Question

For the following question, enter the correct numerical value upto TWO decimal places. If the numerical value has more than two decimal places, round-off the value to TWO decimal places. (For example: Numeric value 5 will be written as 5.00 and 2.346 will be written as 2.35)
0.2 g of an organic compound on complete combustion produces 0.44 g of CO2, then the percentage of carbon in it is _____ %.

Hint:

For combustion analysis, mass of carbon = \(\frac{12}{44} \times \text{mass of CO}_2\), then %C = \(\frac{\text{Mass of C}}{\text{Mass of compound}} \times 100%\).

Answer 1

Correct Answer: 60

Step 1: Determine the mass of carbon in CO\(_2\). The molar mass of CO\(_2\) is 44 g/mol, and carbon contributes 12 g/mol.
Step 2: Use the ratio of masses to find the mass of carbon: \[ \text{Mass of C} = \frac{12}{44} \times 0.44 \, \text{g} = 0.12 \, \text{g} \]
Step 3: Calculate the percentage of carbon in the compound: \[ % \text{C} = \frac{\text{Mass of C}}{\text{Mass of compound}} \times 100 = \frac{0.12}{0.22} \times 100 \]
Step 4: Correct the mass of compound: given is 0.2 g \[ % \text{C} = \frac{0.12}{0.2} \times 100 = 60.00% \]
Step 5: Check the calculation: mass of carbon in CO2 = \(0.44 \times \frac{12}{44} = 0.12\) g. Percentage = \(0.12/0.22\) ? Wait, mass of compound = 0.2 g (given), so \[ % \text{C} = \frac{0.12}{0.2} \times 100 = 60.00 % \]