Question

when 3.478 grams of a hydrocarbon, cxhy, were burned in a combustion analysis apparatus, 11.76 grams of co2 and 2.407 grams of h2o were produced. in a separate experiment, the molar mass of the compound was found to be 26.04 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

From the mass of $CO_2$, moles of $C$ can be found. Molar mass of $CO_2$ is $44.01\ g/mol$. Moles of $CO_2=\frac{11.76\ g}{44.01\ g/mol}=0.2672\ mol$. Since 1 mole of $CO_2$ contains 1 mole of $C$, moles of $C = 0.2672\ mol$.

Step2: Calculate moles of H

From the mass of $H_2O$, moles of $H$ can be found. Molar mass of $H_2O$ is $18.02\ g/mol$. Moles of $H_2O=\frac{3.478\ g}{18.02\ g/mol}=0.193\ mol$. Since 1 mole of $H_2O$ contains 2 moles of $H$, moles of $H=2\times0.193\ mol = 0.386\ mol$.

Step3: Find the ratio of C and H

Divide the moles of each element by the smaller number of moles. For $C:\frac{0.2672\ mol}{0.2672\ mol}=1$, for $H:\frac{0.386\ mol}{0.2672\ mol}\approx1.44$. Multiply by 2 to get whole - number ratio. So, the ratio of $C:H = 2:3$.

Step4: Determine the empirical formula

The empirical formula is $C_2H_3$.

Step5: Determine the molecular formula

The empirical formula mass of $C_2H_3$ is $2\times12.01\ g/mol+3\times1.01\ g/mol = 27.05\ g/mol$. The molar mass of the compound is $26.04\ g/mol\approx27.05\ g/mol$. So, the molecular formula is the same as the empirical formula, $C_2H_3$.

Answer:

empirical formula = $C_2H_3$
molecular formula = $C_2H_3$