# A "1.85 g" sample of mixture of "CuCl"_2 and "CuBr"_2 was dissolved in water and mixed thoroughly with a "1.8 g" portion of "AgCl". After reaction, the solid, a mixture of "AgCl" and "AgBr", was filtered, washed, and dried, and [ . . . ] ?

## [ . . . ] its mass was found to be $\text{2.052 g}$. What percent by mass of the original mixture was ${\text{CuBr}}_{2}$?

Jun 10, 2017

Let's write down what we know.

${m}_{\text{start" = m_(CuCl_2) + m_(CuBr_2) = "1.85 g}}$ $\text{ "" } \boldsymbol{\left(A\right)}$

${m}_{\text{end" = m_(AgCl) + m_(AgBr) = "2.052 g}}$ $\text{ "" } \boldsymbol{\left(B\right)}$

$2 {\text{AgCl"(s) + "CuBr"_2(aq) -> 2"AgBr"(s) + "CuCl}}_{2} \left(a q\right)$

Notice how the only source of $\boldsymbol{\text{Ag}}$ is from the starting $\boldsymbol{\text{AgCl}}$. So:

$1.8 \cancel{\text{g AgCl" xx cancel"1 mol AgCl"/(143.32 cancel"g AgCl") xx "1 mol Ag"^(+)/cancel"1 mol AgCl}}$

$=$ $\text{0.01256 mols Ag"^(+) = "0.01256 mols starting AgCl}$

As a sidenote, we should check that the $\text{AgCl}$ is in excess, because it remains at the end.

$0.01256 \cancel{\text{mols starting AgCl" xx cancel"1 mol AgBr"/cancel"1 mol AgCl" xx "187.77 g AgBr"/cancel"1 mol AgBr}}$

$=$ $\text{2.358 g AgBr dry yield}$, which is more than the dry yield of both solids combined of $\text{2.052 g}$. So, $\text{AgCl}$ is in excess.

Next, it is also useful to note that the only source of $\boldsymbol{{\text{Br}}^{-}}$ on the reactants side is from the $\boldsymbol{{\text{CuBr}}_{2}}$, so we can denote that as:

${n}_{B {r}^{-}} = \left({m}_{C u B {r}_{2}} \times {\text{2 mols Br"^(-)/"1 mol CuBr"_2)/("223.37 g CuBr"_2"/mol CuBr}}_{2}\right)$ $\text{ "" } \boldsymbol{\left(1\right)}$

Now our goal is to find expressions for the mass of each component in the produced precipitate.

Assume we know what mols of ${\text{Br}}^{-}$ we actually have for now, and we can use that as a variable later. This is $1 : 1$ with the mols of $\text{AgBr}$ on the products side.

${n}_{B {r}^{-}} = {n}_{A g B r}$ $\text{ "" } \boldsymbol{\left(2\right)}$

Knowing the mols of $\text{AgBr}$ (supposedly), we can get the mols of $\text{AgCl}$ that were leftover by subtraction.

Since $\text{AgCl}$ reactant contains all the ${\text{Ag}}^{+}$ (from both the $\text{AgBr}$ product and leftover $\text{AgCl}$) and ${\text{Cl}}^{-}$ (from only leftover $\text{AgCl}$), subtracting ${n}_{A g B r}$ subtracts out ${n}_{A {g}^{+}}$ and ${n}_{B {r}^{-}}$ only in $\text{AgBr}$ and leaves ${n}_{A g C l}$ that was leftover:

$\text{0.01256 mols AgCl reactant} - {n}_{A g B r}$

$= {n}_{\text{leftover AgCl}}$ $\text{ "" } \boldsymbol{\left(3\right)}$

Its mass would then be given by:

${m}_{\text{leftover AgCl") = "143.32 g/mol" xx n_("leftover AgCl}}$ $\text{ "" } \boldsymbol{\left(4\right)}$

We're getting there. Now, we need an expression for the mass of $\text{AgBr}$ product. Use $\left(2\right)$ and $\left(1\right)$ and the molar mass of $\text{AgBr}$ to obtain:

=> m_(AgBr) = overbrace((m_(CuBr_2) xx "2 mols Br"^(-)/"1 mol CuBr"_2)/("223.37 g CuBr"_2"/mol CuBr"_2))^(n_(AgBr)) xx "187.77 g"/"mol"

Now we have expressions for the mass of leftover $\text{AgCl}$ and $\text{AgBr}$ product. This means we can (finally) use $\left(B\right)$:

m_("leftover AgCl") + m_(AgBr) = "2.052 g"

$= \text{143.32 g/mol" xx n_("leftover AgCl") + "187.77 g"/"mol} \times {n}_{A g B r}$

• For mols of leftover $\text{AgCl}$, after plugging $\left(1\right)$ and $\left(2\right)$ into $\left(3\right)$, plug $\left(3\right)$ into our current form of $\left(B\right)$.
• For mols of $\text{AgBr}$, plug in $\left(1\right)$ and $\left(2\right)$.

=> "143.32 g/mol" xx ["0.01256 mols AgCl reactant" - (m_(CuBr_2) xx "2 mols Br"^(-)/"1 mol CuBr"_2)/("223.37 g CuBr"_2"/mol CuBr"_2)] + "187.77 g"/"mol" xx (m_(CuBr_2) xx "2 mols Br"^(-)/"1 mol CuBr"_2)/("223.37 g CuBr"_2"/mol CuBr"_2)

To make this easier to read, let ${m}_{C u B {r}_{2}} = x$, in of course, units of $\text{g}$. Then, evaluate some of this to simplify. We know the units will work out, so we omit most of the units for readability:

$= \left\{143.32 \times \left[0.01256 - 0.008954 x\right]\right\} \text{g" + 1.6812x " g}$

$= \left\{1.8001 - 1.2833 x + 1.6812 x\right\} \text{ g}$

$= \left\{1.8001 + 0.3980 x\right\} \text{ g" = "2.052 g}$

$\implies x = \frac{0.2519}{0.3980} {\text{ g" = "0.6329 g CuBr}}_{2}$

Finally, we have the mass! We can use this mass along with the mass given from $\left(A\right)$ to get the percent by mass:

color(blue)(%"w/w CuBr"_2 ) = "0.6329 g CuBr"_2/"1.85 g copper mixture" xx 100%

~~ color(blue)(34.21%)