the combustion of 135 mg of a hydrocarbon sample produces 440. mg of co₂ and 135 mg h₂o. the molar mass of…

Explanation:

Step1: Calculate moles of carbon

The molar mass of $CO_2$ is $M_{CO_2}=44\ g/mol$. The mass of $CO_2$ is $m_{CO_2} = 440\ mg=0.440\ g$. The number of moles of $CO_2$, $n_{CO_2}=\frac{m_{CO_2}}{M_{CO_2}}=\frac{0.440\ g}{44\ g/mol}=0.01\ mol$. Since 1 mole of $CO_2$ contains 1 mole of carbon, the moles of carbon in the hydrocarbon, $n_C = 0.01\ mol$.

Step2: Calculate moles of hydrogen

The molar mass of $H_2O$ is $M_{H_2O}=18\ g/mol$. The mass of $H_2O$ is $m_{H_2O}=135\ mg = 0.135\ g$. The number of moles of $H_2O$, $n_{H_2O}=\frac{m_{H_2O}}{M_{H_2O}}=\frac{0.135\ g}{18\ g/mol}=0.0075\ mol$. Since 1 mole of $H_2O$ contains 2 moles of hydrogen, the moles of hydrogen in the hydrocarbon, $n_H=2\times n_{H_2O}=2\times0.0075\ mol = 0.015\ mol$.

Step3: Find the empirical - formula ratio

Divide the number of moles of each element by the smaller number of moles. For carbon: $\frac{n_C}{n_{min}}=\frac{0.01\ mol}{0.005\ mol}=2$ (where $n_{min}$ is the smaller of $n_C$ and $n_H$, here we can consider a common - factor operation to get whole - number ratios). For hydrogen: $\frac{n_H}{n_{min}}=\frac{0.015\ mol}{0.005\ mol}=3$. So the empirical formula is $C_2H_3$.

Step4: Calculate the empirical - formula mass

The empirical - formula mass of $C_2H_3$, $M_{empirical}=2\times12\ g/mol+3\times1\ g/mol = 27\ g/mol$.

Step5: Determine the molecular formula

The molar mass of the hydrocarbon is $M = 270\ g/mol$. The ratio $n=\frac{M}{M_{empirical}}=\frac{270\ g/mol}{27\ g/mol}=10$. The molecular formula is $(C_2H_3)n$, so the molecular formula is $C{20}H_{30}$.

Answer:

$C_{20}H_{30}$