Copper Lab
Purpose
To demonstrate that the amount of copper, as well as that of other metals, stays the same regardless of the form it takes.
Qualitative Data
Cu + HNO3
Cu(NO3)2 + NaOH Cu(OH)2 CuO + H2SO4 CuSO4 + Zn Zn + H2SO4 |
Copper melts in to aqueous solution, the solution turns blue
Blue precipitate forms Dark, grey Electric blue, clear teal color with no sign of precipitates Precipitates sit at bottom, gives odor, and bubbles form The dried precipitates are red and the aqueous solution is clear. The precipitates sit at the bottom. |
Quantitative Data
Initial Mass of Copper
Mass of Evaporating Dish Mass of Evaporating Dish and Copper Mass of Recovered Copper |
1.955g
56.39g 62.03g 5.64g |
Net Ionic Equations
Cu(s) + 4HNO3(aq) -> Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)
Cu(s) + 4H+ + 4NO3- -> Cu2+ + 2NO3- + 2NO2(g) + 2H2O(l)
Cu(s) + 4H+ + 2NO3- -> Cu2+ + 2NO2(g) + 2H2O(l)
Cu(NO3)2(aq) + 2NaOH(aq) -> Cu(OH)2(s) + 2NaNO3(aq)
Cu2+ + 2NO3- + 2Na+ + 2OH(aq)- -> Cu(OH)2(s) + 2Na+ + 2NO3-
Cu2+ + 2OH- -> Cu(OH)2(s)
Cu(OH)2(s) -> CuO(s) + H2O(l)
CuO(s) + H2SO4(aq) -> CuSO4(aq) + H2O(l)
CuO(s) + 2H+ + SO4- -> Cu2+ + SO42- + H2O(l)
CuO(s) + 2H+ -> Cu2+ + H2O(l)
CuSO4 + Zn(S) -> ZnSO4(aq) + Cu(s)
Cu2+ + SO42- + Zn(s) -> Zn2+ + SO42- + Cu(s)
Cu2+ + Zn(s) -> Zn2+ + Cu(s)
Cu(s) + 4H+ + 4NO3- -> Cu2+ + 2NO3- + 2NO2(g) + 2H2O(l)
Cu(s) + 4H+ + 2NO3- -> Cu2+ + 2NO2(g) + 2H2O(l)
Cu(NO3)2(aq) + 2NaOH(aq) -> Cu(OH)2(s) + 2NaNO3(aq)
Cu2+ + 2NO3- + 2Na+ + 2OH(aq)- -> Cu(OH)2(s) + 2Na+ + 2NO3-
Cu2+ + 2OH- -> Cu(OH)2(s)
Cu(OH)2(s) -> CuO(s) + H2O(l)
CuO(s) + H2SO4(aq) -> CuSO4(aq) + H2O(l)
CuO(s) + 2H+ + SO4- -> Cu2+ + SO42- + H2O(l)
CuO(s) + 2H+ -> Cu2+ + H2O(l)
CuSO4 + Zn(S) -> ZnSO4(aq) + Cu(s)
Cu2+ + SO42- + Zn(s) -> Zn2+ + SO42- + Cu(s)
Cu2+ + Zn(s) -> Zn2+ + Cu(s)
Calculations
Final Mass of Copper
5.64g
Final Moles of Copper
5.64g = 0.0887 mol Cu
63.55g Cu
Initial moles of Copper
1.955g Cu = 0.0308 mol Cu
63.55 g Cu
Percent Yield
0.0887 mol Cu X 100 % = 288%
0.0308 mol Cu
5.64g
Final Moles of Copper
5.64g = 0.0887 mol Cu
63.55g Cu
Initial moles of Copper
1.955g Cu = 0.0308 mol Cu
63.55 g Cu
Percent Yield
0.0887 mol Cu X 100 % = 288%
0.0308 mol Cu
Conclusion
Through various chemical reactions, recovered mass of copper is found to be larger than initial mass of copper; 1.955g of Cu was given initially; 5.64g of Cu was produced. If the lab was accurately done, the amount of copper that was left over should have theoretically weighed the same as the weight of initial copper.
Discussion of Theory
The Copper Lab demonstrates stoichiometry in chemistry. Stoichiometry is helpful in calculating the amount of an element or compound in chemical reactions. For the lab, stoichiometry was used to predict the amount of copper that would be left over. Using stoichiometry, the experimenter could find out that the product of the copper would have the same amount of copper as the original amount. Of course, the misconduct of experiment led to a 288% percent yield than the theoretical yield.
No matter what form copper, as well as any other elements, is in- whether it be in aqueous or solid state- the same amounts of copper is still present in the compound. This fact lines up with The Law of Conservation and Mass, which states that nothing can ever be created nor destroyed. The copper was melted in nitric acid, but was recovered in a solid state after several chemical reactions. The goal of the lab was to recover copper as close to the original copper amount and to demonstrate the Law of Conservation and Mass.
No matter what form copper, as well as any other elements, is in- whether it be in aqueous or solid state- the same amounts of copper is still present in the compound. This fact lines up with The Law of Conservation and Mass, which states that nothing can ever be created nor destroyed. The copper was melted in nitric acid, but was recovered in a solid state after several chemical reactions. The goal of the lab was to recover copper as close to the original copper amount and to demonstrate the Law of Conservation and Mass.
Sources of Error
The yield of copper was too high in this experiment. 1.955g of Cu was given, but 5.46g of Cu was recovered in the conclusion. The percent yield turn out to be 288%, 188% away from the theoretical yield. This could be explained by the reaction between zinc and sulfuric acid.
Zn(S) + H2SO4(aq) -> ZnSO4(aq) + H2(g)
The equation shows that solid zinc would turn form with sulfuric acid to make an aqueous solution of zinc sulfate. This would allow the experimenter to dispose of zinc sulfate, leaving copper behind. However, when the reaction is incomplete, some of the zinc would be left over. When the data was measured, some of the zinc measured along with the copper, resulting in a seemingly higher yield of copper. The mistake could be lessened by adding sulfuric acid, decanting the solution, and repeating it repeatedly.
Zn(S) + H2SO4(aq) -> ZnSO4(aq) + H2(g)
The equation shows that solid zinc would turn form with sulfuric acid to make an aqueous solution of zinc sulfate. This would allow the experimenter to dispose of zinc sulfate, leaving copper behind. However, when the reaction is incomplete, some of the zinc would be left over. When the data was measured, some of the zinc measured along with the copper, resulting in a seemingly higher yield of copper. The mistake could be lessened by adding sulfuric acid, decanting the solution, and repeating it repeatedly.
Questions
1. Why is the product of the reaction between copper and nitric acid in step 2 placed on ice?
The product is placed in ice because its temperature, or the amount of heat in the solution, will change its ph. To make sure the real ph of the solution is 13, the temperature would have to be lowered and remove excess energy in the solution. Also, cold temperatures work as catalyst, speeding up the reaction between the copper and nitric acid.
2. What type of reaction (synthesis, decomp, single rep., double rep., redox, dehydration) occurred in steps 4. 7. And 9?
Step 4: Cu(NO3)2(aq) + 2NaOH(aq) -> Cu(OH)2(s) + 2NaNO3(aq) – Double Replacement
Step 7: CuO(s) + H2SO4(aq) -> CuSO4 + H2O(l) – Double Replacement
Step 9: Zn(S) + H2SO4(aq) -> ZnSO4(aq) + H2(aq) – Single Replacement
3. The reaction of excess zinc with sulfuric acid is a critical step in this investigation. Write the balanced equation for this reaction. What problems would arise from an incomplete reaction?
Zn(S) + H2SO4(aq) -> ZnSO4(aq) + H2(g)
The incomplete reaction of the reaction above would leave some zinc in a premature, solid form. When weighing the copper, the weight of zinc could be included in the copper weight data, causing higher percent yield that the theoretical yield.
4. What ions did you remove when you washed the CuO?
OH ions were removed when CuO was washed.
5. What form of copper is present in the beaker after you added the H2SO4?
CuO(s) + H2SO4(aq) -> CuSO4(aq) + H2O(l)
CuO(s) + 2H+ + SO4- -> Cu2+ + SO42- + H2O(l)
CuO(s) + 2H+ -> Cu2+ + H2O(l)
The copper compound, CuO(s), reacted with H2SO4(aq) to form Cu2+.
6. What ions did you remove when you washed the precipitated copper?
Zinc ions were removed when the precipitated copper was washed.
The product is placed in ice because its temperature, or the amount of heat in the solution, will change its ph. To make sure the real ph of the solution is 13, the temperature would have to be lowered and remove excess energy in the solution. Also, cold temperatures work as catalyst, speeding up the reaction between the copper and nitric acid.
2. What type of reaction (synthesis, decomp, single rep., double rep., redox, dehydration) occurred in steps 4. 7. And 9?
Step 4: Cu(NO3)2(aq) + 2NaOH(aq) -> Cu(OH)2(s) + 2NaNO3(aq) – Double Replacement
Step 7: CuO(s) + H2SO4(aq) -> CuSO4 + H2O(l) – Double Replacement
Step 9: Zn(S) + H2SO4(aq) -> ZnSO4(aq) + H2(aq) – Single Replacement
3. The reaction of excess zinc with sulfuric acid is a critical step in this investigation. Write the balanced equation for this reaction. What problems would arise from an incomplete reaction?
Zn(S) + H2SO4(aq) -> ZnSO4(aq) + H2(g)
The incomplete reaction of the reaction above would leave some zinc in a premature, solid form. When weighing the copper, the weight of zinc could be included in the copper weight data, causing higher percent yield that the theoretical yield.
4. What ions did you remove when you washed the CuO?
OH ions were removed when CuO was washed.
5. What form of copper is present in the beaker after you added the H2SO4?
CuO(s) + H2SO4(aq) -> CuSO4(aq) + H2O(l)
CuO(s) + 2H+ + SO4- -> Cu2+ + SO42- + H2O(l)
CuO(s) + 2H+ -> Cu2+ + H2O(l)
The copper compound, CuO(s), reacted with H2SO4(aq) to form Cu2+.
6. What ions did you remove when you washed the precipitated copper?
Zinc ions were removed when the precipitated copper was washed.