Question

On combustion of 0.21 g of an organic compound containing C, H, and O, it gave 0.127 g of H$_2$O and 0.307 g of CO$_2$. The percentage of H and O in the given organic compound respectively are:

Hint:

When performing combustion analysis, ensure to account for the masses of carbon, hydrogen, and oxygen in the compound. Use stoichiometry to find the moles of each element and calculate the percentages.

Answer 1

Correct Answer: 53 - 54

Step 1: Calculate the moles of carbon in CO_2.
The molar mass of CO\(_2\) is 44 g/mol, and it gives 0.307 g of CO\(_2\) on combustion.
The number of moles of carbon in CO\(_2\) is given by: \[ \text{Moles of C} = \frac{\text{Mass of CO}_2}{\text{Molar mass of CO}_2} = \frac{0.307 \, \text{g}}{44 \, \text{g/mol}} = 0.006977 \, \text{mol} \] Since each mole of CO\(_2\) contains 1 mole of carbon, the moles of carbon in the compound are 0.006977 mol.
Step 2: Calculate the mass of carbon. The molar mass of carbon is 12 g/mol, so the mass of carbon is: \[ \text{Mass of C} = \text{Moles of C} \times \text{Molar mass of C} = 0.006977 \, \text{mol} \times 12 \, \text{g/mol} = 0.0837 \, \text{g} \]
Step 3: Calculate the moles of hydrogen in H_2\text{O}.
The molar mass of H\(_2\)O is 18 g/mol, and it gives 0.127 g of H\(_2\)O on combustion.
The number of moles of hydrogen in H\(_2\)O is given by: \[ \text{Moles of H} = \frac{\text{Mass of H}_2\text{O}}{\text{Molar mass of H}_2\text{O}} = \frac{0.127 \, \text{g}}{18 \, \text{g/mol}} = 0.007056 \, \text{mol} \] Each mole of H\(_2\)O contains 2 moles of hydrogen, so the moles of hydrogen in the compound are 0.007056 mol × 2 = 0.014112 mol.
Step 4: Calculate the mass of hydrogen. The molar mass of hydrogen is 1 g/mol, so the mass of hydrogen is: \[ \text{Mass of H} = \text{Moles of H} \times \text{Molar mass of H} = 0.014112 \, \text{mol} \times 1 \, \text{g/mol} = 0.014112 \, \text{g} \]
Step 5: Calculate the mass of oxygen. The mass of the compound is 0.21 g, so the mass of oxygen is given by: \[ \text{Mass of O} = \text{Total mass of compound} - (\text{Mass of C} + \text{Mass of H}) = 0.21 \, \text{g} - (0.0837 \, \text{g} + 0.014112 \, \text{g}) = 0.21 \, \text{g} - 0.097812 \, \text{g} = 0.112188 \, \text{g} \]
Step 6: Calculate the percentage of hydrogen and oxygen. The percentage of hydrogen is: \[ \text{Percentage of H} = \frac{\text{Mass of H}}{\text{Mass of compound}} \times 100 = \frac{0.014112 \, \text{g}}{0.21 \, \text{g}} \times 100 = 6.72% \] The percentage of oxygen is: \[ \text{Percentage of O} = \frac{\text{Mass of O}}{\text{Mass of compound}} \times 100 = \frac{0.112188 \, \text{g}}{0.21 \, \text{g}} \times 100 = 53.41% \]