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Chem 330 week 3 notes

by: Leslie Pike

Chem 330 week 3 notes Chem 330

Leslie Pike
GPA 3.9

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Beer-lambert law
Quantitative Analysis Chemistry
Dr. Darwin Dahl
Class Notes
Chemistry, Beer-lambert
25 ?




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This 3 page Class Notes was uploaded by Leslie Pike on Friday June 24, 2016. The Class Notes belongs to Chem 330 at Western Kentucky University taught by Dr. Darwin Dahl in Fall 2015. Since its upload, it has received 7 views. For similar materials see Quantitative Analysis Chemistry in Chemistry at Western Kentucky University.


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Date Created: 06/24/16
From longest wavelength to shortest wavelength: radio waves, microwaves, infrared waves, visible light, ultraviolet light, x-rays, gamma rays. Longer wavelengths have lower frequency and therefore lower energy. Imagine that your hand is holding a string that is tied to a hook on the wall. You move your hand up and down very fast, making waves on the string that have small wavelengths. If you keep up for a while, your hand will get tired (you run out of energy) and start moving more slowly and the waves on the string will be longer. Long wavelength, low energy. Energy is Planck’s constant times wave frequency, or Planck’s constant times the speed of light divided by the light’s wavelength. Speed is frequency times wavelength. Total energy of a molecule is the sum of its translational energy, rotational energy, vibrational energy, and electronic energy. Energized molecules emit radiation, and we can identify them from this. Atoms by themselves cannot rotate or vibrate as a whole molecule can, their emission is caused only by electrons changing energy states. This produces line spectra. Whole molecules, with more different types of radiation, produce band spectra. When something is heated and it gives off heat as a result, this produces continuous black body radiation. Multiplicity = 2s + 1, where s is the spin on an electron. M=1 is a singlet (+1/2 electrons are balanced by -1/2 electrons, thus no net s). M=2 is a doublet and it means that you have an unpaired electron. M=3 is a triplet, which means that one of the electrons in a pair has been excited so that it changes its spin (say instead of a +1/2 and -1/2 we have two +1/2 electrons, thus a net s of 1). A resonance emission is when the wavelength emitted by an electron is equal to the same wavelength absorbed by the electron. This does not normally happen; for instance, if you excite an organic compound with UV light, the compound emits light that you can see (visible light is lower energy than UV light, so lower-energy photons are being emitted than what was received, this is called fluorescence). This is because photons can lose energy in a variety of ways before they are re-emitted. Vibrational relaxation: excited electron “vibrates down” to a lower energy level before emitting a photon. Internal conversion: electron crosses from a higher energy level to a lower energy level, such as from s2 to s1. Inter system crossing: electron flips its spin. Vibration is preferred over inter system crossing when possible; therefore, glow in the dark toys work by making vibration impossible (the material is held in a matrix; think glow in the dark plastic Frisbee). A triplet-to-singlet crossing is called phosphorescence. Unlike fluorescence, phosphorescence can continue long after the source of excitation has been removed. Cool things that glow under UV light stop glowing when the light is turned off, but phosphorescing plastic Frisbees that have been left in the sun continue to glow after the sun goes down. The three types of electron pairs we will consider in this class are sigma, pi, and nonbonding. Bonding electrons can be present as bonds or antibonds, antibonds have higher energy. Antibond is symbolized with a star (π and π*, σ and σ*). Sigma antibonds have the highest energy, followed by pi antibonds, nonbonding electrons, pi bonds, and sigma bonds. A nonbonding pair, when energized, can jump to a pi antibond or sigma antibond. Pi bond can jump to pi antibond, and sigma antibond can jump to sigma antibond. For a compound to be analyzed, it must contain either lone pairs or pi bonds, because of the large energy required for σ/σ* transitions and vice versa. Calculating %T: (intensity of light that passed through cuvette)/(intensity of light before passing through cuvette) * 100% Converting from %T to absorbance: log(100%/%T) A=Ebc where E is molar absorptivity in units of L/(mol*cm), b is width of cuvette in cm, and c is concentration of solution in M. If using equation A=abc, a is absorptivity (not molar absorptivity) and things can be in any units. In the homework problems, b is in cm and c is in ppm. If you have a solution with an unknown concentration, you can determine its concentration by measuring the absorbance of the unknown, adding a known amount of a standard of the same analyte in solution, and re-measuring the absorbance. Use this equation: Absorbance −Absorbance standard unknown Volume unknown Absorbance ∗Concentration ∗Volume Concentrationof Unknown= unknown standard standard ¿ For instance, you have two 50 mL samples of copper, unknown concentration. You put them in 100mL volumetric flasks. The first you dilute to 100mL without doing anything to it. The second, you add 4 mL of 2 ppm copper, and then you dilute to 100 mL. Absorbance of the first is .2, and of the second is .31. Using the equation above: Concentration = (0.2*2ppm*4mL)/(50mL(0.31-0.2))=0.29ppm. It is not necessary to use any particular unit for this calculation; just be sure that the volume units cancel (absorbance is unitless) and your concentration of your unknown will be in whatever units you used for the concentration of the standard (ppm, ppb, M, it doesn’t matter). We don’t have to convert mass to moles because our standard is the same solute as our unknown and thus has the same molar mass. Sample problem: you have a sample of iron(II) in solution, concentration unknown. You have a standard of iron(II), 0.0002 M. You are using absorbance to determine concentration, and you adjust the sizes of your cuvettes until both solutions have the same absorbance. When the standard cuvette has a width of 2 cm and the unknown cuvette has a width of 4.5 cm, absorbance is the same. Calculate the concentration of the unknown. Use A=Ebc. Absorbance was the same for both samples, so A=A. Both samples are of iron, and E is an intrinsic property of a material, so E=E. We are left with bunknownunknownb standastandardPlugging in numbers, we get 8.9*10^-5 M for the concentration of the unknown.


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