A saturated open-chain aliphatic hydrocarbon, When 50 ml of it is Burned, it produces 200 ml of carbon dioxide \& 250 ml of water vapour. What is its molecular formula?

To determine the molecular formula of the saturated open-chain aliphatic hydrocarbon (alkane), we start with the combustion reaction: \[ \text{C}_n\text{H}_{2n+2} + O_2 \rightarrow CO_2 + H_2O \] Given that burning 50 ml of the hydrocarbon produces 200 ml of \( CO_2 \) and 250 ml of \( H_2O \), we can make some calculations based on stoichiometry. 1. The volume of \( CO_2 \) (200 ml) indicates the number of moles of carbon in the hydrocarbon since each carbon atom produces one molecule of \( CO_2 \). Thus, there are 200 ml of \( CO_2 \), meaning there are 200 ml of carbon, which corresponds to 200/22.4 L or approximately 8.93 moles (using the gas law at standard temperature and pressure). 2. Since each mole of water vapor \( H_2O \) contains 2 hydrogen atoms, then the volume of water (250 ml) indicates the number of moles of hydrogen in the hydrocarbon. This gives us roughly 11.16 moles of \( H_2 \) from \( H_2O \) (250/22.4 L). Now, we set the equations from the combustion stoichiometry: - For carbon atoms: \( n = 200 ml\) - For hydrogen atoms: \( 2n + 2 \) corresponds to \( H \) = 2(200) ml. Now, we have: - From \( CO_2 \): \( n = 200 \) - From \( H_2O \): \( H = 2n + 2 = 11.16 \to 2n + 2 = 11.16 \) => \( 11.16 \rightarrow n \) which seems to not relate to our calculation. Summarizing \( n : H atoms = 200 : 400 ml \). Thus, the molecular formula comes down to: Overall for C_xH_y; 200:400 (n equations still being obstructive). Thus the molecular formula of the hydrocarbon is approximately \( C_8H_{18} \)(Octane).