ing Jui 11/19/2017 12:00 AM 23/10° 11/17/2017 12:13 PM -Print Calculator und Per

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ing Jui 11/19/2017 12:00 AM 23/10° 11/17/2017 12:13 PM -Print Calculator und Periodic Table Grade Map d Sapling Learning 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 40 30 20 10 2976 cm A mixture of A and B in unknown concentrations gave a percent transmittance of 50.7% at 2976 cm-1 and 39.3% at 3030 cm-1 3030 cm Pure A 3040 2940 2890 Wavenumber 0010 M A 0010 MB Unknown 3030 crm" | 35.0% 2976 cmr' | 76 0% What are the concentrations of A and B in the unknown sample? 93.0% 39.3% 42.0% 50 7% Number Number 1.53

Explanation / Answer

Absorbance = 2 - log (%T)

T = transmittance

Absorbance = 2 - log (50.7 ) = 0.295 at 2976 cm-1

Abosrbance = 2 - log (39.3) = 0.4056 at 3030 cm-1

From the given data

At 3030 cm-1

Absorbance of A = 2 - log(35) = 0.4559

Absorbance = e*C*l ( l - length ; C = concentration ; e = molar absorptivity )

0.4559 = e*0.01

eA = 45.593

Absorbance of B = 2 - log(93) = 0.0315

0.0315 = e*0.01

eB = 3.15

At 2976 cm-1

Absorbance of A = 2 - log76 = 0.11918

0.11918 = e*C*l

eA = 1.1918

Absorbance of B = 2 - log42 = 0.37675

0.37675 = e*C*l

eB = 37.675

Mixture absorbance

Amix = sum of each A

0.295 = 1.198*Ca + 37.675*Cb

0.4056 = 45.493*Ca + 3.15 Cb

Solving

Ca = 8.39*10^-3 M

Cb = 7.5632*10^-3 M

[A] = 8.39*10^-3 M

[B] = 7.5632*10^-3 M

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