JAMB Chemistry 2019 Paper

Concept:Combustion analysis uses the volumes and masses of products to find the mole ratio of carbon to hydrogen in the hydrocarbon.Explanation:At s.t.p., $1$ mole of any gas occupies $22.4 \, \mathrm{dm}^3$.Moles of $\mathrm{CO}_2 = \frac{89.6}{22.4} = 4$ moles.Thus the hydrocarbon contains $4$ moles of carbon atoms, so $x = 4$.Molar mass of water is $18 \, \mathrm{g \, mol}^{-1}$.Moles of $\mathrm{H}_2\mathrm{O} = \frac{54}{18} = 3$ moles.Each mole of water contains $2$ moles of hydrogen, so hydrogen moles $= 6$.Hence $y = 6$.Therefore the hydrocarbon formula is $\mathrm{C}_4\mathrm{H}_6$.Answer:D. $\mathrm{C}_4\mathrm{H}_6$