# CHEM 410 GC Unknown Structure Determination & Value of The Intercept Lab Report

Unknown #2IR
Proton
MS
M+ 94
Chem 410A
Lab Quiz 8 – 1
Fall 2020
Lab 8: Lab Quiz
Part I. Testing the Value of the Intercept.
Put your work in Worksheet 1
A. According to Beerβs Law, the absorbance, A, of a solution of an absorbing species, B, is
proportional to its concentration [B]:
π΄ = ππΏ[π΅]
The molar absorptivity, ο₯, is a characteristic property of the solute and L is the path length of
the sample in the spectrometer. The slope of a graph of A vs [B] will be equal to ο₯L, and as
long as there are no absorbing impurities in the sample, the intercept will be zero (A = 0 when
[B] = 0). You will be given absorbance vs concentration data in class. Using this data,
calculate the slope, intercept, and uncertainties in both. Type the results with uncertainties and
units in a text box. From the slope and the path length (which you will be given), determine
the value of ο₯ with uncertainty. Type the result with units and uncertainty in the text box.
B. To test whether the intercept is actually zero, one computes the t-statistic,
π‘=
π β π0
ππ
b is the intercept, bo is the expected
(true) value for the intercept and ο³b
is the uncertainty in the intercept.
and compares it with the t distribution for f degrees of freedom (f = N-2 for a straight line).
The distribution for 95% probability is given in the table. If the calculated t-statistic is less
than that given in the table, then the intercept is considered to be equal to the true value at the
95% confidence level.
t-table for 95% probability
level. Compare the t-statistic
calculated for the intercept and
compare with this number.
df
1
2
3
4
5
6
7
t-statistic
12.7062
4.302653
3.182446
2.776445
2.570582
2.446912
2.364624
Using the results you obtained in part A, calculate t. Can you say that A = 0 when [B] = 0
1
Chem 410A
Lab Quiz 8 – 2
Part II. Numerical Integration of a Spectral Band
Fall 2020
Put your work in Worksheet 2
Information about a compound can be obtained by measuring absorption spectra or emission
spectra. In each case, the peak wavelength of the spectral band is related to the energy
difference between the ground and excited electronic state. The area under the spectral band
is related to the strength of the transition in the absorption spectrum, and the quantum yield in
the emission spectrum. Both of these topics will be addressed in the lecture class. In this part
of the project you will be given data for either an absorption spectrum or emission spectrum.
Plot the data and determine the area under a spectral band using numerical integration. Type
the result with units in a text box.
Part III. Using Solver to Evaluate an Equation without an Analytical Solution.
We have used Solver to fit a calculated curve to data, but it can also be used to solve an
equation that is difficult to solve analytical. For example, it is straightforward to calculate the
pressure using the van der Waals equation when you know the molar volume, temperature,
and the constants a and b:
π=
ππ
π
β 2
ππ β π ππ
ππ =
π
π
On the other hand, calculating Vm when p, T, a, and b are known is more complicated, but
Solver will do this for you quite easily. Place the known pressure in one cell. Then type the
equation for p in the next cell (referring to the values for Vm, R, T, a, and b in other cells). In
the third cell, type the squared deviation of the known and calculated pressures. Use Solver to
minimize the squared deviation with Vm allowed to vary. Here is a problem to try before the
lab class class:
Practice: Find Vm for argon when p = 1 atm, T = 298 K, R = 0.08206 atm L/K
mol. The constants for argon are a = 1.337 atm L2/mol and b = 0.0320 L/mol.
You should get an answer of Vm = 24.43 L/mol.
You encountered one-dimensional tunneling in Chapter 7 of Atkins. The transmission
probability T of a particle of mass m with energy E through a rectangular barrier of energy Vo
and width W is given by the equation
(π ππ β π βππ )2
π = {1 +
}
16π(1 β π)
β1
π=
πΈ
ππ
π=
{2π(ππ β πΈ)}1/2
β
In this project, you will be asked to find the energy E required to obtain a specific
transmission probability T. Values for T, Vo, W and m will be given. Place your final answer
with units in a text box.
ο Save your project as lastname_8.xlsx and send your file to kpeterson@sdsu.edu
2
Lab 8, Part 1
[dye]/mM
0.23
0.76
1.25
1.86
2.68
A
0.0252
0.112
0.208
0.289
0.431
L = 10.00 Β± 0.02 cm
Lab 8, part 2
wavenumber (cm-1)
16025.50465
16052.00442
16078.59828
16101.46913
16128.24157
16155.11115
16182.07681
16209.14226
16232.41843
16259.66559
16287.01247
16314.45962
16338.06553
16365.69824
16393.43278
16421.26971
16449.2096
16473.23883
16501.37095
16529.60599
16557.94619
16582.32125
16610.85612
16639.49782
16668.24695
16692.97359
16721.92161
16750.97877
16780.14568
16805.23153
16834.60341
16864.08335
16893.67543
16919.13009
16948.93162
16978.84536
17008.8737
17034.70353
17064.94417
17095.3013
17121.41417
17151.98708
17182.67838
e (M-1cm-1)
8.40001868
36.709867
19.3525531
47.5509964
48.9978388
39.4259574
-33.551587
22.9460519
24.9406664
6.76992557
75.9071806
41.935649
22.8630925
56.5388233
36.1465464
59.403418
47.3955399
46.741923
67.0672591
90.1113943
62.6588947
38.7018779
117.614186
33.3883813
59.5200599
108.096201
146.016347
124.687373
119.271643
157.449043
155.427751
103.784372
207.014005
160.598292
204.767322
288.790214
255.38654
285.101963
314.316077
352.48196
364.239322
404.60988
466.476759
17213.48695
17239.98962
17271.02062
17302.17084
17328.96584
17360.34217
17391.83972
17423.45918
17450.65934
17482.51002
17514.48467
17541.99066
17574.19854
17606.53247
17638.9951
17666.92244
17699.62365
17732.45381
17760.69929
17793.77209
17826.97799
17855.54511
17888.99781
17922.58393
17951.48016
17985.31733
18019.29221
18048.52126
18082.74955
18117.11792
18146.68757
18181.31377
18216.0825
18245.99666
18281.02977
18316.20586
18346.47059
18381.91571
18417.50634
18448.12989
18483.99432
18520.00684
18550.99349
18587.28257
18623.72449
18655.08285
18691.80629
512.237396
558.300745
639.865442
731.031306
775.790975
810.046831
945.148401
1072.94606
1133.17996
1331.43772
1427.71678
1544.35888
1757.87246
1929.89505
2155.92434
2433.45055
2632.68285
2970.65744
3175.41158
3527.34776
3929.8854
4420.42656
4829.81402
5416.4223
5872.08867
6514.08086
7219.699
8006.41093
8879.71187
9828.46469
10697.9253
11716.1607
12921.7982
14192.0986
15697.7994
17116.7934
18440.7645
20103.8263
21968.4309
23899.5057
25728.5065
27914.3815
29680.5488
31836.8122
34026.1757
36560.9449
39118.3894
18728.68532
18760.41776
18797.58339
18834.90739
18867.02284
18904.63865
18937.00787
18974.92112
19012.99531
19045.75911
19084.13396
19122.67269
19155.83831
19194.68547
19228.11511
19267.27213
19306.60034
19340.44411
19380.08566
19419.90156
19454.16718
19494.3046
19528.84748
19569.31151
19609.95296
19644.93083
19685.90386
19721.16768
19762.47755
19803.97043
19839.68171
19881.51785
19917.52414
19959.70659
20002.07785
20038.54583
20081.27007
20118.0437
20161.12534
20204.40186
20241.65202
20285.29288
20322.85606
20366.86588
20411.0769
20449.13331
20493.71927
41390.1305
43099.1681
45525.5564
47612.2611
49843.98
51513.5842
52854.3521
54252.8568
55246.1803
55936.8698
56433.8118
56313.1236
56059.5478
55442.5919
54676.8937
53376.7532
51598.9609
50083.8658
47958.1703
46429.0603
44826.6797
42988.718
41126.1826
39490.976
38069.7454
37166.3714
36118.4097
35261.8481
34548.2215
34022.0172
33770.2283
33560.8434
33520.9085
33631.1047
33804.0268
34040.3307
34475.3733
34722.033
35180.6625
35708.3227
36062.3663
36401.6519
36597.8807
36706.7795
36645.9927
36384.25
36027.786
20532.10021
20577.06717
20615.77679
20661.12967
20706.6943
20745.91596
20791.8729
20831.43543
20877.79108
20917.69626
20964.45581
21011.43298
21051.87687
21099.2661
21140.06415
21187.87225
21229.02914
21277.25889
21318.78077
21367.43777
21416.32858
21458.42084
21507.74631
21550.21605
21599.98486
21642.83575
21693.05244
21736.28962
21786.96171
21830.59041
21881.72264
21933.10523
21977.34834
22029.20079
22073.84918
22126.17796
22171.23722
22224.0474
22269.52473
22322.8245
22368.7242
22422.5204
22468.84683
22523.14795
22569.90865
22624.71869
22671.91979
35451.4289
34599.956
33731.3269
32439.9949
31169.3868
29819.9383
28510.6847
27215.6408
25730.269
24387.4299
23075.5731
21659.4127
20782.5436
19709.005
18713.6266
17902.9597
17222.5448
16685.7745
16109.9951
15618.957
15108.1758
14797.4794
14471.9052
14230.4741
13966.3262
13696.9141
13435.011
13237.1821
12924.7871
12700.0574
12352.3806
11772.3746
11326.489
10814.8559
10459.1584
9910.08898
9271.49644
8686.8182
8203.26825
7712.29153
7308.38048
6800.52426
6306.06971
5915.06183
5608.75248
5258.82325
5005.70635
22727.24751
22782.85543
22830.7464
22886.88336
22935.23079
22991.90273
23040.71318
23097.92943
23147.20792
23204.97468
23254.72963
23313.05488
23363.29155
23422.18677
23472.91221
23532.38238
23583.60542
23643.6573
23695.38529
23756.02917
23808.26788
23869.51429
23922.27131
23984.1275
24037.41057
24099.88567
24153.70448
24216.80604
24271.16689
24334.90798
24389.81728
24454.20583
24509.67559
24565.40494
24630.75663
24687.0572
24753.0792
24809.96166
24876.66434
24934.13585
25001.53169
4676.61011
4444.70236
4248.24812
4060.25089
3902.94322
3728.02099
3550.33645
3431.78194
3287.61908
3148.82957
3016.77517
2842.7738
2688.41298
2531.75585
2429.45061
2246.72597
2100.33331
1969.89687
1832.62013
1689.45593
1596.79951
1475.46783
1354.98976
1253.32681
1193.00052
1116.08154
1018.69742
939.254523
920.981361
868.448789
847.335196
774.480123
744.985666
698.473882
650.864963
651.780871
638.329961
587.122073
597.007834
554.621885
553.822227
Lab 3, Part 3. Use Solver to find the solution to this problem. Show all of your work in Worksheet 3.
In class, you may have encountered the one-dimensional tunnelling of a particle of mass m with energy
E through a rectangular barrier of energy Vo and width W. The equation for the transmission probability
T through this barrier is given by the equation
β1
π ππ β π βππ 2
πΈ
2π(ππ β πΈ) 1/2
π = 1+
π=
π=
16π(1 β π)
ππ
β
Find the energy E required to obtain a transmission probability of 0.75 when Vo = 2.0 eV, W = 1.5 x 10
m), and the mass of the particle is 9.109 x 10-31 kg. Recall that Δ§ = h/2p = 1.055 x 10-34 J-s. Be sure to
include the eVβ joules conversion constant in the equation to make sure the units work out (1 eV =
1.6022 x 10-19 J). Solver does not do well varying a very small number, so have it vary the energy before
the conversion to joules. To check your calculations, note that kW should be between about 0.1 and 1.
work in Worksheet 3.
e of mass m with energy
e transmission probability
= 2.0 eV, W = 1.5 x 10-10
s. Be sure to
units work out (1 eV =
it vary the energy before
Format Shape
2.
1
7:01
10:30
1x

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