Chemical Reactions of Copper Lab

Tara FaggioliPd. 5

10/19/09

IntroductionIn this lab, solid copper metal is going to be reacted through a series of reactions using the

cation Cu+2. In order to be sure a reaction has actually taken place, a precipitate or gas will

be formed, there will be a significant temperature change, or a color change will occur. At

the end of the lab, the copper will be precipitated and compared with the starting point to

try and verify the Law of Conservation of Mass.

TheoryThe Law of Conservation of Mass states that matter cannot be created nor destroyed. Our

lab is set up to help prove this theory by starting with a set mass of copper, reacting it in

excess solutions, and then precipitating the copper back out and comparing the start and

end masses.

For part A, copper (II) nitrate formed from solid copper metal. The balanced

oxidation-reduction reaction is Cu(s) + 5HNO3(aq) -> 2NO2(g) + Cu(NO3)2(aq) + 2H2O + H+.

Nitrogen dioxide, NO2, is toxic and must be collected under a fume hood, and it also is a very

strong oxidizing agent. There are many different uses for copper (II) nitrate, Cu(NO3)2,

among them being for light-sensitive papers, insecticide for vines, electroplating, and in

paints.

For part B, copper (II) hydroxide was made from the copper (II) nitrate. The balanced

double displacement reaction is Cu(NO3)2(aq) + 2NaOH(aq) -> Cu(OH)2(s.aq) + 2NaNO3(aq). The net

ionic reaction, including the slightly soluble copper hydroxide, is Cu+2(aq) + 2OH-

(aq) ->

Cu(OH)2(s.aq). Copper hydroxide, Cu(OH)2, can be mixed with latex paint to make a product

that controls root growth in potted plants. Sodium nitrate, NaNO3, is an ingredient in

fertilizers, pyrotechnics, a rocket propellant, and glass and pottery enamels.

For part C, copper (II) oxide came from the copper (II) hydroxide through the

balanced double displacement reaction Cu(OH)2(s) + H2O(l) —-> CuO(s) + 2H2O(l), with the net

ionic being the same. Copper (II) oxide, CuO, can be used to dispose of hazardous

materials, including cyanide, hydrocarbons, and dioxins through oxidation. It is also used as

a pigment in ceramics, producing blue, red, and green, and sometimes gray, pink, or black

glazes. Occasionally, CuO is used as a dietary supplement in animals against copper

deficiency. In addition, it is used in dry cell batteries and some wet cell batteries as the

cathode (using lithium as an anode).

For part D, the copper (II) oxide reacted with sulfuric acid to produce copper (II)

sulfate. The balanced double displacement reaction is CuO(s) + H2SO4(aq) -> CuSO4(aq) + H2O(l).

The net ionic reaction excludes the copper because it would be written as 2H+(aq) + OH-

(aq) ->

H2O(l). However, this is useful to us because we are then left with only copper sulfate in our

beaker. Copper sulfate, CuSO4, is apparently extremely useful. In can be used as a herbicide,

fungicide, a pesticide, and an analytical reagent along with being used in organic synthesis,

and being included in many beginner chemistry sets. Previously, copper sulfate was used as

an emetic, but it is now considered too toxic for this use. One artist, Roger Hiorns, filled a

flat with copper sulfate, allowed it to crystallize for several weeks, then drained off the extra

water to reveal blue crystal-coated walls and floors. The work is titled Seizure.

For part E, zinc was added to the copper sulfate to regenerate the copper metal in

the balanced oxidation-reduction reaction Zn(s) + Cu+2(aq) -> Zn+2

(aq) + Cu(s). The excess zinc

was then dissolved out with hydrochloric acid in the single displacement reaction Zn (s) +

2HCl(aq) -> ZnCl2(aq) + H2(g), net ionic reaction of 2H+(aq) -> H2(g). At this point, the copper from

part A has been retrieved and the compound is rinsed with methanol and acetone to ensure

the other ions will not reappear once the liquid is dried off.

A stoichiometry problem relies on the balanced reactions. Using stoichiometry, you

can calculate how much of a product you can expect to yield based on how much reactant

you start with. Every problem uses a mole to mole ratio of compounds based on the

coefficients of the reaction.

A percent yield calculation is used to determine how much of product you made

compared with how much you expected to make. This is done by first doing a stoichiometry

problem to calculate how much reactant you should have formed, then divide the amount of

product you actually formed by the amount of product you should have formed and multiply

by 100 to get a percent.

Materials and Procedure250mL Erlenmeyerbalancecopperdeionized water16M nitric acidtongs

hot plate6M sodium hydroxidered litmus paperstirring rodbeakerdrying oven

zinc metalevaporating dish12M hydrochloric acidacetonemethanol

Part A: Weigh an Erlenmeyer and add 1.00g copper metal to it. Put the flask under the fume

hood and add 5mL nitric acid. Use tongs to swirl on a hot plate and drive off all the gas by

holding the flask at an angle since the gas is denser than the air. Allow to cool.

Part B: Add 50mL deionized water to the copper (II) nitrate solution, then slowly add 6M

sodium hydroxide and swirl between each addition. Add until red litmus paper turns blue.

Rinse sides first and use a stirring rod to put the solution on the paper.

Part C: Transfer to beaker and gently warm with a hot plate. Allow the precipitate to settle

and decant. Heat 200mL deionized water. Wash the copper oxide with the warm water.

Allow it to settle and decant again. The compound should be washed twice.

Part D: Add 15mL 6M sulfuric acid and swirl until everything is dissolved.

Part E: Add 3.5g zinc metal to the beaker, swirling until the liquid is colorless. Stir in 20mL

12M hydrochloric acid to dissolve any remaining metal. Add another 5mL of acid if the metal

still isn’t dissolved. Pour off the liquid and wash with 20mL deionized water. Transfer the

copper to a pre-weighed evaporating dish and wash with 10mL methanol. Decant the liquid

carefully so none of it touches your skin, and then rinse with acetone and decant. Place the

evaporating dish in the drying oven. Once the liquid is evaporated off, weigh the dried

copper to compare with the starting amount.

Data and ObservationsMass of evaporating dish and copper (end)Mass of evaporating dish (empty)Mass of flask and copper (start)Mass of flask (empty)Mass of copper (start)Mass of copper recovered% recovery of copper

27.87g26.71g80.58g79.62g.96g1.16g120%

Calculations:Mass of copper (start)

80.58g – 79.62g = .96gMass of copper (end)

27.87g – 26.71g = 1.16g% recovery of copper

1.16g / .96g x 100 = 120%% error

|.96 – 1.16| / .96 x 100 = 20%

Questions:

1. A. Copper (II) nitrate - Cu(NO3)2 – blue

B. Copper (II) hydroxide – Cu(OH)2 – pale blue

C. Copper (II) oxide – CuO – black

D. Copper (II) sulfate – CuSO4 – blue

E. Copper metal – Cu – “pinkish,” “peachy,” also metallic, “orangish,” slightly “brownish”

–> just think of a penny and you’ve got it. A clean, shiny penny. =D

2. Considering our percent error of 20%, I’d say we overshot the Law of Conservation of

Mass just a bit. Possible sources of error besides human ones include a bit of greenish-

bluish precipitate remaining when we weighed the copper and weighing dish for the final

mass, or excess reactants not being fully washed off or reacted away at any other point.

Observations:

A.

B.

C.

D.

E.

Summary

Post-lab Questions1. Redox: Cu + NO3- -> Cu+2 + NO

red: (4H+ + 3e- + N(+5)O3- -> N(+2)O + 2H2O)2

ox: (Cu(0) -> Cu+2 + 2e-)3

combined: 8H+ + 6e- + 2NO3- + 3Cu -> 2NO + 4H2O + 3Cu+2 + 6e-

simplified: 8H+ + 2NO3- + 3Cu -> 2NO + 4H2O + 3Cu+2

double displacement: CaCl2(aq) + Na2CO3(aq) -> CaCO3(s) + 2NaCl(aq)

2. A double displacement reaction occurs when a solid (precipitate), gas, or a liquid is

formed. Two aqueous products haven’t really been reacted because most double

displacement reactions happen in solution to start with, and compounds in solution break

up into ions.

3. Generally, percent yield is the percent value of how much product is formed from a

certain amount of reactant. The maximum value a percent yield can have is 100%.

Percent yield added to percent error equals 100.

4. reaction: Cu(s) + 5HNO3(aq) -> 2NO2(g) + Cu(NO3)2(aq) + 2H2O + H+

1.00gCu x 1molCu/63.55gCu x 1molCu(NO3)2/1molCu x 185.75gCu(NO3)2/1molc = 2.95g

1.40g/2.95g x 100 = 47.5% yield

5. a. S + O2 -> SO2, 98.0% yield

.980(1.00kgS x 103gS/1kgS x 1molS/32.07gS x 1molSO2/1molS) = 30.6molSO2

b. 2SO2 + O2 -> 2SO3, 96.0% yield

.960(30.6molSO2 x 2molSO3/2molSO3) = 29.3molSO3

c. SO3 + H2SO4 -> H2S2O7, 100.% yield

1.00(29.3molSO3 x 1molH2S2O7/1molSO3) = 29.3molH2S2O7

d. H2S2O7 + H2O -> 2H2SO4, 97.0% yield

.970(29.3molH2S2O7 x 2molH2SO4/1molH2S2O7 x 98.09gH2SO4/1molH2SO4 x 1kg/103g)

= 5.85kgH2SO4

6.