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

##### 1 Answer

#### Explanation:

In order to figure out how many grams of bromine you get in that many grams of *calcium bromide*, **percent composition**.

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

one moleof calcium cations,#"Ca"^(2+)# two molesof 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*

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**

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

#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**.