A student weighed 0.2440 grams of sample of a recently synthesized cobalt amine bromide compound. After dissolving in 120mL of acidified water, she added 2.55 grams of AgNO3 to precipitate any free Br-. The precipitate was found to weigh 0.3700 grams. Calculate the %Br - in the sample. MW Cl - 35.45 g/mol; MW Br - 79.91 g/mol; MW Ag 107.87 g/mol

The bromine percentage is 49.42 %

Explanation:

Mass of AgBr precipitated = 0.31 g

Molecular weight of AgBr = 187.77 g mol - 1

moles of AgBr = 0.31 g by 187.77 g mol - 1 = 0.0016509 moles

moles of Br ion = 0.0016509 moles

Mass of Br ion = 0.0016509 moles into 79.904 g mol - 1 = 0.1319 g

The Total mass of compound = 0.2669 g

Percentage of bromine = 0.1319 by 0.2669 = 49.42 percent

Hence the bromine percentage is 49.42 %