CIE iGCSE Co-Ordinated Science C3.1 Formulas Exam Style Questions Paper 4
Question
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic C11.1 — Formulas and terminology / Hydrocarbons (Part (a))
• Topic C11.3 — Fuels / Natural gas (Part (b))
• Topic C11.2 — Naming organic compounds (Part (c))
• Topic C11.7 — Polymers / Addition polymerisation (Part (d)(i) & (d)(ii))
• Topic C3.1 — Formulas / Balancing equations (Part (e))
▶️ Answer/Explanation
(a) Compound C is not a hydrocarbon because it contains oxygen; hydrocarbons contain only hydrogen and carbon.
Compound C has an -OH group (alcohol functional group), meaning it contains oxygen atoms in addition to carbon and hydrogen, so it is not a hydrocarbon.
(b) Natural gas
Compound D is methane (CH₄), which is the main constituent of natural gas, a fossil fuel formed from ancient organic matter.
(c) Propane
Compound B has the formula C₃H₈, which is the third member of the alkane homologous series, named propane.
(d)(i) The polymer structure is –[–CH₂–CH₂–]– with single bonds between carbon atoms:
Ethene monomers (compound A with C=C bond) join by breaking the double bond and forming single bonds between carbon atoms in a long chain.
(d)(ii) Difference 1: Addition polymerisation only occurs in unsaturated monomers (with C=C bonds), while condensation polymerisation involves different types of monomers.
Difference 2: Addition polymerisation only forms the polymer molecule, whereas condensation polymerisation forms the polymer and a small molecule (like water) per linkage.
Addition polymerisation uses monomers with double bonds; condensation polymerisation joins monomers with functional groups, releasing a by-product.
(e) \( \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} \)
Propane combustion requires 5 molecules of oxygen to produce 3 molecules of carbon dioxide and 4 molecules of water. The equation is balanced with 3 carbon, 8 hydrogen, and 10 oxygen atoms on both sides.
Question
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic C3.1 — Formulas (Part (a))
• Topic C12.5 — Qualitative analysis (Part (b)(i), (b)(ii))
• Topic C6.3 — Redox (Part (c)(i))
• Topic C3.2 — Relative masses of atoms and molecules (Part (c)(ii))
• Topic C9.6 — Extraction of metals (Part (c)(i), (c)(ii))
▶️ Answer/Explanation
(a) \( \text{Fe}_2(\text{SO}_4)_3 \)
Iron(III) ions have a charge of \( 3+ \) (\( \text{Fe}^{3+} \)) and sulfate ions have a charge of \( 2- \) (\( \text{SO}_4^{2-} \)). To form a neutral compound, the total positive charge must balance the total negative charge. The lowest common multiple of 3 and 2 is 6. Therefore, we need 2 iron(III) ions (total charge \( 2 \times 3+ = 6+ \)) and 3 sulfate ions (total charge \( 3 \times 2- = 6- \)). Hence, the formula is \( \text{Fe}_2(\text{SO}_4)_3 \).
(b)(i) Red-brown.
When aqueous sodium hydroxide is added to a solution containing iron(III) ions, a red-brown precipitate of iron(III) hydroxide is formed. This is a characteristic test for \( \text{Fe}^{3+} \) ions.
(b)(ii) \( \text{Fe}^{3+}(\text{aq}) + 3\text{OH}^-(\text{aq}) \rightarrow \text{Fe(OH)}_3(\text{s}) \)
The ionic equation shows only the ions that participate in the reaction. The \( \text{Fe}^{3+} \) ions from the iron(III) sulfate solution react with \( \text{OH}^- \) ions from the sodium hydroxide solution to form solid iron(III) hydroxide precipitate. The sodium and sulfate ions are spectator ions and are not included in the ionic equation. The state symbols are (aq) for aqueous ions and (s) for the solid precipitate.
(c)(i) Iron(III) oxide is reduced because it loses oxygen / \( \text{Fe}_2\text{O}_3 \) loses oxygen.
In the blast furnace reaction, iron(III) oxide (\( \text{Fe}_2\text{O}_3 \)) is converted to iron (\( \text{Fe} \)). The iron(III) oxide loses oxygen atoms (it is reduced from \( \text{Fe}_2\text{O}_3 \) to \( \text{Fe} \)). Loss of oxygen is reduction. Carbon monoxide (\( \text{CO} \)) gains oxygen to become carbon dioxide (\( \text{CO}_2 \)), so carbon monoxide is oxidised. This is a redox reaction where reduction and oxidation occur simultaneously.
(c)(ii) Minimum mass of iron(III) oxide required = 40,000 g
Calculation steps:
- \( M_r \) of \( \text{Fe}_2\text{O}_3 = (2 \times 56) + (3 \times 16) = 112 + 48 = 160 \)
- From the equation, \( 2 \text{Fe} \) atoms are produced from 1 \( \text{Fe}_2\text{O}_3 \) molecule.
- Mass of Fe produced from 160 g of \( \text{Fe}_2\text{O}_3 = 112 \text{g} \)
- Mass of \( \text{Fe}_2\text{O}_3 \) required = \( \frac{160}{112} \times 28000 = 40,000 \text{g} \)
OR using moles:
- Moles of Fe = \( 28000 \div 56 = 500 \) moles
- Mole ratio \( \text{Fe}_2\text{O}_3 : \text{Fe} = 1 : 2 \)
- Moles of \( \text{Fe}_2\text{O}_3 = 500 \div 2 = 250 \) moles
- Mass of \( \text{Fe}_2\text{O}_3 = 250 \times 160 = 40,000 \text{g} \)