Mad percent of CuS and Cu2S in an ore sample

posted by .

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?

  • Mad percent of CuS and Cu2S in an ore sample -

    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 = 102.5-11.28 = 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 -

    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

  • Mad percent of CuS and Cu2S in an ore sample -

    or how about i make the first equation
    x + y =91.21

  • Mad percent of CuS and Cu2S in an ore sample -

    i have tried this, and apparently this is not the method, is there another way?

  • Mad percent of CuS and Cu2S in an ore sample -

    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

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Chemistry

    An ore contains Fe3O4 and no other iron. The iron in a 51.98 gram sample of the ore is all converted by a series of chemical reactions to Fe2O3. The mass of Fe2O3 is measured to be 32.8 grams. What was the percent Fe3O4 in the sample …
  2. Chem

    3. Chalcocite is an ore from which copper can be extracted for commercial use. The formula for this important ore is Cu2S. In the laboratory it is formed by the following reaction: 2Cu + S ¨ Cu2S The reaction takes place in laboratory …
  3. Chem

    Chalcocite is an ore from which copper can be extracted for commercial use. The formula for this important ore is Cu2S. In the laboratory it is formed by the following reaction: 2Cu + S → Cu2S The reaction takes place in laboratory …
  4. chemistry

    An ore contains Fe3O4 and no other iron. The iron in a 27.07 gram sample of the ore is all converted by a series of chemical reactions to Fe2O3. The mass of Fe2O3 is measured to be 21.9 grams. What was the percent Fe3O4 in the sample …
  5. Ap Chemistry

    An ore contains Fe3O4 and no other iron. The iron in a 60.67 gram sample of the ore is all converted by a series of chemical reactions to Fe2O3. The mass of Fe2O3 is measured to be 36.1 grams. What was the percent Fe3O4 in the sample …
  6. AP Chemistry

    An ore contains Fe3O4 and no other iron. The iron in a 40.93 gram sample of the ore is all converted by a series of chemical reactions to Fe2O3. The mass of Fe2O3 is measured to be 30.7 grams. What was the percent Fe3O4 in the sample …
  7. chemistry

    An ore contains Fe3O4 and no other iron. The iron in a 65.46 gram sample of the ore is all converted by a series of chemical reactions to Fe2O3. The mass of Fe2O3 is measured to be 37.9 grams. What was the percent Fe3O4 in the sample …
  8. AP Chemistry

    An ore contains Fe3O4 and no other iron. The iron in a 39.3-gram sample of the ore is all converted by a series of chemical reactions to Fe2O3. The mass of Fe2O3 is measured to be 10.6 g. What was the mass of Fe3O4 in the sample of …
  9. stoichiometry

    An ore contains Fe3O4 and no other iron.The iron in a 39.5-gram sample of the ore is all converted by a series of chemical reactions to Fe2O3. The mass of Fe2O3 is measured to be 26.7 g. What was the mass of Fe3O4 in the sample of …
  10. Chemistry

    Identify the equation for which Keq = [Cu+]2[S2-]. A) CuS(s) ⇌ Cu2+(aq) + S2- (aq) B) 1/2Cu2S(s) ⇌ Cu+(aq) + 1/2 S2- (aq) C) Cu2S(s) ⇌ 2Cu+(aq) + S2- (aq) D) Cu2S(s) ⇌ Cu+(aq) + 2 S2- (aq) E) none of the answers

More Similar Questions