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

Mar 11, 2016

$\text{156 g Br}$

#### Explanation:

In order to figure out how many grams of bromine you get in that many grams of calcium bromide, ${\text{CaBr}}_{2}$, 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, ${\text{Ca}}^{2 +}$
• two moles of bromide anions, $2 \times {\text{Br}}^{-}$

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 $\text{100 g}$ of calcium bromide.

The two molar mass are

${\text{For CaBr"_2: " "M_M = "199.89 g mol}}^{- 1}$

${\text{For Br:" " " " " " " M_M = "79.904 g mol}}^{- 1}$

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 $\text{100 g}$ of calcium bromide will contain $\text{79.95 g}$ 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 $\text{195-g}$ sample of calcium bromide

$195 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g CaBr"_2))) * overbrace("79.95 g Br"/(100color(red)(cancel(color(black)("g CaBr"_2)))))^(color(purple)("79.95% Br")) = color(green)(|bar(ul(color(white)(a/a)"156 g Br} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to three sig figs.