beer's law Report
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Beer’s Law Lab Report
Alisha Varney
09/13/2023
Marshall University
CHM 218 section 104- Bill Price
I.
Introduction In this lab the relationship between absorbance and concentration of a sample, known as beer’s law, was tested. The sample used to test the law comes from aspirin
(ASA), which is SA. To test beers law, SA was diluted to five different concentrations and with different dilutants. This was done to prove that beer’s law is a linear relationship. As concentration increased, absorption also increased, making a linear relation in the data. II.
Data Part 1
Concentration of SA in stock solution A= 8.68 x 10^-6 g/mol
Concentration of SA in stock solution B= 1.74 x 10^-7 g/mol
Part 2
concentration of SA (M)
Absorbance 4.3 x 10 ^-5
0
8.7 x 10^-5
0.1
1.3 x 10^-4
0.15
1.74 x 10^-4
0.25
2.17 x 10^-4
0.55
Part 3
Aspirin tablet weight= 0.36g
Absorbency of diluted and dissolved aspirin tablet= 0.3
Concentration of aspirin in the 250ML solution is 7.99 x 10^-6.
III.
Example Calculations Concentration=molar mass g/m moles (given g/ x moles) =moles
Moles/ L = concentration amount 138.129g/1mol (0.3g/x mol)
Moles= 0.00217
0.00217/250L = 8.68 x 10^-6 m/mL (1000) = 8.68 x 10^-9 m/mL
Concentration of sample= 8.68 x 10^-6 m/ml (amount SA in sample (ML))
8.68 x 10^-6m/mL (10mL) = 8.7 x 10 ^-5 m/ML IV.
Results and Discussion
In this lab it was found that as the concentration of SA in the sample increased so did
the absorption of the sample. This coincides with the linear relationship stated in Beer’s Law. However, the data collected is not perfectly linear, this could be due to inaccurate measurements when SA was added to a solution. The problem could lie in
the fact that the 5ml sample was measured with a 5 ml pipette, the 10ml sample with a 10ml pipette, and 25ml sample with a 25ml pipette, but 15ml and 20ml samples were measured by eye using the 25ml pipette. Although there is some flaw in accuracy here, the data is still a positive slope with a linear trend line that proves Beer’s law of relationship between concentration and absorbance. Visually, it was seen that with concentration increase, the purple color of the samples deepened. In art it is learned that the darkness of color has to do with how much light is absorbed.
Considering this, it makes since that darker color of the sample will equal more absorption. Overall, Beer’s law was observed by proving that the increase of SA concentration in a solution correlates with the increase of the absorbance.
Concentration(M
)
absorbanc
e 4.3*10^-5
0
8.7*10^-5
0.1
1.3*10^-4
0.15
1.74*10^-4
0.25
2.17*10^-4
0.55
0.0000000
0.0000500
0.0001000
0.0001500
0.0002000
0.0002500
0.00
0.10
0.20
0.30
0.40
0.50
0.60
f(x) = 2868.44 x − 0.16
Figure 3.1 Beer's Law: SA Concentration vs. Absorbance Concentration(M) Absorbance
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V.
Questions Prelab:
1.)
The following concentrations vs. absorbance data were collected for Substance A.
Conduct a Beer’s Law plat of this data and estimate the extinction coefficient, given the path length of the sample was 1.17cm.
concentration(M)
Absorbance
1.0*10^-4
0.58
8.0*10^-5
0.48
6.0*10^-5
0.35
4.0*10^-5
0.239
2.0*10^-5
0.15
0
0
0.00000
0.00002
0.00004
0.00006
0.00008
0.00010
0.00012
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
Beer's Law Plot: prelab Q1 Concentration (M)
Absorbance
A= Ebc
0.150= E (1.17) (2.0*10^-5)
E= 6,410
A solution of Substance A with an unknown concentration was found to have an absorbance of 0.325. What is the. Concentration of substance A in the solution? A= 0.325
0.325= (6410) (1.17) c
0.325= 7499.7c
C= 4.3*10^-5
2.)
Why is it important that a Beer’s law plot be linear?
This is important because for A=Ebc to be true, there must be a linear relationship between absorbance and concentration, or the equation would be inaccurate.
Post-Lab:
1.)
Calculate the concentration of both stock solution A and B (show both calculations)
Stock A= 138.12g/1m =0.36g/0.001998m
0.001998m/250ml= 8.68*10^-6 m/ml
Stock B= 8.68*10^-6/ 50ml= 1.74*10^-7m/ml
2.)
Calculate the concentration of the five standard solution and report those concentrations in a table with the absorbances measured (show one sample calculation
)
Concentration(
M)
absorbanc
e 4.3*10^-5
0
8.7*10^-5
0.1
1.3*10^-4
0.15
1.74*10^-4
0.25
2.17*10^-4
0.55
Ex calculation: 8.68*10^-6 (10ml) = 8.7*10^-5m/ml
3.)
Create a Beer’s Law plot from your absorbance and concentration data for the 5 standard solutions. Use the plot to find the concentration of Fe^3+ -SA in your unknown cuvette, both graphically and mathematically. *Graph as seen In figure 3.1 in Results and Discussions
Graphically when absorbency of Fe^3+ -SA= 0.3 then concentration would be approximately 1.8*10^-4
Mathematically y=2868.4x+0.164
0.3=2868.4x-0.164
X=1.6*10^-4
4.)
Using the data from your Beer’s Law plot, what is the value of the molar extinction coefficient (also called the molar absorptivity coefficient) for the Fe^+3 -SA ion? ( the diameter of the cuvette is 1.00cm)
A=Ebc
0.3=E(1.00)(1.6*10^-4)
E=1875
5.)
Using the data from your Beer’s Law plot, calculate a.
The exact mass (in grams) of ASA in your aspirin unknown 0.36g
b.
The mass percent of ASA in your aspirin unknown.
ASA= 180.157g/mol
0.36g= 0.00198mol
0.1998% ASA in aspirin unknown
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Table II
Cuvette #
Concentration
Absorbance
0.50 M
1
1.049
2
0.40 M
0.926
3
0.20 M
0.740
0.298
4
0.10 M
5
0.05 M
0.202
Unknown # 9
0.427
Unknown # 10
7
0.646
Blank (DI water)
0
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The value of L is normally 1 cm so E = A/C in this experiment.
Table III
Solution #
Calculated Values of E
1
2
3
4
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Sample No.
Water Samples
pH
1
Tap water from household
6.59
2
Water from creek
7.35
3
Water from river
7.35
4
Unknown sample
7.00
Table 2. Absorbance of Samples
Table 3. Calibration Curve Data*
Sample No.
Absorbance
Absorbance
3-
Concentration of PO4
Ions, ppm
(1 pt each)
1
1.33
1.0
0.050
2
3.26
2.0
0.093
0.89
3.0
0.137
4
0.099
4.0
0.185
5.0
0.230
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0.192
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0.248
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0.460
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Law Tutorial Part 2
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Exp #9 - Laboratory Practical►
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em1.png
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Part 3: Q9 Average absorbance
Flask 8
Flask 9
Flask 10
Flask 11
Part 3: Q9
Flask 8
Flask 9
Flask 10
Flask 11
Part 3: Q10
Flask 8
Flask 9
Flask 10
Flask 11
Q
Part 3
0.146
[Fe] -0.002mol/L
0.166
volume 5.00mL
0.215
0.294
[FESCN mol/L)
✪
4.59* 10
5.18*10*
6.64 x 10
9.08*10*
[Fe(mol/L)
5.00 x 10
5.00*10*
Y=3362x -0.0083 R¹ 0.9993
5.00 x 10
5.00*10*
(Amounts of flask 8 (9 - 11's values are in the spreadsheet)
1:1 Ratio so,
Molarity of Fe(NO3))3 = 0.2M
Volume of Fe(NO3))3 = 5.00mL
71%
Moles of Fe³= Molarity x volume(L)
Moles of Fe = 0.2M × 0.005L = 1× 10-³mole
68%
[SCN ](mol/
4 x 10-4
5 x 10-4
7*10*
1*10*
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Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
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Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY