Question

0.48 g of an organic compound on complete combustion produced 0.22 g of CO₂. The percentage of C in the given organic compound is:

• A. 25
• B. 50
• C. 12.5
• D. 87.5

Hint:

The percentage of carbon is calculated using the ratio of carbon weight in CO₂ to the total weight of organic carbon.

Answer 1

The Correct Option is C

To determine the percentage of carbon in the given organic compound, we need to first use the data provided from the combustion reaction.

Given:

• Mass of the organic compound = 0.48 g
• Mass of CO₂ produced = 0.22 g

We know that during combustion, the carbon in the organic compound is converted into carbon dioxide (CO₂). Therefore, we start by calculating the mass of carbon in the 0.22 g of CO₂ produced.

The molar mass of CO₂ is 44 g/mol, which consists of 12 g/mol of carbon and 32 g/mol of oxygen. Thus, the mass fraction of carbon in CO₂ can be expressed as:

\(\frac{\text{Mass of C in CO₂}}{\text{Molar Mass of CO₂}} = \frac{12}{44}\). 

Calculate the amount of carbon in 0.22 g of CO₂:

\(\text{Mass of C} = 0.22 \times \frac{12}{44}\).

Simplifying the equation:

\(\text{Mass of C} = 0.22 \times \frac{12}{44} = 0.06 \text{ g}\).

Now, we calculate the percentage of carbon in the organic compound:

\(\text{Percentage of C} = \left( \frac{\text{Mass of C}}{\text{Mass of organic compound}} \right) \times 100 = \left( \frac{0.06}{0.48} \right) \times 100\).

Calculating the percentage:

\(\text{Percentage of C} = \left( \frac{0.06}{0.48} \right) \times 100 = 12.5\%\).

Thus, the percentage of carbon in the organic compound is 12.5%, which corresponds to the correct option given.