The IR (infrared) spectra of two pure compounds (0.020 M compound A in solvent,
ID: 528168 • Letter: T
Question
The IR (infrared) spectra of two pure compounds (0.020 M compound A in solvent, and 0.020 M compound B in solvent) are shown to the right. The pathlength of the cell is 1.00 cm. Note that the y axis in the spectra is transmittance rather than absorption, so that the wavenumbers at which there is a dip in the curve correspond to absorption peaks. A mixture of A and B in unknown concentrations gave a percent transmittance of 49.2% at 2976 cm–1 and 40.0% at 3030 cm–1.
5/12/2017 11:55 PM A 82.1/100 G 5/5/2017 10:49 AM Gradebook Print Calculator Periodic Table Question 27 of 28 Sapling Learning The IR (infrared) spectra of two pure compounds (0.020 M compound A in solvent, and 0.020 M compound B in solvent) are shown to the right. The pathlength of the cell is 1.00 cm. Note that the y axis in the spectra is transmittance rather than absorption, so that the wavenumbers at which there is a dip in the curve correspond to absorption peaks. A mixture of A and B in unknown concentrations 2976 cm 3030 cmH gave a percent transmittance of 49.2% at 2976 Pure A cm and 40.0% at 3030 cm 1 Pure B 3040 2990 2940 Wavenumber 0,020 MA 0020 MB Unknown Wavenumber (cm 3030 cm 1 35.0% 93.0% 2976 cm-1 76.0% 42.0% What are the concentrations of A and B n the unknown sample? Number M A M B O Previous Give Up & View Solution Check Answer Next HExt AExplanation / Answer
Ans. Beer-Lambert’s Law, A = e C L - equation 1,
where,
A = Absorbance
e = molar absorptivity at specified wavelength (M-1cm-1)
L = path length (in cm)
C = Molar concentration of the solute
Absorbance, A = 2 - log (% transmittance) = 2- log (T)
Or, A = 2 – log (42.8) = 2 - 1.63144 = 0.3685
Wavenumber
[A] = 0.02 M
[B] = 0.02 M
Unknown
3030 cm-1
0.4559
0.0315
0.4283
2976 cm-1
0.1192
0.3767
0.3116
In the unknown solution, let [A] = A molar , and [B] = B molar
#1. At 3030cm-1 ,
Extinction coefficient, A = A /LC = 0.4559 /(1.0 cm x 0.02 M) = 22.795 M-1cm-1
Extinction coefficient, B = A /LC = 0.0315 /(1.0 cm x 0.02 M) = 1.575 M-1cm-1
Total absorbance of the mixture = Abs of A + Abs of B
Or, 0.4283 = (22.795 M-1cm-1) x A M x 1.0 cm + (1.575 M-1cm-1) x B M x 1.0 cm
Or, 0.4283 = 22.795 A + 1.575 B
Hence, 22.795 A + 1.575 B = 0.4283 - equation 1
#2. At 2976 cm-1
Extinction coefficient, A = A /LC = 0.1192 /(1.0 cm x 0.02 M) = 5.96 M-1cm-1
Extinction coefficient, B = A /LC = 0.3767 /(1.0 cm x 0.02 M) = 18.835 M-1cm-1
Total absorbance of the mixture = Abs of A + Abs of B
Or, 0.3116 = (5.96 M-1cm-1) x A M x 1.0 cm + (18.835 M-1cm-1) x B M x 1.0 cm
Or, 0.3116 = 5.96 A + 18.835 B
Hence, 5.96 A + 18.835 B = 0.3116 - equation 2
#3. Comparing (equation 1 x 5.96) – (equation 2 x 22.795)-
136.67882 A + 9.4437 B = 2.5680868
(-)136.67882 A + 429.343825 B = 7.102922
--------------------------------------------------------------------------------------------------------------------------------------------------------- - 419.900125 B = - 4.5348352
Or B = 4.5257172 / 419.900125 = 0.0108
Thus, [B] in the unknown = B molar = 0.0108 M
Now, Putting the values of B in equation 1-
22.795 A + 1.575 (0.0108) = 0.4283
Or, 22.795 A = 0.4283 - 0.01701 = 0.41129
Or, A = 0.41129 / 22.795 = 0.0180
Thus, [A] in the unknown = A molar = 0.0180 M
Wavenumber
[A] = 0.02 M
[B] = 0.02 M
Unknown
3030 cm-1
0.4559
0.0315
0.4283
2976 cm-1
0.1192
0.3767
0.3116
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