Question

On complete combustion $0.12\ \text{g}$ of an organic compound gives $0.11\ \text{g}$ of $CO_{2}$. What is the percentage of carbon in the organic compound?

• A. 15%
• B. 20%
• C. 25%
• D. 17.5%
• E. 21.5%

Hint:

The ratio $\frac{12}{44}$ represents the mass fraction of carbon in carbon dioxide.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This is a problem in quantitative elemental analysis. When an organic compound is combusted, all the carbon in the compound is converted into carbon dioxide (CO\(_2\)). By measuring the mass of CO\(_2\) produced, we can determine the mass of carbon in the original sample and hence its percentage.
Step 2: Key Formula or Approach:
1. Molar Masses: We need the molar mass of Carbon (C) and Carbon Dioxide (CO\(_2\)). - Molar mass of C = 12 g/mol - Molar mass of CO\(_2\) = 12 + 2(16) = 44 g/mol 2. Mass of Carbon: The mass of carbon in a given mass of CO\(_2\) can be found using the ratio of their molar masses. \[ \text{Mass of C} = \left(\frac{\text{Molar mass of C}}{\text{Molar mass of CO}_2}\right) \times \text{Mass of CO}_2 \] 3. Percentage of Carbon: \[ % \text{C} = \frac{\text{Mass of C in sample}}{\text{Total mass of sample}} \times 100% \] Step 3: Detailed Explanation:
We are given: - Mass of organic compound = 0.12 g - Mass of CO\(_2\) produced = 0.11 g First, calculate the mass of carbon in the 0.11 g of CO\(_2\). \[ \text{Mass of C} = \frac{12}{44} \times 0.11 \text{ g} \] \[ \text{Mass of C} = \frac{3}{11} \times 0.11 \text{ g} = 3 \times 0.01 = 0.03 \text{ g} \] This mass of carbon came from the original 0.12 g sample of the organic compound. Now, calculate the percentage of carbon in the compound. \[ % \text{C} = \frac{\text{Mass of C}}{\text{Mass of compound}} \times 100% \] \[ % \text{C} = \frac{0.03 \text{ g}}{0.12 \text{ g}} \times 100% \] \[ % \text{C} = \frac{3}{12} \times 100% = \frac{1}{4} \times 100% = 25% \] Step 4: Final Answer:
The percentage of carbon in the organic compound is 25%.