The IR (infrared) spectra of two pure compounds (0.020 M compound A in solvent,

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 45.1% at 2976 cm^-1 and 45.7% at 3030 cm^-1. What are the concentrations of A and B in the unknown sample?

Explanation / Answer

In order to do this, with the first data given, we calculate The molar absorptivity for both concentrations:
Aa1 = 2 - log(35) = 0.4559; Ab1 = 2 - log93 = 0.0315
Aa2 = 2 - log(76) = 0.1192; Ab2 = 2 - log42 = 0.3768

1 and 2 are the wavelengths of 3030 and 2976 (in that order)

The general beer law is: A = EbC --> solving for E = A/bC
Ea1 = 0.4559 / 0.02 = 22.795 L/mol cm; Eb1 = 0.0315 / 0.02 = 1.575 L/mol cm
Ea2 = 0.1192 / 0.02 = 5.96 L/mol cm; Eb2 = 0.3768 / 0.02 = 18.84 L/mol cm

We also know that Absorbance in a mixture is: A = A1 + A2
A1 = Aa1 + Ab1 (1)
A2 = Aa2 + Ab2 (2)

The absorbance of the unknown are:
A1 = 2 - log45.7 = 0.3401
A2 = 2 - log45.1 = 0.3458

With this, we can rewrite (1) and (2) in terms or A and B so:
0.3401 = 22.795Ca + 1.575Cb (3)
0.3458 = 5.96Ca + 18.84Cb (4)

Solving for Cb first from (3):
Cb = 0.3401 - 22.795Ca / 1.575
Cb = 0.2159 - 14.473Ca (5)

Replacing in (4):
0.3458 = 5.96Ca + 18.84(0.2159 - 14.473Ca)
0.3458 = 5.96Ca + 4.0676 - 272.6713Ca
0.3458 - 4.0674 = (5.96 - 272.6713)Ca
Ca = -3.7216 / -266.7113
Ca = 0.01395 M

To get Cb, we replace this value in (5):
Cb = 0.2159 - 14.473(0.01395)
Cb = 0.014 M

Hope this helps

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