KCSE 2025 Chemistry P2 Q2 — Organic Naming and the Extraction of Copper
Published
The Question
“Question 2 has two parts. (a) Name the following organic compounds: (i) CH₃(CH₂)₅CH₃ and (ii) CH₃(CH₂)₃C≡CH. (b) The flow chart shows the extraction of copper from copper pyrites, CuFeS₂. The ore enters chamber 1 with a blast of air; the product passes to chamber 2 where substance B is added and a molten slag is drained off; in chamber 3 more air is blown through and impure copper is tapped off. A gas C leaves chambers 1 and 3. (i) Identify gas C. (ii) Write the equation for the reaction in chamber 1. (iii) Identify substance B. (iv) Write the equation for the formation of the slag in chamber 2. (v) Write the equations for the reactions in chamber 3 that produce copper. (vi) Name the method used to purify the copper and describe the electrodes and electrolyte used. (vii) Copper articles left in air for a long time acquire a green coating; name the substance responsible.”
(a)(i) Name CH₃(CH₂)₅CH₃
To name an organic compound, count the carbon atoms and check the bonds between them. The shorthand (CH₂) repeated five times, written between the two CH₃ groups, opens out into a straight chain of seven carbon atoms joined only by single bonds. Only single carbon-to-carbon bonds means the compound is an alkane, and alkane names end in -ane. Seven carbons give the stem 'hept-', so the compound is heptane.
(a)(ii) Name CH₃(CH₂)₃C≡CH
Counting again, this chain has six carbon atoms, but the last two are joined by a carbon-to-carbon triple bond. A triple bond between carbons means the compound is an alkyne, and alkyne names end in -yne. Six carbons give the stem 'hex-'. To locate the triple bond, number the chain from the end that reaches it first so it gets the lowest possible number; counting from the right, the triple bond starts at carbon 1. The name is therefore hex-1-yne.
(b)(i) Identify gas C
In chamber 1 the sulphide ore is roasted in a blast of air. The sulphur in the ore burns in the oxygen, exactly as sulphur itself burns to a choking gas, so gas C — the gas leaving both chamber 1 and chamber 3 — is sulphur(IV) oxide, that is, sulphur dioxide.
(b)(ii) Equation for chamber 1
Copper pyrites, CuFeS₂, reacts with oxygen when roasted to give copper(I) sulphide, iron(II) oxide and sulphur(IV) oxide. Checking the balance confirms two coppers, two irons, four sulphurs and eight oxygens on each side.
(b)(iii) Identify substance B
The product entering chamber 2 still contains iron as iron(II) oxide, which must be removed. Substance B is added to react with that iron oxide and carry it off as slag. Substance B is silica, silicon(IV) oxide.
(b)(iv) Formation of the slag
Iron(II) oxide is a basic oxide and silica is an acidic oxide; a basic oxide reacting with an acidic oxide gives a salt. They combine to form iron(II) silicate, which floats off as the molten slag. The equation is already balanced — one oxygen from the iron oxide plus two from the silica make the three in FeSiO₃, so the main compound in the slag is FeSiO₃.
(b)(v) Reactions in chamber 3
In chamber 3 a fresh blast of air completes the extraction. Part of the copper(I) sulphide is first oxidised to copper(I) oxide, which then reacts with the remaining copper(I) sulphide in a self-reduction to give molten impure copper, releasing more sulphur(IV) oxide.
(b)(vi) Purifying the copper
The copper from chamber 3 is still impure, so it is purified by electrolysis. The impure copper is made the anode (the thick positive electrode) and a thin sheet of pure copper is the cathode; the electrolyte is copper(II) sulphate solution. When the current flows the impure anode dissolves into the solution as copper(II) ions, which drift across and are deposited as pure copper on the cathode, while the impurities fall to the bottom as anode sludge. So the anode slowly shrinks and the pure cathode grows.
(b)(vii) The green coating on old copper
Over a long time copper corrodes slowly, reacting little by little with three things present in the air together — oxygen, carbon dioxide and moisture. These build up a green layer of basic copper(II) carbonate on the surface, the patina seen on old copper articles.
Final Result
(a) The compounds are (i) heptane and (ii) hex-1-yne. (b) Gas C is sulphur(IV) oxide, SO₂. Chamber 1: 2CuFeS₂ + 4O₂ → Cu₂S + 2FeO + 3SO₂. Substance B is silica, SiO₂, which forms the slag FeO + SiO₂ → FeSiO₃. Chamber 3: 2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂, then 2Cu₂O + Cu₂S → 6Cu + SO₂. The copper is purified by electrolysis, using the impure copper as the anode, pure copper as the cathode and copper(II) sulphate solution as the electrolyte. The green coating on old copper is basic copper(II) carbonate, CuCO₃·Cu(OH)₂.
Why this method works
Two different skills run through this question. Naming is a fixed routine: the number of carbon atoms fixes the stem (hept-, hex-), the type of carbon-carbon bond fixes the family and suffix (single bonds → -ane, a triple bond → -yne), and the position of a multiple bond is given the lowest locant by numbering from the nearer end. The copper extraction is really a story about oxides. Roasting any sulphide ore in air burns its sulphur to sulphur dioxide, which is why the same gas leaves both roasting chambers. The iron impurity is stripped out by exploiting acid-base behaviour — basic FeO reacts with acidic SiO₂ to make a fusible silicate slag that floats clear of the denser copper. Chamber 3 is a neat self-reduction: some copper(I) sulphide is oxidised to copper(I) oxide, which then oxidises the sulphur of the remaining sulphide while its own copper is reduced to the metal, so no external reducing agent is needed. Finally, electrolytic refining works because copper dissolves selectively from an impure anode and re-deposits pure at the cathode, leaving the impurities behind as sludge.
Chamber 1 balances as 2 Cu, 2 Fe, 4 S and 8 O on each side. Adding the two chamber-3 equations cancels the copper(I) oxide and returns 6 mol of copper metal, confirming the ore is fully reduced to the free metal.