Question

0.1435 g of silver chloride was obtained from 0.0945 g of an organic compound by Carius method. The percentage of chlorine by weight in the compound is (molar mass of AgCl = 143.5 g mol\(^{-1}\))

• A. 18.9 %
• B. 37.6 %
• C. 24.9 %
• D. 56.7 %

Hint:

To calculate the percentage composition, use the formula: \[ \text{Percentage of element} = \frac{\text{Mass of element}}{\text{Total mass of compound}} \times 100 \] In this case, chlorine's mass was determined from the moles of AgCl and then used to find the percentage.

Answer 1

The Correct Option is B

Step 1: Let the weight of chlorine in the compound be \( x \). The weight of AgCl is 0.1435 g, and the molar mass of AgCl is 143.5 g mol\(^{-1}\). Now, from the given data, we know that 0.0945 g of organic compound gives 0.1435 g of AgCl. \[ \frac{0.1435}{143.5} = \text{moles of AgCl} \] The number of moles of AgCl formed is: \[ \text{moles of AgCl} = \frac{0.1435}{143.5} = 0.001 \text{ mol of AgCl} \]

 Step 2: Since the molar ratio of AgCl to chlorine is 1:1, we know that 1 mole of AgCl corresponds to 1 mole of chlorine. Therefore, the number of moles of chlorine is also 0.001 mol. Now, the mass of chlorine is: \[ \text{Mass of chlorine} = 0.001 \text{ mol} \times 35.5 \text{ g mol}^{-1} = 0.0355 \text{ g} \] 

Step 3: The percentage of chlorine by weight in the compound is: \[ \text{Percentage of chlorine} = \frac{0.0355}{0.0945} \times 100 = 37.6 \% \] Thus, the percentage of chlorine in the organic compound is 37.6 %.