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Spectroscopic Method and Chromatographic Separations

See the attached file.
1) When measured in a 1.00 cm cuvet, a 9.45×10-4 M solution of species A exhibited absorbances of 0.155 and 0.687 at 450 nm and 725 nm, respectively. A 2.65×10-3 M solution of species B gave absorbances of 0.613 and 0.078 at 450 nm and 725 nm, respectively. Both species were dissolved in the same solvent, and the solvent's absorbance was 0.007 and 0.002 at 450 nm and 725 nm, respectively.

a) Calculate the molar absorptivities for species A.

b) Calculate the molar absorptivities for species B.

c) Calculate the concentrations of A and B in a solution that yielded the following
absorbance data in a 1.00 cm cuvet: 0.496 at 450 nm and 0.875 at 725 nm.

2) Listed below are retention times and peak widths for an HPLC chromatographic separation on a 10 cm C12 column.
Compound Time (min) Width (min) Unretained 0.954
Peak 1 9.423 0.1317
Peak 2 10.548 0.1433
Peak 3 11.143 0.1517
Peak 4 11.727 0.1550
Peak 5 15.049 0.1883

a) Calculate the theoretical number of plates for Peak 2.

b) Calculate the capacity factors for Peak 3 and Peak 4.

c) Calculate the relative retention between Peak 4 and Peak 5.

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1) When measured in a 1.00 cm cuvet, a 9.45×10-4 M solution of species A exhibited absorbances of 0.155 and 0.687 at 450 nm and 725 nm, respectively. A 2.65×10-3 M solution of species B gave absorbances of 0.613 and 0.078 at 450 nm and 725 nm, respectively. Both species were dissolved in the same solvent, and the solvent's absorbance was 0.007 and 0.002 at 450 nm and 725 nm, respectively.

a) Calculate the molar absorptivities for species A.

According to Beer Lambert's Law,

A = eCl
Where A = absorption
e= molar absoptivity, c = concentration (M) and l = path length (length of Cuvet)(in cm)

so, at 450nm

since absorption of solvent is 0.007, therefore, absorption of A at 450nm = 0.155-0,007 = 0.148
A = ecl
Or e = A/cl = 0.148/ (9.45×10-4 M x ...

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