# Question 8fb2b

Jun 15, 2016

31.2%

#### Explanation:

In order to find the percent concentration by mass of fluorine, $\text{F}$, in krypton difluoride, ${\text{KrF}}_{2}$, you essentially need to determine how many grams of fluorine you get per $\text{100 g}$ of compound.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {\text{% F" = "mass of fluorine"/"mass of KrF}}_{2} \times 100 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

You know that one mole krypton difluoride contains

• one mole of krypton, $1 \times \text{Kr}$
• two moles of fluorine, $2 \times \text{F}$

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)

${M}_{\text{M F" = "18.998 g mol}}^{- 1}$

Since one mole of krypton difluoride contains two moles of fluorine, you can say that $\text{121.79 g}$ of krypton difluoride, the mass of 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.