517 A sample of an antihistamine brompheniramine maleate wei

5-17. A sample of an antihistamine, brompheniramine maleate, weighing 4.6330 g was dissolved in alcohol and decomposed with metallic sodium. The resulting solution was treated with 10.00 mL 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 14.42 mL to reach the end point: Ag+ +SCN- AgSCN (s) Calculate the % Br in the sample. what additional piece of information is needed to deter- mine the percentage of the antihistamine in the sample?

Solution

Ag⁺ reacts with SCN^- on a 1:1 molar basis.

Millimoles of Ag⁺ added (as AgNO₃) = (10.00 mL)*(0.2500 M) = 2.50 mmole.

Millimoles of Ag⁺ reacted with SCN^- = millimoles of SCN^- added (as KSCN) = (14.42 mL)*(0.1214 M) = 1.750588 mmole.

Millimoles of Ag⁺ that has reacted with the bromine in antihistamine = (2.50 – 1.750588) mmole = 0.749412 mmole.

Ag⁺ reacts with bromine (most notably Br^-) on a 1:1 molar basis to form AgBr. Moreover, assume that the bromine in antihistamine is the only source of Br^-; hence, millimoles of Br^- present in antihistamine = 0.749412 mmole.

The atomic mass of bromine is 79.904 g/mol; therefore, mass of bromine corresponding to 0.749412 mmole = (0.749412 mmole)*(1 mole/1000 mmole)*(79.904 g/1 mole) = 0.05988 g 0.0599 g.

Percentage of bromine in antihistamine = (0.0599 g)/(4.6330 g)*100 = 1.29289% 1.2929% (ans).

A crucial bit of information that will help us determine the percentage of antihistamine in the sample is the chemical structure of antihistamine; more specifically, the chemical formula which will give us the number of atoms of bromine per mole of antihistamine.