Question

0.5 g of an organic compound on combustion gave 1.46 g of $ CO_2 $ and 0.9 g of $ H_2O $. The percentage of carbon in the compound is ______ (Nearest integer) $\text{(Given : Molar mass (in g mol}^{-1}\text{ C : 12, H : 1, O : 16})$

Hint:

During combustion of an organic compound, all the carbon present in the compound is converted to \( CO_2 \). Use the molar mass ratio to find the mass of carbon in the \( CO_2 \) produced and then calculate the percentage of carbon in the organic compound.

Answer 1

Correct Answer: 80

Organic Compound → CO₂

Applying POAC on ‘C’

Option (mole) of ‘C’ in compound = $n_{\text{CO}_2} \times 1$

So mass of ‘C’ in compound = $\dfrac{1.46}{44} \times 12$

So, % of ‘C’ in compound =
\[ \dfrac{1.46}{44} \times \dfrac{12}{0.5} \times 100 \]
= 79.63

Answer 2

To determine the percentage of carbon in the compound, we'll first find the amount of carbon in the given combustion products.

Step 1: Calculate the moles of $CO_2$ produced

The molar mass of $CO_2$ is 44 g/mol (C:12, O:16 × 2 = 32, so 12 + 32 = 44).

Using the given mass of $CO_2$ (1.46 g), calculate the moles:

$\text{Moles of } CO_2 = \frac{1.46 \text{ g}}{44 \text{ g/mol}} \approx 0.0332 \text{ mol}$

Step 2: Calculate the mass of carbon in the $CO_2$

Each mole of $CO_2$ contains 1 mole of carbon. Therefore, the mass of carbon is:

$\text{Mass of Carbon} = 0.0332 \text{ mol} \times 12 \text{ g/mol} \approx 0.3984 \text{ g}$

Step 3: Calculate the percentage of carbon in the compound

The percentage of carbon is given by the formula:

$\text{Percentage of Carbon} = \left(\frac{\text{Mass of Carbon}}{\text{Mass of the compound}}\right) \times 100$

Substituting the known values:

$\text{Percentage of Carbon} = \left(\frac{0.3984 \text{ g}}{0.5 \text{ g}}\right) \times 100 \approx 79.68\%$

Rounding to the nearest integer, the percentage of carbon is 80%.

Verification

The computed value of 80% fits within the expected range (80,80).