Question

when 5.682 grams of a hydrocarbon, cxhy, were burned in a combustion analysis apparatus, 17.83 grams of co2 and 7.300 grams of h2o were produced. in a separate experiment, the molar mass of the compound was found to be 28.05 g/mol. determine the empirical formula and the molecular formula of the hydrocarbon. enter the elements in the order presented in the question. empirical formula = molecular formula =

Explanation:

Step1: Calculate moles of C

The mass of $CO_2$ is $17.83$ g. The molar mass of $CO_2$ is $44.01$ g/mol. Moles of $CO_2=\frac{17.83\ g}{44.01\ g/mol}=0.4051$ mol. Since 1 mole of $CO_2$ contains 1 mole of C, moles of C in the hydrocarbon is also $0.4051$ mol.

Step2: Calculate mass of C

The molar - mass of C is $12.01$ g/mol. Mass of C = moles of C×molar mass of C = $0.4051\ mol\times12.01\ g/mol = 4.865\ g$.

Step3: Calculate moles of H

The mass of $H_2O$ is $7.300$ g. The molar mass of $H_2O$ is $18.02$ g/mol. Moles of $H_2O=\frac{7.300\ g}{18.02\ g/mol}=0.4051$ mol. Since 1 mole of $H_2O$ contains 2 moles of H, moles of H in the hydrocarbon is $2\times0.4051\ mol = 0.8102$ mol.

Step4: Find the mole - ratio of C and H

Divide the moles of each element by the smaller number of moles. For C: $\frac{0.4051\ mol}{0.4051\ mol}=1$. For H: $\frac{0.8102\ mol}{0.4051\ mol}=2$. So the empirical formula is $CH_2$.

Step5: Determine the molecular formula

The molar mass of the empirical formula $CH_2$ is $12.01 + 2\times1.01=14.03$ g/mol. The molar mass of the compound is $28.05$ g/mol. The ratio $n=\frac{28.05\ g/mol}{14.03\ g/mol}\approx2$. So the molecular formula is $C_2H_4$.

Answer:

empirical formula = $CH_2$
molecular formula = $C_2H_4$