Chapter 17 spectroscopy
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Transcript of Chapter 17 spectroscopy
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Chapter 17-18: Spectrophotometry
Spectroscopy – the interaction of radiation and matter
Spectroscopic methods measure the amt of radiation produced or absorbedElucidation of molecular structureQual./Quant. Detn. of inorganic and organic compds
Classify
Region of the electromagnetic spectrumX-rayUVVisibleIR
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Electromagnetic radiation – as a wave
Waves – properties of wavelength, frequency, velocity, amplitude
Particles – discrete packets of energy called photons
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Important Equations
E = h
= c
V = 1/
E = hc/
E = hcV
c = 2.998 x 1010 cm/s h = 6.626 x 10-34 Js
UV: 180- 380 nmVis: 380 – 780 nmNear IR: 0.78 – 2.5 mFar IR: 2.5 – 50 m
Memorize
Memorize
wavenumber
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What happens when a molecule absorbs a photon of light?
Energy increases
GS
ES
E = hc/
M + h M*
Energy absorbed = exactly the energy difference betweenthose states
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Three Basic Transitions1- rotational (lower energy)2- vibrational3- electronic (higher energy)
Pure Vibrational IR regionPure Rotational Microwave
UV-Vis: move bonding (outervalence electrons)
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* E large (<150 nm) n* (halogens, N, O, S) E smaller (=150-250 nm)* n* E small (=200-700 nm)
UV-Vis: move bonding (outer valence electrons)
GS
ES
E = hc/
M + h M*
Know this.
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OrganicChromophores
From Skoog, West, Holler
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Clicker questions
If E = 600 KJ/mol , what is the wavelength in nm?
If E = 160 KJ/mol , what is the wavelength in nm?
Hint: 6.023 x 1023 photons/mol
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Instrumentation
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BasisRadiation goes through the sample, certain frequenciesare removed via absorption
Plot A vs determine what frequencies are absorbed
A
Broad: why?
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From Skoog, West, Holler
Why are lines narrower in vapor phase?
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When radiation interacts with matter
1. Some transmitted through sample2. Some absorbed by the sample3. Some reflected at each surface4. Some scattered by dust….
Absorption - Transmission
P0 P
T = P/P0 %T = P/Po x 100%
A = -log T
A = 2 – Log %TMemorize
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Quantitative Chemical Analysis
Beer’s Law: A = abc
a = absorptivityb = pathlengthc = concentration
If C has units of M and b has units of cm,A = bc
= molar absorptivity (M-1cm-1)
Memorize
Beer’s Law is additive:
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1. Chemical
A. Dilute Solutions
B. Analyte dissociates/associates….HIn = H+ + In-
color 1 color 2
Limitations of Beer’s Law1. Instrumental2. Chemical
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2. Instrumental
A. Polychromatic radiation
Beer’s law valid for monochromatic radiation
From Skoog, West, Holler
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B. Stray lightScattered radiation… “stray”Usually a different and may not have passedthrough the sample
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ExampleA solution contains 1.00 mg of K3Fe(CN)6 (FW 328.26)
in 100.0 mL. It transmits 70.0% of incident light comparedto a blank in a 1.00 cm cell. Calculate molar absorptivity?
Clicker Question:
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Applications and Fluorescence
Qualitative AnalysisSupplemental to other techniques (lacks structure)Detect certain chromophoric groupCompare to other spectra
Quantitative AnalysisMolecule must absorb UV-Vis radiationBeers Law must be obeyed
Moderately sensitive, 10-4 – 10-6 MModerately selectiveGood accuracyRelatively easy, convenient, rapid
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Details of Analysis
Properly select wavelengthProperly clean and handle sample cells
Chose standard solutions carefullyProper concentration and composition
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Standard Addition Method
Match the overall composition of the sampleskeep the matrix constant
Impt for solns with complex composition
C
ASingle Point MethodMultiple additions
Extension of the calibration curve methodInvolves addition of known quantity of std to unk
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Photometric Titrations
S + t = Ps t p
s = 0 t > 0 p = 0
A
Vol
Used to locate equiv. PointBeers Law must be obeyedCorrect A for volume changes
Know how sketch thesefor different species
S= substrate, analyteT = titrantP = product
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Fluorescence
Emission processMolecules excited by absorption of electromagnetic radiation,lose excess energy via photon emission
Very sensitive (ppb)Limited number of compds fluoresce (aromatic)
M + h M*
M* M + heatM* M + h
Emits at a longer wavelength than it absorbs
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Energy level Diagram Vibrational relaxationInternal conversionIntersystem crossingFluorescencePhosphorescence
Fluorescence
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Know these terms…
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Longer wavelength- Why?
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Instrument componentsSource (mercury arc lamp)MonochromatorSample (right angles)Photomultiplier tube
900: why?
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Quantitative Chemical AnalysisF = kC (at constant P0)Linear at low concentrations
Why drop in I ?
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A compound with a molecular weight of 125.0 has a molar absorptivity of 2.5 x 105 M-1cm-1. How many grams of thiscompound should be dissolved in 1.00 L such that after a 200-fold dilution the resulting solution will give an absorbanceof 0.60 in a 1.0 cm cell
Clicker question
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Cytochrome c has a molar absorptivity of 106,000 M-1cm-1. 100 uL of a solution cyt. C is diluted to 1.00 mL. The Absorbance of the diluted solution is 0.30 in a 1.0 mm cell. Calculate the concentration of cyt. C in the original soln.
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A 2.00 mL specimen was treated with reagents togenerate color with phosphate following which thesample was diluted to 100.0 mL. Photometric measurement for the phosphate in a 25.0 mL aliquot yielded an absorbance of 0.428. Addition of 1.00mL of a solution containing 0.0500 mg of phosphateto a second 25.0 mL aliquot resulted in an absorbanceof 0.517. Calculate the mgs of phosphate in each milliliter of the specimen.
Clicker question
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Quinine in a 1.664 g antimalarial tablet was dissolved insufficient 0.10 M HCL to give 500.0 mL of solution. A 15.00 mL aliquot was then diluted to 100.0 mL with the acid. Thefluorescent intensity for the diluted sample at 347 nm provideda reading of 288 on an arbitrary scale. A standard 100.0-ppmQuinine solution registered 180 when measured under identical conditions. Calculate the mgs of quinine in the tablet
Clicker question (fluorescence)