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 wave numbers 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 52.4% at 2976 cm^-1 and 39.1 % at 3030 cm^-1. What are the concentrations of A and B in the unknown sample? [A] = [B] =

Explanation / Answer

We need to change transmittance into absorbance.

By applying formula: A = 2- log(%T)

The table becomes:

by applying Beer Lambert law, A = ecl

A = absorbance

e = extinction coefficient

l = path length

absorbance is due to A and B.

Extinction coefficient of A is:

At 3030 cm-1, e1 = cl/A = 0.020*1.00/0.4559 = 0.044

At 2976 cm-1, e2 = 0.020*1.00/0.1192 = 0.168

Extinction coefficient of B is:

At 3030 cm-1, e3 = cl/A = 0.020*1.00/0.0315 = 0.635

At 2976 cm-1, e4 = 0.020*1.00/0.3768 = 0.053

To find concentrations in unknown sample:

let concentration od A = x M and B = y M

At 3030 cm-1, absorbance is due to A and B, so,

A = 0.4078 = e1xl + e3yl = 0.044x + 0.635y

Similarly at 2976 cm-1,

A = 0.2807 = e2xl + e4yl = 0.168x + 0.053y

Solving these two equations,

x = concentration of A = 1.501 M

y = concentration of B = 0.538 M

Wavenumber (cm-1) 0.020 M A 0.020 M B Unknown compound 3030 0.4559 0.0315 0.4078 2976 0.1192 0.3768 0.2807

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