Aluminum to Alum
Assessment Criteria: DCP, CE
Data Collection and Processing
Raw Data
sType of measurement
Value
Mass of KOH (g)
Uncertainty
4.00
±0.01
250.0
±0.05
Mass of aluminum piece (g)
0.39
±0.01
Volume of sulfuric acid used (mL)
13.0
±0.05
Volume of water used to dilute
acid (mL)
12.0
±0.05
Mass of dried crystals and
paper towel used to dry them
(g)
15.34
±0.01
Mass of paper towel (g)
12.65
±0.01
Volume of water to dissolve KOH
(mL)
The uncertainties in mass measurements are based on the smallest digit of the digital readout. Uncertainties in
volume measurements are based half the smallest increment on the measuring cylinders used.
Referenced Data
Value
Uncertainty
Source
26.982
±0.001
www.unit5.org/chemistry/christjs/2MolarMass_element.doc
32.066
±0.001
"
39.098
±0.001
"
1.0079
±0.0001
www.wikipedia.org/wiki/hydrogen
15.999
±0.001
www.unit5.org/chemistry/christjs/2MolarMass_element.doc
Molar mass of
aluminum (g/mol)
Molar mass of
sulfur (g/mol)
Molar mass of
potassium (g/mol)
Molar mass of
hydrogen (g/mol)
Molar mass of
oxygen (g/mol)
Formula of alum
(potassium
aluminum sulfate)
KAl(SO4)2*12H2O
http://antoine.frostburg.edu/chem/senese/101/moles/faq/yieldalum-from-al.shtml
Reaction Formula
for synthesizing
Al3+(aq) + K1+(aq) + 2 SO42-(aq) +
12 H2O ---> KAl(SO4)2.12H2O(s)
http://web.lemoyne.edu/~giunta/chm151L/alum.html
Alum from
Aluminum
Data Processing and Calculations
Calculation of Yield
Moles of Aluminum = Mass Al. Consumed /Molar mass
= 0.39g/ 26.982 g/mol
=0.014mol
Mole ratio of Aluminum to Alum in Reaction = 1:1
Moles of Alum theoretically yielded= moles of Al used x mole ratio
= 0.014mol x 1/1
=0.014mol
Molar mass of Alum = MK + MAl + 2MS + 8MO +12 MH2O
= 474.384 g/mol
Theoretical mass of Alum yielded = moles of Alum theoretically yielded x molar mass of Alum
= 0.014mol x 474.384 g/mol
= 6.9g
Actual Mass yield = Combined mass of crystals and paper towel – mass of paper towel
= 15.34g – 12.65 g
=2.69g
% yield = (Actual yield/Theoretical yield) x 100%
= (2.69g/6.9g)100%
=39%
Processed Data Table
Quantity
Moles of AL used (mol)
Mole Ratio of Al:Alum
Theoretical mole yield of Alum (mol)
Molar mass of Alum (g/mol)
Theoretical mass yield of Alum (mol)
Actual Mass yield op Alum (g)
Percent yield (%)
Value
0.014
1 to 1
0.014
474.384
6.9
2.69
39
Uncertainty Calculations
Unc. In mol of Al = ((0.01/ 0.39) + (0.001/26.982))*100%
=2.56781%
Unc. In Molar Mass of Alum = 0.001+ 0.001 + 0.001 + 0.0001 + 0.001
= ±0.0041g/mol
% err.
= (0.0041 / 474.384)*100%
=0.0008%
Uncertainty in Theoretical mass yield=2.56781% + .0005%
= 2.5687%
Uncertainty Actual mass yield= 0.01 + 0.01
=0.02g
% err.
=(0.02/2.69)*100%
= 0.74349%
Uncerainty in Final % Yield = 2.5687% + 0.74349%
= 3.312% * 39.2
=±1.30%
To proper significant figures = ±1%
Final Value : 39 % Yield, ± 1%
Conclusion and Evaluation
Conclusion
The final crystals formed were of a very wide range of sizes, with some barely visible, to others as large as
approximately 1 cm3. Most of the crystals were quite clear, showing few signs of impurities.
The percent yield of the experiment was 39% ± 1%. This was based on the mole ratio of Aluminum to alum
in the reactions used, the moles of aluminum used in the reaction, calculated from the mass used, the
theoretical value of moles and mass of alum supposedly to be produced. The actual mass of crystals
measured was then compared to the theoretical value to acquire the final percent value. The yield is
far below the perfect yield of 100%, but the difference can, and will, be explained in the evaluation.
Evaluating Procedure
The measurements taken were precise in this experiment. The final uncertainty was within 1% error, a strong
point of this procedure.
However, the procedure lacked the accuracy and high yield desired. Firstly, allowing the aluminum to dissolve
in the Potassium hydroxide solution for only 15 minutes as per indicated in the procedure was not sufficient.
Even though the aluminum piece used was torn into minute pieces in order to speed up the reaction, by the
end of the 15 minutes, small pieces of aluminum remained.
There were also several problems with the filtration process. The paper used needed to be folded in order to fit
into the funnel to be used. This crease in turn was a weak point. Sometime during filtration, the paper tore,
and some of the black precipitate from the aluminum, potassium and hydroxide solution was able to pass
through.
Moreover, the collection and decanting of the liquid from the crystal did not occur as effectively as possible. No
matter how quickly the liquid was poured out of the beaker, some of the tinier crystal could not be retrieved,
and when removing the crystals from the paper towel, not all could be removed for certain. Some residue
could well have remained on the paper. Many crystals were lost in transport as well, when placing the crystal
to dry, when carrying them over to the scale, etc. The container used to carry them around was unstable.
Improving the Investigation
First of all, instead of arbitrarily estimating the reaction time of the for Al to dissolve in the KOH solution,
allowing all of the aluminum to dissolve, and ensuring that it does by step like stirring, should improve the
yield.
Instead of folding the filter paper into the cone to pour our liquid in to, taking the time to properly make a
cone, using some adhesive tape to cover any crease, would reduce our risk of another contamination.
Using a larger cone would increase the surface area of paper for the cone to be used during both filtration and
decanting of the liquid from the crystals. This would be especially useful in the decanting step as all the liquid
could be poured in at once, and none of the crystal would settle at the bottom of the beaker, unable to be
poured out. Furthermore, placing the dried crystals and filter paper in a solid container could reduce the loss of
crystal during handling when compared to just the paper.