Question #f4ff2

1 Answer
Oct 2, 2015

Answer:

#"BrO"_2#

Explanation:

So, you know that you're dealing with an unknown bromine oxide, let's say #"Br"_x"O"_y#, which is converted to silver bromide, #"AgBr"#.

If you assume that all the bromine tht was initially in the oxide is now in the silver bromide, then you can use silver bromide's percent composition to find how many grams of bromine you get in 4.698 g of silver bromide.

To do that, use silver and silver bromide's respective molar amsses

#(79.904color(red)(cancel(color(black)("g/mol"))))/(187.78color(red)(cancel(color(black)("g/mol")))) xx 100 = "42.55% Br"#

This means that every 100 grams of silver bromide will contain 42.55 g of bromine. Therefore, your sample contains

#4.698color(red)(cancel(color(black)("g AgBr"))) * "42.55 g Br"/(100color(red)(cancel(color(black)("g AgBr")))) = "1.999 g Br"#

The bromine oxide will thus contain #"1.999 g"# bromine and

#m_(O) = 2.80 - 1.999 = "0.801 g O"#

FInd the number of moles of each element in the original oxide

#1.999color(red)(cancel(color(black)("g Br"))) * "1 mole Br"/(79.904color(red)(cancel(color(black)("g Br")))) = "0.02501 moles Br"#

and

#0.801color(red)(cancel(color(black)("g O"))) * "1 mole Br"/(16.0color(red)(cancel(color(black)("g O")))) = "0.05006 moles O" #

Divide both values by the smallest one to get the two elements' mole ratio in the unknown oxide

#"For Br: " (0.02501color(red)(cancel(color(black)("g/mol"))))/(0.02501color(red)(cancel(color(black)("g/mol")))) = 1#

#"For O: " (0.05006color(red)(cancel(color(black)("g/mol"))))/(0.02501color(red)(cancel(color(black)("g/mol")))) = 2.002 ~~ 2#

This means that the empirical formula of the oxide is

#"Br"_1"O"_2 -> color(green)("BrO"_2)#