Chemistry Q73 – JEE-MAIN 2026

Home Chemistry Basic Organic ChemistryQuestion #108

2.0 g of a bromo hydrocarbon (X) was subjected to Carius analysis, gave 3.36 g of AgBr. The percentage of carbon in the compound (X) is 26.7\%. Total number of carbon atoms in the empirical formula for compound (X) is _______. (Given molar mass in g mol

^{- 1}

H:1, C:12, Br:80, Ag:108)

Enter Calculated Numerical Value:

Clue 1 of 3: Conceptual Principle

Carius method is used for the quantitative estimation of halogens. The halogen in the organic compound is converted to its silver halide, which allows for the calculation of the halogen's mass percentage.

🧠 Full Solution Path

Step 1: Calculate the percentage of Bromine in compound (X)✦ Active

First, calculate the molar mass of $A g B r$ using the given atomic masses ( $A g = 108$ , $B r = 80$ ). Then, determine the mass of bromine present in the $3.36 g$ of $A g B r$ obtained from the Carius analysis. Finally, calculate the percentage of bromine in the original $2.0 g$ sample of compound (X).

Molar mass of AgBr = 108 + 80 = 188 g/mol

Mass of Br in 3.36 g of AgBr = \frac{80}{188} \times 3.36 g = 1.430 g

Percentage of Br in compound (X) = \frac{1.430 g}{2.0 g} \times 100 = 71.5 %

Step 2: Calculate the percentage of Hydrogen and the relative moles of C, H, and Br○ Expand

Since the compound is a bromo hydrocarbon, it contains only Carbon, Hydrogen, and Bromine. Use the given percentage of Carbon and the calculated percentage of Bromine to find the percentage of Hydrogen. Then, assume a $100 g$ sample of the compound to convert these percentages into masses and subsequently into moles for each element.

Percentage of H in compound (X) = 100 % - (% C + % B r)

Percentage of H = 100 % - (26.7 % + 71.5 %) = 100 % - 98.2 % = 1.8 %

Assuming 100 g of compound (X):

Moles of C = \frac{26.7 g}{12 g/mol} = 2.225 mol

Moles of H = \frac{1.8 g}{1 g/mol} = 1.8 mol

Moles of Br = \frac{71.5 g}{80 g/mol} = 0.89375 mol

Step 3: Determine the empirical formula and the number of carbon atoms○ Expand

To find the simplest whole number ratio for the empirical formula, divide the moles of each element by the smallest number of moles calculated. If the ratios are not whole numbers, multiply by a suitable integer to obtain the smallest whole number ratio. Finally, identify the number of carbon atoms in this empirical formula.

Divide by the smallest number of moles (0.89375 mol for Br):

Ratio of C = \frac{2.225}{0.89375} \approx 2.489 \approx 2.5

Ratio of H = \frac{1.8}{0.89375} \approx 2.014 \approx 2

Ratio of Br = \frac{0.89375}{0.89375} = 1

The ratio C:H:Br is 2.5:2:1. Multiply by 2 to get whole numbers:

C:H:Br = 5 : 4 : 2

The empirical formula is C_{5} H_{4} B r_{2}

The total number of carbon atoms in the empirical formula is 5.

💡 Teacher's Secret Hint

Remember to multiply all ratios by the same integer to obtain the smallest whole numbers for the empirical formula.

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Chemistry Question 73 – JEE-MAIN 2026

Clue 1 of 3: Conceptual Principle

🧠 Full Solution Path