Chemistry Q74 – JEE-MAIN 2025

Home Chemistry Basic ChemistryQuestion #70

0.5 g of an organic compound on combustion gave 1.46 g of

C O_{2}

and 0.9 g of

H_{2} O

. The percentage of carbon in the compound is _______. (Nearest integer) [Given : Molar mass (in g

m o l^{- 1}

) C : 12, H : 1, O : 16]

Enter Calculated Numerical Value:

Clue 1 of 3: Combustion Analysis

In the complete combustion of an organic compound, all the carbon is converted to carbon dioxide ( $C O_{2}$ ) and all the hydrogen is converted to water ( $H_{2} O$ ).

🧠 Full Solution Path

Step 1: Calculate the mass of carbon from

C O_{2}

produced✦ Active

The molar mass of $C O_{2}$ is $12 + (2 \times 16) = 44 g/mol$ . The mass of carbon in $1.46 g$ of $C O_{2}$ is calculated using the ratio of molar masses:

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

Mass of C = \frac{12 g/mol}{44 g/mol} \times 1.46 g = 0.39818 g

Step 2: Calculate the percentage of carbon in the organic compound○ Expand

The percentage of carbon in the organic compound is found by dividing the mass of carbon by the total mass of the organic compound and multiplying by $100 %$ . The mass of the organic compound is $0.5 g$ .

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

Percentage of C = \frac{0.39818 g}{0.5 g} \times 100 % = 79.636 %

Step 3: Round to the nearest integer○ Expand

Rounding $79.636 %$ to the nearest integer gives $80 %$ .

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Chemistry Question 74 – JEE-MAIN 2025

Clue 1 of 3: Combustion Analysis

🧠 Full Solution Path