Question

A sample of 0.3257 g of an ionic compound 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 0.7165 g, what is the percent by mass of Br in the original compound?

Answer 1

Step-by-step explanation:

Given parameters:

Mass of ionic compound = 0.3257g

Mass of AgBr precipitate = 0.7165g

Unknown:

Percent mass of Br in the original compound.

Solution

The percent mass of Br in original compound =
`(mass of  Br  in the sample)/(mass of sample)`

Now we have to find the mass of Br⁻:

We must note that the same mass of Br⁻ would move through the ionic sample to form the precipitate.

Mass of Br in AgBr =
`(Atomic mass of Br)/(Molar mass of AgBr)  x mass of precipitate`

Mass of Br =
`(80)/(80 + 108)` x 0.7165

Mass of Br = 0.426 x 0.7165 = 0.305g

Percent mass of Br =
`(0.305)/(0.3257)` x 100 = 93.7%