Many organic compounds contain carbon and hydrogen only.
(a) (i) Organic compound A has the following composition by mass.
C, 82.76%; H, 17.24%
Calculate the empirical formula of compound A.
(ii) Compound B has the empirical formula \(CH_2\) and a relative molecular mass of 70. Determine the molecular formula of compound B.
(b) Fig. 7.1 shows a section of polymer Q
(c) Propene, \(C_3H_6\), can be produced by heating \(C_{11}H_{24}\). The products of the reaction are propene, hydrogen and one other product in a 1:1:1 mole ratio. Complete the symbol equation for this reaction 
(d) Carboxylic acids and esters contain carbon, hydrogen and oxygen only. An ester X and a carboxylic acid Y both contain 3 carbon atoms. X and Y have the same molecular formula.
(i) State the name given to compounds with the same molecular formula but different structural formulae.
(ii) Esters are made by the reaction between carboxylic acids and alcohols. Ester X is methyl ethanoate. Name the carboxylic acid and the alcohol used to make methyl ethanoate.
carboxylic acid ……………..
alcohol ………………….
(iii) Draw the displayed formula of carboxylic acid Y. Name the carboxylic acid.
Most-appropriate topic codes (Cambridge IGCSE Chemistry 0620):
• Topic 3.3 — The mole and the Avogadro constant (Part (a))
• Topic 11.5 — Alkenes & addition polymerisation (Part (b))
• Topic 11.3 — Fuels & cracking (Part (c))
• Topic 11.1 — Formulae, functional groups and terminology / Structural isomers (Part (d)(i))
• Topic 11.6 — Alcohols & 11.7 — Carboxylic acids (Parts (d)(ii), (d)(iii))
(a)(i) M1: Divide the percentage by the atomic mass: C = 82.76 / 12 = 6.90; H = 17.24 / 1 = 17.24.
M2: Divide by the smaller number (6.90): C = 1; H = 17.24 / 6.90 = 2.5.
M3: Multiply to get whole numbers (×2): C = 2, H = 5. Hence the empirical formula is C₂H₅.
(a)(ii) The empirical formula mass of CH₂ is 12 + (2×1) = 14. The molecular mass is 70. The multiplier (n) = 70 / 14 = 5. Therefore the molecular formula is (CH₂)₅ = C₅H₁₀.
(b) The polymer structure shows a carbon chain with a methyl side group, indicating it is poly(propene) made from propene monomers. The monomer must have a double bond.
Monomer displayed formula: 
Monomer name: propene.
(c) C₁₁H₂₄ → C₃H₆ + H₂ + another product. Balancing carbons: 11 – 3 = 8 carbons in the other product. Balancing hydrogens: Left = 24, Right = 6 (propene) + 2 (hydrogen) = 8, so remaining product needs 16 H. The other product is C₈H₁₆ (an alkene). The balanced equation is:
C₁₁H₂₄ → C₃H₆ + H₂ + C₈H₁₆.
(d)(i) Compounds with the same molecular formula but different structural formulae are called structural isomers.
(d)(ii) Methyl ethanoate is formed from the condensation reaction of ethanoic acid (carboxylic acid) and methanol (alcohol).
(d)(iii) Both X and Y have the same molecular formula (C₃H₆O₂). Since Y is a carboxylic acid with 3 carbons, it is propanoic acid (C₂H₅COOH).
Displayed formula of propanoic acid: 
Name of carboxylic acid Y: propanoic acid.
