Chemistry 110 - Experiment 4 Addenda (2024)

Experiment 4

Chemical Changes: Reactions of Copper

Optional protocol using a penny as the source of copper (at the discretion of your instructor).

Instead of using a piece of copper wire, as described in the printed protocol, you will use a penny (you must use a 1970 or later penny). Current pennies are made from a piece of zinc coated with a thin layer of copper. The modern penny is not 100% pure copper as it was prior to the 1960's. You will dissolve the entire penny in the concentrated nitric acid.

To do this experiment follow the modified protocol shown below:

1. Cu (s) + 4 HNO3 (aq) Cu(NO3)2 (aq) + 2 NO2 (g) + 2 H2O (l)

  • Place a single penny (you should have determined its mass) into a 250-mL Erlenmeyer flask.
  • Add 20 mL concentrated nitric (HNO3) acid.
  • Let the penney completely dissolve, or if you believe all the copper has dissolved, you can remove the zinc core using a pair of forceps.
  • Add about 20 mL of DI water before proceeding with the next step.

2. Cu(NO3)2 + 2 NaOH (aq) Cu(OH)2 (s) + 2 NaNO3 (aq)

  • Add about 50 mL of 6 M NaOH in small increments, with stirring.
  • The bluish Cu(OH)2 solid also contains the Zn(OH)2, which is white.
  • Using litmus paper, check the pH to make certain the liquid is basic (red litmus turns blue).

3. Cu(OH)2 (s) CuO (s) + H2O (l)

  • Heat your sample, with stirring, until all the blue color disappears, leaving a grey slurry (your mixture will contain both the black CuO and the white ZnO).
  • Collect your sample using vacuum filtration and a Büchner funnel (you must use #2 filter paper, not #3 filter paper).
  • Remove the filter from your funnel and place in the bottom of a 250-mL beaker (with the solids on the top side).

4. CuO (s) + H2SO4 (aq) CuSO4 (aq) + H2O (l)

  • Add about 20 mL of the 3 M H2SO4 and let it dissolve, with occassional swirling of the beaker to completely dissolve the solid material.
  • When all the solid has been dissolved, the solution should be a light blue in color.
  • Add about 40 mL of DI water, swirl, and then remove the filter paper.

Cover your sample with a layer of parafilm and place on the cart for storage until the next lab period. The blue color of the liquid is due to the Cu2+ ion. The Zn2+ ion is colorless, but is also present. (If your liquid has pieces of the filter paper in it, you must do a filtration prior to doing the Mg reduction step.)

During the second day of your lab, you will then reduce your Cu2+ to produce Cu metal using Mg. Don't worry about the Zn2+ ion which also gets reduced (producing Zn metal), because any Zn metal produced will either react with the sulfuric acid or it will help to reduce the Cu2+ into the Cu metal. By doing this reduction step (starting with Mg and your solution of ions) you will produce only Cu metal, and all the zinc and magnesium will be converted, or stay, as ions.

5. CuSO4 (aq) + Mg (s) Cu (s) + MgSO4 (aq)

  • If your liquid has pieces of the filter paper in it, you must do a Büchner funnel vacuum filtration prior to adding the Mg filings.
  • Pour the clear filtrate into either a 400-mL or 600-mL beaker.
  • Add the required amount of Mg filings (see published protocol for amount) to your beaker, a few pieces at a time.
  • Continue to add more Mg until all the blue color in the liquid is gone (Cu2+ is blue in solution; therefore, when it is all reduced to Cu, the blue color will disappear).
  • If any bubbles are apparent, then all the Mg has not yet been reacted. If in doubt, add up to 10 mL of 3 M H2SO4 to assist in the dissolving. Do not filter until no more gas (bubbles) is produced.

To collect your copper do the following:

  • Filter the final mixture (the liquid must be colorless, but will continue the reddish or black-looking copper) to collect your Cu metal via Büchner funnel and vacuum filtration.
  • Wash the solid material with copious amounts of water to remove any acid from the filter paper (residual acid will cause the filter paper to turn black when heated).
  • Keep the vacuum on for at least 5 min after your final wash to remove as much water as possible.
  • Dry your Cu metal (along with a preweighed piece of filter paper) in the drying oven (use a pre-weighed evaporating dish or watch glass to put your filter in).

Weigh your dry Cu sample and determine the amount of copper present (remember to subtract the mass of the pre-weighed evaporating dish and filter paper). Determine the percent yield of Cu based on the mass of the original penny using the %Yield formula given below.

To obtain a value for %Yield, you need to know the Theoretical Value you should obtain, based on balanced equations, and if 100% of the reactant(s) is converted to product. After obtaining your Experimental Value, divide it by the Theoretical Value, then multiply by 100 to get %Yield.

Chemistry 110 - Experiment 4 Addenda (1)

Questions:

  1. Why does the copper not dissolve in the H2SO4 mixture in the last step?
  2. Practice doing the oxidation-reduction reaction for Cu combining with concentrated HNO3, which is the first reaction. (You will see this reaction again on quizzes, the lab exam and the regular exam.)
  3. If you used dilute HNO3, instead of the concentrated acid used in this experiment, you would produce NO (g) instead of the NO2 (g). Show the correct oxidation-reduction equation for the production of NO.

Reagents needed for use of a penny (per group)

  • 20 mL concentrated nitric acid
  • 50 mL 6 M NaOH
  • 40 mL H2SO4 (20 mL to dissolve solid CuO and 20 mL to dissolve excess Mg after last step)
  • Use only #2 Büchner filter paper, do not use #3 paper as it is too thick and gets clogged during the CuO filtration

Go To Experiment: 1 2 3 4 5 6 7 8 9 10 11 12
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Chemistry 110 - Experiment 4 Addenda (2024)

FAQs

When you have finished adding the necessary amount of NaOH to your beaker, what color will the litmus paper turn? ›

If you think you have added enough NaOH, dip a glass stirring rod into your mixture and touch this stirring rod to a piece of red litmus paper. If the liquid is basic, it will turn the red litmus blue.

What ions are present in the filtrate when the product of reaction 3 is filtered using vacuum filtration? ›

Final answer: After the filtration of reaction 3, the filtrate contains sodium ions (Na+) and water molecules, with the sodium ions being the primary ions present in the filtrate.

How do you do the oxidation reaction experiment at home? ›

Squeeze the juice from the lemon and add a few drops of water. Write a message on the paper using a cotton swab that has been dipped in juice. Once dry, the message will be invisible. Hold the paper up to the lightbulb, and the juice will turn brown due to oxidation, and the message will appear.

What happened to the color when you added NaOH? ›

When N a O H is added to solution, ions from the solution will get consumed. This will shift the equilibrium towards the reactant side. Equilibrium shift will raise the concentration of C r O 4 2 − ions. Presence of increased C r O 4 2 − ions makes the solution color intense yellow.

What is the color change of NaOH? ›

Phenolphthalein is often used as an indicator in acid–base titrations. For this application, it turns colorless in acidic solutions and pink in basic solutions. And sodium hydroxide is a base. So, when the sodium hydroxide, which is a base, is mixed with phenolphthalein, it turns pink.

What 4 substances are filtered out of the glomerulus? ›

The glomerulus filters water and small solutes out of the bloodstream. The resulting filtrate contains waste, but also other substances the body needs: essential ions, glucose, amino acids, and smaller proteins. When the filtrate exits the glomerulus, it flows into a duct in the nephron called the renal tubule.

What substance is left after filtration? ›

During filtration the substance left behind on the filter paper is called residue.

Which substance is present in the filtrate? ›

The glomerular filtrate contains urea, water, vitamins, fatty acids, water, glucose, nitrogen, salts, etc. The glomerular filtrate enters the nephrons for further filtration termed ultrafiltration.

Does vinegar and baking soda experiment carbon dioxide? ›

When you combine the solid (baking soda) and the liquid (vinegar), the chemical reaction creates a gas called carbon dioxide. Carbon dioxide is invisible, except as the bubbles of gas you may have noticed when the vinegar and baking soda mixture began to fizz. This gas is what made the balloon inflate.

What are two examples of oxidation reactions from your everyday life? ›

Hint: The two major oxidation reactions taking place in our day to day life are corrosion of metals and rancidity of food items.

Which reaction shows oxidation? ›

Oxidation–reduction reactions, commonly known as redox reactions, are reactions that involve the transfer of electrons from one species to another. The species that loses electrons is said to be oxidized, while the species that gains electrons is said to be reduced.

What did you observe in the beaker when preparing the NaOH solution? ›

NaOH present in the form of pellets which react with water readily because it is a base. It is basic and hence, turns red litmus paper blue. The beaker when touched should be hot and not cold as the reaction is exothermic and releases heat.

What is the color change when sodium hydroxide is added to blue litmus paper and red litmus paper? ›

The base turns red litmus into blue. Thus Sodium hydroxide turns red litmus into blue. Sodium hydroxide turns blue litmus red. P and Q are aqueous solutions of sodium chloride and sodium hydroxide, respectively.

Why does the color of water change after adding a few drops of NaOH solution? ›

The initial addition of sodium hydroxide solution is not enough to neutralize the acid present, and the indicators remain colorless. With the second addition of NaOH solution, an excess of base results, the solutions become alkaline, and the indicators become colored.

How the colour of the solution will change as the sodium hydroxide is added to the beaker? ›

As NaOH is added, the solution would become blue (pH 8 - 11) and then purple (pH 13 – 14). Because the water was neutral to start with, as more OH– ions are added, the solution becomes more basic as the OH– ions are immediately in excess.

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