of a gaseous hydrocarbon is completely burnt in excess oxygen, producing of and of vapour under simi — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question
Question
$10\text{ mL}$ of a gaseous hydrocarbon is completely burnt in excess oxygen, producing $40\text{ mL}$ of $CO_2(g)$ and $50\text{ mL}$ of $H_2O$ vapour under similar conditions of temperature and pressure. The molecular formula of the hydrocarbon is:
💡 Solution & Explanation
Step 1: Write the general combustion equation. $C_xH_y(g) + (x+y/4)O_2(g) \rightarrow xCO_2(g) + (y/2)H_2O(g)$. Step 2: Apply volume relationships (since water is in vapour phase, Gay-Lussac's law applies directly). Volume of $CO_2 = 10x = 40\text{ mL} \implies x = 4$. Volume of $H_2O = 10(y/2) = 50\text{ mL} \implies 5y = 50 \implies y = 10$. Step 3: The formula of the hydrocarbon is $C_4H_{10}$.