Question

A sample of antihistamine, bropheniramine maleate, weighing 4.6330g was dissolved in alcohol and decomposed with metallic sodium. The resulting solution was treated with 10.00mL of 0.2500 M AgNO₃, an amount sufficient to precipitate all of the liberated bromide ion as AgBr. The excess AgNO₃ was titrated with 0.1214 M KSCN, requiring 14.42mL to reach an end point:

Ag⁺  + SCN^-   --> AgSCN_{(s) ,} calculate the % Br in the sample. what additional piece of information is needed to determine the percentage of antihistamine in the sample?

(answer is 1.29%Br, but need to know how to get that)

Answer 1

Number of moles of KSCN , n = Molarity x volume in L

= 0.1214 M x 14.42 mL x 10^-3 L/mL

= 1.75x10^-3 mol

Ag+  + SCN-   --> AgSCN(s)

1 mole of AgNO3 reacts with 1 mole of KSCN

1.75x10^-3 mol of AgNO3 reacts with 1.75x10^-3 mol of KSCN

Number of moles of AgNO3 titrated originally , N = Molarity x volume in L

= 0.2500M x 10.00 mL x 10^-3 L/mL

= 2.5x10^-3 mol

So Number of moles of AgNO3 reacted with  bropheniramine maleate are 2.5x10^-3 mol - 1.75x10^-3 mol

= 7.5x10^-4 moles

Mass of bropheniramine maleate , m = Number of moles x molar mass

= 7.5x10^-4 moles x 319.24(g/mol)

= 0.239 g

So percentage of antihistamine = ( mass of antihistamine / mass of sample ) x 100

= ( 0.239 / 4.6330) x 100

= 5.17 %

Molar mass of antihistamine is 319.4 g/mol

Molar mass of Bromine 79.9 g/mol

319.4 g of antihistamine contains 79.9 g of Br

0.239 g of antihistamine contains (0.239 x 79.9) / 319.4 = 0.0598 g of Br

So % Bromine = ( mass of Br / mass of sample ) x 100

= ( 0.0598 / 4.6330) x 100

= 1.29 %