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Posted by **Rose Bud** on Monday, September 24, 2012 at 2:59pm.

- Mad percent of CuS and Cu2S in an ore sample -
**DrBob222**, Monday, September 24, 2012 at 3:36pmI 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 = 102.5-11.28 = 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 sample-mass 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.

- Mad percent of CuS and Cu2S in an ore sample -
**Anonymous**, Monday, September 24, 2012 at 4:39pmi 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

- Mad percent of CuS and Cu2S in an ore sample -
**Anonymous**, Monday, September 24, 2012 at 4:46pmor how about i make the first equation

x + y =91.21

- Mad percent of CuS and Cu2S in an ore sample -
**rose bud**, Monday, September 24, 2012 at 4:58pmi have tried this, and apparently this is not the method, is there another way?

- Mad percent of CuS and Cu2S in an ore sample -
**rose bud**, Monday, September 24, 2012 at 5:02pmmy 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 didnt work, maybe i will try again, any suggestions