# How many grams are in 3 mol of KBr?

Mar 24, 2016

$\text{360 g}$

#### Explanation:

Your strategy here will be to use the molar mass of potassium bromide, $\text{KBr}$, as a conversion factor to help you find the mass of three moles of this compound.

So, a compound's molar mass essentially tells you the mass of one mole of said compound. Now, let's assume that you only have a periodic table to work with here.

Potassium bromide is an ionic compound that is made up of potassium cations, ${\text{K}}^{+}$, and bromide anions, ${\text{Br}}^{-}$. Essentially, one formula unit of potassium bromide contains a potassium atom and a bromine atom.

Use the periodic table to find the molar masses of these two elements. You will find

${\text{For K: " M_M = "39.0963 g mol}}^{- 1}$

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

To get the molar mass of one formula unit of potassium bromide, add the molar masses of the two elements

${M}_{\text{M KBr" = "39.0963 g mol"^(-1) + "79.904 g mol"^(-1) ~~ "119 g mol}}^{-}$

So, if one mole of potassium bromide has a mas of $\text{119 g}$m it follows that three moles will have a mass of

3 color(red)(cancel(color(black)("moles KBr"))) * overbrace("119 g"/(1color(red)(cancel(color(black)("mole KBr")))))^(color(purple)("molar mass of KBr")) = "357 g"

You should round this off to one sig fig, since that is how many sig figs you have for the number of moles of potassium bromide, but I'll leave it rounded to two sig figs

"mass of 3 moles of KBr" = color(green)(|bar(ul(color(white)(a/a)"360 g"color(white)(a/a)|)))