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 100 90 80 70 g6o 50 2976 cm A mixture of A and B in unknown concentrations D gave a percent transmittance of 47.8% at 2976 cm-1 and 37.9% at 3030 cm-1 30 20 10 3030 cm Pure A Pure B 3040 2990 2940 2890 Wavenumber 0.010 M A 0.010 M B 3030 cm 2976 cm1 What are the concentrations of A and B in the unknown sample? Unknown 319% 47.8% Wavenumber (cm -1) 1 35.0% 930% 1 76.0% 42.0% Number Number M A M B

Explanation / Answer

absorbance = molar absorptivity (M-1.cm-1) x path length (cm) x concentration (M)

Absorbance = 2 - log(%T)

For A at 2976 cm-1,

molar absorptivity (e1) = [2 - log(76)]/0.01 = 11.92 M-1.cm-1

For B at 2976 cm-1,

molar absorptivity (e2) = [2 - log(42)]/0.01 = 37.67 M-1.cm-1

For A at 3030 cm-1,

molar absorptivity (e3) = [2 - log(35)]/0.01 = 45.6 M-1.cm-1

For B at 3030 cm-1,

molar absorptivity (e4) = [2 - log(93)]/0.01 = 3.15 M-1.cm-1

absorbance of mixture = sum total of all absorbances

So,

at 2976 cm-1,

absorbance of unknown = 2 - log(47.8) = 0.32

0.32 = 11.92[A] + 37.67[B]

at 3030 cm-1,

absorbance of unknown = 2 - log(37.9) = 0.42

0.42 = 45.6[A] + 3.15[B]

Solving the two equations,

concentration of [A] in unknown = 8.82 x 10^-3 M

concentration of [B] in unknown = 5.70 x 10^-3 M

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