Experiment #5 – Aspirin Synthesis Lab Report
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Wilfrid Laurier University *
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Course
110
Subject
Chemistry
Date
Feb 20, 2024
Type
Pages
10
Uploaded by DeaconSnow11815
1
Abstract
The purpose of this lab is to synthesize aspirin from the mixture of acetic anhydride and
salicylic acid. This resulted in creating crystals and used to filter in the Hirsh funnel. Creating a
vacuum when pouring the liquid will allow the solid to be trapped when the liquid gets drained.
This filtered product was impure and solid acetylsalicylic acid, with a mass of 0.12g. The
melting point of this ASA was determined and compared with the standard ASA. This was done
using a Digimelt melting apparatus which involved placing the solid in a capillary tube into the
machine to find an accurate melting range. The initial melting point of Asa was 102.3
℃
and
final readings of 105.7
℃
. The prepared ASA was involved in the Iron (III) salicylate dianion
solution with NaOH. The diluted sample was mixed with FeCl. A vernier Go Direct calorimeter
was used to measure the absorbance of 5 standard solutions and construct the Beer’s law plot
graph. The mass of pure ASA and concentration values were determined. Also, the percentage
purity was calculated as 99% and percentage yield of 34%. This means that only 34% of aspirin
was synthesized out of 100% and the aspirin was 99% pure. In conclusion, this experiment was
not successful due to low percentage yield.
Procedure
For the procedure, see lab manual (CH110 Lab Manual, Fall 2023) pages 121-124. No deviations
were made to this aspirin synthesis experiment. Wilfrid Laurier University Chemistry
Department. Fall 2023. Pages 117-127 in the Chemistry 110 Lab Manual. Wilfrid Laurier
University, ON, Canada
Observations/Results
Table 1: Before, During, After Qualitative Observations of Solutions From Part A-E
Part
Solutions
2
A
Salicylic Acid: white, odourless powder
Phosphoric Acid: clear, yellow, transparent liquid
Acetic Anhydride: clear, liquid, vinegar odour
DI water (4 drops): some liquid but solid particles formed
DI water (7 drops): white crystal-like solid participles at bottom of tube
B
Prepared ASA Before: wet, cloudy particles, viscous
Prepared ASA After: sand like texture, odourless, white, solid
C
Standard ASA sample
Before heat: small, dry solid, white, very grainy,
After heat: clear, colourless, liquid
Prepared ASA sample
Before heat: powder texture, grainy, opaque, solid, white
After heat: clear, colourless, liquid
D
NaOH: water like viscosity, odourless, colourless, clear
Prepared ASA sample: white, solid, grainy
Mixture of NaOH and prepared ASA sample: clear, colourless, liquid
FeCl
3
: liquid, yellow, clear, odourless
Mixture of NaOH + ASA + FeCl
3
: dark brown, liquid, opaque, odourless
E
Curvette A: dark purple/brown liquid
Curvette B: dark purple/brown liquid
Curvette C: dark reddish brown liquid
Curvette D: light brown liquid
Curvette E: yellow liquid
Prepared ASA sample: dark brown liquid
Table 2: Quantitative Data Collected From Part A-C
Mass of salicylic acid (g)
0.18
Volume of acetic anhydride (mL)
0.34
Mass of filter paper (g)
0.62
Mass of “impure” ASA + filter paper (g)
0.80
Mass of prepared “impure” ASA (g)
0.12
Melting Point range of prepared ASA (
℃
)
102.3
℃
- 105.7
℃
Melting Point range of standard ASA (
℃
)
138.9
℃
- 143.5
℃
Table 3: Results from Part D-E
3
Sample
Absorbance Reading
Standard A
1.30
Standard B
1.02
Standard C
0.80
Standard D
0.55
Standard E
0.25
Prepared ASA
0.88
Questions
1.
Construct a beer’s law plot of Absorbance versus Concentration and add a computer
generated line of best fit. Show the slope intercept form of the best fit line.
2.
a)
Determine the concentration of your prepared iron (III) Salicylate Dianion solution
(pure ASA) in the 10.0 mL of 1.0 M in mg/mL (part D) using the Beer’s law plot
completed in Question 1.
y = 7.2873x + 0.0115
0.88 = 7.2873x + 0.0115
0.88 - 0.0115 = 7.2873x
x =
0.8685
7.2873
x = 0.119 mg/mL
b)
Then, calculate the mass (in mg) of pure ASA knowing that the two consecutive dilutions
were done to prepare the solution. Show all of your work.
C
3
x V
3
= C
2
x V
2
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= C
2
𝐶3 ? 𝑉3
𝑉2
C
2
=
(0.119 ??/?𝐿)(25?𝐿)
(5?𝐿)
C
2
= 0.595 mg/mL
C
1
x V
1
= C
2
x V
2
C
1
=
𝐶2 ? 𝑉2
𝑉1
C
1
=
(0.595 ??/?𝐿)(100?𝐿)
(5?𝐿)
C
1
= 11.9 mg/mL
m = CV
m = (11.9 mg/mL)(10mL)
m = 119 mg of pure ASA
3.
A standard “Bayer Aspirin'' tablet contains 325 mg of acetylsalicylic acid by weight. If
the absorbance of this commercial tablet was measured to be 2.57, is the claim made by
the drug company accurate? Explain your answer and show your work.
y = 7.2873x + 0.0115
(2.57) = 7.2873x + 0.0115
2.57 - 0.0115 = 7.2873x
x =
2.5585
7.2873
x = 0.351 mg/mL
C
3
x V
3
= C
2
x V
2
= C
2
𝐶3 ? 𝑉3
𝑉2
C
2
=
(0.351 ??/?𝐿) (25?𝐿)
(5?𝐿)
C
2
= 1.755 mg/mL
5
C
1
x V
1
= C
2
x V
2
C
1
=
𝐶2 ? 𝑉2
𝑉1
C
1
=
(1.755 ??/?𝐿) (100?𝐿)
(5?𝐿)
C
1
= 35.1 mg/mL
m = CV
m = ( 35.1 mg/mL) (10mL)
m = 351 mg
4.
Calculate the percent purity of ASA. Show your calculations, including units.
Include
your statement of academic integrity in this question
.
Percent purity =
x 100%
?𝑎?? ?? ???? ??????? ?𝑎?? ?? 𝑖????? ???????
Percent purity =
x 100%
119 ??
120 ??
Percent purity = 99%
“I declare that this lab report was done individually. I did not take pictures of my work or
send my own documents with other students. I understand the academic integrity policies and
penalties at WLU.”
5.
Calculate the percentage yield of pure ASA by including a reaction table as shown in the
introduction section of this lab manual. Show all your calculations, including how you
determined theoretical yield and units.
Theoretical yield = 0.18g C
7
H
6
O
3
x
180.16 ?/???
138.12 ?/???
= 0.235g
Percentage yield =
x 100%
𝑎???𝑎? ?𝑖???
?ℎ?????𝑖?𝑎? ?𝑖???
=
x 100%
0.119?
0.235?
= 34%
Table 4: Reaction table for the production of ASA
H
3
PO
4
C
7
H
6
O
3
+ C
4
H
6
O
3
→
C
9
H
8
O
4
+ CH
3
COOH
(L.R)
6
MW (g/mol)
138.12
102.09
180.16
60.05
Volume (mL)
-
0.34
-
-
Density (g/mL)
-
1.08
-
-
Mass (g)
0.18
0.37
-
-
Mmol
1.3
3.60
-
-
Complete calculation #1: calculation of molar weight of all molecules in the reaction
C
7
H
6
O
3
= (7 mol x 12.0107 g/mol) + (6 mol x 1.00794 g/mol) + (3 mol x 15.9994 g/mol)
= 138.12 g/mol
C
4
H
6
O
3
= (4 mol x 12.0107 g/mol) + (6 mol x 1.00794 g/mol) + (3 mol x 15.9994 g/mol)
= 102.09 g/mol
C
9
H
8
O
4
= (9 mol x 12.0107 g/mol) + (8 mol x 1.00794 g/mol) + (4 mol x 15.9994 g/mol)
= 180.16 g/mol
CH
3
COOH = (2 mol x 12.0107 g/mol) + (4 mol x 1.00794 g/mol) + (2 mol x 15.9994 g/mol)
= 60.05 g/mol
Complete calculation #2: calculation of mmol of C
7
H
6
O
3
C
7
H
6
O
3 mmol
=
?𝑎??
???𝑎? ?𝑎??
=
0.18?
138.12 ?/???
= 1.3032 x 10
-3
mol x
1000 ????
1 ???
= 1.3032 mmol
≃
1.3 mmol
Complete calculation #3: calculation of density, mass and mmol of C
4
H
6
O
3
C
4
H
6
O
3 density
=
?𝑎??
??????
=
0.37?
0.34?𝐿
= 1.08 g/mL
C
4
H
6
O
3 mass
= DV
= 1.08 g/mL x 0.34mL
= 0.3672g
≃
0.37g
C
4
H
6
O
3 mmol
=
?𝑎??
???𝑎? ?𝑎??
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7
=
0.3672?
102.09 ?/???
= 3.596x10
-3
mol x
1000 ????
1 ???
= 3.596 mmol
≃
3.60 mmol
Complete calculation #4: determination of limiting reagent
1.3032 x 10
-3
mol C
7
H
6
O
3
x
= 1.3032x10
-3
mol C
9
H
8
O
4
1 ??? 𝐶9𝐻8𝑂4 1 ??? 𝐶7𝐻6𝑂3
3.596x10
-3
mol C
4
H
6
O
3
x
= 3.596x10
-3
mol C
9
H
8
O
4
1 ??? 𝐶9𝐻804
1 ??? 𝐶4𝐻603
C
7
H
6
O
3
< C
4
H
6
O
3
, therefore the limiting reagent is C
7
H
6
O
3
Discussion
The objective of this experiment is to produce aspirin and determine its purity using the melting
point and absorbance measurements with Go Direct. The acetylsalicylic acid was not
successfully made in this experiment since the percentage yield was 34%, meaning it suffered
over 50% of product loss. The aspirin was synthesized by liquid acetic anhydride and powder
salicylic acid in the presence of phosphoric acid. A Hirsch funnel was used to filter the mixture
into producing acetylsalicylic solid. Parts C-E was completed with the use of a calorimeter to
determine absorbance, concentration and mass of pure ASA. A beer’s law plot was constructed
based on absorbance versus concentration of ASA in standards.. The dilution formulas were used
to figure out the mass of aspirin and the percentage purity ASA produced. The percentage purity
resulted in 99% and can further be analyzed by observing the melting point of the prepared ASA
of 143.5
℃
. In part C, we concluded that the melting point of standard ASA was 105.7
℃
. There
is only a 3.4
℃
difference while the prepared ASA was 4.6
℃
, proving that ASA is pure enough.
The small difference in temperature could be sources of errors during the lab, including product
loss during the dilution step. A source of error in this experiment would be wetting the filter and
8
obtaining the remaining crystals from the test tube with water. Water is known to be a good
solvent so it would dissolve some of the crystals and contaminate the solution, which would
affect the amount of precipitate that was dried. This would result in reducing the percent yield
due inaccurately performing this experiment. One way to improve the procedure would be to use
a different solvent when wetting the filter paper and removing the crystals from the test tube.
Ethanol can be used as it will not contaminate the precipitate as it is a good disinfectant. Also, it
would prevent the percent yield from reducing as ethanol would only dissolve a smaller amount
of precipitate since it is not as soluble as water.
Conclusion
The synthesis of aspirin was not successful as the percent yield was low being 34% and the
synthesized mass was 119 mg. The percent purity was 99%, indicating that the sample that was
synthesized was accurate. Also, the absorbance of the prepared sample was 0.88.
Therefore, the lab was ineffective but the criteria for this experiment was met.
References
Refer to page 117-127 in the lab manual
9
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