Answer: C. Combustion of hydrocarbon Z: all C ends up in \(\text{CO}_2\), all H ends up in \(\text{H}_2\text{O}\). Work backwards.
Moles of \(\text{CO}_2\) = \(4.4 / 44 = 0.10\) mol → moles of C in Z = 0.10.
Moles of \(\text{H}_2\text{O}\) = \(2.7 / 18 = 0.15\) mol → moles of H in Z = \(0.15 \times 2 = 0.30\).
Mole ratio C : H = 0.10 : 0.30 = 1 : 3. Empirical formula = \(\text{CH}_3\) → C.
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