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 46.0% at 2976 cm–1 and 41.4% at 3030 cm–1.

What are the concentrations of A and B in the unknown sample?

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 100 90 80 70 60 50 40 2976 cm1 A mixture of A and B in unknown concentrations gave a percent transmittance of 46.0% at 2976 cm-1 and 41.4% at 3030 cm-1. 3030 cm1 20 Pure A 10 Pure B 3040 2990 2940 2890 Wavenumber 0.010 M A 3030cm-1 135.0% 2976 cm-1 176.0% 0.010 M B 93.0% 42.0% Unknown 41.4% 460% Wavenumber (cm-1)

Explanation / Answer

According to Beer-Lambert's law,

A = (ecl)A + (ecl)B

A = 2-log(%T)

Path length l = 1 cm

So, A = (el)A + (el)B

Now, lets look at the data table give,

0.01 M A at 3030 cm^-1 has a transmittance 35% and 0.01 M B 93%

So, Absorbance at 3030 cm^-1 would be,

A = 2-log(%T)

comes out to be, for A = 0.456 & B = 0.03152

similarly absorbance at, 2976 cm^-1 would be,

A = 0.12 & B = 0.3768

Absorption for unknown would be,

at 3030 cm^-1 A = 0.383

at 2976 cm^-1 A = 0.337

Now calculate molar absorbtivity e,

At 3030 cm^-1 or 3300 nm

e = A/bC, with b = 1.00 cm, so e = A/C

A: e = 0.456 / 0.01 = 45.6
B: e = 0.03152 / 0.01 = 3.152

e at, 2976 cm^-1 or 3360.2 nm would be,

A: e = 0.12 / 0.01 = 12
B: e = 0.3768 / 0.01 = 37.68

You will now have 2 equations:

A = eC[x] + eC[y] at 3030 cm^-1
A = eC[x] + eC[y] at 2976 cm^-1

Plug in the molar absorptivity and the mixture absorbances the A at their respective wavelengths. We'll say [x] is A and [y] is B. This will give you 2 equations with 2 unknowns.

0.383 = 45.6C[x] + 3.152C[y]
0.337 = 12C[x] + 37.68C[y]

Solve the linear system of equations,

C[x] = 0.008 M and C[y] = 0.0064 M

So the mixture contains 0.008 M of A and 0.0064 M of B.

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