# Question 59c70

Mar 15, 2016

#### Answer:

"% Br" = 24.91%

#### Explanation:

The trick here is to realize that the number of moles of chloride anions, ${\text{Cl}}^{-}$, and bromide anions, ${\text{Br}}^{-}$, will be equal to the number of moles of silver(I) cations, ${\text{Ag}}^{+}$, present in the sample.

Notice that one mole of silver chloride contains one mole of silver(I) cations and one mole of chloride anions

${\text{AgCl" -> "Ag"^(+) + "Cl}}^{-}$

Likewise, one mole of silver bromide contains one mole of silver(I) cations and one mole of bromide anions

${\text{AgBr" -> "Ag"^(+) + "Br}}^{-}$

color(red)(!)color(white)(a)From this point on, I will refer to the silver(I) cations as silver, and o the chloride and bromide anions as chlorine and bromine, respectively.

If you take ${n}_{x}$ to be the number of moles of silver chloride and ${n}_{y}$ to be the number of moles of silver bromide present in the mixture, you can say that the mixture will contain

${n}_{A g} = {n}_{x} + {n}_{y} \to$ the total number of moles of silver

${n}_{B r} = {n}_{y} \to$ the number of moles of bromine

${n}_{C l} = {n}_{x} \to$ the number of moles of chlorine

Now, use silver's molar mass to determine how many moles you get in that sample

0.6635color(red)(cancel(color(black)("g"))) * "1 mole Ag"/(107.868color(red)(cancel(color(black)("g")))) = "0.006151 moles Ag"

This means that you have

n_(Ag) = n_y + n_x = "0.006151 moles"" " " "color(orange)("(*)")

Use the total mass of the sample to find the combined mass of bromine and chlorine present in the mixture

${m}_{\text{mixture}} = {m}_{A g} + {m}_{B r} + {m}_{C l}$

${m}_{C l} + {m}_{B r} = \text{1.0234 g" - "0.6635 g}$

m_(Cl) + m_(Br) = "0.3599 g" " " " "color(orange)("(* *)")

As you know, the molar mass of a element tells you the mass of one mole of said element. This means that you can write the number of moles of an element in terms of the mass and the molar mass

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{M}_{M} = \frac{m}{n} \implies n = \frac{m}{M} _ M} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

This means that the number of moles of chlorine present in the sample will be

n_(x) = (m_(Cl)color(red)(cancel(color(black)("g"))))/("35.453"color(red)(cancel(color(black)("g"))) "mol"^(-1)) = m_(Cl)/35.453color(white)(a)"moles Cl"

Likewise, the number of moles of bromine present in the sample will be

n_y = (m_(Br)color(red)(cancel(color(black)("g"))))/(79.904color(red)(cancel(color(black)("g")))"mol"^(-1)) = m_(Br)/79.904color(white)(a)"moles Br"

This means that equation $\textcolor{\mathmr{and} a n \ge}{\text{(*)}}$ becomes

${m}_{C l} / 35.453 + {m}_{B r} / 79.904 = 0.006151$

Use equation $\textcolor{\mathmr{and} a n \ge}{\text{(* *)}}$ to find the mass of elemental chlorine

${m}_{C l} = 0.3599 - {m}_{B r}$

Plug this into the above equation to find the mass of elemental bromine present in the sample

$\frac{0.3599 - {m}_{B r}}{35.453} + {m}_{B r} / 79.904 = 0.006151$

This will be equivalent to

$0.01015 - 0.01569 \cdot {m}_{B r} = 0.006151$

${m}_{B r} = \frac{0.01015 - 0.006151}{0.01569} = 0.2549$

The mixture contains

${m}_{B r} = \text{0.2549 g Br}$

This means that the percent composition of bromine in the sample will be

"% Br" = (0.2549 color(red)(cancel(color(black)("g"))))/(1.0234color(red)(cancel(color(black)("g")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)"24.91%"color(white)(a/a)|)))#

The answer is rounded to four sig figs, the number of sig figs you have for the mass of silver present in the sample.