Free notes · Secondary 3 and 4
O-Level Chemistry notes
Free O-Level Chemistry notes on one page, for the 2026 O-Level (6092) and the 2027 SEC G3 Chemistry (K324). Full notes for 10 of the 12 syllabus chapters, each with the key idea, the facts and formulas to know, a worked exam question, the common mistakes and exam tips. The other chapters have a checklist of what the syllabus asks for, so the page maps the whole syllabus.
Free to read, print or save as a PDF, with no sign-up. The print layout keeps every chapter and drops the menus. After each chapter, practise on the O-Level Chemistry prelim papers with answers.
How to use these notes
- Pick one chapter. Read the key idea and the must-know points.
- Close the page and write down what you remember, then check and mark what you missed.
- Try the worked exam question before reading its steps, and read the common mistakes.
- Do questions on that chapter from a school paper and mark them. Redo the wrong ones a few days later.
Syllabus edition
These notes follow SEAB's O-Level Chemistry syllabus 6092, examined in 2026, and SEC G3 Chemistry K324, examined from 2027. The two documents list the same 12 topics and the same three papers. Sources: SEAB, O-Level Chemistry syllabus 6092 (2026), SEAB, SEC G3 Chemistry syllabus K324 (2027).
Exam format
| Paper | What it is | Time | Marks | Weighting |
|---|---|---|---|---|
| Paper 1 | 40 compulsory multiple-choice questions. The Periodic Table is printed in the paper. | 1 h | 40 | 30% |
| Paper 2 | Structured and free-response questions. Section A, 70 marks: compulsory structured questions. The last two carry 20 marks between them, and one is a data-based question worth 8 to 12 marks. Section B, 10 marks: answer one of two questions. The Periodic Table is printed in the paper. | 1 h 45 min | 80 | 50% |
| Paper 3 | Practical: compulsory practical questions. The Notes for Qualitative Analysis are printed in the paper. | 1 h 50 min | 40 | 20% |
- No notes or textbooks are allowed in the practical.
The whole syllabus, chapter by chapter
All 12 chapters of the syllabus in order. 10 have full notes on this page. The rest have a checklist of what the syllabus asks you to do.
| Chapter | What it covers | On this page |
|---|---|---|
| 1. Experimental Chemistry | Apparatus, collecting and drying gases, separation, chromatography and purity. | Checklist |
| 2. The Particulate Nature of Matter | Kinetic particle theory, diffusion and atomic structure. | Checklist |
| 3. Chemical Bonding and Structure | Ionic, covalent and metallic bonding, and how structure decides properties. | Full notes |
| 4. Chemical Calculations | Formulae and equations, the mole and stoichiometry. | Full notes |
| 5. Acid-Base Chemistry | Acids, bases, pH, oxides, making salts, and ammonia. | Acids and Bases, Salts and Their Preparation, plus checklist |
| 6. Qualitative Analysis | Tests for cations, anions and gases. | Full notes |
| 7. Redox Chemistry | Oxidation and reduction, electrolysis, electroplating and simple cells. | Redox Reactions, Electrolysis |
| 8. Patterns in the Periodic Table | Periodic trends, Groups 1, 17 and 18, transition elements and the reactivity series. | Checklist |
| 9. Chemical Energetics | Exothermic and endothermic reactions, energy profiles and bond energies. | Full notes |
| 10. Rate of Reactions | Factors that change the rate, collisions and catalysts. | Full notes |
| 11. Organic Chemistry | Fuels and crude oil, hydrocarbons, alcohols, carboxylic acids, esters and polymers. | Full notes |
| 12. Maintaining Air Quality | Air, pollutants and their effects, the ozone layer and greenhouse gases. | Checklist |
O-Level Chemistry at a glance
| Chapter | The one thing to remember |
|---|---|
| Chemical Bonding | Ionic: metal and non-metal, electrons transferred, strong attraction between oppositely charged ions |
| Mole Concept and Stoichiometry | Moles = mass ÷ molar mass, n = m/Mr |
| Acids and Bases | Acid + metal gives salt + hydrogen |
| Salts and Their Preparation | All sodium, potassium and ammonium salts and all nitrates are soluble |
| Qualitative Analysis | Add NaOH(aq) or NH₃(aq) dropwise, then in excess, and record both results |
| Redox Reactions | OIL RIG: oxidation is loss of electrons, reduction is gain of electrons |
| Electrolysis | The anode is positive and oxidation happens there, the cathode is negative and reduction happens there |
| Energy Changes | Exothermic: ΔH negative, temperature of the surroundings rises |
| Rate of Reaction | Average rate = change in amount ÷ time taken |
| Organic Chemistry | Alkanes CₙH₂ₙ₊₂, alkenes CₙH₂ₙ, alcohols CₙH₂ₙ₊₁OH |
Notes 1 of 10
Chemical Bonding
Key idea. Chemical bonding is the attraction that holds atoms or ions together. Metals transfer electrons to non-metals to form ions held by ionic bonds, non-metals share pairs of electrons in covalent bonds, and metals are held by metallic bonding between positive ions and delocalised electrons. The type of bonding and structure decides melting point, conductivity and hardness.
Must know
- Ionic: metal and non-metal, electrons transferred, strong attraction between oppositely charged ions
- Covalent: non-metals share pairs of electrons, and a double bond is two shared pairs
- Metallic: a lattice of positive ions in a sea of delocalised electrons
- Melting a simple molecular substance overcomes weak intermolecular forces, not covalent bonds
What the syllabus asks you to do
- Describe the formation of ions by electron loss or gain, giving the electronic configuration of a noble gas
- Describe, with dot-and-cross diagrams, ionic bonding between metals and non-metals, such as NaCl and MgCl₂
- Describe covalent bonding as the sharing of a pair of electrons, with dot-and-cross diagrams for molecules such as H₂, O₂, H₂O, CH₄ and CO₂
- Describe metals as a lattice of positive ions in a sea of electrons and relate this to their physical properties
- Explain why alloys have different physical properties to their constituent elements
- Compare simple molecular substances with giant covalent substances such as diamond, graphite and silicon dioxide to deduce their properties
- Relate the physical properties, including electrical conductivity, of ionic and covalent substances to their structure and bonding
Structure and properties compared
| Structure | Melting point and why | Conducts electricity? |
|---|---|---|
| Giant ionic, such as NaCl | High, strong attraction between many oppositely charged ions | Only when molten or aqueous, when the ions can move |
| Simple molecular, such as H₂O and CO₂ | Low, only weak intermolecular forces are overcome | No, there are no mobile ions or electrons |
| Giant covalent, such as diamond and SiO₂ | Very high, many strong covalent bonds must break | No, all outer electrons are used in bonding |
| Graphite | Very high, strong covalent bonds within the layers | Yes, delocalised electrons move along the layers |
| Metallic, such as copper | Usually high, strong attraction between ions and delocalised electrons | Yes, as a solid and as a liquid |
Worked exam question
Magnesium reacts with chlorine to form magnesium chloride. Describe the dot-and-cross diagram for magnesium chloride and explain why it has a high melting point and conducts electricity only when molten or in aqueous solution.
- Write the electronic configurations: Mg is 2.8.2 and Cl is 2.8.7.
- Magnesium loses its two outer electrons, one to each of two chlorine atoms, so the formula is MgCl₂.
- Draw Mg²⁺ in square brackets with the 2.8 configuration and a 2+ charge at the top right.
- Draw two Cl⁻ ions, each in square brackets with eight outer electrons, seven dots of its own and one cross from magnesium, and a minus charge.
- High melting point: MgCl₂ is a giant ionic lattice, and a lot of energy is needed to overcome the strong electrostatic attraction between the Mg²⁺ and Cl⁻ ions.
- Conductivity: in the solid the ions are held in fixed positions, but when molten or dissolved the ions are free to move and carry charge.
Answer: MgCl₂ contains one Mg²⁺ ion and two Cl⁻ ions in a giant ionic lattice. It melts at a high temperature because of the strong attraction between oppositely charged ions, and it conducts only when molten or aqueous because only then are the ions mobile.
Common mistakes
- Writing that molten sodium chloride conducts because electrons move.For an ionic compound, always name the ions as the moving charged particles.
- Writing that boiling water breaks the O-H bonds.For simple molecular substances, say that weak intermolecular forces are overcome and the molecules stay intact.
- Drawing water with only the two shared pairs around oxygen.Count outer electrons after drawing. Oxygen should have eight: two bond pairs and two lone pairs.
- Drawing a shared pair of electrons between Na and Cl in sodium chloride.For a metal with a non-metal, draw separate ions in square brackets with charges and no shared electrons.
- Saying graphite conducts because it has mobile ions.Graphite contains only carbon atoms, so its moving charges are delocalised electrons along the layers.
- Describing metallic bonding as attraction between metal atoms.Say the attraction is between positive metal ions and the sea of delocalised electrons.
Exam tips
- In structure and properties answers, always name the structure (giant ionic, simple molecular, giant covalent or metallic), the particles and the force being overcome. Marks are usually given for each.
- Count the outer electrons in every dot-and-cross diagram before moving on. Each atom in a covalent molecule should end with a full outer shell, two for hydrogen.
- For conductivity, always name the mobile charged particle: ions for ionic compounds, delocalised electrons for metals and graphite.
Notes 2 of 10
Mole Concept and Stoichiometry
Key idea. The mole concept counts particles by mass. One mole contains 6.02 × 10²³ particles and has a mass equal to the Mr in grams. To solve a calculation, change the given quantity into moles, use the mole ratio from the balanced equation, then change moles into the mass, gas volume or concentration the question asks for.
Must know
- Moles = mass ÷ molar mass, n = m/Mr
- Gas volume at r.t.p. = moles × 24 dm³
- Moles in solution = concentration in mol/dm³ × volume in dm³
- Percentage yield = actual yield ÷ theoretical yield × 100%
What the syllabus asks you to do
- Calculate relative molecular mass and relative formula mass as the sum of relative atomic masses
- Define the mole in terms of the Avogadro constant
- Calculate empirical and molecular formulae from relevant data
- Calculate reacting masses and volumes of gases, with one mole of gas occupying 24 dm³ at room temperature and pressure, including limiting reactant problems
- Apply solution concentration in mol/dm³ or g/dm³ to titration results and simple problems
- Calculate percentage yield and percentage purity
Mole formula sheet for O-Level Chemistry
| Quantity | Formula | Unit |
|---|---|---|
| Moles from mass | n = mass ÷ Mr | mol, with mass in g |
| Moles of gas at r.t.p. | n = volume ÷ 24 | mol, with volume in dm³ |
| Moles in solution | n = concentration × volume | mol, with volume in dm³ |
| Mass concentration | g/dm³ = mol/dm³ × Mr | g/dm³ |
| Percentage yield | actual ÷ theoretical × 100 | % |
| Percentage purity | mass of pure substance ÷ mass of sample × 100 | % |
Start with what is given
5.00 g of CaCO₃ and 50.0 cm³ of 1.00 mol/dm³ HCl
Convert each to moles
5.00 ÷ 100 = 0.0500 mol CaCO₃, 1.00 × 0.0500 = 0.0500 mol HCl
Find the limiting reactant
CaCO₃ needs 0.100 mol HCl, only 0.0500 mol is present, so HCl limits
Apply the mole ratio
2HCl gives 1CO₂, so 0.0500 ÷ 2 = 0.0250 mol CO₂
Convert to the answer
0.0250 × 24 = 0.600 dm³ of CO₂ at r.t.p.
Worked exam question
5.00 g of calcium carbonate is added to 50.0 cm³ of 1.00 mol/dm³ hydrochloric acid. CaCO₃ + 2HCl gives CaCl₂ + H₂O + CO₂. Which reactant is limiting, and what volume of carbon dioxide forms at r.t.p.? (Ar: Ca = 40, C = 12, O = 16)
- Mr of CaCO₃ = 40 + 12 + 3 × 16 = 100, so moles of CaCO₃ = 5.00 ÷ 100 = 0.0500 mol.
- Moles of HCl = 1.00 × 50.0/1000 = 0.0500 mol.
- The equation needs 2 mol of HCl for every 1 mol of CaCO₃, so 0.0500 mol of CaCO₃ would need 0.100 mol of HCl.
- Only 0.0500 mol of HCl is present, so the hydrochloric acid is the limiting reactant and some calcium carbonate is left over.
- From the ratio 2HCl to 1CO₂, moles of CO₂ = 0.0500 ÷ 2 = 0.0250 mol.
- Volume of CO₂ = 0.0250 × 24 = 0.600 dm³.
Answer: Hydrochloric acid is limiting, and 0.600 dm³ (600 cm³) of carbon dioxide forms at r.t.p.
Common mistakes
- Calculating moles in 25.0 cm³ of 0.100 mol/dm³ NaOH as 0.100 × 25.0 = 2.50 mol.Divide cm³ by 1000 first, so 25.0 cm³ becomes 0.0250 dm³ and the answer is 0.00250 mol.
- Taking moles of CO₂ as equal to moles of HCl in CaCO₃ + 2HCl.Write the ratio in words, such as 2 mol HCl gives 1 mol CO₂, before using it.
- Using 16 as the Mr of oxygen gas or 35.5 for chlorine gas.Write the formula of every gas first, then add the Ar values: O₂ is 32 and Cl₂ is 71.
- Converting 18 cm³ of liquid water to moles by dividing by 24 000.Use 24 dm³ only for gases at r.t.p. For liquids and solids, use mass ÷ Mr.
- Rounding a ratio of 1 : 1.5 to 1 : 2 for an empirical formula.If a ratio ends in .5 or .33, multiply every number by 2 or 3, so 1 : 1.5 becomes 2 : 3.
- Choosing the reactant with the smaller mass in grams as the limiting reactant.Convert both amounts to moles and compare them with the ratio in the balanced equation.
Exam tips
- Write a short label for every line, such as 'moles of HCl' or 'mole ratio 2 : 1', so that a marker can award method marks even if a later number slips.
- Keep at least four significant figures in your working and round only the final answer, usually to three significant figures.
- When two amounts are given in the question, expect a limiting reactant step and do it before calculating any product.
- Check the units the question asks for. Convert 0.600 dm³ to 600 cm³ or mol/dm³ to g/dm³ only at the end.
Notes 3 of 10
Acids and Bases
Key idea. An acid produces hydrogen ions, H⁺, in water, and an alkali is a soluble base that produces hydroxide ions, OH⁻. Acids react with metals to give a salt and hydrogen, with bases to give a salt and water, and with carbonates to give a salt, water and carbon dioxide. The lower the pH, the higher the concentration of H⁺ ions.
Must know
- Acid + metal gives salt + hydrogen
- Acid + base gives salt + water, with ionic equation H⁺ + OH⁻ gives H₂O
- Acid + carbonate gives salt + water + carbon dioxide
- Strong acids ionise completely in water, weak acids such as ethanoic acid only partly
What the syllabus asks you to do
- Describe acids and alkalis in terms of the ions they produce in aqueous solution and their effect on Universal Indicator
- Describe neutrality, acidity and alkalinity in terms of H⁺ and OH⁻ concentrations, Universal Indicator colour and the pH scale
- Describe qualitatively the difference between strong and weak acids in terms of the extent of ionisation
- Describe the reactions of acids with metals, bases and carbonates to form salts
- Describe the reaction of hydrogen ions with hydroxide ions to produce water as neutralisation
- Describe the reactions of bases with acids and with ammonium salts
- Describe why soil pH is controlled and how excess acidity is treated with calcium hydroxide
- Classify oxides as acidic, basic, amphoteric or neutral based on metallic or non-metallic character
Strong and weak acids of the same concentration and volume compared
| Property | Hydrochloric acid (strong) | Ethanoic acid (weak) |
|---|---|---|
| Ionisation in water | Complete | Partial, most molecules stay un-ionised |
| Concentration of H⁺ ions | Higher | Lower |
| pH | Lower | Higher |
| Rate of reaction with magnesium | Faster | Slower |
| Total hydrogen with excess magnesium | Same as ethanoic acid | Same as hydrochloric acid |
A line graph of pH against volume of sodium hydroxide added, from 0 to 50 cm³. The pH starts at 1, rises slowly to about 3 by 24.5 cm³, then jumps steeply through pH 7 at 25.0 cm³, where the acid is exactly neutralised. With excess alkali it levels off near pH 12.5.
Worked exam question
0.24 g of magnesium ribbon is added to excess dilute hydrochloric acid. Write the balanced equation and the ionic equation, and calculate the volume of hydrogen produced at room temperature and pressure. (Ar: Mg = 24, molar gas volume = 24 dm³ at r.t.p.)
- Balanced equation: Mg + 2HCl gives MgCl₂ + H₂.
- Cl⁻ ions appear unchanged on both sides, so remove them: Mg + 2H⁺ gives Mg²⁺ + H₂.
- Moles of Mg = 0.24 ÷ 24 = 0.010 mol.
- The ratio of Mg to H₂ is 1 : 1, so 0.010 mol of H₂ forms, because the acid is in excess and magnesium is limiting.
- Volume of H₂ = 0.010 × 24 = 0.24 dm³ = 240 cm³.
Answer: Mg + 2HCl gives MgCl₂ + H₂, ionic equation Mg + 2H⁺ gives Mg²⁺ + H₂, and 240 cm³ (0.24 dm³) of hydrogen forms at r.t.p.
Common mistakes
- Writing that a concentrated ethanoic acid solution is a strong acid.Use strong or weak for ionisation, and concentrated or dilute for moles per dm³.
- Calling copper(II) oxide an alkali.Check solubility. An alkali is a base that dissolves to give OH⁻ ions, and copper(II) oxide is insoluble.
- Saying a solution of pH 2 has fewer hydrogen ions than one of pH 5.Remember that the lower the pH, the higher the H⁺ concentration.
- Classifying aluminium oxide as a basic oxide.Learn Al₂O₃ and ZnO as amphoteric, since they react with both acids and alkalis.
- Writing Mg + 2HCl gives MgCl₂ + H, or leaving Cl⁻ in the ionic equation.Write H₂ for hydrogen gas and remove ions that appear unchanged on both sides.
- Predicting that copper fizzes in dilute hydrochloric acid.Check the reactivity series. Metals below hydrogen, such as copper, do not react with dilute acids.
Exam tips
- Learn the three acid reactions as word patterns (metal gives salt and hydrogen, base gives salt and water, carbonate gives salt, water and carbon dioxide). Then any new equation is a matter of filling in names and balancing.
- When asked to compare a strong and a weak acid, mention the degree of ionisation and the concentration of H⁺ ions in the same answer.
- For oxide classification questions, test the oxide against both an acid and an alkali in your answer rather than relying on whether the element is a metal.
Notes 4 of 10
Salts and Their Preparation
Key idea. A salt forms when the hydrogen ions of an acid are replaced by metal or ammonium ions. The preparation method depends on solubility: react an acid with an excess of an insoluble metal, base or carbonate, titrate an acid with an alkali for sodium, potassium and ammonium salts, or mix two solutions to precipitate an insoluble salt.
Must know
- All sodium, potassium and ammonium salts and all nitrates are soluble
- Insoluble chlorides: AgCl and PbCl₂. Insoluble sulfates: BaSO₄ and PbSO₄, with CaSO₄ slightly soluble
- Soluble salt from an insoluble solid: add excess, filter, crystallise
- Insoluble salt: precipitate, filter, wash the residue, dry
What the syllabus asks you to do
- Describe the preparation, separation and purification of salts by reacting acids with metals, insoluble bases and insoluble carbonates
- Describe the preparation of salts by titration
- Describe the preparation of insoluble salts by precipitation
- Describe the general solubility rules for nitrates, chlorides, sulfates, carbonates, hydroxides, Group 1 salts and ammonium salts
- Suggest a method of preparing a given salt from suitable starting materials
- Construct chemical equations with state symbols, including ionic equations
Solubility rules
| Compounds | Soluble | Insoluble or exceptions |
|---|---|---|
| Sodium, potassium and ammonium salts | All soluble | None |
| Nitrates | All soluble | None |
| Chlorides | Most | Silver chloride and lead(II) chloride |
| Sulfates | Most | Barium sulfate and lead(II) sulfate, calcium sulfate slightly soluble |
| Carbonates | Sodium, potassium and ammonium carbonates | All others |
| Oxides and hydroxides | Sodium and potassium hydroxides, calcium hydroxide slightly | Most other metal oxides and hydroxides |
Add excess zinc oxide
Warm the acid and add ZnO until some stays undissolved
Filter
Remove the unreacted zinc oxide and keep the filtrate
Heat to saturation
Evaporate some water until crystals form on a cooled glass rod
Cool to crystallise
Zinc sulfate crystals form as the saturated solution cools
Filter, wash and dry
Wash with a little cold distilled water, then dry between filter papers
Worked exam question
Describe how to prepare pure, dry crystals of zinc sulfate starting from zinc oxide and 50.0 cm³ of 1.00 mol/dm³ sulfuric acid. Show that 5.0 g of zinc oxide is enough to be in excess. ZnO + H₂SO₄ gives ZnSO₄ + H₂O. (Ar: Zn = 65, O = 16)
- Choose the method: zinc sulfate is soluble and zinc oxide is an insoluble base, so react the acid with excess zinc oxide.
- Check the excess: moles of H₂SO₄ = 1.00 × 50.0/1000 = 0.0500 mol, and the ratio is 1 : 1, so 0.0500 mol of ZnO is needed.
- Mr of ZnO = 65 + 16 = 81, so 0.0500 × 81 = 4.05 g reacts, and 5.0 g is therefore in excess.
- Warm the acid, add the zinc oxide a little at a time with stirring until some solid remains undissolved.
- Filter off the excess zinc oxide and collect the filtrate, which is aqueous zinc sulfate.
- Heat the filtrate until it is saturated, then leave it to cool so that crystals form.
- Filter off the crystals, wash them with a little cold distilled water and dry them between filter papers.
Answer: React the warm acid with excess zinc oxide (4.05 g is needed, so 5.0 g is an excess), filter, heat the filtrate to saturation, cool to crystallise, then filter, wash with a little cold distilled water and dry.
Common mistakes
- Adding excess sodium carbonate to an acid and filtering to make a sodium salt.Use titration for sodium, potassium and ammonium salts.
- Evaporating the filtrate to dryness to collect the salt.Heat only to saturation, then cool, so crystals keep their water of crystallisation and impurities stay in solution.
- Planning to make copper(II) sulfate from copper metal and dilute sulfuric acid.Use copper(II) oxide or copper(II) carbonate, since copper is below hydrogen in the reactivity series.
- Keeping the filtrate after precipitating barium sulfate.Ask whether the product is soluble. If it is insoluble, keep the residue.
- Rinsing zinc sulfate crystals under a running tap.Wash with a small amount of cold distilled water only.
- Choosing barium carbonate and sodium sulfate to precipitate barium sulfate.Pick two soluble starting compounds, such as barium chloride or nitrate with sodium sulfate.
Exam tips
- Start every preparation answer by stating whether the salt is soluble, and whether it is a sodium, potassium or ammonium salt. That one sentence decides the method and earns credit.
- Give a reason for each key step, such as 'add excess so that all the acid is used up' or 'wash to remove soluble impurities'. Planning questions reward the reasons as well as the steps.
- In precipitation questions, write the ionic equation with only the two ions that form the solid and add the state symbol (s) to the salt if asked for state symbols.
Notes 5 of 10
Qualitative Analysis
Key idea. Qualitative analysis identifies the ions in a salt and the gases released in a reaction using standard tests. Cations are identified by the colour of the precipitate formed with aqueous sodium hydroxide and aqueous ammonia, and whether it dissolves in excess. Anions are identified with acid, silver nitrate, barium nitrate or aluminium foil, and gases with litmus, limewater and splints.
Must know
- Add NaOH(aq) or NH₃(aq) dropwise, then in excess, and record both results
- Acidify with dilute nitric acid before the chloride, iodide and sulfate tests
- A lighted splint pops for hydrogen, a glowing splint relights for oxygen
- Ammonia turns damp red litmus paper blue
What the syllabus asks you to do
- Use aqueous sodium hydroxide and aqueous ammonia to identify Al³⁺, NH₄⁺, Ca²⁺, Cu²⁺, Fe²⁺, Fe³⁺ and Zn²⁺ from the precipitates formed and their solubility in excess
- Describe tests to identify carbonate, chloride, iodide, nitrate and sulfate ions
- Describe tests to identify ammonia, carbon dioxide, chlorine, hydrogen, oxygen and sulfur dioxide
- Construct ionic equations, including those for the formation of precipitates
Cation tests with aqueous sodium hydroxide and aqueous ammonia
| Cation | With aqueous sodium hydroxide | With aqueous ammonia |
|---|---|---|
| Al³⁺ | White precipitate, soluble in excess giving a colourless solution | White precipitate, insoluble in excess |
| Zn²⁺ | White precipitate, soluble in excess giving a colourless solution | White precipitate, soluble in excess giving a colourless solution |
| Ca²⁺ | White precipitate, insoluble in excess | No precipitate |
| Cu²⁺ | Light blue precipitate, insoluble in excess | Light blue precipitate, soluble in excess giving a dark blue solution |
| Fe²⁺ | Green precipitate, insoluble in excess | Green precipitate, insoluble in excess |
| Fe³⁺ | Red-brown precipitate, insoluble in excess | Red-brown precipitate, insoluble in excess |
| NH₄⁺ | No precipitate, ammonia given off on warming turns damp red litmus blue | Not used for this ion |
Add NaOH(aq) dropwise
A green precipitate forms, pointing to Fe²⁺
Add NaOH(aq) in excess
The precipitate is insoluble in excess, as expected for Fe²⁺
Confirm with NH₃(aq)
Green precipitate, insoluble in excess, gives the same result
Acidify with dilute HNO₃
Removes carbonate ions that would also give a white precipitate
Add Ba(NO₃)₂(aq)
White precipitate of barium sulfate shows SO₄²⁻, so the salt is FeSO₄
Worked exam question
Solution X gives a green precipitate with aqueous sodium hydroxide that is insoluble in excess. A separate sample is acidified with dilute nitric acid, and aqueous barium nitrate is added, giving a white precipitate. Identify X and write ionic equations for both precipitates.
- A green precipitate insoluble in excess sodium hydroxide is the standard result for Fe²⁺.
- Ionic equation: Fe²⁺ + 2OH⁻ gives Fe(OH)₂.
- The nitric acid removes any carbonate ions, so a white precipitate with barium nitrate must be barium sulfate, showing SO₄²⁻.
- Ionic equation: Ba²⁺ + SO₄²⁻ gives BaSO₄.
- Combine the cation and anion to name the salt.
Answer: X is iron(II) sulfate, FeSO₄. Fe²⁺ + 2OH⁻ gives Fe(OH)₂ and Ba²⁺ + SO₄²⁻ gives BaSO₄.
Common mistakes
- Recording 'white solution' or 'cloudy' for a precipitate, or 'clear' for a colourless solution.Write 'white precipitate' and 'colourless solution', since a clear solution can still be coloured.
- Adding only a few drops of sodium hydroxide and stopping there.Add the reagent dropwise and then in excess, and record both results.
- Acidifying with hydrochloric acid before adding silver nitrate.Use dilute nitric acid, because hydrochloric acid adds Cl⁻ ions and always gives a white precipitate.
- Holding dry or blue litmus paper over the tube to test for ammonia.Use damp red litmus paper, which turns blue.
- Testing for hydrogen with a glowing splint.A lighted splint pops for hydrogen, and a glowing splint relights for oxygen, the gas that helps things burn.
- Recording Cu²⁺ with excess ammonia as 'blue precipitate remains'.Remember that the light blue precipitate dissolves in excess ammonia to give a dark blue solution.
Exam tips
- Learn the cation results as a two-column table (sodium hydroxide and aqueous ammonia) for the seven cations, and practise writing it out from memory until every colour and excess result is exact.
- In practical answers, record what you see before you conclude. Observations and conclusions are often marked separately.
- Name the acid you use to acidify. Dilute nitric acid is the safe choice for the chloride, iodide and sulfate tests.
- 6092 states that the Notes for Qualitative Analysis are printed in the practical paper, so practise reading results from them quickly. Still learn the results, since the syllabus does not say the notes appear in other papers.
Notes 6 of 10
Redox Reactions
Key idea. Redox reactions are reactions in which oxidation and reduction happen together. Oxidation is the gain of oxygen, loss of hydrogen, loss of electrons or an increase in oxidation state, and reduction is the reverse. To analyse a reaction, assign oxidation states to every element on both sides and see which increase and which decrease.
Must know
- OIL RIG: oxidation is loss of electrons, reduction is gain of electrons
- Oxidation states add up to 0 in a compound and to the charge in an ion
- The oxidising agent is reduced, and the reducing agent is oxidised
- Acidified KMnO₄ goes purple to colourless with a reducing agent, KI goes colourless to brown with an oxidising agent
What the syllabus asks you to do
- Define oxidation and reduction in terms of oxygen and hydrogen gain or loss
- Define redox in terms of electron transfer and changes in oxidation state
- Identify redox reactions in terms of oxygen or hydrogen gain or loss, electron gain or loss and changes in oxidation state
- Describe the use of aqueous potassium iodide and acidified potassium manganate(VII) to test for oxidising and reducing agents from their colour changes
Common oxidising and reducing agents and what they become
| Substance | Acts as | What it becomes |
|---|---|---|
| Acidified potassium manganate(VII), MnO₄⁻ | Oxidising agent | Mn²⁺, so purple turns colourless |
| Chlorine, Cl₂ | Oxidising agent | Chloride ions, Cl⁻ |
| Oxygen, O₂ | Oxidising agent | Oxide ions, O²⁻, in metal oxides |
| Aqueous potassium iodide, I⁻ | Reducing agent | Iodine, I₂, so colourless turns brown |
| Iron(II) ions, Fe²⁺ | Reducing agent | Iron(III) ions, Fe³⁺ |
| Sulfur dioxide, SO₂ | Reducing agent | Sulfate ions, SO₄²⁻ |
| Reactive metals such as zinc and magnesium | Reducing agents | Metal ions such as Zn²⁺ and Mg²⁺ |
Write the balanced equation
Cl₂ + 2KI gives 2KCl + I₂
Assign oxidation states
Cl goes from 0 to −1, I from −1 to 0, K stays +1
Find the increase
Iodide rises from −1 to 0, so iodide is oxidised
Find the decrease
Chlorine falls from 0 to −1, so chlorine is reduced
Name the agents
Chlorine is the oxidising agent, iodide is the reducing agent
Worked exam question
Chlorine gas is bubbled into aqueous potassium iodide. Cl₂ + 2KI gives 2KCl + I₂. State what you would observe, use oxidation states to show which species is oxidised and which is reduced, name the oxidising agent and write the two half-equations.
- Observation: iodine is produced, so the colourless solution turns brown.
- Chlorine changes from 0 in Cl₂ to −1 in KCl, a decrease, so chlorine is reduced.
- Iodine changes from −1 in KI to 0 in I₂, an increase, so iodide is oxidised.
- Chlorine causes the iodide to be oxidised and is itself reduced, so chlorine is the oxidising agent and iodide is the reducing agent.
- Reduction half-equation: Cl₂ + 2e⁻ gives 2Cl⁻.
- Oxidation half-equation: 2I⁻ gives I₂ + 2e⁻. Potassium ions keep an oxidation state of +1 and are spectators.
Answer: The solution turns from colourless to brown. Chlorine (0 to −1) is reduced and is the oxidising agent, iodide (−1 to 0) is oxidised. Half-equations: Cl₂ + 2e⁻ gives 2Cl⁻ and 2I⁻ gives I₂ + 2e⁻.
Common mistakes
- Writing that chlorine, the oxidising agent, is oxidised when it reacts with iodide.Remember that the oxidising agent takes electrons, so it is itself reduced.
- Working out sulfur in SO₄²⁻ as +8 by setting the total to zero.For any ion, set the sum equal to its charge: x + 4(−2) = −2, so sulfur is +6.
- Saying sulfate is reduced when zinc reacts with copper(II) sulfate.Check each species' oxidation state. Sulfate stays the same, so it is a spectator ion.
- Writing Zn + 2e⁻ gives Zn²⁺ for the oxidation of zinc.Balance charge on both sides. In an oxidation the electrons are products: Zn gives Zn²⁺ + 2e⁻.
- Describing the potassium iodide test result as colourless to purple.Iodine formed in aqueous potassium iodide makes the solution brown.
- Writing the oxidation state of iron in Fe³⁺ as 3+.Write oxidation states with the sign first, so iron in Fe³⁺ is +3.
Exam tips
- When asked whether a reaction is redox, write the oxidation state above each element on both sides. If any element changes, it is redox, and you have your evidence written down.
- When naming the oxidising or reducing agent, name the whole species (for example 'chlorine' or 'Cu²⁺ ions'), not just 'it'.
- For the manganate(VII) and iodide tests, always give both the starting colour and the final colour, such as 'purple to colourless'.
Notes 7 of 10
Electrolysis
Key idea. Electrolysis is the decomposition of a molten or aqueous ionic compound by an electric current. Cations move to the negative cathode and are reduced, while anions move to the positive anode and are oxidised. In aqueous solutions, hydrogen forms at the cathode unless the metal is less reactive than hydrogen, and oxygen forms at the anode unless a halide is concentrated.
Must know
- The anode is positive and oxidation happens there, the cathode is negative and reduction happens there
- Cathode: a metal less reactive than hydrogen is deposited, otherwise hydrogen forms
- Anode: a concentrated halide gives the halogen, otherwise OH⁻ gives oxygen
- A copper anode dissolves instead of releasing oxygen
What the syllabus asks you to do
- Describe electrolysis as the conduction of electricity through a molten or aqueous ionic compound, leading to chemical changes at the electrodes
- Describe the electrolysis of molten sodium chloride and predict the products for a molten binary ionic compound, using inert electrodes
- Apply selective discharge of cations by reactivity and of halide, hydroxide and sulfate anions, including concentration effects, using inert electrodes
- Construct ionic equations for the reactions at the electrodes
- Describe the electrolysis of aqueous copper(II) sulfate with copper electrodes as a means of purifying copper
- Describe the electroplating of metals, such as copper plating, and state one use
- Describe the production of electrical energy from simple cells, linked to the reactivity series and electron transfer
Products of electrolysis
| Electrolyte (electrodes) | Cathode product | Anode product |
|---|---|---|
| Molten lead(II) bromide (inert) | Lead | Bromine |
| Molten sodium chloride (inert) | Sodium | Chlorine |
| Dilute sulfuric acid (inert) | Hydrogen | Oxygen |
| Concentrated aqueous sodium chloride (inert) | Hydrogen | Chlorine |
| Aqueous copper(II) sulfate (inert) | Copper | Oxygen |
| Aqueous copper(II) sulfate (copper) | Copper | No gas, the copper anode dissolves as Cu²⁺ |
List all ions present
Na⁺ and Cl⁻ from the salt, H⁺ and OH⁻ from water
Decide the cathode product
Sodium is more reactive than hydrogen, so hydrogen gas forms
Decide the anode product
Chloride is concentrated, so chlorine forms instead of oxygen
Write the half-equations
2H⁺ + 2e⁻ gives H₂ and 2Cl⁻ gives Cl₂ + 2e⁻
Check what is left
Na⁺ and OH⁻ remain, so the solution becomes sodium hydroxide
Worked exam question
Concentrated aqueous sodium chloride is electrolysed using inert electrodes. Name the ions present, state the product at each electrode with a half-equation and a test, and explain why the solution becomes alkaline.
- Ions present: Na⁺ and Cl⁻ from the salt, and H⁺ and OH⁻ from water.
- Cathode: sodium is more reactive than hydrogen, so H⁺ ions are discharged, 2H⁺ + 2e⁻ gives H₂.
- Test for hydrogen: it extinguishes a lighted splint with a 'pop'.
- Anode: chloride ions are at high concentration, so they are discharged in preference to OH⁻, 2Cl⁻ gives Cl₂ + 2e⁻.
- Test for chlorine: it bleaches damp litmus paper.
- H⁺ and Cl⁻ ions are removed, leaving Na⁺ and OH⁻ ions in solution, which is aqueous sodium hydroxide.
Answer: Hydrogen forms at the cathode (2H⁺ + 2e⁻ gives H₂) and chlorine at the anode (2Cl⁻ gives Cl₂ + 2e⁻). The remaining Na⁺ and OH⁻ ions make the solution sodium hydroxide, so it becomes alkaline.
Common mistakes
- Writing that electrons flow through the electrolyte from the anode to the cathode.Say that electrons flow in the external wires, while ions carry the charge through the electrolyte.
- Predicting sodium at the cathode for aqueous sodium chloride.For aqueous solutions, compare the metal with hydrogen. Sodium is more reactive, so hydrogen forms.
- Predicting sulfur dioxide at the anode for copper(II) sulfate solution.Sulfate and nitrate are not discharged, so OH⁻ ions give oxygen at an inert anode.
- Predicting oxygen at a copper anode in copper(II) sulfate solution.Check the electrode material first. A copper anode dissolves as Cu²⁺ ions.
- Writing Cl⁻ gives Cl₂ + e⁻ at the anode.Balance atoms, then charge: 2Cl⁻ gives Cl₂ + 2e⁻.
- Calling the cathode positive in electrolysis.In electrolysis the cathode is negative and reduction happens there. Cations move to the cathode.
Exam tips
- For every electrolysis question, first list all the ions present, including H⁺ and OH⁻ from water for aqueous solutions. Then decide at each electrode in turn.
- Check whether the electrodes are inert or copper before predicting the anode product. That single word changes the answer.
- When asked about the electrolyte afterwards, look at which ions are left behind. That tells you whether it becomes acidic, alkaline or stays the same.
Notes 8 of 10
Energy Changes
Key idea. Energy changes describe the heat given out or taken in by a reaction. Exothermic reactions release heat, so the surroundings get warmer and ΔH is negative. Endothermic reactions absorb heat, so the surroundings cool and ΔH is positive. Bond breaking absorbs energy and bond making releases it, so ΔH is negative when more energy is released making bonds than is absorbed breaking them.
Must know
- Exothermic: ΔH negative, temperature of the surroundings rises
- Endothermic: ΔH positive, temperature of the surroundings falls
- Bond breaking is endothermic and bond making is exothermic, and their balance sets the sign of ΔH
- A catalyst lowers the activation energy but does not change ΔH
What the syllabus asks you to do
- Describe the meaning of enthalpy change in terms of exothermic (ΔH negative) and endothermic (ΔH positive) reactions
- Represent energy changes by energy profile diagrams, including enthalpy changes and activation energies
- Describe bond breaking as an endothermic process and bond making as an exothermic process
- Explain qualitatively overall enthalpy changes in terms of the energy changes when covalent bonds are broken and made
- Explain how a pathway with a lower activation energy, provided by a catalyst, increases the rate of reaction
Exothermic and endothermic reactions compared
| Feature | Exothermic | Endothermic |
|---|---|---|
| Heat energy | Given out to the surroundings | Taken in from the surroundings |
| Temperature of surroundings | Rises | Falls |
| Sign of ΔH | Negative | Positive |
| Products on an energy profile | Lower than the reactants | Higher than the reactants |
| Bonds broken and made | More energy released forming bonds than absorbed breaking them | Less energy released forming bonds than absorbed breaking them |
| Examples | Combustion, neutralisation, respiration | Thermal decomposition, photosynthesis, dissolving ammonium nitrate |
A line graph of energy against progress of reaction. The reactants start at 0, the curve rises over a hump representing the activation energy, then falls to the products at −184 kJ/mol. A dashed curve with a lower hump shows the catalysed pathway. It starts and ends at the same levels, so ΔH is unchanged.
Worked exam question
An energy profile diagram for H₂ + Cl₂ gives 2HCl shows the reactants at 0 kJ/mol and the products at −184 kJ/mol, relative to the reactants. (a) Is the reaction exothermic or endothermic? (b) State ΔH. (c) Explain the sign of ΔH in terms of the bonds broken and made. (d) State how the diagram changes if a catalyst is used.
- The products are below the reactants, so the chemicals lose energy to the surroundings and the reaction is exothermic.
- ΔH is the difference between the product level and the reactant level: −184 − 0 = −184 kJ/mol.
- Bonds broken: one H-H bond and one Cl-Cl bond, which absorbs energy.
- Bonds made: two H-Cl bonds, which releases energy.
- ΔH is negative, so more energy is released making the two H-Cl bonds than is absorbed breaking the H-H and Cl-Cl bonds.
- A catalyst gives a lower hump, so a lower activation energy, but the reactant and product levels, and therefore ΔH, stay the same.
Answer: (a) Exothermic. (b) ΔH = −184 kJ/mol. (c) More energy is released making two H-Cl bonds than is absorbed breaking one H-H and one Cl-Cl bond. (d) The hump is lower, but ΔH is unchanged.
Common mistakes
- Explaining an exothermic reaction only as 'bonds are made'.Compare the two: more energy is released making bonds than is absorbed breaking them.
- Saying that burning methane needs no energy to start because ΔH is negative.Mention the activation energy, which is why fuels need a spark or flame.
- Writing that the temperature falls during an exothermic reaction.Exothermic means heat passes to the surroundings, so the thermometer reading rises.
- Writing that breaking bonds releases energy.Learn the pair: breaking bonds absorbs energy, making bonds releases energy.
- Drawing a catalysed energy profile with a different product level.Keep both levels fixed and lower only the hump, so ΔH is unchanged.
- Measuring the activation energy from the product level or from the bottom of the graph.Draw the Ea arrow from the reactant level up to the top of the hump.
Exam tips
- When asked to explain ΔH, name the bonds broken and the bonds made, then state which process involves more energy, for example more energy is released forming the bonds in CO₂ and H₂O than is absorbed breaking the bonds in CH₄ and O₂.
- Always write the sign in front of ΔH, including the plus sign for endothermic values, and include the unit kJ/mol.
- On energy profile diagrams, label the axes, both levels with their formulae, ΔH and Ea, and draw the arrows in the correct direction.
Notes 9 of 10
Rate of Reaction
Key idea. Rate of reaction is the change in the amount of a reactant or product per unit time, read from the gradient of a graph. Particles react only when they collide with at least the activation energy. Higher concentration, pressure, surface area or temperature, or a catalyst, increase the frequency of effective collisions and so increase the rate.
Must know
- Average rate = change in amount ÷ time taken
- The rate at any moment is the gradient of the curve, steepest at the start
- The final volume of gas depends only on the limiting reactant
- In timed experiments such as the disappearing cross, rate is proportional to 1 ÷ time
What the syllabus asks you to do
- Describe the effect of concentration, pressure, particle size and temperature on rate, and explain these effects in terms of collisions between reacting particles
- Define the term catalyst and describe the effect of catalysts, including enzymes, on rate
- Explain how pathways with lower activation energies account for the increase in rate
- State that some compounds act as catalysts in industrial processes and that enzymes are biological catalysts
- Suggest a suitable method for investigating the effect of a given variable on rate
- Interpret data obtained from experiments on rate of reaction
Factors that increase the rate of reaction
| Factor increased | Effect on rate | Reason |
|---|---|---|
| Concentration of a solution | Increases | More particles per unit volume, so more frequent effective collisions |
| Pressure of a gas | Increases | Particles closer together, so more frequent effective collisions |
| Surface area of a solid | Increases | More particles exposed, so more frequent effective collisions |
| Temperature | Increases | Faster particles, and a greater proportion with energy at least equal to the activation energy |
| Catalyst added | Increases | Alternative pathway with a lower activation energy |
Two curves of gas volume against time from 0 to 450 s. With 1.00 mol/dm³ acid the curve rises steeply, passes 240 cm³ at 60 s and levels off at 600 cm³. With 0.500 mol/dm³ acid the initial gradient is half as steep and the curve levels off at 300 cm³, because only half as much acid was used.
Worked exam question
Excess marble chips (calcium carbonate) are added to 50.0 cm³ of 1.00 mol/dm³ hydrochloric acid. CaCO₃ + 2HCl gives CaCl₂ + H₂O + CO₂. After 60 s, 240 cm³ of gas has been collected at r.t.p. (a) Calculate the final volume of carbon dioxide. (b) Calculate the average rate over the first 60 s. (c) Describe how the final volume and initial rate change if 50.0 cm³ of 0.500 mol/dm³ acid is used instead.
- Moles of HCl = 1.00 × 50.0/1000 = 0.0500 mol, and the acid is limiting because the marble is in excess.
- Moles of CO₂ = 0.0500 ÷ 2 = 0.0250 mol, from the 2 : 1 ratio.
- Final volume = 0.0250 × 24 = 0.600 dm³ = 600 cm³.
- Average rate = 240 cm³ ÷ 60 s = 4.0 cm³/s.
- With 0.500 mol/dm³ acid, moles of HCl = 0.0250 mol, so the final volume halves to 300 cm³.
- The acid is less concentrated, so there are fewer acid particles per unit volume, collisions are less frequent and the initial rate is lower.
Answer: (a) 600 cm³. (b) 4.0 cm³/s. (c) The final volume halves to 300 cm³ and the initial gradient is less steep, because the less concentrated acid gives less frequent effective collisions.
Common mistakes
- Writing 'the particles collide more' as the whole explanation.Write 'more frequent effective collisions' and name the factor that causes them.
- Explaining a temperature rise only by faster particles colliding more often.Add that a greater proportion of particles have energy at least equal to the activation energy.
- Writing that a catalyst gives the particles more energy.Say that a catalyst provides an alternative pathway with a lower activation energy.
- Drawing a higher final volume for powdered marble than for chips.Find the limiting reactant. Only changing its amount changes the final volume.
- Saying the reaction is fastest where the curve is highest.Look at the steepness. The steepest part, usually at the start, is the fastest.
- Writing that a catalyst does not take part in the reaction.Say that the catalyst takes part but is regenerated, so its mass is the same at the end.
Exam tips
- Use the phrase 'frequency of effective collisions' in every rate explanation, and link it to the specific factor in the question.
- When sketching a second curve on the same axes, decide two things separately: is the initial gradient steeper or less steep, and is the final volume the same, higher or lower.
- If a question gives amounts of both reactants, work out the limiting reactant before predicting the final volume of gas.
Notes 10 of 10
Organic Chemistry
Key idea. Organic chemistry is the study of carbon compounds, grouped into homologous series with the same functional group and general formula. At O-Level you study alkanes, alkenes, alcohols, carboxylic acids and esters, how crude oil is separated and cracked, and how alkenes join into addition polymers. For each reaction, learn the reagents, the conditions and the product.
Must know
- Alkanes CₙH₂ₙ₊₂, alkenes CₙH₂ₙ, alcohols CₙH₂ₙ₊₁OH
- Alkenes decolourise reddish-brown bromine water, alkanes do not
- Carboxylic acid + alcohol, warmed with concentrated sulfuric acid, gives an ester + water
- Addition polymers form when C=C bonds open, with no other product
What the syllabus asks you to do
- Describe crude oil as a mixture of hydrocarbons separated by fractional distillation into fractions used as fuels and as a source of chemicals
- Describe a homologous series and the alkanes and alkenes, with general formulae CₙH₂ₙ₊₂ and CₙH₂ₙ, and define and identify isomers
- Describe alkanes as generally unreactive except for combustion and substitution by chlorine
- Describe cracking, the use of aqueous bromine to distinguish saturated and unsaturated hydrocarbons, and the addition of bromine, steam and hydrogen to alkenes
- Describe the formation of ethanol by the catalysed addition of steam to ethene and by fermentation of glucose, and its oxidation to ethanoic acid
- Describe the reaction of a carboxylic acid with an alcohol to form an ester, and deduce ester names and formulae
- Describe the addition polymerisation of ethene, deduce a polymer from its monomer, and describe nylon and Terylene as condensation polymers
Homologous series at O-Level
| Series | General formula and group | First members |
|---|---|---|
| Alkanes | CₙH₂ₙ₊₂, single bonds only | Methane CH₄, ethane C₂H₆ |
| Alkenes | CₙH₂ₙ, C=C double bond | Ethene C₂H₄, propene C₃H₆ |
| Alcohols | CₙH₂ₙ₊₁OH, -OH group | Methanol CH₃OH, ethanol C₂H₅OH |
| Carboxylic acids | CₙH₂ₙ₊₁COOH, -COOH group | Methanoic acid HCOOH, ethanoic acid CH₃COOH |
| Esters | -COO- linkage between two carbon chains | Ethyl ethanoate CH₃COOC₂H₅ |
Crude oil
A mixture of hydrocarbons, mostly alkanes
Fractional distillation
Separates fractions by boiling point in a fractionating column
Cracking
Long-chain alkanes heated with a catalyst give shorter alkanes and alkenes
Ethene
An unsaturated monomer with a C=C double bond
Addition polymerisation
Double bonds open and monomers join into long chains
Poly(ethene)
Repeat unit -CH₂-CH₂-, used for bags and films
Worked exam question
Compound A has the formula C₃H₆ and decolourises bromine water. (a) Name A and its homologous series. (b) Name the product with bromine and write its formula. (c) Calculate the mass of product formed when 4.2 g of A reacts completely with bromine. (Ar: C = 12, H = 1, Br = 80)
- C₃H₆ fits CₙH₂ₙ, and decolourising bromine water shows a C=C bond, so A is propene, an alkene.
- Bromine adds across the double bond: C₃H₆ + Br₂ gives C₃H₆Br₂, which is 1,2-dibromopropane.
- Mr of C₃H₆ = 3 × 12 + 6 × 1 = 42, so moles of A = 4.2 ÷ 42 = 0.10 mol.
- The ratio of A to product is 1 : 1, so 0.10 mol of C₃H₆Br₂ forms.
- Mr of C₃H₆Br₂ = 42 + 2 × 80 = 202, so mass = 0.10 × 202 = 20.2 g.
Answer: A is propene, an alkene. It forms 1,2-dibromopropane, C₃H₆Br₂, and 4.2 g of propene gives 20.2 g of product.
Common mistakes
- Stating that hexane decolourises bromine water.Link decolourising bromine water only to a C=C bond. Alkanes leave it reddish-brown.
- Naming the ester from ethanol and methanoic acid as methyl ethanoate.Name the alcohol part first (ethyl), then the acid part (methanoate), giving ethyl methanoate.
- Drawing the repeat unit of poly(ethene) with a C=C bond.Open the double bond. The repeat unit is -CH₂-CH₂-, with single bonds and a bond out at each end.
- Leaving out 'absence of oxygen' when describing fermentation.State glucose solution, yeast, about 37 °C and no oxygen.
- Writing propanoic acid as C₃H₇COOH.Count every carbon, including the one in -COOH, so propanoic acid is C₂H₅COOH.
- Describing cracking as a way to make more fuel from nothing.Say cracking turns less useful long-chain fractions into shorter alkanes and alkenes that are in demand.
Exam tips
- Make a one-page map of the reactions: alkane to chloroalkane, alkene to alkane, alcohol, dibromo compound or polymer, alcohol to acid, acid plus alcohol to ester. Write the conditions beside each arrow on your own notes.
- Always state conditions when a reaction needs them, such as ultraviolet light, a nickel catalyst, concentrated sulfuric acid or yeast. They often carry a mark of their own.
- Check every organic equation by counting carbon, hydrogen and oxygen atoms on both sides.
Syllabus checklist for the rest
What the syllabus asks you to do in the chapters, and parts of chapters, that do not have full notes yet. Each point is the syllabus learning outcome in plain words. Tick each one you can do without looking.
Chapter 1. Experimental Chemistry
4 points to check
- Name apparatus for measuring time, temperature, mass and volume, including burettes, pipettes, measuring cylinders and gas syringes.
- Suggest apparatus to collect and dry gases (drying agents are limited to calcium oxide, concentrated sulfuric acid and fused calcium chloride) and to measure rates of reaction.
- Describe and choose separation methods: a solvent with filtration and crystallisation or evaporation, sublimation, distillation, fractional distillation, a separating funnel and paper chromatography.
- Interpret chromatograms using known samples and Rf values, explain why locating agents are used, and use melting and boiling points to identify a substance and judge its purity.
Chapter 2. The Particulate Nature of Matter
4 points to check
- Describe solids, liquids and gases and the changes between them using the kinetic particle theory and energy changes.
- Explain everyday diffusion in terms of particles, and how molecular mass and temperature affect its rate.
- State the relative charges and approximate masses of protons, neutrons and electrons, and describe an atom with electrons in shells.
- Define proton number, nucleon number and isotopes, use nuclide notation, and work out the protons, neutrons and electrons in atoms and ions.
Chapter 5. Acid-Base Chemistry: also in the syllabus
1 point to check
- Describe how ammonia is made from nitrogen and hydrogen, state that some reactions are reversible, and interpret data on the conditions used in the Haber process.
Chapter 8. Patterns in the Periodic Table
5 points to check
- Describe the Periodic Table as elements in order of proton number, and link an element's position to its electronic configuration, valence electrons and ionic charge.
- Describe the change from metal to non-metal across a period, and predict the properties of Group 1 and Group 17 elements.
- Describe the trends in Group 1 (lithium, sodium, potassium) and Group 17 (chlorine, bromine, iodine, including displacement reactions), and why the Group 18 noble gases are unreactive.
- Describe typical transition elements and their use as catalysts.
- Place metals in a reactivity series from their reactions with water, steam and dilute hydrochloric acid and the reduction of their oxides, and deduce an order from experimental results.
Chapter 12. Maintaining Air Quality
4 points to check
- Describe dry air as about 78% nitrogen and 21% oxygen, with the rest mainly noble gases (mostly argon) and carbon dioxide.
- Name common pollutants (carbon monoxide, methane, nitrogen oxides, ozone, sulfur dioxide, unburned hydrocarbons) and the sources of carbon monoxide, nitrogen oxides and sulfur dioxide.
- Describe catalytic converters and the use of calcium carbonate against acid rain and in flue gas desulfurisation, and discuss the effects of pollutants on health and buildings.
- Discuss the ozone layer and its depletion by CFCs, the carbon cycle, and carbon dioxide and methane as greenhouse gases.
If your child is struggling
- Find the exact chapter and point. The checklists and common mistakes on this page show where an answer breaks down.
- Go back to that chapter's worked question and do it again without the steps.
- Ask the school teacher which chapters the class has covered and which are tested next.
- If several chapters are weak, work on one at a time, with short timed questions, rather than rereading everything.