Question

10 mL of gaseous hydrocarbon on combustion gives 40 mL of CO\(_2\)(g) and 50 mL of water vapour. The total number of carbon and hydrogen atoms in the hydrocarbon is ______ .

Answer 1

Correct Answer: 14

Let the formula of the hydrocarbon be \( \text{C}_x\text{H}_y \). On complete combustion, the reaction can be represented as:
\[\text{C}_x\text{H}_y + \left(x + \frac{y}{4}\right) \text{O}_2 \rightarrow x \text{CO}_2 + \frac{y}{2} \text{H}_2\text{O}\]
Step-by-step Calculation:
According to the given data, 10 mL of hydrocarbon produces 40 mL of \(\text{CO}_2\) and 50 mL of water vapor (\(\text{H}_2\text{O}\)).
From the stoichiometry of the reaction:
\[x \times 10 \, \text{mL} = 40 \, \text{mL} \implies x = 4\]
Similarly, for water:
\[\frac{y}{2} \times 10 \, \text{mL} = 50 \, \text{mL} \implies y = 10\]
Therefore, the hydrocarbon is \( \text{C}_4\text{H}_{10} \).
Total number of carbon and hydrogen atoms:
\[\text{Total atoms} = x + y = 4 + 10 = 14\]
Conclusion: The total number of carbon and hydrogen atoms in the hydrocarbon is 14.

Answer 2

Given:

The combustion reaction of a hydrocarbon \( C_xH_y \) is given by: 

\[ C_xH_y (10 \, \text{ml}) + O_2 \rightarrow CO_2 + H_2O \]

Step 1: The balanced combustion equation can be written as:

\[ C_xH_y + \left( x + \frac{y}{4} \right) O_2 \rightarrow xCO_2 + \frac{y}{2} H_2O \]

Step 2: Equating the coefficients of the elements gives the following relationships:

\[ 10x = 40 \quad \text{(for carbon)} \] Solving for \( x \), we get: \[ x = 4. \]

Step 3: For hydrogen, we have:

\[ 5y = 50 \quad \text{(for hydrogen)} \] Solving for \( y \), we get: \[ y = 10. \]

Step 4: The hydrocarbon formula is \( C_4H_{10} \).