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 n solvent) are shown to the right. The path length 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 46.2% at 2976 cm and 44.4% at 3030 cm^-1. Wavenumber 0.010 MA 0.010 M B Unknown 3030 cm^-1 35.0% 93.0 % 44.4 % 2976 cm^-1 76.0 % 42.0 % 46.2% What are the concentrations of A and B in the unknown sample?

Explanation / Answer

According to Beer Lambert law,

A = e c l

e-extinction co efficient

c- concentration

l-path length

First convert transmittance to absorbance and wavenumber to wavelength,

Absorbance = - log (%T/100)

Find extinction co efficient for each pure compound A and B at the two wavelengths,

A3300 = eA3300 l cA

substitute the value , l = 1 cm = 1 *10-2 m, cA = 0.01M

eA3300 = 4560

similarly, eA3360 = 1190 , eB3300 = 310, eB3360 = 3770

Now for finding the unknown conc in the mixture A and B,

(Atotal)3300 = A3300 + A3360 = eA3300 l cA + eB3300 l cB

0.353 = 4560 * 10-2 cA + 310 * 10-2 cB ----- 1

similarly, for wavelength 3360

0.335 = 1190 * 10-2 cA + 3770 * 10^-2 cB --------2

Solving eqn 1 & 2 we get,

CA = 7.3 * 10-3 M

CB = 6.6 * 10-3 M in the unknown mixture.

Wavelength = 1/wavenumber Absorbance of A Abs of B Abs of unknown 3300 nm 0.456 0.031 0.353 3360 nm 0.119 0.377 0.335

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