Question

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

Answer 1

`"156 g Br"`

Explanation:

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.