
Concept explainers
Copper metal can be prepared by roasting copper ore, which can contain cuprite (Cu2S) and copper(II) sulfide.
Cu2S(s) + O2(g) → 2 Cu(s) + SO2(g)
CuS(s) + O2(g) → Cu(s) + SO2(g)
Suppose an ore sample contains 11.0% impurity in addition to a mixture of CuS and Cu2S. Heating 100.0 g of the mixture produces 75.4 g of copper metal with a purity of 89.5%. What is the weight percent of CuS in the ore? The weight percent of Cu2S?

Interpretation:
The weight percentage of CuS and Cu2S in the given sample of ore has to be determined.
Concept introduction:
- For chemical reaction balanced chemical reaction equation written in accordance with the Law of conservation of mass.
- Law of conservation of mass states that for a reaction total mass of the reactant and product must be equal.
Number of mole =Given mass of the substanceMolar mass
- The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
- Stoichiometric factor is a relationship between reactant and product which is obtained from the balanced chemical equation for a particular reaction.
- Weight percent of elements of a compound is the ratio of weight of element to the weight of whole compound and multiplied with hundred.
Answer to Problem 103GQ
The weight percentage of CuS and Cu2S in the given ore is 26.55 % and 62.45 % respectively.
Explanation of Solution
Balanced chemical equation for the reaction occurred in the preparation of copper from cuprite (Cu2S) and copper (II) sulfide is,
Cu2S(s) +O2(g) →2Cu(s) + SO2(g)CuS(s) + O2(g)→ Cu(s) + SO2(g)
Mass of impurity in the ore= 11100 ×100=11 g
Then the mass of CuS and Cu2S is 100 −11 = 89 g
Mass of Cu from CuS = mass of CuSMolar mass of CuS ×1 mol Cu in CuS1 mol CuS × molar mass of CuS
Considering A as the mass of CuS and B as the mass of Cu2S,
Mass of Cu from CuS is,
A95.61×11×63.55 g/mol = 0.6626 A
Similarly mass of Cu from Cu2S is,
mass of Cu2SMolar mass of Cu2S ×2 mol Cu in CuS1 mol Cu2S × molar mass of Cu2S
Mass of Cu from Cu2S is,
B159.1×21×63.55 = 0.7989 B
Total mass of copper metal produced,
Cu from CuS+Cu from Cu2S+ impurities
Thus,
Total mass ofCumetal produced=0.6626 A+ 0.7989 B+ impurities
But here total Cu metal produced with impurity = 75.4 g
Therefore mass of impurity is,
10.5100 ×75.4 = 7.917 g
Total mass of Cu metal produced = mass of Cu from CuS + mass of Cu from Cu2S+ impurities
That is,
75.4 = 0.6626 A + 0.7989 B + 7.917
Calculated mass of CuS and Cu2S is 89 g. By using this value the mass of CuS and the mass of Cu2S can be calculated as follows,
A+ B = 89 , A = 89−B
Substituting this in the above equation,
75.4 = 0.6626 (89-B) + 0.7989 B + 7.917
Then the value of B is 62.45 g
A= 89-62.45 = 26.55 g
Weight percent ofCuSin the ore=26.55100 ×100 %= 26.55 %
Weight percent ofCu2Sin the ore=62.45100 ×100 %= 62.45 %
The weight percentage of CuS and Cu2S in the given sample of ore was determined.
Want to see more full solutions like this?
Chapter 4 Solutions
Chemistry & Chemical Reactivity
- Part C IN H N. Br₂ (2 equiv.) AlBr3 Draw the molecule on the canvas by choosing buttons from the Tools (for bonds and + e (×) H± 12D T EXP. L CONT. דarrow_forward9. OA. Rank the expected boiling points of the compounds shown below from highest to lowest. Place your answer appropriately in the box. Only the answer in the box will be graded. (3) points) OH OH بر بد بدید 2 3arrow_forwardThere is an instrument in Johnson 334 that measures total-reflectance x-ray fluorescence (TXRF) to do elemental analysis (i.e., determine what elements are present in a sample). A researcher is preparing a to measure calcium content in a series of well water samples by TXRF with an internal standard of vanadium (atomic symbol: V). She has prepared a series of standard solutions to ensure a linear instrument response over the expected Ca concentration range of 40-80 ppm. The concentrations of Ca and V (ppm) and the instrument response (peak area, arbitrary units) are shown below. Also included is a sample spectrum. Equation 1 describes the response factor, K, relating the analyte signal (SA) and the standard signal (SIS) to their respective concentrations (CA and CIS). Ca, ppm V, ppm SCa, arb. units SV, arb. units 20.0 10.0 14375.11 14261.02 40.0 10.0 36182.15 17997.10 60.0 10.0 39275.74 12988.01 80.0 10.0 57530.75 14268.54 100.0…arrow_forward
- A mixture of 0.568 M H₂O, 0.438 M Cl₂O, and 0.710 M HClO are enclosed in a vessel at 25 °C. H₂O(g) + C₁₂O(g) = 2 HOCl(g) K = 0.0900 at 25°C с Calculate the equilibrium concentrations of each gas at 25 °C. [H₂O]= [C₁₂O]= [HOCI]= M Σ Marrow_forwardWhat units (if any) does the response factor (K) have? Does the response factor (K) depend upon how the concentration is expressed (e.g. molarity, ppm, ppb, etc.)?arrow_forwardProvide the structure, circle or draw, of the monomeric unit found in the biological polymeric materials given below. HO OH amylose OH OH 행 3 HO cellulose OH OH OH Ho HOarrow_forward
- OA. For the structure shown, rank the bond lengths (labeled a, b and c) from shortest to longest. Place your answer in the box. Only the answer in the box will be graded. (2 points) H -CH3 THe b Нarrow_forwardDon't used hand raitingarrow_forwardQuizzes - Gen Organic & Biological Che... ☆ myd21.lcc.edu + O G screenshot on mac - Google Search savings hulu youtube google disney+ HBO zlib Homework Hel...s | bartleby cell bio book Yuzu Reader: Chemistry G periodic table - Google Search b Home | bartleby 0:33:26 remaining CHEM 120 Chapter 5_Quiz 3 Page 1: 1 > 2 > 3 > 6 ¦ 5 > 4 > 7 ¦ 1 1 10 8 ¦ 9 a ¦ -- Quiz Information silicon-27 A doctor gives a patient 0.01 mC i of beta radiation. How many beta particles would the patient receive in I minute? (1 Ci = 3.7 x 10 10 d/s) Question 5 (1 point) Saved Listen 2.22 x 107 222 x 108 3.7 x 108 2.22 x 108 none of the above Question 6 (1 point) Listen The recommended dosage of 1-131 for a test is 4.2 μCi per kg of body mass. How many millicuries should be given to a 55 kg patient? (1 mCi = 1000 μСi)? 230 mCiarrow_forward
- Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage LearningChemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage LearningChemistry for Engineering StudentsChemistryISBN:9781337398909Author:Lawrence S. Brown, Tom HolmePublisher:Cengage Learning
- ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage LearningChemistry: An Atoms First ApproachChemistryISBN:9781305079243Author:Steven S. Zumdahl, Susan A. ZumdahlPublisher:Cengage Learning





