Question

3.(a)(i) Draw the structure of the sixth member of the alkene series.
(ii) Calculate the relative molecular mass of the sixth member of the alkene series. [H=1, C=12]
(iii) State one difference between cracking and reforming in the petroleum industry.

Accepted Answer

(i) The sixth member of the alkene series is heptene, C₇H₁₄ (structure: a 7-carbon chain with one C=C double bond).
(ii) Molecular mass = (12×7)+(1×14) = 98.
(iii) Cracking involves breaking large hydrocarbon molecules into smaller ones, while reforming involves the re-arrangement of atoms within molecules without necessarily changing their size.

WAEC Chemistry 2018 Theory — Question 15

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