# Question #8fb2b

##### 1 Answer

#### Explanation:

In order to find the **percent concentration by mass** of fluorine, *krypton difluoride*, **per** **of compound**.

#color(blue)(|bar(ul(color(white)(a/a)"% F" = "mass of fluorine"/"mass of KrF"_2 xx 100color(white)(a/a)|)))#

You know that **one mole** krypton difluoride contains

,one moleof krypton#1 xx "Kr"# ,two molesof fluorine#2 xx "F"#

You can thus use the **molar mass** of krypton difluoride and the **molar mass** of fluorine to determine how many *grams* of fluorine you get *per mole* of krypton difluoride.

You have

#M_("M KrF"_2) = "121.79 g mol"^(-1)#

#M_"M F" = "18.998 g mol"^(-1)#

Since **one mole** of krypton difluoride contains **two moles** of fluorine, you can say that *one mole* of the compound, contains

#2 color(red)(cancel(color(black)("moles F"))) * "18.998 g"/(1color(red)(cancel(color(black)("mole F")))) = "37.996 g F"#

This means that the percent concentration by mass of fluorine in krypton difluoride is

#"% F" = (37.996 color(red)(cancel(color(black)("g"))))/(121.79color(red)(cancel(color(black)("g")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)color(black)(31.2%)color(white)(a/a)|)))#

I'll leave the answer rounded to three **sig figs**.