Chemistry Q74 – JEE-MAIN 2025

Home Chemistry Basic Organic ChemistryQuestion #170

An organic compound weighing

500 mg

, produced

220 mg

C O_{2}

, on complete combustion. The percentage composition of carbon in the compound is _______ \%. (nearest integer) (Given molar mass in

{g mol}^{- 1}

C : 12, O : 16

)

Enter Calculated Numerical Value:

Clue 1 of 3: Principle of Conservation of Mass

In combustion analysis, all the carbon from the organic compound is converted into carbon dioxide.

🧠 Full Solution Path

Step 1: Calculate Molar Mass of

C O_{2}

✦ Active

The molar mass of carbon ( $C$ ) is $12 {g mol}^{- 1}$ and oxygen ( $O$ ) is $16 {g mol}^{- 1}$ . The molar mass of carbon dioxide ( $C O_{2}$ ) is calculated as:

Molar mass of C O_{2} = 12 + (2 \times 16) = 12 + 32 = 44 {g mol}^{- 1}

Step 2: Determine Mass of Carbon in

C O_{2}

○ Expand

From the molar mass, $44 g$ of $C O_{2}$ contains $12 g$ of carbon. We produced $220 mg$ of $C O_{2}$ . The mass of carbon in this $C O_{2}$ is:

Mass of C = (\frac{Atomic mass of C}{Molar mass of C O_{2}}) \times Mass of C O_{2}

Mass of C = (\frac{12}{44}) \times 220 mg = 60 mg

💡 Teacher's Secret Hint

Ensure units are consistent (mg in this case) throughout the calculation.

Step 3: Calculate Percentage Composition of Carbon○ Expand

The $60 mg$ of carbon came from the $500 mg$ organic compound. The percentage composition of carbon in the compound is:

Percentage of C = (\frac{Mass of C}{Mass of organic compound}) \times 100 %

Percentage of C = (\frac{60 mg}{500 mg}) \times 100 % = \frac{60}{5} % = 12 %

The percentage composition of carbon in the compound is $12 %$ . Rounded to the nearest integer, it is $12$ .

💡 Teacher's Secret Hint

Remember to express the final answer as a percentage and round to the nearest integer as requested.

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Chemistry Question 74 – JEE-MAIN 2025

Clue 1 of 3: Principle of Conservation of Mass

🧠 Full Solution Path