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 46.6% at 2976 cm–1 and 40.8% at 3030 cm–1.

ea Sapling Learning macmilan 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 y axis in the spectra is 70 transmittance rather than absorption, so that the 60 wavenumbers at which there is a dip in the 50 curve correspond to absorption peaks 40 A mixture of A and B in unknown concentrations F 30 3030 cm gave a percent transmittance of 46.6% at 2976 cm and 40.8% at 3030 cm 1 20 10 3040 Wavenumber 0,010 MA C.010 M B Unknown 3030 cm-1 35.0% 93.0% 40.8% 2976 cm-1 76.0% 46.6% 42.0% What are the concentrations of A and B in the unknown sample? Number Number M A 2976 cm 2990 2940 Wavenumber (cm-1) MI B Pure A Pure B 2890

Explanation / Answer

absorbance = 2 - log(%T)

absorbance = molar absorptivity x path length x concentration

So,

molar absorptivity for A

at 2976 cm-1 = (2 - log(76))/0.01 = 11.92 M-1.cm-1

at 3030 cm-1 = (2 - log(35))/0.01 = 45.60 M-1.cm-1

molar absorptivity for B

at 2976 cm-1 = (2 - log(42))/0.01 = 37.67 M-1.cm-1

at 3030 cm-1 = (2 - log(93))/0.01 = 3.152 M-1.cm-1

For mixture

Absorbance at 2976 cm-1 = 2 - log(46.6) = 0.332

Absorbance at 3030 cm-1 = 2 - log(40.8) = 0.389

Absorbance of solution is sum total of absorbances of all components in solution

0.332 = 11.92[A] + 37.67[B]

0.389 = 45.60[A] + 3.152[B]

Solving,

[A] = 0.0081 M

[B] = 0.00625 M

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