Expt: F Compounds Of Copper

Question 1

Over all reaction schemes and sequences

Answer 1

Part 1:
Cu(s) —> Cu(NO3)2(aq) —> CuCO3(s) —> CuCl2(aq) —> Cu(s)

Part 2:
CuSO4(aq) —> Cu(OH)2(s) —> CuO(s) —> CuSO4(aq)

Part 1 sequence:
Cu(s)
| react with HNO3(aq) in fume hood
V (gas evolved)
Cu(NO3)2(aq)
| add excess Na2CO3(aq) (gas evolved)
V
CuCO3(s)
| recover ppt by vacuum filtration
V Add HCl(aq) (gas evolved)
CuCl2(aq)
| add aluminum (gas evolved)complete
V when colourless and bubbling stops
Cu(s)

Part 2 sequence:
CuSO4(aq)
| react with NaOH(aq)
V
Cu(OH)2(s)
| heat
V
CuO(s)
| add H2SO4(aq)
V
CuSO4(aq)

Part 1 A:
Copper metal —> copper(||) nitrate
3 Cu(s) + 8 HNO3(aq) —> 3 Cu(NO3)2(aq) + 4 H2O(L) + 2 NO(g)

Part 1 B:
copper(||) nitrate —> copper(||) carbonate
Cu(NO3)2(aq) + 2 Na2CO3(aq) —> CuCO3(s) + 2 NaNO3(aq) + CO3^2+(aq)

Part 1 C:
Copper(||) carbonate —> copper(||) chloride
CuCO3(s) + HCl(aq) —> CuCl2(aq) + H^+ + CO^2- (H + CO) —> H2CO3(aq) —> H20(l) + CO2(g)

Part 1 D:
Copper(||) chloride —> copper metal
2 Al(s) + 3 Cu^2+(aq) —> 2 Al^3+(aq) + 3 Cu(s)

Question 2

Copper (||) entities

Answer 2

•Can exist as either copper(|) or copper(||); the more stable Cu^2+ one.

•Can vary considerably in characteristics such as colour and solubility. Since the cation is the same (Cu^2+) in each case, any difference can be attributed to the counter ions bonded to the Cu^2+ ion.

•Copper metal does not readily react with Sulfuric or hydrochloric acid. However, copper does react with nitric acid, a strong oxidizing agent.

•copper is limiting Reagent in all reactions

•Several methods are used to drive reactions to completion use of excess reagent(s)
-Formation of gas
-Formation of precipitate
-Thermal extremes (heat or cold)

•In part 1 C, Because of the excess HCl, Excess proteins React with CO3 and form some CO2 gas and H2O. Which is evolved from via the reaction:
2H3O^+(aq) + CO3^2-(aq) —> H2CO3(aq) + 2 H2O(l). And from
H2CO3(aq) —> H2O(l) + CO2(g).

Question 3

Ligands

Answer 3

Any molecule or Atom that irreversibly binds to a receiving protein molecule, otherwise known as a receptor. When ligand binds to receptors, shape and/or activity is altered to initiate several different types of cellular responses. 

Question 4

What is the Precipitate, and what is the Filtrate?

Answer 4

Filtrate is the liquid that passes through the paper into the filter flask.

Precipitate is the solid left over after the filtrate has gone through. 

Question 5

Distinguish between the terms reagent and reactant.

Answer 5

Reagent is any substance added to a chemical system, which may or may not precipitate in a chemical reaction.

Reactant is a molecule or ion that is converted (via chemical reaction) into product(s).

Question 6

Equation for the formation of copper (||) hydroxide from copper (||) sulfate

Equation for the formation of copper (||) Oxide from Copper (||) hydroxide

Equation for the formation of copper (||) sulfate from copper (||) oxide

Answer 6

CuSO4(aq) + 2 NaOH(aq) —> Cu(OH)2(s) + Na2SO4(aq)

Cu(OH)2(s) + heat —> CuO(s) + H2O(l)

CuO(s) + H2SO4(aq) —> CuSO4(aq) + H2O(l)

Question 7

Write the balanced molecular equation for

Reaction of copper (||) nitrate to Copper (||) carbonate

Reaction of copper (||) carbonate to copper (||) chloride

Answer 7

Cu(NO3)2(s) + Na2CO3(aq) —> CuCO3(s) + 2 Na(NO3)(aq) + CO^2+(g)

CuCO3(s) + HCl(aq) —> CuCl2(aq) + H^+(g) + CO^2+(g)

H^+(g) + CO^2+(g) —> H2CO3(aq) —> H2O(l) + CO2^-

Question 8

Identify the gases produced in parts 1 A, B, and C of this experiment.

Write the balanced molecular equations for the reactions that produce:

The red brown gas in part 1 A:

The gas in part 1 B:

Answer 8

Part 1 A: NO2(g), NO(g)

Part 1 B: CO2(g)

Part 1 C: CO2(g)

Part 1 A:
NO(aq) + 1/2 O2(g) —> No2(g)

Part 1 B:
H2CO3(aq) —> H2O(l) + CO2(g)


Question 9

If the filtrate in part 1B was a clear, dark blue Solution, explain how and why the percent yield of copper (||) carbonate solid would be affected.

Answer 9

The % yield would be lower, this is because some of the copper (||) carbonate has been lost in the filtrate. It is meant to be colourless because the blue colour comes from copper (||) carbonate and therefore we don’t want to lose it in the filtrate.

Question 10

Most reactions do not yield 100% recovery of the product. Suggest two or more reasons, along with a brief explanation, that would result in a percent yield of less than 100%.

Also suggest one reason why someone might obtain a yield of greater than 100%. 

Answer 10

1. Some of the Cu^2+ was lost in the filtrate during part B because not all of the Cu^2+ reacted with CO3.
2. When transferring the Cu(s) to the vial it is likely some was lost due to being left on the funnel or missing the vial.
3. Could have been lost when decanting the final solid in part E.
4. The Cu(s) was still wet, therefore altering the weight.
5. There could have been left over Al(s) that hadn’t fully reacted with Cl^-(aq).

Question 11

Write the balanced molecular equation for the oxidation reduction reaction that occurs between copper (||) chloride and aluminum metal



Answer 11

3 CuCl2(aq) + 2 Al(s) —> 3 Cu(s) + 2 AlCl3(aq)

Question 12

During the final step of this experiment to recover copper metal, colourless gas bubbles continue to be produced in the reaction solution after the solution turned colorless. Explain why these bubbles continue to produce even though the reaction between copper ions and aluminum metal was complete as indicated by the colourless solution. Indicate the appropriate balanced molecular equation to support your answer. 

Answer 12

The bubbles continued because the Al could still react with the H3O^+(aq) that was formed from HCl(aq) left over from the pervious step, forming H^+ gas.

2 Al(aq) + 6 HCl(aq) —> 3 H2(g) + 2 AlCl3(aq)

Question 13

Copper oxide, CuO, may also be formed when copper metal is gently heated in air or oxygen gas. Copper metal surfaces that have been tarnished by a layer of black CuO can be cleaned by treating the surface with dilute solutions of hydrobromic acid, HBr(aq). This process forms of water soluble salt that can be washed away.

Write the balanced molecular equation of the reaction that is responsible for the removal of CuO(s) from the copper metal surface. 

Is CuO an acidic or basic oxide? Justify your answer.

Answer 13

CuO(s) + 2 HBr(aq) —> CuBr2(aq) + H2O(l)

CuO(s) + H2O(l) —> Cu(aq) + 2 OH^-(aq)

It is basic because it produces OH^- ions.

Question 14

Table
Compound. Colour. Solubility in water

Answer 14

Compound Colour Solubility in water
Cu rust red Insoluble
Cu(NO3)2 Bright blue Soluble
CuCO3 Light blue Insoluble
CuCl2 Blue/green Soluble
CUSO4 Bright blue Soluble
Cu(OH)2 Pale blue insoluble
CuO Brown/Black Insoluble