a Sapling Learning macmillan learning 100 The IR (infrared) spectra of two pure

Question

a Sapling Learning macmillan learning 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 yaxis in the spectra is 70 transmittance rather than absorption, so that the 8 60 wavenumbers at which there is a dip in the E 50 curve correspond to absorption peaks 2 40 2976 cm A mixture of A and B in unknown concentrations 3030 cm 1 H- 30 gave a percent transmittance of 52.8% at 2976 cm-1 and 43.6% at 3030 cm 20 10 2940 3040 2990 Wave number 0.010 MA 0.010 M B Unknown Wavenumber (cm-1) 3030 cm 1 35.0% 93.0% 43.6% 2976 cm1 76.0% 42.0% 52.8% What are the concentrations of A and B in the unknown sample? Number Number M A MI B Pure A Pure B 2890

Explanation / Answer

At 2976 cm-1

% of T for 0.010 M of pure A = 76

We have equation Aborbance A =E c l

Where E = molar extinction coefficient , c = concntration , l = path length in cm

A = log 100 - log % T = 2 - log 76 = 2- 1.8808 = 0.1192

So E = A / c l  = 0.1192 / (0.010 mol L-1 x 1 cm ) = 11.92 L mol-1 cm-1

For unknown [A]

% of transmittance at 2976 cm-1 = 52.8

A = 2- log 52.8 = 2 - 1.7226 = 0.2774

c = A/El = 0.2774 / 11.92 L mol-1 cm-1 x1 cm = 0.023 M

At 3030 cm-1

% of T for 0.010 M of pure B = 93

We have equation Aborbance A =E c l

Where E = molar extinction coefficient , c = concntration , l = path length in cm

A = log 100 - log % T = 2 - log93 = 2- 1.9685 = 0.0315

So E = A / c l  = 0.0315 / (0.010 mol L-1 x 1 cm ) = 3.15 L mol-1 cm-1

For unknown [B]

% of transmittance at 3036 cm-1 = 43.6

A = 2- log 43.6 = 2 - 1.6395 = 0.3605

c = A/El = 0.3605 / 3.15 L mol-1 cm-1 x1 cm = 0.114 M

Answer

[A] = 0.023 M

[B] = 0.114 M

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