Exactly of an unknown gaseous hydrocarbon is exploded with a massive excess of pure oxygen gas. Upon — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question
Question
Exactly $10\text{ mL}$ of an unknown gaseous hydrocarbon is exploded with a massive excess of pure oxygen gas. Upon cooling to room temperature, a volumetric contraction of exactly $25\text{ mL}$ is experimentally observed. When the residual gaseous mixture is subsequently treated with concentrated aqueous $KOH$, a further contraction of $20\text{ mL}$ is observed. Deduce the exact number of hydrogen atoms present in the molecular formula of this hydrocarbon.
💡 Solution & Explanation
Step 1: Write general combustion equation: $C_xH_y(g) + (x + y/4)O_2(g) \rightarrow xCO_2(g) + (y/2)H_2O(l)$. Step 2: Analyze the second contraction. $KOH$ specifically absorbs $CO_2$. Therefore, the volume of $CO_2$ produced is $20\text{ mL}$. Using Gay-Lussac's law, $10x = 20 \implies x = 2$. Step 3: Analyze the first contraction. Contraction = (Volume of gaseous reactants) - (Volume of gaseous products). Water is liquid at RT, so its volume is negligible. Contraction = $[10 + 10(x + y/4)] - 10x = 10 + 10x + 2.5y - 10x = 10 + 2.5y$. Step 4: Equate and solve for $y$. $10 + 2.5y = 25 \implies 2.5y = 15 \implies y = 6$. The hydrocarbon is $C_2H_6$ (Ethane), containing exactly 6 Hydrogen atoms.