How many grams of #Br# are in 195 g of #CaBr_2#?

In order to figure out how many grams of bromine you get in that many grams of calcium bromide, #"CaBr"_2#, you must find the compound's percent composition.

To do that, use the fact that one mole of calcium bromide contains

• one mole of calcium cations, #"Ca"^(2+)#
• two moles of bromide anions, #2 xx "Br"^(-)#

You can thus us the molar mass of calcium bromide and the molar mass of bromine to determine how many grams of bromine you get per #"100 g"# of calcium bromide.

The two molar mass are

#"For CaBr"_2: " "M_M = "199.89 g mol"^(-1)#

#"For Br:" " " " " " " M_M = "79.904 g mol"^(-1)#

So, two moles of bromide anions for every one mole of calcium bromide will give you a percent composition of

#(2 xx 79.904 color(red)(cancel(color(black)("g mol"^(-1)))))/(199.89color(red)(cancel(color(black)("g mol"^(-1))))) xx 100 = "79.95% Br"#

This means that every #"100 g"# of calcium bromide will contain #"79.95 g"# of elemental bromine in the form of bromide cations.

All you have to do now is use this percent composition as a conversion factor to determine how many grams of bromine you get in that #"195-g"# sample of calcium bromide

#195color(red)(cancel(color(black)("g CaBr"_2))) * overbrace("79.95 g Br"/(100color(red)(cancel(color(black)("g CaBr"_2)))))^(color(purple)("79.95% Br")) = color(green)(|bar(ul(color(white)(a/a)"156 g Br"color(white)(a/a)|)))#

The answer is rounded to three sig figs.