0620/42 Oct/Nov 2025 Worked Answers (IGCSE Chemistry 0620 Extended)

52 questions · 80 marks · 75 minutes

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1. Question 1(a)

   3 marksParticle separation, arrangement and motion in the three states
   Step 1: In a solid the particles are touching in a regular arrangement and can only vibrate about fixed positions. Step 2: In a liquid the particles are still close together (touching) but in a random arrangement, and they can move around and slide past each other. Step 3: In a gas the particles are far apart in a random arrangement and move quickly in all directions.
   Method:

   Recall the kinetic particle model for each state across separation, arrangement and motion.
   Examiner tips

   • Separation decreases solid≈liquid (touching) but gas is far apart; only solids are regular.

2. Question 1(b)

   3 marksNaming physical changes
   Step 1: A solid turning into a liquid is melting. Step 2: A gas turning into a liquid is condensing (condensation). Step 3: A solid becoming an aqueous solution is dissolving, so the three changes are melting, condensing and dissolving.
   Method:

   Match each change of state to its standard name, checking the direction carefully.
   Examiner tips

   • Watch the direction: melting and condensing are opposites of freezing and evaporating.

3. Question 1(c)(i)

   1 marksWhy gases diffuse
   Step 1: Gas particles are in constant, random motion and move quickly. Step 2: Because they are far apart and free to move, they spread out and mix with other particles. Step 3: This movement of particles from where they are concentrated to fill the available space is diffusion.
   Method:

   Link diffusion to the random movement of gas particles.
   Examiner tips

   • Diffusion happens because particles are in constant random motion.

4. Question 1(c)(ii)

   1 marksWhat determines the relative rate of diffusion
   Step 1: At a given temperature, lighter gas particles move faster than heavier ones. Step 2: The mass of each molecule is its relative molecular mass, $M_r$. Step 3: So the relative rate of diffusion is determined by the relative molecular mass: gases with a lower $M_r$ diffuse faster.
   Method:

   Recall the link between molecular mass and how fast a gas diffuses.
   Examiner tips

   • Rate of diffusion depends on relative molecular mass, Mr.

5. Question 1(d)

   2 marksAdverse effects of oxides of nitrogen
   Step 1: Oxides of nitrogen react in sunlight to form photochemical smog. Step 2: They are also toxic and irritate the lungs, causing respiratory (breathing) problems. Step 3: So two further adverse effects are photochemical smog and respiratory problems.
   Method:

   Recall the harmful effects of nitrogen oxide pollutants beyond acid rain.
   Examiner tips

   • Oxides of nitrogen: acid rain, photochemical smog, respiratory problems.

6. Question 1(e)(i)

   1 marksToxic product of incomplete combustion
   Step 1: Incomplete combustion happens when there is not enough oxygen, producing carbon monoxide instead of carbon dioxide. Step 2: Carbon monoxide is a toxic gas. Step 3: So the toxic gaseous product removed in the catalytic converter is carbon monoxide.
   Method:

   Recall the toxic carbon-containing gas formed when fuels burn in too little oxygen.
   Examiner tips

   • Incomplete combustion of fuels gives toxic carbon monoxide.

7. Question 1(e)(ii)

   2 marksNon-toxic gases formed in a catalytic converter
   Step 1: In the catalytic converter the toxic carbon monoxide reduces the oxides of nitrogen. Step 2: The nitrogen oxides are converted to harmless nitrogen gas, and the carbon monoxide is oxidised to carbon dioxide. Step 3: So the two non-toxic gases formed are nitrogen and carbon dioxide.
   Method:

   Work out the harmless products when carbon monoxide reduces the nitrogen oxides.
   Examiner tips

   • Catalytic converter: CO + NOx -> N2 + CO2.

8. Question 2(a)(i)

   1 marksName of the main iron ore
   The main iron ore used in the blast furnace, which contains iron(III) oxide, is called hematite.
   Method:

   Recall the name of the iron(III) oxide ore used in the blast furnace.
   Examiner tips

   • Hematite = iron(III) oxide ore.

9. Question 2(a)(ii)

   1 marksSource of heat in the blast furnace
   Step 1: Hot air is blasted into the furnace where it reacts with the coke (carbon). Step 2: The carbon burns in the oxygen of the air, and this combustion is strongly exothermic. Step 3: So the main heat comes from the burning of carbon (coke).
   Method:

   Identify the exothermic reaction that warms the blast furnace.
   Examiner tips

   • Burning carbon (coke) in the blast of hot air supplies the heat.

10. Question 2(a)(iii)

    1 marksGaseous reducing agent in the blast furnace
    Step 1: Coke reacts with limited oxygen and carbon dioxide to form carbon monoxide. Step 2: Carbon monoxide is the gas that reduces iron(III) oxide to iron. Step 3: So the gaseous reducing agent is carbon monoxide.
    Method:

    Recall which gas reduces the iron oxide to iron.
    Examiner tips

    • Carbon monoxide is the gaseous reducer in the blast furnace.

11. Question 2(a)(iv)

    2 marksEquation for reduction of iron(III) oxide
    Step 1: Each carbon monoxide gains an oxygen to become carbon dioxide, while iron(III) oxide loses its oxygen to form iron. Step 2: $\text{Fe}_2\text{O}_3$ has 3 oxygen atoms, so 3 CO molecules are needed and 3 $\text{CO}_2$ molecules form. Step 3: Balancing the iron gives 2 Fe, so the equation is $3\text{CO} + \text{Fe}_2\text{O}_3 \rightarrow 2\text{Fe} + 3\text{CO}_2$.
    Method:

    Balance iron and oxygen between the reducing agent and the oxide.
    Examiner tips

    • Balance the 3 oxygens of Fe2O3 with 3 CO and 3 CO2.

12. Question 2(a)(v)

    2 marksRole of limestone in removing silicon(IV) oxide
    Step 1: The limestone first decomposes in the heat: $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$. Step 2: The basic calcium oxide then reacts with the acidic silicon(IV) oxide impurity: $\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3$ (slag). Step 3: So the two equations are the thermal decomposition of limestone and the reaction of calcium oxide with silicon(IV) oxide.
    Method:

    Write the decomposition of limestone, then the acid-base reaction that forms slag.
    Examiner tips

    • Two steps: thermal decomposition of CaCO3, then CaO + SiO2 -> slag.

13. Question 2(b)

    2 marksWhy different iron atoms are isotopes
    Step 1: They are all iron, so they have the same number of protons (26), which defines the element. Step 2: Their nucleon numbers differ (54, 56, 57, 58), which means they have different numbers of neutrons. Step 3: Atoms of the same element with the same protons but different neutrons are isotopes.
    Method:

    State what is the same (protons) and what differs (neutrons) for isotopes.
    Examiner tips

    • Isotopes: same proton number, different neutron number.

14. Question 2(c)(i)

    3 marksSubatomic particles in iron ions
    Step 1: Both ions are iron, so each has 26 protons. Step 2: Neutrons = nucleon number - protons, so $^{54}\text{Fe}$ has 54 - 26 = 28 neutrons and $^{58}\text{Fe}$ has 58 - 26 = 32 neutrons. Step 3: A 2+ ion has lost 2 electrons (26 - 2 = 24) and a 3+ ion has lost 3 electrons (26 - 3 = 23).
    Method:

    Use proton number, nucleon number and charge to count each particle.
    Examiner tips

    • Charge changes only the electron count, never the protons or neutrons.

15. Question 2(c)(ii)

    2 marksTest for aqueous Fe2+ ions
    Step 1: Aqueous metal ions are tested by adding aqueous sodium hydroxide (or aqueous ammonia). Step 2: Iron(II) ions give a green precipitate of iron(II) hydroxide. Step 3: So the test is to add sodium hydroxide and look for a green precipitate.
    Method:

    Recall the sodium hydroxide test and the colour for iron(II).
    Examiner tips

    • Fe2+ -> green precipitate; Fe3+ -> red-brown precipitate with NaOH(aq).

16. Question 2(d)(i)

    2 marksDefinition of an oxidising agent
    Step 1: An oxidising agent causes another substance to be oxidised (to lose electrons). Step 2: In doing so it gains those electrons, so it is itself reduced. Step 3: Hence an oxidising agent oxidises another substance and is itself reduced.
    Method:

    Define it in terms of what it does to others and what happens to itself.
    Examiner tips

    • Oxidising agent = electron acceptor; it is reduced.

17. Question 2(d)(ii)

    1 marksWhy Fe2+ to Fe3+ is oxidation
    Step 1: The charge increases from 2+ to 3+, which means the ion has become more positive. Step 2: An ion becomes more positive by losing a (negative) electron. Step 3: Loss of electrons is the definition of oxidation, so $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+}$ is oxidation.
    Method:

    Link the rise in positive charge to the loss of an electron.
    Examiner tips

    • Oxidation Is Loss of electrons (OIL).

18. Question 2(d)(iii)

    1 marksCondition for oxidation by potassium manganate(VII)
    Step 1: Potassium manganate(VII) acts as an oxidising agent in acidic conditions. Step 2: The reaction needs hydrogen ions, so the solution must be acidified. Step 3: So the condition required is that the solution is acidified.
    Method:

    Recall the standard condition for potassium manganate(VII) oxidations.
    Examiner tips

    • Acidify the manganate(VII) for it to act as the oxidiser.

19. Question 2(e)

    1 marksProduct when iodide reduces Fe3+
    Step 1: The iodide ions act as the reducing agent, so they are themselves oxidised. Step 2: Iodide ions lose electrons and are oxidised to iodine. Step 3: So while $\text{Fe}^{3+}$ becomes $\text{Fe}^{2+}$, the other product is iodine.
    Method:

    Identify what the reducing agent (iodide) turns into when it is oxidised.
    Examiner tips

    • If iodide is the reducer, it is oxidised to iodine.

20. Question 3(a)

    2 marksPrecipitation equation with state symbols
    Step 1: Barium sulfate is insoluble, so it forms as a solid precipitate, $\text{BaSO}_4(\text{s})$. Step 2: Sodium chloride is soluble and stays in solution as $\text{NaCl}(\text{aq})$. Step 3: Balancing the chlorine and sodium needs 2 NaCl, giving $\text{BaSO}_4(\text{s}) + 2\text{NaCl}(\text{aq})$.
    Method:

    Decide which product is insoluble, then balance and add state symbols.
    Examiner tips

    • Precipitation: the insoluble salt is (s), the soluble one stays (aq).

21. Question 3(b)

    2 marksMass from moles and Mr
    Step 1: The relative formula mass of $\text{BaCl}_2$ = 137 + (2 x 35.5) = 208. Step 2: Mass = moles x $M_r$ = 0.100 x 208. Step 3: This gives a mass of 20.8 g.
    Method:

    Find the relative formula mass, then multiply by the moles.
    Examiner tips

    • Work out Mr first, then multiply by the number of moles.

22. Question 3(c)

    1 marksVolume from moles and concentration
    Step 1: Volume in $\text{dm}^3$ = moles / concentration = 0.100 / 1.25 = 0.0800 $\text{dm}^3$. Step 2: Convert to $\text{cm}^3$ by multiplying by 1000. Step 3: 0.0800 x 1000 = 80 $\text{cm}^3$.
    Method:

    Divide moles by concentration, then convert to cm^3.
    Examiner tips

    • volume in cm^3 = (moles / concentration) x 1000.

23. Question 3(d)

    1 marksColour of the barium sulfate precipitate
    Step 1: Barium sulfate is an insoluble salt that precipitates out of the mixture. Step 2: Barium sulfate is a white solid. Step 3: So the colour of the solid formed is white.
    Method:

    Recall the colour of the insoluble barium sulfate precipitate.
    Examiner tips

    • Barium sulfate = white insoluble solid.

24. Question 3(e)

    1 marksTerm for the solid left after filtration
    The solid left behind in the filter paper after filtration is called the residue (the liquid that passes through is the filtrate).
    Method:

    Recall the term for the trapped solid in filtration.
    Examiner tips

    • Residue stays in the paper; filtrate passes through.

25. Question 3(f)

    2 marksWhy the residue mass is too high and how to correct it
    Step 1: When the mixture is filtered, the residue is wet with the solution, which contains dissolved sodium chloride. Step 2: If the residue is not washed, this sodium chloride dries onto it and adds extra mass. Step 3: Rinsing the residue with distilled water washes the soluble sodium chloride away, so the additional step is to rinse with water and the mass was high because of the leftover sodium chloride.
    Method:

    Identify the soluble impurity left on the residue and the washing step that removes it.
    Examiner tips

    • Unwashed residue keeps soluble impurities; rinse with distilled water.

26. Question 3(g)

    1 marksName of the salt-preparation method
    Making an insoluble salt by mixing two soluble solutions so the insoluble salt forms as a solid is called precipitation.
    Method:

    Name the method used to make an insoluble salt from two soluble solutions.
    Examiner tips

    • Insoluble salt from two solutions = precipitation.

27. Question 3(h)

    3 marksAlternative soluble and insoluble salts for the precipitation
    Step 1: A different soluble barium salt to replace barium chloride is barium nitrate (nitrates are soluble). Step 2: A different soluble sulfate to replace sodium sulfate is potassium sulfate (or ammonium sulfate). Step 3: A different insoluble barium salt that precipitation could make is barium carbonate, which is also insoluble.
    Method:

    Use solubility rules to pick a soluble barium salt, a soluble sulfate and an insoluble barium salt.
    Examiner tips

    • Soluble barium salt (nitrate), soluble sulfate (potassium/ammonium), insoluble barium salt (carbonate).

28. Question 4(a)(i)

    1 marksCommon cation produced by acids in water
    Step 1: An acid is a substance that releases hydrogen ions when dissolved in water. Step 2: Nitric, sulfuric and ethanoic acid all release the hydrogen ion. Step 3: So the common cation produced is $\text{H}^+$.
    Method:

    Recall the ion that all acids release in aqueous solution.
    Examiner tips

    • Acids in water release H+ ions.

29. Question 4(a)(ii)

    2 marksStrong vs weak acid in terms of dissociation
    Step 1: A strong acid is one that dissociates (ionises) completely in water, releasing all its hydrogen ions. Step 2: Nitric acid does this, so it is a strong acid. Step 3: A weak acid only partially dissociates, releasing few hydrogen ions; ethanoic acid does this, so it is a weak acid.
    Method:

    Define strong and weak acids by how completely they dissociate.
    Examiner tips

    • Strength = degree of dissociation, not concentration.

30. Question 4(a)(iii)

    1 marksAcid with the highest pH
    Step 1: The higher the concentration of hydrogen ions, the lower the pH. Step 2: Ethanoic acid is weak and only partially dissociates, so it has the lowest hydrogen ion concentration. Step 3: The lowest hydrogen ion concentration gives the highest pH, so ethanoic acid has the highest pH.
    Method:

    Link the weakest acid to the lowest H+ concentration and hence the highest pH.
    Examiner tips

    • Weak acid -> fewer H+ ions -> higher pH.

31. Question 4(a)(iv)

    1 marksColour of thymolphthalein in acids
    Step 1: Thymolphthalein is colourless in acidic and neutral solutions and blue in alkaline solutions. Step 2: All three solutions here are acidic. Step 3: So the thymolphthalein is colourless in each of the dilute acids.
    Method:

    Recall the acidic colour of thymolphthalein.
    Examiner tips

    • Thymolphthalein: colourless (acid) -> blue (alkali).

32. Question 4(a)(v)

    2 marksAnions formed when acids dissociate
    Step 1: Sulfuric acid, $\text{H}_2\text{SO}_4$, loses two $\text{H}^+$ to leave the sulfate ion, $\text{SO}_4^{2-}$. Step 2: Ethanoic acid, $\text{CH}_3\text{COOH}$, loses one $\text{H}^+$ from the -COOH group to leave the ethanoate ion, $\text{CH}_3\text{COO}^-$. Step 3: So the anions are $\text{SO}_4^{2-}$ and $\text{CH}_3\text{COO}^-$.
    Method:

    Remove the acidic hydrogen ions to deduce each anion.
    Examiner tips

    • Take off the acidic hydrogens to find the anion left behind.

33. Question 4(a)(vi)

    2 marksProducts of electrolysis of dilute sulfuric acid
    Step 1: Dilute sulfuric acid is essentially acidified water, so electrolysis splits the water. Step 2: At the cathode (negative electrode) hydrogen ions are reduced to hydrogen gas. Step 3: At the anode (positive electrode) the water/hydroxide is oxidised to oxygen, so the two gases are hydrogen and oxygen.
    Method:

    Treat dilute sulfuric acid electrolysis as splitting water into hydrogen and oxygen.
    Examiner tips

    • Dilute sulfuric acid -> hydrogen (cathode) + oxygen (anode).

34. Question 4(a)(vii)

    1 marksSalt formed from calcium and ethanoic acid
    Step 1: A metal reacting with a carboxylic acid forms a salt and hydrogen. Step 2: Ethanoic acid forms ethanoate salts, and the metal is calcium. Step 3: So the salt formed is calcium ethanoate.
    Method:

    Combine the metal name with the ethanoate ending of the acid.
    Examiner tips

    • Acid name -> salt name: ethanoic acid -> ethanoate.

35. Question 4(b)(i)

    1 marksDefinition of a base
    A base is a substance that is a proton (hydrogen ion) acceptor.
    Method:

    Recall the proton-transfer definition of a base.
    Examiner tips

    • Acid = proton donor; base = proton acceptor.

36. Question 4(b)(ii)

    1 marksSolubility of a base that is not an alkali
    Step 1: An alkali is defined as a soluble base, one that dissolves in water. Step 2: Aluminium hydroxide is a base but is not an alkali. Step 3: Therefore it must be insoluble in water.
    Method:

    Use the definition of an alkali as a soluble base to deduce the solubility.
    Examiner tips

    • Alkali = soluble base. Not an alkali = insoluble base.

37. Question 4(b)(iii)

    1 marksCompound forming aluminium nitrate with nitric acid
    Step 1: Nitric acid reacts with bases such as metal oxides and hydroxides to form the metal nitrate. Step 2: Aluminium oxide is a base containing aluminium, so it reacts with nitric acid to give aluminium nitrate and water. Step 3: So aluminium oxide also forms aluminium nitrate with dilute nitric acid.
    Method:

    Pick an aluminium-containing base that reacts with nitric acid to give the nitrate.
    Examiner tips

    • Acid + metal oxide/hydroxide/carbonate -> metal salt; use an aluminium base.

38. Question 4(b)(iv)

    2 marksOxidation number of nitrogen in a nitrate
    Step 1: $\text{Al(NO}_3)_3$ has 9 oxygen atoms, contributing 9 x (-2) = -18, and aluminium contributes +3. Step 2: The compound is neutral, so the three nitrogen atoms must total +18 - 3 = +15. Step 3: Sharing +15 across the 3 nitrogen atoms gives +15 / 3 = +5 for each nitrogen.
    Method:

    Set the sum of oxidation numbers to zero and solve for nitrogen, then divide by 3.
    Examiner tips

    • Total oxidation numbers = 0; divide the nitrogen total by 3.

39. Question 5(a)

    2 marksGeneral characteristics of a homologous series
    Step 1: A homologous series shares one general formula; for the alkanes this is $\text{C}_n\text{H}_{2n+2}$. Step 2: Each member also differs from the next by a $\text{CH}_2$ unit. Step 3: So two further characteristics are the same general formula and a $\text{CH}_2$ difference between consecutive members.
    Method:

    Recall two more defining features of a homologous series for the alkanes.
    Examiner tips

    • Homologous series: one general formula + CH2 difference between members.

40. Question 5(b)(i)

    1 marksTrend in boiling points of alkanes
    Step 1: As the carbon chain gets longer, the alkane molecules get larger. Step 2: Larger molecules have stronger intermolecular forces, which need more energy to overcome. Step 3: So the boiling points increase as the chain length increases.
    Method:

    Relate longer chains to stronger forces and higher boiling points.
    Examiner tips

    • Boiling point increases down the alkane series.

41. Question 5(b)(ii)

    2 marksAnother physical property showing a trend
    Step 1: Viscosity (how thick or sticky a liquid is) is another physical property that shows a trend in the alkanes. Step 2: As the carbon chain gets longer, the molecules are larger and the forces between them stronger. Step 3: This makes the alkanes more viscous, so viscosity increases as the chain length increases.
    Method:

    Name a valid physical property and state how it changes along the series.
    Examiner tips

    • Viscosity increases (and volatility/flammability decrease) with chain length.

42. Question 5(c)

    1 marksWhy alkanes are saturated
    Step 1: A saturated compound contains only single bonds between its carbon atoms. Step 2: In alkanes every carbon-carbon bond is a single bond, with no double bonds. Step 3: So alkanes are described as saturated because all their carbon-carbon bonds are single.
    Method:

    Define saturation in terms of single carbon-carbon bonds.
    Examiner tips

    • Saturated hydrocarbons have only single bonds.

43. Question 5(d)(i)

    1 marksCondition for substitution of propane with chlorine
    Step 1: The substitution of an alkane by chlorine needs energy to start the reaction. Step 2: This energy is provided by ultraviolet light (sunlight). Step 3: So the condition needed is ultraviolet light.
    Method:

    Recall the condition that starts an alkane substitution with a halogen.
    Examiner tips

    • Substitution of alkanes is driven by ultraviolet light.

44. Question 5(d)(ii)

    4 marksProducts of monosubstitution of propane
    Step 1: Monosubstitution replaces just one hydrogen atom of propane with a chlorine atom. Step 2: Replacing a hydrogen on an end carbon gives $\text{CH}_3\text{CH}_2\text{CH}_2\text{Cl}$, which is 1-chloropropane. Step 3: Replacing a hydrogen on the middle carbon gives $\text{CH}_3\text{CHClCH}_3$, which is 2-chloropropane.
    Method:

    Replace a single hydrogen with chlorine in each possible position and name the products.
    Examiner tips

    • One Cl on the end = 1-chloropropane; one Cl in the middle = 2-chloropropane.

45. Question 5(d)(iii)

    1 marksDefinition of structural isomers
    Structural isomers are compounds with the same molecular formula but different structural formulae (different arrangements of the atoms).
    Method:

    State what is the same (molecular formula) and what differs (structure) for isomers.
    Examiner tips

    • Same molecular formula + different structure = structural isomers.

46. Question 6(a)

    1 marksAtoms in the ester linkage
    Step 1: An ester linkage is formed between a carboxylic acid and an alcohol. Step 2: It contains the carbonyl carbon, the oxygen double-bonded to it (C=O) and the oxygen single-bonded between the two carbon chains. Step 3: So the ester linkage is the -C(=O)-O- group, taking in the carbon and both of its oxygen atoms.
    Method:

    Identify the carbon and the two oxygen atoms that form the ester linkage.
    Examiner tips

    • Ester linkage = -COO- = carbon + both oxygen atoms.

47. Question 6(b)

    1 marksNaming an ester
    Step 1: An ester is named as alkyl + ...oate: the alcohol gives the alkyl part and the acid gives the -oate part. Step 2: The acid part $\text{CH}_3\text{CH}_2\text{CH}_2\text{COO}$ comes from butanoic acid (4 carbons), giving butanoate. Step 3: The $\text{OCH}_2\text{CH}_3$ comes from ethanol, giving the ethyl part, so the ester is ethyl butanoate.
    Method:

    Split the ester into its acid and alcohol parts and combine the names.
    Examiner tips

    • Name = alkyl (from alcohol) + acid stem + "oate".

48. Question 6(c)

    1 marksEmpirical formula of an ester
    Step 1: The molecular formula is $\text{C}_6\text{H}_{12}\text{O}_2$ (6 C, 12 H, 2 O). Step 2: The empirical formula is the simplest whole-number ratio, so divide each number by the highest common factor, 2. Step 3: This gives $\text{C}_3\text{H}_6\text{O}$.
    Method:

    Simplify the molecular formula to the smallest whole-number ratio.
    Examiner tips

    • Empirical formula = molecular formula divided by the highest common factor.

49. Question 6(d)(i)

    2 marksReactants that make an ester
    Step 1: An ester is made by reacting a carboxylic acid with an alcohol. Step 2: The butanoate part comes from butanoic acid and the ethyl part comes from ethanol. Step 3: So ethyl butanoate is made from butanoic acid and ethanol.
    Method:

    Split the ester name into the acid and the alcohol that formed it.
    Examiner tips

    • Match the acid to the -oate part and the alcohol to the alkyl part.

50. Question 6(d)(ii)

    1 marksCatalyst for ester formation
    Step 1: Esterification of a carboxylic acid with an alcohol is catalysed by an acid. Step 2: Concentrated sulfuric acid is the usual catalyst. Step 3: So a suitable catalyst is an acid catalyst such as concentrated sulfuric acid.
    Method:

    Recall the type of catalyst that speeds up ester formation.
    Examiner tips

    • Use an acid catalyst (concentrated sulfuric acid) to make esters.

51. Question 6(e)

    1 marksNumber of unbranched esters with a given formula
    Step 1: An ester is an acid part joined to an alcohol part, and together they must total 5 carbons. Step 2: The unbranched possibilities are: methyl butanoate (1+4), ethyl propanoate (2+3), propyl ethanoate (3+2) and butyl methanoate (4+1). Step 3: That gives 4 different unbranched esters with the formula $\text{C}_5\text{H}_{10}\text{O}_2$.
    Method:

    Systematically split the 5 carbons between the acid and alcohol parts.
    Examiner tips

    • Count the ways to share the 5 carbons between acid and alcohol (1+4, 2+3, 3+2, 4+1).

52. Question 6(f)

    1 marksComplete combustion equation of an ester
    Step 1: Each $\text{C}_5\text{H}_{10}\text{O}_2$ gives 5 $\text{CO}_2$ and 5 $\text{H}_2\text{O}$. Doubling avoids a fraction of oxygen, giving 10 $\text{CO}_2$ and 10 $\text{H}_2\text{O}$. Step 2: The right-hand side has (10 x 2) + 10 = 30 oxygen atoms, and the two ester molecules supply 4 oxygens, so $\text{O}_2$ must supply 30 - 4 = 26 oxygen atoms = 13 $\text{O}_2$. Step 3: The balanced equation is $2\text{C}_5\text{H}_{10}\text{O}_2 + 13\text{O}_2 \rightarrow 10\text{CO}_2 + 10\text{H}_2\text{O}$.
    Method:

    Balance carbon and hydrogen, then balance oxygen allowing for the oxygen in the ester.
    Examiner tips

    • Remember the ester already contains 2 oxygen atoms when balancing the O2.