i need a lab report about Chemical conversion- Alum

Discussion – Explain your results, referring to the chemistry theory where appropriate. Did the dataconform to what was expected? What does it show in relation to the experiment? If the results are notwhat was expected explain why. Possible sources of error (estimate experimental error in numericalresults). Conclusion – briefly state any final result(s), or conclusions that can be drawn from the experiment.References – any sources consulted for the writing of the report. These should be referenced (cited) inthe body of the text.  CHEMICAL CONVERSION (Stoichiometry): ALUM FROM ALUMINUM
anhydrous alum to form the
SCRAP
hydrated salt.
OBJECTIVES:
•
To understand how to perform stoichiometry calculations in
determining the theoretical yield
•
To be able to calculate percent yield from experimental data
•
To learn some new experimental techniques by synthesizing alum
from aluminum metal.
•
To learn the atom economy concept.
INTRODUCTION
Aluminum is a very useful metal in our lives. It is extremely
lightweight and durable, but is very energy intensive to produce.
Aluminum metal is quite reactive, yet it resists corrosion. This is
because forms a protective Al2O3(s) oxide coating that prevents the
rest of the aluminum metal from further reaction with oxygen. On
the other hand, iron forms a non-protective oxide, Fe2O3 (rust). As
a result, iron is susceptible to corrosion. Thus, unlike iron, metallic
aluminum products do not need to be painted.
While aluminum can be recycled or reused, we are going to use
it as the starting material in a chemical synthesis. Through a series
of reactions, we are going to chemically transform metallic
aluminum into “alum”. Alum is the ionic compound KAl(SO4)2 •
12H2O, potassium aluminum sulfate dodecahydrate with a melting
point of 92.5oC.
Alum was known in antiquity and has many practical uses.
Hydrated alum was used as a flame retardant by the ancient
Egyptians and Romans. It is still used in fire extinguishers today.
Its flame suppressant properties are related to the release of
waters of hydration – a phenomenon we studied in last week’s lab.
When alum is heated to a high temperature, it decomposes as
follows:
2 [KAl(SO4)2 • 12H2O(s)] + Heat ® K2SO4(s) + Al2O3(s) + 3SO3(g)
+ 24H2O(g)
The release of the 24 waters of hydration serves to help cool the
fire. SO3 is a noncombustible gas which is heavier than air and
helps smother the fire by reducing the amount of available oxygen.
Unfortunately, it is a smelly gas which is also toxic and dangerous
for humans to breathe. In addition, K2SO4(s) and Al2O3(s) form a
noncombustible solid coating on the fuel source which further
inhibits combustion. Hydrated alum is also used in water
purification to coagulate fine suspended particulates into large
particles which fall to the bottom and can be removed. Finally, alum
has historically been used as a deodorant in many cultures and is
currently marketed in US as a “natural deodorant crystal”.
Anhydrous alum can be
used for preserving food, such
as pickles. It preservative
action arises from hydration –
the reverse of the process we
studied last week – by
removing water that would
normally spoil the product.
Any water present in the food
would be attracted to the
Dr. H.
In addition to referring to
the compound we will make,
the term “alum” is also used for
a whole class of related
compounds in which other
metal cations are substituted
for K+ or Al3+.
These
“substituted alums” have the
general formula AB(SO4)2 •
12H2O where A is a
monovalent (+1) cation and B
is a trivalent (+3) cation. Here
are some examples:
sodium alum
NaAl(SO4)2 •
12H2O chromium alum
KCr(SO4)2 • 12H2O
sodium ferric alum
NaFe(SO4)2 • 12H2O
One of the Sapling postlab questions will
involve a substituted alum compound.
The Synthesis Scheme
The series of reaction
steps we will be performing in
this synthesis are given in Box
1 on the next page. In Step 4
of the procedure, aluminum
metal will be converted into a
basic
salt
solution
of
potassium
aluminum
hydroxide. As you see from
the equation below, hydrogen
gas is formed in this process.
This can create a fire and
explosion hazard. If too much
heat is generated as that could
potentially ignite the hydrogen
gas being produced during the
process. We will reduce the
hazard by performing the
reaction in the hood to
ventilate the hydrogen gas.
Unfortunately, this dangerous
reaction could accidentally be
created at home.
Oven
cleaners commonly contain
strong bases as a means to
convert fats into soap,
whereby you can remove the
cooking mess using hot water.
Warning labels on the oven
cleaner container will alert you
to the fact that it will react with
aluminum metal and can
cause a major hazard.
Once this reaction has
occurred, it will be filtered to
remove the Al2O3(s) coating
that is present on the surface
Page 2 of 7
of the aluminum metal. After
filtration, in Step 6 the basic
potassium
aluminum
hydroxide solution will then be
slowly neutralized with sulfuric
acid to form first aluminum
hydroxide and then the neutral
aluminum sulfate salt. In Step
7, alum is formed by heating
the reaction solution which
contains
a
mixture
of
aluminum
sulfate
and
potassium sulfate.
If you
carefully examine the overall
net reaction, you will notice
that there are
more waters of hydration in the alum product than
the number of waters on the reactant side. The
difference is due to a net of two more water
molecules being produced than are consumed
during the alum synthesis process.
In our synthesis, small crystals of alum will be
produced. Large crystals of alum can be grown by
dissolving the product in a minimum amount of
deionized water and then slowly evaporating the
water from an open container over time. Alum
crystals weighing several hundreds of pounds each
have been grown this way in laboratories using
large fish tanks as the growing containers.
Reference
For information on stoichiometry and percent
yield calculations, consult Sections 4.2 and 4.3 of
Chemistry: A Molecular Approach, (3rd Edition),
Nivaldo J. Tro, Pearson, 2014.
EXPERIMENTAL PROCEDURE
Common Equipment (located in fume hoods)
Buchner funnel with rubber stopper
250 mL side arm vacuum flask
Reagents Used Per Pair:
Litmus paper (blue and red)
Filter paper (#1 and #4)
0.5 grams Aluminum foil
20 mL 3 M KOH(aq)
~5 mL (100 drops) 6 M H2SO4(aq)
10 mL 95% Ethanol
Strong acids and bases will be used during this
lab. Wear your goggles and aprons and use them
carefully to prevent contact with these materials.
Additions of reagents should be in a fume hood to
prevent breathing in toxic fumes. Hydrogen gas will
be produced in the beginning steps of this
experiment, so proper venting in a fume hood is
required. Make sure all glassware is clean and free
of cracks or chips prior to usage in the experiment.
Directions:
Work in pairs throughout this experiment.
1. Weigh out a piece of aluminum foil (between
0.4 to 0.6 grams) and record the actual mass in
your lab report.
2. Carefully tear the aluminum foil into very small
pieces and place them in a 250 mL Erlenmeyer
flask. Reaction time is dependent upon surface
area so smaller pieces of aluminum will react
faster with the KOH than larger ones.
5. In this step the basic solution is neutralized
then made slightly acidic by adding 6 M
sulfuric acid, H2SO4(aq). Be careful as this
is a strong acid solution and the resulting
acid-base reaction is very exothermic.
During this step a white precipitate will likely
form with the initial addition of H2SO4(aq),
and may dissolve near the end of the
addition when the solution becomes acidic.
Since accurate volumes are not critical in
this step we will count drops: 20 drops ≈ 1
mL.
Commented [EO3]: This seems to have already been done in
the prep process. Not needed here?
3. In a hood, use a graduated cylinder to transfer
20 mL of 3 M KOH(aq) to the flask containing
the aluminum foil pieces. Be careful as this is
a strong base solution. Swirling the flask will
help in the reaction process, but be careful not
to spill the liquid. The container will get hot.
The mixture will most likely be cloudy due to the
formation of insoluble Al2O3(s) which is an
unwanted byproduct of the reaction.
4. The Al2O3(s) is removed using simple gravity
filtration. While the mixture is cooling label a
250 mL Erlenmeyer flask with your name – this
flask will be used to collect the filtrate. Place a
piece of Whatman #1 filter paper in your glass
funnel. Once the mixture has cooled to room
Hazards:
temperature slowly transfer the contents of
the reaction flask into the filter funnel and
collect the filtrate in the labelled receiving
flask.
When finished the filter paper
containing waste Al2O3 can be disposed of
in the trash.
Commented [EO1]: Use Bunsen burners. Be sure to revise
prep notes Deleted: ¶
Commented [EO2]: Needs to be added to prep notes
Box 1: Reaction Steps in Synthesis of Alum from Aluminum
Step 4 of the experiment
2 Al(s) + 2 KOH(aq) + 6 H2O(l) → 2 KAl(OH)4(aq) + 3 H2(g)
Step 6, initial addition of H2SO4
2 KAl(OH)4(aq) + H2SO4(aq) → 2 Al(OH)3(s) + 2 H2O(l) + K2SO4(aq)
Step 6, further addition of H2SO4
2 Al(OH)3(s) + 3 H2SO4(aq) → Al2(SO4)3(aq) + 6 H2O(l)
Al2(SO4)3(aq) + K2SO4(aq) + 24 H2O(l) → 2 KAl(SO4)2 • 12H2O(s)
________________________________________________________________________________________________________________________ Overall net
Step 7 upon heating
reaction 2 Al(s) +2 KOH(aq) + 4 H2SO4(aq) + 22 H2O(l) →2 KAl(SO4)2 • 12H2O(s) + 3 H2(g)
Dr. H.
Use a pipet to slowly add approximately 4 mL of
6 M H2SO4(aq) by counting drops: since 20
drops ≈ 1 mL add 80 drops. Swirl the solution
to ensure complete mixing. Test the pH of the
solution (acidic solutions turn blue litmus red) if it is acidic you are finished with this step. If
not, continue adding the sulfuric acid in 5 drop
increments, swirl, and test the pH after each
increment. Stop when the solution is acidic.
heating, continuously stir your flask by swirling. Try to avoid boiling. If the solution begins to boil (starts
bubbling) immediately shut off the Bunsen burner and let sit the remainder of the 5 minutes.
7. Remove the flask from the Bunsen burner and allow the mixture to gradually cool to near room
temperature at your station. Crystals should begin to grow as the solution cools, but alum will often
super saturate. Crystal growth can be encouraged by using a glass stir rod to scratch the walls of the
flask below the liquid surface.
6. Place the flask containing the mixture on a ring
stand with a wire gauze for support over a
Bunsen burner. Adjust your Bunsen burner to
GENTLY heat your flask for 5 minutes. While
9. Clean and dry a watch glass. Together weigh it and a piece of Whatman #4 filter paper and record
the mass in your lab report. Set them aside until after the vacuum filtration.
8. Place the nearly cooled flask into an ice bath to finish the crystallization process. Allow it to sit for at
least thirty minutes. A fine, white, fluffy precipitate should form during this step. While you are waiting,
complete the next two steps
CHEMICAL CONVERSION (Stoichiometry): ALUM FROM ALUMINUM SCRAP
Page 3 of 7
is required >=90 drops.
10. Carefully clean all of your test tubes with soap,
water, and test tube brush. Rinse with tap water
until all soap suds are gone and then rinse three
times with deionized water. (You will need clean
test tubes for next week’s experiment.)
11. Use vacuum assisted filtration to separate the
solid alum product from the remaining solution.
Set up the Buchner funnel and filter flask as
shown in Figure 1. Make sure you place the
piece of Whatman #4 filter paper, weighed in
Step 1, in the funnel above the holes and wet it
with a minimum amount of DI water.
12. Connect the rubber hose from the flask to a
vacuum trap. The rubber hose from the trap is
connected to the vacuum port on the bench top
(the tap with the yellow knob labelled VAC).
Gently open the vacuum port knob so that air is
removed slowly at first. The pressure differential
across the filter paper increases the rate at
which the liquid is drawn through the paper
relative to simple gravity filtration
13. Transfer the mixture from the flask into the
Buchner funnel. Use a minimum amount of
Commented [EO5]: Note that numbering is off. #6 is
somehow skipped/omitted.
water to wash any remaining alum from the
flask into the filter. Do not allow the water level
in the vacuum flask to reach the level of the
side-arm level.
14. Wash the white crystals of alum in your filter with 10 mL of cold ethanol. Alum is soluble in water and
ethanol so you need to make sure you limit your use of either of them when washing the crystals.
Ethanol is added to help remove water that adheres to the surface of the crystals; however, as a drying
agent it is not strong enough to remove the waters of hydration.
15. Allow your product to dry under vacuum suction for at approximately 20 minutes to achieve a dried
solid.
C
o
16.
Carefully
transfer the
reaction product, KAl(SO4)2•12H2O(s), and the filter paper
m
onto the watch glass (from Step 1) via a spatula or scoopula.
m
e
17.
Weigh the combined watch glass, filter paper, and product, and record this mass in your lab report.
n
The difference between this mass and the mass of the watch glass/filter paper from Step 1 will give
t
mass of the product.
e
d
18.
Dispose of the weighed alum product by placing in the regular trash can at the end of the bench.
[
E
19.
O Dispose of the filtrate solution you collected in the vacuum flask according to the direction of your
laboratory instructor.
4
]
20.
Clean up your bench area before checking out of lab.
:
B
a
Green
Chemistry Connections
s
e
Aluminum
Recycling
d
o
Aluminum metal is a valuable resource that is
n
often
taken for granted. Most of the aluminum found
S
e earth is in the oxide form, which requires energy
on
c
to
reduce down to metallic aluminum. Pure natural
t
occurring
aluminum metal was so rare that at one
A
2
time
it was worth 10 times its weight in gold. It was
0
not
1 until the mass production of electricity in the late
1800s
that aluminum became used by the average
6
e
person.
Sadly, we have trained ourselves to look at
x
itp as a disposable metal with little value.
e
Many people think that aluminum metal will stay the same when buried underneath soil for long periods
r
of
i time. This is simply not true, as aluminum is a reactive metal. When an aluminum foil or can is disposed
e in a landfill, the aluminum metal slowly reacts with oxygen to produce bauxite – a mixture of Al(OH)3,
of
n
AlO(OH)
and Al2Si2O5(OH)4. This process can be enhanced by the pH of the soil around it. Once the
c
e
aluminum
has been converted back to the oxide, it will require much more energy to form aluminum metal
,
than
would be required to simply melt the same sample in metallic form. The same can be said of the alum
t
sample
that you produced in this laboratory experiment. This is the reason why recycling of aluminum
h
Dr. H.
Page 4 of 7
metal is important to do. It saves energy to do the
recycling conversion when it is in the metallic form.
So the next time you are about to throw away an
aluminum item, please remember this process…
Atom Economy†
A main focus of green chemistry is to reduce the
amount of pollution created in chemical syntheses.
Reactions in which a large proportion of the reactant
atoms end up in waste products contribute to
pollution, make ineffective use of resources, and
raise the costs of production. One way to measure
the relative efficiency or inefficiency of a reaction is
to look at its atom economy. Atom economy lets us
know what percent of the reactants’ atoms make
their way into the desired product. Atom economy
of is defined as:
FW of Desired Product ´100 =
% Atom Economy FW of All Products where FW
stands for formula weight. The concept of atom
economy was developed by Barry Trost, a professor
of chemistry at Stanford University, who received a
1998 Presidential Green Chemistry Challenge
Award for his work.
†This introduction to atom economy is an excerpt adapted
from Introduction to Green Chemistry; American
Chemical Society: Washington, DC, 2002.
Atom Economy versus Percent Yield
Traditionally, chemists have used percent yield
as a measure of how efficient a chemical reaction is.
Percent yield measures the fraction of desired
reactant atoms which end up in the desired product.
CHEMICAL CONVERSION (Stoichiometry): ALUM FROM ALUMINUM SCRAP
What percent yield ignores is the fraction of reactant
Dr. H.
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Page 6 of 7
atoms which ends up in the undesired products (i.e.,
waste). In contrast, atom economy measures the
fraction of all reactant atoms which end up in the
desired product.
The difference is perhaps best illustrated by an
example. Suppose that we are going to prepare
ZnSO4(s) from aqueous solutions of ZnCl2 and
K2SO4:
ZnCl2(aq) + K2SO4(aq) ® ZnSO4(s) + 2 KCl (aq)
The percent yield calculation focuses only on the
atoms in highlighted in bold underline font. Let’s
suppose that we had a solution containing 0.136 g
of ZnCl2 which was combined with an excess of
K2SO4 to produce 0.152 g of ZnSO4(s). As shown in
Box 2 below, the theoretical yield (expected amount
of product if all ZnCl2 is converted to product) is
0.161 g. From a percent yield perspective, our
chemical process appears to be a very efficient
method of producing ZnSO4(s):
152 g actual ´100 = 94.4%
0.161 g theoretical
However, this calculation ignores the two moles of KCl produced. The KCl is unneeded waste matter
which must be disposed of. In the atom economy calculation, the KCl mass is included in the calculation
below. (Note that in the denominator we multiplied the KCl molar mass by two because two moles of KCl
are produced in the balanced chemical equation.)
´100 = 51.9%
161 g/mol ZnSO4
161 g/mol ZnSO4 +(2´74.6 g/mol KCl)
The result of 51.9% indicates that, even under the best of circumstances, half of our reactant mass is
winding up as waste product. Thus, from an atom economy perspective, this reaction is a relatively
inefficient method of producing ZnSO4(s)
Note that atom economy doesn’t replace percent yield. Both concepts are necessary to express the
overall efficiency of an experimental yield. The overall efficiency of a reaction can be defined as:
% Overall Efficiency = Yield ´ Atom Economy ´ 100%
In this equation, “Yield” and “Atom Economy” are expressed as fractional values, not percent values.
Thus, the overall efficiency of our example experimental result is:
% Overall Efficiency = 0.944´ 0.519´100 = 49.0%
Note that less than half of the reactants actually made it into the desired product. This is inefficient from
both an environmental and an economic
0
.
Box 2
Theoretical Yield = 0.136 g ZnCl2 ´
161 g ZnSO4
´
= 0.161 g ZnSO4 1 mol ZnCl
Dr. H.
136 g ZnCl2 1 mol ZnSO4
1 mol ZnSO4
2
´
1 mol ZnCl2
CHEMICAL CONVERSION (Stoichiometry): ALUM FROM ALUMINUM SCRAP
standpoint.
Dr. H.
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CHEMICAL CONVERSION (Stoichiometry): ALUM FROM ALUMINUM SCRAP
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Dr. H.
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Name:
Instructor:
CHEMICAL CONVERSION (Stoichiometry): ALUM FROM ALUMINUM SCRAP
DATA (values must be recorded in ink! Include units and proper precision where appropriate.)
Mass of aluminum:
__________________________________
Mass of watch glass + filter paper:
__________________________________
Mass of alum product + watch glass + filter paper: __________________________________
CALCULATIONS (complete this section in pencil!)
1. Calculate the mass of your alum product – this is the actual yield for your synthesis.
2. Calculate the molar mass of alum [KAl(SO4)2 • 12H2O] – round the result to the nearest 0.1 gram/mol. Use
the elemental molar masses listed in your text book for this calculation.
3. Calculate the theoretical yield in grams for this reaction. Assume that aluminum is the limiting reagent and
use the molar mass of Al, the reaction stoichiometry, and the molar mass of alum in your calculation. An
example calculation is shown in Box 2 on Page 4.
4. Use the actual yield (#1) and the theoretical yield (#3) to calculate the % yield for your synthesis.
Dr. H.
Page 2 of 5
Dr. H.
CHEMICAL CONVERSION: ALUMINUM SCRAP TO ALUM
Page 3 of 5
5. Calculate the % atom economy for this synthesis – use the example given on Page 4 of the experiment
instructions as a guide. Use the overall net reaction shown in Box 1 of the instructions.
6. Calculate the % overall efficiency for your yield in this synthesis.
REFLECTION QUESTIONS
1. For many chemical reactions, even under the best of circumstances, it is difficult to achieve reaction yields
close to 100%. Which step in this alum synthesis procedure is most responsible for a low yield? Justify
your answer.
2. Although it is theoretically impossible to achieve a yield greater than 100%, occasionally students do report
yields above 100%. Give an example of a mistake (other than simply misreading/recording a mass value)
that would lead to a yield greater than 100%.
Dr. H.
Page 4 of 5
3. Why is proper handling of samples (quantitative transfer) important in calculating percent yield?
4. A cleaned, dried aluminum can (Cherry Pepsi) was recently weighed. The recorded mass was 13.2518
grams. Assuming that 0.5000 g of the weight was other materials (paint, aluminum oxide), what would
the theoretical yield be (on a mass basis) if all of the aluminum in the can was converted into alum?
Remember to report your result with the correct units and precision.
Dr. H.
CHEMICAL CONVERSION: ALUMINUM SCRAP TO ALUM
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Dr. H.
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