appear on Paper 1 or Paper 2.

Links: 4CH1 Chemistry course hub, printable 4CH1 checklist, Principles of chemistry overview and the free 4CH1 all-topics diagnostic.

Paper 1 (2 hours, 110 marks) assesses content without a C reference. Paper 2 (1 hour 15 minutes, 70 marks) assesses all content. A calculator may be used on both.

What this unit covers

Spec point	What you must be able to do	Paper
1.1	Describe solids, liquids and gases by arrangement, movement and energy of particles	1 and 2
1.2	Name each change of state, say how it is achieved and what happens to the particles	1 and 2
1.3	Explain dilution of coloured solutions and diffusion of gases using particles	1 and 2
1.4	Define solvent, solute, solution and saturated solution	1 and 2
1.5C	Define solubility in g per 100 g of solvent	Paper 2 only
1.6C	Plot and interpret solubility curves	Paper 2 only
1.7C	Practical: find the solubility of a solid in water at a set temperature	Paper 2 only
1.8	Classify a substance as an element, compound or mixture	1 and 2
1.9	Use melting and boiling points to tell a pure substance from a mixture	1 and 2
1.10	Describe five separation techniques	1 and 2
1.11–1.13	Interpret chromatograms, calculate Rf values; practical with inks or food colourings	1 and 2

(a) States of matter

The three states (1.1)

	Solid	Liquid	Gas
Arrangement	Close together, regular pattern	Close together, irregular	Far apart, random
Movement	Vibrate about fixed positions	Move around each other	Move quickly in all directions
Energy of particles	Lowest	Higher	Highest

Forces of attraction between particles are strongest in a solid and weakest in a gas.

Changes of state (1.2)

Change	Name	How it is achieved
Solid → liquid	Melting	Heating
Liquid → solid	Freezing	Cooling
Liquid → gas	Boiling or evaporating	Heating
Gas → liquid	Condensing	Cooling
Solid → gas directly	Sublimation	Heating (for example, solid carbon dioxide)

When a solid melts, the particles gain energy and vibrate more until they overcome some of the forces holding them in place; they stay close but can now move around each other. When a liquid boils, the particles gain enough energy to overcome the remaining forces and move far apart. Freezing and condensing are the reverse: particles lose energy and move closer together.

Evaporation happens only at the surface, below the boiling point. Boiling happens throughout the liquid, at the boiling point.

Why the temperature stops rising. Heat a pure solid steadily and its temperature stays constant while it melts, and again while it boils. The energy is being used to overcome forces between particles, not to make them move faster.

Evidence for particles: dilution and diffusion (1.3)

Dilution of a coloured solution. Dissolve a few crystals of potassium manganate(VII) in water. Take 1 cm³ of the purple solution and add 9 cm³ of water; repeat several times. The colour gets paler each time but is still visible after several dilutions. So the crystals must contain a very large number of very small particles. Each dilution spreads them among more water particles, so each cm³ holds fewer coloured particles.

Worked example 1. Each step takes 1 cm³ and makes it up to 10 cm³. What fraction of the original concentration is left after three steps?

each step: concentration × 1/10
after 3 steps: 1/10 × 1/10 × 1/10 = 1/1000

Answer: one thousandth of the original concentration.

Diffusion is the random movement of particles from a region where they are at higher concentration to a region where they are at lower concentration, until they are evenly spread.

• Bromine in gas jars. A jar of brown bromine vapour sits under a jar of air; the cover between them is removed. After a while both jars are the same pale brown, because bromine and air particles move randomly and mix until evenly spread. If the upper jar is a vacuum, the colour fills it almost at once: there are no air particles to collide with.
• Ammonia and hydrogen chloride. The white ring forms nearer the hydrochloric acid end because ammonia particles are lighter and diffuse faster (see the Principles of chemistry revision notes).

Diffusion is faster at a higher temperature, because the particles have more energy and move faster.

Solutions (1.4)

• Solvent: the liquid in which a substance dissolves (water is the most common).
• Solute: the substance that dissolves in the solvent.
• Solution: the mixture formed when a solute dissolves in a solvent.
• Saturated solution: a solution in which no more solute can dissolve at that temperature.

Solubility (1.5C) – Paper 2 only

Solubility is the mass of solute that dissolves in 100 g of solvent to form a saturated solution at a stated temperature. Its unit is g per 100 g of solvent. Always give the temperature, because solubility changes with temperature.

Measuring solubility (1.7C) – Paper 2 only

1. Stir excess solid into water in a water bath at the chosen temperature until some solid stays undissolved (saturated).
2. Pour some clear solution into a weighed evaporating basin. Weigh basin and solution.
3. Evaporate the water by gentle heating. Weigh basin and dry solid.
4. Reheat, cool and reweigh until the mass stops changing (constant mass), so all the water has gone.

Worked example 2.

Measurement	Mass / g
Empty basin	32.50
Basin + saturated solution	58.90
Basin + dry solid (constant mass)	39.10

mass of solid  = 39.10 − 32.50 = 6.60 g
mass of water  = 58.90 − 39.10 = 19.80 g
solubility     = 6.60 / 19.80 × 100 = 33.3 g per 100 g of water

The mass of water is the solution minus the solid, not the solution minus the basin.

Solubility curves (1.6C) – Paper 2 only

Plot solubility (g per 100 g water) on the y-axis against temperature (°C) on the x-axis and draw a smooth curve. For most solids, solubility rises with temperature. Points on the curve are saturated solutions; points below it are unsaturated.

Worked example 3. A solid X has these solubilities:

Temperature / °C	0	20	40	60	80
Solubility / g per 100 g water	12	20	34	55	82

A saturated solution of X in 50 g of water at 60 °C is cooled to 20 °C. What mass of crystals forms?

at 60 °C, 100 g water holds 55 g; 50 g water holds 27.5 g
at 20 °C, 100 g water holds 20 g; 50 g water holds 10.0 g
crystals = 27.5 − 10.0 = 17.5 g

Or in one line: (55 − 20) × 50/100 = 17.5 g.

(b) Elements, compounds and mixtures

Classifying substances (1.8)

• Element: only one type of atom; cannot be broken down into anything simpler by chemical reactions. Examples: iron (Fe), oxygen (O₂).
• Compound: two or more elements chemically combined in fixed proportions; only a chemical reaction splits it up. Examples: water (H₂O), sodium chloride (NaCl).
• Mixture: two or more substances not chemically combined; proportions can vary, each keeps its own properties, and physical methods separate them. Examples: air, sea water, ink.

On a particle diagram: identical units of one kind of atom (single atoms or molecules like O₂) = element; identical units containing different atoms bonded together = compound; two or more kinds of unit = mixture.

Worked example 4. Classify (a) a sample containing only N₂ molecules, (b) a sample containing only NH₃ molecules, (c) a sample containing N₂ and O₂ molecules.

Answers: (a) element; (b) compound: N and H atoms bonded in a fixed ratio; (c) mixture: two substances not bonded to each other.

Purity, melting and boiling (1.9)

A pure substance (one element or one compound) has a fixed melting point and boiling point; pure water melts at 0 °C and boils at 100 °C at normal atmospheric pressure. A mixture may melt or boil over a range of temperatures. A sharp melting point matching the data-book value suggests a pure sample; melting over a range shows it is impure.

Separation techniques (1.10)

Technique	Separates	Key points
Filtration	Insoluble solid from a liquid	Solid stays on filter paper (residue); liquid passes through (filtrate)
Crystallisation	Soluble solid from its solution	Evaporate some water until saturated (crystals form on a glass rod dipped in and removed); cool; filter off the crystals and dry them between filter papers
Simple distillation	A solvent from a solution (e.g. water from salt water)	The solvent boils; its vapour cools in a condenser (water in at the bottom) and is collected; the solute stays behind
Fractional distillation	Liquids that mix, with different boiling points (e.g. ethanol and water)	A fractionating column sits above the flask; the lower-boiling liquid (ethanol, about 78 °C) distils first; the thermometer at the top shows the boiling point of the liquid coming over
Paper chromatography	Soluble substances in a mixture, such as dyes in an ink	Substances move different distances up the paper with the solvent

Chromatography and Rf values (1.11–1.13)

Practical method (1.13). Draw a pencil baseline about 1–2 cm from the bottom (pencil does not dissolve and run). Put small, concentrated spots of each ink or food colouring on it, with reference dyes alongside. Stand the paper in solvent with the level below the baseline, so the spots do not wash into the solvent. Cover with a lid. Remove the paper before the solvent reaches the top, mark the solvent front and let it dry.

Reading a chromatogram (1.11).

• A pure substance gives one spot.
• A mixture gives two or more spots: the minimum number of substances present (two could travel the same distance).
• A spot level with a reference spot, in the same solvent, is probably the same substance.
• A spot that stays on the baseline is insoluble in that solvent.

Rf values (1.12).

Rf = distance moved by the substance / distance moved by the solvent

Measure both distances from the baseline. Rf has no units and is less than 1. A substance has the same Rf in the same solvent and paper, so matching Rf values identifies it.

Worked example 5. The solvent front is 8.0 cm above the baseline. A food colouring gives spots at 3.2 cm and 5.6 cm. Known dyes have Rf: yellow 0.40, red 0.55, blue 0.70.

spot 1: Rf = 3.2 / 8.0 = 0.40  → yellow
spot 2: Rf = 5.6 / 8.0 = 0.70  → blue

Answer: the colouring contains the yellow and blue dyes, and not the red.

Common errors

• Saying particles in a solid “do not move”. They vibrate about fixed positions.
• Saying the temperature rises during melting. It stays constant while forces between particles are overcome.
• Defining a saturated solution without “at that temperature”.
• Using the mass of the solution, not the mass of water, in a solubility calculation (Paper 2 only).
• Calling a mixture of two elements a compound. Ask: are they chemically bonded?
• Drawing the chromatography baseline in ink, or putting the solvent above it.
• Measuring Rf distances from the bottom of the paper instead of the baseline, or dividing the solvent distance by the spot distance.
• Using filtration to recover a dissolved solid. It only removes insoluble solids.

Next steps

After the revision notes and practice questions for this unit, try the Principles of chemistry practice set and the formulae, equations and calculations practice set.

Official syllabus

Pearson Edexcel International GCSE in Chemistry (4CH1) specification, Issue 3, September 2024, Pearson Education Limited. Topic 1 Principles of chemistry, sub-topics (a) States of matter and (b) Elements, compounds and mixtures, points 1.1–1.13.