Sum of Carbon and Hydrogen Atoms in Combustion of Alkane

To find the sum of the number of carbon and hydrogen atoms in the alkane $(\text{x})$, we begin by writing the general formula for an alkane, which is $\text{C}_n\text{H}_{2n+2}$, where $n$ is the number of carbon atoms.

The general balanced chemical equation for the complete combustion of one mole of an alkane in oxygen is given by: $\text{C}_n\text{H}_{2n+2} + \left( n + \frac{2n+2}{4} \right) \text{O}_2 \longrightarrow n \text{CO}_2 + (n+1) \text{H}_2\text{O}$

Simplifying the coefficient of oxygen ($\text{O}_2$): $n + \frac{2n+2}{4} = n + \frac{n+1}{2} = \frac{3n+1}{2}$

According to the question, $1\text{ mole}$ of the alkane requires $8\text{ moles}$ of oxygen for complete combustion. Therefore, we set the stoichiometric coefficient of $\text{O}_2$ equal to $8$: $\frac{3n+1}{2} = 8$

Solving for $n$: $3n + 1 = 16$ $3n = 15$ $n = 5$

Thus, the molecular formula of the alkane is: $\text{C}_5\text{H}_{2(5)+2} = \text{C}_5\text{H}_{12}$

The alkane contains:

• Number of Carbon atoms ($N_{\text{C}}$) = $5$
• Number of Hydrogen atoms ($N_{\text{H}}$) = $12$

The sum of the number of carbon and hydrogen atoms in the alkane is: $\text{Sum} = N_{\text{C}} + N_{\text{H}} = 5 + 12 = 17$