WAEC Chemistry 2019 Paper

Concept:Molecular mass is the sum of the atomic masses of all atoms in a molecule.Compounds with the same total atomic mass will have equal molecular masses.Explanation:Calculate the molecular mass of each compound using approximate atomic masses: $C = 12$, $H = 1$, $O = 16$.Ethene is $\mathrm{C_2H_4}$.Its molecular mass is $2(12) + 4(1) = 24 + 4 = 28$.Ethyne is $\mathrm{C_2H_2}$.Its molecular mass is $2(12) + 2(1) = 24 + 2 = 26$.Carbon(II) oxide is $\mathrm{CO}$.Its molecular mass is $12 + 16 = 28$.Carbon(IV) oxide is $\mathrm{CO_2}$.Its molecular mass is $12 + 2(16) = 12 + 32 = 44$.Comparing the results, $\mathrm{C_2H_4}$ and $\mathrm{CO}$ both have a molecular mass of $28$.Thus, the pair having the same molecular mass is Carbon(II) oxide and Ethene.Answer:Option C — Carbon(II) oxide and Ethene.