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) $\text{(Given : Molar mass (in g mol}^{-1}\text{ C : 12, H : 1, O : 16})$
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) $\text{(Given : Molar mass (in g mol}^{-1}\text{ C : 12, H : 1, O : 16})$
Show Hint
Correct Answer: 80
Approach Solution - 1
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
Approach Solution -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).
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