Question

combustion analysis

1. 33.658 g of oxygen was used to completely react with a sample of a hydrocarbon (cxhy) in a combustion reaction. the reaction products were 33.057 g of carbon dioxide and 10.816 g of water. ascertain the empirical formula of the compound. what is the molecular formula of the compound if its molar mass is 204 g/mol?

Explanation:

Step1: Calculate moles of carbon

The molar mass of $CO_2$ is $M_{CO_2}=44.01\ g/mol$. The moles of $CO_2$ is $n_{CO_2}=\frac{33.057\ g}{44.01\ g/mol}=0.7511\ mol$. Since 1 mole of $CO_2$ contains 1 mole of carbon, the moles of carbon $n_C = 0.7511\ mol$.

Step2: Calculate moles of hydrogen

The molar mass of $H_2O$ is $M_{H_2O}=18.02\ g/mol$. The moles of $H_2O$ is $n_{H_2O}=\frac{10.816\ g}{18.02\ g/mol}=0.6\ mol$. Since 1 mole of $H_2O$ contains 2 moles of hydrogen, the moles of hydrogen $n_H=2\times0.6\ mol = 1.2\ mol$.

Step3: Find the ratio of carbon to hydrogen for empirical - formula

Divide the number of moles of each element by the smaller number of moles. For carbon: $\frac{n_C}{n_C}=\frac{0.7511\ mol}{0.7511\ mol}=1$. For hydrogen: $\frac{n_H}{n_C}=\frac{1.2\ mol}{0.7511\ mol}\approx1.6$. Multiply both by 5 to get whole - number ratios. So the empirical formula is $C_5H_8$.

Step4: Calculate the empirical - formula mass

The empirical - formula mass of $C_5H_8$ is $M_{empirical}=5\times12.01\ g/mol + 8\times1.01\ g/mol=60.05\ g/mol+8.08\ g/mol = 68.13\ g/mol$.

Step5: Determine the molecular formula

The molar mass of the compound is $M = 204\ g/mol$. The ratio $n=\frac{M}{M_{empirical}}=\frac{204\ g/mol}{68.13\ g/mol}\approx3$. The molecular formula is $(C_5H_8)_3 = C_{15}H_{24}$.

Answer:

Empirical formula: $C_5H_8$; Molecular formula: $C_{15}H_{24}$