Question

a 5.366 gram sample of an organic compound containing c, h and o is analyzed by combustion analysis and 13.54 grams of co₂ and 2.376 grams of h₂o are produced. in a separate experiment, the molar mass is found to be 122.1 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 $CO_2$, the molar mass of $CO_2$ is $M_{CO_2}=44.01\ g/mol$. The moles of $CO_2$ is $n_{CO_2}=\frac{13.54\ g}{44.01\ g/mol}=0.3077\ mol$. Since 1 mole of $CO_2$ contains 1 mole of C, moles of C, $n_C = 0.3077\ mol$.

Step2: Calculate moles of H

From the mass of $H_2O$, the molar mass of $H_2O$ is $M_{H_2O}=18.02\ g/mol$. The moles of $H_2O$ is $n_{H_2O}=\frac{2.376\ g}{18.02\ g/mol}=0.1319\ mol$. Since 1 mole of $H_2O$ contains 2 moles of H, moles of H, $n_H=2\times0.1319 = 0.2638\ mol$.

Step3: Calculate mass of C and H

Mass of C, $m_C=n_C\times12.01\ g/mol=0.3077\ mol\times12.01\ g/mol = 3.696\ g$. Mass of H, $m_H=n_H\times1.008\ g/mol=0.2638\ mol\times1.008\ g/mol=0.265\ g$.

Step4: Calculate mass of O

Mass of O, $m_O=5.366\ g-(3.696\ g + 0.265\ g)=1.405\ g$. Moles of O, $n_O=\frac{1.405\ g}{16.00\ g/mol}=0.0878\ mol$.

Step5: Find mole - ratio of C, H and O

Divide each number of moles by the smallest number of moles ($n_O = 0.0878\ mol$). For C: $\frac{0.3077\ mol}{0.0878\ mol}\approx3.5$. For H: $\frac{0.2638\ mol}{0.0878\ mol}\approx3$. For O: $\frac{0.0878\ mol}{0.0878\ mol}=1$. Multiply by 2 to get whole - numbers. So, the mole - ratio of C:H:O is 7:6:2.

Answer:

Empirical formula: $C_7H_6O_2$
Molecular formula:
The empirical formula mass, $M_{empirical}=7\times12.01+6\times1.008 + 2\times16.00=84.07+6.048+32.00 = 122.118\ g/mol$. Since the molar mass of the compound is 122.1 g/mol and the empirical formula mass is approximately 122.1 g/mol, the molecular formula is the same as the empirical formula.