Full set of Notes with Fill-Ins - San Jose State University

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Chem 155 Unit 10 Page 178 of 316The BlankThere are two problems with calculating theabsorbance of a liquid sample in a container.These are that losses of light other thanabosrobance by analyte molecules.b. Absorption by solvent or non-analyte solutein the sample.a. Reflections at each interface (Fresnel eqn).Scattering particles in the sample.Slightly overblown…Reflection losses are about 7% for glass, air andwater. So – even without analyte, the transmittedpower for the sample, P, is less than the incidentpower.How do we compensate for this loss in intensity in thesample power “P” that is not due to the analyte?Measure Po with a blank in place that has nearlythe same reflection and scattering losses as thecuvette – so the only difference in P and Po isdue to absorbance.Page 178 of 316
Chem 155 Unit 10 Page 179 of 316Theory of light absorbance:Absorbance theory is based on the assumption thatmolecules have a cross section through which nolight can pass.Each “slice” of the sample removes a fixed fraction ofthe light entering it – so an exponential relationshipbetween light power and distance results:P(x) = Po⋅exp(-aNx)a = the cross section of the absorber in cm 2N = number of absorbers per cm 3 (# concentration)x = distancePage 179 of 316
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Full set of Notes with Fill-Ins - San Jose State University

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