Question

0.378g of an organic compound containing carbon and hydrogen was subjected to combustion by Leibig's method, the CO2 and H2O formed were passed through KOH tube and anhydrous CaCl₂ tube. At the end of the experiment, the increase in the respective weights of KOH and CaCl₂tubes were 0.26334g and 0.162g. The percentage of carbon in the given organic compound is (Nearest integer)

Answer 1

19%

Answer 2

Solution:

Given:

• Mass of organic compound = 0.378 g
• Increase in KOH tube = 0.26334 g (mass of CO₂ produced)
• Increase in CaCl₂ tube = 0.162 g (mass of H₂O produced)

Steps:

1. Find moles of CO₂ produced:

   $n(\text{CO}_2) = \frac{0.26334\,\text{g}}{44\,\text{g/mol}} \approx 0.00598\,\text{mol}$

2. Determine mass of Carbon:

   Each mole of CO₂ contains 1 mole of C (12 g/mol):

   $\text{Mass of C} = 0.00598\,\text{mol} \times 12\,\text{g/mol} \approx 0.07176\,\text{g}$

3. Calculate percentage of Carbon in the compound:

   $\%\,\text{C} = \left(\frac{0.07176\,\text{g}}{0.378\,\text{g}}\right) \times 100 \approx 18.99\%$

Rounded to the nearest integer, the percentage of Carbon is 19%.