Sapling Learning macmillan learning 100 CoDe IR (infrared) spectra of two pure c

Question

Sapling Learning macmillan learning 100 CoDe IR (infrared) spectra of two pure compounds (0.020 M Compound A in solvent, 90 and 0.020 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 2976 cm A mixture of A and B in unknown concentrations 3030 cm gave a percent transmittance of 47.8% at 2976 30 cm and 45.9% at 3030 cm 1 Pure A 20 Pure B 10 2940 3040 2990 2890 Wavenumber 0.020 M A 0.020 M B Unknown Wave number (cm-1) 3030 cm-1 35.0% 45.9% 93.0% 2976 cm 1 76.0% 42.0% 478% What are the concentrations of A and B in the unknown sample? Number Number Al- .0146 M A B .0104 MI B Incorrect Previous Give Up & View Solution Try Again Next Exit Explanation

Explanation / Answer

Calculate molar absorptivities for A (e1 and e2) and B (e3 and e4) at the two wavelengths

at 3030 cm-1

For A, e1 = absorbance/concentration = (2 - log(35))/0.02 = 22.80 M-1.cm-1

For B, e3 = (2 - log(93))/0.02 = 1.576 M-1.cm-1

at 2976 cm-1

For A, e2 = (2 - log(76))/0.02 = 5.96 M-1.cm-1

For B, e4 = (2 - log(42))/0.02 = 18.84 M-1.cm-1

For solution, total absorbance = sum of individual absorbances

at 3030 cm-1

absorbance = 2 - log(45.9) = 0.34

at 2976 cm-1

absorbance = 2 - log(47.8) = 0.32

Let C1 be concentration of A in unknown and C2 be concentration of B in the unknown

0.34 = 22.8C1 + 1.576C2

0.32 = 5.96C1 + 18.84C2

Solving the two equations for to unknowns,

[A] = 0.01404 M

[B] = 0.01254 M

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