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#?

1 Answer
Jun 10, 2017

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%)#