Question

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 `"2.052 g"`. What percent by mass of the original mixture was `"CuBr"_2`?

Answer 1

DISCLAIMER: COMPLICATED ANSWER!

Let's write down what we know.

`m_"start" = m_(CuCl_2) + m_(CuBr_2) = "1.85 g"` `" "" "bb((A))`

`m_"end" = m_(AgCl) + m_(AgBr) = "2.052 g"` `" "" "bb((B))`

`2"AgCl"(s) + "CuBr"_2(aq) -> 2"AgBr"(s) + "CuCl"_2(aq)`

Notice how the only source of `bb"Ag"` is from the starting `bb"AgCl"`. So:

`1.8 cancel"g AgCl" xx cancel"1 mol AgCl"/(143.32 cancel"g AgCl") xx "1 mol Ag"^(+)/cancel"1 mol AgCl"`

`=` `"0.01256 mols Ag"^(+) = "0.01256 mols starting AgCl"`

As a sidenote, we should check that the `"AgCl"` is in excess, because it remains at the end.

`0.01256 cancel"mols starting AgCl" xx cancel"1 mol AgBr"/cancel"1 mol AgCl" xx "187.77 g AgBr"/cancel"1 mol AgBr"`

`=` `"2.358 g AgBr dry yield"`, which is more than the dry yield of both solids combined of `"2.052 g"`. So, `"AgCl"` is in excess.

Next, it is also useful to note that the only source of `bb("Br"^(-))` on the reactants side is from the `bb("CuBr"_2)`, so we can denote that as:

`n_(Br^(-)) = (m_(CuBr_2) xx "2 mols Br"^(-)/"1 mol CuBr"_2)/("223.37 g CuBr"_2"/mol CuBr"_2)` `" "" "bb((1))`

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

Assume we know what mols of `"Br"^(-)` we actually have for now, and we can use that as a variable later. This is `1:1` with the mols of `"AgBr"` on the products side.

`n_(Br^(-)) = n_(AgBr)` `" "" "bb((2))`

Knowing the mols of `"AgBr"` (supposedly), we can get the mols of `"AgCl"` that were leftover by subtraction.

Since `"AgCl"` reactant contains all the `"Ag"^(+)` (from both the `"AgBr"` product and leftover `"AgCl"`) and `"Cl"^(-)` (from only leftover `"AgCl"`), subtracting `n_(AgBr)` subtracts out `n_(Ag^(+))` and `n_(Br^(-))` only in `"AgBr"` and leaves `n_(AgCl)` that was leftover:

`"0.01256 mols AgCl reactant" - n_(AgBr)`

`= n_("leftover AgCl")` `" "" "bb((3))`

Its mass would then be given by:

`m_("leftover AgCl") = "143.32 g/mol" xx n_("leftover AgCl")` `" "" "bb((4))`

We're getting there. Now, we need an expression for the mass of `"AgBr"` product. Use `(2)` and `(1)` and the molar mass of `"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 `"AgCl"` and `"AgBr"` product. This means we can (finally) use `(B)`:

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

`= "143.32 g/mol" xx n_("leftover AgCl") + "187.77 g"/"mol" xx n_(AgBr)`

• For mols of leftover `"AgCl"`, after plugging `(1)` and `(2)` into `(3)`, plug `(3)` into our current form of `(B)`.
• For mols of `"AgBr"`, plug in `(1)` and `(2)`.

`=> "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_(CuBr_2) = x`, in of course, units of `"g"`. Then, evaluate some of this to simplify. We know the units will work out, so we omit most of the units for readability:

`= {143.32 xx [0.01256 - 0.008954x]} "g" + 1.6812x " g"`

`= {1.8001 - 1.2833x + 1.6812x} " g"`

`= {1.8001 + 0.3980x} " g" = "2.052 g"`

`=> x = 0.2519/0.3980 " g" = "0.6329 g CuBr"_2`

Finally, we have the mass! We can use this mass along with the mass given from `(A)` 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%)`