Recycling: Alum From Aluminum
Learning Goals:
1. To conduct the synthesize of alum from aluminum: recycle aluminum.
2. To use recrystallization as a technique to purifying alum.
3. To calculate the percent yield of the alum synthesis reaction.
Recycling Aluminum. In the United States, 1500 aluminum cans are recycled per
second with an energy savings of 95 percent over refining and smelting bauxite ore. Though
aluminum cans are currently recycled to make more aluminum products, scrap aluminum metal
can also be used to product alum. Alum is a chemical used in a myriad of applications including
water purification, marble cement, explosives, tanning, hardening gelatin, baking powders,
clarifying sugar, hardening plaster casts, and as a medicinal astringent. In this experiment, you
will produce alum from an aluminum can in a multi-step synthesis reaction summarized by
equation 1. You will then purify the alum through a process called recrystallization and
determine the percent yield of the alum synthesis reaction. %
2 Al(s) + 2 KOH(aq) + 4 H2SO4(aq) + 22 H2O ! 2 [KAl(SO4)2•12 H2O](s) + 3 H2(g)
Table 1. Chemicals in Equation 1
Al(s)
KOH(aq)
H2SO4(aq)
H2 O
[KAl(SO4)2•12 H2O](s)
H2(g)
!
(1)
aluminum
potassium hydroxide
sulfuric acid
water
alum
hydrogen
Alum. Alum is the common name given to several chemical compounds that consist of a
positively charged ion (K+, Na+, NH4+ ect.), an aluminum (Al3+) ion, several sulfate ions (SO4-2)
and twelve water of hydration. The compound to be synthesized in this procedure is Potassium
aluminum sulfate dodecahydrate, KAl(SO4)2•12 H2O. When compounds crystallize out of an
aqueous solution, water molecules can fit into the crystalline structure and become bound up in
the solid. This water is included in the formula weight of the compound and is represented by a
dot (•) and then the number of water molecules for each molecule of solid. This adds up for the
above alum compound to a formula weight of 474.37.
Lecture Connections: In Moore et al. (4th ed.), there is a discussion of
aluminum and its properties in chapter 21 (pg. 1016). In Shultz, there is a discussion of
aluminum and its properties in chapter 9 (pg. 338).
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Instructions before going into lab
You can work in pairs for this experiment. If you work in pairs, please remember to
write the name of your partner and contact information in your lab notebooks. This is
critical, if you will be sharing any spectra that should be reproduced for your lab report.
• Each pair of students should discuss and answer questions as they encounter them in this
experiment. These questions are set off in the experiment (i.e. “Q:”). The questions offer
guidance during the lab. Answers to these questions do not need to be reproduced in
your lab report.
• Each student is to record notes and observations and all other data in their own lab notebook.
• Each student is responsible for collecting the information needed to complete the Lab Report
and Post Lab Portions of the Experiment.
• Goggles are required at all times in the lab. There are no exceptions. %
•
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Chemicals and Equipment
Chemicals
1.0 g aluminum can
1.5 M Potassium hydroxide (KOH)
9.0 M sulfuric acid
Ice for ice bath
50%:50% Ethanol:water mixture
Equipment and Supplies
steel wool
Buchner funnel
Vacuum flask
scissors
Bunsen burner and ring stand
150 and 250 mL beakers
100 mL graduated cylinder
large beaker for ice bath
balance
Prelab Assignment.
View the video clips on Using the Balance, Using the Bunsen
Burner and Using a Vacuum filtration flask (Buchner Funnel). You will need Quick Time video
player to see them.
In your lab notebook, prepare the following information:
1.
A brief (2-3 sentence) introduction to the lab.
2.
A table of safety information including the chemicals used in the lab and any safety
handling precautions. This information can be obtained from the MSDS safety sheets.
3.
To prepare for lab, we’d like you to answer three questions. A worked example is
presented at the end of this section to help answer questions 2 and 3.
Q1. Please define “waters of hydration” and indicate there are any present in equation 1. Hint:
this may be in your textbooks index or found with a simple Google search.
Q2. What volume of potassium hydroxide (KOH), using a 1.5M KOH solution, is needed to
react with 1.0 g of aluminum (Al) according to equation 2?
Q3 What volume of potassium hydroxide (KOH) would be needed to react with 1.0 g of
aluminum (Al) with a two-fold excess of KOH?
2 Al(s) + 6 H2O(l) + 2 KOH(aq) + heat ! 2 KAl(OH)4(aq) + 3 H2 (g) (2)
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The reason for the two-fold excess of potassium hydroxide, is that many reaction proceed slowly
if the exact amounts of reactants are used. To promote all of the aluminum reacting, a two-fold
excess of hydroxide product is used to drive the reaction to completion. This is a strategy that is
frequently used by chemists to synthesize specific products. It should be no surprise that the
aluminum will be the limiting reactant, if a two-fold excess of potassium hydroxide is used.
Strategy for solving question 1. You will note that for every 2 moles of aluminum an equal
number of moles of KOH is needed in the reaction. Since we are given 1.0 g of aluminum, we
can convert the mass (in grams) to moles. Then, using equation 2, we can calculate the number
of moles of potassium hydroxide needed. However, the question is not answered, because we
need to know what volume of a 1.5M KOH solution will be equal to the number of moles
needed. This issue becomes easier, if we recognize what is the definition of a 1.5M KOH
solution. In other words, that there are 1.5 moles of KOH for every liter, or there are 1.5 moles
for every 1,000 mL. The conversion factor of 1.5 moles = 1,000 mL of a 1.5 M solution can then
be used to find the volume of KOH needed.
Strategy for solving question 2. Once we know the volume of a 1.5 M KOH solution we need
to react with 1.0 g of aluminum, a two-fold excess will be having twice the number of moles of
KOH present at the start of the reaction. This will simply be a doubling of the volume needed in
the answer to question 1.
Worked Problem. Starting with approximately 1 g of iron, how much of a 1.55 M
solution of H2SO4 would be needed to react according to equation 3?
Fe(s) + H2SO4(aq) !FeSO4(aq) + H2 (g)
(3)
Step 1. The piece of iron actually weighed 1.060 g.
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Step 2. For every 1 mole of Fe we need 1 mole of H2SO4, therefore we need 0.0190 moles of
H2SO4
Step 3.
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Answer: we would need about 12 mL of a 1.5 M H2SO4 solution.
Give the information to your Instructor at the beginning of the lab. You will not be allowed
to work in the lab without this information.
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Procedure:
Q: Companies apply a polymer coating to the inside of the aluminum soda can? Can you guess
why?
1. Obtain a strip of aluminum from a soda can. Using steel wool, scrub both sides of the strip to
remove the paint from the outside and the polymer coating from the inside.
2. Cut the scrap into small pieces and weigh out approximately 1 g of the metal. [Notes: (a.) It is
not important to have exactly 1.000g of aluminum, but it is important to record the mass to
± 0.001g; (b.) remember to record the mass of aluminum in your lab notebooks.]
2 Al(s) + 6 H2O(l) + 2 KOH(aq) + heat ! 2 KAl(OH)4(aq) + 3 H2 (g)
(2)
3. Using the actual mass of aluminum and concentration of of KOH(aq), calculate the volume of
KOH(aq) needed to add a two-fold mole excess.
4. Place the aluminum pieces in a 250 mL beaker and SLOWLY add a two-fold mole excess of
KOH(aq) solution. [Note: see prelab worked problem for an example calculation.]
Note: You should not have to add more than 65 mL of KOH. If this occurs, see your lab
instructor.
Note: Make sure you record the initial mass of the aluminum metal, the amount of KOH added,
and observations from adding KOH to Al.
5. When the initial bubbling and foaming has slowed, heat the solution very gently. DO NOT
BOIL. If the level of the liquid drops to less than one quarter of the original level, add distilled
water. After the fizzing has completely ceased (10-15 minutes), remove the beaker from the
flame.
Q: How can you tell that hydrogen gas is produced?
6. While the solution is still hot, filter it by gravity filtration to remove insoluble impurities.
7. Cool the solution to room temperature and then slowly add 15 mL of 9 M sulfuric acid
(H2SO4). Stir the mixture continuously. Record your observations. After all of the sulfuric acid
has been added, Al2(SO4)3(aq) will be present in solution.
8. Using a Bunsen Burner heat the mixture gently for 10 minutes. Watch carefully for splattering.
The solution should now be clear. If any solid is seen, filter the warm mixture again. Please
record your observations in your lab notebook.
9. Cool the resulting clear solution in an ice bath until crystals form. The crystals will be alum,
KAl(SO4)2•12 H2O. If no crystals form after 15-20 minutes, try scratching the inside of the
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beaker with a glass stirring rod. The scratching forms small grooves in the glass, and the crystals
can adhere to the rough surface. If you are still unsuccessful in obtaining crystals, boil the
solution to reduce the volume of liquid by 25 % and then cool in the ice bath. When crystals
begin to form, leave the beaker in ice for another 10 minutes.
10. Filtering and washing crystals. Place a flask containing 20 mL of 50%:50% ethanol: water
mixture into the ice bath to cool. Collect your crystals by vacuum filtration. Wash them with
approximately 20 mL of the cooled ethanol/water mixture. Continue to apply the vacuum to the
crystals until the filter paper is dry and the crystals do not stick to a stirring rod if touched. To
obtain any product remaining in the funnel and to make sure the crystals are dry, make a paper
sandwich by removing the filter paper from the funnel and placing it product side down on top of
a piece of paper towel. Press down on the top of the sandwich with another paper towel to
remove excess moisture. Transfer the crystals by scrapping from the paper to a small beaker
(previously tared) and weigh the mass of the crystals. This is the mass you will use to calculate
your yield in your lab report.
11. Recrystallization. Recrystallization is a process to purify the alum product. This process
results in increased purity because impurities are left in solution. Large crystals often indicate
high purity. Transfer approximately 5 g of your alum into a 150 mL beaker for recrystallization.
Add 30 mL distilled water and heat with stirring to dissolve the solid. When all of the alum has
dissolved, remove the solution from the heat. Suspend a piece of thread into the solution to begin
nucleation. Loop some thread over a glass rod and rest the rod across the top of the beaker. Make
sure the thread is about 1-2 inches below the surface of the liquid but is not touching the sides or
bottom of the beaker.
12. Place the solution uncovered in your drawer and allow it to stand until the next laboratory
period. This product does not have to be reported on in your lab report.
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Lab Report
Include the following information in a lab report:.
1. Data table containing the following:
•
•
initial mass of aluminum metal
mass of the alum crystals (from step 10 in your procedure)
2. The chemical formula for this type of alum is KAl(SO4)2•12 H2O. What is the molecular
weight?
3. Based on the balanced chemical equation for the production of alum (see equation 1), how
much alum could you synthesized from your initial mass of aluminum, assuming a complete
reaction.
4. The actual amount of product produced in an experiment or industrial process is called the
actual yield. The amount of product that could be theoretically produced from a reaction if every
molecule of the limiting reagent is included in the product is called the theoretical yield.
Calculate the theoretical yield, your actual yield, and the percent yield of alum obtained in the
experiment.
Q: Can you make a claim to have synthesized alum from aluminum? Do you need to make any
assumptions? What evidence do you have?
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Q: Can you make any claims about the yield of the synthesis?
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