The Nature of Substances and Chemical Reactions — WJEC GCSE Chemistry Revision Notes & Topical Questions

Section 1 of 38Elements

Element = a substance made of only one type of atom

Listing

• Listing all elements are listed in the Periodic Table

Forms

• Forms elements exist as single atoms or as molecules of joined atoms

Building blocks

• Building blocks elements cannot be broken down into simpler substances by chemical means, but they react to make more complex substances

Section 2 of 38Compounds

Compound = a substance made of two or more different types of atom chemically joined

Formation

• Formation in chemical reactions the atoms of elements rearrange to form compounds

Properties

• Properties compounds can differ greatly from their elements, as sodium chloride differs from sodium and chlorine

Molecules

• Molecules a compound of non-metal atoms is a molecule containing different types of atom

Section 3 of 38Chemical symbols and diatomic elements

Symbols

• Symbols each element has a symbol from the Periodic Table, such as Li, W, Al, Ca, Au and Si

Seven diatomic elements

• Seven diatomic elements they exist as molecules of two atoms with formula X₂
• Hydrogen H₂, nitrogen N₂, oxygen O₂.
• Fluorine F₂, chlorine Cl₂, bromine Br₂, iodine I₂.

Section 4 of 38Simple molecules and their formulae

Simple molecules

• Simple molecules they are made of two or more non-metal atoms chemically joined, and can be elements or compounds

Covalent

• Covalent simple molecules are covalent substances and contain no metal atoms

Formula

• Formula the symbols show which atoms are present and the subscript number shows how many of each

Substance	Formula	Type	Atoms present
Hydrogen	H2	Element	2 H
Chlorine	Cl2	Element	2 Cl
Water	H2O	Compound	2 H, 1 O
Methane	CH4	Compound	1 C, 4 H
Ammonia	NH3	Compound	1 N, 3 H
Sulfuric acid	H2SO4	Compound	2 H, 1 S, 4 O

Section 5 of 38Representing molecules in diagrams

Information shown

• Information shown the number of atoms and how they are arranged

Style 1

• Style 1 use chemical symbols for the atoms, and no key is needed

Style 2

• Style 2 use coloured circles for atoms, and a key shows the colour for each element

Patterns

• Patterns different patterns can replace colours when none are available

Arrangement

• In water, the two hydrogen atoms are joined to either side of the oxygen atom, not in a row.
• In carbon dioxide, an oxygen atom is on either side of the central carbon atom.
• In methane, four hydrogen atoms surround the central carbon atom.

Section 6 of 38Ions used in formulae

• Ionic compounds
  • a metal and a non-metal react, with the metal losing electrons and the non-metal gaining electrons to form ions, so they are not simple molecules

Ion	Formula
Hydrogen	H+
Group 1	Li+, Na+, K+
Group 7	F-, Cl-, Br-
Copper(II)	Cu2+
Iron(II)	Fe2+
Iron(III)	Fe3+
Carbonate	CO32-
Sulfate	SO42-
Hydroxide	OH-
Ammonium	NH4+

• Data
  • a table of common ions is given in the exam, but the knowledge must be applied to any ion given

Section 7 of 38Formulae of ionic compounds

Overall charge

• Overall charge an ionic compound has no overall charge, so the positive and negative charges cancel

Direct comparison

• Direct comparison balance the charges by choosing numbers of each ion
• Iron(II) sulfate has Fe²⁺ and SO₄^2-, so one of each is needed, giving FeSO₄.
• Iron(III) chloride has Fe³⁺ and Cl^-, so three chlorides are needed, giving FeCl₃.

Swap-and-drop

• Swap-and-drop swap the charge numbers and drop them as subscripts
• Remove the signs and omit the number 1.
• Copper(II) chloride, Cu²⁺ and Cl^-, gives CuCl₂.
• If the charges are equal, reduce to the simplest whole-number ratio.

Polyatomic ions

• Polyatomic ions put brackets around the ion when more than one is needed, so ammonium sulfate is (NH₄)₂SO₄

Naming

• Naming the metal keeps its name and the non-metal ending changes, so iron and chlorine give iron chloride

Section 8 of 38Relative atomic mass, A_r

Symbol and location

• A_r is the larger number above the symbol in the Periodic Table, and it is always larger than the atomic number except for hydrogen

Standard

• Standard carbon-12 is the standard atom with a mass of exactly 12 units

Comparison

• Mg with A_r 24 is twice as heavy as carbon, and H with A_r 1 is one twelfth of a carbon-12 atom

Definition

• A_r is the average mass of an element’s isotopes compared with one twelfth of the mass of a carbon-12 atom

Section 9 of 38Relative formula mass, M_r

Definition

• M_r is the sum of the relative atomic masses of all atoms in the formula

Term

• Term relative formula mass is used for ionic compounds

Substance	Calculation	Mr
H2	2 × 1	2
H2O	(2 × 1) + 16	18
K2CO3	(2 × 39) + 12 + (3 × 16)	138
Ca(OH)2	40 + (2 × 16) + (2 × 1)	74
(NH4)2SO4	(2 × 14) + (8 × 1) + 32 + (4 × 16)	132
NaCl	23 + 35.5	58.5
Mg(NO3)2	24 + (2 × 14) + (6 × 16)	148

Section 10 of 38Mass number and relative atomic mass

Mass number

• Mass number the sum of protons and neutrons in the nucleus, always a whole number

Relative atomic mass

• Relative atomic mass the ratio of the mass of an element to one twelfth of a carbon atom, and it can be a decimal

Section 11 of 38Percentage composition

Definition

• Definition the mass of each element in a compound expressed as a percentage of the total mass

Equation

• Equation percentage composition = × 100
  • (total mass of the element ÷ total mass of the compound)

1

Find the relative atomic masses from the Periodic Table.

2

Add them to find the M_r of the compound.

3

Multiply the A_r by the number of atoms of the element.

4

Divide by M_r and multiply by 100.

Water

• M_r is 18
• Hydrogen is (2 × 1) ÷ 18 × 100 = 11.1%.
• Oxygen is 16 ÷ 18 × 100 = 88.9%.

Check

• Check the percentages of all elements add up to 100%

Glucose, C₆H₁₂O₆

• M_r = 180.
• Carbon is (6 × 12) ÷ 180 × 100 = 40%.

Ammonium nitrate, NH₄NO₃

• Ammonium nitrate, NH₄NO₃ count every atom of the element, even in different parts of the formula
• M_r = 80 and there are two nitrogen atoms.
• Nitrogen is (2 × 14) ÷ 80 × 100 = 35%.

Common error

• Common error an answer above 100% means the fraction was written upside down

Section 12 of 38Mixtures

Mixture = two or more substances that are not chemically joined, which can be elements, compounds or both

Separation

• Separation mixtures are separated by physical processes: filtration, evaporation, distillation and chromatography

Basis

• Basis every method uses a difference in a physical property, such as boiling point, solubility or magnetism

Section 13 of 38Filtration

• Use
  • it separates an insoluble solid from a liquid or solution, such as sand from water

1

Place filter paper in a funnel above a beaker.

2

Pour in the mixture.

3

Liquid and dissolved solids pass through the paper as the filtrate.

4

Insoluble solid stays behind as the residue.

Section 14 of 38Evaporation and crystallisation

Evaporation

• Evaporation it separates a dissolved solid from a solution by heating to boil off the solvent

Crystallisation

• Crystallisation it is used when the solid is more soluble in hot solvent than cold, such as copper(II) sulfate

1

Heat the solution to evaporate about half of the solvent, leaving a saturated solution.

2

Test for saturation by dipping a cold dry glass rod in, because crystals form on the rod if it is saturated.

3

Cool the saturated solution slowly so that crystals grow as solubility decreases.

4

Collect the crystals by filtration.

5

Wash them with distilled water to remove impurities.

6

Dry them between filter papers or in a drying oven.

Section 15 of 38Simple and fractional distillation

• Use of simple distillation
  • it separates a liquid from a solution with a dissolved solid, such as pure water from salt water, or a pure liquid from a mixture of liquids

1

Heat the solution so that the liquid evaporates.

2

The vapour rises through the neck of the flask into the condenser.

3

The vapour cools and condenses into the pure liquid, which is collected in a beaker.

4

The dissolved solid is left behind in the flask.

Fermentation products

• Fermentation products simple distillation can separate alcohol and water

Fractional distillation

• Fractional distillation it is more effective when boiling points are close or higher purity is needed, as for crude oil

Section 16 of 38Chromatography method

• Use
  • it separates substances with different solubilities in a solvent, such as the dyes in black ink

1

Draw a pencil line on chromatography paper and place spots of the samples on it.

2

Lower the paper into the solvent so that the pencil line is above the solvent level.

3

The solvent rises by capillary action and carries the substances with it.

4

More soluble substances travel further.

• Why pencil
  • ink would run into the chromatogram along with the samples

Section 17 of 38Interpreting chromatograms and R_f values

Interpreting chromatograms and R_f values

Identical substances

• Identical substances they give identical chromatograms, and a known compound is often run as a reference spot

Purity

• Purity a pure substance gives one spot, and an impure substance gives more than one spot

Retention factor

• R_f = distance travelled by the substance ÷ distance travelled by the solvent

Features

• R_f is a ratio with no units, and a given compound always has the same R_f under the same conditions

Example

• Example a spot that moves 3 cm while the solvent moves 6 cm has R_f = 0.5

Section 18 of 38Rearrangement of atoms

Reactions

• Reactions atoms of the reactants rearrange to form products, which changes the properties

Same atoms

• Same atoms the number of atoms of each element is the same at the start and at the end, which is why equations must be balanced

Example

• Example sodium reacting with chlorine has the same sodium and chlorine atoms on both sides of the equation

Section 19 of 38Law of conservation of mass

1Law

• Law no matter is lost or gained in a chemical reaction, so mass is always conserved

2Consequence

• Consequence the total mass of reactants equals the total mass of products, so equations must be balanced

3Calculation

• Calculation the sum of the relative formula masses of reactants equals that of the products

4What changes

• What changes the arrangement of atoms and the properties of the substances, not the total amount of matter

Section 20 of 38Demonstrating conservation of mass

Demonstrating conservation of mass

Closed system

• Closed system a precipitation reaction in a closed flask shows constant total mass
• CaCl₂(aq) + Na₂SO₄(aq) → CaSO₄(s) + 2NaCl(aq) forms a precipitate of calcium sulfate.
• Nothing enters or leaves, so the mass before and after is the same.

Open system

• Open system a reaction that releases a gas shows an apparent mass loss
• 2HCl(aq) + CaCO₃(s) → CaCl₂(aq) + H₂O(l) + CO₂(g).
• Carbon dioxide escapes to the surroundings, so the flask mass decreases unless it is closed.

Section 21 of 38Colour change as evidence

Colour change as evidence

Ways to monitor a reaction

• Ways to monitor a reaction colour change, temperature change and effervescence, plus others

Copper and silver nitrate

• Copper and silver nitrate orange-brown copper is added to colourless silver nitrate solution
• The solid changes from orange-brown to silver.
• The solution changes from colourless to blue as copper displaces silver.

Halide ion test

• Halide ion test the solution changes from colourless to a cream, white or yellow precipitate, so a solid also forms

Hydrochloric acid with sodium thiosulfate

• Hydrochloric acid with sodium thiosulfate colourless turns to yellow
• A solid precipitate forms and a pungent smell is released.

Thermal stability of carbonates

• Thermal stability of carbonates green copper(II) carbonate forms black copper(II) oxide on decomposition

Section 22 of 38Temperature change and effervescence

Exothermic

• Exothermic heat is given out and the temperature rises
• Calcium oxide (quicklime) with water forming calcium hydroxide is highly exothermic.
• Sodium with water is exothermic and also effervesces.

Endothermic

• Endothermic heat is absorbed and the temperature falls
• Photosynthesis absorbs light energy to make glucose and oxygen from carbon dioxide and water.
• Dissolving solid ammonium chloride in water absorbs heat, and this is used in cold packs.

Effervescence

• Effervescence fizzing often accompanies reactions of acids
• Alkali metals with water release hydrogenmetal + water → metal hydroxide + hydrogen.

Other signs

• Other signs a change of state such as liquid to solid, light being produced, a smell being produced, or a change in pH

Section 23 of 38Word equations

Layout

• Layout reactants → products, with reactants on the left, products on the right, and the arrow meaning "forms" or "produces"

Conditions

• Conditions they or a catalyst name can be written above the arrow

Example

• Example sodium hydroxide + hydrochloric acid → sodium chloride + water

1

Identify the substances that react, which are the reactants.

2

Identify the substances formed, which are the products.

3

Write reactants, an arrow, then products, even if the question lists the products first.

4

Add any knowledge not given, such as "burns in air" meaning reacts with oxygen.

• Examples
  • ammonia + nitric acid → ammonium nitrate, and carbon + oxygen → carbon dioxide

Section 24 of 38Symbol equations

Symbol equation

• Symbol equation it uses formulae of the reactants and products

Diatomic molecules

• Diatomic molecules write H₂, N₂, O₂, F₂, Cl₂, Br₂ and I₂ as molecules

State symbols

• (s) solid, (l) liquid, (g) gas and (aq) aqueous, which should be known and are good practice to include

Balance

• Balance there must be the same number of atoms of each element on both sides, by the law of conservation of mass

Example

• S(s) + O₂(g) → SO₂(g) shows one sulfur atom reacting with one oxygen molecule to form one sulfur dioxide molecule

Section 25 of 38Balancing equations

1

Write the correct formulae for all reactants and products.

2

Work from left to right, balancing one element at a time.

3

Count an unchanged group such as NO₃^- as a single unit.

4

Change only the numbers in front of formulae (coefficients).

5

Balance elements that appear on their own last.

6

Check each element on both sides.

1Neutralisation

• NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)

2Redox

• 2Fe₂O₃ + 3C → 4Fe + 3CO₂

3Aluminium and copper(II) oxide

• 2Al(s) + 3CuO(s) → Al₂O₃(s) + 3Cu(s)

4Magnesium oxide and nitric acid

• MgO(s) + 2HNO₃(aq) → Mg(NO₃)₂(aq) + H₂O(l)

Section 26 of 38Checking a balanced equation

1Method

• Method count atoms or unchanged groups on each side, keeping groups such as NO₃ as one unit

2Magnesium oxide and nitric acid

• Magnesium oxide and nitric acid each side has 1 Mg, 2 H, 2 NO₃ groups and 1 O outside the group

3Never change formulae

• Never change formulae adding or removing a subscript changes the substance, so H₂O must not become H₂O₂

4Strategy

• Strategy draw the atoms out if balancing is difficult

Section 27 of 38Percentage yield

Yield

• Yield the amount of product obtained from a reaction

Actual yield = the recorded amount of product obtained

Theoretical yield = the amount expected under perfect conditions, calculated from the balanced equation

Equation

• Equation percentage yield = × 100
  • (actual yield ÷ theoretical yield)

Example

• 1.6 g obtained against a theoretical 2.0 g gives 80%

Check

• Check a value above 100% means the numbers are the wrong way round

Industry aim

• Industry aim a high percentage yield increases profit and reduces cost and waste

Section 28 of 38Why yield is less than 100%

Left-behind reactants

• Left-behind reactants some reactants remain in the equipment

Reversible reactions

• Reversible reactions products keep turning back into reactants, so a high yield is impossible

Losses in separation

• Losses in separation some product is lost during filtration, distillation or purification

Side reactions

• Side reactions a reactant may react with a gas in the air or an impurity

Transfer losses

• Transfer losses product is lost when moved between containers

Section 29 of 38Simplest (empirical) formula

Simplest formula = the whole-number ratio of atoms of each element in a compound

Organic compounds

• Organic compounds the simplest formula often differs from the molecular formula, as ethanoic acid C₂H₄O₂ has simplest formula CH₂O

Ionic compounds

• Ionic compounds the chemical formula is always the simplest formula, as for NaCl

Exam wording

• Exam wording questions usually say "find the simplest formula" rather than "empirical formula"

Section 30 of 38Calculating the simplest formula

1

Write the mass (or percentage) of each element, taking percentages as grams in 100 g.

2

If percentages do not add to 100%, calculate the missing element by difference.

3

Divide each mass by its A_r to get moles.

4

Divide all by the smallest number to get the ratio.

5

If a value is not whole, such as 1.5, multiply all by the smallest number that gives whole numbers.

Example

• 10 g H and 80 g O give 10 mol and 5 mol, so ratio 2 : 1 and H₂O

Example

• 31.58% C, 5.26% H and 63.16% O give 2.63, 5.26 and 3.95 mol, ratio 1 : 2 : 1.5, doubled to C₂H₄O₃

Caution

• Caution show working, since experimental data can give an unfamiliar answer such as Mg₂O for incomplete combustion of magnesium

Section 31 of 38Molecular formula

• Molecular formula = the actual number of atoms of each element in a molecule

Compound	Simplest formula	Molecular formula
Methane	CH4	CH4
Ethane	CH3	C2H6
Ethene	CH2	C2H4
Benzene	CH	C6H6

Molecular formula

Work out the relative formula mass of the simplest formula.

Divide the M_r of the molecular formula by that value.

Multiply every number of atoms in the simplest formula by the result.

• Example
  • C₄H₁₀S has M_r 90 and the compound has M_r 180, so the factor is 2 and the molecular formula is C₈H₂₀S₂

Section 32 of 38Calculating reacting masses

• Use
  • a balanced equation gives the moles and masses of reactants and products

Calculating reacting masses

Convert the given mass to moles using moles = mass ÷ M_r.

Use the molar ratio from the balanced equation to find the moles of the other substance.

Convert moles to mass using mass = moles × M_r.

Magnesium example

• Magnesium example in 2Mg + O₂ → 2MgO, 6.0 g Mg is 0.25 mol, the ratio is 1 : 1, so MgO is 0.25 × 40 = 10 g

Aluminium example

• Aluminium example in 2Al₂O₃ → 4Al + 3O₂, 51 tonnes (5.1 × 10⁷ g) is 500 000 mol, the ratio is 1 : 2, giving 1 000 000 mol and 27 tonnes of aluminium

Units

• 1 tonne = 10⁶ g

Section 33 of 38Balancing equations from reacting masses

• Idea
  • masses of reactants and products can be converted to moles, and the simplest mole ratio gives the coefficients

1

Calculate the M_r of each substance.

2

Convert each mass to moles using moles = mass ÷ M_r.

3

Divide by the smallest number of moles to get the simplest whole-number ratio.

4

Write the ratio numbers in front of the formulae.

Zinc example

• 1.2 g C, 16.2 g ZnO, 4.4 g CO₂ and 13.0 g Zn give 0.1, 0.2, 0.1 and 0.2 mol, ratio 1 : 2 : 1 : 2, so C + 2ZnO → CO₂ + 2Zn

Units

• Units reacting masses stay in proportion, so any mass unit works if it is used consistently

Wording

• "minimum mass" and "maximum mass" questions follow the same mole method

Section 34 of 38The mole and the Avogadro constant

Mole = the SI unit of amount of substance, containing the same number of the stated particles

Avogadro constant

• Avogadro constant one mole contains 6.02 × 10²³ particles, which can be atoms, molecules or ions

Examples

• One mole of Na contains 6.02 × 10²³ atoms.
• One mole of H₂ contains 6.02 × 10²³ molecules.
• One mole of NaCl contains 6.02 × 10²³ formula units.

Molar mass = the mass of one mole of a substance in g mol^-1

Value

• Value for an element it equals the A_r in grams, and for a compound it equals the M_r in grams

Syllabus note

• Syllabus note the value of the Avogadro constant need not be remembered, but how to use and rearrange the equation must be known

Section 35 of 38Avogadro’s law and molar gas volume

Avogadro’s law

• Avogadro’s law equal amounts of gases at the same temperature and pressure occupy the same volume

Molar gas volume

• Molar gas volume one mole of any gas occupies 24 dm³ (24 000 cm³) at room temperature and pressure

RTP conditions

• 20 °C and 1 atmosphere

Equation

• Equation moles = volume ÷ molar volume, and volume = moles × molar volume

Units

• Units use 24 for volumes in dm³ and 24 000 for volumes in cm³

Section 36 of 38Converting between moles and gas volume

Gas	Moles	Volume
Hydrogen	3	72 dm3
Carbon dioxide	0.25	6 dm3
Oxygen	5.4	129.6 dm3
Ammonia	0.02	0.48 dm3 (480 cm3)

Gas	Volume	Moles
Methane	225.6 dm3	9.4
Carbon monoxide	7.2 dm3	0.3
Sulfur dioxide	960 dm3	40
Oxygen	1200 cm3	0.05

• Formula triangle
  • cover the wanted quantity to see how to calculate it

Section 37 of 38Key terms

Term	Meaning
Element	A substance made of only one type of atom.
Compound	A substance made of two or more different atoms chemically joined.
Diatomic molecule	A molecule made of two atoms, such as H2 or Cl2.
Simple molecule	A small covalent molecule of non-metal atoms.
Relative atomic mass	The average mass of an element’s atoms compared with one twelfth of carbon-12.
Relative formula mass	The sum of the relative atomic masses of the atoms in a formula.

Term	Meaning
Mixture	Substances not chemically joined, separable by physical processes.
Filtrate	The liquid that passes through filter paper.
Residue	The insoluble solid left on filter paper.
Crystallisation	Forming crystals from a saturated solution as it cools.
Rf value	The distance moved by a substance divided by the distance moved by the solvent.
Percentage composition	The mass of an element as a percentage of the total mass of the compound.

Term	Meaning
Law of conservation of mass	No matter is created or destroyed in a chemical reaction.
Precipitate	An insoluble solid formed when two solutions react.
Effervescence	Fizzing caused by the release of a gas.
Word equation	An equation showing reactants and products by name.
State symbol	A symbol showing whether a substance is solid, liquid, gas or aqueous.
Balanced equation	An equation with the same number of each type of atom on both sides.

Term	Meaning
Yield	The amount of product obtained in a reaction.
Theoretical yield	The maximum product expected from the balanced equation.
Simplest (empirical) formula	The simplest whole-number ratio of atoms in a compound.
Molecular formula	The actual number of atoms of each element in one molecule.
Molar ratio	The ratio of moles of substances shown by the coefficients in an equation.

Term	Meaning
Mole	The SI unit of amount of substance, containing 6.02 × 1023 particles.
Avogadro constant	The number of particles in one mole, 6.02 × 1023.
Molar mass	The mass of one mole of a substance in g mol-1.
Molar gas volume	The volume of one mole of gas, 24 dm3 at RTP.
RTP	Room temperature and pressure, 20 °C and 1 atm.

Section 38 of 38Exam technique

1

Writing the formula of an ionic compound with unbalanced charges — the total positive and negative charge must cancel.

2

Forgetting brackets for repeated polyatomic ions — write (NH₄)₂SO₄, not NH₄2SO₄.

3

Treating ionic compounds as simple molecules — simple molecules form only between non-metals.

4

Missing atoms inside brackets in M_r — multiply every atom in the bracket by the subscript.

1

Percentage above 100% — the formula has been inverted, because the element mass must be on top.

2

Counting only one atom of an element — multiply the A_r by the number of atoms.

3

Putting the pencil line below the solvent — the samples would wash into the solvent.

4

Using evaporation to separate a solid that decomposes — crystallisation and filtration are used instead.

1

Saying mass is lost when a gas escapes — the mass is conserved but the gas has left the open flask.

2

Changing a subscript to balance an equation — only coefficients can change.

3

Writing oxygen as O in an equation — oxygen is the molecule O₂.

4

Putting products on the left because they are mentioned first in a question — reactants always go on the left.

1

Inverting the percentage yield — actual yield goes on top.

2

Not turning 1.5 into whole numbers — multiply all ratios to get whole numbers.

3

Giving the empirical formula as the molecular formula — compare M_r values to find the multiplier.

4

Ignoring the mole ratio — use the coefficients in the balanced equation.

1

Using 24 with a volume in cm³ — divide by 24 000 for cm³.

2

Treating the mole as a mass — a mole is an amount, and molar mass gives the mass of one mole.

3

Using the molar volume for liquids or solids — 24 dm³ applies only to gases at RTP.

4

Using unsimplified mole numbers as coefficients — divide by the smallest number first.