Question

An organic compound weighing 500 mg, produced 220 mg of CO$_2$ on complete combustion. The percentage composition of carbon in the compound is ............... % (nearest integer).
(Given molar mass in g mol$^-$ of C: 12, O: 16)

Hint:

When calculating the percentage composition, use the ratio of the mass of the element to the total mass of the compound, then multiply by 100.

Answer 1

Correct Answer: 12

Step 1: Determine the amount of carbon in CO\(_2\):
When the compound undergoes complete combustion, carbon from the compound reacts with oxygen to produce CO\(_2\). The molar mass of CO\(_2\) is: \[ \text{Molar mass of CO}_2 = 12 + 2 \times 16 = 44 \text{ g/mol} \] The molar mass of carbon (C) is 12 g/mol. In 44 g of CO\(_2\), 12 g is carbon. Therefore, the mass of carbon in 220 mg of CO\(_2\) can be calculated as: \[ \text{Mass of C} = \left(\frac{12}{44}\right) \times 220 = 60 \text{ mg} \]
Step 2: Calculate the percentage of carbon in the compound:
The mass of carbon in the original organic compound is 60 mg. The total mass of the compound is 500 mg. The percentage of carbon in the compound is given by: \[ \text{Percentage of C} = \left(\frac{60}{500}\right) \times 100 = 12% \]