The IR (infrared) spectra of two pure compounds (0.010 M compound A in solvent,

Question

The IR (infrared) spectra of two pure compounds (0.010 M compound A in solvent, and 0.010 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 50.7% at 2976 cm–1 and 40.0% at 3030 cm–1.

100 The IR (infrared) spectra of two pure compounds (0.010 M compound A in solvent, 90 and 0.010 M compound B in solvent) are shown 80 to the right. The pathlength of the cell is 1.00 cm. Note that the yaxis in the spectra is 70 transmittance rather than absorption, so that the 8 60 wavenumbers at which there is a dip in the 50 curve correspond to absorption peaks. 2 40 A mixture of A and B in unknown concentrations 30 3030 cm H gave a percent transmittance of 50.7% at 2976 cm and 40.0% at 3030 cm 20 10 3040 Wavenumber 0.010 M A 0.010 MB Unknown 3030 cm-1 35.0% 93.0% 40.0% 2976 cm 1 76.0% 42.0% 50.7% What are the concentrations of A and Bin the unknown sample? Number Number M A 2976 cm 2940 2990 Wave number (cm- M B Pure A Pure B 2890

Explanation / Answer

Ans. Beer-Lambert’s Law, A =e C L                - equation 1,               where,

                                    A = Absorbance

                                    e = molar absorptivity at specified wavelength

                                    L = optical path length (in cm)

                                    C = Molar concentration of the solute

Note: The value of % transmittance is preferred at the wavenumber that gives maximum absorption. So,

Preferred wavenumber for compound A = 3030 cm-1

Preferred wavenumber for compound A = 2976 cm-1

# Absorbance of a solution can be calculated using given value of transmittance as follow-

Absorbance, A = 2 - log (% transmittance) = 2- log (T)

#1. Compound A:

                Absorbance, A = 2 – log = 2 – log 35.0 = 0.45593

el = Al / (CL)    , Where el = molar absorptivity at specified wavenumber

Or, el = 0.45593 / (0.01 M x 1 cm) = 45.593 M-1 cm-1

Thus, e3030 (compound A) = 45.593 M-1 cm-1

The “unknown” with 40% transmittance is compound A because it gives maximum absorptivity at wavenumber 3030cm-1.

Now,

Absorbance of A in unknown solution = 2 – log 40 = 0.39794

Concentration of (unknown A) =

                                                C = A / (e3030 x L)

Or, C = 0.39794 / (45.593 M-1 cm-1 x 1 cm)

Or, C = 0.008728 M

Hence, concentration of A in the unknown = 0.008728 M

#2. Compound B:

                Absorbance, A = 2 – log = 2 – log 42.0 = 0.37675

el = Al / (CL)    , Where el = molar absorptivity at specified wavenumber

Or, el = 0.37675 / (0.01 M x 1 cm) = 37.675 M-1 cm-1

Thus, e2976 (compound B) = 37.675 M-1 cm-1

The “unknown” with 50.7 % transmittance is compound B because it gives maximum absorptivity at wavenumber 2976cm-1.

Now,

Absorbance of B in unknown solution = 2 – log 50.7 = 0.29499

Concentration of (unknown B) =

                                                C = A / (e2976 x L)

Or, C = 0.29499 / (37.675 M-1 cm-1 x 1 cm)

Or, C = 0.00782 M

Hence, concentration of B in the unknown = 0.00782 M

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