$10 \mathrm{~mL}$ of gaseous hydrocarbon on combustion gives $40 \mathrm{~mL}$ of $\mathrm{CO}_2(\mathrm{~g})$ and $50 \mathrm{~mL}$ of water vapour. Total number of carbon and hydrogen atoms in the hydrocarbon is _________ .

<p>To solve this problem, we will use the fact that the volume ratio of gases in a reaction at the same conditions of temperature and pressure represents their mole ratio according to Avogadro's law. Thus, since the gas volumes given are at the same conditions, we can directly relate them to their stoichiometric coefficients in the balanced equation for combustion of a hydrocarbon.</p> <p>The general equation for complete combustion of a hydrocarbon with a formula $\mathrm{C}_x\mathrm{H}_y$ can be represented as:</p> $$ \mathrm{C}_x\mathrm{H}_y + (x + \frac{y}{4})\mathrm{O}_2 \rightarrow x\mathrm{CO}_2 + \frac{y}{2}\mathrm{H}_2\mathrm{O} $$ <p>Given:</p> <ul> <li>$10 \mathrm{~mL}$ of hydrocarbon ($\mathrm{C}_x\mathrm{H}_y$)</li> <li>$40 \mathrm{~mL}$ of $\mathrm{CO}_2$</li> <li>$50 \mathrm{~mL}$ of $\mathrm{H}_2\mathrm{O}$</li> </ul> <p>Since the volume ratios represent the mole ratios for gases, we can write the following ratios for the coefficients:</p> <p>$$ \frac{x}{1} = \frac{40 \mathrm{~mL} \mathrm{CO}_2}{10 \mathrm{~mL} \mathrm{C}_x\mathrm{H}_y} = 4 $$</p> <p>$$ \frac{y}{2} = \frac{50 \mathrm{~mL} \mathrm{H}_2\mathrm{O}}{10 \mathrm{~mL} \mathrm{C}_x\mathrm{H}_y} = 5 $$</p> <p>From the first ratio, we see that:</p> <p>$x = 4$</p> <p>This tells us that there are four carbon atoms in the hydrocarbon molecule.</p> <p>From the second ratio, by multiplying both sides by 2, we get:</p> $y = 5 \times 2 = 10$ <p>This means there are ten hydrogen atoms in the hydrocarbon molecule.</p> <p>So, the total number of carbon and hydrogen atoms in the hydrocarbon is:</p> $\text{Total atoms} = x + y = 4 + 10 = 14$ <p>Therefore, the total number of carbon and hydrogen atoms in the hydrocarbon is 14.</p>