Question

A sample of .3301 g of an ionic comound containing the bromide ion (Br^-) is dissolved in water and treated with an excess of AgNO3. If the mass of the AgBr precipitate that forms is .07045 g, what is the percent by mass of Br in the original compound?

Answer 1

no ofmoles of AgNO3 = weight of AgNO3 in grams / molar mass of AgNO3

= 0.07045 / 169.87

= 4.15 x 10^-4 moles

let me write the equation

AgNO3 + XBr ----> AgBr + XNO3

from this equation it clear that no ofmoles of AgBr formed = no ofmoles of Br- reacted

no of moles of Br- = 4.15 x 10^-4 moles

weight of Br- = moles x molar mass of Br-

= 4.15 x 10^-4 moles x 79.90 g/mol

= 0.033 grams

total mass of the sample given = 0.3301gr

weight of Br- = 0.033 gr

% of Br- = (0.033 / 0.33) x (100)

= 10%