The figure at the right shows the absorption spectra of two pure compounds, A an

Question

The figure at the right shows the absorption spectra of two pure compounds, A and B, each at a concentration of 1.50 × 10–4 M. The table below shows the measured absorbance of these standards in a 1.000-cm pathlength cell at five different wavelengths, along with the absorbance of an unknown mixture of the two compounds at the same wavelengths.

Compound A Standard

Use a spreadsheet to calculate the concentration of each compound in the mixture.

[A]=                M

[B]=                M

Wavelength (nm)

Compound A Standard

Compound B Srandard Mixture 440 0.430 0.213 0.263 470 0.642 0.367 0.427 500 0.609 0.444 0.474 530 0.392 0.457 0.429 560 0.231 0.334 0.299

Explanation / Answer

At 440 nM:

For compound A, molar absorptivity, EA(440 nm) = 0.430 / (1.50x10-4x1 cm) = 2867 M-1cm-1

For compound B, molar absorptivity, EB(440 nm) = 0.213 / (1.50x10-4x1 cm) = 1420 M-1cm-1

For the mixture, absrbance = 0.263 = EA(440 nm)xCAx1.00 cm + EB(440 nm)xCBx1.00 cm

=> 0.263 = 2867 M-1cm-1 x CAx1.00 cm + 1420 M-1cm-1xCBx1.00 cm

=> 0.263 = 2867 M-1x CA + 1420 M-1xCB  ----- (1)

At 470 nm:

For compound A, molar absorptivity, EA(470 nm) = 0.642 / (1.50x10-4x1 cm) = 4280 M-1cm-1

For compound B, molar absorptivity, EB(470 nm) = 0.367 / (1.50x10-4x1 cm) = 2447 M-1cm-1

For the mixture, absorbance = 0.427 = EA(470 nm)xCAx1.00 cm + EB(470 nm)xCBx1.00 cm

=> 0.427 = 4280 M-1cm-1 x CAx1.00 cm + 2447 M-1cm-1xCBx1.00 cm

=> 0.427 = 4280 M-1x CA + 2447 M-1xCB  ----- (2)

By solving eqn(1) and (2) we get

[A] = 3.97x10-5 M (answer)

[B] = 1.05x10-4 M (answer)

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