Question

A 0.8838-g sample of an ionic compound containing bromide ions and an unknown metal cation is dissolved in water and treated with an excess of AgNO3. If 1.573 g of a AgBr precipitate forms, what is the percent by mass of Br in the original compound?

Answer 1

%
`Br=75.7`%

Step-by-step explanation:

Hello,

From the AgBr precipitate, one could compute one can compute the bromine grams as follows:

`1.573gAgBr*(80gBr)/(188gAgBr)=0.669gBr`

As long as there was an excess of silver nitrate, one knows that into the 1.573 g of AgBr, 0.669 g correspond to the bromine that was initially contained into the 0.8838-g sample, thus, the percent is computed as follows:

%
`Br=(0.669gBr)/(0.8838g)*100`%

%
`Br=75.7`%

Best regards.