Question

a 7.044 - gram sample of an organic compound containing c, h and o is analyzed by combustion analysis and 10.32 grams of co₂ and 4.227 grams of h₂o are produced. in a separate experiment, the molar mass is found to be 60.05 g/mol. determine the empirical formula and the molecular formula of the organic compound. (enter the elements in the order c, h, o.) empirical formula: molecular formula:

Explanation:

Step1: Calculate moles of C

From the mass of $CO_2$, find moles of C. Molar mass of $CO_2$ is $44.01\ g/mol$. Moles of $CO_2=\frac{10.32\ g}{44.01\ g/mol}\approx0.2345\ mol$. Since 1 - mole of $CO_2$ has 1 - mole of C, moles of C is also $0.2345\ mol$.

Step2: Calculate moles of H

From the mass of $H_2O$, find moles of H. Molar mass of $H_2O$ is $18.02\ g/mol$. Moles of $H_2O=\frac{4.227\ g}{18.02\ g/mol}\approx0.2346\ mol$. Since 1 - mole of $H_2O$ has 2 - moles of H, moles of H is $2\times0.2346\ mol = 0.4692\ mol$.

Step3: Calculate mass of C and H

Mass of C = moles of C×molar mass of C = $0.2345\ mol\times12.01\ g/mol\approx2.816\ g$. Mass of H = moles of H×molar mass of H = $0.4692\ mol\times1.008\ g/mol\approx0.473\ g$.

Step4: Calculate mass of O

Mass of the sample is $7.044\ g$. Mass of O = mass of sample−(mass of C + mass of H) = $7.044\ g-(2.816\ g + 0.473\ g)=3.755\ g$.

Step5: Calculate moles of O

Molar mass of O is $16.00\ g/mol$. Moles of O = $\frac{3.755\ g}{16.00\ g/mol}\approx0.2347\ mol$.

Step6: Find the empirical - formula ratio

Divide the moles of each element by the smallest number of moles (0.2345 mol in this case). For C: $\frac{0.2345\ mol}{0.2345\ mol}=1$. For H: $\frac{0.4692\ mol}{0.2345\ mol}\approx2$. For O: $\frac{0.2347\ mol}{0.2345\ mol}\approx1$. So the empirical formula is $CH_2O$.

Step7: Calculate the empirical - formula mass

Empirical - formula mass of $CH_2O$ is $12.01\ g/mol+(2\times1.008\ g/mol)+16.00\ g/mol = 30.03\ g/mol$.

Step8: Determine the molecular formula

Molar mass of the compound is $60.05\ g/mol$. $n=\frac{\text{Molar mass of compound}}{\text{Empirical - formula mass}}=\frac{60.05\ g/mol}{30.03\ g/mol}\approx2$. So the molecular formula is $C_2H_4O_2$.

Answer:

Empirical formula: $CH_2O$
Molecular formula: $C_2H_4O_2$