Chemistry 155 Introduction to Instrumental Analytical Chemistry

Transcription

1 Chem 155 Unit 1 Page 1 of 316 Chemistry 155 Introduction to Instrumental Analytical Chemistry Unit 1 Spring 2010 San Jose State University Roger Terrill Page 1 of 316

2 Chem 155 Unit 1 Page 2 of Overview and Review Propagation of Error Introduction to Spectrometric Methods Photometric Methods and Spectroscopic Instrumentation Radiation Transducers (Light Detectors): Monochromators for Atomic Spectroscopy: Photometric Issues in Atomic Spectroscopy Practical aspects of atomic spectroscopy: Atomic Emission Spectroscopy Ultraviolet-Visible and Near Infrared Absorption UV-Visible Spectroscopy of Molecules Intro to Fourier Transform Infrared Spectroscopy Infrared Spectrometry: Infrared Spectrometry - Applications Raman Spectroscopy: Mass Spectrometry (MS) overview: Chromatography Page 2 of 316

3 Chem 155 Unit 1 Page 3 of Overview and Review Tools of Instrumental Analytical Chem Instrumental vs. Classical Methods Vocabulary: Basic Instrumental Vocabulary: Basic Statistics Review Statistics Review Calibration Curves and Sensitivity Vocabulary: Properties of Measurements Detection Limit Linear Regression Experimental Design: Validation Assurance of Accuracy: Spike Recovery Validates Sample Prep Reagent Blanks for High Accuracy: Standard additions fix matrix effects: Internal Standards Propagation of Error Introduction to Spectrometric Methods Electromagnetic Radiation: Energy Nomogram Diffraction Properties of Electromagnetic Radiation: Photometric Methods and Spectroscopic Instrumentation General Photometric Designs for the Quantitation of Chemical Species Block Diagrams Optical Materials Optical Sources Continuum Sources of Light: Line Sources of Light: Laser Sources of Light: Radiation Transducers (Light Detectors): Desired Properties of a Detector: Photoelectric effect photometers Limitations to photoelectric detectors: Operation of the PMT detector: PMT Gain Equation: Noise in PMT s and Single Photon Counting: Semiconductor-Based Light Detectors: Charge Coupled Device Array Detectors: Monochromators for Atomic Spectroscopy: Adjustable Wavelength Selectors Monochromator Designs: The Grating Equation: Dispersion Angular dispersion: Page 3 of 316

4 Chem 155 Unit 1 Page 4 of Effective bandwidth Bandwith and Atomic Spectroscopy Factors That Control EFF Resolution Defined Grating Resolution Grating Resolution Exercise: High Resolution and Echelle Monochromators Photometric Issues in Atomic Spectroscopy Practical aspects of atomic spectroscopy: Nebulization (sample introduction): Atomization Flame Chemistry and Matrix Effects Flame as sample holder : Optimal observation height: Flame Chemistry and Interferences: Matrix adjustments in atomic spectroscopy: Atomic Emission Spectroscopy AAS / AES Review: Types of AES: Inert-Gas Plasma Properties (ICP,DCP) Predominant Species are Ar, Ar +, and electrons Inductively Coupled Plasma AES: ICP-AES ICP Torches Atomization in Ar-ICP Direct Current Plasma AES: DCP-AES Advantages of Emission Methods Accuracy and Precision in AES Ultraviolet-Visible and Near Infrared Absorption Overview The Blank Theory of light absorbance: Extinction Cross Section Exercise: Limitations to Beer s Law: Noise in Absorbance Calculations: Deviations due to Shifting Equilibria: Monochromator Slit Convolution in UV-Vis: UV-Vis Instrumentation: Single vs. double-beam instruments: UV-Visible Spectroscopy of Molecules Spectral Assignments Classification of Electronic Transitions Spectral Peak Broadening Aromatic UV-Visible absorptions: UV-Visible Bands of Aqeuous Transition Metal Ions Charge-Transfer Complexes Page 4 of 316

5 Chem 155 Unit 1 Page 5 of Lanthanide and Actinide Ions: Photometric Titration Multi-component Analyses: Intro to Fourier Transform Infrared Spectroscopy Overview: molecular vibrations IR Spectroscopy is Difficult! Monochromators Are Rarely Used in IR Interferometers measure light field vs. time The Michelson interferometer: How is interferometry performed? Signal Fluctuations for a Moving Mirror Mono and polychromatic response Interferograms are not informative: Transforming time frequency domain signals: The Centerburst: Time vs. frequency domain signals: Advantages of Interferometry Resolution in Interferometry Conclusions and Questions: Answers: Infrared Spectrometry: Absorbance Bands Seen in the Infrared: IR Selection Rules Rotational Activity Normal Modes of Vibration: Group frequencies: a pleasant fiction! Summary: Infrared Spectrometry - Applications Strategies used to make IR spectrometry work Solvents for IR spectroscopy: Handling of neat (pure no solvent) liquids: Handling of solids: pelletizing: Handling of Solids: mulling: A general problem with pellets and mulls: Group Frequencies Examples Fingerprint Examples Diffuse Reflectance Methods: Quantitation of Diffuse Reflectance Spectra: Attenuated Total Reflection Spectra: Raman Spectroscopy: What a Raman Spectrum Looks Like Quantum View of Raman Scattering Classical View of Raman Scattering The classical model of Raman: The classical model: catastrophe! Page 5 of 316

6 Chem 155 Unit 1 Page 6 of Raman Activity: Some general points regarding Raman: Resonance Raman Raman Exercises Mass Spectrometry (MS) overview: Example: of a GCMS instrument: Block diagram of MS instrument Information from ion mass Ionization Sources Mass Analyzers: Mass Spec Questions: Chromatography Chapter General Elution Problem / Gradient Elution T-gradient example in GC of a complex mixture High Performance Liquid Chromatography Types of Liquid Chromatography Normal Phase: HPLC System overview: Example of Reverse-phase HPLC stationary phase: Ideal qualities of HPLC stationary phase: Page 6 of 316

7 Chem 155 Unit 1 Page 7 of Overview and Review Skoog Ch 1A,B,C (Lightly) 1D, 1E Emphasized Analytical Chemistry is Measurement Science. Simplistically, the Analytical Chemist answers the following questions: What chemicals are present in a sample? QUALITATIVE ANALYSIS At what concentrations are they present? QUANTITATIVE ANALYSIS Additionally, 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 in solution dissociate? At what temperature does this chemical decompose? Myriad questions about chemical states Page 7 of 316

8 Chem 155 Unit 1 Page 8 of Tools of Instrumental Analytical Chem Spectroscopy w/ Electromagnetic (EM) Radiation Name of EM Wavelength Predominant Name of regime: Excitation Spectroscopy Gamma ray 0.1 nm Nuclear Mossbauer X-Ray 0.1 to 10 nm Core electron x-ray absorption, fluorescence, xps Vacuum nm Valence Vuv Ultraviolet electron Ultraviolet Valence Uv or uv-vis electron Visible Valence Vis or uv-vis electron Near Infrared 800-2,500 Vibration Near IR or NIR (overtones) Infrared m Vibration IR or FTIR Microwave 40 m 1 mm rotations Microwave 30 mm Electron spin in mag field Radiowave 1 m Nuclear spin in mag field Rotational or microwave ESR or EPR NMR Page 8 of 316

9 absorbance Chem 155 Unit 1 Page 9 of Chromatography Chemical Separations Different chemicals flow through separation medium (column or capillary) at different speeds plug of mixture goes in chemicals come out of column one-by-one (ideally) Gas Chromatography GC Powerful but Suitable for Volatile chemicals only Liquid Chromatography High Performance (pressure), HPLC in it s many forms Electrophoresis -Liquids, pump with electric current, capillary, gel, etc. Chromatogram time / s Page 9 of 316

10 Chem 155 Unit 1 Page 10 of Mass Spectrometry Detection method where sample is: volatilized, injected into vacuum chamber, ionized, usually fragmented, accelerated, ions are weighed as M/z mass charge. Often coupled to: chromatograph laser ablation atmospheric sniffer. Very sensitive (pg) quantitation Powerful identification tool Page 10 of 316

11 Chem 155 Unit 1 Page 11 of Electrochemistry Simple, sensitive, limited to certain chemicals Ion selective electrodes (ISE s): e.g. ph, pcl, po 2 etc. ISE s measure voltage across a selectively permeable membrane (e.g. glass for ph) E α log[concentration] ISE s have incredible dynamic range! ph 4 ph 10 [H+] = M Dynamic electrochemistry measure current (i) resulting from redox reactions at an driven by a controlled voltage at an electrode surface i(e,t) α [concentration] Gravimetry Precipitate and weigh products very precise, very limited Thermal Analysis Thermogravimetric Analysis TGA Mass loss during heating loss of waters of hydration, or decomposition temperature Differential Scanning Calorimetry DSC Heat flow during heating or cooling Page 11 of 316

12 Chem 155 Unit 1 Page 12 of Instrumental vs. Classical Methods. Separation Classical Methods of Analytical Chemistry # Chemicals Isolated / hr and amount (g) Extraction 1-2 g Distillation 1-2 g Precipitation 1-2 g Crystallization 1-2 g Instrumental High Performance Liquid Chromatrography Gas Chromatography Relax, you don t need to memorize this table just humor Dr. Terrill while he talks about it. # Chemicals Isolated / hr and amount (g) 10 ng 100 ng Electrophoresis 50 pg Qualitative Speciation Estimated Number of uniquely identifiable molecules by method Combination of Color / Smell Melt / Boiling Point, Solubility Wetting Density Hardness 100 s UV-Vis Infrared 1,000 s 100,000 s Mass Spectrometry > 10 6 NMR Spectroscopy Best Quantitative Precision and Sensitivity Titration 0.1 % 1 ppm > 10 6 Optical Spectroscopy 0.1% M Quantitation Precision Gravimetry 0.01 % 1 ppm Mass Spectrometry 0.1% amount 10-4 % mass M Colorimetry 10% 1 ppm NMR Spectroscopy 1% 100 pppm What are the more precise measurements that you have made and what were they? Page 12 of 316

13 Chem 155 Unit 1 Page 13 of Vocabulary: Basic Instrumental Analyte The chemical species that is being measured. Matrix The liquid, solid or mixed material in which the analyte must be determined. Detector Device that records physical or chemical quantity. Transducer The sensitive part of a detector that converts the chemical or physical signal into an electrical signal. Sensor Device that reversibly monitors a particular chemical e.g. ph electrode Analog signal A transducer output such as a voltage, current or light intensity. Digital signal When an analog signal has been converted to a number, such 3022, it is referred to as a digital signal. Analog signals are susceptible to distortion, and so are usually converted into digital signals (numbers) promptly for storage, transmission or readout. Page 13 of 316

14 Chem 155 Unit 1 Page 14 of Vocabulary: Basic Statistics Review Precision Random Error Accuracy Systematic Error (bias) Histogram Probability Distribution If repeated measurements of the same thing are all very close to one another, then a measurement is precise. Note that precise measurements may not be accurate (see below). Random error is a measure of precision. Random differences between sequential measurements reflect the random error. If a measurement of something is correct, i.e. close to the true value, then that measurement is accurate. Systematic error is the difference between the mean of a population of measurements and the true value. A graph of the number or frequency of occurences of a certain measurement versus the measurement value. A theoretical curve of the probability of a certain measured value occurring versus the measured value. A histogram of a set of data will often look like a Gaussian probability distribution. Average The sum of the measured. x values divided by the number n n of measurements. Median Variance ( 2 ) Standard Deviation ( ) Relative Standard Deviation Propagaion of Error x MEAN Half of the measurements fall above the median value, and half fall below. A measurement of precision. The sum of the 2 n squares of the random measurement errors. n 1 A widely accepted standard measurement of precision. The square root n of the variance. n 1 The standard deviation divided by the mean, and often expressed as : %RSD= /x MEAN 100%. When the mean of a set of measurement (x) has a random error ( x ), it is reported as x± x. If we wish to report the result of a calculation y=f(x) based on x, we propagate the error through the calculation using a mathematical method. n x MEAN x n 2 x MEAN x n 2 Page 14 of 316

15 Number of times it was observed Chem 155 Unit 1 Page 15 of Statistics Review Precision and Accuracy Histogram of normally distribut ed event s a Mean Mean - one standard deviation Mean + one standard deviation Value observed His togram of 1024 events Basic Formulae Mean : Average: lim x N N N N x N N x avg N Population Standard Deviation: Sample Standard Deviation: s x x 2 x N N lim N N 2 x N x avg N N 1 Bias or absolute systematic error = x avg s Relative standard deviation = x avg Page 15 of 316

16 Chem 155 Unit 1 Page 16 of 316 When you make real measurements of things you generally don t know the true value of the thing that you are measuring. (Call this the true mean,, for now. For the purposes of this discussion let us assume that there is no systematic (accuracy) error (i.e. no bias) Confidence Interval WHAT DO YOU DO TO ENSURE THAT YOUR ANSWER IS AS CLOSE AS POSSIBLE TO THE TRUTH? TAKE THE AVERAGE x AVG But, 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 within this 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: x The standard deviation: or s The number of measurements (observations) made: N This interval is called a confidence interval (CI). Which is better,a bigger or a smaller CI? Smaller is better How can you improve your CI? Make more measurements (N) Page 16 of 316

17 Chem 155 Unit 1 Page 17 of 316 Confidence Interval (continued). If you know the standard deviation,, (less common case), then: The x% confidence interval for = x AVG ± z / N ½ If you don t know the standard deviation,, (more commonly the case), then: If you have only a rough estimate of x AVG, then you are less confident that it is close to, hence you divide by N ½. The x% confidence interval for In this case t is a function of N = x AVG ± ts / N ½ This leaves only z and t what are they? These numbers represent the multiple of one standard deviation ( or s) that correspond to the confidence interval. In the second case, s is only an estimate of, so the error in s needs to be taken into account, so t is a function of the number of degrees of freedom. For our purposes, i.e. averaging multiple identical measurements, the number of degrees of freedom is simply N-1. Page 17 of 316

18 Chem 155 Unit 1 Page 18 of 316 An example: Assume that we do our best to measure the concentration of basic amines in a fish tank. Our answers are 4.2, 4.6, 4.0. N = 3 X AVG = ( ) / 3 = 4.27 s = ( ) 2 +( ) 2 +( ) = 0.31 Number of degrees of freedom for confidence interval = 3-1=2 95% confidence limits for = ± (4.3*0.31) / (3 1/2 ) = 4.27±0.93 or = 4.3 ± 0.9 or = 3.4 to Page 18 of 316

19 Chem 155 Unit 1 Page 19 of Confidence Interval (CI) in Words Consider this experiment. You have a camera device that measures the temperature of objects from a distance by measuring their infrared light emission. It is very convenient but somewhat 97.5, 96.0, imprecise. Assume for the moment that the camera is perfectly accurate that is, if you measure the same object with it many times the average temperature result will equal the true temperature. In order to evaluate the precision of your camera thermometer, you measure the temperature of each item three times. In each case you get an average and a standard deviation. PROBLEM: The average camera reading is sometimes higher than the true value, and sometimes lower, but you don t know how to evaluate this fluctuation. In just a few words, how can you characterize this fluctuation? SOLUTION: Calculate the sample standard deviation. QUESTION: A series of experiments, each of three measurements each yields a set of sample standard deviations that also different each time! If you repeat the whole experiment, but this time you measure each sample ten times, then the standard deviations are much closer, but still not equal each time. Why does the sample standard deviation calculation give a different result each time? Assume for the moment that the camera performance (precision) is not changing. ANSWER: Realize this fact: Sample standard deviations are only estimates. Page 19 of 316

20 Chem 155 Unit 1 Page 20 of CI PROBLEM: This imperfect camera thermometer is going to be used to screen passengers boarding an airliner. Passengers with a high temperature may have avian flu. Our criterion is this: if there is more than a 90% probability that a given passenger s temperature exceeds 102, then we will take him aside and test him for bird flu. We have only moments to acquire three measurements per passenger, so precision is low. Also, the precision is not the same each time. For three passengers we get the following results: Passenger 1: 100.3, 101.1, Passenger 2: 98.8, 98.5, 98.4 Passenger 3: 104.0, 103.9, How do we answer the question: does this person s temperature exceed our 90% / 102 criterion? To answer this, we must accept the following: Assuming that measurements are unbiased (accurate) we can state, for the 80% CI, that there is a 10% probability that the true mean lies below the lower limit of the CI, an 80% probability that the true mean lies within this CI, and a 10% probability that the true mean lies above this CI. So, there is a 90% probability that the true mean lies within or above the 80% CI. For example, if we took some measurements and then computed the 80% CI to be to then we could say that the probability that the true temperature is or higher is 90% % chance is or lower 80% chance < < % chance is or higher Page 20 of 316

21 Chem 155 Unit 1 Page 21 of 316 Use this table: % confidence interval freedom Use this formula: CI ± t s N Note the following: For a straight average of N points, the number of degrees of freedom is N-1. Complete the following table: Passenger T AVG ( F) St dev ( F) Lower boundary of 80% CI Upper boundary of 80% CI Is the probability that the passenger s Temp is > % or more? What is another way to word the conclusion above? Assuming our equipment is accurate, then averaged over many passengers, and using this criterion, our conclusion that the person s temperature is > 102 will be wrong less than 10% of the time! Page 21 of 316

22 Chem 155 Unit 1 Page 22 of t EXP PROBLEM: Another way of approaching this type of problem is to calculate an experimental value of t called t EXP. In the example below, we will compare a measured result with an exact one. The question one answers with t EXP is this: Am I confident that the observed value (x) differs from the expected value ( )? Our threshold temperature, exactly 102 was tested, so we can make measurements and test the hypothesis that the true temperature is greater than 102. Given the following three measurements of a passenger s temperature: , , calculate an experimental value of the t statistic for this population relative to the true value of 102. Average = , std dev = 1.34 % confidence interval freedom t exp x av N s = test value ( ) 3 t exp 1.34 t exp Can you state with the given confidence that this person s temperature differs from the expected value of 102? 99%? No 95%? No 80%? Yes 50%? Yes Page 22 of 316

23 Instrument Signal (S) Chem 155 Unit 1 Page 23 of Calibration Curves and Sensitivity Calibration Sensitvity: S = mc + S b S = instrument signal C = analyte conc. m = slope calibration Sb = signal for blank m = calibration sensitivity 10 A highly sensitive instrument can discriminate between small differences in analyte concentration. S C = m S i 0 S S 5 But, can we really distinguish C between small changes in concentration? 0 C i Analyte Concentration (C) 100 = Analytical Sensitvity = m / S = ( S/ C) / S = 1/ C for the C corresponding to S m and S so 1 / = noise in concentration C. So small is? Good Bad Page 23 of 316

24 Chem 155 Unit 1 Page 24 of Vocabulary: Properties of Measurements Sensitivity Selectivity Specimen Sample Analyte Calibration Curve Interferant Detection Limit C MIN or C M Limit of Quantitation (LOQ) Limit of Linearity (LOL) Linear Dynamic Range (LDR) A detector or instrument that responds to only a small change in analyte concentration is sensitive. Numerically, the sensitivity of an instrument is the slope of the calibration curve in untis of: signal / unit concentration often given as m. Sometimes the detection limit (see below) is called the instrument sensitivity but this is not correct. The ratio of the sensitivity of an instrument to an analyte to that of an interferant. The material removed for analysis e.g. a given tablet from an assembly line. The mixture that contains the chemical to be measured e.g. the blood sample that we want to measure iron in. The chemical that is being measured e.g. the iron in the blood sample. A linear or non-linear function relating instrument response (signal) to analyte concentration. Another chemical in the sample that either affects the instrument s sensitivity to the sample or gives a signal of it s own that may be indistinguishable from the analyte signal. The minimum detectable concentration of analyte. Usually defined as that concentration that gives a signal of magnitude equal to three times the standard deviation in the blank signal. The minimum concentration of analyte for which an accurate determination of concentration can be made. The LOQ is typically that concentration for which the signal is 10x the standard deviation of the blank signal. The largest concentration for which a calibration curve remains linear. The range of concentrations (or signal strengths) between the LOQ and the LOL. An instrument is most useful within its LDR. Page 24 of 316

25 Chem 155 Unit 1 Page 25 of Detection Limit The 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 change as: S M Therefore: S M = mc M must be a multiple (n) of the noise level in the blank: ( b ). S M 3 b = mc M Minimum Detectable Signal Signal due to blank Minimum detectable concentration By convention, m=3. Derive a formula for C M based on B and m: Derive a formula for C M based on : Page 25 of 316

26 Chem 155 Unit 1 Page 26 of 316 A Graphical Look at Detection Limit (C MIN ) Consider the minimum detectable signal in the context of the confidence interval: S MIN = 3 b + S b To what confidence interval does 3 correspond? Assume N=2 (two replicate measurements of S b ) Another way of saying this (crudely) is that we consider a signal detected when it falls outside of the boundaries corresponding to the 99% confidence interval for the blank signal. Page 26 of 316

27 Chem 155 Unit 1 Page 27 of Minimum Detectable Temperature Change? Using the thinking that we developed for the general case of signals with random error what do you think is the probability that the following signal change is due to random fluctuations? Page 27 of 316

28 Chem 155 Unit 1 Page 28 of 316 Dynamic Range LOQ = Limit of Quantitation ( S /S 0.3) LOL = Limit of Linearity Between LOQ and LOL your instrument is most useful! This is called: Linear Dynamic Range Page 28 of 316

29 Chem 155 Unit 1 Page 29 of 316 Selectivity Sometimes another chemical species will add to or subtract from your analyte signal S = m A C A + m B C B + m C C C + m D C D + + S B Analyte Interferants Blank Selectivity coefficients determine how serious an interferant is to your determination of analyte C A : k B,A = m B /m A, k C,A = m C /m A, etc So: S = m A (C A + k B,A C B + k C,A C C + ) + S B Page 29 of 316

30 Chem 155 Unit 1 Page 30 of Direct Interference In order to measure a concentration directly, without corrections, k B,A, k C,A must be approximately: If there is interference, it is necessary to know both: and before one can determine the desired quantity C A! Page 30 of 316

31 Chem 155 Unit 1 Page 31 of Linear Regression Least-Squares Regression or Linear Regression Assuming that a set of x,y data pairs are well described by the linear function y = a+bx, and assuming that most error is in y (x is more precisely known) and the errors in y are not a function of x, then the coefficients that minimize the residual function: 2 = (y I (a + bx I )) 2 can be found from the following equations: Non-linear Regression Standard Additions Plot Sample Matrix / Standards Matrix Matrix Effect Internal Standards from Data Reduction and Error Analysis for the Physical Sciences, Philip R. Bevington, cw Mc Graw Hill. Methods for fitting arbitrary curves to data sets. A nearly matrix-effect free form of analysis. A standard additions plot is a linear plot of instrument signal versus quantity of a standard analyte solution spiked or added to the unknown analyte sample. The unknown analyte concentration is derived from the concentration axis-intercept of this plot. The matrix is the solution, including solvent(s) and all other solutes in which an analyte is dissolved or mixed A matrix effect refers to the case where the instrumental sensitivity is different for the sample and standards because of differences in the matrix. A calibration method in which fluctuation in the instrument signals due to matrix effects are, ideally, cancelled out by monitoring the fluctuations in the instrument sensitivity to chemicals, internal standards, that are chemically similar to the analyte. Page 31 of 316

32 Instrument Signal (S) Chem 155 Unit 1 Page 32 of Linear Least Squares Calibration The most common method for determining the concentration of an unknown analyte is the simple calibration curve. In the calibration curve method, one measures the instrument signal for a range of analyte concentrations (called standards) and develops an approximate relationship (mathematically or graphically) between some signal S and analyte concentration C. If the signal-concentration relationship is linear, then: S = S B + mc y = a + bx But, one can not just draw the line between any two points because all the points have some error. So, one mathematically attempts to minimize the residuals. 2 = (y i (a + bx i )) S i 1 S i F i F i C i 10 Analyte Concentration (C) Page 32 of 316

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

34 Chem 155 Unit 1 Page 34 of 316 See 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 XX Also it is a somewhat more subtle problem to calculate the error in a concentration determined from a calibration curve of signal (y) versus concentration (x). You will need to follow Skoog appendix-a calculations to deal with this problem in your lab reports where you determine an unknown concentration. M = the number of replicate analyses, N = the number of data points. Equation a1-37 Skoog 5th edition. s c s y m 1 M 1 N yc avg y avb 2 m 2 S xx Note: when computing a confidence interval using this s C value, the degrees of freedom are N-2. (See for example Salter C., Error Analysis Using the Variance-Covariance Matrix J. Chem. Ed. 2000, 77, Page 34 of 316

35 Chem 155 Unit 1 Page 35 of Experimental Design: Designing an experiment involves planning how you will: make calibration and validation standards; prepare the sample for analysis; and perform the measurements Making a set of calibration standards. 1. You need to know the dynamic range of your instrument. 2. You need to know the sample size requirement of your instrument. 3. You need to know the estimated expected concentration of your sample. 4. You need to prepare and dilute your sample until it is a. within the dynamic range of your instrument and b. such that there is enough solution to measure. 5. You need to choose target standard concentrations that bracket the expected sample concentration generously e.g. by a factor of 2 to 3. For example, if your expected concentration is 5.3 ppm, you may wish to make a calibration set that consists of standards that are about 2,4,8,10 and 12 ppm. Page 35 of 316

36 Chem 155 Unit 1 Page 36 of You need to make a primary stock solution, the concentration of which you know accurately and precisely. This solution will usually be more than twice as concentrated as your most concentrated calibration standard. You will dilute this primary stock solution to make the calibration standards. You need to have enough to make all of your calibration standards. 7. You need to choose the pipets and volumetric flasks that you will use to perform the dilutions. This means that you plan the preparation of each standard. This takes some planning and compromising and many choices there are many ways to do this correctly there is more than one right answer! 8. Decide on and record a labeling system in your notebook, collect the glassware, and do the work. I have a labeling system for your caffeine, benzoic acid, iron and zinc standards I need you to use these labels so that we can sort things out in the class. Page 36 of 316

37 Chem 155 Unit 1 Page 37 of Exercise in planning an analysis: Assume that you will be analyzing sucrose in corn syrup sweetened ketchup packets. The packets are thought (i.e. expected) to contain about 0.8 grams of corn syrup that is about 70% sucrose by weight. The HPLC instrument that you will be using can detect sucrose by refractive index in the parts per thousand (ppth) range (this is the instrument s dynamic range for sucrose). You have pure sucrose for standards, and will be making five calibration standard solutions. The instrument requires between 250 and 1000 L of sample. You have the following glassware at your disposal: Pipets Volumetric Flasks Volume Relative Precision Volume Relative Precision L 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

38 Chem 155 Unit 1 Page 38 of Plan the analysis! (start with guesses ) 1. If you dissolve the packet in water, dilute it to ml and filter it, what will the approximate sucrose concentration be in ppth (parts per thousand)? 0.8 g syrup 0.7 g sucrose ml g syrup water 1000 mg sucrose g sucrose 1 ml water = 5.6 ppth 1 g water Is this within the dynamic range? Would it be better to use 10, 25 or 50 ml of water? 2. You need to prepare a stock solution of sucrose to make the calibration standards. What concentration 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? 1.0 ml 100 mg sucrose = 1 ml soution 100 mg sucrose = g sucrose 10 ml => 1.00 g sucrose 100 ml => 10.0 g sucrose Page 38 of 316

39 Chem 155 Unit 1 Page 39 of 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. without any loss, in the desired solvent. 3. Transfer this liquid quantitatively into the desired volumetric flask. 4. Dilute to volume with the desired solvent this process is important! It is often poor practice to add 5 ml of a to 5 ml of b and anticipate that the final volume will be exactly 10 ml! Remember the words DILUTE TO VOLUME! Page 39 of 316

40 Chem 155 Unit 1 Page 40 of What concentrations bracket the 5.6 ppth target? 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 , 0.50, 5.0, 50, 500 ppth Do you need to make standards on an even spacing? No but calibration points above and below the std. are needed. Does the analyte have to fall right in the middle of the calibration standards? No, but it minimizes error! 5. What volumes of standards should you prepare? How much is needed by the instrument? What is the smallest volume that you can conveniently 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 ml Page 40 of 316

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

42 Chem 155 Unit 1 Page 42 of How to plan a set of standard preparations in the MS Excel spreadsheet program: Calibration Standard Preparation: Target Conc: Stock Conc: Final Volume: Volume to Pipet: C2 / ppt C1 / ppt V2 / ml V1=C2*V2/C ml ml ml ml ml Calibration Stand Target Conc: Stock Conc: Final Volume: Volume to Pipet: C2 / ppt C1 / ppt V2 / ml V1=C2*V2/C =A4*C4/B4 ml =A5*C5/B5 ml =A6*C6/B6 ml =A7*C7/B7 ml =A8*C8/B8 ml These are formulas that you type into Excel normally only the result of the formula calculation is displayed. Page 42 of 316

43 Chem 155 Unit 1 Page 43 of Validation Assurance of Accuracy: Calibration and linear regression optimizes the precision of a calibration curve determination but a calibration curve can only be said to be accurate if: the analysis has been validated of saying this is that a method is valid if: all significant sources of bias have been removed Another way Validation addresses the various aspects of an analysis that can go wrong and give you a wrong answer. The table below lists some of the aspects of an analysis that can be invalid, and suggests ways to validate them. Source of bias: Analyst erratic Error in analyst technique Calibration standards are in error (are not what they say the are) Calibration method Instrument function erratic Instrument drift Possible Solution(s) or Diagnostic: Analyst can repeat experiment Different analyst does same analysis and gets same result. Entire analysis Independent standard repeated with measured periodically - indpendent called validation standard calibration or QC standard (quality standard set control) Different calibration method used (standard additions) Different instrument used (can be a different kind of instrument) Periodically measure validation standard or internal standard Page 43 of 316

44 Chem 155 Unit 1 Page 44 of 316 In general validation of an analysis is done using some kind of independent analysis of the sample. For the purposes of this class, if the two methods agree (t-test does not show a difference) then the method is said to be validated. If the two methods disagree (i.e. there is evidence of bias), then there is evidence for systematic error like a matrix effect, an interferant, or a mistake in the preparation of standards or samples. If an analysis is repeated as described below, what aspects of the analysis are validated (i.e. what must have been good for the answers to have come out the same)? Validation Approach Completely independent measurements of the same sample using a different instrument or technique. Use the same method, but measure independently prepared standards as samples. Aspect(s) validated Calibration method, including matrix effects. Instrument, including distortion and drift. You! The analyst, did it right You! The analyst made no mistakes in the implementation of the procedure. Page 44 of 316

45 Chem 155 Unit 1 Page 45 of Spike Recovery Validates Sample Prep. To do a spike recovery analysis, one takes replicate samples and to a subset of them adds a spike of analyte before the sample prep begins. So, for example, one could take four vitamin tablets, and divide them into two groups of two. To one group one could add some Fe, say half the amount originally expected. For example, if there is supposed to be 15 mg of iron in the tablet, one could spike two samples each with 5 mg of iron and leave two unspiked. Acids etc. digest filter Dilute to 100 ml volume Analyze 140 ppm 14.0 mg found Sample Sample mg spike digest filter Dilute to 100 ml volume Analyze 187 ppm 18.7 mg found = 4.7 mg of spike found: spike recovery percent = 4.7 / % What does this say about our sample preparation method? We may be losing about 6% of the analyte during sample prep. Page 45 of 316

46 Chem 155 Unit 1 Page 46 of Reagent Blanks for High Accuracy: A reagent blank is a blank that is made by doing everything for the sample prep etc. but without the sample: Acids etc used in sample prep. digest filter Dilute to 100 ml volume Analyze 2.3 ppm 0.23 mg Fe found No Sample sample. Ultrapure water blank In this example the reagent blank is analyzed against ultrapure water. The 0.23 mg Fe found in the reagent blank may be due to Fe impurities in the acids, but it also may be a matrix effect. In either case, it suggests that we should do what in order to arrive at a more accurate result? Either: a. analyze all standards and samples against reagent blanks or b. obtain higher purity acids c. subtract the signal from the reagent blank (with regulatory approval ) Page 46 of 316

47 Chem 155 Unit 1 Page 47 of Standard additions fix matrix effects: Use the method of standard additions when matrix effects degrade the accuracy of the calibration curves. There is one important assumption built into the calibration curve idea. That assumption is the following: the sensitivity of the instrument to the analyte in the standards is EQUAL TO the sensitivity of the instrument to the analyte in the sample matrix Why would the sensitivity be different? 1. Many Instruments are sensitive to things like: 1. ph 2. ionic strength 3.organic components of the solvent matrix 2. Sometimes other chemicals (interferants) can change the calibration sensitivity by: chemically binding to or interacting with the analyte atom/molecule These two things are examples of: Matrix Effects Page 47 of 316

48 Chem 155 Unit 1 Page 48 of How does one deal with matrix effects? 1. Make the sample and standard matrix as nearly identical as possible: matrix matching But matrix matching requires that you already know a lot about your unknown often not the case. So, the matrix-immune alternative to the calibration curve is to: 2. Use the method of: standard additions: In other words, the assumption built into the calibration curve method is that the matrix effects are negligible or identical for standards and samples. If this can not be assumed, one must match the sample and standard matrices. The way that the method of standards additions does this is to dilute both standards and samples: in the same matrix (solution) Page 48 of 316

49 Chem 155 Unit 1 Page 49 of An Example of Standard Additions: 1. The analyte sample is split up into e.g. 6 aliquots of identical, known volume e.g 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 dissolved in known matrix like water or 0.1M ph 7 phosphate etc. 3. Each aliquot is diluted to a total volume. 1. Add sample 2. Add standard 3. Dilute to volume, and mix mix mix! Note: you split and dilute your sample how does this impact the precision of your measurement? How does this process impact the accuracy of your measurement? Page 49 of 316 It may decrease it If dilution is too great!. It increases it!

50 Chem 155 Unit 1 Page 50 of Calculating Conc. w/ Standard Additions: a' a b' b One way of analyzing this uses similar triangles: a / b = a /b The y-axis absorbance signal (S) is proportional to the moles of analyte (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 the same amount of analyte as the sample. a is proportional to moles of analyte in the sample = V X C X a is proportional to moles of std added = V S C S b is (V S ) 0 the x-intercept of the graph b is V S the spike volume Substitute for a,a,b,b : V X C X (V S ) 0 = V S C S V S Solve for C X : C X = C S (V S ) 0 /V X Page 50 of 316

51 Chem 155 Unit 1 Page 51 of Standard Additions by Linear Regression: Let: V X = volume of unknown analyte solution added to each flask C X = concentration of unknown solution V T = final, diluted volume V S = volume of spike added to unknown soln. before dilution C S = concentration of analyte in spike solution Dilution 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 T Dilution Calc 2: Contribution to concentration of analyte from spike: V S C S /V T Total Signal (sensitivity = k) given that x variable is V S. S = k V X C X /V T + k V S C S /V T Slope = m = kc S /V T intercept = b = kv X C X /V T we can get b and m from linear regression and we want C X so b/m = = so : C X = bc S mv X kv X C X /V T kc S /V T V X C X /C S Page 51 of 316

52 Chem 155 Unit 1 Page 52 of Internal Standards Internal standards can correct for sampling, injection, optical path length and other instrument sensitivity variations. An internal standard is a substance added (or simply present) in constant concentration in all samples, and standards. When something unexpected decreases the sensitivity of the instrument (m) so the signal drops (S = mc + S B ) it can be impossible to distinguish this from a change in analyte concentration without an internal standard. Consider the ratio of the blank corrected analyte ( S' internal standard (IS) signals (S): S' S' AN IS m m AN IS C C AN IS C k C AN AN IS S AN s B ) to the Assumes for the moment that k is a constant, i.e. invariant to factors affecting overall instrumental sensitivity. As an example, let s consider k to be a correction for injection volume in a chromatographic system. It is perfectly reasonable to assume that an accidentally low or high injection volume would affect the internal standard and analyte signals identically e.g. if a given injection were 6% high, then both analyte and internal standard peaks would be 6% larger than expected. If k is invariant to instrument fluctuations, then the true analyte concentration can always be derived from the ratio of the corrected signals so long as the internal standard concentration remains constant. Anlalyte conc. in a sample. C AN S' S' AN IS C k IS C where IS is easily derived from a previously measured k S' A STD) and C ) of the analyte and internal standard ( calibration standard for which the analyte signal ( concentration ( S ', C IS STD IS STD A STD ) are known: k S' S' A STD IS STD C C IS STD A STD m m AN IS Page 52 of 316 From a calibration standard.

53 Chem 155 Unit 1 Page 53 of 316 In other words, if C IS is held constant in all experiments, then the ratio of the analyte to internal standard signals will be independent of instrument sensitivity. When to use an internal standard? When substantial influence on instrument sensitivity is expected due to variation in things like: Sample matrix Temperature Detector sensitivity Injection volume Amplifier (electronics) drift Flow rate An internal standard must: a. not interfere with your analyte b. 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 of analyte is sandwiched between two salt plates and this is placed in the IR beam. The path-length, b, is highly variable from experiment to experiment. The signal A = bc where is characteristic of the molecule, b is pathlength and C is concentration. If you are doing an experiment to measure the increase in amide formation versus time by the intensity of the amide bands near 1700 cm -1. You could take samples and measure them periodically, but the variability of the pathlength would distort the results. On the other hand, if all the samples were spiked with the same amount of acetonitrile then the sharp nitrile stretch at 2250 cm -1 could be used as an internal standard. Assuming you always spiked with the same amount of benzonitril, what simple metric, based on A 1700 and A 2250 would always be proportional to the amide concentration? In other words, simply by taking the ratio of the analyte to the internal standard peaks while holding the internal standard concentration constant, one can cancel a variable such as path-length! Page 53 of 316

54 absorbance absorbance absorbance Chem 155 Unit 1 Page 54 of 316 Internal Standards Example: In HPLC the sample is injected into a flowing stream, and the signal is 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 the internal standard. 100 ppm NaNO 3, 480 All mau. s Peaks 2, 3 and 4 are other ingredients in the sample. 1 Peak 5 is the analyte, a caffeine standard, 25 ppm, 530 mau. s 100 ppm NaNO 3, 520 mau. s time / s time / s Chromatogram 1 is a standard. Did the caffeine concentration increase from chromatogram 1 to 2? Did the caffeine concentration increase in chromatogram 1 to 3? Unknown caffeine conc. 630 mau. s time / s Page 54 of 316

55 Chem 155 Unit 1 Page 55 of Internal Standards Calculation How about calculating the actual concentration of the sample in 3 above? S' A STD 530 mau. s C A STD 25 ppm Standard S' IS STD 480 mau. s C IS STD 100 ppm k S S' ' A STD C C IS STD m m AN unitles s IS STD A STD IS S' AN 630 mau. s S' IS 520 mau. s Sample C IS 100 ppm C AN S S' ' AN IS IS C k ppm Page 55 of 316

56 Chem 155 Unit 2 Page 56 of Propagation of Error Skoog Chapters Covered: Appendix a1b-4 a1b-5 and eqn. a1-28 table a1-5 There is a general problem in experimental science and engineering: How to estimate the error in calculated results that are based on measurements that have error? Let s consider the sum of two measurements a and b that both have some 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 lb b = 5.0 lb Let s say that we are weighing a and b and putting them into a box for shipment. 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 at the same time because it has a limited capacity. So, we have to first weigh a, then b and then calculate the total weight. But we know that there is a problem with fluctuations, so we repeatedly weigh the same items a and b and do the following experiment: Characteristics of numbers in sum a and b Characteristics of sum c Trial # Weight of: Deviation: Total Deviation a b da db weight: from avg: average This is somewhat artificial and is not quite right, but it gives you the general idea. Errors in a and b propagate into c. Page 56 of 316

57 Real distribution theoretical theoretical Chem 155 Unit 2 Page 57 of 316 From this example above, one would conclude that the fluctuation in the sum is equal to the fluctuation in the numbers summed but this is not right. Which is bigger? a. the fluctuations in the individual numbers summed b. the fluctuations in the sum Graphically we consider here a similar case, for clarity we let a have a slightly larger fluctuation than b. a = 5 2, b = 3 1 a+sa+b+sb a+sa+b-sb a-sa+b-sb c a a+sa a-sa a-sa+b+sb Page 57 of 316

58 Chem 155 Unit 2 Page 58 of 316 error propagation through sums : a rnorm( ) b rnorm( ) c a b a b c mean( a) stdev ( a) mean( b) stdev ( b) mean( c) 6 stdev ( c) a_dist b_dist c_dist histogram ( 40 a) histogram ( 40 b) histogram ( 40 c) a_dist j 1 b_dist j c_dist j a_dist j 0 b_dist j 0 c_dist j 0 Page 58 of 316

59 Chem 155 Unit 2 Page 59 of 316 error propagation through products : c a b i i i mean( c) c stdev ( c) a_dist b_dist c_dist a_dist j 1 b_dist j c_dist j 1 hist ogram ( 40 a) hist ogram ( 40 b) hist ogram ( 40 c) a_dist j 0 b_dist j 0 c_dist j 0 Page 59 of 316

60 Chem 155 Unit 2 Page 60 of 316 For a simple example if you construct a simple calibration curve of instrument 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

61 Chem 155 Unit 2 Page 61 of 316 Let s consider an example wherein the relationship between the measured signal and the desired quantity is non-linear : Absorbance A = -log (P/Po) Absorbance is a nonlinear function of light power - the measured quantity in a spectrophotometric experiment. A log P P l A i P i Obviously the same error in P can give rise to different errors in A! This is a propagation of error problem. How can you calculate the error in A that should result from a particular error in P? Page 61 of 316

62 Chem 155 Unit 2 Page 62 of 316 Obviously, the answer has to do with the way that the dependent variable (A) in this case changes as a function of the independent variable (P). Consider a calculated value S that depends on the measured quantites, 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± a b± b c± c 2 S S a 2 2 S a b 2 2 S b c 2 2 c So let s take an example: S = a+b-c find S = f(s,a, a, b, b,c, c) Note that the larger terms dominate! Page 62 of 316

63 Chem 155 Unit 2 Page 63 of 316 Exercise: Prove the multiplication / division rule Show that for S = a*b/c S 2 a 2 b 2 c 2 S 2 a 2 b 2 c 2 Page 63 of 316

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

65 Chem 155 Unit 3 Page 65 of Introduction to Spectrometric Methods Skoog Chapter 6 all sections Electromagnetic Radiation (EMR) is: Light! EMR described as a classical, electromagnetic wave: distance In vacuum (air): n=1 In matter: n>1 = c = wavelength = frequency c = light speed MEDIUM = c MEDIUM MEDIUM = VAC / c MEDIUM = c VAC / Note: When the refractive index ( ) changes Stays same changes Page 65 of 316

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

67 Chem 155 Unit 3 Page 67 of 316 Energy Nomogram Energy nomogram - Hard UV 100 nm to Far IR 1mm nm cm -1 s E / ev E / kcal /mol 10-1 E / kj /mol 10 1 T / K Page 67 of 316

68 Chem 155 Unit 3 Page 68 of Diffraction Diffraction is the basis of wavelength selection in most spectrometers: d Page 68 of 316

69 Chem 155 Unit 3 Page 69 of Diffraction Exercises: 1. Derive a formula for the path difference, as a function of the spacing d, and the angle of incidence. (Use the diagram on the following page.) first note that = sin( ) = /d = dsin( ) 2. Under what conditions is the light intensity at the detector high or bright? = n n = -2,-1,0,1,2 i.e. any integer 3. Calculate the wavelengths of light that are diffracted at 48 for d = 1 micrometer (1 ). Page 69 of 316

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

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

72 Chem 155 Unit 3 Page 72 of Polarization: Most light is unpolarized Electric vector randomly oriented Special filters, called polarizers, can remove all E-field components except those falling in a given plane. The result is plane polarized light: All E-field lies in one plane Elliptically polarized light: E-field vector rotates around direction of propagation. Page 72 of 316

73 Chem 155 Unit 3 Page 73 of Refractive index: ratio of speed of light in vacuum to speed of light in medium. = c VAC / c MEDIUM = c / Refractive index, Wavelength Frequency s -1 Energy J nm 3x10 8 ms -1 / 500x10-9 m = 6x10 15 s -1 6x10 15 s -1 x 6.626x10-34 Js = 4x10-20 J /1.5 = 333 nm 6x10 15 s -1 4x10-20 J Page 73 of 316

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

75 Chem 155 Unit 3 Page 75 of Reflection: Fresnel Equation for a special case: Normal Incidence No Absorption Incident beam Reflected beam Transmitted beam Calculate the total reflection loss due to the two reflections (air glass and glass water) when a light beam passes through one side of a cuvette ( = 1.5), containing water ( = 1.3). Page 75 of 316

76 Chem 155 Unit 3 Page 76 of Scattering: Incident Transmitted Scattered Scattering Process Raleigh (elastic) Tyndall (Mie) Elastic normally Scatterer atoms and molecules Colloids (bigger / nm) Raman (visible) Molecules vibrational spec Features weak, favors blue Strong (easy to image) very weak, inelastic Page 76 of 316

77 Electron Energy Chem 155 Unit 3 Page 77 of The Photoelectric Effect and the Photon: Electrons are emitted from metal surfaces that are irradiated with light of sufficiently high frequency. Max Planck and Albert Einstein discovered that Kinetic energy of the emitted electrons depends on: Light frequency and Type of Metal Kinetic energy of emitted electrons is independent of: Light intensity - more intensity means more electrons, but not more energetic electrons! Light current electrons Metal Slope = h Free e - 0 E F Page 77 of 316 Light Frequency

78 Chem 155 Unit 3 Page 78 of 316 This leads to the concept of the light as a particle and an expression for the energy of a photon. Electron energy relationship to light frequency: = Work function of metal E F = Fermi energy of metal = Light frequency (s -1 ) E = h - h = slope = Formula of line Planck s constant Also the photoelectric effect is the basis of many light detectors because it converts light energy into: Electricity current or single-photon pulses Page 78 of 316

79 Chem 155 Unit 3 Page 79 of 316 The discovery of photon energy correlated with the theory of atomic and molecular orbital energy. Phase gas phase atom molecular liquid molecular solid Quantum States electronic electronic electronic vibrational vibrational phonon rotational Resulting Spectra lines bands broad Page 79 of 316

80 Chem 155 Unit 3 Page 80 of Spectra typical of gas, liquid and solid. Absorber Phase Notes / transition types Atom Gas Extremely narrow lines / electronic Molecule Gas Molecule / small Solution Fine structure due to electronic+ rotational+ vibrational Broad bands + some fine structure / electronic + vibrational Molecule / larger Solution Broad bands electronic only resolved Page 80 of 316

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

82 Chem 155 Unit 3 Page 82 of Typical fluorphore Jablonski Diagram Photophysical Processes Page 82 of 316

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

84 Emission or Heated Mat. Emission or Absorbance Emission or Absorbance Chem 155 Unit 3 Page 84 of Energy Level Diagram on its Side Line spectra, band spectra and continuum spectra of atoms, molecules and solids: The relationship between energy states, photon energies and spectra. Light Frequency Energy Light Frequency Energy Light Frequency Page 84 of 316 Energy

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

86 Chem. 155 Unit 4 Page 86 of Photometric Methods and Spectroscopic Instrumentation Skoog Chapters Covered: Review: Quantitation in Absorbance and Emission 7A Optical Designs Absorbance Emission 7A Optical Materials (lightly!) 7B Light Sources Continuum and Line 7B Lasers! Page 86 of 316

87 Chem. 155 Unit 4 Page 87 of General Photometric Designs for the Quantitation of Chemical Species Generalized Detector Response: S = kp + S DARK S = Signal k = proportionality P = light power S DARK = detector response in absence of light Absorbance More Analyte Less Signal All absorbance methods! IR, VIS, UV, Xray! Quantitation A = bc = -log(p/p0) Emission and Fluorescence More Analyte More Signal = molar absorptivity b = pathlength (cm) c = concentration in moles/l P = Light Power at Detector P O = Light Power for Blank c = concentration in moles/l All light emission methods! Fluorescence (Xray - UV), Scattering, Luminescence, even NMR! Quantitation P = P O + mc P = Light Power at Detector P O = Light Power for Blank c = concentration in any unit Page 87 of 316

88 Chem. 155 Unit 4 Page 88 of Block Diagrams Instruments for Analytical Spectrometry: 1. Absorbance Cuvette Flame Gas Cell Photomultiplier Tube, Photodiode Light beam is absorbed by analytes, not re-emitted Amount of Absorption Amount of Analyte 2.1 Fluorescence 2.2 Raman Scattering Light beam stimulates analytes to emit light Emitted light is collected at right angles to stimulating Amount of Emission Amount of Analyte 3. Chemiluminescence Analyte reacted with chemical that makes it emit light Emitted light is collected Amount of Emission Amount of Analyte Page 88 of 316

89 Chem. 155 Unit 4 Page 89 of Optical Materials Cost Trade Off LiF Sapphire Al 2 O 3 Crystal $$ Water Soluble! $$$$ Hardness! $$ UV-Vis $ vis only $ Water Soluble! $$$$ Red and IR only $$$$ IR Only $$ non-linear dispersion $$- $$$$ most common tool order-overlap $$ low-res short range $ only one per filter $-$$ very low resolution Time Domain Interferometers $$$$ hard to optimize for both visible and IR at once Page 89 of 316

90 Chem. 155 Unit 4 Page 90 of Optical Sources Cost Trade Off $$ VUV Only? $$ Short Lifetimes $$ Popular UV-source used with W-Halogen for UV-Vis $ W-Halogen Visible only, long lifetimes $$ NIR and IR Only $ IR only $ IR Only $$ Low intensity, limited s $$- $$$$ Exellent Intensity, limited s $ - $$ very high resolution, poor dynamic range $$-$$$ PMT v. fast, v.v. sensitive, delicate, limited in IR $-$$ low sensitvity, limited in IR but cheap $ - $$$ sensitive and fast, much tougher than PMT $$-$$$$ v. sensitive, slow, tough, $$-$$$ detector of choice most IR More exotic energy detectors don t know much about these Page 90 of 316

91 Chem. 155 Unit 4 Page 91 of Continuum Sources of Light: Broadband or White Light Sources Also called: 1. Simplest Design: Blackbody Sources a. Tungsten b. Quartz-Tungsten-Halogen c. Nernst Glower Visible / Near IR / IR IR / Far IR 2. Gas Emission Designs: a. H 2 / D 2 UV Only H 2 + e - H 2 * H + H + h Kinetic Energy of H atoms = Continuum Therefore h can be: Continuum b. Ar, Xe, Hg UV-Vis-Near IR Heavy (High-Z) atoms Many atomic states + High Pressure (extensive broadening of lines) Quasi-Continuum Page 91 of 316

92 Chem. 155 Unit 4 Page 92 of Line Sources of Light: 1. Low Pressure Gas Emission a. Hg b. Ar c. Xe d. Na Many Lines can be filtered to emit only one predominant line. Na-D Line Predominates and nm doublet 2. Hollow Cathode Lamps: a. Metals (Cathode!) b. Used in atomic spectroscopy (absorbance and fluorescence) Sputtering 300V DC Electrical Discharge Ne + Ne + Fe* + Fe + h Atomic Emission Fe Metal Cathode Page 92 of 316

93 Chem. 155 Unit 4 Page 93 of Laser Sources of Light: Partially transmitting mirror An Introduction to Lasers: LASER is an acronym for a light amplification process: L A S E R Light Amplification by Stimulated Emission of Radiation Fully reflective mirror Pumping Energy Source: Intense Light Electrical Discharge Gain Medium atoms or molecules that undergo lasing transitions What happens in the Gain Medium? 1. relatively long lived Page 93 of 316

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

95 Chem. 155 Unit 4 Page 95 of A laser is a light amplifier Some of the above processes degrade the light in the cavity Spontaneous Emission Absorption Some of the above processes amplify the light in the cavity Pumping Stimulated Emission gain! stimulated emission pump power in losses absorption, spontaneous emission beam out Pump Fast Decay Lasing! Two common laser configurations: 3-state 3 2 Fast Decay Pump 0 1 Lasing! Fast Decay 4-state (or more) Page 95 of 316

96 Chem. 155 Unit 4 Page 96 of Polulation Inversion and laser amplification absorption population in state Ex lasing amplification population in state Ey Roughly speaking lasing is possible when: population in upper state 1 is greater than the population in lower state 2 This is called a population inversion. Page 96 of 316

97 Chem. 155 Unit 4 Page 97 of Necessity of 3 or more states Why are three or more levels (states) necessary for lasing? recall: N EXCITED N GROUND e kt E E = ENERGY DIFFERENCE us-ls k = Boltzmann const. 1.3x10-23 J/K T = temperature (K) N EXCITED = population (concentration) US N GROUND = population (concentration) LS What happens when T infinity? N EXCITED N GROUND Page 97 of 316

98 Chem. 155 Unit 4 Page 98 of Common lasers categorized by lasing medium. 1. Solid Red - Near Infrared 1.1. Lanthanide-and transition metal ion lasers Nd-YAG Neodymium ions in a crystal called a garnet made of yttrium oxide and aluminum oxide. Other lanthanide ions are substituted for other wavelengths in the near IR Ho-ZBLA Holmium (Ho +3 ) doped glasses made from ZrF 3, BaF 3, La F 3, Al F 3 can be fabricated into optical fibers that lase and can amplify optical signals of certain wavelengths 1.2. Ti-sapphire tunable ( nm) CW or pulsed Mode locked: fs, (MHz repetition) Chirped-pulse amplification: fs, 5, mj, (khz rep) gw power densities Fundamental or second harmonic can be used in OPA optical parametric amplifier lasers that are freely tunable, normally pulsed 1.3. Semiconductor Lasers Silicon, gallium arsenide and other light-emitting diodes can be made to lase when many electrons are promoted to excited states within microfabricated cavities in the semiconductor crystal. Blue (very new) Near Infrared 2. Liquid dye-lasers (blue-red) are made from solutions of many different fluorescent dye molecules. The dye molecules have multiple excited-states that can be induced to lase usually by pumping with other lasers (Ar-ion). Tunable! 3. Gas-Phase lasers (UV-near infrared) are very common, and typically 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

99 Chem. 155 Unit 4 Page 99 of nm - excitation of Hoechst stain, Calcium Blue, and other fluorescent dyes in fluorescence microscopy 405 nm - InGaN blue-violet laser, in Blu-ray Disc and HD DVD drives 473 nm - Bright blue laser pointers, still very expensive, output of DPSS systems 485 nm - excitation of GFP and other fluorescent dyes 532 nm - AlGaAs Bright green laser pointers, frequency doubled 1064 nm IR lasers (SHG) 593 nm - Yellow-Orange laser pointers, DPSS 635 nm - AlGaInP better red laser pointers, same power subjectively 5 times as bright as 670 nm one 640 nm nm - AlGaInP DVD drives, laser pointers 660 nm nm - AlGaInP cheap red laser pointers 785 nm - GaAlAs Compact Disc drives 808 nm - GaAlAs pumps in DPSS Nd:YAG lasers (e.g. in green laser pointers or as arrays in higher-powered lasers) 848 nm - laser mice 980 nm - InGaAs pump for optical amplifiers, for Yb:YAG DPSS lasers 1064 nm - AlGaAs fiber-optic communication 1310 nm - InGaAsP fiber-optic communication 1480 nm - InGaAsP pump for optical amplifiers 1550 nm - InGaAsP fiber-optic communication 1625 nm - InGaAsP fiber-optic communication, service channel Page 99 of 316

100 Chem. 155 Unit 4 Page 100 of Excimer Lasers: Kr * + F KrF * Excimer! Kr + F KrF What is unusual about this molecule? Because of fast dissociation, [KrF * ] > or < [KrF] This favors: population inversion! Page 100 of 316

101 Chem. 155 Unit 4 Page 101 of Some laser pointers. LS = lower state US = upper state GS = ground state hν + US 2hν + LS stimulated emission gain hν + LS US absorption loss US LS + hν spontaneous emisson loss Based on this, should the LS be: short lived or long lived? Based on this, should the US be: short lived or long lived? Page 101 of 316

102 Chem. 155 Unit 4 Page 102 of 316 Laser Questions: Indicate all that apply: 1. For a laser gain medium with three or more states, a population inversion is / means: a. More electrons in the upper than lower states. b. More molecules, atoms or ions in a given excited than ground state. c. Excited state energy is greater than ground state energy. d. Concentration of molecules / atoms or ions in upper state is greater than concentration in lower state. e. Concentration in ground state is zero. f. Concentration in upper state is greater than half of the total. 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] Spontaneous emission Absorption Stimulated Emission 3. For gain to occur in a laser [upper states] must be relative to [lower states]: a. > b. < c. = Page 102 of 316

103 Chem 155 Unit 5 Page 103 of Radiation Transducers (Light Detectors): Chapter 7, Skoog Holler & Nieman 7E Photomultipliers Photodiodes Charge-Coupled Device Arrays 5.1 Desired Properties of a Detector: High Quantum Efficiency: A large fraction of photons that strike it result in a response few are lost. High Gain: For each photon that strikes the detector, a large signal is generated. Low Noise: Constant light flux gives a constant signal - for measurement is low. Low Dark Count: Low or no signal is present in the dark. Page 103 of 316

104 Chem 155 Unit 5 Page 104 of Photoelectric effect photometers Phototube: High UVtransparency fused silica window. Evacuated tube in which electrons can travel if emitted from cathode. + - Readout Amplifier Photocathode transducer - emits electrons via the photoelectric effect Low work function metal: Na, K, GaAs Anode positive bias, collects photocurrent. Page 104 of 316

105 Chem 155 Unit 5 Page 105 of 316 What is the maximum gain of a phototube expressed as electrons per incident photon? 1! What is a typical gain for a phototube expressed as electrons per incident photon? 0.01 What limitation is built in to all photoelectric detectors? h must be Page 105 of 316

106 Photoelectron energy / j Chem 155 Unit 5 Page 106 of Limitations to photoelectric detectors: Recall Einstein et al. used the photoelectric effect to discover what particle: The photon! Slope of this line = Plank s Constant: h = 6.626x10-34 Js Light frequency / s -1 Peak quantum efficiency: About 10% or 1 photoelectron 10 photons Page 106 of 316

107 Chem 155 Unit 5 Page 107 of Operation of the PMT detector: Faceplate material: UV-transparent Silica Photocathode potential: From pmtconstruct.pdf -100 to 1000V A typical PMT may have 12 dynodes, each of which gives off somewhere between 4 and 12 secondary electrons depending on the applied voltage. Question: Derive a formula for PMT gain: How many electrons are collected for each incident photon? (Use outline on next page ) Page 107 of 316

108 Chem 155 Unit 5 Page 108 of PMT Gain Equation: 1 Photon Photocathode Quantuum Efficiency - How many photoelectrons per incident photon are emitted? How many secondary electrons? Secondary electron yield - Dynode 1 Dynode 2 PMT Gain = G = = Quantum efficiency = Secondary electron yield n = Number of dynodes n Dynode 12 Page 108 of 316

109 Chem 155 Unit 5 Page 109 of 316 How many dynodes do you want or need? Ultimate sensitivity = Detect a single photon! Consider the arrival packet of electrons: h e - at photocathode e - strikes n- dynodes n electrons strike anode 3x10-9 s What is the average current during the 10ns pulse? Assume = 4, n = 10 F = coulombs/mole e - Mol = 6.023x10 23 Current (amperes) = # coulombs / s 4 10 = Current (amperes) = # coulombs / s i = e - * coulombs / 6.023e23 e - / 10-8 s = 1.7x10-5 amperes easily measureable Page 109 of 316

110 current / microamperes Chem 155 Unit 5 Page 110 of Noise in PMT s and Single Photon Counting: Two major sources: Thermal: Individual electrons are thermionically emitted from photocathode or one of the dynodes. Cosmic Ray Background: Energetic particles ( and particles, -rays etc. from space) strike photocathode or dynodes and cause a large current spike. time / s Reject: cosmic ray e- cascade. Accept: photonelectron from photocathode Reject: Thermal e - from dynodes Muon: created by very high energy cosmic ray interaction with upper atmosphere, similar to electron (-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 charged particles 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 matter before attenuating as a result of absorption or deflection by other atoms. Mark Wolvertron, science writer, Scientific American magazine, September 2007, page 26 "Muons for Peace" Page 110 of 316

111 Chem 155 Unit 5 Page 111 of 316 In single photon counting: Accept only those pulses originating at the photocathode. Retain signal Reject noise Recall: Detection limit S MIN = S BLANK + 3 BLANK Large and small pulses are all noise! These contribute to both S BLANK and BLANK! Limitations to the PMT and to single photon counting: 1. Wavelength limitations: Same as phototube - h 2. Intensity limitations: In general, the PMT is used in LOW LIGHT ONLY ambient light will destroy the tube. In Single Photon Counting, light levels must be low enough that light pulses do not pile up or overlap. Page 111 of 316

112 Chem 155 Unit 5 Page 112 of Semiconductor-Based Light Detectors: Photodiodes: p-type Si p-n junction n-type Si - + P + B - + B - P + - The only carriers of current in p-type Si are: The only carriers of current in n-type Si are: Holes! Electrons! Why does B (or Al) become (-) and P become (+)? Both B and P are forced to become tetravalent by the bounding atoms in the lattice this leaves a curious formal charge! B (p-dopant) Mobile holes. P (n-dopant) Mobile electrons Page 112 of 316

113 Chem 155 Unit 5 Page 113 of 316 Depletion, anhiliation, injection and rectification: 1. Forward Bias: + p-n junction Injection of holes 2. Reverse Bias: Anhiliation of holes and electrons + Injection of electrons Removal of holes Depletion of holes and electrons Removal of electrons Under reverse bias: Page 113 of 316 No current flows: Rectification

114 Chem 155 Unit 5 Page 114 of 316 Photodiode: Steps in Photoconduction: Photon absorbed by Si, promotes e- into concution band. This leaves a mobile hole and electron. Hole and electron are swept apart by the applied field and: h current transduction Properties of the photodiode: 1. Gain: = 1 but is often near unity (1) 2. Dynamic Range: 3. Ease of fabrication: Page 114 of 316 Largest measureable signal >> smallest Excellent! Can be made very small. Can be made into 1D and 2D

115 Chem 155 Unit 5 Page 115 of Charge Coupled Device Array Detectors: -10V -10V -10V Al contact SiO2 n-si e- are repelled from Al contact 2. h strikes depletion region 3. h+ accumulate at Al contact -10V V -10V V Page 115 of

116 Chem 155 Unit 5 Page 116 of 316 CCD Array detectors are very useful for spectroscopy. 1. Maximum gain per pixel 1 Not as sensitive as a PMT. 2. Readout is relatively slow: Ca. 100 frames per second 10-2 s. 3. But, conventional spectrometers measure the light spectrum (power versus wavelength) work by limiting the light incident on the detector to one wavelength at a time. With a CCD Array detector: You can have a different detector for every wavelength CCD s can acquire an entire spectrum much more quickly than a PMT-based system. 4. The light spectrum falls in a 2-dimensional image or picture. 5. Thermal noise is lower than in PMT, but cannot be completely rejected. 6. Cosmic ray background can be dealt with statistically: If there is a single particular pixel (spot on the picture) that is anomalously bright (i.e has an unusually high reading) it can be rejected as cosmic radiation noise using statistical analysis. Page 116 of 316

117 Chem 155 Unit 6 Page 117 of Monochromators for Atomic Spectroscopy: Chapter 7, Skoog Holler & Nieman 7C-2 Monochromators Dispersion Resolution Speed 7C-3 Monochromator Slits and Spectral Resolution Effective Bandwidth Line Convolution (Effect of Slit Width on Resolution) Czerny-Turner and Echelle designs Why do we care about monochromators? Monochromatic light is useful! An analysis of the wavelength dependence of the absorbance or emission of light is called: Spectroscopy This technique is used to: analyze the structure of atoms and molecules identify the atoms and molecules observe interactions between atoms and molecules Measurement of the amount of monochromatic light absorbed or emitted by atoms or molecules is called: Spectrometry This technique can tell you: the concentration of atoms or molecules in a sample Page 117 of 316

118 Chem 155 Unit 6 Page 118 of Adjustable Wavelength Selectors Adjustable wavelength selectors are called: Monochromators Light Input Polychromatic White Broadband Light Output Monochromatic Narrow-band Wavelength / nm Two important characteristics of monochromators are: The amount of light that makes it through at a given wavelength: The range of wavelengths that exit the monochromator: throughput bandwidth - Page 118 of 316

119 Chem 155 Unit 6 Page 119 of Monochromator Designs: Czerney-Turner blue red Prism Wavelength Dispersion: Linear Nonlinear with UV-Absorption Nonlinear Page 119 of 316

120 Chem 155 Unit 6 Page 120 of The Grating Equation: Grooves or blazes: Lines or rulings etched into grating that scatter or diffract the light. Light emitted from the grating surface is called: Diffracted light For a single, polychromatic input beam: Infinite monochrmatic output beams differing in angle For a particular wavelength ( ) to be diffracted at a particular angle, the corresponding pathlength difference (DBC) must be equal to: n must = DBC where n is any integer Page 120 of 316

121 Chem 155 Unit 6 Page 121 of 316 The Grating Equation (cont): For outgoing rays 1 and 2 to interfere constructively: pathlength difference DBC must = n i. b + i =? ii. b =? iii. a+b+90 =? i 180 iv. substitute ii into iii a + (90 i) + 90 = 180 a i +180 = 180 a = i Page 121 of 316

122 Chem 155 Unit 6 Page 122 of 316 The Grating Equation (cont): For a particular wavelength to appear bright at incident angle i, the following must be true: n = DB + BC We 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 / d opposite / hypotenuse = DB / AB = DB / d d sin(i) + d sin (r) = d ( sin(i) + sin(r) ) For constructive interference (i.e. a bright condition): Page 122 of 316 n = d ( sin(i) + sin(r) )

123 Chem 155 Unit 6 Page 123 of Dispersion Dispersion characterizes the extent to which a monochromator separates (disperses) the different wavelengths of light. Angular dispersion: For a given incident angle, i, how quickly does the color change as you change the viewing angle r? In other words what is the derivative of with respect to r? n = (r) = d / dr = d / dr d ( sin(i) + sin(r) ) d ( sin(i) + sin(r) )/ n d/dr[dsin(i)] /n + d/dr[dsin(r)] / n d cos(r) / n Page 123 of 316

124 Chem 155 Unit 6 Page 124 of Angular dispersion: dr n d = d cos(r) Reciprocal linear dispersion: dy n = Fdr d = r = order Groove spacing on grating Angle of diffraction = wavelength sin(r) sin(r) r y/f d D -1 dy = d Fdr D -1 = d cos(r) / nf Page 124 of 316

125 Chem 155 Unit 6 Page 125 of 316 For monochromators the important factor is: Reciprocal Linear Dispersion D -1 = d / dy = d cos(r) / nf What is dispersion? Consider a typical monochromator: d = 1000 lines / mm r = 45 n = 1 F = 1m First let s get the units straight! d= F= d = 1/1000 mm = 10-3 mm = 1 m = 1000 nm F = 1m = 1000 mm D -1 = 1000 nm x 0.7 / 1 x 1000 mm D -1 = nm / mm Exit Slit Focal Plane y / mm Page 125 of 316 / nm

126 Chem 155 Unit 6 Page 126 of Effective bandwidth The combination of exit slit width and dispersion determine EFF the effective bandwidth of the monochromator. EFF is a number, in units of nm, equal to the range of wavelengths that exit the monochromator at any one time. EFF = wd -1 W = slit width in mm D -1 = dispersion in nm / mm EFF is controlled in part by: the exit slit width - w W = slit width in mm D -1 = dispersion in nm / mm EFF 15 nm EFF 2 nm Page 126 of 316 / nm

127 Chem 155 Unit 6 Page 127 of Bandwith and Atomic Spectroscopy Consider the case of a Ni atomic spectroscopy experiment. Ni atoms emit at 231.7, and nm. However, Ni atoms only absorb light at nm. (Two emission lines are non-resonant.) So, it is important in atomic absorption spectroscopy to remove Ni emission at and 232.2, but efficiently transmit light at nm. Page 127 of 316

128 Chem 155 Unit 6 Page 128 of Factors That Control EFF In general, one wants a monochromator with as small an effective bandwidth is practical. Such a monochormator is called: high resolution To achieve low effective bandwith / high resolution EFF should be: small so EFF = wd -1 = wd cos(r) / nf, d should be: F should be: w should be: small large small The essential tradeoff is that very small EFF usually equate to low light levels. For example, achieving very small EFF by making w very small is often unsuccessful because the source is imaged onto the focal plane. Unless the source is a very bright, point source, then making w is smaller than the source will often cause light levels to drop dramatically. Page 128 of 316

129 Light throughput Chem 155 Unit 6 Page 129 of Resolution Defined Resolution is defined as R / Resolution is the ability to just separate two adjacent spectral lines. For example, consider the output of a monochromator with 0.1 nm versus 1 nm resolution. In this case two purely monochromatic sources of 501 and 502 nm are focused into the entrance slit. Grating Res olution Limted Throughput Low resolution (1 nm) Wavelength / nm R=0.1 nm R=1 nm x 10 Note that the low resolution case has much lower intensity as well. Page 129 of 316

130 Chem 155 Unit 6 Page 130 of Grating Resolution Resolution may be limited by the size of the grating as well. The grating resolution, R, is given by: R / = nn n = N = Diffraction order Number of grooves of the grating that are illuminated Grating resolution imposes a kind of effective bandwidth limitation just like the EFF defined above Please note these three things: 1. Both grating resolution and effective bandwith define a minimum resolvable for the monochromator. 2. A given monochromator will be limited by the worse of the two parameters. 3. This is usually EFF Page 130 of 316

131 Chem 155 Unit 6 Page 131 of Grating Resolution Exercise: How large must a 1000 groove/mm grating be to give a resolution of 1nm at 500nm? (Assume that the monochromator is functioning in first order.) R = / = nn N = / n = 500 nm /1*1nm = 500 grooves 500 grooves / 1000 grooves/mm = 0.5 mm! Page 131 of 316

132 Chem 155 Unit 6 Page 132 of 316 If a Czerny-Turner monochromator has the following specifications: Holographically-ruled diffraction grating with 1582 grooves per mm. 250 mm focal length Grating position such that the diffracted angle is 45 degrees Operation in first order A slit width of 0.5 mm a. What is the reciprocal linear dispersion effective bandwidth of this monochromator (use appropriate units)? b. What is the effective bandwidth of this monochromator (use appropriate units)? A Czerny-Turner monochromator is set to nm and the slits are set so that the effective bandwidth is 1.0 nm. If a broadband UV light source is directed into the 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 length c. Grating position such that the refracted angle is 45 degrees d. Operation in first order What slit width is required to give this monochromator an effective bandwidth of 0.5 nm? How big must the grating be in order to achieve this resolution (R). Page 132 of 316

133 Chem 155 Unit 6 Page 133 of High Resolution and Echelle Monochromators Atomic spectroscopy demands high resolution because atomic absorption / emission lines are: 1. narrow 2. numerous So we want effective bandwidth EFF = wd -1 = w d cos(r) / nf to be: We also want resolution (R = / = nn) to be: small large EFF R w should be small - d should be small small F should be large large n should be large large But there are limitations to the above: w If w is too small 1 no light 2 exceeds the ability to focus light d F Limitations in the fabrication of gratings How big of a spectrometer is tolerable? What about n? n is the key Page 133 of 316

134 Chem 155 Unit 6 Page 134 of 316 A monochromator that operates in very high order is called: An Echelle monochromator The grooves are machined to reflect light at high incident angles i and r are nearly identical and are called An echelle monochromator outperforms a Czerny- Turner monochromator in high-resolution applications: Czerny-Turner Echelle F / m Groove density 1200 / mm 79 / mm i,r or R (300nm) 60, ,000 D -1 (nm / mm) Page 134 of 316

135 Chem 155 Unit 6 Page 135 of 316 But Echelle monochromators have a strange handicap. At a given angle, many wavelengths may simultaneously be striking the detector! n = 2dsin( ) 63x300 = 2dsin( ) but 62x??? = 2dsin( ) also! n 1 1 = n = n 1 1 /n 2 n / nm This problem is called: Order overlap Is this a problem for Czerny-Turner monochromators? If n 1 =1 and 1 =300nm then for n 2 =2, 2 = 150 nm If n 1 =1 and 1 =800nm then for n 2 =2, 2 = 400 nm Czerny-Turner monochromators handle this problem with: Echelle Monochromators handle the order-sorting problem with cross dispersion usually a silica SiO 2 prism. UV-absorption is a problem with glass Page 135 of 316

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137 Chem 155 Unit 6 Page 137 of 316 This means that the Echelle monochromator has a 2- dimensional focal plane! What kind of detector is tailor-made for an Echelle mnochromator The CCD array detector! Page 137 of 316

138 Chem 155 Unit 7 Page 138 of Photometric Issues in Atomic Spectroscopy Skoog Chapters Covered: 9B, 9C Question: How can one quantify low concentrations of atoms that have extremely narrow band absorption spectra? How can one do this when the atoms are contained in a glowing flame that contains many broadband emitting species. Answer: Interrogate the sample of atoms with an equally narrow band and very bright source of radiation. Where do you find such a source? Hollow Cathode Lamp Cathode is made of analyte metal e.g. Fe or Zn Ne + ions in plasma accelerate into cathode, sputter metals atoms, excite them and emission collected out lamp end. Page 138 of 316

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151 Chem 155 Unit 7 Page 151 of 316 Three photometric problems: 1. EFF >> ATOMIC so stray light is a major issue. The maximum absorbance that one could realistically expect with a 0.25 nm EFF is roughly 0.01 using a broadband source, and this is going to be noisy so: a. Use hollow cathode lamp with emission identically matched to absorption. This makes maximum absorbance much higher, and less noisy. For the remaining source and flame stray light, one can: b. Modulate the source to help to distinguish it from flame and other radiation that leaks into detector around atomic line. 2. Flame and molecules in flame present a broadband absorption interference. These are direct interferants, so they must be measured in some way that is not sensitive to the analyte atoms: a. Use broadband (D 2 ) source that is approximately blind to analyte to approximate the broadband flame absorbance. b. Use Zeeman splitting to polarization select between resonant absorption 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 power and concentration. #2 Understand how stray light can affect computed values of absorption. #3 Understand how narrow atomic bands, that are narrower than the attainable effective bandwidth, yield a difficult stray light problem. #4 Understand how the hollow cathode lamp addresses the problem of atomic absorption lines that are narrower than the effective bandwidth of the monochromator by matching the spectum of the source (hollow cathode lamp) to that of the absorber (free atoms in flame). #5 Understand how the absorbance of broadband D 2 radiation actually approximates the blank absorbance of the flame even when there is analyte present in the flame. #6 Understand how Zeeman splitting samples the flame background but not the analyte absorbance. Page 151 of 316

152 Chem 155 Unit 8 Page 152 of Practical aspects of atomic spectroscopy: Sample introduction and plasma chemistry. Skoog chapters covered: 8C, 9A Flame or Plasma Monochromator Atomic Emission from Hollow Cathode Lamp Light source for absorption measurement only. Flame or Plasma Nebulizer / Spray Chamber Atomic Absorption and / or Atomic Emission in Flame/Plasma Molecular gas Dry solids Aerosol Analytes in Solution PMT Absorbance: A Log Emission: C ] k( P P P O [ PO ) P P 0 b[c] In AAS and AES solutions of analytes are a. aerosolized, b. measured as they 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 in front of the detector Why is this tradeoff worthwhile? Intense and narrow emission and absorption bands make atomic spectroscopy sensitive and selective compared to most comparable solution methods. Page 152 of 316

153 Chem 155 Unit 8 Page 153 of Nebulization (sample introduction): Sample introduction in AAS and AES is usually through a nebulizer. The nebulizer makes a very fine mist of the analyte solution. The finest droplets in this mist are then selected by the spray chamber. These droplets are then swept into the flame or plasma. This sample introduction process is necessary but: a. inefficient b. problematic c. hard to do reproducibly Page 153 of 316

154 Chem 155 Unit 8 Page 154 of 316 Nebulizer types: 1. concentric 2. cross flow 3. fritted disk 4. Babbington 5. ultrasonic pneumatic Concentric: Page 154 of 316

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

156 Chem 155 Unit 8 Page 156 of 316 Ultrasonic: Page 156 of 316

157 Atoms Chem 155 Unit 8 Page 157 of Atomization Atoms and ions diluted Optimal AES region Optimal AAS region Molecules Particles Droplets Entire process happens in: milliseconds So large droplets are: bad Page 157 of 316

158 Chem 155 Unit 8 Page 158 of Flame Chemistry and Matrix Effects Our understanding of what happens is based on the idea of: Local Thermodyamic Equilibrium (LTE) This means that we theorize and analyze the process occurring in the flame/plasma as though they were equilibrium processes. Some flame / plasma properties (Table 9-1): Fuel Oxidant Temp. C Burn velocity (cm/s) Natural Air Gas Natural Gas O Acetylene Air Acetylene O Acetylene N 2 O Page 158 of 316

159 Chem 155 Unit 8 Page 159 of Flame as sample holder : 1. For best sensitivity: 2. Page 159 of 316 Interrogate with hollow cathode radiation through atom rich region. Changes in observation height: Change sensitivity so don t move observation region during experiment.

160 Chem 155 Unit 8 Page 160 of Optimal observation height: Optimize: Observation region Flame stoichiometry 1. Increased absorbance corresponds to: More gas phase atoms 2. Absorbance normally starts low because: Molecules have not been broken down into atoms yet 3. Absorbance normally ends low because: Gas in flame expands and atoms are diluted. 4. Cr profile decreases because: Cr atoms form stable oxides and this increases with height. 5. Ag profile increases because: Ag atoms do not form stable oxides Page 160 of 316

161 Chem 155 Unit 8 Page 161 of Flame Chemistry and Interferences: 1. Chemical equilibria can occur in flames. Common oxyanions can react with atoms: H 2 PO Fe FePO4 + 2H + + 3e - SO Fe FeSO4 + e - 2. Electrons can be a reagent. Ionization can be suppressed by addition of KCl: KCl K + + Cl + e - Fe + + e - Fe 3. Oxides can be suppressed by: a. Higher Temp: Fe + Ox FeOx b. Competition: Ti + Ox TiOx a. Ti absorbs O from flame c. richer flame: more C 2 H 2, less O 2 4. Hydroxides can be suppressed by: Higher Temp: CaOH Ca + OH Page 161 of 316

162 Chem 155 Unit 8 Page 162 of Matrix adjustments in atomic spectroscopy: Releasing agents: Compete with analyte for interferant: Adding Sr +2 to react with PO 4 and release Ca +2 and Mg +2 for analysis. Protecting agents: Form stable but volatile complex with analyte e.g. EDTA Radiation buffering (when sensitivity is not an issue): Add interferant to samples and standards Standard Additions: Match matrix of sample and standards Page 162 of 316

163 Chem 155 Unit 9 Atomic Emission Page 163 of Atomic Emission Spectroscopy Responsibilities in the Chapter: Introduction and all of 10A Overview Nomenclature Plasma Types Pro s and Con s Inductively Coupled Plasma Torch Design Operating Principles Plasma Properties Plasma Optimization Direct Current Plasma Design Properties Spectrometers Bandwidth Considerations Multi-channel Designs Quantitation and Accuracy Comparisons with Atomic Absorption Page 163 of 316

164 Chem 155 Unit 9 Atomic Emission Page 164 of AAS / AES Review: M+ - ions -e - +e - h M* Emission [M] = k (P-Po) HCL h Atomic gas M Absorption [M] = -k log(p/po) Molecular gas Solid particles Nebulize Liquds Solids Gases Page 164 of 316

165 Chem 155 Unit 9 Atomic Emission Page 165 of Types of AES: PLASMA Plasma is ionized gas typically Ar, Ar +, e - Ionization is maintained in one of three ways: A DC electrical discharge called Direct Current Plasma, DCP or Plasma Jet An induction coil operating at 27 or 41 MHz Inductively Coupled Plasma, ICP A Microwave Cavity called Microwave Induced Plasma or MIP FLAME Flame Emission Spectrometry or FES Flame is chemical plasma typically comprising 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 as Na, Li, K ARC / SPARK Arc = Continuous Discharge Spark = Pulsed Discharge These two are like DCP, but are done in air, on solids (e.g. pressed powders) or metal surfaces Page 165 of 316

166 Chem 155 Unit 9 Atomic Emission Page 166 of Inert-Gas Plasma Properties (ICP,DCP) Ar mostly, also Ne and He (MIP) Chemically Inert 9.4 Predominant Species are Ar, Ar +, and electrons Very Hot 5,000 to 10,000 K vs. 2,000 to 3,000 K for flames Superior Atomizers B 2 O 3, PO 4, WO x, VO x, ZrO x and other REFRACTORY compounds can be analyzed Refractory compounds are extremely thermally stable, and therefore hard to atomize. Nonmetals Can Be Analyzed Cl, Br, I, S for example: Try making a hollow cathode lamp out of a non-metal! Plasma Emission Simultaneous Detection Flame Atomic Absorption One or Two at a Time Optimization Simpler Easier to Use Excellent Dynamic Range Higher Precision Wide Range of Elements Costly to Use / Buy Metals Only Significantly Cheaper Page 166 of 316

167 Chem 155 Unit 9 Atomic Emission Page 167 of Inductively Coupled Plasma AES: ICP-AES Induction and Eddy Currents: Switch on electromagnet & eddy currents (circular) flow in conductor around changing magnetic field lines: Page 167 of 316

168 Chem 155 Unit 9 Atomic Emission Page 168 of ICP Torches Intensely luminescent, 5000K plasma tauroidal core. Water-cooled induction coils with 27 or 41 MHz large AC currents flowing through them. Quartz envelope Sample aerosol Tangential Ar flow pushes plasma away from quartz housing Page 168 of 316

169 Chem 155 Unit 9 Atomic Emission Page 169 of Atomization in Ar-ICP Excitation = Optimal observation region. Desolvation / Atomization Inert Properties: Hot Electron rich Inert Analyte = low Atomization oxide formation, / Ionization: chemical interference low Hot = efficient atomization Electron rich = low ionization Excellent atomizer Page 169 of 316

170 Chem 155 Unit 9 Atomic Emission Page 170 of Direct Current Plasma AES: DCP-AES Ar flow: Argon Plasma Jet: Argon: Relative to ICP: Page 170 of 316

171 Chem 155 Unit 9 Atomic Emission Page 171 of Advantages of Emission Methods One of the main advantages of emission methods is: Simultaneous multielement determinations With the appropriate spectrometer design, it is possible to measure many elements at once using emission. This means: Simultaneous light detection at different wavelengths i.e. different points along the focal plane. With conventional spectrometers, this means many photomultiplier tubes (PMT s) placed along the focal plane of the Czerny-Turner, or similar monochromator. But, PMT s are neither small, nor cheap. So, PMT-based multi-channel spectrometers are both BIG, and EXPENSIVE. With the echelle design, there is a 2-dimensional focal plane that happens to be extremely well suited to the lithographically designed semiconductor charge-transfer array detector such as the CCD and CID. Page 171 of 316

172 Chem 155 Unit 9 Atomic Emission Page 172 of 316 P P 0 Page 172 of 316

173 Chem 155 Unit 9 Atomic Emission Page 173 of Accuracy and Precision in AES The Good News: Plasmas of Inert Gas are Excellent Atomizers so: Atomization is High Chemical Interference is Low Oxidation is Low The Bad News: AES relies on the existence of excited states, but excited state populations: Are small relative to ground states Depend strongly on temperature N Excited Excited g Recall: e N o g o The Solution: (when very high accuracy or precision is needed) Internal Standards kt E Page 173 of 316

174 Chem 155 Unit 9 Atomic Emission Page 174 of 316 A perfect internal standard will have: Identical Activation Energy ( E) E A S A Q A N A g g 0 e kt S IS Q IS N IS g g 0 e E IS kt Exercise: Derive an equation for the ratio of the analyte to IS signals as a function of temperature: Recall: e a e b e a b The answer: S A S IS Q A N A Q IS N IS e E A kt E IS So: Page 174 of 316

175 Chem 155 Unit 9 Atomic Emission Page 175 of 316 A good internal standard can compensate for drift i.e. changes that appear between measurements due to: Sample introduction rate Instrument Gain Atomization Efficiency Temperature Do you need one? Do the experiment Check your precision and accuracy and see! Page 175 of 316

176 Chem 155 Unit 9 Atomic Emission Page 176 of 316 Two examples of precision / accuracy assessments of ICP-AES: Page 176 of 316

177 Chem 155 Unit 9 Atomic Emission Page 177 of 316 Page 177 of 316

178 Chem 155 Unit 10 Page 178 of Ultraviolet-Visible and Near Infrared Absorption Responsibilities in the Chapter 13 Measurement of Transmittance / Absorbance (13A) Beer s Law (13B) Mixtures (13B-1) Limitations (13B-2) Effect of Instrumental Noise on Spectrophotometric Analysis (13C) Instrumentation (13D, 13D-1, 13D-2) 10.1 Overview Ultraviolet Visible Near Infrared nm nm nm We group these wavelengths together for two reasons. Analytes: This wavelength range covers many atomic and molecular electronic absorptions. Optics: Pure SiO 2, called fused silica or quartz is transparent in this range. Even air absorbs strongly at <180 nm. Page 178 of 316

179 Chem 155 Unit 10 Page 179 of The Blank There are two problems with calculating the absorbance of a liquid sample in a container. These are that losses of light other than abosrobance by analyte molecules. b. Absorption by solvent or non-analyte solute in the sample. a. Reflections at each interface (Fresnel eqn). Scattering particles in the sample. Reflection losses are about 7% for glass, air and water. So even without analyte, the transmitted power for the sample, P, is less than the incident power. How do we compensate for this loss in intensity in the sample power P that is not due to the analyte? Measure Po with a blank in place that has nearly the same reflection and scattering losses as the cuvette so the only difference in P and Po is due to absorbance. Page 179 of 316

180 Chem 155 Unit 10 Page 180 of Theory of light absorbance: Absorbance theory is based on the assumption that molecules have a cross section through which no light can pass. Each slice of the sample removes a fixed fraction of the light entering it so an exponential relationship between light power and distance results: P(x) = Po exp(-anx) a = the cross section of the absorber in cm 2 N = number of absorbers per cm 3 (# concentration) x = distance Page 180 of 316

181 Chem 155 Unit 10 Page 181 of Cross section a and molar extinction : The cross-section a can be related to the molar extinction coefficient,, in Beer s law. Use this to answer the question: can the absorption cross section of a molecule be bigger than the molecule itself? Beer s law: M max 10 5 M -1 cm -1 A = bc = -Log(P/Po) a is about 1 nm theoretical cross section is about the same size as the molecule! Page 181 of 316

182 Chem 155 Unit 10 Page 182 of Limitations to Beer s Law: 1. Fundamental Deviations: 1.1. Electrostatic or other interactions between the absorbing ions or molecules can change the energy levels and symmetries of molecules, and therefore change ( ). These appear at high (>0.01 M) absorber concentration Refractive Index depends on refractive index (n) so, in cases where comparisons between absorbance in solutions of greatly differing n must be made, then you can re-write Beer s Law as: A = n/(n 2 +2) 2 b c. 2. Apparent Chemical Deviations: Concentration dependent equilibria may occur such as HA H + + A -. If for HA is different from for A - then cal curve may be nonlinear! 3. Instrumental Deviations: Several instrumental factors are known to cause non-beer s law behavior: 3.1. The monochromator effective bandwidth is large compared to the spectral bandwidths analyzed Stray light is getting into the detector The light levels are too low for the detector to accurately represent. Page 182 of 316

183 Chem 155 Unit 10 Page 183 of Instrumental Limitations to Beer s Law: The Stray Radiation Problem: Light reaching the detector that a. does not come from the source, b. does not go through the sample. or c. is not the intended wavelength contributes a background power, P STRAY that, in addition to making the absorbance deviate from Beer s law. Does stray radiation make a positive or a negative deviation from Beer s law (A = bc) Stray radiation can come from the source! In a conventional Czerny-Turner monochromator operating in first order, if the monochromator setting is 700nm, 350nm radiation diffracted in second order will also fall on the exit slit! Order-sorting filters that are moved into the light beam according to the wavelength normally remove this but if a filter fails to remove all of the 350nm light then there will be a P STRAY component. P stray 0.05 A truei log P i A P appi log P i P stray o P o P stray P i A truei A appi A truei C i Page 183 of 316

184 Chem 155 Unit 10 Page 184 of Instrumental Noise: Consider the signals P and Po you actually measure these, and they always contain noise. P Po P Po Po P P A = A = A = Case 1. Hard to distinguish P and Po For example: P=0.99±0.01 Po=1.00±0.01 Possible : Possible: P/Po 1.00 A=-Log(P/Po) NEGATIVE Case 2. Hard to distinguish P and zero For example: P=0.01±0.01 Possible : A = -Log(P/Po) Infinite! P = 0.00 Page 184 of 316

185 Chem 155 Unit 10 Page 185 of Apparent Deviations due to Equilibria: For many molecules, especially weak acids, only one of the conjugate acid / base pair may absorb at a given wavelength. 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 an indicator like phenolphthalein (HA) in pure water look like? HA H + + A - 1. K EQ = [H + ][A - ] / [HA] absorbing Concentration of HA in solution, the undissociated fraction. 2. Conservation of mass: [HA 0 ] = [HA] + [A - ] so: [HA] = [HA 0 ] [A - ] Number of moles of HA 4. [A - ] = [H + ] (if >> 10-7 M) divided by volume of in solution. Total or formal concentration. 5 = K EQ = [A - ][A - ] / [HA] 6 = K EQ = [A - ] 2 / [HA 0 ] [A - ] Page 185 of 316

186 Chem 155 Unit 10 Page 186 of 316 K EQ = [A - ] 2 / [HA 0 ] [A - ] try to solve for [A - ] [A - ] 2 + K EQ [A - ] K EQ [HA 0 ] = 0 a x 2 b x c 0 x b b 2 4 a c 2 a a = 1 b = K EQ c = K EQ [HA 0 ] [A - ] = x( K HAo) K K 2 4 K HAo 2 Page 186 of 316

187 Chem 155 Unit 10 Page 187 of 316 Let K EQ = 10-5 [HA] in solution: HA( K HAo) HAo x( K HAo) x K HA o HA o Fraction ionized: HA o 1 x K HA o z HA oz HA oz Page 187 of 316

188 Chem 155 Unit 10 Page 188 of Monochromator Slit Convolution in UV-Vis: A monochromator may qualitatively distort absorbance peaks as well as quantitatively distort them if the EFF is too large. Page 188 of 316

189 Chem 155 Unit 10 Page 189 of 316 So if monochromator slits are too wide (relative to peaks): Peaks are broadened Non-Beer s law behavior occurs But if the monochromator slits are too narrow: Low light power leads to noise Low light power can lead to spurious positive spikes positive deviations from Beer s law! Page 189 of 316

190 Chem 155 Unit 10 Page 190 of UV-Vis Instrumentation: Light Sources a. D 2 Lamps nm, excellent UV sources Electrical Discharge in D 2 gas in a sealed, SiO 2 bulb. D 2 D 2 * (dissociative molecular state) D 2 * D a + D b + h E = KE(D a ) + KE(D b ) + h Kinetic Energy Continuum Wavelength continuum b. Tungsten(W)-Halogen nm, excellent Vis-NIR (visible-near Infrared) sources blackbody radiators (e.g. 2500K) halogen helps to re-deposit W onto filament and improves lifetime for very hot filaments c. Xe- or Hg-arc lamps nm, excellent UV-Vis sources high-atomic number noble gas in electrical discharge many states many wavelengths that coalesce into a continuum. Materials: silica = SiO 2 ( nm) glass = SiO 2 + NaBO 4, etc. ( nm) sapphire = Al 2 O 3 ( nm) $$ Page 190 of 316

191 Chem 155 Unit 10 Page 191 of Single vs. double-beam instruments: Single-beam design: Measure blank signal (Po), digitize and record the number, replace blank with 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 a particular noise source is prominent: i.e. slow change in: o source intensity o detector sensitivity gives Page 191 of 316 drift absorbance error.

192 Chem 155 Unit 10 Page 192 of 316 Double-beam instruments continually monitor or alternate between P and Po using a second, blank cuvette Advantages of double beam designs: Accuracy: P and Po are measured: almost simultaneously so source and detector drift are: minimized! Hence absorbances are more: accurate. Disadvantages of double beam designs: Higher Cost: More complicated design Often: Often: Two detectors More optics Page 192 of 316

193 Chem 155 Unit 10 Page 193 of 316 Photodiode 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 grating all detectors are measuring the signal at once The spectrometers are LESS EXPENSIVE because: they have few moving parts and CCD detectors 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 accuracy Page 193 of 316

194 Chem 155 Unit of UV-Visible Spectroscopy of Molecules Skoog Ch. 14A all, 14B all, 14E-1 and 14E-3 Consider the simplest carbonyl: H H C O Lewis Structure of Formaldehyde 3 types of valence electrons are important: n non-bonding sigma-bonding pi-bonding Orbital geometries * * n Page 194 of 316

195 Absorbance Energy Chem 155 Unit of Spectral Assignments Wavelength / nm UV spectrum & energy levels of a fictitious carbonyl: If: * * n * n * are allowed, assign peaks 1, 2 and 3 to specific transitions: Transition Energy From Diagram From Spectrum Possible Energy / Peak # / nm / cm -1 Transitions arbitrary units * 10-0 = ,000 n * 10-4 = ,000 * 9-1 = ,000 n * 9-4 = 5 E (spectrum) E (digram) / nm Assignment 50, * 40, n * 30, * n * Page 195 of * 9 * n

196 Chem 155 Unit of Classification of Electronic Transitions The following simplified classifications apply to organic molecules: 1. * 1.1. Highest energy! 1.2. Shortest wavelength < 200 nm 1.3. O 2, N 2, all molecules have -bonds so: this radiation is absorbed by air and SiO 2 so spectroscopy must be done in vacuum and is very difficult called VUV or vacuum ultraviolet spectroscopy 2. n * nm most all transitions are found in the ultraviolet 2.2. Molecules with lone (nonbonding) pairs of electrons participate such as: O, N, Halogens to 1000 medium strength absorbers 3. n *, * 3.1. Longest wavelength transitions nm 3.2. Strongest absorbers 100 to 100,000 M -1 cm Limited to molecules with -bonds e.g. alkenes, alkynes, aromatics, carbonyls, azides etc. Page 196 of 316

197 Chem 155 Unit of Spectral Peak Broadening Why are solution-phase electronic spectra hundreds of nm wide if atomic emission and absorption peaks are only ca nm wide? Page 197 of 316

198 Chem 155 Unit of 316 They are molecules, not atoms so electronic transitions, vibrational transitions, rotational transitions all happen simultaneously Solution phase interactions perturb the energy levels of each molecule differently. The spectrum averages over many molecules, creating a continuum of energy levels for the transition Page 198 of 316

199 Chem 155 Unit of Influence of Solvent Spectral Fine Structure: Temperature and Spectral Fine Structure: Page 199 of 316

200 Chem 155 Unit of Solvatochromism: The shift in wavelength of an absorbance band with changes in polarity (dielectric constant) of the solvent. Conjugation: The presence of alternating single and double bonds in organic molecules. Page 200 of 316

201 Chem 155 Unit of Aromatic UV-Visible absorptions: Three sets of * bands are prominent in the spectrum of benzene: Primary E2 B (nm) MAX (M -1 cm -1 ) 60,000 7, V. Strong Strong Intermed -OH and NH 2 red-shift and intensify the B-band Based on this, does the nonbonding pair stabilize or destabilize the orbitals? (Also, assuming that it acts predominantly on as opposed to *) Page 201 of 316

202 Chem 155 Unit of UV-Visible Bands of Aqeuous Transition Metal Ions Transitions are known as d-d because electronic states correspond to changes in the population of: Page 202 of 316

203 Chem 155 Unit of 316 d-d transitions imply that d-orbitals have different energies. When ligands such as H 2 O coordinate (bond) to the metal ion, they interact with the d-orbitals and change their energies predictable ways depending on the bonding geometries. They split the d-orbitals by and amount, the ligand field strength. Page 203 of 316

204 Chem 155 Unit of 316 Ligand field splitting energy varies with the type of ligand 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 < < 1, but seldom used for detecting or determining the concentration of metal ions why? Page 204 of 316

205 Chem 155 Unit of Charge-Transfer Complexes Strongly Absorbing Metal Complexes: Example: Fe(III) 3+ -SCN - + h Fe(II) 2+ SCN 0 electron transfer from SCN - to Fe(III) 5,000 Fe(II) 2+ (o-phenanthroline) 3 + hv Fe(III)3 + (o-phenanthroline) -1 (o-phenanthroline) 2 electron transfer from Fe(II) to (o-phenanthroline) 12,000 Assuming that the absorption noise, σ A = 0.001, what is the best detection limit for Fe +2 aquo and FeSCN +2 in the nm range? Recall, Cm=3sb/m relate sb to sa and m to eb! Page 205 of 316

206 Chem 155 Unit of Lanthanide and Actinide Ions: Lanthanide and Actinice ions have unusual spectral properties. Lots of weak visible and infrared lines Very narrow bands! The 4f and 5f orbitals are smaller than the filled 6s and 7s orbitals so the f orbitals are not split or broadened very much by solvent ligands Electronic configurations differ in energy due primarily to orbital angular momentum, i.e. magnetic interactions described by Russell- Saunders and other coupling schemes. Page 206 of 316

207 Chem 155 Unit of Photometric Titration Trace metals may be analyzed by by UV-Vis absorption using a method known as photometric titration. For the complexation reaction: S + T P S = 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? Why might titration be a more accurate and precise way to compute C than using C=A/ b (i.e. from an absorbance calibration curve). Page 207 of 316

208 Chem 155 Unit of Multi-component Analyses: Consider a cell containing two analytes with overlapping absorbance spectra: Consider two absorbing species A and B: 6 B i ( 0.25A ) i ( 0.75B ) i 4 ( 0.5A) i ( 0.5B) i ( 0.75A ) i ( 0.25B ) i A i w i MAX(A) = 233 nm, FWHM = 60 nm A(233) = 5.3 A(263) = 3.2 A(293) = 0.72 MAX(B) = 293 nm, FWHM = 60 nm B(233) = 0.72 B(263) = 3.2 B(293) = 5.3 Page 208 of 316

209 Absorbance Chem 155 Unit of 316 Given: A(233) = 5.3 A(293) = 0.72 B(233) = 0.72 B(293) = 5.3 A(233) = A (233)bC A + B (233)bC B A(293) = A (293)bC A + B (293)bC B A(233) = 2.55 A(293) = Unknown mixture of A and B Wavelength / nm What are the concentrations of A and B? Answer: C A = 0.4, C B = 0.6 A1 C A B1 C B A 1 A2 C A B2 C B A 2 I II 5.3 C A 0.72C B C A 5.3 C B 3.47 III II III III C A C B C A 0.72C B Substitute CA into I and II (to check): I III 5.3 C A 0.72C B 2.55 I C B C A 0.72C B C B C A II C B 3.47 C A C B Page 209 of 316

210 Chem 155 Unit of 316 Matrix approach: A1 C A B1 C B A 1 A1 A2 C A B2 C B A 2 A2 A3 C A B3 C B A 3 A3 B1 B2 B3. C A C B A 1 A 2 A 3 A4 C A B4 C B A 4 A4 B4 A 4 Ai Bi C A C B A i = the extinction coefficient of s pecies A at wavelength i = the extinction coefficient of s pecies B at wavelength i = the concentration of species A = the concentration of species B = the absorbance at wavelength i Use computers to solve matrix for C A and C B, C C etc. Overdetermine matrices to average out noise. Page 210 of 316

211 Chem 155 Unit of Intro to Fourier Transform Infrared Spectroscopy Skoog Chapters Covered Fourier Transform Optical Measurements 16A1 Vibrational Transitions 16A2 Heteronuclear Diatomics - Classical 16A3 Heteronuclear Diatomics - Quantum 16A4 Vibrational Modes 16A5 16C1 Vibrational Coupling Fourier Transform Instruments Page 211 of 316

212 Chem 155 Unit of Overview: Infrared (IR) spectroscopy deals mainly with: 1 molecular vibrations IR radiation is visible light. low energy relative to IR wavelengths range (approximately) from: Wavelength (nm) Frequency (cm -1 ) Near IR 1,000 2,500 10,000 4,000 Mid IR 2,500 20,000 4, Far IR 20, , Page 212 of 316

213 Chem 155 Unit of Why is IR spectroscopy important? 1. UV-Vis spectra are useful, but: all molecules: Vibrate So most all molecules have: Vibrational absorption spectra And most vibrational modes absorb: Strongly So IR spectroscopy is a very general and sensitive analytical tool. 2. Vibrational spectra have a lot of information: a. Fingerprinting / Identification Each molecule has a complex but unique signature. b. Quantitation Beer s law holds for vibrational absorption too! c. Measurement of Interactions / Configurations Vibrational frequencies (spectra) are sensitive to intermolecular interactions so one can identify phenomena such as hydrogen bonding, trans or gauche interactions that may control reactivity of molecules. Page 213 of 316

214 Chem 155 Unit of Some Applications of IR spectroscopy Identification of organic molecules Identification of functional groups Elucidation of molecular structure Elucidation of intermolecular interactions Monitoring of atmospheric pollutants. Some breathalizers use IR absorption of ethanol. Quantitation of amide nitrogen for the determination of protein content in food and other materials. IR Spectroscopy applications will follow in a later lecture section. Page 214 of 316

215 Chem 155 Unit of IR Spectroscopy is Difficult! IR beams are invisible to the eye. IR bands are narrow, so SiO 2 absorbs IR radiation. EFF must be small. Water and other solvents absorb IR radiation. IR photons are low energy, so they are hard to detect! Why can t you use a photomultiplier tube in the IR? Infrared radiation has a longer wavelength than visible radiation: IR > VIS so IR radiation has a lower frequency: IR = c/ IR so IR radiation has a lower energy: so Photomultiplier tubes don t work because: E IR = h IR, For a PMT to work, the photon energy must be greater than the work function of the metal! E But E IR Page 215 of 316

216 Chem 155 Unit of Monochromators Are Rarely Used in IR Because: 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 effective bandwidth, Grating or prism-monochromator based IR spectrometers are usually: slow noisy What is done instead? Interferometry In interferometry all light frequencies strike the detector at once and the information comes from signal oscillation as a function of: time! Could you measure the frequency of light by measuring the time between maxima in the electric field? c m s m c s Page 216 of 316

217 Signal / Volts Chem 155 Unit of Interferometers measure light field vs. time Imagine you were a light-speed gremlin with a voltmeter in hand and a light beam passed by: c 1 c c Time / femtoseconds This can be done with an optical device called an: Interferometer Take this on faith for now, and let s explore what that might look like Page 217 of 316

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

219 Chem 155 Unit of How is interferometry performed? Recall light interference is how gratings work. cos( ) 6 cos( ) 4 cos( ) cos( ) Light Electric Fields Sum Constructive Interference cos( ) 6 cos( ) 4 cos( ) cos( ) Destructive Interference Constructive interference: n = f( sin(i)+sin(r) ) for a grating For gratings constructive interference occurs when the path length difference between the rays,, is an integral multiple of the wavelength, : = d(sin(i)+sin(r)) = n n = f( time ) for an interferometer Page 219 of 316

220 Chem 155 Unit of Signal Fluctuations for a Moving Mirror Consider the case of mirror 2 moving at constant velocity, v (cm/s). AC(t) = Distance from beamsplitter to mirror 2 AB = Distance from beamsplitter to mirror 1 AC(0) = AB AC(t) = AB + vt Constant Velocity Mirror We can define the retardation - - as the difference in pathlength between the rays going to the fixed and moving mirrors respectively: (t) = 2(AC(t) - AB) (0) = 2(AC(0) - AB) = 2(AB - AB) = 0 What is the time interval, (s) between the conditions = 0 and =? Find for which ( ) = = 2(AC( ) - AB) =? ( hint: substitute AB + vt for AC(t) ) ( ) = = 2(AC( )-AB) = 2( (AB-v ) -AB) = 2v = 2v = / 2v Page 220 of 316

221 Chem 155 Unit of 316 So at every time interval the condition true and the detector sees a bright light: = n is What is the frequency (s -1 )of the detector signal in terms of, and in terms of the instrumental parameters v and? f (s -1 ) = 1/ (s) = 2v(cm/s) / (cm) For v = 1 cm/s and = 1000 nm, what will be the detector frequency, f? f (s -1 ) = 1/ (s) = 2 1(cm/s) / 1000x10-7 (cm) = 20,000Hz Page 221 of 316

222 Chem 155 Unit of Mono and polychromatic response The interferometer makes a low frequency oscillating signal from a high frequency light signal. But we still need to see the signal as a function of frequency (or wavelength) to understand it. Source: interferogram Spectrum monochromaticone frequency only. Two frequencies (lines) only. Polychromatic many frequencies. Page 222 of 316

223 Amplitude Chem 155 Unit of Interferograms are not informative: IR and other spectra are typically presented in the frequency domain i.e. as a function of frequency or wavelength not as a function of time. How can one transform: x T ime Into something informative like: In other words, the frequency or wavelength spectrum of the light is where the information is, how can one get this information from time domain signals? Page 223 of 316

224 Chem 155 Unit of Transforming time frequency domain Amplitude1 0 Amplitude time_seconds Frequency_Hz Amplitude time_seconds Amplitude Frequency_Hz Amplitude Amplitude time_seconds signals: Frequency_Hz Time Domain Frequency Domain conversion is done by computer using an algorithm called the: In what domain does a conventional monochromator operate? Discrete Fourier Transform Frequency or Wavelength Domain A vs. Page 224 of 316

225 Chem 155 Unit of The Centerburst: For a broad spectrum of frequencies going into a Michelson interferometer, for what value of is the constructive interference condition, n = satisified for all wavelengths? n = AB-BC = is always true if: = 0! Page 225 of 316

226 Amplitude Amplitude Amplitude Amplitude Chem 155 Unit of Time vs. frequency domain signals: It is simple to imagine deciphering the E(t) signal of a monochromatic light beam, but, what would a broadband light signal look like in the time domain? Broadband spectrum: Line spectrum: Frequency Frequency 2000 Interferogram T ime T ime Page 226 of 316

227 Chem 155 Unit of Advantages of Interferometry. In the IR, interferometers have advantages over monochromators: Advantage 1: Interferometers are fast: Consider a grating onochromator: If your source and detector are good from 200nm to 1200nm (1000nm total) and your effective bandwidth is 1 nm, how much light is being measured or lost at any one time? Multiplex Advantage 1 / (1000) is being measured so 1-[1/1000], or 99.9% is being thrown away! You only get one at a time! Advantage 2: Interferometers are high resolution. Recall: If EFF > the width of the spectral peak: Deviation from Beer s law Peak distortion (convolution) Page 227 of 316

228 Chem 155 Unit of Resolution in Interferometry Interferometers are easily made into highresolution spectrometers. Consider two spectral lines differing in frequency by only 2%: How do these lines appear in the time domain? cos( ) 6 cos( 1.02) 3 cos( ) cos( 1.02) Page 228 of 316

229 Chem 155 Unit of 316 Let us define resolution as: just be separated from 2. = 1-2 when 1 can I assert that to separate, or resolve these lines in the frequency domain spectrum, we must scan the retardation,, from 0 (where 1 and 2 are in phase) to a new value, let s just call it where the two signals are again in phase. The frequency of oscillation of the signal in time is: f = 2v(cm/s) / (cm) = 2 v where is the light frequency in wavenumbers (cm -1 ). The light power function will be sinusoidal: P( ) = P i cos(2 f t) = P i cos(2 (2v ) t) If we substitute v, the mirror velocity, with /2t = v we get intensity versus retardation: P( ) = B cos(2 ) Page 229 of 316

230 Chem 155 Unit of 316 For two light rays of identical power P i but different frequency the total signal will be: P( ) = P i cos(2 1) + P i cos(2 2) For identical B P( ) = P ( cos(2 1) + cos(2 2) ) What is P( ) when = 0? B( cos(0)+cos(0) ) = 2B So, when will P( ) be 2B again? In other words, at what next value >0 will the two different wavelengths 1 and 2 both give bright signals again? 1 = some integer 2 = some integer ± 1 1 = 1 ± = ±1 i.e. they have slipped one wavelength and are both bright again so, in this case: 1-2 = = 1/ Page 230 of 316

231 Chem 155 Unit of 316 For a large, is small and resolution is high! Consider an interferometer with a mirror travel d = 0.5 cm that is operating in the mid visible at 500 nm: = 10 7 /500 nm = 20,000 cm -1 = 2d = 1 cm = 1/ = 1/1cm, so 20,001 cm -1 can be resolved from 20,000 cm / 20,001 cm -1 = nm. So, 500 nm can be resolved from nm. = = 0.025nm EFF = 0.025nm is very difficult to achieve with a monochromator! Remember that displacement. is equal to 2d twice the mirror Page 231 of 316

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

233 Chem 155 Unit of Answers: f 2 v 2 1 cm s 3000cm s -1 effective bandwidth in cm 1 : EFF d cm 0.5cm 1 effective bandwidth in nm: 10 7 nm cm 3000cm nm cm cm nm EFF w d cos ( r) n F F w d cos ( r) n EFF 0.4 mm 1000nm cos nm 0.514m The detector frequency would be 6 khz not a problem. 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 this bandwidth with a monochromator would be 0.5 m, a fairly large system, especially when compared to a 1 cm interferometer drive. Page 233 of 316

234 Chem 155 Unit 13 Page 234 of Vibrational Spectroscopy: Selection Rules, Normal Modes and Group Frequencies Skoog Chapters Covered: 16A all, 16B Sources and Transducers 16C-1, FT-Instruments Rough overview of the optical spectroscopy ranges: (nm) (cm -1 ) T (K) Designation Phenomena , ,000 UV - *, n- *, VIS n- *, - * Valence electron excitation 1,000 10,000 14,000 Near IR low-e electronic Hi-E vibrations OH, C O, NH Mid IR Group Frequencies 10,000 1,000 1,400 Fingerprint Inorganics, High-Z, Low-k Far IR Lattice vibration, Pure rotations 100, wave 1,000, Page 234 of 316

235 Chem 155 Unit 13 Page 235 of Absorbance Bands Seen in the Infrared: : Vibrations Rotations Solid broad (phonon) rare (C 60 ) Liquid broad unresolved Gas sharp sharp Page 235 of 316

236 Chem 155 Unit 13 Page 236 of IR Selection Rules How does electromagnetic radiation make a molecule vibrate? Consider the diatomic molecule H-Cl. E( ) sin Electric Field Vector of Light E( ) + H H H - Cl Cl Cl In Out In Force Page 236 of 316

237 Chem 155 Unit 13 Page 237 of 316 Now consider the molecule N 2 : E( ) sin Electric Field Vector of Light E( ) N N N N N N When will a molecular vibration absorb light? when vibration causes change in dipole moment dynamic dipole needed Will interaction with electromagnetic radiation change the FREQUENCY or the AMPLITUDE of the vibration? Page 237 of 316

238 E( ) sin 2 Chem 155 Unit 13 Page 238 of Rotational Activity Indicate the forces on the molecule below: 360 Electric Field Vector of Light E( ) ` When can a molecular rotor absorb light? Example: Cl H H Cl When a molecule has a permanent dipole moment it is rotationally active. J = v = J = v = Page 238 of 316 Energy

239 Chem 155 Unit 13 Page 239 of Normal Modes of Vibration: Consider a diatomic molecule translations rotations vibration! For linear molecules, there are: modes of vibration. Non-linear molecules have: modes of vibration. 3N-5 3N-6 Page 239 of 316

240 Chem 155 Unit 13 Page 240 of 316 Modes of vibration: For example CO 2 : Linear or Non-linear? # of Modes: 3( ) - ( ) = S = 1340 cm -1 AS = 2340 cm -1 symmetric stretch anti-symmetric stretch S = 666 cm -1 S = cm -1 anti-symmetric stretch, IR active symmetric stretch, IR inactive bend modes, degenerate, IR active For example H 2 O: Linear or Non-linear? # of Modes: 3( ) - ( ) = S(OH) = AS(OH) = S(OH) = 3562 cm cm cm -1 Page 240 of 316

241 Chem 155 Unit 13 Page 241 of One may see more than 3N-5/6 bands: Overtones: Combinations: v = ±2, ±3 C = A + B and A - B One may see fewer than 3N-5/6 bands: not in 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 no change in dipole moment during the vibrational mode. Page 241 of 316

242 Chem 155 Unit 13 Page 242 of Group 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 group modes and show up in IR spectra. Examples of groups : C-H C=O C N NH 3 OH etc. 3N-6 rule irrelevant to this analysis is somewhat variable because: environment other bonds other modes influence the group vibration. Page 242 of 316

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

244 Chem 155 Unit 13 Page 244 of 316 Group frequencies can be calculated approximately: Interatomic forces are approximated by Hooke s Law: F = k y Since F = ma (Newton s Law) F = ma = m d 2 y dt 2 = ky(t) y(t) is proportional to its own second derivative. This differential equation has a sinusoid solution: y(t) = A cos(2 m t) m = 1 2 k = m 1 m 2 m 1 + m 2 reduced mass Units of are: units of mass Page 244 of 316

245 Chem 155 Unit 13 Page 245 of 316 Classical calculations (above) tell you the natural frequency of a classical oscillator. Quantuum mechanical calculations tell you the allowed energy levels (E). From Planck and Einstein s equation E=h we can calculate the light frequencies ( ). E = (v + ½ )h m v = vibrational quantuum number = 0,1,2 v = 0,±1 = the selection rule for vibrational transitions again: m = 1 2 k so: E = v h m = ±1 h 2 k = h note: m is the oscillator frequency and is the light frequency What is the relationship between and m? the light frequency equals the oscillator frequency What is the relationship between light frequency,, and light wavenumber, (cm -1 )? ( = 1/ (cm)). = c Page 245 of 316

246 Chem 155 Unit 13 Page 246 of Summary: Molecules vibrate their atoms move back and forth relative to each other. Vibrations come in certain modes that involve the whole molecule. Modes wherein the molecule has a change in dipole moment 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, some individual parts of the molecule behave as though they were independent. The absorption bands corresponding to these group modes are known as group frequencies and allow one to identify functional groups such as aldehydes, amides, olefins etc. on a given molecule. Page 246 of 316

247 Chem 155 Unit of Infrared Spectrometry - Applications Skoog Chapters Covered 17A1 Sample Handling in Mid-IR 17A2 Group Frequencies 17A3 Quantitation in Mid IR 17B Reflection Methods 17D Near IR applications 17E Far IR applications The major difficulty in infrared spectroscopy is related to the IR selection rule: during vibration => IR active Nearly everything absorbs IR radiation! optics and solvents will have strong absorption Other difficulties relative to, e.g. visible spectroscopy: Beam is invisible to the eye hard to align IR Sources are often weak (blackbodies) Detectors are not very sensitive Page 247 of 316

248 Chem 155 Unit of Strategies used to make IR spectrometry work - 1. Poor sources and detectors: use interferometer instead of monochromator 1. Optics absorb IR light: use mirrors instead of lenses 2. Materials for transmissive optics: use high-atomic-mass materials with weak inter atomic bonds typical materials are: 1. CaF 2 2. ZnSe 3. KBr 4. NaCl 3. Sample Handling: use solvents transparent in selected spectral regions avoid solvents completely use reflection-absorption instead of transmission Page 248 of 316

249 Chem 155 Unit of Solvents for IR spectroscopy: 14.3 Handling of neat (pure no solvent) liquids: Sandwich your sample between IR-transparent plates: e.g. NaCl: Q. If your sample is 10 microns thick, and the concentration of molecules in the neat sample is 10 M, and a typical band gives an absorbance of 0.1 a. what is? A= bc = A/bC = 0.1/0.001*10 = 10 b. why can t you use a 1-cm pathlength NaCl cuvette? Page 249 of 316 A= bc = 10*1*10 = 100 T= =

250 Chem 155 Unit of Handling of solids: pelletizing: grind put into die 5 10 tons of pressure a few minutes vacuum to reduce voids 14.5 Handling of Solids: mulling: sample 2-5 mg 2 m Nujol = mineral (aliphatic ) oil 1. Grind sample finely 2. Mix with mulling oil 3. Sandwich w/ KBr/NaCl plates or Fluorolube = fluorocarbon oil Page 250 of 316

251 Chem 155 Unit of A general problem with pellets and mulls: A = -log[p/po] How do you measure Po? You use an open beam to approximate Po So the spectra you obtain, are a sum of absorbances the spectra contain: Sample plus window plus mulling oil absorbance plus reflection losses. IR spectra obtained with these sampling methods are useful for identification, not quantitation: qualitative The two pieces of information that are useful: Functional group identification Spectral fingerprinting Page 251 of 316

252 Chem 155 Unit of Group Frequencies Examples : C-O-H versus C-Cl Which stretching frequency is higher: Why? O-H or C-Cl Reduced mass of C-Cl much larger so (k/ ) 1/2 for C-Cl smaller Page 252 of 316

253 Chem 155 Unit of Fingerprint Examples Mid-IR Group Frequencies and Fingerprints Example Branched Alkane Structural Isomers Page 253 of 316

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

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

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

257 Chem 155 Unit of 316 Attenuated Total Reflectance Sampling Volume and Quantitation: Penetration depth Even though Snell s Law says the light should totally internally reflect, the light actually behaves as though it were reflecting from a plane situated at: A point just above the surface, slightly penetrating the sample. Page 257 of 316

258 Chem 155 Unit of 316 The reflected beam interacts with the sample on the crystal surface. It behaves as though it passes through the sample and is reflected back by a plane inside the sample some fraction of a wavelength above the crystal surface (i.e. the n 1 -n 2 interface). Page 258 of 316

259 Chem 155 Unit of Raman Spectroscopy: vibrational spectroscopy with near UV, visible or near infrared light is called: Raman Scattering Spectroscopy Intense monochromatic light beam usually a laser H 2 O A tiny fraction of the incident radiation scatters off of the molecules in the solution: Rayleigh An even tinier fraction of the incident radiation scatters off of the molecules and lose or gain energy. Raman SCATTERED = INCIDENT SCATTERED INCIDENT elastic scattering inelastic scattering Page 259 of 316

260 Chem 155 Unit of 316 A Raman scattering experiment: Scattered Beam High resolution monochromator / interferometer + sensitive detector Laser Transmitted Beam Incident Beam The sample may be a concentrated solution, a neat liquid analyte, a transparent solid or a surface. Raman is not usually used for dilute solutions. If the laser is, e.g. a HeNe laser ( =632.8 nm), most of the scattered radiation will be: cm -1 nm or 1/(632.8x10-7 cm) = This scattered radiation is monochromatic, and is known in the spectrum as the: Rayleigh Line Page 260 of 316

261 Light Power Chem 155 Unit of What a Raman Spectrum Looks Like Identify 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 differences between the Stokes and anti-stokes peaks.. Hypothetical Raman Scattering Spectrum Wavelength / nm Page 261 of 316

262 Light Power Energy Chem 155 Unit of Quantum View of Raman Scattering. S1 So Vibrational coordinate Hypothetical Raman Scattering Spectrum Wavelength / nm Page 262 of 316

263 Chem 155 Unit of Classical View of Raman Scattering Visible frequency radiation is higher frequency than IR radiation and vibrational frequencies. Molecular Vibration (e.g. C=O stretch, 2000 cm -1 ) r r eq cos 2 vib t time Visible Frequency Light (e.g. 500 nm / 20,000 cm -1 / blue-green) E E 0 cos 2 ex t time Page 263 of 316

264 Chem 155 Unit of 316 Radiation interacts with molecules by inducing a dipole moment. The induced dipole m is proportional to the polarizability,. m E E 0 cos 2 ex t But the polarizability is itself oscillating! 0 d dr r r e 0 d dr r A cos 2 vib t So the induced dipole moment is oscillating in a more complex way: m E 0 cos 2 ex t 0 d dr r A cos 2 vib t The dynamic induced dipole that gives rise to Raman scattering: Incident E-field Oscillating polarizability Amplitude modulation in induced dipole Page 264 of 316

265 Light Power Chem 155 Unit of The classical model of Raman: The induced dipole moment has a complex description, but how does this relate to Raman scattering? Recall that: cos( x) cos( y) So m becomes: m E 0 0 cos 2 ex t 1 2 E 0 1 cos( x y) 1 cos( x y) 2 2 d dr r A cos 2 ex vib cos 2 ex vib Both Stokes and AntiStokes peaks are predicted by classical theory! Classical Model 'Catastrophe' Wavelength / nm but Page 265 of 316

266 Light Power Chem 155 Unit of The classical model: catastrophe! Real intens ity dis tributions Frequency / cm-1 The classical model does not predict the Stokes / anti-stokes intensity ratios. With what does an equal intensity distribution violate? Boltzmann Distribution between v=0 and v=1 vibrational levels. So, the answer is: the quantum model wins, and we accept the virtual state. I m told that time-dependent solutions to the molecular structure allow for these evanescent states. more The Stokes peaks are anti Stokes peaks because: intense than the there are more molecules in the ground vibrational state, that can give rise to Stokes peaks, than in the excited vibrational state, that can give rise to anti-stokes peaks. Page 266 of 316

267 Chem 155 Unit of Raman Activity: mode Raman active? IR active? O C Symmetric stretch NO YES YES NO O O C Asymmetric stretch YES NO NO YES O O C Bend YES YES YES YES O Page 267 of 316

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

269 Chem 155 Unit of Some general points regarding Raman: Pro: Raman uses visible light: Optics Solvents Biological samples Imaging with Raman has better spatial resolution because of the shorter wavelengths used. Microscopes can be used in Raman microprobe mode. Silver and gold particles or rough surfaces can very strongly enhance Raman scattering. Different vibrational modes can be seen (e.g. centrosymmetric) vib 1 2 k Very low frequency modes are accessible (e.g. inorganics): Depolarization ratio can distinguish between symmetric and asymmetric stretches. Page 269 of 316

270 Chem 155 Unit of 316 Con: Raman signals are often very weak. High-powered lasers can damage samples, so this is not always a good solution. Fluorescent molecules can give bad spectral interference for Stokes lines. Page 270 of 316

271 Chem 155 Unit of Resonance Raman S1 So Pro s 1. Intensity greatly enhanced ( ) 2. Selective for vibrations of chromophore Con s 1. Sample degradation because of absorption. 2. Fluorescence can swamp Raman signals. Fluorescence can be rejected with a very short pulsed laser source: typical fluorophore - FLUOR ~ 10-9 s. Raman - RAMAN ~ S Page 271 of 316

272 Light Power Chem 155 Unit of Raman Exercises Given the spectrum below, compute the following: 1. The laser wavelength. 2. The vibrational frequencies of the modes in the molecule giving rise to the following spectrum. 3. The effective bandwidth is needed in the visible wavelength monochromator to achieve 1 cm -1 resolution with Raman given the above laser system and chromophore.. Hypothetical Raman Scattering Spectrum Wavelength / nm las er frequency: cm 1 Stokes frequency: cm cm 1 Effective bandwidth: Must resolve and cm nm high res olution needed! 2759 cm Page 272 of 316

273 Chem 155 Unit of Using and Ar-ion laser at 514.5nm, at what wavelength would you expect to see the Stokes peak for the NO2 resonance of nitrobenzene given that it is normally observed at 1520 cm -1 in IR absorption spectroscopy? cm nm 5. Does the CO bond in CO 2 have a dipole moment? YES 6. 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 symmetric stretching mode. O C O O C O O C O 8. Is the CO 2 symmetric stretch IR active? Why? No, because at no time during the vibration does the dipole moment change. 9. Is the CO 2 symmetric stretch Raman active? Why? Yes, because during the vibration the polarizability changes. Page 273 of 316

274 Chem 155 Unit of Draw a series of CO 2 molecules to illustrate the asymmetric stretching mode. O C O O C O O C O 11. Is the asymmetric stretch IR active? Why? Yes, because at during the vibration the dipole moment changes. 12. Is the asymmetric stretch Raman active? Why? No, because at during the vibration the polarizability does not change. (increase in one CO bond polarizability is offset by decrease in the other bond). 13. Why is N 2 not a greenhouse gas? Homonuclear diatomics have no dipole moment change during their only vibrational mode: symmetric stretch. Page 274 of 316

275 Chem 155 Unit of Cyclohexanone has a strong absorption peak at 5.86 m and at this wavelength a linear relationship exists between absorbance and concentration. a. Calculate the frequency of this mode in cm -1. b. Identify the part of the molecule responsible (the group frequency) for the absorbance at this wavelength. This is the carbonyl or ketone symmetric stretch. c. Suggest a solvent that would be suitable for a quantitative analysis of cyclohexanone at this wavelength. CCl 4, CHCl 3 or tetrachloroethylene d. A solution of cyclohexanone (2.0 mg/ml) in the selected solvent has an absorbance of 0.40 in a cell with a path length of mm. What is the detection limit for this compound under these conditions, if the noise associated with the spectrum of the solvent is absorbance units. A=εbC calibration slope = εb cm 1 b A C mg ml 0.2 mg ml 1 C m 3 s b m mg ml mg ml Page 275 of 316

276 Chem 155 Unit of Calculate the ratio of HCl molecules in the first vibrational excited state at 25 C relative to the ground state. The IR absorption appears at 2885 cm -1. N1 g1 a. The Boltzmann equation is: N0 g0 kt e where g 0 and g 1 are the degeneracies of the states 0 and 1 and for HCl 35 may be considered equal, and k is Boltzmann s constant, T is absolute temperature and E is the k J K T 298 K energy of state cm J s cm J s relative to 0. E 2885cm J s cm s E k T J N1 N0 e kt E E k T e b. Assuming that the bond strengths are equal, calculate the vibrational frequency that you expect for the isotope DCl 35. DCl HCl VIB k k DCl DCl k HCl HCl since m 1 m 2 m 1 m 2 k HCl k DCl DCl HCl HCl DCl HCl DCl HCl DCl DCl DCl 2069 Page 276 of 316

277 Chem 155 Unit of Draw three simple Jablonski diagrams, e.g. for a simple molecule and indicating just two electronic states S0 and S1 (no T state is needed). a. On the first diagram, draw arrows corresponding to light absorption followed by fluorescent light emission. b. On the second, draw arrows corresponding to the Stokes Raman peaks. c. On the third, draw arrows corresponding to the Anti- Stokes Raman process. S1 Virtual state S0 Fluorescence Raman-Stokes Raman-AntiStokes Page 277 of 316

278 Chem 155 Unit of A peak is observed in a fluorescence spectrometer when analyzing trace impurities in water. With the monochromator excitation source set to 250 nm, the mysterious weak peak appears at 274 nm, unusually close to the excitation band. When the excitation wavelength was changed to 300 nm, the mystery bands moved to 335 nm. a. Calculate the Stokes shift for the mystery band in cm -1 for the 250 nm excitation experiment. b. Calculate the Stokes shift for the mystery band in cm -1 for the 300 nm excitation experiment. c. What might this mystery peak be? The band moves with the excitation it is the Raman peak for water OH stretch modes. Page 278 of 316

279 Chem 155 Unit of Mass Spectrometry (MS) overview: Skoog: sections 20A and B In MS one a. often initially performs b. gets analyte molecules into the c. converts analyte molecules into d. that are both e. exposes the analyte ions to f. separates them based on g. detects them based on GC or LC separation gas phase in vacuum ions intact and in fragments E and/or B fields mass / charge current (they are ions) 16.1 Example: of a GCMS instrument: Page 279 of 316

280 Chem 155 Unit of Block diagram of MS instrument. Sample Introduction Ionization Source Mass Analyzer Computer for Display / storage Pattern recognition to ID molecule Page 280 of 316

281 Chem 155 Unit of Information from ion mass Mass spectra can distinguish o Different isotopes of the same element. o Small mass differences between molecules with the same nominal 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 Cl But this can be informative: Isobaric interferants Asparagine 13 CH 2 CONH 2 (133.12) Aspartic acid CH 2 CO 2 H (133.10) Distinguishable only by high resolution MS Source: Quantitative Chemical Analysis 6 th Ed. By Daniel Harris Page 281 of 316

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

283 Chem 155 Unit of 316 Example of CI vs EI MS of 1-octanol: Electron impact ionization chemical ionization (CH 4 ) Which is a better MS? CI: EI: more easily interpretable complex, but highly unique Page 283 of 316

284 Chem 155 Unit of Matrix Assisted Laser Desorption-Ionization (MALDI): o Soft o Pulsed introduction of solids o Large molecules proteins / biomolecules o Gives predominance of M + sample in vacuum chamber proteins in matrix matrix ionizes supersonic expansion Page 284 of 316

285 Chem 155 Unit of Electrospray Ionization (ESI): o Soft o Continuous introduction of solids o Large molecules proteins / biomolecules o 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

286 Chem 155 Unit of 316 Which is a better ionization source for large molecules, MALDI or ESI? MALDI: ESI: spectra are more easily interpretable spectra are complex, but ESI easily interfaced to LC Page 286 of 316

287 Chem 155 Unit of Mass Analyzers: One important quality of mass analyzers is: Resolution: R M/ M Small 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 Mass 12 C2 1 H C 1 H2 14 N + 14 N C 16 O What resolution would be needed to distinguish these? Smallest M = = m / M = 28 / = 2300 What resolution would one need to calculate the nominal mass of immunoglobulin-g (M = )? m / M = 149,190 / 1 = 150,000 Page 287 of 316

288 Chem 155 Unit of Magnetic Sectors: Resolution of a single magnetic sector is ca The properties that limit R of the ions entering it, mainly: speed distribution directional distribution Page 288 of 316

289 Chem 155 Unit of Double-Focusing Electric Magnetic Sectors Electric Sector Focuses: Speed distribution Magnetic Sector Re-focuses: Directional distribution At double-focus point, all ions have narrow speed and direction distributions, so only ions of a given M/z exit at that point. What determines the M/z values that exit? E-field on electric sector B-field on magnetic sector R attainable: 100,000 Page 289 of 316

290 Chem 155 Unit of Quadrupole Mass Filters: 10,000 Resolution up to: Masses limted to: < 4000 AMU Stable trajectory is a function of: AC amplitude DC field A given AC amplitude + DC field yields a stable trajectory for one: Like E and B sectors this is a: One scans m/z: M/z mass filter scanning mode (see one at a time) Page 290 of 316

291 Chem 155 Unit of Time of Flight (TOF): Ion kinetic energy E = zv PULSE But kinetic energy E = ½mv 2 So, a drift length L gives a the flight time t F = L/v Solve for m/z as a function of V PULSE and distance L: E = zv PULSE = ½mv 2 = ½m(L/t F ) 2 2V PULSE (t F /L) 2 = m/z = (2V/L²) t F 2 So, one measures ion m/z by measuring the ion s Arrival time at the detector: t F Page 291 of 316

292 Chem 155 Unit of 316 TOF analyzers: 1. High mass capable. 2. Fast 100 of spectra per second. 3. Efficient / sensitive all ions are collected. 4. Good Resolution up to Triple Quadrupole Analyzers: 16.6 Mass Spec Questions: 1. What might you label the axes on a mass spectrum? 2. Consider a fragment of nominal mass 180 AMU in a mass spectrometer. What peaks might arise from this one fragment? 3. Consider a high resolution mass spectrum of the singly charged fragment CHCl 3 +. Though varying greatly in intensity, it will be possible to resolve 12 different peaks for this fragment. Assign the isotopomers and predict the masses. 4. Forensic evidence was gathered against U.S. cyclist Floyd Landis was based on a high testosterone level seen in a blood sample. Since testosterone is a naturally occurring hormone, it was necessary to analyze this further. Testosterone made H by biosynthesis in the human body differs in only one way H H from that prepared in a laboratory 13 C/ 12 O C isotopic abundances. What resolution is required to measure this ratio? Page 292 of 316 testosterone OH C 19 H 28 O 2 Exact Mass: Mol. Wt.: m/e: (100.0%), (20.6%), (2.1%) C, 79.12; H, 9.78; O, 11.09

293 Chem 155 Unit of What type of mass analyzer might be useful for this? 6. Would it be necessary to perform a separation step before 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 of testosterone above, can you distinguish between the exact mass and the molecular weight? 9. To what do the 20% and 2% fragments correspond? C 19 H 28 O 2 Exact Mass: Mol. Wt.: m/e: (100.0%), (20.6%), (2.1%) C, 79.12; H, 9.78; O, Element Iso % Carbon 12 C 98.90% 13 C 1.10% Hydrogen 1 H 99.99% 2 H 0.01% Oxygen 16 O O 0.038% 18 O 0.2% Page 293 of 316

294 Chem 155 Unit 17 Chromatography Page 294 of Chromatography Chapter 26 General and descriptive aspects of chromatographic retention and separation: phenomenological k, efficiency, selectivity. Quantitative description of zone migration in partition chromatography: migration velocity, partition coefficient and theoretical k. Theoretical description of efficiency: zone broadening and 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 normal phase or flash ), gel permeation (GPC), gel filtration, ion exchange. Electrophoresis. Page 294 of 316

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

296 Chem 155 Unit 17 Chromatography Page 296 of 316 Some definitions: Mobile Phase is the gas, liquid or supercritical fluid that passes through separation column or slab. Stationary Phase is the solid or immobilized liquid into which the solutes or analytes partition, adsorb or bind. Band or Zone describes the region (stripe, plug) of the separation column or slab that contains the solute or analyte. Retention is the word used to refer to the delay of the solute or analyte in reaching the detector relative to the mobile phase. Page 296 of 316

297 Chem 155 Unit 17 Chromatography Page 297 of 316 Page 297 of 316

298 Chem 155 Unit 17 Chromatography Page 298 of 316 Page 298 of 316

299 Chem 155 Unit 17 Chromatography Page 299 of 316 Page 299 of 316

300 Chem 155 Unit 17 Chromatography Page 300 of 316 Page 300 of 316

301 Chem 155 Unit 17 Chromatography Page 301 of 316 Page 301 of 316

302 Chem 155 Unit 17 Chromatography Page 302 of 316 x Page 302 of 316

303 signal Chem 155 Unit 17 Chromatography Page 303 of 316 Theoretical Plates and Plate Count: x i i x 1 1 S x exp N: N Illustration of different p late counts plates plates 6400 plates plates retention time or column length Page 303 of 316

304 Chem 155 Unit 17 Chromatography Page 304 of 316 Plate count depends on retention time: Why does the plate count go up for the more retained peaks? N: N Illustration of different p late counts signalx retention time or column length Page 304 of 316

305 Chem 155 Unit 17 Chromatography Page 305 of 316 The van Deemter Equation: H = plate height A = multipath term B = longitudinal diffusion term C = resistance to mass transfer H A B u Plate height as a function of linear flow rate: H( A B C u) A B u C u C u H( u) u Rationalize why is there an optimum flow rate: Page 305 of 316

306 Chem 155 Unit 17 Chromatography Page 306 of 316 A 2 d p B 2 D M H A B C u C = com plex function of particle u size, coating, diffusion coefficient, favored in general by large diffusivity and small particle size Page 306 of 316

307 Chem 155 Unit 17 Chromatography Page 307 of General Elution Problem / Gradient Elution A mixture of molecules with strongly differing k values is hard to separate because: If the solvent is weak enough to separate 1 and 2: a. b. If the solvent is strong enough to elute 5 and 6 in a reasonable time: x Page 307 of 316

308 Chem 155 Unit 17 Chromatography Page 308 of 316 Isocratic (LC) and isothermal (GC) separations are vulnerable to the general elution problem. The solution to the general elution problem: o solution gradient elution (liquid x chromatography) o temperature gradient elution (gas chromatography) The gradient methods: o begin with weakly eluting conditions and this helps to separate the weakly retained species o finish with strongly eluting conditions to expedite the elution of strongly retained species and to limit diffusional broadening In LC gradient elution is a bit more difficult because: o the columns require a fairly significant equilibration time between runs o the detector baseline can drift significantly if the different solvents have different UV absorbance at the detection wavelength if this is a problem, then one can use o a different wavelength or o a different set of solvents Page 308 of 316

309 Chem 155 Unit 17 Chromatography Page 309 of T-gradient example in GC of a complex mixture. x Page 309 of 316

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

311 Chem 155 Unit 17 Chromatography Page 311 of Types of Liquid Chromatography Most can be performed as HPLC in high performance instrumentation x if necessary Normal Phase: stationary phase: hydrophilic (SiO 2 particles) mobile phase: hydrophobic mobile phase (hexane, methanol). weak solvent: hexane strong solvent: methanol retained: polar compounds eluted first: non-polar compounds Reverse Phase: stationary phase: hydrophobic (SiO 2 particles coated with hexadecane monolayer) mobile phase: hydrophilic mobile phase (water, methanol) weak solvent: water strong solvent: methanol retained: eluted first: non-polar compounds polar compounds Page 311 of 316

312 Chem 155 Unit 17 Chromatography Page 312 of Exclusion Gels GPC Gel permeation chromatography: porous polymer gel particles x small molecules penetrate and are retained, large molecules elute more quickly. PAGE polyacrlamide gel electrophoresis: Aqueous gel: -CH 2 -CONH- used for electrophoresis, large molecules experience more drag and are retained. Page 312 of 316

313 Chem 155 Unit 17 Chromatography Page 313 of 316 Ion Exchange: Stationary phase is an ionic polymer e.g. polystyrenesulfonic acid: SO 3 x - * * n The solid or gel phase polymer transiently binds cations from solution. Cations that have a higher affinity for the resin are retained longer and viceversa. Page 313 of 316

314 Chem 155 Unit 17 Chromatography Page 314 of HPLC System overview: Injection valve x Page 314 of 316

315 Chem 155 Unit 17 Chromatography Page 315 of Example of Reverse-phase HPLC stationary phase: x Page 315 of 316