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

Chem 155 Unit 1 Page 3 of 3161 Overview and Review ........................................................................................ 71.1 Tools of Instrumental Analytical Chem. ................................................ 81.2 Instrumental vs. Classical Methods. ................................................... 121.3 Vocabulary: Basic Instrumental .......................................................... 131.4 Vocabulary: Basic Statistics Review .................................................. 141.5 Statistics Review ................................................................................ 151.6 Calibration Curves and Sensitivity ...................................................... 231.7 Vocabulary: Properties of Measurements .......................................... 241.8 Detection Limit ................................................................................... 251.9 Linear Regression .............................................................................. 311.10 Experimental Design: ....................................................................... 351.11 Validation – Assurance of Accuracy: ................................................ 431.12 Spike Recovery Validates Sample Prep. .......................................... 451.13 Reagent Blanks for High Accuracy: .................................................. 461.14 Standard additions fix matrix effects:................................................ 471.15 Internal Standards ............................................................................ 522 Propagation of Error ......................................................................................... 563 Introduction to Spectrometric Methods ............................................................ 653.1 Electromagnetic Radiation: ................................................................ 663.2 Energy Nomogram ............................................................................. 673.3 Diffraction ........................................................................................... 683.4 Properties of Electromagnetic Radiation: ........................................... 714 Photometric Methods and Spectroscopic Instrumentation ............................... 864.1 General Photometric Designs for the Quantitation of Chemical Species................................................................................................................. 874.2 Block Diagrams .................................................................................. 884.3 Optical Materials ................................................................................ 894.4 Optical Sources .................................................................................. 904.5 Continuum Sources of Light: .............................................................. 914.6 Line Sources of Light: ........................................................................ 924.7 Laser Sources of Light: ...................................................................... 935 Radiation Transducers (Light Detectors): ...................................................... 1025.1 Desired Properties of a Detector: ..................................................... 1025.2 Photoelectric effect photometers ...................................................... 1035.3 Limitations to photoelectric detectors: .............................................. 1055.4 Operation of the PMT detector: ........................................................ 1065.5 PMT Gain Equation: ......................................................................... 1075.6 Noise in PMT’s and Single Photon Counting: ................................... 1095.7 Semiconductor-Based Light Detectors: ............................................ 1115.8 Charge Coupled Device Array Detectors: ........................................ 1146 Monochromators for Atomic Spectroscopy: ................................................... 1166.1 Adjustable Wavelength Selectors ..................................................... 1176.2 Monochromator Designs: ................................................................. 1186.3 The Grating Equation: ...................................................................... 1196.4 Dispersion ........................................................................................ 122Page 3 of 316

Chem 155 Unit 1 Page 4 of 3166.5 Angular dispersion: .......................................................................... 1236.6 Effective bandwidth .......................................................................... 1256.7 Bandwith and Atomic Spectroscopy ................................................. 1266.8 Factors That Control Δλ EFF ............................................................... 1276.9 Resolution Defined ........................................................................... 1286.10 Grating Resolution ......................................................................... 1296.11 Grating Resolution Exercise: .......................................................... 1306.12 High Resolution and Echelle Monochromators .............................. 1327 Photometric Issues in Atomic Spectroscopy .................................................. 1378 Practical aspects of atomic spectroscopy: ..................................................... 1518.1 Nebulization (sample introduction): .................................................. 1528.2 Atomization ...................................................................................... 1568.3 Flame Chemistry and Matrix Effects ................................................ 1578.4 Flame as ‘sample holder’: ................................................................ 1588.5 Optimal observation height: .............................................................. 1598.6 Flame Chemistry and Interferences: ................................................ 1608.7 Matrix adjustments in atomic spectroscopy: ..................................... 1619 Atomic Emission Spectroscopy ...................................................................... 1629.1 AAS / AES Review: .......................................................................... 1639.2 Types of AES: .................................................................................. 1649.3 Inert-Gas Plasma Properties (ICP,DCP) .......................................... 1659.4 Predominant Species are Ar, Ar + , and electrons .............................. 1659.5 Inductively Coupled Plasma AES: ICP-AES .................................. 1669.6 ICP Torches ..................................................................................... 1679.7 Atomization in Ar-ICP ....................................................................... 1689.8 Direct Current Plasma AES: DCP-AES ........................................... 1699.9 Advantages of Emission Methods .................................................... 1709.10 Accuracy and Precision in AES ...................................................... 17210 Ultraviolet-Visible and Near Infrared Absorption .......................................... 17710.1 Overview ........................................................................................ 17710.2 The Blank ....................................................................................... 17810.3 Theory of light absorbance ............................................................. 17910.4 Extinction Cross Section Exercise: ................................................. 18010.5 Limitations to Beer’s Law: .............................................................. 18210.6 Noise in Absorbance Calculations: ................................................. 18510.7 Deviations due to Shifting Equilibria: .............................................. 18610.8 Monochromator Slit Convolution in UV-Vis: ................................... 18910.9 UV-Vis Instrumentation: ................................................................. 19110.10 Single vs. double-beam instruments: ........................................... 19211 UV-Visible Spectroscopy of Molecules ........................................................ 19511.1 Spectral Assignments ..................................................................... 19611.2 Classification of Electronic Transitions ........................................... 19711.3 Spectral Peak Broadening .............................................................. 19811.4 Aromatic UV-Visible absorptions: ................................................... 20111.5 UV-Visible Bands of Aqeuous Transition Metal Ions ...................... 20211.6 Charge-Transfer Complexes .......................................................... 205Page 4 of 316

Chem 155 Unit 1 Page 5 of 31611.7 Lanthanide and Actinide Ions: ........................................................ 20611.8 Photometric Titration ...................................................................... 20711.9 Multi-component Analyses: ............................................................ 20812 Intro to Fourier Transform Infrared Spectroscopy ........................................ 21112.1 Overview: ....................................................................................... 2121 molecular vibrations ....................................................................................... 21212.2 IR Spectroscopy is Difficult! ............................................................ 21512.3 Monochromators Are Rarely Used in IR ......................................... 21612.4 Interferometers measure light field vs. time .................................... 21712.5 The Michelson interferometer: ........................................................ 21812.6 How is interferometry performed? .................................................. 21912.7 Signal Fluctuations for a Moving Mirror .......................................... 22012.8 Mono and polychromatic response ................................................ 22212.9 Interferograms are not informative: ................................................ 22312.10 Transforming time frequency domain signals: ......................... 22412.11 The Centerburst: .......................................................................... 22512.12 Time vs. frequency domain signals: ............................................. 22612.13 Advantages of Interferometry. ...................................................... 22712.14 Resolution in Interferometry ......................................................... 22812.15 Conclusions and Questions: ......................................................... 23212.16 Answers: ...................................................................................... 23313 Infrared Spectrometry: ................................................................................. 23413.1 Absorbance Bands Seen in the Infrared: ........................................ 23513.2 IR Selection Rules .......................................................................... 23613.3 Rotational Activity ........................................................................... 23813.4 Normal Modes of Vibration: ............................................................ 23913.5 Group frequencies: a pleasant fiction! ............................................ 24213.6 Summary: ....................................................................................... 24614 Infrared Spectrometry - Applications ............................................................ 24714.1 Strategies used to make IR spectrometry work - ............................ 24814.2 Solvents for IR spectroscopy: ......................................................... 24914.3 Handling of neat (pure – no solvent) liquids: .................................. 24914.4 Handling of solids: pelletizing: ........................................................ 25014.5 Handling of Solids: mulling: ............................................................ 25014.6 A general problem with pellets and mulls: ...................................... 25114.7 Group Frequencies Examples ........................................................ 25214.8 Fingerprint Examples ..................................................................... 25314.9 Diffuse Reflectance Methods: ........................................................ 25414.10 Quantitation of Diffuse Reflectance Spectra: ................................ 25514.11 Attenuated Total Reflection Spectra: ............................................ 25615 Raman Spectroscopy: .................................................................................. 25915.1 What a Raman Spectrum Looks Like ............................................. 26115.2 Quantum View of Raman Scattering. ............................................. 26215.3 Classical View of Raman Scattering .............................................. 26315.4 The classical model of Raman: ...................................................... 26515.5 The classical model: catastrophe! .................................................. 266Page 5 of 316

Chem 155 Unit 1 Page 7 of 316Overview and ReviewSkoog Ch 1A,B,C (Lightly) 1D, 1E EmphasizedAnalytical Chemistry is Measurement Science.Simplistically, the Analytical Chemist answers thefollowing questions:What chemicals are present in a sample?QUALITATIVE ANALYSISAt what concentrations are they present?QUANTITATIVE ANALYSISAdditionally, Analytical Chemists are asked:• Where are the chemicals in the sample?• liver, kidney, brain• surface, bulk• What chemical forms are present?• Are metals complexed?• Are acids protonated?• Are polymers randomly coiled or crystalline?• Are aggregates present or are molecules insolution dissociate?• At what temperature does this chemicaldecompose?• Myriad questions about chemical states…Page 7 of 316

Chem 155 Unit 1 Page 8 of 316Tools of Instrumental Analytical Chem.1.1.1 Spectroscopy w/ Electromagnetic (EM)RadiationName of EM Wavelength Predominant Name ofregime:Excitation SpectroscopyGamma ray ≤ 0.1 nm Nuclear MossbauerX-Ray 0.1 to 10 nm Coreelectronx-ray absorption,fluorescence, xpsVacuum 10 - 180 nm Valence VuvUltravioletelectronUltraviolet 180 - 400 Valence Uv or uv-viselectronVisible 400-800 Valence Vis or uv-viselectronNear Infrared 800-2,500 Vibration Near IR or NIR(overtones)Infrared 2.5-40 μm Vibration IR or FTIRMicrowave 40 μm – 1mmrotationsMicrowave ≈30 mm Electron spinin mag fieldRadiowave ≈1 m Nuclear spinin mag fieldRotational ormicrowaveESR or EPRNMRPage 8 of 316

Chem 155 Unit 1 Page 9 of 3161.1.2 Chromatography – Chemical SeparationsDifferent chemicals flow through separationmedium (column or capillary) at different speeds‘plug’ of mixture goes in chemicals come out ofcolumn one-by-one (ideally)Gas Chromatography ‘GC’Powerful but Suitable for Volatile chemicals onlyLiquid Chromatography High Performance(pressure), ‘HPLC’ in it’s many forms –Electrophoresis -Liquids, pump with electriccurrent, capillary, gel, etc.Chromatogramabsorbancetime / sPage 9 of 316

Chem 155 Unit 1 Page 10 of 3161.1.3 Mass SpectrometryDetection method where sample is:volatilized,injected into vacuum chamber,ionized,usually fragmented,accelerated,ions are ‘weighed’ as M/z – mass charge.Often coupled to:chromatographlaser ablationatmospheric “sniffer”.Very sensitive (pg) quantitationPowerful identification toolPage 10 of 316

Chem 155 Unit 1 Page 13 of 316Vocabulary: Basic InstrumentalAnalyteThe chemical species that isbeing measured.MatrixThe liquid, solid or mixedmaterial in which the analytemust be determined.DetectorDevice that records physical orchemical quantity.Transducer The sensitive part of a detectorthat converts the chemical orphysical signal into an electricalsignal.SensorDevice that reversibly monitors aparticular chemical – e.g. pHelectrodeAnalog signal A transducer output such as avoltage, current or light intensity.Digital signal When an analog signal has beenconverted to a number, such‘3022’, it is referred to as a digitalsignal.Analog signals are susceptible to distortion, and soare usually converted into digital signals (numbers)promptly for storage, transmission or readout.Page 13 of 316

Chem 155 Unit 1 Page 14 of 316Vocabulary: Basic Statistics ReviewPrecisionRandom ErrorAccuracySystematic Error(bias)HistogramProbabilityDistributionIf repeated measurements of the same thing areall very close to one another, then ameasurement is precise. Note that precisemeasurements may not be accurate (see below).Random error is a measure of precision.Random differences between sequentialmeasurements reflect the random error.If a measurement of something is correct, i.e.close to the true value, then that measurement isaccurate.Systematic error is the difference between themean of a population of measurements and thetrue value.A graph of the number or frequency ofoccurences of a certain measurement versus themeasurement value.A theoretical curve of the probability of a certainmeasured value occurring versus the measuredvalue.A histogram of a set of data will often look like a Gaussian probability distribution.AverageThe sum of the measured .xvalues divided by the numbernnof measurements.MedianVariance (σ 2 )Standard Deviation(σ)Relative StandardDeviationPropagaion ofErrorx MEANHalf of the measurements fall above the medianvalue, and half fall below.A measurement ofprecision. The sum of thesquares of the randommeasurement errors.n 1A widely accepted‘standard’ measurement ofprecision. The square root nσof the variance.n 1The standard deviation divided by the mean, andoften expressed as : %RSD=σ/x MEAN 100%.σ 2 n x MEAN x n2When the mean of a set of measurement (x) hasa random error (σ x ), it is reported as x±σ x . If wewish to report the result of a calculation y=f(x)based on x, we propagate the error through thecalculation using a mathematical method.nx MEAN x n2Page 14 of 316

Chem 155 Unit 1 Page 15 of 316Statistics Review1.1.7 Precision and AccuracyHistogram of normally distributed events80MeanNumber of times it was observed604020Mean -onestandarddeviationMean +onestandarddeviation00 1 2 3 4 5 6Value observed1.1.8 Basic FormulaeHistogram of 1024 eventsMean : μlimN → ∞N∑ x NNPopulationStandardDeviation:σ xlimN → ∞x N − μ∑ ( )2NNAverage:x avg∑ x NNNSampleStandardDeviation:s xx N − x∑ ( avg )2NN−1Bias or absolute systematic error = x avg − μRelative standard deviation =sx avgPage 15 of 316

Chem 155 Unit 1 Page 16 of 316When you make real measurements of things you generally don’tknow the ‘true’ value of the thing that you are measuring. (Call thisthe true mean, μ, for now. For the purposes of this discussion let usassume that there is no systematic (accuracy) error (i.e. no bias).1.1.9 Confidence IntervalWHAT DO YOU DO TO ENSURE THAT YOUR ANSWER IS ASCLOSE AS POSSIBLE TO THE TRUTH?TAKE THE AVERAGEx AVGBut, you still don’t know the exact answer…SO WHAT DO YOU REALLY WANT TO SAY?I am highly confident that the true mean lies withinthis interval (e.g between 92 and 94 grams).In fact, there is only a 1 in 20 chance that I am wrong!How do you calculate what that interval is?You need to know:The average of the data set: xThe standard deviation: σ or sThe number of measurements (observations) made: NThis interval is called a confidence interval (CI). Which is better,abigger or a smaller CI?Smaller is better…How can you improve your CI?Make more measurements (N)Page 16 of 316

Chem 155 Unit 1 Page 19 of 3161.1.10 Confidence Interval (CI) in WordsConsider thisexperiment. You havea camera device thatmeasures thetemperature of objectsfrom a distance bymeasuring theirinfrared light emission.It is very convenientbut somewhat97.5, 96.0,99.197.32398.05599.051imprecise. Assume for the moment that the camera is perfectlyaccurate – that is, if you measure the same object with it many timesthe average temperature result will equal the true temperature.In order to evaluate the precision of your camera thermometer, youmeasure the temperature of each item three times. In each case youget an average and a standard deviation.PROBLEM 1: The average camera reading is sometimes higher thanthe true value, and sometimes lower, but you don’t know how toevaluate this fluctuation. In just a few words, how can youcharacterize this fluctuation?SOLUTION 1:Calculate the sample standard deviation.PROBLEM 2: A series of experiments, each of three measurementseach yields a set of sample standard deviations that also differenteach time! If you repeat the whole experiment, but this time youmeasure each sample ten times, then the standard deviations aremuch closer, but still not equal each time.Why does the sample standard deviation calculation give a differentresult each time? Assume for the moment that the cameraperformance (precision) is not changing.SOLUTION 2:Realize this fact: Sample standard deviations areonly estimates.Page 19 of 316

Chem 155 Unit 1 Page 21 of 316Use this table:% confidence interval°freedom 50 80 95 991 1.00 3.08 12.71 63.662 0.82 1.89 4.30 9.923 0.76 1.64 3.18 5.844 0.74 1.53 2.78 4.605 0.73 1.48 2.57 4.036 0.72 1.44 2.45 3.717 0.71 1.41 2.36 3.508 0.71 1.40 2.31 3.369 0.70 1.38 2.26 3.2510 0.70 1.37 2.23 3.1720 0.69 1.33 2.09 2.8550 0.68 1.30 2.01 2.68100 0.68 1.29 1.98 2.63Use this formula:CIμ± ⋅ts ⋅ NNote the following:For a straight average of Npoints, the number ofdegrees of freedom is N-1.Complete the following table:Passenger T AVG (°F) Stdev(°F)1 101.5 1.1Lowerboundaryof 80%CIUpperboundaryof 80%CIIs the probability thatthe passenger’sTemp is > 102°90% or more?2 98.57 0.173 103.93 0.047Page 21 of 316

Chem 155 Unit 1 Page 22 of 3161.1.12 t EXP PROBLEM:Another way of approaching this type of problem is to calculate anexperimental value of ‘t’ called t EXP.In the example below, we will compare a measured result with anexact one. The question one answers with t EXP is this: Am I confidentthat the observed value (x) differs from the expected value (μ)?Our threshold temperature, exactly 102° was tested, so we can makemeasurements and test the hypothesis that ‘the true temperature isgreater than 102°’.Given the following three measurements of a passenger’stemperature: 103.76°, 102.11°, 105.38° – calculate an experimentalvalue of the ‘t’ statistic for this population relative to the true value of102°. Average = 103.75, std dev = 1.34% confidence interval°freedom 50 80 95 991 1.00 3.08 12.71 63.662 0.82 1.89 4.30 9.923 0.76 1.64 3.18 5.84( x av − μ) ⋅ Nt expsμ = test value( 103.76−102) ⋅ 3t exp :=1.34t exp = 2.275Can you state with the given confidence that this person’stemperature differs from the expected value of 102°?99%? No95%? No80%? Yes50%? YesPage 22 of 316

Chem 155 Unit 1 Page 23 of 316C = analyte conc.m = slope calibration 1513.166Sb = signal for blankInstrument Signal (S)Calibration Curves and SensitivityCalibration Sensitvity:S = mC + S bS = instrument signalm = calibrationsensitivity10A highly sensitive instrument candiscriminate between small differences inanalyte concentration.ΔSΔC= mS ≈σ SiΔS5But, can wereallydistinguishΔCbetween smallchanges in0 00 20 40 60 80 100concentration?1000 C iAnalyte Concentration (C)γ = Analytical Sensitvityγ = m / σ Sγ = (ΔS/ΔC) / σ Sγ = 1/ΔC for the “ΔC”corresponding to σ Sm ≈andσ S ≈soγ ≈1 / γ = ‘noise’ inconcentration ≈σ C .So small γ is?GoodBadPage 23 of 316

Chem 155 Unit 1 Page 24 of 316Vocabulary: Properties of MeasurementsSensitivitySelectivitySpecimenSampleAnalyteCalibration CurveInterferantDetection Limit C MIN orC MLimit of Quantitation(LOQ)Limit of Linearity (LOL)Linear Dynamic Range(LDR)A detector or instrument that responds to only asmall change in analyte concentration issensitive. Numerically, the sensitivity of aninstrument is the slope of the calibration curve inuntis of: signal / unit concentration – often givenas ‘m’. Sometimes the detection limit (see below)is called the instrument sensitivity – but this is notcorrect.The ratio of the sensitivity of an instrument to ananalyte to that of an interferant.The material removed for analysis – e.g. a giventablet from an assembly line.The mixture that contains the chemical to bemeasured – e.g. the blood sample that we wantto measure iron in.The chemical that is being measured – e.g. theiron in the blood sample.A linear or non-linear function relating instrumentresponse (signal) to analyte concentration.Another chemical in the sample that either affectsthe instrument’s sensitivity to the sample or givesa signal of it’s own that may be indistinguishablefrom the analyte signal.The minimum detectable concentration ofanalyte. Usually defined as that concentrationthat gives a signal of magnitude equal to threetimes the standard deviation in the blank signal.The minimum concentration of analyte for whichan accurate determination of concentration canbe made. The LOQ is typically that concentrationfor which the signal is 10x the standard deviationof the blank signal.The largest concentration for which a calibrationcurve remains linear.The range of concentrations (or signal strengths)between the LOQ and the LOL. An instrument ismost useful within its’ LDR.Page 24 of 316

Chem 155 Unit 1 Page 25 of 316Detection LimitThe detection limit is denoted “C M ”C M is the minimum concentration that can be“detected,” or distinguished confidently from a blank.Let us define the minimum detectable signal changeas: ΔS MTherefore: ΔS M = mC M must be a multiple (n) of thenoise level in the blank: (σ b ).ΔS M ≡ 3σ b = mC MMinimumDetectable SignalSignal due toblankMinimumdetectableconcentrationBy convention, m=3.Derive a formula for C M based on σ B and m:mC M = 3σ bsoC M = 3σ b /mDerive a formula for C M based on γ :γ = m/σ so 1/γ = σ/mC M = 3σ/m = 3/γPage 25 of 316

Chem 155 Unit 1 Page 26 of 3161.1.13 A Graphical Look at Detection Limit (C MIN )Consider the minimum detectable signal in thecontext of the confidence interval:S MIN = 3σ b + S bTo what confidence interval does 3σ correspond?Assume N=2 (two replicate measurements of S b )zσ/N 1/2 = 3σ z/N 1/2 = 3 z = 3*1 1/2 = 3z = 3 corresponds to the 99.7% confidence intervalAnother way of saying this (crudely) is that weconsider a signal ‘detected’ when it falls outside of theboundaries corresponding to the 99.7% confidenceinterval for the blank signal.Page 26 of 316

Chem 155 Unit 1 Page 27 of 3161.1.14 Minimum Detectable Temperature Change?Using the thinking that we developed for the generalcase of signals with random error – what do you thinkis the probability that the following signal change isdue to random fluctuations?Page 27 of 316

Chem 155 Unit 1 Page 28 of 3161.1.15 Dynamic RangeLOQ = Limit of Quantitation (σ S /S ≈ 0.3)LOL = Limit of LinearityBetween LOQ and LOL your instrument is mostuseful!This is called:Linear Dynamic RangePage 28 of 316

Chem 155 Unit 1 Page 30 of 3161.1.17 Direct InterferenceIn order to measure a concentration directly, withoutcorrections, k B,A , k C,A must be approximately:Zero!If there is interference, it is necessary to know both:k B,A , k C,AandC B , C Cbefore one can determine the desired quantity C A !Page 30 of 316

Chem 155 Unit 1 Page 31 of 316Linear RegressionLeast-SquaresRegression orLinear RegressionAssuming that a set of x,y data pairs are welldescribed by the linear function y = a+bx, andassuming that most error is in y (x is moreprecisely known) and the errors in y are not afunction of x, then the coefficients that minimizethe residual function:χ 2 = Σ(y I – (a + bx I )) 2can be found from the following equations:Non-linearRegressionStandard AdditionsPlotSample Matrix /Standards MatrixMatrix EffectInternal Standardsfrom Data Reduction and Error Analysis for thePhysical Sciences, Philip R. Bevington, cw. 1969Mc Graw Hill.Methods for fitting arbitrary curves to data sets.A nearly matrix-effect free form of analysis. Astandard additions plot is a linear plot ofinstrument signal versus quantity of a standardanalyte solution ‘spiked’ or added to the unknownanalyte sample. The unknown analyteconcentration is derived from the concentrationaxis-intercept of this plot.The matrix is the solution, including solvent(s)and all other solutes in which an analyte isdissolved or mixedA matrix effect refers to the case where theinstrumental sensitivity is different for the sampleand standards because of differences in thematrix.A calibration method in which fluctuation in theinstrument signals due to matrix effects are,ideally, cancelled out by monitoring thefluctuations in the instrument sensitivity tochemicals, internal standards, that are chemicallysimilar to the analyte.Page 31 of 316

Chem 155 Unit 1 Page 32 of 3161.1.18 Linear Least Squares CalibrationThe most common method for determining theconcentration of an unknown analyte is the simplecalibration curve.In the calibration curve method, one measures theinstrument signal for a range of analyteconcentrations (called standards) and develops anapproximate relationship (mathematically orgraphically) between some signal ‘S’ and analyteconcentration ‘C’.If the signal-concentration relationship is linear, then:S = S B + mC y = a + bxBut, one can not just draw the line between any twopoints because all the points have some error. So,one mathematically attempts to minimize theresiduals.χ 2 = Σ(y i – (a + bx i )) 21.51.5Instrument Signal (S)S iS iF i10.5F i10000 2 4 6 8 10Page 32 of 3160 C iAnalyte Concentration (C)

Chem 155 Unit 1 Page 33 of 316The values of a and b for which χ 2 is a minimum arethe following:The errors in the coefficients, σ a and σ b can be found,similarly, using:These quantities are best found using a computerprogram. Modern versions of Microsoft Excel willcalculate a and b for you (use ‘display equation’option on the ‘trend line’), and σ a and σ b if you havethe data analysis ‘toolpack’ option installed. Excel isalso fairly well suited to doing the sums and formulas.Page 33 of 316

Chem 155 Unit 1 Page 34 of 316See also, appendix a1C in Skoog, Holler and Nieman.But, be advised that there is a typo in older editions.Equation a1-32 should read:m = S XY /S XXAlso – it is a somewhat more subtle problem tocalculate the error in a concentration determined froma calibration curve of signal (y) versus concentration(x).You will need to follow Skoog appendix-a calculationsto deal with this problem in your lab reports whereyou determine an unknown concentration. M = thenumber of replicate analyses, N = the number of datapoints.Equation a1-37Skoog 5th edition.s cs y⋅m1M+1N+( ) 2yc avg− y avbm 2 ⋅S xxNote: when computing a confidence interval using thiss C value, the degrees of freedom are N-2. (See forexample Salter C., “Error Analysis Using theVariance-Covariance Matrix” J. Chem. Ed. 2000, 77,1239.Page 34 of 316

Chem 155 Unit 1 Page 35 of 316Experimental Design:Designing an experiment involves planning how youwill: make calibration and validation standards;prepare the sample for analysis; and perform themeasurements.1.1.19 Making a set of calibration standards.1. You need to know the dynamic range of yourinstrument.2. You need to know the sample sizerequirement of your instrument.3. You need to know the estimated expectedconcentration of your sample.4. You need to prepare and dilute your sampleuntil it is a. within the dynamic range of yourinstrument and b. such that there is enoughsolution to measure.5. You need to choose target standardconcentrations that bracket the expectedsample concentration generously – e.g. by afactor of 2 to 3. For example, if your expectedconcentration is 5.3 ppm, you may wish to makea calibration set that consists of standards thatare about 2,4,8,10 and 12 ppm.Page 35 of 316

Chem 155 Unit 1 Page 36 of 3166. You need to make a primary stock solution,the concentration of which you know accuratelyand precisely. This solution will usually bemore than twice as concentrated as your mostconcentrated calibration standard. You willdilute this primary stock solution to make thecalibration standards. You need to have enoughto make all of your calibration standards.7. You need to choose the pipets and volumetricflasks that you will use to perform the dilutions.This means that you plan the preparation ofeach standard. This takes some planning andcompromising and many choices – there aremany ways to do this correctly – there is morethan one right answer!8. Decide on and record a labeling system inyour notebook, collect the glassware, and dothe work. I have a labeling system for yourcaffeine, benzoic acid, iron and zinc standards –I need you to use these labels so that we cansort things out in the class.Page 36 of 316

Chem 155 Unit 1 Page 37 of 3161.1.20 Exercise in planning an analysis:Assume that you will be analyzing sucrose in cornsyrup sweetened ketchup packets. The packets arethought (i.e. expected) to contain about 0.8 grams ofcorn syrup that is about 70% sucrose by weight. TheHPLC instrument that you will be using can detectsucrose by refractive index in the 0.1-20 parts perthousand (ppth) range (this is the instrument’sdynamic range for sucrose). You have pure sucrosefor standards, and will be making five calibrationstandard solutions. The instrument requires between250 and 1000 μL of sample.You have the following glassware at your disposal:PipetsVolumetric FlasksVolume RelativePrecisionVolume RelativePrecision20-200 μL 5-1% 1 mL 1%1 mL 1% 5 mL 1%5 mL 1% 10 mL 1%10 mL 1% 25 mL 1%15 mL 1% 50 mL 1%20 mL 1% 100 mL 0.5%25 mL 0.5% 250 mL 0.5%50 mL 0.5% 1000 mL 0.25%Page 37 of 316

Chem 155 Unit 1 Page 38 of 3161.1.21 Plan the analysis!1. If you dissolve the packet in water, dilute it to100.0 mL and filter it, what will the approximatesucrose concentration be (ppth)?0.8 gsyrup0.7 gsucrose 1100 mLg syrup water1000 mgsucroseg sucrose1 mLwater1 gwater= 5.6 ppthIs this within the dynamic range?Would it be better to use 10, 25 or 50 mL ofwater?2. You need to prepare a stock solution of sucroseto make the calibration standards. Whatconcentration should this stock solution be?1, 10, 100 or 1000 ppth?3. How much sucrose would be required to make:1, 10, 100 or 1000 mL of this solution?1.0mL100 mgsucrose =1 mLsoution100 mgsucrose =0.100 gsucrose10 mL =>1.00 gsucrose100mL =>10.0 gsucrosePage 38 of 316

Chem 155 Unit 1 Page 39 of 3161.1.22 How to make primary stock solution:1. Weigh out the desired quantity of pure (e.g. dry,or oxide-free) analyte material.2. Dissolve this amount quantitatively, i.e. withoutany loss, in the desired solvent.3. Transfer this liquid quantitatively into thedesired volumetric flask.4. Dilute to volume with the desired solvent – thisprocess is important! It is often poor practice toadd 5 mL of ‘a’ to 5 mL of ‘b’ and anticipate thatthe final volume will be exactly 10 mL!Remember the words DILUTE TO VOLUME!Page 39 of 316

Chem 155 Unit 1 Page 40 of 3164. What concentrations ‘bracket’ the 5.6 ppthtarget?For example:5.2, 5.4, 5.6, 5.8, 6.0 ppth--- or ---1, 2, 4, 6, 8 ppth--- or ---1, 2, 5, 10, 15 ppth--- or ---0.050, 0.50, 5.0, 50, 500 ppthDo you need to makestandards on an evenspacing?No – but calibrationpoints above andbelow the std. areneeded.Does the analytehave to fall right in themiddle of thecalibration standards?No, but itminimizeserror!5. What volumes of standards should you prepare?How much is needed by the instrument?What is the smallest volume that you canconveniently and precisely measure?How expensive is the analyte and solvent?How expensive is it to dispose of the waste?0.1 or 1 or 10 or 100 or 1000 mLPage 40 of 316

Chem 155 Unit 1 Page 41 of 3166. How do you prepare the calibration series fromthe primary stock solution?Primary Stock (1) is diluted to make Calibration Standard (2)C 1 V 1 = C 2 V 2First calibration standard:10 mL of 1 ppth sucrose from 100 ppth stocksolution.C 1 = 100 ppthC 2 = 1 ppthV 2 = 10.0 mLV 1 = ? = volume to pipet overV 1 = C 2 V 2 /C 1Page 41 of 316

Chem 155 Unit 1 Page 42 of 3167. How to plan a set of standard preparations in theMS Excel spreadsheet program:Calibration Standard Preparation:Target Conc: Stock Conc: Final Volume: Volume to Pipet:C2 / ppt C1 / ppt V2 / mL V1=C2*V2/C11.00 100.0 10.00 0.100 mL2.00 100.0 10.00 0.200 mL5.00 100.0 10.00 0.500 mL10.00 100.0 10.00 1.000 mL15.00 100.0 10.00 1.500 mLCalibration StandTarget Conc: Stock Conc: Final Volume: Volume to Pipet:C2 / ppt C1 / ppt V2 / mL V1=C2*V2/C11 100 10 =A4*C4/B4 mL2 100 10 =A5*C5/B5 mL5 100 10 =A6*C6/B6 mL10 100 10 =A7*C7/B7 mL15 100 10 =A8*C8/B8 mLThese are formulas that you type intoExcel – normally only the result of theformula calculation is displayed.Page 42 of 316

Chem 155 Unit 1 Page 43 of 316Validation – Assurance of Accuracy:Calibration and linear regression optimizes theprecision of a calibration curve determination but acalibration curve can only be said to be accurate if:the analysis has been validatedAnother way of saying this is that a method is valid if:all significant sources of biashave been removedValidation addresses the various aspects of ananalysis that can ‘go wrong’ and give you a wronganswer.The table below lists some of the aspects of ananalysis that can be invalid, and suggests ways tovalidate them.Source of bias:Analyst erraticError in analysttechniqueCalibrationstandards are inerror (are not whatthey say the are)CalibrationmethodInstrumentfunction erraticInstrument driftPossible Solution(s):Analyst can repeat experimentDifferent analyst does same analysis andgets same result.Entire analysis Independent standardrepeated with measured periodically -indpendent called validation standardcalibration or QC standard (qualitystandard set control)Different calibration method used (standardadditions)Different instrument used (can be a differentkind of instrument)Periodically measure validation standard orinternal standardPage 43 of 316

Chem 155 Unit 1 Page 44 of 316In general validation of an analysis is done usingsome kind of independent analysis of the sample.• For the purposes of this class, if the two methodsagree (t-test does not show a difference) then themethod is said to be validated.• If the two methods disagree (i.e. there isevidence of bias), then there is evidence forsystematic error like a matrix effect, aninterferant, or a mistake in the preparation ofstandards or samples.If an analysis is repeated as described below, whataspects of the analysis are validated (i.e. what musthave been ‘good’ for the answers to have come outthe same)?Validation ApproachCompletely independentmeasurements of thesame sample using adifferent instrument ortechnique.Use the same method,but measureindependently preparedstandards as samples.Aspect(s) validatedCalibration method,including matrix effects.Instrument, includingdistortion and drift.You! The analyst, did itrightYou! The analyst made nomistakes in theimplementation of theprocedure.Page 44 of 316

Chem 155 Unit 1 Page 45 of 316Spike Recovery Validates Sample Prep.To do a spike recovery analysis, one takes replicatesamples and to a subset of them adds a spike ofanalyte before the sample prep begins. So, forexample, one could take four vitamin tablets, anddivide them into two groups of two. To one group onecould add some Fe, say half the amount originallyexpected. For example, if there is supposed to be 15mg of iron in the tablet, one could spike two sampleseach with 5 mg of iron and leave two unspiked.Acids etc.digestfilterDilute to100 mLvolumeAnalyze 140ppm14.0 mgfoundSampleSample+ 5.00mg spikedigestfilterDilute to100 mLvolumeAnalyze 187ppm18.7 mgfound18.7-14.0 = 4.7 mg of spike found: spikerecovery percent = 4.7 / 5.00 94%What does this say aboutour sample preparationmethod?We may be losing about6% of the analyte duringsample prep.Page 45 of 316

Chem 155 Unit 1 Page 46 of 316Reagent Blanks for High Accuracy:A reagent blank is a blank that is made by doingeverything for the sample prep etc. but without thesample:Acids etc usedin sample prep.digestfilterDilute to100 mLvolumeAnalyze 2.3ppm0.23 mgFe found“No Sample” sample.Ultrapurewater blankIn this example the reagent blank is analyzed againstultrapure water. The 0.23 mg Fe found in the reagentblank may be due to Fe impurities in the acids, but italso may be a matrix effect. In either case, itsuggests that we should do what in order to arrive ata more accurate result?Either:a. analyze all standards and samplesagainst reagent blanks orb. subtract the signal from the reagentblank against result from sample resultsmade with pure water blanks!Page 46 of 316

Chem 155 Unit 1 Page 47 of 316Standard additions fix matrix effects:Use the method of standard additions when matrixeffects degrade the accuracy of the calibration curves.There is one important assumption built into thecalibration curve idea. That assumption is thefollowing:the sensitivity of the instrument to the analyte in the standards isEQUAL TOthe sensitivity of the instrument to the analyte in the sample matrixWhy would the sensitivity be different?1. Many Instruments are sensitive to things like:1. pH 2. ionic strength 3.organiccomponents of the solvent matrix2. Sometimes other chemicals can change thecalibration sensitivity by:chemically binding to or interacting with theanalyte atom/moleculeThese two things are examples of:Matrix EffectsPage 47 of 316

Chem 155 Unit 1 Page 48 of 3161.1.23 How does one deal with matrix effects?1. Make the sample and standard matrix as nearlyidentical as possible:matrix matchingBut – matrix matching requires that you already knowa lot about your ‘unknown’ – often not the case. So,the matrix-immune alternative to the calibration curveis to:2. Use the method of: standard additions:In other words, the assumption built into thecalibration curve method is that the matrix effects arenegligible or identical for standards and samples. Ifthis can not be assumed, one must match the sampleand standard matrices.The way that the method of standards additions doeswith this is to dilute both standards and samples:in the same matrix (solution)Page 48 of 316

Chem 155 Unit 1 Page 49 of 3161.1.24 An Example of Standard Additions:1. The analyte sample is split up into e.g. 6 aliquots ofidentical, known volume – e.g. 1.00 ml.2. To each of these, a known quantity of standard(known as a spike) is added – e.g.0, 0.1, 0.2, 0.3, 0.4, 0.5 ml – standard is dissolvedin known matrix like water or 0.1M pH 7 phosphateetc.3. Each aliquot is diluted to a total volume.1. Add sample2. Add standard3. Dilute to volume,and mix mix mix!Note: you split and diluteyour sample – how doesthis impact the precision ofyour measurement?How does this processimpact the accuracy ofyour measurement?Page 49 of 316It decreases it!If volume change is big.It increases it!

Chem 155 Unit 1 Page 50 of 3161.1.25 Calculating Conc. w/ Standard Additions:a'ab'bOne way of analyzing this uses similar triangles:a / b =a’/b’The y-axis absorbance signal (S) is proportional to the moles ofanalyte (V X C X ) and standard (V S C S ).The x-axis is simply the standard ‘spike’ volume.The x-intercept is the hypothetical spike volume (V S ) 0 containing thesame amount of analyte as the sample.a is proportional to moles of analyte in the sample = V X C Xa’ is proportional to moles of std added = V S C Sb is (V S ) 0 – the x-intercept of the graphb’ is V S – the spike volumeSubstitute for a,a’,b,b’:V X C X(V S ) 0=V S C SV SSolve for C X :C X = C S (V S ) 0 /V XPage 50 of 316

Chem 155 Unit 1 Page 51 of 3161.1.26 Standard Additions by Linear Regression:Let:V X = volume of unknown analyte solution added to each flaskC X = concentration of unknown solutionV T = final, diluted volumeV S = volume of ‘spike’ added to unknown soln. before dilutionC S = concentration of analyte in spike solutionDilution Calc 1: (V 1 C 1 = V T C T C T = V 1 C 1 /V T )Contribution to concentration of analyte from sample:V X C X /V TDilution Calc 2:Contribution to concentration of analyte from spike:V S C S /V TTotal Signal (sensitivity = k) given that ‘x’ variable is C S .S = kV X C X /V T+ kV S C S /V TSlope = m =kC S /V Tintercept = b =kV X C X /V Twe can get b and m from linear regressionand we want C X … so … b/m = =so : C X =bC SmV XkV X C X /V TkC S /V TV X C X /C SPage 51 of 316

Chem 155 Unit 1 Page 52 of 316Internal StandardsInternal standards can correct for sampling, injection, optical pathlength and other instrument sensitivity variations.An internal standard is a substance added (or simply present) inconstant concentration in all samples, and standards.When something unexpected decreases the sensitivity of theinstrument (m) so the signal drops (S = mC + S B ) – it can beimpossible to distinguish this from a change in analyte concentrationwithout an internal standard.Consider the ratio of the blank corrected analyte ( S'S'ANmANCANCinternal standard (IS) signals (S): = = kS'm C CISISISANANIS= SAN− sB) to theAssumes for the moment that k is a constant, i.e. invariant to factorsaffecting overall instrumental sensitivity. As an example, let’sconsider k to be a correction for injection volume in achromatographic system. It is perfectly reasonable to assume that anaccidentally low or high injection volume would affect the internalstandard and analyte signals identically – e.g. if a given injection were6% high, then both analyte and internal standard peaks would be 6%larger than expected.If k is invariant to instrument fluctuations, then the true analyteconcentration can always be derived from the ratio of the correctedsignals so long as the internal standard concentration remainsconstant.Anlalyte conc. in a sample.S CC = 'AN ISANS'ISkwhere C ISis easily derived from a previously measuredkcalibration standard for which the analyte signal ( S'A − STD) andconcentration ( CA − STD) of the analyte and internal standardS'A−STDCIS−STDmAN( S'IS− STD, CIS−STD) are known: k ==S'C mIS−STDA−STDISPage 52 of 316From a calibration standard.

Chem 155 Unit 1 Page 53 of 316In other words, if C IS is held constant in all experiments, then theratio of the analyte to internal standard signals will beindependent of instrument sensitivity.When to use an internal standard?When substantial influence on instrument sensitivity is expected dueto variation in things like:Sample matrixTemperatureDetector sensitivityInjection volumeAmplifier (electronics) drift Flow rateAn internal standard must:a. not interfere with your analyteb. ideally have the same dependence on the chemical matrix,temperature (or other troublesome variable) as the analyte.Consider the ubiquitous ‘salt plate’ IR sampling method. A drop ofanalyte is sandwiched between two salt plates and this is placed inthe IR beam. The path-length is highly variable from experiment toexperiment. The signal A = εbC where ε is characteristic of themolecule, b is pathlength and C is concentration. If you are doing anexperiment to measure the increase in amide formation versus timeby the intensity of the amide bands near 1700 cm -1 . You could takesamples and measure them periodically, but the variability of thepathlength would distort the results.On the other hand, if all the samples were spiked with the sameamount of acetonitrile then the sharp nitrile stretch at 2250 cm -1 couldbe used as an internal standard.Page 53 of 316

Chem 155 Unit 1 Page 54 of 3161.1.27 Internal Standards Example:In HPLC the sample is injected into a flowing stream, and the signalis a peak in a plot of absorbance versus time (a chromatogram).Often the volume of the injection will vary from injection to injection,so the peaks will vary in size!Peak 1 is theinternal standard.100 ppm NaNO 3 ,480 All mAu . sabsorbance10.5Peaks 2, 3 and 4 are otheringredients in the sample.1Peak 5 is theanalyte, a caffeinestandard, 25 ppm,530 mAu . sabsorbance100 ppm NaNO 3 ,520 mAu . sabsorbance00 400 800 1200 1600 2000 2400 2800 3200 3600 4000time / s10.500 400 800 1200 1600 2000 2400 2800 3200 3600 4000time / s10.523Chromatogram1 is a standard.Did the caffeineconcentrationincrease fromchromatogram1 to 2?Did the caffeineconcentrationincrease inchromatogram1 to 3?Unknowncaffeine conc.630 mAu . s00 400 800 1200 1600 2000 2400 2800 3200 3600 4000time / sPage 54 of 316

Chem 155 Unit 1 Page 55 of 3161.1.28 Internal Standards CalculationHow about calculating the actual concentration of the sample in 3above?StandardS'A − STDCA − STDS'IS − STDCIS− STD530 mAu . s25 ppm480 mAu . s100 ppmSCA−STDIS−STDk ==S'IS−STDCA−STDS'ANmm' 530⋅100ANIS480⋅25= 4.417 unitless630 mAu . sSampleS'ISCISCANS'S'ISCk520 mAu . s100 ppmAN IS=630⋅100520⋅4.417= 27.4 ppmPage 55 of 316

Chem 155 Unit 2 Page 56 of 316Propagation of ErrorSkoog Chapters Covered:Appendix a1B-4 a1B-5 and – eqn. a1-28 table a1-5There is a general problem in experimental science and engineering:How to estimate the error in calculated results that are based onmeasurements that have error?Let’s consider the sum of two measurements a and b that both havesome fluctuation s a = 0.5 lb and s b = 0.5 lb.Let’s also pretend that we know the true values of a and b.True value of:a = 5.0 lbb = 5.0 lbLet’s say that we are weighing a and b and putting them into a box forshipment. We need to know the total weight. Our scale is really bad(poor precision, lots of fluctuation), and it can’t weigh both a and b atthe same time because it has a limited capacity. So, we have to firstweigh a, then b and then calculate the total weight. But we know thatthere is a problem with fluctuations, so we repeatedly weigh the sameitems a and b and do the following experiment:Characteristics of numbers insum ‘a’ and ‘b’Characteristics of sum‘c’Trial # Weight of: Deviation: Total DeviationA b da db weight: from avg:1 4.5 4.5 -0.5 -0.5 9.0 -12 5.5 4.5 +0.5 -0.5 10.0 03 4.5 5.5 -0.5 +0.5 10.0 04 5.5 5.5 +0.5 +0.5 11.0 +1average 5.0 5.0 0.5 0.5 10 0.5This is somewhat artificial and is not quite right, but it gives you thegeneral idea. Errors in a and b propagate into c.Page 56 of 316

Chem 155 Unit 2 Page 57 of 316From this example above, one would conclude that the fluctuation inthe sum is equal to the fluctuation in the numbers summed – but thisis not right.Which is bigger?a. the fluctuations in the individual numbers summedb. the fluctuations in the sumGraphically we consider here a similar case, for clarity we let a havea slightly larger fluctuation than b.a = 5 ± 2, b = 3± 1a+sa+b+sbtheoreticala+sa+b-sba-sa+b-sbcaa+saa-saa-sa+b+sb“Real”distributiontheoreticalPage 57 of 316

Chem 155 Unit 2 Page 58 of 316error propagation through sums:a := rnorm( 10000, 3,0.5)b := rnorm( 10000, 3,0.5)c:=a + b00002.78103.55406.334a=122.662.763b=123.3311.98c=125.9914.74432.52433.48236.00642.15743.84846.005mean( a) = 3.004stdev ( a) = 0.498mean( b) = 2.996stdev ( b) = 0.504mean( c) = 6stdev ( c) = 0.712a_distb_distc_dist:=:=:=histogram( 40,a)histogram( 40,b)histogram( 40,c)0.5 2 + 0.5 2 = 0.7071000a_dist j,1b_dist j,1500c_dist j 1,a dist j 0 b dist j 000 2 4 6 8 10, , c dist j 0Page 58 of 316

Chem 155 Unit 2 Page 59 of 316error propagation through products:c =c := i0123456789101112131415a ⋅b i i09.8818.8615.4738.7898.3018.0577.5619.9093.5897.4719.2066.0541.9440.6435.7738.27mean( c) = 9.003stdev ( c) = 2.156a_dist :=b_dist :=c_dist :=a_dist j,1b_dist j,1histogram( 40,a)histogram( 40,b)histogram( 40,c)1000500c_dist j, 100 5 10 15 20 25a_dist j, 0 , b_dist j,0 , c_dist j,0Page 59 of 316

Chem 155 Unit 2 Page 60 of 316For a simple example – if you construct a simple calibration curve ofinstrument signal versus concentration, and then use a real (i.e.noisy) signal to determine concentration.Errors propagate from S to C according to the sensitivity.Page 60 of 316

Chem 155 Unit 2 Page 61 of 316Let’s consider an example wherein the relationship between themeasured signal and the desired quantity is non-linear :AbsorbanceA = -log (P/Po)Absorbance is a nonlinear function oflight power - the measured quantity in aspectrophotometric experiment.Alog P P l21.5A i10.500 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1P iObviously – the same error in P can give rise to different errors in A!This is a propagation of error problem. How can you calculate theerror in A that should result from a particular error in P?Page 61 of 316

Chem 155 Unit 2 Page 62 of 316Obviously, the answer has to do with the way that the dependentvariable (A) in this case changes as a function of the independentvariable (P).Consider a calculated value ‘S’ that depends on the measuredquantites, e.g. instrument signals, a,b,c…S = f(a,b,c…)In fact – the variance in S is proportional to the variance in a,b,c –and the proportionality is the partial derivative squared:(Skoog appendix a1B-4 has a derivation if you are curious)for: S f( a±σ a , b±σ b , c±σ c )σ S2⎛⎜⎝δSδa⎞ ⎟⎠2⋅σ a2+⎛⎜⎝δSδb⎞ ⎟⎠2⋅σ b2+⎛⎜⎝δSδc⎞ ⎟⎠2⋅σ c2So – let’s take an example:S = a+b-cfind σ S = f(S,a,σ a , b,σ b ,c,σ c )Note that the larger terms dominate!Page 62 of 316

Chem 155 Unit 2 Page 63 of 316Exercise: Prove the “multiplication / division rule”Show that for S = a*b/cσ S2S 2 σ a2a 2 σ b22σ cb 2 c 2Page 63 of 316

Chem 155 Unit 2 Page 64 of 316The propagation of error rules for special cases:Page 64 of 316

Chem 155 Unit 3 Page 65 of 316Introduction to Spectrometric MethodsSkoog Chapter 6 all sectionsElectromagnetic Radiation (EMR) is:Light!EMR described as a classical, electromagnetic wave:In vacuum (air): n=1 In matter: n>1λν = cλ = wavelengthν = frequencyc = light speedNote: When the refractive index (η) changesStays sameν ___________λ ___________changesdistanceλ MEDIUM ν = c MEDIUMλ MEDIUM = λ VAC / ηc MEDIUM = c VAC /ηPage 65 of 316

Chem 155 Unit 3 Page 66 of 316Electromagnetic Radiation:A vast range of energies covers many physicalprocesses. These processes are the basis ofspectroscopy.Page 66 of 316

Chem 155 Unit 3 Page 67 of 316Energy NomogramEnergy nomogram -Hard UV 100 nm to Far IR 1mm10 310 2 10 1210 1510 410 5 10 -310 1410 1 10 -110 010 2 10 -110 110 3 10 110 510 210 410 410 310 1310 -110 010 110 310 510 210 -210 010 210 610 1λ / nm ν / cm -1 ν / s −1 E / eV E / kCal/molE / kJ/molT / KPage 67 of 316

Chem 155 Unit 3 Page 68 of 316DiffractionDiffraction is the basis of wavelength selection in mostspectrometers:dφθδλPage 68 of 316

Chem 155 Unit 3 Page 69 of 3161.1.29 Diffraction Exercises:1. Derive a formula for the path difference δ, as afunction of the spacing d, and the angle ofincidence θ. (Use the diagram on the following page.)first note that θ = φsin(θ) = δ/dδ = dsin(θ)2. Under what conditions is the light intensity at thedetector high or bright?δ = nλn = … -2,-1,0,1,2 … i.e. any integer3. Calculate the wavelengths of light that arediffracted at 48° for d = 1 micrometer (1μ).nλ = dsin(θ)λ = dsin(θ) / nλ = 1.00 μm sin(48) / 1 (assumes order = 1, common)0.743 μm = 743 nm (red)Page 69 of 316

Chem 155 Unit 3 Page 70 of 316Diagram for deriving a formula for the path lengthdifference between diffracted rays:Page 70 of 316

Chem 155 Unit 3 Page 71 of 316Properties of Electromagnetic Radiation:Most EM Radiation is polychromatic, i.e. the beam isa mixture of rays of different:frequenciesphasesE.g. incandescent light sources are polychromatic.Another term for polychromatic light: is white light.EM Radiation is monochromatic if all rays haveidentical:E.g. Na-atomic emission lamps are nearlymonochromatic because nearly all of the light is froma single atomic transition that emits at 590nm.White light can be filtered so that it is nearlymonochromatic.Coherence:EM Radiation is coherent if all rays have identical:frequencywavelengthphaseCoherent radiation comes from lasers and exotic lightsources called synchrotrons.Page 71 of 316

Chem 155 Unit 3 Page 72 of 3161.1.30 Polarization:Most light is unpolarizedElectric vector randomly orientedSpecial filters, called polarizers, can remove allE-field components except those falling in a givenplane. The result is plane polarized light:All E-field lies in one planeElliptically polarized light:E-field vector rotates around direction ofpropagation.Page 72 of 316

Chem 155 Unit 3 Page 73 of 3161.1.31 Refractive index:ratio of speed of light in vacuum to speed of light inmedium.η = c VAC / c MEDIUMη = c / ν 1 ≤ η ≤ ∼2.5Wavelength Frequency s -1 EnergyJRefractiveindex, η1.00 500 nm 3x10 8 ms -1 /500x10 -9 m =6x10 15 s -11.50 500/1.5 =333 nm6x10 15 s -16x10 15 s -1 x6.626x10 -34 Js=4x10 -20 J4x10 -20 JPage 73 of 316

Chem 155 Unit 3 Page 74 of 3161.1.32 Refraction and Snell’s Law:Index = f(wavelength)Blue Red RedBluePage 74 of 316

Chem 155 Unit 3 Page 75 of 3161.1.33 Reflection:Fresnel Equation for a special case:• Normal Incidence• No AbsorptionIncident beamReflected beamTransmitted beamCalculate the total reflection loss due to the tworeflections (air | glass and glass | water) when a lightbeam passes through one side of a cuvette (η = 1.5),containing water (η = 1.3).Page 75 of 316

Chem 155 Unit 3 Page 76 of 3161.1.34 Scattering:IncidentTransmittedScatteredScatteringProcessRaleighScattereratoms andmoleculesFeaturesweak, favorsblueTyndall (Mie) colloids strongRaman molecules very weak,inelasticPage 76 of 316

Chem 155 Unit 3 Page 77 of 3161.1.35 The Photoelectric Effect and the Photon:Electrons are emitted from metal surfaces that areirradiated with light of sufficiently high frequency.Max Planck and Albert Einstein discovered thatKinetic energy of the emitted electrons depends on:Light frequencyandType of MetalKinetic energy of emitted electrons is independent of:Light intensity - more intensity means moreelectrons, but not more energetic electrons!LightcurrentelectronsMetalPage 77 of 316

Chem 155 Unit 3 Page 78 of 316This leads tothe concept ofthe light as aparticle andan expressionfor the energyof a photon.Electron EnergySlope =“h”0 ϕLight FrequencyFree e -E FElectron energy relationship to light frequency:Φ =E F =Work function of metalFermi energy of metalν = Light frequency (s -1 )E = hν - Φh = slope =Formula of linePlanck’s constantAlso – the photoelectric effect is the basis of manylight detectors because it converts light energy into:Electricity – current or single-photon pulsesPage 78 of 316

Chem 155 Unit 3 Page 79 of 316The discovery of photon energy correlated with thetheory of atomic and molecular orbital energy.Phasegas phaseatommolecularliquidmolecularsolidQuantumStateselectronicelectronicelectronicvibrationalphononrotationalResultingSpectralinesbandsbroadPage 79 of 316

Chem 155 Unit 3 Page 80 of 3161.1.36 Spectra typical of gas, liquid and solid.Absorber Phase Notes /transitiontypesAtom Gas Extremelynarrow lines/ electronicMolecule GasMolecule/ smallSolutionFinestructuredue toelectronic+rotational+vibrationalBroadbands +some finestructure /electronic +vibrationalMolecule/ largerSolutionBroadbandselectroniconlyresolvedPage 80 of 316

Chem 155 Unit 3 Page 81 of 3161.1.37 Energy levels photon absorption emission.Page 81 of 316

Chem 155 Unit 3 Page 82 of 3161.1.38 Typical fluorphore Jablonski Diagram.1.1.39 Photophysical ProcessesPage 82 of 316

Chem 155 Unit 3 Page 83 of 3161.1.40 Typical organic electronic spectrum.Page 83 of 316

Chem 155 Unit 3 Page 84 of 3161.1.41 Energy Level Diagram on its SideLine spectra, band spectra and continuum spectra ofatoms, molecules and solids: The relationshipbetween energy states, photon energies and spectra.Light FrequencyLight FrequencyEmission orHeated Mat.Emission orAbsorbanceEnergyEmission orAbsorbanceEnergyLight FrequencyPage 84 of 316Energy

Chem 155 Unit 3 Page 85 of 3161.1.42 Quantitation by interaction with light:Page 85 of 316

Chem. 155 Unit 4 Page 86 of 316Photometric Methods and SpectroscopicInstrumentationSkoog Chapters Covered:Review: Quantitation in Absorbance and Emission7A Optical Designs – Absorbance Emission7A Optical Materials (lightly!)7B Light Sources Continuum and Line7B Lasers!Page 86 of 316

Chem. 155 Unit 4 Page 87 of 316General Photometric Designs for the Quantitation ofChemical Species1.1.43 Generalized Detector Response:S = kP + S DARKS = Signalk = proportionalityP = light powerS DARK = detector response in absence of light1.1.44 AbsorbanceMore AnalyteLess SignalAll absorbance methods!IR, VIS, UV, Xray!1.1.45 QuantitationA = εbc = -log(P/P0)1.1.46 Emission and FluorescenceMore AnalyteMore Signalε = molar absorptivityb = pathlength (cm)c = concentration in moles/LP = Light Power at DetectorP O = Light Power for Blankc = concentration in moles/LAll light emission methods!Fluorescence (Xray - UV),Scattering, Luminescence,even NMR!1.1.47 QuantitationP = P O + mCP = Light Power at DetectorP O = Light Power for Blankc = concentration in any unitPage 87 of 316

Chem. 155 Unit 4 Page 88 of 316Block DiagramsInstruments for Analytical Spectrometry:1. Absorbance • Cuvette• Flame• Gas CellPhotomultiplierTube,PhotodiodeLight beam is absorbed by analytes, not re-emittedAmount of Absorption Amount of Analyte2.1 Fluorescence2.2 RamanScatteringLight beam stimulates analytes to emit lightEmitted light is collected at right angles to stimulatingAmount of Emission Amount of Analyte3. ChemiluminescenceAnalyte reacted with chemical that makes it emit lightEmitted light is collectedAmount of Emission Amount of AnalytePage 88 of 316

Chem. 155 Unit 4 Page 89 of 316Optical MaterialsCostTrade OffLiFSapphire – Al 2 O 3Crystal$$ Water Soluble!$$$$ Hardness!$$ UV-Vis$ vis only$ Water Soluble!$$$$ Red and IR only$$$$ IR Only$$ non-linear dispersion$$- $$$$ most common toolorder-overlap$$ low-res short λ range$ only one λ per filter$-$$ very low resolutionTimeDomainInterferometers$$$$ hard to optimize for bothvisible and IR at oncePage 89 of 316

Chem. 155 Unit 4 Page 90 of 316Optical SourcesCostTrade Off$$ VUV Only?$$ Short Lifetimes$$ Popular UV-source – usedwith W-Halogen for UV-Vis$ W-HalogenVisible only, long lifetimes$$ NIR and IR Only$ IR only$ IR Only$$ Low intensity, limited λ’s$$- $$$$ Exellent Intensity, limitedλ’s$ - $$ very high resolution, poordynamic range$$-$$$ PMT v. fast, v.v. sensitive,delicate, limited in IR$-$$ low sensitvity, limited in IRbut cheap$ - $$$ sensitive and fast, muchtougher than PMT$$-$$$$ v. sensitive, slow, tough,$$-$$$ detector of choice most IRMore exotic ‘energy detectors’ – don’tknow much about thesePage 90 of 316

Chem. 155 Unit 4 Page 91 of 316Continuum Sources of Light:Broadband or White Light SourcesAlso called:1. Simplest Design:Blackbody Sourcesa. Tungstenb. Quartz-Tungsten-Halogenc. Nernst GlowerVisible / Near IR / IRIR / Far IR2. Gas Emission Designs:a. H 2 / D 2UV OnlyH 2 + e - H 2 * H + H + hνKinetic Energy of H atoms =ContinuumTherefore hν can be:Continuumb. Ar, Xe, HgUV-Vis-Near IRHeavy (High-Z) atoms Many atomic states+High Pressure (extensive broadening of lines)Quasi-ContinuumPage 91 of 316

Chem. 155 Unit 4 Page 92 of 316Line Sources of Light:1. Low Pressure Gas Emissiona. Hg Many Lines – can beb. Ar filtered to emit onlyc. Xe one predominant line.d. Na Na-D Line Predominates –589.00 and 589.59 nm doublet2. Hollow Cathode Lamps:a. Metals (Cathode!)b. Used in atomic spectroscopy (absorbanceand fluorescence)Sputtering300V DC Electrical DischargeNe + Ne +Fe* + Fe + hνAtomic EmissionFe Metal CathodePage 92 of 316

Chem. 155 Unit 4 Page 93 of 316Laser Sources of Light:PartiallytransmittingmirrorAn Introduction to Lasers:LASER is an acronym fora light amplificationprocess:LASERLightAmplification byStimulatedEmission ofRadiationFullyreflectivemirrorPumping EnergySource:• Intense Light• Electrical DischargeGain Medium – atoms ormolecules that undergo‘lasing’ transitionsWhat happens in the Gain Medium?1.“relativelylong lived”Page 93 of 316

Chem. 155 Unit 4 Page 94 of 316What happens in the Gain Medium?2.3.4.Stimulated Emission is:CoherentMonochromaticPage 94 of 316

Chem. 155 Unit 4 Page 95 of 3161.1.48 A laser is a light amplifier –Some of the above processes degrade the light in the cavity• Spontaneous Emission• AbsorptionSome of the above processes amplify the light in the cavity• Pumping• Stimulated Emissionbeam outPump201Fast DecayLasing!Two common laserconfigurations:3-state32Fast DecayPump01Lasing!Fast Decay4-state (or more)Page 95 of 316

Chem. 155 Unit 4 Page 96 of 3161.1.49 Polulation Inversion and laser amplificationabsorption ∝population in state Exlasing amplification ∝population in state EyRoughly speaking lasing is possible when:population in upper state 1 is greater than thepopulation in lower state 2This is called a population inversion.Page 96 of 316

Chem. 155 Unit 4 Page 97 of 3161.1.50 Necessity of 3 or more statesWhy are three or more levels (states) necessary forlasing?recall:N EXCITEDN GROUNDeΔEkTΔE =k =T =N EXCITED =N GROUND =What happens when T infinity?N EXCITED N GROUNDPage 97 of 316

Chem. 155 Unit 4 Page 98 of 3161.1.51 Common lasers categorized by lasing medium.1. Solid Red - Near Infrared1.1. Lanthanide-and transition metal ion lasers1.1.1. Nd-YAG – Neodymium ions in a crystal called a ‘garnet’made of yttrium oxide and aluminum oxide. Otherlanthanide ions are substituted for other wavelengths in thenear IR1.1.2. Ho-ZBLA – Holmium (Ho +3 ) doped glasses made fromZrF 3 , BaF 3 , La F 3 , Al F 3 can be fabricated into optical fibersthat lase and can amplify optical signals of certainwavelengths1.2. Semiconductor Lasers –1.2.1. Silicon,1.2.2. gallium arsenide and other light-emitting diodes can bemade to lase when many electrons are promoted to excitedstates within microfabricated cavities in the semiconductorcrystalBlue (very new) – Near Infrared2. Liquid dye-lasers (blue-red) are made from solutions of manydifferent fluorescent dye molecules. The dye molecules havemultiple excited-states that can be induced to lase usually bypumping with other lasers3. Gas-Phase lasers (UV-near infrared) are very common, andtypically pumped by electrical discharge. Examples include:3.1. He-Ne lasers (633 nm)3.2. Ar-ion, Kr-ion (514, 488, 325 nm)3.3. N 2 (337 nm)3.4. CO 2 (10,600nm and many other IR wavelengths)3.5. XeF (351) and KrF (248) and ArF (193) excimer lasers.Page 98 of 316

Chem. 155 Unit 4 Page 99 of 3161.1.52 What is an excimer laser?Kr * + FKrF *Excimer!Kr + FKrFWhat is unusualabout thismolecule?Because of fast dissociation,[KrF * ] > or < [KrF]This favors:population inversion!Page 99 of 316

Chem. 155 Unit 4 Page 100 of 3161.1.53 Some laser “pointers”.LS = lower stateUS = upper stateGS = ground statehν + US 2hν + LSstimulatedemissiongainhν + LS US absorption lossUS LS + hνspontaneousemissonlossBased on this, should the LS be:short lived orlong lived?Based on this, should the US be:short lived orlong lived?Page 100 of 316

Chem. 155 Unit 4 Page 101 of 316Laser Questions:Indicate all that apply:1. For a laser gain medium with three or more states, a populationinversion is / means:a. More electrons in the upper than lower states.b. More molecules, atoms or ions in a given excited thanground state.c. Excited state energy is greater than ground state energy.d. Concentration of molecules / atoms or ions in upper stateis greater than concentration in lower state.e. Concentration in ground state is zero.f. Concentration in upper state is greater than half of thetotal.2. For efficient laser gain:a. Upper state should be long-lived.b. Lower state should be short lived.c. Ground state should be unstable.d. Ground state should not be lower state.e. Lower state should be long lived.f. Upper state should be short lived.Contributes to: Proportional To:Phenomenon Gain Loss [upper states] [lower states]SpontaneousemissionAbsorptionStimulatedEmission3. For gain to occur in a laser [upper states] must be _____relative to [lower states]:a. >b.

Chem 155 Unit 5 Page 102 of 316Radiation Transducers (Light Detectors):Chapter 7, Skoog Holler & Nieman7E PhotomultipliersPhotodiodesCharge-Coupled Device ArraysDesired Properties of a Detector:High Quantum Efficiency:A large fraction of photons that strike itresult in a response – few are lost.High Gain:For each photon that strikes thedetector, a large signal isgeneratedLow Noise:Constant light flux gives a constantsignal - σ for measurement is low.Low Dark Count:Low or no signal is present in thedark.Page 102 of 316

Chem 155 Unit 5 Page 103 of 316Photoelectric effect photometersPhototube:High UVtransparencyfusedsilica window.Evacuated tube inwhich electrons cantravel if emitted fromcathode.+-ReadoutAmplifierPhotocathode –transducer - emitselectrons via thephotoelectric effectLow work functionmetal: Na, K, GaAsAnode – positivebias, collectsphotocurrent.Page 103 of 316

Chem 155 Unit 5 Page 104 of 316What is the maximum gain of aphototube expressed as electrons perincident photon?1!What is a typical gain for a phototubeexpressed as electrons per incidentphoton?0.01What limitation is built in to allphotoelectric detectors?hν must be ≥ φPage 104 of 316

Chem 155 Unit 5 Page 105 of 316Limitations to photoelectric detectors:Recall Einstein et al. used the photoelectric effect todiscover what particle: The photon!Photoelectron energy / jSlope ofthis line =Plank’s Constant:h = 6.626x10 -34 JsLight frequency ν / s -1Peak quantum efficiency:About 10%or1 photoelectron10 photonsPage 105 of 316

Chem 155 Unit 5 Page 106 of 316Operation of the PMT detector:Faceplate material:UV-transparentSilicaPhotocathode potential:From www.hamamatsu.com pmtconstruct.pdf-100 to –1000VA typical PMT may have 12 dynodes, each of whichgives off somewhere between 4 and 12 secondaryelectrons depending on the applied voltage.Question:Derive a formula for PMT gain:How many electrons are collected for each incidentphoton? (Use outline on next page…)Page 106 of 316

Chem 155 Unit 5 Page 107 of 316PMT Gain Equation:1 PhotonPhotocathodeQuantuumEfficiency - φHow many photoelectronsper incident photon areemitted?How many secondaryelectrons?Secondaryelectron yield - αDynode 1φαDynode 2φα 2Dynode 12PMT Gain = G = φα nφ = Quantum efficiencyα = Secondary electron yieldn = Number of dynodesPage 107 of 316

Chem 155 Unit 5 Page 108 of 316How many dynodes do you want or need?Ultimate sensitivity =Detect a single photon!Consider the arrival packet of electrons:hν e - atphotocathodee - strikes n-dynodesα n electronsstrike anode3x10 -9 sWhat is the average current during the 10nspulse? Assume α = 4, n = 10F = 96485 coulombs/mole e -Mol = 6.023x10 23Current (amperes) = # coulombs / s4 10 = 1048576Current (amperes) = # coulombs / si = 1048576 e - * 96485 coulombs / 6.023e23 e - / 10 -8 s= 1.7x10 -5 amperes – easily measureablePage 108 of 316

Chem 155 Unit 5 Page 109 of 316Noise in PMT’s and Single Photon Counting:Two major sources:Thermal:Individual electrons are ‘thermionically’emitted from photocathode or one of thedynodes.Cosmic Ray Background:Energetic particles (α and β particles, γ-raysetc. from space) strike photocathode ordynodes and cause a large current spike.current / microamperestime / sReject: cosmicray – e- cascade.Accept: photonelectronfromphotocathodeReject:Thermal e - fromdynodesMuon: created by very high energy cosmic ray interaction with upper atmosphere, similar toelectron (-1 charge, spin 1/2) but about 200 times heavier.About 10,000 muons reach every square meter of the earth's surface a minute; these chargedparticles form as by-products of cosmic rays colliding with molecules in the upper atmosphere.Traveling at relativistic speeds, muons can penetrate tens of meters into rocks and other matterbefore attenuating as a result of absorption or deflection by other atoms.—Mark Wolvertron, science writer, Scientific American magazine, September 2007, page 26 "Muons forPeace"Page 109 of 316

Chem 155 Unit 5 Page 110 of 316In single photon counting:Accept only those pulses originating at thephotocathode.RetainsignalRejectnoiseRecall: Detection limitS MIN = S BLANK + 3σ BLANKLarge and small pulses are all noise!These contribute to both S BLANK and σ BLANK !Limitations to the PMT and to single photon counting:1. Wavelength limitations:Same as phototube - hν ≥ φ2. Intensity limitations:In general, the PMT is used inLOW LIGHT ONLY – ambientlight will destroy the tube.In Single Photong Counting,light levels must be lowenough that light pulses donot pile up or overlap.Page 110 of 316

Chem 155 Unit 5 Page 111 of 316Semiconductor-Based Light Detectors:Photodiodes:p-type Sip-n junctionn-type Si+P + -B - +B -P + -The only carriers ofcurrent in p-type Si are:The only carriers ofcurrent in n-type Si are:Holes!Electrons!Why does B (or Al) become (-) and P become (+)?Both B and P are forced to become tetravalentby the bounding atoms in the lattice – this leavesa curious formal charge!B(p-dopant)Mobileholes.P(n-dopant)MobileelectronPage 111 of 316

Chem 155 Unit 5 Page 112 of 316Depletion, anhiliation, injection and rectification:1. Forward Bias:++-+ +- -Injection ofholes2. Reverse Bias:Anhiliation ofholes andelectrons+Injection ofelectrons+++---Removal ofholesDepletion ofholes andelectronsRemoval ofelectronsUnder reverse bias:Page 112 of 316No current flows:Rectification

Chem 155 Unit 5 Page 113 of 316Photodiode:++++---Steps in Photoconduction:1.2.3.Photon absorbed by Si, promotes e- intoconcution band.This leaves a mobile hole and electron.Hole and electron are swept apart by theapplied field and:hν currenttransductionProperties of the photodiode:1. Gain:= φ ≤ 1 but φ is often near unity2. Dynamic Range:3. Ease of fabrication:Page 113 of 316Largest measureable signal >>smallest Excellent!Can be made very small.Can be made into 1D and 2D arrays

Chem 155 Unit 5 Page 114 of 316Charge Coupled Device Array Detectors:-10V-10V-10VAl contactSiO2n-Si--- ----- ----- --1. e- arerepelledfrom Al2. hν strikesdepletionregion3. h+accumulateat Al-10V+ + ++ + +-10V-10V+ + ++ + +-10V+ + +Page 114 of 316+ + +

Chem 155 Unit 5 Page 115 of 316CCD Array detectors are very useful forspectroscopy.1. Maximum gain per pixel φ 1Not as sensitive as a PMT.2. Readout is relatively slow:Ca. 100 frames per second 10 -2 s.3. But, conventional spectrometers measure thelight spectrum (power versus wavelength) workby limiting the light incident on the detector toone wavelength at a time. With a CCD Arraydetector:You can have a different detector for everywavelength CCD’s can acquire an entire spectrummuch more quickly than a PMT-based system.4. The light spectrum falls in a 2-dimensional imageor picture.5. Thermal noise is lower than in PMT, but cannotbe completely rejected.6. Cosmic ray background can be dealt withstatistically:If there is a single particular pixel (spot on thepicture) that is anomalously bright (i.e has anunusually high reading) – it can be rejected ascosmic radiation noise using statistical analysis.Page 115 of 316

Chem 155 Unit 6 Page 116 of 316Monochromators for Atomic Spectroscopy:Chapter 7, Skoog Holler & Nieman7C-2 Monochromators DispersionResolutionSpeed7C-3 Monochromator Slits and Spectral ResolutionEffective BandwidthLine Convolution (Effect of Slit Width on Resolution)Czerny-Turner and Echelle designsWhy do we care about monochromators?Monochromaticlight is useful!An analysis of the wavelength dependence of theabsorbance or emission of light is called: SpectroscopyThis technique is used to:analyze the structure of atoms and moleculesidentify the atoms and moleculesobserve interactions between atoms and moleculesMeasurement of the amount of monochromatic lightabsorbed or emitted by atoms or molecules is called:SpectrometryThis technique can tell you:the concentration of atoms or molecules in a samplePage 116 of 316

Chem 155 Unit 6 Page 117 of 316Adjustable Wavelength SelectorsAdjustable wavelength selectors are called:MonochromatorsLight InputPolychromatic“White”“Broadband”Light Output“Monochromatic”Narrow-bandWavelength / nm455.3Two important characteristics of monochromators are:The amount of light thatmakes it through at a givenwavelength:The range of wavelengthsthat exit themonochromator:throughputbandwidth - ΔλPage 117 of 316

Chem 155 Unit 6 Page 118 of 316Monochromator Designs:Czerney-TurnerblueredPrismWavelengthDispersion:LinearNonlinear withUV-AbsorptionNonlinearPage 118 of 316

Chem 155 Unit 6 Page 119 of 316The Grating Equation:Grooves or blazes:Lines or rulings etched into grating that scatter ordiffract the light.Light emitted from the grating surface is called:Diffracted lightFor a single, polychromatic input beam:Infinite monochrmatic output beams differing inangleFor a particular wavelength (λ) to be diffracted at aparticular angle, the corresponding pathlengthdifference (DBC) must be equal to:nλ must = DBC where n is any integerPage 119 of 316

Chem 155 Unit 6 Page 120 of 316The Grating Equation (cont):For outgoing rays 1 and 2 to interfere constructively:pathlength difference DBC must = nλi. b + i = ?ii. b = ?iii. a+b+90° = ?90°90° - i180°iv. substitute ii into iiia + (90 – i) + 90 = 180 a – i +180 = 180 a = iPage 120 of 316

Chem 155 Unit 6 Page 121 of 316The Grating Equation (cont):For a particular wavelength λ to appear bright atincident angle i, the following must be true:nλ = DB + BCWe know that angle a = angle i. We can measure i,so we use it in calculations.sin(i) = sin(a) =sin(r) = sin(a’) =BC + DB =opposite / hypotenuse = BC / AB = BC / dopposite / hypotenuse = DB / AB = DB / dd sin(i) + d sin (r) = d ( sin(i) + sin(r) )For constructive interference (i.e. a bright condition):Page 121 of 316nλ = d ( sin(i) + sin(r) )

Chem 155 Unit 6 Page 122 of 316DispersionDispersion characterizes the extent to which amonochromator separates (disperses) the differentwavelengths of light.Angular dispersion:For a given incident angle, i, how quickly does thecolor change as you change the viewing angle r?In other words – what is the derivative of λwith respect to r?nλ =λ(r) =dλ / dr =dλ / drd ( sin(i) + sin(r)d ( sin(i) + sin(r) / nd/dr[sin(i)] + d/dr[sin(r)] / nd cos(r) / nPage 122 of 316

Chem 155 Unit 6 Page 123 of 316Angular dispersion:dr nn =dλ=d cos(r)d =Reciprocal linear dispersion: dy ≈ Fdrr =orderGroovespacing ongratingAngle ofdiffractionλ =wavelengthsin(r) ≈ rsin(r) ≈ y/FdλD -1 ≡ =dydλFdrD -1 =d cos(r) / nFPage 123 of 316

Chem 155 Unit 6 Page 124 of 316For monochromators the important factor is:Reciprocal Linear DispersionD -1 = dλ / dy = d cos(r) / nFWhat is dispersion?Consider a typical monochromator:d = “1000 lines / mm” r = 45° n = 1 F = 1mFirst – let’s get the units straight!d=F=d = 1/1000 mm = 10 -3 mm = 1 μm = 1000 nmF = 1m = 1000 mmD -1 = 1000 nm x 0.7 / 1 x 1000 mmD -1 = 0.700 nm / mmExit SlitFocal Plane0 100 200 300 400y / mm300 370 440 510 580Page 124 of 316λ / nm

Chem 155 Unit 6 Page 125 of 316Effective bandwidthThe combination of exit slit width and dispersiondetermine Δλ EFF – the effective bandwidth of themonochromator. Δλ EFF is a number, in units of nm,equal to the range of wavelengths that exit themonochromator at any one time.Δλ EFF = wD -1W = slit width in mmD -1 = dispersion in nm / mmΔλ EFF is controlled in part by:the exit slit width - wW = slit width in mmD -1 = dispersion in nm / mmΔλ EFF ≈15 nm300 400 500 600 700Δλ EFF ≈2 nm300 400 500 600 700Page 125 of 316λ / nm

Chem 155 Unit 6 Page 126 of 316Bandwith and Atomic SpectroscopyConsider the case of a Ni atomic spectroscopyexperiment. Ni atoms emit at 231.7, 232.0 and232.2 nm. However, Ni atoms only absorb light at232.0 nm. (Two emission lines are non-resonant.)So, it is important in atomic absorption spectroscopyto remove Ni emission at 231.7 and 232.2, butefficiently transmit light at 232.0 nm.Page 126 of 316

Chem 155 Unit 6 Page 127 of 316Factors That Control Δλ EFFIn general, one wants a monochromator with as smallan effective bandwidth is practical. Such amonochormator is called:high resolutionTo achieve low effective bandwith / high resolutionΔλ EFF should be:smallΔλ EFF = wD -1 = wd cos(r) / nF ,sod should be:F should be:w should be:smalllargesmallThe essential tradeoff is that very small Δλ EFF usuallyequate to low light levels. For example, achievingvery small Δλ EFF by making w very small is oftenunsuccessful because the source is imaged onto thefocal plane. Unless the source is a very bright, pointsource, then making w is smaller than the source willoften cause light levels to drop dramatically.Page 127 of 316

Chem 155 Unit 6 Page 128 of 316Resolution DefinedResolution is defined as R ≡ λ / ΔλResolution is the ability to just separate two adjacentspectral lines. For example, consider the output of amonochromator with 0.1 nm versus 1 nm resolution.In this case two purely monochromatic sources of 501and 502 nm are focused into the entrance slit.Grating Resolution Limted ThroughputLow resolution(1 nm)Light throughput500 500.5 501 501.5 502 502.5 503Wavelength / nmR=0.1 nmR=1 nm x 10Note that the low resolution case has much lowerintensity as well.Page 128 of 316

Chem 155 Unit 6 Page 129 of 316Grating ResolutionResolution may be limited by the size of the grating aswell. The grating resolution, R, is given by:R ≡ λ / Δλ = nNn =N =Diffraction orderNumber of grooves of the grating that areilluminatedGrating resolution imposes a kind of effectivebandwidth limitation Δλ just like the Δλ EFF definedabovePlease note these three things:1. Both grating resolution and effective bandwithdefine a minimum resolvable Δλ for themonochromator.2. A given monochromator will be limited by theworse of the two parameters.3. This is usually Δλ EFFPage 129 of 316

Chem 155 Unit 6 Page 130 of 316Grating Resolution Exercise:How large must a 1000 groove/mmgrating be to give a resolution of 1nm at500nm? (Assume that themonochromator is functioning in firstorder.)R = λ / Δλ = nNN = n λ / Δλ = 1*500/0.1 = 5000 grooves5000 grooves / 1000 grooves/mm = 5 mm!Page 130 of 316

Chem 155 Unit 6 Page 131 of 316If a Czerny-Turner monochromator has the following specifications:Holographically-ruled diffraction grating with 1582 grooves per mm.250 mm focal lengthGrating position such that the diffracted angle is 45 degreesOperation in first orderA slit width of 0.5 mma. What is the reciprocal linear dispersion effective bandwidth of thismonochromator (use appropriate units)?b. What is the effective bandwidth of this monochromator (useappropriate units)?A Czerny-Turner monochromator is set to 300.00 nm and the slits are set so thatthe effective bandwidth is 1.0 nm. If a broadband UV light source is directed intothe monochromator, what wavelengths of light will exit?What type of light source might be appropriate for this experiment?If a Czerny-Turner monochromator has the following specifications:a. Holographically-ruled diffraction grating with 940 grooves per mm.b. 250 mm focal lengthc. Grating position such that the refracted angle is 45 degreesd. Operation in first orderWhat slit width is required to give this monochromator an effective bandwidth of0.5 nm?How big must the grating be in order to achieve this resolution (R).Page 131 of 316

Chem 155 Unit 6 Page 132 of 316High Resolution and Echelle MonochromatorsAtomic spectroscopy demands high resolutionbecause atomic absorption / emission lines are:1. narrow 2.So we want effective bandwidth Δλ EFF =wD -1 = w d cos(r) / nF to be:We also want resolution (R = λ / Δλ = nN)to be:smalllargeΔλ EFFRw should be small -d should be small smallF should be large largen should be large largeBut there are limitations to the above:If w is too small 1 – no light 2 – exceedsw the ability to focus lightdFLimitations in the fabrication of gratingsHow big of a spectrometer is tolerable?What about n?n is the key…Page 132 of 316

Chem 155 Unit 6 Page 133 of 316A monochromator that operates in very high order iscalled:An Echelle monochromatorThe grooves are machined to reflect light at highincident angles – i and r are nearly identical and arecalled βAn echelle monochromator outperforms a Czerny-Turner monochromator in high-resolution applications:Czerny-Turner EchelleF / m 0.5 0.5Groove density 1200 / mm 79 / mmi,r or β 10° 63°R (300nm) 60,000 700,000D -1 (nm / mm) 1.6 0.15Page 133 of 316

Chem 155 Unit 6 Page 134 of 316But – Echelle monochromators have a strangehandicap. At a given angle β, many wavelengths maysimultaneously be striking the detector!nλ = 2dsin(β)63x300 = 2dsin(β)but…62x??? = 2dsin(β) also!n 1 λ 1 = n 2 λ 2 λ 2 = n 1 λ 1 /n 2n……61 309.8362 304.8463 30064 295.3165 290.77…………λ / nmThis problem is called:Order overlapIs this a problem for Czerny-Turner monochromators?If n 1 =1 and λ 1 =300nm then for n 2 =2, λ 2 = 150 nmIf n 1 =1 and λ 1 =800nm then for n 2 =2, λ 2 = 400 nmCzerny-Turner monochromators handle this problemwith:Order sorting filters – when long λ’s aredesired, short λ’s are absorbed by filtersPage 134 of 316

Chem 155 Unit 6 Page 135 of 316Echelle Monochromators handle the order-sortingproblem with cross dispersion – usually a silica SiO 2prism. (Why SiO 2 and not other glass?)UV-absorption is a problem with glassPage 135 of 316

Chem 155 Unit 6 Page 136 of 316This means that the Echelle monochromator has a 2-dimensional focal plane!What kind of detector is tailor-made for an EchellemnochromatorThe CCD array detector!Page 136 of 316

Chem 155 Unit 7 Page 137 of 316Photometric Issues in Atomic SpectroscopySkoog Chapters Covered: 9B, 9CQuestion: How can one quantify low concentrations ofatoms that have extremely narrow band absorptionspectra? How can one do this when the atoms arecontained in a glowing flame that contains manybroadband emitting species.Answer: Interrogate the sample of atoms with anequally narrow band and very bright source ofradiation.Where do you find such a source?Hollow Cathode LampCathode is made of analyte metal – e.g. Fe or ZnNe + ions in plasma accelerate into cathode, sputtermetals atoms, excite them and emission collected outlamp end.Page 137 of 316

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Chem 155 Unit 7 Page 150 of 316Three photometric problems:1. Δλ EFF >> Δλ ATOMIC so stray light is a major issue. The maximumabsorbance that one could realistically expect with a 0.25 nm Δλ EFF isroughly 0.01 using a broadband source, and this is going to be noisy so:a. Use hollow cathode lamp with emission identically matched toabsorption. This makes maximum absorbance much higher, andless noisy. For the remaining source and flame stray light, one can:b. Modulate the source to help to distinguish it from flame and otherradiation that ‘leaks’ into detector ‘around’ atomic line.2. Flame and molecules in flame present a broadband absorptioninterference. These are direct interferants, so they must be measured insome way that is not sensitive to the analyte atoms:a. Use broadband (D 2 ) source that is approximately ‘blind’ to analyteto approximate the broadband flame absorbance.b. Use Zeeman splitting to polarization select between resonantabsorption and off-peak background absorption.c. Use Smith-Hiefje ‘splitting’ to approximate the same effect.Major Learning Objectives:#1 Understand the exponential relationship between transmitted light powerand concentration.#2 Understand how stray light can affect computed values of absorption.#3 Understand how narrow atomic bands, that are narrower than theattainable effective bandwidth, yield a difficult stray light problem.#4 Understand how the hollow cathode lamp addresses the problem ofatomic absorption lines that are narrower than the effective bandwidth ofthe monochromator by matching the spectum of the source (hollowcathode lamp) to that of the absorber (free atoms in flame).#5 Understand how the absorbance of broadband D 2 radiation actuallyapproximates the ‘blank’ absorbance of the flame even when there isanalyte present in the flame.#6 Understand how Zeeman splitting samples the flame background but notthe analyte absorbance.Page 150 of 316

Chem 155 Unit 8 Page 151 of 316Practical aspects of atomic spectroscopy:Sample introduction and plasma chemistry.Skoog chapters covered: 8C, 9AFlame or PlasmaMonochromatorAtomic EmissionfromHollowCathodeLampLight sourcefor absorptionmeasurementonly.Flame orPlasmaNebulizer /Spray ChamberAtomicAbsorptionand / orAtomicEmission inFlame/PlasmaMolecular gasDry solidsAerosolAnalytes inSolutionAbsorbance:⎡ P ⎤A = −Log⎢⎥ = ε b[C]⎢⎣PO⎥⎦Emission:C]= k(P − P )[OPMTPP 0In AAS and AES solutions of analytes are a. aerosolized, b. measured asthey race past the detector as a very dilute gas and in a flame or plasma.This a. very strongly dilutes the analyte and b. limits the time it spends infront of the detector. ---- Why is this tradeoff worthwhile?Intense and narrow emission and absorption bands makeatomic spectroscopy sensitive and selective compared tomost comparable solution methods.Page 151 of 316

Chem 155 Unit 8 Page 152 of 316Nebulization (sample introduction):Sample introduction in AAS and AES is usuallythrough a nebulizer.The nebulizer makes a very fine mist of the analytesolution.The finest droplets in this mist are then selected bythe spray chamber.These droplets are then swept into the flame orplasma.This sample introduction process is necessary but:a.inefficientb.problematicc.hard to do reproduciblyPage 152 of 316

Chem 155 Unit 8 Page 153 of 316Nebulizer types:1. concentric2. cross flow3. fritted disk4. Babbington5. ultrasonicpneumaticConcentric:Page 153 of 316

Chem 155 Unit 8 Page 154 of 316Nebulizers, cont:Cross flow:Babbington:Fritted disk:Page 154 of 316

Chem 155 Unit 8 Page 155 of 316Ultrasonic:Page 155 of 316

Chem 155 Unit 8 Page 156 of 316AtomizationAtoms andions dilutedAtomsOptimal AESregionOptimal AASregionMoleculesParticlesDropletsEntire process happens in:millisecondsSo large droplets are:badPage 156 of 316

Chem 155 Unit 8 Page 157 of 316Flame Chemistry and Matrix EffectsOur understanding of what happens is based on theidea of:Local Thermodyamic Equilibrium(LTE)This means that we theorize and analyze the processoccurring in the flame/plasma as though they wereequilibrium processes.Some flame / plasma properties (Table 9-1):Fuel Oxidant Temp. °C Burn velocity(cm/s)Natural Air 1800 40GasNatural O 2 2800 400GasAcetylene Air 2400 230Acetylene O 2 3100 2000Acetylene N 2 O 2700 300Page 157 of 316

Chem 155 Unit 8 Page 158 of 316Flame as ‘sample holder’:1.For best sensitivity:2.Page 158 of 316Interrogate with hollow cathoderadiation through ‘atom rich’ region.Changes in observation height:Change sensitivity – so don’t moveobservation region during experiment.

Chem 155 Unit 8 Page 159 of 316Optimal observation height:Optimize:Observation regionFlame stoichiometry1. Increased absorbance corresponds to:More gas phase atoms2. Absorbance normally starts low because:Molecules have not been broken downinto atoms yet3. Absorbance normally ends low because:Cr atoms form stable oxides and thisincreases with height.4. Cr profile decreases because:Cr atoms form stable oxides and thisincreases with height.5. Ag profile increases because:Ag atoms do not form stable oxidesPage 159 of 316

Chem 155 Unit 8 Page 160 of 316Flame Chemistry and Interferences:1. Chemical equilibria can occur in flames.Common oxyanions can react with atoms:-H 2 PO 4 + Fe FePO4 + 3H + + 4e -SO 4-2 + Fe FeSO4 + e -2. Electrons can be a reagent.Ionization can be suppressed by addition of KCl:KCl K + + Cl + e -Fe + + e - Fe3. Oxides can be suppressed by:a. Higher Temp: Fe + Ox ↔ FeOxb. Competition: Ti + Ox TiOxa. Ti ‘absorbs’ O from flamec. ‘richer’ flame: more C 2 H 2 , less O 24. Hydroxides can be suppressed by:Higher Temp: CaOH ↔ Ca + OHPage 160 of 316

Chem 155 Unit 8 Page 161 of 316Matrix adjustments in atomic spectroscopy:Releasing agents:Compete with analyte for interferant:Adding Sr +2 to react with PO 4 and releaseCa +2 and Mg +2 for analysis.Protecting agents:Form ‘stable but volatile’ complex withanalyte – e.g. EDTARadiation buffering (when sensitivity is not an issue):Add interferant to samples and standardsStandard Additions:Match matrix of sample and standardsPage 161 of 316

Chem 155 Unit 9 Atomic Emission Page 162 of 316Atomic Emission SpectroscopyResponsibilities in the Chapter:Introduction and all of 10A• OverviewNomenclaturePlasma TypesPro’s and Con’s• Inductively Coupled PlasmaTorch DesignOperating PrinciplesPlasma PropertiesPlasma Optimization• Direct Current PlasmaDesignProperties• SpectrometersBandwidth ConsiderationsMulti-channel Designs• Quantitation and Accuracy• Comparisons with Atomic AbsorptionPage 162 of 316

Chem 155 Unit 9 Atomic Emission Page 163 of 316AAS / AES Review:M+ - ionsM*-e - +e -Emission[M] = k’·(P-Po)HCLΔAtomic gas“M”ΔMoleculargasΔSolidparticlesΔNebulizehνAbsorptionΔ[M] = -k·log(P/Po)LiqudsSolidsGasesPage 163 of 316

Chem 155 Unit 9 Atomic Emission Page 164 of 316Types of AES:1.1.54 PLASMAPlasma is ionized gas – typically Ar, Ar + , e -Ionization is maintained in one of three ways:• A DC electrical discharge called “Direct CurrentPlasma”, DCP or Plasma Jet• An induction coil operating at 27 or 41 MHz“Inductively Coupled Plasma” , ICP• A Microwave Cavity called Microwave InducedPlasma or MIP1.1.55 FLAME• Flame Emission Spectrometry or “FES”• Flame is “chemical plasma” – typicallycomprising CO 2 , H 2 O, other Molecules, e - e -• Flame types include: Air - C 2 H 2 , N 2 O - C 2 H 2 ,O 2 – C 2 H 2• FES is limited to easily excited species such asNa, Li, K1.1.56 ARC / SPARK• Arc = Continuous Discharge• Spark = Pulsed Discharge• These two are like DCP, but are done in air, onsolids (e.g. pressed powders) or metalsurfacesPage 164 of 316

Chem 155 Unit 9 Atomic Emission Page 165 of 316Inert-Gas Plasma Properties (ICP,DCP)Ar mostly, also Ne and He (MIP)• Chemically InertPredominant Species are Ar, Ar + , and electrons• Very Hot5,000 to 10,000 K vs. 2,000 to 3,000 K for flames• Superior AtomizersB 2 O 3 , PO 4 , WO x , VO x , ZrO x and otherREFRACTORY compounds can be analyzedRefractory compounds are extremely thermallystable, and therefore hard to atomize.• Nonmetals Can Be AnalyzedCl, Br, I, S for example: Try making a hollowcathode lamp out of a non-metal!Plasma EmissionSimultaneous DetectionFlame Atomic AbsorptionOne or Two at a TimeOptimization SimplerExcellent Dynamic RangeWide Range of ElementsCostly to Use / BuyEasier to UseHigher PrecisionMetals OnlySignificantly CheaperPage 165 of 316

Chem 155 Unit 9 Atomic Emission Page 166 of 316Inductively Coupled Plasma AES:ICP-AESInduction and Eddy Currents:Switch ‘on’ electromagnet & eddy currents (circular)flow in conductor around changing magnetic fieldlines:Page 166 of 316

Chem 155 Unit 9 Atomic Emission Page 167 of 316ICP TorchesIntensely luminescent, 5000Kplasma “tauroidal core”.Water-cooled inductioncoils with 27 or 41 MHzlarge AC currentsflowing through them.Quartz envelopeSampleaerosolTangential Arflow pushesplasma awayfrom quartzhousingPage 167 of 316

Chem 155 Unit 9 Atomic Emission Page 168 of 316Atomization in Ar-ICPExcitation =Optimalobservationregion.Desolvation /AtomizationInert Properties:HotElectron richInert Analyte = low Atomization oxide formation, / Ionization: chemical interference lowHot = efficient atomizationElectron rich = low ionizationExcellent atomizerPage 168 of 316

Chem 155 Unit 9 Atomic Emission Page 169 of 316Direct Current Plasma AES: DCP-AESIntense emissionfrom electricaldischargeObservationregionAr flow:Cool anodesMake plasmaArgon Plasma Jet:Argon: hot e- richInertRelative to ICP:Less expensiveLess reliableLess sensitivePage 169 of 316

Chem 155 Unit 9 Atomic Emission Page 170 of 316Advantages of Emission MethodsOne of the main advantages of emission methods is:Simultaneous multielement determinationsWith the appropriate spectrometer design, it ispossible to measure many elements at once usingemission. This means:Simultaneous light detection at differentwavelengths – i.e. different points along the focalplane.With conventional spectrometers, this means manyphotomultiplier tubes (PMT’s) placed along the focalplane of the Czerny-Turner, or similarmonochromator. But, PMT’s are neither small, norcheap. So, PMT-based multi-channel spectrometersare both BIG, and EXPENSIVE.With the echelle design, there is a 2-dimensionalfocal plane that happens to be extremely well suitedto the lithographically designed semiconductorcharge-transfer array detector such as the CCD andCID.Page 170 of 316

Chem 155 Unit 9 Atomic Emission Page 171 of 316PP 0Page 171 of 316

Chem 155 Unit 9 Atomic Emission Page 172 of 316Accuracy and Precision in AES• The Good News:Plasmas of Inert Gas are Excellent Atomizers so:Atomization is HighChemical Interference is LowOxidation is Low• The Bad News:AES relies on the existence of excited states, butexcited state populations:Are small relative to ground statesDepend strongly on temperatureN Excitedg ExcitedRecall:. eN o g o• The Solution: (when very high accuracy or precisionis needed)Internal StandardsΔEkTPage 172 of 316

Chem 155 Unit 9 Atomic Emission Page 173 of 316A perfect internal standard will have:Identical Activation Energy (ΔE)Exercise: Derive an equation for the ratio of theanalyte to IS signals as a function of temperature:Recall:e ae be a−bTheanswer:S AS ISQ A ⋅N AQ IS ⋅N IS⋅e( ΔE A−ΔE IS )kTSo:Internal Standards makemeasurements of C Analyte moreindependent of temperature if ΔE A ≈ΔE ISPage 173 of 316

Chem 155 Unit 9 Atomic Emission Page 174 of 316• A good internal standard can compensate for “drift”i.e. changes that appear between measurementsdue to:Sample introduction rateInstrument GainAtomization EfficiencyTemperature• Do you need one?Do the experiment…Check your precision and accuracy and see!Page 174 of 316

Chem 155 Unit 9 Atomic Emission Page 175 of 316Two examples of precision / accuracy assessments ofICP-AES:Page 175 of 316

Chem 155 Unit 9 Atomic Emission Page 176 of 316Page 176 of 316

Chem 155 Unit 10 Page 177 of 316Ultraviolet-Visible and Near Infrared AbsorptionResponsibilities in the Chapter 13Measurement of Transmittance / Absorbance (13A)Beer’s Law (13B)Mixtures (13B-1)Limitations (13B-2)Effect of Instrumental Noise on SpectrophotometricAnalysis (13C)Instrumentation (13D, 13D-1, 13D-2)OverviewUltravioletVisibleNear Infrared190-400 nm400-750 nm750-2500 nmWe group these wavelengths together for tworeasons.Analytes:This wavelength range covers many atomicand molecular electronic absorptions.Optics:Pure SiO 2 , called fused silica or quartz istransparent in this range. Even airabsorbs strongly at λ

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

This came out a bit tedious again.Chem 155 Unit 10 Page 180 of 316We need a coherent section with the layer-bylayerapproach.The whole derivation needs to be scrapped.Extinction Cross Section Exercise:Relate the cross-section “a” to the molar extinctioncoefficient, ε, in Beer’s law. Use this to answer thequestion: can the absorption cross section of amolecule be bigger than the molecule itself?Beer’s law: A = εbC = -Log(P/Po)Basic theory: P = Po⋅exp(-aNx)Recall: Ln(x) = 2.303Log(x)Molar : C = N(cm -3 )(1000cm 3 /L)/(6.023⋅10 23 /mol) =C = 1000 N / N A1. Let x = b (the full path length), solve for aNb,convert to Log from Ln.P = Po exp(-aNb)ln(P/Po) = -aNbaNb = -ln(P/Po) = -2.303Log(P/Po)2. Substitute εbC for –Log(P/Po)aNb = 2.303εbC3. Substitute 1000 N / N A for C, cancel terms, solvefor ε.aNb = 2.303εb (1000 N / N A )aNb = 2.303εb (1000 N / N A )a = 2303 ε / N APage 180 of 316

Chem 155 Unit 10 Page 181 of 3164. Is the light absorption cross section “a” bigger orsmaller than the physical size of the proteinβ-phycoerythrin, a 240,000g/mol protein withε = 2.4x10 6 M -1 cm -1 at λ = 540nm?The physical size of this protein is probably 10 nm(diameter).Recall that a has the dimensions of cm 2a = 2303 ε / N Aa = 2303 (2.4x10 6 ) / 6.02x10 23 = 9.2x10 -15 cm²size ~ a 1/2 = 9.5x10 -8 cm = 0.95 nmPage 181 of 316

Chem 155 Unit 10 Page 182 of 316Limitations to Beer’s Law:1. Fundamental Deviations:1.1. Charge-charge, or simple electrostaticinteractions between atoms/molecules canchange the energy levels and symmetries ofmolecules, and therefore change ε(λ). Theseappear at high absorber concentration.1.2. Refractive Index – ε depends on refractiveindex (n) – so, in cases where comparisonsbetween absorbance in solutions of greatlydiffering n must be made, then you can re-writeBeer’s Law as: A = ε⋅n/(n 2 +2) 2 ⋅b⋅c.Why is charge-charge so different from Van derWaals?2. Chemical Deviations: Van der Waals bonds,hydrogen bonds, or other non-covalent bonds or“chemical interactions” can change the energylevels and symmetries of molecules, and thereforechange ε(λ). These appear at high concentration.3. Instrumental Deviations: Several instrumentalfactors are known to cause non-Beer’s lawbehavior:3.1. The monochromator effective bandwidth islarge compared to the spectral bandwidthsanalyzed.3.2. Stray light is getting into the detector.3.3. The light levels are too low for the detector toaccurately represent.Page 182 of 316

Chem 155 Unit 10 Page 183 of 316Limitations to Beer’s Law cont.:Too large of a monochromator effective bandwidthmeans that polychromatic radiation will be incident on yourdetector at any given supposedly monochromatic setting.This is depicted below. Analyzing for the moment just two ofthe wavelengths of light that strike the detector, we cancalculate the apparent aggregate absorbance assuming thatthe individual absorbances obey Beer’s law:Way overblown…Page 183 of 316

Chem 155 Unit 10 Page 184 of 316Limitations to Beer’s Law cont.:Bigger graph, that’s all. Keep it simple.The Stray Radiation Problem:Light reaching the detector that a. does not come from thesource, b. does not go through the sample. or c. is not theintended wavelength contributes a background power,P STRAY that, in addition to making the absorbance deviatefrom Beer’s law.Does stray radiation make a positive or a negativedeviation from Beer’s law (A = εbC)Stray radiation can come from the source!In a conventional Czerny-Turner monochromator operatingin first order, if the monochromator setting is 700nm, 350nmradiation diffracted in second order will also fall on the exitslit! Order-sorting filters that are moved into the light beamaccording to the wavelength normally remove this but if afilter fails to remove all of the 350nm light then there will be aP STRAY component.P stray 0.05 A truei log P iP oA appi log P i P strayP o P stray0 1 2 3 4P i =A truei =A appi =1000.10.011·10 -31·10 -412340.8451.2431.3141.321A truei43210C iPage 184 of 316

Chem 155 Unit 10 Page 185 of 316Noise in Absorbance Calculations:A complex problem to treat exactly – hence ourmerciful, qualitative approach!Consider the signals P and Po – you actuallymeasure these, and they always contain noise.PBigger graph, that’s all. Keep it simple.Consider an alternative picture and simple statement.Po P PoPoPPA = A = A =Case 1.Hard to distinguish P and PoFor example:P=0.99±0.01Po=1.00±0.01Possible : Possible:1.000.99P/Po 1.01A=-Log(P/Po)NEGATIVECase 2.Hard to distinguish P and zeroFor example:P=0.01±0.01Possible :P = 0.00A = -Log(P/Po) Infinite!Page 185 of 316

Chem 155 Unit 10 Page 186 of 316Deviations due to Shifting Equilibria:For many molecules, especially weak acids, only one of theconjugate acid / base pair may absorb at a givenwavelength.This is how acid-base indicators work!Suppose only the conjugate base of a weak acid is colored.(think phenolphthalein!) What would a Beer’s law plot of anindicator like phenolphthalein (HA) in pure water look like?HA ⇔ H + + A -1. K EQ = [H + ][A - ] / [HA]absorbingGod, please just dump this one, please!Vastly simplify or dump.Concentration of HA insolution, theundissociated fraction.2. Conservation of mass: [HA 0 ] = [HA] + [A - ]so: [HA] = [HA 0 ] – [A - ]Number of moles of HA4. [A - ] = [H + ] (if >> 10 -7 M) divided by volume of insolution. Total or formalconcentration.5 = 1 + 4 K EQ = [A - ][A - ] / [HA]6 = 5 + 2 K EQ = [A - ] 2 / [HA 0 ] – [A - ]Page 186 of 316

Chem 155 Unit 10 Page 187 of 316K EQ = [A - ] 2 / [HA 0 ] – [A - ]try to solve for [A - ]…[A - ] 2 + K EQ [A - ] – K EQ [HA 0 ] = 0ax ⋅2+ bx ⋅ + c 0 x:=( )−b+ b 2 − 4a ⋅ ⋅c2a ⋅a = 1b = K EQc = K EQ [HA 0 ][A - ] =x( K , HAo):=−K+ K 2 + 4K ⋅ ⋅HAo2Page 187 of 316

Chem 155 Unit 10 Page 188 of 316Let K EQ= 10 -5 …[HA] in solution:HA( K , HAo) := HAo − x( K , HAo)4 . 10 4( )xKHA , oHA o3 . 10 42 . 10 41 . 10 4Fraction ionized:00 5 . 10 5 1 . 10 4 1.5 . 10 4 2 . 10 4 2.5 . 10 4 3 . 10 4HA o1xKHA ( , o )zHA oz0.10.011 . 10 7 1 . 10 6 1 . 10 5 1 . 10 4 1 . 10 3 0.01 0.1HA ozPage 188 of 316

Chem 155 Unit 10 Page 189 of 316Somehow I lost the forest for the trees hereagain…Monochromator Slit Convolution in UV-Vis:A monochromator may qualitatively distort absorbancepeaks as well as quantitatively distort them if the Δλ EFF is toolarge.Page 189 of 316

Chem 155 Unit 10 Page 190 of 316Somehow I lost the forest for the trees here again…So – if monochromator slits are too wide (relative topeaks):Peaks are broadenedNon-Beer’s law behavior occursBut – if the monochromator slits are too narrow:Low light power leads to noiseLow light power can lead to spuriouspositive deviations from Beer’s law!Page 190 of 316

Chem 155 Unit 10 Page 191 of 316UV-Vis Instrumentation:Light Sourcesa. D 2 Lamps – 190-400nm,• excellent UV sources• Electrical Discharge in D 2 gas in a sealed, SiO 2bulb.D 2 D 2 * (dissociative molecular state)D 2 * D a + D b + hνE = KE(D a ) + KE(D b ) + hνWavelengthKinetic Energy Continuum continuumb. Tungsten(W)-Halogen – 350-2500nm ,• excellent Vis-NIR (visible-near Infrared) sources• blackbody radiators (e.g. 2500K)• halogen helps to re-deposit W onto filament andimproves lifetime for very hot filamentsc. Xe- or Hg-arc lamps – 200-1000nm,• excellent UV-Vis sources• high-atomic number noble gas in electricaldischarge – many states many wavelengthsthat coalesce into a continuum.Materials:silica = SiO 2 (190-2500nm)glass = SiO 2 + NaBO 4 , etc. (350-1500nm)sapphire = Al 2 O 3 (180-5500nm) $$Page 191 of 316

Chem 155 Unit 10 Page 192 of 316Single vs. double-beam instruments:Single-beam design: Measure blank signal (Po),digitize and record the number, replace blankwith sample cuvette, measure P, calculate A.Advantages of single beam designs:Simple design, relatively inexpensive.Disadvantages of single beam designs:P and Po are measured at different times so aparticular noise source is prominent:i.e. slow change in:o source intensityo detector sensitivitygivesPage 192 of 316driftabsorbance error.

Chem 155 Unit 10 Page 193 of 316Double-beam instruments continually monitoror alternate between P and Po using a second,blank cuvetteAdvantages of double beam designs:Accuracy:P and Po are measured:almost simultaneouslyso source and detector drift are:minimized!Hence absorbances are more:accurate.Disadvantages of double beam designs:Higher Cost: More complicated designOften:Often:Two detectorsMore opticsPage 193 of 316

Chem 155 Unit 10 Page 194 of 316Photodiode Array and Charge Coupled Device(CCD) Array Multichannel Spectrometers (13D-3)Advantages of Multichannel UV-Vis:The spectrometers are FAST because:no need to mechanically rotate a gratingall detectors are measuring the signal at onceThe spectrometers are LESS EXPENSIVE because:they have few moving parts and CCDdetectors are a mass produced technology.The Wavelengths are VERY ACCURATE because:the grating position never changes.Disadvantage of Multichannel UV-Vis:Single-beam design limits photometric accuracyPage 194 of 316

Chem 155 Unit 11 195 of 316UV-Visible Spectroscopy of MoleculesSkoog Ch. 14A all, 14B all, 14E-1 and 14E-3Consider the simplest carbonyl:HHCOLewis Structure ofFormaldehyde3 types of valence electrons are important:• “n” – non-bonding• “σ” – sigma-bonding• “π” – pi-bondingOrbital geometriesσ*π*nπσPage 195 of 316

AbsorbanceChem 155 Unit 11 196 of 316.Spectral Assignments15 . 10 4 2 30200 250 300 350 400Wavelength / nmUV spectrum & energy levels of a fictitiouscarbonyl:If: π π* σ σ* n π* n σ* are allowed,assign peaks 1, 2 and 3 to specific transitions:Transition EnergyFrom DiagramFrom SpectrumPossible Energy / Peak # λ / nm ν / cm -1Transitions arbitrary unitsσ σ* 10-0 = 10 1 200 50,000n σ* 10-4 = 6 2 250 40,000π π* 9-1 = 8 3 330 30,000n π* 9-4 = 5ΔE (spectrum) ΔE (digram) λ / nm Assignment50,000 10 200 σ σ*40,000 8 250 n σ*30,000 6 330 π π*n π*Page 196 of 316Energy10 σ*9 π*87654 n321 π0 σ

Chem 155 Unit 11 197 of 316Classification of Electronic TransitionsThe following simplified classifications apply toorganic molecules:1. σ σ*1.1. Highest energy!1.2. Shortest wavelength λ < 200 nm1.3. O 2 , N 2 , all molecules have σ-bonds so: thisradiation is absorbed by air and SiO 2 sospectroscopy must be done in vacuum and isvery difficult – called VUV or vacuumultraviolet spectroscopy2. n σ*2.1. 150-250nm – most all transitions are foundin the ultraviolet2.2. Molecules with lone (nonbonding) pairs ofelectrons participate such as: O, N, Halogens2.3. ε ∼ 100 to 1000 “medium” strength absorbers3. n π*, π π*3.1. Longest wavelength transitions 200-700 nm3.2. Strongest absorbers ε ∼ 100 to 100,0003.3. Limited to molecules with π-bonds – e.g.alkenes, alkynes, aromatics, carbonyls, azidesetc.Page 197 of 316

Chem 155 Unit 11 198 of 316Spectral Peak BroadeningWhy are solution-phase electronic spectra hundredsof nm wide if atomic emission and absorption peaksare only ca. 0.005 nm wide?They aremolecules, notatoms – soelectronictransitions,vibrationaltransitions,rotationaltransitions allhappensimultaneouslySolution phaseinteractionsperturb the energylevels of eachmoleculedifferently. Thespectrumaverages overmany molecules,creating acontinuum ofenergy levels forthe transitionPage 198 of 316

Chem 155 Unit 11 199 of 3161.1.57 Influence of Solvent Spectral Fine Structure:1.1.58 Influence of Temperature on Spectral FineStructure:Page 199 of 316

Chem 155 Unit 11 200 of 3161.1.59 Solvatochromism:The shift in wavelength of an absorbance band withchanges in polarity (dielectric constant) of the solvent.Conjugation: The presence of alternating single anddouble bonds in organic molecules.Page 200 of 316

Chem 155 Unit 11 201 of 316Aromatic UV-Visible absorptions:• Three sets of ππ* bands are prominent in thespectrum of benzene:Primary E2 Bλ (nm) 184 204 256ε MAX (M -1 cm -1 ) 60,000 7,900 200V. Strong Strong Intermed• -OH and –NH 2 red-shift and intensify the B-bandDoes the nonbonding pair stabilize or destabilize theπ orbitals? (assuming it acts predominantly on π asopposed to π*)Page 201 of 316

Chem 155 Unit 11 202 of 316UV-Visible Bands of Aqeuous Transition Metal IonsTransitions are known as “d-d” because electronicstates correspond to changes in the population of:d-electron orbitalsPage 202 of 316

Chem 155 Unit 11 203 of 316“d-d” transitions imply that d-orbitals have differentenergies.When ligands such as H 2 O coordinate (bond) to themetal ion, they interact with the d-orbitals and changetheir energies predictable ways depending on thebonding geometries. They “split” the d-orbitals by andamount Δ, the ligand field strength.Page 203 of 316

Chem 155 Unit 11 204 of 316Ligand field splitting energy Δ varies with the type ofligand in the approximate following series:I - < Br - < Cl - < F- < OH - < C 2 O 4 -2 < H 2 O < SCN < NH 3< Ethylenediamine < o-phenanthroline < NO 2 - < CN -“d-d” transitions are common, 0.1 < ε < 1, but seldomused for detecting or determining the concentration ofmetal ions – why?A = εbC εb is the sensitivity and it is small!Page 204 of 316

Chem 155 Unit 11 205 of 316Charge-Transfer ComplexesStrongly Absorbing Metal Complexes:Example:Fe(III) 3+ -SCN - + hν Fe(II) 2+ SCN 0electron transfer from SCN - to Fe(III)ε ∼ 5,000Fe(II) 2+ (o-phenanthroline) 3 + hv Fe(III)3 + (o-phenanthroline) -1 (o-phenanthroline) 2electron transfer from Fe(II) to (o-phenanthroline)ε ∼ 12,000Page 205 of 316

Chem 155 Unit 11 206 of 316Lanthanide and Actinide Ions:Lanthanide and Actinice ions have unusual spectralproperties.• Lots of weak visibleand infrared lines• Very narrow bands!The 4f and 5f orbitalsare smaller than thefilled 6s and 7s orbitalsso the f orbitals are not‘split’ or broadenedvery much by solventligands• Electronicconfigurations differ inenergy due primarily toorbital angularmomentum, i.e.magnetic interactionsdescribed by Russell-Saunders and othercoupling schemes.Page 206 of 316

Chem 155 Unit 11 207 of 316Photometric TitrationTrace metals may be analyzed by by UV-Visabsorption using a method known as photometrictitration.For the complexation reaction:S + T PS = Analyte metal ion – e.g. Fe 3+T = Titrant – e.g. SCN -P = Product – e.g. FeSCN 2+Consider a big, stirred cuvette that you titrate with T.To which of the above schemes would the FeSCN 2+titration above correspond?Page 207 of 316

Chem 155 Unit 11 208 of 316Multi-component Analyses:Consider a cell containing two analytes with overlapping absorbancespectra:Consider two absorbing species A and B:6B i( 0.25A) i+( 0.75B) i4( 0.5A) i+( 0.5B) i( 0.75A) i+( 0.25B) iA i20200 250 300 350 400w iλ MAX (A) = 233 nm, FWHM = 60 nmε A (233) = 5.3 ε A (263) = 3.2 ε A (293) = 0.72λ MAX (B) = 293 nm, FWHM = 60 nmε B (233) = 0.72 ε B (263) = 3.2 ε B (293) = 5.3Page 208 of 316

Chem 155 Unit 11 209 of 316Given:ε A (233) = 5.3 ε A (293) = 0.72ε B (233) = 0.72 ε B (293) = 5.3A(233) = ε A (233)bC A + ε B (233)bC BA(293) = ε A (293)bC A + ε B (293)bC BA(233) = 2.55A(293) = 3.47What are the concentrations of A and B?Answer: C A = 0.4, C B = 0.6ε A1 ⋅C Aε A2 ⋅C AUnknown mixture of A and B42Wavelength / nm+ ε B1 ⋅C A B 1+ ε B2 ⋅C A B 2Absorbance0200 220 240 260 280 300 320 340 360 380 400III5.3⋅C A0.72⋅C A+ 0.72⋅C B 2.55+ 5.3⋅C B 3.47III II⋅0.725.3III0.72⋅0.72 ⋅C A + 5.3⋅0.72 ⋅C B 3.47⋅0.725.3 5.35.3III0.098⋅C A+ 0.72⋅C B 0.471Substitute CA into I and II (to check):I − III5.3⋅C A+ 0.72⋅C B 2.55−( )0.098⋅C A + 0.72⋅C B 0.4715.202⋅ C A 2.079C A2.0795.2020.40III5.3⋅0.42.55 − 5.3⋅0.4C B0.720.72⋅0.4+ 0.72⋅C B 2.55+ 5.3⋅C B 3.473.47 − 0.72⋅0.4C B5.30.600.60Page 209 of 316

Chem 155 Unit 11 210 of 316Matrix approach:ε A1 ⋅C Aε A2 ⋅C Aε A3 ⋅C Aε A4 ⋅C A+ ε B1 ⋅C BA 1⎛ ε⎜ A1+ ε B2 ⋅C BA 2 ⎜ ε A2+ ε B3 ⋅C A B 3+ ε B4 ⋅C A B 4⎜⎜⎜⎝ε A3ε A4ε B1ε B2ε B3ε B4⎞ ⎟⎟ ⎛⎟⎟ .⎜⎜⎝⎟ ⎠C AC B⎞⎟⎟⎠⎛⎜⎜⎜⎜⎜⎝A 1A 2A 3A 4⎞⎟⎟⎟⎟⎟⎠ε Aiε BiC AC BA i= the extinction coefficient of species A at wavelength i= the extinction coefficient of species B at wavelength i= the concentration of species A= the concentration of species B= the absorbance at wavelength iUse computers to solve matrix for C A and C B , C C etc.Overdetermine matrices to average out noise.Page 210 of 316

Chem 155 Unit 12 211 of 316Intro to Fourier Transform Infrared SpectroscopySkoog Chapters CoveredFourier Transform OpticalMeasurements16A1 VibrationalTransitions16A2 HeteronuclearDiatomics - Classical16A3 HeteronuclearDiatomics - Quantum16A4 Vibrational Modes16A516C1Vibrational CouplingFourier TransformInstrumentsPage 211 of 316

Chem 155 Unit 12 212 of 316Overview:Infrared (IR) spectroscopydeals mainly with:molecular vibrationsIR radiation islight.lowenergy relative to visibleIR wavelengths range (approximately) from:Wavelength (nm) Frequency (cm -1 )Near IR 1,000 – 2,500 10,000 – 4,000Mid IR 2,500 – 20,000 4,000 - 500Far IR 20,000 – 200,000 500 - 50Page 212 of 316

Chem 155 Unit 12 213 of 316Why is IR spectroscopy important?1. UV-Vis spectra are useful, but: all molecules:VibrateSo most all molecules have:Vibrational absorption spectraAnd most vibrational modes absorb:StronglySo IR spectroscopy is a very general and sensitiveanalytical tool.2. Vibrational spectra have a lot of information:a. Fingerprinting / IdentificationEach molecule has a complex but uniquesignature.b. QuantitationBeer’s law holds for vibrationalabsorption too!c. Measurement of Interactions / ConfigurationsVibrational frequencies (spectra) aresensitive to intermolecular interactions –so one can identify phenomena such ashydrogen bonding, trans or gaucheinteractions that may control reactivity ofmolecules.Page 213 of 316

Chem 155 Unit 12 214 of 316Some examples of IR spectroscopy• Identification of organic molecules• Identification of functional groups• Elucidation of molecular structure• Elucidation of intermolecular interactionsRoutine uses of IR spectroscopy• Confirmation of functional groups• Confirmation of molecule identity• Monitoring of atmospheric pollutants.• Some ‘breathalizers’ use IR absorption of ethanol.• Quantitation of amide nitrogen for the determinationof protein content in food and other materials.IR Spectroscopy applications will follow in a laterlecture section.Page 214 of 316

Chem 155 Unit 12 215 of 316IR Spectroscopy is Difficult!• IR beams are invisible to the eye.• IR bands are narrow, so Δλ EFF must be small.• SiO 2 absorbs IR radiation.• Water and other solvents absorb IR radiation.• IR photons are low energy, so they are hard todetect!Why can’t you use a photomultiplier tube in the IR?Infrared radiation has a longer wavelengththan visible radiation:λ IR > λ VIS– so –IR radiation has a lower frequency:ν IR = c/λ IR– so –IR radiation has a lower energy:– so –Photomultiplier tubes don’t work because:E IR = hν IR ,For a PMT to work, the photon energy must be greaterthan the work function of the metal!E ≥ φButE IR ≤ φPage 215 of 316

Chem 155 Unit 12 216 of 316Monochromators Are Rarely Used in IRBecause:• IR sources are somewhat weak and• IR radiation is somewhat hard to detect, and• IR radiation is hard to focus and• IR spectroscopy must be done at very low effectivebandwidth,Grating or prism-monochromator based IRspectrometers are usually:slownoisyWhat is done instead?InterferometryIn interferometry all light frequencies strike thedetector at once and the information comes fromsignal oscillation as a function of:time!Could you measure the frequency of light bymeasuring the time between maxima in theelectric field?c 3⋅10 8 − 1:= ⋅m⋅s λ := 10⋅10 − 6 ⋅mλ=c3.333× 10 − 14 sPage 216 of 316

Chem 155 Unit 12 217 of 316Interferometers measure light field vs. timeImagine you were a light-speed gremlin with avoltmeter in hand and a light beam passed by:τcSignal / Volts0 5 10 15 20 25 30Time / femtosecondsThis can be done with an optical device called an:InterferometerTake this on faith for now, and let’s explore what thatmight look like…Page 217 of 316

Chem 155 Unit 12 218 of 316The Michelson interferometer:Consider that S, A, B, and D are fixed – i.e.unmoving, but Mirror 2, C, can move. In terms of thevariable distance AC, and the fixed distance AB,When will there be a bright light at the detector asopposed to darkness?When 2AB - 2AC = nλPage 218 of 316

Chem 155 Unit 12 219 of 316How is interferometry performed?Recall – light interference is how gratings work.cos( θ ) 6cos( θ ) 4cos( θ ) cos( θ )cos( θ ) 6cos( θ π ) 4cos( θ ) cos( θ π )105050 0.5 1 1.5 2 2.5 3θ2 . π1050Light ElectricFields SumConstructiveInterferenceDestructiveInterference50 0.5 1 1.5 2 2.5 3θ2 πConstructive interference:nλ = f( sin(i)+sin(r) ) for a gratingFor gratings constructiveinterference occurs when the pathlength difference between the rays,δ, is an integral multiple of thewavelength, λ:δ = d(sin(I)+sin(r)) = nλnλ = f( time) for an interferometerPage 219 of 316

Chem 155 Unit 12 220 of 316Signal Fluctuations for a Moving MirrorConsider the case of mirror 2 moving at constantvelocity, v (cm/s).AC(t) = Distance from beamsplitter to mirror 2AB = Distance from beamsplitter to mirror 1AC(0) = ABAC(t) = AB + vt Constant Velocity MirrorWe can define the retardation - δ - as the differencein pathlength between the rays going to the fixed andmoving mirrors respectively:δ(t) = 2(AC(t) - AB)δ(0) = 2(AC(0) - AB) = 2(AB - AB) = 0What is the time interval, τ (s) between the conditionsδ = 0 and δ = λ?Find τ for which δ(τ) = λ = 2(AC(τ) - AB)τ = ? ( hint: substitute AB + vt for AC(t) )δ(τ) = λ = 2(AC(τ)-AB) = 2( (AB-vτ) -AB) = 2vτλ = 2vτ τ = λ / 2vPage 220 of 316

Chem 155 Unit 12 221 of 316So – at every time interval τ the condition δ = nλ istrue and the detector “sees” a bright light:What is the frequency (s -1 )of the detector signal interms of τ, and in terms of the instrumentalparameters v and λ?f (s -1 ) = 1/τ(s) = 2v(cm/s) / λ (cm)For v = 1 cm/s and λ = 1000 nm, what will be thedetector frequency, f?f (s -1 ) = 1/τ(s) = 2 1(cm/s) / 1000x10 -7 (cm) =20,000HzPage 221 of 316

Chem 155 Unit 12 222 of 316Mono and polychromatic responseThe interferometer makesa low frequency oscillatingsignal from a highfrequency light signal.But we still need to see thesignal as a function offrequency (or wavelength)to understand it.Source: interferogram Spectrummonochromaticonefrequencyonly.Two frequencies(lines) only.Polychromatic –manyfrequencies.Page 222 of 316

Chem 155 Unit 12 223 of 316Interferograms are not informative:IR and other spectra are typically presented in the“frequency domain” i.e. as a function of frequency orwavelength – not as a function of time.How can one transform:2000Amplitude1000010000 20 40 60 80 100TimeInto something informative like:In other words, the frequency or wavelength spectrumof the light is where the information is, how can oneget this information from time domain signals?Page 223 of 316

Chem 155 Unit 12 224 of 316Transforming time frequency domain signals:121.5Amplitude10Amplitude10.510 0.5 1 1.5time_seconds00 1 2 3 4Frequency_Hz121.5+Amplitude2010 0.5 1 1.5time_secondsAmplitude10.500 1 2 3 4Frequency_Hz------------------------------------------------------------------------------------------221.5Amplitude30Amplitude10.520 0.5 1 1.5time_seconds00 1 2 3 4Frequency_HzTime Domain Frequency Domain conversion isdone by computer using an algorithm called the:Discrete Fourier TransformIn what domain does a conventional monochromatoroperate?Frequency or Wavelength Domain – A vs. λPage 224 of 316

Chem 155 Unit 12 225 of 316The Centerburst:For a broad spectrum of frequencies going into aMichelson interferometer, for what value of δ is theconstructive interference condition, nλ = δ satisifiedfor all wavelengths?nλ = AB-BC = δ is always true if:δ = 0!Page 225 of 316

Chem 155 Unit 12 226 of 316Time vs. frequency domain signals:It is simple to imagine deciphering the E(t) signal of amonochromatic light beam, but, what would abroadband light signal look like in the time domain?Broadband spectrum:Line spectrum:10020Amplitude50Amplitude10000 0.5 1 1.5Frequency100 20 40Frequency2000InterferogramAmplitude020000 20 40 60 80 100Time10Amplitude01020 25 30 35 40TimePage 226 of 316

Chem 155 Unit 12 227 of 316Advantages of Interferometry.In the IR, interferometers have advantages overmonochromators:• Advantage 1: Interferometers are fast:Consider a grating onochromator:If your source and detector aregood from 200nm to 1200nm(1000nm total) and your effectivebandwidth is 1 nm, how much light is beingmeasured or lost at any one time?MultiplexAdvantage1 / (1000) is being measured so1-[1/1000], or 99.9% is being thrown away!You only get one λ at a time!• Advantage 2: Interferometers are high resolution.Recall: If Δλ EFF > the width of the spectral peak:Deviation fromBeer’s lawPeak distortion(convolution)Page 227 of 316

Chem 155 Unit 12 228 of 316Resolution in InterferometryInterferometers are easily made into highresolutionspectrometers.Consider two spectral lines differing in frequency byonly 2%:How do these lines appear in the time domain?786cos( θ ) 6cos( θ . 1.02 ) 3cos( θ ) cos( θ . 1.02 )24200.99901320 10 20 30 40 50 60 70 80 90 1000 θ2 π100Page 228 of 316

Chem 155 Unit 12 229 of 316Let us define resolution as: Δν = ν 1 - ν 2 when ν 1 canjust be separated from ν 2 .I assert that to separate, or resolve these lines in thefrequency domain spectrum, we must scan theretardation, δ, from 0 (where ν 1 and ν 2 are in phase)to a new value, let’s just call it “δ” where the twosignals are again in phase.The frequency of oscillation of the signal in time is:f = 2v(cm/s) / λ(cm) = 2 v νwhere ν is the light frequency in wavenumbers (cm -1 ).The light power function will be sinusoidal:P(δ) = P i cos(2 π f t) = P i cos(2 π (2vν) t)If we substitute v, the mirror velocity, with δ/2t = v weget intensity versus retardation:P(δ) = B cos(2 π δ ν)Page 229 of 316

Chem 155 Unit 12 230 of 316For two light rays of identical power P i but differentfrequency the total signal will be:P(δ) = P i cos(2 π δ ν 1 ) + P i cos(2 π δ ν 2 )For identical BP(δ) = P ( cos(2 π δ ν 1 ) + cos(2 π δ ν 2 ) )What is P(δ) when δ = 0?B( cos(0)+cos(0) ) = 2BSo, when will P(δ) be 2B again?In other words, at what next δ value >0 will the twodifferent wavelengths ν 1 and ν 2 both give ‘bright’signals again?δ ν 1 = some integerδ ν 2 = some integer ± 1δ ν 1 = δ ν 1 ± 1δν 1 - δ ν 2 = ±1i.e. they have ‘slipped’ one wavelength and are bothbright again –so, in this case: ν 1 - ν 2 = Δν = 1/δPage 230 of 316

Chem 155 Unit 12 231 of 316For a large δ, Δν is small and resolution is high!Consider an interferometer with a mirror traveld = 0.5 cm that is operating in the mid visible at500 nm:ν = 10 7 /500 nm = 20,000 cm -1δ = 2d = 1 cmΔν = 1/δ = 1/1cm,so 20,001 cm -1 can be resolved from 20,000 cm -1 .10 7 / 20,001 cm -1 = 499.975 nm.So, 500 nm can be resolved from 499.975 nm.Δλ = 500 - 499.765 = 0.025nmΔλ EFF = 0.025nm is very difficult to achieve with amonochromator!Remember that δ is equal to 2d – twice the mirrordisplacement.Page 231 of 316

Chem 155 Unit 12 232 of 316Conclusions and Questions:Infrared spectroscopy is important because:1. Virtually all molecules have a unique vibrationalspectroscopic signature in the infrared.2. Most molecules absorb strongly! So IR can bequite sensitive for both identification andquantitation.3. But IR radiation is harder to manage with amonochromator because sources are weak,detectors are less sensitive and high resolution isoften needed.4. So interferometry is done instead, which is bothinherently fast and high reseolution.Interferometry converts optical frequency directly intoa frequency on the detector:1. What frequency would a 3000 cm -1 sourceproduce in an interferometer operating with amirror velocity of 1 cm/s?2. If the mirror travel were 1 cm, what would be theresolution (Δλ EFF ) in cm -1 ?3. What will this value be in nm?4. What focal length would be needed with amonochromator with a 1000 nm grating spacingand slits at 0.4 mm?Page 232 of 316

Chem 155 Unit 12 233 of 316Answers:fcm2⋅v⋅ν21 ⋅ ⋅ ⋅3000⋅ cm − 1 = 6000s -1seffective bandwidth in cm − 11: Δν EFFδ12d ⋅121 ⋅ ⋅cm=0.5cm − 1effective bandwidth in nm:10 7 ⋅nmcm3000⋅cm − 1( )−10 7 ⋅nmcm3000.5cm ⋅− 1= 0.555nmΔλ EFFwd ⋅ ⋅cos () rnF ⋅Fwd ⋅⋅cos () rn⋅Δλ EFF⎛⎝0.4⋅ mm⋅1000⋅ nm⋅cos⎜45⋅π1801⋅0.55⋅nm⎞ ⎟⎠= 0.514mThe detector frequency would be 6 kHz – not aproblem.The effective bandwidth for a 1 cm drive is 0.5 cm -1 .At 3000 cm -1 – this equates to 0.55 nm.The minimum focal length required to achieve thisbandwidth with a monochromator would be 0.5 m, afairly large system, especially when compared to a1 cm interferometer drive.Page 233 of 316

Chem 155 Unit 13 Page 234 of 316Infrared Spectrometry:Selection Rules, Normal Modes and GroupFrequenciesSkoog Chapters Covered:16A all, 16B – Sources and Transducers16C-1, FT-InstrumentsRough overview of the optical spectroscopy ranges:λ (nm) ν (cm -1 ) T (K) Designation Phenomena100 100,000 140,000 UV σ-σ*, n-σ*,VISn-π*, π-π*Valence electronexcitation1,000 10,000 14,000 Near IR low-E electronicHi-E vibrationsOH, C≡O, NHMid IR GroupFrequencies10,000 1,000 1,400 FingerprintInorganics,High-Z, Low-kFar IRLattice vibration,Pure rotations100,000 100 140μ-wave1,000,000 10 14Page 234 of 316

Chem 155 Unit 13 Page 235 of 316Absorbance Bands Seen in the Infrared::VibrationsRotationsSolidbroad(phonon)rare(C 60 )Liquid broad unresolvedGas sharp sharpPage 235 of 316

Chem 155 Unit 13 Page 236 of 316IR Selection RulesHow does electromagnetic radiation make a moleculevibrate? Consider the diatomic molecule H-Cl.Electric Field Vector of LightE( θ )δ+HHθHδ-Cl Cl ClIn Out InForcePage 236 of 316

Chem 155 Unit 13 Page 237 of 316Now consider the molecule N 2 :Electric Field Vector of LightE( θ )θδ0NNNδ0NNN0 0 0ForceWhen will a molecular vibration absorb light?when vibration causes change in dipole momentdynamic dipole neededWill interaction with electromagnetic radiation changethe FREQUENCY or the AMPLITUDE of thevibration?Page 237 of 316

Chem 155 Unit 13 Page 238 of 316Rotational ActivityIndicate the forces on the molecule below:Electric Field Vector of LightE ( θ )`θWhen can a molecular rotor absorb light?When a molecule has a permanent dipolemoment it rotationally active.Example:Cl HH ClΔJ =Δv =ΔJ =Δv =Page 238 of 316Energy

Chem 155 Unit 13 Page 239 of 316Normal Modes of Vibration:Consider a diatomic molecule –translationsrotationsvibration!For linear molecules, there are:modes of vibration.Non-linear molecules have:modes of vibration.3N-53N-6Page 239 of 316

Chem 155 Unit 13 Page 240 of 316Modes of vibration:For example – CO 2 : Linear or Non-linear?# of Modes: 3( ) - ( ) =ν S = 1340 cm -1 ν AS = 2340 cm -1symmetric stretch anti-symmetric stretchδ S = 666 cm -1 δ S = cm -1anti-symmetric stretch, IR activesymmetric stretch, IR inactivebend modes, degenerate, IR activeFor example – H 2 O: Linear or Non-linear?# of Modes: 3( ) - ( ) =ν S (OH) = ν AS (OH) = δ S (OH) =3562 cm -1 3756 cm -1 1596 cm -1Page 240 of 316

Chem 155 Unit 13 Page 241 of 3161.1.60 One may see more than 3N-5/6 bands:• Overtones:• Combinations:Δv = ±2, ±3ν C = ν A + ν B and ν A - ν B1.1.61 One may see fewer than 3N-5/6 bands:• ν not in 400-4000 cm -1 range of typical FTIR• Band may be weak• Band may be broad and overlap another band• Bands may be degenerate• Bands may be ‘forbidden’, i.e. if there may be nochange in dipole moment during the vibrationalmode.Page 241 of 316

Chem 155 Unit 13 Page 242 of 316Group frequencies: a pleasant fiction!Most molecules have many atoms.That means many modes! Some very complex!But –some functional groups behave like small molecules!Characteristic groupmodes and show up inIR spectra.Examples of ‘groups’:• C-H• C=O• C≡N• NH 3• OH etc.3N-6 rule irrelevant tothis analysisν is somewhatvariable because:• environment• other bonds• other modesinfluence the groupvibration.Page 242 of 316

Chem 155 Unit 13 Page 243 of 316p. 136 of Spectrometric Identification of Organic MoleculesR.M. Silverstein, F.M. Webster, Wiley 1998Page 243 of 316

Chem 155 Unit 13 Page 244 of 316Group frequencies can be calculated approximately:Interatomic forces are approximated by Hooke’s Law:F =k · ySince F = mA (Newton’s Law)d 2 yF = mA = m = ky(t)y(t) is proportional to its own second derivative.This differential equation has a sinusoid solution:dt 2y(t) = A cos(2πν m t) ν m =1 k2π μμ =m 1 m 2m 1 + m 2reduced massUnits of μ are:Page 244 of 316units of mass

Chem 155 Unit 13 Page 245 of 316Classical calculations (above) tell you the natural frequencyof a classical oscillator.Quantuum mechanical calculations tell you the allowedenergy levels (E). From Planck and Einstein’s equationE=hν we can calculate the light frequencies (ν).E = (v + ½ )h ν mv = vibrational quantuum number = 0,1,2…Δv = 0,±1 = the selection rule for vibrational transitionsagain: ν m =12πkμso:ΔE = Δv h ν m = ±1h2πkμ= hνnote:ν m is the oscillator frequency andν is the light frequencyWhat is the relationship between ν and ν m ?the light frequency equals the oscillatorfrequencyWhat is the relationship between light frequency, ν, and lightwavenumber, ν (cm -1 )? (ν = 1/λ(cm)).ν = cνPage 245 of 316

Chem 155 Unit 13 Page 246 of 316Summary:• Molecules vibrate – their atoms move back and forthrelative to each other.• Vibrations come in certain modes that involve the wholemolecule.• Modes wherein the molecule has a change in dipolemoment may be excited by IR radiation.• The higher the mass, the lower the mode frequency.• The stronger the bond, the higher the mode frequency.• In complex molecules with many atoms, someindividual parts of the molecule behave as though theywere independent.• The absorption bands corresponding to these groupmodes are known as group frequencies and allow oneto identify functional groups such as aldehydes,amides, olefins etc. on a given molecule.Page 246 of 316

Chem 155 Unit 14 247 of 316Infrared Spectrometry - ApplicationsSkoog Chapters Covered17A1 Sample Handling in Mid-IR17A2 Group Frequencies17A3 Quantitation in Mid IR17B Reflection Methods17D Near IR applications17E Far IR applicationsThe major difficulty in infrared spectroscopy is relatedto the IR selection rule:Δμ during vibration => IR activeNearly everything absorbs IR radiation!optics and solvents will have strongabsorptionOther difficulties relative to, e.g. visible spectroscopy:Beam is invisible to the eye – hard to alignIR Sources are often weak (blackbodies)Detectors are not very sensitivePage 247 of 316

Chem 155 Unit 14 248 of 316Strategies used to make IR spectrometry work -1. Poor sources and detectors:use interferometer instead of monochromator1. Optics absorb IR light:use mirrors instead of lenses2. Materials for transmissive optics:use high-atomic-mass materials with weakinter atomic bonds – typical materials are:1. CaF 2 2. ZnSe 3. KBr 4. NaCl3. Sample Handling:use solvents transparent in selected spectralregionsavoid solvents completelyuse reflection-absorption instead oftransmissionPage 248 of 316

Chem 155 Unit 14 249 of 316Solvents for IR spectroscopy:Handling of neat (pure – no solvent) liquids:Sandwich your sample between IR-transparentplates: e.g. NaCl:Q. If your sample is 10 microns thick, and theconcentration of molecules in the neat sample is10 M, and a typical band gives an absorbance of 0.1a. what is ε?A=εbC ε = A/bC = 0.1/0.001*10ε = 10b. why can’t you use a 1-cm pathlength NaCl cuvette?Page 249 of 316A=εbC = 10*1*10 = 100T=10 -100 = 0.00000000000000…….01

Chem 155 Unit 14 250 of 316Handling of solids: pelletizing:grindput into die5 – 10 tons of pressurea few minutes…vacuum to reduce voidsHandling of Solids: mulling:sample2-5 mg2 μm“Nujol” =mineral(aliphatic ) oil1. Grind sample finely2. Mix with mulling oil3. Sandwich w/ KBr/NaClplatesor“Fluorolube” =fluorocarbon oilPage 250 of 316

Chem 155 Unit 14 251 of 316A general problem with pellets and mulls:A = -log[P/Po]How do you measure Po?You use an open beam to approximate PoSo – the spectra you obtain, are a sum ofabsorbances – the spectra contain:Sample plus window plus mulling oilabsorbance plus reflection losses.IR spectra obtained with these sampling methods areuseful for identification, not quantitation:qualitativeThe two pieces of information that are useful:Functional group identificationSpectral ‘fingerprinting’Page 251 of 316

Chem 155 Unit 14 252 of 316Group Frequencies Examples:C-O-H versus C-ClWhich stretching frequency is higher:Why?O-H or C-ClReduced mass of C-Cl muchlarger so (k/μ) 1/2 for C-Cl smallerPage 252 of 316

Chem 155 Unit 14 253 of 316Fingerprint ExamplesMid-IR Group Frequencies and FingerprintsExample – Branched Alkane Structural IsomersPage 253 of 316

Chem 155 Unit 14 254 of 316Diffuse Reflectance Methods:A way to do IR of solids without mulling or pelletizing:Page 254 of 316

Chem 155 Unit 14 255 of 316Quantitation of Diffuse Reflectance Spectra:Page 255 of 316

Chem 155 Unit 14 256 of 316Attenuated Total Reflection Spectra:Page 256 of 316

Chem 155 Unit 14 257 of 316Attenuated Total Reflectance Sampling Volume andQuantitation:Penetration depth ≈ λEven though Snell’s Law says the light should totallyinternally reflect, the light actually behaves as thoughit were reflecting from a plane situated at:A point just above the surface, slightlypenetrating the sample.Page 257 of 316

Chem 155 Unit 14 258 of 316The reflected beam interacts with the sample on thecrystal surface. It behaves as though it passesthrough the sample and is reflected back by a planeinside the sample some fraction of a wavelengthabove the crystal surface (i.e. the n 1 -n 2 interface).Page 258 of 316

Chem 155 Unit 15 259 of 316Raman Spectroscopy:vibrationalspectroscopy withnear UV, visible or near infraredlight is called:Raman Scattering SpectroscopyIntense monochromatic lightbeam – usually a laserH 2 OA tiny fraction ofthe incidentradiation scattersoff of the moleculesin the solution:RayleighAn even tinierfraction of theincident radiationscatters off of themolecules and loseor gain energy.Ramanν SCATTERED = ν INCIDENTν SCATTERED ≠ ν INCIDENTelasticscatteringinelasticscatteringPage 259 of 316

Chem 155 Unit 15 260 of 316A Raman scattering experiment:Scattered BeamHigh resolutionmonochromator /interferometer +sensitive detectorLaserTransmitted BeamIncident BeamThe sample may be aconcentrated solution, a neatliquid analyte, a transparentsolid or a surface.Raman is not usuallyused for dilutesolutions.If the laser is, e.g. a HeNe laser (λ=632.8 nm), most of the scatteredradiation will be:632.8 15803 cm -1nm or 1/(632.8x10 -7 cm) =This scattered radiation is monochromatic, and is known in the spectrumas the:Rayleigh LinePage 260 of 316

Chem 155 Unit 15 261 of 316What a Raman Spectrum Looks LikeIdentify in the following:1. The Rayleigh peak.2. The elastically scattered light peaks.3. The inelastically scattered light peaks.4. The Stokes peaks.5. The anti-Stokes peaks.6. Rationalize the relative intensity differencesbetween the Stokes and anti-Stokes peaks..Hypothetical Raman Scattering Spectrum .Light Power400 425 450 475 500 525 550 575 600 625 650Wavelength / nmPage 261 of 316

Chem 155 Unit 15 262 of 316Quantum View of Raman Scattering.S1EnergySoVirtualstatesVibrational coordinateHypothetical Raman Scattering SpectrumLight Power400 425 450 475 500 525 550 575 600 625 650Wavelength / nmPage 262 of 316

Chem 155 Unit 15 263 of 316Classical View of Raman ScatteringVisible frequency radiation is higher frequency than IRradiation and vibrational frequencies.Molecular Vibration (e.g. C=O stretch, 2000 cm -1 )r( )r eq ⋅cos 2⋅π⋅ν vib ⋅ttimeVisible Frequency Light (e.g. 500 nm / 20,000 cm -1 / blue-green)E( )E 0 ⋅cos 2⋅π⋅ν ex ⋅ttimePage 263 of 316

Chem 155 Unit 15 264 of 316Radiation interacts with molecules by inducing adipole moment.The induced dipole ‘m’ is proportional to thepolarizability, α.m( )α⋅Eα⋅E 0 ⋅cos 2⋅π⋅ν ex ⋅tBut the polarizability is itself oscillating!dαα α 0 dr r r dα+ ( − e)α 0 +dr⋅( ( ))r A ⋅cos 2⋅π⋅ν vib ⋅tSo the induced dipole moment is oscillating in a morecomplex way:m E 0 ⋅α 0( )⎛⎝( )dα⋅cos 2⋅π⋅ν ex ⋅t⋅⎜α 0 +dr ⋅rA⋅cos 2⋅π⋅ ν vib⋅tThe dynamic induced dipole that gives rise to Raman scattering:⎞⎟⎠Incident E-fieldOscillatingpolarizabilityAmplitudemodulation ininduced dipolePage 264 of 316

Chem 155 Unit 15 265 of 316The classical model of Raman:The induced dipole moment has a complexdescription, but how does this relate to Ramanscattering?Recall that:1cos( x) ⋅cos( y)⋅cos( x + y)+1⋅cos(x − y)2 2So m becomes:m E 0 ⋅α 0 ⋅cos( 2⋅π⋅ν ex ⋅t)...1E dα+ ⋅2 0⋅⋅r dr A cos 2⋅π⋅( ν ex + ν vib )⋅( )⋅⎡ ⎣ ⎡ ⎣ ⎡⎣ ⎤ ⎦+ cos 2 ⎡⎣⋅πν ex − ν vib ⎤ ⎦ ⎤ ⎦Both Stokes and AntiStokes peaks are predicted by classical theory!Light Power1.6 .10 4 1.68 .10 4 1.76 .10 4 1.84 .10 4 1.92 .10 4 2 .10 4 2.08 .10 4 2.16 .10 4 2.24 .10 4 2.32 .10 4 2.4 .10 4Wavelength / nmbut…Page 265 of 316

Chem 155 Unit 15 266 of 316The classical model: catastrophe!Real intensity distributionsLight Power1.6 .10 4 1.68 .10 4 1.76 .10 4 1.84 .10 4 1.92 .10 4 2 .10 4 2.08 .10 4 2.16 .10 4 2.24 .10 4 2.32 .10 4 2.4 .10 4Frequency / cm-1The classical model does not predict the Stokes /anti-Stokes intensity ratios.With what does an equal intensity distribution violate?Boltzmann Distribution between v=0 andv=1 vibrational levels.So, the answer is: the quantum model wins, and weaccept the ‘virtual state’. I’m told that time-dependentsolutions to the molecular structure allow for theseevanescent states.moreThe Stokes peaks areanti Stokespeaksbecause:intense than thethere are more molecules in theground vibrational state, that cangive rise to Stokes peaks, than inthe excited vibrational state, thatcan give rise to anti-stokes peaks.Page 266 of 316

Chem 155 Unit 15 267 of 316Raman Activity:modeΔμ ΔαRamanactive?IR active?OCSymmetricstretchNOYES YES NOOOCAsymmetricstretchYESNO NO YESOOCBendYESYES YES YESOPage 267 of 316

Chem 155 Unit 15 268 of 316Also note that:Purely centro-symmetric modes are only Ramanactive, but IR inactive.Most modes are both Raman and IR allowed.Many allowed modes may be too weak to detect ineither Raman or IR or both.Intensity distributions between Raman and IR oftendiffer substantially.Page 268 of 316

Chem 155 Unit 15 269 of 316Some general points regarding Raman:Pro: Raman uses visible light:OpticsSolventsBiological samplesImaging with Raman has better spatial resolutionbecause of the shorter wavelengths used.Microscopes can be used in ‘Raman microprobe’mode.Silver and gold particles or rough surfaces can verystrongly enhance Raman scattering.Different vibrational modes can be seen (e.g.centrosymmetric)1ν vib ⋅2πkμVery low frequency modes are accessible (e.g.inorganics):Depolarization ratio can distinguish betweensymmetric and asymmetric stretches.Page 269 of 316

Chem 155 Unit 15 270 of 316Con: Raman signals are often very weak.High-powered lasers can damage samples, so this isnot always a good solution.Fluorescent molecules can give bad spectralinterference for Stokes lines.Page 270 of 316

Chem 155 Unit 15 271 of 316Resonance RamanS1SoPro’s1. Intensity greatly enhanced (10 2 – 10 6 )2. Selective for vibrations of chromophoreCon’s1. Sample degradation because of absorption.2. Fluorescence can swamp Raman signals.Fluorescence can be rejected with a very short pulsed laser source:‘typical’ fluorophore - τ FLUOR ~ 10 -9 s.Raman -τ RAMAN ~ 10 -14 SPage 271 of 316

Chem 155 Unit 15 272 of 316Raman ExercisesGiven the spectrum below, compute the following:.1. The laser wavelength.2. The vibrational frequencies of the modes in the molecule givingrise to the following spectrum.3. The effective bandwidth is needed in the visible wavelengthmonochromator to achieve 1 cm -1 resolution with Raman giventhe above laser system and chromophore.Hypothetical Raman Scattering Spectrum .Light Power400 425 450 475 500 525 550 575 600 625 650Wavelength / nmlaser frequency:Stokes frequency:10 7500= 20000 cm − 110 7560= 17857 cm − 110 7 10 7− = 2143 cm − 1 Effective bandwidth: Must resolve 20000 and 20001 cm-1.500 56010 7 10 710 7 10 7− = 0.025 nm high resolution needed!− = 2759 cm − 120000 20001500 580Page 272 of 316

Chem 155 Unit 15 273 of 3164. Using and Ar-ion laser at 514.5nm, at what wavelength wouldyou expect to see the Stokes peak for the NO2 resonance ofnitrobenzene given that it is normally observed at 1520 cm -1 inIR absorption spectroscopy?10 7514.5− 1520 = 17916 cm − 1 10717916= 558.2 nm5. Does the CO bond in CO 2 have a dipole moment? YES6. Does the CO 2 molecule have a dipole moment? Why?No, because individual C=O bond dipoles exactly cancel.7. Draw a series of CO 2 molecules to illustrate the symmetricstretching mode.O C O O C O O C O8. Is the CO 2 symmetric stretch IR active? Why?No, because at no time during the vibration does the dipolemoment change.9. Is the CO 2 symmetric stretch Raman active? Why?Yes, because during the vibration the polarizabilitychanges.Page 273 of 316

Chem 155 Unit 15 274 of 31610. Draw a series of CO 2 molecules to illustrate theasymmetric stretching mode.O CO O C OOCO11. Is the asymmetric stretch IR active? Why?Yes, because at during the vibration the dipole momentchanges.12. Is the asymmetric stretch Raman active? Why?No, because at during the vibration the polarizability doesnot change. (increase in one CO bond polarizability isoffset by decrease in the other bond).13. Why is N 2 not a greenhouse gas?Homonuclear diatomics have no dipole moment changeduring their only vibrational mode: symmetric stretch.Page 274 of 316

Chem 155 Unit 15 275 of 31614. Cyclohexanone has a strong absorption peak at 5.86μmand at this wavelength a linear relationship exists betweenabsorbance and concentration.a. Calculate the frequency of this mode in cm -1 .b. Identify the part of the molecule responsible (the groupfrequency) for the absorbance at this wavelength.This is the ‘carbonyl’ or ketone symmetric stretch.c. Suggest a solvent that would be suitable for a quantitativeanalysis of cyclohexanone at this wavelength.CCl 4 , CHCl 3 or tetrachloroethylened. A solution of cyclohexanone (2.0 mg/mL) in the selectedsolvent has an absorbance of 0.40 in a cell with a pathlength of 0.025 mm. What is the detection limit for thiscompound under these conditions, if the noise associatedwith the spectrum of the solvent is 0.001 absorbanceunits.A=εbC calibration slope = εb10 45.86= 1706 cm − 1εbAC− 1⎞ ⎟⎠0.402.00 mg0.2 mg3s ⋅⎛b= ⎜C⎝ mLmm⋅mL3⋅0.001⎛⎝0.2⋅⎜mgmL⎞⎟⎠=− 10.015 mgmLPage 275 of 316

Chem 155 Unit 15 276 of 31615. Calculate the ratio of HCl molecules in the first vibrationalexcited state at 25°C relative to the ground state. The IRabsorption appears at 2885 cm -1 .N1 g1kTa. The Boltzmann equation is:N=g 0⋅ e where g00 and g 1are the degeneracies of the states 0 and 1 and for HCl 35may be considered equal, and k is Boltzmann’s constant,T is absolutetemperatureand ΔE is theenergy ofstate 1relative to 0.k 1.38⋅10 − 23 J:= ⋅K−ΔET :=298⋅KkT ⋅ = 4.112× 10 − 21 J2885⋅ cm − 1 ⋅6.626⋅10 − 34 ⋅J⋅s⋅3.0010 10 cm⋅ ⋅ = 5.735× 10 − 20 JsΔE 2885⋅ cm − 1 ⋅6.626⋅10 − 34 ⋅J⋅s⋅3.00⋅10 10 cm:=⋅sN1N0− ΔEkTe− ΔEkT ⋅e=8.784×10 − 7b. Assumingthat the bondstrengths areequal,calculate thevibrationalfrequencythat youexpect for theisotope DCl 35 .1ν VIB ⋅2⋅π1⋅ν DCl 2⋅πν HCl1⋅2⋅πkμk DClμ DClk HClμ HClsinceμk HClm 1 ⋅m 2m 1 + m 2k DClν DClν HClμ HClμ DCl35⋅1μ HCl :=35 + 135⋅2μ DCl :=35 + 2μ HCl= 0.717μ DClν DCl := 2885⋅0.717ν DCl = 2069Page 276 of 316

Chem 155 Unit 15 277 of 31616. Draw three simple Jablonski diagrams, e.g. for a simplemolecule and indicating just two electronic states S0 and S1(no T state is needed).a. On the first diagram, draw arrows corresponding to lightabsorption followed by fluorescent light emission.b. On the second, draw arrows corresponding to the StokesRaman peaks.c. On the third, draw arrows corresponding to the Anti-Stokes Raman process.S1Virtual stateS0Fluorescence Raman-Stokes Raman-AntiStokesPage 277 of 316

Chem 155 Unit 15 278 of 31617. A peak is observed in a fluorescence spectrometer whenanalyzing trace impurities in water. With the monochromatorexcitation source set to 250 nm, the mysterious weak peakappears at 274 nm, unusually close to the excitation band.When the excitation wavelength was changed to 300 nm, themystery bands moved to 335 nm.a. Calculate the Stokes shift for the mystery band in cm -1 forthe 250 nm excitation experiment.b. Calculate the Stokes shift for the mystery band in cm -1 forthe 300 nm excitation experiment.c. What might this ‘mystery peak’ be?10 7 10 7− = 3504250 27410 7 10 7− = 3483300 335The band moves with the excitation – it is the Ramanpeak for water OH stretch modes.Page 278 of 316

Chem 155 Unit 16 279 of 316Mass Spectrometry (MS) overview:Skoog: sections 20A and BIn MS onea. often initially performsb. gets analyte molecules into thec. converts analyte molecules intod. that are bothe. exposes the analyte ions tof. separates them based ong. detects them based onGC or LC separationgas phase in vacuumionsintact and in fragmentsE and/or B fieldsmass / chargecurrent (they are ions)Example: of a GCMS instrument:Page 279 of 316

Chem 155 Unit 16 280 of 316Block diagram of MS instrument.SampleIntroductioIonizationSourceMassAnalyzerPage 280 of 316

Chem 155 Unit 16 281 of 316Information from ion massMass spectra can distinguisho Different isotopes of the same element.o Small mass differences between molecules with the samenominal molecular weight.This makes mass spectra complex because:There are peaks for every isotopomer!1H 35 Cl, 2 H 35 Cl, 1 H 37 Cl, 2 H 37 ClBut this can be informative: Isobaric interferantsAsparagine – 13 CH 2 CONH 2 (133.12)Aspartic acid – CH 2 CO 2 H (133.10)Distinguishable only by high resolution MSPage 281 of 316Source: Quantitative Chemical Analysis 6 th Ed. By Daniel Harris

Chem 155 Unit 16 282 of 316Ionization Sources1.1.62 Electron Impact (EI):‘Hard’ ionization:extensive fragmentation1.1.63 Chemical Ionization:Derivative of EI but with 1000:1 excess of:reagent gas e.g.CH 4‘Soft’ ionization:limited fragmentationCH 4 collision with electron CH 4+ , CH 5+ , CH3 +CH 4+ , CH 5+ , CH3 + recombination, rxn w/ CH 4 C 2 H 5+CH 5+ + MH MH 2+ Proton transfer M+1+ CH 4C 2 H 5+ + M M C 2 H 5+ethyl transfer M+15Page 282 of 316

Chem 155 Unit 16 283 of 316Example of CI vs EI MS of 1-octanol:Electron impactionizationchemicalionization (CH 4 )Which is a ‘better’ MS?CI:EI:more easily interpretablecomplex, but highly uniquePage 283 of 316

Chem 155 Unit 16 284 of 3161.1.64 Matrix Assisted Laser Desorption-Ionization(MALDI):o ‘Soft ‘o Pulsed introduction of solidso Large molecules – proteins / biomoleculeso Gives predominance of M +sample invacuumchamberproteins inmatrixmatrixionizessupersonicexpansionPage 284 of 316

Chem 155 Unit 16 285 of 3161.1.65 Electrospray Ionization (ESI):o ‘Soft ‘o Continuous introduction of solidso Large molecules – proteins / biomoleculeso Gives many different charge states: M + , M +2 , M +3 ,M +4 , M +5 etc.What happens when charged droplets dry out?Page 285 of 316

Chem 155 Unit 16 286 of 316Which is a ‘better’ ionization source for large molecules,MALDI or ESI?MALDI:ESI:spectra are more easilyinterpretablespectra are complex, but ESI easilyinterfaced to LCPage 286 of 316

Chem 155 Unit 16 287 of 316Mass Analyzers:One important quality of mass analyzers is:Resolution: R ≡M/ΔMSmall molecule MS may demand high resolution:For example, can one distinguish C 2 H 4+ , CH2 N + , N 2+ , CO +in most mass spectrometers?Ion Nominal Mass Exact Mass12 C21 H4+28 28.031312 C 1 H214 N +14 N2+12 C 16 O +28 28.018728 28.006128 27.9949What resolution would be needed to distinguish these?Smallest ΔM =28.0061-27.9949 = 0.012m / ΔM =28 / 0.012 = 2300What resolution would one need to calculate the nominalmass of immunoglobulin-G (M = 149190)?m / ΔM =149,190 / 1 = 150,000Page 287 of 316

Chem 155 Unit 16 288 of 3161.1.66 Magnetic Sectors:Resolution of a single magnetic sector is ca.2000The properties that limit R of the ions entering it, mainly:speeddistributiondirectionaldistributionPage 288 of 316

Chem 155 Unit 16 289 of 3161.1.67 Double-Focusing Electric Magnetic SectorsElectric Sector Focuses:Speed distributionMagnetic Sector Re-focuses:Directional distributionAt double-focus point, all ions have narrow speed anddirection distributions, so only ions of a given M/z exit atthat point. What determines the M/z values that exit?E-field on electricsectorB-field on magneticsectorR attainable:100,000Page 289 of 316

Chem 155 Unit 16 290 of 3161.1.68 Quadrupole Mass Filters:10,000Resolution up to:Masses limted to:< 4000 AMUStable trajectory is a function of:AC amplitudeDC fieldA given AC amplitude + DC fieldyields a stable trajectory for one:Like E and B sectors this is a:One ‘scans’ m/z:M/zmass filterscanning mode (seeone at a time)Page 290 of 316

Chem 155 Unit 16 291 of 3161.1.69 Time of Flight (TOF):Ion kinetic energy E = zV PULSEBut kinetic energy E =½mv 2So, a drift length L gives a the “flight” time t F =L/vSolve for m/Z as a function of V PULSE and distance L:E = zV PULSE = ½mv 2 = ½m(L/t F ) 22V PULSE (t F /L) 2 = m/z = (2V/L²) t F2So, one measures ion m/z by measuring the ion’sArrival time at the detector: t FPage 291 of 316

Chem 155 Unit 16 292 of 316TOF analyzers:1. High mass capable.2. Fast – 100 of spectra per second.3. Efficient / sensitive – all ions are collected.4. Good Resolution up to 25000.Mass Spec Questions:1. What might you label the axes on a mass spectrum?2. Consider a fragment of nominal mass 180 AMU in amass spectrometer. What peaks might arise from thisone fragment?3. Consider a high resolution mass spectrum of the singlycharged fragment CHCl 3 + . Though varying greatly inintensity, it will be possible to resolve 12 differentpeaks for this fragment. Assign the isotopomers andpredict the masses.4. Forensic evidence was gathered against U.S. cyclistFloyd Landis was based on a high testosterone levelseen in a blood sample. Since testosterone is anaturally occurring hormone, itwas necessary to analyze thisfurther. Testosterone madeHby biosynthesis in the humanbody differs in only one wayH Hfrom that prepared in alaboratory – 13 C/ 12 OC isotopicabundances. Whatresolution is required tomeasure this ratio?testosteroneOHC 19 H 28 O 2Exact Mass: 288.21Mol. Wt.: 288.42m/e: 288.21 (100.0%), 289.21 (20.6%), 290.22 (2.1%)C, 79.12; H, 9.78; O, 11.09Page 292 of 316

Chem 155 Unit 16 293 of 3165. What type of mass analyzer might be useful for this?6. Would it be necessary to perform a separation stepbefore analyzing Landis’ blood for testosterone by MS?7. Chemists are very sloppy in the way we use the word‘resolution’ – for example, the GCMS that we use has‘unit mass resolution’ on the quadrupole analyzer.What does this mean?8. In the description oftestosterone above, can youdistinguish between the exactmass and the molecularweight?9. To what do the 20% and 2% fragments correspond?C 19 H 28 O 2Exact Mass: 288.21Mol. Wt.: 288.42m/e: 288.21 (100.0%), 289.21 (20.6%), 290.22 (2.1%)C, 79.12; H, 9.78; O, 11.09Element Iso %Carbon12 C 98.90% 13 C 1.10%Hydrogen1 H 99.99% 2 H 0.01%Oxygen16 O 99.7617 O 0.038% 18 O 0.2%Page 293 of 316

Chem 155 Unit 17 Chromatography Page 294 of 316Chromatography – Chapter 26General and descriptive aspects of chromatographicretention and separation: phenomenological k’,efficiency, selectivity.Quantitative description of zone migration in partitionchromatography: migration velocity, partitioncoefficient and theoretical k’.Theoretical description of efficiency: zone broadeningand the Van Deemter equation.Solving the general elution problem: Gradient vs.isocratic elutions.HPLC Instrumentation: pump, injector, column,detector, data collection.Chromatographic modes: HPLC (reverse and normalphase or ‘flash’), gel permeation (GPC), gel filtration,ion exchange.Electrophoresis.Page 294 of 316

Chem 155 Unit 17 Chromatography Page 295 of 316Chromatography: separation of chemicals in a mobilephase by differential flow rates through stationaryphase.Skoog Holler and Nieman Principles of Instrumental Analysis 5 th ed.Page 295 of 316

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Chem 155 Unit 17 Chromatography Page 303 of 316Theoretical Plates and Plate Count:x i := i100 21000( )S x, μ , σ1 ⎡ −1exp⎢⋅(2⋅π⋅σ2⋅σ 2 x − μ) 2 ⎤:= ⋅50⎥2⎣⎦1000= 10= 2.525 2= 0.625100012.5 2= 0.1561000N:2⎛ 1000 ⎞⎜ ⎟ = 100⎝ 100 ⎠2⎛ 1000 ⎞⎜ ⎟ = 1600 N⎝ 25 ⎠2⎛ μ ⎞⎜ ⎟⎝ σ ⎠2⎛ 1000 ⎞⎜ ⎟ = 400⎝ 50 ⎠2⎛ 1000 ⎞⎜ ⎟ = 6400⎝ 12.5 ⎠Illustration of different plate counts0.030.0256400 platessignal0.020.0151600 plates0.01400 plates0.005100 plates00 200 400 600 800 1000 1200 1400 1600 1800 2000retention time or column lengthPage 303 of 316

Chem 155 Unit 17 Chromatography Page 304 of 316Plate count depends on retention time:Why does the plate count go up for the more retainedpeaks?Because more retained peaks are expected to getbroader by longitudinal diffusion – but these donot.N:2⎛ 400 ⎞⎜ ⎟⎠ = 1024⎝ 12.52⎛ 1200 ⎞⎜ ⎟ = 9216 N⎝ 12.5 ⎠2⎛ μ ⎞⎜ ⎟⎝ σ ⎠2⎛ 800 ⎞⎜ ⎟⎠ = 4096⎝ 12.52⎛ 1600 ⎞⎜ ⎟ = 16384⎝ 12.5 ⎠Illustration of different plate counts0.030.0250.02signal0.0150.010.00500 200 400 600 800 1000 1200 1400 1600 1800 2000retention time or column lengthPage 304 of 316

Chem 155 Unit 17 Chromatography Page 305 of 316The van Deemter Equation:H = plate heightA = ‘multipath term’B = longitudinal diffusion termC = resistance to mass transferPlate height as a function of linear flow rate:H A+ B + C⋅uu12108H12 ( , , 0.5,u)6420 2 4 6 8 10 12uRationalize why is there an optimum flow rate:Excessive flow rate can exceed the rate ofpartitioning into and out of the stationary phase.Very slow flow allows too much time fordiffusional broadening.Page 305 of 316

Chem 155 Unit 17 Chromatography Page 306 of 316H A+ B + C⋅uuA2λ ⋅d pmultipath termB2γ ⋅D Mlongitudinal diffusionC = complex function of particlesize, coating, diffusion coefficient,favored in general by largediffusivity and small particle sizeresistance tomass transferPage 306 of 316

Chem 155 Unit 17 Chromatography Page 307 of 316General Elution Problem / Gradient ElutionA mixture of molecules with strongly differing k’ valuesis hard to separate because:If the solvent is weak enough to separate 1 and 2:a. Separation takes too longb.Peaks 5 and 6 broaden into baselineIf the solvent is strong enough to elute 5 and 6 in areasonable time:Peaks 1 and 2 and 3 and 4 co-elutePage 307 of 316

Chem 155 Unit 17 Chromatography Page 308 of 316Isocratic (LC) and isothermal (GC) separations arevulnerable to the general elution problem.The solution to the general elution problem:o solution gradient elutionchromatography)(liquido temperature gradient elution (gaschromatography)The gradient methods:o begin with weakly eluting conditions and thishelps to separate the weakly retained specieso finish with strongly eluting conditions toexpedite the elution of strongly retained speciesand to limit diffusional broadeningIn LC gradient elution is a bit more difficult because:o the columns require a fairly significantequilibration time between runso the detector baseline can drift significantly if thedifferent solvents have different UV absorbanceat the detection wavelength – if this is a problem,then one can useo a different wavelengthoro a different set of solventsPage 308 of 316

Chem 155 Unit 17 Chromatography Page 309 of 316T-gradient example in GC of a complex mixture.Isothermal:45°CIsothermal:145°CT-programmed:30° - 180° CPage 309 of 316

Chem 155 Unit 17 Chromatography Page 310 of 316High Performance Liquid ChromatographySkoog Holler and Nieman Principles of Instrumental Analysis 5 th ed.Page 310 of 316

Chem 155 Unit 17 Chromatography Page 311 of 316Types of Liquid ChromatographyMost can be performed as HPLC in ‘highperformance’ instrumentation if necessary.Normal Phase:stationary phase: hydrophilic (SiO 2 particles)mobile phase: hydrophobic mobile phase(hexane, methanol).weak solvent: hexanestrong solvent: methanolretained: polar compoundseluted first: non-polar compounds1.1.70 Reverse Phase:stationary phase: hydrobic (SiO 2 particlescoated with hexadecane monolayer)mobile phase: hydrophilic mobile phase (water,methanol)weak solvent: waterstrong solvent: methanolretained: non-polar compoundseluted first: polar compoundsPage 311 of 316