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)
(Given molar mass in g mol$^-$ of C: 12, O: 16)
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Correct Answer: 12
Approach Solution - 1
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: \[ \text{Percentage of C} = \left(\frac{60}{500}\right) \times 100 = 12% \]
Approach Solution -2
Step 1:
When an organic compound undergoes complete combustion, it produces CO₂ and H₂O. The carbon in the compound is converted into CO₂. Therefore, the amount of carbon in the CO₂ produced will help in calculating the percentage composition of carbon in the compound.
Step 2:
We are given that 220 mg of CO₂ is produced. To find the mass of carbon in CO₂, we use the molar masses:
- Molar mass of CO₂ = 12 (C) + 32 (O₂) = 44 g/mol.
- Molar mass of carbon (C) = 12 g/mol.
Now, the mass of carbon in 220 mg of CO₂ can be calculated using the ratio of the molar masses of carbon and CO₂:
Mass of carbon = (12 / 44) × 220 = 60 mg
Step 3:
The total mass of the compound is 500 mg. To find the percentage of carbon in the compound, we use the formula:
Percentage of carbon = (Mass of carbon / Mass of compound) × 100
Substituting the values:
Percentage of carbon = (60 / 500) × 100 = 12%
Final Answer:
The percentage composition of carbon in the compound is 12%.
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