A substance on analysis gave the following data: 0.3112 g gave 0.4291 g of `CO_2` and 0.0585 g of water. 0.2293 g of the substance when heated with nitric acid and silver nitrate gave 0.3969 g of bromide and chloride of silver. 0.2202 g of this mixture of halides was found to contain 0.1435 g of silver. Find the empirical formula of the original substance.

Options

A. `C_6H_4ClBr`
B. `C_6H_3Cl_2Br`
C. `C_6H_3ClBr_2`
D. `C_6H_4Cl_2`

Answer

A

Explanation

Let xg of AgCl be present in the 0.2202 g of the mixture of AgBr and AgCl.
So, `(108x)/(143.5) +(108xx(0.2202-x))/(188) = 0.1435`
Determine the amounts of AgCl and AgBr in 0.3969 g of the mixture.
`AgCl=x/(0.2202)xx0.3969= 0.1720g`
`AgBr = (0.3969- 0.1720)= 0.2249g`
Thus, percentage of CI = `(35.5)/(143.5) xx(0.1720)/(0.2293) xx100= 18.55`
and percentage of Br = `80/(188) xx(0.2249)/(0.2293) xx100=41.73`
Percentage of carbon = `12/44xx(0.4291)/(0.3112)xx100=37.6%`
Percentage of hydrogen `=2/18xx(0.0585)/(0.3112) xx100 = 2.09%`