Elements, compounds and mixtures

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Chemistry · Models of the particulate nature of matter · SL · Guide 2025

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Elements, compounds and mixtures

Particle comparison: oxygen molecules contain one kind of atom; water contains hydrogen and oxygen bonded in a fixed two-to-one ratio; a mixture contains separate oxygen and hydrogen molecules in variable proportions.

Every sample of matter is an element, a compound or a mixture.

Elements and compounds are pure substances: each has one fixed composition.

A mixture does not.

An element contains atoms of one kind only (all with the same number of protons).

It cannot be broken down into anything simpler by a chemical reaction.

The atoms need not be separate: oxygen gas is made of $\ce{O2}$ molecules and diamond is a giant network of carbon atoms, yet each is an element because only one kind of atom is present.

A compound contains atoms of two or more elements chemically bonded in a fixed ratio.

That fixed ratio is why a compound has a formula: water is always $\ce{H2O}$ , wherever it comes from.

Its properties are new.

Sodium is a reactive metal and chlorine a toxic gas, yet sodium chloride is a stable white solid.

A compound can be split into its elements only by a chemical reaction, such as electrolysis or heating until it decomposes.

A mixture contains two or more elements or compounds that are not chemically bonded, in no fixed ratio.

Each component keeps its own properties, which is exactly why a mixture can be separated by physical methods.

Alloys such as brass or steel are mixtures, even though they look uniform and contain metallic bonding, because their composition can vary.

Telling the three apart

A question usually gives you properties and asks you to classify, or gives the class and asks for the properties.

These are the contrasts to use.

Element

Compound

Mixture

Kinds of atom

one

two or more, bonded

two or more substances, not bonded to each other

Composition

fixed

fixed ratio, shown by a formula

variable

Properties

its own

different from those of its elements

each component keeps its own

How to split it

cannot be split chemically

chemical reaction only

physical methods

Melting and boiling

sharp, fixed temperatures

usually over a range of temperatures

The last row is a practical test.

A pure substance melts sharply at one temperature.

An impure solid, which is a mixture, starts to melt at a lower temperature and melts over a range.

A measured melting point that is sharp and matches the accepted value is evidence that a product is pure.

Compare a mixture of hydrogen and oxygen gases with water.

The gases can be mixed in any proportion, the hydrogen in the mixture still burns, and the two can be separated physically.

In water the same two elements are bonded in a fixed 2 : 1 ratio of atoms, the properties are entirely different, and only a chemical reaction will release the elements.

Homogeneous and heterogeneous mixtures

A homogeneous mixture has the same composition throughout and a single visible phase: salt water, air, brass.

Its components are mixed at the level of individual particles, so every sample taken from it is the same.

A solution is a homogeneous mixture of a solute dissolved in a solvent.

A heterogeneous mixture has regions of different composition, often ones you can see: sand in water, oil on water, granite.

It contains more than one phase, with boundaries between them, and two samples taken from different places can differ.

"Homogeneous" does not mean "pure".

Air is uniform, but it is still a mixture, because its composition can change and its components can be separated physically.

Whether two substances form a homogeneous or a heterogeneous mixture depends on the attractions between their particles, which you study with intermolecular forces.

Choosing a separation method

Pick the method from the property that differs between the components.

Method

Property it uses

Example

Solvation (dissolving)

one component dissolves in a solvent, the other does not

salt dissolves in water, sand does not

Filtration

an insoluble solid is trapped by filter paper

sand from salt solution

Evaporation

a dissolved solid is left when the solvent evaporates

salt from salt solution

Recrystallization

a solid is more soluble in hot solvent than cold

purifying an impure solid product

Distillation

liquids (or a solvent and a solute) have different boiling points

pure water from sea water; fractional distillation for ethanol and water

Chromatography

components spread differently between a mobile and a stationary phase

dyes in an ink

Ask two questions.

What are the states of the components, and are they dissolved?

Which component do you want to keep?

An insoluble solid in a liquid is filtered.

A dissolved solid is recovered by evaporation, but the solvent is lost.

To keep the solvent, or to separate two liquids that mix, you distil.

Small amounts of several dissolved substances are separated and identified by chromatography.

Solvation, filtration and evaporation

Solvation is what happens when a substance dissolves: solvent particles surround the solute particles and hold them in solution.

It separates a soluble component from an insoluble one.

Filtration then removes the insoluble solid.

The solid left on the paper is the residue and the liquid that passes through is the filtrate.

Evaporation recovers the solute from the filtrate.

Heating to dryness can decompose the solid or make it spit.

The usual method is to evaporate until crystals just begin to form, then leave the solution to cool so that crystals grow.

The crystals are filtered off and dried.

Recrystallization

Recrystallization purifies an impure solid.

It works because most solids are much more soluble in a hot solvent than in the same solvent when cold.

1. Dissolve the impure solid in the minimum volume of hot solvent. This gives a solution that is close to saturated while hot.

2. If insoluble impurities are present, filter the solution while it is still hot. They stay on the paper.

3. Let the filtrate cool slowly, then stand it in ice. The solubility falls, so the wanted substance crystallizes.

4. Soluble impurities are present in small amounts, so the cold solvent can still hold them. They stay in solution.

5. Filter off the crystals, rinse them with a little cold solvent and dry them.

Three points explain the details.

Too much solvent keeps more of the product dissolved when cold, so less is recovered.

Hot rinsing solvent would redissolve the crystals.

Some product always stays in the cold solution, so the recovery is never 100%.

A good solvent dissolves a lot of the solid when hot and very little when cold.

It must not react with the solid.

Distillation

Distillation separates by boiling point.

The mixture is heated in a flask, the vapour passes into a water-cooled condenser, and the condensed liquid, the distillate, is collected.

Simple distillation separates a solvent from a dissolved solid, or two liquids with very different boiling points. Sea water gives pure water, and the salts stay in the flask.

Fractional distillation separates liquids that mix and have closer boiling points, such as ethanol (78 °C) and water (100 °C). A fractionating column sits between the flask and the condenser. Vapour condenses and boils again many times as it rises, so the vapour at the top is rich in the component with the lower boiling point.

The thermometer bulb sits level with the side arm that leads to the condenser.

There it reads the temperature of the vapour that is about to be collected, which is the boiling point of that liquid.

Cooling water enters the condenser at its lower end, so the jacket stays full of water.

Anti-bumping granules in the flask make the liquid boil smoothly.

Chromatography

In chromatography a mobile phase (a solvent) moves through a stationary phase.

In paper chromatography the stationary phase is the paper.

In thin layer chromatography (TLC) it is a thin coating of a solid such as silica on a plate.

Each component of the mixture is shared between the two phases.

A component that is more soluble in the mobile phase, and less strongly attracted to the stationary phase, is carried further.

Different components therefore travel different distances and separate into spots.

The method has details that examiners test:

The baseline is drawn in pencil, because graphite does not dissolve in the solvent. Ink would run and add its own spots.

The spots are small and are allowed to dry, so they do not spread into each other.

The solvent level starts below the baseline. Otherwise the samples dissolve into the pool of solvent instead of travelling up the paper.

A lid keeps the container full of solvent vapour, so the solvent does not evaporate from the paper.

The paper is removed before the solvent reaches the top, and the solvent front is marked at once.

A substance that gives two or more spots is a mixture.

A single spot is consistent with a pure substance, but does not prove it: two components may happen to travel the same distance in that solvent.

A spot that travels the same distance as a known substance on the same paper is likely to be that substance.

The position of a spot is reported as its retardation factor, $R_f$ , which you meet with intermolecular forces.

✓

EXAM FOCUS

How the IB examines this

Paper 1A, classification: you pick the element, compound or mixture from a list, or a table whose rows classify several substances. The mark is lost by calling an alloy a compound, or by treating a uniform appearance as proof of a pure substance.

Paper 1A, choice of method: a mixture is described and you choose the technique that isolates a named component. Read which component must be kept. Evaporation loses the solvent, and filtration cannot remove a dissolved solid.

Paper 1B, procedure: a preparation ends with a purification step and you explain a detail of it: why the minimum volume of hot solvent, why a cold rinse, what a melting point shows. Each mark needs the specific reason. A general phrase such as "to make it more accurate" earns nothing.

Paper 1B, data: you interpret a chromatogram or a set of solubility or melting-point data, and judge whether a sample is pure. The conclusion must be tied to the evidence given.

Paper 2, distinguish or explain: a short part asks you to distinguish between two classes, or to explain why a named material is a mixture. Distinguish needs a statement about each side. For two marks, give two separate contrasts.

Paper 2, outline a separation: steps in a workable order, each naming the technique and what it removes. A missing step, such as drying the crystals, usually costs a mark.

◆

Key terms

Element: a substance made of one type of atom that cannot be broken down chemically into simpler substances.

Compound: two or more elements chemically bonded together in a fixed ratio.

Mixture: two or more substances, not chemically bonded, in no fixed ratio, which can be separated by physical methods.

Homogeneous mixture: a mixture with uniform composition and a single phase throughout.

Heterogeneous mixture: a mixture whose composition is not uniform, with more than one phase.

Solvation: the surrounding of solute particles by solvent particles as a substance dissolves.

Recrystallization: purifying a solid by dissolving it in the minimum volume of hot solvent and letting it crystallize as the solution cools.

Filtrate and residue: the liquid that passes through a filter, and the solid that stays on it.

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Worked example 1: pure salt from rock salt

Rock salt is a heterogeneous mixture of sodium chloride and insoluble grit.

Outline how to obtain dry salt crystals.

1. Add water and stir: the salt dissolves (solvation), the grit does not.

2. Filter: the grit stays on the paper as the residue; salt solution passes through as the filtrate.

3. Evaporate some of the water, then leave the solution to crystallize, and dry the crystals.

Answer: dissolve, filter, evaporate and crystallize.

Each step uses a physical difference, so no chemical change takes place.

↗

Worked example 2: classifying matter

Classify each of these as an element, a compound or a mixture, with a reason.

For a mixture, state whether it is homogeneous or heterogeneous.

(a) Ozone, $\ce{O3}$ .

(b) Ethanol, $\ce{C2H5OH}$ .

(d) A salad dressing of oil and vinegar that has been shaken.

1. (a) Ozone contains only oxygen atoms. Being made of molecules does not make it a compound. It is an element.

2. (b) Ethanol contains carbon, hydrogen and oxygen atoms bonded in a fixed ratio, shown by its formula. It is a compound.

3. (c) Stainless steel is iron with chromium and other elements in proportions that vary from one grade to another. It is a mixture. It is uniform throughout, so it is homogeneous.

4. (d) Oil and vinegar are not bonded to each other and can be mixed in any proportion. Droplets of one liquid are visible in the other, so there are two phases. It is a heterogeneous mixture.

Answer: (a) element; (b) compound; (c) homogeneous mixture; (d) heterogeneous mixture.

↗

Worked example 3: a structured question on distillation (Paper 2 style)

Propanone and water mix completely.

Propanone boils at 56 °C and water at 100 °C. A mixture of the two is separated by fractional distillation.

(a) State the property of the two liquids that allows them to be separated by distillation. [1]

(b) Identify the first liquid to collect as the distillate, giving a reason. [1]

(c) Explain why the thermometer bulb is placed level with the side arm leading to the condenser, not in the liquid. [1]

(d) State where the cooling water enters the condenser and give a reason. [2]

(e) Suggest why a fractionating column gives a better separation than simple distillation. [1]

(f) Propanone is highly flammable. Suggest one precaution, specific to this hazard, for heating the flask. [1]

1. (a) They have different boiling points. [1]

2. (b) Propanone, because it has the lower boiling point. [1]

3. (c) It measures the temperature of the vapour entering the condenser, which is the boiling point of the liquid being collected. [1]

4. (d) At the lower end of the condenser [1], so that the jacket fills completely with water and the vapour is cooled along its whole length [1].

5. (e) The vapour condenses and vaporizes again many times in the column, so the vapour reaching the top is richer in propanone. [1]

6. (f) Heat with an electric heating mantle or a hot water bath, with no flame nearby. [1]

Answer: the marks are for the specific reason in each part.

In (f), "wear goggles" is good practice but does not address flammability, so it would not score.

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Worked example 4: interpreting a thin layer chromatogram

A student prepares a solid product P from a starting material S and purifies it by recrystallization.

Three samples are spotted on one TLC plate.

The solvent front moves 6.0 cm.

The data were written for this example.

Sample

Distance moved by each spot / cm

Pure S (reference)

4.2

Crude P

1.8 and 4.2

Recrystallized P

1.8

Deduce what the chromatogram shows about the crude and the recrystallized product.

1. Crude P gives two spots, so it is a mixture of at least two substances.

2. One of its spots moved 4.2 cm, the same as pure S on the same plate. The crude product contains unreacted starting material.

3. The spot at 1.8 cm appears in both product samples, so it is P. It moved less far than S, so P is more strongly attracted to the stationary phase, or less soluble in the solvent, than S.

4. Recrystallized P gives one spot at 1.8 cm. The starting material has been removed.

Answer: the crude product is contaminated with S, and recrystallization removed it.

One spot is consistent with pure P, but a melting point would be needed to confirm it.

↗

Worked example 5: evaluating a recrystallization (Paper 1B style)

A student purifies 2.50 g of crude benzoic acid by recrystallization from water.

The crude solid is dissolved in 50 cm³ of boiling water, the solution is filtered hot, and the filtrate is cooled to 25 °C. The crystals are filtered off, rinsed and dried.

The results were written for this example.

Value

Solubility of benzoic acid in water at 100 °C

5.6 g per 100 cm³

Solubility of benzoic acid in water at 25 °C

0.34 g per 100 cm³

Mass of dry crystals obtained

2.05 g

Melting range of crude solid

115 to 119 °C

Melting range of crystals

121 to 122 °C

Melting point of pure benzoic acid

122 °C

(a) Calculate the minimum volume of boiling water needed to dissolve 2.50 g of benzoic acid. [1]

(b) Calculate the mass of benzoic acid that stays dissolved in 50 cm³ of water at 25 °C. [1]

(d) Explain, using a calculation, why dissolving the crude solid in 150 cm³ of boiling water would give a lower recovery. [2]

(e) Explain what the melting ranges show about the crude solid and the crystals. [2]

(f) State why the crystals are rinsed with cold water and not hot water. [1]

1. (a) $\dfrac{2.50}{5.6} \times 100 = 45$ cm³. [1]

2. (b) $0.34 \times \dfrac{50}{100} = 0.17$ g. [1]

3. (c) $\dfrac{2.05}{2.50} \times 100 = 82.0\%$ . [1]

4. (d) At 25 °C, 150 cm³ of water holds $0.34 \times \dfrac{150}{100} = 0.51$ g of benzoic acid [1]. That is three times as much product left in solution, so less crystallizes [1].

5. (e) The crude solid melts below 122 °C and over a range of 4 °C, so it is impure [1]. The crystals melt over a narrow range that reaches the value for the pure substance, so they are much purer [1].

6. (f) Benzoic acid is far more soluble in hot water, so a hot rinse would dissolve some of the crystals. [1]

Answer: (a) 45 cm³; (b) 0.17 g; (c) 82.0%; (d) 0.51 g would stay dissolved.

The 50 cm³ used is just above the minimum, which is why the loss to the solution is small.

✦

Exam tip

A Distinguish question wants a contrasting point for each substance, not a description of one.

"A compound has a fixed ratio of elements chemically bonded; a mixture has no fixed ratio and its components are not bonded" earns the mark in one sentence.

Common mistakes

Calling an alloy a compound. It is a mixture, because its composition is not fixed.

Saying a mixture must be heterogeneous. Air and salt water are mixtures, and both are homogeneous.

Calling a substance made of molecules a compound. $\ce{O2}$ , $\ce{O3}$ and $\ce{S8}$ are elements, because each contains one kind of atom.

Using evaporation to separate two liquids. Evaporation throws away the solvent; to keep both liquids you distil.

Writing "dissolve in hot solvent" for recrystallization. The mark is for the minimum volume of hot solvent.

Concluding that one spot on a chromatogram proves a substance is pure. It is consistent with purity, no more.

Check your understanding

1. Which is a homogeneous mixture? [1]
A. Distilled water
B. Dry, filtered air
C. Sand mixed with iron filings
D. Copper wire

2. Which row classifies all three substances correctly? [1]

Diamond

Ammonia

Brass

element

compound

mixture

compound

mixture

element

mixture

compound

element

compound

3. Which statements about a mixture are correct? [1]

I. Its components keep their own chemical properties.
II. Its composition can vary.
III. Its components can be separated only by a chemical reaction.

A. I and II only
B. I and III only
C. II and III only
D. I, II and III

4. Which method gives pure water from copper(II) sulfate solution? [1]
A. Filtration
B. Evaporation
C. Simple distillation
D. Recrystallization

5. Which row describes the best solvent for recrystallizing a solid? [1]

Solubility of the solid in hot solvent

Solubility of the solid in cold solvent

high

low

high

low

6. A green ink gives a blue spot and a yellow spot on a paper chromatogram. What can be deduced about the ink? [1]
A. It is a compound of two elements.
B. It contains at least two substances.
C. It contains exactly two substances.
D. It is insoluble in the solvent used.

7. Distinguish between a homogeneous and a heterogeneous mixture. [2]

8. Bronze contains copper and tin atoms held together by metallic bonding. Explain why bronze is classified as a mixture and not as a compound. [2]

9. Iron filings are stirred with powdered sulfur. A magnet lifts the iron out of the powder. When the powder is heated strongly instead, it glows and forms a black solid, iron(II) sulfide, $\ce{FeS}$ , that a magnet does not attract. Explain, using this evidence, why the powder is a mixture and the black solid is a compound. [3]

10. Copper(II) sulfate crystals have been contaminated with powdered charcoal, which is insoluble in water. Outline how to obtain pure, dry crystals of copper(II) sulfate from the mixture. [3]

11. A student uses paper chromatography to find out which dyes are in the coating of a green sweet. Spots of the coating and of three known dyes, E, F and G, are placed on a pencil baseline. The solvent front moves 9.0 cm. The results were written for this question.

Sample

Distance moved by each spot / cm

Sweet coating

2.7 and 6.3

Dye E

2.7

Dye F

4.5

Dye G

6.3

(a) State why the baseline is drawn in pencil and not in ink. [1]
(b) Explain why the solvent level must start below the baseline. [1]
(c) Deduce which of the dyes are present in the coating, giving your reasoning. [2]
(d) Identify the dye that is most strongly attracted to the paper, giving a reason. [1]
(e) The student claims: "Dye F gives one spot, so it must be a pure substance." Comment on this claim and suggest how it could be tested further. [2]

12. A chemist needs to recrystallize an impure solid Z. The table shows the solubility of Z in three solvents. Z reacts with none of them. The data were written for this question.

Solvent

Solubility of Z when cold / g per 100 cm³

Solubility of Z when hot / g per 100 cm³

0.40

12.0

9.5

14.0

0.10

0.30

(a) Identify the most suitable solvent and justify your choice by referring to the other two. [3]
(b) A 6.00 g sample contains 5.85 g of Z and 0.15 g of a soluble impurity Y. It is dissolved in 60 cm³ of hot solvent P and cooled. Calculate the mass of Z that stays in solution and the mass of Z that crystallizes. [2]
(c) The solubility of Y in cold solvent P is 2.0 g per 100 cm³. Explain, using a calculation, why the crystals contain no Y. [2]
(d) Calculate the percentage of the Z in the sample that is recovered. [1]

Try the questions before checking your work.

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