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

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1 Overview and Review ...... 7 2 Propagation of Error ...... 56 3 Introduction to Spectrometric Methods ...... 65 4 Photometric Methods and Spectroscopic Instrumentation ...... 86 5 Radiation Transducers (Light Detectors): ...... 103 6 Monochromators for Atomic Spectroscopy: ...... 117 7 Photometric Issues in Atomic Spectroscopy ...... 138 8 Practical aspects of atomic spectroscopy: ...... 152 9 Atomic Emission Spectroscopy ...... 163 10 Ultraviolet-Visible and Near Infrared Absorption ...... 178 11 UV-Visible Spectroscopy of Molecules ...... 194 12 Intro to Fourier Transform Infrared Spectroscopy ...... 211 13 Infrared Spectrometry: ...... 234 14 Infrared Spectrometry - Applications ...... 247 15 Raman Spectroscopy: ...... 259 16 Mass Spectrometry (MS) overview: ...... 279 17 Chromatography ...... 294

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1 Overview and Review ...... 7 1.1 Tools of Instrumental Analytical Chem...... 8 1.2 Instrumental vs. Classical Methods...... 12 1.3 Vocabulary: Basic Instrumental ...... 13 1.4 Vocabulary: Basic Statistics Review ...... 14 1.5 Statistics Review ...... 15 1.6 Calibration Curves and Sensitivity ...... 23 1.7 Vocabulary: Properties of Measurements ...... 24 1.8 Detection Limit ...... 25 1.9 Linear Regression ...... 31 1.10 Experimental Design: ...... 35 1.11 Validation – Assurance of Accuracy: ...... 43 1.12 Spike Recovery Validates Sample Prep...... 45 1.13 Reagent Blanks for High Accuracy: ...... 46 1.14 Standard additions fix matrix effects: ...... 47 1.15 Internal Standards ...... 52 2 Propagation of Error ...... 56 3 Introduction to Spectrometric Methods ...... 65 3.1 Electromagnetic Radiation: ...... 66 3.2 Energy Nomogram ...... 67 3.3 Diffraction ...... 68 3.4 Properties of Electromagnetic Radiation: ...... 71 4 Photometric Methods and Spectroscopic Instrumentation ...... 86 4.1 General Photometric Designs for the Quantitation of Chemical Species ...... 87 4.2 Block Diagrams ...... 88 4.3 Optical Materials ...... 89 4.4 Optical Sources ...... 90 4.5 Continuum Sources of Light: ...... 91 4.6 Line Sources of Light: ...... 92 4.7 Laser Sources of Light: ...... 93 5 Radiation Transducers (Light Detectors): ...... 103 5.1 Desired Properties of a Detector: ...... 103 5.2 Photoelectric effect photometers ...... 104 5.3 Limitations to photoelectric detectors: ...... 106 5.4 Operation of the PMT detector: ...... 107 5.5 PMT Gain Equation: ...... 108 5.6 Noise in PMT‟s and Single Photon Counting: ...... 110 5.7 Semiconductor-Based Light Detectors: ...... 112 5.8 Charge Coupled Device Array Detectors: ...... 115 6 Monochromators for Atomic Spectroscopy: ...... 117 6.1 Adjustable Wavelength Selectors ...... 118 6.2 Monochromator Designs: ...... 119 6.3 The Grating Equation: ...... 120 6.4 Dispersion ...... 123 6.5 Angular dispersion:...... 124

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6.6 Effective bandwidth ...... 126 6.7 Bandwith and Atomic Spectroscopy ...... 127 6.8 Factors That Control EFF ...... 128 6.9 Resolution Defined ...... 129 6.10 Grating Resolution ...... 130 6.11 Grating Resolution Exercise: ...... 131 6.12 High Resolution and Echelle Monochromators ...... 133 7 Photometric Issues in Atomic Spectroscopy ...... 138 8 Practical aspects of atomic spectroscopy:...... 152 8.1 Nebulization (sample introduction): ...... 153 8.2 Atomization ...... 157 8.3 Flame Chemistry and Matrix Effects...... 158 8.4 Flame as „sample holder‟: ...... 159 8.5 Optimal observation height: ...... 160 8.6 Flame Chemistry and Interferences: ...... 161 8.7 Matrix adjustments in atomic spectroscopy: ...... 162 9 Atomic Emission Spectroscopy ...... 163 9.1 AAS / AES Review: ...... 164 9.2 Types of AES: ...... 165 9.3 Inert-Gas Plasma Properties (ICP,DCP) ...... 166 9.4 Predominant Species are Ar, Ar+, and electrons ...... 166 9.5 Inductively Coupled Plasma AES: ICP-AES ...... 167 9.6 ICP Torches ...... 168 9.7 Atomization in Ar-ICP ...... 169 9.8 Direct Current Plasma AES: DCP-AES ...... 170 9.9 Advantages of Emission Methods ...... 171 9.10 Accuracy and Precision in AES ...... 173 10 Ultraviolet-Visible and Near Infrared Absorption ...... 178 10.1 Overview ...... 178 10.2 ...... 178 10.3 The Blank ...... 179 10.4 Theory of light absorbance: ...... 180 10.5 Extinction Cross Section Exercise: ...... 181 10.6 Limitations to Beer‟s Law: ...... 182 10.7 Noise in Absorbance Calculations: ...... 184 10.8 Deviations due to Shifting Equilibria: ...... 185 10.9 Monochromator Slit Convolution in UV-Vis: ...... 188 10.10 UV-Vis Instrumentation: ...... 190 10.11 Single vs. double-beam instruments: ...... 191 11 UV-Visible Spectroscopy of Molecules ...... 194 11.1 Spectral Assignments ...... 195 11.2 Classification of Electronic Transitions ...... 196 11.3 Spectral Peak Broadening ...... 197 11.4 Aromatic UV-Visible absorptions: ...... 201 11.5 UV-Visible Bands of Aqeuous Transition Metal Ions ...... 202 11.6 Charge-Transfer Complexes ...... 205

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11.7 Lanthanide and Actinide Ions: ...... 206 11.8 Photometric Titration ...... 207 11.9 Multi-component Analyses: ...... 208 12 Intro to Fourier Transform Infrared Spectroscopy ...... 211 12.1 Overview: ...... 212 1 molecular vibrations ...... 212 12.2 IR Spectroscopy is Difficult! ...... 215 12.3 Monochromators Are Rarely Used in IR...... 216 12.4 Interferometers measure light field vs. time ...... 217 12.5 The Michelson interferometer: ...... 218 12.6 How is interferometry performed? ...... 219 12.7 Signal Fluctuations for a Moving Mirror ...... 220 12.8 Mono and polychromatic response ...... 222 12.9 Interferograms are not informative: ...... 223 12.10 Transforming time  frequency domain signals: ...... 224 12.11 The Centerburst: ...... 225 12.12 Time vs. frequency domain signals: ...... 226 12.13 Advantages of Interferometry...... 227 12.14 Resolution in Interferometry ...... 228 12.15 Conclusions and Questions: ...... 232 12.16 Answers: ...... 233 13 Infrared Spectrometry: ...... 234 13.1 Absorbance Bands Seen in the Infrared: ...... 235 13.2 IR Selection Rules ...... 236 13.3 Rotational Activity ...... 238 13.4 Normal Modes of Vibration: ...... 239 13.5 Group frequencies: a pleasant fiction! ...... 242 13.6 Summary: ...... 246 14 Infrared Spectrometry - Applications ...... 247 14.1 Strategies used to make IR spectrometry work - ...... 248 14.2 Solvents for IR spectroscopy: ...... 249 14.3 Handling of neat (pure – no solvent) liquids: ...... 249 14.4 Handling of solids: pelletizing: ...... 250 14.5 Handling of Solids: mulling: ...... 250 14.6 A general problem with pellets and mulls: ...... 251 14.7 Group Frequencies Examples ...... 252 14.8 Fingerprint Examples ...... 253 14.9 Diffuse Reflectance Methods: ...... 254 14.10 Quantitation of Diffuse Reflectance Spectra: ...... 255 14.11 Attenuated Total Reflection Spectra: ...... 256 15 Raman Spectroscopy: ...... 259 15.1 What a Raman Spectrum Looks Like ...... 261 15.2 Quantum View of Raman Scattering...... 262 15.3 Classical View of Raman Scattering ...... 263 15.4 The classical model of Raman: ...... 265 15.5 The classical model: catastrophe! ...... 266

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15.6 Raman Activity: ...... 267 15.7 Some general points regarding Raman: ...... 269 15.8 Resonance Raman ...... 271 15.9 Raman Exercises ...... 272 16 Mass Spectrometry (MS) overview: ...... 279 16.1 Example: of a GCMS instrument: ...... 279 16.2 Block diagram of MS instrument...... 280 16.3 Information from ion mass ...... 281 16.4 Ionization Sources ...... 282 16.5 Mass Analyzers: ...... 287 16.6 Mass Spec Questions: ...... 292 17 Chromatography – Chapter 26 ...... 294 17.1 General Elution Problem / Gradient Elution ...... 307 17.2 T-gradient example in GC of a complex mixture...... 309 17.3 High Performance Liquid Chromatography ...... 310 17.4 Types of Liquid Chromatography ...... 311 17.5 Normal Phase: ...... 311 17.6 HPLC System overview: ...... 314 17.7 Example of Reverse-phase HPLC stationary phase: ...... 315 17.8 Ideal qualities of HPLC stationary phase: ...... 316

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1 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…

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1.1 Tools of Instrumental Analytical Chem. 1.1.1 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 x-ray absorption, electron fluorescence, xps Vacuum 10 - 180 nm Valence Vuv Ultraviolet electron Ultraviolet 180 - 400 Valence Uv or uv-vis electron Visible 400-800 Valence Vis or uv-vis electron Near Infrared 800-2,500 Vibration Near IR or NIR (overtones) Infrared 2.5-40 m Vibration IR or FTIR

Microwave 40 m – 1 rotations Rotational or mm microwave Microwave 30 mm Electron spin ESR or EPR in mag field Radiowave 1 m Nuclear spin NMR in mag field

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1.1.2 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 absorbance

time / s

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1.1.3 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

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1.1.4 Electrochemistry Simple, sensitive, limited to certain chemicals

Ion selective electrodes (ISE‟s): e.g. pH, pCl, pO2 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+] = 0.0001 0.0000000001 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]

1.1.5 Gravimetry Precipitate and weigh products – very precise, very limited

1.1.6 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

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Relax, you don‟t need 1.2 Instrumental vs. Classical Methods. to memorize this table – just humor Dr. Terrill while he talks about it. Methods of Analytical Chemistry # Chemicals # Chemicals Isolated / hr Isolated / hr Classical Instrumental and amount and amount (g) (g) High Performance Extraction 1-2 g Liquid 10 ng Chromatrography Gas Distillation 1-2 g 100 ng Chromatography Separation Precipitation 1-2 g Electrophoresis 50 pg Crystallization 1-2 g

Estimated Number of uniquely identifiable molecules by method

Combination of UV-Vis 1,000‟s Color / Smell Melt / Boiling Infrared 100,000‟s Qualitative Point, 100‟s Speciation Solubility Mass Spectrometry > 106 Wetting Density NMR > 106 Hardness Spectroscopy Best Quantitative Precision and Sensitivity Optical Titration 0.1 Spectroscopy 0.1% 10-23 M % 1 ppm

0.1% Mass Quantitation Gravimetry 0.01 amount -13 1 ppm Spectrometry 10 M Precision % -4 10 % mass Colorimetry NMR 100 10% 1 ppm Spectroscopy 1% pppm

What are the more precise measurements that you have made and what were they?

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1.3 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.

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1.4 Vocabulary: Basic Statistics Review

Precision 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 Random error is a measure of precision. Random differences between sequential measurements reflect the random error. Accuracy If a measurement of something is correct, i.e. close to the true value, then that measurement is accurate. Systematic Error Systematic error is the difference between the mean of a population of measurements and the (bias) true value. Histogram A graph of the number or frequency of occurences of a certain measurement versus the measurement value. Probability A theoretical curve of the probability of a certain measured value occurring versus the measured Distribution 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. xMEAN n Median Half of the measurements fall above the median value, and half fall below. 2 A measurement of x x 2 Variance ( ) precision. The sum of the MEAN n squares of the random 2 n measurement errors. n 1 A widely accepted Standard 2 „standard‟ measurement of xMEAN xn Deviation ( ) precision. The square root n of the variance. n 1 Relative Standard The standard deviation divided by the mean, and often expressed as : %RSD= /x 100%. Deviation MEAN Propagaion of When the mean of a set of measurement (x) has a random error ( x), it is reported as x± x. If we Error 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.

Page 14 of 316 0 02.781 1 2.66 22.763 32.524 42.157 53.022 6 2.94 a 73.278 84.096 Chem 155 Unit 1 Page 15 of93.404 316 103.493 113.431 1.5 Statistics Review 123.458 133.337 142.478 1.5.1 Precision and Accuracy 153.035 Histogram of normally distributed events 80 Mean

60 Mean - one Mean + 40 standard one deviation standard

Number of times it times of was Number observed deviation 20

0 0 1 2 3 4 5 6 Value observed

Histogram of 1024 events

1.5.2 Basic Formulae

2 x Population N lim xN Standard N N x lim N Deviation: N Mean : N N 2 x x x N Sample N avg N Standard N xavg sx Average: N Deviation: N 1

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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).

1.5.3 Confidence Interval

WHAT DO YOU DO TO ENSURE THAT YOUR ANSWER IS AS CLOSE AS POSSIBLE TO THE TRUTH?

TAKE THE AVERAGE xAVG

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)

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Confidence Interval (continued).

If you know the standard deviation, , (less common case), then: ½ The x% confidence interval for = xAVG ± z / N

If you have only a rough If you don‟t know the standard estimate of x , then you are deviation, , (more commonly the AVG less confident that it is close to case), then: , hence you divide by N½.

½ The x% confidence interval for = xAVG ± ts / N In this case t is a function of N

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.

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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

XAVG = (4.2+4.6+4.0) / 3 = 4.27

2 2 2 s = (4.2-4.27) +(4.6-4.27) +(4.0-4.27) = 0.31 3-1

Number of degrees of freedom for confidence interval = 3-1=2

1/2 95% confidence limits for = 4.27± (4.3*0.31) / (3 ) = 4.27±0.93 or 5.4 = 4.3 ± 0.9 or

= 3.4 to 5.2 5.0

4.5

3.5

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1.5.4 Confidence Interval (CI) in Words Consider this experiment. You have a camera device that 97.5, 96.0, measures the 99.1 temperature of objects 98.055 from a distance by measuring their 97.323 infrared light emission. 99.051 It is very convenient but somewhat 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.

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1.5.5 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°, 103.0°. Passenger 2: 98.8°, 98.5°, 98.4° Passenger 3: 104.0°, 103.9°, 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 101.8° to 102.6° then we could say that the probability that the true temperature is 101.4° or higher is 90%.

101° 102° 103°

10% chance 80% chance 10% chance is 101.8° or lower 101.8° < < 102.6° is 102.6° or higher

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Use this table: Use this formula: % confidence interval t s °freedom 50 80 95 99 CI ± 1 1.00 3.08 12.71 63.66 N 2 0.82 1.89 4.30 9.92 3 0.76 1.64 3.18 5.84 Note the following: 4 0.74 1.53 2.78 4.60 5 0.73 1.48 2.57 4.03 6 0.72 1.44 2.45 3.71 For a straight average of N 7 0.71 1.41 2.36 3.50 points, the number of 8 0.71 1.40 2.31 3.36 degrees of freedom is N-1. 9 0.70 1.38 2.26 3.25 10 0.70 1.37 2.23 3.17 20 0.69 1.33 2.09 2.85 50 0.68 1.30 2.01 2.68 100 0.68 1.29 1.98 2.63

Complete the following table:

Lower Upper Is the probability that Passenger TAVG(°F) St boundary boundary the passenger‟s dev of 80% of 80% Temp is > 102° (°F) CI CI 90% or more? 1 101.5 1.1 2 98.57 0.17 3 103.93 0.047

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!

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1.5.6 tEXP PROBLEM:

Another way of approaching this type of problem is to calculate an experimental value of „t‟ called tEXP. In the example below, we will compare a measured result with an exact one. The question one answers with tEXP 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: 103.76°, 102.11°, 105.38° – calculate an experimental value of the „t‟ statistic for this population relative to the true value of 102°. Average = 103.75, std dev = 1.34

% confidence interval x N °freedom 50 80 95 99 av 1 1.00 3.08 12.71 63.66 texp 2 0.82 1.89 4.30 9.92 s 3 0.76 1.64 3.18 5.84 = test value

(103.76 102) 3 t exp 1.34

t 2.275 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

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1.6 Calibration Curves and Sensitivity

A highly sensitive instrument can discriminate between small differences in Calibration Sensitvity: analyte concentration. S = mC + Sb S = instrument signal C = analyte conc. m = slope calibration15 13.166 Sb = signal for blank S m = calibration = m sensitivity C 10

S S i S 5 But, can we Instrument Signal (S) Signal Instrument really distinguish C between small changes in 0 0 0 20 40 60 80 100concentration? 0 C 100 i Analyte Concentration (C)

= Analytical Sensitvity m 1 / = „noise‟ in = m / S and concentration S C. = ( S/ C) / S so

= 1/ C for the “ C” So small is? corresponding to S Good

Bad

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1.7 Vocabulary: Properties of Measurements

Sensitivity 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. Selectivity The ratio of the sensitivity of an instrument to an analyte to that of an interferant. Specimen The material removed for analysis – e.g. a given tablet from an assembly line. Sample The mixture that contains the chemical to be measured – e.g. the blood sample that we want to measure iron in. Analyte The chemical that is being measured – e.g. the iron in the blood sample. Calibration Curve A linear or non-linear function relating instrument response (signal) to analyte concentration. Interferant 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. Detection Limit CMIN or The minimum detectable concentration of analyte. Usually defined as that concentration CM that gives a signal of magnitude equal to three times the standard deviation in the blank signal. Limit of Quantitation The minimum concentration of analyte for which (LOQ) 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. Limit of Linearity (LOL) The largest concentration for which a calibration curve remains linear. Linear Dynamic Range The range of concentrations (or signal strengths) (LDR) between the LOQ and the LOL. An instrument is most useful within its‟ LDR.

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1.8 Detection Limit The detection limit is denoted “CM” CM is the minimum concentration that can be “detected,” or distinguished confidently from a blank.

Let us define the minimum detectable signal change as: SM

Therefore: SM = mCM must be a multiple (n) of the noise level in the blank: ( b).

SM ≡ 3 b = mCM

Minimum Signal due to Minimum Detectable Signal blank detectable concentration

By convention, m=3.

Derive a formula for CM based on B and m:

Derive a formula for CM based on : 1.8.1

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A Graphical Look at Detection Limit (CMIN)

Consider the minimum detectable signal in the context of the confidence interval:

SMIN = 3 b + Sb To what confidence interval does 3 correspond? Assume N=2 (two replicate measurements of Sb)

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.

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1.8.2 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?

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Dynamic Range

LOQ = Limit of Quantitation ( S/S 0.3)

LOL = Limit of Linearity

1.8.3 Between LOQ and LOL your instrument is most useful! This is called:

Linear Dynamic Range

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Selectivity

Sometimes another chemical species will add to or subtract from your analyte signal –

S = mACA + mBCB + mCCC + mDCD + … + SB

Analyte Interferants Blank

Selectivity coefficients determine how serious an interferant is to your determination of analyte CA: kB,A = mB/mA, kC,A = mC/mA, etc…

So:

S = mA(CA + kB,ACB + kC,ACC + …) + SB

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1.8.4 Direct Interference

In order to measure a concentration directly, without corrections, kB,A, kC,A must be approximately:

If there is interference, it is necessary to know both:

and

before one can determine the desired quantity CA!

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1.9 Linear Regression

Least-Squares Assuming that a set of x,y data pairs are well described by the linear function y = a+bx, and Regression or assuming that most error is in y (x is more Linear Regression precisely known) and the errors in y are not a function of x, then the coefficients that minimize the residual function: 2 2 = (yI – (a + bxI)) can be found from the following equations:

from Data Reduction and Error Analysis for the Physical Sciences, Philip R. Bevington, cw. 1969 Mc Graw Hill. Non-linear Methods for fitting arbitrary curves to data sets. Regression Standard Additions A nearly matrix-effect free form of analysis. A standard additions plot is a linear plot of Plot 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. Sample Matrix / The matrix is the solution, including solvent(s) and all other solutes in which an analyte is Standards Matrix dissolved or mixed Matrix Effect A matrix effect refers to the case where the instrumental sensitivity is different for the sample and standards because of differences in the matrix. Internal Standards 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.

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1.9.1 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 = SB + 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 2 = (yi – (a + bxi)) 1.5 1.5

S i 1 S i F i F 0.5 Instrument Signal (S) Signal Instrument i

0 0 0 2 4 6 8 10 0 C 10 i Analyte Concentration (C) Page 32 of 316 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.

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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 = SXY/SXX

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.

2 Equation a1-37 sy 1 1 ycavg yavb s Skoog 5th edition. c m M N 2 m S xx

Note: when computing a confidence interval using this sC 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, 1239.

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1.10 Experimental Design:

Designing an experiment involves planning how you will: make calibration and validation standards; prepare the sample for analysis; and perform the measurements.

1.10.1 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.

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6. 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.

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1.10.2 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 0.1-20 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 Volume Relative Precision Precision 20-200 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%

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1.10.3 Plan the analysis! (start with guesses…)

1. If you dissolve the packet in water, dilute it to 100.0 mL and filter it, what will the approximate sucrose concentration be in ppth (parts per thousand)?

0.8 g 0.7 g 1000 mg 1 mL syrup sucrose 1 sucrose water = 5.6 ppth 100 mL 1 g g syrup water g sucrose 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?

1.0 100 mg 100 mg 0.100 g mL sucrose = sucrose = sucrose 1 mL soution

1.00 g 10 mL => sucrose

100 10.0 g mL => sucrose

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1.10.4 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!

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4. What concentrations „bracket‟ the 5.6 ppth target?

For example: Do you need to make standards on an even 5.2, 5.4, 5.6, 5.8, 6.0 ppth spacing? --- or --- 1, 2, 4, 6, 8 ppth No – but calibration --- or --- points above and 1, 2, 5, 10, 15 ppth below the std. are --- or --- needed. 0.050, 0.50, 5.0, 50, 500 ppth

Does the analyte No, but it have to fall right in the minimizes middle of the error! calibration standards?

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

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6. How do you prepare the calibration series from the primary stock solution?

Primary Stock (1) is diluted to make  Calibration Standard (2)

C1V1 = C2V2

First calibration standard: Make 10 mL of 1 ppth sucrose from 100 ppth stock solution.

C1 = 100 ppth C2 = 1 ppth V2 = 10.0 mL V1 = ? = volume to pipet over

V1 = C2V2/C1

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7. 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/C1 1.00 100.0 10.00 0.100 mL 2.00 100.0 10.00 0.200 mL 5.00 100.0 10.00 0.500 mL 10.00 100.0 10.00 1.000 mL 15.00 100.0 10.00 1.500 mL

Calibration Standard Preparation: Target Conc: Stock Conc: Final Volume: Volume to Pipet: C2 / ppt C1 / ppt V2 / mL V1=C2*V2/C1 1 100 10 =A4*C4/B4 mL 2 100 10 =A5*C5/B5 mL 5 100 10 =A6*C6/B6 mL 10 100 10 =A7*C7/B7 mL 15 100 10 =A8*C8/B8 mL

These are formulas that you type into Excel – normally only the result of the formula calculation is displayed.

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1.11 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 Another way of saying this is that a method is valid if:

all significant sources of bias

have been removed

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: Possible Solution(s) or Diagnostic: Analyst erratic Analyst can repeat experiment Error in analyst Different analyst does same analysis and technique gets same result. Calibration Entire analysis Independent standard standards are in repeated with measured periodically - error (are not what indpendent called validation standard they say the are) calibration or QC standard (quality standard set control) Calibration Different calibration method used (standard method additions) Instrument Different instrument used (can be a different function erratic kind of instrument) Instrument drift Periodically measure validation standard or internal standard

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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 Aspect(s) validated Completely independent Calibration method, measurements of the including matrix effects. same sample using a different instrument or Instrument, including technique. distortion and drift. You! The analyst, did it right Use the same method, You! The analyst made no but measure mistakes in the independently prepared implementation of the standards as samples. procedure.

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1.12 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. Dilute to Analyze 100 mL  140 digest filter volume ppm 14.0 mg found

Sample

Dilute to Analyze 100 mL  187 digest filter volume ppm 18.7 mg found

Sample+ 5.00 18.7-14.0 = 4.7 mg of spike found: spike mg spike recovery percent = 4.7 / 5.00  94%

What does this say about We may be losing about our sample preparation 6% of the analyte during method? sample prep.

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1.13 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. Dilute to Analyze 100 mL  2.3 digest filter volume 0.23 mg ppm 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…)

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1.14 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

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1.14.1 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)

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1.14.2 An Example of Standard Additions: 1. The analyte sample is split up into e.g. 6 aliquots of identical, known volume – e.g. 1.00 ml. 2. To each of these, a known quantity of standard (known as a spike) is added – e.g. 0, 0.1, 0.2, 0.3, 0.4, 0.5 ml – standard is 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 It may decrease it this impact the precision of If dilution is too great!. your measurement?

How does this process impact the accuracy of It increases it! your measurement? Page 49 of 316 Chem 155 Unit 1 Page 50 of 316

1.14.3 Calculating Conc. w/ Standard Additions:

a' b' a

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 (VXCX) and standard (VSCS). The x-axis is simply the standard „spike‟ volume. The x-intercept is the hypothetical spike volume (VS)0 containing the same amount of analyte as the sample. a is proportional to moles of analyte in the sample = VXCX a‟ is proportional to moles of std added = VSCS b is (VS)0 – the x-intercept of the graph b‟ is VS – the spike volume VXCX VSCS

Substitute for a,a‟,b,b‟: = (VS)0 VS

Solve for C : X CX = CS(VS)0/VX

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1.14.4 Standard Additions by Linear Regression: Let: VX = volume of unknown analyte solution added to each flask CX = concentration of unknown solution VT = final, diluted volume VS = volume of „spike‟ added to unknown soln. before dilution CS = concentration of analyte in spike solution

Dilution Calc 1: (V1C1 = VTCT  CT = V1C1/VT) Contribution to concentration of analyte from sample:

VXCX/VT

Dilution Calc 2: Contribution to concentration of analyte from spike:

VSCS/VT

Total Signal (sensitivity = k) given that „x‟ variable is VS.

S = k VXC X/VT + k VSCS/VT

Slope = m = kCS/VT

intercept = b = kVXCX/VT we can get b and m from linear regression

kVXCX/VT and we want CX … so … b/m = = VXCX/CS kCS/VT bCS so : C = X mVX

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1.15 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 + SB) – 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'AN S AN sB ) to the S' m C C internal standard (IS) signals (S): AN AN AN k AN S'IS mISCIS CIS

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.

S'AN CIS CIS CAN where is easily derived from a previously measured S'IS k k calibration standard for which the analyte signal ( S'A STD ) and concentration (CA STD ) of the analyte and internal standard (

S'A STD CIS STD mAN S'IS STD ,CIS STD) are known: k S'IS STD CA STD mIS

From a calibration standard.

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In other words, if CIS 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 A1700 and A2250 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!

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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 Peaks 2, 3 and 4 are other internal standard. ingredients in the sample. 100 ppm NaNO3, 480All mAu .s Peak 5 is the 1 analyte, a caffeine 1 standard, 25 ppm, 530 mAu.s

0.5 absorbance Chromatogram

0 1 is a standard. 0 400 800 1200 1600 2000 2400 2800 3200 3600 4000 time / s Did the caffeine

1 concentration increase from 2 chromatogram 1 to 2? 0.5

absorbance Did the caffeine concentration increase in 0 0 400 800 1200 1600 2000 2400 2800 3200 3600 4000 chromatogram time / s 1 to 3? 100 ppm NaNO3, 520 mAu.s 1 3 Unknown caffeine conc. . 0.5 630 mAu s

absorbance

0 0 400 800 1200 1600 2000 2400 2800 3200 3600 4000 time / s Page 54 of 316 Chem 155 Unit 1 Page 55 of 316

1.15.1 Internal Standards Calculation How about calculating the actual concentration of the sample in 3 above?

. S'A STD 530 mAu s

CA STD 25 ppm

. S' Standard IS STD 480 mAu s

CIS STD 100 ppm

S'A STD CIS STD mAN 530 100 k 4.417 unitless S' C m 480 25 IS STD A STD IS S' . AN 630 mAu s

S' . IS 520 mAu s Sample C IS 100 ppm

S'AN CIS 630 100 C 27.4 ppm AN S' k IS 520 4.417

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2 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 sa = 0.5 lb and sb = 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 Characteristics of sum sum „a‟ and „b‟ „c‟ Trial # Weight of: Deviation: Total Deviation a b da db weight: from avg: 1 4.5 4.5 -0.5 -0.5 9.0 -1 2 5.5 4.5 +0.5 -0.5 10.0 0 3 4.5 5.5 -0.5 +0.5 10.0 0 4 5.5 5.5 +0.5 +0.5 11.0 +1 average 5.0 5.0 0.5 0.5 10 0.5

This is somewhat artificial and is not quite right, but it gives you the general idea. Errors in a and b propagate into c.

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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 c

a-sa+b-sb

theoretical

a+sa a-sa+b+sb

a

theoretical

a-sa

distribution

“Real”

Page 57 of 316 Chem 155 Unit 2 Page 58 of 316

error propagation through sums :

a rnorm(10000 3 0.5) b rnorm(10000 3 0.5) c a b

0 0 0 0 2.781 0 3.554 0 6.334 1 2.66 1 3.331 1 5.991 a b c 2 2.763 2 1.98 2 4.744 3 2.524 3 3.482 3 6.006 4 2.157 4 3.848 4 6.005

mean(a) 3.004 mean(b) 2.996 mean(c) 6 stdev(a) 0.498 stdev(b) 0.504 stdev(c) 0.712

a_dist histogram(40 a) 2 2 0.5 0.5 0.707 b_dist histogram(40 b) c_dist histogram(40 c)

1000

a_dist j 1

b_dist j 1 500

c_dist j 1

0 0 2 4 6 8 10

a_dist j 0 b_dist j 0 c_dist j 0

Page 58 of 316 Chem 155 Unit 2 Page 59 of 316

error propagation through products :

c a b i i i mean(c) 9.003 stdev(c) 2.156 0 0 9.881 a_dist histogram(40 a) 1 8.861 b_dist histogram(40 b) 2 5.473 c_dist histogram(40 c) 3 8.789 4 8.301 1000 5 8.057 6 7.561 c 7 9.909 a_dist j 1

813.589 b_dist j 1 500 9 7.471 c_dist j 1 10 9.206 11 6.054 1211.944 0 0 5 10 15 20 25 1310.643 a_dist j 0 b_dist j 0 c_dist j 0 14 5.773 15 8.27

Page 59 of 316 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 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 P light power - the measured quantity in a A log spectrophotometric experiment. P l

2

1.5

A i 1

0.5

0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 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?

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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 2 2 2 S 2 S 2 S 2 S a b c a b 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 Chem 155 Unit 2 Page 63 of 316

Exercise: Prove the “multiplication / division rule” Show that for S = a*b/c

2 2 2 2 S a b c 2 2 2 2 S a b c

Page 63 of 316 Chem 155 Unit 2 Page 64 of 316

The propagation of error rules for special cases:

Page 64 of 316 Chem 155 Unit 3 Page 65 of 316 3 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 MEDIUM = cMEDIUM

= wavelength MEDIUM = VAC / = frequency c = light speed cMEDIUM = cVAC /

Note: When the refractive index ( ) changes

______Stays same

______changes

Page 65 of 316 Chem 155 Unit 3 Page 66 of 316 3.1 Electromagnetic Radiation: A vast range of energies covers many physical processes. These processes are the basis of spectroscopy.

Page 66 of 316 Chem 155 Unit 3 Page 67 of 316 Energy Nomogram

Energy nomogram - Hard UV 100 nm to Far IR 1mm 2 5 10 10 1 3 10 10 105 2 1015 10

103 104 0 2 10 10 104 1 1014 10

4 3 -1 1 10 10 10 10 103 100 1013

5 2 10 10 10-2 100 102 10-1 1012

106 101 10-3 10-1 101 nm cm-1 s E / eV E / kCal E / kJ T / K /mol /mol

Page 67 of 316 Chem 155 Unit 3 Page 68 of 316 3.2 Diffraction Diffraction is the basis of wavelength selection in most spectrometers:

d

Page 68 of 316 Chem 155 Unit 3 Page 69 of 316 3.2.1 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 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 Chem 155 Unit 3 Page 71 of 316 3.3 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 Chem 155 Unit 3 Page 72 of 316 3.3.1 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 Chem 155 Unit 3 Page 73 of 316 3.3.2 Refractive index:

ratio of speed of light in vacuum to speed of light in medium.

= c / c VAC MEDIUM

= c / 1 2.5

Refractive Wavelength Frequency s-1 Energy J index, 1.00 500 nm 3x108 ms-1 / 6x1015 s-1 x 500x10-9 m = 6.626x10-34 Js 6x1015 s-1 = 4x10-20 J 1.50 500/1.5 = 6x1015 s-1 4x10-20 J 333 nm

Page 73 of 316 Chem 155 Unit 3 Page 74 of 316 3.3.3 Refraction and Snell‟s Law:

Index = f(wavelength)

Blue Blue Red Red

Page 74 of 316 Chem 155 Unit 3 Page 75 of 316 3.3.4 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 Chem 155 Unit 3 Page 76 of 316 3.3.5 Scattering:

Incident Transmitted

Scattered

Scattering Scatterer Features Process Raleigh (elastic) atoms and weak, favors molecules blue Tyndall (Mie) Colloids Strong (easy to Elastic normally (bigger / nm) image) Raman (visible) Molecules – very weak, vibrational spec inelastic

Page 76 of 316 Chem 155 Unit 3 Page 77 of 316 3.3.6 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 Type of Metal and

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

(KE) ElectronEnergy EF

Page 77 of 316 Light Frequency

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

EF = Fermi energy of metal = Light frequency (s-1) E = h - Formula of line h = slope = 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 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 molecular molecular atom liquid solid

Quantum electronic States electronic electronic

vibrational

vibrationalctronic phonon

rotational

Resulting lines bands broad Spectra

Page 79 of 316 Chem 155 Unit 3 Page 80 of 316 3.3.7 Spectra typical of gas, liquid and solid.

Absorber Phase Notes / transition types Atom Gas Extremely narrow lines / electronic

Molecule Gas Fine structure due to electronic+ rotational+ vibrational Molecule Solution Broad / small bands + some fine structure / electronic + vibrational

Molecule Solution Broad / larger bands electronic only resolved

Page 80 of 316 Chem 155 Unit 3 Page 81 of 316 3.3.8 Energy levels and photon absorption and emission.

Page 81 of 316 Chem 155 Unit 3 Page 82 of 316 3.3.9 Typical fluorphore Jablonski Diagram. 3.3.10 Photophysical Processes

Page 82 of 316 Chem 155 Unit 3 Page 83 of 316 3.3.11 Typical organic electronic spectrum.

Page 83 of 316 Chem 155 Unit 3 Page 84 of 316 3.3.12 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.

Emission or or Emission Absorbance Light Frequency

Energy

Emission or or Emission Absorbance Light Frequency

Energy

Emission or or Emission Mat. Heated

Light Frequency

Page 84 of 316

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

Page 85 of 316 Chem. 155 Unit 4 Page 86 of 316

4 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 Chem. 155 Unit 4 Page 87 of 316

4.1 General Photometric Designs for the Quantitation of Chemical Species 4.1.1 Generalized Detector Response: S = kP + SDARK S = Signal k = proportionality P = light power S = detector response in absence of light DARK

All absorbance methods! 4.1.2 Absorbance IR, VIS, UV, Xray! More Analyte

Less Signal

4.1.3 Quantitation = molar absorptivity b = pathlength (cm) A = bc = -log(P/P0) c = concentration in moles/L P = Light Power at Detector PO = Light Power for Blank c = concentration in moles/L

4.1.4 Emission and Fluorescence All light emission methods! Fluorescence (Xray - UV), More Analyte Scattering, Luminescence, even NMR! More Signal

4.1.5 Quantitation

P = Light Power at Detector P = PO + mC PO = Light Power for Blank c = concentration in any unit

Page 87 of 316 Chem. 155 Unit 4 Page 88 of 316

4.2 Block Diagrams

Instruments for Analytical Spectrometry:

1. Absorbance Cuvette Photomultiplier Tube, Flame Photodiode Gas Cell 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

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4.3 Optical Materials

Cost Trade Off

$$ Water Soluble! LiF Sapphire – Al2O3 Crystal $$$$ Hardness! $$ 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

$$$$ hard to optimize for both Interferometers visible and IR at once Time Domain

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4.4 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

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4.5 Continuum Sources of Light:

Broadband or White Light Sources Also called:

1. Simplest Design: Blackbody Sources

a. Tungsten Visible / Near IR / IR b. Quartz-Tungsten-Halogen c. Nernst Glower IR / Far IR

2. Gas Emission Designs: UV Only a. H2 / D2

- H2 + e  H2*  H + H + h

Kinetic Energy of H atoms = Continuum

Therefore h can be: Continuum

UV-Vis-Near IR b. Ar, Xe, Hg

Heavy (High-Z) atoms  Many atomic states + High Pressure (extensive broadening of lines) Quasi-Continuum

Page 91 of 316 Chem. 155 Unit 4 Page 92 of 316

4.6 Line Sources of Light: 1. Low Pressure Gas Emission a. Hg Many Lines – can be filtered to b. Ar emit only one predominant line. c. Xe d. Na Na-D Line Predominates – 589.00 and 589.59 nm doublet

2. Hollow Cathode Lamps: a. Metals (Cathode!) b. Used in atomic spectroscopy (absorbance and fluorescence)

300V DC  Electrical Discharge Sputtering

+ + Ne + Fe* + h Ne Fe

Atomic Emission

Fe Metal Cathode Page 92 of 316 Chem. 155 Unit 4 Page 93 of 316

4.7 Laser Sources of Light:

L Light An Introduction to Lasers: LASER is an acronym for A Amplification by a light amplification S Stimulated process: E Emission of Partially transmitting R Radiation mirror Fully reflective mirror

Pumping Energy Gain Medium – atoms or Source: molecules that undergo Intense Light „lasing‟ transitions Electrical Discharge

What happens in the Gain Medium?

1.

“relatively long lived”

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What happens in the Gain Medium?

2.

3.

4.

Stimulated Emission is: Coherent

Monochromatic

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4.7.1 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 losses – Stimulated Emission absorption, gain! – stimulated spontaneous pump emission emission power in beam out

2 Two common laser configurations: Fast Decay 1 Pump Lasing! 0 3-state

3 Fast Decay 2

Lasing! Pump 1 Fast Decay 4-state (or more) 0

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4.7.2 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.

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4.7.3 Necessity of 3 or more states

Why are three or more levels (states) necessary for lasing?

E E = ENERGY DIFFERENCE us-ls N EXCITED -23 recall: e kT k = Boltzmann const. 1.3x10 J/K T = temperature (K) N GROUND NEXCITED = population (concentration) US NGROUND = population (concentration) LS

What happens when T  infinity? NEXCITED  NGROUND

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4.7.4 Common lasers categorized by lasing medium.

1. Solid Red - Near Infrared 1.1. Lanthanide-and transition metal ion lasers 1.1.1. 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 1.1.2. Ho-ZBLA – Holmium (Ho+3) doped glasses made from ZrF3, BaF3, La F3, Al F3 can be fabricated into optical fibers that lase and can amplify optical signals of certain wavelengths 1.2. Ti-sapphire – tunable (650-1100 nm) CW or pulsed 1.2.1. Mode locked: 5-2000 fs, (MHz repetition) 1.2.2. Chirped-pulse amplification: 10-100 fs, 5, mJ, (kHz rep)  gW power densities… 1.2.3. Fundamental or second harmonic can be used in „OPA‟ optical parametric amplifier lasers that are freely tunable, normally pulsed

1.3. Semiconductor Lasers – 1.3.1. Silicon, 1.3.2. 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. N2 (337 nm) 3.4. CO2 (10,600nm and many other IR wavelengths) 3.5. XeF (351) and KrF (248) and ArF (193) excimer lasers

Page 98 of 316 Chem. 155 Unit 4 Page 99 of 316

. 375 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 - . 650 nm - AlGaInP DVD drives, laser pointers . 660 nm - . 670 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 Chem. 155 Unit 4 Page 100 of 316

4.7.5 Excimer Lasers:

* Kr + F * KrF Excimer!

KrF What is unusual Kr + F about this molecule?

Because of fast dissociation,

[KrF*] > or < [KrF]

This favors:

population inversion!

Page 100 of 316 Chem. 155 Unit 4 Page 101 of 316

4.7.6 Some laser “pointers”.

LS = lower state US = upper state GS = ground state

stimulated hν + US  2hν + LS gain emission

hν + LS  US absorption loss

spontaneous US  LS + hν loss emisson

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 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 Chem 155 Unit 5 Page 103 of 316

5 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 Chem 155 Unit 5 Page 104 of 316

5.2 Photoelectric effect photometers Phototube:

High UV- Evacuated tube in transparency fused which electrons can silica window. travel if emitted from cathode.

+ - Readout

Amplifier

Photocathode – Anode – positive transducer - emits bias, collects electrons via the photocurrent. photoelectric effect Low work function metal: Na, K, GaAs

Page 104 of 316 Chem 155 Unit 5 Page 105 of 316

What is the maximum gain of a phototube expressed as electrons per 1! incident photon?

What is a typical gain for a phototube expressed as electrons per incident 0.01 photon?

What limitation is built in to all photoelectric detectors? h must be

Page 105 of 316 Chem 155 Unit 5 Page 106 of 316

5.3 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

j / energy Photoelectron

Light frequency / s-1

Peak quantum efficiency:

About 10% or 1 photoelectron 10 photons

Page 106 of 316 Chem 155 Unit 5 Page 107 of 316

5.4 Operation of the PMT detector:

Faceplate material:

UV-transparent Silica

Photocathode potential:

From www.hamamatsu.com 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 Chem 155 Unit 5 Page 108 of 316

5.5 PMT Gain Equation:

1 Photon Photocathode

Quantuum Efficiency - How many photoelectrons per incident photon are emitted? Secondary electron yield -

How many secondary Dynode 1 electrons?

Dynode 2

Dynode 12

PMT Gain = G = n = Quantum efficiency = Secondary electron yield n = Number of dynodes

Page 108 of 316 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:

- - n h  e at e strikes n- electrons photocathode dynodes strike anode

-9 3x10 s

What is the average current during the 10ns pulse? Assume = 4, n = 10

F = 96485 coulombs/mole e- Mol = 6.023x1023 Current (amperes) = # coulombs / s

410 = 1048576

Current (amperes) = # coulombs / s i = 1048576 e- * 96485 coulombs / 6.023e23 e- / 10-8 s = 1.7x10-5 amperes – easily measureable

Page 109 of 316 Chem 155 Unit 5 Page 110 of 316

5.6 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.

Reject: cosmic ray – e- cascade. Accept: photon- electron from

current current microamperes / photocathode Reject: time / s 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 Chem 155 Unit 5 Page 111 of 316

In single photon counting:

Accept only those pulses originating at the photocathode.

Retain signal

noise Reject

Recall: Detection limit

SMIN = SBLANK + 3 BLANK

Large and small pulses are all noise!

These contribute to both SBLANK 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 Chem 155 Unit 5 Page 112 of 316

5.7 Semiconductor-Based Light Detectors: Photodiodes:

p-type Si p-n junction n-type Si

-

+ + P - B

- + B P + -

The only carriers of The only carriers of current in p-type Si are: current in n-type Si are:

Electrons! Holes!

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 Mobile P Mobile (p-dopant) holes. (n-dopant) electrons

Page 112 of 316 Chem 155 Unit 5 Page 113 of 316

Depletion, anhiliation, injection and rectification: 1. Forward Bias:

+

p-n junction s + - + + - -

Injection of Anhiliation of Injection of

holes holes and electrons

electrons 2. Reverse Bias: +

+ - - + - +

Removal of Depletion of Removal of holes holes and electrons electrons

Under reverse bias: No current flows: Rectification Page 113 of 316 Chem 155 Unit 5 Page 114 of 316

Photodiode: +

+ - - + + -

Steps in Photoconduction:

1. Photon absorbed by Si, promotes e- into concution band.

2. This leaves a mobile hole and electron.

Hole and electron are swept apart by the 3. applied field and:

h  current transduction

Properties of the photodiode:

1. Gain: = 1 but is often near unity (1)

2. Dynamic Range: Largest measureable signal >> smallest  Excellent! 3. Ease of fabrication: Can be made very small. Can be made into 1D and 2D Page 114 of 316 arrays Chem 155 Unit 5 Page 115 of 316

5.8 Charge Coupled Device Array Detectors: -10V -10V -10V

Al contact

SiO2 ------n-Si ------

1. e- are 2. h strikes 3. h+ repelled depletion accumulate from Al region at Al contact contact -10V

+ + + + + +

-10V -10V

+ + + + + +

-10V

+ + + + + + Page 115 of 316 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: -2 Ca. 100 frames per second 10 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 Chem 155 Unit 6 Page 117 of 316

6 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

Monochromatic Why do we care about monochromators? 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 Chem 155 Unit 6 Page 118 of 316

6.1 Adjustable Wavelength Selectors Adjustable wavelength selectors are called:

Monochromators

Light Input Light Output

Polychromatic “Monochromatic” “White” Narrow-band

“Broadband”

Wavelength / nm

455.3

Two important characteristics of monochromators are:

The amount of light that throughput makes it through at a given wavelength:

The range of wavelengths bandwidth - that exit the monochromator:

Page 118 of 316 Chem 155 Unit 6 Page 119 of 316

6.2 Monochromator Designs:

Czerney-Turner

blue red

Prism

Wavelength Dispersion:

Linear

Nonlinear with UV-Absorption Nonlinear

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6.3 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 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 = ? 90°

ii. b = ? 90° - i

iii. a+b+90° = ? 180°

iv. substitute ii into iii

a + (90 – i) + 90 = 180  a – i +180 = 180  a = i

Page 121 of 316 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) = opposite / hypotenuse = BC / AB = BC / d sin(r) = sin(a‟) = opposite / hypotenuse = DB / AB = DB / d

BC + DB = d sin(i) + d sin (r) = d ( sin(i) + sin(r) )

For constructive interference (i.e. a bright condition): n = d ( sin(i) + sin(r) )

Page 122 of 316 Chem 155 Unit 6 Page 123 of 316

6.4 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?

d ( sin(i) + sin(r) ) n =

d ( sin(i) + sin(r) )/ n (r) = d / dr = d/dr[dsin(i)] /n + d/dr[dsin(r)] / n d / dr d cos(r) / n

Page 123 of 316 Chem 155 Unit 6 Page 124 of 316

6.5 Angular dispersion: order dr n d d cos(r) n = Groove = spacing on Reciprocal linear dispersion: dy Fdrd = grating

Angle of r = diffraction

wavelength =

sin(r) r sin(r) y/F

d d dy Fdr D-1 =

d cos(r) / nF D-1 =

Page 124 of 316 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= d = 1/1000 mm = 10-3 mm = 1 m = 1000 nm

F= F = 1m = 1000 mm

D-1 = 1000 nm x 0.7 / 1 x 1000 mm Focal Plane D-1 = 0.700 nm / mm Exit Slit

0 100 200 300 400

y / mm

300 370 440 510 580 Page 125 of 316 / nm Chem 155 Unit 6 Page 126 of 316

6.6 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.

W = slit width in mm -1 -1 EFF = wD D = 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

300 400 500 600 700

EFF 2 nm

300 400 500 600 700 / nm Page 126 of 316

Chem 155 Unit 6 Page 127 of 316

6.7 Bandwith and Atomic Spectroscopy

Consider the case of a Ni atomic spectroscopy experiment. Ni atoms emit at 231.7, 232.0 and 232.2 nm. However, Ni atoms only absorb light at 232.0 nm. (Two emission lines are non-resonant.) So, it is important in atomic absorption spectroscopy to remove Ni emission at 231.7 and 232.2, but efficiently transmit light at 232.0 nm.

Page 127 of 316 Chem 155 Unit 6 Page 128 of 316

6.8 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

small EFF should be:

-1 EFF = wD = wd cos(r) / nF , so d should be: small

F should be: large

w should be: 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 Chem 155 Unit 6 Page 129 of 316

6.9 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 Resolution Limted Throughput Low resolution

(1 nm) Light throughput Light

500 500.5 501 501.5 502 502.5 503 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 Chem 155 Unit 6 Page 130 of 316

6.10 Grating Resolution

Resolution may be limited by the size of the grating as well. The grating resolution, R, is given by:

R / = nN n = Diffraction order

N = 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 Chem 155 Unit 6 Page 131 of 316

6.11 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 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 300.00 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 Chem 155 Unit 6 Page 133 of 316

6.12 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 = small wD-1 = w d cos(r) / nF to be:

We also want resolution (R = / = nN) large to be:

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: If w is too small 1 – no light 2 – exceeds w the ability to focus light d Limitations in the fabrication of gratings

F How big of a spectrometer is tolerable?

What about n?

n is the key…

Page 133 of 316 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 0.5 0.5 Groove density 1200 / mm 79 / mm i,r or 10° 63° R (300nm) 60,000 700,000 D-1 (nm / mm) 1.6 0.15

Page 134 of 316 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( ) n1 1 = n2 2  63x300 = 2dsin( ) 2 = n1 1/n2 but… 62x??? = 2dsin( ) also!

n / nm … … 61 309.83 62 304.84 63 300 64 295.31 65 290.77 … … … …

This problem is called: Order overlap

Is this a problem for Czerny-Turner monochromators?

If n1=1 and 1=300nm then for n2=2, 2 = 150 nm If n1=1 and 1=800nm then for n2=2, 2 = 400 nm Czerny-Turner monochromators handle this problem with: Echelle Monochromators handle the order-sorting problem with cross dispersion – usually a silica SiO2 prism. UV-absorption is a problem with glass

Page 135 of 316 Chem 155 Unit 6 Page 136 of 316

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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 Chem 155 Unit 7 Page 138 of 316

7 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.

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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 (D2) 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 D2 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 Chem 155 Unit 8 Page 152 of 316

8 Practical aspects of atomic spectroscopy: Sample introduction and plasma chemistry. Skoog chapters covered: 8C, 9A

Flame or Plasma Monochromator Atomic Emission Atomic PMT from Absorption P Hollow and / or Cathode Lamp Atomic P0 Emission in Light source Flame/Plasma for absorption measurement Molecular gas Absorbance: only. P Dry solids A Log b[C] PO Flame or Plasma Aerosol Emission:

Nebulizer / [C] k(P PO ) Spray Chamber Analytes in Solution

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 Chem 155 Unit 8 Page 153 of 316

8.1 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

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Nebulizer types:

1. concentric 2. cross flow 3. fritted disk pneumatic 4. Babbington 5. ultrasonic

Concentric:

Page 154 of 316 Chem 155 Unit 8 Page 155 of 316

Nebulizers, cont:

Cross flow:

Babbington:

Fritted disk:

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

Ultrasonic:

Page 156 of 316 Chem 155 Unit 8 Page 157 of 316

8.2 Atomization

Atoms and

ions diluted

Optimal AES

region Atoms

Optimal AAS region

Molecules

Particles

Droplets

Entire process happens in: milliseconds

So large droplets are: bad

Page 157 of 316 Chem 155 Unit 8 Page 158 of 316

8.3 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 1800 40 Gas Natural O2 2800 400 Gas Acetylene Air 2400 230

Acetylene O2 3100 2000

Acetylene N2O 2700 300

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8.4 Flame as „sample holder‟:

1. For best sensitivity:

Interrogate with hollow cathode radiation through „atom rich‟ region.

Changes in observation height:

2. Change sensitivity – so don‟t move observation region during experiment. Page 159 of 316 Chem 155 Unit 8 Page 160 of 316

8.5 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

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8.6 Flame Chemistry and Interferences:

1. Chemical equilibria can occur in flames. Common oxyanions can react with atoms:

- + - H2PO4 + Fe  FePO4 + 2H + 3e -2 - SO4 + 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 C2H2, less O2 4. Hydroxides can be suppressed by:

Higher Temp: CaOH ↔ Ca + OH

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8.7 Matrix adjustments in atomic spectroscopy:

Releasing agents: Compete with analyte for interferant: +2 Adding Sr to react with PO4 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

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9 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

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9.1 AAS / AES Review:

M+ - ions -e- +e- Emission M*h [M] = k‟·(P-Po)

h Absorption HCL Atomic gas [M] = -k·log(P/Po) “M”

Molecular gas

Solid particles

Nebulize

Liquds Solids Gases

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9.2 Types of AES:

9.2.1 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

9.2.2 FLAME

Flame Emission Spectrometry or “FES”

Flame is “chemical plasma” – typically - - comprising CO2, H2O, other Molecules, e e Flame types include: Air - C2H2 , N2O - C2H2 , O2 – C2H2 FES is limited to easily excited species such as Na, Li, K

9.2.3 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

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9.3 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

B2O3, PO4, WOx, VOx, ZrOx 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 Flame Atomic Absorption Simultaneous Detection One or Two at a Time

Optimization Simpler Easier to Use

Excellent Dynamic Range Higher Precision

Wide Range of Elements Metals Only

Costly to Use / Buy Significantly Cheaper

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9.5 Inductively Coupled Plasma AES: ICP-AES

Induction and Eddy Currents:

Switch „on‟ electromagnet & eddy currents (circular) flow in conductor around changing magnetic field lines:

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9.6 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

Tangential Ar flow pushes Sample plasma away aerosol from quartz housing

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9.7 Atomization in Ar-ICP

Excitation = Optimal observation region.

Desolvation / Atomization

InertProperties:

Hot

Electron rich

InertAnalyte = low Atomization oxide formation, / Ionization: chemical interference low

Hot = efficient atomization

Electron rich = low ionization

Excellent atomizer

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9.8 Direct Current Plasma AES: DCP-AES

Ar flow:

Argon Plasma Jet: Argon:

Relative to ICP:

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9.9 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.

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P

P0

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9.10 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

E Excited Excited N g kT Recall: e N o g o The Solution: (when very high accuracy or precision is needed) Internal Standards

Page 173 of 316 Chem 155 Unit 9 Atomic Emission Page 174 of 316

A perfect internal standard will have: Identical Activation Energy ( E)

EA

g kT SA QA NA e g0

EIS

g kT SIS QIS NIS e g0

Exercise: Derive an equation for the ratio of the analyte to IS signals as a function of temperature:

a e a b e Recall: b e

E E The A IS answer: S A Q A N A kT e S Q N IS IS IS

So:

Page 174 of 316 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 Chem 155 Unit 9 Atomic Emission Page 176 of 316

Two examples of precision / accuracy assessments of ICP-AES:

Page 176 of 316 Chem 155 Unit 9 Atomic Emission Page 177 of 316

Page 177 of 316 Chem 155 Unit 10 Page 178 of 316

10 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 190-400 nm

Visible 400-750 nm

Near Infrared 750-2500 nm

We group these wavelengths together for two reasons. Analytes: This wavelength range covers many atomic and molecular electronic absorptions.

Optics:

Pure SiO2, called fused silica or quartz is transparent in this range. Even air absorbs strongly at <180 nm.

Page 178 of 316 Chem 155 Unit 10 Page 179 of 316 10.2 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 Chem 155 Unit 10 Page 180 of 316 10.3 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 cm2 N = number of absorbers per cm3 (# concentration) x = distance

Page 180 of 316 Chem 155 Unit 10 Page 181 of 316

10.4 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: A = bC = -Log(P/Po)

M max 105 M-1 cm-1

 a is about 1 nm  theoretical cross section is about the same size as the molecule!

Page 181 of 316 Chem 155 Unit 10 Page 182 of 316 10.5 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. 1.2. 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/(n2+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. 3.2. Stray light is getting into the detector. 3.3. The light levels are too low for the detector to accurately represent.

Page 182 of 316 Chem 155 Unit 10 Page 183 of 316 10.5.1 Instrumental Limitations to Beer‟s Law:

10.5.1.1 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, PSTRAY 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 PSTRAY component.

Pi Pi P stray P stray 0.05 A true log A app log i P o i P o P stray

P A true A i i appi 4 0 1 0 1 0.1 0.845 3 2 0.01 1.243 3 A 1·10 -3 1.314 truei 2 4 1·10 -4 1.321 1

0 0 1 2 3 4 C i

Page 183 of 316 Chem 155 Unit 10 Page 184 of 316 10.5.1.2 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. Case 2.

Hard to distinguish P and Po Hard to distinguish P and zero For example: For example: P=0.01±0.01 P=0.99±0.01 Po=1.00±0.01 Possible : Possible : Possible: P = 0.00

0.99 P/Po  1.00 A = -Log(P/Po)  1.01 A=-Log(P/Po) Infinite! NEGATIVE

Page 184 of 316 Chem 155 Unit 10 Page 185 of 316 10.5.2 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?

+ - absorbing HA H + A Concentration of HA in solution, the + - 1. KEQ = [H ][A ] / [HA] undissociated fraction.

- 2. Conservation of mass: [HA0] = [HA] + [A ] - so: [HA] = [HA0] – [A ] Number of moles of HA 4. [A-] = [H+] (if >> 10-7 M) divided by volume of in solution. Total or formal concentration.

K = [A-][A-] / [HA] 5 = 1 + 4  EQ

- 2 - 6 = 5 + 2  KEQ = [A ] / [HA0] – [A ]

Page 185 of 316 Chem 155 Unit 10 Page 186 of 316 - 2 - KEQ = [A ] / [HA0] – [A ]

try to solve for [A-]…

- 2 - [A ] + KEQ[A ] – KEQ[HA0] = 0 a = 1 2 2 b b 4 a c a x b x c 0 x 2 a b = KEQ

c = KEQ[HA0]

2 K K 4 K HAo [A-] = x(K HAo) 2

Page 186 of 316 Chem 155 Unit 10 Page 187 of 316

-5 Let KEQ = 10 …

[HA] in solution: HA(K HAo) HAo x(K HAo)

4 10 4

3 10 4

x K HAo 2 10 4 HAo

1 10 4

0 0 5 10 5 1 10 4 1.5 10 4 2 10 4 2.5 10 4 3 10 4 HAo

Fraction ionized:

1

x K HAo z 0.1 HAo z

0.01 1 10 7 1 10 6 1 10 5 1 10 4 1 10 3 0.01 0.1 HAo z

Page 187 of 316 Chem 155 Unit 10 Page 188 of 316 10.6 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 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 Chem 155 Unit 10 Page 190 of 316 10.7 UV-Vis Instrumentation:

Light Sources a. D2 Lamps – 190-400nm, excellent UV sources

Electrical Discharge in D2 gas in a sealed, SiO2 bulb. D2  D2* (dissociative molecular state) a b D2*  D + D + h E = KE(Da) + KE(Db) + h Wavelength Kinetic Energy Continuum  continuum

b. Tungsten(W)-Halogen – 350-2500nm, excellent Vis-NIR (visible-near Infrared) sources blackbody radiators (e.g. 2500K) halogen helps to re-deposit W onto filament and improves lifetime for very hot filaments c. Xe- or Hg-arc lamps – 200-1000nm, excellent UV-Vis sources high-atomic number noble gas in electrical discharge – many states  many wavelengths that coalesce into a continuum. Materials: silica = SiO2 (190-2500nm) glass = SiO2 + NaBO4, etc. (350-1500nm) sapphire = Al2O3 (180-5500nm) $$

Page 190 of 316 Chem 155 Unit 10 Page 191 of 316 10.8 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:

drift

i.e. slow change in: o source intensity o detector sensitivity

gives

absorbance error.

Page 191 of 316 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: Two detectors Often: More optics

Page 192 of 316 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 Chem 155 Unit 11 194 of 316

11 UV-Visible Spectroscopy of Molecules Skoog Ch. 14A all, 14B all, 14E-1 and 14E-3

Consider the simplest carbonyl:

H C O Lewis Structure of Formaldehyde H

3 types of valence electrons are important: “n” – non-bonding “ ” – sigma-bonding “ ” – pi-bonding

Orbital geometries

*

*

n

Page 194 of 316 Chem 155 Unit 11 195 of 316

11.1 Spectral Assignments 10 * 9 * 8

. 7

1 6 5 10 4 5

Energy 4 n

Absorbance 2 3 3

0 200 250 300 350 400 2 Wavelength / nm 1 0 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 = 10 1 200 50,000 n  * 10-4 = 6 2 250 40,000  * 9-1 = 8 3 330 30,000 n  * 9-4 = 5 E (spectrum) E (digram) / nm Assignment 50,000 10 200  * 40,000 8 250 n  * 30,000 6 330  * n  *

Page 195 of 316 Chem 155 Unit 11 196 of 316

11.2 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. O2, N2, all molecules have -bonds so: this radiation is absorbed by air and SiO2 so spectroscopy must be done in vacuum and is very difficult – called VUV or vacuum ultraviolet spectroscopy

2. n * 2.1. 150-250nm – most all transitions are found in the ultraviolet 2.2. Molecules with lone (nonbonding) pairs of electrons participate such as: O, N, Halogens 2.3. 100 to 1000 “medium” strength absorbers

3. n  *,  * 3.1. Longest wavelength transitions 200-700 nm 3.2. Strongest absorbers 100 to 100,000 M-1cm-1 3.3. Limited to molecules with -bonds – e.g. alkenes, alkynes, aromatics, carbonyls, azides etc.

Page 196 of 316 Chem 155 Unit 11 197 of 316

11.3 Spectral Peak Broadening Why are solution-phase electronic spectra hundreds of nm wide if atomic emission and absorption peaks are only ca. 0.005 nm wide?

Page 197 of 316 Chem 155 Unit 11 198 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 Chem 155 Unit 11 199 of 316

11.3.1 Influence of Solvent Spectral Fine Structure:

11.3.2 Temperature and Spectral Fine Structure:

Page 199 of 316 Chem 155 Unit 11 200 of 316

11.3.3 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 Chem 155 Unit 11 201 of 316

11.4 Aromatic UV-Visible absorptions:

Three sets of  * bands are prominent in the spectrum of benzene:

Primary E2 B (nm) 184 204 256 -1 -1 MAX (M cm ) 60,000 7,900 200 V. Strong Strong Intermed

-OH and –NH2 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 Chem 155 Unit 11 202 of 316

11.5 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 Chem 155 Unit 11 203 of 316

“d-d” transitions imply that d-orbitals have different energies.

When ligands such as H2O 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 Chem 155 Unit 11 204 of 316

Ligand field splitting energy varies with the type of ligand in the approximate following series:

- - - - -2 I < Br < Cl < F- < OH < C2O4 < H2O < SCN < NH3 - - < Ethylenediamine < o-phenanthroline < NO2 < 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 Chem 155 Unit 11 205 of 316

11.6 Charge-Transfer Complexes Strongly Absorbing Metal Complexes:

Example: Fe(III)3+-SCN- + h  Fe(II)2+SCN0 electron transfer from SCN- to Fe(III) 5,000

2+ Fe(II) (o-phenanthroline)3 + hv  + -1 Fe(III)3 (o-phenanthroline) (o-phenanthroline)2 electron transfer from Fe(II) to (o-phenanthroline) 12,000

Assuming that the absorption noise, σA = 0.001, what is the best detection limit for Fe+2 aquo and FeSCN+2 in the 300-800 nm range? Recall, Cm=3sb/m relate sb to sA and m to eb!

Page 205 of 316 Chem 155 Unit 11 206 of 316

11.7 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 Chem 155 Unit 11 207 of 316

11.8 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. Fe3+ T = Titrant – e.g. SCN- P = Product – e.g. FeSCN2+

Consider a big, stirred cuvette that you titrate with T.

To which of the above schemes would the FeSCN2+ 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 Chem 155 Unit 11 208 of 316

11.9 Multi-component Analyses: Consider a cell containing two analytes with overlapping absorbance spectra: Consider two absorbing species A and B:

6

Bi (0.25A) (0.75B) 4 i i

(0.5A) (0.5B) i i (0.75A) (0.25B) i i 2 Ai

0 200 250 300 350 400 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 Chem 155 Unit 11 209 of 316

Unknown mixture of A and B

Given: 4 A(233) = 5.3 A(293) = 0.72 B(233) = 0.72 B(293) = 5.3

2 Absorbance A(233) = A(233)bCA + B(233)bCB

A(293) = A(293)bCA + B(293)bCB 0 200220240260280300320340360380400 Wavelength / nm A(233) = 2.55 A(293) = 3.47

What are the concentrations of A and B?

Answer: CA = 0.4, CB = 0.6 C C A A1 A B1 B 1 C C A A2 A B2 B 2

I 5.3 CA 0.72CB 2.55

II 0.72CA 5.3 CB 3.47

0.72 III II 5.3 0.72 0.72 0.72 III 0.72 CA 5.3 CB 3.47 5.3 5.3 5.3

III 0.098CA 0.72CB 0.471 Substitute CA into I and II (to check):

I III 5.3 CA 0.72CB 2.55 I 5.3 0.4 0.72CB 2.55 2.55 5.3 0.4 0.098CA 0.72CB 0.471 CB 0.60 0.72 5.202CA 2.079 II 0.720.4 5.3 CB 3.47 2.079 3.47 0.720.4 CA 0.40 CB 0.60 5.202 5.3

Page 209 of 316 Chem 155 Unit 11 210 of 316

Matrix approach:

A1 C B1 C A A A B 1 A1 B1 1 C C A C A A2 A B2 B 2 A2 B2 A 2 . C C A A3 B3 C A A3 A B3 B 3 B 3 A4 B4 A C C A 4 A4 A B4 B 4

Ai = the extinction coefficient of species A at wavelength i

Bi = the extinction coefficient of species B at wavelength i

C = the concentration of species A A C = the concentration of species B B A = the absorbance at wavelength i i

Use computers to solve matrix for CA and CB, CC etc.

Overdetermine matrices to average out noise.

Page 210 of 316 Chem 155 Unit 12 211 of 316 12 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 Vibrational Coupling 16C1 Fourier Transform Instruments

Page 211 of 316 Chem 155 Unit 12 212 of 316 12.1 Overview: Infrared (IR) spectroscopy 1 molecular deals mainly with: vibrations

IR radiation is low energy relative to visible light.

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 4,000 - 500 Far IR 20,000 – 200,000 500 - 50

Page 212 of 316 Chem 155 Unit 12 213 of 316 12.2 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 Chem 155 Unit 12 214 of 316 12.3 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 Chem 155 Unit 12 215 of 316 12.4 IR Spectroscopy is Difficult!

IR beams are invisible to the eye.

IR bands are narrow, so EFF must be small.

SiO2 absorbs IR radiation. 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: EIR = h IR, – so – Photomultiplier tubes don‟t work because: For a PMT to work, the photon energy must be greater than the work function of the metal! E But

EIR

Page 215 of 316 Chem 155 Unit 12 216 of 316 12.5 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? 8 1 6 14 c 3 10 m s 10 10 m 3.333 10 s c

Page 216 of 316 Chem 155 Unit 12 217 of 316 12.6 Interferometers measure light field vs. time

Imagine you were a light-speed gremlin with a voltmeter in hand and a light beam passed by:

1 c c c

SignalVolts /

0 5 10 15 20 25 30

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 Chem 155 Unit 12 218 of 316 12.7 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 Chem 155 Unit 12 219 of 316 12.8 How is interferometry performed?

Recall – light interference is how gratings work.

10 Light Electric

cos( ) 6 Fields Sum 5 cos( ) 4 Constructive cos( ) cos( ) 0 Interference

5 0 0.5 1 1.5 2 2.5 3

2 10 Destructive cos( ) 6 5 Interference cos( ) 4 cos( ) cos( ) 0

5 0 0.5 1 1.5 2 2.5 3

2

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

Constructive interference:

n = f( sin(i)+sin(r) ) for a grating

n = f( time ) for an interferometer

Page 219 of 316

Chem 155 Unit 12 220 of 316 12.9 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 Chem 155 Unit 12 221 of 316

So – at every time interval the condition = n is true and the detector “sees” a bright light:

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 Chem 155 Unit 12 222 of 316 12.10 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 monochromatic- one frequency only.

Two frequencies (lines) only.

Polychromatic – many frequencies.

Page 222 of 316 Chem 155 Unit 12 223 of 316 12.11 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:

2000

1000

x

Amplitude 0

1000 0 20 40 60 80 100 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 Chem 155 Unit 12 224 of 316

12.12 Transforming time  frequency domain

1 2

1.5 Amplitude1 0 Amplitude 1

0.5 1 0 0.5 1 1.5 0 time_seconds 0 1 2 3 4 Frequency_Hz

1 2

1.5

Amplitude2 0 Amplitude 1 + 0.5 1 0 0 0.5 1 1.5 0 1 2 3 4 time_seconds Frequency_Hz

------

2 2

1.5 Amplitude3 0 Amplitude 1

0.5 2 0 0.5 1 1.5 0 0 1 2 3 4 time_seconds Frequency_Hz signals:

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 Chem 155 Unit 12 225 of 316 12.13 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 Chem 155 Unit 12 226 of 316 12.14 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:

100 20

10

50 Amplitude Amplitude 0

0 10 0 0.5 1 1.5 0 20 40 Frequency Frequency

2000 Interferogram

0 Amplitude

2000 0 20 40 60 80 100 T ime

10

0 Amplitude

10 20 25 30 35 40 T ime Page 226 of 316 Chem 155 Unit 12 227 of 316 12.15 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 Multiplex good from 200nm to 1200nm Advantage (1000nm total) and your effective bandwidth is 1 nm, how much light is being measured or lost at any one time?

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 Peak distortion Beer‟s law (convolution)

Page 227 of 316 Chem 155 Unit 12 228 of 316 12.16 Resolution in Interferometry

Interferometers are easily made into high- resolution spectrometers.

Consider two spectral lines differing in frequency by only 2%:

How do these lines appear in the time domain?

8 7

6 cos( ) 6 4 cos( 1.02) 3 cos( ) cos( 1.02) 2 2

0

0.999013 2 0 10 20 30 40 50 60 70 80 90 100 0 100 2

Page 228 of 316 Chem 155 Unit 12 229 of 316

Let us define resolution as: = 1 - 2 when 1 can just be separated from 2.

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( ) = Pi cos(2 f t) = Pi 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 Chem 155 Unit 12 230 of 316

For two light rays of identical power Pi but different frequency the total signal will be:

P( ) = Pi cos(2 1) + Pi 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 1 - 2 = ±1 i.e. they have „slipped‟ one wavelength and are both bright again – so, in this case: 1 - 2 = = 1/

Page 230 of 316 Chem 155 Unit 12 231 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:

= 107/500 nm = 20,000 cm-1 = 2d = 1 cm = 1/ = 1/1cm, so 20,001 cm-1 can be resolved from 20,000 cm-1.

107 / 20,001 cm-1 = 499.975 nm.

So, 500 nm can be resolved from 499.975 nm.

= 500 - 499.765 = 0.025nm

EFF = 0.025nm is very difficult to achieve with a monochromator!

Remember that is equal to 2d – twice the mirror displacement.

Page 231 of 316 Chem 155 Unit 12 232 of 316 12.17 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 -1 resolution (ΔλEFF) in cm ? 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 Chem 155 Unit 12 233 of 316 12.18 Answers:

cm 1 -1 f 2 v 2 1 3000cm 6000s s

1 1 1 1 1 effective bandwidth incm : EFF 0.5cm 2 d 2 1 cm

7 nm 7 nm 10 10 cm cm 0.555nm effective bandwidth in nm: 1 1 3000cm 3000.5cm

0.4 mm 1000nm cos 45 w d cos (r) w d cos (r) 180 EFF F 0.514m n F n EFF 1 0.55nm

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 Chem 155 Unit 13 Page 234 of 316

13 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 100 100,000 140,000 UV - *, n- *, n- *, - * VIS 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

100,000 100 140 -wave

1,000,000 10 14

Page 234 of 316 Chem 155 Unit 13 Page 235 of 316

13.1 Absorbance Bands Seen in the Infrared: :

Vibrations Rotations

Solid broad rare (phonon) (C60) Liquid broad unresolved

Gas sharp sharp

Page 235 of 316 Chem 155 Unit 13 Page 236 of 316

13.2 IR Selection Rules

How does electromagnetic radiation make a molecule vibrate? Consider the diatomic molecule H-Cl. E( ) sin 2 360 Electric Field Vector of Light

E( )

+ H H H

- Cl Cl Cl

In Out In Force

Page 236 of 316 Chem 155 Unit 13 Page 237 of 316

Now consider the molecule N2: E( ) sin 2 360 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 Chem 155 Unit 13 Page 238 of 316

13.3 Rotational Activity E( ) sin 2 Indicate360 the forces on the molecule below: Electric Field Vector of Light

E( )

`

Cl H H Cl

When can a molecular rotor absorb light?

When a molecule has a permanent dipole

moment it is rotationally active.

Example:

J = J = v = v =

Energy Page 238 of 316 Chem 155 Unit 13 Page 239 of 316

13.4 Normal Modes of Vibration: Consider a diatomic molecule –

translations

rotations

vibration!

For linear molecules, there are: 3N-5 modes of vibration.

Non-linear molecules have: 3N-6 modes of vibration.

Page 239 of 316 Chem 155 Unit 13 Page 240 of 316

Modes of vibration: For example – CO2: Linear or Non-linear? # of Modes: 3( ) - ( ) =

= 1340 cm-1 = 2340 cm-1 S AS symmetric stretch anti-symmetric stretch

-1 -1 S = 666 cm S = cm

anti-symmetric stretch, IR active

symmetric stretch, IR inactive

bend modes, degenerate, IR active

For example – H2O: Linear or Non-linear? # of Modes: 3( ) - ( ) =

S(OH) = AS(OH) = S(OH) = 3562 cm-1 3756 cm-1 1596 cm-1

Page 240 of 316 Chem 155 Unit 13 Page 241 of 316

13.4.1 One may see more than 3N-5/6 bands:

Overtones: v = ±2, ±3

Combinations: C = A + B and A - B

13.4.2 One may see fewer than 3N-5/6 bands:

not in 400-4000 cm-1 range of typical FTIR

Band may be weak

Band may be broad and overlap another band

Bands may be degenerate

Bands may be „forbidden‟, i.e. if there may be no change in dipole moment during the vibrational mode.

Page 241 of 316 Chem 155 Unit 13 Page 242 of 316

13.5 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 NH3 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 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 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) d2y F = mA = m = ky(t) dt2 y(t) is proportional to its own second derivative. This differential equation has a sinusoid solution: 1 k y(t) = A cos(2 mt) m = 2

m1 m2 = reduced mass m1 + m2

Units of are: units of mass Page 244 of 316 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

1 k again: m = 2

so: h k E = v h m = ±1 = h 2

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 Chem 155 Unit 13 Page 246 of 316

13.6 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 Chem 155 Unit 14 247 of 316

14 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 Chem 155 Unit 14 248 of 316

14.1 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. CaF2 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 Chem 155 Unit 14 249 of 316

14.2 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?

A= bC = 10*1*10 = 100 T=10-100 = 0.00000000000000…….01 Page 249 of 316 Chem 155 Unit 14 250 of 316

14.4 Handling of solids: pelletizing:

grind put into die

5 – 10 tons of pressure a few minutes…

vacuum to reduce voids

2-5 mg 14.5 Handling of Solids: mulling: sample 2 m

“Nujol” = mineral (aliphatic ) oil

or 1. Grind sample finely “Fluorolube” = fluorocarbon oil 2. Mix with mulling oil 3. Sandwich w/ KBr/NaCl plates

Page 250 of 316 Chem 155 Unit 14 251 of 316

14.6 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 Chem 155 Unit 14 252 of 316

14.7 Group Frequencies Examples : C-O-H versus C-Cl

Which stretching frequency is higher:

O-H or C-Cl Why? Reduced mass of C-Cl much larger so (k/ )1/2 for C-Cl smaller

Page 252 of 316 Chem 155 Unit 14 253 of 316

14.8 Fingerprint Examples

Mid-IR Group Frequencies and Fingerprints Example – Branched Alkane Structural Isomers

Page 253 of 316 Chem 155 Unit 14 254 of 316

14.9 Diffuse Reflectance Methods: A way to do IR of solids without mulling or pelletizing:

Page 254 of 316 Chem 155 Unit 14 255 of 316

14.10 Quantitation of Diffuse Reflectance Spectra:

Page 255 of 316 Chem 155 Unit 14 256 of 316

14.11 Attenuated Total Reflection Spectra:

Page 256 of 316 Chem 155 Unit 14 257 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 Chem 155 Unit 14 258 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 n1-n2 interface).

Page 258 of 316 Chem 155 Unit 15 259 of 316

15 Raman Spectroscopy:

vibrational spectroscopy with

near UV, visible or near infrared light is called:

Raman Scattering Spectroscopy

Intense monochromatic light H2O beam – usually a laser

A tiny fraction of An even tinier the incident fraction of the radiation scatters incident radiation off of the molecules scatters off of the in the solution: molecules and lose or gain energy.

Rayleigh Raman

SCATTERED = INCIDENT SCATTERED ≠ INCIDENT

elastic scattering inelastic scattering Page 259 of 316 Chem 155 Unit 15 260 of 316 A Raman scattering experiment:

High resolution monochromator / Scattered Beam interferometer + sensitive detector

Laser Transmitted Beam

Incident Beam

The sample may be a Raman is not usually concentrated solution, a neat used for dilute liquid analyte, a transparent solutions. solid or a surface.

If the laser is, e.g. a HeNe laser ( =632.8 nm), most of the scattered radiation will be:

632.8 nm or 1/(632.8x10-7 cm) = 15803 cm-1

This scattered radiation is monochromatic, and is known in the spectrum as the:

Rayleigh Line

Page 260 of 316 Chem 155 Unit 15 261 of 316 15.1 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 . Light Power Light

400 425 450 475 500 525 550 575 600 625 650 Wavelength / nm

Page 261 of 316 Chem 155 Unit 15 262 of 316 15.2 Quantum View of Raman Scattering.

S1

So Energy

Vibrational coordinate

Hypothetical Raman Scattering Spectrum Light Power Light

400 425 450 475 500 525 550 575 600 625 650 Wavelength / nm

Page 262 of 316 Chem 155 Unit 15 263 of 316 15.3 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 cos 2 t eq vib

time

Visible Frequency Light (e.g. 500 nm / 20,000 cm-1 / blue-green)

E E0 cos 2 ex t

time

Page 263 of 316 Chem 155 Unit 15 264 of 316 Radiation interacts with molecules by inducing a dipole moment. The induced dipole „m‟ is proportional to the polarizability, . m E E0 cos 2 ex t

But the polarizability is itself oscillating!

d d r r r cos 2 t 0 dr e 0 dr A vib

So the induced dipole moment is oscillating in a more complex way: d m E cos 2 t r cos 2 t 0 ex 0 dr A vib

The dynamic induced dipole that gives rise to Raman scattering:

Incident E-field

Oscillating polarizability

Amplitude modulation in induced dipole

Page 264 of 316 Chem 155 Unit 15 265 of 316 15.4 The classical model of Raman: The induced dipole moment has a complex description, but how does this relate to Raman scattering?

Recall that: 1 1 cos(x) cos(y) cos(x y) cos(x y) 2 2 So m becomes: m E0 0 cos 2 ex t 1 d E0 rA cos 2 cos 2 2 dr ex vib ex vib

Both Stokes and AntiStokes peaks are predicted by classical theory!

Classical Model 'Catastrophe' Light Power Light

4 4 4 4 4 4 4 4 4 4 4 1.6 10 1.68 101.76 101.84 101.92 10 2 10 2.08 102.16 102.24 102.32 10 2.4 10 Wavelength / nm but…

Page 265 of 316 Chem 155 Unit 15 266 of 316 15.5 The classical model: catastrophe!

Real intensity distributions Light Power Light

4 4 4 4 4 4 4 4 4 4 4 1.6 10 1.68 101.76 101.84 101.92 10 2 10 2.08 102.16 102.24 102.32 10 2.4 10 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. The Stokes peaks are more intense than the anti Stokes peaks there are more molecules in the ground vibrational state, that can because: give rise to Stokes peaks, than in the excited vibrational state, that can give rise to anti-stokes peaks.

Page 266 of 316 Chem 155 Unit 15 267 of 316 15.6 Raman Activity:

mode Raman IR active? active? Symmetric O stretch

C NO YES YES NO

O

Asymmetric O stretch

YES NO NO YES C

O

Bend O

YES YES YES YES C

O

Page 267 of 316 Chem 155 Unit 15 268 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 Chem 155 Unit 15 269 of 316 15.7 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)

1 k vib 2

Very low frequency modes are accessible (e.g. inorganics):

Depolarization ratio can distinguish between symmetric and asymmetric stretches.

Page 269 of 316 Chem 155 Unit 15 270 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 Chem 155 Unit 15 271 of 316 15.8 Resonance Raman S1

So

Pro‟s 1. Intensity greatly enhanced (102 – 106) 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:

-9 „typical‟ fluorophore - FLUOR ~ 10 s. -14 Raman - RAMAN ~ 10 S

Page 271 of 316 Chem 155 Unit 15 272 of 316 15.9 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 . Light Power Light

400 425 450 475 500 525 550 575 600 625 650 Wavelength / nm

7 10 1 laser frequency: 20000 cm 500

7 10 1 Stokes frequency: 17857 cm 560

7 7 10 10 1 2143 cm Effective bandwidth: Must resolve 20000 and 20001 cm-1. 500 560 7 7 7 7 10 10 10 10 1 0.025 nm high resolution needed! 2759 cm 20000 20001 500 580

Page 272 of 316 Chem 155 Unit 15 273 of 316

4. 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?

7 7 10 1 10 1520 17916 cm 558.2 nm 514.5 17916

5. Does the CO bond in CO2 have a dipole moment? YES

6. Does the CO2 molecule have a dipole moment? Why? No, because individual C=O bond dipoles exactly cancel.

7. Draw a series of CO2 molecules to illustrate the symmetric stretching mode.

O C O O C O O C O

8. Is the CO2 symmetric stretch IR active? Why? No, because at no time during the vibration does the dipole moment change.

9. Is the CO2 symmetric stretch Raman active? Why? Yes, because during the vibration the polarizability changes.

Page 273 of 316 Chem 155 Unit 15 274 of 316

10. Draw a series of CO2 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 N2 not a greenhouse gas? Homonuclear diatomics have no dipole moment change during their only vibrational mode: symmetric stretch.

Page 274 of 316 Chem 155 Unit 15 275 of 316

14. Cyclohexanone has a strong absorption peak at 5.86 m and at this wavelength a linear relationship exists between absorbance and concentration. 4 10 1 -1 1706 cm a. Calculate the frequency of this mode in cm . 5.86 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. CCl4, CHCl3 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 0.025 mm. What is the detection limit for this compound under these conditions, if the noise associated with the spectrum of the solvent is 0.001 absorbance units. A=εbC  calibration slope = εb

1 A 0.40 mg 3 sb 3 0.001 mg b 0.2 C 0.015 C mg mL m m 1 mL 2.00 mg mL 0.2 mL

Page 275 of 316 Chem 155 Unit 15 276 of 316

15. 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. E a. The Boltzmann equation is: N1 g1 e kT where g and g N0 g0 0 1 are the degeneracies of the states 0 and 1 and for HCl35 may be considered equal, and k is Boltzmann‟s constant, T is absolute temperature 23 J 21 k 1.3810 T 298 K k T 4.112 10 J and E is the K energy of 1 34 10 cm 20 state 1 2885cm 6.62610 J s 3.0010 5.735 10 J relative to 0. s 1 34 10 cm E 2885cm 6.62610 J s 3.0010 s

E E N1 kT k T 7 e e 8.784 10 N0

b. Assuming that the bond strengths are 1 k m1 m2 VIB equal, 2 m1 m2 calculate the vibrational 1 kDCl frequency DCl 2 DCl DCl HCl that you since kHCl kDCl k expect for the HCl 1 HCl HCl DCl isotope DCl35. 2 HCl

35 1 35 2 HCl HCl DCl 0.717 35 1 35 2 DCl

DCl 28850.717 DCl 2069

Page 276 of 316 Chem 155 Unit 15 277 of 316

16. 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 Chem 155 Unit 15 278 of 316

17. 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?

7 7 7 7 10 10 10 10 3504 3483 250 274 300 335

The band moves with the excitation – it is the Raman peak for water OH stretch modes.

Page 278 of 316 Chem 155 Unit 16 279 of 316 16 Mass Spectrometry (MS) overview: Skoog: sections 20A and B In MS one a. often initially performs GC or LC separation b. gets analyte molecules into the gas phase in vacuum c. converts analyte molecules into ions d. that are both intact and in fragments e. exposes the analyte ions to E and/or B fields f. separates them based on mass / charge g. detects them based on current (they are ions)

16.1 Example: of a GCMS instrument:

Page 279 of 316 Chem 155 Unit 16 280 of 316 16.2 Block diagram of MS instrument.

Sample

Introduction

Ionization Source

Mass Analyzer

Computer for Display / storage

Pattern recognition to ID molecule

Page 280 of 316 Chem 155 Unit 16 281 of 316 16.3 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! 1H35Cl, 2H35Cl, 1H37Cl, 2H37Cl

But this can be informative: Isobaric interferants 13 Asparagine – CH2CONH2 (133.12) Aspartic acid – CH2CO2H (133.10) Distinguishable only by high resolution MS

Source: Quantitative Chemical Analysis 6th Ed. By Daniel Harris Page 281 of 316 Chem 155 Unit 16 282 of 316 16.4 Ionization Sources

16.4.1 Electron Impact (EI):

„Hard‟ ionization: extensive fragmentation

16.4.2 Chemical Ionization: reagent gas e.g. Derivative of EI but with 1000:1 excess of: CH4

„Soft‟ ionization: limited fragmentation

+ + + CH4  collision with electron  CH4 , CH5 , CH3 + + + + CH4 , CH5 , CH3  recombination, rxn w/ CH4  C2H5

+ + Proton transfer  M+1 CH5 + MH  MH2 + CH4

+ + ethyl transfer  M+15 C2H5 + M  M C2H5

Page 282 of 316 Chem 155 Unit 16 283 of 316

Example of CI vs EI MS of 1-octanol:

Electron impact ionization

chemical ionization (CH4)

Which is a „better‟ MS?

CI: more easily interpretable

EI: complex, but highly unique

Page 283 of 316 Chem 155 Unit 16 284 of 316 16.4.3 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 proteins in matrix supersonic vacuum matrix ionizes expansion chamber

Page 284 of 316 Chem 155 Unit 16 285 of 316

16.4.4 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 Chem 155 Unit 16 286 of 316

Which is a „better‟ ionization source for large molecules, MALDI or ESI?

MALDI: spectra are more easily interpretable

spectra are complex, but ESI easily ESI: interfaced to LC

Page 286 of 316 Chem 155 Unit 16 287 of 316 16.5 Mass Analyzers:

One important quality of mass analyzers is:

16.5.1.1 Resolution: R M/ M

Small molecule MS may demand high resolution: + + + + For example, can one distinguish C2H4 , CH2N , N2 , CO in most mass spectrometers?

Ion Nominal Mass Exact Mass 12 1 + 28 28.0313 C2 H4 12 1 14 + 28 28.0187 C H2 N 14 + 28 28.0061 N2 12C16O+ 28 27.9949

What resolution would be needed to distinguish these?

Smallest M = 28.0061-27.9949 = 0.012

m / M = 28 / 0.012 = 2300

What resolution would one need to calculate the nominal mass of immunoglobulin-G (M = 149190)? m / M = 149,190 / 1 = 150,000

Page 287 of 316 Chem 155 Unit 16 288 of 316

16.5.2 Magnetic Sectors:

Resolution of a single magnetic sector is ca. 2000

The properties that limit R of the ions entering it, mainly:

speed directional distribution distribution

Page 288 of 316 Chem 155 Unit 16 289 of 316 16.5.3 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 B-field on magnetic sector sector

R attainable: 100,000

Page 289 of 316 Chem 155 Unit 16 290 of 316

16.5.4 Quadrupole Mass Filters:

10,000 Resolution up to:

< 4000 AMU Masses limted to:

Stable trajectory is a function of: AC amplitude

DC field

M/z A given AC amplitude + DC field yields a stable trajectory for one: mass filter Like E and B sectors this is a:

One „scans‟ m/z: scanning mode (see one at a time)

Page 290 of 316 Chem 155 Unit 16 291 of 316 16.5.5 Time of Flight (TOF):

Ion kinetic energy E = zVPULSE

But kinetic energy E = ½mv2

So, a drift length L gives a the “flight” time tF = L/v

Solve for m/Z as a function of VPULSE and distance L:

2 2 E = zVPULSE = ½mv = ½m(L/tF)

2 2 2VPULSE(tF/L) = m/z = (2V/L²) tF

So, one measures ion m/z by measuring the ion‟s

Arrival time at the detector: t F

Page 291 of 316 Chem 155 Unit 16 292 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 25000.

16.5.6 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 CHCl3 . 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 OH further. Testosterone made by biosynthesis in the human H body differs in only one way from that prepared in a H H 13 12 O laboratory – C/ C isotopic testosterone abundances. What C19H28O2 Exact Mass: 288.21 resolution is required to Mol. Wt.: 288.42 m/e: 288.21 (100.0%), 289.21 (20.6%), 290.22 (2.1%) measure this ratio? C, 79.12; H, 9.78; O, 11.09

Page 292 of 316 Chem 155 Unit 16 293 of 316

5. 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 C19H28O2 testosterone above, can you Exact Mass: 288.21 Mol. Wt.: 288.42 distinguish between the exact m/e: 288.21 (100.0%), 289.21 (20.6%), 290.22 (2.1%) C, 79.12; H, 9.78; O, 11.09 mass and the molecular weight? 9. To what do the 20% and 2% fragments correspond?

Element Iso % Carbon 12C 98.90% 13C 1.10% Hydrogen 1H 99.99% 2H 0.01% Oxygen 16O 99.76 17O 0.038% 18O 0.2%

Page 293 of 316 Chem 155 Unit 17 Chromatography Page 294 of 316

17 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 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 5th ed.

Page 295 of 316 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 Chem 155 Unit 17 Chromatography Page 297 of 316

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Page 301 of 316 Chem 155 Unit 17 Chromatography Page 302 of 316 x

Page 302 of 316 Chem 155 Unit 17 Chromatography Page 303 of 316 Theoretical Plates and Plate Count:

2 2 x i 100 25 i 10 0.625 1000 1000 1 1 2 2 2 S x exp x 50 12.5 2 2.5 0.156 2 2 1000 1000

2 2 2 1000 1000 N: 100 1600 N 100 25

2 2 1000 1000 400 6400 50 12.5

Illustration of different plate counts

0.03 6400 plates 0.025

0.02

signal 1600 plates 0.015

0.01 400 plates

0.005 100 plates

0 x 0 200 400 600 800 1000 1200 1400 1600 1800 2000 retention time or column length

Page 303 of 316 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?

2 2 2 400 1200 N: 1024 9216 N 12.5 12.5

2 2 800 1600 4096 16384 12.5 12.5

Illustration of different plate counts

0.03

0.025

0.02 signal 0.015

0.01 x

0.005

0 0 200 400 600 800 1000 1200 1400 1600 1800 2000 retention time or column length

Page 304 of 316 Chem 155 Unit 17 Chromatography Page 305 of 316 The van Deemter Equation: H = plate height B A = „multipath term‟ H A C u B = longitudinal diffusion term u C = resistance to mass transfer B PlateH(A B C heightu) A asC u a function of linear flow rate: u 12

10

8

H(1 2 0.5 u)

6

4

2 0 2 4 6 8 10 12 u

Rationalize why is there an optimum flow rate:

Page 305 of 316 Chem 155 Unit 17 Chromatography Page 306 of 316

A 2 dp

B 2 DM B H A C u C = com plex function of particleu size, coating, diffusion coefficient, favored in general by large diffusivity and small particle size

Page 306 of 316 Chem 155 Unit 17 Chromatography Page 307 of 316 17.1 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. x

If the solvent is strong enough to elute 5 and 6 in a reasonable time:

Page 307 of 316 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 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 differentx wavelength or o a different set of solvents

Page 308 of 316 Chem 155 Unit 17 Chromatography Page 309 of 316 17.2 T-gradient example in GC of a complex mixture.

x

Page 309 of 316 Chem 155 Unit 17 Chromatography Page 310 of 316 17.3 High Performance Liquid Chromatography

Skoog Holler and Nieman Principlesx of Instrumental Analysis 5th ed.

Page 310 of 316 Chem 155 Unit 17 Chromatography Page 311 of 316 17.4 Types of Liquid Chromatography

Most can be performed as HPLC in „high performance‟ instrumentation if necessary.

17.5 Normal Phase: stationary phase: hydrophilic (SiO2 particles) mobile phase: hydrophobic mobile phase (hexane, methanol). weak solvent: hexane strong solvent: methanol retained: polar compounds eluted first: non-polar compounds

17.5.1 Reverse Phase: stationary phase: hydrophobic (SiO2 particles coated with hexadecane monolayer) mobile phase: hydrophilic mobile phase (water, methanol) weak solvent: water strong solvent:x methanol retained: non-polar compounds eluted first: polar compounds

Page 311 of 316 Chem 155 Unit 17 Chromatography Page 312 of 316 17.5.2 Exclusion Gels

„GPC‟ Gel permeation chromatography: porous polymer gel particles – small molecules penetrate and are retained, large molecules elute more quickly.

PAGE polyacrlamide gel electrophoresis:

Aqueous gel: -CH2-CONH- used for electrophoresis, large molecules experience more drag and are retained.x

Page 312 of 316 Chem 155 Unit 17 Chromatography Page 313 of 316

Ion Exchange: Stationary phase is an ionic polymer – e.g. polystyrenesulfonic acid:

- SO3

* * 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 vice- versa. x

Page 313 of 316 Chem 155 Unit 17 Chromatography Page 314 of 316 17.6 HPLC System overview:

Injection valvex

Page 314 of 316 Chem 155 Unit 17 Chromatography Page 315 of 316 17.7 Example of Reverse-phase HPLC stationary phase:

x

Page 315 of 316 Chem 155 Unit 17 Chromatography Page 316 of 316 17.8 Ideal qualities of HPLC stationary phase: Very fast partitioning between mobile and stationary phase. 1. Ultimate thin film: single molecular layer. 2. „Perfect‟ – defect free film with no sites for strong adsorption.

Mechanically strong to withstand high pressure w/o collapse.

Chemically inert.

Partition coefficients of solute analytes are all different in single solvent – i.e. the SP provides a measure of selectivity without being limited to just a few chosen analytes. x

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