Mad percent of CuS and Cu2S in an ore sample
posted by Rose Bud .
Suppose an ore sample contains 11.01 % impurity in addition to a mixture of Cu2S and CuS. Heating 102.5 g of the mixture produces 74.90 g of solid which is 88.18 % copper. What is the mass percent CuS in the ore?

I would approach the problem this way.
102.5 g Cu2S + CuS + 11.01% impurity; therefore, mass of impurity is 102.5 x 0.1101 = 11.28 which means
mass Cu2S + mass CuS = 102.511.28 = 91.21g.
The problem isn't clear what happens to the impurity when heated; I suppose we are to assume it is inert.
Let x = mass Cu2S
and y = mass CuS

x + y = 91.21
x(2*atomic mass Cu/molar mass Cu2S) + y(atomic mass Cu/molar mass CuS) = 74.90*0.8818

Two equations in two unknowns. Solve for x and y
Then %CuS = (mass CuS/mass sample)*100 = ?
Post your work if you get stuck.
Note: x + y = 91.21 is just mass Cu2S + mass CuS = mass samplemass impurity.
x(2*atomic mass Cu/molar mass Cu2S)= mass Cu contributed by Cu2S in the sample.
y(atomic mass Cu/molar mass CuS) = mass Cu contributed by CuS in the sample.
The sum of Cu contributed by Cu2S and Cu = mass Cu in the heated sample = 74.90*0.8818. 
i tried to solve for x and y, but when i put them in that form, and then solved for only x, and then plugged it in to the equation, it just cancels everything out.
am i suppose to make one equation
x(Cu2/Cu2S)+y(Cu/CuS)= 91.21
and the other equation
x(Cu2/Cu2S)+y(Cu/CuS)= 74.90(.8818)
i havent tried these 2 equations yet 
or how about i make the first equation
x + y =91.21 
i have tried this, and apparently this is not the method, is there another way?

my answers are a little different because the numbers changed,
i got 12.7% for CuS and 75.83% for Cu2S. it almost added to my needed percent of 88.56%.
i even tried to recalculate, incause i left out some numbers, but it didn't work, maybe i will try again, any suggestions