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As the S/N of an analytical signal decreases, so does the accuracy and precision of that signal. The pair of plots below illustrate this point. The plot to the left contains three analyte peaks with a peak-to-peak noise level of 0.19 μV. The S/N for each peak is 52, 26, and 10 respectively. Increasing the peak-to-peak noise level ten-fold (1.9 μV) decreases the S/N of each peak by a factor of ten. (5.2, 2.6, 1.0 respectively)
Note that the signal at 2 minutes, with a S/N ratio of ~3, is at a level commonly known as the detection limit, which is defined as the magnitude at which the signal is statistically distinguishable from the noise. The signal at 3 minutes, which has a S/N equal to 1, is indistinguishable from the baseline noise. This comparison illustrates the need to reduce noise to a level at which chemical information is not compromised. A spreadsheet has been designed to illustrate the relationship between signal and noise. Click here to perform these exercises.
Can Noise be Reduced After the Data has been Recorded?
In the examples above, the frequency of the signal is less than the frequency of the noise. In all cases, if the signal frequency and the noise frequency are not equal, then there should be at least one suitable approach to noise reduction.
Overview of S-N Enhancement Techniques
This module will describe two general categories of noise reduction techniques:
1. Analog Filtering (Hardware-Based)
2. Digital Filtering (Software-Based)
Most of these S/N enhancement methods, whether analog or digital, are based on either:
1. Bandwidth Reduction (i.e. decreasing Δf).
2. Signal Averaging (i.e. decreasing Δf or averaging out random noise fluctuations)
Bandwidth reduction is important --- Remember, if fsignal ≠ fnoise, we have a chance of isolating the signal from the noise. This results in an enhanced signal-to-noise ratio and more reliable information about the chemical sample of interest. We will see that there are limitations to how much bandwidth reduction can be applied before distorting the instrumental signal. Nevertheless, these can be effective approaches to improving the quality of the instrumental signal.
04 Signal-to-Noise Enhancement
Introduction
Exercise #1 is designed to familiarize the student with the effect of noise on the detectability of a signal. This exercise is designed to be completed with the Signal Noise Exercise spreadsheet. This spreadsheet allows the user to create an ideal chromatographic separation containing up to three peaks, which represent three different compounds.
The height of each peak is proportional to the amount of analyte being separated. A noise component may be added to the ideal separation in order to simulate data that could be acquired in an actual separation.
Table of Spreadsheet Parameters
The table below describes all of the parameters on the spreadsheet needed to complete the exercise below. Parameters with a light yellow background may be adjusted. Parameters with a light green background may not be adjusted.
Parameter
Definition
S/N (signal-to-noise ratio)
This figure of merit indicates the magnitude of the signal with respect to the noise level at the 95% or 99% confidence level.
Peak Intensity
This value controls the height of the analyte peak. A value of zero indicates no analyte present.
Mean
This value controls the position of the maximum peak intensity. Values are restricted between 0.5 to 4.5 minutes
Standard Deviation (Std Dev)
This value controls the width of the peak. Values are restricted between 0.001 and 0.250 minutes
Noise
This value controls the amount of noise added to the plot. Since it is based on Excel's RAND function, it appears as high-frequency noise superimposed on the analyte peaks. This is also known as the peak-to-peak noise.
Offset
This value is used to raise or lower the baseline of this plot. It is most effective when the plot has large peaks on a noiseless baseline.
RMS Noise
This value is the magnitude of the noise (Max Signal — Min Signal) divided by 5.16 or 3.92 (±zσ), where z = 2.58 or 1.96 (99% or 95% confidence level). Dividing the RMS noise into the peak intensity provides the S/N for that analyte.
Peak + Noise
This cell represents the sum of the peak intensity and the superimposed noise.
Part 1: Spreadsheet Orientation
1. Familiarize yourself with the Signal Noise Exercise spreadsheet. Observe changes to the plot when:
1. The peak parameters are adjusted (peak intensity, mean, standard deviation)
2. The magnitude of the noise is increased from zero
3. The magnitude of the offset is increased from zero
2. Answer the following questions
1. Which parameter(s) control the signal level?
2. Which parameter(s) control the noise level?
3. Which parameter(s) or concept(s) control the character of the instrumental response?
Part 2: Evaluating Baseline Noise
1. Start with a flat baseline by eliminating all traces of signal and noise.
1. How would you accomplish this?
2. Which parameter would you adjust if you can’t see the flat baseline?
2. Calculating Noise Magnitude
1. Based on the discussion of noise in this e-module, if the peak-to-peak noise (VN) is 1.0 μV, calculate the RMS Noise at the 99% confidence level.
2. Increase the noise level to 1.0 μV on the spreadsheet and look at the RMS Noise. Does your answer agree with the spreadsheet?
3. Repeat the RMS Noise calculation at the 95% confidence level. Is the RMS Noise smaller or larger? Explain this difference based on your knowledge of population distributions.
Part 3: Evaluating Signal-to-Noise Ratios
1. Enter the following signal parameters into the Signal Noise Exercise spreadsheet
1. Peak 1
1. Peak Intensity = 10 μV
2. Mean = 1 minute
3. Standard Deviation = 0.1 minute
2. Peak 2
1. Peak intensity = 5 μV
2. Mean = 2 minutes
3. Standard Deviation = 0.1 minute
3. Peak 3
1. Peak Intensity = 1 μV
2. Mean = 3 minutes
3. Standard Deviation = 0.1 minute
2. Change the peak-to-peak noise level on the spreadsheet to 1.00 μV.
1. Look at the S/N ratio for each peak and determine which peaks are below the detection limit (S/N < 3)
2. Record the "Peak+Noise" magnitude for each peak.
3. Obtain replicate data by pressing F9 to refresh the spreadsheet. Record the new magnitude for each peak.
4. For each analyte peak, calculate the percent relative error range and average percent relative error introduced by adding 1.00 μV peak-to-peak noise to the simulated signal.
5. As the S/N decreases, comment on any trends regarding the effect of noise on the accuracy and precision of the signal
3. Increase the peak-to-peak noise level on the spreadsheet to 5.00 μV and determine which peaks are now below the detection limit.
4. Based on your results, what role would signal-to-noise enhancement have in analyte detection? | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Introduction_to_Signals_and_Noise/04_Signal-to-Noise_Enhancement/01_Signal-to-Noise_Enhancement_Exercise_1.txt |
Signals and noise are almost always expressed as electrical quantities. The electrical quantities you should be familiar with are:
Voltage
Voltage is a measure of energy available when an electron moves from a point of higher potential to a point of lower potential. The SI Unit for voltage is the Volt (V). 1V = 1 Joule/Coulomb.
Physicochemical phenomena that generate voltage include:
• Chemical Reactions, such as those that take place in a battery
• Electromagnetic Induction, such as moving a coil of wire through a magnetic field (i.e. an electric generator)
• Photovoltaic Cells, which convert light energy into electrical work
Current
Current is a measure of the amount of electronic charge flowing per unit time past a given point. The SI Unit for current is the Ampere (A). 1A = 1 Coulomb/second. Types of current include
02 Analog Filtering
In 1827, Georg Ohm published his work Die galvanische Kette mathematisch bearbeitet, indicating that the current flowing through a conductor is proportional to the voltage across the conductor. All conductors of electricity obey Ohm’s Law, which is mathematically expressed as:
$\mathrm{R = \dfrac{V}{I}}$
• V = Voltage across the conductor (in Volts, V)
• I = Current through the conductor (in Amperes, A)
• R = Resistance of the conductor (in Ohms, Ω)
Simple applets to test out Ohm’s Law:
02 Passive Electronic Com
Resistor
• A resistor is a component that resists electron flow.
• The unit of resistance is called an ohm (Ω). 1Ω = 1V/A
• In an electronic circuit schematic, a resistor is represented by:
Capacitor
• A capacitor is an electronic component that stores charge
• It consists of two conductive plates separated by an insulating medium
• The unit of capacitance is called a farad (F). 1 F = 1 C/V
• In an electronic circuit schematic, a capacitor is represented by:
A simple applet used to illustrate the principle of resistance
http://micro.magnet.fsu.edu/electromag/java/filamentresistance/index.html
Simple applets used to illustrate the principle of capacitance
http://micro.magnet.fsu.edu/electromag/java/capacitance/index.html
http://micro.magnet.fsu.edu/electromag/java/capacitor/
03 Passive Electronic Cir
Remember that signal-to-noise ratios can be enhanced if the signal frequency is different than the noise frequency. You will be introduced to these frequency-dependent analog filters at the end of this section. For now, let’s start very simply...
Resistor Fundamentals
The simplest circuit involving a resistor and a voltage source is shown below. The dotted lines are just there to represent where a high-quality voltmeter would be connected if we wished to measure the voltage across the resistor. Calculating the current flowing through this resistor requires the use of Ohm’s Law.
Circuit #1
According to Ohm's Law $\mathrm{I = \dfrac{V}{R} = \dfrac{1.0\: Volt}{20\: Ohms} = 0.050\: Amperes}$
• V = 1.0 Volt
• R = 20 Ohms
Resistors in Series
Practically speaking, we are not limited to a single resistor. Circuit #1 could also be represented by Circuit #2 below:
• Resistors placed in a "head-to-tail" configuration are in series.
• The total resistance is the sum of all the individual resistances
• Putting resistors together in series gives a larger total resistance
$\mathrm{R_T = (R_1 + R_2) = (10\,Ω + 10\,Ω) = 20\,Ω}$
Resistors in Parallel
Resistors placed in a “side-to-side” configuration are in parallel.
The total resistance is the reciprocal of the sum of each reciprocal resistance. So for a pair of resistors as shown in Circuit #3 above:
$\mathrm{R_T = \dfrac{1}{\dfrac{1}{R_1} + \dfrac{1}{R_2}} = \dfrac{R_1R_2}{R_1 + R_2}}$
Applying this to Circuit #3:
$\mathrm{R_T = \dfrac{1}{\dfrac{1}{40} + \dfrac{1}{40}} = \dfrac{40 * 40} {40 + 40} = 20\: ohms}$
Putting resistors together in parallel always gives a smaller total resistance. Note that Circuit #3 has the same current as Circuits #1 and #2.
Voltage Divider
Sometimes, the output of an instrument is too large for a readout device. One circuit used to reduce a voltage is a voltage divider
Note that:
• A representation for a voltmeter has been added to the schematic
• The voltage is only being accessed across one of the two resistors
Assuming that the meter resistance is much larger than R2 (i.e. no loading error occurs), then according to Ohm’s Law
$\mathrm{V_{in} = I(R_1 + R_2)}$
For a discussion of loading errors, click here.
If the readout device (i.e. a meter) is placed across R2, than the voltage read by the meter is
Or in other words, the divider output equals the instrument output multiplied by R2 over the total resistance
$\mathrm{V_{out} = V_{in} \left(\dfrac{R_2}{R_1 + R_2}\right)}$
In this case, the divider output is:
$\mathrm{V_{out} = 1.0\: V \left(\dfrac{10\, Ω}{10\, Ω + 10\, Ω}\right) = 0.5\; V}$
RC Voltage Dividers (Analog Filters)
Although voltage dividers are extremely useful, they are unable to selectively filter signal voltages from noise voltages. That is:
Voltage dividers are frequency independent.
However, the impedance of a capacitor is frequency dependent, as shown by the following equation:
$\mathrm{X_C = \dfrac{1}{2πfC}}$
• XC is the impedance of the capacitor (impedance is the generalized form of resistance that applies to AC signals)
• f is the frequency of the voltage source in Hertz
• C is the capacitance in Farads
As the frequency increases, the impedance of a capacitor decreases!
When a capacitor is charging or discharging, the voltage across the capacitor lags behind (i.e. is out-of-phase) by 90°. If this phase difference did not exist, one could simply insert the value of the impedance into the voltage divider equation and use the same mathematics for the DC voltage divider circuit to calculate the output of the RC voltage divider circuit at a given frequency.
However, if each impedance is treated as a phasor, the total impedance of a RC series (ZRC) is calculated using the following relationship
$\mathrm{Z_{RC} = \sqrt{R^2 + \left(\dfrac{1}{2πfC}\right)^2}}$
An excellent overview of the mathematics of AC circuits can be accessed by clicking on the following URL:
http://www.animations.physics.unsw.edu.au/jw/AC.html
(Accessed June 6, 2014).
Low-Pass Filters
• Used when the signal frequency < noise frequency
• The relationship between Vin and Vout is analogous to a frequency independent voltage divider
• The desired filter output is obtained across the frequency dependent component (capacitor)
$\mathrm{V_{out} = V_{in}\left(\dfrac{X_C}{Z_{RC}}\right) = V_{in} \left(\dfrac{\dfrac{1}{2πfC}}{\sqrt{R^2 + \left(\dfrac{1}{2πfC}\right)^2}}\right) = V_{in} \left(\dfrac{1}{\sqrt{1+(2πfC)^2}}\right)}$
High-Pass Filters
• Used when the signal frequency > noise frequency
• The relationship between Vin and Vout is analogous to a frequency independent voltage divider
• The desired filter output is obtained across the frequency independent component (resistor)
$\mathrm{V_{out} = V_{in}\left(\dfrac{X_C}{Z_{RC}}\right) = V_{in} \left(\dfrac{R}{\sqrt{R^2 + \left(\dfrac{1}{2πfC}\right)^2}}\right) = V_{in} \left(\dfrac{2πfRC}{\sqrt{1 + (2πfRC)^2}}\right)}$
04 Decibel Scale
Expressing Signal Attenuation of RC filters
• Because an ideal analog filter would not attenuate the signal but only the noise, the decibel scale is used to express the degree of electrical attenuation (or gain) attributable to an electronic device, such as a RC filter.
• A decibel is defined as: $\mathrm{dB = 20 \log_{10} \left(\dfrac{V_{out}}{V_{in}}\right)}$
• So 0 dB represents no signal attenuation, and -20 dB represents an order of magnitude decrease in the RC filter output compared with the input.
• Remember that S/N enhancement is possible if the frequency of the signal and the noise are different. We can express the attenuation of the RC filter response as a function of frequency using a Bode plot.
• Bode plots are log-log plots: decibels are a logarithmic quantity and frequency is plotted on a logarithmic scale. They are quite frequently used to illustrate the frequency response of electronic circuits.
Bode Plots
• Below is a Bode plot of the low-pass RC filter frequency response shown a few sections back. Notice that low frequencies are unattenuated, but attenuation increases with higher frequencies.
• Every Bode plot has two straight lines: the relatively flat response where little attenuation occurs and a linear response of -20 dB/decade at higher frequencies. The intersection point of these two lines coincides with the rounded section of the plot. This is the cutoff frequency, f0, of the RC filter, which is expressed by the following relationship: f0 = 1/(2πRC)
The cutoff frequency, which is 1592 Hz for this particular circuit, corresponds to a 3 dB attenuation, and can be used as a figure-of-merit for the response of the filter.
• Below is a Bode plot of the high-pass RC filter frequency response a few sections back. Note that because the same resistor and capacitor were used, the cutoff frequency has not changed. The filter output is simply accessed across the resistor instead of the capacitor.
05 Analog Filter Exercise
Analog Filter Demo
A lecture demonstration of how an RC filter isolates noise from signal can be obtained as a MS Word document by clicking here or as a web page by clicking here.
Bode Plot Exercise
An exercise on interpreting the frequency response of RC filters using a Bode plot can be accessed by clicking here.
Analog Filter Exercises
A couple of exercises have been included to reinforce your understanding about the design and application of analog filters. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Introduction_to_Signals_and_Noise/04_Signal-to-Noise_Enhancement/02_Analog_Filtering/01_Ohms_Law.txt |
A digital filter is a noise reduction technique that is software-based. It is an approach that was popularized once personal computers became widely available.
Digitally-based signal-to-noise enhancement techniques described in this e-module include:
• Ensemble Averaging
• Boxcar Averaging
• Moving Average (Weighted & Unweighted)
03 Digital Filtering
Ensemble averaging is a data acquisition method that enhances the signal-to-noise of an analytical signal through repetitive scanning. Ensemble averaging can be done in real time, which is extremely useful for analytical methods such as:
• Nuclear Magnetic Resonance Spectroscopy (NMR)
• Fourier Transform Infrared Spectroscopy (FTIR)
• Near-Infrared (NIR) Spectrophotometry
• UV-Visible Spectrophotometry
Ensemble averaging also works well with multiple datasets once data acquisition is complete. In either case, this method of S/N enhancement requires that:
• The analyte signal must be stable
• The source of noise is random
How Ensemble Averaging Works
• Repeated experiments (scans) are performed on the chemical system in question. The scans are averaged either in real-time or after the data acquisition is complete. A visualization of this process is shown below for five spectra of 8.8 μg/mL 1,1'-ferrocenedimethanol in water.
Pros of Ensemble Averaging
• Ensemble averaging filters out random noise, regardless of the noise frequency
• Ensemble averaging is effective, even if the original signal has a S/N<1
• Ensemble averaging is straightforward to implement
• Improvement in S/N is proportional to:
$\sqrt{\#\textrm{ of datasets averaged together}}$
Cons of Ensemble Averaging
• Requirement of a stable signal
• Ensemble averaging will not work if noise is not random (e.g. 60 Hz electrical noise)
Example of Ensemble Averaging
These simulated 5-μV gaussian signals illustrate S/N improvement of ensemble averaging. The bottom dataset represents a S/N of 2 (single dataset), the middle dataset represents a S/N of 8 (average of 16 datasets), and the top dataset represents a S/N of 20 (average of 100 datasets).
Click here to work on an ensemble averaging exercise.
02 Ensemble Averaging Ex
An ensemble averaging spreadsheet similar to the Signal Noise Exercise spreadsheet can be accessed by clicking here.
1. Adjust the following parameters in the ensemble averaging spreadsheet:
• Peak #1 Intensity = 2 μV (Peaks 2 & 3 should have zero intensity)
• Peak #1 Mean = 1 minute
• Peak #1 Standard Deviation = 0.01 minute
• Noise = 6 μV
2. Adjust the number of datasets averaged in the spreadsheet
• Top = 1 dataset
• Middle = 16 datasets
• Bottom = 100 datasets
3. How does the S/N improve with the number of datasets averaged?
4. Compare the appearance of the original signal with the ensemble averaged signal.
03 Boxcar Averaging
Boxcar averaging is a data treatment method that enhances the signal-to-noise of an analytical signal by replacing a group of consecutive data points with its average. This treatment, which is called smoothing, filters out rapidly changing signals by averaging over a relatively long time but has a negligible effect on slowly changing signals. Therefore, boxcar averaging mimics a software- based low-pass filter. Boxcar averaging can be done both in real time and after data acquisition is complete.
How Boxcar Averaging Works
During Data Acquisition:
• The signal is sampled several times. Theoretically, any number of points may be used.
• The samples are summed together and an average is calculated.
• The average signal (dependent variable) is stored in the smoothed data set as the y-coordinate, and the average value of the independent variable (e.g. time, wavelength) is used as the x-coordinate.
After Data Acquisition (see figure below):
• Sum the data points within the boxcar
• Divide by the number of points in the boxcar
• Plot the average y-value at the central x-value of the boxcar
• Repeat with Boxcar 2, etc until the last full boxcar is smoothed
Main Points about Boxcar Averaging
• Boxcar averaging is equivalent to software-based low-pass filtering.
• Boxcar averaging is straightforward to implement.
• Improvement in S/N is proportional to:
$\sqrt{\textrm{# of data points in boxcar}}$
• (N-1) points are lost from each boxcar in the smoothed data set, where N is the boxcar length. The data density of the smoothed data set will be reduced by (N-1)/N
• Significant loss of information can occur if the length of the boxcar is comparable to the data acquisition rate. It is best to implement boxcar averaging with a sufficient data acquisition rate.
Example of Boxcar Averaging
• There are two 5 μV signals below
• Peak at 1.00 minutes with a width of 0.04 minutes
• Peak at 2.00 minutes with a width of 0.40 minutes
• Levels of boxcar averaging are as follows
• Bottom dataset: Theoretical S/N of 13 (no smoothing)
• Middle dataset: Theoretical S/N of 29 (Five-point boxcar, 0.05 min long)
• Top dataset: Theoretical S/N of 39 (Nine-point boxcar, 0.09 min long)
• Notice that little distortion occurs if the peak width is much larger than the boxcar and significant S/N enhancement is possible.
• Signals with frequencies similar to the rate of data acquisition are quickly attenuated, analogous to a low-pass RC filter.
• Click here to work on a boxcar averaging exercise.
04 Boxcar Averaging Exer
Boxcar Averaging Exercise #1
A boxcar averaging spreadsheet similar to the Ensemble Averaging spreadsheet can be accessed by clicking here.
1. Adjust the following parameters in the boxcar averaging spread sheet:
• Peak Intensity = 5 μV (All 3 peaks)
• Peak #1 Mean & Standard Deviation (1.000 ± 0.005) min
• Peak #2 Mean & Standard Deviation = (2.000 ± 0.025) min
• Peak #3 Mean & Standard Deviation = (3.000 ± 0.250) min
• Noise = 2 μV
2. Adjust the number of boxcar elements for each dataset
• Top = 9 data points
• Middle = 3 data points
• Bottom = 1 data point (raw data)
3. Which of the peak parameters is the most significant when using boxcar averaging to increase S/N? Support your choice based on what you observe in the boxcar averaging spreadsheet.
Boxcar Averaging Exercise #2
1. Adjust the following parameters in the boxcar averaging spreadsheet:
• Peak Intensity = 4 μV (Peak #1), 2.5 μV (Peak #2), 1 μV (Peak #3)
• Peak #1 Mean & Standard Deviation = (1.00 ± 0.15) min
• Peak #2 Mean & Standard Deviation = (2.00 ± 0.15) min
• Peak #3 Mean & Standard Deviation = (4.00 ± 0.15) min
• Noise = 2.0 μV
2. Adjust the number of boxcar elements for each dataset
• Top = 9 data points
• Middle = 5 data points
• Bottom = 1 data point (raw data)
3. Discuss the ability of boxcar filters to clearly extract signals from noise at or below the detection limit based on S/N enhancement. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Introduction_to_Signals_and_Noise/04_Signal-to-Noise_Enhancement/03_Digital_Filtering/01_Ensemble_Averaging.txt |
Overview
Digital filtering is a data treatment method that enhances the signal-to-noise ratio of an analytical signal through the convolution of a data set with an appropriate filter. This treatment method is another smoothing technique. If the filter is unweighted, it will perform in a similar manner to the boxcar filter. That is, it filters out rapidly changing signals by averaging over a relatively long time but has a negligible effect on slowly changing signals, and it too behaves as a software-based low-pass filter. However, a weighted filter may be constructed to mimic a low-pass, high-pass or even a bandpass filter. This module will focus on a weighted filter application based on least-squares quadratic smoothing that was popularized by Savitzky and Golay in the 1960’s.
Convolution
Before we explore the differences in the meaning and construction of unweighted versus weighted filters, the concept of convolution needs to be addressed. Let’s start with an analytical signal sampled every second for ten seconds. The raw data in this ideal case, which is represented in the figure below, consists of a slowly changing peak-shaped function.
For the moment, let’s ignore the independent variable (i.e. x-axis) and treat this instrumental response as a vector. We can represent the data above by the following matrix:
x = [ 0 0 1 3 6 7 6 3 1 0 0 ]
and a three-point unweighted filter to convolve the raw data
f = [ 1 1 1 ]
The result will be a smoothed data matrix, x’
The convolution process involves the following steps:
1. Matrix multiplication of the first raw data segment with the same number of array elements as the appropriate filter function, f. The filter function has the same sampling rate as the raw data.
1. This operation is called the dot product.
$\mathrm{f\cdot x = \sum\limits_{i = 1}^{n}f_ix_i \:\:\:(n = length\: of\: filter)}$
2. So for the first set of three raw data points:
$\mathrm{f\cdot x=[f_1\:\:\: f_2\:\:\: f_3] \cdot \begin{bmatrix}\ce x_1 \ \ce x_2 \ \ce x_3 \end{bmatrix} =(f_1x_1 + f_2x_2 + f_3x_3)}$
2. Normalizing the dot product with the sum of the filter elements and placing the result in the smooth data matrix with an x-value equivalent to the x-value of the center of the filter function.
$\mathrm{x'_2 = \dfrac{f \cdot x}{\sum\limits_{i = 1}^{n}f_i}= \dfrac{(f_1x_1 + f_2x_2 + f_3x_3)}{(f_1 + f_2 + f_3)}}$
1. So in this case, x’2 has the same time as x2 (i.e. time = 2 s).
3. Slide the filter function over one data point and repeat the matrix multiplication process, placing the next normalized dot product as the next array element in the smoothed data matrix. Therefore,
$\mathrm{x'_3 = \dfrac{f \cdot x}{\sum\limits_{i = 1}^{n}f_i} = \dfrac{\sum\limits_{i = 1}^{n}f_ix_{(i+1)}}{\sum\limits_{i = 1}^{n}f_i} = \dfrac{(f_1x_2 + f_2x_3 + f_3x_4)}{(f_1 + f_2 + f_3)}}$
1. x’3 has the same time as x3 (i.e. time = 3 s).
4. Repeat step 3 until the leading edge of the filter has the same x-value as the last point in the raw data matrix. This means that (n-1)/2 data points will be lost from each side of x’
Because the filter function is unweighted, we call this convolution process the moving window averaging technique, as shown in the figure below.
Convolving the filter function with the original response in the previous figure results in the smoothed response below.
Effect of Unweighted Filter Width
In the unweighted moving window averaging approach, we assume that each data point is equally important in the instrumental response above. This works well if the peak width is much larger than the filter width. However, if the width of the filter is comparable to the peak width of the signal, applying an unweighted filter distorts the signal, decreasing the signal intensity and increasing its width. In the figure below, the raw data is smoothed by a 3-point, 5-point, and 7-point unweighted filter.
Weighted (Savitzky-Golay) Filters
In order to avoid distorting the signal significantly, one convolves the raw data with a filter that looks more like the signal itself. A weighted filter that emphasizes the response at the central filter element and de-emphasizes the response at the outer filter elements is used. This approach, which is called least-squares polynomial smoothing, was popularized in analytical chemistry by Savitzky and Golay. Savitzky and Golay used the least-squares approach to derive a set of convolution integers for a given filter width. Below is a list of Savitzky-Golay coefficients for 5, 9, and 13-point quadratic smoothing of instrumental responses.
Filter Points 13 9 5
-6 -11
-5 0
-4 9 -21
-3 16 14
-2 21 39 -3
-1 24 54 12
0 25 59 17
1 24 54 12
2 21 39 -3
3 16 14
4 9 -21
5 0
6 -11
Normalizing Factor 143 231 35
If we use a five-point filter function, instead of the unweighted function below
f = [ 1 1 1 1 1 ]
we would use the Savitzky-Golay coefficients
f = [ -3 12 17 12 -3 ]
using the original raw data, the normalized dot product for the first smoothed data point would be
$\mathrm{x'_3 = \dfrac{f \cdot x}{\sum\limits_{i = 1}^{n}f_i} = \dfrac{(f_1x_1 + f_2x_2 + f_3x_3 + f_4x_4 + f_5x_5)}{(f_1 + f_2 + f_3 + f_4 + f_5)}\ = \dfrac{((-3*0)+(12*0)+(17*1)+(12*3)+(-3*6))}{((-3)+12+17+12+(-3))}}$
Just like the unweighted moving average smooth, the raw data would be convolved with the weighted moving average smooth using the appropriate Savitzky-Golay coefficients. A comparison of the 5-point unweighted and weighted moving average smoothing functions on a "noisy" version of the raw data set is shown below. Notice that the polynomial filter (smoothed response in black) distorts the signal to a lesser extent than the unweighted filter (smoothed response in red).
Points to Consider Using Moving Average Filtering
• The moving average technique retains greater data density than boxcar averaging.
• The moving average technique is straightforward to implement.
• Improvement in S/N is proportional to (# filter elements)1/2 if the noise is normally distributed.
• (N-1)/2 points are lost on either end of the smoothed data set, where N is the filter length.
• Significant distortion and loss of resolution may occur if the length of the filter is comparable to the peak width. It is best to implement a moving average with a filter width much smaller than the narrowest peak to be smoothed.
• Optimal filter choices are typically chosen in an empirical fashion.
Click here to work on a moving average exercise
06 Moving Average Exerci
A moving average spreadsheet similar to the Ensemble Averaging spreadsheet can be accessed by clicking here.
1. Remove all noise from the data, and adjust the following parameters in the moving average spreadsheet:
• Peak Intensity = 5 μV (All 3 peaks)
• Peak #1 Mean & Standard Deviation = (1.00 ± 0.02) min
• Peak #2 Mean & Standard Deviation = (2.00 ± 0.05) min
• Peak #3 Mean & Standard Deviation = (3.00 ± 0.10) min
• Offset= 1 μv
2. Select unweighted (MA) as the filter type for Dataset 1 and set the length of the smoothing window to 1. Select MA as the filter type for Dataset 2 and observe what happens to the smoothed data as you increase the size of the filter window.
3. Select weighted (SG) as the filter type for Dataset 2 and observe what happens to the smoothed data as the size of the filter window increases.
4. Compare and contrast unweighted and weighted filters and their effect on the distortion of the peak width and height of the original signal.
5. Adjust the number of elements in both moving average filters to 5 points. Make one filter unweighted (MA) and one weighted (SG) Compare and contrast the effect of unweighted and Savitzky-Golay smoothing filters with equal lengths on peak distortion.
6. Repeat this exercise with filter lengths of 9, 13, 17 and 21 elements. Under what conditions does minimal peak distortion occur? Will an unweighted filter always distort the original signal? Will a weighted signal always provide an undistorted signal?
7. Remove all noise from the data, and adjust the following parameters in the moving average spreadsheet:
• Peak Intensity 1 μV (Peak #1), 1 μV (Peak #2), 1 μV (Peak #3).
• Peak #1 Mean & Standard Deviation = (1.00 ± 0.02) min
• Peak #2 Mean & Standard Deviation = (2.00 ± 0.04) min
• Peak #3 Mean & Standard Deviation = (4.00 ± 0.10) min
• Offset= 1 μV
8. Select unweighted (MA) as the filter type for Dataset 1 and set the length of the smoothing window to 1. Select MA as the filter type for Dataset 2 and set the length of the smoothing window to 5. The peak distortion for the widest peak will be ~1%. Gradually increase the noise until Peak #3 in Dataset #1 has a S/N of ~3. Then select weighted (SG) as the filter type for Dataset #2 and repeat the gradual increase in the noise level. Note the appearance of the raw and smoothed signals in both cases as the noise increases.
9. Compare and contrast the ability of properly sized unweighted and weighted filters to clearly extract signals from noise based on S/N enhancement. Which of the S/N enhancement approaches best enhances signals at or below the detection limit?
05 References
1. Adams, M. J. Acquisition and Enhancement of Data. Chemometrics in Analytical Spectroscopy; The Royal Society of Chemistry: Cambridge, 1995; pp 27 - 53.
2. Binkley, D.; Dessy, R. J. Chem. Educ. 1979, 56, 148.
3. Savitzky, A.; Golay, M. J. E. Anal. Chem. 1964, 36, 1627. (Errors in reported equations corrected in Steinier, J.; Termonia, Y.; Deltour, J. Anal. Chem. 1972, 44, 1906.)
4. Sharaf, M. A.; lllman, D. L.; Kowalski, B. R. Signal Detection and Manipulation. Chemometrics; John Wiley and Sons: New York, 1986; pp 65 - 117.
5. Skoog, D. A.; Holler, F. J.; Nieman, T. A. Principles of Instrumental Analysis; Harcourt Brace: Philadelphia, 1998.
Instructor’s Resources
Click here for suggested approaches to solving the exercises.
06 Acknowledgements
Author
Steven C. Petrovic
Department of Chemistry
Southern Oregon University
1250 Siskiyou BIvd., Ashland, OR 97520
Email: [email protected]
Acknowledgements
This module was first published in 2008 and revised in 2014. The citation for the original module is:
Petrovic, S. C. Introduction to Signals and Noise: eLearning Module. J. Anal. Sci. Digital Lib. [Online] 2008, Article 10055. http://jasdl.asdlib.org/2008/10/introduction-to-signals-and-noise-elearning-module/ (accessed Dec 3, 2014)
• Updated content in the revised (2014) module:
• Clarification of the voltage divider equations for low-pass filters and high-pass filters
• The author would like to thank participants from the following ASDL Curriculum Development Workshops for their contributions to this module
• University of Kansas, Lawrence, KS, June 18-22, 2007
• University of California at Riverside, Riverside, CA, June 9-13, 2008
This work is licensed under a Creative Commons Attribution Noncommercial-Share Alike 2.5 License | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Introduction_to_Signals_and_Noise/04_Signal-to-Noise_Enhancement/03_Digital_Filtering/05_Convolution-Based_Smo.txt |
Learning Objectives
After Completing the Module the Student will be able to:
• Describe Diffraction and Scattering
• Explain how variables in Bragg's Law relate to generic lattice of points
• Predict relationships between cell and powder diffraction pattern
• Draw a diagram of a powder XRD
• Interpret XRD diffraction patterns
This module provides an introduction to X-ray Diffraction (XRD), which is a versatile, non-destructive technique that reveals detailed information about the chemical composition and crystallographic structure of materials. It is utilized in a variety of settings ranging from chemistry and materials to geology and biological sciences. This module is aimed at presenting the basic theory and applications of X-ray Diffraction to a novice user. An interactive tutorial is also provided to help in obtaining an understanding of diffraction.
Introduction to X-ray Diffraction (XRD)
1912: Maxwell von Laue first discovered X-ray diffraction
• determined that X-rays would be scattered by atoms in a crystalline solid if there was similarity in the wavelength of X-rays and the interatomic distances of the crystalline material.
1913: Sir William H. Bragg and and his son Sir W. Lawrence Bragg derived the equation known as Bragg’s Law to define diffraction as a function of the angle of incidence
$\mathrm{2d \sinθ = nλ}$
λ = wavelength of the x-ray
θ = scattering angle
n = integer representing the order of the diffraction peak
d = inter-plane distance of (i.e atoms, ions, molecules)
• determined why the cleavage faces of crystals appeared to reflect X-ray beams at certain angles of incidence (θ). This is due to constructive interference.
• simulated the experiment, using visible light and tiny arrays of dots and pinholes to mimic atomic arrangements on a much larger scale. These experiments provided similar patterns to X-rays but were safer to work with than X-rays.
1914: von Laue awarded Nobel Prize for discovery of X-ray diffraction by crystals and showing X-rays are electromagnetic waves 1915: Braggs awarded Nobel Prize for their work determining the crystal structure of diamond, NaCl and ZnS.
01 X-rays
X-rays are electromagnetic radiation with very short wavelengths and high energy.
X-rays have high energy and they penetrate opaque material, but are absorbed by materials containing heavy elements. As an x-ray beam travels through a substance its intensity decreases with distance traveled through the matter.
Exercise \(1\)
1. Is water or Cr a better absorber of X-rays?
2. What effect does the wavelength of the X-ray sources have on penetration of X-rays?
Try the following demonstration below to help in answering these questions:
Click on Image (Note this will open in a new window)
Applet from: matter.org.uk (www.matter.org.uk/diffraction...iffraction.htm)
In addition to adsorption, X-rays can also be scattered and diffracted by a material.
Exercise \(2\)
1. What do you think makes something scatter vs. diffract?
Let’s take a look at the next section on Diffraction to learn more.
02 Diffraction and Braggs Law
Take a look at the diagram below:
1. When X-rays interact with a single particle, it scatters the incident beam uniformly in all directions.
2. When X-rays interact with a solid material the scattered beams can add together in a few directions and reinforce each other to yield diffraction. The regularity of the material is responsible for the diffraction of the beams.
You have likely seen diffraction before. For example if you look at a CD when exposed to white light you can see it diffracted into varies wavelengths of color. The pits (or grooves) in the CD are the regularity of the material that causes the diffraction.
Google images: outreach.atnf.csiro.au
Diffraction can occur when any electromagnetic radiation interacts with a periodic structure. The repeat distance of the periodic structure must be about the same wavelength of the radiation. For example, light can be diffracted by a grating having scribed lines arranged on the order of the wavelength of light.
Exercise $1$
So if all electromagnetic radiation can diffract, why are X-rays used in crystallography?
X-ray Diffraction and Bragg’s Law
X-rays have wavelengths on the order of a few angstroms (1 Angstrom = 0.1 nm). This is the typical inter-atomic distance in crystalline solids, making X-rays the correct order of magnitude for diffraction of atoms of crystalline materials.
How are Diffractions Patterns Made?
When X-rays are scattered from a crystalline solid they can constructively interfere, producing a diffracted beam. What does this mean?
Constructive vs. Destructive Interference
Interference occurs among the waves scattered by the atoms when crystalline solids are exposed to X-rays. There are two types of interference depending on how the waves overlap one another.
Constructive interference occurs when the waves are moving in phase with each other. Destructive interference occurs when the waves are out of phase.
This constructive interference results in diffraction patterns.
Bragg's Law and Diffraction
The relationship describing the angle at which a beam of X-rays of a particular wavelength diffracts from a crystalline surface was discovered by Sir William H. Bragg and Sir W. Lawrence Bragg and is known as Bragg’s Law
$\mathrm{2d\sin θ= nλ}$
λ = wavelength of the x-ray
θ = scattering angle
n = integer representing the order of the diffraction peak.
d = inter-plane distance of (i.e atoms, ions, molecules)
Click on the following image below to get to an Applet where you can explore this relationship of Bragg’s Law
www.eserc.stonybrook.edu/ProjectJava/Bragg/
Guide to how to use Applet: There are 2 rays incident on two atomic layers of a crystal (d). At the beginning the scattered rays are in phase and interfering constructively. Bragg’s Law is satisfied and diffraction is occurring. If you click on the details button you can see the detector, which measures how well the phases of the two rays match. When the meter is green it indicates that Bragg’s law is satisfied. You can change three variables (d, λ, and θ) to see how they effect the diffraction.
Exercise $2$: Unit Cell Size from Diffraction Data
The diffraction pattern of copper metal was measured with X-ray radiation of wavelength of 1.315Å. The first order Bragg diffraction peak was found at an angle 2theta of 50.5 degrees. Calculate the spacing between the diffracting planes in the copper metal. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Introduction_to_X-ray_Diffraction_(XRD)/02_Historical_Perspective.txt |
Utilization of X-Ray Diffraction
Crystallography: Diffraction data provides information on the structures of crystalline solids
• The symmetry in structures shows itself in the diffraction by the appearance and disappearance of characteristic reflections.
• The intensity of the diffracted beams depends on the arrangement and atomic number of the atoms in the repeating motif, called the unit cell. Unit cells describe the symmetry of all structures.
What are some of the key crystallography concepts?
The emphasis of this learning module is not on obtaining a detailed understanding of crystal structures. However a few key concepts are useful in understanding crystallography and results obtained using XRD.
Key Concepts
• Crystal structures are the periodic arrangement of atoms in a crystal
• Structures are defined using lattices (infinite array of points in space)
• Common lattices are shown below (note the spheres represent lattice points, NOT atoms)
www.chem.ox.ac.uk/icl/heyes/s...1.html#anchor2
• Unit cells are used to describe the smallest repeating structure in a lattice of a crystalline material.
• These repetitions are what give rise to the constructive interference defined by Bragg’s Law
Note: It is important to recognize that simple examples shown in this module have atoms existing within lattice sites. This is rarely the case beyond metals and simple salts.
Let’s examine in the next section how different unit cells and atoms influence diffraction using a program called Powdercell.
Powdercell is a freeware program that can be downloaded.
Werner, K. Nolze, G. Powder Cell 2.3
http://www.ccp14.ac.uk/ccp/web-mirro...er/e_cell.html
POWDER CELL Analysis
The following activity allows you to evaluate factors that influence diffraction, creating unique diffraction patterns for a material.
It requires a free ware program (Powdercell) that can be downloaded, but does not have to be installed on a computer to be operational.
Files for Powdercell Activity
Tungsten
Potassium 1
Potassium 2
Tungsten heated
Fe BCC
Fe FCC
Initial Analysis
1. Open Powdercell 2.3
2. Under File select Tungsten
Notes: If you do not immediately see a diffraction pattern, go under diffraction (top menu) and select Diffraction On.
Hints:
• The diffraction from the crystal structure is denoted at different scattering angles (2θ). The y-axis is the intensity of the diffraction at a given angle.
• Above the diffraction peaks are miller indices (h,k,l). These are a shorthand notation used to describe the crystallographic planes and directions in a material. Depending on the crystal structure different planes will constructively diffract.
Questions:
1. What do you notice in the diffraction pattern for tungsten? (i.e. what diffraction peaks are present and at what angle of diffraction)
2. Tungsten exhibits a body centered cubic (BCC) structure. What would the diffraction pattern look like for Potassium, which also exhibits a BCC structure?
3. In actuality the unit cell for Potassium is much larger a=b=c = 5.247Å vs. the 3.180 Å for Tungsten. Open up the file Potassium 2. Compare this to Potassium 1. What do you notice?
4. Temperature will also change a unit cell dimension for a compound. Let’s look at the Tungsten structure again. Under elevated temperatures the dimensions of the unit cell change from a=b=c = 3.180 Å to a=b=c= 3.200 Å. Do you think this will change the diffraction pattern?
Examine the diffractions of Tungsten room temperature and tungsten heated using Powder cell. (Compare the files Tungsten and Tungsten heated). What do you observe?
5. Sometimes temperature can also cause phase transitions in a material. For example α-Fe exhibits a BCC structure at room temperatures. However, above 912°C and up to 1401°C alpha iron undergoes a phase transition from BCC to the FCC configuration of γ-Fe, called austenite. Compare the differences between the patterns for BCC (alpha) and FCC (gamma) iron. (Use files Fe BCC and Fe FCC.)
What is different in the crystal structures and diffraction patterns between FCC and BCC?
XRD Analysis-Summary
From diffraction patterns one can find the crystal structure of an unknown material. In addition one can also determine factors such as the orientation of single crystals, or measure the size and shape of crystalline regions. There are several X-ray diffraction techniques. Two of the most common are:
• Single crystal X-ray diffraction: used to solve structure of crystalline materials ranging from inorganic compounds to complex macromolecules such as proteins or polymers. You can learn everything about a crystal structure, but requires a single crystal. Although obtaining single crystals is difficult, single crystal X-ray crystallography is a primary method for determining the molecular conformations of biological interest such as DNA, RNA and proteins.
• Power X-ray diffraction: used to characterize crystallographic structure, grain size, and preferred orientation in polycrystalline or powder solid samples. This is a preferred method of analysis for characterization of unknown crystalline materials. Compounds are identified by comparing diffraction data against a database of known materials. It can be used to follow phase changes as a function of variable such as temperature, pressure.
This module focuses on providing an introduction to X-ray diffraction Powder X-ray analysis to identify crystalline materials.
04 Instrumental Design
Let’s reexamine the simple description of Bragg’s law. In the example shown below, diffraction is only being measured from the horizontal planes of a crystal as function of 2θ. For a complete analysis of a material, the diffraction from all the possible lattice planes should be examined.
1. Based on this above statement what do you think are important considerations in the instrumental design for XRD analysis?
2. In Powder XRD analysis, the assumption is that all orientations are present in the sample and interact with the X-ray source simultaneously. What effect does this have on the collection of the diffraction pattern?
Answer: Diffraction occurs at all the angles of 2θ simultaneously in powder samples. In order to obtain a diffraction pattern, the detector (in most designs) rotates to various 2θ angles to measure diffraction from the sample.
Below is a schematic diagram for a powder X-ray diffractometer, showing the rotating detector.
The source shown is an X-ray tube, which is the most common source of X-rays. Filters are used to provide a narrow wavelength range for analysis.
3. Why is a monochromatic source desirable for XRD analysis?
Typical detectors are scintillation counters. These are materials that release photons of energy when X-ray radiation passes through the scintillation counter. The photons of energy are measured using a photomultiplier tube.
1. What effect does detector collection time have on the analysis?
Hint: Think about S/N
04 Instructor Notes
This module provides an introduction to X-ray Diffraction (XRD).
XRD is a versatile, non-destructive technique that reveals detailed information about the chemical composition and crystallographic structure of materials. It is utilized in a variety of settings ranging from chemistry and materials to geology and biological sciences.
Objective
This module is aimed at presenting the basic theory and applications of X-ray Diffraction to a novice user. It focuses on the application of Powder XRD to identify unknown crystalline samples.
The module is designed with learning activities for the students. The activities are interactive and inquiry based. Activities also include web-links to interactive applets to aid in student understanding.
The activities can be completed independently or in a classroom/laboratory setting. The Learning activity on Crystallography utilizes a freeware program called Powdercell. Crystal structure files have been created for this specific module and are available to open in this program for student analysis.
Note: In addition to the learning activities there is a section on the module that gives a historical perspective on X-ray diffraction.
Learning Activities
The learning activities are web-based but have also been broken up into separate .pdf files that could be printed or utilized online. They are designed to build off of each other but could be used independently depending on student background.
1. X-rays
2. Diffraction and Bragg’s Law
3. Crystallography
4. Instrument Design
X-ray Learning Activity: The activity is designed to get students to think about why one would use X-rays to probe a material. It leads into the second activity about diffraction.
Diffraction and Bragg’s Law Learning Activity: This is an interactive tutorial that walks through diffraction, constructive/deconstructive interference and Bragg’s Law. The activity includes a useful applet on Bragg’s law that allows students to modify wavelength, theta, and inter-atomic distances of a lattice to see how that effect diffraction.
Crystallography Learning Activity: This is an interactive tutorial that provides an overview of how X-rays can provide information on the structure of material. It does not go into a detailed understanding of crystal structures, but does provide an introduction into some common lattices and the concept of the unit cell. The activity uses a freeware program Powdercell to examine powder patterns of materials. The activity is designed so crystal structures are already created to be opened up in the program (concepts of unit cells, and point groups not needed). By working through the module students are able to understand how symmetry and intensity effect diffraction and can predict relationships between cell and powder diffraction pattern.
The Powdercell files needed for this activity can be accessed on the homepage for Crystallography.
Instrument Design Learning Activity: This activity is designed to get students to think about important factors in instrument design for XRD analysis rather than providing a list of detectors and sources. It is focused on the design commonly seen for powder XRDs. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Introduction_to_X-ray_Diffraction_(XRD)/03_Basic_Theory/03_Crystallography.txt |
Introduction to Energy-Dispersive X-Ray Fluorescence (XRF) – An Analytical Chemistry Perspective
Dr. Pete Palmer
Professor
Department of Chemistry & Biochemistry
San Francisco State University
Science Advisor
San Francisco District Laboratory
U.S. Food and Drug Administration
What is XRF?
1. X-ray Fluorescence Spectrometry
2. An elemental analysis technique
3. Another acronym to remember
4. A new scientific gadget to play with
5. The closest thing we have to a tricorder
6. An advanced, highly automated, portable analytical tool that can be used by scientists, lab staff, field investigators, and even non-experts to support their job functions
7. All of the above
Typical Applications of XRF
XRF is currently used in many different disciplines:
Geology
• Major, precious, trace element analysis
• Characterization of rocks, ores, and soils
Environmental Remediation
• Pb in paint
• Heavy metals in soil (EPA method 6200)
Recycling
• Alloy identification
• Waste processing
Miscellaneous
• Art and archeology
• Industrial hygiene
• Forensics
“Ownership” of XRF Within Academia
• Although XRF is a physical phenomena involving the interaction of X-rays with matter, most of the applications of XRF are in areas outside of physics (chemistry, environmental sciences, food and product quality monitoring, etc.)
• Although XRF requires specialized knowledge in chemistry (spectral interpretation, calibration, sample prep, etc.), it is not even mentioned in 99% of undergraduate chemistry programs in the U.S.
• These materials will hopefully encourage wider dissemination and use of XRF in undergraduate chemistry and biochemistry programs and demonstrate its potential as a means for teaching concepts such as spectroscopy, sampling, qualitative and quantitative analysis, and elemental composition in
• Analytical Chemistry (Quantitative & Instrumental Analysis)
• Environmental Chemistry
• Independent student research projects
Intended Audience & Objectives
These materials were specifically designed for undergraduate chemistry and biochemistry majors
They are also appropriate for novices to the field of XRF and assume only a basic knowledge of chemistry (i.e., general chemistry)
By the end of this presentation, students should understand the following:
1. The basic theory of XRF
2. How to interpret XRF spectra
3. How to do quantitative analysis via XRF
4. Typical applications of XRF
The CSI Syndrome:
The growing popularity of forensic sciences as evidenced by TV series on this subject has attracted many young people to this discipline
Unfortunately, these shows often trivialize the science and rigor needed to derive reliable results on “real world” samples
Science does not always give yes/no answers (and real world problems are usually not solved in a 60-minute episode)
Forensic science requires careful work and is a lot harder than it looks on TV
Nothing is more useless than an powerful tool that is not used properly
https://xkcd.com/
Acknowledgements
FDA & its San Francisco District Laboratory for applications, collaborations, and funding for and access to a variety of XRF instrumentation
• Dr. Richard Jacobs (Toxic Element Specialist)
• Sally Yee (Chemistry Supervisor)
• Tom Sidebottom and Dr. Rod Asmundson (Lab Directors)
• Dr. George Salem, Carl Sciaccitano, and Dr. Selen Stromgren (DFS)
XRF vendors for freely providing their knowledge and expertise as well as the loan of several XRF analyzers to San Francisco State University
• Bruker (Dr. Bruce Kaiser, Dr. Alexander Seyfarth)
• Innov-X (Jack Hanson, Kim Russell, Innov-X University Equipment grant)
• Thermo-Niton (Rich Phillips, Peter Greenland)
Dr. Palmer would like to extend a special thanks to the several generations of San Francisco State University students, who acquired XRF data, participated in the research and development of various XRF methods, and were intimately involved in numerous XRF case studies
Contributors and Attributions
• Dr. Pete Palmer (San Francisco State University). This work is licensed under the Creative Commons Attribution-ShareAlike 3.0 Unported License.
Introduction to XRF- An Analytical Perspective
The Electromagnetic Spectrum
How does light affect molecules and atoms?
D.C. Harris, Quantitative Chemical Analysis, 7th Ed., Freeman, NY, 2007.
X-Ray Interactions With Matter
When X-rays encounter matter, they can be:
• Absorbed or transmitted through the sample (Medical X-Rays – used to see inside materials)
www.seawayort.com/hand.htm
• Diffracted or scattered from an ordered crystal (X-Ray Diffraction – used to study crystal structure)
http://commons.wikimedia.org/wiki/Fi...ern_3clpro.jpg
• Cause the generation of X-rays of differentcolors” (X-Ray Fluorescence – used to determine elemental composition)
Atomic Structure
• An atom consists of a nucleus (protons and neutrons) and electrons
• Z is used to represent the atomic number of an element (the number of protons and electrons)
• Electrons spin in shells at specific distances from the nucleus
• Electrons take on discrete (quantized) energy levels (cannot occupy levels between shells
• Inner shell electrons are bound more tightly and are harder to remove from the atom
Adapted from Thermo Scientific Quant’X EDXRF training manual
Electron Shells
Shells have specific names (i.e., K, L, M) and only hold a certain number of electrons
X-rays typically affect only inner shell (K, L) electrons
Adapted from Thermo Scientific Quant’X EDXRF training manual
Moving Electrons to/from Shells
Binding Energy versus Potential Energy
• The K shell has the highest binding energy and hence it takes more energy to remove an electron from a K shell (i.e., high energy X-ray) compared to an L shell (i.e., lower energy X-ray)
• The N shell has the highest potential energy and hence an electron falling from the N shell to the K shell would release more energy (i.e., higher energy X-ray) compared to an L shell (i.e., lower energy X-ray)
Adapted from Thermo Scientific Quant’X EDXRF training manual
XRF – A Physical Description
Step 1: When an X-ray photon of sufficient energy strikes an atom, it dislodges an electron from one of its inner shells (K in this case)
Step 2a: The atom fills the vacant K shell with an electron from the L shell; as the electron drops to the lower energy state, excess energy is released as a Kα X-ray
Step 2b: The atom fills the vacant K shell with an electron from the M shell; as the electron drops to the lower energy state, excess energy is released as a Kβ X-ray
www.niton.com/images/XRF-Excitation-Model.gif
XRF – Sample Analysis
www.niton.com/images/fluoresc...tal-sample.gif
• Since the electronic energy levels for each element are different, the energy of X-ray fluorescence peak can be correlated to a specific element
Simple XRF Spectrum
~10% As in Chinese supplement
• The presence of As in this sample is confirmed through observation of two peaks centered at energies very close (within ±0.05 keV) to their tabulated (reference) line energies
• These same two peaks will appear in XRF spectra of different arsenic-based materials (i.e., arsenic trioxide, arsenobetaine, etc.)
~10% Pb in imported Mexican tableware
• The presence of Pb in this sample is confirmed through observation of two peaks centered at energies very close (within ±0.05 keV) to their tabulated (reference) line energies
• These same two peaks will appear in XRF spectra of different lead-based materials (i.e., lead arsenate, tetraethyl lead, etc.)
Box Diagram of XRF Instrument
X-ray Source Detector Sample Digital Pulse Processor XRF Spectrum (cps vs keV) Results (elements and conc’s) software
• High energy electrons fired at anode (usually made from Ag or Rh)
Can vary excitation energy from 15-50 kV and current from 10-200 µA
Can use filters to tailor source profile for lower detection limits
• Energy-dispersive multi-channel analyzer – no monochromator needed, Peltiercooled solid state detector monitors both the energy and number of photons over a preset measurement time
The energy of photon in keV is related to the type of element
The emission rate (cps) is related to the concentration of that element
• Concentration of an element determined from factory calibration data, sample thickness as estimated from source backscatter, and other parameters
Different Types of XRF Instruments
• EASY TO USE (“point and shoot”)
• Used for SCREENING
• Can give ACCURATE RESULTS when used by a knowledgeable operator
• Primary focus of these materials
• COMPLEX SOFTWARE
• Used in LAB ANALYSIS
• Designed to give ACCURATE RESULTS
(autosampler, optimized excitation, report generation) | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Introduction_to_XRF-_An_Analytical_Perspective/1._Introduction.txt |
XRF Spectra
Consecutive elements in periodic table
• Plotting only a portion of the XRF spectra of several different elements
• Note periodicity - energy is proportional to Z2 (Moseley’s law)
Periodic Table of XRF Fluorescence Data
Including K and L line energies & detection limits
XRF Energies for Various Elements
Generalizations based on use of field portable analyzers
• ORGANIC ELEMENTS (i.e., H, C, N, O) DO NOT GIVE XRF PEAKS
Fluorescence photons from these elements are too low in energy to be transmitted through air and are not efficiently detected using conventional Sibased detectors
• LOW Z ELEMENTS (i.e., Cl, Ar, K, Ca) GIVE ONLY K PEAKS
L peaks from these elements are too low in energy (these photons are not transmitted through air and not detected with conventional Si-based detectors)
• HIGH Z ELEMENTS (i.e., Ba, Hg, Pb, U) GIVE ONLY L LINES
K peaks from these elements are too high in energy (these electrons have high binding energies and cannot be removed with the limited voltage available in field portable analyzers)
• MIDDLE Z ELEMENTS (i.e., Rh through I) MAY GIVE BOTH K AND L LINES
XRF – More Detailed Description
Note energy level diagrams are not drawn to scale
www.niton.com/images/fluoresc...tal-sample.gif
• Since XRF affects inner shell and not bonding electrons, the XRF spectrum of an element is independent of its chemical form (i.e., spectra of lead, lead arsenate, and tetraethyl lead will ALL show peaks at 10.61 and 12.55 keV)
K Line Series
~10% As in Chinese supplement
• L lines not observed (1.28 and 1.32 keV - too low in energy to be excited)
• Kα and Kβ peak energies are often close together (1.2 keV apart for As)
• K lines observed for low to medium Z elements (i.e., Cl, Fe, As)
• Kα and Kβ peaks have typical ratio of ~ 5 to 1
L Line Series
~10% Pb in imported Mexican tableware
• K lines not observed (75.0 and 94.9 keV - too high in energy to be excited)
• Lα and Lβ peak energies are often further apart (2.1 keV apart for Pb)
• L lines observed for high Z elements (i.e., Hg, Pb, Th)
• Lα and Lβ peaks have typical ratio of ~ 1 to 1 Pb Lγ line
More Complex XRF Spectrum
Chinese supplement containing 4% As and 2% Hg
• Line overlaps are possible and users must evaluate spectrum to confirm the presence or absence of an element
Effect of Detector Resolution
Spectra of 900 ppm Pb added into Pepto-Bismol
Older Si(PIN) detector
• Resolution ~0.2 keV (FWHM)
• Cannot resolve Pb and Bi peaks
Newer SDD
• Resolution ~0.15 keV (FWHM)
• Can resolve Pb and Bi peaks
Adapted from Bruce Kaiser, Bruker AXS
Artifact Peaks
Arising from X-ray tube source
• Electrons with high kinetic energy (typically 10-50 kV) strike atoms in the X-ray tube source target (typically Rh or Ag) and transfer energy
• The interaction of X-ray source photons with the sample generates several characteristic features in an XRF spectrum which may include the following:
• Bremsstrahlung
• Rayleigh peaks
• Compton peaks
Bremsstrahlung
Continuum/backscatter from cellulose sample
E0 = initial energy of electron in X-ray tube source
E1, E2 = energy of X-ray
• Very broad peak due to backscattering of X-rays from sample to detector that may appear in all XRF spectra
• Maximum energy of this peak limited by kV applied to X-Ray tube, maximum intensity of this peak is ~ 2/3 of the applied keV
• More prominent in XRF spectra of less dense samples which scatter more of X-ray source photons back to the detector
Rayleigh Peaks
Elastic scattering from metal alloy sample
E0 = initial energy of X-ray from target element in x-ray tube source
E1 = energy of X-ray elastically scattered from (typically dense) sample
• Peaks arising from target anode in X-ray tube source (Rh in this case) that may appear in all XRF spectra acquired on that instrument
• No energy is lost in this process so peaks show up at characteristic X-ray energies (Rh Lα and Lβ at 20.22 and 22.72 keV in this case)
• Typically observed in spectra of dense samples as weak peaks (due to increased absorption of X-ray source photons by sample)
Compton Peaks
Inelastic scattering from cellulose sample
E0 = initial energy of X-ray from target element in x-ray tube source
E1 = energy of X-ray inelastically scattered from (typically non-dense) sample
• Peaks arising from target element in X ray tube (again, Rh in this case) that may appear in all XRF spectra acquired on that instrument
• Some energy is lost in this process so peaks show up at energies slightly less than characteristic X-ray tube target energies
• Typically observed in spectra of low density samples as fairly intense peaks (note these peaks are wider than Rayleigh peaks)
Artifact Peaks
Arising from detection process
• The interaction of X-ray fluorescence photons from the sample with the detector can generate several different types of artifact peaks in an XRF spectrum which may include the following:
• Sum peaks
• Escape peaks
Sum Peaks
Example from analysis of Fe sample
• Artifact peak due to the arrival of 2 photons at the detector at exactly the same time (i.e., Kα + Kα, Kα + Kβ)
• More prominent in XRF spectra that have high concentrations of an element
• Can be reduced by keeping count rates low
Escape Peaks
Example from analysis of Pb sample
• Artifact peak due to the absorption of some of the energy of a photon by Si atoms in the detector (Eobserved = Eincident – ESi where ESi = 1.74 keV)
• More prominent in XRF spectra that have high concentrations of an element and for lower Z elements
• Can be reduced by keeping count rates low
Artifact Peaks Due to Blank Media
• May observe peaks due to contaminants in XRF cups, Mylar film, and matrix
• In this case, the cellulose matrix is highly pure and the peaks are due to trace elements in the XRF analyzer window and detector materials
• This can complicate interpretation (false positives)
Summary of Factors That Complicate Interpretation of XRF Spectra
Elements in the sample may produce 2 or more lines
• Kα, Kβ, Lα, Lβ, (we use simplified nomenclature and discussed only α and β lines)
• Lγ, Lα1, Lβ1, Lβ2 (can also have α1 and α2 lines, β1 and β2 lines, γ lines, etc.)
Peak overlaps arising from the presence of multiple elements in the sample and limited detector resolution
Peaks from X-ray source
• Bremsstrahlung (more prominent in less dense samples)
• Rayleigh peaks from X-ray source target (typically Ag Lα, Lβ)
• Compton peaks from X-ray source target (typically at energies < Ag Lα, Lβ)
Sum peaks (two X-ray photons arriving at the detector at the same time)
• E = Kα + Kα
• E = Kα + Kβ
Escape peaks (Si in the detector absorbing some of the energy from a X-ray)
• E = Kα – Kα for Si (where Si line energy = 1.74 keV)
• E = Lα – Kα for Si
Other artifact peaks
• Product packaging, XRF cup, Mylar film, (measure what you want to measure)
• Contaminants on XRF window or trace levels of elements in XRF window or detector materials (analyze blanks to confirm source of these artifacts) | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Introduction_to_XRF-_An_Analytical_Perspective/2._Interpretation_of_XRF_Spectra.txt |
Qualitative Analysis
Issues to consider
Question: What is the GOAL of the analysis and WHAT ELEMENTS do we want to look for (toxic elements such as As, Cd, Hg, Pb; nutrient elements such as Ca, Fe)?
Answer: Define the problem (what to measure, typical concentration range, required detection limit, accuracy, precision, etc.)
Question: Are there any potential SPECTRAL OVERLAPS with other elements in sample?
Answer: Compare line energies of target elements and other elements to identify any possible interferences
Question: If we get a “positive” (detection of a toxic element), do we know for certain that it is IN THE SAMPLE and not in the product packaging or the background materials used to hold the sample?
Answer: Measure what you want to measure and be sure to do “blanks”
Question: How do we know that the analyzer software is not giving ERRONEOUS RESULTS (false positives or false negatives)?
Answer: Users must evaluate the spectrum to verify the reported results – positive identification of an element requires observation of two peaks at energies close to their tabulated values
Spectra for positive, tentative, and negative identifications
• As and Hg clearly present in blue spectrum (see both α and β peaks)
• As and Hg possibly present in purple spectrum (β peaks barely > blank)
• As and Hg not present in black spectrum (no visible peaks)
False positive for Pb in baby food cap
Spectrum
Closeup of Pb lines
• User acquired sample spectrum near lid (>10% Fe), which gave Fe sum peak at 6.40 keV * 2 photons = 12.80 keV
• Vendor algorithm incorrectly identified Pb in this sample at over 2000 ppm (detection and quantitation based on signal at the Pb Lβ line at 12.61 keV, zero intensity of Pb Lα line at 10.55 keV not considered by algorithm)
• Be wary of analyzer software and be sure to avoid potential false positives such as this by evaluating the spectrum to confirm the presence of an element
False negative for U in tableware
• Vendor algorithm did not identify U in this sample (algorithm not intended to attempt this identification of this and other relatively uncommon elements)
• Lack of manual interpretation of the spectrum of a product containing only U would have led to the assumption that it was safe
• Be wary of analyzer software and be sure to avoid potential false negatives such as this by evaluating the spectrum to identify unexplained peaks
Conclusions on Qualitative Analysis
Vendor software on commercial XRF analyzers are usually reliable in identifying which elements are present in a sample, but are not foolproof and an occasional false positive or false negative is possible
FALSE POSITIVES (element detected when not present)
• Due to limitations in the vendor software, which make not take into account line overlaps, sum peaks, escape peaks
• Users must confirm positive detection of an element based on the observation of two peaks centered within ±0.05 keV of the tabulated line energies for that element at the proper intensity ratio (5:1 for K lines, 1:1 for L lines)
FALSE NEGATIVES (element not detected when present)
• Due to limitations in the analyzer software, which may not be set up to detect all possible elements in the periodic table
• Unlikely occurrence for toxic elements such as As, Hg, Pb, and Se, more common for rare elements such as U, Th, and Os
• Users must identify “non-detected” elements through manual interpretation of the spectrum
Quantitative Analysis
Issues to consider
Question: Are the element CONCENTRATIONS within the detection range of XRF (% to ppm levels)?
Answer: Define the problem, research sample composition, or take a measurement
Question: What sort of SAMPLE PREP is required (can samples be analyzed as is or do they need to be ground up)?
Answer: Consider sample - is it homogeneous?
Question: For SCREENING PRODUCTS, are semi-quantitative results good enough? For example, if percent levels of a toxic element are found in a supplement, is this sufficient evidence to detain it or to initiate a regulatory action?
Question: For ACCURATE QUANTITATIVE ANALYSES, what is the most appropriate calibration model to use for the samples of interest (Compton Normalization, Fundamental Parameters, empirical calibration, standard additions)?
Types of Calibration Models
VISUAL OBSERVATION (rough approximation, depends on many variables)
• Peak intensity >100 cps corresponds to concentrations >10,000 ppm (% levels)
• Peak intensity of 10-100 cps corresponds to concentrations of ~100-1000 ppm
• Peak intensity of 1-10 cps corresponds to concentrations ~10-100 ppm
• Peak intensity < 1 cps corresponds to concentrations ~1-10 ppm
FUNDAMENTAL PARAMETERS (aka FP or alloy mode)
• Uses iterative approach to select element concentrations so that modeled spectrum best matches samples spectrum (using attenuation coefficients, absorption/enhancement effects, and other known information)
• Best for samples containing elements that can be detected by XRF (i.e., alloys, well characterized samples, and samples containing relatively high concentrations of elements)
COMPTON NORMALIZATION (aka CN or soil mode)
• Uses “factory” calibration based on pure elements (i.e., Fe, As2O3) and ratioing the intensity of the peak for the element of interest to the source backscatter peak to account for differences in sample matrices, orientation, etc.
• Best for samples that are relatively low density (i.e., consumer products, supplements) and samples containing relatively low concentrations of elements (i.e., soil)
OTHER MODES – thin film/filters, RoHS/WEEE, pass/fail, etc.
• Beyond scope of these training materials
EMPIRICAL CALIBRATION
• Involves preparation of authentic standards of the element of interest in a matrix that closely approximates that of the samples
• Provides more accurate results than factory calibration and Compton Normalization
• Note that the XRF analyzer can be configured and used with this type of calibration to give more accurate results for the elements and matrices of interest
• Usually reserved for laboratory analyses by trained analysts, using a high purity metal salt containing the element of interest, an appropriate matrix, homogenization via mixing or grinding
STANDARD ADDITIONS
• Involves adding known amounts of element of interest into the sample
• Provides most accurate results as the standards are prepared in the sample matrix as the sample
• Usually reserved for laboratory analyses by trained analysts, and even then used only as needed as this is labor intensive and time consuming
Effect of Concentration
Spectra of As standards in cellulose
• Intensity is proportional to concentration
• Detection limits depend on element, matrix, measurement time, etc.
• Typical detection limits are as low as 1 part per million (ppm)
Peak Intensity vs Concentration
Linearity falls off at high concentrations
• Response becomes nonlinear between 1000-10,000 ppm
• Use of Compton Normalization will partially correct for this
Compton Normalized Intensity vs Conc.
Linearity improves through use of “internal standard”
• Use of Compton Normalization (X-ray tube source backscatter from sample) partially corrects for self absorption and varying sample density
Quantitative Analysis at High Conc’s
Cr standards in stainless steel for medical instrument analysis
• Although Fundamental Parameters based quantitation gives fairly accurate results, it also gives determinate error (consistently negative errors)
• Determination of Cr in surgical grade stainless steel samples using an XRF analyzer calibrated with these standards gave results that were statistically equivalent to flame atomic absorption spectrophotometry
• For determining % levels of an element, use Fundamental Parameters mode
Quantitative Analysis at Low Conc’s
As, Hg, Pb, and Se standards in cellulose for supplement analysis
• Although Compton Normalization based quantitation gives fairly accurate results, it can also give significant determinate error (slopes > 1)
• Determination of Pb in supplements using an XRF analyzer calibrated with these standards gave results that were statistically equivalent to ICP-MS
• For determining ppm levels of an element, use Compton Norm. mode
Standard Additions Method
Determination of As in grapeseed sample
• Typically gives more reliable quantitative results as this method involves matrix matching (the sample is “converted” into standards by adding known amounts of the element of interest)
• This process is more time consuming (requires analysis of sample “as is” plus two or more samples to which known amounts of the element of interest have been added)
Effect of Measurement Time
Longer analysis times give better precision and lower LODs
• S/N = mean signal / standard deviation of instrument response (noise)
• As per theory, S/N is proportional to square root of measurement time
• 1-2 min measurement gives a good compromise between speed and precision
• Longer measurement times give better S/N and lower LODs
Conclusions on Quantitative Analysis
For field applications, the sample is often analyzed “as is” and some accuracy is sacrificed in the interest of shorter analysis times and higher sample throughput, as the more important issue here is sample triage (identifying potential samples of interest for more detailed lab analysis)
• Use FP mode to analyze samples that contain % levels of elements
• Use CN mode to analyze samples that contain ppm levels of elements and have varying densities
For lab applications, more accurate quantitative results are obtained by an empirical calibration process
• Grind/homogenize product to ensure a representative sample
• Calibrate the analyzer using standards and/or SRMs
• Use a calibration curve to compute concentrations in samples
• When suitable standards are not available or cannot be readily prepared, consider using the method of standard additions
For either mode of operation, getting an accurate number involves much more work than implied in the “point and shoot” marketing hype of some XRF manufacturers | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Introduction_to_XRF-_An_Analytical_Perspective/3._Qualitative_and_Quantitative_Analysis.txt |
Four Key Advantages of XRF for Many Applications
SIMPLICITY
• Relatively simple theory, instrument, and spectra (versus IR, MS, NMR)
MINIMAL SAMPLE PREP
• For many screening applications, samples can often be analyzed “as is” with minimal sample processing
• For accurate quantitative analysis, samples must be ground up and homogenized (faster and easier than acid digestion required for conventional atomic spectrometry methods)
TYPICAL ANALYSIS TIMES ON THE ORDER OF 1 MINUTE
• For determining % levels of an element (which typically gives high count rates), measurement times can be as short as a few seconds
• For ppm-level detection limits, measurement times on the order of 1-10 minutes are needed
PORTABILITY
• Instrument can be brought to the samples
Analytical Process Stream
Typical
analysis protocol
More intelligent
analysis protocol
• Use XRF for “sample triage” (sort into “detects” and “non-detects”)
• Avoid wasting time trying to quantify non-detectable levels of a toxic element with more time consuming methods such as ICP-MS
• Avoid problems trying to quantify % levels of a toxic element with a very sensitive technique such as ICP-MS (contaminating digestion vessels, glassware, instrument, etc. in low-level process stream)
• Perform accurate quantitative analysis (via XRF or ICP-MS) where warranted Typical analysis protocol More intelligent analysis protocol
Toxic Elements in Tableware
Pb and other elements are still causing problems
• Ceramic plates may contain toxic elements that can leach into food
• XRF can be used to quickly identify elements and their concentrations in tableware, glazes, and base ceramic material, and food
Pb in Imported Tableware and Food Products
“The prevalence of elevated blood lead levels was significantly higher in 1 of the 3 clinics (6% among screened children and 13% among prenatal patients)”
Consumption of foods imported from Oaxaca was identified as a risk factor for elevated blood lead levels in Monterey County, California.”
Handley et al, Am J Public Health, May 2007, Vol 97, No. 5, pp 900-906
“…the source was found to be related to contamination of foods in Mexico that was inadvertently transported to California through a… practice, called ‘envios’ (Spanish for send or transport) … the frequent transport of prepared foods from Mexico to California. Envios in fact are ‘mom and pop’ express air transport businesses in which foods are sent from home in Oaxaca to home in California, often on a daily basis. Unfortunately, it was discovered that some of the foods contained lead. The as yet unidentified sources of the lead are currently undergoing investigation.”
Handley et al, Intl J of Epidemiology, 2007, 36, pp 1205–1206
“An interdisciplinary investigation…was undertaken to determine the contamination source and pathway of an on-going outbreak of lead poisoning among migrants originating from Zimatlán, Oaxaca, Mexico and living in Seaside, California, and among their US-born children…
The focus in the present work concentrates on the Oaxacan area of origin of the problem in Mexico, and two potential sources of contamination were investigated: wind-borne dusts from existing mine residues as potential contaminants of soil, plant, and fauna; and food preparation practices using lead-glazed ceramic cookware
The results indicated significant presence of lead in minewastes, in specific foodstuffs, and in glazed cookware, but no extensive soil contamination was identified. In-situ experiments demonstrated that lead incorporation in food is made very efficient through grinding of spices in glazed cookware, with the combination of a harsh mechanical action and the frequent presence of acidic lime juice, but without heating, resulting in high but variable levels of contamination.”
Villalobos et al, Science of the Total Environment, in press
Pb in Tableware
Samples from Monterey County, CA
Analysis via handheld XRF calibrated with Pb standards
H. Gregory, P.T. Palmer, manuscript in prep
Pb in Food and New Tableware
Samples from Monterey County, CA
Analysis via handheld XRF calibrated with Pb standards
H. Gregory, P.T. Palmer, manuscript in prep
Museum Artifacts Preserved With As and Hg
Ideal for nondestructive testing via handheld XRF
Results From Basket Collection
Handheld XRF calibrated with Hg and As standards
Detectable Hg contamination on 17% of the baskets
K. Cross, P.T. Palmer, manuscript in prep
Results From Bird Collection
Handheld XRF calibrated with Hg and As standards
Significant As contamination on most of the birds
K. Cross, P.T. Palmer, manuscript in prep
Determination of Cr in Stainless Steel
Handheld XRF analysis of Kervorkian-designed biopsy forceps
• Atomic absorption method gave 12.7% Cr (difficult prep and digestion, >1-day effort)
• XRF analysis gave 12.8% Cr and correctly identified alloy (no sample prep, FP mode, empirical calibration with Cr standards, <1 min reading)
• Results used to confirm labeling requirements for Cr content in surgical products used in medical applications
P.T. Palmer et al, “Rapid Determination of Cr in Stainless Steel via XRF”, FDA Lab Information Bulletin, July 2006.
XRF vs Atomic Absorption for Cr in Stainless Steel
• t test indicates no significant differences at the 95% confidence level between handheld XRF and conventional Atomic Absorption Spectrophotometry method
• Such data demonstrate that XRF can give accurate quantitative results
P.T. Palmer et al, FDA Lab Information Bulletin, August 2010.
Chinese Herbal Medicine - Niuhuang Jiedu Pian
• Product manufactured in China (Cow yellow detoxification tablet), Intended to treat mouth ulcers, relieve tooth aches, reduce fever, and “release toxins”, product import document indicated that As in the form of realgar (As4S4 )
• ICP-MS showed 6.85% As (note low value here versus XRF may be due to inability of acid digestion procedures to dissolve realgar)
• Handheld XRF showed 11.7% As in product (Compton Normalization mode, empirical calibration with As standards, diluted sample into range of standards)
• Recommended max dose of 9 tablets per day is equivalent to consumption of 0.173 g of As (minimum lethal dose ~0.130 g*)
P.T. Palmer et al, J Ag. Food Chem, 57 (2009) 2605. *http://www.atsdr.cdc.gov/toxprofiles/tp2.pdf, p. 60, 127.
Toxic Elements in Supplements
• Dietary supplement sales in the U.S. surpassed \$21 billion in 2006 and 60% of people use them on a daily basis
• The Dietary Supplement Health and Education Act (DSHEA) does not require manufacturers to perform any efficacy or safety studies on dietary supplements
• FDA’s Current Good Manufacturing Practice (cGMP) requirements for Dietary Supplements provides no recommended limits for specific contaminants
• Numerous studies have reported the presence of toxic elements in a large numbers of domestic and imported supplement products
• Concerns for consumer safety have led to a Canadian ban on imports of Ayurvedic medicines in 2005 and a call for more testing and better regulation of these products
• Clearly XRF is an ideal tool for this application
Ayurvedic Medicines – Pushpadhanwa
• Ayurvedic medicine Pushpadhanwa (ironically, a fertility drug), label information indicates that it contains the following:
Rasasindoor = Pure mercury and sulfur
Nag Bhasma = Lead oxide (ash)
Loha Bhasma = Grom oxide
Abhrak Bhasma = Mica oxide
• Santa Clara County Health Dept issued a press release (Aug 2003) regarding this product which caused two serious illnesses and a spontaneous abortion
• Atomic absorption analysis by private lab showed 7% Pb in this product
• Handheld XRF analysis showed 8% Pb and 7% Hg (Compton Normalization mode, empirical calibration with authentic standards, diluted sample into range of standards)
P.T. Palmer et al, J Ag. Food Chem, 57 (2009) 2605.
Imported and Domestic Supplements
• Dolan, Capar, et al (FDA/CFSAN) reported on determination of As, Hg, and Pb in dietary supplements via microwave digestion followed by high resolution ICP-MS Dolan et al, J Ag & Food Chem, 2003, 51, 1307.
• A subset of these samples (28) were the focus of a study to compare and evaluate several different XRF analysis methods
• This represents a very challenging application for XRF due to
• Low levels of toxic elements in these samples (highest was 50 ppm)
• Tremendous variability of sample matrices and preparation of appropriate standards for an empirical calibration (cellulose was used to approximate the predominantly organic content of the samples)
• As and Pb spectral overlaps and co-occurrence of both in some samples
• Our goal was to evaluate XRF in two different modes of operation
• Screening products “as is” using an empirically calibrated handheld XRF (results not included in this presentation)
• Accurate quantitative analysis of homogenized products using an empirically calibrated lab-based XRF (completely automated data acquisition, calibration, quantitative analysis, and report generation)
XRF vs ICP-MS for Toxic Elements in Supplements
• t test indicates no significant differences at the 95% confidence level between lab-grade XRF and conventional ICP-MS method
• Such data demonstrate that XRF can give accurate quantitative results (impressive considering most samples contain these elements at concentration that are very close to the detection limit)
P.T. Palmer et al, FDA Lab Information Bulletin, August 2010.
5. Conclusions
Advantages of XRF
Selectivity:
True multi-element analysis (from S to U, ~80 different elements)
Measures total element concentration (independent of chemical form)
LODs:
1 to 10 ppm at best (depends on source, element, matrix, etc.)
Linearity:
Linear response over 3 orders of magnitude (1-1000 ppm)
Accuracy:
Relative errors ~ 50% with factory calibrated instrument
Relative errors < 10% using authentic standards for calibration
Precision:
RSDs < 5% (must have homogeneous sample)
Speed:
Minimal sample prep (analyze “as is” or homogenize and transfer to cup)
Fast analysis times (typically seconds to minutes)
Cost:
\$25,000-\$50,000 for field portable instrument
Far less expensive per sample than FAAS, GFAAS, ICP-AES, and ICP-MS
Miscellaneous:
Simple (can be used by non-experts in the field)
Nondestructive (sample can be preserved for follow up analysis)
Field-portable instruments can operate under battery power for several hours
Limitations of XRF
Selectivity:
Interferences between some elements (high levels of one element may give a false positive for another due to overlapping emission lines and limited resolution of ~0.2 keV FWHM)
No info on chemical form of element (alternate technique required for speciation)
Detection Limits:
Must use alternate technique to measure sub-ppm levels (TXRF, GFAAS, ICP-AES, ICP-MS)
Accuracy:
XRF is predominantly a surface analysis technique (X-rays penetrate few mm into sample)
To get more accurate results, one must homogenize the samples and calibrate instrument response using authentic standards
Trends in Elemental Analysis Techniques
XRF and ICP-MS are complementary
These techniques are replacing conventional atomic spectroscopy techniques such as FAAS and GFAAS
Technique XRF ICP-MS
Elements Na-U Li-U
Interferences spectral overlaps, limited resolution isobaric ions
Detection Limit ~1-10 ppm
~10 ppt (liquids)
~10 ppb (solid-0.1 g into 100 mL)
Sample prep minimal (homogenization) significant (digestion/filtration)
Field work yes not possible
Capital cost \$25-50K \$170-250K
Safety Considerations
• XRF X-ray tube sources are far less intense than medical and dental X-ray devices
• When an XRF analyzer is used properly, users will be exposed to nondetectable levels of radiation
Scenario/situation exposure units
Exposures from normal operation of XRF analyzer in sampling stand
Left/right/behind analyzer << 0.1 mREM/hour
Exposures from background radiation sources
Chest X-ray 100 mREM/X-ray
Grand Central Station 120 mREM/year
Airline worker 1000 mREM/year
Exposure limits set by regulatory agencies
Max Permissible Limit during pregnancy 500 mREM/9 months
Max Permissible Limit for entire body 5000 mREM/year
Max Permissible Limit for an extremity (i.e., finger) 50,000 mREM
Exposures from unauthorized and unacceptable use of XRF analyzer outside sampling stand
4 feet directly in front of analyzer window 14 mREM/hour
1 foot in front of analyzer window 186 mREM/hour
Directly in front of analyzer window 20,000 mREM/hour | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Introduction_to_XRF-_An_Analytical_Perspective/4._Applications_of_XRF.txt |
Separation Science - Chromatography Unit
Thomas Wenzel
Department of Chemistry
Bates College, Lewiston ME 04240
[email protected]
The following textual material is designed to accompany a series of in-class problem sets that develop many of the fundamental aspects of chromatography.
02 Text
Before examining chromatographic separations, it is useful to consider the separation process in a liquid-liquid extraction. Certain features of this process closely parallel aspects of chromatographic separations. The basic procedure for performing a liquid-liquid extraction is to take two immiscible phases, one of which is usually water and the other of which is usually an organic solvent. The two phases are put into a device called a separatory funnel, and compounds in the system will distribute between the two phases. There are two terms used for describing this distribution, one of which is called the distribution coefficient (DC), the other of which is called the partition coefficient (DM).
The distribution coefficient is the ratio of the concentration of solute in the organic phase over the concentration of solute in the aqueous phase (the V-terms are the volume of the phases). This is essentially an equilibration process whereby we start with the solute in the aqueous phase and allow it to distribute into the organic phase.
$\mathrm{solute_{aq} = solute_{org}}$
$\mathrm{D_C = \dfrac{[solute]_{org}}{[solute]_{aq}} = \dfrac{mol_{org}/V_{org}}{mol_{aq}/V_{aq}} = \dfrac{mol_{org}\times V_{aq}}{mol_{aq}\times V_{org}}}$
The distribution coefficient represents the equilibrium constant for this process. If our goal is to extract a solute from the aqueous phase into the organic phase, there is one potential problem with using the distribution coefficient as a measure of how well you have accomplished this goal. The problem relates to the relative volumes of the phases. For example, suppose the volume of the organic phase was very small compared to the volume of the aqueous phase. (Imagine using 100 mL of organic solvent relative to a volume of water equal to that in an Olympic-sized swimming pool). You could have a very high concentration of the solute in the organic phase, but if we looked at the amount of solute in the organic phase relative to the amount still in the water, it might only be a small portion of the total solute in the system. Since we really want as much of the solute in the organic phase as possible, this system has not yet achieved that outcome.
The partition coefficient is the ratio of the moles of solute in the two phases, and is a more effective means of measuring whether you have achieved the desired goal. The larger the value of DM, the more of the solute we have extracted or partitioned into the organic phase.
$\mathrm{D_M = \dfrac{mol_{org}}{mol_{aq}}}$
Note as well how we can relate DC to DM:
$\mathrm{D_C = \dfrac{mol_{org}\times V_{aq}}{mol_{aq}\times V_{org}} = D_M\left(\dfrac{V_{aq}}{V_{org}} \right )}$
From experience you have probably had in your organic chemistry lab, you know that the approach that is often used in liquid-liquid extraction is to add some organic phase, shake the mixture, and remove the organic phase. A fresh portion of the organic phase is then added to remove more of the solute in a second extraction. As we will see shortly, this distribution of a solute between two immiscible phases forms the basis of chromatographic separations as well.
Next we want to examine some general types of extraction procedures that are commonly used. The first is a classic example of an extraction procedure that can be used to separate acids, bases, and neutrals.
An aqueous sample contains a complex mixture of organic compounds, all of which are at trace concentrations. The compounds can be grouped into broad categories of organic acids, organic bases and neutral organics. The desire is to have three solutions at the end, each in methylene chloride, one of which contains only the organic acids, the second contains only the organic bases, and the third contains only the neutrals. Devise an extraction procedure that would allow you to perform this bulk separation of the three categories of organic compounds.
Two things to remember:
• Ionic substances are more soluble in water than in organic solvents.
• Neutral substances are more soluble in organic solvents than in water.
The key to understanding how to do this separation relates to the effect that pH will have on the different categories of compounds.
Neutrals – Whether the pH is acidic or basic, these will remain neutral under all circumstances.
Organic acids – RCOOH
At very acidic pH values (say a pH of around 1) – these are fully protonated and neutral
At basic pH values (say a pH of around 13) – these are fully deprotonated and anionic
Organic bases – R3N
At very acidic pH values (say a pH of around 1) – these are protonated and cationic
At very basic pH values (say a pH of around 13) – these are not protonated and neutral
Step 1: Lower the pH of the water using concentrated hydrochloric acid.
Neutrals – neutral
Acids – neutral
Bases – cationic
Extract with methylene chloride – the neutrals and acids go into the methylene chloride, the bases stay in the water.
Step 2: Remove the water layer from step (1), adjust the pH back to a value of 13 using a concentrated solution of sodium hydroxide, shake against methylene chloride, and we now have a solution of the organic bases in methylene chloride. (Solution 1 – ORGANIC BASES IN METHYLENE CHLORIDE)
Step 3: Take the methylene chloride layer from step (1) and shake this against an aqueous layer with a pH value of 13 (adjusted to that level using a concentrated solution of sodium hydroxide).
Neutrals – neutral
Organic acids – anionic
The neutrals stay in the methylene chloride layer. (Solution 2: NEUTRALS IN METHYLENE CHLORIDE) The acids go into the water layer.
Step 4. Take the water layer from Step (3), lower the pH to a value of 1 using concentrated hydrochloric acid, shake against methylene chloride, and the neutral organic acids are now soluble in the methylene chloride (Solution 3: ORGANIC ACIDS IN METHYLENE CHLORIDE).
Devise a way to solubilize the organic anion shown below in the organic solvent of a two-phase system in which the second phase is water. As a first step to this problem, show what might happen to this compound when added to such a two-phase system.
This compound will align itself right along the interface of the two layers. The non-polar C18 group is hydrophobic and will be oriented into the organic phase. The polar carboxylate group is hydrophilic and will be right at the interface with the aqueous phase.
One way to solubilize this anion in the organic phase is to add a cation with similar properties. In other words, if we added an organic cation that has a non-polar R group, this would form an ion pair with the organic anion. The ion pair between the two effectively shields the two charged groups and allows the pair to dissolve in an organic solvent. Two possible organic cations that could be used in this system are cetylpyridinium chloride or tetra-n-butylammonium chloride.
A somewhat similar procedure can often be used to extract metal complexes into an organic phase. Metal salts with inorganic anions (halide, sulfate, phosphate, etc.) are generally water-soluble but not organic-soluble. It is possible to add a relatively hydrophobic ligand to the system. If the ligand complexes with the metal ion, then the metal complex may be organic-soluble. Usually it helps to form a neutral metal complex. Also, remember back to our examination of the effect of pH on the complexation of metal ions with ligands. The extraction efficiency of a metal ion in the presence of a ligand will depend on the pH of the aqueous phase. Adjustment of the pH is often used to alter the selectivity of the extraction, thereby allowing different metal ions to be separated. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Separation_Science/02_Text/01_Liquid-Liquid_Extraction.txt |
Chromatography refers to a group of methods that are used as a way of separating mixtures of compounds into their individual components. The basic set up of a chromatographic system is to have two phases, one stationary and one mobile. A compound, which we will usually call the solute, is introduced into the system, and essentially has a choice. If it is attracted to the mobile phase (has van der Waal attractions to the mobile phase), it will move through the system with the mobile phase. If it is attracted to the stationary phase, it will lag behind. It’s easy to imagine some solute compounds with some degree of attraction for both phases, such that these move through the system with some intermediate rate of travel.
The first report of a chromatographic application was by Mikhail Tswett, a chemist from Estonia, in 1903. (A list of important literature articles is provided at the end). Undoubtedly other people had observed chromatography taking place, but no one until Tswett recognized its applicability for the separation of mixtures in chemistry. Tswett was interested in separating the pigments in plants. He packed a glass column that might have been comparable to a buret with starch, mashed up the plant and extracted the pigments into a solvent, loaded the solution onto the top of the starch column, and ran a mobile phase through the starch. Eventually he saw different color bands separate on the column, hence the term chromatography. Rather than eluting the colored bands off of the column, he stopped it, used a rod to push out the starch filling, divided up the bands of color, and extracted the individual pigments off the starch using an appropriate solvent. Twsett used a solid stationary phase (starch) and a liquid mobile phase.
Chromatographic systems can use either a gas or liquid as the mobile phase. Chromatographic systems can use either a solid or liquid as the stationary phase. A solid stationary phase is easy to imagine, as we have already seen for Tswett’s work. Another common example is to use paper as the solid (actually, paper is made from a compound called cellulose). The sample is spotted onto the paper as shown in Figure 1a, and the bottom of the piece of paper is dipped into an appropriate liquid (note that the spot is above the liquid – Figure 1b). The liquid mobile phase moves up the paper by capillary action, and components of the mixture can separate into different spots depending on their relative attraction to the cellulose or chemicals that make up the liquid (Figure 1c).
Figure 1. Paper chromatography
Another type of chromatography that you may be familiar with is thin layer chromatography (TLC). This is very similar to paper chromatography, although the stationary phase is usually a coating of small particles of a material known as silica gel (silica gel is a polymer with the formula SiO2, although a bit later we will talk in more detail about the exact chemical nature of this material) on a glass or plastic plate. Similarly, we could take these silica gel particles and pack them into a glass column, and then flow a liquid through the column. If we used some gas pressure to force the liquid through the silica gel column more quickly, we would have a common technique known as flash chromatography that synthetic chemists use to separate relatively large amounts of materials they have prepared. Another common solid to use in column chromatography is alumina (Al2O3 is the general formula, although we will say more later about its exact chemical nature).
These methods we have just been talking about are examples of liquid-solid chromatography. If we used gas as the mobile phase, injected either a gaseous sample or liquid sample into a high temperature zone that flash volatilized it, with some solid as the stationary phase (there are a range of possible materials we could use here), we could perform gas-solid chromatography. If we thought about how a solute compound would interact with such a solid stationary phase, we would realize that it must essentially “stick” to the surface by some intermolecular (van der Waal) forces. This sticking process in chromatography is known as adsorption.
Earlier I said that it is also possible to use a liquid as the stationary phase. This might seem odd at first, because it would seem as if the flowing mobile phase would somehow push along a liquid stationary phase. The way to use a liquid as the stationary phase is to coat a very thin layer of it onto a solid support, as illustrated in Figure 2.
Figure 2. Liquid phase coated onto a solid support.
Alternatively, it is possible to coat a liquid phase onto the inside walls of a small-diameter, open-tubular column (known as a capillary column) as shown in Figure 3.
Figure 3. Liquid phase coated onto the inside walls of a capillary column.
The liquid coating has some attractive forces for the underlying solid surface, such that it sticks to the surface and stays in place. We could probably see how we could easily have a gas flowing around a packed bed of such particles and, so long as the liquid coating did not evaporate, it would stay in place on the particles. Flowing a liquid around it might be a bit more problematic, and we will examine that in length later in this unit. Suffice it to say, though, that it can be done. This would lead to gas-liquid and liquid-liquid chromatographic methods. The interaction of a solute compound with a liquid stationary phase is different than with the solid stationary phases discussed above. It will still depend on having attractive intermolecular forces, but instead of sticking to the solid surface by adsorption, the solute molecule will now dissolve into the liquid stationary phase. If a solute molecule dissolves into a liquid stationary phase in a chromatographic system, we say that the molecule partitions between the two phases.
These two terms are so important in chromatographic systems that it is worth summarizing them again:
Adsorption – describes the process of a solute molecule adhering to a solid surface
Partition – describes the process of a solute molecule dissolving into a liquid stationary phase
There are two terms we can describe in chromatography that are analogous to those we already described for liquid-liquid extraction. One of these is the distribution constant (KC) that is the ratio of the concentration of the solute in the stationary phase (CS) to that in the mobile phase (CM).
$\mathrm{K_C = \dfrac{C_S}{C_M} = \dfrac{mol_S/V_S}{mol_M/V_M} = \dfrac{mol_S\times V_M}{mol_M \times V_S}}$
But just like in liquid-liquid extraction, a problem with this ratio is that the volumes of the stationary and mobile phases might be significantly different from each other. With a solid stationary phase, VS is comparable to the surface area of the particles. With a liquid stationary phase, the coating is usually so thin that VM would be much larger than VS. In many cases, a more useful term is the partition coefficient (KX), which is the ratio of the moles of solute in each of the two phases.
$\mathrm{K_X = \dfrac{mol_S}{mol_M}}$
Notice that we can relate KC to KX because the mole ratio shows up in both. That allows us to write the following:
$\mathrm{K_C = K_X\left(\dfrac{V_M}{V_S}\right)}$
There are also some other fundamental figures of merit that we often use when discussing chromatographic separations. The first is known as the selectivity factor (α). In order to separate two components of a mixture, it is essential that the two have different distribution or partition coefficients (note, it does not matter which one you use since the volume terms will cancel if KC is used). The separation factor is the ratio of these two coefficients, and is always written so that it is greater than or equal to one. K2 represents the distribution coefficient for the later eluting of the two components.
$\mathrm{\alpha = \dfrac{K_2}{K_1}}$
If α = 1, then there will be no separation of the two components. The larger the value of α, the greater the separation. One thing to realize is that there is some limit as to how large we want an α-value to be. For example, the chromatogram in Figure 4b has a much greater α value than that in Figure 4a. But the two components are fully separated so that the extra time we would need to wait for the chromatogram in Figure 4b is undesirable. In most chromatographic separations, the goal is to get just enough separation, so that we can keep the analysis time as short as possible.
Figure 4. Elution of compounds where the α-value in (b) is much greater than the α-value in (a).
Another important figure of merit in chromatographic separations is known as the retention factor (k'), which is a measure of the retention of a solute, or how much attraction the solute has for the stationary phase. The k’ term was formerly known as the capacity factor. The retention factor is actually identical to the partition coefficient.
$\mathrm{k' = \dfrac{mol_S}{mol_M} = \dfrac{C_SV_S}{C_MV_M} = K_C\left(\dfrac{V_S}{V_M} \right )}$
We can determine k' values from a chromatogram using the following equation:
$\mathrm{k' = \dfrac{t_R - t_0}{t_0}}$
Figure 5. Chromatographic information used to calculate the k’ value for a peak.
As seen in Figure 5, the term tR is the retention time of the compound of interest, whereas t0 refers to the retention time of an unretained compound (the time it takes a compound with no ability to partition into or adsorb onto the stationary phase to move through the column). Since t0 will vary from column to column, the form of this equation represents a normalization of the retention times to t0.
Another important figure of merit for a chromatographic column is known as the number of theoretical plates (N). It turns out that one helpful way to think of a chromatographic column is as a series of microscopically thin plates as shown in Figure 6. Using the picture in Figure 6, we could imagine a compound moving down the column as a series of steps, one plate at a time. The compound enters a plate, distributes according to its distribution coefficient between the stationary and mobile phase, and then moves on to the next plate. Now it turns out that chromatographic systems never really reach equilibrium, and in fact are not steady state systems, but things like the distribution and partition coefficients, which are equilibrium expressions, are useful ways to examine the distribution in chromatographic systems. A compound with less favorable solubility in the stationary phase would then move through the series of plates faster. The only way to separate two compounds is to have enough plates, or enough equilibrations, to exploit the difference in partition coefficients between the two compounds. A column with more theoretical plates is more likely to separate two compounds than one with fewer.
Figure 6. Representation of a chromatographic column as a series of microscopically thin plates.
We can determine the number of plates for a compound as shown in Figure 7.
$\mathrm{N = 16\left( \dfrac{t_R}{W}\right )^2}$
Figure 7. Chromatographic information used to determine N for a column.
W is the width of the peak where it intersects with the baseline. The important thing to remember is to use the same units when measuring W and tR (e.g., distance in cm on a plot, time in seconds, elution volume in mL – which is common in liquid chromatography). It should be pointed out that a column will have a set number of plates that will not vary much from compound to compound. The reason for this is that a compound with a longer retention time will exhibit a larger peak width, such that the ratio term is correcting for these two effects. If we determine the number of plates for a column, dividing the column length (L) by the number of plates leads to the height equivalent to a theoretical plate (H). Note that the smaller the value of H the better. It actually turns out the H is the important figure of merit for a column.
$\mathrm{H = \dfrac{L}{N}}$
The chromatograms in Figure 8 show the distinction between a column with a smaller value of H (Figure 8a) and one with a much larger value of H (Figure 8b).
Figure 8. Chromatograms for a column with (a) a smaller value of H and (b) a larger value of H.
Usually today, though, people refer to something called the reduced plate height (h) for a column. The reduced plate height is defined as shown below, in which dp is the particle diameter of the packing material used in the column (or the internal tube diameter if an open tubular capillary column is used).
$\mathrm{h = \dfrac{H}{d_p}}$
You will also come across a term known as the reduced velocity ($\nu$) in the chromatographic literature. The reduced velocity is defined as follows, in which v is the mobile phase velocity, and DM the diffusion coefficient of the solute in the mobile phase.
$\mathrm{\nu = \dfrac{d_p\times v}{D_M}}$
Finally, we can define an equation for the resolution (RS) of two compounds. This will be a measure of how much two compounds in a chromatogram are separated from each other.
$\mathrm{R_S = \dfrac{2(t_2 - t_1)}{W_1 + W_2}}$
The terms t2 and t1 refer to the retention time of the two compounds, and W1 and W2 to the width of each peak at baseline.
Figure 9 shows the separation of two peaks at resolution values of 0.75 (Figure 9a), 1.0 (Figure 9b) and 1.5 (Figure 9c). The better separation of A and B that occurs with increasing resolution is obvious. It is also worth noting that the analysis time is longer the higher the resolution. Chromatographic analyses are always a compromise between the degree of separation of the peaks and how long it takes to perform the analysis. In many cases, the goal is to get just enough separation in the shortest period of time.
Figure 9. Separation of two peaks with resolution values of (a) 0.75, (b) 1.0 and (c) 1.5. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Separation_Science/02_Text/02_Chromatography__Background/01_Introduction.txt |
Distribution Isotherms (Isotherm means constant temperature):
We have used the distribution and partition coefficients as ways to express the distribution of a molecule between the mobile and stationary phase. Note that these equations take the form of equilibrium expressions, and that KC is a constant for the distribution of a solute compound between two particular phases. (KX depends on the relative volumes of the two phases.)
Suppose that we plotted CS versus CM, as shown on the coordinate system in Figure 10.
Figure 10. Coordinate system with the concentration of analyte in the stationary phase (CS) shown on the y-axis and the concentration of analyte in the mobile phase (CM) shown on the x-axis.
Notice that CM plus CS would give the total amount of the solute injected. So as you move to the right on the CM axis, it means that more sample is being injected into the chromatograph. If we have the expression for the distribution coefficient:
$\mathrm{K_C = \dfrac{C_S}{C_M} = constant}$
The idealized form of the plot of the distribution isotherm is shown in Figure 11. If we inject more sample into the column, it distributes according to a set ratio of the distribution coefficient. The result would be a straight line, with the slope being the distribution coefficient.
Figure 11. Idealized plot of the distribution isotherm.
If we examine this in more detail, though, we will realize that the volume of stationary phase is some fixed quantity, and is usually substantially less than the volume of the mobile phase. It is possible to saturate the stationary phase with solute, such that no more can dissolve. In that case, the curve would show the behavior shown in Figure 12, a result known as the Langmuir isotherm (Langmuir was a renowned surface scientist and a journal of the American Chemical Society on surface science is named in his honor). If we get a region of Langmuir behavior, we have saturated the stationary phase or overloaded the capacity of the column.
Figure 12. Plot of the Langmuir isotherm.
We might also ask whether you could ever get the following plot, which is called anti-Langmuir behavior.
Figure 13. Plot of the anti-Langmuir isotherm.
There are actually two ways this could happen. One is if the solute dissolves in the stationary phase, creating a mixed phase that then allows a higher solubility of the solute. This behavior is not commonly observed. Another way that this can occur is in gas chromatography if too large or concentrated a sample of solute is injected. At a fixed temperature, a volatile compound has a specific vapor pressure. This vapor pressure can never be exceeded. If the vapor pressure is not exceeded, all of the compound can evaporate. If too high a concentration of compound is injected such that it would exceed the vapor pressure if all evaporates, some evaporates into the gas phase but the rest remains condensed as a liquid. If this happens, it is an example of anti-Langmuir behavior because it appears as if more is in the stationary phase (the condensed droplets of sample would seem to be in the stationary phase because they are not moving).
The last question we need to consider is what these forms of non-ideal behavior would do to the shape of a chromatographic peak. From laws of diffusion, it is possible to derive that an “ideal” chromatographic peak will have a symmetrical (Gaussian) shape. Either form of overloading will lead to asymmetry in the peaks. This can cause either fronting or tailing as shown in Figure 14. The Langmuir isotherm, which results from overloading of the stationary phase, leads to peak tailing. Anti-Laugmuir behavior leads to fronting.
Figure 14. Representation of a chromatographic peak exhibiting ideal peak shape, fronting, and tailing. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Separation_Science/02_Text/02_Chromatography__Background/02_Distribution_Isotherms.txt |
If we go back now and consider Tswett’s first separation, we see that he used solid starch as the stationary phase, and so solutes exhibited an adsorption to the surface of the starch. A useful thing to consider is the nature of the chemical groups on the surface of starch. Starch is a carbohydrate comprised of glucose units. The glucose functionality of starch is shown below, and we see that the surface is comprised of highly polar hydroxyl groups.
Glucose
Silica gel (SiO2) and alumina (Al2O3) were mentioned previously as two other common solid phases. If we examine the structure of silica gel, in which each silicon atom is attached to four oxygen atoms in a tetrahedral arrangement (this corresponds to two silicon atoms sharing each of the oxygen atoms), we run into a problem when you try to create a surface for this material. If we have an oxygen atom out on the edge, we would then need to attach a silicon atom, which necessitates more oxygen atoms. We end up with a dilemma in which we can never wrap all of these around on each other and only make something with the formula SiO2. Some of these groups are able to do this and some of the surface of silica gel consists of what are known as siloxane (Si–O–Si) groups. But many of the outer oxygen atoms cannot attach to another silicon atom and are actually hydroxyl or silanol (Si–OH) groups.
–Si–O–Si– –Si–OH
Siloxane units Silanol units
What we notice is that the surface of silica gel consists of very polar hydroxyl groups. It turns out that the same thing will occur for the surface of alumina.
If we consider a solute molecule (S), and have it adsorb to the surface of silica gel, we could write the following equation to represent the process of adsorption.
Si–OH + S \(\leftrightarrow\) Si–OH - - - S
With any reaction we can talk about its enthalpy and entropy. So in this case, we could talk about the enthalpy of adsorption (ΔHADS). Suppose now we took one solute molecule and millions of surface silanols, and went through one by one and measured ΔHADS for this one solute molecule with each of these millions of surface silanols. The question is whether we would get one single ΔHADS value for each of the measurements. Hopefully it might seem intuitive to you that we would not get an identical value for each of these individual processes. Instead, it seems like what we would really get is a distribution of values. There would be one value that is most common with other less frequent values clustered around it. For some reason, some silanol sites might be a bit more active because of differences in their surrounding microenvironment, whereas others might be a bit less active. Suppose we entered all these measurements in a spreadsheet and then plotted them as a histogram with number of measurements of a particular value versus ΔHADS. But another important question involves the nature of this distribution of ΔHADS values. Would it be symmetrical as shown in Figure 15, or would it have some asymmetry?
Figure 15. Representation of a symmetrical distribution for ΔHADS
The easiest way to think of this is to examine the nature of the silanol groups on the surface of silica gel. One thing we could ask is whether there are different types of silanols, as shown below (are there also disilanols and trisilanols?).
–Si–OH –Si–(OH)2 –Si(OH)3
It turns out that we will get some di- and trisilanols, although these are far fewer in number than the monosilanols. It should make sense that even if the distribution about a monosilanol were symmetrical, that the di- and trisilanols will have different distributions of ΔHADS. Assuming that ΔHADS is larger for the di- and trisilanols, that might lead to the plot shown in Figure 16 for each of the individual distributions and then the composite drawing for the overall distribution.
Figure 16. Representation of the distribution of ΔHADS values for a surface that contains silanol, disilanol and trisilanol groups.
Notice the asymmetry in the plot in Figure 16 with a smaller number of highly active sites. Something to realize is that even if we had only monosilanols, there is still experimental data that shows a small proportion of highly active sites that produced an asymmetric distribution. It is important to realize that not everything in nature occurs in a symmetrical manner.
The key question for us to ask is what a chromatographic peak would look like for a solute traveling through this stationary phase. Hopefully it seems reasonable to predict that the peak for the solute would also be asymmetric. Those solute molecules adsorbed at the highly active sites would essentially get stuck on the column and take much longer to elute. The asymmetric peak observed in Figure 17 exhibits tailing. Note that tailed peaks are undesirable in chromatographic separations because they are more likely to overlap with each other and interfere with the separation. Unless the solid surface is highly deactivated, all chromatographic separations based on adsorption will exhibit peak tailing. This causes an inherent inefficiency in adsorption methods.
Figure 17. Tailed peak shape in adsorption chromatography
One last thing before we move on is to understand that strong adsorption by molecules on active solid surfaces has a profound implication in environmental chemistry. Consider a leaking, underground tank of gasoline at a service station. The soil around the tank gets contaminated by the leaking gas and it is desirable to clean it up. One way is to remove all the contaminated soil, heat it in a very high temperature oven that combusts all the hydrocarbon components of the gasoline, and recycle the soil. This is a common remediation procedure. But suppose the soil exists over a very large area (say the soil surrounding an old hazardous waste dump). This may be too large a volume of soil to use the removal and combustion method. The chemicals may have leached far away from the original site, contaminating nearby drinking wells.
Based on our knowledge of chromatography, and essentially using the ground as something similar to a chromatographic column, we could imagine pumping lots of clean water in from wells outside the contaminated area, and removing contaminated water from wells internal to the contaminated region. After some time of pumping, all the contaminants ought to migrate by a chromatographic-like process from the outside to the inside and be removed. If you do this, you eventually see that the level of contaminants in the water coming out of the inner wells drop considerably (maybe even to acceptable levels). But if you allow the system to sit for quite a while, you start to discover elevated levels of contaminants in the water again. What has happened is that some of the contaminants adsorbed quite strongly to the solid surfaces in the ground and were unavailable to the water. As the system sat, these slowly began to come off the surface into the water, raising the concentration. This kind of treatment process has generally proven ineffective as a way of completely treating such ground water contamination. It does get rid of a lot of the contaminants, but the strong adsorption requires enormous lengths of time before the levels would drop low enough.
Chromatographic methods languished under these inefficient methods for many years until Martin and Synge reported the first application of a partition separation in chromatography in 1941. Recognizing that WWII was taking place in 1941, there was a considerable need for wool clothing for soldiers from England fighting in the war. Wool is rather unique as a fabric since it retains much of its warmth even if wet. In fact, so much wool was needed that England did not have enough sheep to provide the volume of wool clothing that was necessary. Martin and Synge were interested in seeing whether it was possible to make artificial wool, and sought to examine the amino acid content of the proteins that make up wool fibers as a first step in understanding the chemical nature of wool. What they also realized, though, was that separating amino acids using adsorption chromatographic methods available at that time was going to be a difficult process. They therefore investigated whether it would be possible to take a solid particle and coat it with a liquid stationary phase, and perform the separation based on partitioning. What they found was that it was possible, and that the chromatographic efficiency was improved considerably when compared to methods based on adsorption.
The important thing to realize here is that we no longer have an adsorption process, but the dissolution of the solute into another liquid solvent. The relevant enthalpy to consider here is the enthalpy of solvation (ΔHSOLV). It turns out that ΔHSOLV of a particular solute in a particular solvent will also show a distribution of values; however, in this case, the distribution is a symmetrical one. Therefore, a chromatographic peak for a solute being separated entirely by a partitioning mechanism between the mobile and stationary phases ought to be symmetrical as well. This will greatly enhance the efficiency of the chromatography. This work was so important that it was recognized with the Nobel Prize in 1952. The only problem was that there were still other important issues with liquid chromatography such that it really could not flourish as an analytical method (we will examine these other issues later in our unit on chromatography).
One intriguing aspect of Martin and Synge’s 1941 paper is the last sentence, which predicted that it should be possible to use gas as the mobile phase (i.e., gas chromatography), and that some of the limitations that restrict the efficiency of liquid chromatography should not occur in gas chromatography. What was also interesting is that it was not until 1951, that James and Martin published the first report in which gas chromatography was described as an analysis method. Much of the delay was because of WWII and the need for scientists to devote their research to areas of immediate national concern related to the war effort.
The introduction of gas chromatography revolutionized the entire field of chemical analysis. One thing, as we will develop in the next substantial portion of this unit, is that gas chromatographic methods had certain fundamental advantages over liquid chromatographic methods when it came to column efficiency. It was possible, using gas chromatography, to separate very complex mixtures of volatile chemicals in very short periods of time. Before gas chromatography, no one even considered that it might be possible to separate as many as 50 to 100 constituents of a sample in only an hour. The other thing that prompted an explosion of interest in gas chromatography as an analysis method was the development of some highly sensitive methods of detection in the 1950s and 1960s. People were now able to sense levels of molecules that were not detectable in other, conventional solution-phase systems. Part of the problem with solution-phase analysis is the overwhelming volume of solvent that can interfere with the technique used to perform the measurement. Gas phase measurements, where the overall density of molecules is very low, do not have as much potential for interference from gases other than those being measured.
What we now need to do is develop an understanding of what created the inherent advantage in efficiency of gas chromatographic columns, and then understand what took place to improve the efficiency of liquid chromatographic columns. When we talk about the efficiency of a chromatographic column, what we really refer to is the width of the chromatographic bands of solutes as they migrate through the column. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Separation_Science/02_Text/02_Chromatography__Background/03_Adsorption_Compared_to_Partition_as_a_Separation_M.txt |
One of the most important occurrences in chromatographic systems is the broadening of peaks as compounds move through the chromatographic column. For example, if we examine Figure 18 in which both sets of peaks have the same retention times but different extents of broadening, we see that the set of conditions that produce the narrower set of peaks (Figure 18b) resulted in “better” or more efficient chromatography. What we observe in the chromatogram with less peak broadening is that the peaks are fully resolved and that we could fit more peaks into a similar window of time in the chromatogram. An ideal chromatographic system would therefore produce peaks that were straight line spikes in which no broadening occurred (Figure 18a).
Figure 18. Representation of chromatographic peaks in which the broadening increases going from (a) to (c).
There are specific processes that occur in chromatographic systems that cause peaks to broaden. It also turns out that these processes are often influenced by experimental variables that we may have some control over. Contributions to broadening in chromatographic systems can be divided into two broad areas of concern. One is the contribution from what is known as dead volume. Dead volume refers to all the volume in a chromatographic system from the injector to the detector other than the column. Remember, the separation only occurs in the column. All other volumes (tubing used to connect components, volume within the detector cell, etc.) have the ability to contribute to peak broadening but not to the separation. One general goal then is to try to reduce the total dead volume to as small a quantity as possible. In liquid chromatography this involves using very narrow internal diameter tubing and short lengths of tubing to connect the components, using small-volume detector cells, etc. The fittings that are used to connect pieces of tubing together or components to each other have been designed specifically to reduce the dead volume to minimal levels and in ways that do not promote mixing and broadening.
The other source of broadening is within the column. If you were to examine state-of-the-art columns that are used today in gas and liquid chromatography, it turns out that there are several features of their design that lead to significant reductions in peak broadening. In other words, these columns represent the best we can do today to reduce the broadening of peaks, and therefore represent the most efficient column technology. It is worth taking the time to understand the various contributions to peak broadening that occur within the chromatographic column and to examine the ways in which current gas and liquid chromatographic columns have been designed to minimize these effects.
There are four general contributions to broadening within chromatographic columns. These are known as:
• longitudinal diffusion
• eddy diffusion
• mass transport broadening in the stationary phase
• mass transport broadening in the mobile phase
Before moving on, it is worth remembering back to two fundamental criteria we talked about with regards to chromatographic columns, the number of theoretical plates (N) and the reduced plate height (h). Remember that a column with more plates, or better yet a column with a small reduced plate height, was more efficient and provided better separations. We can therefore use the reduced plate height as a determining measure of the efficiency of a chromatographic column. The smaller the value of h, the more efficient the column. What we will develop as we analyze the four contributions to broadening above is an equation, which was first known as the van Deemter equation (J. J. van Deemter described the first treatment of this for chromatographic systems in 1956), that relates these four terms to the reduced plate height.
03 Broadening of Chromatographic Peaks
Consider a band of a compound in a chromatographic column. The band has the concentration profile shown in Figure 19.
Figure 19. The representation on the left shows a band of a compound on a chromatographic column. The representation on the right shows the concentration profile of the band. The concentration is greatest at the center of the band.
The first thing to consider is what would happen to this profile if the flow of the column was stopped and the column was allowed to sit. We know from the random process of diffusion that there is a statistical preference in which more molecules diffuse away from regions of high concentration to regions of low concentration. If we were to allow this band to sit stopped in the column, it would slowly diffuse out from its central region and lead to a broadening of the concentration profile as shown in Figure 20.
Figure 20. Representation of band or peak broadening that results from longitudinal diffusion. Compare to the concentration profile in Figure 19.
This observation of diffusion from a region of high concentration to one of low concentration will occur whether the band of material is sitting stationary in a column or is flowing through the column. A molecule flowing through a column has two means of movement. One is the physical flow that is taking place. But the other is still its ability to diffuse in a random manner from one point to another. All compounds moving through a chromatographic column must exhibit some degree of longitudinal diffusion broadening. Therefore, we can never get the idealized chromatogram shown in Figure 18a.
An important thing to consider is whether this phenomenon is more significant (i.e., happens faster and therefore causes more broadening, everything else being equal) in gas or liquid chromatography. To consider this, we would need to know something about the relative rates of diffusion of gases and liquids. A substance with a faster rate of diffusion will broaden more in a certain amount of time than something with a slower rate of diffusion. So the relevant question is, which diffuses faster, gases or liquids? I suspect we all know that gases diffuse appreciably faster than liquids. Just imagine yourself standing on the opposite side of a room from someone who opens a bottle of a chemical with the odor of a skunk. How fast do you smell this odor? Compare that with having the room full of water, and someone adds a drop of a colored dye to the water at one side of the room. How fast would that color make its way across the water to the other side of the room? In fact, gases have diffusion rates that are approximately 100,000 times faster than that of liquids. The potential contribution of longitudinal diffusion broadening to chromatographic peaks is much more serious in gas chromatography than in liquid chromatography. In liquid chromatography, the contribution of longitudinal diffusion broadening is so low that it’s really never a significant contribution to peak broadening.
Finally, we could ask ourselves whether this phenomenon contributes more to band broadening at higher or lower flow rates. What we need to recognize is that longitudinal broadening occurs at some set rate that is only determined by the mobile phase (gas or liquid) and the particular molecule undergoing diffusion. In the gas or liquid phase, it would be reasonable to expect that a small molecule would have a faster rate of diffusion than a large molecule. If we are doing conventional gas or liquid chromatography using organic compounds with molecular weights from about 100 to 300, the differences in diffusion rates are not sufficient enough to make large differences here. If we were comparing those molecules to proteins with molecular weights of 50,000, there might be a significant difference in the rate of diffusion in the liquid phase. If the longitudinal diffusion occurs at a set, fixed rate, then the longer a compound (solute) is in the column, the more time it has to undergo longitudinal diffusion. The compound would be in the column a longer time at a slower flow rate. This allows us to say that the contribution of longitudinal diffusion to overall peak broadening will be greater the slower the flow rate. If we use the term B to represent longitudinal diffusion broadening and v to represent flow rate, and want to relate this to h, we would write the following expression:
$\mathrm{h = \dfrac{B}{v}}$
Remember that the smaller the reduced plate height the better. At high flow rates, B/v gets smaller, h is smaller, and the contribution of longitudinal diffusion to peak broadening is smaller.
We can also write the following expression for B: B = 2φDM
In this case, DM refers to the diffusivity (diffusion coefficient) of the solute in the solvent. Notice that this is a direct relationship: the faster the rate of diffusion of the solute, the greater the extent of longitudinal diffusion. φ is known as the obstruction factor, and occurs in a packed chromatographic column. In solution, a molecule has an equal probability of diffusion in any direction. In a packed column, the solid packing material may restrict the ability of the solute to diffuse in a particular direction, thereby hindering longitudinal diffusion. This term takes this effect into account. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Separation_Science/02_Text/03_Broadening_of_Chromatographic_Peaks/01_Longitudinal_Diffusion_Broadening.txt |
Consider a group of molecules flowing through a packed bed of particles (Figure 21). Another way to think of this is to imagine you and a group of friends following a river downstream in a set of inner tubes. The river has a number of rocks in the way and a variety of different flow paths through the rocks. Different people go in different channels either because they paddle over to them or get caught in different flows of the current as they bounce off of obstacles.
Figure 21. Representation of a chromatographic column packed with particles.
The question we need to consider is whether different molecules would have different path lengths as they passed through the bed. Would each person traveling down the river travel a slightly different distance, or would everyone travel the exact same distance? I suspect we can see that different molecules would end up traveling paths with different lengths as shown in Figure 22. One molecule (Molecule 1) might find a relatively straight shot through the packed bed, whereas another might encounter more particles that needed to be circumvented (Molecules 2 and 3). If we watched a group of molecules pass through a packed bed, and then mapped out each path with a length of string, we would see if we then stretched out all the strings, that there was a distribution of path lengths with some being shorter and some being longer. If that were the case, the molecule with the shortest path length would move through the column more quickly (Molecule 1). The molecule with the longest path length would move through the column more slowly (Molecule 3). If we have a distinction between the time it takes a set of molecules to move through the column based only on different path lengths, we have broadened the peak. This is known as eddy diffusion.
Figure 22. Representation of the different paths of three different molecules traveling through a packed bed of particles.
A key factor to consider when examining eddy diffusion is to ask whether the difference in length between the shortest and longest path depends at all on the diameter of the particles. If it is, we could then ask which particles (smaller or larger) would lead to a greater difference in path length?
Almost everyone who is asked the first question seems to intuitively realize that the size of the particles must somehow make a difference. It just seems too coincidental to think that the difference between the shortest and longest path would be identical if the particle sizes are different. But interestingly enough, almost everyone, when they first consider this, seems to select the wrong answer when figuring out which particle (small or large) would lead to a greater difference in path length between the shortest and longest path. Remember, the important distinction is the size of the difference between the shortest and longest path, not whether one column would uniformly have longer path lengths than the other.
To help with this assessment, two columns are pictured in Figure 23, one with small particles, the other with larger particles.
Figure 23. Representation of two columns with different size particles.
Note that the greatest distinction will occur between two molecules, one of which has found a relatively straight shot through the column, the other of which has encountered a lot of particles and so has to travel around them. Also note that the distance a molecule needs to travel around a larger particle is larger. Now you might be inclined to say that if we have to travel around lots of small particles, wouldn’t that eventually add up to the larger distance of travelling around one larger particle? Turns out that it doesn’t! The conclusion is that smaller particles will reduce the contribution of eddy diffusion to peak broadening (although we will add a proviso in just a bit). In other words, the distribution of path lengths of a set of molecules travelling through a packed bed is more uniform for a bed containing smaller particles than it is for a bed made up of larger particles. Thus, regarding eddy diffusion, there is a theoretical advantage to using smaller particles. That is the reason we use the reduced plate height (h = H/dp) instead of the plate height for an overall assessment of column efficiency. If we had two people pack a column with the idea of seeing who packed a better one, if one person used smaller particles they would have a competitive advantage if we only compared the plate height (H).
It is also worth realizing that the flow profile shown in Figure 24 for a molecule would not occur (okay, there’s probably some infinitesimally small possibility that this could occur, but it’s so small that we could ignore it) in a chromatographic column. There is a physical flow pushing the material through the column and when a molecule reaches a flowing area, it will generally be swept downstream.
Figure 24. Unrealistic representation of the flow profile for a molecule through a packed chromatographic column.
Some packed columns exhibit channeling, and channeling leads to a significant and undesirable amount of eddy diffusion. The illustration in Figure 25 shows a column with a channel and compares two flow paths, one of which goes through the channel.
Figure 25. Representation of a packed chromatographic column with a channel.
Channels provide a straight path through a portion of the column and molecules in a channel avoid any need to move around any particles. A molecule moving through a channel will have a much more streamlined or shorter path than a molecule in an adjacent part of the column that has to move through the packed bed. Channels occur when the particles stick together in some way and separate from each other, instead of nesting together in a packing arrangement with every particle closest together. A liquid chromatographic column that has dried out (all of the mobile phase is allowed to evaporate) will likely develop channels during the drying process that will never close back up if it is rewetted. Channels in chromatographic columns are undesirable and introduce a lot of broadening into the system.
Another key thing to ask then is whether channeling is more likely to occur with smaller or larger particles. Channeling will occur if a column is poorly packed. There are very specific procedures that have been developed for packing gas and liquid chromatographic columns that are designed to minimize the chance that channeling will occur. One key to packing a good column is to slowly lay down a bed of particles so that they nest into each other as well as possible. In packing a gas chromatographic column, this will usually involve slowly adding the particles to the column while vibrating it so that the particles settle in together. Liquid chromatographic columns are usually packed slowly under high pressure. A column is packed efficiently when the particles are in a uniform bed with the minimum amount of voids. Given a particular particle size, the goal is to fit as many of them as possible into the column. We can then ask which is more difficult to pack efficiently, larger or smaller particles.
One way to think about this is if you were asked to fill a large box (say a refrigerator box) with basketballs or tennis balls. The goal is to fit as many of either one in as possible. You could imagine readily taking the time to carefully lay down each basketball into the refrigerator box, layer by layer, and fitting in as many basketballs as possible. You might also be able to imagine that you would start slowly with the tennis balls, laying in one in a time, and quickly lose patience at how long this would take to fill the entire box. If you then sped up, say by slowing dumping in balls from a pail while a helper shook the box, you would probably create more voids in the box. Also, because the interstitial volumes between the tennis balls will be smaller than that with the basketballs, any channels become more significant. The result is that it is more difficult to avoid the formation of channels with smaller particles. Recapping, smaller particles have a theoretical advantage over larger particles, but more care must be exercised when packing smaller particles if this theoretical advantage in column efficiency is to be realized.
Do open tubular capillary columns exhibit eddy diffusion? Capillary columns do not have packing material. Instead, they are long, narrow diameter tubes that have a coating of a liquid stationary phase on the internal walls of the column. A representation of a capillary column with a uniform coating of a liquid on the walls is shown in Figure 26.
Figure 26. Representation of a coated capillary column.
Because there is no packing material to move around, there would not be any eddy diffusion in such a column. The absence of eddy diffusion broadening is one advantage that capillary columns have when compared to packed columns.
Finally, we could ask whether eddy diffusion exhibits any dependence on the flow rate. This actually turns out to be a difficult question to answer with conflicting opinions and data about whether there is a flow dependence, and if so, exactly what the dependency is. If we go back to van Deemter’s initial development of peak broadening in chromatography in 1956, we would see that van Deemter believed that the contribution of eddy diffusion to peak broadening did not depend in any way on the flow rate. This is a reasonable argument if we thought that we could draw a variety of different flow paths through a packed bed, and the difference in length between the shortest and longest flow path would be fixed irrespective of how fast the molecules were moving through the path.
But let’s return to our river analogy to see how the flow rate might get involved in this. Suppose the river had a relatively fast flow rate, such that different people in different inner tubes got locked into particular flow channels and stayed in those all the way down the river. Under that situation, each path would have a preset length and even if we slowed down the flow, so long as you were locked into a particular path, the difference in distance would be invariant. But suppose now we slowed down the flow so much that there were opportunities to drift around at points and sample a variety of flow paths down the river. This might lead to some averaging and the amount of sampling of different flow paths would be greater the slower the flow. This same thing can happen in a chromatographic column and leads to conflicting data and opinions about the nature of any flow rate dependence on eddy diffusion. Also, the exact point at which there is a crossover between a rapid flow that locks in a flow path versus a slow one that allows each molecule to sample many different flow paths is impossible to determine. This point is still not fully resolved, and literature on band broadening show different terms for eddy diffusion. We usually denote eddy diffusion by the term A. If you look at van Deemter’s initial treatment of peak broadening, you would see (note, van Deemter also did not use the reduced plate height, but we could write h = A as well):
H = A
(where A = 2λdp)
Note how the particle diameter (dp) is included in this equation for A, so that the smaller the particle diameter, the smaller the contribution of eddy diffusion to the reduced plate height.
Another common conclusion today is that the contribution of eddy diffusion broadening does exhibit a slight dependence on flow rate. The usual form of this is that the dependence is v1/3, and you might often see books or articles that include the following term in the overall band broadening equation.
h = A v1/3
Still other people throw up their hands at all the confusion regarding eddy diffusion and show overall band broadening equations that do not have any A-term in them. We will develop another broadening term that has to do with processes going on in the mobile phase (and note that eddy diffusion only involves the mobile phase - none of what we talked about even requires the presence of a liquid stationary phase), so some people lump the eddy diffusion term into this other mobile phase term. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Separation_Science/02_Text/03_Broadening_of_Chromatographic_Peaks/02_Eddy_Diffusion_%28Multipath%29_Broadening.txt |
Consider a compound that has distributed between the mobile and stationary phase within a plate in a chromatographic column. Figure 27 might represent the concentration distribution profiles in the two phases (note that the compound, as depicted, has a slight preference for the mobile phase).
Figure 27. Representation of the concentration profiles for a compound distributed between the stationary (left) and mobile (right) phases of a chromatographic column. Note that the compound has a preference for the mobile phase.
What we want to do is consider what would happen to these two concentration profiles a brief instant of time later. Since the mobile phase is mobile and the solute molecules in it are moving, we could anticipate that the profile for the mobile phase would move ahead a small amount. The figure below illustrates this.
Figure 28. Representation of the two concentrations profiles in Figure 27 a brief instant of time later. Note that the mobile phase profile has moved ahead of the stationary phase profile.
What about the concentration profile for the solute molecules in the stationary phase? Consider the picture in Figure 29 for two solute molecules dissolved in the stationary phase of a capillary column and let’s assume that these are at the trailing edge of the stationary phase distribution.
Figure 29. Two molecules dissolved in the liquid stationary phase of a capillary column.
What we observe is that the molecule labeled 1 is right at the interface between the stationary and mobile phase and provided it is diffusing in the right direction, it can transfer out into the mobile phase and move along. The molecule labeled 2, however, is “trapped” in the stationary phase. It cannot get out into the mobile phase until it first diffuses up to the interface. We refer to this process as mass transport. The solute molecules in the stationary phase must be transported up to the interface before they can switch phases. What we observe is that the solute molecules must spend a finite amount of time in the stationary phase. Since the mobile phase solute molecules are moving away, molecules stuck in the stationary phase lag behind and introduce a degree of broadening.
If we then consider the leading edge of the mobile phase distribution, we would observe that the molecules are encountering fresh stationary phase with no dissolved solute molecules and so these start to diffuse into the stationary phase when they encounter the surface. We can illustrate this in Figure 30 with arrows showing the direction of migration of solute molecules out of the stationary phase at the trailing edge and into the stationary phase at the leading edge.
Figure 30. Representation of the movement of analyte molecules at the leading and trailing edge of the concentration distribution.
Hopefully it is obvious from Figure 30 that the finite time required for the molecules to move out of the stationary phase leads to an overall broadening of the concentration distribution and overall broadening of the peak.
A critical question to ask is whether the contribution of stationary phase mass transport broadening exhibits a dependence on the flow rate. Suppose we go back to the small amount of time in the first part above, but now double the flow rate. Comparing the first situation (solid line in the figure above) to that with double the flow rate (dashed line) leads to the two profiles. Hopefully it is apparent that the higher flow rate leads to a greater discrepancy between the mobile and stationary phase concentration distributions, which would lead to more broadening.
The term used to express mass transport broadening in the stationary phase is CS (not to be confused with the CS that we have been using earlier to denote the concentration of solute in the stationary phase). If we wanted to incorporate this into our overall band broadening equation, recognizing that higher flow rates lead to more mass transport broadening and reduced column efficiency (higher values of h), it would take the following form:
h = CS v
Something we might ask is whether the flow rate dependency of the stationary phase mass transport term has any troublesome aspects. An alternative way to phrase this question is to ask whether we would like to use slow or fast flow rates when performing chromatographic separations. The advantage of fast flow rates is that the chromatographic separation will take place in a shorter time. Since “time is money”, shorter analysis times are preferred (unless you like to read long novels, and so prefer to inject a sample and then have an hour of reading time while the compounds wend their way through the column). If you work for Wenzel Analytical, we’re going to try to perform analyses as fast as possible and maximize our throughput. The problem with speeding up the flow rate too high is that we begin to introduce large amounts of stationary phase mass transport broadening. The shortening of the analysis time begins to be offset by broad peaks that are not fully separated. Ultimately, stationary phase mass transport broadening forces us to make a compromise between adequate efficiency and analysis time. You cannot optimize both at the same time.
Another thing we need to think about is what effect the thickness of the stationary phase has on the magnitude of stationary phase mass transport broadening. The pictures in Figure 31 for one wall of a coated capillary column serve to illustrate this point.
Figure 31. Representation for one wall of a coated capillary column with a thicker (left) and thinner (right) stationary phase coating.
Remember that the key point is that solute molecules spend a finite amount of time in the stationary phase, and since solute molecules in the mobile phase are moving away, the longer this finite time the worse. Therefore the thicker the phase, the more broadening will occur from stationary phase mass transport processes. This says that the ideal stationary phase coating ought to be microscopically thin, so that molecules rapidly diffuse into and out of the stationary phase, thereby reducing how far ahead mobile phase molecules can move in this finite amount of time. We should also be able to realize that the optimal coated phase ought to have a uniform thickness. If we have thin and thick regions as shown in Figure 32, we see that the time spent in the stationary phase by solute molecules will vary considerably, an undesirable situation. Compound 1 will likely spend less time in the stationary phase than compound 2.
Figure 32. Representation of one wall of a coated capillary column with non-uniform thickness of the coating.
While microscopically thin coatings reduce stationary phase mass transport broadening, there are two problems with microscopically thin coated phases. If we consider the picture in Figure 31, where we have two capillary columns with different thickness coatings, the capillary column with the thinner coating will have much lower capacity than the one with the thicker coating. This means that there is much less weight of stationary phase over a theoretical plate for the column with the thinner coating and much less analyte dissolves into the stationary phase. Increasing the thickness of the coating or capacity of the column has several advantages. One is that it helps in the separation of the mixture (something we will learn more about later in the course). The other is that it is easy to overload or saturate a column that has a very low capacity.
This raises the question of whether you could design a column that has a thin stationary phase coating but high capacity. For coated capillary columns, this is not possible. A thinner coating in a capillary column means less capacity. For a coated packed column; however, it is possible to retain a high capacity while thinning the coating. Accomplishing this involves using smaller particles but the same weight of coating. Imagine taking large solid support particles and crushing them into a bunch of smaller particles. What you should realize is that the smaller particles have a much larger surface area. If we then coated the same amount of stationary phase (e.g., 5% by weight) relative to the weight of solid support, because of the larger surface area a thinner coating results. What we see is that the use of smaller particles has a theoretical advantage over the use of larger particles for coated stationary phases.
Another problem with coating microscopically thin phases is the risk of leaving some of the surface of the underlying solid support uncoated. These exposed solid surfaces (Figure 33) provide highly active sites for adsorption of solute molecules, and we have already seen how adsorption is an inefficient process that leads to peak tailing. While thin coatings have an advantage, great care must be taken in coating these phases to insure a complete coverage with uniform thickness of the surface.
Figure 33. Representation of the wall of a coated capillary column showing an exposed solid surface.
Open tubular capillary columns are common in gas chromatography because it is possible to coat their inside walls with an exceptionally thin, uniform stationary phase. Today, it is also possible to chemically bond the liquid phase onto the interior surface of the capillary column. In the early days of capillary gas chromatographic columns, these were made of glass tubing that was approximately the same diameter as a melting point capillary that you are familiar with from organic chemistry. These columns were stretched from thick-walled glass tubes that were heated in an oven. As the capillary tube was stretched out, it was coiled in a coiling oven. It was common to use 30-meter lengths, essentially a 30-meter long glass slinky. The most common process for coating a capillary column involves what is known as the “moving plug” technique. As illustrated in Figure 34, the liquid stationary phase is dissolved in a plug of solvent that is pushed through the column using pressure from an inert gas. As the plug moves, it coats a very thin layer of liquid onto the inside walls of the column. As the solvent evaporates, a very thin layer of liquid stationary phase remains on the walls of the column. A systematic process is used to treat the inside walls of the column prior to coating to ensure that the liquid wets the surface well and is deposited uniformly over all of the interior surfaces.
Figure 34. Moving plug technique for coating a capillary column.
One problem with these glass columns was their fragility. Many frustrated workers broke the columns trying to mount them into a gas chromatograph with leak-proof fittings. The fittings used with these glass columns are usually made of graphite, a soft substance that often can be molded around the tube without breaking it (but if you’re not careful, it’s easy to break it). Another problem is caused by the chemical nature of glass. We think of glass as a silicate material (SiO2), but it actually turns out that most silicate glasses contain other metal ions as constituents (aluminum, magnesium, calcium, and iron oxides are some of the other metals present). In some glasses, these other metals can be as much as 50% of the glass. These metals are positively charged centers, and if some of the surfaces are not coated by the liquid phase (an inevitable occurrence), these metal ions provide active sites for adsorption that cause tailing of compounds (especially oxygen- and nitrogen-containing compounds that have dipoles).
The capillary columns used in gas chromatography today are known as fused silica columns. Fused silica is pure silicon dioxide (SiO2) and lacks the metal ions in regular glass. The surface of fused silica is considerably less active than the surface of regular glass. Fused silica is widely used in the production of devices known as fiber optics. Fiber optics are thin, solid glass fibers. Light is shined into one end of the fiber at an angle that causes complete internal reflection of the light as shown in Figure 35.
Figure 35. Complete internal reflection of a light beam inside a fiber optic.
Light going in one end (e.g., New York City) exits out the other end (e.g., Los Angeles, CA). The light can be pulsed (sent in small bursts) and the speed of light allows for very rapid communication. It turns out to be easy to make fiber optic-like devices with a hole in the center (in fact, it took people a while to learn how to make glass fibers without the hole since the fibers tend to cool from the outside to the inside, leading to contraction and a hole in the center). The hole in the center of fused silica capillary columns is so small you cannot see it with the naked eye (we will see later that this very small opening has advantages in chromatographic applications). Also, these capillary columns are incredibly flexible. They can be tied into knots, and more importantly for chromatographic applications, leak-tight fittings can be attached without breaking the columns. The deactivated surface, flexible nature making them easy to install and use, and chromatographic efficiency (partly because of the deactivated surface, partly because of how well they can be coated with thin phases, and partly because of the small diameter) make them the column of choice for most gas chromatographic applications today. The only drawback to these columns is that they have very small capacities. Gas chromatographs built to use fused silica capillary columns usually have what are known as split injection systems. A typical injection size for a gas chromatographic sample is 1 µL. Even this amount is too much for a fused silica capillary column, but reproducibly injecting smaller volumes is very difficult. Instead, the flow from the injector is split, and only a small part (often 1 in 50 to 1 in 100) is actually sent into the capillary column. The rest is vented away and never enters the column.
Another thing we need to examine under the topic of stationary phase mass transport broadening is the nature of the stationary phase used today in liquid chromatography. In 1963, Calvin Giddings published a significant paper titled “Liquid Chromatography with Operating Conditions Analogous to Those of Gas Chromatography” (Analytical Chemistry, 1963, 25, 2215). At this point in time, gas chromatography was the method of choice when performing analyses because existing gas chromatographic columns were far more efficient than the liquid chromatographic columns that were available. In fact, gas chromatography was so preferable to liquid chromatography that many publications of the 1960s and 1970s described ways of preparing volatile derivatives of non-volatile compounds so that they could be analyzed by gas chromatography. The extra steps involved in the derivatization were worthwhile since liquid chromatographic methods were not good enough to separate most compounds in complex mixtures.
What Giddings did in this paper was show that it was theoretically possible to perform liquid chromatography at the efficiency of gas chromatography. The key feature, which we have not yet fully developed because we have one more contribution to broadening to examine, was to use exceptionally small particles with exceptionally thin coatings in liquid chromatographic columns. Of course, these exceptionally thin coatings cause several problems. One is how to coat them so that none of the surface of the solid support is exposed. If you can coat them uniformly, another problem in a liquid chromatographic system is that there might be locations in the column where the flowing liquid mobile phase can physically wash away regions of the coated phase thereby exposing the underlying solid surface. Finally, there is no such thing as two liquids that are completely immiscible in each other. It’s true that oil and water don’t mix, but it’s also true that a little bit of oil will dissolve in water. Over time, the coated stationary phase will gradually dissolve away in the mobile phase, eventually exposing the underlying solid surface. Even though Giddings showed in 1963 how to perform liquid chromatography at the efficiency of gas chromatography, the phases that were needed could not be coated in a way that it was a widely practical method.
It was not until the late 1970s, when bonded liquid chromatographic stationary phases were introduced, that it was finally practical to do liquid chromatography with the same efficiency as gas chromatography. Bonded phases relied on silica gel, something we already encountered when we first talked about solid phase, adsorption chromatography. Silica gel is an excellent support for liquid chromatographic phases. It is physically robust, stable at pH values from about 2 to 8, and can be synthesized in a range of particle sizes including the very small particle sizes needed for liquid chromatography. The chemistry used in preparing bonded phases is shown in Figure 36. Basically this involves the surface derivatization of the silanol groups on the surface of the silica gel with chlorodimethyloctadecylsilane. The C18 or octadecylsilane (ODS) phase shown in the scheme are the most common ones used. The pyridine is added to remove the HCl produced in the reaction.
Figure 36. Scheme for bonding C18 groups to the surface of silica gel.
Other common bonded phases use C8 or C1 surface groups. You can also purchase bonded phases with phenyl groups, C3H6NH2 (aminopropyl) groups, and C3H6CN groups. But the C18 phases are so common, and so versatile, that we will examine these in more detail.
The first thing to realize is that attaching the C18 groups converts a highly polar surface (the silanol groups can form hydrogen bonds) into a non-polar surface. In effect, this is like attaching a one-molecule thick, oily skin to the surface of silica gel. There is some question about the exact conformation of the attached C18 groups. Figure 37 shows two forms, one with the C18 groups extended, the other with the C18 groups collapsed. There is substantial evidence to support the idea that if the non-polar C18 phase is in contact with a mobile phase that is very polar (e.g., water), that the C18 groups are in the collapsed form. Putting the C18 phase in contact with a less polar mobile phase (e.g, methanol) leads to more extension of the C18 groups.
Extended Form Collapsed Form
Figure 37. Extended and collapsed forms of C18 groups bonded to the surface of silica gel.
If we examine the extended form of the C18 phase, it is interesting to consider the nature of this material at the outer regions. Even though the C18 groups are attached to a solid, are they long enough such that the outer edges are “fluid” enough to behave more like a liquid than a solid? If so, then molecules might distribute into these phases by a partitioning mechanism. If not, then molecules adsorb to the surface. We previously learned that adsorption was undesirable when compared to partitioning, but that was adsorption on a polar surface. The non-polar, deactivated nature of the C18 surface does not provide strong adsorption sites so peak tailing is quite minimal with C18 bonded phases.
If you look at most liquid chromatograms, though, you will see that the peaks tail more than in gas chromatography. The likely reason for this is that some of the surface silanol groups remain underivatized. Crowding of the C18 groups during the derivatization process makes it unlikely that all of the silanols can be reached and deactivated. In an effort to minimize the number of unreacted silanol groups, some commercially available C18 phases are end-capped. End capping involves exhaustively reacting the C18 phase with a much smaller silane such as chlorotrimethylsilane (ClSi(CH3)3).
Because of the covalent bonds, the C18 groups cannot wash or dissolve off of the solid support. With regards to mass transport in the stationary phase, we said that we want the stationary phase as thin as possible. C18 bonded phases are essentially one-molecule thick. We could never coat a one-molecule thick chromatographic phase that was uniform and had no exposed solid surfaces. Therefore, we could never really make a liquid chromatographic phase with faster stationary phase mass transport properties than the bonded silica phases.
Until recently, the common particle sizes to find in commercially available C18 columns had diameters of 3, 5, and 10 µm. Columns using particles of these sizes are commonly referred to as high performance liquid chromatography (HPLC). Recently columns with 1.7 or 1.8 um particles have become commercially available and there use is referred to as ultra-high pressure liquid chromatography (UPLC). Smaller particles will require higher pressures to force a liquid through the column and UPLC columns can run at pressures up to 15,000 psi. In gas chromatography, we observe that smaller particles usually have thinner coatings because they have more surface area. Note that for these bonded phases the thickness of the stationary phase with the 1.7, 3, 5, and 10 µm particles is always identical. What does change is the surface area. A column packed with 3 µm particles will have more surface area than a column of the same length packed with 5 or 10 µm particles. This increased surface area leads to an increase in the column’s capacity. If the columns are the same length, the increased capacity will lead to increased separation of compounds and lengthen the retention times. The usual approach is to use shorter columns with the smaller particles. The shorter column reduces the analysis time because there is less volume of mobile phase in the column. Typically C18 columns packed with 10 µm particles are 25 cm long; 5 µm particles are used in columns that are 15 cm long; and 3 µm particles are used in columns that are 3 cm long. These 3×3 columns can have very short analysis times. A negative aspect is that they are more susceptible to fouling by contaminants in the sample. Also, a difficulty with ultra-small particles involves packing the columns so that there are no channels. Because of the small particles, liquid chromatographic columns require highly specialized packing procedures. Finally, the particles need to have uniform sizes (the smaller the particle, the smaller the tolerance on the range of particle sizes that can be used). A mixture of particles of different sizes is known as an aggregate. Concrete is an excellent example of an aggregate. The problem with an aggregate is that the smaller particles fill in the voids between the larger particles, making it very difficult to get a liquid to flow through the bed.
One last thing we can consider is whether stationary phase mass transport broadening is more significant in gas or liquid chromatography. We have to be careful here, because it’s a somewhat subtle distinction. What we need to consider is the diffusion rate of the solute in the stationary phase. In gas chromatography, we have a gaseous molecule dissolved in a liquid coating. The important thing is that it is a dense liquid with a molecule dissolved in it. The diffusion rate is therefore rather comparable in this stationary phase to that of a liquid. Of course, a liquid chromatographic column is usually at room temperature whereas a gas chromatographic column is usually at some elevated temperature that is often higher than 100oC. We know that diffusion rates are temperature dependent, so that the rate ought to be faster in gas chromatography, and mass transport broadening in the stationary phase ought to be more significant in liquid chromatography, assuming everything else is equal. Just realize that this is not a 100,000-fold difference, because the stationary phase in gas chromatography is not a gas, but a heated liquid. It is also possible to operate liquid chromatographic columns at higher temperatures to speed up the rate of diffusion and improve the mass-transport broadening. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Separation_Science/02_Text/03_Broadening_of_Chromatographic_Peaks/03_Stationary_Phase_Mass_Transport_Broadenin.txt |
It’s going to turn out that a similar mass transfer effect occurs within the mobile phase. The thing to consider here is that we want solute molecules to encounter the surface of the stationary phase as quickly as possible so that they can immediately undergo a distribution between the two phases.
For example, if we reconsider the picture for mass transport broadening in the stationary phase (Figure 38), we realize that we want the molecules at the leading edge of the mobile phase profile to quickly encounter and enter the stationary phase, otherwise they will move further ahead and broaden the distribution.
Figure 38. Representation of the movement of analyte molecules at the leading and trailing edge of the concentration distribution.
Another way to examine mobile phase mass transport broadening is to consider a capillary column as shown below.
Figure 39. Capillary column showing a molecule (black dot) that has just left the stationary phase.
The dot represents a molecule that has just left the stationary phase and is about to diffuse across the mobile phase and re-encounter stationary phase on the other side of the column. It is helpful to draw a line representing the path of the molecule, and then draw a second line for the path of the molecule if the flow rate were doubled. We could also imagine a situation in which the flow was so fast that the molecule never re-encountered the stationary phase. This would be a problem since it’s important for the molecule to encounter the stationary phase if we are to ever have a distribution occurring that leads to a separation of two compounds.
Figure 40. Representation of the movement of a molecule as it diffuses through the mobile phase. The longer the arrow, the higher the flow rate.
If we are using capillary columns in a chromatographic system, what does this observation above suggest about the desirable diameter for such a column? Hopefully it would be apparent that a much smaller column diameter would lead to much faster encounters with the stationary phase. So we want capillary columns of very small diameter. That is one of the reasons why the fused silica columns used today in gas chromatography are so efficient. We can not even see the opening in these columns with the naked eye. They really approach the limit of how narrow we are able to make chromatographic capillary columns.
Next we need to ask whether this effect occurs in a packed column. If we think about a packed column, we should realize that there are voids, or interstitial volume, between the particles that make up the packing. Solute molecules need to diffuse across these voids to encounter the stationary phase. The larger the void, the more time it takes to diffuse across, and the more significant the mobile phase mass transport broadening. How can we reduce the voids? If we use smaller particles, they will pack closer together and reduce the interstitial volume between the particles. Once again, we see how small particles offer a theoretical advantage when compared to large particles. Note that this has happened in every instance when particle size made a difference. Using a packing material with smaller particles is better, provided we pack a good column with minimal to no channeling.
Figure 41. Representation of a packed column showing the interstitial regions.
Another critical question to consider is whether mobile phase mass transport broadening is more significant in gas or liquid chromatography. To answer this, we need to realize that we want the solute compound to diffuse across regions of mobile phase as fast as possible. If we remember that gases diffuse about 100,000 times faster than liquids, we realize that the impact of mobile phase mass transport broadening is significantly lower for gases than it is for liquids. In fact, mobile phase mass transport broadening is the most important distinction between gas and liquid chromatography. Gases diffuse far more quickly across regions of mobile phase than do liquids. That means in liquid chromatography we need far smaller mobile phase voids to reduce the magnitude of this term. When Giddings published his important paper in 1963 on doing liquid chromatography with the efficiency of gas chromatography, this was a critical realization put forward in this paper.
What does this mean for practical applications of liquid chromatography? If we were to try to perform capillary liquid chromatography, it would mean that we ought to have columns with much smaller internal diameters than we use for gas chromatography. But gas chromatography already uses fused silica columns, which essentially reach the limit of practical internal diameters (remember, if we reduced this diameter even smaller, we introduce a problem of not having enough sample capacity in the column and would have a very difficult time actually introducing sample into the column), so capillary liquid chromatography is not a practical method and does not offer the same advantages of capillary gas chromatography. There are researchers who investigate aspects of capillary liquid chromatography, but it is not a commercial method and is not used widely by practitioners in the field. As far as packed column liquid chromatography, we have already seen how we end up using very small particles (1.7, 3, 5, and 10 µm). The real value of these small particles is that they reduce the interstitial volume of mobile phase between the particles, thereby reducing the time required for mobile phase mass transport. Gas chromatography can be done with much larger particles because of the much faster mobile phase mass transport.
The last thing we could consider about mobile phase mass transport broadening is whether its overall contribution to band broadening depends on flow rate. If we go back to our initial picture of the capillary column, we know that we want the solute molecules to encounter stationary phase as often as possible. The faster the flow, the fewer encounters, so a faster flow makes this contribution worse (this is analogous to the leading edge of the mobile phase distribution moving ahead more per unit time, thereby causing more broadening of the overall distribution). If we needed to write a term to go into our equation for h, and we represent mobile phase mass transport broadening as CM (be careful not to confuse this with CM the concentration of solute in the mobile phase that we have used earlier), the term would take the following form:
h = CM v
Notice how the relationship is analogous to what we saw with mass transport broadening in the stationary phase. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Separation_Science/02_Text/03_Broadening_of_Chromatographic_Peaks/04_Mobile_Phase_Mass_Transport_Broadening.txt |
Concluding Comments – Peak broadening in Chromatographic Systems
Can you now draw a generalized plot of h versus flow rate?
We have now developed all four contributions to peak broadening in chromatographic systems, compared their significance in gas and liquid chromatography, and seen how current state-of-the-art columns have been designed to minimize band broadening effects. We can now pull together all four of the contributions into one equation that reflects overall contributions to peak broadening. It is shown below, using one common way of expressing the eddy diffusion term:
h = Av1/3 + B/v + CS v + CM v
It is worthwhile examining the overall form of this expression. First note how the longitudinal diffusion term (B/v) will become very large at very low flow rates. Note how the mass transport terms (CS v and CM v) will become very large at very high flow rates. This implies that this equation will have a minimum at some intermediate flow rate. A generalized plot of this equation for a gas chromatographic packed column is shown below. The general contributions of the longitudinal diffusion and mass transport terms to the equation are also represented. The optimal flow rate for a gas chromatographic packed column is around 15 to 20 mL/minute. A lot of times people actually run these at higher flow rates of about 30 mL/minute. The reason for this is to shorten the analysis time. Some column efficiency is sacrificed, but the entire separation process is a balance between chromatographic efficiency and analysis time, where there is a desirability to shorten this if possible.
Figure 42. Generalized van Deemter plot for a packed gas chromatographic column.
A similar plot is also shown below for liquid chromatography. Notice in this case that extremely low flow rates must be used before longitudinal diffusion becomes a significant contribution to broadening. This means that the optimal flow rate for liquid chromatographic separations is very slow, which is undesirable because of the long analysis times this would create. Instead, a common flow rate for a packed C18 liquid chromatographic column is in the 1.0 to 1.5 mL/minute range.
Figure 43. Generalized van Deemter plot for a packed liquid chromatographic column.
For capillary GC columns, which are more commonly used today than packed GC columns, instead of setting a column flow rate you often set the column head pressure to a value recommended by the instrument manufacturer. By adjusting the pressure of the carrier gas (usually helium) at the beginning of the column, this insures an adequate flow of gas through the column. The desired head pressure depends on the length and internal diameter of the capillary column. Longer or narrower columns will require higher pressures to maintain adequate flow. One observation is that the pressure of the gas drops from the beginning to the end of the column, such that the flow rate changes over the column as well. Instead of measuring a flow rate for a fused silica capillary column in mL/min, it is more common to measure the carrier gas velocity (u) in cm/sec. Since this value changes as well over the column, an average value is obtained by dividing the column length in cm by the retention time in seconds of an unretained compound.
$\mathrm{u\: (cm/sec) = \dfrac{column\: length\: (cm)}{RT\: of\: unretained\: peak\: (sec)}}$
A plot of the plate height versus carrier gas velocity for a column that is 25 m long and has a 0.25 mm internal diameter is shown below. The characteristic observation of an optimum velocity as which to minimize peak broadening is observed.
Figure 44. van Deemter plots for a capillary gas chromatographic column.
Why is the plot different for nitrogen versus helium as the carrier gas? The key difference between nitrogen and helium is the viscosity of the two gases. Nitrogen, as a heavier molecule, will have a higher viscosity than helium. The higher viscosity of nitrogen slows down the diffusion rates and increases the contribution of mobile phase mass transport broadening. Using helium as the carrier gas leads to improved efficiency because of the reduction in mobile phase mass transport broadening relative to nitrogen. Also, the uopt for helium is at a higher flow rate meaning that the analysis time can be shortened. This plot shows that hydrogen would actually be a better carrier gas than helium. Hydrogen is rarely used as a carrier gas because of its explosion risk. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Separation_Science/02_Text/03_Broadening_of_Chromatographic_Peaks/05_Concluding_Comments.txt |
Another important equation in chromatographic separations is the fundamental resolution equation. The equation relates the resolution of two compounds (RS) to the number of theoretical plates (N), retention or capacity factor (k'), and selectivity factor (α). Appendix I shows the derivation of the fundamental resolution equation. Since we would usually be interested in separating two compounds that are close to each other in a chromatogram, and therefore possibly not resolved, we usually think of applying this equation when we want to improve the resolution of two adjacent components.
$\mathrm{R_S = \left(\dfrac{\sqrt{N}}{4} \right)\left(\dfrac{\alpha - 1}{\alpha} \right)\left(\dfrac{k_2'}{1+k_2'} \right)}$
The interesting part about this equation is that it is possible to examine a chromatogram with poor resolution, decide which of the three terms is causing the most serious problem, and then make systematic experimental changes to improve the results. It is worth examining the types of experimental changes that can be made to influence each of these terms. Notice that in each case, making the term larger will improve the resolution.
04 Fundamental Resolution Equation
One obvious way to increase the number of plates is to increase the length of the column. Doubling the length doubles the number of theoretical plates. One cautionary note about this is to consider the square root dependency on the number of plates in the equation. Doubling the column length will double the analysis time, but will not double the resolution. We would need a column four times longer to achieve a doubling of the resolution. Increasing the number of plates will always give better resolution, but there are diminishing returns when the increased analysis time is considered.
Another thing to examine is factors that influence the band broadening equation. That equation related broadening to h, and the smaller the value of h, the more plates in a column. One thing we could do is optimize the flow rate of the system. We mentioned how it was common to use a flow rate higher than the optimal one to shorten the analysis time. Slowing down the flow rate may provide enough gains in efficiency to separate two compounds that are not fully resolved.
Another change to make is to use smaller particles. In gas chromatography, this will lead to a thinner coating, improving mass transport broadening in the stationary phase. In both gas chromatography and liquid chromatography, smaller particles will reduce the interstitial volume and reduce mobile phase mass transport broadening. Smaller particles, provided they are packed well, also will reduce broadening contributions from eddy diffusion.
02 Selectivity Factor
The first thing to notice about this term is that there are diminishing returns to resolution as it is made excessively large. Table 1 provides values of α and (α – 1)/ α.
Table 1. Values of α and (α – 1)/ α.
α (α – 1)/α
1 0
2 \(\dfrac{1}{2}\)
3 \(\dfrac{2}{3}\)
4 \(\dfrac{3}{4}\)
5 \(\dfrac{4}{5}\)
6 \(\dfrac{5}{6}\)
Notice that the (α – 1)/α term approaches a value of 1 at high values of α. Optimal α-values for most separations are between 2 and 5.
Remember that the α-value refers to the ratio of the two partition coefficients for the two components. If we wanted to make a substantive change in α, we need to change the partition coefficient of one component but not the other (or we need to change both appreciably but differently to impact the ratio). There is really only one way to do this and it involves changing the identity of the stationary phase. This would involve using a different liquid phase in gas chromatography. As we will see, there are many available liquid phases for use in gas chromatography, so changing the nature of this phase and hence the α-values is a common procedure.
In liquid chromatography, this would mean switching from a C18 to some other bonded phase. Since the range of bonded phases is more limited and changes of questionable utility in many instances, this is not done that frequently. One thing you might wonder about is whether changing the nature of the mobile phase in liquid chromatography would be viewed as constituting a change in α. This is easily done and is the most common way in liquid chromatography to improve the resolution of two substances that are not fully resolved. It turns out that changes to the mobile phase are generally regarded as changes to the retention or capacity factor. Usually this might involve altering some aspect of the mobile phase so that all the compounds exhibit a higher capacity, stay on the column longer, and exhibit better resolution. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Separation_Science/02_Text/04_Fundamental_Resolution_Equation/01_N__Number_of_Theoretical_Plates.txt |
Similar to the α-value term in the fundamental resolution equation, the term with k' also approaches 1 at higher values of k', as seen by the data provided in Table 2. Therefore we can say that optimal k' values are also between 2 and 5. Also, larger k' values will always lead to longer retention and analysis times.
Table 2. Values of k2’ and k2'/(1 + k2').
k2' k2'/(1 + k2')
0 0
1 $\dfrac{1}{2}$
2 $\dfrac{2}{3}$
3 $\dfrac{3}{4}$
4 $\dfrac{4}{5}$
5 $\dfrac{5}{6}$
Remember that we defined k' earlier in the course as:
$\mathrm{k_2' = K_2\left(\dfrac{V_S}{V_M} \right )}$
So anything that alters any of the terms of this equation will lead to a change in k'. As we already mentioned above, the most common way to change k' in liquid chromatography is to alter the constituents of the mobile phase. This will change the distribution constant (K2), and if the correct change is made, make K2 larger and k' larger. The analog to this in gas chromatography is to change the temperature. Recall that:
$\mathrm{K_2 = \dfrac{C_S}{C_M}}$
We could consider what happens to this ratio as the temperature is changed. The stationary phase is a liquid. The mobile phase is a gas. Hotter temperatures reduce the solubility of a gas in a liquid (warm soda will go flatter faster). Thermal pollution refers to the reduction in the concentration of dissolved oxygen gas when warm water from a power plant or industrial cooling process is added to a river or lake. So hotter temperatures will reduce CS, reduce K2 and reduce k'. This would actually make the resolution worse. So by cooling down a gas chromatographic column, we increase the retention factor and improve the resolution. Of course, this also lengthens the analysis time, which may be undesirable.
If we examine the volume terms, we realize that it is difficult to change VM, the volume of the mobile phase. VM is the interstitial volume and is essentially fixed for a particular particle size. There is no way to crunch the particles closer together to reduce VM. In gas chromatography, it is easy to alter VS, the volume of the stationary phase. This would involve coating a thicker loading (e.g., instead of 3% liquid coating, raise it to 5%). Changing VS in liquid chromatography really is not an option since we have bonded C18 groups and do not coat different thickness phases. If we change the particle size from 5 µm to 3 µm, and keep everything else, we do create more capacity in the column (because the 3 µm particles have more surface area), but we create more plates as well.
Consider the chromatogram in Figure 45. We observe two problems with this separation. The first few compounds come out of the column very quickly and are not fully resolved. The latter compounds are fully resolved but stay in the column too long and are very broad.
Figure 45. Chromatogram with non-optimal k’ values for the earlier and later eluting compounds.
What problem exists in this separation? If we were to enlarge N to improve the resolution of the first few compounds, we would lengthen the retention time of the latter ones as well, which would be unacceptable. If we changed α, it might resolve some of the overlapped constituents but might cause some that are already resolved to now overlap with each other. We have a capacity problem (k') in this chromatogram. The first few compounds do not have enough capacity and would benefit by spending more time in the stationary phase. The latter few have too high a retention factor and are spending too much time in the stationary phase. What we say is that a chromatogram has a limited peak capacity. In other words, it is only possible to separate a certain amount of compounds within a fixed period of time in the chromatogram. There is only some number of peaks that can be fit in side by side before they start to overlap with each other.
In gas chromatography, there is a way to address these problems during the chromatogram so that all of the constituents are chromatographed at a k' value that is more optimal. If we lowered the temperature during the early portion of the chromatogram, we would raise the retention factor of these constituents of the sample and they would stay on the column longer. If we raised the temperature during the latter portion of the chromatogram, we would lower the retention factor of these constituents and they would come off of the column faster. This systematic raise in temperature during the chromatogram is known as a temperature program. Most gas chromatograms are obtained using temperature programming rather than isothermal conditions. The temperature program chromatogram for the same mixture is shown in Figure 46.
Figure 46. Chromatogram with more optimal k’ values.
There is an analogous procedure in liquid chromatography that is known as gradient elution. In this technique, you start with a mobile phase that causes the constituents of the sample to have a high retention factor and then systematically vary the mobile phase during the run to lower the capacity of the constituents. The exact nature of the changes that might be made to accomplish this will be discussed later in our unit. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Separation_Science/02_Text/04_Fundamental_Resolution_Equation/03_k__Retention_or_Capacity_Factor.txt |
Liquid chromatographs are fairly simple pieces of equipment. There is a solvent reservoir that holds the mobile phase, a high pressure pump, an injection valve, a column, and a detector (Figure 47).
Figure 47. Components of a liquid chromatograph.
The pumps are usually reciprocating small-volume pumps that use inlet and outlet check valves (Figure 48). There is a small cylinder with a sapphire piston. The inlet and outlet valves essentially have a moveable ball in them. When the cylinder is full of fluid and pushing it into the chromatograph, notice how the ball in the inlet check valve will move down and block the opening to the solvent reservoir. Also notice how the ball in the outlet check valve will pop up, allowing flow into the column. When the cylinder is almost empty and the piston recycles to fill (this recycle step is very fast), the ball in the inlet check valve pops up and allows flow from the solvent reservoir to the cylinder. The ball in the outlet check valve moves down, blocking the hole, and prohibiting flow back from the column. The flow out of a reciprocating small volume pump such as this is pulsed (when the piston recycles to fill, flow is stopped). Many liquid chromatographs employ some form of a pulse-dampening device to reduce these pulses.
Figure 48. Diagram of the pump head of a reciprocating small-volume pump.
The injection valve is usually a six-port, two-way valve with a sample loop (Figure 49). In one position (the load position), the flow comes in from the pump through one port and goes out another port to the column. The other two ports connect the syringe adapter to a sample loop and a vent. A common sample loop size is 20 µL. Usually you inject about 100 µL of sample into the loop. The first 80 µL cleans the loop and goes out the vent. The last 20 µL are retained in the loop and ready for injection in the system.
In the second position (the inject position), the flow from the pump now goes through the sample loop and then to the column. Moving the valve to this position allows the mobile phase to sweep the sample into the column for analysis.
Figure 49. Flow diagram of a six-port injection valve.
We have already discussed most of the important features of liquid chromatographic columns. The column lengths are usually 3, 15, or 25 cm depending on the particle size used. The internal diameter of conventional liquid chromatographic columns is 4.5 mm (about ¼ inch).
Many methods have been used for detection in liquid chromatography. The most common detection method is to use ultraviolet absorption. Fluorescence spectroscopy is also used for certain classes of compounds. More recently people have figured out ways to adapt mass spectrometers to liquid chromatographs. The large volume of solvent in liquid chromatography is incompatible with conventional mass spectrometric methods, so these techniques are quite specialized. Also, micro columns with smaller diameters and less solvent are often used with liquid chromatographic-mass spectrometers. One advantage of mass spectrometry over most other detection methods is that the mass spectrum provides information that may allow you to identify the chemical structure of an eluting compound.
05 Liquid Chromatographic Separation Methods
Steric exclusion chromatography is a technique that separates compounds solely on the basis of size. In order for the results of steric exclusion separations to be meaningful, there can be no directed forces between the compounds being separated and the surface of the particles used as the stationary phase. Instead, the particles are prepared with well-characterized pore sizes. Figure 50 shows a particle with one pore in it. Figure 50 also shows representations for several molecules of different size (note, the pore size and molecule size are not representative of the scale that would exist in real steric exclusion phases – the pore is bigger than would actually occur and the molecules would never have sizes on the order of that of the particle).
Figure 50. Representation of a particle with a pore.
The smallest molecule is small enough to fit entirely into all regions of the pore. The largest molecule is too big to fit into the pore. The intermediate sized one can only sample some of the pore volume. Provided these molecules have no interaction with the surface of the particle and are only separated on the basis of how much of the pore volume they can sample, the largest molecule would elute first from the column and the smallest one last.
Suppose we took a molecule that was even larger than the biggest one in the picture above. Note that it would not fit into any of the pores either, and would elute with the biggest in the picture. If we wanted to separate these large species, we would need a particle with even larger pores. Similarly, if we took a smaller molecule than the smallest pictured above, it would sample all of the pore volume and elute with the smallest one pictured above. To separate these two compounds, we would need particles with smaller pores.
Steric exclusion chromatography requires large compounds, and is not generally effective on things with molecular weights of less than 1000. It is commonly used in biochemistry for the separation of proteins and nucleic acids. It is also used for separation or characterization purposes in polymer chemistry (a polymer is a large compound prepared from repeating monomeric units – polyethylene is a long, linear polymer made of repeating ethylene units).
One critical question is how to remove the possibility of an attractive force between the compounds being separated and the surface of the porous particles. The way this is done is to use a particle with very different properties than the compound being separated, and to use a solvent that the solutes are very soluble in. For example, if we want to separate proteins or nucleic acids, which are polar and very soluble in water, we would need to use porous particles made from an organic material that had a relatively non-polar surface. If we wanted to separate polyethylenes of different chain lengths, which are non-polar, we would dissolve the polyethylene in a non-polar organic solvent and use porous particles made from a material that had a highly polar surface. This might involve the use of a polydextran (carbohydrate), which has hydroxyl groups on the surface of the pores.
The classic organic polymer that has been used to prepare porous particles for the steric exclusion separation of water-soluble polymers is a co-polymer of styrene and divinylbenzene.
Styrene Divinylbenzene
If we just had styrene, and conducted a polymerization, we would get a long, single-bonded chain of carbon atoms with phenyl rings attached to it. Linear polystyrenes are soluble in certain non-polar organic solvents. The divinylbenzene acts as a cross-linking agent that bridges individual styrene chains (Figure 51). A typical ratio of styrene to divinylbenzene is 12:1. The cross-linking serves to make the polymer into particles rather than linear chains. The cross-linked polymer is an insoluble material with a very high molecular weight. By controlling the reaction conditions including the amount of cross-linker and rate of reaction, it is possible to make polystyrenes with a range of pore sizes.
Figure 51. Cross-linked styrene-divinylbenzene copolymer.
If we think about what is inside the column, there are four important volume terms to consider. One is the total volume of the column (VT). The second is the interstitial volume between the particles (VI). The third is the volume of all of the pores (VP). The last one is the actual volume occupied by the mass of the material that makes up the particles.
It turns out that VI is usually about 0.3 of VT for a column. There is simply no way to crunch the particles closer together to reduce this term appreciably.
VI = 0.3 VT
It turns out that the maximum pore volume that can be achieved is about 0.4 of VT. If you make this larger, the particles have more pores than structure, become excessively fragile, and get crushed as you try to push a liquid mobile phase through the column.
VP = 0.4 VT
Totaling this up, we can write that:
VI + VP = 0.7 VT
Since there are no directed forces in a steric exclusion column, this means that all of the separation must happen in one column volume, or in 0.7 VT. If we then showed a chromatogram for a series of compounds being separated on a steric exclusion, it might look like that shown in Figure 52.
Figure 52. Representation of a chromatogram on a steric exclusion column.
The peak labeled 1, which has the shortest retention time (0.3 VT), corresponds to all molecules in the sample that had a size that was too large to fit into the pores. It cannot be known whether this is one compound or a mixture of large compounds. The peak labeled 5, which has the longest retention time (0.7 VT), corresponds to all molecules in the sample that had a size small enough to sample all of the pore volume. It cannot be known whether it is one compound or a mixture of two or more small compounds. The peaks labeled 2 through 4 sample some of the pore volume depending on their size. The peak labeled 6 elutes with a retention volume that is greater than 0.7 VT. How would we explain the elution volume of this peak? The only way we can do so is if there are attractive forces between this molecule and the surface of the particles. This is a serious problem and we would need to examine this system and eliminate the cause of this attractive force.
Note that this method separates things on the basis of their size. What people try to do when using steric exclusion chromatography is equate size with molecular weight. The two representations for the shape of a molecule shown in Figure 53 point out a potential problem with equating size with molecular weight. One molecule is spherical. The other is rod shaped. The spherical molecule probably has a larger molecular weight, because it does occupy more volume. However, the size of a molecule is determined by what is known as its hydrodynamic radius. This is the volume that the molecule sweeps out in space as it tumbles. If you took the rod shaped molecule below and allowed it to tumble, you would see that its size is essentially comparable to that of the spherical molecule.
Figure 53. Representation of the molecular size of spherical (left) and rod-shaped (right) molecules.
When performing steric exclusion chromatography, you need to use a series of molecular weight standards. It is essential that these standards approximate the molecular features and shapes of the molecules you are analyzing. If you are analyzing proteins, you need to use proteins as standards. The same would apply to nucleic acids, or to particular organic polymers. Figure 54 shows the typical outcome of the calibration of elution time with molecular weight (always plotted on a log scale) for a steric exclusion column.
Figure 54. Generalized calibration curve for a steric exclusion column.
Note that excessively large molecules never enter the pores and elute together at 0.3 VT. Excessively small molecules sample all of the pore volume and elute at 0.7 VT. There is then some range of size (or molecular weight) where the compounds sample some of the pore volume and elute at volumes between 0.3 and 0.7 VT.
The plots that would occur for particles with the next larger (dashed line) and smaller pore sizes (dotted line) are shown in Figure 55.
Figure 55. Calibration curves for a steric exclusion column. The dashed line is the column with the largest pore sizes. The dotted line is the column with the smallest pore sizes. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Separation_Science/02_Text/05_Liquid_Chromatographic_Separation_Methods/01_Steric_Exclusion_Chromatography.txt |
A second sub-category of liquid chromatography is known as ion-exchange chromatography. This technique is used to analyze ionic substances. It is often used for inorganic anions (e.g., chloride, nitrate, and sulfate) and inorganic cations (e.g., lithium, sodium, and potassium). It can also be used for organic ions, although this is less common with the advent of reversed phase liquid chromatographic methods that will be described later. Another significant application of ion-exchange chromatography is as a step in the purification of proteins. Some of the substituent groups of amino acids are charged (the total charge for a particular protein is a function of the pH of the solution), which makes ion exchange a suitable method for protein purification.
The approach is to attach fixed ionic groups to the surface of a solid support. One common support used in the formation of ion exchange resins is polystyrene-divinylbenzene copolymers. The fixed ions are attached through a derivatization of the phenyl rings of the polystryene. Common fixed ions involve either sulfonate groups or quaternary amines as shown in Figure 56. The aromatic sulfonate groups are strong enough acids that they are deprotonated at all but highly acidic pH values (pH < 1).
Sulfonate group Quaternary amine
Figure 56. Polystyrene polymer derivatized with fixed ionic groups.
Since we must preserve neutrality in such a system, there must also be an exchangeable counterion associated with each of these fixed groups. In the case of the sulfonate group, the counterion is a cation and this is a cation exchange resin. With the quarternary amine phases, the exchangeable counterion is an anion and this is an anion exchange resin. These counterions can be exchanged with each other. For example, this would enable you to have a cation exchange resin in the sodium form or the hydrogen form as seen in Figure 57
Sodium form Hydrogen form
Figure 57. Sulfonate cation exchange resin with sodium and hydrogen counterions.
An anion exchange resin could be in the hydroxide form (OH) or chloride form (Cl). One common use of ion exchange resins is in the deionization of water. It is useful to consider a scheme using ion exchange resins that would enable you to deionize water. If for example, we had water with sodium chloride in it (Na+Cl), we would need a way of removing both cations and anions. If we first passed the water through a cation exchange resin in the hydrogen form, the sodium ions would exchange with the hydrogen ions as shown in the top picture of Figure 58. If we then passed the water through an anion exchange resin in the hydroxide form, the chloride ions would exchange with the hydroxide ions as shown in the bottom picture of Figure 58.
Figure 58. Use of a cation (top) and anion (bottom) exchange resin to deionize water.
The H+ and OH given off by the two resins would combine to form water. Eventually the resin will fill up with impurity ions (Na+ and Cl in this case), and we would either need to replace the resin or reactivate it. We can reactivate the cation exchange resin by passing relatively concentrated hydrochloric acid through it to remove all the Na+ and replace it with H+. We can reactivate the anion exchange resin by passing a relatively concentrated solution of sodium hydroxide through it to remove all the Cl and replace it with OH. To minimize how often we need to replace these resins or how frequently we need to recharge them, it’s best to have resins with as high an ion exchange capacity as possible (the capacity is determined by the number of phenyl rings that have been derivatized with fixed ions).
The capacity of ion exchange resins used for deionizing water is too high for analytical purposes. Presumably, the concentrations of ions in samples that we want to analyze are relatively low. If we use high capacity resins, the retention times will be much too long. One of the early impediments to the use of ion exchange as an analytical method was the lack of methods to reproducibly synthesize resins with low capacities. One of the first groups of people to figure out a way to do this was chemists at Dow Chemical. They were motivated by a need to measure inorganic anions and cations at low levels and realized that ion-exchange chromatography would be an ideal method for doing so.
With any chromatographic method, it helps to know some basic rules for predicting retention order. An interesting example to consider in ion-exchange chromatography is the retention order for the ions Li+, Na+, and K+ on a cation exchange resin. One thing we can consider is whether these ions have different strength interactions with the fixed anion in the resin. All three ions have the same charge. What should also be apparent is that lithium is the smallest ion with the densest charge, potassium the largest with the most diffuse or fluffiest charge. We therefore might predict that the attraction of the lithium ion for the sulfonate groups in the resin is the strongest of the three ions. It turns out that the equation for ionic attraction has a distance term in it, and the closer the ions, the stronger the attraction. Lithium being the smallest is effectively closer to the sulfonate and has the strongest attractive force of the three for the resin. This would suggest a retention order with potassium eluting first, sodium second, and lithium last as shown in Figure 59.
Figure 59. Retention order of Li+, Na+ and K+ on a cation-exchange column based on stationary phase effects.
But there is also a mobile phase in chromatographic separations, and it is useful to examine whether these ions have different relative stabilities in the mobile phase. If one of the three is most stable in the mobile phase, it might be expected to stay in the mobile phase more than the others and elute first. What might affect the stability of these ions in the mobile phase? The mobile phase in this separation would be an aqueous phase. If we consider water, we know that it is unusual in having a very elaborate network of hydrogen bonds. Any ion that dissolves in water will cause some disruption of this network. The question to consider is which ion would be the most disruptive of the three to the hydrogen bond network? Since the charges of these three ions are identical, this decision will be based on the size. The larger potassium ion will be more disruptive of the hydrogen bonds, and more difficult for water to accommodate. You could then say that Li+ is more stable in the water, stays in the mobile phase more, and elutes first. You could phrase this another way by saying that the K+ is least stable in the aqueous mobile phase, and so is “forced” into the resin by the mobile phase and elutes last. Sometimes the chromatographic literature refers to this as the solvophobic effect. The K+ fears the solvent and so spends more time in the resin. The relative retention order based on mobile phase effects would be lithium first and potassium last, as shown in Figure 60, the exact opposite of what was predicted above based on stationary phase effects.
Figure 60. Retention order of Li+, Na+ and K+ on a cation-exchange column based on mobile phase effects.
With conflicting predictions, the only way to know which one is more important is to inject the ions and determine the retention order. In this case, the measurements show that the Li+ elutes first and the K+ elutes last. The mobile phase effects are more significant in determining the retention order.
Now suppose we had two ions with exactly the same size, but one had a charge of 1+ and the other a charge of 2+. Examining stationary phase effects, the equation that describes the attraction between two ions has the charges of both ions in it. The greater the charge of the ions, the greater the attraction. We would therefore expect the 2+ ion to show greater attraction for the fixed sulfonate anions and elute later than the 1+ ion as shown in Figure 61.
Figure 61. Retention order on a cation exchange column for ions of the same size but different charge based on stationary phase effects.
Examining mobile phase effects, both ions would have a similar extent of disruption of the hydrogen bond network since both have the same size. What we then need to consider is the strength of the attraction between the positive ion and the sphere of water molecules that surround it (remember, these cations would be surrounded by the slightly negative oxygen atoms of water molecules). This represents an electrostatic attraction, and again is dependent on charge. Therefore the 2+ ion is more stable in the water and ought to elute first as shown in Figure 62, the exact opposite of what was predicted based on stationary phase effects.
Figure 62. Retention order on a cation exchange column for ions of the same size but different charge based on mobile phase effects.
Faced with conflicting predictions, it’s necessary to perform the experiment and see which comes out first. In this case, the measurements show that the 1+ ion elutes first, the 2+ ion elutes second. The stationary phase effects are more important in this case in determining the retention order. The likely reason why the stationary phase effects are more significant is that the 2+ ion can actually bind simultaneously at two adjacent sulfonate sites as shown in Figure 63.
Figure 63. Representation of a +2 ion associating at two anionic sites on the resin.
The likelihood that this occurs can be observed if a lanthanide(III) ion is added to these resins. In this case, the resin particles actually shrink in size as the lanthanide ion is added. The reason for the compression is that the binding of three sulfonate groups to the lanthanide causes the polymer to collapse in a bit to fit these groups around the lanthanide.
The separation of Li+, Na+, and K+ described in the prior problem would often be done on a polystyrene resin using a fairly dilute solution of hydrochloric acid (perhaps 0.1 M) as the mobile phase. The bound ions would be sulfonate groups and the mobile counter ion would be the H+ ion. An important issue is how to detect these ions. They do not absorb ultraviolet or visible light in the accessible portion of the spectrum. They do not absorb infrared light. Conductivity is one possibility for performing the measurement. The conductivity of a solution is a measure of the extent to which the solution conducts electricity. Dissolved ions are needed for a solution to conduct electricity, and the higher the concentration of ions, the higher the conductivity. We can measure the conductivity of a solution quite sensitively. In fact, this is the reading that is performed on water that has been purified by passage through a MilliQ water purification device to see just how well the water has been deionized. The only problem with trying to apply a direct conductimetric measurement is that the hydrochloric acid in the mobile phase produces too high a background signal. The chemists at Dow who had developed low capacity ion exchange resins recognized this problem as well and devised an ingeneous way to remove the conductivity of the eluent ions (HCl) but retain the conductivity of the alkali ions they wanted to detect.
What they did was use a device called a suppressor column. If we imagine measuring Na+ in a solution containing sodium chloride (Na+Cl), we first start with a cation exchange resin in the H+ form and would have Na+Cl and H+Cl eluting out the end of the column when the sodium band comes off as shown in Figure 64.
Figure 64. Effluent from the analytical column when analyzing a sample containing sodium. H+Cl is the eluent ion.
Note that there is still a high concentration of HCl mixed with the NaCl as it elutes from the column. This HCl will interfere with the measurement of conductivity. Suppose we took the column eluent and then passed it through an anion exchange resin in the hydroxide form as shown in Figure 65. The Cl ion of the Na+Cl would exchange with the hydroxide, converting this into sodium hydroxide (Na+OH), a conducting electrolyte because it stays ionized. The Cl ion of the HCl would exchange with the hydroxide, converting this into H+ and OH, which is non-conducting water. We can therefore measure a conductivity that only relates to the amount of sodium ion in the original sample. An analogous scheme, which had the columns reversed, could be used to measure the conductivity of anions that were separated on an anion exchange column in the hydroxide form.
Figure 65. Effect of the suppressor column on the eluent from the column in Figure 64.
Eventually, the hydroxide counterions in the suppressor column will all become replaced with chloride ions and the device would not work anymore. The suppressor column must be periodically regenerated in the hydroxide form.
Today, instead of using suppressor columns to remove the conductivity of the eluent ions, membrane-based electrolytic neutralization devices are employed. The electrolysis of water can be used to generate hydronium and hydroxide ions, and by proper design, the desired ion can be generated in such a way to pass through a membrane and suppress the conductivity of the eluent ions. In some instruments, similar electrolytic strategies are used prior to the analytical column to generate the eluent ions as well. The use of this eluent generation technology leads to less background conductivity and better sensitivity, making it especially useful for the analysis of low levels of ions.
One other way of detecting the ions, and especially anions, was also developed by the chemists at Dow. It involved the use of indirect spectrophotometric detection. In this procedure, an anion such as the phthalate anion is used as the counterion in the mobile phase. The material is usually added as potassium phthalate, and maintained at a concentration of 0.001 M. The phthalate ion absorbs ultraviolet light. The background measurement in the system then consists of a high absorption reading of UV light.
phthalate anion
Suppose we were separating Cl in a sample containing sodium chloride. Figure 66 shows the end of the column with the chloride ions in the resin, just about to elute in the column. At this point, the concentration of phthalate ion leaving the column is 0.001 M. But what happens when a chloride ion is displaced from the resin to exit the column? This will only happen if a phthalate ion replaces it in the resin. The replacement that occurs will cause the concentration of phthalate ion to drop below 0.001 M.
Figure 66. Representation of an anion exchange column when using phthalate ion for indirect spectrophotometric detection.
In other words, when the chloride elutes from the column, what you observe in the mobile phase is that the total concentration of anions equals 0.001 M (this has to happen in order to preserve neutrality in the system).
[Cl] + [pht ] = 0.001 M
If the concentration of phthalate ion drops proportionally to the concentration of chloride eluting from the column, the UV absorption drops proportionally as well, as shown in Figure 67. The name indirect spectrophotometric detection is aptly chosen since the value of the UV absorption drops as analyte anions elute from the column.
Figure 67. The negative peak for the chloride ion in indirect spectrophotometric detection.
The scientists at Dow Chemical who developed the ion exchange methods for some analyses they needed to perform realized that there were others who would want to be able to measure the same species, and that the device has commercial potential. A company called Dionex, which is still the leading vendor of ion chromatographs, was created to market their invention. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Separation_Science/02_Text/05_Liquid_Chromatographic_Separation_Methods/02_Ion-Exchange_Chromatography.txt |
In our development of band broadening, we talked extensively about the C18 columns that are the most common in use today. We already mentioned that less common bonded phases including those with C8, C1, aminopropyl (C3H6NH2) and cyanopropyl (C3H6CN) groups are also commercially available. Prior to the introduction of the C18 columns, the normal way to do liquid chromatography involved the use of a polar stationary phase like silica gel and a non-polar mobile phase. When the C18 columns were first reported, they constituted a non-polar stationary phase, and so a polar mobile phase could be employed. Since this configuration of polarities of the phases was the reverse of what was normally done, the use of C18 columns became known as reversed-phase liquid chromatography. Even though today almost everyone uses these reversed-phase methods, the name has stuck.
We have already seen how the deactivation of the surface with the C18 groups has a very positive effect on the chromatographic efficiency. Another significant advantage of using C18 stationary phases is that aqueous-based mobile phases are used with them. Water is the ideal mobile phase for liquid chromatography because so many substances are soluble in water. Water is the solvent of the environment. Water is the primary solvent of living systems. Many pharmaceutical agents need to possess water solubility to enter the body. Since many organic compounds are not that soluble in water, organic pharmaceuticals that are not water-soluble usually have ionic groups incorporated into them to render them water-soluble.
It’s rare to use only water as the mobile phase with a C18 column. Instead, an organic modifier (methanol, acetonitrile, and tetrahydrofuran are the three most common) is added in some amount to the mobile phase. Compared to water, the chemical properties of the organic modifier are more like that of the C18 phase. The presence of the organic modifier therefore shortens the retention time compared to a mobile phase that was only water. Raising the concentration of the organic modifier will further shorten the retention time of almost all compounds. We mentioned the concept of a gradient elution earlier in the unit when we discussed the problem of having a limited peak capacity in a chromatogram. Remember the goal was to start with a high retention factor and systematically change the system so that the retention factor got shorter as the chromatogram progressed. One possible gradient procedure might be to start with a mobile phase that is 80:20 water-methanol and gradually change is to 20:80 water-methanol during the chromatogram. The increased concentration of methanol will reduce the retention factor of the later eluting components.
Experimentally, it is easy to carry out a gradient. Using two pump heads, one pumps the initial phase and the other the final phase. A special mixing chamber exists to adequately mix the output from the two pumps. During the chromatogram, a computer systematically slows down the pump rate of the initial phase and ramps up the pump rate of the final phase.
Another important variable with an aqueous mobile phase is the pH of the system. C18 silica gel phases are stable over the range of 2 to 8 (although 3 to 7 really represents a safer window). If the pH is too acidic, the surface C18 groups will be hydrolyzed off. If the pH is too basic, the silica gel itself will decompose. This pH range is one that can be used to change the nature of many acids and bases. For example, carboxylic acids at pH 3 might mostly be in their protonated (neutral) form, whereas at pH 7 they are likely to be in their deprotonated (anionic) form. Since we would expect the neutral form to have more solubility in a C18 phase then the anionic form, the retention time for most carboxylic acids would be longer at pH 3 than at pH 7. If we consider nitrogen bases, we note that at pH 3 they are likely to be in their protonated (cationic) form, whereas at pH 7 they are likely to be in their deprotonated (neutral) form. Organic bases therefore would have a longer retention time at pH 7 than at pH 3. You might also recollect how we used a similar process to distinguish acids and bases in an acid/base/neutral extraction procedure discussed at the beginning of this unit. This is the same behavior and rationale going in liquid chromatography. Another gradient procedure that could be used to advantage in liquid chromatography is a pH gradient.
There is some debate in the literature on liquid chromatography about the exact nature of the separation procedure (adsorption on a deactive surface versus partitioning), and whether stationary phase or mobile phase (solvophobic) effects are most important in determining retention properties. It actually turns out that either one usually leads to the same conclusion as to retention order, and for our purposes, thinking about which compound of a pair has a greater attraction for the stationary phase is sufficient to predict which will come out first and which second. We have already examined the two pairs below as a function of pH.
RCOOH RCOO
pH = 3 (longer tR) pH = 7 (shorter tR)
R3NH+ R3N
pH = 3 (shorter tR) pH = 7 (longer tR)
How about a comparison of the retention order of pyridine and tert-butylpyridine on a reversed-phase column? If you thought that the presence of the tert-butyl group on the pyridine would make that compound more attractive to the C18 groups, thereby causing it to have the longer retention time of the two, you are correct. In fact, in all likelihood, there is some preference of the association of these two compounds with a C18 phase as shown in Figure 68. The non-polar portion of the molecule is likely oriented toward the C18 phase, whereas the polar nitrogen atom is oriented toward the mobile phase.
Figure 68. Representation of the association of tert-butylpyridine (left) and pyridine (right) with a C18 bonded phase.
Lastly, picking up on another technique we learned with regards to liquid-liquid extractions, we could ask how we might lengthen the retention time of organic cations on a C18 column. For example, there are a series of neurotransmitters such as dopamine that are based on aromatic amine derivatives.
dopamine
Dopamine and other similar derivatives may well be protonated under the separation conditions, and because of the protonation, have short retention times that do not facilitate the separation of the similar analogues. Can you think of a way to lengthen the retention time that does not involve adjustment of the pH? A way to do this is to add a “greasy” anion such as heptylsulfonate to the mobile phase (note, since this is added to the mobile phase the anion is continuously pumped through the system).
$\mathrm{C_7H_{15}SO_3^- = heptylsulfonate}$
One way to think of this is that, just like in liquid-liquid extraction, the sulfonate anion can form an ion pair with the cation and the ion pair would have more solubility in a C18 phase than the cation by itself. An alternative way to view this is that some of the sulfonate salt distributes into the C18 phase as shown in Figure 69. This serves to create a pseudo ion exchange material that the organic cations are attracted to. Detailed studies have shown that the latter mechanism (pseudo ion exchange resin) is the one that actually operates.
Figure 69. Formation of a pseudo ion exchange material with heptylsulfonate on a C18 column.
One other important category of bonded phases involves the attachment of optically pure groups to silica gel. The compound dinitrobenzoyl-L-leucine is one example of such a phase. These types of phases are used for the separation of enantiomers.
Dinitrobenzoyl-L-leucine
The distinction of enantiomers is analogous to the distinction that would occur if you considered putting your feet into only one of your shoes. If you put your left foot into your left shoe, it feels fine. If you put your right foot into your left shoe, it does not fit quite as well, and even in the dark, you would notice that something isn’t right. If we expressed this as a reaction with an equilibrium constant, we could say that the association of your left foot with your left shoe is higher than the association of your right foot with your left shoe. If we had bonded the left shoes to silica gel, your right foot would elute faster from the column, your left foot slower. The compound pictured above has a number of groups that can form intermolecular interactions with enantiomeric solutes. The amide functionality is a site with dipoles and the potential for hydrogen bonding. The electron-deficient dinitrobenzoyl ring can undergo what is known as pi-stacking with electron-rich aromatic rings. The isobutyl group of the leucine provides steric hindrance. Other chiral liquid chromatographic phases are designed to have similar groups that provide sites for intermolecular attraction, π-stacking, and steric constraints. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Separation_Science/02_Text/05_Liquid_Chromatographic_Separation_Methods/03_Bonded-Phase_Liquid_Chromatography.txt |
A gas chromatograph typically has a carrier gas (it is common to refer to the mobile phase in gas chromatography as the carrier gas since the mobile phase has no influence at all over the separation – its only purpose is to carry the solutes through the column), heated injection port, column in a heated oven, and detector.
The carrier gas must be inert and helium or nitrogen is used for most gas chromatographic applications. Sample injection is most commonly accomplished through the use of a small-volume syringe (1 μL is a typical injection volume). The liquid sample is injected into a hot zone that is sealed with a polymer septum. The syringe pierces the septum, and on injection the liquid sample is flash-volatilized in the heated injection port and flushed into the column by the carrier gas. With fused silica capillary columns, 1 μL of liquid sample will overload the column so the injector is designed with a splitter. Adjusting the flow of the splitter gas allows you to calibrate the split ratio. Ratios of 50:1 to 100:1 (only the one part is injected) are common. Most injection ports of this type have glass liners. Different glass liners are inserted under the septum depending on whether you are doing split or splitless injections.
Another gas chromatographic injection technique that is used in certain applications involves thermal desorption from a polymer trap. The technique is illustrated in Figure 70 for the analysis of organic chemicals in air. Assume that we have a small piece of tubing filled with a polymer. The polymer is a material that organic chemicals will adsorb to. Water in the air will pass through unretained. A polymer that is used quite frequently for this application is Tenax. Tenax is a p-polyphenylene oxide polymer that is stable at very high temperatures. In the adsorption step, we use a pump to draw some large volume of air (this might be as high as 15 liters of air) through the Tenax trap, which is held at room temperature. The organic chemicals from the air will adsorb onto the polymer.
Figure 70. Use of a sorbent trap to remove organic compounds from an air sample.
The organic chemicals can be desorbed from the Tenax trap by heating the gas under a flow of helium. The illustration in Figure 71 shows how the Tenax trap is desorbed in a backflush mode (the end that had the higher concentrations of organic chemicals is put closer to the gas chromatograph). Switching valves comparable to liquid chromatographic injection valves make it easy to redirect the flow of gases through these Tenax traps between the adsorption and backflushing orientation.
Figure 71. Desorption of a sorbent trap in a backflush mode.
Desorption might typically be done at a temperature of 225oC over a period of five minutes. One thing to realize is that we would never be able to tolerate a five-minute injection of a sample, since this would mean that all of the peaks might be as long as five minutes. It is essential to maintain a “plug” injection, which is achieved by thermal focussing. Thermal focussing is accomplished by cooling a region in the injector (or in older gas chromatographs, the entire oven) to a temperature of –50oC. This cooling is accomplished using either liquid carbon dioxide or liquid nitrogen. At –50oC, the organic compounds that desorb from the Tenax trap freeze in a small band at the head of the column. Rapid heating of this frozen band injects the sample at a comparable rate to a normal syringe injection.
Sorbent traps are frequently used for the analysis of trace levels of volatile organic chemicals in water as well. In this case, the method is known as a purge and trap technique and is shown in Figure 72. A sample of water is taken (usually 5 mL – too large to ever inject into a gas chromatograph) and purged with helium gas. After exiting the water, the helium gas flows through a Tenax trap. The helium gas bubbling through the water displaces the dissolved volatile organic chemicals, which adsorb onto the Tenax trap. The Tenax trap is desorbed as described above for the analysis of volatile organic chemicals in air.
Figure 72. Purge and trap device used to analyze volatile organic compounds in water.
Information on a variety of commercially available gas chromatographic stationary phases is provided in Table 3. The distinction between different stationary phases is based on a comparison of their polarity. Five different compounds are typically used to represent different types of functional groups. Individual indices for these five compounds are measured, and a composite value (P) is determined as well. These polarity indices are referred to as Rohrschneider constants. The higher the number the more polar the phase. In reality, a sampling of four or five stationary phases would be good enough to span the range of polarities that is needed for gas chromatographic separations.
Table 3. Rohrschneider constants for gas chromatographic liquid phases
Liquid phase Max T
oC
X' Y' Z' U' S' P
Squalane 100 0 0 0 0 0 0
Apiezon L 250 32 22 15 32 42 29
SE-30 300 15 53 44 64 41 43
OV-1, methyl gum 350 16 55 44 65 42 44
OV-3, 10% phenyl 350 44 86 81 124 88 85
OV-7, 20% phenyl 350 69 113 111 171 128 118
Dioctyl sebacate 125 72 168 108 180 123 130
Dilauryl phthalate - 79 158 120 192 158 141
Dinonyl phthalate 150 83 183 147 231 159 161
OV-17, 50% phenyl 375 119 158 162 243 202 177
Versamid 930 150 109 313 144 211 209 197
Trimer acid 150 94 271 163 182 328 218
OV-25, 75% phenyl 350 178 204 208 305 280 235
Polyphenylether 225 182 233 228 313 293 250
Triton X-305 200 262 467 314 488 430 392
Carbowax 20M 225 322 536 368 572 510 462
Carbowax 4000 200 317 545 378 578 521 468
Reoplex 400 200 364 619 449 647 671 550
Carbowax 1540 175 371 639 453 666 641 554
Diglycerol 100 371 826 560 676 854 657
EGSS-X 200 484 710 585 831 778 678
Ethylene glycol phthalate 200 453 697 602 816 872 688
Diethylene glycol succinate 200 496 746 590 837 835 701
Tetrahydroxyethylenediamine 150 463 942 626 - 893 731
Hexakis(2-cyanoethoxycyclohexane) 150 567 825 713 978 901 797
N,N-bis(2-cyanoethyl)formamide 125 690 991 853 1110 1000 929
X' = benzene; Y' = butanol; Z' = 2-pentanone; U' = nitropropane; S' = pyridine
P = (X' + Y' + Z' + U' + S')/5
If we consider what governs retention order in gas chromatography, there are two important parameters. One is the volatility of the compound, with the observation that more volatile compounds (those with a higher vapor pressure or lower boiling point) elute first. The other is the attractive forces between the compound and the stationary phase. If we consider the homologous series of alcohols listed in Table 4, it’s interesting to note that the boiling point goes up by approximately 20oC for each additional methylene group (CH2) in the chain. The increase in boiling point with each additional methylene group is a little higher for the series of alkanes. The alcohols have higher boiling points than the corresponding alkanes because they can hydrogen bond with each other.
Table 4. Boiling points for homologous series of alkanes and primary alcohols.
Hydrocarbon Boiling Point (oC) Alcohol Boiling Point (oC)
Pentane 36 1-Pentanol 137
Hexane 69 1-Hexanol 157
Heptane 98 1-Heptanol 176
Octane 126 1-Octanol 196
Nonane 151 1-Nonanol 215
Decane 174 1-Decanol 231
If we consider a homologous series, it turns out that the boiling points are actually determined by the molar volumes of the molecules. Just as we observed with the size of molecules in steric exclusion chromatography, the molar volume of a molecule is the volume swept out by the molecule as it tumbles. Molecules with larger molar volumes have higher boiling points, provided the molecules being compared have identical intermolecular forces (you cannot compare hydrocarbons to alcohols). An interesting comparison is observed by looking at the boiling point of n-octane and iso-octane (2,2,4-trimethylpentane).
n-octane (bp = 126oC) iso-octane (bp = 98oC)
The branched iso-octane would have a smaller molar volume than the linear n-octane, and this is clearly reflected in the boiling points of the two compounds.
When predicting retention order in gas chromatography, the overriding factor is a comparison of the boiling points. The compound with the lowest boiling point elutes first. Only when two compounds have very close boiling points (within 5oC or less) does it become important to consider the polarity of the compounds and the polarity of the stationary phase. Remember that like dissolves like, so a polar stationary phase will show more retention of a polar compound. The Rohrschneider values are used to determine the polarity of the stationary phase.
06 Gas Chromatographic Separation Methods
As mentioned previously, highly sensitive detection methods were developed for gas chromatography in the 1950s and 1960s that facilitated its use as an analytical method.
One of the most useful detection methods is known as the flame ionization detector (FID). The flame ionization detector is highly sensitive and involves burning the sample in an air-hydrogen flame. Molecules with a CH bond will form CH+ ions in the flame, and these are measured using a negatively charged collector above the flame. When these positive ions strike the negatively charged collector, a current proportional to the amount of ions is measured. This detector is essentially universal, as it is able to measure all organic compounds. It is not sensitive toward compounds like water and carbon dioxide, which is advantageous.
Another important detector in the development of gas chromatography is the thermal conductivity detector. This is a universal detector, but not nearly as sensitive as the FID. A resistance circuit called a Wheatstone bridge is used. The bridge has two halves, and the resistance of these halves is compared. Instruments with a thermal conductivity detector need two matched columns. One column has only carrier gas flowing through it. The other has carrier gas and sample. Changes in the thermal conductivity of the gas change the resistance reading over the Wheatstone bridge. Some representative values for themal conductivities of gases are listed below. The value for butane is representative of most organic compounds. Since we need to measure a difference in thermal conductivity, we need a carrier gas with as different a thermal conductivity as possible from most organic compounds. This would suggest the use of hydrogen gas as a carrier gas, however hydrogen is too much of an explosion risk for this purpose. Helium is therefore the carrier gas of choice. Thermal conductivity detectors tend to be used on less expensive pieces of equipment like the gas chromatographs used for the analysis of the purity of compounds prepared in organic chemistry instructional labs.
\begin{align} \ce{&Thermal\: conductivity\: values: &&Hydrogen &471\ & &&Helium &376\ & &&Nitrogen &66\ & &&\mathit{n}-Butane &43} \end{align}
One of the most important gas chromatographic detectors ever developed was the electron capture detector. This device, which is illustrated in Figure 73, was invented by James Lovelock. The device has a radioactive foil (usually containing tritium or radioactive nickel) that emits beta particles. Beta particles are high-energy electrons emitted by a nuclear decay process (a neutron decays into a beta particle and a proton). The opposite side of the device is positively charged so the beta particles stream across the tube. As the high-energy beta particles strike carrier gas molecules (usually either nitrogen or an argon/methane mixture), they create a cascade of lower energy electrons, which generate a current that can be measured. If we had a compound coming out of the gas chromatograph that could capture some of these low-energy electrons, the current would drop. The drop in current can be related to the concentration of electron capturing compound in the sample. What is especially impressive about this detector is the level of sensitivity that can be achieved for certain classes of compounds.
Figure 73. Representation of an electron capture detector.
What types of compounds ought to have the ability to capture electrons? Remembering back to your knowledge of periodic properties, you should predict that halogen-containing compounds (compounds with a chlorine, fluorine, or bromine atom) ought to be very effective at capturing electrons. What are especially significant are the classes of compounds that contain halogens. These include the following (with their respective application or concern):
• Chlorofluorocarbons (or freons) – ozone layer destruction
• Chloroform – potential carcinogen formed in drinking water from water chlorination
• Dioxin – by-product of combustion and certain industrial processes
• Chlorinated pesticides such as DDT
• Polychlorinated biphenyls – used in electronic devices (transformers)
These are only some examples, but they form an extensive array of compounds of environmental significance. The invention of the electron capture detector facilitated the discovery that freons were making their way to the stratosphere and leading to the destruction of the ozone layer. Without this detector, this discovery would have been delayed by many years.
Mass spectrometric detection is especially important because it provides information that can be used for compound identification. Most gas chromatograph-mass spectrometers (GC-MS) use fused silica capillary columns, and the column effluent can go straight into the mass spectrometer. What happens in the mass spectrometer is that the column effluent is bombarded with a high-energy beam of electrons. As these electrons strike the sample, they generate ions. These can be anions or cations, but most mass spectrometers are designed to draw off only the cations for analysis. Usually almost all the cations have a +1 charge. If we consider the molecule acetophenone, we can illustrate what will take place (three functional groups within the molecule are labeled A, B, and C).
Acetophenone
If an electron struck this molecule and knocked out one other electron, we would have the entire molecule with a +1 charge. This ion is referred to as the molecular ion, and we would show this as ABC+. It’s usually very desirable to have a molecular ion in the spectrum. The other valuable observation in mass spectrometry when you use an electron beam for ionization is that you get fragment ions as well. Possible fragments we could get from acetophenone are listed in Table 5, along with their masses. Actually, what is really measured is the mass-to-charge ratio, denoted as m/e. If all of the ions have a charge of +1, the mass equals the mass-to-charge ratio.
Table 5. Possible fragment ions from acetophenone using A, B and C to denote pieces of the molecule.
Ion m/e
ABC+ 120
AB+ 105
AC+ 92
A+ 77
BC+ 43
B+ 28
C+ 15
A good question to ask is whether we would ever get AC+ as a fragment? This would require an intramolecular reaction to take place. This does happen with some molecules, but not the one we are using as an example. Similarly, it might be very unlikely to get a B+ ion, which would require the loss of two groups. If you were to examine a book on reaction mechanisms that occur in a mass spectrometer, you would notice that it would look very similar to an organic chemistry textbook. The two things we observe is that different fragments have different weights, but they also have different intensities. For example, if a molecule has a good leaving group, it will tend to leave in the mass spectrometer. Depending on which fragment is more stable as a positive ion, the relative intensity of the two ions will vary. The most intense ion is given a value of 100 and the intensity of each other ion is reported relative to it. A possible mass spectrum for acetophenone is provided in Figure 74. It is worth noting that there will be many more ions in the spectrum than these. For one thing, isotope effects show up. It turns out that 1% of all carbon is carbon-13 so one out of every hundred molecules weighs one more (meaning that we will see a small peak at m/e = 121). Note that the mass spectrometer does not measure an average molecular weight based on isotopic abundance, but the exact weight of each ion. For another thing, the molecule might lose a hydrogen atom and show a fragment with a mass of 119.
Figure 74. Representation of the mass spectrum for acetophenone
Notice how we could use these masses to determine a possible molecular formula for the compound, and the masses of fragments to determine possible groups that are found in the molecule. The compound below (2-phenylethanal) has the exact same molecular weight (120) as acetophenone, but you might appreciate that it would probably have a different mass spectrum.
2-phenylethanal
Mass spectra are relatively difficult to interpret and to assign an unequivocal structure to. What is usually done instead is that the mass spectrometers with a GC-MS come with a library of spectra of known compounds. The computer will compare your measured mass spectrum to those in the library and report the ten best matches. The top match does not confirm the assignment. To do that, you would need an authentic sample of that compound and would have to show that it has the same retention time and mass spectrum on your instrument. With those two matches, you can be assured that you have identified the compound. GC-MS is commonly used for drug testing at sports events like the Olympics.
07 Appendix 1: Derivation of the Fundamental Resolution Equation
$\mathrm{R_S = \dfrac{2(t_2-t_1)}{W_1 + W_2}}\tag{1}$
$\mathrm{W_1 \cong W_2}$
$\mathrm{W = 4\sigma}$
$\mathrm{R_S = \dfrac{2(t_2-t_1)}{8\sigma_2} = \dfrac{t_2-t_1}{4\sigma_2}}\tag{2}$
$\mathrm{N = \left(\dfrac{t_2}{\sigma_2} \right )^2 \hspace{60px} \sigma_2 = \dfrac{t_2}{\sqrt{N}}}$
$\mathrm{R_S = \dfrac{t_2-t_1}{4\left( \dfrac{t_2}{\sqrt{N}}\right )}}\tag{3}$
$\mathrm{R_S = \left(\dfrac{\sqrt{N}}{4}\right )\left(\dfrac{t_2-t_1}{t_2}\right ) = \left(\dfrac{\sqrt{N}}{4} \right) \left(1 - \dfrac{t_1}{t_2} \right )}\tag{4}$
Write an expression for the fraction of material in the mobile phase (φM):
$\mathrm{\varphi_M = \dfrac{C_MV_M}{C_MV_M + C_SV_S}}\tag{5}$
$\mathrm{\varphi_M = \dfrac{\dfrac{C_MV_M}{C_MV_M}}{\dfrac{C_MV_M}{C_MV_M} + \dfrac{C_SV_S}{C_MV_M}} = \dfrac{1}{1 + k'}}\tag{6}$
Express the average migration velocity of component 2:
$\mathrm{v_{S_2} = \varphi_M v}\tag{7}$
(where v is the mobile phase velocity)
$\mathrm{v_{S_2} = \dfrac{L}{t_2}\hspace{20px}(8) \hspace{60px} v = \dfrac{L}{t_0} \hspace{20px} (9)}$
$\mathrm{\dfrac{L}{t_2} = \dfrac{L}{t_0}\varphi_M}\tag{10}$
$\mathrm{t_2 = \dfrac{t_0}{\varphi_{M_2}} \hspace{60px} t_1 = \dfrac{t_0}{\varphi_{M_1}}}\tag{11}$
$\mathrm{t_2 = \dfrac{t_0}{\left(\dfrac{1}{1+k_2'}\right)} \hspace{60px} t_1 = \dfrac{t_0}{\left(\dfrac{1}{1+k_1'}\right)}}\tag{12}$
$\mathrm{t_2 = t_0(1+k_2') \hspace{60px} t_1=t_0(1+k_1')}\tag{13}$
$\mathrm{\dfrac{t_1}{t_2} = \dfrac{t_0(1+k_1')}{t_0(1+k_2')} =\dfrac{1+k_1'}{1+k_2'}}\tag{14}$
Substitute (14) into (4):
$\mathrm{R_S = \left(\dfrac{\sqrt{N}}{4}\right) = \left(1 - \dfrac{1+k_1'}{1+k_2'}\right) }\tag{15}$
Consider the $\mathrm{\left(1 - \dfrac{1+k_1'}{1+k_2'}\right) }$ term:
$\mathrm{\left(1 - \dfrac{1+k_1'}{1+k_2'}\right) =\dfrac{1+k_2'}{1+k_2'} - \dfrac{1+k_1'}{1+k_2'}}\tag{15a}$
$\mathrm{=\dfrac{k_2' - k_1'}{1+k_2'}}\tag{15b}$
$\mathrm{=\dfrac{k_1'\left( \dfrac{k_2'}{k_1'} - 1\right )}{1+k_2'}}\tag{15c}$
$\mathrm{=\dfrac{\left(\dfrac{k_2'}{k_1'}\right)k_1'\left( \dfrac{k_2'}{k_1'} - 1\right )}{\left(\dfrac{k_2'}{k_1'}\right)(1+k_2')} = \dfrac{k_2'\left( \dfrac{k_2'}{k_1'} - 1\right )}{\left(\dfrac{k_2'}{k_1'}\right)(1+k_2')}}\tag{15d}$
$\mathrm{=\left(\dfrac{k_2'}{1+k_2'}\right)\left(\dfrac{\dfrac{k_2'}{k_1'}-1}{\dfrac{k_2'}{k_1'}} \right )}\tag{15e}$
$\mathrm{\alpha=\dfrac{k_2'}{k_1'}}$
$\mathrm{=\left( \dfrac{k_2'}{1+k_2'}\right )\left(\dfrac{\alpha - 1}{\alpha}\right )}\tag{15f}$
Substitute (15f) into (15):
$\mathrm{R_S=\left( \dfrac{\sqrt{N}}{4}\right )\left(\dfrac{\alpha - 1}{\alpha}\right )\left(\dfrac{k_2'}{1+k_2'} \right )}$
Relationship to retention time:
$\mathrm{v_{S_2} = \dfrac{L}{t_2} \hspace{60px} t_2 = \dfrac{L}{v_{S_2}}}$
$\mathrm{v_{S_2} = \varphi_M v = \left(\dfrac{1}{1+k_2'} \right )v}\tag{16}$
$\mathrm{H = \dfrac{L}{N} \hspace{60px} L = HN}$
$\mathrm{t_2 = \dfrac{HN(1 + k_2')}{v}}\tag{17}$
Rearrange the fundamental resolution equation to solve for N:
$\mathrm{N = 16{R_S}^2 \left( \dfrac{\alpha}{\alpha - 1}\right )^2 \left( \dfrac{1 + k_2'}{k_2'}\right )^2}$
$\mathrm{t_2 = \left(\dfrac{16{R_S}^2H}{v}\right ) \left( \dfrac{\alpha}{\alpha - 1}\right )^2 \left( \dfrac{(1 + k_2')^3}{(k_2')^2}\right )}\tag{18}$ | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Separation_Science/02_Text/06_Gas_Chromatographic_Separation_Methods/01_Detection_Methods.txt |
The problem sets on chromatography can be used in at least two different manners. The primary intent is to use these as a set of in-class, collaborative learning exercises. Groups of 3-4 students work together in discussing and working through the problems. When using the problem sets in this manner, the instructor must actively facilitate and guide students through the material. This manual will guide instructors through each of the problem sets, identifying possible student responses to the questions and the response and activities of the instructor during the progression of the problem.
An alternative to the use of the problems in class is to assign them as out-of-class activities, preferably done as a group activity among students or as a peer-led learning activity (REF). The accompanying text that goes with each problem provides a detailed discussion of each step of the thought process of solving it, such that students could work back and forth between the problem and text on an iterative basis to gain an understanding of the material.
There is no perfect way to assemble groups for such collaborative learning activities. I gather information on the first day of class (year in college, major, prior chemistry courses) and then use this to set groups of 3-4 students that start on the second day of class. I try to make the groups as heterogeneous as possible and they work together for the entire semester. Another strategy is to assign groups for a shorter period of time that might encompass completion of a specific topic or unit, and to then create new groups for the next unit. One other possibility is to have different groups every day of class. Since it is important for groups to work well together, having new groups every day may be less successful than allowing groups to work together for more extended periods of time. I would recommend that the instructor assign groups rather than allowing the students to pick their own. This avoids the potential problem of friends who want to be in the same group but who then do not work well together or stay focused on the assigned task. It also avoids the problem of the student who is left without a group at the end of the selection process, something that can be especially problematic if it is a member of a minority group. When using collaborative groups, it is also important for the instructor to monitor the functioning of the groups and to step in to address either dysfunctional groups or the recalcitrant individual within a group. (Ref articles on group learning). Peer-evaluation processes are often used by instructors who employ group activities as a way of assessing how well groups are working (REF).
I also expect the groups to meet outside of class for any homework assignments, something that is aided because I am at a residential college. An alternative to this is to schedule a room on the evening before a homework assignment is due and encourage them to come to this place and work in any arrangement they wish on the homework. I have run such sessions for several years now and attend them as a facilitator (one result is that it has cut down considerably the individual traffic to my office seeking help on the homework problems) and it has been an excellent way to promote collaboration among the students.
The instructor has an especially important role to fulfill during such group activities. I have observed that the more engaged that I am in the process in helping to guide the students through the material, the more effective the learning that occurs. In most instances, it seems that the students are initially stumped by the question, that they begin to explore things that they do know that might apply to answering the question, and that help from the instructor either by letting them know that they are on the right track or by suggesting another direction in which to take their thinking is necessary. As they begin a question, I roam around the room listening in on conversations and looking over their shoulders at what might be written in their notebook. If I hear something interesting, I indicate that to the group. If I see that someone has written something interesting and relevant in their notebook, I tell other group members that they ought to talk with this individual about what they have written, and that the individual should explain to the other group members why they wrote that down. If I hear a group going entirely in the wrong direction, I probe them on why they are heading in that way and then offer suggestions about things to consider that will set them off in the right direction. When all groups have realized an important point, I call time out and summarize the concept at the board. Then I send them back to continue with the next part of the problem. Most of the problems are handled in such an iterative manner where the students work through some important part of the problem, I summarize it at the board when they have developed the concept, and then they return to the next part of the problem. Occasionally a group will just not see something, whereas every other group has gotten the point, and it may require a direct intervention from the instructor with that group to explain the concept. Similarly, there are times when I call their attention to the board to summarize a point when one of the groups still has not gotten the concept but waiting would slow down the remainder of the class to an unacceptable level.
When using these materials, I want the students to discuss and discover the concepts inherent in the problems, so they do not have the text when working on the problems. After they have completed a particular problem, I then give them a copy of that portion of the text (everyone is instructed to have a three-ring loose-leaf binder of a certain minimum thickness that will accommodate the entire text that will be passed out in increments as the semester develops). The text thoroughly goes through the thought process for solving each problem and I encourage the students to read it over that evening to reinforce the concepts developed in class that day. I also give homework problems designed to reinforce the concepts developed in class.
04 Instructor's Manual
Before providing students with the problem set, I spend about twenty minutes introducing extraction. This includes a discussion of how extractions are carried out experimentally (most students have probably encountered a separation funnel, although lately I have often had a few first-year students in the course). I also introduce distribution and partition coefficients and talk about how extraction is used for bulk separations of chemicals with similar properties.
1. Devise a way to separate the materials in the following sample by performing an extraction.
The sample consists of water with a complex mixture of trace levels of organic compounds. The compounds can be grouped into broad categories of organic acids, organic bases, and neutral organics. The desire is to have three solutions at the end, each in methylene chloride, one of which contains only the organic acids, the second contains only the organic bases, and the third contains only the neutrals.
Students are also given the following hint to aid in thinking about a solution to this problem.
Remember - Ions are more soluble in water than in organic solvents.
- Neutrals are more soluble in organic solvents than in water.
I point out how the separation of acids, bases and neutrals is a common bulk separation scheme that is often used in areas like the analysis of environmental samples. Groups are then allowed to think about the problem. Often the students think they can just add methylene chloride and extract the neutrals without extracting anything else. Groups often realize that the presence of acids and bases in the same sample then means that neutralization has likely occurred to some degree. At this point I prompt them to write down answers to the following questions.
What do organic acids and bases look like?
After a few minutes the students can identify carboxylic acid groups as acids and amine groups as bases. I then ask them to think about the nature of these chemicals as a function of pH, and more specifically at extremes of pH.
What would these groups look like at a pH of 1? What about at a pH of 14?
It should take the students just a few minutes to correctly draw each of the four cases. They should realize that the key point is that at a low pH amines are protonated and carry a positive charge while at a high pH carboxylic acids are deprotonated and carry a negative charge. At this point I ask them
Can you now devise a scheme for separating these organic molecules?
Give the students five minutes to come up with the scheme. I ask individual groups as I’m circulating through the class what acid and base they would use to adjust the pH, and they immediately respond with hydrochloric acid and sodium hydroxide. Once each group has an acceptable scheme I spent a few minutes going through a scheme at the board to show where each class of molecules is at each step. I also point out that the usual way this is done in practice involves a separation of the acidic compounds from the base/neutral components so only involves two solutions instead of three.
I then ask the students the following question.
Suppose you had metal ions in water. Can you think of a way to extract them into the organic phase?
After completing the equilibrium unit, students talking with their group usually recognize quickly that complexing the metal with a ligand to make a neutral complex will move the metal into the organic layer. I then talk a little about how we can selectively complex metals by varying the pH, so that in some cases it is possible to adjust the pH of the aqueous phase to extract one metal ion in the presence of others.
How would you get the metal ions back into the aqueous phase?
Students should all suggest decreasing the pH in order to protonate the ligand to drive the equilibrium back toward the uncomplexed species.
2. Devise a way to solubilize the organic anion shown below in the organic solvent of a two phase system in which the second phase is water. As a first step to this problem, show what might happen to this compound when added to such a two phase system.
C18H37C(O)O-Na+
What would happen to this molecule in the two phase system?
The groups often initially think that the species will enter the aqueous layer because it is an ion. I then ask them to consider the non-polar nature of the long carbon chain and where this group would prefer to reside in such a two-phase system. Groups usually then wonder if it is possible for the species to lie right at the interface of the system with the ionic end in the aqueous phase and carbon chain in the organic phase. I indicate that this is what would happen and then briefly talk about the formation of micelles if this salt were to be solely dissolved in water and a reverse micelle if dissolved in an organic phase.
Can you now think of a way to move it to the organic phase?
Students immediately propose lowering the pH as one method, and I challenge the students to think of others. Usually they are stumped by this and I ask them to think about the solubility properties of sodium relative to other possible cations. After a few minutes, I usually have to call the groups’ attention to me and discuss the concept of ion pairing and how the use of a lipophillic organic cation (e.g., quaternary amine with bulky aliphatic groups) would create an organic-soluble ion pair. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Separation_Science/04_Instructor's_Manual/01_In-class_Problem_Set_-_Extraction.txt |
At the beginning of this unit, I spend essentially an entire class in a lecture format providing background information on chromatography. This includes some of the history of chromatography beginning with the initial work of Mikhail Tswett, and the introduction of key concepts within chromatography such as the difference between adsorption and partitioning, the distribution constant, the partition coefficient, the selectivity factor, the concept of capacity and the retention factor, the idea of dividing a column into a set of theoretical plates, and retention time. With this background, I then give them the first problem set.
02 Chromatography Unit
1. Consider a plot that has the concentration of analyte in the stationary phase on the y-axis and the concentration of analyte in the mobile phase on the x-axis.
a) Draw an idealized plot as greater concentrations of analyte are injected into the chromatographic column.
If this is all students are presented with, most are confused as to what is being asked. I spend a few minutes thoroughly describing the experiment that will be performed (a series of consecutive injections in which the total amount of analyte is increased for each subsequent injection). I then give the groups about five minutes to consider this problem, but most students will have no idea how to proceed. Some students may draw shapes resembling parabolas; others may draw lines with negative slope. Some may realize that the concentration in the stationary phase divided by the concentration in the mobile phase must be a constant (the distribution coefficient), but not know how to represent this on the graph. After they have had some time to think it through, I then draw the idealized plot on the board and allow them some time to consider it.
b) Draw what you suspect would really happen.
Again, many students probably will not know how to approach this problem, although some usually realize that at some high enough concentration of analyte the stationary phase will become saturated and are able to draw a correct plot. I make sure to point out correct plots to other members of the groups and other groups, hearing that someone has the correct answer, usually try to listen in to see if they can figure out what would occur. We then spend a few minutes talking about what happens when you exceed the capacity of the stationary phase. Introduce the Langmuir isotherm and anti-Langmuir and talk about why they might occur in both liquid chromatography and gas chromatography.
c) What might the peaks look like in the real versus ideal situations?
I have usually mentioned previously mentioned that an ideal chromatographic peak is Gaussian in shape, but if it has not been mentioned, it may be worth asking the students what they think an ideal peak would look like. They usually respond quickly with a Gaussian and then realize that these non-deal behaviors likely introduce some asymmetry in the peak. We then have a discussion of fronting and tailing and which would be observed for the two different types of non-ideal behavior.
2. What term would we use to describe the movement of a molecule in a liquid stationary phase?
Students will probably arrive at the word “diffusion” very quickly, but some students may be under the impression that diffusion must occur across something such as a membrane. I also ask the students what we commonly say about diffusion as it relates to concentration and they readily offer that compounds diffuse from regions of high concentrations to regions of low concentrations. When probed, it is clear that the students have a misconception about this process, believing that the diffusion is almost purposeful (go from high to low) rather than the result of purely random motion and a statistical consideration and comparison of how many molecules are moving from high-to-low (more) versus low-to-high (fewer). Diagramming this on the board with a system restricted to movement in only two directions usually gets the point across, although it make take a few minutes for every student to be comfortable with this concept.
3. What processes would account for the movement of a molecule through a region of interstitial volume in a mobile phase?
Students should all agree on diffusion and flow as the processes that would account for the movement of a molecule through the interstitial regions.
4. Tswett used starch as his stationary phase.
a) What is the dominant surface functionality of starch?
Allow students a few minutes for this question. Some students may know immediately, but many have probably never considered the chemical composition of starch. Spend a few minutes talking about glucose and hydroxyl groups. Discuss the intermolecular forces relevant to this type of molecule.
b) The two other common solid stationary phases are silica gel (SiO2) and alumina (Al2O3). What do you think are the surface functionalities of these materials?
Draw the structure of a few units of silica gel to give the students an idea of where to start. Some students may think that silica gel is made into a sphere in order to avoid having surface functionality. Some may suggest capping off molecule by protonating the oxygen groups.
I spend a few minutes talking about the siloxane and silanol groups found within silica gel. I discuss the polarity of the solid support and why an organic liquid would make a good mobile phase. I also ask them
What are some problems with using a non-polar organic mobile phase?
Most groups realize that many molecules we are interested in from environmental and biological samples are water-soluble and not organic-soluble.
c) Draw a plot of the distribution of enthalpies of adsorption for a molecule on the surface of starch, silica, or alumina.
Students are completely confused by this question. I begin this problem by talking about what enthalpy of adsorption is. I also indicate that what we are going to consider is the enthalpy of adsorption of a single molecule attaching to a single silanol group (we may not be able to measure this, but we can pretend we can to see what happens). And then that we are going to measure this value over and over again many times and prepare a histogram with the number of measurements at a particular value (y-axis) versus the measured enthalpy of adsorption (x-axis). Even though students may now understand what is being asked, they rarely draw the plots correctly. Many draw exponential decays either going from high to low or low to high. Some draw linear curves with either negative or positive slopes. At this point I encourage them to think about what would happen at the two extremes. Would there be an infinite number of molecules with low energy? Would there be an infinite number of molecules with high energy? Usually by this point at least one member of each group realizes that a Gaussian distribution might be a likely observation (in reality, the distribution is unlikely to be a true Gaussian but at this point, if they can come up with a Gaussian that is good enough). I then point out that the surface of silica gel would also have some disilanol and trisilanol groups, with the expectation that silanol > disilanol > trisilanol, and I then ask them:
On the same plot, draw in two additional distributions of the enthalpy of adsorption for a molecule associating at the disilanol and trisilanol groups.
It is worth reminding them that since there are fewer of these groups than silanols, the two distributions should be smaller. Most groups have some disagreement about whether the two new distributions will occur at a higher or lower average energy. It is worth discussing at the board how a disilanol group can form two simultaneous interactions with an adsorbing molecule that is likely to make the value higher on average. Then point out how the overall enthalpy of adsorption would be the sum of the three different distributions, which is not an asymmetric distribution.
d) What would a chromatographic peak look like on such a phase with such a plot of adsorption enthalpies?
Seeing the overall distribution, most of the groups readily appreciate that the peak should show tailing.
e) What is the problem with this peak?
Concluding this problem set, I discuss some examples of adsorption that occur in the natural world. I especially talk about why oil spills in soil are so hard to clean up since active sites in the soil do not readily release the oil molecules. Also, I point out how most solid surfaces seem to have a small concentration of some especially active sites, such that the distribution of the enthalpy of adsorption is rarely symmetrical and often skewed toward the high energy side of the distribution.
I then ask them to consider a similar plot where the stationary phase is now a coated liquid (partition mechanism) rather than a solid surface (adsorption).
I indicate that we are now considering the enthalpy of solvation. While students are not necessarily sure what to think about this, when prompted they cannot think of any reason why this would be something other than a Gaussian. I then summarize why chromatographic methods based on partitioning are going to be more efficient than those based on adsorption (especially if it involves a polar surface), and remind them why the first work of Martin and Synge on the use of partitioning in chromatography was significant enough to merit a Nobel Prize. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Separation_Science/04_Instructor's_Manual/02_Chromatography_Unit/In-class_Problem_Set_1.txt |
Before beginning this problem set, I spend about five minutes introducing peak broadening and reminding them about the ideal chromatographic peak. I also remind them of the concept of theoretical plates, the idea that we want a plate height that is as small as possible, and propose that we will spend the next section of the course examining the question: “What contributes to peak broadening and how can we reduce it?”
Longitudinal Diffusion Broadening
1. Consider a “band” of a compound in a chromatographic column. The band had the following concentration profile:
a) What would happen to this profile if the flow of the column were stopped and the column was allow to sit?
Students may be confused about the diagram that accompanies this question. Some students may be tempted to say that band will shrink, others might say it will spread out but not realize why, others may even think that since the flow has stopped it will stay exactly the same. After allowing the students some time to consider this question, summarize how diffusion will cause this peak to become broader.
b) Would this phenomena be more significant (i.e. happen faster) in a gas or a liquid?
Students should realize right away that diffusion occurs faster in a gas, but ask them how much faster. You may get answers ranging from ten times to ten million times faster.
c) Does this phenomenon contribute more to band broadening at higher or lower flow rates?
Most groups realize that a slower flow rate means that the compounds will have more time in the column and then more time for longitudinal diffusion to occur. Students should realize that this means that there is a greater contribution to peak broadening from longitudinal diffusion at lower flow rates. I summarize this conclusion, set up the concept of how we will be developing something known as the van Deemter equation that will eventually include terms for all of the different contributions to peak broadening, and then ask them:
Should the term for the contribution to peak broadening from longitudinal broadening be multiplied or divided by the mobile phase velocity.
Even though they understand that there is a greater contribution at lower flow rates, they may have trouble understanding how to incorporate that into the equation. With a few minutes time, each group arrives at the correct form of the expression to include in the equation.
Eddy Diffusion Broadening
Consider molecules flowing through a packed bed of particles.
a) Would different molecules have different path lengths as they passed through the bed?
Students should have no trouble with this question.
b) Is the difference in path length between the shortest and longest path dependent at all on the diameter of the particle? If so, which particles (smaller or larger) would lead to a greater difference?
Students will immediately acknowledge that there is a dependence on particle size, but may not know which will lead to a greater difference. Some may argue that with larger particles there are fewer paths that lead straight out of the column and with smaller particles there is the possibility of zigzagging all the way across the column. If anything over the years I have taught this course, more students think that smaller particles will be worse than larger particles. I encourage them to only consider realistic paths; with the flow pushing the molecule down the column, the molecule will not zigzag across the column. Eventually we get to the point of agreement that large particles lead to the greatest differences in path length.
c) Some packed columns exhibit channeling. What do you think is meant by this term?
Some students think that channeling has to do with the large interstitial regions resulting from using large particles. Other may recognize that it has something to do with improperly packed columns. Allow them just a few minutes to think about this and then discuss streamlined paths and the problem this introduces with regards to broadening of peaks.
d) Would channeling be more likely to occur with smaller or larger particles? In other words, which is more difficult to pack efficiently, larger or smaller particles? A column is packed efficiently when the particles are in a uniform bed with the minimum amount of voids.
A majority of the students will probably argue that it is harder to pack large particles efficiently. Remind them that efficient packing has to do with minimizing the voids; just because large particles result in large voids, it doesn’t mean that the voids can’t be minimized. Encourage them to imagine trying to pack large and small objects efficiently (fitting in the most number of particles possible in the allotted space implies the tightest and most efficient packing). How easy is it to get the smaller particles perfectly settled in relative to larger particles? It may take a while for the students to be convinced that smaller particles are harder to pack efficiently.
e) Do open tubular capillary columns exhibit eddy diffusion?
Students should have no trouble with this question. They should realize that having no particles means that there is no eddy diffusion.
f) Does this phenomenon exhibit any dependence on flow rate?
This is actually a highly debated question with conflicting data and representations in the literature (see text for a further discussion). Students may initially be tempted to say that there is no dependence on flow rate, which is consistent with the initial form of the van Deemter equation. They will argue that since eddy diffusion just has to do with the difference in path length, the time that it takes to travel those paths is insignificant. Encourage them to think about this question in more depth. Won’t a slower flow rate allow the molecules to sample more of the column? Won’t a faster flow rate cause molecules to be more likely to stay in one path or channel? Allow them to debate both arguments and then discuss how this is still a greatly debated topic. Introduce the eddy diffusion term and include it in the van Deemter equation.
Stationary Phase Mass Transport Broadening
Consider a compound that has distributed between the mobile and the stationary phase within a plate in a chromatographic column. The following diagram might represent the concentration distribution profiles in the two phases (note that the compound depicted has a preference for the mobile phase).
a) What would happen to these concentration profiles a brief instant of time later?
Allow students a few minutes to think about this problem. Some students may think that it will just stay the same. Some students may think that both concentration profiles will shift down. Remind them of the key difference between a mobile phase and a stationary phase. Encourage them to think about what would happen to molecules in the stationary phase at the top of the diagram once the molecules in the mobile phase have shifted down. What about the molecules in the mobile phase at the bottom of the diagram?
b) Will what happens in (a) contribute to band broadening? Explain.
Students should realize that this will contribute to peak broadening because the overall Gaussian distribution becomes wider. Discuss how molecules must spend a finite amount of time in the stationary phase. This is a good time to discuss that the lag in mass transport that effectively traps molecules in the stationary phase for a finite amount of time means that a chromatographic column is constantly at disequilibrium (even though we often present it as if the compound equilibrates over the section of the column where it is distributing between the two phases).
c) Does the contribution of this phenomenon to band broadening exhibit a dependence on flow rate? If so, are there any troublesome aspects to this dependence?
Students should realize that the faster the flow rate, the more broadening there will be. You may ask them to draw two profiles for the distribution in the mobile phase a brief instant of time later for one flow rate and then double that flow rate. Once students realize the correct answer, spend a few minutes summarizing the conclusions and talking about how with slower flow rates the system is able to reach a state that is closer to equilibrium. Introduce the stationary phase mass transport term for the van Deemter equation. Allow students some time to realize that they now have one term in the equation that recommends use of a faster flow rate and one term that recommends use of a slower flow rate to optimize efficiency.
d) What happens to this effect as the stationary phase coating is made thicker?
Students should realize right away that with a thicker coating, it will take longer for particles to diffuse out of the stationary phase and therefore broadening will be increased. At this point, I then ask them:
How could you create a packed column with a thinner coating of stationary phase?
Students usually decide to apply a lower percent coating, but talk about the disadvantages (reduced capacity) of doing that. Ask them to think of a method that retains the same percent loading but does not decrease the capacity. The students should eventually think of using smaller particles in order to keep the same weight of the liquid phase and solid support but increase the surface area. Discuss some of the drawbacks to using smaller particles (need to take more care in coating to insure uniform surface coverage, more care in packing to avoid channeling).
e) Capillary GC columns have very thin coatings. Describe one advantage of these columns.
After completing part (d), students realize that a thinner coating will minimize the broadening due to stationary phase mass transport. I usually spend a few minutes talking about how liquid coatings are applied to capillary columns. This is also a good time to talk about the history of capillary gas chromatography including how glass capillary columns used to be made. Also discuss the use of fused silica capillary columns and how they compare to regular glass. It would also be a good idea to mention the low capacity of these types of columns and the use of a split injection system. Talk about the role that fiber optics played in the development of capillary gas chromatography.
f) Compare the effect of this phenomenon on a uniform versus non-uniform stationary phase coating.
The groups are usually quick to realize that a non-uniform coating will be undesirable. I summarize this, spend a few minutes talking about the methods for coating particles, and remind them of the deleterious effect if the coating process leads to regions of exposed surface that will result in adsorption.
g) Is this effect of more concern in gas or liquid chromatography?
Allow the students a few minutes to consider this question. The students often focus on the difference in the mobile phases of gas and liquid chromatography, and assume that since gases diffuse faster than liquids, are tempted to say that this is more of a concern with liquid chromatography. Remind them that stationary phase mass transport broadening just has to do with the amount of time a molecule stays in the stationary phase. Once a molecule has entered the stationary phase, it doesn’t matter whether the mobile phase is a gas or a liquid. In our current understanding, the stationary phases in gas and liquid chromatography are both liquids. Challenge them to think of some other key differences between gas and liquid chromatography. Ask them what temperature a gas chromatograph oven is usually at compared to a liquid chromatography column. With this information, you can ask them the effect that the different temperatures of gas and liquid chromatographic columns will have on diffusion rates. They should now see that stationary phase mass transport broadening is likely to be more of a concern in liquid chromatography, but not as much as they originally thought when they only considered diffusion rates within the mobile phase.
At this point, I discuss the explosion of gas chromatography as an analysis method in the 1950s-70s and how that led to efforts to volatilize non-volatile compounds.
I also introduce Calvin Giddings and his paper titled, Liquid Chromatography with Operating Conditions Analogous to Those of Gas Chromatography, which concludes that ultra-small particles will be needed if LC is to have an efficiency comparable to GC (although we have not yet discussed mobile phase mass transport broadening).
What issues might there be in coating ultra-small particles with a very thin liquid coating?
What issues arise trying to keep a liquid coating on ultra-small particles when a liquid is the mobile phase?
I usually pose these two questions to the entire class rather than the groups and solicit individual student responses. We then discuss issues such as coating uniformity and degradation of the liquid coating due to physical removal from the solid support by the flowing liquid and because the liquid coating and liquid mobile phase will still have some slight solubility in each other even if they are regarded as immiscible (while oil and water don’t mix, they actually do to some small extent). I then spend some time talking about the development of bonded phases for liquid chromatography (especially C-18 phases) and the significance of this discovery to the field of liquid chromatography.
Mobile Phase Mass Transport Broadening
Consider a capillary column as shown below.
The dot represents a molecule that has just left the stationary phase is about to diffuse across the mobile phase and re-encounter the stationary phase on the other side of the column.
a) Draw a line representing the path of this molecule.
Most students realize right away to draw a line that traverses from one side of the channel to the other. Some need prompting (usually to look at what another student in the group has drawn).
b) What would this path look like if the flow rate were doubled?
Students recognize that this would result in the molecule being pulled further down the column before encountering the stationary phase again.
c) Is it important for the molecule to encounter the stationary phase? Think about a situation in which the flow was so fast that the molecule never re-encountered the stationary phase.
Students realize that without a molecule encountering the stationary phase, there is no separation and therefore no chromatography. Therefore the contribution of mobile phase mass transport broadening is more significant at higher flow rates.
d) If using capillary columns, what does this suggest about the desirable diameter for such a column?
Students realize right away that the narrower the column the better.
e) Is this phenomenon worse in gas or liquid chromatography?
Having fully developed the effect for the stationary phase, they realize right away that the much slower diffusion in liquids means that mobile phase mass transport broadening is much more significant in liquid chromatography.
f) How does the contribution to band broadening depend on flow rate?
Again, they usually realize it is exactly analogous to stationary phase mass transport broadening.
g) Would this effect be observed in a packed column? If so, how?
Generally, students realize immediately that the interstitial volume in a packed column is analogous to the open space in a capillary column.
h) How could you reduce this effect in a packed column?
Students usually immediately propose using smaller particles.
i) Say something about the length of a column needed if you undertake what you have suggested in (h), assuming that you have maintained a constant thickness of stationary phase.
I first point out that this is exactly the situation that occurs with a bonded phase liquid chromatographic column, where the thickness is set by the nature of the bonded C-18 groups and does not depend on the particle size they are bonded to. Students immediately realize that the smaller particle has a larger surface area, and then usually soon realize that this means that the capacity of a bonded phase LC column with smaller particles is greater than that with larger particles. With greater capacity they realize that, when using smaller particles in LC, you can maintain the same capacity by using a shorter column. When asked what else they will observe with a shorter column, most immediately realize that it will take less time to run the chromatogram.
I then spend a few minutes discussing plots of h versus flow rate and talk about the important differences between the plots for liquid chromatography and gas chromatography. Talk about the optimal flow rate and realistic flow rates. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Separation_Science/04_Instructor's_Manual/02_Chromatography_Unit/In-class_Problem_Set_2.txt |
Fundamental Resolution Equation
I introduce the fundamental resolution equation, spend a few minutes reminding them about the variables in the equation (each of which we have seen before), and provide them with a photocopy of a derivation of the equation that we quickly go through. I do not want them to know how to do the derivation, but do want them to appreciate that it is a derivable expression.
1. Describe ways in which the number of plates on a chromatographic column can be increased. Are there any tradeoffs associated with these changes?
Allow students a few minutes to consider this question. They may think of some methods such as decreasing the plate height (remind them to go back to all of the terms in the van Deemter equation and think of experimental variables they can optimize), packing the column more efficiently, using a longer column, using smaller particles, or optimizing the flow rate. Challenge them to think about what each of these changes would mean experimentally.
2. Describe ways in which the separation factor can be increased. Are there limits to the effect that increasing the separation factor has on chromatographic resolution?
Most students have probably had some experience with thin-layer chromatography. If so, they may think of changing the eluent identity to separate compounds. Spend time talking about how you might change the identity of the stationary phase in both liquid and gas chromatography. Mention that changing the identity of the mobile phase in liquid chromatography is considered changing the retention factor.
Is there is any disadvantage to having too large a selectivity factor.
If they are stumped by this, drawing a chromatogram on the board for two components with a large selectivity factor will make apparent that too much time is being spent waiting for the second peak and that the separation is too good (unless one wanted to scale this up for a preparative separation where a higher selectivity means that more compound can be isolated in each injection).
3. Describe ways in which the retention or retention or capacity factor can be increased. Are there any tradeoffs associated with those changes? Are there limits to the effect that increasing the retention factor has on chromatographic resolution?
Since we have just discussed that changing the mobile phase in liquid chromatography is considered a change in capacity, ask students to think of ways in which the mobile phase can be changed. Reminding them that the mobile phase is aqueous-based, they will usually come up with pH. Since some groups are using LC in their lab project, they also mention varying the percent organic modifier. I then spend some time talking about the effect of pH on retention time, and how it is often possible to reverse the retention order of organic acids and bases by changing the pH. We also spend a few minutes talking about the effect of changing the percent organic modifier.
It is then worth reminding them of (or having them look up) the equation we had earlier in this section that defined the retention factor. I then asking for other ways they might change the capacity in gas chromatography. Some groups immediately see that the volumes of the stationary and mobile phases are in the expression and think of changing them. At some point we have a discussion about whether it is practical to change the volume of the stationary phase (yes) or mobile phase (no) in GC. Similarly, we think about bonded phase LC materials and realize that the use of smaller particles leads to both an improvement in the number of plates and the retention factor. When prompted about GC, and variables that might affect the distribution coefficient, students talking in their groups eventually think of temperature as a variable in gas chromatography.
In order to increase the capacity of the column, would you need to increase or decrease the temperature?
While the students usually seem to want to raise the temperature as a way of “improving” the chromatography, if asked to think and reason their way through the effect that the temperature has on the retention factor and resolution (and I remind them that presumably we have two compounds that are not fully resolved and want to improve this), they usually determine that they will want to lower the temperature of the GC oven.
Is there is any disadvantage to having too large a retention factor.
The groups usually realize immediately that too large a value means a much longer analysis time.
4. Consider the following chromatogram.
The early eluting peaks and the later eluting peaks exhibit a problem.
a) Describe the chromatographic nature (there is a particular chromatographic term that describes each) of the problem.
Let the students spend a few minutes on this problem. Students will probably be tempted to say that the early peaks exhibit an issue with the separation factor. Point out to them that the peaks have started to separate, which is a good sign. It is likely worth indicating that presumably we want to focus on terms within the fundamental resolution equation and that they may wish to go through them one by one and think about what adjusting each would do. It may be worth pointing out that the later two eluting peaks are well resolved but stay on the column too long. Eventually the groups usually get to the realization that the early eluting peaks have too small a retention factor, and that the later eluting peaks have too large a retention factor. This conclusion should be summarized for the group.
b) Propose a way in gas chromatography to eliminate both problems.
Students will most likely not think of solutions on their own. Remind them of variables that affected the retention factor that were discussed when considering the fundamental resolution equation. Eventually someone in a group usually brings up the role of temperature and I ask them to consider how temperature might be used to solve this problem. Groups eventually propose changing the temperature during the chromatogram (it is worth asking them what they would start and finish with to addresses the problem) and we then discuss running a temperature program in gas chromatography.
c) Propose a way in liquid chromatography to eliminate both problems.
When prompted as to whether there is something essentially equivalent in LC to the role of temperature in GC, most groups realize that altering the mobile phase composition (either pH or organic modifier) during the chromatogram could have a similar effect. I mention the concept of a gradient elution and ask them to determine what they would do with the percentage of organic modifier during the gradient. They are usually able to reason this out correctly, and we have a brief discussion of the difference between isocratic elution to gradient elution.
At this point in the course I use a lecture format to introduce basic aspects of liquid chromatography and the topic of steric exclusion chromatography. See the learning objectives and textbook for the topics that are covered. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Separation_Science/04_Instructor's_Manual/02_Chromatography_Unit/In-class_Problem_Set_3.txt |
Ion Exchange Chromatography
I spend a few minutes introducing ion exchange chromatography, showing how it is possible to attach fixed cations or anions to polymeric resins and how these then have an exchangeable counterion.
1. Describe a scheme using ion exchange chromatography that would enable you to deionize water. Say something about the capacity of the ion exchange resins you would use for this purpose.
Allow students about few minutes to work on this problem. Students should have no trouble recognizing that you would need to run the water through a pair of columns in order to remove the anions and the cations.
What ions should be used as the counter-ions in the column?
Within a few minutes the groups can usually figure out that hydronium and hydroxide ions are needed if the goal is to deionize the water. They also realize that high capacity resins will be the best for deionizing water. I then spend time summarizing what would happen to each ion in the two-column system. I make sure that the students understand that the capacity is limited by the number of derivatized aromatic rings. I discuss the water purification systems that we use in the department and how the measurement of conductivity is used as a way to determine how well the water has been deionized.
2. Would ion exchange resins that are useful for deionizing water be useful for analytical separations?
The students will need to know what is meant by analytical separations (trace levels of ions). I also ask them to consider what it would take to be able to actually have the ions elute from the column. The groups can usually reason out that the high capacity resins used for deionizing water would lead to exceptionally long analysis times if used for trace analyses. They also propose including eluent ions in the mobile phase. I ask them to specify what ions they might use as their mobile phase counterions and they can usually come up with hydronium (hydrochloric acid) for cations or hydroxide (sodium hydroxide) for anions. I then summarize these concepts at the board.
3. What would be the order of retention for the ions Li(I), Na(I), and K(I) on a cation exchange resin? Justify your answer.
I allow students about ten minutes to consider this problem on their own. They always address this by considering how strongly the ions might associate with the resin. They may find compelling arguments for both sides including the charge density of lithium or the steric hindrance of potassium. I ask them to think about the equation that describes the electrostatic attraction of two ions, which has the two radii in it, and most eventually conclude that the lithium will associate more strongly with the resin and elute last.
After summarizing this as a reasonable prediction, I then ask them to consider mobile phase effects. In particular, I ask them to think about what they know about the structure of water and what would happen to this structure when ions dissolve in water. I also ask them to think about drawing a picture for the environment around a lithium, sodium or potassium ion in water. Students may realize that lithium is more stable in the eluent water because it causes the least disruption to the physical network of hydrogen bonds in the liquid water and has the strongest electrostatic attraction with the negative ends of the water molecules. They eventually conclude that a consideration of mobile phase effects would lead one to predict the opposite retention order.
With two opposite predictions, I suggest that we try the experiment. We don’t actually try the experiment, but I indicate that experimental data shows that for ions of the same charge, the mobile phase effects are more important and smaller ions elute first. I also introduce the concept of the solvophobic effect.
What if you had ions of varying charge but the same size (a +1 and +2 ion of similar size)? What is the retention order based on mobile phase effects? What about the retention order based on stationary phase effects?
Students can reason out that, again, the two different considerations lead to different retention orders. Experimental data shows that the +2 ion elutes last, and I discuss how ions of higher charge have much stronger association with the resin because they can bind simultaneously to two of the fixed ion sites and require two mobile phase counterions to be pushed out of the resin and migrate down the column.
4. Consider the case of separating the alkali ions in (3) on a polystyrene resin using a fairly dilute solution of hydrochloric acid as the mobile phase.
a) What is the bound ion and the mobile counter ion?
Groups can readily answer this.
b) One problem is how to detect these ions. They do not absorb ultraviolet or visible light in the accessible portion of the spectrum. They do not absorb infrared light. Conductivity might work except that the hydrochloric acid in the mobile phase produces too high of a background signal. Devise a way to remove the conductivity of the eluent ions (HCl) but retain the conductivity of the alkali ions you wish to detect.
Students understand the problem but usually struggle to come up with a way to solve it. I usually draw a representation of the column on the board and show the NaCl and HCl eluting simultaneously and indicate that the goal is to remove the conductivity of the HCl while retaining the conductivity of the NaCl. Many of the groups start to think that there should be some way of neutralizing the acid but when asked what they would use (and the typical response is NaOH), they realize that this would eliminate the conductivity of the H+ (by reacting it with OH- and converting it to water) but replace it with Na+ leaving NaCl, which has not solved the problem. Through prompting they come to the realization that they need a way to selectively replace the Cl- with OH-, which converts the HCl to water and the NaCl being analyzed to NaOH, which is still conducting. Groups then realize that running the sample eluting from the analytical column through an anion exchange column in the hydroxide form will solve the problem. I then summarize the use of suppressor columns in ion chromatography at the board. I also explain the possibility of using an indirect spectrophotometric detection method as well. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Separation_Science/04_Instructor's_Manual/02_Chromatography_Unit/In-class_Problem_Set_4.txt |
(See also: Affinity Chromatography)
07 Specialty Topics
Ion chromatography (IC) is a subset of liquid chromatography methods: ion exchange, ion exclusion, ion pair chromatography. IC is a useful tool for determining the presence and concentration of ions in samples and is utilized in numerous industrial and research settings including a test for authentic tequila and for environmental analyzes such as the determination of anions (\(PO_4^{3-}\), \(Cl^-\), \(NO_3^-\), etc) in surface waters. Ion-exchange is the basic principle behind the removal of cations and anions from drinking water using most commercial, such as Brita, water filters. Ion-exchange is also a natural process that occurs with clay substrates, resulting in the mobility of cations in soils.
This module provides an introduction to Ion Chromatography. In this module the basic theory and applications of IC will be presented at a level that assumes a basic general chemistry background. It is designed to move through the module sequentially using the links below. This module presents the basic history, theory, and applications of IC. The module focuses on ion exchange chromatography, as it is the most common use of IC in environmental analysis, but includes the related ion exclusion and ion pair methods.
Ion-exchange Chromatography
Upon completion of this module you should be able to
1. Explain the basic principles, operation and application of IC.
2. Differentiate between IC and other chromatographic methods.
3. Explain the chemical basis for stationary phase effects and mobile phase effects.
4. Predict retention order given the relative dominance between stationary phase effects and mobile phase effects.
5. Differentiate between stationary phases used in anion exchange and cation exchange.
6. Explain the basis for the common IC detection methods.
7. Describe the general process of analyzing a sample by IC.
8. Explain the rationale for using a suppressor cartridge and how it works.
9. Predict the effects of overloading, eluting too quickly, eluting too slowly, baseline drift.
02 History
Ion Chromatography (IC) methods were first reported around 1850 when H.Thomson and J.T. Way used various clays as an ion exchange and extracted labile calcium, magnesium, and ammonium ions. In 1927, the first zeolite column was used to remove Mg2+ and Ca2+ from water. Cation exchange using a sulfonated polystyrene/divinylbenzene column was developed in the 1940s as part of the Manhattan project. Very large columns were used to concentrate and purify the radioactive nucleotides required for the atom bomb. In the late 1940s anion exchange was performed with the attachment of a quaternary ammonia on the polystyrene/divinylbenzene support. The industrialization of the technique occurred in the 1970s when Small et al (Small, H.; Stevens, T.S.; Bauman, W.C. Anal Chem, 1975, 47, 1801) developed a suppressor column that enabled conductivity detection.
For a more thorough description of the history see Small, H J Chem Ed, 2004, 81(9), 1277-1284.
03 Basic Principles of Ion Chromatography
Basic process of IC
The basic process of chromatography using ion exchange can be represented in 5 steps: eluent loading, sample injection, separation of sample, elution of analyte A, and elution of analyte B, shown and explained below. Elution is the process where the compound of interest is moved through the column. This happens because the eluent, the solution used as the solvent in chromatography, is constantly pumped through the column. The chemical reactions below are for an anion exchange process. (Eluent ion = , Ion A= , Ion B = )
Step 1: The eluent loaded onto the column displaces any anions bonded to the resin and saturates the resin surface with the eluent anion.
(key: Eluent ion = , Ion A= , Ion B = )
This process of the eluent ion (E -) displacing an anion (X -) bonded to the resin can be expressed by the following chemical reaction:
Resin + -X - + E - <=> Resin + -E - + X -
Step 2: A sample containing anion A and anion B are injected onto the column. This sample could contain many different ions, but for simplicity this example uses just two different ions ready to be injected onto the column.
Step 3: After the sample has been injected, the continued addition of eluent causes a flow through the column. As the sample elutes (or moves through the column), anion A and anion B adhere to the column surface differently. The sample zones move through the column as eluent gradually displaces the analytes.
Question to consider: How would you write the chemical reaction for elution process with respect to anion A and anion B. How would you write the Kf expression for the two reactions? How would you sketch the elution process at this step using a figure similar to the figure in Step 1 if the Kf for anion A() is larger than the Kf for anion B()?
Step 3: The continued addition of the eluent causes a flow through the column. As sample elutes, anion A and anion B adhere to the column surface differently. The sample zones move through the column as eluent gradually displaces the analytes.
In reality not every eluent ion is removed from the surface of the column. It depends on the amount of analyte loaded. A better representation of the column can be seen by just looking at a slice of the column where the separation is occurring, as shown in the figure below.
Step 4: As the eluent continues to be added, the anion A moves through the column in a band and ultimately is eluted first.
This process can be represented by the chemical reaction showing the displacement of the bound anion (A -) by the eluent anion (E -).
Resin + -A - + E - <=> Resin + -E - + A -
Question to consider: If ion B had a very strong affinity for the resin, how would the elution time for ion B be affected? If it takes forever to come off, would this be useful in trying to determine the quantity of that ion present? When might this be useful? (Hint: go back to the introduction to the module and look at where ion-exchange is used...)
(Answer: As the affinity ion B has for the resin increases, the elution time would increase. If the affinity becomes large enough, in essence anion B will stay on the column. This phenomena is utilized in water filtration where ion exchange is used to remove particular ions from the sample.)
Step 5: The eluent displaces anion B, and anion B is eluted off the column.
Resin + -B - + E - <=> Resin + -E - + B -
The overall 5 step process can be represented pictorally:
Stationary phase (or resin) composition
There are a number of different resins or stationary phases that have been developed for use in IC. The main classes of substances used are: modified organic polymer resins, modified silica gels, inorganic salts, glasses, zeolites, metal oxides, and cellulose derivatives. The most commonly used resins are the silica gels and polymer resins. As the sample is injected onto the column, the two different analytes briefly displace the eluent as the counter -ion to the charged resin. The analyte is briefly retained at the fixed charge on the resin surface. The analytes are subsequently displaced by the eluent ions as the eluent is added to the column. The different affinities (see the chemical reactions in the basic process section) are the basis for the separation. The Kf values of each reaction is also known as the selectivity coefficient. The greater the difference between the Kf values for the two analytes, the more the two analytes will be separated during the ion chromatography process. In reality, the interaction between the solvent and the analyte can also have an impact on the order each analyte is eluted. For a more in-depth analysis of predicting the retention order see the material by Dr. Thomas Wenzel. (http://www.bates.edu/x65385.xml)
The common cation exchange resins are based on either polystyrenedivinylbenzene (PS-DVB) or methacrylate polymers. The surface of these polymers (Figure 1) is functionalized with a negatively charged sulfonated group (-SO3-). The cation in the eluent or the analyte of interest is the counter-ion in the vicinity of the charged functional group.
Figure 1: cation exchange surface
The surface of the polymer is functionalize with a quaternary amine (-N+R3) for anion exchange (see Figure 2). The quaternary amine provides a positive charge to the surface, attracting negatively charged anions in the liquid phase. Just like the cation exchange resin, the anion of the eluent or the analyte of interest exists as the counter-ion in the vicinity of the positive charge residing on the amine.
Figure 2: anion exchange surface. The R stands for some organic (C and H) chain. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Separation_Science/07_Specialty_Topics/Ion-exchange_Chromatography/01_Learning_Objectives.txt |
In order to have useful information, you need to be able to detect what comes out of the column. The analytes are ions that come out in separate sample bands, which means there is a small part of the solution coming out with a higher concentration of ions.
Based on your knowledge, how could you possibly detect ions or changes in ion concentrations coming off of the column?
There are many different possibilities of detecting ions. Due to its simplicity, most instruments use conductivity.
Conductivity
Conductivity is the measure of a material’s ability to conduct electricity. Since conductivity is proportional to the number of ions in solution, it is the primary method of detection for ion chromatography. One problem with measuring conductivity is the high conductivity that may be present in the eluent. Conductivity became common with the use of a suppressor.
The suppressor is a cation or anion exchanger after the ion exchange column that replaces the eluent ions with either H+ or OH-. If you are performing cation anlysis, the eluent is acid, and the exchanger replaces the eluent counterion with OH-. This then converts much of the eluent to neutral H2O. Thus the suppressor greatly reduces the conductivity contribution from the eluent, enabling the signal from the analyte of interest to be more readily detected.
Other Methods
Other detection methods have been coupled with IC, including mass spectrometry, atomic spectroscopy, fluorescence, luminescence, UV-Vis, and potentiometric. Most require post-column reactions to generate the signal or are so selective they are not useful in detecting multiple analytes simultaneously. Of these methods, the most likely to be broadly used is mass spectrometry-(i.e. the determination of ionic compounds in toothpaste Cavalli, S; Herrmann, H; Höfler, F; LC GC Europe, 2004, 17(3), 160).
05 Chromatograms
The time to elute an analyte is a function of how long the analyte is retained on the column, therefore the output of IC is a graph of conductivity as a function of time, called a chromatogram. Based on the previous discussion of elution, you may expect a chromatogram to look something like:
However, we need to consider another process that impacts shape of the elution peaks. As an analyte flows through the column, some of the analyte molecules may pass by through the length of the stationary phase faster or slower than would be expected due to diffusion processes or the formation of channels. An example of this can be seen in the figure with the path of each ion shown in the figure.
If the two ions are traveling at the same speed, set by the flow of the eluent, what can you say about when they will emerge? Will they emerge at the same time?
From the figure, you should be able to see that the path in red is much shorter than the path in blue. Since the path is shorter, and the ions are traveling at the same speed, the ion following the red path will emerge first. Thus a normal chromatogram peak will have a gaussian distribution, symmetric around the mean, as seen in the figure on the right. You can also peruse a more extensive mathematical modeling of the chromatogram peak here (link takes you to a different website http://www.chem.uoa.gr/applets/AppletChrom/Appl_Chrom2.html ).
Since we are concerned with the concentration of ions present in the solution, how will the chromatogram change as you increase the amount of analyte loaded onto the column?
After considering the prior question: Since we are concerned with the concentration of ions present in the solution, how will the chromatogram change as you increase the amount of analyte loaded onto the column?
The chromatogram peak will increase in height and concamitantly in area. Therefore when quantifying data, the peak height or area is used. In order to determine actual concentrations, a series of standards must be analyzed to calibrate the response between peak area and actual concentration for each ion.
An example chromatogram of Poland Springs bottled water. Each separate peak is due to a different cation.
The area under each peak is used to calculate the concentration of each ion. What information would you need in order to determine the relationship between peak area and concentration?
Just like any other quantitative method, you need to calibrate the response, in this case by analyzing a series of known standards and plotting a calibration curve.
If you notice, the last peak is not completely guassian, there are other factors that do have an impact on peak shapes. For more in-depth information on peak shape, see the following material. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Separation_Science/07_Specialty_Topics/Ion-exchange_Chromatography/04_Detection_Methods.txt |
You now have an outline of the basic ion chromatography process. The loading of the sample onto the column varies with the instrument. The sample is eluted off of the column, through the detector. The signal from the detector is converted into the chromatograph. Lastly, considering you have the eluent flowing through a column with very fine particles. How can you force water through a column containing very fine particles?
When a water filter is used in a pitcher, the force of gravity is used, and when a water filter is used on a tap, the water pressure in the pipes is used. In our situation, a much higher pressure is required, since the solid phase particles are much smaller that a water filter on a household tap.
The pressures required for most IC instruments is at least 600 psi. In order to achieve this pressure a double piston high pressure pump is used, one such example is shown in the picture.
The injection of the sample onto the column is performed using a multiport valve that is inline with the eluent tubing. Different instruments have slightly different styles, with the most common a direct port for injection.
Another example is where an autosampler is used. In the photo to the right, an autosampler is shown, where a peristaltic pump (image below) pulls the sample up the autosampler tube and into the sample loop on a six-port valve. This is the setup used in the Lachat 8500 QuickChem series with an ion chromatography channel.
Software controls the six-port valve and at the appropriate time the valve switches to have the sample flow onto the column.
The six port valve works by first having the flow of the sample go through the sample loop (figure on the left) and then to the waste. This fills the sample loop with the sample. At a set period, the valve turns connecting different lines coming in. The eluent now forces the sample from the sample loop onto the column.
One instrument, a Lachat 8500 QuickChem with IC, is shown below illustrating a guard column, analytical column, and the suppressor cartridge. The guard column is used to protect the analytical ion column from contamination.
07 Experiments
There are a number of different possible lab experiments that utilize IC analysis, from one week experiments to semester long projects.
1) An experiment to cover one four hour lab period is the analysis of the ion content of bottled water. If you wish to use this as a dry lab exercise, sample data can be provided to students with the sample data set (in Excel), and chromatograms (in pdf) collected on a suite of bottled water samples in 2009.
2) An experiment that would cover 2-3 four hour lab periods:
-evaluating whether a stream is contaminated by road salt applications.
3) An project that could cover multiple weeks with more sample preparation due to more complex samples. This could easily be adapted for project based learning experiences.
- anions in fruit juices see Whelan et al, J Chem Ed 81(9), Sep 2004, 1299-1302.
08 Troubleshooting
There are troubleshooting guides for HPLC, a related chromatographic technique. In many instances the troubleshooting is similar. The HPLC troubleshooter (www.dq.fct.unl.pt/QOF/hplcts.html) online interactive guide can be useful for IC as well.
There are several common issues that show up in IC, see if you can think about these chromatograms and what might cause the problem.
Sample Chromatogram 1
If you look at the first big peak, the shape is quite a bit different than the other peaks. It is very broad and has fronting to the peak. This is due to overloading the column, thus there is not a great separation sodium peak, and it dwarfs any other peak that would have a shorter elution time. This sample was from a tidal river, thus the high sodium concentration.
Several other common issues that show up in IC are if the suppressor column starts to fail, the eluent solution changes either in pH or ion concentration, or the analytical column gets contaminated. These tend to cause baseline drifts or very poor peaks. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Separation_Science/07_Specialty_Topics/Ion-exchange_Chromatography/06_Basic_Instrumentation.txt |
Goal:
This module is designed to provide theory and practical use of size exclusion chromatography (SEC) for separation of proteins.
In-Class Questions:
1. What is an obvious physical difference between an immunoglobulin (antibody), which is a protein, and glucose, which is a monosaccharide?
2. When would it be useful to separate immunoglobulins from glucose in blood?
3. How could you devise an experiment to separate immunoglobulins from glucose utilizing the physical difference between the molecules determined in the first question?
Introduction:
Size exclusion chromatography (SEC), also called gel filtration, gel permeation, molecular sieve, and gel exclusion chromatography, is a separation technique used to separate molecules on the basis of size and shape (hydrodynamic radius). Size exclusion chromatography is called gel filtration chromatography because the gel essentially allows for the filtering of molecules from a sample based upon molecular size. However, unlike other techniques, the larger molecules elute first. This technique is widely applicable to purification or desalting of proteins in complex samples such as blood, which contain molecules of widely different sizes.
Theory:
In-class question:
4. Depicted below are three different sized molecules. Next to them is shown a pore. What would happen as each molecule interacted with the pore? How would this effect the retention time of each molecule?
Molecules of different sizes can be separated by this technique because of differential time spent inside a solid phase particle which excludes entrance of relatively larger molecules, allows some entrance of medium-sized molecules, and allows free accessibility of the smallest molecules. The particles contain pores with tunnels in which the size can be controlled depending on the size of molecules to be separated. Smaller molecules experience a more complex pathway (like a maze) to exit the particle than do larger molecules. Because molecules that have a large size compared to the pore size of the stationary phase have very little entrance into the pores, these larger sized molecules elute first from the column. Medium sized molecules are relatively large compared to the pore size of the solid phase and therefore may find some pores in which they enter and spend some time. Smaller-sized molecules have more pores that are accessible to them and therefore spend more time inside the pores relative to larger-sized molecules. Therefore, smaller molecules elute last and larger molecules elute first in Size Exclusion Chromatography.
Figure depicting the separation of three molecules by Size Exclusion Chromatography (SEC):
a.The black molecule is much smaller in size compared to the pore size of the solid particle and has total access to the pores (not excluded).
b.The Green molecule is somewhat smaller in size compared to the pore size of the solid particle and has some access to the pores (partially excluded).
c.The purple molecule is larger in size than the pore size of the solid particle and does not have any access to the pores (totally excluded).
The figure below shows a schematic representation of the operation of a size exclusion chromatographic column:
The link below shows an animation of a SEC separation:
www.wiley.com/college/fob/quiz/quiz05/5-6.html
The Solid Phase Material:
Pore Size
Solid phase materials used in SEC are usually classified based on their ability to separate different sizes of proteins. Since size is a difficult item to accurately measure for a large molecule, the solid phase materials are identified with a molecular weight range instead and the weight is equated with size. All compounds with a molecular weight less than or equal to the lower number in the range will see the entire internal volume of the beads resulting in no selection and therefore no separation. All compounds with a molecular weight greater than or equal to the higher number in the range are completely excluded from the inside of a bead and therefore no separation is achieved. Molecules with weights or sizes between these two extremes of the range can be separated. This is the numerical pore size range reported for each solid phase material used in SEC. The pore size used for a separation is dependent on the size range of the particular set of molecules to be separated. Smaller pore sizes are used for rapid desalting of proteins or for protein purification. Intermediate pore sizes are used to separate relatively small proteins. Very large pore sizes are used for purification of biological complexes.
In-class Question:
5. Other chromatographic methods depend on specific interactions between the molecules being separated and the surface of the solid support. What effect might such specific interactions have on the method you came up with to separate glucose from immunoglobulins? Can you think of how you might limit specific interactions between the molecules and the surface?
Types of Material
Protein separations are generally performed using materials composed of dextrose, agarose, polyacrylamide, or silica which have different physical characteristics. Polymer combinations are also used. Ideally the materials will have no interaction with the proteins or maybe modified so that there is little interaction between the material and protein. The dextran sephadex is most commonly used in protein separations. Typical separation ranges that can be achieved using sephadex are given below. Molecules ranging from 100 to 600,000 Da can be separated depending on the type of sephadex chosen.
G10 (100-800Da), G15 (500-1500 Da), G25 (1000-5,000 Da), G50 (1,500-30,000 Da), G75 (3,000-80,000 Da), G100 (4,000-150,000 Da), G150 (5,000-300,000 Da), and G200 (5,000-600,000 Da).
Therefore, the solid phase packing material is selected based upon some knowledge of the size of the range of molecules to be separated.
Applications of SEC:
SEC is not a high resolution technique; generally the molecules to be separated must differ by at least two-fold in molecular weight. Therefore, SEC is useful for desalting or removal of small molecule contaminants from protein samples, determination of the solution subunit composition of a multimeric protein, and to isolate different multimers from each other.
Size exclusion chromatography is generally used near the end of the purification process for a protein of interest, for example to desalt the protein or separate the correctly folded native protein from the denatured protein.
Other Important Aspects of SEC:
The ability to correctly pack a column is very important in obtaining the desired separation. Overpacking a column can lead to inaccessible pores and poor resolution. Underpacking a column can lead to surface area accessibility problems and cause poor resolution.
In-class Question:
6. Would you expect glucose or immunoglobulins to produce a broader peak profile when dissolved in buffer solution at a similar concentration? Which of the peaks below is most likely to belong to glucose? Which peak is most likely to belong to an immunoglobulin? Explain.
Obtaining Size and Molecular Weight Data from SEC:
Calibrations must be performed in order to get information about the size or molecular weight of a protein separated by SEC. Standard protein samples of known molecular size are separated on a SEC column and the retention volumes or elution volumes (i.e. the volume of eluting buffer necessary to remove a particular analyte from a packed column) are recorded. A calibration plot of log molecular mass (Y axis) versus elution volume (X axis) is prepared. The calibration plot can be used to estimate the molecular weight of a protein(s) by separating the protein(s) on the same column as the standards and recording its retention volume. The molecular weight can then be extrapolated from the calibration plot as shown below.
If standard proteins 1, 2, and 3 have molecular masses of 60,000, 30,000, and 15,000 Da, respectively, then an unknown protein which has an elution volume of 9mL will have an estimated log molecular mass of 4.7 and a molecular mass of 45,000 Da.
In-class Question:
7. Would it be easier to separate glucose from an immunoglobulin or two immunoglobulins from each other? Explain. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Separation_Science/07_Specialty_Topics/Size_Exclusion_Chromatography/01_SEC_Separations_Module.txt |
Inquiry-based in class Questions: (Designed to be presented to the students preceding the textual materials).
(1) What is an obvious physical difference between an immunoglobulin (antibody), which is a protein, and glucose, which is a monosaccharide?
(2) When would it be useful to separate immunoglobulins from glucose in blood?
(3) How could you devise an experiment to separate immunoglobulins from glucose utilizing the physical difference between the molecules determined in (1)?
(4) Depicted below are three different sized molecules. Next to them is shown a pore. What would happen as each molecule interacted with the pore? How would this effect the retention time of each molecule?
(5) Other chromatographic methods depend on specific interactions between the molecules being separated and the surface of the solid support. What effect might such specific interactions have on the method you came up with to separate glucose from immunoglobulins? Can you think of how you might limit specific interactions between the molecules and the surface?
(6) Would you expect glucose or immunoglobulins to produce a broader peak profile when dissolved in buffer solution at a similar concentration? Which of the peaks below is most likely to belong to glucose? Which peak is most likely to belong to an immunoglobulin? Explain.
(7) Would it be easier to separate glucose from an immunoglobulin or two immunoglobulins from each other? Explain.
03 Out-of-Class Problem Set
(1) Draw a chromatogram depicting separation of proteins 1, 2, and 3 (protein #1, mw 30,000, protein #2, mw 60,000, and protein #3, mw 90,000). Which type of packing material would you use to separate these proteins?
(2) Indicate on the chromatogram in (1) where insulin would most likely elute. Would it have total accessibility, some accessibility or no accessibility to the particle pores on this column?
(3) Referring to the size exclusion chromatogram in (1), Indicate on the chromatogram where you would expect a molecule with a size of 40,000 to elute. Would you expect the molecule to be well resolved, somewhat resolved, or not resolved from the peak for protein #1? Explain.
(4) Use the SEC calibration plot shown below to estimate the molecular weight (MW) of an unknown protein with a retention volume of 3mL.
04 Instructors Guide Resource
This module is designed to be used by instructors teaching undergraduate courses in which the concept of protein separation by size/shape is important (e.g. Biochemistry, Molecular biology, Biotechnology, Immunology, Bioanalytical chemistry, etc.). The students should have some basic knowledge of chemistry and biology. The in class activity portion of this module is designed to be completed by a small group of students in about a 1 hour time period.
Objectives:
After completing the module, the students will be able to:
1. Explain how size exclusion chromatography works
2. Illustrate (by drawing) the SEC separation process
3. Predict elution order of proteins
4. Interpret size exclusion chromatograms
5. Recognize when SEC would be useful
6. Estimate molecular mass using SEC
Inquiry-based in class questions:
(1) Most students will immediately answer this question correctly. They are aware that there is a large difference in size or mass between monosaccharides and proteins. If problems arise, the instructor may decide to ask them to find and compare the structures.
(2) Students will probably make the connection between glucose and diabetes. Therefore, they may say to get rid of the protein so that glucose can be measured in the blood. They should probably know what immunoglobulins do in the body. Guide them to think about what immunoglobulins are and what role they play in the body. Ask them if there may be a need for pure immunoglobulin (examples include use as a pharmaceutical or for development of an immunoassay).
(3) Students will know that small molecules move faster than larger molecules in solution. So, they may say something like put the sample in a flow system and the glucose molecules will move faster and can be separated first. Suggest to them what might happen if the sample is placed in a column which contains only buffer and the sample is added to the top and the solution allowed to flow through the column using a pump to continuously add buffer. They will most likely say that the glucose will separate out first. Now, draw some particles in the column with holes a little larger that glucose, but a lot smaller than an antibody. They will get that glucose can enter the holes (use the term pores) and that the antibody cannot. Now ask them what will happen to the rate at which glucose is able to exit (use elute) from the column. They will understand that glucose will be held up inside the pores. Now ask them what will happen to the antibody. They will probably get that the antibody will not be held up much due to the presence of the particles and will elute before the glucose molecules. It is worth emphasizing at this point that SEC is a method that depends on the molecules having no attractive forces with the surface of the solid phase. Attractive forces, which are exploited in virtually all other chromatographic methods but SEC, could be different for glucose and an immunoglobulin and if present, could interfere with the separation scheme.
(4) The smallest molecule can diffuse through the entire pore. The largest molecule does not fit into any of the pore. The intermediate-sized one can only diffuse through part of the pores. Because of this, the smallest one should take the longest to come through the column. The largest one should come through fastest. Of course, this requires that the molecules have no attractive interaction with the surface of the solid particles.
(5) Dissolving the compounds being separated in an especially good solvent (one they are highly soluble in) and using a solid particle with a surface with properties very different from the nature of glucose and the proteins (so a nonpolar surface) creates an environment where specific interactions with the surface can be eliminated.
(6) Students will generally know that larger molecules will produce broader peak profiles. So, they will most likely understand that peak 1 is most likely to belong to glucose and peak 2 is most likely to belong to an immunoglobulin.
(7) Students will probably not have too much difficulty in realizing that placing a smaller peak and larger peak in the same amount of space has less overlap (like resolution) than placing two larger peaks in the same amount of space. If there is any question, the instructor may have the students perform an exercise to illustrate this concept.
Out-of-class problem set:
(1) Students will refer to the schematic representation of the operation of an SEC column to draw this chromatogram. Ensure that they correctly show protein #1 eluting last and protein #3 eluting first. Based on the information in the module, sephadex G100 would be a good packing material.
(2) Students will look up the mass of insulin and realize that it is a lot smaller in mass (~6000 Da) than the other proteins depicting on the chromatogram, and therefore it will elute a lot later than other proteins depicted. Based on use of G100, insulin would have some accessibility to the pores.
(3) They will understand that for peaks to be well separated (resolved), there must be a two-fold difference in molecular mass. So, a peak of 40,000 Da eluting between proteins 1 and 2 may not be completely separated.
(4) Students should be able to extrapolate the corresponding Log MW associated with 3 mL retention volume:
Retention volume = 3mL yields a corresponding Log MW ~ 4.8
MW ~ 60,000Da | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Separation_Science/07_Specialty_Topics/Size_Exclusion_Chromatography/02_In-Class_Questions.txt |
You are likely familiar with a device called a centrifuge where a sample is put into a suitable holder and spun. The spinning creates a centrifugal force that is greater than gravity, with the result that the centrifugal force causes particles in the sample holder to settle out.
Q1. What are the variables that influence the settling of a particle in a centrifuge?
Q2. What does the name ultracentrifuge imply?
Q3. Suppose you wanted to separate molecules or particles using an ultracentrifugation procedure. What variables could you alter that would allow you to complete the separation?
Q4. How could you use the speed of the centrifuge or time that the sample has been allowed to spin to perform a separation of several different components of the mixture?
A second variable is the density of the solvent in the centrifuge tube.
Q5. Can you think of a procedure based on solvent density that could be used to separate several different components of a mixture in an ultracentrifuge?
Q6. Can you think of a procedure for generating a density gradient in a centrifuge tube?
02 Answers
You are likely familiar with a device called a centrifuge where a sample is put into a suitable holder and spun. The spinning creates a centrifugal force that is greater than gravity, with the result that the centrifugal force causes particles in the sample holder to settle out.
Q1. What are the variables that influence the settling of a particle in a centrifuge?
There are a number of variables that influence the settling of a particle. Obvious ones are the speed or frequency at which the sample rotates and the mass, density and size of the particle. Another one that may not be as obvious is the density or viscosity of the solvent that the sample is in.
Q2. What does the name ultracentrifuge imply?
The name ultracentrifuge implies that samples are rotated at a very high speed. Ultracentrifuges operate at speeds of more than about 20,000 revolutions per minute. The ultrahigh speed of a centrifuge allows you to settle out much smaller particles. In fact, the speed of an ultracentrifuge is high enough that large molecules such as proteins, nucleic acids, and other polymers settle differently based on their molecular weights. Ultracentrifugation is often used to isolate cells, subcellular organelles, and macromolecules.
Q3. Suppose you wanted to separate molecules or particles using an ultracentrifugation procedure. What variables could you alter that would allow you to complete the separation?
One variable is the speed of the centrifuge and the time the sample has been allowed to spin in the centrifuge.
Q4. How could you use the speed of the centrifuge or time that the sample has been allowed to spin to perform a separation of several different components of the mixture?
You could use a process known as differential centrifugation. Larger particles will settle faster and at slower speeds than smaller particles. The materials that settle to the bottom of the centrifuge tube are known as the pellet. It is possible to pellet a subset of material in the tube, remove the supernatant and add it to a second tube, up the spin rate to create a second pellet, and so on. The pellets obtained from a differential centrifugation are often not pure enough for follow-up studies and must be subjected to additional purification.
A second variable is the density of the solvent in the centrifuge tube.
Q5. Can you think of a procedure based on solvent density that could be used to separate several different components of a mixture in an ultracentrifuge?
You could use a process known as density gradient centrifugation. This is the preferred mechanism to purify compounds using ultracentrifugation. The heaviest or densest layer is placed at the bottom of the tube while the lightest layer is at the top. The sample is then placed on top of the gradient and the sample spun in the centrifuge. The most common materials used to generate density gradients are sucrose or cesium iodide.
Q6. Can you think of a procedure for generating a density gradient in a centrifuge tube?
One procedure would be quite similar to the process used to perform a gradient elution separation in liquid chromatography. This procedure would involve preparing solutions at the two extremes of the gradient you want to create in the tube. You would start with the heavier solution, and then using two pumps, systematically reduce the amount of the heavier solution while increasing the amount of lighter solution. The two solutions coming from the pumps must be mixed for homogeneity and then carefully added to the centrifuge tube such that the layers of different densities do not mix with each other. Density gradient centrifugation can be used to separate substances on the basis of size or mass (a rate zonal separation) or density (an isopycnic separation).
A rate zonal separation has certain aspects similar to a differential centrifugation. The sample is introduced at the top of the gradient and allowed to spin for a certain period of time during which the particles migrate at different rates toward the bottom of the tube. There are several important criteria that are needed for a successful rate zonal separation.
• The sample solution must have a density that is lower than the lowest density portion of the gradient.
• The density of the particles being separated must be greater than the highest density portion of the gradient.
• The density gradient must have a sufficiently long pathlength so that the separation can occur.
• The separation cannot be run for too long a time, otherwise all of the particles can form a single pellet at the bottom of the tube.
In an isopycnic separation, the particles will sink into the gradient to the point where the density of the gradient equals the density of the particle (there isopycnic point). Once there, the particles will remain suspended at that point in the gradient irrespective of the length of time that the sample is kept in the centrifuge. There are several criteria that are needed for a successful isopycnic separation.
• The density of the particles must fall between the density limits of the gradient.
• The time must be long enough so that each particle has reached its isopycnic point. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Separation_Science/07_Specialty_Topics/Ultracentrifugation/01_In-class_Questions.txt |
There is strong motivation for instructors of Instrumental Analysis to use online supplements for their course or to possibly function without a textbook. But paring down the ASDL sources and links into one usable document takes quite a bit of time. Hopefully this document will allow those interested to be able to gather the information quickly.
The following contains learning objectives and corresponding web links for topics seen in most Instrumental Analysis courses. Instrumental Analysis is the second course of a twosemester sequence for analytical chemistry at the United States Military Academy. The textbook used in the course was different editions of Principles of Instrumental Analysis, by Skoog, Holler, and Crouch. Dr. Way Fountain, Dr. Dawn Riegner, and Dr. Tom Spudich have taught instrumental analysis at USMA from 1996 to the present. We have assembled this information for anyone to use in their course; all we request is that you acknowledge us and/or others that have provided the links for the work that has been done. Note that there is a plan to include problems (and access to solutions) for all the lessons, but we cannot guarantee a date as to when this will occur. If you have suggestions or comments, please do not hesitate to email us: [email protected] or [email protected]
With the exception of material marked as copyrighted, information presented on these pages is considered public information and may be distributed or copied. Use of appropriate byline/photo/image credits is requested.
The views expressed herein are those of the authors and do not reflect the position of the United States Military Academy, the Department of the Army, or the Department of Defense.
How to use this document:
These topics and learning objectives are currently grouped for a traditional 75 minute class and set up so they can be ported directly into a syllabus. There are general topics listed in the Table of Contents (below) that take you to a listing of possible learning objectives for this topic. There are references, which are mainly online links, for the general topic OR specific learning objective. All online links are underlined and in italics. Links that are highlighted in gray are references that are in the Analytical Sciences Digital Library (www.asdlib.org). Note that other than the Analytical Chemistry 2.0 textbook by David Harvey, and the Atomic Emission Spectroscopy learning module by Alexander Scheeline and Thomas Spudich, the links are not shortened in any way, thus giving the reference for the document. Additionally, there are no links or learning objectives for any electrochemistry, as it was not offered in the instrumental analysis portion of the course.
Contributors and Attributions
• Dr. Augustus W. Fountain III (Edgewood Chemical Biological Center), Dr. Dawn E. Riegner (United States Military Academy), Dr. Thomas M. Spudich (Maryville University)
• Sourced from the Analytical Sciences Digital Library
Teaching Instrumental Analysis without a Textbook
1. Be able to explain how there are physical and chemical characteristics of matter that can be probed by various instrumental techniques.
2. Define transducer.
3. Define data domain.
4. Understand the numerical criteria for selecting analytical methods.
5. Relate the usefulness and differences of calibration curves, standard additions, and internal standards (this means be able to create & use them to identify unknown concentrations as well).
(Data Reduction Section of AES web page, Harvey – Chapter 5)
https://facultystaff.richmond.edu/~rdominey/301/local/Intro_Instrum_Analysis.pdf | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Teaching_Instrumental_Analysis_without_a_Textbook/01_What_is_Instrumental_Analysis___Data_domains.txt |
Note: You should be familiar with computer components and applications. If not, ask about them in class.
1. Convert between base-10 and binary numbers and vise versa.
http://www.asdlib.org/onlineArticles/elabware/Scheeline_ADC/ADC_visual/numrep.html
http://www.purplemath.com/modules/numbbase.htm
www.ling.ohio-state.edu/~adri...mal-binary.pdf
http://en.Wikipedia.org/wiki/Binary_numeral_system
2. Be able to explain how analog to digital conversion (ADC) occurs and be able to compare an analog plot to a digital plot of the same data.
http://www.asdlib.org/onlineArticles/elabware/Scheeline_ADC/ADC_visual/adc.html
www.chem.uic.edu/chem520/adc1.pdf
3. Relate why more bits in an ADC leads to a reduced uncertainty as well as max % error. Relate cost to number of bits (research outside the text for approximate costs).
http://www.asdlib.org/onlineArticles/elabware/Scheeline_ADC/ADC_visual/adc.html
4. Be able to explain the flexibility that is needed using computers and instrument specific hardware and software.
03 Signals and Noise DAQ Lab
General Signal and Noise references:
terpconnect.umd.edu/~toh/spectrum/TOC.html
terpconnect.umd.edu/~toh/spectrum/SignalsAndNoise.html
http://www.asdlib.org/onlineArticles/ecourseware/Petrovic/signals_noise1.htm
1. Be able to explain what the S/N ratio represents; know how to calculate it, and how the relative ratio can be increased.
http://www.asdlib.org/onlineArticles/ecourseware/Petrovic/signals_noise1.htm
2. Know the five types of noise and characteristics in which to reduce noise.
http://www.asdlib.org/onlineArticles/ecourseware/Spudich/SIGNALSANDNOISE.ppt
3. Discuss ways to improve the S/N ratio both via analog and digital signal processing. Be able to use the ensemble averaging formula to determine the number of scans needed to improve the S/N ratio by a known numerical amount.
http://www.asdlib.org/onlineArticles/ecourseware/Petrovic/signals_noise21.htm
4. Be able to discuss potential problems associated with signal processing.
http://www.ee.buffalo.edu/faculty/paololiu/edtech/roaldi/tutorials/labview.htm
http://www.asdlib.org/onlineArticles/elabware/Scheeline_ADC/ADC_visual/why%20digitize1.html
04 Electromagnetic Radiation
1. Define, explain, apply, & use terms associated with wave and particle properties of electromagnetic radiation.
Harvey – Chapter 10.1.1-10.1.2
www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/UVVis/spectrum.htm#uv1
2. Be able to calculate wavelength, frequency, and energy of electromagnetic radiation.
Harvey – Chapter 10.1.2
3. Understand the possible interactions of electromagnetic radiation with matter, i.e. scattering, transmission, diffraction, refraction, reflection, polarization. (AES web page)
Harvey – Chapter 10.1.2, 10.8.1
Scheeline & Spudich – links above
Polarization -- http://en.Wikipedia.org/wiki/Polarization_(waves)
4. Be able to explain the photoelectric effect, how spectra are generated, black-body radiation, absorption and emission of electromagnetic radiation. (AES web page)
Scheeline & Spudich – links above
Harvey 10.1.1, 10.1.3
5. Understand the quantitative aspects of emission and absorption methods. Be able to calculate Beer’s Law values.
Harvey 10.2
05 Introduction to Optical Spectroscopy
1. Know the 5 components necessary for building optical instruments for absorption and emission studies.
Harvey, Figure 10.27
www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/UV-Vis/uvspec.htm#uv1
2. Know the differences between continuum and line sources. Be able to give examples of each.
For blackbody radiation sources, be able to calculate the wavelength maximum at different temperatures.
Harvey 10.1.3
3. Be able to explain the differences between filters and monochromators, i.e. the benefits and inferiorities of each.
Harvey 10.1.3
4. Be able to explain the two types of monochromators. (AES web page goes into significant depth for single detector spectrometers and multiple detector spectrometers)
Harvey 10.3.1
5. Know what factors contribute to choosing the slit width of a monochromator.
AES web page goes into great depth
06 More Optical Spectroscopy
1. Have a general understanding of the characteristics of an ideal transducer. Understand the differences between single channel and multichannel transducers.
Harvey 10.3.1
2. Know the two general types of radiation transducers: photoelectric and heat. Specifically, be able to describe how the phototube, photomultiplier, silicon photodiode, photodiode array, CCD, thermocouple, and pyroelectric transducers operate. (AES web page)
3. Determine which materials would be necessary to construct specific spectroscopic instruments. Figure 7-2 & 7-3 from: D.A. Skoog, F.J. Holler, and S.R. Crouch, Principles of Instrumental Analysis, (Thomson, Brooks/Cole, Belmont, CA 2007), 6th ed., Chap. 7, pp. 166-167.
07 Introduction to Atomic Spectroscopy
1. Be able to draw and explain atomic energy level diagrams. Be able to indicate atomic absorption and atomic emission (fluorescence) on the diagram, AES web page, (History & Theory, energy level diagrams).
2. Understand phenomena contributing to atomic spectral line width, i.e. Doppler and pressure broadening.
http://en.Wikipedia.org/wiki/Spectral_line
3. Be able to explain the effects of temperature on atomic spectra, i.e. Boltzmann equation.
4. Be able to explain the functionality of an atomizer and be able to describe how samples (solid and aqueous) are prepared for eventual atomization. (Introduction of analyte (electrothermal vaporization and nebulizers)).
08 Atomic Absorption
AA Schematic info: slc.umd.umich.edu/slconline/ADVAA/AdvAA.swf
1. Be able to relate the differences between atomic absorption and atomic emission with respect to the transition that is observed. This includes why limits of detection are better in an analysis for some metals with a particular method (such as FAA vs. FAE, for example) as opposed to others.
www.andor.com/learning/applic..._spectroscopy/
www.water800.com/jspx/AConceptBook.pdf
2. Be able to relate specifically why graphite furnace (electrothermal vaporization) AA has a much higher sensitivity than flame AA.
www.water800.com/jspx/AConceptBook.pdf
3. Understand source modulation; be able to relate why there tends to be improved LODs using source modulation as opposed to not using source modulation.
www.water800.com/jspx/AConceptBook.pdf
4. Be able to acquire and process data from an analysis. This includes obtaining a signal for a blank, standards, and unknown; taking into account how the sample was processed (sample blank), which includes dilutions of both the sample & standards; plot the data (Absorbance vs. Concentration) & obtain a calibration curve; calculating a limit of detection for the analyte.
slc.umd.umich.edu/slconline/ADVAA/AdvAA.swf – sample problems in part 9.
http://terpconnect.umd.edu/~toh/mode...tionCurve.html
09 Atomic Emission
1. Know the advantages of using an ICP for atomic emission spectroscopy. (AES web page (sources, ICP))
www.cee.vt.edu/ewr/environmen...r/icp/icp.html
2. Be able to explain the differences between a sequential and a simultaneous multi-channel instrument for ICP spectroscopy. (AES web page (spectrometers))
3. Be able to explain the different sample introduction methods/techniques used in atomic spectroscopy as well as relate the advantages and disadvantages of all. (AES web page, Introduction of Analyte).
www.inorganicventures.com/tec...ambers-torches
4. Be able to relate the functionality of an internal standard with atomic emission spectroscopy. (AES web page, Data Reduction) | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Teaching_Instrumental_Analysis_without_a_Textbook/02_Digital_Electronics.txt |
1. Be able to describe the advantages and disadvantages of atomic X-ray Fluorescence.
http://serc.carleton.edu/research_education/geochemsheets/techniques/XRF.html
2. Be able to explain the five components of the X-ray fluorimeter that you use in the lab. Specifically, it has an X-ray tube source, a sample chamber (for solid, powder, or liquid samples), a monochromator, a scintillation counter with photomultiplier tube for detection, it requires a vacuum pump for operation of X-rays, and computer processing.
http://serc.carleton.edu/research_education/geochemsheets/wds.html
www.innovx.com/products/handheld
www.rigaku.com/edxrf/nex-cg.html
3. Be able to explain how a spectrum is generated from X-ray Fluorescence using an energy level diagram. Also be able to describe the features of the spectrum (i.e. continuous and line features).
http://www.amptek.com/xrf.html or pdf file www.amptek.com/pdf/xrf.pdf
http://en.Wikipedia.org/wiki/X-ray_fluorescence
11 Preparation of Lab Report Critical Journal Reading
Note that this is a class discussion the components needed for a lab report/journal article.
George Whitesides interview --- Publishing your research via the ACS web page, May 11, 2011.
www2.fiu.edu/~collinsl/Article reading tips.htm
http://www.analytictech.com/mb870/Handouts/How_to_read.htm
www.biochem.arizona.edu/class...568/papers.htm
www.examiner.com/x-6378-Balti...ientific-paper
12 Instrumentation for Molecular Absorption Spectroscopy
1. Have a working understanding of Beer’s Law and know the instrumental deviations to it.
terpconnect.umd.edu/~toh/models/BeersLaw.html
http://www.youtube.com/watch?v=O39avevqndU&feature=channel_page
2. Be able to explain the three categories of instrumental noise or uncertainties in transmittance measurements. Know typical sources of each.
Figures 14-1 (pp. 368) & 14-6 (pp. 372) from: D.A. Skoog, F.J. Holler, and S.R. Crouch, Principles of Instrumental Analysis, (Thomson, Brooks/Cole, Belmont, CA 2007), 6th ed., Chap. 14, pp. 367-372.
3. Be able to explain the differences between H2, D2, W, and Xe arc lamps as well as light emitting diodes (LEDs) as sources for UV-Vis spectroscopy.
Harvey 10.1.3
http://en.Wikipedia.org/wiki/Fluorescent_lamp
http://en.Wikipedia.org/wiki/Light-emitting_diode
http://electronics.howstuffworks.com/led.htm
4. Understand (and be able to apply the knowledge of) the general types of UV-Vis instrument (single beam, double beam, multi-channel), and the advantages and disadvantages of each.
Harvey 10.3.1
www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/UV-Vis/uvspec.htm#uv1
13 Applications of Molecular Absorption Spectroscopy
1. Be able to describe the molecular absorption process. Review molecular orbital theory from organic chemistry.
http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/UVVis/spectrum.htm#uv3
http://www.docstoc.com/docs/21697175...V-SPECTROSCOPY
http://www.chemguide.co.uk/analysis/uvvisible/bonding.html
http://www.chemguide.co.uk/analysis/uvvisible/theory.html#top
2. Be able to explain what is happening to molecular electrons using energy level diagrams in molecular absorption spectroscopy.
Harvey 10.2.1
http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/UV-Vis/spectrum.htm#uv3
http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/intro3.htm#strc8a
3. Understand kinetic methods and what happens to analyte concentration during a spectrophotometric titration.
Harvey 13.2.1
Understand the effects of solvents on spectral properties. Figures 14-1 (pp. 368) & 14-6 (pp. 372) from D.A. Skoog, F.J. Holler, and S.R. Crouch, Principles of Instrumental Analysis, (Thomson, Brooks/Cole, Belmont, CA 2007), 6th ed., Chap. 14, pp. 367-372.
14 Molecular Luminescence Spectroscopy
General Reference: http://www.prenhall.com/settle/chapters/ch26.pdf
1. Know the possible deactivation processes available to a molecule which absorbs a photon and undergoes an electronic energy transition.
http://teaching.shu.ac.uk/hwb/chemistry/tutorials/molspec/lumin1.htm
http://teaching.shu.ac.uk/hwb/chemistry/tutorials/molspec/lumin3.htm
2. Be able to explain the impact of molecular structure, temperature, solvent, and concentration on fluorescence.
Harvey 10.6.3
www.springerlink.com/content/...7/fulltext.pdf -- solvent effects
3. Be able to explain the various instrumental components and diagrams for fluorimeters and spectrofluorimeters.
Harvey 10.6.1
http://www.fluorescence-foundation.org/lectures/genova2006/lecture3.ppt (This site asks for user name and password, but opens without them when “cancel” is selected)
http://www.instrumentalchemistry.com/liquidphase/pages/fluoresence.htm
4. Be able to acquire and process data from an analysis. This includes, but is not limited to (quantitative analysis): obtaining a signal for a blank, standards, and unknown; taking into account how the sample was processed (sample blank), which includes dilutions of both the sample and standards; plotting the data (Signal vs. Concentration) and obtaining a calibration curve; calculating a limit of detection for the analyte.
15 Infrared Spectroscopy
General reference: http://www.prenhall.com/settle/chapters/ch15.pdf
1. Understand the quantum treatment of molecular vibrations; be able to apply equation listed in the first reference.
www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#
http://www.youtube.com/watch?v=DDTIJgIh86E&feature=channel_page
http://orgchem.colorado.edu/hndbksup...r/IRtheory.pdf
infrared.als.lbl.gov/content/web-links
2. Be able to explain what kinds of motions molecules are undergoing when they absorb IR radiation.
www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/infrared.htm#ir2
3. Be able to explain the advantages of Fourier transform spectrometers, and how the Michelson interferometer works.
www.files.chem.vt.edu/chem-ed...a/fourier.html
www3.wooster.edu/Chemistry/analytical/ftir/default.html | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Teaching_Instrumental_Analysis_without_a_Textbook/10_X-ray_Fluorescence.txt |
General reference: http://www.prenhall.com/settle/chapters/ch15.pdf
1. Understand attenuated total reflectance and transmissive IR absorption techniques.
las.perkinelmer.com/content/T...CH_FTIRATR.pdf
http://www.niu.edu/ANALYTICALLAB/ftir/index.shtml
2. Be able to discuss the disadvantages of performing quantitative IR analyses.
http://pubs.acs.org/doi/pdf/10.1021/ed062p886
3. Review interpretation of IR spectra = qualitative analysis (including limitations).
www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/infrared.htm#ir3
http://www.chem.ucla.edu/~webspectra/
http://riodb01.ibase.aist.go.jp/sdbs/cgi-bin/cre_index.cgi?lang=eng
http://www.colby.edu/chemistry/JCAMP/IRHelper.html
4. Be aware of the variety of ways to prepare samples for specific instruments
17 Nuclear Magnetic Resonance Spectroscopy (NMR)
General NMR references:
http://www.cis.rit.edu/htbooks/nmr/
www.cem.msu.edu/~reusch/VirtualText/Spectrpy/nmr/nmr1.htm
http://www.asdlib.org/onlineArticles/ecourseware/Larive/qnmr1.htm
http://www.ias.ac.in/initiat/sci_ed/resources/chemistry/James.T.pdf
http://www-keeler.ch.cam.ac.uk/lectures/
www.biophysics.org/Portals/1/PDFs/Education/donne.pdf
1. Understand and be able to explain the quantum and classical descriptions of NMR.
www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/nmr/nmr1.htm
http://teaching.shu.ac.uk/hwb/chemistry/tutorials/molspec/nmr1.htm
http://www.cis.rit.edu/htbooks/nmr/chap-3/chap-3.htm
2. Be able to describe relaxation processes in NMR.
www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/nmr/nmr1.htm
http://www.cis.rit.edu/htbooks/nmr/chap-3/chap-3.htm
3. Understand and be able to explain the differences between continuous wave (CW) NMR and FT-NMR (aka ”pulsed”) instruments.
www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/nmr/nmr1.htm
http://www.cis.rit.edu/htbooks/nmr/chap-3/chap-3.htm
4. Review interpretation of NMR spectra = qualitative analysis (chemical shift and coupling).
http://www.nd.edu/~smithgrp/structure/workbook.html
http://www.chem.ucla.edu/~webspectra/
http://riodb01.ibase.aist.go.jp/sdbs/cgi-bin/cre_index.cgi?lang=eng
18 Applications of NMR
1. Know the components of an FT-NMR instrument.
http://www.cis.rit.edu/htbooks/nmr/inside.htm
2. Be able to describe why we need a frequency lock system, shims, and sample spinning.
http://www.cis.rit.edu/htbooks/nmr/chap-7/chap-7.htm
http://www.cis.rit.edu/htbooks/nmr/chap-8/chap-8.htm
3. Understand and be able to explain the RF source used for excitation.
http://www.cis.rit.edu/htbooks/nmr/chap-7/chap-7.htm
4. Know why we do not need a Michelson interferometer to perform FT-NMR but we do need one for FT-IR.
http://www.cis.rit.edu/htbooks/nmr/chap-5/chap-5.htm
http://www.cis.rit.edu/htbooks/nmr/chap-8/chap-8.htm
19 Mass Spectrometry
General links:
http://www.youtube.com/watch?v=J-wao0O0_qM&feature=channel_page
eu.shimadzu.de/products/chromato/gcms/TutorialGCMS/Default.aspx?page=p1
http://www.waters.com/waters/nav.htm...S&cid=10073244
http://www.shsu.edu/~chm_tgc/primers/gcms.html
spectroscopyonline.findanalyt...86/article.pdf
http://svmsl.chem.cmu.edu/vmsl/default.htm
http://www.prenhall.com/settle/chapters/ch31.pdf
1. Know the similarities and differences between gas-phase ionization sources and desorption ionization sources for mass spectrometry.
http://www.instrumentalchemistry.com/gasphase/pages/elecion.htm
http://www.instrumentalchemistry.com/liquidphase/pages/maldi.htm
http://en.Wikipedia.org/wiki/Electrospray_ionization
http://en.Wikipedia.org/wiki/Chemical_ionization
www.public.iastate.edu/~kamel/ci.html
2. Understand and be able to explain electron impact ionization (EI); chemical ionization (CI); matrix assisted laser desorption (MALDI); and electrospray ionization sources for mass spectrometry.
20 Mass Spectrometry continued
1. Understand how ions are detected in mass spectrometers. Especially know how electron multipliers work.
http://en.Wikipedia.org/wiki/Electron_multiplier
www.sge.com/products/electron...ltipliers-work
http://www.vias.org/simulations/simusoft_emultiplier.html
sales.hamamatsu.com/en/produc...ultipliers.php
2. Understand the role of mass analyzers in mass spectrometry. Be able to describe how quadrupole, time-of-flight, double-focusing, and FTICR analyzers work.
http://www.asms.org/whatisms/p8.html
http://www.instrumentalchemistry.com/gasphase/pages/iontrap.htm
http://www.instrumentalchemistry.com/gasphase/pages/tof.htm
http://www.instrumentalchemistry.com/gasphase/pages/magsecone.htm
21 Re-introduction to Chromatography
1. Understand important chromatographic quantities and relationships, specifically retention time, width at half height, retention factor, linear flow velocity, resolution, number of plates and plate height.
www.separatedbyexperience.com/glossary/
http://web.njit.edu/~kebbekus/analysis/4CHROMAT.htm
http://www.edusolns.com/gc/gctutorial/
http://www.instrumentalchemistry.com/theory/pages/hetp.htm
Harvey 12.1, 12.2
2. Be able to describe the general elution problem.
http://web.njit.edu/~kebbekus/analysis/4CHROMAT.htm
Harvey 12.3
3. Understand how to utilize a Van Deemter plot (ie. kinetic variables affecting zone broadening).
http://en.Wikipedia.org/wiki/File:Van-deemter.jpg
http://en.Wikipedia.org/wiki/Van_Deemter_equation
http://www.edusolns.com/hplc/hplctutorial/
Harvey 12.3
4. Understand the major components of a chromatogram and quantitative analysis in chromatography.
http://www.gmu.edu/depts/SRIF/tutorial/gcd/gc-ms2.htm
www.aerosol.us/ADIweb/ResearchTAG.html
22 Gas Chromatography
1. Be able to describe the major components of a gas chromatograph.
http://www.edusolns.com/gc/gctutorial/
eu.shimadzu.de/products/chromato/gc/default.aspx
http://orgchem.colorado.edu/hndbksupport/GC/GC.html
http://www.instrumentalchemistry.com/gasphase/pages/injector.htm
http://www.instrumentalchemistry.com/gasphase/pages/split.htm
http://www.instrumentalchemistry.com/gasphase/pages/splitless.htm
Harvey 12.4
2. Know the differences between the flame ionization, thermal conductivity, electron capture, and flame photometric detectors.
http://elchem.kaist.ac.kr/vt/chem-ed/sep/gc/gc-det.htm
http://www.edusolns.com/gc/gctutorial/
http://www.instrumentalchemistry.com/gasphase/pages/tcd.htm
http://www.instrumentalchemistry.com/gasphase/pages/ecd.htm
http://www.instrumentalchemistry.com/gasphase/pages/fid.htm
http://www.instrumentalchemistry.com/gasphase/pages/npd.htm
http://www.instrumentalchemistry.com/gasphase/pages/source.htm
Harvey 12.4.5
3. Understand different factors that affect column efficiency.
http://orgchem.colorado.edu/hndbksupport/GC/GC.html
Harvey 12.3
23 Liquid Chromatography Capillary Electrophoresis
Definitions: cid=10049080&locale=en_US
Video of instrument in action: http://www.youtube.com/watch? v=kz_egMtdnL4&feature=channel_page
1. Understand what is going on during partition chromatography. Know how to perform gradient and isocratic elutions.
www.chemistry.nmsu.edu/Instru...qd_Chroma.html
Harvey 12.5.2
2. Know the differences between normal and reversed phase partition chromatography.
www.chemistry.nmsu.edu/Instru...qd_Chroma.html
Harvey 12.5.1
3. Know the basic components of an HPLC instrument. Specifically, know the difference between bulk property detectors (refractive index detector) and solute –property detectors (UV-Vis detectors). Also be familiar with the mass spectral detector.
http://www.instrumentalchemistry.com/liquidphase/pages/sampleloop.htm
http://www.instrumentalchemistry.com/liquidphase/pages/dampner.htm
http://www.instrumentalchemistry.com/liquidphase/pages/diodearray.htm
http://www.instrumentalchemistry.com/liquidphase/pages/fluoresence.htm
http://www.instrumentalchemistry.com/liquidphase/pages/singlewavelength.htm
Harvey 12.5.4 | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Teaching_Instrumental_Analysis_without_a_Textbook/16_Instrumentation_for_Infrared_Spectroscopy.txt |
This module is an introduction to biological mass spectrometry with an emphasis on proteomics, the study of proteins, particularly their structures and functions. The first section provides key background information on proteins and proteomics. The second section describes the use of electrospray ionization to determine the molecular weight of a protein using an ion trap mass analyzer. The third section shows how to identify an unknown protein using peptide mass fingerprinting with matrix-assisted laser desorption ionization time of flight mass spectrometry (MALDI-TOF). The fourth section describes shotgun proteomics and tandem MS experiments for de novo peptide sequencing. An instructor’s manual provides answers to the questions and guidelines for how to implement the activities.
• Instructor’s Manual
• Section 1: Proteins and Proteomics
• Section 2: Electrospray
Electrospray ionization (ESI) is a solution-phase “soft” ionization source which converts ions in solution to gaseous ions. The technique is commonly applied to biomolecules and was developed to permit the investigation of analytes from condensed phase samples, such as liquids and solids
• Section 3: Peptide Mass Mapping
Often the first step in analyzing a complex mixture of proteins is to separate them using two-dimensional (2D) gel electrophoresis. If 2D gel analysis indicates that new protein is being expressed or the expression level of a protein has changed, that protein must be identified. Peptide mass mapping is a method used to determine the identity of a protein spot in a gel. The protein is cut from the gel, destained, and extracted.
• Section 4: MS-MS and De Novo Sequencing
Contributors and Attributions
• Dr. Jill Robinson (Indiana University), Dr. Michelle L. Kovarik (Trinity College)
Biological Mass Spectrometry: Proteomics
Introduction
This module is an introduction to biological mass spectrometry with an emphasis on proteomics, the study of proteins, particularly their structures and functions. The first section provides key background information on proteins and proteomics. The second section describes the use of electrospray ionization to determine the molecular weight of a protein using an ion trap mass analyzer. The third section shows how to identify an unknown protein using peptide mass fingerprinting with matrix-assisted laser desorption ionization time of flight mass spectrometry (MALDI-TOF). The fourth section describes shotgun proteomics and tandem MS experiments for de novo peptide sequencing.
Each section includes some brief explanatory text with accompanying reading questions that emphasize student comprehension. The text and accompanying reading questions are intended to be completed as a pre-class assignment so that students have the background information needed to work on more difficult questions during the class period. Alternatively, the reading questions could be completed at the start of the class period. Each section also includes discussion questions and activities that are intended for in-class work in small groups. The instructor’s manual gives more detailed suggestions for how different portions of the module might be used and the amount of class time needed to complete each section.
Information on how to use this module
What prior knowledge do students need before beginning this module?
It is helpful if students have had at least one class period on the basics of mass spectrometry before starting this module. For example, students should understand that analytes must be ionized for mass spectrometric analysis so that the molecules may be separated by m/z. No prior knowledge of proteomics, specific ion source, or mass analyzers is needed because these topics are covered in the module.
The information in Section 1 is a necessary prerequisite to Sections 2-4, but Sections 2-4 are independent of each other. An individual instructor could complete one or two of these sections without the others and in any order. Students who have taken biochemistry may not need Section 1A, which is an introduction to amino acids, the peptide bond, and proteins.
How much time is required in and out of class to complete each section?
Section 1: Proteins and proteomics
Class Period 1: (Section 1A What is a protein?) Reading questions should be completed prior to class. Discussion questions can be answered in small groups during class.
Class Period 2: (Section 1 B Proteomics) Reading questions should be completed prior to class. It is best to summarize and discuss the answers to this set of reading questions at the start of class. During class, distribute the following comparative proteomics paper and have students work in groups to answer the discussion questions.
Comparative proteomics of oral cancer cell lines: identification of cancer associated proteins, Karsani et al, Proteome Science, 2014, 12:3. doi:10.1186/1477-5956-12-3 (Open access journal)
Section 2: Calculating the molecular weight of a protein from its electrospray ionization mass spectrum
Class Period 1: Reading and video questions should be completed prior to class. Discussion questions can be answered in small groups during class. (If all questions are completed during the class period, then two class sessions are needed.)
Section 3: Identification of a protein: Peptide mass mapping
Class Period 1 and 2: Two full class periods are suggested for the reading and discussion of MALDI and TOF-MS (Sections 3A-B).
Class Period 3: Section 3C Reading questions should be completed prior to class. Discussion questions can be answered in small groups during class.
Class Periods 4 and 5: Section 3D Reading questions should be completed prior to class. Students will need a computer to access the maxtrix science website and mascot database. It is best if students download an excel file of the mass list and then they can simply copy and paste the m/z values into the search query. Students will explore how changing search parameters affects the protein score (statistical significance that a protein has been correctly identified). The varying protein scores must be interpreted.
A series of homework problems accompanies Section 3D. These are meant to be assigned out of class but could also be used during another class period.
Section 4: Sequencing peptides and identifying proteins from tandem MS (MS-MS) data
Class Period 1 and 2: If the reading and video questions are completed prior to class, the discussion questions can be completed in one to two class periods, depending on whether the groups work through sketching fragment ions and calculating masses together or divide the work among group members. To save time, the instructor may allow students to use the Institute for Systems Biology’s peptide calculator. The Challenge Question may be assigned as homework or worked in class during a class session. If readings, videos, and discussions will all occur in class, at least three class periods are recommended.
What is the difference between reading questions, video questions, and discussion questions?
Reading questions and video questions are short answer questions that are relatively straightforward. Individual students should be able to answer these questions on their own without group discussion. Reading questions are based on reading the text portion of the module, and video questions are based on linked videos. Students could be assigned to read and view videos before coming to class. This will ensure that students have the requisite background knowledge to discuss the more open-ended and challenging discussion questions in small groups during class. It is best to briefly re-cap the answers to the reading questions at the start of the class period.
Alternatively, students could complete the reading and video questions independently during class; this works especially well for shorter sections like the introduction to proteins (Section 1). Discussion questions should always be used in-class where students discuss the answers in groups of 3-4 students. A good strategy is to have each group discuss a question for several minutes while the instructor circulates to answer questions, correct misunderstandings, and monitor discussions. Then the instructor can optionally call the class together as a group to report out answers and for large group discussion before moving on to the next question. For the time-of-flight portion (Section 3B), there is an accompanying slide show that includes animations to be shown as a wrap-up to discussions.
Instructors Manual
Proteins are biological macromolecules consisting of long chains of amino acids. A shorter chain of amino acids is called a peptide. Proteins and peptides are biological polymers formed from amino acid monomers. Each amino acid is composed of an amine group, a side chain (R), and a carboxylic acid functionality (Figure 1).
Figure 1. An amino acid.
There are 22 naturally occurring amino acids, 20 of which are encoded by the genome (Figure 2). These amino acids are usually grouped according to the character of their side chains, which may be acidic, basic, neutral, or hydrophobic. In a protein, portions of the linear sequence of amino acids may take on a secondary structure (such as a helix or a sheet) based on intermolecular forces between amino acid residues. The full-length protein will form a tertiary structure or overall shape, and the structure of the protein is intimately linked to the protein’s function.
Figure 2. Table of naturally occurring amino acids sorted by side chain and with their three-letter and one-letter codes. (Reproduced under a Creative Commons license from Dancojocari.)
Reading Question
1. In Figure 1 above, circle the amine group, draw a triangle around the side chain, and draw a square around the carboxylic acid group of the amino acid.
A.
The linear chains of amino acids that form proteins are connected by amide (or peptide) bonds that form by a dehydration reaction between two amino acids (Figure 3). By convention, we write and draw peptides from the N-terminus (the side with the free amine group) to the C-terminus (the side with the free carboxylic acid).
Figure 3. Formation of a peptide bond between two amino acids.
Reading Question
2. In Figure 3 above, label the N-terminus and the C-terminus of the dipeptide product. Circle the amide bond.
A.
While many properties of peptides and proteins can be measured to provide useful information, we will focus on measurements of peptide and protein masses using mass spectrometry. Mass spectrometry measures the mass and charge of molecules in the gas phase. As a result, the mass of individual molecules is important. The monoisotopic mass of a molecule is the mass of that compound when it is composed of the most abundant isotope of each atom in the molecule. For example, if we calculate the monoisotopic mass of CO2, we will use the masses of carbon-12 and oxygen-16 since these are the most abundant isotopes of these elements. This gives us a monoisotopic mass of
12.000 amu + 2 (15.995 amu) = 43.99 amu.
This is different than the molar mass we would calculate using average atomic weights from the periodic table; on the periodic table, we find a molar mass for carbon of 12.011 amu. This represents the weighted average of all carbon atoms, which includes 98.9% carbon-12 and 1.1% carbon-13. Table 1 shows the monoisotopic mass of each amino acid and the mass of that amino acid as a residue in a protein or peptide chain.
Table 1. Monoisotopic molecular weight information for all 20 genetically encoded, naturally occurring amino acids.
Amino Acid Single-Letter Code Residue MW (amu) Amino Acid MW (amu)
glycine G 57.02 75.03
alanine A 71.04 89.05
serine S 87.03 105.04
proline P 97.05 115.06
valine V 99.07 117.08
threonine T 101.05 119.06
cysteine C 103.01 121.02
isoleucine I 113.08 131.09
leucine L 113.08 131.09
asparagine N 114.04 132.05
aspartic acid D 115.03 133.04
glutamine Q 128.06 146.07
lysine K 128.09 146.11
glutamic acid E 129.04 147.05
methionine M 131.04 149.05
histidine H 137.06 155.07
phenylalanine F 147.07 165.08
arginine R 156.10 174.11
tyrosine Y 163.06 181.07
tryptophan W 186.08 204.09
Discussion Questions
1. Consider the data in Table 1. By what value do the residue molecular weight (MW) and the amino acid MW differ? Why is the MW of an amino acid residue in a peptide chain different from the mass of the full amino acid?
A. The mass of the amino acid residue is 18 amu lighter than the mass of the amino acid because the formation of peptide bonds at the N- and C-termini results in the loss of water from the amino acid.
2. Draw the structure for the tetrapeptide G, C, L, W. Refer to Figure 2 for the structure of amino acid side chains.
A.
3. Calculate the monoisotopic molecular weight of the tetrapeptide using the data in Table 1.
A. When computing a peptide or protein molecular weight, sum up all residue masses and add 18.02 (for the H and OH at the termini):
57.02+103.01+113.08+186.08+18.01 = 477.20 amu
Note: In protein mass spectrometry, the Da is a common unit. It is the same as amu.
4. Calculate the molecular weight of the tetrapeptide using molar mass information from the periodic table. Why is this molecular weight different from the monoisotopic mass you calculate in question 3? Which mass is the mass measured in mass spectrometry?
A. Glycine is C2H5NO2. Cysteine is C3H7NO2S. Leucine is C6H13NO2. Tryptophan is C11H12N2O2.
This is a total of 22 carbons, 37 hydrogens, 5 nitrogens, 8 oxygens, and 1 sulfur atom. For the peptide, we must subtract the equivalent of 3 water molecules representing dehydryation to form the peptide bonds (i.e., total atom count should be less 6 hydrogens and 3 oxygens). This gives us 22 C, 31 H, 5 N, 5 O, and 1 S.
22 (12.011) + 31 (1.008) + 5 (14.007) + 5 (15.999) + 1 (32.06) = 477.58 Da.
The periodic table gives the weighted average molar masses of each element. This is the average mass of a population of GCLW peptides. No single GCLW peptide actually has this mass, and it is slightly heavier than the monoisotopic mass since the most abundant isotopes of H, C, N, and O are the lightest isotopes. The monoisotopic mass is the mass of a GCLW peptide composed of atoms that are the most abundant isotope of each element in the peptide. This will be the most intense peak on the mass spectrum. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Biological_Mass_Spectrometry%3A_Proteomics/Instructors_Manual/Section_1%3A_Proteins_and_Proteomics/Se.txt |
Gel electrophoresis gives some indication of the molecular weight of a protein based on its migration in a gel; however, the molecular weights determined from this method are not precise. In contrast, mass spectrometry can provide accurate and precise molecular weights that aid in identification of proteins. Initially, mass spectrometry experiments were limited to small molecules that could be readily volatilized into the gas phase and ionized before entering the mass analyzer. In the late 1980s, several ionization methods were developed and applied to biomolecules that permitted analytes from condensed phase samples, such as liquids and solids, to be volatilized and analyzed by mass spectrometry. Electrospray ionization (ESI) is one such technique.
ESI is a solution-phase “soft” ionization source which converts ions in solution to gaseous ions. The mechanism of ion formation in an ESI source is depicted schematically in Figure 1. Briefly, gaseous ions are formed when a solution containing the analytes of interest (from an LC column or from a syringe infusion pump) is sprayed through a stainless steel capillary to which a high voltage is applied, creating a fine mist of droplets which are charged on their surface. As solvent evaporates from the charged droplets, the charge density on their surface increases to a critical limit, at which point electrostatic repulsion causes the larger droplets to break up into smaller charged droplets. Finally, analyte ions are ejected into the gas phase by electrostatic repulsion, and these ions enter the mass analyzer for subsequent mass analysis. You can watch a video from the Johnson lab showing the formation of a stable electrospray as the voltage applied to the tip is increased.
Figure 1. The mechanism of electrospray ionization (ESI). High voltage is applied to a steel capillary to produce charged droplets containing the analyte molecules. As solvent evaporates, charge accumulates until the Rayleigh limit is reached and the droplet undergoes Coulomb fission into smaller droplets. This process continues until gas phase analyte molecules enter the mass spectrometer. Figure is reproduced with permission from S. Banerjee and S. Mazumdar, Int. J. Anal. Chem., 2012, 2012, 282574 under a Creative Commons Attribution License.
For electrospray, the protein sample is prepared in a solution composed of water with acid and a low surface tension organic solvent, such as methanol. Relatively pure protein samples can be infused directly into the mass spectrometer using a syringe pump, or more complex protein and peptide mixtures may be separated by chromatography first and electrosprayed directly from the column.
Electrospray ion sources are compatible with many types of mass analyzers. For this application, we will consider the quadrupole ion trap as a mass analyzer. The quadrupole ion trap is composed of two end cap electrodes at the entrance and the exit of the trap and a ring electrode (shaped like a donut) in the middle. The voltages and AC frequencies of the applied potentials on the electrodes are varied to control the motion of ions in and through the trap. The ion trap mass analyzer is a small, relatively inexpensive mass analyzer that typically generates mass spectra with unit resolution (Δm ≈ 1) or slightly better. This video shows the operating steps involved in generating a mass spectrum using an ion trap.
Video Question
1. Summarize what you saw in the video on the ion trap. What are the operating steps used to generate a mass spectrum using an ion trap mass analyzer?
A. The ions are formed by electrospray and then enter the trap. In the trap, the ions collide with buffer gas particles (gray) and their motion is damped. The ions are then scanned out of the trap sequentially by m/z value and detected. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Biological_Mass_Spectrometry%3A_Proteomics/Instructors_Manual/Section_2%3A_Electrospray/Section_2A._E.txt |
One of the advantages of ESI is that it is a “soft” ionization technique in which little fragmentation of large, thermally fragile biomolecules occurs. Consequently, molecular weight information is readily obtained with this technique. Additionally, the ions formed are often multiply charged, which enables the analysis of molecular masses exceeding 100,000 Da because the multiple charges bring the m/z values into the mass range of conventional mass analyzers, such as ion traps. With proteins and peptides, the mass spectrum consists of a series of peaks, call the “peak envelope” which represents a distribution of multiply charged analyte ions.
Ubiquitin is a small protein with a monoisotopic molecular weight of 8560 Da. Electrospray ionization of this small protein typically results in major charge states of +8, +9, +10, +11, +12, and +13.
Reading Questions
1. Using this information, complete the table below, assuming that the charges on each ion come from protonation rather than sodium or potassium adducts. Round masses and m/z values to the ones place.
z mass of [M+zH]+z m/z
8
9
10
11
12
13
A.
z mass of [M+zH]+z m/z
8 8568 1071
9 8569 952
10 8570 857
11 8571 779
12 8572 714
13 8573 659
2. Using the data you entered in the table above, sketch an expected ESI-MS spectrum for ubiquitin. Label each peak with its charge state. What do you notice about the spacing of the peaks along the x-axis?
A.
The spacing of peaks widens along the length of the x-axis as the charge state decreases, as seen in the predicted spectrum.
Based on the m/z values and peak spacing observed in the charge envelope, we can determine the charge state, z, for each peak and the molecular weight of the analyte using Equations $\ref{1}$ and $\ref{2}$,
$z = \dfrac{M_2-A}{M_1-M_2} \label{1}$
$MW= \dfrac{(M_1-A)(M_2-A)}{M_1-M_2} \label{2}$
where MW is the molecular weight of the analyte, M1 is the m/z value for the first ion, z is the charge state of the first ion, M2 is the m/z value for a second ion of lower m/z, and A is the mass of the adduct ion, which is usually a proton (H+) but can be sodium (Na+) or potassium (K+) ions from glassware or buffers used in the experiment.
Discussion Questions
1. Why do aqueous samples for electrospray typically include an acid and a low surface tension solvent such as methanol?
A. The acidity of the solution helps to ionize analytes by protonation, and low surface tension solvent promotes droplet fission.
2. Figure 2 shows an experimentally obtained mass spectrum for ubiquitin. Compare this spectrum to the spectrum you predicted in Reading Question 2. Are there any differences? If so, what might cause these differences?
Figure 2. ESI-MS spectrum of bovine ubiquitin from Protea Biosciences (proteabio.com/products/PS-143)
A. The experimental spectrum agrees with the predicted spectrum with respect to the m/z values of the major peaks, which differ only slightly from the predicted values. The spectra differ in two ways: first, the intensity of the peaks varies because some charge states are more favorable than others; second, the experimental spectrum has additional smaller peaks at higher m/z values arising from low abundances of ions with charges of less than +8.
3. Using Equation ($\ref{2}$) and any pair of peaks from Figure 2, calculate the molecular weight of ubiquitin and its percent error compared to the theoretical monoisotopic mass of 8560 amu.
A. Any pair of peaks can be used. An example solution is shown below using the peaks at 714.92 and 779.83.
$MW= \dfrac{(M_1-A)(M_2-A)}{M_1-M_2} =\dfrac{(779.83-1.01)(714.92-1.01)}{779.83-714.92}=8566\: amu$
$\%\: error= \dfrac{8566-8560}{8560}×100\%=0.07\%$
Figure 3. ESI-MS spectrum of cytochrome C. Data obtained at Trinity College.
4. Figure 3 shows the ESI-MS spectrum for cytochrome C electrosprayed from a mixture of water, methanol, and acetic acid with pH of 2.5. What is the charge state of the peak at m/z = 773.36?
A. Using equation ($\ref{1}$), $z= \dfrac{727.94-1.008}{773.36-727.94}=\dfrac{726.932}{45.42}=+16$
If students use the peak at m/z=824.85 the charge comes out negative. It is helpful to remind them that the sample is usually prepared in acid and that they should expect positive charges from protonation.
5. Determine the MW of the analyte, cytochrome C, using the data in Figure 3.
A. Any pair of peaks can be used. An example solution is shown below using the peaks at 773.36 and 727.94 and equation ($\ref{2}$).
$MW= \dfrac{(M_1-A)(M_2-A)}{M_1-M_2} =\dfrac{(773.36-1.01)(727.94-1.01)}{773.36-727.94}= 12360\: amu$
6. How would you expect the mass spectrum to change if the cytochrome C sample was electrosprayed from a solution of higher pH? Explain your answer.
A. Using a higher pH sample results in less protonation of the cytochrome C analyte and therefore lower charge states. This shifts the charge envelope to higher m/z values. For example, at pH 4.2 the z=+9 charge state is the base peak, and at pH 6.4 the +8 charge state is the base peak. See L. Konermann and D. J. Douglas, Biochem., 1997, 36, 12296-12302 for example data.
7. Note that in the ESI-MS spectra for ubiquitin and cytochrome C, each major peak is accompanied by a series of less intense peaks of slightly different m/z. What is the source of these peaks?
A. These peaks represent the isotopic distribution of atoms in the protein. Because the m/z value of a mass spectrum represents the mass-to-charge ratio of an individual molecule, they reflect the exact isotopic composition of each molecule. The most intense peak arises from the most likely composition of isotopes. Less intense peaks arise from molecules that happen to contain a different isotopic distribution, with the relative intensities of these peaks representing the isotopic abundance of each element in the ion and the number of atoms.
Note to instructors: For small molecules, the most intense peak arises from ions composed solely of the most abundant isotopes of each atom, such as carbon-12, oxygen-16, hydrogen-1, and nitrogen-13. For organic molecules composed of these elements, the most abundant isotope is also the lightest isotope, and the most intense peak gives the monoisotopic mass of the ion. For larger molecules, the most intense peak may include heavier isotopes, and for very large molecules the most intense peak may actually present the average mass as calculated from atomic weights for each atom. For example, 1.10% of all carbon atoms are carbon-13, so molecules that contain more than 100 carbon atoms are likely to include at least one carbon-13 atom, making their most intense peak heavier than the monoisotopic mass. More information on isotopic distributions in large molecules, such as proteins, can be found in the 1983 Yergey et al. reference.
8. Compare your description of the ion trap mass analyzer video with your group mates’ descriptions. As a group, write a consensus explanation for how an ion trap works.
A. The video shows that the trap starts out with some gas molecules already in the trap. (This buffer gas is usually helium. Collisions with the buffer gas “cool” the ions as they enter the trap. This reduction in kinetic energy makes them easier to trap.) Ions from the ESI source enter the trap and follow a complex path between the end cap electrodes and the ring electrode. Eventually, the ions are damped to the middle of the trap. Then ions are scanned out of the trap by m/z value and detected to generate a mass spectrum. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Biological_Mass_Spectrometry%3A_Proteomics/Instructors_Manual/Section_2%3A_Electrospray/Section_2B._E.txt |
Matrix-assisted laser desorption ionization (MALDI) is a technique that generates gas-phase peptide and protein ions. Proteins are nonvolatile and cannot simply be heated to make the gaseous form. (Think about a steak on the grill. The steak chars but does not vaporize.) Instead proteins are desorbed (released from a surface) by mixing the protein with a matrix molecule and irradiating the mixture with a pulse from an ultraviolet laser. The following figure shows the structure of various molecules used as the MALDI matrix.
Discussion Question
1. What structural features do these molecules have in common? Why are these common features important for a MALDI matrix?
A. They have aromatic rings so they have strong absorption in the ultraviolet region of the spectrum. The have a carboxylic acid group so they can transfer a proton to the biomolecule giving it a positive charge.
The α-cyano-4-hydroxcinnamic acid (CCA) molecule is a MALDI matrix used for peptide and protein analysis. The matrix is combined with a mixture of peptides and placed on a metal plate. A pulsed nitrogen laser with a wavelength of 337 nm is used to irradiate the peptides and matrix. The absorption spectrum for a peptide and the CCA matrix is shown.
(a)
(b)
Figure. (a) Absorbance spectrum of a peptide. (b) Absorbance spectrum of CCA matrix.
Discussion Questions
1. Does the peptide absorb the laser light? Does the matrix absorb the laser light?
A. No, the peptide does not absorb the laser light. The peptide absorbs strongly at 220 nm and 280 nm. The matrix does absorb the laser light.
2. Propose a reason that the laser is matched to the absorption of the matrix (CCA) and not the peptide. What would happen to the peptide if it absorbed a high energy pulse of laser light?
A. It would break bonds in the peptide causing fragmentation.
In the MALDI process, the laser is directed at the mixture of CCA matrix and peptides. The matrix absorbs the laser light, but the peptides do not. The matrix heats up and “explodes” from the surface carrying the peptide with it. (It is similar to skydiving. The airplane (matrix) carries the people (peptides) into the air and when they jump, they are temporarily flying.) After absorbing light the matrix is in an excited state and transfers a proton (H+) to the peptide giving it a positive charge. The invention of MALDI enabled the mass spectrometric analysis of large nonvolatile biomolecules and was so important that its inventor Koichi Tanaka of Shimadzu Corporation was awarded one-third of the 2002 Nobel Prize in Chemistry. The MALDI process is depicted in the figure below.
Figure. The MALDI process for creating gas-phase ions of peptides. (Figure from Wikimedia Commons)
The purple color represents the MALDI matrix and the green color represents peptides. Energy from the laser is absorbed by the matrix, causing the matrix to desorb from the surface. The matrix carries the peptides into the gas phase and gives them a charge by donating a proton. Molecular ions of the peptides are created because peptides do not absorb laser light which would break bonds within the molecule.
An example of a typical mass spectrum of a mixture of proteins ionized by the MALDI method is shown below. The matrix is 2,5-dihydroxybenzoic acid (DHB) and the mixture contains three proteins; ubiquitin, cytochrome C and equine myoglobin.
Figure. MALDI-TOF mass spectrum of three proteins; ubiquitin, cytochrome C, and equine myoglobin.
Figure adapted from King’s College London
http://www.kcl.ac.uk/innovation/research/corefacilities/smallrf/mspec/cemsw/instr/maldi-tof-ms.aspx
Discussion Questions
1. Examine the mass spectrum of the mixture of three proteins obtained using MALDI as the ionization method. Would you classify MALDI as a “soft” (little to no fragmentation) ionization technique or a “hard” (fragmentation) ionization technique?
A. Soft ionization method. Fragmentation is not evident in this mass spectrum.
2. What is the identity of Peak 1 in the mass spectrum?
A. Cytochrome C+ 2 H with a +2 charge. The m/z ratio of cytochrome C+H with a +1 charge is 12400. If cytochrome C accepts 2 protons from the matrix, then the m/z ratio is 6200. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Biological_Mass_Spectrometry%3A_Proteomics/Instructors_Manual/Section_3%3A_Peptide_Mass_Mapping/Secti.txt |
Learning Objectives
At the end of this assignment, you should be able to:
1. Draw the chemical structure of an amino acid and small peptide.
2. Describe the difference between free and residue amino acid mass.
3. Calculate the monoisotopic and average mass of a peptide and explain why they are different.
Section 1: Proteins and Proteomics
Proteins are biological macromolecules consisting of long chains of amino acids. A shorter chain of amino acids is called a peptide. Proteins and peptides are biological polymers formed from amino acid monomers. Each amino acid is composed of an amine group, a side chain (R), and a carboxylic acid functionality (Figure 1).
Figure 1. An amino acid.
There are 22 naturally occurring amino acids, 20 of which are encoded by the genome (Figure 2). These amino acids are usually grouped according to the character of their side chains, which may be acidic, basic, neutral, or hydrophobic. In a protein, portions of the linear sequence of amino acids may take on a secondary structure (such as a helix or a sheet) based on intermolecular forces between amino acid residues. The full-length protein will form a tertiary structure or overall shape, and the structure of the protein is intimately linked to the protein’s function.
Figure 2. Table of naturally occurring amino acids sorted by side chain and with their three-letter and one-letter codes. (Reproduced under a Creative Commons license from Dancojocari.)
Reading Question
1. In Figure 1 above, circle the amine group, draw a triangle around the side chain, and draw a square around the carboxylic acid group of the amino acid.
The linear chains of amino acids that form proteins are connected by amide (or peptide) bonds that form by a dehydration reaction between two amino acids (Figure 3). By convention, we write and draw peptides from the N-terminus (the side with the free amine group) to the C-terminus (the side with the free carboxylic acid).
Figure 3. Formation of a peptide bond between two amino acids.
Reading Question
2. In Figure 3 above, label the N-terminus and the C-terminus of the dipeptide product. Circle the amide bond.
While many properties of peptides and proteins can be measured to provide useful information, we will focus on measurements of peptide and protein masses using mass spectrometry. Mass spectrometry measures the mass and charge of molecules in the gas phase. As a result, the mass of individual molecules is important. The monoisotopic mass of a molecule is the mass of that compound when it is composed of the most abundant isotope of each atom in the molecule. For example, if we calculate the monoisotopic mass of CO2, we will use the masses of carbon-12 and oxygen-16 since these are the most abundant isotopes of these elements. This gives us a monoisotopic mass of
12.000 amu + 2 (15.995 amu) = 43.99 amu.
This is different than the molar mass we would calculate using average atomic weights from the periodic table; on the periodic table, we find a molar mass for carbon of 12.011 amu. This represents the weighted average of all carbon atoms, which includes 98.9% carbon-12 and 1.1% carbon-13. Table 1 shows the monoisotopic mass of each amino acid and the mass of that amino acid as a residue in a protein or peptide chain.
Table 1. Monoisotopic molecular weight information for all 20 genetically encoded, naturally occurring amino acids.
Amino Acid Single-Letter Code Residue MW (amu) Amino Acid MW (amu)
glycine G 57.02 75.03
alanine A 71.04 89.05
serine S 87.03 105.04
proline P 97.05 115.06
valine V 99.07 117.08
threonine T 101.05 119.06
cysteine C 103.01 121.02
isoleucine I 113.08 131.09
leucine L 113.08 131.09
asparagine N 114.04 132.05
aspartic acid D 115.03 133.04
glutamine Q 128.06 146.07
lysine K 128.09 146.11
glutamic acid E 129.04 147.05
methionine M 131.04 149.05
histidine H 137.06 155.07
phenylalanine F 147.07 165.08
arginine R 156.10 174.11
tyrosine Y 163.06 181.07
tryptophan W 186.08 204.09
Discussion Questions
1. Consider the data in Table 1. By what value do the residue molecular weight (MW) and the amino acid MW differ? Why is the MW of an amino acid residue in a peptide chain different from the mass of the full amino acid?
2. Draw the structure for the tetrapeptide G, C, L, W. Refer to Figure 2 for the structure of amino acid side chains.
3. Calculate the monoisotopic molecular weight of the tetrapeptide using the data in Table 1.
4. Calculate the molecular weight of the tetrapeptide using molar mass information from the periodic table. Why is this molecular weight different from the monoisotopic mass you calculate in question 3? Which mass is the mass measured in mass spectrometry? | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Biological_Mass_Spectrometry%3A_Proteomics/Section_1%3A_Proteins_and_Proteomics/Section_1A._What_is_a.txt |
The central dogma of molecular biology, DNA to RNA to protein, has given us an explanation of how information encoded by our DNA is translated and used to make an organism. It describes how a gene made of DNA is transcribed by messenger RNA and then translated into a protein by transfer RNA in a complex series of events utilizing ribosomal RNA and amino acids. Although in essence the central dogma remains true, studies of genes and proteins are revealing a complexity that we had never imagined. For example, distinct genes are expressed in different cell types and the physiological state of the cell alters which proteins are produced and at what level. Furthermore, chemical changes (i.e. phosphorylation) to proteins occur after translation and are critical to a protein’s function. The importance and diversity of proteins started a whole new field termed proteomics.
Proteomics is the study of proteins, particularly their structures and functions. This term was coined to make an analogy with genomics. The Human Genome Project, started in 1990 and completed in 2003, sequenced three billion bases in genes (the human genome). The entire set of proteins in existence in an organism throughout its life cycle, or on a smaller scale the entire set of proteins found in a particular cell type under a specific set of conditions is referred to as the proteome.
Proteomics is much more complicated than genomics for several reasons. The genome is a rather constant entity while the proteome differs from cell to cell and is constantly changing through its biochemical interactions with the genome and the environment. Consequently, the proteome reflects the particular stage of development or the current environmental condition of the cell or organism. One organism will have radically different protein expression in different parts of the body, in different stages of its life cycle, and in different environmental conditions. For example, when E. coli cells are grown under conditions of elevated temperature a class of proteins known as heat shock proteins are upregulated. Many members of this group perform a chaperone function by stabilizing new proteins to ensure correct folding or by helping to refold proteins that were damaged by the cell stress. Ultimately, the comparison of proteomes of healthy and diseased tissues may identify the molecular nature of a disease and provide potential new targets for drug development. The field of proteomics also presents many analytical challenges when compared to genomics. In DNA there are only four nucleotide bases with similar molecular weights and properties. In a proteome there are thousands of different proteins with a wide range of concentrations, molecular weights, and properties.
Proteomics was initially defined as the effort to catalog all the proteins expressed in all cells at all stages of development. That definition has now been expanded to include the study of protein functions, protein-protein interactions, cellular locations, expression levels, and post-translational modifications of all proteins within all cells and tissues at all stages of development. It is hypothesized that a large amount of the non-coding DNA in the human genome functions to regulate protein production, expression levels, and post-translational modifications. It is regulation of our complex proteomes, rather than our genes, that makes us different from simpler organisms with a similar number of genes. An international collaboration of scientists in the human Proteome Project (HPP) is working to characterize all 20,300 genes of the known genome and generate a map of the protein based molecular architecture of the human body. Completion of this project will enhance understanding of human biology at the cellular level and lay a foundation for development of diagnostic, prognostic, therapeutic, and preventive medical applications.
Reading Questions
1. Define the term proteome.
2. Define the term proteomics.
3. Why is the analysis of proteins in a cell more difficult than sequencing DNA?
4. What types of questions can be answered by studying the proteome?
Read the following research paper to learn how the field of proteomics can be useful in the treatment of cancer.
Comparative proteomics of oral cancer cell lines: identification of cancer associated proteins, Karsani et al, Proteome Science, 2014, 12:3. doi:10.1186/1477-5956-12-3 (Open access journal)
Discussion Questions
1. What was goal of the scientific study reported in the paper?
2. Why would a study of the change in oral cancer proteins add to our understanding of the disease?
3. Refer to the results and discussion section and Figure 1 to answer the following questions.
a. How were the proteins in the healthy and cancerous cells separated and detected?
b. How many individual protein spots were resolved on the silver stained gels?
c. How many protein spots exhibited a significant difference in abundance from normal cells to cancerous cells?
4. Table 1 is a list of proteins with different abundances in the cancer cell line.
(The section in this module on peptide mass mapping describes how the identity of the protein in the gel spot was determined.)
Examine the data for two structural proteins: Stathmin (STMN1) and myosin regulatory light chain-2 (ML12A).
What is the change in abundance for each protein? Can this change be visualized from the image of the spot?
How is the change quantified?
5. The simplified 2D gel shown represents the proteins from a healthy cell line.
Draw a new 2D gel which could represent the changes in protein expression that occur in a cancerous cell line. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Biological_Mass_Spectrometry%3A_Proteomics/Section_1%3A_Proteins_and_Proteomics/Section_1B._What_is_p.txt |
Learning Objectives
At the end of this assignment, you should be able to:
• Explain how the m/z values and spacing of peaks in an ESI charge envelope is used to determine molecular weight.
• Calculate the charge state and mass of a protein from its ESI-MS spectrum.
• Explain how a quadrupole ion trap mass analyzer works.
Calculating the molecular weight of a protein from its electrospray ionization mass spectrum (ESI-MS)
Section 2: Electrospray
Gel electrophoresis gives some indication of the molecular weight of a protein based on its migration in a gel; however, the molecular weights determined from this method are not precise. In contrast, mass spectrometry can provide accurate and precise molecular weights that aid in identification of proteins. Initially, mass spectrometry experiments were limited to small molecules that could be readily volatilized into the gas phase and ionized before entering the mass analyzer. In the late 1980s, several ionization methods were developed and applied to biomolecules that permitted analytes from condensed phase samples, such as liquids and solids, to be volatilized and analyzed by mass spectrometry. Electrospray ionization (ESI) is one such technique.
ESI is a solution-phase “soft” ionization source which converts ions in solution to gaseous ions. The mechanism of ion formation in an ESI source is depicted schematically in Figure 1. Briefly, gaseous ions are formed when a solution containing the analytes of interest (from an LC column or from a syringe infusion pump) is sprayed through a stainless steel capillary to which a high voltage is applied, creating a fine mist of droplets which are charged on their surface. As solvent evaporates from the charged droplets, the charge density on their surface increases to a critical limit, at which point electrostatic repulsion causes the larger droplets to break up into smaller charged droplets. Finally, analyte ions are ejected into the gas phase by electrostatic repulsion, and these ions enter the mass analyzer for subsequent mass analysis. You can watch a video from the Johnson lab showing the formation of a stable electrospray as the voltage applied to the tip is increased.
Figure 1. The mechanism of electrospray ionization (ESI). High voltage is applied to a steel capillary to produce charged droplets containing the analyte molecules. As solvent evaporates, charge accumulates until the Rayleigh limit is reached and the droplet undergoes Coulomb fission into smaller droplets. This process continues until gas phase analyte molecules enter the mass spectrometer. Figure is reproduced with permission from S. Banerjee and S. Mazumdar, Int. J. Anal. Chem., 2012, 2012, 282574 under a Creative Commons Attribution License.
For electrospray, the protein sample is prepared in a solution composed of water with acid and a low surface tension organic solvent, such as methanol. Relatively pure protein samples can be infused directly into the mass spectrometer using a syringe pump, or more complex protein and peptide mixtures may be separated by chromatography first and electrosprayed directly from the column.
Electrospray ion sources are compatible with many types of mass analyzers. For this application, we will consider the quadrupole ion trap as a mass analyzer. The quadrupole ion trap is composed of two end cap electrodes at the entrance and the exit of the trap and a ring electrode (shaped like a donut) in the middle. The voltages and AC frequencies of the applied potentials on the electrodes are varied to control the motion of ions in and through the trap. The ion trap mass analyzer is a small, relatively inexpensive mass analyzer that typically generates mass spectra with unit resolution (Δm ≈ 1) or slightly better. This video shows the operating steps involved in generating a mass spectrum using an ion trap.
Video Question
1. Summarize what you saw in the video on the ion trap. What are the operating steps used to generate a mass spectrum using an ion trap mass analyzer? | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Biological_Mass_Spectrometry%3A_Proteomics/Section_2%3A_Electrospray/Section_2A._Electrospray_and_Qua.txt |
One of the advantages of ESI is that it is a “soft” ionization technique in which little fragmentation of large, thermally fragile biomolecules occurs. Consequently, molecular weight information is readily obtained with this technique. Additionally, the ions formed are often multiply charged, which enables the analysis of molecular masses exceeding 100,000 Da because the multiple charges bring the m/z values into the mass range of conventional mass analyzers, such as ion traps. With proteins and peptides, the mass spectrum consists of a series of peaks, call the “peak envelope” which represents a distribution of multiply charged analyte ions.
Ubiquitin is a small protein with a monoisotopic molecular weight of 8560 Da. Electrospray ionization of this small protein typically results in major charge states of +8, +9, +10, +11, +12, and +13.
Reading Questions
1. Using this information, complete the table below, assuming that the charges on each ion come from protonation rather than sodium or potassium adducts. Round masses and m/z values to the ones place.
z mass of [M+zH]+z m/z
8
9
10
11
12
13
2. Using the data you entered in the table above, sketch an expected ESI-MS spectrum for ubiquitin. Label each peak with its charge state. What do you notice about the spacing of the peaks along the x-axis?
Based on the m/z values and peak spacing observed in the charge envelope, we can determine the charge state, z, for each peak and the molecular weight of the analyte using Equations $\ref{1}$ and $\ref{2}$,
$z = \dfrac{M_2-A}{M_1-M_2} \label{1}$
$MW= \dfrac{(M_1-A)(M_2-A)}{M_1-M_2} \label{2}$
where MW is the molecular weight of the analyte, M1 is the m/z value for the first ion, z is the charge state of the first ion, M2 is the m/z value for a second ion of lower m/z, and A is the mass of the adduct ion, which is usually a proton (H+) but can be sodium (Na+) or potassium (K+) ions from glassware or buffers used in the experiment.
Discussion Questions
1. Why do aqueous samples for electrospray typically include an acid and a low surface tension solvent such as methanol?
2. Figure 2 shows an experimentally obtained mass spectrum for ubiquitin. Compare this spectrum to the spectrum you predicted in Reading Question 2. Are there any differences? If so, what might cause these differences?
Figure 2. ESI-MS spectrum of bovine ubiquitin from Protea Biosciences (proteabio.com/products/PS-143)
3. Using Equation ($\ref{2}$) and any pair of peaks from Figure 2, calculate the molecular weight of ubiquitin and its percent error compared to the theoretical monoisotopic mass of 8560 amu.
Figure 3. ESI-MS spectrum of cytochrome C. Data obtained at Trinity College.
4. Figure 3 shows the ESI-MS spectrum for cytochrome C electrosprayed from a mixture of water, methanol, and acetic acid with pH of 2.5. What is the charge state of the peak at m/z = 773.36?
5. Determine the MW of the analyte, cytochrome C, using the data in Figure 3.
6. How would you expect the mass spectrum to change if the cytochrome C sample was electrosprayed from a solution of higher pH? Explain your answer.
7. Note that in the ESI-MS spectra for ubiquitin and cytochrome C, each major peak is accompanied by a series of less intense peaks of slightly different m/z. What is the source of these peaks?
8. Compare your description of the ion trap mass analyzer video with your group mates’ descriptions. As a group, write a consensus explanation for how an ion trap works. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Biological_Mass_Spectrometry%3A_Proteomics/Section_2%3A_Electrospray/Section_2B._ESI-MS_Data.txt |
Learning Objectives
At the end of this assignment, you will be able to:
1. Describe the technique of Matrix Assisted Laser Desorption Ionization (MALDI) for creating gas phase ions of peptides and proteins.
2. Explain how ions of different m/z ratios separate in a Time of Flight (TOF) mass spectrometer.
3. Define resolution and explain how the reflectron design increases resolution in TOF mass spectrometry.
4. Explain the process of peptide mass mapping for identifying a protein.
5. Understand how variations in search parameters influence the probability score in a peptide mass mapping search.
6. Given MALDI-TOF data for a digested protein, identify the protein using the technique of peptide mass mapping in the MASCOT database.
Overview
Often the first step in analyzing a complex mixture of proteins is to separate them using two-dimensional (2D) gel electrophoresis. (The Introduction section of this module includes a paper on protein analysis using 2D gel electrophoresis.) If 2D gel analysis indicates that new protein is being expressed or the expression level of a protein has changed, that protein must be identified. Peptide mass mapping is a method used to determine the identity of a protein spot in a gel. The protein is cut from the gel, destained, and extracted. Then the protein is digested (cut into pieces) with a proteolytic enzyme such as trypsin. The digest contains a mixture of peptides whose mass is analyzed using Matrix-Assisted Laser Desorption Ionization Time of Flight (MALDI-TOF) mass spectrometry. In this module you will learn the principles of MALDI-TOF mass spectrometry and the method of peptide mass mapping for identifying a protein.
Section 3: Peptide Mass Mapping
Matrix-assisted laser desorption ionization (MALDI) is a technique that generates gas-phase peptide and protein ions. Proteins are nonvolatile and cannot simply be heated to make the gaseous form. (Think about a steak on the grill. The steak chars but does not vaporize.) Instead proteins are desorbed (released from a surface) by mixing the protein with a matrix molecule and irradiating the mixture with a pulse from an ultraviolet laser. The following figure shows the structure of various molecules used as the MALDI matrix.
Discussion Question
1. What structural features do these molecules have in common? Why are these common features important for a MALDI matrix?
The α-cyano-4-hydroxcinnamic acid (CCA) molecule is a MALDI matrix used for peptide and protein analysis. The matrix is combined with a mixture of peptides and placed on a metal plate. A pulsed nitrogen laser with a wavelength of 337 nm is used to irradiate the peptides and matrix. The absorption spectrum for a peptide and the CCA matrix is shown.
(a)
(b)
Figure. (a) Absorbance spectrum of a peptide. (b) Absorbance spectrum of CCA matrix.
Discussion Questions
1. Does the peptide absorb the laser light? Does the matrix absorb the laser light?
2. Propose a reason that the laser is matched to the absorption of the matrix (CCA) and not the peptide. What would happen to the peptide if it absorbed a high energy pulse of laser light?
In the MALDI process, the laser is directed at the mixture of CCA matrix and peptides. The matrix absorbs the laser light, but the peptides do not. The matrix heats up and “explodes” from the surface carrying the peptide with it. (It is similar to skydiving. The airplane (matrix) carries the people (peptides) into the air and when they jump, they are temporarily flying.) After absorbing light the matrix is in an excited state and transfers a proton (H+) to the peptide giving it a positive charge. The invention of MALDI enabled the mass spectrometric analysis of large nonvolatile biomolecules and was so important that its inventor Koichi Tanaka of Shimadzu Corporation was awarded one-third of the 2002 Nobel Prize in Chemistry. The MALDI process is depicted in the figure below.
Figure. The MALDI process for creating gas-phase ions of peptides. (Figure from Wikimedia Commons)
The purple color represents the MALDI matrix and the green color represents peptides. Energy from the laser is absorbed by the matrix, causing the matrix to desorb from the surface. The matrix carries the peptides into the gas phase and gives them a charge by donating a proton. Molecular ions of the peptides are created because peptides do not absorb laser light which would break bonds within the molecule.
An example of a typical mass spectrum of a mixture of proteins ionized by the MALDI method is shown below. The matrix is 2,5-dihydroxybenzoic acid (DHB) and the mixture contains three proteins; ubiquitin, cytochrome C and equine myoglobin.
Figure. MALDI-TOF mass spectrum of three proteins; ubiquitin, cytochrome C, and equine myoglobin.
Figure adapted from King’s College London
http://www.kcl.ac.uk/innovation/research/corefacilities/smallrf/mspec/cemsw/instr/maldi-tof-ms.aspx
Discussion Questions
1. Examine the mass spectrum of the mixture of three proteins obtained using MALDI as the ionization method. Would you classify MALDI as a “soft” (little to no fragmentation) ionization technique or a “hard” (fragmentation) ionization technique?
2. What is the identity of Peak 1 in the mass spectrum? | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Biological_Mass_Spectrometry%3A_Proteomics/Section_3%3A_Peptide_Mass_Mapping/Section_3A._MALDI-TOF_Ma.txt |
This is set of questions guides students through the principle of a TOF mass analyzer. There is very little reading and all questions are for small group discussion.
(Source for figures and animations: Dr. Jon Karty of Indiana University Mass Spectrometry Facility)
Consider the diagram above, which shows 4 ions (the circles), between two electrical plates. The ions have different masses and higher mass is represented by a larger circle. Imagine the entire schematic is contained in a long tube under vacuum. After the ions enter, the left plate is brought to a positive high voltage (+HV) such as +15,000 V. The right plate is a grid, like a screen door, so that ions can pass through, and is held at electrical ground (0 V).
1. In this scenario, the ions will accelerate once the +HV is applied. If the ions are positively charged, which direction will they move?
2. If all of the ions have a charge of +1, which ion will reach the detector first? The larger circles represent ions with higher mass.
3. What factors other than mass will influence the velocity of the ions?
4. $KE = zV \label{1}$
where z is the charge on the ion and V is the magnitude of the high voltage (HV).
1. Does this equation support your answer to number 3?
2. If all of the ions have the same charge, what will be true about their kinetic energies?
5. You may remember another equation for kinetic energy from physics:
where m is the mass of the object and v is its velocity.
1. Consider the four ions at the start of this worksheet. Which ion will have the highest velocity?
2. What is the order of ions reaching the detector?
3. Is the TOF mass analyzer dispersive or scanning?
4. Why do you think it is called a “time-of-flight” mass analyzer?
5. Consider a +1 ion with m/z = 115 which enters the time-of-flight source region (between the two plates).
1 elementary charge = 1.6 × 10-19 C $\mathrm{1\: V = 1\:\: \large{ {}^J/_C} }$
6. $\mathrm{1\: J = 1\: \dfrac{kg \cdot m^2}{s^2}}$ 1 amu = 1.66 × 10-27 kg
7. If the flight tube is 2 m long, how long will the ion take to reach the detector? Remember that v = d/t, where v is velocity, d is distance traveled, and t is time.
6. The ability of a mass spectrometer to distinguish between two different m/z ions is called resolving power.
$\textrm{Resolving Power} = \dfrac{m}{\Delta m}$
Image adapted from UC Davis Fiehn Metabolomics Lab. fiehnlab.ucdavis.edu/projects...ass_Resolution
1. Calculate the resolving power of the mass analyzer from the peak in the figure.
2. Using the resolving power calculated in part (a), sketch the appearance of the two peaks with m/z ratios of 1500 and 1500.5. Would you say these peaks are resolved?
7. Consider the two instrument schematics for a time of flight mass analyzer below. Which one do you think will be better at resolving small differences in the masses of the ions? Explain.
1. Equation $\ref{1}$ can also be written as:
where E is the electric field (V/d) and ds is the distance traveled in the source region. Use this equation to explain which ion (1 or 2) will leave the source with the highest velocity.
2. How will the position of different ions in the source affect the resolving power of the mass analyzer? Draw a peak for a given m/z similar to the one in question 7 assuming that the position of the ion in the source did not affect kinetic energy. Draw a new peak taking into account how different positions of ions in the source affect the kinetic energy of the ions.
8. Consider the instrument diagram below, which adds a “reflectron” to the mass analyzer. The reflectron consists of a series of grids held at increasingly high positive potentials from ground (0 V) up to +HV2. Note: +HV2 > +HV1.
1. On the diagram above, sketch the flight paths for Ion 1 and Ion 2 (these ions have the same m/z ratio and are the same ions in question 8). Note that the reflectron is angled toward the detector.
2. Which ion penetrates farther into the reflectron?
3. How does this arrangement correct for the ions’ different starting positions in the source?
9. Examine the diagram of a reflectron TOF mass analyzer.
1. What is the order of ions reaching the detector in the reflectron time of flight mass spectrometer? (all ions have a +1 charge)
2. Why is the ionization method MALDI frequently coupled with a time of flight mass analyzer?
3. Examine the two mass spectra below. Identify which spectrum is from a reflectron TOF instrument and which is from a linear TOF instrument? | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Biological_Mass_Spectrometry%3A_Proteomics/Section_3%3A_Peptide_Mass_Mapping/Section_3B._Time_of_Flig.txt |
Peptide mass mapping is a technique that uses powerful search engines (e.g. Mascot) to identify a protein from mass spectrometry data and primary sequence databases. The general approach is to take a small sample of the protein and digest it with a proteolytic enzyme, such as trypsin. Trypsin cleaves the protein after lysine and arginine residues. The resulting mixture of peptides is analyzed by MALDI-TOF mass spectrometry.
The experimental mass values of the peptides are then compared with theoretical peptide mass values. Theoretical mass values of peptides are obtained by using the genome sequence for an organism and predicting all the proteins that can be expressed. Once all the proteins are predicted then the cleavage rules for the digest enzyme are applied and the masses of the resulting peptides calculated by the computer (in-silico digest). By using an appropriate scoring algorithm, the closest match or matches can be identified. If the "unknown" protein is present in the sequence database, then the aim is to pull out that precise entry. If the sequence database does not contain the “unknown” protein, then the aim is to pull out those entries which exhibit the closest homology, often equivalent proteins from related species. The steps in peptide mass mapping are outlined in the flow chart.
The analysis of a complex mixture of proteins from an organism always involves some type of separation step to isolate a certain protein. The separation methods frequently used are two dimensional (2D) gel electrophoresis or liquid chromatography. The figure shows the experimental workflow used to identify a protein spot from a 2D gel.
Reading Questions
1. In your own words, describe the general principle of peptide mass mapping for protein identification.
The enzyme trypsin is frequently used to digest proteins in the peptide mass mapping technique. Trypsin cleaves the amide bond after lysine (K) and arginine (R) residues. K and R make up about 10% of the amino acids in a protein and digesting with trypsin typically results in peptides in a useful range mass range for mass spectrometry (500-3,000 Da). (Remember: The mass unit amu is the same as Da.)
The structures of lysine and arginine are shown.
Discussion Questions
1. What is the purpose of digesting the protein with trypsin? (Hint: Think about how mass spectrometry is used to identify and/or elucidate the structure of small organic molecules such as caffeine).
2. Why is a well annotated genome for the organism of interest needed in the peptide mass mapping technique for identifying a protein?
3. Classify the side chains of R and K as acidic or basic.
4. Digesting the protein with trypsin ensures that there is an R or K residue in each peptide. Why is this helpful for MALDI-TOF analysis? (Hint: Think about the function of the matrix in MALDI).
5. a. Can the following two peptides with the same amino acid composition be distinguished using MALDI-TOF mass spectrometry?
Peptide 1: GASPVRTCILKMHFY
Peptide 2: GMFHRATIKYPVCSL
b. Calculate the expected (monoisotopic) masses if the enzyme trypsin was used to digest peptides 1 and 2. A table of amino acid masses is provided. Can the peptides be distinguished after digestion with trypsin?
(Refer to the Introduction section of this module if you need assistance in calculating the mass of a peptide or defining the difference between a monoisotopic mass and average mass.)
6. There are other enzymes that could be used to digest the proteins.
Pepsin is most efficient in cleaving peptide bonds between hydrophobic amino acids (leucine) and aromatic amino acids such as phenylalanine, tryptophan, and tyrosine. Pepsin is less specific and results in many small peptides. Would pepsin be a good choice for digesting proteins for peptide mass mapping? Explain your reasoning.
There are also enzymes that are highly specific and result in only a few cleavage sites in a protein. Is a highly specific enzyme that creates a few large peptides be a good choice for peptide mass mapping? Explain your reasoning.
Table 2. Molecular weight information for all twenty naturally occurring amino acids.
Amino Acid
Single-Letter Code
Residue MW (amu)
Amino Acid MW (amu)
glycine
G
57.02
75.03
alanine
A
71.04
89.05
Serine
S
87.03
105.04
proline
P
97.05
115.06
Valine
V
99.07
117.08
threonine
T
101.05
119.06
cysteine
C
103.01
121.02
isoleucine
I
113.08
131.09
leucine
L
113.08
131.09
asparagine
N
114.04
132.05
aspartic acid
D
115.03
133.04
glutamine
Q
128.06
146.07
Lysine
K
128.09
146.11
glutamic acid
E
129.04
147.05
methionine
M
131.04
149.05
histidine
H
137.06
155.07
phenylalanine
F
147.07
165.08
arginine
R
156.10
174.11
tyrosine
Y
163.06
181.07
tryptophan
W
186.08
204.09 | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Biological_Mass_Spectrometry%3A_Proteomics/Section_3%3A_Peptide_Mass_Mapping/Section_3C._Peptide_Mass.txt |
The peptide mass fingerprint search in the MASCOT database is used to identify a protein from mass spectrometry data. The following MALDI-TOF mass spectrum shows the masses detected after digesting a protein with trypsin. This protein is from the organism Escherichia coli (E. coli) and was cut from a 2D gel and analyzed by students in Analytical Chemistry at Indiana University. .coli (Escherichia coli)
m/z
S/N
832.312 10.3
842.53 23
1045.541 3.7
1179.58 2.3
1210.545 5.2
1247.598 5.7
1283.741 74.7
1307.679 2.1
1401.67 3.1
1403.711 2.3
1473.801 43.2
1521.775 37.2
1537.764 6
1618.801 7.3
1626.907 10.4
1648.88 2.6
1811.976 53
2141.078 8.5
2391.068 2.3
2753.496 12.3
Some peaks in the mass spectrum do not come from peptides from the protein we are trying to identify. Peaks that should be eliminated before performing the Mascot search are:
Trypsin autolysis peaks: Trypsin cleaves the protein of interest after lysine and arginine residues. However, trypsin will also cleave other trypsin molecules. The peaks due to trypsin autolysis are: 514.63, 842.5, 906.05, 1006.15, 1045.12, 1736.97, 1768.99, 2158.48, 2211.4, 2239.1
Matrix Clusters: The MALDI matrix may also combine with Na+ and K+ during ionization. These peaks are 855.1, 861.1, 871.1, 877.1, 1060.1. If the scientist had good lab technique, these ions would be eliminated or greatly reduced in a sample preparation step.
Reading Question
1. Examine the m/z data in the previous table and figure. Which peaks should be removed before entering m/z data into the MASCOT database?
Peptide Mass Mapping Search Parameters
Before m/z information is entered into the MASCOT database, there are also a number of search parameters that must be set appropriately. The image shows a data entry page for a peptide mass fingerprint search. Let’s examine the meaning of each of the following search parameters; Database, Enzyme, Taxonomy, Fixed Modifications, Variable Modifications, Protein Mass, Peptide Tolerance, Mass Values, Monoisotopic or Average Mass, and Data Input.
The following section on search parameters has been adapted from the Peptide Mass Fingerprint Tutorial in the MASCOT Database.
Database and Taxonomy: The first choice you have to make is which database to search. Some databases contain sequences from a single organism. Others contain entries from multiple organisms, but usually include the taxonomy for each entry, so that entries for a specific organism can be selected during a search using a taxonomy filter.
If your target organism is well characterized, such as human or mouse or yeast, Swiss-Prot is the recommended choice. The entries are all high quality and well annotated. Because Swiss-Prot is non-redundant, it is relatively small, which makes it easier to get a statistically significant match. If you know what is in the sample, you can restrict the search to an organism or family by means of the taxonomy filter, but remember that you can never rule out contaminants.
If you are interested in a bacterium or a plant, you may find that it is poorly represented in Swiss-Prot, and it would be better to try one of the comprehensive protein databases, which aim to include all known protein sequences. The two best known are NCBInr and UniRef100. These are very large databases, and you will almost certainly want to select a limited taxonomy. But, never choose a narrow taxonomy without looking at the counts of entries and understanding the classification. In the current Swiss-Prot, for example, there are 26,139 entries for rodentia, of which all but 1,602 are for mouse and rat. So, even if your target organism is hamster, it isn’t a good idea to choose ‘other rodentia’. Better to search rodentia and hope to get a match to a homologous protein from mouse and rat.
Enzyme: Choose the enzyme used to digest the protein. Trypsin is commonly used and will be the enzyme utilized in all the data in this module.
Missed Cleavages: The number of missed cleavages refers to the completeness of the enzyme digest. Did the enzyme trypsin cleave after every lysine and arginine residue in the protein? Or were some cleavages missed? The number of allowed missed cleavages should be set empirically, by running a standard and/or trying different values to see which gives the best score.
Modifications in database searching are handled in two ways.
First, there are the fixed modifications. The most common example is the reduction and alkylation of cysteine. This reaction is performed to break disulfide bonds and prevent them from reforming. In the absence of disulfide bonds, the protein will be unfolded and the enzyme will be more effective in digesting the protein. Since all cysteines are modified, this is effectively just a change in the mass of cysteine. It carries no penalty in terms of search speed or specificity.
The alkylation agent used is iodoacetamide (select modification carbamidomethyl). In proteins, the reduced thiol group in cysteine is alkylated with iodoacetamide in the reaction shown:
In contrast, most post-translational modifications do not apply to all instances of a residue. For example, phosphorylation might affect just one serine in a protein containing many serines and threonines. These variable or non-quantitative modifications are expensive in the sense that they increase the search space. This is because the software has to permute out all the possible arrangements of modified and unmodified residues that fit to the peptide molecular mass. As more and more modifications are considered, the number of combinations and permutations increases geometrically, and we get a so-called combinatorial explosion.
One common variable modification is the oxidation of methionine shown:
Protein Mass: If the protein mass is known from its position in a 2D gel, this value can be entered. Usually, this adds little to the score, and the general advice is to leave this field blank.
Peptide Tolerance: Making an estimate of the mass accuracy doesn’t have to be a guessing game. The Mascot Protein View report includes graphs of mass errors.
One way to evaluate the mass accuracy of the mass spectrometer is to run a standard and look at the error graphs for the correct match. Another method of evaluating mass accuracy is to compare the experimental value of a trypsin autolysis peak with the theoretical value.
In the data set provided, one trypsin autolysis peak had a measured mass of 1045.54 and the theoretical mass is 1045.12. The measurement indicates the mass spectrometer has mass error of approximately 0.42 Da.
(Note: Da is the same as amu).
Mass values: Most frequently MALDI produces the singly charged molecular ion (MH+). Your peak list will only contain Mr values (relative molecular mass) if the peak picking software has ‘de-charged’ the measured m/z values. Peak picking software may be programmed to do this because the data contained a mixture of charge states.
Most modern instruments produce monoisotopic mass values. You will only have average masses if the entire isotope distribution has been centroided into a single peak, which usually implies very low resolution.
The following MALDI-TOF mass spectrum of a protein digest zooms in on the mass region of different peptides near m/z 1500. The isotope distribution in a peptide with m/z 1515.7 is shown. The natural abundance of carbon-12 is 98.90% and carbon-13 is 1.10%. Therefore, peptides with a large number of carbon atoms will contain significant contributions to the M+1 peak and M+2 peak from carbon-13 atoms. The monoisotopic peptide contains all carbon-12 atoms. The M+1 peak has one carbon-13 atom and the M+2 peak has two carbon-13 atoms.
Data Input: The first requirement for a Peptide Mass Fingerprint (PMF) search is a peak list (a list of m/z values). Peak lists are text files and come in various different formats. You can also copy and paste a list of values into the query area of the search form, or even type them in. Each m/z value goes on a separate line. If you also have an intensity value for the peak, this follows the m/z value, separated by a space or a tab.
Reading Questions
1. You are analyzing a protein from E. coli.
a. What is the advantage of setting the taxonomy to E. coli.?
b. What is a disadvantage of setting the taxonomy to E. coli. instead of a more general class of bacteria to which E. coli. belongs (Proteobacteria).
2. a. What is meant by the search parameter “missed cleavages?”
b. How will one missed cleavage affect the number of peptides created after digestion with trypsin?
3. A common fixed modification is carbamidomethyl. Why is a protein chemically modified in this way?
4. Briefly describe one method for determining the peptide tolerance (or the mass accuracy) of the mass spectrometer.
Performing a Peptide Mass Mapping Search: Now that you understand the various search parameters, you are now ready to perform a peptide mass fingerprint search in MASCOT.
1. Go to www.matrixscience.com and choose “Mascot search database” “Peptide mass fingerprint”, and “Perform search”
2. A good set of search parameters to start with are:
Database: SwissProt
Taxonomy: Escherichia Coli
Enzyme: Trypsin
Missed Cleavages: 1
Fixed Modification: Carbamidomethyl
Variable Modification: Oxidation of M
Protein Mass: leave blank
Peptide Tolerance: ±1 Da
Mass Values: MH+
Monoisotopic
Report Top 5 Hits
3. We will start with the MALDI-TOF data for the protein from E. Coli cut from a 2D gel. Copy and paste the m/z values in the table. Don’t forget to remove the trypsin autolysis peaks or matrix clusters from the data set.
m/z
832.312
842.53
1045.541
1179.58
1210.545
1247.598
1283.741
1307.679
1401.67
1403.711
1473.801
1521.775
1537.764
1618.801
1626.907
1648.88
1811.976
2141.078
2391.068
2753.496
4. Record the search results.
What protein has the highest score?
What is its protein score? What score is needed for significance?
Click on the identity of the protein for information about sequence and which peptide masses were found experimentally.
How many mass values were searched?
How many mass values were matched?
What was the percent sequence coverage?
Instructor Note: Students can run the search on computers and then compare with the results shown here.
Results Summary: The first results screen identifies the protein as cysteine synthase A with a protein score of 120. Scores outside the green region (>56) are significant. A score of 120 indicates that there is a high probability that the protein has been correctly identified.
The optimum data set for a peptide mass fingerprint is, of course, all of the correct peptides and none of the wrong ones. By correct, we mean that the textbook enzyme cleavage rules were followed, and only specified modifications are present. Sadly, real life data are generally far from ideal, and it is almost unknown to get every single experimental mass value matching and 100% sequence coverage. However, it is not always recognized that having too many peptide mass values can create similar difficulties to having too few.
Imagine a tryptic digest of a 20 kDa protein. We would expect something around 20 perfect cleavage peptides. If the digest was incomplete, or there was a non-quantitative modification, we might expect to double the number of peptides observed.
If 100 peaks are taken from the mass spectrum of this digest and submitted to Mascot then either 60 to 80 peaks are noise or there are extensive non-quantitative modifications. Either possibility is bad news for search specificity. The peaks which cannot be matched correctly will still contribute to the population of random matches.
The Mowse Scoring Algorithm is described in [Pappin, 1993]. (Reference available on MASCOT database)
The first stage of a Mowse search is to compare the calculated peptide masses for each entry in the sequence database with the set of experimental data. Each calculated value which falls within a given mass tolerance of an experimental value counts as a match.
Rather than just counting the number of matching peptides, Mowse uses empirically determined factors to assign a statistical weight to each individual peptide match.
Probability Based Scoring
Mascot incorporates a probability based implementation of the Mowse algorithm. The Mowse algorithm is an excellent starting point because it accurately models the behavior of a proteolytic enzyme. By casting the Mowse score into a probabilistic framework a simple rule can be used to judge whether a result is significant or not.
Matches using mass values are always handled on a probabilistic basis. The total score is the absolute probability that the observed match is a random event. Reporting probabilities directly can be confusing because they encompass a very wide range of magnitudes, and also because a "high" score is a "low" probability. For this reason, we report scores as -10*LOG10(P), where P is the absolute probability. A probability of 10-20 thus becomes a score of 200.
Significance Level
Given an absolute probability that a match is random, and knowing the size of the sequence database being searched, it becomes possible to provide an objective measure of the significance of a result. A commonly accepted threshold is that an event is significant if it would be expected to occur at random with a frequency of less than 5%. This is the value which is reported on the master results page.
The master results page for typical peptide mass fingerprint search reports that "Scores greater than 56 are significant (p<0.05).” The protein with the score of 120 is a nice result because the highest score is highly significant, leaving little room for doubt.
After clicking on the protein with the top score, additional information from the search is displayed (as shown in the screen capture below). The molecular weight (34,525 Da) and pI value (5.83) are provided. If the protein was cut from a 2D gel, the position in the gel should correlate with the molecular weight of pI value of the protein identified. The protein sequence coverage was 49% and 13 of the 18 mass values that were searched matched the protein of interest.
Discussion Questions
1. The protein identified has a very high score; however, less than half of the sequence was matched.
Why can a protein have a high score even with low sequence coverage?
2. What are some experimental reasons for low sequence coverage? In other words, why are some peptides not found in the MALDI-TOF data?
Peptide Mass Mapping for Protein Identification
In this computer exercise, Mascot search parameters will be varied to explore their effect on protein score. The following MALDI-TOF data for an E. coli protein cut from a 2D gel will initially give a low protein score using default search parameters. The parameters will then be changed in a systematic way to see if a significant protein score can be achieved.
Open the Excel File: Proteomics Data AST4
The data is shown here as well.
m/z
Intensity
842.589
1962
864.585
483
874.496
99
886.559
93
936.473
91
976.436
87
993.491
1285
1015.477
93
1045.643
571
1067.664
99
1145.477
125
1155.598
142
1202.521
520
1254.658
153
1316.769
246
1333.777
186
1428.78
1282
1450.796
69
1525.82
119
1675.941
862
1753.983
1171
1804.047
307
1884.98
508
2013.05
271
2094.001
147
2508.378
155
2530.349
141
2662.519
162
2691.481
333
2807.501
476
3337.949
307
3794.978
221
3809.024
338
1. Use the following default search parameters to run a Peptide Mass Fingerprint search in Mascot search to identify the protein with the file name AST4.
Database: SwissProt
Taxonomy: All
Enzyme: Trypsin
Missed Cleavages: 1
Fixed Modification: Carbamidomethyl
Variable Modification: Oxidation of M
Protein Mass: leave blank
Peptide Tolerance: ±1 Da
Mass Values: MH+
Monoisotopic
Report Top 5 Hits
What protein has the highest score?
What is its protein score? What score is needed for significance?
How many mass values were searched?
How many mass values were matched?
What was the percent sequence coverage?
2. Use the same masses and initial search parameters and change the taxonomy to Metazoa (Animals) because the sample was from chicken.
What protein has the highest score?
Protein score? What score is needed for significance?
How many mass values were searched?
How many mass values were matched?
What was the percent sequence coverage?
The score for significance changed between all taxonomies and Metazoa. What do you think is the reason for the change?
3. Retain the taxonomy as Metazoa and vary the number missed cleavages.
Zero missed cleavages:
What protein has the highest score?
Protein score? What score is needed for significance?
How many mass values were searched?
How many mass values were matched?
What was the percent sequence coverage?
Two missed cleavages:
What protein has the highest score?
Protein score? What score is needed for significance?
How many mass values were searched?
How many mass values were matched?
What was the percent sequence coverage?
Summarize how the number of missed cleavages parameter affects protein score. Can you explain why the score changes?
4. Vary the mass tolerance.
Repeat the search choosing 1 missed cleavage and a peptide tolerance of 0.5 Da.
What protein has the highest score?
Protein score? What score is needed for significance?
How many mass values were searched?
How many mass values were matched?
What was the percent sequence coverage?
Repeat the search choosing 1 missed cleavage and a peptide tolerance of 0.3 Da.
What protein has the highest score?
Protein score? What score is needed for significance?
How many mass values were searched?
How many mass values were matched?
What was the percent sequence coverage?
Repeat the search choosing 1 missed cleavage and a peptide tolerance of 0.1 Da.
What protein has the highest score?
Protein score? What score is needed for significance?
How many mass values were searched?
How many mass values were matched?
What was the percent sequence coverage?
Summarize how the mass tolerance affects score. Can you explain why the score changes?
5. Set the number of missed cleavages to 1 and mass tolerance to 0.30 Da. Choose only carbamidomethyl (C) as a fixed modification and no variable modification.
What protein has the highest score?
Protein score? What score is needed for significance?
How many mass values were searched?
How many mass values were matched?
What was the percent sequence coverage?
Summarize how the selection of modifications affects score. Can you explain why the score changed?
Homework (Peptide Mass Mapping)
1. To turn in for each protein:
1. Record the search parameters that were used to generate the highest probability score.
2. What masses were used and which were discarded? Explain your reasoning.
3. Record the protein identity, probability score, molecular weight, pI, number of mass values searched and matched, and percent sequence coverage.
4. Interpret the results.
***Note: The reduction and alkylation procedure was not efficient in this set of data. It may help scores to not use choose carbamidomethyl as a fixed modification. (Do not select any fixed modifications if you cannot get a significant protein score.)
2. Imagine that you have performed a 2D gel separation of proteins from healthy and cancerous cells and have identified a protein implicated in the cancerous state by cutting out the spot, digesting with trypsin, and performing peptide mass mapping. What else could you do experimentally to increase your level of certainty that the protein from the gel spot was identified correctly?
3. The following data were obtained by analytical chemistry students at Indiana University. The students grew E. coli samples at two different temperatures (37°C and 46°C). The cells were lysed, proteins isolated, and 2D gel electrophoresis performed. Based upon differences in the 2D gel pattern between the high and low temperature E coli samples, one protein spot (from high temp. sample) was analyzed as it was suspected to be a heat shock protein.
Change search parameters for the following data set to see if you can achieve a significant score for a heat shock (chaperone protein).
High temp. E. coli gel
Between MW band 5 (50 kDa)
MW band 6 (75 kDa)
Sample
m/z
Signal
719.3664
262.6387
842.4825
4195.775
855.0103
204.8235
861.033
273.0225
864.4673
410.5345
877.0083
428.2205
892.9921
220.0347
1045.548
444.506
1567.886
1874.742
1694.763
476.5625
1707.792
276.5812
1845.939
1238.335
1939.982
243.3293
2402.261
209.4211 | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Biological_Mass_Spectrometry%3A_Proteomics/Section_3%3A_Peptide_Mass_Mapping/Section_3D._MASCOT_Datab.txt |
Learning Objectives
At the end of this assignment, you should be able to:
1. Explain the purpose and methodology of a tandem MS experiment.
2. Identify b and y ions and use them to determine the sequence of a short peptide by de novo sequencing
Section 4: MS-MS and De Novo Sequencing
While enzymes, such as trypsin, can be used to cleave proteins and peptides at specific amino acid linkages, we can also fragment peptides inside of a mass spectrometer to obtain additional information. These types of experiments are called tandem MS or MS-MS experiments. These experiments are particularly helpful for “shotgun” proteomics or bottom-up proteomics. In these experiments, protein mixtures are first digested with enzymes (such as trypsin), then separated by one or more chromatography steps, and then electrosprayed into a mass spectrometer. A mass analyzer is then used to select a precursor ion with a specific m/z value for fragmentation. Fragmentation requires that some energy be added to the system. The most common method of fragmentation in MS-MS experiments is collision-induced dissociation (CID). In CID, the precursor ion is accelerated into an interaction cell that contains a collision gas, such as helium or nitrogen. When the precursor ion collides with the collision gas, the ion can fragment into two fragments, an ion and a neutral. The fragment ions are then analyzed to produce the MS/MS spectrum.
To better understand the operation of a commonly used triple quadruple or “triple quad” MS/MS, watch an animation, then answer the questions below.
Video Questions
1. What is the purpose of the skimmer in the instrument?
2. The animation uses color to indicate difference m/z value ions. Which “color” of ion is selected as the precursor ion?
3. Although this style of instrument is commonly called a “triple quad,” the collision cell is not actually a quadrupole. What is it?
4. The last quadrupole selected fragment ions to be sent to the detector. Neutrals also pass through this quadrupole. Why don’t they produce a signal at the detector?
MS/MS experiments are useful because the fragment m/z values give information about the analyte’s molecular structure. Low energy CID often produces small neutral losses, such as H2O, CH3OH, CO, CO2, NH3, and CN. Higher energy collisions can lead to retrosynthetic reactions, in which characteristic bonds in the precursor ion are broken. For example, CID of peptides often results in cleavage of peptide bonds along the backbone.
Reading Question
1. Complete the table below to summarize the expected mass differences from common neutral losses. Round expected masses to the nearest amu.
Table 1. Common neutral losses produced by CID of peptide ions.
Neutral Loss Mass (amu)
NH3
H2O
CN
CO
CH3OH
CO2
The identity of the precursor ion can be scanned through the mass range so that a mass spectrum of each precursor ion’s corresponding fragments is obtained. These experiments produced large quantities of data extremely rapidly. Interpreting these large data sets usually involves specialized software programs that identify peptides from their fragments and then identify proteins from their peptides; however, for simple mixtures, the data may be interpreted manually since peptide fragmentation in tandem MS experiments is well-characterized. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Biological_Mass_Spectrometry%3A_Proteomics/Section_4%3A_MS-MS_and_De_Novo_Sequencing/Section_4A._Tand.txt |
Peptide fragments produced in tandem MS experiments are named using a letter-number scheme that identifies which bond was broken and which side of the peptide (the N-terminus or the C-terminus) became the charged fragment (Figure 1). As noted above, CID typically results in cleavage of the peptide bond and therefore produces b and y ions. When the N-terminal side of the peptide is the charged fragment and the C-terminus is a neutral loss, the result is a b ion. When the C-terminal side is the charged fragment, the result is a y ion. In either case, the ions are numbered sequentially from the charged terminus, meaning that the first N-terminal residue gives the b1 fragment, and the first C-terminal residue gives the y1 ion. Other fragmentation methods, such as photodissociation, electron capture, and electron transfer dissociation can cleave other bonds within the peptide, giving rise to a, x, c, and z ions. In some cases, CID may also produce a ions, which differ from b ions by the absence of a carbonyl, making them 28 amu lighter than the corresponding b ion.
For appropriate CID conditions, it is often possible to obtain a series of b and y ions for a peptide (Figure 1). The mass differences between b ions (or between y ions) are characteristic of the amino acid residues which have been lost, and careful examination of the mass spectrum can yield the peptide sequence from the MS-MS data. For example, Figure 3 shows model data for fragmentation of the MRFA peptide. The mass difference between peaks from right to left are 71, 147, 156 amu, corresponding in mass to sequential losses of A, F, and R amino acid residues (Table 2), indicating that these amino acids form a series in the peptide backbone of the analyte.
Figure 1. Fragment ions are named based on the bond cleaved and the location of the charge on the resulting fragments. CID most commonly produces b and y ions, which result from cleavage of the peptide bond with the charged fragment occurring on the N- or C-terminal fragment, respectively. Figure is adapted with permission from S. Banerjee and S. Mazumdar, Int. J. Anal. Chem., 2012, 2012, 282574 under a Creative Commons Attribution License.
Figure 2. Structure of the tetrapeptide MRFA.
Figure 3. A model mass spectrum for MS-MS of the peptide, MRFA, which has a precursor m/z of 524.27.
Table 2. Molecular weight information for all 20 naturally occurring amino acids.
Amino Acid Single-Letter Code Residue MW (amu) Amino Acid MW (amu)
glycine G 57.02 75.03
alanine A 71.04 89.05
serine S 87.03 105.04
proline P 97.05 115.06
valine V 99.07 117.08
threonine T 101.05 119.06
cysteine C 103.01 121.02
isoleucine I 113.08 131.09
leucine L 113.08 131.09
asparagine N 114.04 132.05
aspartic acid D 115.03 133.04
glutamine Q 128.06 146.07
lysine K 128.09 146.11
glutamic acid E 129.04 147.05
methionine M 131.04 149.05
histidine H 137.06 155.07
phenylalanine F 147.07 165.08
arginine R 156.10 174.11
tyrosine Y 163.06 181.07
tryptophan W 186.08 204.09
Reading Questions
2. Sketch the a2, b2, and y2 ions for the tetrapeptide MRFA shown in Figure 2. What are the expected masses of these fragments?
3. What is the mass difference between the y3 and the y2 ion in Figure 3?
4. What amino acid residue corresponds to this mass difference? Does this make sense given the sequences of these two ions?
In practice, a complete b and y ion series may not be obtained, but it is often possible to deduce the peptide sequence from MS-MS data without referring to external databases or genomic data. This method of determining peptide structure is called de novo sequencing since the sequence is determined without reference to outside data. Because de novo sequencing does not rely on an external database, peptides can be identified even if they have unexpected post-translational modifications or arise from organisms with unsequenced genomes. Several research groups have developed algorithms to automate de novo sequencing from MS-MS data. These algorithms are designed to account for missing b and y ions, identify post-translational modifications, and address other challenges, including identification of oxidation and other chemical changes to proteins and peptides. Database searching is an alternative to de novo sequencing for longer peptides and proteins. Similarly to database searching for peptide mass fingerprinting, MS-MS database searching relies on genomic data to predict the expected spectra for MS-MS analysis. The experimental spectra are then matched to the predicted spectra to make peptide and protein assignments.
Discussion Questions
1. Consider the data in Table 2. By what value do the residue MW and the amino acid MW differ?
2. If needed, review the module introduction to peptides and proteins. Why is the MW of an amino acid residue in a peptide chain different from the mass of the full amino acid? Sketch a reaction to support your answer.
3. Are there any amino acids that could not be distinguished from one another using a mass analyzer with unit resolution (e.g., a quadrupole ion trap)?
4. Leu-enkephalin is a pentapeptide with the sequence YGGFL that is involved in neurotransmission. Tabulate the amino acid sequence and the expected monoisotopic MWs, and fragments of leu-enkephalin. Work in groups, dividing the required calculations among members.
Table 3. Expected precursor and fragment ions for CID MS-MS of leu-enkephalin.
Ion Peptide Sequence Expected m/z
[M+H]+
b1
b2
b3
b4
b5
y1
y2
y3
y4
5. Using your completed Table 3 above, identify as many of the peaks in the leu-enkephalin MS-MS spectrum shown in Figure 4 below as possible, remembering that CID can result in the small neutral losses that you noted in Table 1 in addition to retrosynthetic fragmentation.
Figure 4. ESI-MS-MS spectrum of leu-enkephalin. Data obtained at Trinity College.
6. Imagine researchers take a sample of blood serum from a patient and perform a shotgun proteomics experiment with a trypsin digestion and tandem MS detection. Would you recommend a separation step in between the tryptic digest of the serum proteins and the mass spectrometric detection? Why or why not?
Challenge Question
To the best of your ability, use the MS-MS spectrum in Figure 5 below to de novo sequence an unknown tetrapeptide. To support your proposed sequence, make a table showing the expected mass of each fragment and the actual mass you observe for the peak.
Figure 5. MS-MS mass spectrum of an unknown peptide. Data from F. Klink, Separation Science. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Biological_Mass_Spectrometry%3A_Proteomics/Section_4%3A_MS-MS_and_De_Novo_Sequencing/Section_4B._CID_.txt |
A series of collaborative learning activities and accompanying text that develop the concept of concentration calibration, utilizing the external standard, internal standard, and standard addition methods. The module is based primarily on flavonoids, particularly quercetin, as an example analyte. The activities are designed as in-class, small group exercises. An additional out-of-class activity is also available.
Contributors and Attributions
• Dr. Sandra L. Barnes (Alcorn State University), Dr. David Thompson (Sam Houston State)
Concentration Calibration
Learning Objectives
By the end of this module, students should be able to:
• Use chemical information to obtain concentration and manipulate concentration units
• Prepare standard solutions and an external calibration curve, determine R2 value and linear regression equation
• Determine the concentration of an analyte given raw data gathered by any of the following methods that incorporate:
1. external standards
2. internal standards
3. standard additions
• Use standard addition and internal standard methods for determining concentration
Introduction
Accurately knowing the amount, in terms of concentration, of a particular substance (analyte) is important in fundamental research and also in many applied fields of study such as medicine, environmental studies, and the food industry.
In medicine, many drug dosages are effective within a narrow concentration range. If too little is administered, the benefits of the drug will not be obtained. However, if too much is administered in a particular time frame, overdose or toxicity can occur. Therefore, many patients must have their blood tested on a regular basis to determine the amount of such drugs in their system. Also, many common analytes such a metals are required by the human body at low levels, but are toxic to the cells at high levels.
In environmental studies, it is important to ensure that levels of environmental contaminants are monitored. In 2010 the BP Oil spill devastated the wildlife on the Gulf Coast of Mexico in the United States. It was one of the worst environmental accidents in the history of the United States. Years and vast amounts of money was required to clean up the spill. The Gulf waters have since been deemed safe to use; however, four years later, scientists still monitor the concentration of oil contaminants to ensure safety of Gulf water for patron use.
In the food industry, most products that are commercially available for human consumption are required to disclose the nutritional content of the product on the label (sodium, potassium, sugar, fat, etc.); most products in your kitchen cabinet will provide this information, which can be useful for health reasons. For example, individuals with hypertension need to monitor and limit the amount of sodium, present in common table salt, they consume on a daily basis. Therefore, they avoid food products that have a high sodium content.
The previous examples provide important reasons for knowing the concentration of a particular analyte in a sample (biological fluid, water, food etc.). The substance that we want to know the amount of in a sample is called the analyte. Common analytes include the pesticide alachlor in water, the antioxidant quercetin in plant foods, the carbohydrate glucose in blood, oxygen in air, and ethanol in water (rubbing alcohol). In order to determine the concentration of a particular analyte in a sample we must perform a procedure called “concentration calibration”. This module discusses the three most common types of concentration calibration procedures.
Q1. What is meant by the term “concentration” and what is the most common unit for concentration in a typical analytical measurement?
Q2. To answer this question, refer to M. Bener, M. Ozyurek, K. Guclu, R. Apak, Polyphenolic contents of natural dyes produced from industrial plants assayed by HPLC and novel spectrophotometric methods, Industrial Crops and Products, 2010, vol. 32, 499-506.
(a) find the concentration used to prepare all stock phenolic compounds in ethanol (b) Convert this unit to mM (c) Compare the numerical values in (a) and (b). What might be the advantage of reporting the value in mM?
Other common units of concentration contain prefixes to represent relatively small amounts of analyte (milli, m, 10-3; centi, c, 10-2; and micro, μ, 10-6); for example, 3 X 10-3 M = 3 mM. When a very small amount of analyte is present in a solution, it is convenient to represent the analyte concentration as the mass of analyte to the volume of solution (analyte mass:solution volume) in units such as parts per million (ppm, 1ppm=1mg/1L) or parts per billion (ppb, 2ppb=2mg/1L). For example, in the reference: J.A. Conquer, G. Maiani, E. Azzini, A. Raguzzini, and B.J. Holub, Supplementation with quercetin markedly increases plasma quercetin concentration without effect on selected risk factors for heart disease in healthy subjects, J. Nutr. March 1, 1998 vol. 128 no. 3, 593-597, a supplement tablet contains 250 mg quercetin along with other components.
Q3. If the supplement tablet is dissolved in 10 mL of solution (remember, 10 mL of water weighs 10 grams), what will be the concentration of quercetin in ppm?
Analytes themselves cannot be measured directly; however, specific properties of the analyte can. Many analysis techniques provide a response to a solution containing some concentration of an analyte. Two common ways to get a response for an analyte is to measured light absorbance or an electrochemical property. In some cases, a particular analyte of interest may only have one useful property available for measurement. However, some analytes have more than one property that can be measured. For example, the quercetin flavonoid molecule absorbs light and is electrochemically active. Hence, one might measure the light absorption properties or the electrochemical properties of quercetin in order to determine its concentration in a sample.
However, the absorption or electrochemical response of a sample of quercetin can rarely be used alone to determine the concentration in the sample. For example, the absorption or electrochemical measurement might be slightly different from one day to the next for the exact same sample due to a variety of uncontrollable variables including background noise from the instrument.
Q4. If an instrument response to the same concentration of an analyte varies from day-to-day, can you devise a procedure to determine the concentration of an analyte in a solution despite this issue?
Finding suitable standard solutions
A primary standard analyte solution is a solution that contains a known amount of the analyte called the standard. The standard may be a pure analyte, a solution containing the analyte, or a solution containing the analyte along with other solutes; in either case, the analyte concentration is accurately known. The solution may be a solid, liquid, or gas phase solution and the standard analyte may exist in either of these three phases as well. Standards are referred to as Standard Reference Materials (SRM) because they have been tested by The National Institute of Standards and Technology (NIST). This agency ensures the analyte concentration in many types of samples is accurate. The picture below is a SRM for Gulf of Mexico Crude Oil, which may be used to test for crude oil contaminants, for example to monitor the safety of the gulf waters after the BP oil spill.
Q5. What are common characteristics of a primary standard such as the one shown above?
Q6. Do you think that standards are available for all analytes? Why or why not? Elaborate.
Preparation of standard solutions
A primary standard solution is usually made into a stock solution (a solution that contains a relatively high amount of analyte). Then, the stock solution is diluted (mixed with solvent) to give a series of lower concentrations of the analyte; this is most commonly done using the serial dilution method as shown below for the dilution of quercetin with methanol in which an aliquot of the stock solution is transferred into a new volumetric flask, diluted and mixed well to produce diluted solution 1. Then, the same aliquot amount from diluted solution 1 is transferred into a new volumetric flask, diluted to the mark, and mixed well to prepare diluted solution 2. The process is repeated in this manner until a series of solutions have been prepared in which each successive solution is more dilute than the previous. The figure below illustrates preparation of three dilutions starting with a solid primary standard; It is common practice to dilute each successive solution by half of the previous solution concentration. The aliquot removed from the final solution is discarded to waste.
Calculations are performed using the Dilution Equation:
$\mathrm{C_1V_1 = C_2V_2}$
Where C is concentration, V is solution volume, and the subscripts 1 and 2 refer to the initial solution and the more dilute final solution, respectively.
Beer’s Law
Spectrophotometric measurements are one of the most frequently used methods in analytical laboratories; therefore, the Beer-Lambert Law, commonly referred to simply as Beer’s Law, is one of the most frequently utilized equations. This law shows the relationship between solution absorbance (A) and concentration of absorbing species (C) at a specific wavelength. The equation is given below:
$\mathrm{A = εbC}$
Where A = absorbance, ε = molar absorptivity, and C = concentration of absorbing species
In many cases, there is a proportional or linear relationship between A and C; therefore, when the concentration of the absorbing species increases, the absorbance of the solution increases and when the concentration of the absorbing species decreases, the absorbance decreases.
For example, if a solution of an analyte at 0.12 M produces an absorbance signal (corrected for blank response) of 0.50, the expected approximate absorbance of a 0.24M and 0.06M solution of the analyte would give an absorbance reading of approximately 1.0 and 0.25, respectively, assuming a linear relationship between the absorbance and concentration.
Quercetin is an antioxidant molecule that appears yellow in solution. Below are standard quercetin solutions in methanol which range in concentration from 1000 mM to 9.16X10-3 mM, along with a solution blank.
Q7. Why do you think the stock solution has a stronger yellow color than the 9.374 mM quercetin solution?
Q8. To the human eye, the 9.16X10-3 mM quercetin standard solution depicted above is indistinguishable from the blank solution. But in the UV-VIS absorbance spectrophotometer, the absorbance measured for this solution is significantly larger than that measured from the blank. Explain why this might be, and reference Beer’s law in your answer.
Q9. Refer to Bener et al., 2010 section 2.4. CUPRAC assay of total antioxidant capacity. The authors use a reagent blank solution in their measurements. What do you think the blank solution is composed of and why do you think there is a need for a blank solution?
Q10. Refer to reference Bener et. al., 2010. Use the phenolic acid stock solution concentration and the dilution equation to calculate the volume of the stock solution required to prepare a 1 mL solution containing .25 mM quercetin.
Q11. For the series of quercetin standard solutions below, plot the concentration values on the x-axis and the absorbance values on the y-axis using excel.
If you need a refresher on plotting graphs using Excel, visit the following website:
http://www.ncsu.edu/labwrite/res/gt/gt-reg-home.html
Quercetin standard concentration (mM)
Absorbance at 270nm
500
0.81
375
0.62
250
0.40
125
0.19
The resulting plot is called a calibration curve, and the method used to obtain it is called the external calibration method.
Linear Regression and the Method of Least Squares
Regression lines are a visual way to depict the relationship between independent (x) and dependent (y) variables in a graph like the one generated in question 10 . For linear regression, the equation that best represents the relationship between x and y is called the linear regression equation; it is the equation that is a best fit to a straight line, y = mx + b, where y values represent the detector responses to some physical property of the analyte at different concentrations, m is the slope of the line, x values represent the different concentrations of the standard analyte, and b is the y intercept. The linear regression equation is commonly referred to as the calibration equation. Linear regression models are often fitted using the Least Squares method. This method is used to draw the best straight line through experimental data points that have some scatter and do not lie perfectly on a straight line; some points will lie above the line and some points will lie below the line. This procedure assumes that the error in the y values are greater than the error in the x values and that the standard deviation in the y values are similar. Because the goal is to minimize the magnitude of the deviations regardless of the signs, all deviations are squared so that all numbers are positive; hence, the method is called “Method of Least Squares”. How well the resulting regression equation describes the data is expressed as a correlation coefficient, R2 (R-squared). The closer R2 is to 1.00, the better the fit because the data will more closely fit the equation for a straight line in which R2 is 1.00 .
Q12. Determine the regression equation and R2 value for the graph in Q10 .
Visit the website below for a refresher on how to determine this information.
http://www.ncsu.edu/labwrite/res/gt/gt-reg-home.html
Q13. Use the equation information generated in Q11 to calculate the concentration of quercetin that would yield an absorbance of 0.30
The general steps involved in using standard analyte concentration and detector response data to calculate an unknown analyte concentration using linear regression and the method of least squares is as follows:
1. Create a scatter plot of the data
2. Create a linear regression line (trendline)
3. Obtain a regression equation and R-squared value
4. Rearrange the regression equation to calculate the unknown analyte concentration (software can automatically perform this calculation)
Q14. Referring to the calibration curve you determined for quercetin in Q10, what could be done to obtain the concentration of quercetin in a solution that has an absorbance reading beyond the linear range of the calibration curve (for example, when a sample absorbance value is 3)?
Accounting for changes in extraction efficiency
Suppose you wanted to measure the quercetin concentration in a plant food such as Prunus serotina. Before performing the measurement, you would first need to remove quercetin from the plant sample because the plant itself will not be compatible with the measurement technique. For example, most measurement techniques require the sample in liquid form. A likely procedure for removing the quercetin from the plant is to use an extraction process. A sample of the plant might be mixed with a suitable solvent in a blender, homogenized and filtered. One concern in this process is whether all of the quercetin has been extracted from the plant. If a lesser amount is extracted, the concentration of quercetin in Prunus serotina will be underestimated using an external standard curve.
Q15.Can you think of a way to determine the extraction efficiency of an analyte such as quercetin?
Internal Standard Method
An internal standard (I.S.) in analytical chemistry is a substance that is similar to the analyte that is added in a constant amount to the blank, the standards, and the samples. Internal standards should be primary standards just as external standards should. They are useful to compensate for changes in extraction efficiency, detector response due to sample loss during other sample preparation steps, fluctuations in sample analyzed, or changes in detector response due to different flow rates. Internal standards are widely used in chromatography because of differences in the reproducibility of sample injected into the chromatograph. All of these changes should affect the internal standard to the same degree as the analyte so that the ratio of the standard to analyte remains constant.
Refer to the following reference to answer Q16: J. Zhang, M.B. Satterfield, J. S. Brodbelt, S.J. Britz, B. Clevidence, J.A. Novotny, Structural characterization and detection of kale flavonoids by electrospray ionization mass spectrometry, Anal. Chem., 2003, vol.75, 6401-6407.
Q16. What internal standard did the author’s use in the paper? Based on the internal standard used, identify the characteristics of an appropriate I.S. for quercetin?
In a high performance liquid chromatography experiment, a standard solution containing 0.0750 M of quercetin, represented by X, and 0.0600M of internal standard kaempferol, represented by S, gave peak areas of Ax=300 and As=200 (areas in arbitrary units), respectively. The response factor or response ratio (F) can be calculated using the following equation and rearranging the equation to solve for F:
Standard Mixture:
$\dfrac{Ax}{[X]}=F\left(\dfrac{As}{[S]} \right )$
Where [X] and [S] are the concentrations of X and S in moles per liter.
The concentration of quercetin [X] in a sample extract can then be determined by spiking the extract with a known amount of the internal standard [S] kaempferol, measuring both the quercetin and kaempferol peak areas, and using the equation above and the F value obtained from the standards to calculate the concentration of quercetin [X] in the extract.
Q17. Consider the data provided above: a standard solution containing 75 ppm of quercetin and 60 ppm of internal standard kaempferol gave peak areas of 300 and 200, respectively. A plant sample is spiked such that the extract to be analyzed should have 60 ppm of kaempferol. Analysis of the sample gives a peak area for the kaempferol of 163. The quercetin peak in the same extract has an area of 407. What is the concentration of quercetin in the extract?
As shown in the chromatogram below, the area under each peak is proportional to the concentration of the species injected into the column. However, the detector generally has a different response to each component. For example, if both the quercetin (X) and kaempferol (S) have concentrations of 10.0 mM, the area under the analyte peak (AX) might be 2.30 times greater than the area under the standard peak (AS). We say that the detector response ratio or response factor, F, is 2.30 times greater for X than for S.
Q18. Refer to reference: M. Olszewska, Quantitative HPLC analysis of flavonoids and chlorogenic acid in the leaves and inflorescences of Prunus serotina EHRH, Acta Chromatographica no. 19 2007, 253-269. (www.us.edu.pl/uniwersytet/jednostki/wydzialy/chemia/acta/ac19/zrodla/23_AC19.pdf). What extraction efficiency (recovery) do the authors report for quercetin in the leaves of the cherry tree (Prunus serotina) sample? How do the authors determine the percent quercetin that is extracted from this sample?
Accounting for matrix effects
Another common problem that can occur in chemical analysis is known as a matrix effect. The matrix is everything else in the sample except the analyte. Components of the matrix may interact with the analyte and alter its response in the measurement technique. Matrix effects can sometimes enhance the response, whereas other times they can decrease the response. For example, interactions of metals such as aluminum in the matrix with quercetin may cause variations in the absorbance at the wavelength used for the measurement. Since the components that make up the matrix are often complex, not completely known, and possibly vary from sample to sample, an internal standard may not exhibit an identical matrix effect to that of the analyte.
Q19. Can you devise a procedure that would allow you to account for the matrix effect on the absorbance measurement of quercetin?
Standard Addition Method
The standard addition method is similar to the external calibration method in that the concentration of an analyte is determined by comparison to a set of standard solutions of the analyte. However, in the standard addition method, increasing quantities of the standard are added to samples to correct for ‘matrix effects’ (a change in the analytical signal caused by anything in the sample other than the analyte). This is called “spiking.” The concentration of the analyte is obtained by extrapolating back to the x-intercept; the absolute value of the x-intercept is the concentration of the unknown as shown below:
Q20. Refer to reference Berner et. al., 2010. Determine the concentration of quercetin in Rubia tinctorum L. extract from figure 2.
Q21. In Q13, the sample had an absorbance value of 0.30. The same sample is now subjected to a standard addition experiment. The table below shows the concentrations of quercetin spiked into the sample and the resulting absorbances. Plot the calibration curve and determine the amount of quercetin in the original (unspiked) sample. How does the concentration of quercetin in the sample compare with the value in Q13? Did the matrix have an effect on the measurement? If so, what effect did it have?
Quercetin Concentration (mM) Detector Signal
0 0.300
50 0.367
100 0.444
150 0.512
200 0.577
REFERENCES
Module questions:
M. Olszewska, Quantitative HPLC analysis of flavonoids and chlorogenic acid in the leaves and inflorescences of Prunus serotina EHRH, Acta Chromatographica no. 19 2007, 253-269. (www.us.edu.pl/uniwersytet/jednostki/wydzialy/chemia/acta/ac19/zrodla/23_AC19.pdf)
M. Bener, M. Ozyurek, K. Guclu, R. Apak, Polyphenolic contents of natural dyes produced from industrial plants assayed by HPLC and novel spectrophotometric methods, Industrial Crops and Products, 2010, vol. 32, 499-506.
J. Zhang, M.B. Satterfield, J. S. Brodbelt, S.J. Britz, B. Clevidence, J.A. Novotny, Structural characterization and detection of kale flavonoids by electrospray ionization mass spectrometry, Anal. Chem., 2003, vol.75, 6401-6407.
J.A. Conquer, G. Maiani, E. Azzini, A. Raguzzini, and B.J. Holub, Supplementation with quercetin markedly increases plasma quercetin concentration without effect on selected risk factors for heart disease in healthy subjects, J. Nutr. March 1, 1998 vol. 128 no. 3, 593-597.
Background Reading:
A.J. Larson, J.D. Symons, T. Jalili, Therapeutic potential of quercetin to decrease blood pressure:review of efficacy and mechanisms, Adv. Nutr., 2012, vol. 3, 39-46.
F. Perez-Vizcaino, J. Duarte, R. Andriantsitohaina, Endothelial function and cardiovascular disease: Effects of quercetin and wine polyphenols, Free Radical Research, October 2006, vol. 40 no. 10, 1054-1065.
Additional Resources:
High Pressure Liquid Chromatographic Determination of Nitrendipine in Human Plasma After Solid Phase Extraction, J. Liq. Chromatogr. Rel. Technol. 1999, 22 (9), 1381 – 1390.
Quantitative Determination of Deferiprone in Human Plasma by Reverse Phase High Performance Liquid Chromatography and its Application to Pharmacokinetic Study, Pak. J. Pharm. Sci. 2012, 25 (2), 343 – 348.
Determination of Biochemical Oxygen Demand of Area Waters: A Bioassay Procedure for Environmental Monitoring, J. Chem. Educ. 2012, 89, 807 – 811.
Simultaneous Determination of Methylxanthines in Coffees and Teas by UV-Vis Spectrophotometry and Partial Least Squares, Analytica Chimica Acta. 2003, 493 (1), 83 – 94.
Lab Exercises for Internal Standard Method:
Determination of the Antibiotic Oxytetracycline in Commercial Milk by Solid-Phase Extraction: A High-Performance Liquid Chromatography (HPLC) Experiment for Quantitative Instrumental Analysis, J. Chem. Educ. 2012, 89, 656 – 659.
Lab Exercise for External Calibration and Standard Addition Methods:
Undergraduate Analytical Chemistry: Method Development and Results of the Analysis of Bismuth in Pharmaceuticals, Spectr. Lett. 2010, 43, 545 – 549.
Francisco Perez-Vizcaino, Juan Duarte, and Ramaroson Andriantsitohaina, Endothelial function and cardiovascular disease: Effects of quercetin and wine polyphenols, Free Radical Research, October 2006; 40(10): 1054-1065.
The authors thank Dr. Tom Wenzel, Bates College, Dr. Sheila A. Sanders, New York University, and the ASDL group members for valuable feedback on this module. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Concentration_Calibration/01_In-Class_Modular_Unit.txt |
Using Flavonoid Content as a way of Predicting Health Benefits of Plant-based Food
Developed by Sandra Barnes
Instructions: Read the following information and answer the questions at the end of the module.
Basic information about flavonoids
• Flavonoids are polyphenolic secondary plant metabolites.They serve a variety of functions in plants, such as protection from UV radiation, signals for pollinators, selective admittance of blue-green and red light for photosynthesis, and plant defense.Thousands of individual flavonoid compounds are known, but the core structure is based on three cyclic rings A, B, and C, as shown below.The A ring is synthesized from the malonate pathway and the B ring and majority of the C ring are synthesized from the Shikimic acid pathway via phenylalanine.
• Flavonoids can be classified as flavonols, flavones, isoflavones, flavanals, procyanidins, tannins, and anthocyanins based mainly on the number andplacement of OH and OCH3 groups to the core ring structure. The flavonol compound quercetin is the most commonly found and widespread flavonoid in plant-based food; therefore, it is commonly used as a standard compound when analyzing and comparing flavonoid content in plants.
Structure of quercetin indicating the A, B, and C rings
Health benefits of plant flavonoids
Chronic diseases such as cancer and cardiovascular disease are major health issues and are leading causes of death and economic strain for people of the United States and the world. Food scientists and nutritionists have taken great interest in flavonoids because consumption of these compounds have been found to infer long term health benefits to humans. Most flavonoids are reducing agents that act as antioxidants and free-radical scavengers due to their ability to be easily oxidized yet remain stable, while preventing oxidation of biomolecules in the human body. The antioxidant properties of flavonoids protect the human body against oxidative reactions which may damage proteins, DNA, cell membranes and other biocomponents, which may lead to chronic diseases. Flavonoids cannot be synthesized by humans. Therefore, they must be obtained from food sources. Plants offer a wide variety of flavonoids, which are mostly concentrated in the outer parts of plants such as flowers, leaves, fruit peels, seed coats, and bark. The table below shows a representative list of food groups which contain different classes of flavonoids.
FLAVONOID
FOOD GROUP
Flavonols
Vegetables, fruits, herbs, wine, tea
Flavones
Vegetables, herbs, berries
Isoflavones
Legumes (nuts and cereals)
Flavanones
Vegetables, fruits, berries, citrus fruits
Flavanols
Fruits, berries, nuts, cider, cocoa, cereals
Procyanidins
Fruits, berries, nuts, cocoa, wine, cider
Anthocyanins
Colored vegetables, berries, red wine, red onion, red cabbage, red beans
tannins
Nuts and carrot
Information taken from: Laura Bravo and Raquel Mateos, Analysis of Functional Foods and Nutraceuticals, P. 147-206. In Methods of Analysis of Functional Foods and Nutraceuticals, W. Jeffrey Hurst editor, CRC Press second edition, 2008.
General mechanism of action of flavonoids against free-radicals
• Hydroxyl (OH) groups attached to benzene rings become oxidized and donate a hydrogen (H) to the free radical molecule to stabilize the free radical and prevent oxidative damage to biomolecules and cells.The flavonoid compound is stabilized due to resonance hybridization of the benzene ring.
Figure adapted from Pieta P. J. Nat. Prod., 2000, 63, 1035-1042.
Flavonoid Analysis by High Performance Liquid Chromatography (HPLC)
HPLC with UV-Vis detection is the most commonly used method for quantifying individual flavonoid compounds in food*. Flavonoid compounds are first separated using HPLC and then detected spectroscopically. Different HPLC separation methods are required for analysis of compounds in different flavonoid classes.
The HPLC chromatogram shown above is of collard (Brassica oleracea) and corn (Zea mays) samples separated and analyzed by Barnes S. L. et al. 2011 (unpublished).
*Hertog, M.G.L., Hollman, P.C.H., and Venema, D.P., Optimization of a quantitative HPLC determination of potentially anticarcinogenic flavonoids in vegetables and fruits, J. Agric. Food Chem., 40, 1591, 1992.
Flavonoid Analysis by Aluminum Chloride Method
The Aluminum Chloride Method is a method commonly used to quickly determine the total flavonoid content of food. Flavonoid molecules react with Al3+ to form a complex which can be detected spectroscopically at 367nm.
Complexation of the flavonoid quercetin with aluminum
Structure taken from: Jiang Liuyun, Liu Yuming, Preparation, biological activity and quantum chemistry calculation of quercetin-Aluminum complex, Chemical Journal on Internet. Dec. 1, 2004, vol. 6 no. 12 P. 87.
Module Questions
• Your employer, Veggy ways, is seeking to produce the healthiest vegetables possible. Therefore, the compnay has brought one hundred food samples to your lab and as the food chemist, your job is to determine which foods have the greatest health benefits by measuring the flavonoid content of the samples. Discuss the advantages and disadvantages of using HPLC and Aluminum Chloride method for these tests.
• Based on the introductory information provided in this module, which one of the following plant-based foods would you predict to contain the highest amount of total flavonoids: Corn (Zea mays), squash (Cucurbita spp.), or collard greens (Brassica oleracea)?
• Based on the aluminum chloride data provided for each food sample below, prepare a concentration calibration curve, determine the concentration calibration equation, R2 value, calculate the concentration of total flavonoid for each food, and determine which food contains the highest flavonoid content.
Corn Squash Collard
Quercetin Calibration Standard (µg/mL)
Standard Solution Absorbance @ 367nm (blank corrected)
Food Sample
*Flavonoid Analyte Solution Absorbance @ 367nm (not blank corrected)
12.63
0.2001
Squash
0.166
31.25
0.4198
Corn
0.312
62.50
0.8244
Collard
1.226
125
1.6622
*Analyte solutions were diluted by ½ to obtain readings within the range of the calibration standards. The blank reading was 0.001 | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Concentration_Calibration/02_Out-of-Class_Modular_Unit.txt |
A series of collaborative learning activities and accompanying text that develop fundamental aspects of electrochemistry and electrochemical methods of analysis. These activities are intended to be done in class by students working in groups, but can be modified for use as out-of-class exercises. Learning objectives and an instructor’s manual are provided as well. The instructor’s manual provides tips for how to use the in-class exercises, the types of responses that students often provide to the questions in the exercises, and how the instructor can build from these responses to develop the concepts. Analytical methods developed in this unit include ion-selective electrodes, electrodeposition, coulometry, electrochemical titrations, and voltammetric methods including anodic stripping voltammetry, linear sweep voltammetry, differential pulse linear sweep voltammetry, and cyclic voltammetry.
Contributors and Attributions
• Dr. Thomas Wenzel (Bates College)
Electrochemical Methods of Analysis
In-class Questions – Electrochemistry
1. Define what is meant by oxidation and reduction.
2. Define what is meant by an oxidizing and reducing agent. Give a good example of each.
3. Define what is meant by a half-reaction.
4. Give an example of a half-reaction and determine whether a half-reaction can be an equilibrium expression. Why or why not?
Chemical Energy
Now let's think about chemical energy (otherwise known as the Gibbs energy and denoted by G). A bottle of pure A has some amount of chemical energy. A bottle of pure B has some amount of chemical energy. A solution of A in water that is 2 Molar has some amount of chemical energy as does a solution of B in water that is 2 Molar.
1. Do you think that those four examples have the same or different chemical energy?
2. How would you measure or determine the absolute chemical energy of those four systems?
3. $\mathrm{A \Leftrightarrow B}$
When will the chemical energy of that system achieve its lowest value?
4. $\mathrm{[A] + [B] = 2\: Molar}$
is always satisfied, and the reaction of A to produce B has a fairly large equilibrium constant.
5. What would the plot look like if the reaction had a fairly small equilibrium constant?
6. Why do we need to define something called the standard state?
7. What does it mean for the difference in chemical energy (DG) to be positive or negative?
Electrochemical Cells
1. Describe what you know about an electrochemical cell.
2. What processes are responsible for conduction of electricity in an electrochemical cell?
3. What is the purpose of a salt bridge? What would you put inside a salt bridge?
4. Describe two types of situations that would result in the irreversibility of an electrochemical process.
Use of the Nernst Equation
Potassium dichromate reacts with Fe(II) to produce Cr(III) and Fe(III).
1. What is the standard state potential and K for this reaction?
2. What is the cell potential if one half-cell is made up with 1.50 M potassium dichromate and 0.30 M chromium(III)nitrate hexahydrate in 1.00 M nitric acid and the other half cell is made up with 0.050 M iron(III)chloride hexahydrate and 0.10 M iron(II)chloride tetrahydrate?
3. What is the cell potential if the chromium half-cell were operated at a pH of 7 instead of using 1 M nitric acid?
Electrodeposition/Electrogravimetry
Will the presence of Fe(II) at 0.0800 M interfere with the electroplating of 99.9% of the cadmium(II) in a solution in which the cadmium is expected to be present at a concentration of no less than 0.0500 M? Calculate potential values relative to a standard hydrogen electrode.
Suppose the solution had 0.0800 M Cr(III) as a possible interference. Is it possible to plate out 99.9% of the cadmium(II) without any interference from the chromium.
Coulometry
1. Draw the plot you would obtain for current (y-axis) versus time (x-axis) if you applied a constant potential high enough to carry the reduction of Cd(II) to cadmium metal. The system has an electrode with a large surface area.
2. How would you relate the outcome of your plot above to the concentration of Cd(II) in the sample?
3. What advantages does coulometry have over electrogravimetry?
Coulometric Titration
What are some advantages of using a coulometric titration?
Amperometric Titration
Draw the plot that would be obtained in an amperometric titration if each of the following occurred (assume that the system is set up with a reducing potential):
1. Only the analyte undergoes a reduction at the applied potential.
2. Only the titrant undergoes a reduction at the applied potential.
3. Both the analyte and titrant undergo a reduction at the applied potential.
Potentiometric Titration
Consider a solution of iron(II) that is to be titrated with cerium(IV). The products of the reaction are iron(III) and cerium(III). The course of the titration will be monitored potentiometrically.
$\mathrm{Fe(II) + Ce(IV) = Fe(III) + Ce(III)}$
If 20 ml of an 0.10 M solution of Fe(II) is to be titrated with 0.10 M Ce(IV), calculate the junction potential that would be measured at a platinum electrode when 5 ml, 10 ml, 20 ml, and 30 ml of titrant have been added.
The following steps will allow you to work through the calculation of the junction potentials.
1. Write an expression for the total electrochemical potential of this system (this is the Nernst equation for the complete reaction).
2. What is the total electrochemical potential of the titration reaction at any point during the titration? Think carefully about this because your first intuition may not be correct.
3. Which half reaction is easier to use to calculate the junction potential before the equivalence point? Which half reaction is easier to use after the equivalence point?
Voltammetric Methods
1. How can electrostatic migration be eliminated in an electrochemical cell?
2. Samples subjected to voltammetric methods are usually purged with nitrogen or another inert gas before the analysis and maintained under an inert atmosphere during the analysis. Why is this done?
Anodic Stripping Voltammetry
1. What feature of the plot can be related to the concentration of the metal?
2. What advantage(s) does anodic stripping voltammetry offer over coulometry or electroplating?
Linear Sweep Voltammetry
1. What feature of the plot could be related to concentration?
2. What feature of the plot can be used for species identification?
3. For the exact same solution of Cd(II) and Zn(II), draw the current that would be observed if the solution is not stirred.
4. What advantage(s) does the method in question 5 have over regular linear sweep measurements?
Cyclic Voltammetry
1. Draw a plot that could be obtained for an electrochemical reaction that is chemically irreversible and forms an electrochemically inactive product.
2. Draw the plot of current vs. potential that would be obtained for a reversible chemical reaction in which only the reverse reaction has an overpotential, but the potential eventually is sufficient to complete the reverse reaction.
3. Propose a reaction mechanism that would explain the following cyclic voltammogram. The first voltage cycle is shown as a solid line. The second voltage cycle is shown as a dotted line. A third voltage cycle gives the same output as the dotted line. Assume that a reducing potential is applied. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Electrochemical_Methods_of_Analysis/01_In-class_Problems.txt |
The following textual material is designed to accompany a series of in-class problem sets that develop many of the fundamental aspects of electrochemical analytical methods.
02 Text
Electrochemical processes are commonly used for analytical measurements. There are a variety of electrochemical methods with different degrees of utility for quantitative and qualitative analysis that are included in this unit. The coverage herein is not exhaustive and methods that are most important or demonstrate different aspects of electrochemical measurements are included. Also, in most cases the coverage is designed to provide a broad overview of how the method works and not delve deeply into all of the associated subtleties. There are two free sources of information available through the Analytical Sciences Digital Library for those desiring a more in-depth coverage of particular methods. One is a general textbook on analytical chemistry written by David Harvey.
community.asdlib.org/activele...line-textbook/
The other is a module on electrochemistry written by Richard Kelly.
community.asdlib.org/activele...asic-concepts/
The methods we will explore in this unit rely on one of two different electrochemical phenomena. The first is that many chemical species have the ability to transfer electrons through an oxidation-reduction process. With appropriate design of an electrochemical system, this transfer of electrons can be measured as a current. Since we also know that different species have different oxidation or reduction abilities, electrochemical measurements relying on electron transfer can often be used for the purpose of species identification.
The second method of using electrochemical processes for measurement purposes relies on the measurement of a potential. In particular, we will focus on some methods that rely on something called a junction potential. You likely have some familiarity with electrochemical cells. Such a device consists of electrodes and the design of electrodes creates interfaces or junctions (e.g., a metal electrode in contact with a solution represents a junction). Any junction in an electrochemical system will have a potential associated with it and in certain cases, the magnitude of this junction potential can be related to the concentration of a species in solution. For example, a pH electrode is the best known example of the use of a junction potential for determining the concentration of a species. The key feature of a pH electrode is a thin glass membrane. When placed into an aqueous solution, a junction potential occurs at the glass-solution interface and the magnitude of this potential is determined by the concentration of H+ in solution.
Define what is meant by oxidation and reduction.
In a chemical reaction involving a transfer of electrons, one species gains one or more electrons while another species loses one or more electrons. Oxidation refers to the species that loses electrons and reduction to the species that gains electrons. If you have trouble remembering which is which, using the pneumonic “LEO the lion goes GER” can be helpful (LEO = Loss of Electrons is an Oxidation; GER = Gain of Electrons is a Reduction). It is important to remember that both processes must occur simultaneously.
Define what is meant by an oxidizing and reducing agent. Give a good example of each.
An oxidizing reagent promotes the oxidation of another substance so is reduced in the overall electrochemical reaction. Good or strong oxidizing agents are species that really want to be reduced. Fluorine and chlorine are strong oxidizing agents because they very much want to be the fluoride or chloride ion.
A reducing agent promotes the reduction of another substance so is oxidized in the overall electrochemical reaction. Good or strong reducing agents are species that really want to be oxidized. Since alkali metals such as lithium, sodium or potassium want to be oxidized into their cationic forms, they would be good reducing agents.
Define what is meant by a half-reaction.
Overall electrochemical reactions consist of both a reduction and oxidation. Each half of this overall process is represented by an appropriate half reaction. A half-reaction shows the reduced and oxidized form of the species and these two forms are referred to as a redox couple.
Give an example of a half-reaction and determine whether a half-reaction can be an equilibrium expression. Why or why not?
One of many possible half reactions is shown below for the reduction of cadmium ion to cadmium metal.
$\mathrm{Cd^{2+}(aq) + 2e^- = Cd(s)}$
If we were to try to write an equilibrium constant expression for this reaction, it would need to have the concentration of free electrons in the expression. Since we really cannot weigh out a mass of electrons to use in a reaction and cannot dissolve free electrons into a solution, we cannot write a true equilibrium constant expression for a half reaction.
However, an interesting aspect of electrochemical reactions is that we can design a device known as an electrochemical cell that has each of the half reactions isolated from each other in separate halves of the cell. As we begin to examine electrochemical cells and processes in more detail, we will often focus our attention on only one half of the overall process and will write expressions for half reactions that are essentially an equilibrium constant expression. The expression will not have a term for electrons in it. For the half reaction shown above the expression would be as follows:
$\mathrm{\dfrac{1}{[Cd^{2+}(aq)]}}$
Just like in equilibrium constant expressions, there is no term for the Cd(s) because a solid will not have a concentration. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Electrochemical_Methods_of_Analysis/02_Text/1._Basic_Concepts_in_Electrochemistry.txt |
Before examining something known as the electrochemical potential (E), it is useful to explore some aspects of what is known as the chemical energy that you have likely learned about before in general or physical chemistry. The chemical energy of a system is known as the Gibbs energy and is denoted by G.
Suppose we wanted to think about the simplest chemical reaction possible – a situation in which A reacts to produce B.
$\mathrm{A \Leftrightarrow B}$
One example of an actual chemical reaction of A reacting to produce B would be something known as a keto-enol tautomerism.
A bottle of pure A has some amount of chemical energy. A bottle of pure B has some amount of chemical energy. A solution of A in water that is 2 Molar has some amount of chemical energy. A solution of B in water that is 2 Molar has some amount of chemical energy.
Do all four examples listed above (pure A, pure B, [A] = 2 M, B = [2 M]) have the same or different chemical energy?
Hopefully it will seem reasonable or intuitive to think that all four of these systems will have different chemical energies. Since the two compounds are different from each other, in their pure forms they likely have different chemical energies. Similarly, a chemical species dissolved in a solvent is now different than the pure compound and the two will have different chemical energies.
How would you measure or determine the absolute chemical energy of those four systems?
This is actually a trick question. We do not have methods available to determine the absolute chemical energy for a system. In reality the best that we can do is measure the difference in chemical energies between two systems. This difference is denoted as ΔG, which is likely something you are familiar with from prior classes you have taken.
Let’s continue to examine the situation in which A reacts to produce B.
$\mathrm{A \Leftrightarrow B}$
When will the chemical energy (G) of this system achieve its lowest value?
The lowest value of chemical energy for a reaction occurs when it is at equilibrium. For the example above, this means that pure A and pure B must have a higher chemical energy than the equilibrium state of the reaction. It is important to note that chemical systems strive to have the lowest chemical energy as that will be their most stable state.
We will next consider what a plot of G versus the mole fraction of A and B in which the situation
$\mathrm{[A] + [B] = 2\: Molar}$
is always satisfied. We can consider what the plot would look like for two different situations:
The reaction of A to produce B has a fairly large equilibrium constant.
The reaction of A to produce B has a fairly small equilibrium constant.
Remember that the equilibrium constant expression is:
$\mathrm{K= \dfrac{[B]}{[A]}}$
Since the equilibrium situation will have some concentration of both A and B, the system at equilibrium must have a lower chemical energy than either a 2 M solution of A or a 2M solution of B. That would allow us to draw two possible representations for the situation in which K is large (there is a much higher concentration of B than A) and K is small (there is a much higher concentration of A than B) as seen in Figure 1.
Figure 1. Plot of G versus molar quantities of reactant and product for a reaction with a large (left) and small (right) equilibrium constant (K).
Remember that we do not know how to calculate the value of G. What we can possibly do is measure the difference in chemical energy (ΔG) between two different states (e.g., we could possibly measure the difference in chemical energy between a solution in which [A] = 2 M and equilibrium). The plot in Figure 2 indicates ΔG for such a situation when K is large.
Figure 2. Magnitude of ΔG between [A] = 2 M and equilibrium.
Why do we need to define something called the standard state?
Consider an example where every student in a class was asked to perform a laboratory measurement of ΔG for the reaction of A to produce B. If we consider the example where K is large, you could imagine a situation in which one student starts with a solution where the concentration of A is 2 M and B is 0 while another starts with a solution where the concentration of B is 2 M and A is 0. Looking at the plot in Figure 2, both students may measure accurate and correct values for ΔG but the values that the students report will be different. As seen by the plot in Figure 2, the value one would measure for ΔG is dependent on the particular starting point, since each starting point has a different value of G. In order to record ΔG values that can be tabulated and compared, everyone must agree in advance to a uniform set of starting conditions. These agreed upon starting conditions are known as the standard state.
What conditions constitute the standard state?
The standard state is the situation where every reactant and product in the reaction has an activity of 1. Since there are many conditions where we are unable to determine the exact activity of a substance, an approximation is made where the standard state has the concentration of all soluble species in the reaction at 1 Molar, gases at a pressure of 1 atmosphere and a temperature of 25oC (298 K). For A reacting to produce B, the standard state would be a solution in which [A] = 1 M and [B] = 1 M. The plot in Figure 3 shows the difference in chemical energy between the standard state and equilibrium for the situation in which K is large. The difference in chemical energy between the standard state and equilibrium is given a special designation: ΔGo.
Figure 3. Difference in chemical energy between the standard state and equilibrium (ΔGo) for a reaction with a large equilibrium constant.
You can imagine that it might be difficult to think about how you could actually make a standard state solution and somehow hold it in a suspended state before allowing it to react and achieve equilibrium, thereby allowing a measurement of ΔGo. As we will see further on in this unit, electrochemical cells are actually ideal systems for measurements of ΔGo because it is possible to prepare the two half reactions in separate chambers but prevent the reaction from taking place by not completing all of the circuitry between the two chambers.
What does it mean for the difference in chemical energy (ΔG of ΔGo) to be positive or negative?
In considering this, it is important to examine the two different plots of G for the reaction with a large K and small K. These are shown in Figure 4 and the value of ΔGo is indicated in each plot.
Figure 4. ΔGo for a reaction with a large (left) and small (right) equilibrium constant.
Note that in both cases the chemical energy drops as the system goes from the standard state to equilibrium. However, when we think about carrying out a chemical reaction, our usual goal is to have the reaction form products. For this reason, we adopt a convention that says that ΔGo is less than 0 (negative) for a reaction that favors the formation of products. ΔGo is greater than 0 (positive) for a reaction that favors the formation of reactants.
The following equation is something you should have seen before in a unit on thermodynamics in either a general or physical chemistry course.
$\mathrm{∆G^o= -RT \ln K}$
This equation provides the difference in chemical energy between the standard state and equilibrium. Note that if K is large (favors products), lnK is positive and ΔGo is negative. Similarly if K is small (favors reactants), lnK is negative and ΔGo is positive.
Suppose the Starting Conditions are not at the Standard State
There are many situations where you do not want to examine reactions that begin with standard state conditions but instead want to start with a set of non-standard state conditions. The difference in chemical energy between non-standard state conditions and equilibrium is designated as ΔG. The equation that allows us to calculate ΔG is as follows:
$\mathrm{∆G= -RT\ln K + RT\ln Q\: or\: \Delta G= \Delta G^o + RT\ln Q}$
The term Q is an expression that looks exactly like the equilibrium constant expression except that it is evaluated using the starting non-standard state concentrations. It is important to recognize what each part of this equation is providing.
–RTLnK (denoted as Go) is the difference in chemical energy between the standard state and equilibrium.
RTlnQ is the difference in chemical energy between the non-standard state starting conditions and the standard state.
The plot in Figure 5 shows the placement of these two terms for the example of A reacting to produce B that we have been examining. In this case the non-standard state conditions are further removed from chemical equilibrium than the standard state. Note that with [A] = 1.50 M and [B] = 0.50 M, the value of Q is 0.33.
$\mathrm{Q= \dfrac{[B]}{[A]}= \dfrac{(0.50)}{(1.50)}=0.33}$
With the value of Q below 1, lnQ will be negative and the value of ΔG is more negative than the value of Go.
Figure 5. Representation of RTlnQ and –RTlnK for a reaction starting in non-standard state conditions. Non-standard state conditions are further from equilibrium than the standard state conditions.
The plot in Figure 6 shows starting non-standard state conditions that are closer to equilibrium than those of the standard state: [A] = 0.50 M; [B] = 1.50 M. In this case the value of Q is 3.0.
$\mathrm{Q= \dfrac{[B]}{[A]}= \dfrac{(1.50)}{(0.50)}=3.0}$
With the value of Q above 1, lnQ is positive and the value of G is less negative than the value of Go.
Figure 6. Representation of RTlnQ and –RTlnK for a reaction starting in non-standard state conditions. Non-standard state conditions are closer to equilibrium than the standard state conditions. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Electrochemical_Methods_of_Analysis/02_Text/2._The_Chemical_Energy_of_a_System.txt |
Electrochemical reactions have a similar drive toward the lowest possible energy. Instead of referring to G for electrochemical reactions, we refer to the electrochemical potential (E). Similar to chemical energy, Eo refers to the difference in electrochemical potential between the standard state and equilibrium. E refers to the difference in electrochemical potential between non-standard state conditions and equilibrium. An interesting facet of electrochemical reactions is that with proper design the change in chemical energy can be converted to electrical energy in the form of an electrical current.
The standard state electrochemical potential (Eo) can be related to ΔGo by the following equation:
$\mathrm{ΔG^o = -nFE^o}$
In this equation, n is the number of electrons transferred in the overall balanced electrochemical reaction and F is Faraday’s Constant. Faraday’s Constant relates the total charge in Coulombs (C) of a reaction to the amount of product that forms. For a reaction in which n = 1, there are 96,485 C/mole.
The non-standard state electrochemical potential (E) can be related to ΔG by the following equation:
$\mathrm{ΔG = -nFE}$
There are several important outcomes of the relationship between chemical energy and electrochemical potential. Setting equal and rearranging the two expressions for ΔGo leads to the following:
$\mathrm{\Delta G^o= -RT\ln K\: and\: \Delta G^o= -nFE^o}$
$\mathrm{E^o= \dfrac{(RT)}{(nF)} \ln K}$
Substituting in the standard state temperature, gas constant and Faraday’s constant and converting the natural log to a base ten log leads to the following expression:
$\mathrm{E^o= \dfrac{(0.059)}{(n)} \log K}$
Setting equal and rearranging the two expressions for ΔG leads to the following:
$\mathrm{\Delta G= -RT\ln K + RT\ln Q\: and\: \Delta G= -nFE}$
$\mathrm{E= \dfrac{(RT)}{(nF)} \ln K - \dfrac{(RT)}{(nF)}\ln Q}$
Again, substituting in the standard state temperature, gas constant and Faraday’s constant and converting the natural log to a base ten log leads to the following expression:
$\mathrm{E= \dfrac{(0.059)}{(n)} \log K - \dfrac{(0.059)}{(n)}\log Q}$
Substituting in for the first expression in this provides:
$\mathrm{E= E^0 - \dfrac{(0.059)}{(n)} \log Q}$
This last equation is especially important in electrochemistry and is known as the Nernst Equation. In the Nernst Equation:
$\mathrm{\dfrac{0.059}{n} \log K}$ (denoted as Eo) is the difference in electrochemical potential between the standard state and equilibrium.
$\mathrm{\dfrac{0.059}{n}\log Q}$ is the difference in electrochemical potential between the non-standard state starting conditions and the standard state.
One last thing worth pointing out is the sign convention of electrochemical potentials and how they relate to whether the reaction favors products or reactants. An examination of the Nernst Equation shows that electrochemical reactions that favor products will have positive values of electrochemical potential. Electrochemical reactions that favor reactants will have negative values of electrochemical potential.
4. Table of Standard State Electrochemical Potentials
It is possible to measure the standard state electrochemical potential for individual half reactions. Doing so requires setting one particular half reaction as a reference point to which all other potentials are compared. The half reaction used as the reference involves reduction of the hydrogen ion (H+). This half reaction is arbitrarily assigned a standard state reduction potential of 0.00 Volts.
$\mathrm{2H^+(aq) + 2e^- = H_2(g) \hspace{40px} E^o = 0.00\: V}$
Tables of standard state electrochemical potentials are freely available on the internet. By convention, all of the half reactions are written as reductions. Earlier we mentioned how alkali metals are strong reducing agents as they have a strong driving force to be oxidized. The half reaction and standard state potential for the reduction of Li+ is shown below.
$\mathrm{Li^+(aq) + e^- = Li(s) \hspace{40px} E^o = -3.045\: V}$
Note that this reaction has a very large negative standard state potential. Remember from earlier that electrochemical reactions that favor the reactants have negative potentials. An examination of the standard state potentials indicates that the reduction of Li+ has the highest value in the table. Therefore it should not be surprising that lithium batteries are in common use today. Also, for those readers who are fans of Star Trek, we can now understand when the Enterprise needed more fuel the call from the bridge was for more dilithium crystals.
We also used fluorine as an example of a powerful oxidizing agent meaning that it has a strong driving force to be reduced. The half reaction and standard state potential for the reduction of fluorine gas is shown below.
$\mathrm{F_2(g) + 2e^- = 2F^-(aq) \hspace{40px} E^o = 2.87\: V}$
Note that this reaction has a very large positive standard state potential. Remember that electrochemical reactions that favor the products have positive potentials.
Since an overall electrochemical reaction has a reducing and oxidizing half to it, we often work with systems in which two half reactions are paired up. When considering any pair, the one with the more positive Eo value proceeds as a reduction and the one with the less positive value proceeds as an oxidation.
It is worth noting that reactions are usually run under non-standard state conditions. It is possible to take the conditions so far away from the standard state used to generate the values in a table of Eo values that the reaction may actually proceed in the reverse direction from what occurs in the standard state. For example, looking back at the plot of G for a reaction with a large K shown in Figure 1, starting at a very high concentration of B and a very low concentration of A that is to the right of the equilibrium state in the plot would mean that the reaction proceeds toward reactants instead of products. This would be the reverse of the reaction direction predicted by comparing the ΔGo or Eo values of A and B.
One final thing to note is that Eo values do vary with other conditions of the solution. For example, electrochemical reactions with H+ in one of the half reactions are highly influenced by the pH. The standard state will have [H+] = 1 M (note, this constitutes a pH of 0) and the measured Eo value will often have slightly different values if different acids (e.g., nitric, hydrochloric, perchloric) are used to make the 1 M solution. Another common condition with electrochemical reactions involves the ionic strength of the solution. The ionic strength (μ) is defined as follows:
$\mathrm{\mu = \dfrac{1}{2} \sum [C_i]Z_i^2}$
Where Ci is the concentration of each ion and Zi is the charge of each ion. Note that both cations and anions are included in the summation term.
In some cases with electrochemical reactions it is desirable to have a concentration of unreactive ions in the solution as a background electrolyte (e.g., alkali cations paired with halide anions might be such a background electrolyte). Measured Eo values often vary slightly at different ionic strengths.
In some cases it is more common to use formal potentials (Eo’). A formal potential is the reduction potential that applies to a half reaction under a specified set of conditions (e.g., pH, ionic strength, concentration of complexing agents). One common example is that the formal potential of important biological electrochemical reactions are often measured at pH 7, which is much closer to physiological pH than the standard state pH of 0.
Because Eo values vary slightly with conditions, calculated values for a system you wanted to study obtained using the Nernst equation are often only close approximations of what you would actually obtain as a measured value. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Electrochemical_Methods_of_Analysis/02_Text/3._Relationship_of_Chemical_Energy_to_Electrochemical_Pot.txt |
Before developing analytical methods based on electrochemistry, it is worth exploring aspects about electrochemical cells. Concepts needed to comprehend the nature of an electrochemical cell are informative in understanding some of the analytical methods we will develop. From a more practical standpoint, batteries are examples of electrochemical cells.
Describe what you know about an electrochemical cell.
The components of an electrochemical cell are shown in Figure 7.
Figure 7. Diagram of the components in an electrochemical cell.
The particular cell shown involves a half reaction with zinc and a half reaction with copper.
$\mathrm{Zn^{2+}(aq) + 2e^- = Zn(s) \hspace{40px} E^o = -0.763\: V}$
$\mathrm{Cu^{2+}(aq) + 2e^- = Cu(s) \hspace{40px} E^o = 0.337\: V}$
Based on the two Eo values, the copper ion will be reduced and zinc metal will be oxidized. In an electrochemical cell, the reduction half reaction is referred to as the cathode and the oxidation half reaction is referred to as the anode. By convention, the anode is always put on the left and the cathode on the right in the diagram.
The zinc half-cell consists of a piece of zinc metal in a solution containing zinc ion. The copper half-cell consists of a piece of copper metal in a solution containing copper ion. If a half reaction does not form a solid metallic species (e.g., Fe3+ + e = Fe2+) an inert metal such as platinum is used in the cell.
The two half-cells need to be connected to complete the circuitry and allow the reaction to proceed. Two connections are needed for a complete circuit. One is a metal wire that connects the two pieces of metal. The other is something known as a salt bridge that connects the two solutions.
What processes are responsible for conduction of electricity in an electrochemical cell?
The processes responsible for the current flow in an electrochemical cell depend on which part of the cell you are in. For the metallic components (zinc, copper, copper connecting wire), electrons are responsible for the current flow. In the solution, conduction of electricity is caused by migration of ions.
The ability of ions to conduct electricity is the reason why someone should never use a hairdryer while sitting in a bathtub full of water. If a hairdryer is dropped into the water, the water conducts electricity because of ions in it with the end result that the person will be electrocuted. Conductivity is a measurement of the ability of a solution to conduct electricity. The conductivity of a solution directly correlates with the ionic strength of the solution. Many science buildings have a device that is designed to generate highly purified water. One of the goals of these purification systems is to deionize the water. With these systems the conductivity is measured to determine the degree to which the water has been deionized (the reading is reported as a resistance and the higher the resistance, the less conductive the solution).
It is also important to consider the portions of the cell where the metal interfaces with the solution. In the cathode where reduction occurs, electrons must “jump” from the metal to a species in solution. In the anode of the cell represented in Figure 7, zinc atoms need to give up two electrons and a zinc ion is released into the solution. For an anodic half-cell with two water-soluble species (e.g., Fe2+ = Fe3+ + e), an electron would need to “jump” from a species in solution to the platinum electrode.
What is the purpose of the salt bridge?
In order to understand the purpose of the salt bridge it is necessary to consider the process taking place in each of the half cells in Figure 7. If each half cell started at standard state conditions, the cathode would begin with a 1 M concentration of a copper salt such as copper sulfate ([Cu2+] = 1 M; [SO42-] = 1 M) and the anode would have a zinc salt such as zinc sulfate ([Zn2+] = 1 M; [SO42-] = 1 M). Note that in both half cells, the sulfate ion is a spectator ion that is not involved and does not change in the electrochemical reaction. As the electrochemical reaction proceeds, Cu2+ in the cathode gets reduced and plates out as copper metal. In the other half call, zinc metal gets oxidized to form Zn2+. Without any form of intervention, this means that over time [Cu2+] < [SO42-] in the cathode and [Zn2+] > [SO42-] in the anode. The buildup of charge in both of the half cells is an undesirable situation because nature wants to maintain systems that are neutral. If this charge continued to build up, it will inhibit the electrochemical reaction and prevent it from going to its full extent. The purpose of the salt bridge is to act as a source of spectator ions that can migrate into each of the half cells to preserve neutrality. Any charge buildup in the solutions of the two half cells is known as a junction potential. Therefore, the purpose of a salt bridge is to reduce the junction potential between the solution interface of the two half cells.
What would you put inside a salt bridge?
First, it is important to put ionic species into the salt bridge that will not be reduced or oxidized in either of the half cells. Alkali cations and halide anions would be ideal for this purpose. It is also important that the charge balance in each of the half cells facilitated by the ions in the salt bridge occurs at the same rates. That means that the halide anions moving from the salt bridge into the anode to balance out the excess Zn2+ ions do so at the same rate as the alkali cations moving from the salt bridge into the cathode to balance out the depletion of Cu2+ ions. Ions have a property known as mobility and the mobility of an ion depends on its size. Smaller ions have a higher mobility than larger ions. That means that the ideal species for a salt bridge should have a cation and anion of the same size and charge. Potassium chloride is the ideal species for incorporation into a salt bridge, as K+ and Cl have the same number of electrons and are approximately the same size. Potassium nitrate (K+NO3) can also be used in a salt bridge. Amazingly, the nitrate ion, which has atoms with second shell electrons, has approximately the same size as a chloride ion, which has atoms with third shell electrons.
Another thing to consider is the concentration of KCl in the salt bridge. It is desirable to have a salt bridge that can overcome the possibility of a large charge buildup. To achieve this and not deplete the ions in the salt bridge over the course of the reaction, the KCl is typically at a high concentration, usually 4 M.
Describe two types of situations that would result in the irreversibility of an electrochemical process
An interesting aspect of an electrochemical cell is that it can be operated in two directions. If the circuitry is completed and the reaction proceeds in its spontaneous direction toward equilibrium, it is referred to as a voltaic or galvanic cell. In this case a current is drawn from the cell and can be used to perform some sort of work (e.g., light a bulb in a flashlight or operate a cell phone). At some point the cell or battery will “die” as the reaction reaches equilibrium and no more current can be drawn from it. Next time your car battery dies, you may feel better about the situation by remembering that it has just reached equilibrium.
If the cell instead is attached to an external source of power (e.g., plugged into a wall outlet), the reaction can be forced it its reverse direction back away from equilibrium. This is what happens when a battery is recharged. Rechargeable batteries require the use of a reversible reaction. An electrochemical cell being forced in its non-spontaneous direction is referred to as an electrolytic cell.
There are two situations that factor into the reversibility of reactions used in electrochemical cells. One involves chemical reversibility, which relates to the stability of the reactants and products. The other involves electrochemical reversibility, which involves the kinetics of electron transfer and relates to the ability to regenerate or recharge the cell to its initial conditions.
One common misconception is that an electrochemical reaction that produces a gas such as the reduction of hydrogen ion to hydrogen gas is chemically irreversible because the gas escapes.
$\mathrm{2H^+(aq) + 2e^- \rightarrow H_2(g)}$
However, if the cell is designed properly and is sealed, the gas can be trapped and reversing the potential through the use of an external power source can drive the reaction in the reverse direction. Therefore, electrochemical reactions that produce a gas are not necessarily chemically irreversible.
An example of an electrochemical process that is chemically irreversible occurs if the product rapidly decomposes to something else. In this case, when an external power source is applied to reverse the process, the appropriate species is no longer in solution. In the example below, if the A species degrades rapidly to B and C, there is no remaining A for the chemical regeneration of A.
$\ce{A + e^-} \rightleftharpoons \ce{A^-} \rightarrow \ce{B^- + C}$
The second example of an irreversible electrochemical reaction occurs when there is something known as an overvoltage or overpotential. Since electrons must transfer from one species to another in an electrochemical reaction, the kinetics of the electron transfer must be considered. In cases of slow kinetics, it is possible to have an electrochemically irreversible reaction.
If we consider the reduction of H+ to hydrogen gas shown above, there is the key step where the electron must “jump” from the electrode to the hydrogen ion in solution. With some electrochemical reactions, there is a resistance of the electron to making the jump. If one were to apply a potential that in theory was suitably large such that the electrons should complete the jump, it would still not happen. The electron can be forced to “jump” by applying a higher voltage – an overpotential – to the electrode (electrons of higher energy are put onto the electrode until a point is reached where it becomes favorable for an electron to leave the electrode and go to the ion in solution). You are likely familiar with the concept of activation energies in chemical reactions. The occurrence of an overpotential indicates the presence of an activation energy barrier for an electrochemical reaction.
Whether or not a particular half reaction has an overpotential is determined in part by the nature of the electrode material. The reduction of hydrogen ion to hydrogen gas has almost no overpotential with a platinum electrode but has a very high overpotential with mercury and many other electrodes.
As an aside, it is worthwhile to examine two particular half reactions that have practical applications and potential future implications for society. These two reactions are shown below.
$\mathrm{2H^+(aq) + 2e^- = H_2(g) \hspace{40px} E^o = 0.00\: V}$
$\mathrm{O_2(g) + 4H^+(aq) + 4e^- = 2H_2O \hspace{40px} E^o = 1.229\: V}$
Based on the Eo values, the spontaneous reaction involves the oxidation of hydrogen gas and reduction of oxygen gas, as shown in the balanced reaction below.
$\mathrm{2H_2(g) + O_2(g) = 2H_2O \hspace{40px} E^o = 1.229\: V}$
A device that electrochemically combines hydrogen and oxygen gas and uses the transfer of electrons as a source of electricity is known as a fuel cell. One example of the use of this technology as a source of electricity is aboard the International Space Station.
If one considers the reverse reaction shown below, through the application of an external source of power it will be possible to electrochemically convert water to hydrogen and oxygen gas.
$\mathrm{2H_2O = 2H_2(g) + O_2(g)}$
The intriguing aspect of the electrolytic splitting of water into hydrogen and oxygen gas is that hydrogen is a useful fuel for a fuel cell or through combustion. If someone were able to economically carry out this reaction, it would mean that our fuel would come from water, thereby providing a limitless source of fuel. Also, because the reaction is best done under conditions with a relatively high ionic strength, it could be done using ocean water.
There are two problems with economically carrying out the electrolytic splitting of water. One is that it takes energy to do it, since it’s the reverse of the spontaneous direction. Because of losses of efficiency when combusting hydrogen, it would take more energy to split water then you would get in return by using hydrogen as a fuel. A second problem is that both half reactions have overpotentials with many different electrodes meaning that it will cost even more to carry out the reaction.
An active area of research is an attempt to devise electrodes that have two particular features. One is that they are made of materials that do not have high overpotentials toward the two relevant half reactions (Note that two different electrodes are needed: one for the hydrogen half reaction and the other for the oxygen half reaction). The second feature would be a system where sunlight could be used as a source of power to assist the splitting reaction. In the two electrodes, the energy from the sun would be used to promote an electron on the electrode into a higher molecular orbital. With that extra energy it would be much easier for the electron to transfer to the species in solution and then much less costly to split the water. Many of the electrodes being examined in this application are transition metal complexes and while some advances have been made, it is also critical that the electrode be inexpensive enough to make the entire process cost effective. No one has yet to find electrodes with low enough overpotentials and the ability to harness the energy of the sun to facilitate the reactions in a cost effective manner.
Shorthand Notation for an Electrochemical Cell
There is a shorthand notation used to specify the conditions of an electrochemical cell. The notation shows the anode on the left and cathode on the right. The concentrations of important species in each of the half reactions are included. Spectator ions are not included in the notation. Phase boundaries are shown with a single line (|) and a salt bridge with a double line (||). The electrochemical cell described above would have the following notation:
$\ce{Zn \:|\: Zn^2+ (1\: M) \:||\: Cu^2+ (1\: M) \:|\: Cu}$ | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Electrochemical_Methods_of_Analysis/02_Text/5._Electrochemical_Cells.txt |
Potassium dichromate (K2Cr2O7) reacts with Fe(II) to produce Cr(III) and Fe(III). What is the standard state potential and K for this reaction?
The first step in solving this is to identify the two appropriate half reactions that make up the cell. From a table of Eo values we find the following two reactions:
$\mathrm{Fe^{3+}(aq) + e^- = Fe^{2+}(aq) \hspace{40px} E^o = 0.77\: V}$
$\ce{Cr2O7^2- (aq) + 14H+(aq) + 6e-} = \ce{2Cr^3+(aq) + 7H2O} \hspace{40px} \mathrm{E^o = 1.33\: V}$
An examination of the Eo values indicates that the iron reaction proceeds as an oxidation and is the anode and the chromium reaction proceeds as a reduction and is the cathode.
Standard State Potential
The following equation is used when calculating the standard state potential of an electrochemical cell.
$\mathrm{E^o_{CELL} = E^o_{CAT} - E^o_{AN}}$
When using this equation, the Eo values are used directly from the table without changing the sign and without using any coefficients in the balanced overall electrochemical reaction. So even though the anodic reaction is reversed in direction from that in the table, the values in the table are all ranked relative to each other and the minus sign in the equation is accounting for the fact that the anodic reaction occurs in a reverse direction. There is no use of coefficients because Eo values are measured for standard state starting conditions (i.e., all reactants and products beginning at 1 Molar). With the pair in this problem, inequivalent molar amounts of the iron and chromium species will be present when the system achieves equilibrium because six mole equivalents of the iron are used up for each mole equivalent of the chromium.
One final point to note is that EoCELL will always be positive.
$\mathrm{E^o_{CELL} = 1.33 - 0.77 = 0.56\: V}$
Equilibrium Constant
Earlier in the unit we had developed the following equation relating Eo to the equilibrium constant.
$\mathrm{E^o = \dfrac{0.059}{n}\log K}$
This expression can be rewritten as follows:
$\mathrm{K = 10^{(n)(E^o)/0.059}}$
n is the number of electrons that are transferred in the balanced electrochemical reaction (6 in this example). Putting numbers in and evaluating this term gives the following:
$\mathrm{K = 10^{(6)(0.56)/0.059} = 8.9 \times 10^{56}}$
Note that this is an exceptionally large equilibrium constant meaning that the reaction goes nearly toward completion leaving only tiny amounts of reactants at equilibrium.
Cell Potential with the Non-standard State Conditions Given in the Problem
In this example, one half-cell is made up with 1.50 M potassium dichromate and 0.30 M chromium(III)nitrate hexahydrate in 1.00 M nitric acid and the other half cell is made up with 0.050 M iron(III)chloride hexahydrate and 0.10 M iron(II)chloride tetrahydrate?
Using the shorthand notation for an electrochemical cell, we could write the above cell as follows:
$\mathrm{Pt \:|\: Cr_2O_7^{2-} (1.50\: M),\: Cr^{3+} (0.30\: M),\: H^+ (1.00\: M) \:||\: Fe^{2+} (0.050\: M),\: Fe^{3+} (0.10\: M) \:|\: Pt}$
There are two ways to solve this problem. One is to use the Nernst equation in the form we previously defined.
$\mathrm{E_{CELL}= E_{CELL}^0 - \dfrac{0.059}{n} \log Q}$
When using this method, we first need the complete electrochemical reaction for the cell. Note that n = 6 for this reaction as six electrons were needed to balance out the two half reactions.
$\ce{Cr2O7^2- (aq) + 14H+(aq) + 6Fe^2+(aq)} = \ce{2Cr^3+(aq) + 6Fe^3+(aq) + 7H2O}$
We can then write the term for Q:
$\mathrm{Q= \dfrac{[Cr^{3+}]^2 [Fe^{3+}]^6}{[Cr_2 O_7^{2-}][H^+ ]^{14} [Fe^{2+}]^6}}$
$\mathrm{Q= \dfrac{(0.30)^2 (0.050)^6}{(1.50)(1.00)^{14} (0.10)^6} =9.4 \times 10^{-4}}$
We can now substitute values into the Nernst equation and solve for ECELL.
$\mathrm{E_{CELL}= 0.56 -\dfrac{0.059}{6} \log(9.4 \times 10^{-4})=0.56+0.03=0.59 \:V}$
Instead of using the Nernst equation, is also possible to use the following equation to calculate the cell potential.
$\mathrm{E_{CELL} = E_{CAT} - E_{AN}}$
When using this equation, each of the half reactions is written and evaluated as a reduction. The minus sign accounts for the fact that the anodic reaction is reversed in the complete, balanced electrochemical reaction. When evaluating each of the two terms, use the associated value of n for each of the individual half reactions.
$\mathrm{E_{CAT}= E_{CAT}^0 - \dfrac{0.059}{n}\log Q_{CAT}}$
$\mathrm{Q_{CAT}=\dfrac{[Cr^{3+}]^2}{[Cr_2 O_7^{2-} ][H^+ ]^{14}} =\dfrac{(0.30)^2}{(1.50)(1.00)^{14}}=0.060}$
The number of electrons in the cathode reaction is six so n = 6.
$\mathrm{E_{CAT}= 1.33-\dfrac{0.059}{6} \log(0.060)=1.33+0.01=1.34\: V}$
The number of electrons in the anode reaction is one so n = 1.
$\mathrm{E_{AN}= E_{AN}^0 - \dfrac{0.059}{n} \log Q_{AN}}$
$\mathrm{Q_{AN}=\dfrac{[Fe^{2+}]}{[Fe^{3+}]} =\dfrac{(0.10)}{(0.050)}=2.0}$
$\mathrm{E_{AN}= 0.77-\dfrac{0.059}{1} \log(2.0)=0.77-0.02=0.75\: V}$
Now we can evaluate the cell potential.
$\mathrm{E_{CELL} = E_{CAT} - E_{AN} = 1.342 - 0.75 = 0.59\: V}$
As expected, the use of these two possible methods provides the exact same value for the cell potential.
What is the cell potential if the chromium half-cell were operated at a pH of 7 instead of using 1 M nitric acid?
The point of this problem is to realize that some electrochemical reactions are critically dependent on the pH. The particular reaction in this problem has a stoichiometric coefficient of 14 for the H+, meaning that the reaction will be highly dependent on pH. Since only the cathodic half reaction depends on pH, we can evaluate the overall cell potential using the second of the two methods from above.
$\mathrm{E_{CAT}= E_{CAT}^0 - \dfrac{0.059}{n} \log Q_{CAT}}$
$\mathrm{Q_{CAT}=\dfrac{[Cr^{3+}]^2}{[Cr_2 O_7^{2-}][H^+]^{14}} =\dfrac{(0.30)^2}{(1.50)(1.00 \times 10^{-7})^{14}} =6.0 \times 10^{96}}$
$\mathrm{E_{CAT}= 1.33 - \dfrac{0.059}{6} \log(6.0 \times 10^{96} )=1.33- -.95=0.38\: V}$
Remember that:
$\mathrm{E_{CELL} = E_{CAT} - E_{AN}}$
and that EAN calculate previously was 0.75 V.
$\mathrm{E_{CELL} = 0.38 - 0.75 = -0.37\: V}$
It is important to note that ECELL in this case is negative. The cell potential is positive for a reaction that proceeds in the forward direction toward products. The fact that this value is negative means that the reaction actually goes in the reverse direction under the conditions provided. Looking again at the overall reaction for this cell provided below, with the concentration of H+ so low (1.00 x 10-7 M) and the coefficient of 14 for the H+, it is reasonable to think that the overall reaction will actually proceed toward reactants to establish equilibrium than proceeding toward products.
$\ce{Cr2O7^2- (aq) + 14H+(aq) + 6Fe^2+(aq)} = \ce{2Cr^3+(aq) + 6Fe^3+(aq) + 7H2O}$
Similar calculations at pH 1 ([H+] = 0.10 M) and pH 3 ([H+] = 0.0010 M) give cell potentials of 0.46 V and 0.18 V, respectively. The importance of pH on this particular electrochemical cell is apparent from these data. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Electrochemical_Methods_of_Analysis/02_Text/6._Potential_of_an_Electrochemical_Cell.txt |
There are a wide variety of analytical methods that are based on the use of electrochemical reactions or processes. Only a subset of all the electrochemical methods will be discussed herein. The simplest electrochemical cell has two electrodes. One of them is sensitive toward the analyte and is referred to as either the working or indicator electrode. The second electrode is referred to as the counter or reference electrode, with the potential of the working electrode measured relative to it. The most common reference electrode is comprised of a silver-silver chloride (Ag/AgCl) half-cell, with a fill solution that is saturated with KCl and AgCl (Figure 8).
$\mathrm{AgCl (s) + e^- \Leftrightarrow Ag(s) + Cl^-(sat) \hspace{20px} E = +0.197\: V}$
Figure 8. Diagram of a silver-silver chloride reference electrode. (Figure from Analytical Chemistry 2.0, David Harvey, community.asdlib.org/activele...line-textbook/).
The measurement uses a device known as a potentiostat, which is an electronic component that can run a three electrode cell. The potentiostat maintains the potential of the working electrode at a constant level with respect to the reference electrode. In some cases, it will be important that there be no current running through the reference electrode, which is achieved by having the system at a high impedance.
While a Ag/AgCl electrode is commonly used as a reference so that potentials are measured relative to it, for the purposes of the methods discussed herein, we will use the standard hydrogen electrode (SHE) as the reference electrode. Since the potential of the SHE is 0.00 Volts, this will allow us to use the numbers directly from a table of standard electrode potentials without having to correct for the potential of the reference electrode.
7: Electrochemical Analytical Methods
As the name implies, ion-selective electrodes are devices that are selective for a particular ionic species. The most well-known and highly used ion-selective electrode is a pH electrode, which is sensitive toward the H+ ion. A desirable aspect of ion selective electrodes is their ease of use. Using pH electrodes as an example, after appropriate calibration, it is quite trivial to measure the pH of a solution. pH meters are relatively small devices so that it is easy to transport a battery operated device to remote sites and measure the pH of samples like surface waters.
Ion-selective electrodes are used in a configuration where there is a completed circuit but the system is at equilibrium so there is no current flow. When there is no current flow it is possible to measure a potential. The basis of all ion-selective electrodes involves the measurement of a junction potential. The system is designed in such a way that the magnitude of a particular junction potential in the device depends only on the ion being measured. The design of a pH electrode will be discussed in detail and will exemplify the method by which other ion selective electrodes function.
7.1.1. pH Electrode
The components necessary for the measurement of pH involve internal and external Ag/AgCl reference electrodes and a thin glass membrane (about 50 um thick) (Figure 9). You are probably familiar with pH electrodes and have likely used one at some point to measure the pH of a solution. In all likelihood you used a single electrode system with a glass bulb on the bottom. In actuality, the device you used, and which is commonly in use today, is known as a combination electrode. The design of a combination electrode is shown in Figure 9. The system is an ingenious design that has the two Ag/AgCl reference electrodes incorporated into the single electrode system with appropriate connections to allow for a complete circuit.
Figure 9. Diagram of a combination pH electrode.
(Figure from Analytical Chemistry 2.0, David Harvey, community.asdlib.org/activele...line-textbook/).
The inside of the glass membrane is filled with a solution consisting of 0.1 M hydrochloric acid (HCl) saturated with silver chloride (AgCl). The external layer of the glass membrane is immersed in the solution whose pH you wish to measure. The complete circuit of a pH electrode has multiple junction potentials, but only one – the potential between the glass membrane and the external solution whose pH is being measured – is important. All the other junction potentials in the system are constant or zero and combined into one cell constant. The membrane potential with the external solution varies with the concentration of H+ in the solution, and the two reference electrodes measure the potential difference across the glass membrane.
We need to look in more detail at the nature of the glass to understand how a variable membrane potential forms (Figure 10).
Figure 10. Diagram of the glass membrane in a pH electrode showing the dry glass and hydrated gel layers.
There are many ways to formulate glass. While glasses are based on a silicate framework (i.e., repeating SiO2 units), there are other impurities within the glass that give it different properties. The glass used in a pH electrode contains sodium (Na+) ions. When placed in contact with a solution, a microscopically thin (10 nm thick) hydrated gel layer forms on the surface of the glass. Within this gel layer, hydrogen ions from the solution can migrate a very short distance into the glass and displace sodium ions. The number of H+ ions that migrate into the gel layer depends directly on the concentration of H+ in the solution. For the internal solution, the concentration of H+ is constant so a fixed number of Na+ ions are displaced from the gel layer by H+ ions. For the external solution, the number of H+ ions changes with the pH so the number of Na+ ions displaced from the gel layer varies as a function of the pH. The ratio of [H+] to [Na+] in the outside gel layer influences the magnitude of the membrane potential. The reason the membrane potential varies with the ratio of the two ions is because the smaller H+ ions have a higher mobility than the larger Na+ ions.
It is important to note that a pH electrode does not inherently know the pH of a solution and must be calibrated against a buffer solution with a known pH. The electrode provides a measurement in mV and the measurement can be related to pH through an appropriate calibration. Therefore, the accuracy of pH measurements is only as reliable as the accuracy of the calibration buffers. Also, the electrode should be calibrated with a buffer that has a pH close to the pH you expect for the final solution. Common pH calibration buffers with pH values of 4, 7 and 10 are commercially available.
Another important consideration with a pH and any other ion-selective electrode is the possibility of interferences. Considering that H+ ions can migrate from the external solution into the hydrated gel layer, it should seem reasonable to conclude that other cations can similarly migrate into the gel layer and influence the junction potential. The degree to which other ions may influence the membrane potential depends on the nature of the glass and the concentration of the interfering ions. Many ion-selective electrodes come with instructions that list known interferences and the concentrations at which they become problematic to accurate measurements. For pH electrodes, values below pH 0.5 are unreliable because of what is known as an acid error. Values above pH 9 are less reliable because of alkaline error that is usually due to the relatively high concentration of other cations in solution that migrate into the glass.
The response of a pH electrode is also sensitive to the ionic strength of the solution. In solutions with exceptionally low ionic strength and low conductivity, the reading on a pH meter often takes a long time to stabilize after the pH electrode is inserted into the solution (rainwater is an excellent example of a solution with a low ionic strength – a pH measurement is a convenient way to monitor the acidity in acid rain). In solutions of higher ionic strength, the stabilization of the membrane potential and pH reading occur much faster. Some pH electrodes are designed purposely for the measurement of solutions with low ionic strengths. Such an electrode is designed to leak out some of the ions from the reference electrode to raise the ionic strength and conductivity of the external solution and promote faster establishment of the membrane potential.
Allowing the external gel layer of the glass membrane to completely dry out may prove detrimental to the performance of the electrode. pH electrodes are often capped for storage with a small volume of a pH 7 buffer in the cap. Before using a pH electrode removed from storage for measurements, it is advisable to soak it in a pH 7 buffer for several hours before use.
7.1.2. Other Glass Electrodes
Varying the composition of the glass enables the construction of ion selective electrodes that are sensitive for cations other than H+. Glass electrodes sensitive for Na+ and Ag+ are examples. When using these electrodes, one must always be concerned about interferences from other cations in solution and instructions with the electrode will describe possible interferences and the concentrations at which they become important.
7.1.3. Membrane Electrodes with Organic Polymers
In addition to membranes formed of glass, many membrane electrodes employ a hydrophobic organic polymeric material with appropriate functional groups designed to selectively associate with a certain ion. These functional groups can be anion exchangers, cation exchangers and neutral ionophores that are cavity compounds with the correct size to selectively bind certain metal ions. The ion can migrate into or bind to the polymeric membrane and the membrane potential depends on the concentration of that ion. A noteworthy feature of these membrane electrodes is that they can be designed to measure anionic species such as nitrate (NO3) and chloride (Cl) ions and cationic species such as NH4+ and Ca2+. As with other ion-selective electrodes, other anions may interfere with the measurement so care must be taken when using these devices.
7.1.4. Enzyme Electrodes
Enzyme electrodes are constructed similarly to the membrane electrodes described in the previous section except that the active material in the membrane is an enzyme. Typically the desired analyte reacts with the enzyme to produce a product whose concentration is monitored. A virtue of these electrodes is that they are highly selective for the particular substrate that binds to the enzyme. These electrodes are often constructed for use in clinical labs where there is a need to monitor a particular substrate on a regular basis as an indicator of a health issue. Examples of the use of an enzyme electrode are to measure urea or glucose.
7.1.5. Solid-State Electrodes
There are two common solid-state ion-selective electrodes where the active surface consists of a polished crystal of lanthanum fluoride (LaF3) or silver sulfide (Ag2S). The electrode is sensitive toward either of the ions that make up the crystal. While there is seldom interest in measuring the concentration of lanthanum, fluoride is often added to municipal water supplies to reduce tooth decay. Fluoride selective electrodes are easy to use on a sample by sample or continuous monitoring basis to insure that the fluoride levels of the water are in the proper range. Similarly, there is usually not much need to measure silver ion in solution but sulfide ion is indicative of the ability to form hydrogen sulfide (H2S), a toxic, foul-smelling gas. The silver sulfide electrode can also be doped with cadmium, copper and lead to create ion-selective electrodes for these metals.
7.1.6. Gas-Sensing Electrodes
Gas-sensing electrodes are designed to measure water-soluble gases such as carbon dioxide (CO2), sulfur dioxide (SO2), nitrogen dioxide (NO2), ammonia (NH3). The electrode has a semi-permeable membrane that allows the gas of interest to pass through it into an internal solution. These gases can all react with water to form acids that liberate hydronium ions (CO2, NO2 and SO2) or a base that liberates hydroxide ions (NH3). The electrode incorporates a pH electrode in contact with the internal solution and the measured pH can be related back to the amount of gas in solution. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Electrochemical_Methods_of_Analysis/02_Text/7%3A_Electrochemical_Analytical_Methods/7.1%3A_Ion-Select.txt |
Electrodeposition or electrogravimetry are two terms used to describe the same analysis method. The general procedure is to use something like a platinum electrode and apply a constant reducing potential that is sufficient to plate out a solid metal. For example, this method could be used to plate out cadmium metal from solution by the reaction shown below.
$\mathrm{Cd^{2+}(aq) + 2e^- \rightleftharpoons Cd(s) \hspace{40px} E^o = - 0.403\: V}$
The platinum is weighed before and after the plating step. The difference in weight is solid cadmium and this amount can be related back to the concentration of cadmium in the original solution.
Because it is essential that all the Cd2+ in solution get plated out onto the electrode, it is necessary use large surface area electrodes (on the order of 50 cm2) and to stir the solution to bring the Cd2+ up to the electrode. The plating step typically occurs at a potential well above the Eo value and over 30-60 minutes to insure that virtually all the Cd2+ has been plated onto the electrode. The plating potential must be well above the Eo value because, as the plating proceeds, the concentration of Cd2+ in solution will diminish, thereby raising the reducing potential needed for plating.
Example $1$
Will the presence of Fe2+ at 0.0800 M interfere with the electroplating of 99.9% of the Cd2+ in a solution in which the cadmium is expected to be present at a concentration of no less than 0.0500 M? Calculate potential values relative to a standard hydrogen electrode.
Solution
The first thing to consider is the two relevant half reactions for the metal ions in this problem.
$\mathrm{Cd^{2+}(aq) + 2e^- \rightleftharpoons Cd(s) \hspace{40px} E^o = - 0.403\: V}$
$\mathrm{Fe^{2+}(aq) + 2e^- \rightleftharpoons Fe(s) \hspace{40px} E^o = - 0.440\: V}$
Because the reduction potential of Cd2+ is less negative than that of Fe2+, it is easier to plate out the cadmium. This means that it may be possible to apply a reducing potential that is sufficient to plate out cadmium but not large enough to plate out any of the iron.
The next step is to focus on the plating of cadmium. We only need to consider the half reaction for cadmium for this calculation.
$\mathrm{E= E^0 - \dfrac{0.059}{n} \log \dfrac{1}{[Cd^{2+}]}}$
In this equation, n = 2. If we substitute in the initial concentration of Cd2+ (0.0500 M), we get the following value for E.
$\mathrm{E= -0.403- \dfrac{0.059}{2} \log \dfrac{1}{0.0500} = -0.403-0.038= -0.441\: V}$
Note, this is the potential needed to start the plating of cadmium. As the cadmium plates out, the concentration will be lower than 0.0500 M and that will impact the value of E. What we really need to consider is the concentration of Cd2+ when 99.9% of it has plated out. That would mean that only 0.1% of it is left in solution. 0.1% of 0.0500 M is 0.0000500 M (5.00 x 10-5 M). Let’s now reevaluate the potential for this solution.
$\mathrm{E= -0.403- \dfrac{0.059}{2} \log \dfrac{1}{0.0000500}= -0.403-0.126= -0.529\: V}$
So the potential needed to plate out 99.9% of the cadmium is –0.529 V. Note that as the concentration of Cd2+ drops, the potential needed to plate the cadmium becomes more negative (i.e,. the plating of cadmium becomes more difficult and requires a higher potential). Hopefully it seems reasonable that it would take a higher potential to force the cadmium to plate from solution as the concentration of Cd2+ becomes smaller.
The question we now need to answer is whether the potential needed to plate out 99.9% of the Cd2+ will begin to plate out the Fe2+. In this case, we only need to consider the potential that would be needed to start the plating of Fe2+, since plating of any iron will interfere with the measurement of cadmium. This requires using the half reaction for iron:
$\mathrm{E= -0.440- \dfrac{0.059}{2} \log \dfrac{1}{0.0800}= -0.440-0.032= -0.472\: V}$
Since the potential needed to begin plating out iron (– 0.472 V) is less negative than that needed to plate out 99.9% of the cadmium (– 0.529 V), it is not possible to selectively analyze for the cadmium in the presence of iron. The iron will plate out as well and interfere with the method.
Example $2$:
Suppose the solution had 0.0800 M Cr3+ as a possible interference. Is it possible to plate out 99.9% of the cadmium without any interference from the chromium?
Solution
We first need to consider the appropriate half reaction for the reduction of Cr3+.
$\mathrm{Cr^{3+}(aq) + 3e^- \rightleftharpoons Cr(s) \hspace{40px} E^o = - 0.744\: V}$
Then we can solve for the potential that would just start the plating of chromium.
$\mathrm{E= -0.744- \dfrac{0.059}{3} \log \dfrac{1}{0.0800}= -0.744-0.022= -0.766\: V}$
Since the potential needed to begin plating out the chromium (– 0.766 V) is more negative than that needed to plate out 99.9% of the cadmium (– 0.529 V), it is possible to selectively plate out the cadmium. Setting the potential somewhere in the window between – 0.529 V and – 0.766 V relative to a standard hydrogen electrode would be necessary for the analysis.
7.3: Coulometry
Coulometry is similar to electrogravimetry in that a constant potential is applied that is sufficient to carry out a particular electrochemical reaction. In this case, instead of plating out a metal and measuring its weight, the current generated through the electrochemical reaction is measured as a function of time. As with electrogravimetry, it is essential that all of the species in solution undergo the electrochemical reaction. To accomplish this, the potential is typically applied to the solution for 30-60 minutes, an electrode with a large area is used and the solution is stirred. The charge (Q) can be related to the number of electrons using Faraday’s Constant, which can then be related to the moles of substance being measured. We can use the reduction of Cd2+ to cadmium metal to examine the technique of coulometry.
Example \(1\):
Draw the plot you would obtain for current (y-axis) versus time (x-axis) if you applied a constant potential high enough to carry out the reduction of Cd2+ to cadmium metal.
Solution
Just like in electrogravimetry, the concentration of Cd2+ is high at the beginning of the analysis and diminishes as it is reduced to cadmium metal. This means that the current starts out high and will diminish with time. That leads to the plot in Figure 11.
Figure 11. Plot of current versus time in a coulometric analysis. (Figure from Analytical Chemistry 2.0, David Harvey, community.asdlib.org/activele...line-textbook/).
Example \(1\):
How would you relate the outcome of the plot in Figure 11 to the concentration of Cd2+ in the sample?
Solution
What is important to determine here is the total amount of charge that flowed as that can be related using Faraday’s constant to the total number of electrons that were used in reducing Cd2+. The total number of electrons can be related to the total moles of cadmium in solution. Determining the total charge would involve integrating the area under the curve in Figure 11.
Advantages of Coulometry over Electrogravimetry
There are several advantages that coulometry has over electrogravimetry.
Since coulometry measures the charge needed to complete the electrochemical reaction instead of the weight of the substance plated out, reactions in which both species are water soluble can be examined (e.g., the half reaction \(\ce{Fe^{3+}(aq) + e^{–} -> Fe^{2+}(aq)}\)).
Electrogravimetry is only useful for reduction processes involving the plating of a metal. Coulometry can be used in either a reduction or oxidation mode, increasing its versatility. The direction in which the electrons flow will be different depending on whether the reaction involves a reduction or oxidation, but that is not a hindrance to measuring the current and obtaining a plot like that in Figure 11.
One last advantage is that coulometry is more sensitive than electrogravimetry. The sensitivity and detection limits in electrogravimetry are limited by the minimum weight that can be measured on a balance – analytical balances typically measure down to 0.0001 gram but a much higher weight is needed for suitable accuracy and precision. We have the ability to measure very small quantifies of current and can accurately measure time. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Electrochemical_Methods_of_Analysis/02_Text/7%3A_Electrochemical_Analytical_Methods/7.2%3A_Electrodep.txt |
7.4.1. “Classical” Redox Titration
A “classical” redox titration is one in which a titrant of a known concentration is prepared and standardized versus some appropriate primary standard. Then an accurate volume of an analyte solution with an unknown concentration is measured out. To be suitable for a titration, the analyte and titrant must undergo a rapid redox reaction that has a very large equilibrium constant so that formation of products is highly favored. The titrant is added from a burette and the volume needed to reach the endpoint is determined. The endpoint is identified through the use of an appropriate color-changing indicator. Besides an obvious color change, the criteria for selection of a suitable indicator is that it undergoes a redox reaction with the titrant that is less favored (has a lower equilibrium constant) than that of the analyte so that it only reacts after all of the analyte has reacted away. If the reaction between the titrant (TITR) and analyte (AN) is one-to-one, the following equation can be used to calculate the concentration (C) of the analyte.
$\mathrm{C_{TITR} \times V_{TITR} = C_{AN} \times V_{AN}}$
If the titrant and analyte react in something other than a one-to-one equivalency, then the above equation needs to be modified to reflect the proper stoichiometry.
7.4.2. Coulometric Titration (Controlled Current Coulometry)
A coulometric titration is an interesting variation on a classical redox titration. Consider the reaction below in which Fe2+ is the analyte and Ce4+ is the titrant. An examination of the two standard potentials for the two half reactions confirms that in this pair cerium will proceed as the reduction and iron as the oxidation.
$\ce{Fe^2+(aq) + Ce^4+(aq) <=> Fe^3+(aq) + Ce^3+(aq)}$
$\ce{Ce^{4+}(aq) + e^- <=> Ce^{3+}(aq)} \mathrm{\hspace{40px} E^o = 1.61\: V}$
$\ce{Fe^{3+}(aq) + e^- <=> Fe^{2+}(aq)} \mathrm{\hspace{40px} E^o = 0.77\: V}$
Rather than performing the titration by adding the Ce4+ from a burette, an excess of Ce3+ is added to an accurately measured volume of the unknown analyte solution. A constant electrochemical potential suitable to convert Ce3+ to Ce4+ is applied to the solution. As Ce4+ is produced, it immediately reacts with Fe2+ and is converted back to Ce3+. The result is that the concentration of Ce3+ is constant. A plot of the current measured by the generation of Ce4+ versus time would look like that in Figure 12.
Figure 12. Plot of the current versus time in a coulometric titration.
There is more than one way to determine the endpoint of a coulometric titration. One is to add an indicator that reacts only when all the analyte is used up. In this case, you would measure the time until the color change and integrate the current versus time plot to determine the number of electrons and moles of analyte in the sample. Another possibility is that after the equivalence point, the “titrant” that is electrochemically generated no longer reacts away and there are other ways to measure that both members of the redox couple (Ce3+ and Ce4+ in the example above) are present.
What are some advantages of using a coulometric titration?
There are a number of advantages of a coulometric titration when compared to a classical redox titration.
• A coulometric titration has no burette.
• There is no need to standardize the titrant in a coulometric titration.All one has to do is add sufficient amounts of the titrant to the analyte solution (remember, we don’t actually add the titrant but add the appropriate species that will be converted into the titrant). Faraday’s constant is essentially the standard as it relates charge to the number of electrons and we know Faraday’s constant to a high degree of precision.
• Since we can accurately and precisely measure current and time, coulometric titrations are highly accurate and precise.
• Coulometric titrations are much more sensitive and can often measure lower concentrations than conventional titrations.
• Finally, because the titrant immediately reacts with the analyte after it is produced, it is possible to use an unstable titrant in a coulometric titration.Note that the species added to the solution that is converted into the titrant must be stable.
7.4.3. Amperometric Titration
An amperometric titration is done analogously to a classical titration in which a known volume of an analyte is measured out and a standardized titrant is added using a burette. The difference is that instead of using a color-changing indicator to determine the end point, the ability of the solution to generate a current is measured. It should be pointed out that you do not generate a large amount of current in an amperometric titration so the current generation does not deplete the solution of significant levels of the chemical species. Instead, you merely measure whether and how much current is produced at any given point during the titration.
$\mathrm{Analyte + Titrant = Products}$
A stipulation for monitoring a particular titration reaction amperometrically is that the products can never generate a current at the potential being applied to the system. What is also necessary is that the analyte and/or titrant are able to generate a current. This means that there are the three possible scenarios to consider for an amperometric titration.
Example $1$:
Draw the plot of current (y-axis) versus ml of titrant (x-axis) if only the analyte generates a current.
Solution
In this case, at the beginning before any titrant is added there is a high concentration of analyte and therefore the current would be high. As titrant is added, it reduces the concentration of analyte leading to the formation of products and the current would drop. Since every similar increment of titrant reacts with the same amount of analyte, the current should decay in a linear manner and go to zero at the endpoint. The result is the plot seen in Figure 13.
Figure 13. Plot of current versus ml of titrant in an amperometric titration when only the analyte produces a current.
Example $2$:
Draw the plot of current (y-axis) versus ml of titrant (x-axis) if only the titrant generates a current.
Solution
In this case, there is no current at the beginning because the analyte does not generate one. As small amounts of titrant are added, all of the titrant reacts with the analyte to form products so there is still no measureable current. In fact, up until the equivalence point, any added titrant reacts away. Only after the equivalence point and all of the analyte is used up is there extra unreacted titrant in the solution that produces a current. The result is the plot seen in Figure 14.
Figure 14. Plot of current versis ml of titrant in an amperometric titration when only the titrant produces a current.
Example $3$:
Draw the plot of current (y-axis) versus ml of titrant (x-axis) if both the analyte and titrant generate a current.
Solution
The plot in this case is the combination of those in Figures 13 and 14. Prior to the equivalence point, there is analyte and a current is observed but diminishes as more titrant is added. After the equivalence point the concentration of unreacted titrant goes up so the current gets larger. The only point where there is no species in solution capable of generating a current is at the equivalence point. The result is the plot seen in Figure 15.
Figure 15. Plot of current versis ml of titrant in an amperometric titration when both the analyte and titrant produce a current.
7.4.4. Potentiometric Titration
The final titrimetric method we will examine is known as a potentiometric titration. This is similar to a classical titration as titrant is added from a burette. Instead of determining the endpoint using a color-changing indicator, a junction potential is continuously monitored during the titration. An example of a potentiometric titration that you are likely familiar with is a pH titration. In this case, a pH electrode is placed in the solution and the pH (determined by a membrane potential at the external glass membrane as described earlier in this unit) is measured continuously as titrant is added. The pH is plotted versus the ml of titrant added and there is a characteristic “break” or sharp rise in the plot that occurs at the equivalence point as seen in Figure 16. The plot has an inflection point that occurs at the equivalence point, although it is easier to identify the location of the inflection point by taking either the first or second derivative of the plot shown in Figure 16.
Figure 16. Plot of pH of a strong acid versus ml NaOH in a pH titration. (Figure from Analytical Chemistry 2.0, David Harvey, community.asdlib.org/activele...line-textbook/).
It is also possible to measure the progress of other redox reactions potentiometrically. For example, we could monitor the progress of the titration of a solution of Fe2+ using Ce4+ as the titrant. The products of the reaction are Fe3+ and Ce3+.
$\mathrm{Fe^{2+}(aq) + Ce^{4+}(aq) = Fe^{3+}(aq) + Ce^{3+}(aq)}$
The two relevant half reactions are shown below and indicate that the cerium reaction is the cathode and iron reaction is the anode.
$\mathrm{Ce^{4+}(aq) + e^- = Ce^{3+}(aq) \hspace{40px} E^o = 1.61\: V}$
$\mathrm{Fe^{3+}(aq) + e^- = Fe^{2+}(aq) \hspace{40px} E^o = 0.77\: V}$
Note that the Eo for the complete reaction is 0.84 V. Using the equation on p. 10 that relates Eo to K, the equilibrium constant for the reaction is greater than 1014. As with other electrochemical devices, a reference and working electrode is needed to perform the measurement. In this case, a platinum working electrode could be used. Let’s consider the following example to examine what will happen during the course of the titration and how one goes about calculating the potential during the titration.
Example $4$:
20 ml of an 0.10 M solution of Fe2+ is titrated with 0.10 M Ce4+. Calculate the potential that would be measured when 5 ml, 10 ml, 20 ml, 30 ml and 40 ml of titrant have been added.
Solution
The following steps will allow us to work through the calculation of the junction potentials.
Write an expression for the total electrochemical potential of the titration reaction (this is the Nernst equation for the complete reaction).
$\mathrm{E_{TOTAL}= E_{TOTAL}^0- \dfrac{0.059}{1} \log \dfrac{[Fe^{3+} ][Ce^{3+}]}{[Fe^{2+} ][Ce^{4+}]}}$
Note, the reaction has one electron transferred between the cerium and iron so n = 1.
What is the total electrochemical potential of the titration reaction at any point during a titration? Think carefully about this because your first intuition may not be correct.
One of the criteria for any reaction to be useful in a titration is that the system must rapidly achieve equilibrium. What that means is that the titration above is always at equilibrium. In other words, when a drop of the cerium titrant is added, it immediately reacts with the iron and the system achieves equilibrium. Once past the equivalence point, there is some continuing distribution of species that occurs to establish equilibrium after each increment of titrant is added, but the system is always at equilibrium during all phases of the titration. That means that the total electrochemical potential of the reaction between Fe2+ and Ce4+ is zero at all points of the titration.
If we consider the other way we developed for determining the total electrochemical potential of a system.
$\mathrm{E_{TOTAL} = E_{CATHODE} - E_{ANODE}}$
Remember that when using this form of the equation, both the cathode and anode values are evaluated as reductions. Since we determined that ETotal = 0 at all steps for the titration reaction, the following relationship is valid throughout the titration:
$\mathrm{E_{CATHODE} = E_{ANODE}}$
However, there are two reactions to consider. One is the titration reaction and the other is the overall cell reaction. The overall cell reaction includes species in the reference electrode. Whereas the titration reaction goes to completion, the extent of the overall cell reaction is negligible because the voltmeter has high impedance that prevents current flow. The result is that the potential difference measured between the platinum and reference electrode during the titration is either the value ECATHODE or EANODE developed above. Since these will both be equal, in theory either one of them could be used to calculate the potential at any point in the titration. However, there are advantages to using one of the half reactions over the other at certain points of the titration.
Which half reaction is easier to use to calculate the junction potential before the equivalence point? Which half reaction is easier to use after the equivalence point?
A way to consider this is to examine the amounts of the different species present at different points in the titration. The chart below shows the amounts of the different species in moles as the titration progresses.
Titrant (ml) Fe2+ Ce4+ Fe3+ Ce3+
0 0.0020 m 0 m 0 m 0 m
5 ml 0.0015 m 0 m 0.0005 m 0.0005 m
10 ml 0.0010 m 0 m 0.0010 m 0.0010 m
20 ml 0 m 0 m 0.0020 m 0.0020 m
30 ml 0 m 0.0010 m 0.0020 m 0.0020 m
40 ml 0 m 0.0020 m 0.0020 m 0.0020 m
It is important to note that except for the first line when no titrant has been added, the concentrations listed as “0 m” are not actually zero. Because the equilibrium constant for the reaction has a finite value, the concentration of each species must be a finite value. However, the equilibrium constant for this reaction is large (>1014), so the redistribution or back reaction of species that occurs involves only a small amount. As an example, let’s consider the values at 5 ml.
5 ml Titrant Fe2+ Ce4+ Fe3+ Ce3+
Initial 0.0015 m 0 m 0.0005 m 0.0005 m
Back reaction 0.0015 + X m X m 0.0005 – X m 0.0005 – X m
Approximation 0.0015 X m 0.0005 0.0005 m
For the line labeled “back reaction”, a small amount (X) of the Fe3+ and Ce3+ are removed to generate some Ce4+ and Fe2+. However, the value of X is so small relative to 0.00050 and 0.0015 moles (again because K is so large) that it is insignificant and can be ignored. Therefore, the amounts shown in the line labeled “approximation” can be used for any calculations.
An important outcome that arises by examining the amounts in the chart above is that prior to the equivalence point (5 and 10 ml of titrant) there are appreciable amounts of both Fe2+ and Fe3+. That means it will be preferable to use the iron half reaction in calculating the junction potential before the equivalence point.
Also note that after the equivalence point (30 and 40 ml of titrant) there are appreciable amounts of both Ce3+ and Ce4+. That means it will be preferable to use the cerium half reaction in calculating the junction potential after the equivalence point.
We can now calculate the values before and after the equivalence point.
5 ml
$\mathrm{E_{Fe}= E_{Fe}^0- \dfrac{0.059}{1} \log \dfrac{[Fe^{2+}]}{[Fe^{3+}]}}$
$\mathrm{E_{Fe}=0.77- \dfrac{0.059}{1} \log \dfrac{[0.0015\: m]}{[0.0005\: m]} =0.74\: V}$
There is one other important thing to note in this calculation. Because we have a ratio of [Fe2+] to [Fe3+] in the log term, the volumes associated with the two concentrations cancel each other out. So using the number of moles of the two species when there are appreciable amounts of both gives the same value as using the concentration.
10 ml
$\mathrm{E_{Fe}= E_{Fe}^0- \dfrac{0.059}{1} \log \dfrac{[Fe^{2+}]}{[Fe^{3+}]}}$
$\mathrm{E_{Fe}=0.77- \dfrac{0.059}{1} \log \dfrac{[0.0010\: m]}{[0.0010\: m]} =0.77\: V}$
Note that in this case, with the concentrations of Fe2+ and Fe3+ being equal, the log of 1 is zero and the potential is equal to EoFe.
30 ml
$\mathrm{E_{Ce}= E_{Ce}^0- \dfrac{0.059}{1} \log \dfrac{[Ce^{3+}]}{[Ce^{4+}]}}$
$\mathrm{E_{Fe}=1.61- \dfrac{0.059}{1} \log \dfrac{[0.0010\: m]}{[0.0020\: m]} =1.59\: V}$
40 ml
$\mathrm{E_{Ce}= E_{Ce}^0- \dfrac{0.059}{1}\log\dfrac{[Ce^{3+}]}{[Ce^{4+}]}}$
$\mathrm{E_{Fe}=1.61- \dfrac{0.059}{1} \log \dfrac{[0.0020\: m]}{[0.0020\: m]} =1.61\: V}$
Note that in this case, with the concentrations of Ce3+ and Ce4+ being equal, the log of 1 is zero and the potential is equal to EoCe.
The only item remaining potential to be calculated is at the equivalence point where we cannot easily use either of the two half reactions to calculate the potential. Using the two half reactions and knowledge of the stoichiometric equivalence of the different species at the equivalence point, It is possible to derive the following generalizable equation for the potential that will be measured at the equivalence point. A and B refer to the two half reactions and nA and nB are the number of electrons in each of the half reactions, respectively.
$\mathrm{E_{Eq.Pt.}= \dfrac{(n_A E_A^0+ n_B E_B^0)}{(n_A+ n_B)}}$
For the reaction in our problem, nA and nB are both 1 so the result is that the equivalence point potential is the average of the two Eo values.
$\mathrm{E_{Eq.Pt.}= \dfrac{(0.77+ 1.61)}{(2)}=1.19\: V}$
The chart below summarizes the potential measured at different points in the titration.
Titrant (ml) Fe2+ Ce4+ Fe3+ Ce3+ Potential
5 ml 0.0015 m 0 m 0.0005 m 0.0005 m 0.74 V
10 ml 0.0010 m 0 m 0.0010 m 0.0010 m 0.77 V (EoFe)
20 ml 0 m 0 m 0.0020 m 0.0020 m 1.19 V
30 ml 0 m 0.0010 m 0.0020 m 0.0020 m 1.59 V
40 ml 0 m 0.0020 m 0.0020 m 0.0020 m 1.61 V (EoCe)
A plot of the potential (y-axis) versus the ml of titrant (x-axis) is shown in Figure 17.
Figure 17. Plot of potential versus ml titrant for the potentiometric titration of Fe2+ with Ce4+. (Figure from Analytical Chemistry 2.0, David Harvey, community.asdlib.org/activele...line-textbook/).
The break in potential that occurs at the equivalence point is apparent in this plot. Note how the general form of this plot is similar to what is observed in a pH titration. This is not surprising since a pH titration is also a potentiometric titration.
One last thing worth examine is the equivalence point potential for a redox reaction where the number of electrons is different in each of the half reactions. For example, suppose Sn2+ was titrated with Ce4+. The two relevant half reactions are shown below.
$\mathrm{Ce^{4+}(aq) + e^- = Ce^{3+}(aq) \hspace{40px} E^o = 1.61\: V}$
$\mathrm{Sn^{4+}(aq) + 2e^- = Sn^{2+}(aq) \hspace{40px} E^o = 0.154\: V}$
In this case, the balanced reaction needs two equivalents of Ce4+ for each Sn2+ to balance out the electrons being transferred.
$\mathrm{Sn^{2+}(aq) + 2Ce^{4+}(aq) = Sn^{4+}(aq) + 2Ce^{3+}(aq)}$
The equivalence point potential is calculated as follows:
$\mathrm{E_{Eq.Pt.}= \dfrac{(n_A E_A^0+ n_B E_B^0)}{(n_A+ n_B)}}$
$\mathrm{E_{Eq.Pt.}= \dfrac{(E_{Ce}^0+ 2E_{Sn}^0)}{(1+2)}}$
$\mathrm{E_{Eq.Pt.}= \dfrac{(1.61+ 2(0.154))}{(3)}=0.639\: V}$
The plot of potential versus ml titrant that would be obtained for this titration is shown in Figure 18.
Figure 18. Plot of potential versus ml titrant for the potentiometric titration of Sn2+ with Ce4+.
It is important to note that the equivalence point in Figure 18 is not symmetrically placed between the two potential plateaus surrounding the Eo value of tin and Eo value of cerium as it was in the reaction of cerium and iron we examined earlier. Instead, the equivalence point potential is weighted toward the Eo value of tin because the tin half reaction requires two electrons whereas the cerium half reaction only requires one. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Electrochemical_Methods_of_Analysis/02_Text/7%3A_Electrochemical_Analytical_Methods/7.4%3A_Titrimetic.txt |
Voltammetry refers to electrochemical methods in which a specific voltage profile is applied to a working electrode as a function of time and the current produced by the system is measured. This is commonly done with an instrument called a potentiostat, which for these measurements is capable of applying variable potentials to the working electrode relative to a reference electrode (like Ag/AgCl) while measuring the current that flows as a result of the electrode reaction. Depending on the particular method, it is possible to apply reducing and/or oxidizing potentials. When a reduction occurs, the current is called a cathodic current. When an oxidation occurs, the current is called an anodic current. Different voltammetric methods involve different voltage profiles. Voltammetric methods are among some of the most common electrochemical methods in use today. There are a variety of voltammetric methods. This unit will only explore three of these methods: anodic stripping voltammetry (ASV), linear sweep voltammetry, and cyclic voltammetry (CV).
Voltammetric methods typically involve the use of microelectrodes that frequently have areas on the order of 0.3-10 cm2. Originally it was common to use mercury electrodes often as a hanging mercury drop (HMDE) or as drops through a glass capillary (DME) for voltammetric methods. Mercury had several desirable properties in electrode applications. One advantage of mercury is that it has a high overvoltage toward the reduction of H+ so it can be used at high reducing potentials in water without leading to the electrochemical splitting of water into hydrogen and oxygen gas. A second advantage of mercury electrodes is that metals dissolve in mercury by forming amalgams, which improves the measurement of low concentrations of analytes. A concern with electrodes is that mercury can become fouled or the surface of a solid electrode can become poisoned, which significantly alters their properties. This can occur if species in the matrix adsorb to the surface of the electrode. Solid electrodes usually are put through a prescribed polishing procedure before used for measurement purposes, while mercury drops can easily be replaced through a glass capillary.
The use of mercury electrodes has fallen into disfavor today because of the toxicity of the metal and the difficulty of controlling spills of the material when used in electrode applications. Other possible electrodes for use in voltammetric methods include carbon paste, glassy carbon, platinum and gold. In addition, with a glassy carbon electrode it is possible to create a thin film of mercury on the electrode by reducing Hg2+. This provides the advantages of mercury electrodes described earlier without the necessity for using large quantities of mercury.
Today, ultramicroelectrodes have been fabricated with dimensions on the um scale or smaller. These are so small they can be inserted into single cells and used in certain voltammetric methods to probe chemical processes taking place inside the cell.
An important consideration in voltammetric methods is the process by which species in solution move up to the surface of the electrode, where if electroactive, they can be oxidized or reduced. There are three general processes by which ions move in solution: diffusion, convection and electrostatic migration. Diffusion involves the random motion of species in solution. Convection is physical movement created by something like a magnetic stir bar. Electrostatic migration results from the attraction of a positive species towards a negatively charged electrode or alternatively the attraction of a negative species toward a positively charged electrode. Diffusion will always occur in a solution. In some methods, it may or may not be desirable to have convective movement or electrostatic migration.
Example $1$
How can electrostatic migration be eliminated in an electrochemical cell?
Solution
The key to answering this question is to realize that ions in solution must somehow be shielded from the charge on the electrode. The addition of a large concentration of an inert electrolyte such as potassium chloride will accomplish this goal. For example, for a negatively charged cathode, K+ ions will be adjacent to the electrode and throughout the solution. Analyte cations that are in the solution and away from the surface of the electrode will be shielded from the charge on the electrode by all the K+ ions in between.
Example $2$
Samples subjected to voltammetric methods are usually purged with nitrogen or other inert gas before the analysis and maintained under an inert atmosphere during the analysis. Why is this done?
Solution
Most aqueous solutions have reasonable levels of dissolved oxygen gas (O2). Oxygen gas has a relatively low reduction potential and if not removed from the solution will generate a current that will interfere with the measurement. An inert gas used to purge the solution will displace the dissolved oxygen. Maintaining the solution under an inert atmosphere will prevent oxygen from the air from dissolving back into the solution.
7.5.1. Anodic Stripping Voltammetry (ASV)
In ASV, a potential sufficient to reduce the metal of interest is applied to the working electrode for 1-30 minutes in the solution being analyzed to plate out or deposit the metals onto the surface of the working electrode. The lower time is used for samples with metal concentrations of 10-7 M or more. The longer time is used with samples with concentrations as low as 10-11 to 10-12 M. After the deposition step, the voltage is scanned towards more positive potentials. The voltage profile for ASV as a function of time is shown in Figure 19. The solution is stirred during the deposition period in ASV to bring the species in solution up to the surface of the electrode to be reduced, but not stirred during the stripping step.
Figure 19. Potential versus time profile used in anodic stripping voltammetry.
Suppose you are analyzing a solution that contains Cd2+ and Pb2+ using ASV. You apply a potential of –1.0 V for several minutes to plate out both metals. You then scan to more positive potentials in the stripping step. Draw a generalized plot of the current you would observe as a function of time during the deposition step and during the stripping step.
The first thing to consider is the relevant half reactions for the metals being deposited out of solution.
$\ce{Cd^{2+}(aq) + 2e^{–} <=> Cd(s)} \mathrm{\hspace{40px} E^o = - 0.403\: V}$
$\ce{Pb^{2+}(aq) + 2e^{–} <=> Pb(s)} \mathrm{\hspace{40px} E^o = - 0.126\: V}$
Note that cadmium, by virtue of its more negative Eo value, is more difficult to reduce and plate out or deposit than lead. However, during the deposition period, the reducing potential should be set negative enough to deposit both metals onto the electrode. In considering what the plot of current versus time would look like during the deposition period, it might first be tempting to consider a situation similar to what we observed in coulometry (Figure 13). In the coulometric experiment, the current starts high and diminishes as the metal ions are deposited from solution. However, coulometry involves the use of large electrodes and a complete deposition or reaction of the species being analyzed. ASV involves microelectrodes so only a small fraction of the total metal ions in solution will be deposited onto the electrode. Provided the solution is stirred, the result is that a constant current occurs during the deposition step of ASV.
At the initial stage of the potential sweep towards more oxidizing values, the potential is still negative enough to deposit both metals. At some point the voltage will approach the Eo value of – 0.403 Volts and which point deposition of the cadmium ends and the cadmium that has been reduced and deposited onto the electrode will be oxidized. In the deposition step, electrons flowed from the electrode to the metal ions in the solution. In the oxidation of cadmium, the electrons flow in the opposite direction. Whereas in the deposition step, metal ions needed to be brought to the electrode to be reduced, in the oxidation step all of the metal is already deposited onto the electrode. This means that all of the deposited cadmium oxidizes almost instantaneously (and is said to be stripped from the electrode). The result is to produce a large peak of current in the oxidation step as seen in Figure 20. As the reducing potential continues to ramp down, it eventually approaches – 0.126 Volts and the lead that has been deposited onto the electrode will oxidize and be stripped from the electrode. This produces a second peak of current as seen in Figure 20.
Figure 20. Plot of current versus potential for the stripping portion of an anodic stripping voltammogram of a mixture containing cadmium and lead.
What feature of the plot can be related to the concentration of the metal?
The area or height of each peak can be related to concentration using a set of standards.
What advantage(s) does anodic stripping voltammetry offer over coulometry or electroplating?
The most significant advantage of ASV over methods such as coulometry or electroplating is its unprecedented sensitivity. Looking at the plot of current versus time in Figure 20, it is important to recognize that the total area under the two peaks resulting from the two oxidation steps must be equal to the total area under the curve that resulted during the reduction step (since all the electrons that went into reducing Pb2+ and Cd2+ must be released back during the oxidation step). This means that the current peaks for the two metals will be quite intense, meaning that the method can be used to detect low concentrations of metal ions. The sensitivity can be further increased by using larger electrodes and longer deposition times, effectively concentrating more metal atoms onto the electrode. ASV can be used to measure metal concentrations as low as 10-12 M, which is an exceptionally low level (superior to flame atomic absorption spectroscopy and comparable to graphite furnace atomic absorption spectroscopy). ASV can therefore be used for ultratrace level analysis of certain metal ions.
A second advantage of ASV over coulometry or electroplating is that more than one metal species can be analyzed in a sample. The reducing step used in coulometry, electroplating or ASV does not distinguish between different species unless you are able to select the right potential to only reduce a single species. However, the oxidation step in ASV produces a separate peak for each metal plated onto the electrode.
An advantage that coulometry has over ASV is that it can be used in either a reducing or oxidizing mode and can be used with half reactions where both the reactants and products are soluble species. ASV requires a product that plates onto the electrode so is not as versatile as coulometry.
7.5.2. Linear Sweep Voltammetry
The voltage profile in linear sweep voltammetry involves a linear scan of either a reducing or oxidizing voltage (Figure 21). We will illustrate the features of this method by examining the current that is measured when a reducing potential is applied to a two component mixture.
Figure 21. Potential versus time profile used in linear sweep voltammetry.
Draw the current that would be measured for a solution consisting of Cd2+ and Zn2+ in linear sweep voltammetry. Assume that the concentration of Cd2+ is about twice as large as that of Zn2+ and that the solution is stirred.
First we need to write the two relevant half reactions that will occur.
$\mathrm{Cd^{2+}(aq) + 2e^– = Cd(s) \hspace{40px} E^o = - 0.403\: V}$
$\mathrm{Zn^{2+}(aq) + 2e^– = Zn(s) \hspace{40px} E^o = - 0.763\: V}$
Comparing the Eo values indicates that cadmium is more easily reduced than zinc. As the voltage ramps to more negative values, it will reach a point where any Cd2+ species at the surface of the electrode will be reduced and a current will be observed. Because the solution is being stirred, fresh Cd2+ will be forced up to the electrode at a constant rate, thereby sustaining the current. Since this technique uses microelectrodes, as previously discussed in ASV, there will be no diminishment of current due to the depletion of Cd2+. As the reducing voltage continues to increase, eventually it will reach a value sufficient to reduce the Zn2+. Beyond this point, both Cd2+ and Zn2+ are being reduced at a constant rate and the overall current is the sum of both reductions. The plot of current versus time for the linear sweep voltammogram of this solution is shown in Figure 22.
Figure 22. Plot of current versus potential for a linear sweep voltammogram of a mixture containing cadmium and zinc. The solution is stirred.
Something to note in the plot in Figure 22 is that baseline rises slightly as potential is increased before the reduction of Cd2+ begins. One reason for the rising baseline is something known as the capacitive current. Capacitive current occurs because cations in solution form a layer adjacent to the negatively charged electrode as shown in Figure 23. The separated layers of charge act as an electronic device known as a capacitor. The more negative the electrode potential becomes, the more positive charge that can be stored at its surface. Another reason a slight rise in the baseline may occur is if there are trace amounts of impurities in solution that undergo an electrochemical reaction.
Figure 23. Alignment of cations in solution adjacent to the surface of a negatively charged electrode. The two separated layers of charge form something analogous to a capacitor.
What feature of the plot can be related to concentration?
The magnitude of the current generated for each of the reduction steps, which is seen as the height of the plateau, can be related to the concentration. Because the concentration of Cd2+ and Zn2+ in the solution was about twice that of zinc and because both are two electron reductions, the height of the plateau for cadmium in Figure 22 is higher than that of zinc. However, the relationship between the height of the plateau and the concentration for each particular species would need to be calibrated against a series of standards with known concentrations of each metal.
What feature of the plot can be used for species identification?
What is important here is that the different species are reduced at different applied potentials. The value used for species identification is known as E1/2, which is the potential at the half-way point up the reduction step. The E1/2 values for the reduction of cadmium and zinc are shown in Figure 22 and occur at an inflection point in the plot.
For the exact same solution of Cd2+ and Zn2+, draw the current that would be observed if the solution is not stirred.
Stirring the solution insures that ions continue to reach the electrode surface to maintain the current. If the solution is not stirred, when the reduction potential for an ion such as Cd2+ is reached, any Cd2+ at the electrode surface will be reduced so a rise in current comparable to that in Figure 22 will occur. Once these species are reduced, there are two ways additional Cd2+ can make it to the electrode to be reduced. One is through the process of diffusion. The other is that the electrode does have a negative charge and there can be an electrostatic attraction of Cd2+ for the electrode.
We already mentioned earlier that electrostatic attraction of species for the electrode can be eliminated by adding a high concentration of a background electrolyte to the solution (e.g., potassium chloride, K++ and Cl). It is often desirable to run these processes in a manner that is under diffusion control. In such a case, diffusion will not cause enough Cd2+ to strike the electrode to maintain a constant current. Therefore the current will begin to drop after the initial reduction step. It will not drop to zero because diffusion of Cd2+ does occur and some current is maintained. The plot in Figure 24 shows an example of the plot that would result for a solution of Cd2+ and Zn2+ if it were run without stirring the sample.
Figure 24. Plot of current versus potential for a linear sweep voltammogram of a mixture containing cadmium and zinc. The solution is not stirred.
7.5.3. Differential Pulse Linear Sweep Voltammetry
In differential pulse linear sweep voltammetry, a voltage profile similar to a linear sweep experiment is applied but then a small, increased voltage pulse is superimposed over the voltage ramp (Figure 25). In this method the current is sampled twice: (1) just before the pulse is applied and (2) just before the pulse ends. The two measurements are subtracted [(2) – (1)] and the difference is plotted.
Figure 25. Potential versus time profile used in differential pulse linear sweep voltammetry.
Draw the resulting plot for the same sample shown in Figure 22 (a solution consisting of Cd2+ and Zn2+; the concentration of Cd2+ is about twice that of Zn2+, and the solution is stirred).
In order to understand what happens here it is useful to look at the plot in Figure 22 and then consider where the measurements are made for each (1) and (2) in Figure 22. Figure 22 is reproduced as Figure 26a and examples of measurements (1) and (2) for some different pulses are shown on the plot. By applying a short pulse, the measuring of current at point (2) is essentially jumping ahead a little bit on the linear ramp. If we look at the flat portion of the plot early in the ramp, there is no change in current going from (1) to (2) so the difference between the two points is zero. If we look later on during the reduction of Cd2+ at the electrode, since there is a large jump in the current, subtracting (1) from (2) gives a large positive value. If we consider a pulse after the inflection point in the curve (E1/2) there is still a large difference between the current at points (1) and (2) but it begins to diminish. If we examine the plateau of current between the point where Cd2+ is reduced and Zn2+ has not yet started to reduce, there is no change in current between points (1) and (2) for a particular pulse so the difference is now back to zero. The plot shown in Figure 26b shows the result that would be obtained using the voltage profile shown in Figure 25. In effect, the measurement done using this profile is taking the derivative of the curve, hence the name differential pulse linear sweep voltammetry.
Figure 26. Plot of current versus potential for (a) a linear sweep voltammogram and (b) a differential pulse linear sweep voltammogram of a mixture containing cadmium and zinc. The points (1) and (2) indicate some examples of where current readings are taken before and after a pulse.
What advantage(s) does differential pulse linear sweep voltammetry have over conventional linear sweep voltammetry?
There are two important advantages of the differential pulse method over conventional linear sweep voltammetry. The first is that it is much more sensitive. The differential pulse method provides a sharp peak for each species being analyzed and it is much more reliable to measure either the height or area of this peak and relate it to concentration than it is to measure the height of the current in the linear sweep mode. This can be especially appreciated at low concentrations where a small peak in the differential pulse mode would be much easier to measure than only a small change in the height of the current (see Figure 27).
Figure 27. Plot of current versus potential for a linear sweep voltammogram and differential pulse linear sweep voltammogram for a mixture containing a high concentration of one metal and small concentration of another.
The second is that it is much easier to distinguish two species with similar E1/2 values. An example of this is shown by the linear sweep and differential pulse results in Figures 28, respectively. Whereas the linear sweep method requires a difference of about 0.2 V in the E1/2 values to distinguish two species, the differential pulse method only requires a difference of about 0.05 V.
Figure 28. Plot of current versus potential for a linear sweep voltammogram and differential pulse linear sweep voltammogram for a mixture containing two metal ions with similar E1/2 values.
7.5.4. Cyclic Voltammetry
The last method we will develop is known as cyclic voltammetry (CV). In CV, scan rates can typically range from 10 mV/s to 100,000 V/s with an increase in the applied potential followed by a return back to the starting voltage as shown in Figure 29. This can be performed as an anodic or cathodic scan depending on the system being studied. For the examples herein, we will consider a cathodic scan that causes in the reduction of species in the initial solution.
Figure 29. Applied potential vs. time profile used in cyclic voltammetry.
Draw a plot of current (y-axis) versus voltage (x-axis) that you would measure for a solution in which the Fe3+ in ferricyanide (Fe(CN)63–) is reduced to Fe2+ during the reducing phase of the CV voltage profile shown in Figure 29. The half reaction is chemically and electrochemically reversible and the solution is not stirred.
Prior voltammetric methods have given us the ability to understand the plot that will result for CV. The first half of the voltage profile (i.e., the forward sweep of potential) is essentially a linear sweep voltammogram. The second half is similar to the profile used in the latter stripping portion of ASV, although as we will see, the output usually looks quite different than the output in ASV because nothing has plated out of solution. Note, in the example given in the problem, the two species involved in the reaction (Fe2+, Fe3+) are water soluble and neither deposits out onto the electrode.
During the forward sweep, Fe3+ will be reduced to Fe2+ and a current (the forward wave) will be produced as the Fe3+ at the surface of the electrode is reduced (Location 1 in Figure 30). Beyond this point and since the solution is not stirred, the current falls because Fe3+ cannot diffuse up to the electrode fast enough to maintain a constant current (Location 2 in Figure 30). As the potential returns to the initial value it eventually reaches the potential where Fe2+ that remains next to the electrode can be oxidized back to Fe3+. This process results in an anodic current that appears as the reverse wave (Location 3 in Figure 30). For an electrode reaction that is electrochemically reversible (i.e., fast electron transfer meaning no overpotential), the peak of the forward wave will align with the E1/2 value of the reverse wave and the peak of the reverse wave will align with the E1/2 value of the forward wave as seen in Figure 30. The formal potential for a redox couple (Eo) can be calculated by taking the average of the peak potentials of the forward and reverse wave. Provided none of the Fe2+ formed in the reduction step has been lost in a subsequent chemical reaction (Figure 31), under diffusion controlled conditions where the equilibrium concentrations determined by the Nernst equation are maintained, the reverse wave should have the same peak current as the forward wave.
Figure 30. Plot of current versus potential for a cyclic voltammogram of a redox couple with an uncomplicated electron transfer and a diffusion-controlled chemically reversible redox reaction.
Figure 31. Representation of the solution near the electrode before and after the reduction step in a cyclic voltammogram.
The output from a method such as CV does depend significantly on a number of factors and the magnitude of the peak current is described more fully using what is known as the Randles-Sevcik equation. Factors such as the number of electrons transferred in the redox reaction, the electrode area, the diffusion coefficient for the species being analyzed, the concentration of the species being analyzed and the potential scan rate all influence the peak current. The reader is referred to the following resource for a discussion of the Rangles-Sevcik equation:
https://chem.libretexts.org/Core/Analytical_Chemistry/Analytical_Sciences_Digital_Library/JASDL/Courseware/Analytical_Electrochemistry%3A_The_Basic_Concepts
The value of CV is not so much for quantitative analysis but for the analysis of mechanistic properties of electrochemical reactions. Sometimes it is advantageous to do more than one scan when using CV to examine reaction mechanisms. If we consider the chemically and electrochemically reversible electrode reaction discussed above, a second and third voltage scan would produce plots identical to that shown in Figure 30. However, if the product of the initial electrochemical reaction underwent a further reaction, it may produce an electrochemically inactive product that is not observed in the CV, or an active one that is observed in subsequent scans. Let’s consider some other examples of outcomes of CV.
Draw a plot that could be obtained for an electrochemical reaction that is chemically irreversible and forms an electrochemically inactive product.
In this case, there would be no reverse wave. Assuming the process involved applying a reduction potential, the product of the reduction would not be able to be oxidized as the voltage came back down the ramp. The current vs. potential plot obtained in this situation is shown in Figure 32.
Figure 32. Plot of current versus potential for a cyclic voltammogram of a chemically irreversible redox couple in which the product that is formed is electrochemically inactive.
Draw the plot of current vs. potential that would be obtained for a reversible chemical reaction in which only the reverse reaction has an overpotential, but the potential eventually is sufficient to complete the reverse reaction.
In this case, there would be a reverse wave but because of the overvoltage it would be offset from the forward wave as shown in Figure 33. Another difference is that the slower electron transfer occurring in the case of an electrochemically irreversible reaction would lead to a flatter profile on the reverse wave.
Figure 33. Plot of current versus potential for a cyclic voltammogram of a redox couple that is chemically reversible but electrochemically irreversible because of an overpotential on the reverse reaction.
Propose a reaction mechanism that would explain the following cyclic voltammogram. The first voltage cycle is shown as a solid line. The second voltage cycle is shown as a dotted line. A third voltage cycle gives the same output as the dotted line. Assume that a reducing potential is applied.
Figure 34. Plot of current versus potential for a cyclic voltammogram. Solid line is the first scan. Dotted line is the second scan.
The first cycle indicates that a species was reduced. However, the reaction is chemically irreversible as there is no reverse wave. The second cycle shows a new species showing up in the cyclic voltammogram. Presumably the product formed from the reduction either decomposed or further reacted to produce some new species. This new species is electrochemically active so a new forward wave shows up in the second cycle. The electrochemical reaction of the new species is reversible so a reverse wave for it also occurs in the second cycle. The new species likely has a smaller wave than the original species in solution because in this case some of the product, which was produced through a chemical reaction, will diffuse away from the electrode. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Electrochemical_Methods_of_Analysis/02_Text/7%3A_Electrochemical_Analytical_Methods/7.5%3A_Voltammetr.txt |
Overall Unit
After completing this entire unit on electrochemical methods of analysis, a student will be able to:
1. Describe each of the electrochemical methods discussed in the unit.
2. Perform quantitative calculations for each of the methods if provided appropriate information or data.
3. Compare the advantages and disadvantages of the different electrochemical methods of analysis.
4. Select the best method from among those covered in the unit for a particular analysis and justify the choice.
Problem set titled “Electrochemistry”
After completing this problem set, a student will be able to:
1. Define the processes of oxidation and reduction.
2. Define the meaning of an oxidizing and reducing agent.
3. Describe what is meant by a half-reaction.
Problem set titled “Chemical Energy”
After completing this problem set, a student will be able to:
1. Relate chemical energy to electrochemical potential.
2. Relate the sign of the electrochemical potential to the direction of a reaction.
3. Define the standard state and explain why it is necessary to define a standard state.
Problem set titled “Electrochemical Cells”
After completing this problem set, a student will be able to:
1. Describe the components of an electrochemical cell, the purpose of each, and the process that occurs in each.
2. Write a shorthand notation for an electrochemical cell.
3. Describe the processes responsible for conduction of electricity in different part of an electrochemical cell.
4. Describe what it means for a reaction to be chemically and electrochemically irreversible.
Problem set titled “Use of the Nernst Equation”
After completing this problem set, a student will be able to:
1. Calculate the potential of an electrochemical cell under standard and non-standard state conditions.
2. Calculate the equilibrium constant for an electrochemical reaction.
Unit on Ion Selective Electrodes
After completing this problem set, a student will be able to:
1. Describe the functioning of an ion selective electrode.
2. Describe the limitations in the use of ion selective electrodes.
Problem set titled “Electrodeposition/Electrogravimetry”
After completing this problem set, a student will be able to:
1. Describe how the method of electrogravimetry is performed.
2. Determine if one species will interfere with another in electrogravimetry.
Problem set titled “Coulometry”
After completing this problem set, a student will be able to:
1. Describe how the method of coulometry is performed.
2. Compare the advantages and/or disadvantages of electrogravimetry and coulometry.
Problem set titled “Coulometric Titration”
After completing this problem set, a student will be able to:
1. Describe how a coulometric titration is performed.
2. Describe the advantages of a coulometric titration over a conventional redox titration.
Problem set titled “Amperometric Titration”
After completing this problem set, a student will be able to:
1. Describe the process of performing an amperometric titration.
2. Draw the resulting plot of current vs. titrant volume for different scenarios in an amperometric titration.
Problem set titled “Potentiometric Titration”
After completing this problem set, a student will be able to:
1. Calculate the potential at various stages of a potentiometric titration.
Problem set titled “Voltammetric Methods”
After completing this problem set, a student will be able to:
1. Describe how it is possible to eliminate electrostatic migration be eliminated in an electrochemical cell.
2. Describe why it is common to purge samples with an inert gas and conduct the analysis under an inert gas environment when performing voltammetry.
Problem set titled “Anodic Stripping Voltammetry”
After completing this problem set, a student will be able to:
1. Describe how anodic stripping voltammetry (ASV) is performed.
2. Draw the output of an ASV analysis if given the identity of metal species being analyzed.
3. Explain the advantages of ASV over methods such as electrogravimetry and coulometry.
Problem set titled “Linear Sweep Voltammetry”
After completing this problem set, a student will be able to:
1. Describe how linear sweep voltammetry (LSV) is performed.
2. Draw the output of an LSV analysis for samples that are stirred or not stirred if given the identity of the species being analyzed.
3. Explain how concentration and species identication is performed in LSV.
4. Explain how differential pulse LSV is performed.
5. Draw the output obtained from a differential pulse LSV analysis.
6. Explain the advantages of differential pulse LSV over LSV.
Problem set titled “Cyclic Voltammetry”
After completing this problem set, a student will be able to:
1. Describe how cyclic voltammetry (CV) is performed.
2. Draw an example of the output in CV for the analysis of reversible, chemical irreversible and electrochemically irreversible reactions.
3. Provide a reaction mechanism that would explain the output of a cyclic voltammogram. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Electrochemical_Methods_of_Analysis/03_Learning_Objectives.txt |
INTRODUCTION
The problem sets on electrochemistry can be used in at least two different manners. The primary intent is to use these as a set of in-class, collaborative learning exercises. Groups of 3-4 students work together in discussing and working through the problems. When using the problem sets in this manner, the instructor must actively facilitate and guide students through the material. This manual will guide instructors through each of the problem sets, identifying possible student responses to the questions and the response and activities of the instructor during the progression of the problem.
An alternative to the use of the problems in class is to assign them as out-of-class activities, preferably done as a group activity among students or as a peer-led learning activity. The accompanying text that goes with each problem provides a detailed discussion of each step of the thought process of solving it, such that students could work back and forth between the problem and text on an iterative basis to gain an understanding of the material.
There is no perfect way to assemble groups for such collaborative learning activities. I gather information on the first day of class (year in college, major, prior chemistry courses) and then use this to set groups of 3-4 students that start on the second day of class. I try to make the groups as heterogeneous as possible and they work together for the entire semester. Another strategy is to assign groups for a shorter period of time that might encompass completion of a specific topic or unit, and to then create new groups for the next unit. One other possibility is to have different groups every day of class. Since it is important for groups to work well together, having new groups every day may be less successful than allowing groups to work together for more extended periods of time. I would recommend that the instructor assign groups rather than allowing the students to pick their own. This avoids the potential problem of friends who want to be in the same group but who then do not work well together or stay focused on the assigned task. It also avoids the problem of the student who is left without a group at the end of the selection process, something that can be especially problematic if it is a member of a minority group. When using collaborative groups, it is also important for the instructor to monitor the functioning of the groups and to step in to address either dysfunctional groups or the recalcitrant individual within a group. Peer-evaluation processes are often used by instructors who employ group activities as a way of assessing how well groups are working.
I also expect the groups to meet outside of class for any homework assignments, something that is aided because I am at a residential college. An alternative to this is to schedule a room on the evening before a homework assignment is due and encourage them to come to this place and work in any arrangement they wish on the homework. I have run such sessions for several years now and attend them as a facilitator (one result is that it has cut down considerably the individual traffic to my office seeking help on the homework problems) and it has been an excellent way to promote collaboration among the students.
The instructor has an especially important role to fulfill during such group activities. I have observed that the more engaged that I am in the process in helping to guide the students through the material, the more effective the learning that occurs. In most instances, it seems that the students are initially stumped by the question, that they begin to explore things that they do know that might apply to answering the question, and that help from the instructor either by letting them know that they are on the right track or by suggesting another direction in which to take their thinking is necessary. As they begin a question, I roam around the room listening in on conversations and looking over their shoulders at what might be written in their notebook. If I hear something interesting, I indicate that to the group. If I see that someone has written something interesting and relevant in their notebook, I tell other group members that they ought to talk with this individual about what they have written, and that the individual should explain to the other group members why they wrote that down. If I hear a group going entirely in the wrong direction, I probe them on why they are heading in that way and then offer suggestions about things to consider that will set them off in the right direction. When all groups have realized an important point, I call time out and summarize the concept at the board. Then I send them back to continue with the next part of the problem. Most of the problems are handled in such an iterative manner where the students work through some important part of the problem, I summarize it at the board when they have developed the concept, and then they return to the next part of the problem. Occasionally a group will just not see something, whereas every other group has gotten the point, and it may require a direct intervention from the instructor with that group to explain the concept. Similarly, there are times when I call their attention to the board to summarize a point when one of the groups still has not gotten the concept but waiting would slow down the remainder of the class to an unacceptable level.
When using these materials, I want the students to discuss and discover the concepts inherent in the problems, so they do not have the text when working on the problems. After they have completed a particular problem, I then put an electronic copy of that portion of the text on the class site. The text thoroughly goes through the thought process for solving each problem and I encourage the students to read it over that evening to reinforce the concepts developed in class that day. I also give homework problems designed to reinforce the concepts developed in class.
ELECTROCHEMISTRY MODULE
I provide a brief introduction to the idea that two electrochemical processes can be used for analytical purposes. One is on the ability of chemical species to transfer electrons. The other is our ability to measure junction potentials. After this, I introduce the first problem set.
Define what is meant by oxidation and reduction.
Define what is meant by an oxidizing and reducing agent. Give a good example of each.
Define what is meant by a half-reaction.
The answers to these three questions have been thoroughly covered in general chemistry and, in the case of the second one they have experience from organic chemistry, and students readily answer all three.
Give an example of a half-reaction and determine whether a half-reaction can be an equilibrium expression. Why or why not?
I prompt the students by providing a particular half reaction (usually the reduction of Cd(II) to cadmium metal) asking them to write the equilibrium constant. Some of them initially want to put the cadmium metal in the expression and it leads to a question about whether the electrons go into the expression. We then discuss how electrons would need to be included to be correct and we can’t just weigh out a bunch of electrons to add to a reaction. Then I do talk about how we will often write equilibrium-like expressions for half reactions where we ignore the electrons – pointing out that one half reaction can never occur without the presence of another so there is a source of those electrons.
CHEMICAL ENERGY
I find that it helps to review the concept of chemical energy as a way of relating it to electrochemical potential and deriving the Nernst Equation and do so using the following set of problems. The only information I provide them in advance of this problem set is reminding them that in general chemistry they discussed the importance of something known as the chemical or Gibbs energy.
Now let's think about chemical energy (otherwise known as the Gibbs energy and denoted by G). A bottle of pure A has some amount of chemical energy. A bottle of pure B has some amount of chemical energy. A solution of A in water that is 2 Molar has some amount of chemical energy as does a solution of B in water that is 2 Molar.
Do you think that those four examples have the same or different chemical energy?
Intuition tends to tell the students that these four situations are so different from each other that they likely have different chemical energy.
How would you measure or determine the absolute chemical energy of those four systems?
Students are usually stumped by this question. The often remember something called ΔG or more commonly ΔGo and remember looking up these values in tables. I ask about the meaning of Δ and they realize that this must represent a difference of some sort. I point out that the question is not asking about ΔG but is asking about measuring G. I then indicate that we do not actually know how to calculate G for a chemical system and the best we can ever do is measure ΔG.
Now consider a situation in which A reacts to produce B.
$\mathrm{A \rightleftharpoons B}$
When will the chemical energy of that system achieve its lowest value?
I first point out that a keto-enol isomerism is an example of the simple reaction of A producing B. Most groups tend to have one or more students who realize that the lowest chemical energy likely occurs when the system is at equilibrium.
Draw a plot of G versus the mole fraction of A and B in which the situation
$\mathrm{[A] + [B] = 2\: Molar}$
is always satisfied, and the reaction of A to produce B has a fairly large equilibrium constant.
I fully explain these plots at the board so they understand the X-axis notation and then allow them to discuss the question and draw pictures. Sometimes I need to remind them to write the equilibrium constant expression. While they often do not have plots that are exactly correct, they realize that the values of G at [A] = 2 M and [B] = 2 M have to be higher than the value at equilibrium and that the minimum value for the plot should be closer to [B] = 2 M.
What would the plot look like if the reaction had a fairly small equilibrium constant?
After developing the previous question, students can readily answer this one.
Why do we need to define something called the standard state?
The students can usually reason out that the value of ΔG will depend on what you choose as your starting conditions. If not, I prompt them to think of the problem of asking different students to actually go into the lab and conduct a measurement of ΔG for the reaction with no additional instructions about how to do it. That gets them to realize that without specific instructions different people will choose different concentrations of the chemicals.
What constitutes the standard state?
They usually can answer this, although they may not fully realize that it involves all of the chemical species in the solution starting at a concentration of 1 M. We also discuss how it could be challenging to experimentally set up such a solution since the species involved in the reaction will often immediately react to reach equilibrium. I then point out how electrochemical reactions are rather unique because the two half cells can be set up at standard state conditions but no reaction will take place until the circuitry is completed. I also point out that the difference in chemical energy between the standard state and equilibrium is referred to as ΔGo.
What does it mean for the difference in chemical energy (ΔG) to be positive or negative?
Students usually remember that a negative value of ΔG means that the reaction proceeds toward products and a positive value means it proceeds toward reactants. I like to point out that they should notice that the value of G goes down in both cases, and it is by convention that we assign a negative value to ΔG when the reaction proceeds toward products.
At this point I remind them of two equations they developed in general chemistry.
\begin{alignat}{3} &\Delta \ce G^\circ &&= \ce{-RT\ln K}\ &\Delta \ce G &&= \ce{-RT\ln K + RT\ln Q} \end{alignat}
I point out how –RTlnK is the difference in chemical energy between the standard state and equilibrium and RTlnQ is the difference in chemical energy between the non-standard state conditions and the standard state. It is necessary to ask and remind them of what Q is in the second equation.
I also show them using arrows on a diagram (like the one they drew above) representations for the magnitude of ΔGo and several examples of the magnitude of ΔG under different starting conditions (e.g., if the reaction favors products, starting conditions further from equilibrium than the standard state, closer to equilibrium than the standard state, and on the other side of the equilibrium conditions such the reaction would actually proceed in the reverse direction).
I then provide them with two additional equations for electrochemical reactions, indicating that for an electrochemical reaction we can measure something called the electrochemical potential and that this relates to the Gibbs free energy.
\begin{alignat}{3} \ce{&\Delta G^{\circ} && \:=\: -nFE^{\circ}\ &\Delta G &&\:=\: -nFE} \end{alignat}
We then combine the equations relating the Gibbs energy to electrochemical potential to the equations above relating the Gibbs energy to K and Q with the result that we are able to derive an expression relating Eo to K and E to K and Q, the latter of which is the Nernst Equation.
At this point, I introduce the table of standard state potentials. I point out how we can also only measure relative electrochemical potentials and that the reduction of H+ to hydrogen gas has been arbitrarily assigned a value of 0. I ask them to look at both the magnitude and signs of the Eo values for the reduction of Li+ and fluorine gas. Since they know that lithium prefers to be Li+ and fluorine prefers to be F, they realize that values that are negative (as for the reduction of Li+) means that the oxidized form is favored and values that are positive (as for the reduction of fluorine gas) means that the reduced form is favored.
Finally, I mention how things like the ionic strength of a solution and acid used in reactions where H+ is present influence the magnitude of Eo and that in some cases it is common to use something known as the formal potential (Eo‘).
ELECTROCHEMICAL CELLS
The students are given the following questions on the nature of an electrochemical cell without any preliminary information.
Describe what you know about an electrochemical cell.
The students usually remember aspects of the classic electrochemical cell consisting of a copper and zinc half-cell. They can usually draw a representation of the cell, remember that there is an anode and cathode, and that a wire and salt bridge connect the two half cells.
After they have worked on this for a few minutes, I draw a representation for the copper/zinc electrochemical cell on the board and ask them to look at the table of standard state potentials to identify which half-cell is the cathode and which is the anode.
What processes are responsible for conduction of electricity in an electrochemical cell?
They usually are able to identify that it is electrons in the wire. They may need some prompting to ask what is in solution to realize that it is ions in solution that are able to conduct electricity.
What is the purpose of a salt bridge? What would you put inside a salt bridge?
They often remember what is inside a salt bridge (although they may not have the exact right species) but often don’t have a clear recollection of why a cell has a salt bridge other than to complete the circuit. If this is the case, I ask them to consider what happens to the relative concentrations of positive and negative ions as the reaction proceeds. They are able to determine that the copper side will have more negative sulfate ions than position copper ions as the reaction proceeds and that the zinc side will have more positive zinc ions than negative sulfate ions as the reaction proceeds. Often without prompting, they recognize that the cells won’t want to have such a charge buildup. If they don’t state this, I prompt them with a question about whether this is a desirable situation to occur.
With that information, they realize that the salt bridge must be a source of ions that can migrate out into the solutions to balance the charge of each half-cell.
I then describe the nature of the junction potential that would build up in the cell without the salt bridge and how a junction potential is inhibitory to the cell proceeding to equilibrium.
When considering what to put in a salt bridge, I provide them with NaCl and KCl as two possibilities and ask if they could see a reason why one would be preferable to the other. Usually one or more students in each group will suggest that KCl might be preferred because of the similar size of the two ions.
I then ask whether you would want a high or low concentration of the salt in the salt bridge and they quickly conclude that a high concentration would be preferable as it could overcome a larger charge buildup or work for a longer time in an electrochemical cell.
Describe two types of situations that would result in the irreversibility of an electrochemical process.
Before considering this problem, I introduce the concept of a voltaic and electrolytic electrochemical cell and the idea that applying a high enough external voltage allows one to drive an electrochemical cell back away from equilibrium as is done when charging a battery.
Students are usually stumped at figuring out a situation when an electrochemical reaction would be reversible. Sometimes one or more groups will propose a situation where the reaction produces a gas and the gas escapes. I discuss with them how if the cell were designed to trap the gases, then the process could be reversed. I then ask them whether they can envision another situation where the produce of the reaction goes away and they usually propose a reaction where the product either further reacts with something or decomposes.
The idea of an overpotential is completely new to them so to get them thinking about this, I ask what needs to happen at each of the electrode surfaces in our example cell, but to especially consider the cathode. They can usually determine that an electron needs to leave the solid electrode to transfer to an ion in solution. I then talk about how the electron must “jump” from the solid to the ion in solution and that there are cases where there is a resistance of the electron to jumping. I then ask them what they might do to make the electron jump and they usually propose to increase the voltage.
Finally, I introduce the shorthand notation that is used to describe an electrochemical cell.
USE OF THE NERNST EQUATION
Since we have derived this, I give them the following set of questions without any significant introduction and allow the groups to think how to go about solving it.
Potassium dichromate reacts with Fe(II) to produce Cr(III) and Fe(III).
1. What is the standard state potential and K for this reaction?
2. What is the cell potential if one half-cell is made up with 1.50 M potassium dichromate and 0.30 M chromium(III)nitrate hexahydrate in 1.00 M nitric acid and the other half cell is made up with 0.050 M iron(III)chloride hexahydrate and 0.10 M iron(II)chloride tetrahydrate?
3. What is the cell potential if the chromium half-cell were operated at a pH of 7 instead of using 1 M nitric acid?
Working in groups, the students encounter few problems identifying the two appropriate half reactions and writing the complete electrochemical reaction. We have not yet covered the equation for determining the standard state cell potential but usually one or more students remembers it. When we’ve written the equation on the board, I ask the groups to discuss why you use the numbers directly from the table without needing to change the sign of the oxidation half reaction and why there is no need to account for the stoichiometry in the complete, balanced electrochemical reaction. The second part (why no need to account for stoichiometry) usually gives them a bit more trouble so prompting them think about the meaning of Eo is useful to get them on the right track.
After calculating K, I like to have them reflect on the magnitude of the value.
The question in part C gets them to think about the significant effect of pH on the problem. The value of E actually turns out to be negative for the conditions given in the problem, meaning that the reaction goes in the reverse direction from that of standard state conditions.
ELECTROCHEMICAL ANALYTICAL METHODS
Ion-Selective Electrodes
I do a one-day class on ion-selective electrodes in primarily a lecture format. Using the pH electrode as an example, I show pictures from the text material to illustrate how an ion-selective electrode works. I then briefly go over other types of ion-selective electrodes.
Electrodeposition/Electrogravimetry
I first explain the basic concept behind electrogravimetry and then give them the following problem to solve.
Will the presence of Fe(II) at 0.0800 M interfere with the electroplating of 99.9% of the cadmium(II) in a solution in which the cadmium is expected to be present at a concentration of no less than 0.0500 M? Calculate potential values relative to a standard hydrogen electrode.
While doing this problem, they might need help realizing that the final concentration of cadmium will be much lower and that this will change the potential needed for electroplating. They may also need help determining the final concentration after 99.9% has plated, although, one student in each group usually can determine this without any prompting. It is worth developing with them during the problem that the more negative the potential, the less the species wants to be reduced. Once they have completed the calculation for the potential needed to plate out 99.9% of the cadmium, they can see that the iron will interfere.
Suppose the solution had 0.0800 M Cr(III) as a possible interference. Is it possible to plate out 99.9% of the cadmium(II) without any interference from the chromium.
In this case, they see that it should be possible to do the analysis, provide care is taking in setting the plating potential.
Coulometry
Before giving the students the following problem set, I briefly point out how coulometry is similar to electrogravimetry, but instead of weighing a material plated onto an electrode, the current is measured as a function of time.
Draw the plot you would obtain for current (y-axis) versus time (x-axis) if you applied a constant potential high enough to carry the reduction of Cd(II) to cadmium metal. The system has an electrode with a large surface area.
Students in groups usually end up drawing a number of possibilities for this question. Some draw a straight line starting at a high current and going to zero. Others draw the current increasing with time. Others have a fixed current that does not change with time. What I do after seeing the various responses is write the different ideas on the board and then ask groups to see if they can come up with reasons that would rule out any of the possibilities. In that process, the class eventually realizes that the current will diminish with time and then it becomes a matter of determining the shape of the curve. A good question to ask is whether the current would actually ever get to zero, meaning that every ion was reduced. That often sparks the thought that it is likely an exponential decay.
How would you relate the outcome of your plot above to the concentration of Cd(II) in the sample?
Most groups are able to arrive at the conclusion that you need the area under the curve.
What advantages does coulometry have over electrogravimetry?
Some groups realize that coulometry can be used in either an oxidizing or reducing potential. Others may realize that since it doesn’t require plating out of a metal, that it can be used for species where both forms in the redox couple are soluble. I usually have to prompt them about the question of sensitivity, asking what the lowest value we can reliably measure by weight and how that might compare to the lowest currents we can measure.
Titrimetric Methods of Analysis
I briefly go over the process of a classical redox titration where the analyte and titrant undergo an electrochemical reaction, the titrant is dispensed from a buret, and a colorimetric indicator is used. Then I point out that there are other ways to conduct and monitor titrations that involve an electrochemical reaction.
Coulometric Titration
I explain the details involved in performing a coulometric titration and then ask the students the following question. During this, I ask what the plot of current versus time would look like when performing a coulometric titration and then after we have this plot on the board, ask them how you would determine the endpoint. Then I give them the following question.
What are some advantages of using a coulometric titration?
They usually can propose some of the advantages (e.g., no buret, no need to prepare solutions with specific concentrations) and I either ask some leading questions to draw out others and then go over any they missed.
Amperometric Titration
I describe the process of performing an amperometric titration and the need for a system in which the products of the titration cannot produce a current. I then give them the following questions.
Draw the plot that would be obtained in an amperometric titration if each of the following occurred (assume that the system is set up with a reducing potential):
1. Only the analyte undergoes a reduction at the applied potential.
2. Only the titrant undergoes a reduction at the applied potential.
3. Both the analyte and titrant undergo a reduction at the applied potential.
Some groups are able to draw the current diagram for (A) whereas others draw an incorrect picture. I draw representations for all of these on the board and then ask them in their groups to either justify what they wrote or point out the problems with others that are on the board. We then have a discussion that usually results in the class agreeing on the correct plot. Having developed the correct answer to (A), most groups are able to readily draw what would be observed in (B) and (C).
Potentiometric Titration
I briefly mention that the titration will be performed by adding the titrant from a burette and a junction potential will be measured using a suitable electrode during the course of the titration. I then give them the following problem set.
Consider a solution of iron(II) that is to be titrated with cerium(IV). The products of the reaction are iron(III) and cerium(III). The course of the titration will be monitored potentiometrically.
$\mathrm{Fe(II) + Ce(IV) = Fe(III) + Ce(III)}$
If 20 ml of an 0.10 M solution of Fe(II) is to be titrated with 0.10 M Ce(IV), calculate the junction potential that would be measured at a platinum electrode when 5 ml, 10 ml, 20 ml, and 30 ml of titrant have been added.
The following steps will allow you to work through the calculation of the junction potentials.
Write an expression for the total electrochemical potential of this system (this is the Nernst equation for the complete reaction).
The groups are able to write this given our prior work in the class.
What is the total electrochemical potential of the titration reaction at any point during the titration? Think carefully about this because your first intuition may not be correct.
Groups are often confused by this question. If so, I ask them to consider when the titration reaction is at equilibrium. Some students answer that equilibrium is reached at the equivalence point while others in most groups often realize that equilibrium is reached after each drop of titrant is added. I remind them that useful titration reactions must quickly reach equilibrium.
Which half reaction is easier to use to calculate the junction potential before the equivalence point? Which half reaction is easier to use after the equivalence point?
In answering the previous question, we have discussed that with a total electrochemical potential of zero, it means that the potentials of the cathode and anode are equal (ECAT = EAN). I then point out that ECAT (or EAN) is equal to the junction potential measured in the system.
I suggest that the groups write out expressions for ECAT and EAN and then construct a chart that shows the amounts of the different species in the reaction for 5 ml, 10, 20, 30 and 40 ml increments of titrant. After making sure they have the chart written correctly, I then ask if they can answer the questions above. During this process, I often need to remind them that K for the reaction is large, and remind them of the use of a back reaction for calculating the concentration of the Ce(IV) before the equivalence point. Groups can then usually see that it is easy to calculate the junction potential using the iron half reaction before the equivalence point and the cerium half reaction after.
I then ask them to calculate the junction potential for each step in their chart. They realize that they don’t know how to calculate the potential at the equivalence point, so they complete the rest of the chart. I then give them the equation for calculating the junction potential at the equivalence point and also note what happens with the two half reactions have different numbers of electrons in them.
Voltammetric Methods
After a brief introduction to the general process of voltammetry, I give the students the following two questions to answer.
How can electrostatic migration be eliminated in an electrochemical cell?
After allowing the students to think about this, if they are perplexed or don’t have a good answer, I find it helpful to draw a picture on the board of an electrode surface with a negative charge, show a representation of a positive ion in solution but removed from the surface of the electrode, and ask if they can think of something they could do to the system that would eliminate attraction between the negative electrode and positive ion (without changing either the charge on the electrode or ion). Usually the groups are able to think of adding a supporting electrolyte and when asked, reason that a higher concentration would be better.
Samples subjected to voltammetric methods are usually purged with nitrogen or another inert gas before the analysis and maintained under an inert atmosphere during the analysis. Why is this done?
Most groups can reason out that dissolved oxygen is a problem in electrochemical processes.
Anodic Stripping Voltammetry
I explain the features of the applied potential as a function of time used in ASV. I also explain a bit about the common microelectrodes used in ASV and then give them the following problem.
Suppose you are analyzing a solution that contains Cd(II) and Pb(II) using anodic stripping voltammetry. Considering the voltage profile that is applied as a function in time in anodic stripping voltammetry, draw a plot of the current you would observe in the deposition step (applied voltage is –1.00 V) and in the stripping step. Remember, the procedure uses a microelectrode, which is particularly important to consider in the deposition step.
I try to initially focus the students on the deposition process. They may remember the plot in coulometry (or I may prompt them) and draw something similar for this step. We then discuss that because a microelectrode is used, the solution will never be depleted of the species being analyzed. With the realization that the ions are now plated onto the electrode, they can consider the stripping step. They often realize that the species will strip back off the electrode through an oxidation but may not at first realize that it will produce a sharp peak. As with other portions of this unit, when they are asked to draw a result, I draw their different proposals on the board and ask them to examine all the suggestions and rule out some or select which they think is correct. It may require some questions to get them to realize that the current will have a different sign for the deposition and stripping step.
What feature of the plot can be related to the concentration of the metal?
With a final plot complete, groups realize that the height of the peaks can be related to the concentration of the metal.
What advantage(s) does anodic stripping voltammetry offer over coulometry or electroplating?
Groups may need some prompting to get to a conclusion that ASV will be more sensitive. If they don’t come up with it, I ask whether coulometry could be used to analyze the mixture studied in this problem.
Linear Sweep Voltammetry
I explain the potential versus time profile used in LSV and then give them the following problem set.
Draw the current that would be measured for a solution consisting of Cd(II) and Zn(II) for the voltage profile that is applied in a linear sweep experiment. Assume that the concentration of Cd(II) is about twice as large as that of Zn(II) and that the solution is stirred.
I draw the different responses I see among the groups on the board and ask them to provide reasons why some might be ruled out or another defended. Often they realize at some point that the potential will start to reduce one of the metal ions, but may want to draw a peak instead of realizing that the current will plateau at that point because of the solution being stirred. It is helpful to remind them that it is a microelectrode so there will always be a large excess of ions in solution and the bulk concentration effectively does not change.
What feature of the plot could be related to concentration?
The groups realize that the height of a plateau will relate to concentration.
What feature of the plot can be used for species identification?
The groups realize that the location of the wave will vary for different species. I then introduce the idea of a half wave potential.
For the exact same solution of Cd(II) and Zn(II), draw the current that would be observed if the solution is not stirred.
We have discussed the concepts of diffusion, electrostatic attraction and physical convection at several places in this unit so the students realize those are the three things to consider. Assuming a supporting electrolyte has been added to eliminate electrostatic attraction, we can then discuss whether diffusion is fast enough to maintain a plateau for the current. Most of them, when prompted, know that diffusion in solution is fairly slow and are able to reason out that the current will diminish in this case. Sometimes they want to draw a sharp peak instead of a slighter drop.
Consider a voltage profile in a linear sweep experiment where a small, increased voltage pulse is superimposed over a voltage ramp. The current is sampled twice: (1) just before the pulse is applied and (2) just before the pulse ends. The two measurements are substracted [(2) – (1)] and the difference is plotted. Draw the resulting plot for the sample in question 1. What do you think is the name of this method?
It is important to leave the current versus time drawing on the board from the previous questions and ask them to consider that plot as they think about what will occur here. If they remain stumped, I usually draw some of the points (1) and (2) on the board for places where the current is not changing and for places going through the wave, and then tell them to consider the difference they would get in each case and to plot that. With that prompt, they are able to see that this would lead to a sharp peak for a species. I then point out that the process involves taking the derivative of the curve, hence the name differential pulse linear sweep voltammetry.
What advantage(s) does the method in the prior question have over regular linear sweep measurements?
Usually the groups are able to propose that the sharp peaks result in a more sensitive method and when prompted, realize that it will be easier to differentiate two species with closer reduction potentials. I draw pictures of these two situations on the board (Figures 27 and 28 in the text) to further illustrate this conclusion.
Cyclic Voltammetry
The prior topics within voltammetry have created the background needed to answer questions on CV, so after a very brief introduction showing the potential versus time profile, I give them the following questions.
Draw a plot of the current (y-axis) versus voltage (x-axis) that you would measure for a solution in which Fe(III) in ferricyanide (Fe(CN)63-) is reduced to Fe(II) during the reducing phase of the CV voltage profile shown below. The half reaction is chemically and electrochemically reversible and the solution is not stirred.
Given our discussion of LSV, they can readily draw the forward wave. I remind them that the plot is current (Y-axis) versus potential (X-axis), so that the X-axis is not time. This allows them to realize that there will be a reverse plot of the current and that the sign of the current will change for the reverse process. We can discuss the location of the reverse wave and its magnitude.
Draw a plot that could be obtained for an electrochemical reaction that is chemically irreversible and forms an electrochemically inactive product.
They usually can draw the resulting CV without any issue.
Draw the plot of current vs. potential that would be obtained for a reversible chemical reaction in which only the reverse reaction has an overpotential, but the potential eventually is sufficient to complete the reverse reaction.
They usually can draw the resulting CV without any issue.
Propose a reaction mechanism that would explain the following cyclic voltammogram. The first voltage cycle is shown as a solid line. The second voltage cycle is shown as a dotted line. A third voltage cycle gives the same output as the dotted line. Assume that a reducing potential is applied.
The groups can usually determine that the first reaction is irreversible and a new species seems to be showing up, indicating that the first product likely decomposed or further reacted. They can also reason out that the new species is a reversible couple. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Electrochemical_Methods_of_Analysis/04_Instructors_Manual.txt |
These assignments are designed to be capstone activities at the end of units on figures of merit (such as sensitivity and LOD), acid-base equilibria, separations, spectroscopy, mass spectrometry, and electrochemistry. Each assignment consists of an out-of-class reading assignment drawn from the primary literature accompanied by objective questions and a set of open-ended, in-class discussion questions. The assignments are designed to require just one class period and can be used before an exam to review important concepts, examine them from new angles, and apply them to new situations.
Interpreting the Primary Literature
Out-of-Class Questions
Article (Link): D.L. Phillips, I.R. Tebbett, and R.L. Bertholf, “Comparison of HPLC and GC-MS for measurement of cocaine and metabolites in human urine” J. Anal. Toxicol. 1996, 20, 305-308.
1. Using only the information in the abstract, answer the following questions about this work.
1. What did the authors do?
2. Why did they do it?
3. How did they do it?
4. What did they find?
2. Fill in the table below using the information from the introduction.
GC-MS
HPLC-UV
Advantages
Disadvantages
3. Toward the end of the introduction, often in the last paragraph, it is common to transition from background information to a brief summary of the specific question addressed in the manuscript. What question did these authors want to answer?
4. Skim the Materials and Methods section with a focus on what types of details are included. Check the boxes to indicate which pieces of information below are in the methods section, and be sure to note the level of detail.
□ chemicals used
□ glassware used
□ concentrations of solutions
□ instruments used
□ operating parameters for instruments
□ manufacturers of supplies
□ sample calculations for how to make stock solutions or dilutions
□ description of data analysis procedures
□ results of the experiment
5. Why do the authors spike the standards into a urine sample instead of diluting them in water or buffer?
6. Look up the structures of the analytes and bupivacaine and record them.
7. For their statistical analysis, why did the authors use paired t-tests?
8. How did the authors define the limit of quantitation (LOQ)? What is the typical way to define LOQ?
9. What metric was used to compare the precision of the data? What was used to evaluate sensitivity?
In-Class Questions
Article (Link): D.L. Phillips, I.R. Tebbett, and R.L. Bertholf, “Comparison of HPLC and GC-MS for measurement of cocaine and metabolites in human urine” J. Anal. Toxicol. 1996, 20, 305-308.
1. Consider the structures of the analytes and bupivacaine that you looked up outside of class. What was the purpose of the bupivacaine in this analysis? Do you think this compound was a good choice for this purpose? Explain.
2. The authors state that “Overall, GC-MS demonstrated better precision than HPLC, but the methods had generally equivalent sensitivities.” Consider the data in Tables 1 and 2. Do you agree with the authors’ statement? Why or why not? What additional data might be useful in comparing the two methods’ sensitivity and precision? What other information might you want to know when choosing a method for your application? Is this information provided in the manuscript?
3. Do you agree with the authors’ conclusion that HPLC with UV detection is a suitable alternative to GC-MS for analysis of cocaine in urine? Consider this conclusion generally and in the context of the following scenarios. Decide whether you would recommend HPLC-UV or GC-MS for each analysis, and justify your choices.
1. You manage a small business. The owner of the company decides to require drug testing for cocaine for all new hires.
2. You are working on a research project to study how genetic variation in rats affects the metabolism of cocaine.
4. Since this article was published, a method to couple HPLC to MS, called electrospray ionization, has become much more widely available. What would be the advantages of this method for this application? | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Interpreting_the_Primary_Literature/01_Comparing_Analytical_Methods_Assignment.txt |
Out-of-Class Questions
Article (Link): M. Dittrich and S. Sibler, “Cell surface groups of two picocyanobacteria strains studied by zeta potential investigations, potentiometric titration, and infrared spectroscopy,” J. Coll. Int. Sci. 2005, 286, 487-495.
Note that you can skip Sections 2.6 and 3.5 because we will not be discussing the infrared spectroscopy results.
1. Because this article deals with a microbiological and geochemical application, you need to familiarize yourself with a few terms. Match the following terms with their definitions/descriptions.
______ autotropic
______ oligotrophic
______ pelagic
______ calcite
______ electrophoretic mobility
______ zeta potential
______ peptidoglycan
______ picocyanobacteria
______ picoplankton
______ Synechococcus
______ phycocyanin
______ phycoerythrin
1. “self-feeding” using energy from light or chemical reactions
2. the most stable crystal form of CaCO3
3. used to describe living systems in open water
4. a protein complex in blue-green algae
5. a protein complex in red algae
6. the polymer that forms the bacterial cell wall and is composed of sugars and proteins
7. the smallest organisms in the water column that are incapable of swimming against current
8. a general name for photosynthetic bacteria with diameters < 3 µm
9. a specific genus of photosynthetic bacteria commonly found in marine environments
10. capable of living in a low-nutrient environment
11. the electric potential difference between the surface of a particle and the medium around it that arises from net electrical charge on the particle’s surface
12. a measure of how much an electric field affects the migration of a particle
2. In the abstract of the article, mark which sentences correspond to (1) background information and significance, (2) methodological details, (3) results, and (4) conclusions.
3. In your own words, describe why the surface chemistry of picocyanobacteria is an important area of research.
4. Research and summarize the major difference between Gram negative and Gram positive bacteria. How might this difference affect the surface properties of these two types of cells?
5. Re-read the final paragraphs of the introduction. What specific question/problem is being addressed in this article?
6. In Section 2.5, the authors state that the titration data will be plotted with –log[H+] on the x-axis and CA-CB-[H+]+Kw/[H+] on the y-axis. Assuming that =1 (i.e., that activity is equal to concentration), fill in the blanks below.
1. –log[H+] = ________
2. Kw/[H+] = ________
7. In their modeling of the data, what assumption do the authors make about surface charge on the bacteria?
8. On Figure 1, circle the isoelectric point of the bacteria. How do you know that this is the isoelectric point?
9. Based on their modeling results, the authors identify three separate pKa values associated with the picocyanobacteria surface. Fill in the table below to match the range of fitted pKa values with the most likely corresponding functional group.
Approximate Fitted pKa Value
Corresponding Functional Group
~5
~6.5
amine
10. The inflection points at the three pKa values are very weak, making them impossible to identify accurately without modeling. What explanation do the authors give for the weakness of these inflection points?
11. Note that generally the conclusion section of an article should not just summarize the paper. Instead, the conclusion might address (1) questions that remain to be answered about the data, (2) potential future experiments, (3) limitations of the work, and/or (4) the broader significance of the results. Give an example of one of these from the conclusion of this paper.
In-Class Questions
1. Prior to the titrations, the authors washed the cells in a solution of 1 mM EDTA and then resuspended them in NaNO3. Both the NaNO3 and the NaOH used in the titration experiments were degassed with N2 before use. What was the purpose of each of these steps, and why were they necessary?
2. When interpreting Figures 2-4, it will be helpful to consider how these plots differ from typical plots of titrations.
1. Have we typically plotted pH on the x- or y-axis? How does that compare to these plots?
2. What have we plotted on the other axis? How is that different from what is plotted here?
3. We know that ultimately the charges must balance, so where is this “excess” [H+] coming from? What is being deprotonated to release these hydrogen ions?
4. How do the bacteria contribute to the buffering of the system, i.e. how is this possible?
3. For the third pKa value (pK3 in the manuscript), the authors state that either amine or hydroxyl functionality could give rise to the observed pKa value, but they conclude based on the zeta potential measurements that this pKa corresponds to amine groups on the cell surface. Sketch the protonation reactions for a generic amine and a generic hydroxyl group in aqueous solution. Use your sketches to explain the authors’ reasoning.
4. A related article (Aquat. Sci., 2004, 66, 19-26) critiques studies like this one, in which a cell suspension is titrated. In the related article, Claessens et al. argue that because cells are dynamic, living systems, they respond differently to titrant than a chemical solution of weak acid or weak base would respond. For example, in addition to the chemical process of protonation or deprotonation, cells may also respond with metabolic activity or biochemical reactions, including pumping of protons across the cell membrane, unfolding of cell wall proteins, cell lysis, etc. As a result, Claessens et al. suggest that titration data does not necessarily just reflect the acid-base surface chemistry of bacterial assemblies. If you were a program officer at a funding agency, would you provide financial support for further studies like the one you read? Consider the authors’ purpose, as you described it in out-of-class questions 3 and 5, and justify your answer. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Interpreting_the_Primary_Literature/02_Acid-Base_Chemistry_Assignment.txt |
Out-of-Class Questions
Article (Link): B. Wei, D.S. Malkin, and M.J. Wirth, “Plate heights below 50 nm for protein electrochromatography using silica colloidal crystals,” Anal. Chem. 2010, 82, 10216-10221.
1. This abstract is a good example of how to present a large amount of quantitative data concisely in an abstract. Using the abstract and the body of the article as needed, report the following values for these experiments, and compare them to typical values for HPLC and CE from your text, class notes, or lab data. Don’t forget units where needed!
Parameter
This Work
Typical HPLC
Typical CE
dp
L
i.d.
H
N
2. Using a form of the resolution equation, the authors remind the reader that there are two contributors to peak resolution.
1. What are they?
2. Copy equation (1), then circle and label the terms that correspond to each of the contributors to Rs that you listed above.
3. Why do the authors choose to target plate height as a means to improve resolution? What other parameters could they have targeted?
4. Based on the final paragraph of the introduction, what was the objective of this work?
5. Why do the authors use an electric field, rather than pressure, to drive these separations?
6. The authors use silanes to polymerize the packing material and to form what is likely to be a very thin layer of short carbon chains on the surface (i.e., df is extremely low). If the stationary phase is so thin, how do the authors know that chromatography, rather than just electrophoresis, is occurring?
7. Estimate the value of H for the lysozyme peak in Figure 2. Show your work to receive credit. Does your estimate match the authors’?
8. Why is the plate height lower for these separations than for previous separations of dyes using the same packing material?
9. Why does heterogeneous packing increase the plate height? What term(s) in the van Deemter equation is/are affected by packing inhomogeneity like that seen in Figure 3?
10. In Figure 4C, why does peak variance increase linearly with time? In other words, what process causes this peak broadening?
In-Class Questions
Reading Assignment 3 (Link): B.Wei, D.S. Malkin, and M.J. Wirth, “Plate heights below 50 nm for protein electrochromatography using silica colloidal crystals,” Anal. Chem. 2010, 82, 10216-10221.
1. When characterizing plate height for lysozyme in Figures 4 and 5, the authors determine the width of the peak in space rather than in time. They are able to do this because they are using a camera as a detector, but why do they need to do this? How does the width of the peak in space relate to the width of the peak in time? How does the detector contribute to plate height in these experiments?
2. In discussing Figure 6, the authors assert that the A and C terms of the van Deemter equation are negligible for their separations. In Figure 7 and the latter part of the Results & Discussion, the authors address whether the extremely low plate heights observed could be due to focusing rather than to the achievement of a diffusion-limited separation. Why would the A and C terms be negligible under the conditions used in this work? What evidence supports the authors’ assertion that the efficiency of their separations is limited only by diffusion?
3. The authors specifically state that their goal for this work was not to achieve a practical method for protein separations. That would have been outside the scope of this paper because many practical considerations would need to be addressed before this type of packing could be made available in commercial columns. Imagine that an instrument manufacturer wants to use columns like these in a commercial HPLC instrument. What changes to the instrument and practical improvements in the column would be needed? | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Interpreting_the_Primary_Literature/03_Separations_Assignment.txt |
Out-of-Class Questions
Article (Link): C.N. LaFratta, I. Pelse, J.L. Falla, M.A. Palacios, M. Manesse, G.M. Whitesides, and D.R. Walt, “Measuring atomic emission from beacons for long-distance chemical signaling,” Anal Chem. 2013, 85, 8933-8936.
1. Define “infochemistry”.
2. In this paper, the salts are atomized by combustion (i.e., a flame). What other methods for atomization for atomic emission are available?
3. A methanol and air flame like the one used here typically obtains a maximum temperature of 2000 °C. What is the maximum temperature of an ICP? Why are high temperatures necessary for AES?
4. Read section 10C-1 on p. 273 of the textbook. Why did the authors use alkali metals, instead of other elements, for this application?
5. What do the authors mean when they say the signal is “isotropic”? How is isotropic emission an advantage for this application?
6. In the custom telescope in Figure 2, what components act as wavelength selectors? What wavelength selector is typically used in AES instrumentation? Explain the advantages of these two types of wavelength selectors for each application.
7. What signal processing method was used on the data in Figure 3? What are the advantages of this method for this application?
8. Estimate the signal-to-noise ratio of the processed cesium signal in the top right panel of Figure 6.
9. Go to the NIST Handbook of Basic Atomic Spectroscopic Data and look up the most intense persistent strong line(s) between 700-900 nm for Na, Li, and Ca in air. Lines indicated “P” next to the intensity refer to persistent lines, which are detectable even for low concentrations of the element in the presence of other species. http://www.nist.gov/pml/data/handbook/index.cfm#
10. Peruse the bandpass filters available from Spectrofilm.com (the vendor used for parts in this manuscript). What is the narrowest range of wavelengths that can be passed by these commercial filters? How does this compare to the width of an atomic emission line?
In-Class Questions
Article (Link): C.N. LaFratta, I. Pelse, J.L. Falla, M.A. Palacios, M. Manesse, G.M. Whitesides, and D.R. Walt, “Measuring atomic emission from beacons for long-distance chemical signaling,” Anal Chem. 2013, 85, 8933-8936.
1. Consider Figure 4.
1. Show the calculation used to determine the distance limit of detection of 1.7 km.
2. If the authors wanted to double the distance limit of detection, by what factor would the intensity of the atomic emission need to be increased?
3. Is it reasonable to expect that this increase in intensity (needed to double the distance limit of detection) could be achieved by increasing the concentration of the metal salts? Explain your answer.
2. In the top panels of Figure 6, why is it that the cesium signal can be the highest magnitude in the raw data, but is at the “low” level after decoding in the processed data?
3. What is the purpose of the instrument response matrix and what factors affect it?
4. The authors suggest that greater information density could be achieved by adding Na, Li, and Ca to the signal. Do you think that this would be feasible? Support your answer using the information you looked up for out-of-class questions 9 and 10.
5. The authors suggest that this beacon could be used to transmit data in resource-poor environments, such as natural disaster sites. Evaluate the feasibility of this application given the information in the manuscript. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Interpreting_the_Primary_Literature/04_Spectroscopy_Assignment.txt |
Out-of-Class Questions
Article (Link): R.A. Sowell, K.E. Hersberger, T.C. Kaufman, and D.E. Clemmer, “Examining the proteome of Drosophila across organism lifespan,” J. Proteome Res., 2007, 6, 3637-3647.
1. In the paper, the authors focus solely on proteins expressed in the fly’s head.
1. Give one reason for this choice.
2. How were the fly heads separated from the bodies?
2. Based on the details in the sample preparation section, about how many flies were used in this study?
3. Review or research the chromatographic method SCX. What is SCX, and how does it work?
4. What ionization method was used for these experiments? What is the advantage of this ionization method for these experiments?
5. Look up ion mobility spectrometry (IMS). How does this technique work? How does it differ from TOF-MS in the experimental parameters and the separation of analytes?
6. What is the total pressure in the IMS drift tube? Calculate the mean free path of an ion in an IMS drift tube and compare it to the mean free path of an ion in the mass analyzer used for the MS analysis part of the IMS-MS experiments.
7. Consider Figure 1.
1. Sketch the path of ions from the source to the MS detector, and label the IMS and MS regions.
2. What type of mass analyzer is used for the MS analysis on the IMS-MS instrument?
3. What type is on the LC-MS-MS instrument?
4. Which has higher mass resolving power?
8. Briefly and in your own words describe how mass spectra were assigned (matched) to specific peptides.
9. What “semi-quantitative” methods were used to measure changes in protein abundance with fly age in this work?
10. The bottom part of Figure 2 shows the base peak chromatograms for three separations. Define base peak. What is a base peak chromatogram?
11. How effective was the SCX-LC-MS/MS method at identifying peptides compared to the LC-MS/MS method? What do you think is the reason for this difference in performance when an SCX step is added?
12. Which proteins in Figure 6 change with the flies’ age? What biological processes are associated with these proteins?
13. How do the authors’ results compare to results obtained by more traditional methods used to study proteins?
In-Class Questions
1. Consider your answer to out-of-class question 7c. How are tandem MS experiments achieved on the LC-MS-MS system? How does fragmentation occur, and what types of fragments are formed? How do these fragments yield information about the peptides in the sample?
2. Proteomics experiments often produce vast quantities of data.
1. In Figure 4, what is being shown in the panels on the far left of the figure? How does this data relate to the data in the six panels to the right?
2. In your opinion, should authors be required to share all of the raw data used to construct figures in published articles? Why or why not?
3. When data is shared, who should bear the cost of the servers, web development and other infrastructure needed to do so?
Some possibilities include individual researchers through their grants, government organizations like the National Science Foundation and National Institutes of Health, the publishers of the journals the data appears in, or other researchers who choose to access the data. You may think of others.
3. Quantitation by mass spectrometry in proteomics is challenging because the wide variety of peptide chemistries result in variable MS signal between peptides, and ionization efficiency can also change over the course of an LC run.
1. Why didn’t the authors use internal standards to account for these differences? How were they able to compare the results from the different samples in the absence of an internal standard?
2. In Figure 6, what is the difference between the open symbols and the filled symbols?
3. Consider the Figure 6 data and your answer to out-of-class question 13. Do you think the two “semi-quantitative” methods used for analysis are reliable? What limitations do these techniques face? | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Interpreting_the_Primary_Literature/05_Mass_Spectrometry_Assignment.txt |
Out-of-Class Questions
Article (Link): P.E.M. Phillips, G.D. Stuber, M.L.A.V. Heien, R.M. Wightman, and R.M. Carelli, “Subsecond dopamine release promotes cocaine seeking,” Nature, 2003, 422, 614-618.
1. What is the difference between tonic and phasic dopamine signaling?
2. Look up any other unfamiliar words in the abstract. Then summarize this study in one complete sentence.
Skip to the end and read the methods section to orient yourself to the study design.
1. How were the rats taught to self-administer cocaine? How was the cocaine delivered?
2. What stimulus accompanied cocaine delivery?
3. What were the working and reference electrodes in this study? Why wasn’t a counter/auxiliary electrode necessary?
4. Sketch the potential applied to the working electrode as a function of time. Include as much detail in your plot as possible.
5. How many cyclic voltammograms (scans) were acquired per second?
6. How does the scan rate in these experiments compare to a typical CV scan rate? How does scan rate usually affect CV data?
7. How did the authors correct for current from interferents, movement of the animal, and pH changes in the extracellular space?
8. Complete the table below summarizing the means used to demonstrate that their signal came from dopamine release.
Experiment
How did this demonstrate the signal was from dopamine release?
anatomical
physiological
chemical
pharmacological
9. Draw the half-cell reaction for the oxidation of dopamine to dopamine quinone.
10. Consider the CV inset in Figure 1a. Label the oxidation and reduction half-waves. Was the oxidation of dopamine reversible? How can you tell?
11. Calculate the S/N ratio for the trace in Figure 3a.
12. Consider Figures 1-4. In the table, summarize the experimental design used for each experiment and the major results. In the column labeled stimulus/conditions, record what stimulus, if any, was used to elicit dopamine release or drug-seeking behavior. If no stimulus was used, note the conditions of the experiment. In the column labeled dopamine release, estimate the concentration of dopamine released in nM. In the column labeled timing, summarize any important findings about how the timing of the dopamine release related to the timing of the stimulus.
Figure
Stimulus/Conditions
Dopamine Release (nM)
Timing of Response
1
2
3
4
n/a
Note that the small bars labeled 500 nA or 50 nA should be used as a scale bar in evaluating peak height.
In-Class Questions
Reading Assignment 3 (Link): P.E.M. Phillips, G.D. Stuber, M.L.A.V. Heien, R.M. Wightman, and R.M. Carelli, “Subsecond dopamine release promotes cocaine seeking,” Nature, 2003, 422, 614-618.
1. Suggest some advantages of electrochemical detection of dopamine for this application, compared to spectroscopic or mass spectrometric detection.
2. Why was the cyclic voltammetry data not sufficient to identify the signal as coming from dopamine? (Why do the additional experiments that you summarized in out-of-class question #10?)
3. In Figure 1, where do the values for the color scale in the bottom panel come from? What plot would you obtain if you took a slice horizontally through the color plot? What would you obtain for a vertical slice?
4. Suggest one software-based and one hardware-based method to improve the S/N or Figure 3a, and justify your choices.
5. Individuals recovering from drug addiction are often counseled to avoid “triggers”, including locations and situations in which they previously used drugs. Do the findings in this study support this advice? Why or why not? Cite a specific figure or figures in your answer. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Interpreting_the_Primary_Literature/06_Electrochemistry_Assignment.txt |
General Notes
These assignments are designed to be capstone activities at the end of units on figures of merit (such as sensitivity and LOD), acid-base equilibria, separations, spectroscopy, mass spectrometry, and electrochemistry. Each assignment consists of an out-of-class reading assignment accompanied by objective questions and an in-class discussion of more open-ended questions. Each reading assignment is drawn from the peer-reviewed literature. The assignments are designed to require just one class period, and I commonly use them just before an exam as an opportunity for students to review important concepts, examine them from new angles, and apply them to new situations.
The reading and out-of-class questions should be completed before the in-class discussion and are designed to encourage close reading, article comprehension, and application of concepts from class. For example, the out-of-class questions for the fast-scan cyclic voltammetry article ask students to identify the oxidation and reduction half-waves of a representative cyclic voltammogram in the manuscript and to calculate the signal-to-noise for one of the dopamine release peaks. These out-of-class questions can be answered by students individually or in groups, depending on the arrangement that is most effective and feasible at your institution. All of the out-of-class questions are designed to elicit specific objective answers based on the article or material from class or the textbook. A few questions require some additional thought or extra care, and notes on these questions are included in this instructor manual.
The second part of the assignment consists of several open-ended, more conceptually challenging questions designed for in-class discussion. At the start of class, I find it helpful to solicit questions from the students and clarify any confusing points in the paper. In particular, it is useful to make sure that the students understand the objective out-of-class questions before they begin their discussions. Students should then break into small groups of 3-5 students to discuss the in-class questions. (For a discussion of effective methods for creating small groups, see the Instructor’s Guide for the In-Class Module on Separations.) During the discussions, I circulate to answer questions, highlight important points, or keep groups on-task as needed. Discussions for a question typically last from 5-15 minutes. When most of the groups have had time to develop their ideas, it is helpful to have each group report out to the class. If you are unable to check in with each group during their discussion, this is a useful opportunity to check for misconceptions or incomplete responses. Additionally, if you would like a written record of the students’ discussion to be turned in for evaluation or assessment, you may find it helpful to have one student to take notes for each group.
Note: If you do not provide the students with a color copy of the article, be sure to remind them to print the article in color or access it in color online while answering the questions because several of the articles include figures that are uninterpretable in grayscale.
07 Instructors Manual
Article: D.L. Phillips, I.R. Tebbett, and R.L. Bertholf, “Comparison of HPLC and GC-MS for measurement of cocaine and metabolites in human urineJ. Anal. Toxicol. 1996, 20, 305-308.
This article compares two different methodologies, HPLC with UV absorbance detection and GC-MS for the quantification of cocaine and its metabolites in urine. The assignment is designed so that no prior knowledge of the separation mechanisms of HPLC and GC is needed. Similarly, minimal background knowledge about MS is needed. Before the students start their discussions, I check to make sure they have understood that MS detection provides much more detailed structural information about the analyte compared to UV absorbance. The assignment is designed to highlight the figures of merit used to compare two techniques: measures of sensitivity, precision, LOD and LOQ, etc. The methods used also provide an opportunity to discuss matrix effects, use of an internal standard, and statistical analysis of data. Additionally, because these topics are typically covered early in the analytical curriculum, when students might first be encountering the primary literature, some questions are included to draw students’ attention to certain conventions of writing in the scientific literature.
Out-of-Class Questions
Several questions in this assignment (Q1, Q3, Q4) are designed to draw students’ attention to the structure of a scientific article and important conventions in scientific writing for the literature. The remaining questions review important concepts in method calibration and statistics or ensure that students have the necessary background for in-class discussion.
Q1. Using only the information in the abstract, answer the following questions about this work.
(a) What did the authors do?
The authors compared HPLC to GC-MS for detection and determination of cocaine and its metabolites.
(b) Why did they do it?
The aim of the study was to evaluate whether HPLC is a useful alternative to GC-MS for this application.
(c) How did they do it?
The authors used solid phase extraction followed by either HPLC or GC-MS on samples of urine with bupivacaine added as an internal standard. The methods were compared based on sensitivity, precision, and dynamic range.
(d) What did they find?
They found that HPLC is a useful alternative to GC-MS.
Q2. Fill in the table below using the information from the introduction.
(This question prepares students for in-class discussion questions Q4 and Q5.)
GC-MS
HPLC-UV
Advantages
High specificity
Currently the accepted method
Very sensitive methods available
DAD gives spectral information
Low cost
Technology is improving
Disadvantages
Analytes must be volatile and therefore often require derivatization
No UV detection of the metabolite ecgonine methyl ester available, more susceptible to matrix interferents
Q3. Toward the end of the introduction, often in the last paragraph, it is common to transition from background information to a brief summary of the specific question addressed in the manuscript. What question did these authors want to answer?
From p. 305, “The purpose of this work was to determine whether HPLC analysis of urine samples containing cocaine and metabolites produced comparable precision, sensitivity, and reproducibility to GC-MS analysis of the same samples.
Q4. Skim the Materials and Methods section with a focus on what types of details are included. Check the boxes to indicate which pieces of information below are in the methods section, and be sure to note the level of detail.
Discussing this question in class is a useful opportunity to clarify the differences in format and detail between lab manual protocols, procedures recorded in lab notebooks, and the methods sections of journal articles.
■ chemicals used
□ glassware used
■ concentrations of solutions
■ instruments used
■ operating parameters for instruments
■ manufacturers of supplies
□ sample calculations for how to make stock solutions or dilutions
■ description of data analysis procedures
□ results of the experiment
Q5. Why do the authors spike the standards into a urine sample instead of diluting them in water or buffer?
Spiking the standards into a urine sample helps to keep matrix effects constant between samples and calibration standards.
Q6. Look up the structures of the analytes and bupivacaine and record them.
(This question ensures that students have information required to evaluate the suitability of the internal standard in class.)
Q7. For their statistical analysis, why did the authors use paired t-tests?
As noted on p. 306, the same samples were used for each method after being split in half after the SPE step. This is a common experimental design for comparing two methods and means that a paired t-test is appropriate.
Q8. How did the authors define the limit of quantitation (LOQ)? What is the typical way to define LOQ?
(This question asks students to recognize that the LOQ determination in this paper may not match the methodology described in previous classes. If desired this question can be moved to the in-class assignment and used to prompt discussion about practical considerations of false positive and false negatives in determining the best way to assess LOD and LOQ for a specific application.)
On p. 307, the authors defined the LOQ as five times the LOD. [The LOD was determined empirically by running samples of progressively lower concentration until the signal was not consistently detected.] A typical textbook definition of the signal LOQ is 10 standard deviations above the blank. The concentration LOQ is then 10sbl /m. With the standard definition of LOD as 3sbl /m, the LOQ is typically defined as 3.33 times the LOD.
Q9. What metric was used to compare the precision of the data? What was used to evaluate sensitivity?
(This question prompts students to identify the interday variation in the slope as a metric for precision and the slope itself as the metric for sensitivity. As students discuss in-class Q2, I encourage them to consider their responses to this question. Note that the authors use the term “analytical sensitivity” in referring to the interday variation of the slope (calibration sensitivity), which may confuse students who have learned that analytical sensitivity, γ, refers to the calibration sensitivity, m, divided by the standard deviation of the signal for a given calibrator.)
As noted on p. 307, the authors use the interday variability of the average coefficient of variability for all calibrators to evaluate precision. [The coefficient of variation, CV, is the relative standard deviation expressed as a percent.] These values for each calibrator were averaged, and then the average value and standard deviation were used to perform a paired t-test to determine whether there was any difference in precision between the methods. This measure of precision emphasizes day-to-day reproducibility.
Also on p. 307, the authors state that the average slope for each analyte was used to evaluate sensitivity. Again, a t-test was used to determine whether there was a significant difference in sensitivity for each method.
In-Class Questions
Q1. Consider the structures of the analytes and bupivacaine that you looked up outside of class. What was the purpose of the bupivacaine in this analysis? Do you think this compound was a good choice for this purpose? Explain.
Students should consider the presence of acidic and basic groups, hydrophobicity, molecular weight, etc. Because the internal standard is expected to behave similarly to the analytes, students should also consider relative retention times of the analytes and internal standard in Figures 1 and 2. Students may also consider the possibility of including a second internal standard that can be derivatized like BZE and NC for GC analysis to account for incomplete derivatization of these analytes.
If students conclude that bupivacaine is not sufficiently similar to the analytes, it may be helpful to prompt students to consider challenges to identifying an internal standard that is more structurally similar to the analytes. For example, one must be certain that the internal standard cannot possibly occur in the samples due to cocaine metabolism. Time-permitting, you may give the students an opportunity to search the literature for internal standards used in other studies of cocaine metabolites. (They may be surprised to find that for GC-FID studies the most commonly used internal standards may be even less structurally similar to cocaine than bupivacaine.)
Q2. The authors state that “Overall, GC-MS demonstrated better precision than HPLC, but the methods had generally equivalent sensitivities.” Consider the data in Tables 1 and 2. Do you agree with the authors’ statement? Why or why not? What additional data might be useful in comparing the two methods’ sensitivity and precision? What other information might you want to know when choosing a method for your application? Is this information provided in the manuscript?
This question asks students to make a global assessment of the data in Tables 1 and 2. It is important to ensure that the students understand their answers to out-of-class Q9 before this discussion gets underway. The authors provide their interpretation in the text, but this question is actually fairly complex when one considers the standard deviations of the slopes reported in Table 1. Students should be encouraged to consider all the data rather than relying on the authors’ conclusions. Based on Table 2, GC-MS is more precise, as the individual calibrators appear to agree well from day-to-day based on the average CV values reported. However, the slopes for the GC-MS calibration curves had much larger variation than those for HPLC, as seen in Table 1. Students may consider whether reporting the analytical sensitivity, γ, in addition to the slope would be useful, as this metric considers both sensitivity and precision. Some students may bring up the LOD values presented at the end of the paper, giving an opportunity to underscore the distinction made by analytical chemists in discussing sensitivity and LOD.
Q3. Do you agree with the authors’ conclusion that HPLC with UV detection is a suitable alternative to GC-MS for analysis of cocaine in urine? Consider this conclusion generally and in the context of the following scenarios. Decide whether you would recommend HPLC-UV or GC-MS for each analysis, and justify your choices.
1. You manage a small business. The owner of the company decides to require drug testing for cocaine for all new hires.
2. You are working on a research project to study how genetic variation in rats affects the metabolism of cocaine.
This question asks students whether they agree with the paper’s conclusion, which is that HPLC is a useful alternative to GC-MS. Rather than treating this as a question with “correct” answers, students should be encouraged to consider this conclusion critically and use data and experience to support their answer. For example, students who are familiar with derivatization as a sample preparation step may bring up issues of cost and time associated with this technique. I often find it useful to direct students to both the text and the tables and to instruct them to consider all the available data in their decision. The first paragraph of the Results on p. 307 of the article discusses the specificity of each method. The last paragraph of the Results on p. 307 discusses the agreement between the tested methods and the results of an established drug lab. The LOD data are at the end of the Discussion. Students should also be encouraged to consider the ramifications of a false positive in each situation. In summarizing the class discussions, I explain that GC-MS is commonly used as a confirmatory tool in drug testing because of the structural information provided by MS. For this reason, GC-MS is likely the best of the two options for scenario (a), while HPLC-UV may be acceptable or even preferable for scenario (b).
Q4. Since this article was published, a method that readily couples HPLC to MS, called electrospray ionization, has become much more widely available. What would be the advantages of this method for this application?
If needed, students can be referred back to Q2 from the out-of-class questions, which highlights the advantages and disadvantages of each technique. Depending on students’ familiarity with HPLC, GC, and MS, it might be helpful to briefly discuss the difference between a separation technique and a detection method. This should lead into a discussion of how HPLC-MS combines the minimal sample preparation of HPLC with the structural information available by MS. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Interpreting_the_Primary_Literature/07_Instructors_Manual/01_Comparing_Analytical_Methods%3A_Cocaine_.txt |
Article: M. Dittrich and S. Sibler, “Cell surface groups of two picocyanobacteria strains studied by zeta potential investigations, potentiometric titration, and infrared spectroscopy,” J. Coll. Int. Sci. 2005, 286, 487-495.
This article presents an investigation of the surface chemistry of picocyanobacteria, which contribute to calcite precipitation in lakes. The article highlights a unique application of acid-base measurements while also encouraging students to think of the challenges presented by chemical investigations of live cells. Potentiometric titration is used to determine the pKa values of functional groups on the cells’ surface, and infrared spectroscopy is used to confirm identification of these groups. While the potentiometric titration plots look quite different from the titration curves students have seen in class, the same acid-base principles are used to interpret the data. When I use this paper in analytical chemistry, we skip the IR data so no questions on those sections are included; however, if the paper is read in a class that covers both techniques, questions on the second part of the paper could be added.
Out-of-Class Questions
Q1. Because this article deals with a microbiological and geochemical application, you need to familiarize yourself with a few terms. Match the following terms with their definitions/descriptions.
Students should have no trouble finding the definitions of these terms online:
__A__ autotropic
__J___ oligotrophic
__C__ pelagic
__B__ calcite
__L__ electrophoretic mobility
__K__ zeta potential
__F__ peptidoglycan
__H__ picocyanobacteria
__G__ picoplankton
__I__ Synechococcus
__D__ phycocyanin
__E__ phycoerythrin
Q2. In the abstract of the article, mark which sentences correspond to (1) background information and significance, (2) methodological details, (3) results, and (4) conclusions.
There is some flexibility in the correct answer to this question, but a suggested mark-up is shown below: background and significance, methodological details, results, conclusions.
Q3. In your own words, describe why the surface chemistry of picocyanobacteria is an important area of research.
I typically look for some form of the underlined concepts in student responses:
The surface chemistry of picocyanobacteria is an ecologically significant area of research since these organisms play a role in biogeochemistry of lakes by precipitating calcite, potentially through Ca2+ binding to the cell surface followed by hydroxide diffusion through the membrane.
Q4. Research and summarize the major difference between Gram negative and Gram positive bacteria. How might this difference affect the surface properties of these two types of cells?
As student should readily be able to find online or in a biology text, Gram positive bacteria have a simple peptidoglycan cell wall, while Gram negative bacteria have a bilayer membrane around their cell wall.
According to the article, published results show no clear trend in the relative density of reactive sites between Gram positive and Gram negative bacteria; however, one can reasonably hypothesize that a peptidoglycan layer, which is composed of amino acid chains and sugars, would potentially have different acid-base chemistry than a lipid bilayer.
Q5. Re-read the final paragraphs of the introduction. What specific question/problem is being addressed in this article?
The authors want to characterize the surface chemistry of Gram negative picocyanobacteria with respect to “the identity, abundance, and acid-base properties of binding sites” on two strains of Synechoccocus (p. 488).
Q6. In Section 2.5, the authors state that the titration data will be plotted with –log[H+] on the x-axis and CA-CB-[H+]+Kw/[H+] on the y-axis. Assuming that γ =1 (i.e., that activity is equal to concentration), fill in the blanks below.
Assuming that concentration is roughly equal to activity (γ≈1)
1. –log[H+] = pH
2. Kw/[H+] = [OH-]
Q7. In their modeling of the data, what assumption do the authors make about surface charge on the bacteria?
The authors assume that the surface charges are homogeneously distributed across the bacterial surface. While this simplifies the modeling, it may not be entirely justified based on prior published research about bacterial surface chemistry.
Q8. On Figure 1, circle the isoelectric point of the bacteria. How do you know that this is the isoelectric point?
[This question prepares students to discuss in-class question 3.)
The isoelectric point is the pH at which a molecule (or surface in this case) has a net charge of zero. Because the zeta potential arises from a net charge at a surface, the zeta potential will also be zero at the isoelectric point.
Q9. Based on their modeling results, the authors identify three separate pKa values associated with the picocyanobacteria surface. Fill in the table below to match the range of fitted pKa values with the most likely corresponding functional group.
Approximate Fitted pKa Value
Corresponding Functional Group
~5
carboxyl
~6.5
phosphate
~8.8
amine
Q10. The inflection points at the three pKa values are very weak, making them impossible to identify accurately without modeling. What explanation do the authors give for the weakness of these inflection points?
The bacterial surface includes several functional groups with similar pKa values (Section 3.2, p. 490). If the students have looked at titration curves for diprotic weak acids, this is a helpful analogy since as the pKa values of the two acidic groups get closer together the two distinct inflection points are harder to distinguish.
Q11. Note that generally the conclusion section of an article should not just summarize the paper. Instead, the conclusion might address (1) questions that remain to be answered about the data, (2) potential future experiments, (3) limitations of the work, and/or (4) the broader significance of the results. Give an example of one of these from the conclusion of this paper.
Although the section is labeled “Summary,” the authors do extend their comments beyond a mere summary of their results. For example, they address the broader significance of the results to calcite formation by discussing how the functional groups identified could interact with calcium and carbonate ions. They also describe an additional potential application in remediation of metal pollution that would require further study of bacterial surface chemistry as well as studies to identify conditions for culturing picocyanobacteria in the lab.
In-Class Questions
Q1. Prior to the titrations, the authors washed the cells in a solution of 1 mM EDTA and then resuspended them in NaNO3. Both the NaNO3 and the NaOH used in the titration experiments were degassed with N2 before use. What was the purpose of each of these steps, and why were they necessary?
The composition of the cell culture medium is described in Section 2.1 (p.488) as containing numerous divalent metal ions, including Ca2+, Mg2+, Fe2+, Mn2+, Zn2+, Co2+, and Cu2+. These divalent metal ions disproportionately influence ionic strength and can interact with negatively charged groups on the cell surface, influencing measurements of zeta potential. While students may or may not realize that, classes which have covered complexometric titrations should recognize that the EDTA will chelate these metal ions (with varying affinity). The EDTA wash removes most of these ions, minimizing their effect on measurements of bacterial surface charge. The cell culture media is also buffered by dibasic potassium phosphate and boric acid. To titrate functional groups on the cell surface, it is necessary to eliminate the these weakly basic and weakly acidic components of the culture medium. Re-suspending the cells in sodium nitrate, which is a neutral salt, eliminates buffering from the solution so that acid-base behavior is coming only from the cells. Finally, the solutions were degassed with N2 to displace dissolved CO2, which forms carbonic acid in water and affects pH.
Q2. When interpreting Figures 2-4, it will be helpful to consider how these plots differ from typical plots of titrations.
1. Most potentiometric titration plots have pH on the y-axis. In these plots, pH is on the x-axis.
2. Most titration plots give mL of titrant on the x-axis. In this case, pH is on the x-axis and an expression representing “charge excess” (CA-CB-[H+]+Kw/[H+]) is on the y-axis.
3. As the researchers add NaOH to the bacterial cell suspension to increase pH (on the x-axis), the change in pH does not directly correlate with the moles of base added. This occurs because as the NaOH is added, additional H+ is released into the solution as functional groups on the bacteria are deprotonated.
4. Because the functional groups found in biological molecules, such as amines, carboxylic acids, hydroxyl groups, and phosphate groups act as weak acids or bases, biological molecules, including those at the surface of the cells act as buffering agents and influence the pH of their environment.
Q3. For the third pKa value (pK3 in the manuscript), the authors state that either amine or hydroxyl functionality could give rise to the observed pKa value, but they conclude based on the zeta potential measurements that this pKa corresponds to amine groups on the cell surface. Sketch the protonation reactions for a generic amine and a generic hydroxyl group in aqueous solution. Use your sketches to explain the authors’ reasoning.
In the manuscript, pK3 has a value of ~8.8, which could correspond to a hydroxyl or to an amine. If the functional group giving rise to pK3 were an amine, then below pH 5 the functional group would be protonated (because pH < pKa) and positively charged. If this functional group were a hydroxyl, then the protonated form predominant below pH 5 would be neutral. We know that the other two functional groups present, carboxylic acid and phosphate groups, have pKa values of ~5 and ~6.5, respectively. Both of these groups will be negatively charged above pH 5 and neutral below pH 5. If pK3 corresponded to the hydroxyl, all functional groups would be neutral below pH 5, resulting in a zeta potential of zero. However, the data indicate that below pH 5 the zeta potential is positive, corresponding to the presence of amines.
Q4. A related article (Aquat. Sci., 2004, 66, 19-26) critiques studies like this one, in which a cell suspension is titrated. In the related article, Claessens et al. argue that because cells are dynamic, living systems, they respond differently to titrant than a chemical solution of weak acid or weak base would respond. For example, in addition to the chemical process of protonation or deprotonation, cells may also respond with metabolic activity or biochemical reactions, including pumping of protons across the cell membrane, unfolding of cell wall proteins, cell lysis, etc. As a result, Claessens et al. suggest that titration data does not necessarily just reflect the acid-base surface chemistry of bacterial assemblies. If you were a program officer at a funding agency, would you provide financial support for further studies like the one you read? Consider the authors’ purpose, as you described it in out-of-class questions 3 and 5, and justify your answer.
The authors’ goal was to evaluate the surface chemistry of the bacteria to see how it might be involved in calcite precipitation. If the available proton donors are changing during the measurement because of biological effects of the titration then this is a problem for this application. Interestingly, the authors acknowledge that they could not measure the acid-base behavior of the cell surface as a function of ionic strength because the cells may lyse (Section 3.2, p. 491), but they do not consider the possibility of physiological responses of the cells to pH values tested during the titration. Students who have taken a cell biology course may make suggestions to fix the cells (which stops metabolic activity but also causes many of the acidic and basic groups on the cell surface to react during fixation) or to isolate just the membrane components by centrifugation for titration (which could be done, but assumes that the inner and outer membrane leaflets are the same and that all embedded proteins and other biomolecules remain intact, which may not be the case). If students immediately suggest that they would not fund these studies, consider asking if they can suggest other methods to obtain the necessary data. Classes that discuss the FT-IR data may also note that these data were obtained in the absence of any titration, allowing the authors to make some functional group assignments without dramatically perturbing the cells (although there were still suspended in NaNO3). This is a challenging problem, in that the titration of live cells is not an ideal method, but no perfect alternative currently exists, so some useful information may still be obtained by this technique. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Interpreting_the_Primary_Literature/07_Instructors_Manual/02_Acid-Base_Equilibria%3A_Titrations_of_Ba.txt |
Article: B.Wei, D.S. Malkin, and M.J. Wirth, “Plate heights below 50 nm for protein electrochromatography using silica colloidal crystals,” Anal. Chem. 2010, 82, 10216-10221.
This article describes the use of colloidal particles just 330 nm in diameter as a packing material for electrochromatography. Students do not need to have been previously introduced to electrochromatography specifically, but they should have a background in reverse phase HPLC and capillary electrophoresis. The article provides an opportunity for students to apply their conceptual understanding of band broadening and the van Deemter equation to a novel experimental system. To answer several of the questions, students will need to have a detailed understanding of band broadening in separations, such as that provided by the ASDLIB Active Learning In-Class Activity on Separation Science written by Prof. Tom Wenzel. Also of interest: the small size of the particles means that when crystalline packing is achieved Bragg diffraction results in an artificial opal and the column appears blue. This can be seen in images on the website of the corresponding author, Mary Wirth.
Out-of-Class Questions
Q1. This abstract is a good example of how to present a large amount of quantitative data concisely in an abstract. Using the abstract and the body of the article as needed, report the following values for these experiments, and compare them to typical values for HPLC and CE from your text, class notes, or lab data. Don’t forget units where needed!
Parameter
This Work
Typical HPLC
Typical CE
dp
330 nm
1-5 µm
no packing
L
2 cm
15-25 cm
20-50 cm
(to detector)
i.d.
75 µm
4.6 mm
50-100 µm
H
< 50 nm
~10 µm
0.5-5 µm
N
106
104
105-106
The length of the capillaries in this work is found in the Experimental section on p. 10217. Typical data for HPLC and CE should be provided previously during class lectures or laboratories, though some of the values are also provided in commonly used textbooks.
Q2. Using a form of the resolution equation, the authors remind the reader that there are two contributors to peak resolution.
1. Efficiency (peak sharpness/peak width) and selectivity are the two overall contributors to peak resolution. Many experimental variables can be changed to affect these two terms.
2. $R_s=\dfrac{\sqrt{L/H}}{4} \dfrac{\Delta t}{<t>}$
The $\sqrt{L/H}$ term is equal to $\sqrt{N}$ and corresponds to the efficiency of the separation. The $\Delta t\,/<t>$ term corresponds to selectivity.
Q3. Why do the authors choose to target plate height as a means to improve resolution? What other parameters could they have targeted?
The authors choose to target efficiency by improving the plate height because the experimental factors that increase plate height are determined primarily by materials science. Advances in materials science, such as formation of colloidal crystals from nanoparticles, can improve plate height. Improvements to efficiency only increase resolution by the square root, so the authors might have chosen to target selectivity instead. This requires differentially affecting the partition coefficient of each analyte between the mobile and stationary phases. This can be achieved by varying the temperature or by varying the chemical composition of either the mobile phase or the stationary phase.
Q4. Based on the final paragraph of the introduction, what was the objective of this work?
The authors purpose was to explore theoretical possibilities for this new packing, not to develop a practical method. The goal was “to characterize the contributions of [a silica colloidal crystal packing] to plate height for proteins.” (p. 10217)
Q5. Why do the authors use an electric field, rather than pressure, to drive these separations?
The pressure required to drive fluid flow through a column which such a packing would be immense. [This is the extent of the answer provided by my students; however, for those interested, a more detailed discussion is provided below.]
For a given flow rate, u, the pressure difference across the column required, ΔP, is given by
$∆P= \dfrac{ϕηLu}{d_p^2}$
where Φ is a flow resistance factor, η is the viscosity of the mobile phase, L is the length of the column, and dp is the diameter of the packing particles (see, e.g., J.E. McNair, et al. Anal Chem. 1999, 71, 700). Because ΔP increases as the inverse square of the packing particle diameter, much higher pressures are required as diameter decreases. In contrast, electroosmosis is a surface-driven phenomenon that does not depend on particle diameter.
Q6. The authors use silanes to polymerize the packing material and to form what is likely to be a very thin layer of short carbon chains on the surface (i.e., df is extremely low). If the stationary phase is so thin, how do the authors know that chromatography, rather than just electrophoresis, is occurring?
On p. 10218, the authors note that “[t]he retention order (Lyz, RnaseA, CytC) is different from what we observe in a CE separation (Lyz, CytC, RnaseA) by the same as what we observe in a reversed phase separation.” Because these two separation methods resolve analytes based on different chemical characteristics, they result in different retention orders.
Q7. Estimate the value of H for the lysozyme peak in Figure 2. Show your work to receive credit. Does your estimate match the authors’?
To calculate the plate height, H, from experimental data, it is easiest to first calculate the plate number, N, using
$N=16 \left(\dfrac{t_r}{w}\right)^2=5.54 \left(\dfrac{t_r}{w_{1/2}} \right)^2$
where tr is the retention time, w is the width, and w1/2 is the full-width at half-maximum, which is often easier to estimate accurate than the width of the peak at its base. For lysozyme in the Figure 2 inset, tr appears to be approximately 107 s and w1/2 appear to be ~1 s:
$N=5.54 \left(\dfrac{107\: s}{1\: s}\right)^2=\textrm{63,000}$
H is related to N by N = L/H. The caption of Figure 2 states that the separation distance for the data shown is 0.91, so
$H= \dfrac{L}{N}=\dfrac{0.91\: cm}{63000}=1.4 \times 10^{-5}\: cm=140\: nm$
The authors state that the plate heights for the peaks in Figure 2 are all less than 100 nm. Considering that we do not have the raw data, this calculation is in fairly good agreement.
(Note: students may estimate the values of tr and w or w1/2 slightly differently, resulting in varying degrees of agreement between their value and the authors’.)
Q8. Why is the plate height lower for these separations than for previous separations of dyes using the same packing material?
The authors state that the lower plate heights occur “presumably because the injected widths and diffusion coefficients are lower” (p. 10218). Students can be reminded that although we most often consider how to minimize the contribution of the column to peak width, the injection has a finite width that may also appreciably increase plate height. Additionally, proteins have much lower diffusion coefficients than dyes due to their larger size. As a result, the longitudinal diffusion term, B, in the van Deemter equation is smaller, leading to lower plate heights.
Q9. Why does heterogeneous packing increase the plate height? What term(s) in the van Deemter equation is/are affected by packing inhomogeneity like that seen in Figure 3?
Heterogeneous packing can lead to channels in which the packing particles are not well-packed, resulting in larger interstitial spaces in some parts of the column. This means that analyte molecules that pass through the void take a much less tortuous path than molecules that go through a well-packed section of column, a process called channeling. This primarily contributes to band broadening by increasing the A term of the van Deemter equation. Uneven packing also means that some molecules will travel further through the column without encountering the stationary phase than others. This results in increased band broadening due to the mass transport term, C (specifically mass transport in the mobile phase, Cm).
Q10. In Figure 4C, why does peak variance increase linearly with time? In other words, what process causes this peak broadening?
Diffusion causes the peak variance to increase linearly with time. As seen in Figure 4C, the relationship between peak variance and time is given by σ2 = 2Dt. The longer the plug spends on the column, the more diffusion contributes to band broadening through the B and C terms. [Note to instructors: In well-packed columns in this paper, the C term is negligible, but in typical HPLC columns this is not the case.]
In-Class Questions
Q1. When characterizing plate height for lysozyme in Figures 4 and 5, the authors determine the width of the peak in space rather than in time. They are able to do this because they are using a camera as a detector, but why do they need to do this? How does the width of the peak in space relate to the width of the peak in time? How does the detector contribute to plate height in these experiments?
To investigate band broadening in their columns, the authors needed to determine the variance, σ2, of their peaks. The variance is found by fitting the peak to the equation for a Gaussian; however, the authors could not fit the peaks as a function of time because they were only a few points wide. This occurred because the camera needed to use a relatively long exposure time (0.2 s, given as the acquisition time on p. 10218) to capture the image of each peak because the channels are small and the analytes are dim. As a result, by the time the 0.2 s exposure time has passed, the peak is almost passed the detector. The authors can only capture a couple images of the peak as it passes.
However, each picture of the peak is many pixels wide on the camera sensor. This means the authors can fit the Gaussian peak in space very well, as seen in Figure 5. Because the width of the peaks in space is related to their width in time by the velocity of the peaks, the authors can then convert between these two data domains.
[Students can be prompted to sketch the peak as a function of time and as a function of space to help bring them to the conclusion described above.]
One caveat is that the authors must account for the fact that the detector is contributing some variance, $σ_{detector}^2$. This occurs because the peak is moving as the image is acquired. As a result, the peaks look a bit wider than they really are. (Image using a camera with a slow shutter speed to take an image of a sprinter. The picture of the runner looks smeared out wider than it should because the runner is moving as the image is exposed.) The authors account for this using Equation 3 on p. 10219 which relates the variance to the exposure time, τ, and the velocity of the peak, v.
Q2. In discussing Figure 6, the authors assert that the A and C terms of the van Deemter equation are negligible for their separations. In Figure 7 and the latter part of the Results & Discussion, the authors address whether the extremely low plate heights observed could be due to focusing rather than to the achievement of a diffusion-limited separation. Why would the A and C terms be negligible under the conditions used in this work? What evidence supports the authors’ assertion that the efficiency of their separations is limited only by diffusion?
The A term, Eddy diffusion, is negligible because the particle diameter is so small. Recall that A decreases with dp. The C term is negligible because Cs is minimized by the very thin stationary phase since it is inversely related to df and because the Cm term is minimized by the small packing particles which result in very small interstitial spaces for analyte to diffuse across before encountering the stationary phase.
The authors present quite a bit of evidence that their peak widths are diffusion-limited. One of the easiest pieces of evidence for students to appreciate is the shape of the curve in Figure 6. This is a van Deemter curve, but rather than having the characteristic “Nike Swoosh” shape of a curve that includes a C term, it has the shape of the B/u term, which depends only on longitudinal diffusion. The authors also check to make sure that the diffusion coefficient obtained from the data in Figure 6 matches the diffusion coefficient in the absence of an electric field (Figure 7). This rules out the possibility of focusing that could occur if aqueous solution entered during the injection, increasing the retention of the front of the sample plug (since this is more polar than the mobile phase, which contained some acetonitrile). Finally, the authors calculated the expected diffusion-limited peak width based on the diffusion coefficient of lysozyme and the velocity of the peaks (Equation 5) and found good agreement with their experimental data.*
*This calculation is fairly hard to follow. I do not usually go through it in detail in class. We simply discuss how it is related to the authors’ assertion that their separation is diffusion-limited.
Q3. The authors specifically state that their goal for this work was not to achieve a practical method for protein separations. That would have been outside the scope of this paper because many practical considerations would need to be addressed before this type of packing could be made available in commercial columns. Imagine that an instrument manufacturer wants to use columns like these in a commercial HPLC instrument. What changes to the instrument and practical improvements in the column would be needed?
The most important improvement to the column would be to eliminate the channeling issue by improving the packing process. On p. 10218, the authors state that about two-thirds of the columns they packed had gaps in packing at the walls. This would be unacceptable for commercial preparation of columns.
A traditional HPLC instrument would need to be significantly modified to accept these columns. The pressure system would need to be replaced with a high voltage power supply to produce electroosmotic flow. The injection system would also need to be modified for electrically-driven injections. The detector would need to be replaced with a highly sensitive camera, and the software would need to be re-written to accept the output of this detector and convert it into a chromatogram. Ultimately it would probably make more sense to have a dedicated instrument for electrochromatography rather than to retrofit these columns into an existing HPLC design. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Interpreting_the_Primary_Literature/07_Instructors_Manual/03_Separations%3A_Colloidal_Particle_Column.txt |
Article: C.N. LaFratta, I. Pelse, J.L. Falla, M.A. Palacios, M. Manesse, G.M. Whitesides, and D.R. Walt, “Measuring atomic emission from beacons for long-distance chemical signaling,” Anal Chem. 2013, 85, 8933-8936.
This paper describes a method of sending information via atomic emission signals from emergency beacons consisting of alkali-metal doped string soaked in methanol. The authors describe a telescope for reliably quantifying atomic emission from three alkali metals at a distance of up to 1.7 km from the flame. This novel application is engaging to students while also addressing instrument design and fundamentals of atomic emission. Consequently, the assignment is designed to be used after previous in-class discussions of AES theory and instrumentation. Students who have taken linear algebra may appreciate the mathematics used to determine the instrument response matrix on p. 8935; however, the questions are designed so that it is not critical to understand these mathematical sections of the paper as long as students reach some conceptual understanding of why the response matrix is used (in-class question Q2).
Out-of-Class Questions
The out-of-class questions ensure students understand new terms (Q1, Q5) and review concepts from previous class periods on atomic emission (Q2, Q3, Q6).
Q1. Define “infochemistry”.
According to the abstract, infochemistry is “encod[ing] and transmit[ting] information using chemistry instead of electronics.”
Q2. In this paper, the salts are atomized by combustion (i.e., a flame). What other methods for atomization for atomic emission are available?
Individual instructors may vary in their coverage of atomization methods for emission, but common methods presented in textbooks include arcs, sparks, lasers, and of course, inductively coupled plasma (ICP).
Q3. A methanol and air flame like the one used here typically obtains a maximum temperature of 2000 °C. What is the maximum temperature of an ICP? Why are high temperatures necessary for AES?
The maximum temperature of an ICP is approximately 10,000 °C. The high temperature is needed to excite atomic electrons (since atomic emission, unlike atomic fluorescence, does not use light to do this). The high temperatures of an ICP also reduce the formation of complexes, minimizing interference from broad molecular absorption bands.
Q4. Read section 10C-1 on p. 273 of the textbook. Why did the authors use alkali metals, instead of other elements, for this application?
The section and page number reference given are for Skoog, Holler, and Crouch, Principles of Instrumental Analysis, 6th ed., Brooks/Cole, © 2007. Alternatively, students can be referred to "10.7.3 Quantitative Applications: Choice of Atomization and Excitation Source" in Chapter 10 of David Harvey’s freely available text Analytical Chemistry 2.0. Both texts address the fact that most other elements are not excited by the low temperatures of a methanol flame.
Q5. What do the authors mean when they say the signal is “isotropic”? How is isotropic emission an advantage for this application?
Isotropic processes occur in all directions equally. In this application, isotropic emission means that light will be emitted in all directions, so that the person transmitting the signal does not need to consider in which direction the receiver might be located.
Q6. In the custom telescope in Figure 2, what components act as wavelength selectors? What wavelength selector is typically used in AES instrumentation? Explain the advantages of these two types of wavelength selectors for each application.
The telescope uses band pass filters (BP) and dichroic mirrors (DC) as wavelength selectors. Band pass filters pass a range of wavelengths above and below specific thresholds. Dichroic mirrors typically reflect light with a wavelength lower than a threshold and transmit light of longer wavelengths. More conventional atomic emission instrumentation uses a monochromator to disperse all wavelengths so that many different elemental lines can be measured. Filters are good for applications where a small number of specific wavelengths will be measured but tend to be impractical for applications that scan across a wide wavelength range.
[In my course, we discuss both filters and monochromators as wavelength selectors. For students who have not previously discussed the advantages and disadvantages of these options, this question could be moved to the in-class portion of the assignment.]
Q7. What signal processing method was used on the data in Figure 3? What are the advantages of this method for this application?
The signal is processed using boxcar averaging. [This question can be omitted from classes that do not discuss signal processing.] Students who are familiar with these techniques, however, should note that this method can be done in real-time without intensive computational resources; it does not require a repeatable signal (as would ensemble averaging); and although it can result in loss of temporal resolution, this is not an issue for this application since the beacons should burn steadily, and no useful information is encoded in their variation in time.
Q8. Estimate the signal-to-noise ratio of the processed cesium signal in the top right panel of Figure 6.
[This question is a review of signal-to-noise calculations and can be a useful reference for the in-class discussions of the distance LOD (in-class question Q1).]
The average signal, S, is approximately 0.8 V. The peak-to-peak noise is approximately 1.3-0.6 V = 0.7 V. Assuming that the root mean square noise, N, is approximately one fifth the peak-to-peak noise gives N ≈ 0.7 V/5 = 0.14 V.
$\dfrac{S}{N}=\dfrac{0.8\: V}{0.14\: V}=6$
Q9. Go to the NIST Handbook of Basic Atomic Spectroscopic Data and look up the most intense persistent strong line(s) between 700-900 nm for Na, Li, and Ca in air. Lines indicated “P” next to the intensity refer to persistent lines, which are detectable even for low concentrations of the element in the presence of other species. http://www.nist.gov/pml/data/handbook/index.cfm#
[Along with out-of-class Q10, this question ensures that students have the information needed to answer in-class Q4.]
Sodium: 819.4824 nm
Lithium: 812.6453 nm
Calcium: 854.2089 and 866.2140 nm
Q10. Peruse the bandpass filters available from Spectrofilm.com (the vendor used for parts in this manuscript). What is the narrowest range of wavelengths that can be passed by these commercial filters? How does this compare to the width of an atomic emission line?
In addition to the lines above, the elements used in the paper were detected at 766, 852, 780 nm for K, Cs, and Rb, respectively. The closest lines are Cs at 852 nm and Ca at 854 nm. The Spectrofilm bandpass filters are available with bandwidths as narrow as 1 nm. The width of the atomic emission lines is 1.5-5 pm (even after Doppler broadening and other contributions beyond the natural line width), which means the two wavelengths will be well-separated.
In-Class Questions
Q1. Consider Figure 4.
This question addresses a fundamental property of light, LOD calculations for non-linear relationships, and the practical considerations needed in designing beacons based on emission.
1. From Figure 4, the equation for the calibration plot is $y = 50760x^{-1.94}$, an inverse square relationship as expected based on the decrease in intensity of light with distance. The standard deviation of the blank, sbl = 9 mV, which means the signal LOD = 3*9 mV = 27 mV. Plugging this value in for y in the calibration plot gives us the distance at which the signal will drop to LOD:
$0.027=50760x^{-1.94}$
$x=\left(\dfrac{0.027}{50760}\right)^{ \Large{{}^1/_{-1.94} } }$
Solving for x gives 1700 m, the reported LOD.
2. The intensity is related to distance by an inverse square relationship (rounding from -1.94 to ‑2 for the exponent, as predicted by theory). As a result, increasing the distance by a factor of 2 means that the intensity of the signal must be increased by a factor of 4.
[Students may initially try to solve this problem using the calibration curve equation from (a). It is helpful to explain that an increase in the signal means that this calibration curve will change since the same distance will give a higher signal value.]
Some students with good mathematical intuition will readily appreciate the explanation above. Students who value a more concrete mathematical explanation may find the following helpful:
$2x=2*1700\: m=3400\: m$
We need the signal at 3400 m to be 27 mV, assuming that the LOD is improved by increasing the signal rather than decreasing the noise. The current signal at 3400 m is 7 mV, as shown below:
$y=50760(3400)^{-1.94}=0.007$
If the signal LOD is 27 mV, then this means that the signal must increase by a factor of 27/7, or 3.9. (The discrepancy between the 3.9-fold increase in intensity calculated here and the 4-fold increase estimated above arises because the exponent used in these explicit calculations is -1.94 rather than -2.)
3. For this question, I encourage the students to visualize the process of preparing the flares, burning them, and detecting the light. Possible complications could involve solubility limits of the salts being used and self-absorption of the emission at high concentrations.
If desired, students can investigate this question more quantitatively. The supporting information states that between 15-375 µL of 1 M solutions of the nitrate salts were used to prepare the beacons. Students can research the solubility limits for the three salts, which are very near 1 M. At 20 °C, the limits for nitrate salts of cesium, potassium, and rubidium are 23 g, 33 g, and 53 g per 100 g of water respectively. Ignoring changes to solvent volume, these concentrations correspond roughly to 1.2 M, 3.3 M, and 3.6 M. These values mean that increasing the concentration of each salt in the beacon by a factor of 4 would be impossible to do simply by increasing the concentrations of the solutions used to prepare the beacons. Depending on the time allotted for this question, the supporting information and solubility limits can be given to the students so that they can do these calculations themselves, or this information can be presented by the instructor once all the groups have reported on their discussions.
Q2. In the top panels of Figure 6, why is it that the cesium signal can be the highest magnitude in the raw data, but is at the “low” level after decoding in the processed data?
This question is meant to lead into Q3, which asks students to explain the instrument response matrix. The emission of Cs is higher at all concentrations than the emission of the other elements. The high and low levels of Cs relative to the other elements are only apparent after normalization using the instrument response matrix. The higher intensity of the Cs emission compared to the K and Rb emission at equal concentrations (as seen in Figure 3) is one of the more easily grasped difference between the three “channels” used to encode information in the beacons.
Q3. What is the purpose of the instrument response matrix and what factors affect it?
While the mathematics on p. 8935 of the article may not be accessible to students who have not taken linear algebra, the text does provide some guidance here, noting on the same page that “The 3 × 3 instrument response matrix, R, corrects for differences between the three channels of the detector and any crosstalk.” The key is for students to consider what these two items mean and explain them in their own words. Difference between the three channels include differences in the intensity of the emission from the three metals, as highlighted in the previous question, Q2. Other differences might include the sensitivity of the PMTs to specific wavelengths and other imperfections in the instrumentation. In my experience, this may be the first time that students consider, for example, that filters may not be perfect. If students have not looked up the term “crosstalk” or learned about it in a previous class, a short conversation about what this term means is helpful. In particular, students will probably need to be told that crosstalk can be spectral as well as electrical.
Q4. The authors suggest that greater information density could be achieved by adding Na, Li, and Ca to the signal. Do you think that this would be feasible? Support your answer using the information you looked up for out-of-class questions 9 and 10.
See notes for out-of-class Q10 above. Make sure that the students consider both the width of the bandpass filters and the width of the emission lines. Encourage students to consider the added complexity of the instrument with the addition of three more channels. For students who have not taken a physics class with an optics component, it may be helpful to tell them that light is lost at each interface (mirror, filter, etc.) and have them consider the effect of additional signal channels on LOD. Some students may suggest redesigning the telescope to include a monochromator, leading to a discussion of cost and the increased complexity of instruments that involve moving parts.
Q5. The authors suggest that this beacon could be used to transmit data in resource-poor environments, such as natural disaster sites. Evaluate the feasibility of this application given the information in the manuscript.
Students should consider the distance LOD, the information density, the susceptibility of the technique to sunlight and adverse weather conditions. When students have concluded fairly quickly that the telescope is not practical for emergency situations, I often start a conversation about whether this research and the idea of “infochemistry” in general might have other applications or value. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Interpreting_the_Primary_Literature/07_Instructors_Manual/04_Spectroscopy%3A_Infochemistry_using_Atom.txt |
Article: R.A. Sowell, K.E. Hersberger, T.C. Kaufman, and D.E. Clemmer, “Examining the proteome of Drosophila across organism lifespan,” J. Proteome Res., 2007, 6, 3637-3647.
This article describes an in-depth study of the changes in the identity and abundance of the proteins present in fruit fly heads as a function of fly age. Three methodologies, LC-MS/MS, SCX-LC-MS/MS, and LC-IMS-MS, were used, and the article discusses the relative sequence coverage obtained by each technique. The assignment assumes prior knowledge of ESI, common mass analyzers, and peptide fragmentation patterns (i.e., b and y ions). The article by Gary L. Glish and Richard W.Vachet, “The basics of mass spectrometry in the twenty first century,” Nature Rev. Drug Discov. 2003, 2, 140-150 is a good resource. No previous knowledge of IMS is required, as briefly researching this technique is part of the assignment. The LC-IMS-MS is a home-built instrument, which gives students an opportunity to identify components in an instrument block diagram. Additionally, the 2D separations demonstrate the “Big Data” obtainable by shotgun proteomics and provide opportunities for students to practice interpreting complex data plots. The latter part of the discussion requires more biological background knowledge, and the assignment has been designed to focus more heavily on the earlier section of the paper describing the separations and MS instrumentation. However, questions could be added on the later sections in a class for biochemistry majors.
Out-of-Class Questions
Q1. In the paper, the authors focus solely on proteins expressed in the fly’s head.
1. The authors give two reasons to focus on the flies’ heads as opposed to the bodies (p. 3638). First, previous studies of the transcriptome demonstrated an observable difference between fly heads and bodies. Second, studying the head makes the work more applicable to studies of neurology. [On p. 3637, the authors list many reasons to study fruit flies, but none of these explanations are specific to studies of the head.]
2. From p. 3638: “Briefly, heads are obtained as follows: animals were anesthetized by exposure to CO2 gas, transferred to Nalgene tubes, and frozen by submerging the tube in liquid nitrogen; tubes were shaken by hand several times to separate body parts, and heads were collected over a bed of dry ice and stored at -80 °C until further use.”
Q2. Based on the details in the sample preparation section, about how many flies were used in this study?
From p. 3638: The authors pooled 175 heads/time point for 8 time points. This pooling strategy was conducted twice, for a total of 2*175*8 = 2800 flies. The authors state that “time-point 9 includes only 100 heads due to the lack of living animals that remained at later ages.” So the total number of flies can be estimated as 2900.
Q3. Review or research the chromatographic method SCX. What is SCX, and how does it work?
SCX stands for strong cation exchange chromatography, which is a separation method based on the affinity of the cations in the sample for an anionic resin (e.g., with sulfonic acid residues) as the stationary phase. Retention times depend on ionic interactions, and therefore on the charge and size of the ions.
If students have not covered ion exchange chromatography previously, they can be directed to p. 635 in the 8th edition Harris text, p. 839 in the 6th edition Skoog text, or "12.6.2 Ion-Exchange Chromatography" in Chapter 12 of the freely available Harvey text, Analytical Chemistry 2.0.
Q4. What ionization method was used for these experiments? What is one advantage of this ionization method for these experiments?
Electrospray ionization (ESI) was used for the LC-IMS-MS experiments (see Figure 1). Although unspecified, it is most likely that ESI was also used for the LC-MS/MS experiments and SCX-LC-MS/MS experiments.
Advantages of ESI for this application include its easy coupling to condensed phase effluent from LC, soft ionization with minimal fragmentation for easy database searching, and multiple charges for MS analysis of heavier analytes (although singly-charged tryptic peptides would typically be within the mass range of both a quadrupole and a TOF).
Q5. Look up ion mobility spectrometry (IMS). How does this technique work? How does it differ from TOF-MS in the experimental parameters and the separation of analytes?
In IMS, the drift tube is filled with a low pressure of buffer gases (~2.6 Torr He and 0.1 Torr N2 in this work), and the ions are separated in a weak electric field (~11.67 V/cm in this work) based on ion mobility, which depends on mass, cross-sectional area, and charge. Cross-sectional area influences the drift time of the ions because ions with larger cross-sectional areas will experience more collisions with the buffer gas, resulting in longer drift times. The conditions in the drift tube are in contrast to those of a TOF flight tube, which consists of a field-free region kept at high vacuum (~10-6 Torr). As a result, TOF analyzers separate ions based on mass and charge. Cross-sectional area has a negligible influence on flight time because the probability of collisions is low.
Q6. What is the total pressure in the IMS drift tube? Calculate the mean free path of an ion in an IMS drift tube and compare it to the mean free path of an ion in the mass analyzer used for the MS analysis part of the IMS-MS experiments.
The total pressure in the IMS drift tube is 2.6 Torr + 0.1 Torr = 2.7 Torr or 2.7×103 mTorr.
Mean free path, λ, can be determined by the equation
$\lambda= \dfrac{k_B T}{\sqrt{2} \pi d^2 P}\approx \dfrac{5\: cm}{P\: (mTorr)}$ for a small molecule at room temperature
where kB is Boltzmann’s constant, T is the temperature, d is the diameter of the molecule, and P is the pressure. [I present this simplified version of the equation in class before this assignment is due.]
For the IMS, $\lambda \approx 5\: cm/(2.7\times 10^3) = 2\times 10^{-3}\: cm$
For the TOF, $\lambda \approx 5\: cm/(10^{-3}) = 5\times 10^3\: cm$
The much shorter mean free path in the IMS results in collisions which separate ions in drift time based on cross-sectional area.
Q7. Consider Figure 1.
1. The mass analyzer on the IMS-MS instrument is a time-of-flight (see Figure 1).
2. The mass analyzer on the LC-MS-MS is a quadrupole.
3. A TOF has higher mass resolving power than a quadrupole. A TOF typically has a value of R in the thousands or even above 10,000, while a quadrupole usually has unit resolution (Δm = 1 with R = m/Δm depending on mass).
Q8. Briefly and in your own words describe how mass spectra were assigned (matched) to specific peptides.
In evaluating student responses, I look for correct application of the underlined concepts:
The spectra were matched against a database (the National Center for Biotechnology Information Drosophila protein database) using an algorithm (MASCOT). At least one unique peptide hit was required for assignment of a protein, and results from IMS-MS experiments were confirmed visually to reduce the false positive rate. (See p. 3639).
Q9. What “semi-quantitative” methods were used to measure changes in protein abundance with fly age in this work?
Changes in protein abundance were estimated based on (1) the number of unique peptide hits identified for each protein and (2) the integration of peaks from the LC-IMS-MS data sets. In method (1), the number of hits was normalized to the number of hits at time point 2 (ages of 8-14 days). Method (2) was used to validate or further investigate trends identified by method (1), and the integration values were normalized to the total ion signal at each time point.
Q10. The bottom part of Figure 2 shows the base peak chromatograms for three separations. Define base peak. What is a base peak chromatogram?
The base peak is the most intense peak in the mass spectrum. (The other peak intensities are often normalized to this peak, which is assigned an intensity of 100.)
A base peak chromatogram plots the intensity of the base peak in each mass spectrum (y-axis) versus the separation time (x-axis).
Q11. How effective was the SCX-LC-MS/MS method at identifying peptides compared to the LC-MS/MS method? What do you think is the reason for this difference in performance when an SCX step is added?
The addition of the SCX step greatly increased the number of peptides (and therefore proteins) identified from 1102 peptides assigned to 367 proteins to 5430 peptides assigned to 1437 proteins.
The improved performance is likely due to the increased peak capacity with the two-dimensional separation. By further separating the samples before MS analysis the data are cleaner and mass spectra are easier to match.
Q12. Which proteins in Figure 6 change with the flies’ age? What biological processes are associated with these proteins?
Prophenoloxidase and CG4784 both decrease with age. “[P]rophenoloxidase is an enzyme that catalyzes the synthesis of melanin as a defense response and part of wound healing” (p. 4643). CG4784 is a component of the cuticle (exoskeleton).
Q13. How do the authors’ results compare to results obtained by more traditional methods used to study proteins?
[For this section, students who do not have a biology background will benefit from an explanation of the central dogma of biology: DNA $\rightarrow$ RNA $\rightarrow$ protein and the information that Western blotting is a traditional method of assessing protein expression in molecular biology.]
The authors’ data for certain proteins agree with previous Western blotting results. For example, data on heat shock protein 22 and larval serum protein 2 are consistent with previous studies (p. 3644).
The authors’ results show mixed comparisons to previous data on transcript levels. They note significant disagreement with reference 15 in terms of whether changes were observed with age and in which direction (p. 3644-3645). The authors suggest that fly strain, sex, or temperature of growth could account for some of these differences. [There is also precedent in the literature to suggest that transcript levels and protein levels are not always tightly correlated due to additional regulatory components.] In a few cases, the authors’ data on protein levels corresponded to the transcript data. For example, CG15093 and Punch were down-regulated with age in both studies, while Cyp28a5 and S-adenosyl-L-homocysteine hydrolase were up-regulated.
In general, the biological trends observed in this study seem reasonable: down-regulation of metabolic components and reproductive proteins with age and changes to proteins involved in defense response (p. 3645-3646).
In-Class Questions
Q1. Consider your answer to out-of-class question 7c. How are tandem MS experiments achieved on the LC-MS-MS system? How does fragmentation occur, and what types of fragments are formed? How do these fragments yield information about the peptides in the sample?
As noted in out-of-class question 7c, the mass analyzer is a quadrupole. Because this is a single quadrupole (rather than a triple quad), the instrument is being operated as a tandem-in-time MS-MS instrument. The precursor ion is selected and fragmented by collision-induced dissociation (CID) in the mass analyzer. The product ions are then scanned out to obtain the MS-MS spectrum. For peptides, CID typically produces b and y ions. A series of b or y ions can be used to de novo sequence the peptide or matched to a database mass spectrum for identification of the peptide and corresponding protein.
[I spend a class period discussing MS-based proteomics before assigning this article. The article by Ruedi Aebersold & Matthias Mann, “Mass spectrometry-based proteomics,” Nature, 2003, 422, 198-207 is a good resource for this.]
Q2. Proteomics experiments often produce vast quantities of data.
1. The panels on the far left are heat maps showing the LC-IMS results. The false color indicates the peak intensity for each retention time, drift time pair. Each panel on the right is a mass spectrum from the TOF analyzer and corresponds to a single pixel in one of the three left-hand panels. In other words, each pixel in each panel on the right has a corresponding mass spectrum – a huge quantity of data.
2. It is often productive to have students discuss this question along with question (c) since few students think that no raw data should be shared. If the discussion is foundering, I find it helpful to point out how little biological interpretation has been done of the results in this paper (considering the huge number of proteins identified). As a results, future targeted biological studies may wish to cross-reference results with proteomics studies like these.
3. Some possibilities include individual researchers through their grants, government organizations like the National Science Foundation and National Institutes of Health, the publishers of the journals the data appears in, or other researchers who choose to access the data. You may think of others.
The duration of data availability and the need to make data formats compatible with new software versions or programs are additional considerations. I have sometimes assigned an article by Julie Manoharan, “Thank you for sharing,” Biotechniques, 2011 as a follow-up reading.
Q3. Quantitation by mass spectrometry in proteomics is challenging because the wide variety of peptide chemistries result in variable MS signal between peptides, and ionization efficiency can also change over the course of an LC run.
1. For complex mixtures like these protein digests, numerous internal standards would be needed to accurately reflect all the analytes. Additionally, these internal standards would need to be readily separated from the analytes during the chromatographic or mass spectrometric steps. This would be difficult and likely expensive. [Quantitative mass spectrometry experiments for proteomics often used isotope labeling of some kind to introduce internal standards.]
Although internal standards were not used, the authors were still able to compare results semi-quantitatively using peptide hits and peak integration as proxies for protein abundance. To improve the quality of their data interpretation, the authors only looked at changes in abundance with age for proteins that had a difference of at least 5 peptide hits between time points. The authors also separated the proteins out by the magnitude of change in their peptide hits (≥1.5, ≥2.0, or >10) to separate small and large changes in expression. (See p. 3644 for detail.)
2. The open symbols are from the normalized peak intensities (integrated as described on p. 3640). The closed symbols are the normalized number of peptide hits.
3. Based on Figure 6, the semi-quantitative methods agree with one another, which suggests that they are at least reasonably reliable. As discussed in out-of-class question 13, the results from these semi-quantitative method often (but not always) agree with results of more traditional measures of protein expression, further suggesting that the methods are reliable.
The comparisons are limited in that they are semi-quantitative, relative for each protein, and may not capture changes in protein abundance near the detection limit accurately. While imperfect, these methods are practical with current technology and seem to provide useful biological information. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Interpreting_the_Primary_Literature/07_Instructors_Manual/05_Mass_Spectrometry%3A_Fruit_Fly_Proteomic.txt |
Article: P.E.M. Phillips, G.D. Stuber, M.L.A.V. Heien, R.M. Wightman, and R.M. Carelli, “Subsecond dopamine release promotes cocaine seeking,” Nature, 2003, 422, 614-618.
This article describes the results of experiments using fast scan cyclic voltammetry to monitor dopamine release in the brains of drug-seeking rats. While students do not need any specific background in neuroscience to understand the article, they should be reminded to look up unfamiliar biological terms as needed. Additionally, it is important to discuss Q13 from the out-of-class assignment at the start of the in-class portion to ensure that the students understood the experimental design and key results reported in each figure. Q1 of the in-class questions assumes that the students have previously encountered common spectroscopic methods and mass analyzers for MS; if this is not the case, this question can be omitted.
Out-of-Class Questions
Many of the out-of-class questions are designed to encourage the students to familiarize themselves with the study and related neuroscience terminology (Q1-Q4, Q6-Q7, Q9).
Q1. What is the difference between tonic and phasic dopamine signaling?
From the abstract, tonic signaling occurs on a minute-to-minute timescale, while phasic signaling occurs on a subsecond timescale.
Q2. Look up any other unfamiliar words in the abstract. Then summarize this study in one complete sentence.
Two common terms that students need to research are ventral tegmental area and nucleus accumbens:
ventral tegmental area: a section of the midbrain composed of dopamine-containing neurons that are involved in drug-seeking and reward circuitry
nucleus accumbens: a section of the forebrain involved in reward, pleasure, and laughter as well as aggression and fear
I look for some aspect of the four underlined components of the sentence below when evaluating student summaries of the study:
This study investigated subsecond (or phasic) dopamine signaling in the nucleus accumbens using fast scan cyclic voltammetry (or electrochemical measurements) in rats engaged in drug taking.
Q3. How were the rats taught to self-administer cocaine? How was the cocaine delivered?
As described on p. 617, the rats were trained to self-administer cocaine during 2-h sessions during which they were presented with a lever illuminated by a cue light. When the rats pressed the lever, the cocaine dose was administered through a jugular catheter.
Q4. What stimulus accompanied cocaine delivery?
As described on p. 617, cocaine administration was accompanied by a change in lighting and “a continuous auditory tone”.
Q5. What were the working and reference electrodes in this study? Why wasn’t a counter/auxiliary electrode necessary?
As described on p. 617, the working electrode was a carbon fiber microelectrode and the reference electrode was Ag/AgCl.
The latter question is meant for classes that have discussed the use of microelectrodes previously. For the very small currents that pass through ultramicroelectrodes, ohmic losses are negligible, and a two-electrode system does not pose any problems to accurate measurement.
Q6. Sketch the potential applied to the working electrode as a function of time. Include as much detail in your plot as possible.
Students should label the time for each half-wave and the time between cycles in addition to the applied potentials. Some students may not initially realize that they can calculate these times from the maximum and minimum applied potentials (-0.6 V and +1.4V) and the scan rate (400 V/s), but usually at least one student in a group will identify this calculation. The potential on the electrode between scans is not specified in the manuscript. Students often assume that the potential is held at -0.6 V. The electrode was probably allowed to float (no current sourced or sunk) or may have been held at an unspecified potential relative to the reference. The plot below does not indicate the potential between scans.
Q7. How many cyclic voltammograms (scans) were acquired per second?
Scans were acquired at a frequency of 10 per second (one scan every 100 ms, as described on p. 617).
Q8. How does the scan rate in these experiments compare to a typical CV scan rate? How does scan rate usually affect CV data?
The scan rate for these experiments was 400 V/s. If students have performed a laboratory exercise on cyclic voltammetry, they can refer to the range of scan rates used (e.g., 100 mV/s to 1 V/s). Otherwise, this information may come from a previous lecture or their textbook. For example, p. 737 of the Skoog textbook gives a 50 mV/s scan rate as an example.
For reversible reactions, peak current (ip) increases linearly with the square root of the scan rate (v), as described by the Randles-Sevcik equation, $i_p=2.686×10^5 n^{3/2} ACD^{1/2} v^{1/2}$.
Q9. How did the authors correct for current from interferents, movement of the animal, and pH changes in the extracellular space?
On page 617, in the methods section on fast-scan cyclic voltammetry, the authors note that artifacts from pH change or movement were correct by using a differential measurement was made between currents that occurred for dopamine oxidation at +0.7 V vs Ag/AgCl and another potential at which dopamine oxidation does not occur, but the interfering currents did.
This question, along with Q10, should highlight for the students the problem of selectivity at an unmodified electrode. Students should be referred to these questions as needed in their discussions of in-class Q3.
Q10. Complete the table below summarizing the means used to demonstrate that their signal came from dopamine release.
Experiment
How did this demonstrate the signal was from dopamine release?
anatomical
Post-mortem histology was used to verify that the electrode was placed in the nucleus accumbens.
physiological
Electrical pulses were used to stimulate dopamine release and confirm its detection before and after each experiment.
chemical
Cyclic voltammograms recorded in the live animals were compared to CVs from in vitro standards.
pharmacological
An MAOI was administered to suppress the potential interferent, DOPAC, and no difference in signal intensity was observed.
Q11. Draw the half-cell reaction for the oxidation of dopamine to dopamine quinone.
Q12. Consider the CV inset in Figure 1a. Label the oxidation and reduction half-waves. Was the oxidation of dopamine reversible? How can you tell?
The oxidation was not reversible, as evidenced by the lack of current on the reduction half-wave.
Q13. Calculate the S/N ratio for the trace in Figure 3a.
This question is an opportunity for students to practice calculating SNR while also encouraging them to examine the data in a paper closely.
Measuring with a ruler, I estimate that the peak-to-peak noise is approximately 3/16” while the peak height is approximately ½”. Treating the peak-to-peak noise as roughly five times the rms noise gives
$\dfrac{S}{N}=\dfrac{0.5}{0.2\left(\dfrac{3}{16}\right)}=13$
Q14. Consider Figures 1-4. In the table, summarize the experimental design used for each experiment and the major results. In the column labeled stimulus/conditions, record what stimulus, if any, was used to elicit dopamine release or drug-seeking behavior. If no stimulus was used, note the conditions of the experiment. In the column labeled dopamine release, estimate the concentration of dopamine released in nM. In the column labeled timing, summarize any important findings about how the timing of the dopamine release related to the timing of the stimulus.
This question is helpful for the in-class discussion since it collects the most relevant points about each figure into one location. A quick check of student answers to this question is helpful before discussions start to ensure that the students are reading the figures correctly.
Figure
Stimulus/Conditions
Dopamine Release (nM)
Timing of Response
1
electrical stimulus train (24 pulses at 60 Hz)
~700 nM
Appears instantaneous, but not quantified in the text
2
no stimulus, rats were seeking and obtaining cocaine by lever press
64.9 ± 16.1 nM
(post-response)
7.7 ± 0.6 s before lever press, small peak;
3.7 ±0.5 s before lever press is final approach which starts large peak.
3
audiovisual stimulus
(no cocaine)
93.9 ±12.2 nM
peak starts at 0.1 ±0.5 s after stimulus
4
electrical stimulus train (24 pulses at 60 Hz every 120 s)
n/a
animals had highest probability of lever-pressing 5-15 s after stimulus
Notable features of the data summarized in this table are that there was a spike in dopamine before cocaine was actually administered (Figure 2), that a large dopamine response (comparable to the response from cocaine) was observed in the absence of drug simply by recreating the conditions of drug-use (Figure 3) and that dopamine release triggered drug-seeking behavior (Figure 4).
In-Class Questions
Q1. Suggest some advantages of electrochemical detection of dopamine for this application, compared to spectroscopic or mass spectrometric detection.
This question can be omitted in classes that have not covered spectroscopic and mass spectrometric methods. A major advantage is the ability to miniaturize electrodes for electrochemical detection for implantation into the brain of a rat. Because the measurements can be made in situ without derivatization or removing samples via microdialysis, the authors are able to achieve the temporal resolution needed for measurements of phasic dopamine signaling. The sensitivity of electrochemical detection, especially when miniaturized, is also an advantage since the dopamine concentrations are in the nanomolar range. Spectroscopic methods are often less amenable to miniaturization because of dependence of signal strength on path length.
Q2. Why was the cyclic voltammetry data not sufficient to identify the signal as coming from dopamine? (Why do the additional experiments that you summarized in out-of-class question #10?)
Refer students to out-of-class questions Q8 and Q9. Students who are unfamiliar with biological experiments may be surprised at the large number of controls needed. Additionally, this question is a useful opportunity to discuss the selectivity of electrochemistry at unmodified electrodes, which is often poor, and the use of controls or chemically modified electrodes to address this.
Q3. In Figure 1, where do the values for the color scale in the bottom panel come from? What plot would you obtain if you took a slice horizontally through the color plot? What would you obtain for a vertical slice?
This question is an opportunity for students to interpret 2D data plots requiring color mapping as a third dimension. Most students readily identify the color scale as representative of the current, but further thought and occasional prompting are required to answer the subsequent questions. The blue trace above the color map in Figure 1 is a horizontal slice. These slices show the current at a given potential as a function of time. If one chooses the correct potential, i.e., +0.7 V vs Ag/AgCl, this current is correlated with dopamine concentration, yielding a trace that shows the spike and subsequent decline in dopamine concentrations after the stimulus. The red cyclic voltammogram is a vertical slice through the data, folded back on itself. The continuous appearance of the data along the time axis is only possible because the authors used fast-scan cyclic voltammetry and were able to record a CV every 0.1 s (out-of-class Q6).
Q4. Suggest one software-based and one hardware-based method to improve the S/N or Figure 3a, and justify your choices.
This question is best suited for an instrumental class that has covered a variety of hardware and software methods to reduce noise. It may be helpful to quickly review hardware and software methods from earlier in the semester before the students begin their discussion of this question. Students can generate a list of hardware methods, such as grounding, shielding, hardware filters, lock-in amplifiers, etc. and a list of software methods, such as ensemble averaging, boxcar averaging, moving average, Fourier transform based filters, etc. In evaluating these possibilities, students should consider the fact that the goal is to collect data with high temporal resolution, which precludes extreme low-pass filtering. Additionally, students should consider that a rat brain is not a static, homogeneous system. Individual rats and even individual stimuli in the same rat may not consistently produce the same response. Instead, individual responses are likely to vary somewhat both in their timing and their magnitude.
Q5. Individuals recovering from drug addiction are often counseled to avoid “triggers”, including locations and situations in which they previously used drugs. Do the findings in this study support this advice? Why or why not? Cite a specific figure or figures in your answer.
Encourage students to build a logical progression from one dataset to the next, logically linking stimuli and responses, as shown below. Further conversation could include the limitations of animal models for complex human behaviors. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Interpreting_the_Primary_Literature/07_Instructors_Manual/06_Electrochemistry%3A_Fast_Scan_Cyclic_Vol.txt |
This module introduces students to ways of thinking about and working with data using, as a case study, the analysis of 1.69-oz packages of plain M&Ms. The module is divided into six parts:
Part I. Ways to Describe Data
Part II. Ways to Visualize Data
Part III. Ways to Summarize Data
Part IV. Ways to Model Data
Part V. Ways to Draw Conclusions From Data
Part VI. Now It’s Your Turn!
Interspersed within the module’s narrative is a series of investigations, each of which asks students to stop and consider one or more important issues. Many of these investigations draw upon a data set that consists of 30 samples of 1.69-oz packages of plain M&Ms. This case study is meant to serve as an introduction to data and to data analysis and, as with any introduction, it considers a small number of topics; additional resources that provide a deeper introduction to data and to data analysis are listed in Appendix 1 of the case study.
Introduction to Data and the Analysis of Data
At the heart of any analysis is data. Sometimes our data is categorical and sometimes it is numerical; sometimes our data conveys order and sometimes it does not; sometimes our data has an absolute reference and sometimes it is has an arbitrary reference; and sometimes our data takes on discrete values and sometimes it takes on continuous values. Whatever its form, when we gather data our intent is to extract from it information that can help us solve a problem. In this case study we consider how to find meaning in data, including ways to describe data, to visualize data, to summarize data, to model data, and to draw conclusions from data.
If we are to consider how to describe, to visualize, to summarize, to model, and to draw conclusions from data, then we need some data with which we can work. For the purpose of this case study, we need data that is easy to gather and easy to understand, and that allows us to ask interesting questions; it is helpful, as well, if we can find expected results for at least some of our questions so that we can check our analysis. It also is helpful if you can gather your own data so that you can repeat and verify our work, or so that you can extend our analysis. A simple system that meets these criteria is to analyze the contents of bags of M&Ms. There is a rich history of using M&Ms to introduce or to illustrate the analysis of data in a variety of disciplines; Appendix 1 provides examples of such studies. Although this system may seem trivial, keep in mind that reporting the percentage of yellow M&Ms in a bag is analogous to reporting the concentration of Pb2+ in a sample of soil: both express the amount of an analyte present in a unit of its matrix.
Interspersed within the case study’s narrative are a series of investigations, each of which asks you to stop and consider one or more important issues. Some of these investigations include data for you to analyze; you can use this link to access an on-line version of this case study with interactive figures. For additional coverage of the topics in this case study, see Chapter 4 of Analytical Chemistry 2.0, which is available using this link, and the additional resources cited within.
01 Student Handout
In the introduction to this case study we identified four contrasting ways to describe data: categorical vs. numerical, ordered vs. unordered, absolute reference vs. arbitrary reference, and discrete vs. continuous. To give meaning to these descriptive terms, let’s consider the data in Table 1.
Table 1. Distribution of Yellow and Red M&Ms
bag id
year
weight (oz)
type
# yellow M&Ms
% red M&Ms
total M&Ms
rank (total M&Ms)
a
2006
1.74
peanut
2
27.8
18
sixth
b
2006
1.74
peanut
3
4.3
23
fourth
c
2000
0.80
plain
1
22.7
22
fifth
d
2000
0.80
plain
5
20.8
24
third
e
1994
10.0
plain
56
23.0
331
second
f
1994
10.0
plain
63
21.9
333
first
The entries in Table 1 are organized by column and by row, with the first row (shaded here for emphasis) identifying the variables used to describe the data. Each additional row is the record for one sample and each entry in a sample’s record provides information about one of its variables; thus, the data in the table lists the result for each variable and for each sample.
Investigation 1.
Of the variables included in Table 1, some are categorical and some are numerical. Define these terms and assign each of the variables in Table 1 to one of these terms.
Investigation 2.
Suppose we decide to code the type of M&M using 1 for plain and 2 for peanut. Does this change your answer to Investigation 1? Why or why not?
Investigation 3.
Categorical variables are described as nominal or ordinal. Define the terms nominal and ordinal and assign each of the categorical variables in Table 1 to one of these terms.
We also can use a numerical variable to assign samples to groups. For example, we can divide the plain M&Ms in Table 1 into two groups based on the sample’s weight. What makes a numerical variable more interesting, however, is that we can use it to make quantitative comparisons between samples; thus, we can report that there are 14.8× as many plain M&Ms in a 10-oz. bag as there are in a 0.8-oz. bag. Although we can complete meaningful calculations using any numerical variable, the type of calculation we can perform depends on whether or not the variable’s values have an absolute reference.
Investigation 4.
A numerical variable is described as either ratio or interval depending on whether it has (ratio) or does not have (interval) an absolute reference. Explain what it means for a variable to have an absolute reference and assign each of the numerical variables in Table 1 as either a ratio variable or an interval variable. Why might this difference be important?
Finally, the granularity of a numerical variable provides one more way to describe our data.
Investigation 5.
Numerical variables also are described as discrete or continuous. Define the terms discrete and continuous and assign each of the numerical variables in Table 1 to one of these terms. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Introduction_to_Data_and_the_Analysis_of_Data/01_Student_Handout/01_Part_I%3A_Ways_to_Describe_Data.txt |
Suppose we are interested in characterizing 1.69-oz (47.9-g) packages of plain M&Ms. We obtain 30 bags (ten from each of three stores) and, for each bag, report the number of blue, brown, green, orange, red, and yellow M&Ms—for yellow, the number in parentheses is the number of yellow M&Ms in the first five drawn from the bag—and their combined net weight. Table 2 summarizes the data for the last six samples. The full set of data for all 30 samples is available as a separate spreadsheet or R file.
Table 2. Source, Distribution, and Net Weight of Plain M&Ms in 1.69-oz Bags
bag
store
blue
brown
green
orange
red
yellow
net weight (g)
25
CVS
7
13
0
4
15
16 (2)
48.212
26
Target
6
15
1
13
10
14 (1)
51.682
27
CVS
5
17
6
4
8
19 (1)
50.802
28
Kroger
1
21
6
5
10
14 (0)
49.055
29
Target
4
12
6
5
13
14 (2)
46.577
30
Kroger
15
8
9
6
10
8 (1)
48.317
Having collected some data, our next step is to examine it for possible problems, such as missing values or errors introduced when we recorded the data, or to identify important variables and interesting patterns or trends within or between these variables. Although this information is embedded within the data itself, often it is difficult to see it when the data is displayed as a table, particularly if the data set is large in size. Instead, we use one or more simple visualizations of the data.
Two simple visualizations are box and whisker plots and dot plots, examples of which are shown in Figure 1 using the data for yellow M&Ms. Note that neither plot has meaningful information along the y-axis as the vertical dimension simply helps us visualize the data. The vertical distribution of points in the dot plot, for example, is the result of jittering, which offsets samples that share a common number of yellow M&Ms so that, we hope, each appears as a distinct point.
Investigation 6.
Use the dot plot in Figure 1 to deduce the general structure of a box and whisker plot, giving particular attention to the position along the x-axis of the three vertical lines that make up the yellow box and the two vertical lines that make up the whiskers on either side of the yellow box. You might begin by tabulating the number of samples that fall to the left of the box, that fall within the box, including its boundaries, and that fall to the right of the box, and the number of samples that lie to the left and to the right of line inside the box.
As suggested by the next two investigations, one way to use a box and whisker plot is to look for unexpected features in our data that merit attention, such as an oddly shaped distribution of results or an unusually large or an unusually small result for a variable.
Investigation 7.
The box and whisker plot in Figure 1 is perfectly symmetrical in that each side of the box is two units from the box’s middle line, and each whisker is six units from the box’s nearest edge. What does this symmetry suggest about how the results are distributed? Is the actual distribution of the 30 results perfectly symmetrical? If no, is this a problem?
Investigation 8.
In Figure 1 we see that the result for sample 22 falls outside the range of values included within the whiskers. Why might a result that falls outside the whiskers concern us? Does the presence of this particular point suggest a problem? How might your response change if this sample’s reported value is 0 yellow M&Ms? How might your response change if this sample’s reported value is 45 yellow M&Ms?
In addition to providing us with insight into the results for a single variable, we can use box and whisker plots and dot plots to examine differences between variables and differences within a single variable when we can divide that variable into different groups.
Investigation 9.
Figure 2 shows box and whisker plots and dot plots for all six colors of M&Ms included in Table 2 (note: even with jittering, you will not be able to see all 30 samples in these dot plots). Based on these plots, where do you see similarities and where do you see differences in the distribution of M&Ms? What do these similarities and differences suggest to you? For those distributions that do not appear symmetrical, suggest one or more reasons for the lack of symmetry. What do the relative positions of the data for brown and for green M&Ms suggest about their relative abundance in 1.69-oz packages of plain M&Ms?
Investigation 10.
Figure 3 shows box and whisker plots and dot plots for yellow M&Ms grouped by the store where the packages of M&Ms were purchased. Based on these plots, where do you see similarities and where do you see differences in the distribution of yellow M&Ms? What do these similarities and differences suggest to you? In what ways might this data be reassuring to us? Give an example of a result that might suggest we look more closely at our data.
Investigation 11.
Draw a box and whisker plot and an accompanying dot plot for the total number of M&Ms. Compare your plots to those in Figure 2 and discuss any similarities and differences.
Although a box and whisker plot provides some evidence of how a variable’s values are distributed, it is not particularly easy to see the shape of that distribution. For this we use a histogram, which displays the number of results that fall within a sequence of (usually) equally spaced bins. Figure 4, for example, shows histograms for each color of M&Ms in our data set.
Investigation 12.
For the histograms in Figure 4, where do you see similarities and where do you see differences in the distribution of M&Ms? How do the results seen here compare with your interpretation of the box and whisker plots and the dot plots in Figure 2?
One challenge when we draw a histogram is choosing the width for the bins or the number of bins. In Figure 4, for example, the bins for yellow M&Ms are five units wide—the first bin, for example, includes samples with 5, 6, 7, 8, and 9 yellow M&Ms—but the bins are two units wide for all other colors of M&Ms. There are no simple rules for determining the number or the width of bins, so it is a good idea to try several bin sizes before we settle on a final choice.
Investigation 13.
The histograms in Figure 5, from left-to-right, use bins widths of 1, 2, and 3 units, respectively. Note that the x-axis shows the specific results gathered into each bin. How does the choice of bin size affect your understanding of this data? Which of these histograms provides the best representation of the data? As part of your answer, identify what you see as the limitations of the other two histograms.
Investigation 14.
Draw a histogram for the total number of M&Ms and explain the reason(s) for your choice of bin size. Compare your plots to those in Figure 4 and discuss any similarities and any differences. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Introduction_to_Data_and_the_Analysis_of_Data/01_Student_Handout/02_Part_II%3A_Ways_to_Visualize_Data.txt |
Although box and whisker plots, dot plots, and histograms help us see qualitative patterns in our data, they do not allow us to express this information in a quantitative way. For example, in Figure 3 and in Investigation 10 we learned that the distribution of yellow M&Ms in 1.69-oz bags is relatively similar between the three different sources, although the plot for samples purchased from Target has much shorter whiskers and the individual results seem more tightly clustered than is the case for samples purchased at CVS and at Kroger, and the box for the samples purchased from Kroger is quite a bit wider than is the case for the samples from CVS and Target.
Qualitative phrases such as “relatively similar,” “much shorter,” “more tightly clustered,” and “quite a bit wider” are, frankly, fuzzy, but in the absence of a more quantitative way to characterize our data, we have little choice but to adopt such fuzzy terms. When we summarize data, our goal is to report quantitative characteristics, or statistics, that we can use to provide clearer statements about the differences and the similarities between results for different variables, or between the results for a variable and an expected result already known to us. In this part of the case study we consider several useful statistics that we can use to summarize the data for our samples.
Investigation 15.
Before we consider ways to summarize our data, we need to draw a distinction between a sample and a population. We collect and analyze samples with the hope that we can deduce something about the properties of the population. Using our data for M&Ms as an example, define the terms sample and population.
Both a box and whisker plot and a histogram suggests that the distribution of results for a single variable has two important features: its center, which presumably lies somewhere in the middle of the data, and its spread, which is suggested by the length of the whiskers in a box and whisker plot, or how quickly or how slowly the counts in a histogram’s bins decrease as we move away from the bin that has the most counts. For our purposes, we will consider two quantitative measures of central tendency—the mean and the median—and four quantitative measures of spread: the variance, the standard deviation, the range, and the interquartile range.
Central Tendency. The mean, $\bar{x}$, is the arithmetic average of all n of a variable’s results; thus
$\bar{x}=\dfrac{\sum x_i}{n}$
where xi is the result for an individual sample. The median is the middle value when the n results are ordered by rank from smallest-to-largest. If n is odd, then the median is the (n + 1)/2th; if n is even, then the median is the average of the (n / 2)th value and the ((n ⁄ 2 ) + 1)th value.
Investigation 16.
Using the data for yellow M&Ms, calculate the mean and the median for each store and discuss your results. If the mean and the median are equal to each other, what might you reasonably conclude about your data? If the mean is larger than the median, or if the mean is smaller than the median, what might you reasonably conclude about your data? A measure of central tendency is considered robust when it is not changed by one or more results that differ substantially from the remaining results. Which measure of central tendency is more robust? Why?
Spread. A sample’s variance, s 2, provides an estimate of the average squared deviation of its n results relative to its mean; thus
$s^2 =\dfrac{\sum (x_i-\bar{x})^2}{n-1}$
where xi is the result for an individual sample and $\bar{x}$ is the variable’s mean value. The standard deviation, s, is the square root of the variance.
The range is the difference between the sample’s largest value and its smallest value. A variable’s interquartile range, IQR, spans the middle 50% of its values. To find the IQR, we order the data from smallest-to-largest, and separate it into two equal parts; if the sample has an odd number of values, then we do not include the median in either part. Next, we find the median for each of the two parts. The IQR is the difference between these two medians. Note: There actually are several methods for calculating the IQR, which differ in how they divide the data into four parts. As you might expect, different methods may result in different values for the IQR. The method described here was used to create the box and whisker plots in Figures 1–3, where the width of the box is the interquartile range.
Investigation 17.
Using the data for yellow M&Ms, calculate the variance, the standard deviation, the range, and the IQR for each store and discuss your results. Is there a relationship between the standard deviation, the range, or the IQR? A measure of spread is considered robust when its value is not changed by one or more values that differ substantially from the remaining values. Which measure of spread—the variance, the standard deviation, the range, or the IQR—is the most robust? Why? Which is the least robust? Why? | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Introduction_to_Data_and_the_Analysis_of_Data/01_Student_Handout/03_Part_III%3A_Ways_to_Summarize_Dat.txt |
In Part III we made a distinction between a sample and a population, noting that a population is every member of a system that we could analyze and that a sample is the discrete subset of a population that we actually analyze. We collect and analyze samples with the hope that we can use their properties to deduce something about the population’s properties. We accomplish this by using suitable mathematical models.
Investigation 18.
So, what does it mean to build a model? Consider the histograms in Figure 4. What property of the population are we attempting to model? What do your responses imply about the model’s general mathematical form? What does it mean to test a model and how might we accomplish this?
There are a variety of ways in which we might model our data, three of which we consider in this section: the binomial distribution, the Poisson distribution, and the normal distribution.
Binomial Distribution. A binomial distribution describes the probability, P, of a particular event, X, during a fixed number of trials, N, given the probability, p, that the event happens during a single trial. Mathematically, we express the binomial distribution as
$P(X,N)=\dfrac{N!}{X!(N-X)!}\times p^X \times (1-p)^{N-X}$
where ! is the factorial symbol. The theoretical mean, μ, and the theoretical variance, σ 2, for a binomial distribution are
$μ = Np \hspace{40px} σ^2 = Np (1 - p)$
Investigation 19.
The box and whisker plot in Figure 1 includes data from the analysis of 30 samples of 1.69-oz bags of plain M&Ms. Collectively, the samples have 1699 M&Ms, of which 435 are yellow. If you pick one M&M at random from these 1699 M&Ms, what is the probability, p, that it is yellow? Suppose that this probability applies to the population of all plain M&Ms. If we draw a sample of five M&Ms from this population, what is the probability that the sample contains no yellow M&Ms? Repeat for each of 1–5 yellow M&Ms. Construct a histogram of your results and report the mean and the variance. Repeat this analysis for green M&Ms. Compare your two histograms and discuss their similarities and their differences. Using the data in Table 2, comment on the suitability of the binomial distribution for modeling the number of yellow M&Ms in samples of five M&Ms.
Poisson Distribution. The binomial distribution is useful if we wish to model the probability of finding a fixed number of yellow M&Ms in a sample of M&Ms of fixed size, but not the probability of finding a fixed number of yellow M&Ms in a single bag.
Investigation 20.
Explain why we cannot use the binomial distribution to model the distribution of yellow M&Ms in 1.69-oz bags of plain M&Ms.
To model the number of yellow M&Ms in packages of M&Ms, we use the Poisson distribution, which gives the probability of a particular event, X, given an average rate, λ, for that event. Mathematically, we express the Poisson distribution as
$P(X,λ)=\dfrac{e^{-λ} λ^X}{X!}$
The theoretical mean, μ, and the theoretical variance, σ 2, are both equal to λ.
Investigation 21.
The histograms in Figure 4 include data from the analysis of 30 samples of 1.69-oz bags of plain M&Ms. Collectively, the samples have an average of 14.5 yellow M&Ms per bag. Suppose this rate applies to the population of all 1.69-oz bags of plain M&Ms. If you pick a 1.69-oz bag of plain M&Ms at random, what is the probability that it contains exactly 11 yellow M&Ms? Repeat for each of 0–29 yellow M&Ms. Construct a histogram that shows the actual distribution of bags of M&Ms for each of 0–29 yellow M&Ms, using a bin size of 1 unit, and overlay a line plot that shows the predicted distribution of bags; be sure to you use the same scale for each plot’s y-axis. Comment on your results.
Normal Distribution. The binomial distribution and the Poisson distribution are examples of discrete functions in that they predict the probability of a discrete event, such as finding exactly two green M&Ms in the next bag of M&Ms that we open. Not all data we might collect on M&Ms, however, is discrete.
Investigation 22.
Explain why we cannot use the binomial distribution or the Poisson distribution to model data for the net weight of M&Ms in Table 2.
To model the net weight of packages of M&Ms, we use the normal distribution, which gives the probability of obtaining a particular outcome from a population with a known mean, μ, and a known variance, σ 2. Mathematically, we express the normal distribution as
$P(X)=\dfrac{1}{\sqrt{2πσ^2}}e^{\large{-(X-μ)^2 ⁄2πσ^2}}$
Figure 6 shows the normal distribution curves for μ = 0 and for variances of 25, 100, and 400.
Investigation 23.
Using the curves in Figure 6 as an example, discuss the general features of a normal distribution, giving particular attention to the importance of variance. How do you think the areas under the three curves from -∞ to +∞ are related to each other? Why might this be important?
Because the equation for a normal distribution depends solely on the population’s mean, μ, and variance, σ 2, the probability that a sample drawn from a population has a value between any two arbitrary limits is the same for all populations. For example, 68.26% of all samples drawn from a normally distributed population will have values within the range μ ± σ, and only 0.621% will have values greater than μ + 2.5σ; see Appendix 2 for further details.
Investigation 24.
Assuming that the mean, $\bar{x}$, and the standard deviation, s, for the net weight of our samples of M&Ms are good estimates for the population’s mean, μ, and standard deviation, σ, what is the probability that the contents of a 1.69-oz bag of plain M&Ms selected at random will weigh less than the stated net weight of 1.69 oz? Suppose the manufacturer wants to reduce this probability to no more than 5%: How might they accomplish this?
For a binomial distribution, if N × p ≥ 5 and N × (1 - p) ≥ 5, then a normal distribution closely approximates a binomial distribution; the same is true for a Poisson distribution when λ ≥ 20.
Investigation 25.
Suppose we arrange to collect samples of plain M&Ms such that each sample contains 330 M&Ms—an amount roughly equivalent to a 10-oz bag of plain M&Ms—drawn from the same population as the data in Table 2. Can we model this data using a normal distribution in place of the binomial distribution or the Poisson distribution? What advantages are there in being able to use the normal distribution? How might you apply this to more practical analytical problems, such as determining the concentration of Pb2+ in soil? | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Introduction_to_Data_and_the_Analysis_of_Data/01_Student_Handout/04_Part_IV%3A_Ways_to_Model_Data.txt |
In Part IV we noted that when a population is normally distributed, the probability of obtaining a particular result for any single sample is determined by that result’s area under the normal distribution curve defined by the population’s mean and standard deviation. For example, in Investigation 24 we showed that for 1.69-oz bags of plain M&Ms, 22.8% have a net weight less than 1.69 oz if the population’s mean is 48.98 g and its standard deviation is 1.433 g.
Suppose we select a single sample from this population: What can we predict about the net weight of M&Ms in that sample? Rearranging our equation for z, we find that
$x = μ ± zσ$
We call this equation a confidence interval because the value we choose for z defines the probability (our confidence) that the result for a single sample is in the range μ ± .
Investigation 26.
A z of 1.96 corresponds to a 95% confidence interval. Using Appendix 2, show that this is correct. What value of z corresponds to a 90% confidence inteval, and what value of z corresponds to a 99% confidence inteval? Report the 90%, the 95% and the 99% confidence intervals for the net weight of a single 1.69-oz bag of plain M&Ms drawn from a population for which μ is 48.98 g and σ is 1.433 g. For the data in Table 2, how many of the 30 samples have net weights that fall outside of the 90% confidence interval? Does this result make sense given your understanding of a confidence interval?
In Investigation 26 we calculated the confidence interval for a single sample based on the properties of the population from which we obtained the sample. If we draw several replicate samples from this population and calculate their mean, $\bar{x}$, then the confidence interval becomes
$\bar{x} = μ ± \dfrac{zσ}{\sqrt{n}}$
where n is the number of samples.
Investigation 27.
Suppose we draw four 1.69-oz bags of M&Ms from a population for which μ is 48.98 g and σ is 1.433 g. What are the 90%, the 95% and the 99% confidence intervals for the mean, $\bar{x}$, of these samples? Prepare a plot that shows how n affects the width of the 95% confidence interval, expressed as $±zσ/\sqrt{n}$, and discuss the significance of your plot. Suppose we wish to decrease the confidence interval by a factor of 3× solely by increasing the number of samples taken. If the original confidence interval is based on the mean of four samples, how many additional samples must we acquire?
In both Investigation 26 and Investigation 27 we attempt to predict a property of a sample based on a population with known values of μ and σ. For most practical analytical problems, however, we need to work in the opposite direction, using the sample’s mean, $\bar{x}$, and its standard deviation, s, to predict the population’s mean, μ. To do this, we make three modifications to our equation for the confidence interval: we rewrite the equation so that it expresses μ in terms of $\bar{x}$; we replace the population’s standard deviation, σ, with the sample’s standard deviation, s; and we replace z with the variable t, where we define t such that, for any confidence level, tz and the value of t approaches z as the number of samples, n, increases.
$μ=\bar{x}±\dfrac{ts}{\sqrt n}$
Clearly the value of t depends on the confidence interval and the number of samples; see Appendix 3 for further details.
Investigation 28.
Our data for 1.69-oz bags of plain M&Ms includes 30 measurements of the net weight. What are the 90%, the 95% and the 99% confidence intervals for the mean, $\bar{x}$, of these samples? Using the 99% confidence interval as an example, explain the meaning of this confidence interval. Is the stated net weight of 1.69 oz a reasonable estimate of the true mean for the population of 1.69-oz bags of plain M&Ms?
Our approach in Investigation 28 suggests we can use a confidence interval to decide whether a known value is consistent with our results, a process that we call significance testing and that we carry out a bit more formally than suggested by Investigation 28. To illustrate the process, we will use the data from Table 2 for the bags of M&Ms purchased at Target and evaluate whether the mean net weight for these samples is consistent with the stated net weight of 1.69 oz (47.9 g).
To begin, we summarize the experimental results for our sample, which in this case is a mean of 49.52 g and a standard deviation of 1.649 g for n = 10 samples. Next, we state our problem in the form of a yes/no question, the answers to which we define using a null hypothesis (H0) and an alternative hypothesis (HA); for example, for this problem our yes/no question is “Is the mean of the samples consistent with the stated net weight of 1.69 oz?,” which we define as
$H_0\textrm{: }\bar{x}=μ\: \textrm{(yes)}$
$H_A\textrm{: }\bar{x}≠μ\: \textrm{(no)}$
where $\bar{x}$ is 49.52 g and μ is 47.9 g. To evaluate the two hypotheses, we rewrite the equation for the confidence interval so that we can solve for t
$t= \dfrac{|\bar{x}-μ| \sqrt n}{s}=\dfrac{|49.52 - 47.9 | \sqrt{10}}{1.649} =3.087$
Finally, we compare this experimental value of t to the critical values of t for the correct number of degrees of freedom (in this case, $ν = n - 1 = 10 - 1 = 9$). From Appendix 3 we see that $t(α,ν)$ is 1.833 for a 90% confidence interval (an $\alpha$ of 0.10), 2.262 for a 95% confidence interval (an $\alpha$ of 0.05), 2.821 for a 98% confidence interval (an $\alpha$ of 0.02), and 3.250 for a 99% confidence interval (an $\alpha$ of 0.01). Our experimental value for t of 3.087 falls between the critical values for the 98% and the 99% confidence interval; if we are willing to accept an uncertainty of 1–2%, then we can reject the null hypothesis and accept the alternative hypothesis, concluding that the mean of 49.52 g is not consistent with the stated net weight of 1.69 oz. We call this a t-test of $\bar{x}$ vs. μ.
Investigation 29.
In 1996, Mars, the manufacturer of M&Ms, reported the following distribution for the colors of plain M&Ms: 30% brown, 20% red, 20% yellow, 10% blue, 10% green, and 10% orange. Pick any one color of M&Ms and, using the data in Table 2, calculate the percentage of that color in each of the 30 samples. Report the mean and the standard deviation for your color and use a t-test to determine whether your sample’s mean is consistent with the result reported by Mars. Gather results for the remaining five colors from other students and discuss your pooled results. Assuming that the distribution of colors reported by Mars is correct, what can you conclude about the manufacturing process. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Introduction_to_Data_and_the_Analysis_of_Data/01_Student_Handout/05_Part_V%3A_Ways_to_Draw_Conclusion.txt |
Using the data from Table 2, data from the references in Appendix 1, data from one or more of the four data sets listed below, and/or data you collect on your own, pose a question, gather relevant data, complete an analysis of that data, and propose an answer to your question. Although this case study’s introduction to data and to data analysis is limited to just a few types of visualizations, a few ways to summarize data, a few ways to model data, and one way to draw conclusions from data, your background is sufficient to use the texts in Appendix 1 and your textbook to explore other ways to visualize, to summarize, to model, and to draw conclusions from data. Be adventurous!
Data Set 1
source: Math Department, University of Alabama at Huntsville
structure: color distribution and net weight for 30 samples of plain M&Ms in 1.69-oz bags
Data Set 2
source: University of Puget Sound Data Hoard
structure: type, color, diameter, and mass for M&Ms in a 14.0-oz bag of plain M&Ms, a 12.7-oz bag of peanut M&Ms, and a 12.7-oz bag of peanut butter M&Ms
Data Set 3
source: Stats Monkey
website: apstatsmonkey.com/StatsMonkey...ctivities.html
structure: type and color (all data is simulated)
Data Set 4
source: various
Excel spreadsheet: link
structure: color distribution of plain M&Ms attributed to Mars (1996–2008)
Appendix 1: External
The papers and websites gathered here provide examples of studies using M&Ms (and other similar items) to illustrate concepts in the broad area of data analysis.
In addition to the many standard textbooks used in introductory and advanced courses in analytical chemistry, these references introduce additional ways to visualize, to summarize, to model, and to draw conclusions from data.
• Boslaugh, S. Statistics in a Nutshell, 2nd Edition, O’Reilly: Sebastopol, CA, 2013.
• Harvey, D. T. Analytical Chemistry 2.0.
• Larose, D. T.; Larose, C. D. Discovering Knowledge in Data, 2nd Edition, Wiley: Hoboken, NJ, 2014.
• Miller, J. C.; Miller, J.N. Statistics and Chemometrics for Analytical Chemistry, Pearson: Essex, England, 2010.
• Robbins, N. Creating More Effective Graphs, Chart House: Wayne, NJ, 2013.
• van Belle, G. Statistical Rules of Thumb, 2nd Edition, Wiley: Hoboken, NJ, 2008.
Appendix 2: Single-S
The table below gives the proportion, P, of the area under a normal distribution curve that lies to the right of a deviation, z, which is defined as
\[z= \dfrac{x-μ}{σ}\]
where μ and σ are the distribution’s mean and standard deviation, respectively, and where x is the value for which the deviation is defined. For example, the area under a normal distribution to the right of a deviation of +0.04 is 0.4840 (see entry in red), or 48.40% of the total area. The area to the left of the deviation is 1 - P. For a deviation of +0.04, this is 1 – 0.4840, or 51.60%. If x is smaller than μ, then z is negative. In this case, the values in the table give the area to the left of z. For example, if z is –0.04, then 48.40% of the area lies to the left of the deviation.
To use the single-sided normal distribution table, sketch the normal distribution curve for your problem and shade the area that corresponds to your answer. This divides the normal distribution curve into three regions: the shaded area that corresponds to your answer, the area to the right of this, and the area to the left of this. Calculate the values of z for the limits of the area that corresponds to your answer. Use the table to find the areas to the right and to the left of these deviations, subtract these values from 100% and, voilà, you have your answer.
z
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.0
0.5000
0.4960
0.4920
0.4880
0.4840
0.4801
0.4761
0.4721
0.4681
0.4641
0.1
0.4602
0.4562
0.4522
0.4483
0.4443
0.4404
0.4365
0.4325
0.4286
0.4247
0.2
0.4207
0.4168
0.4129
0.4090
0.4502
0.4013
0.3974
0.3396
0.3897
0.3859
0.3
0.3821
0.3783
0.3745
0.3707
0.3669
0.3632
0.3594
0.3557
0.3520
0.3483
0.4
0.3446
0.3409
0.3372
0.3336
0.3300
0.3264
0.3228
0.3192
0.3156
0.3121
0.5
0.3085
0.3050
0.3015
0.2981
0.2946
0.2912
0.2877
0.2843
0.2810
0.2776
0.6
0.2743
0.2709
0.2676
0.2643
0.2611
0.2578
0.2546
0.2514
0.2483
0.2451
0.7
0.2420
0.2389
0.2358
0.2327
0.2296
0.2266
0.2236
0.2206
0.2177
0.2148
0.8
0.2119
0.2090
0.2061
0.2033
0.2005
0.1977
0.1949
0.1922
0.1894
0.1867
0.9
0.1841
0.1814
0.1788
0.1762
0.1736
0.1711
0.1685
0.1660
0.1635
0.1611
1.0
0.1587
0.1562
0.1539
0.1515
0.1492
0.1469
0.1446
0.1423
0.1401
0.1379
1.1
0.1357
0.1335
0.1314
0.1292
0.1271
0.1251
0.1230
0.1210
0.1190
0.1170
1.2
0.1151
0.1131
0.1112
0.1093
0.1075
0.1056
0.1038
0.1020
0.1003
0.0985
1.3
0.0968
0.0951
0.0934
0.0918
0.0901
0.0885
0.0869
0.0853
0.0838
0.0823
1.4
0.0808
0.0793
0.0778
0.0764
0.0749
0.0735
0.0721
0.0708
0.0694
0.0681
1.5
0.0668
0.0655
0.0643
0.0630
0.0618
0.0606
0.0594
0.0582
0.0571
0.0559
1.6
0.0548
0.0537
0.0526
0.0516
0.0505
0.0495
0.0485
0.0475
0.0465
0.0455
1.7
0.0466
0.0436
0.0427
0.0418
0.0409
0.0401
0.0392
0.0384
0.0375
0.0367
1.8
0.0359
0.0351
0.0344
0.0336
0.0329
0.0322
0.0314
0.0307
0.0301
0.0294
1.9
0.0287
0.0281
0.0274
0.0268
0.0262
0.0256
0.0250
0.0244
0.0239
0.0233
2.0
0.0228
0.0222
0.0217
0.0212
0.0207
0.0202
0.0197
0.0192
0.0188
0.0183
2.1
0.0179
0.0174
0.0170
0.0166
0.0162
0.0158
0.0154
0.0150
0.0146
0.0143
2.2
0.0139
0.0136
0.0132
0.0129
0.0125
0.0122
0.0119
0.0116
0.0113
0.0110
2.3
0.0107
0.0104
0.0102
0.00964
0.00914
0.00866
2.4
0.00820
0.00776
0.00734
0.00695
0.00657
2.5
0.00621
0.00587
0.00554
0.00523
0.00494
2.6
0.00466
0.00440
0.00415
0.00391
0.00368
2.7
0.00347
0.00326
0.00307
0.00289
0.00272
2.8
0.00256
0.00240
0.00226
0.00212
0.00199
2.9
0.00187
0.00175
0.00164
0.00154
0.00144
Appendix 3: Critical
The table below gives values of $t(α,ν)$ where $α$ defines the confidence level and $ν$ defines the degrees of freedom. Values for $α$ are defined as follows
$α = 1 - \textrm{confidence level (as fraction)}$
For example, for a 95% confidence level, $α = 1 - 0.95 = 0.05$. The degrees of freedom is the number of independent measurements given any constraints that we place on the measurements. For example, if we have n measurements and we calculate their mean, $\bar{x}$, then we have n - 1 degrees of freedom because the mean, $\bar{x}$, and the values for the first four measurements, x1, x2, x3, and x4, removes the independence of the fifth measurement, x5, whose value is defined exactly as
$x_5=\bar{x} -x_1-x_2-x_3-x_4$
The values of t in this table are two-tailed in that they define a confidence interval that is symmetrical around the mean. For example, for a 95% confidence interval ($α=0.05$), half of the area not included within the confidence interval is at the far right of the distribution and half is at the far left of the distribution. For a one-tailed confidence interval, in which the excluded area is on one side of the distribution, divide the values of $α$ in half; thus, for a one-tailed 95% confidence interval, we use values of t from the column where $α=0.10$.
$ν$
$α=0.10$
$α=0.05$
$α=0.02$
$α=0.01$
1
6.314
12.706
31.821
63.657
2
2.920
4.303
6.965
9.925
3
2.353
3.182
4.541
5.841
4
2.132
2.776
3.747
4.604
5
2.015
2.571
3.365
4.032
6
1.943
2.447
3.143
3.707
7
1.895
2.365
2.998
3.499
8
1.860
2.306
2.896
3.255
9
1.833
2.262
2.821
3.250
10
1.812
2.228
2.764
3.169
12
1.782
2.179
2.681
3.055
14
1.761
2.145
2.624
2.977
16
1.746
2.120
2.583
2.921
18
1.734
2.101
2.552
2.878
20
1.725
2.086
2.528
2.845
30
1.697
2.042
2.457
2.750
50
1.676
2.009
2.311
2.678
$\infty$
1.645
1.960
2.326
2.576 | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Introduction_to_Data_and_the_Analysis_of_Data/01_Student_Handout/06_Part_VI%3A_Now_Its_Your_Turn.txt |
This module introduces students to ways of thinking about and working with data using, as a case study, the analysis of 1.69-oz packages of plain M&Ms. The module is divided into six parts:
Part I. Ways to Describe Data
Part II. Ways to Visualize Data
Part III. Ways to Summarize Data
Part IV. Ways to Model Data
Part V. Ways to Draw Conclusions From Data
Part VI. Now It’s Your Turn!
Interspersed within the module’s narrative is a series of investigations, each of which asks students to stop and consider one or more important issues. Some of these investigations include data sets for students to analyze; for the data in the module’s figures, you may wish to have students use the interactive, on-line version available at http://dpuadweb.depauw.edu/harvey_web/Data_Analysis/index.html.
Many of the investigations draw on a data set that consists of 30 samples of 1.69-oz packages of plain M&Ms, the general structure of which is shown here
Table 2. Source, Distribution, and Net Weight of Plain M&Ms in 1.69-oz Bags
bag
store
blue
brown
green
orange
red
yellow
net weight (g)
27
CVS
5
17
6
4
8
19
50.802
28
Kroger
1
21
6
5
10
14
49.055
29
Target
4
12
6
5
13
14
46.577
30
Kroger
15
8
9
6
10
8
48.317
The counts for the different colors of M&Ms were collected by students at DePauw University in the fall 1996 semester as part of an in-class exercise; this was the only data collected at that time. To allow for a consideration of grouping in this case study, samples were assigned randomly to one of three hypothetical sources. Because the original data did not include a consideration of mass, the net weights included in the data set were generated specifically for this case study by simulating the random sampling of single plain M&Ms from a normally distributed population of weights. The values of μ and σ for this population were derived using the mean, $\bar{x}$, and the standard deviation, s, for the published weights of 462 plain M&Ms available through the Puget Sound Data Hoard (http://stat.pugetsound.edu/hoard/Default.aspx). The appendix to this Instructor’s Guide includes the R code used to generate these net weights, as well as the R code used to generate the figures that accompany some of the suggested responses.
This case study is meant to serve as an introduction to data and to data analysis and, as with any introduction, it considers a small number of topics, principally those covered in Chapter 4 of Analytical Chemistry 2.0; additional resources that provide a deeper introduction to data and to data analysis are listed in Appendix 1 of the case study.
Suggested responses are presented in normal font; additional comments, suggestions, and supplementary materials are in italic font.
Instructors Guide
Investigation 1.
Of the variables included in Table 1, some are categorical and some are numerical. Define these terms and assign each of the variables in Table 1 to one of these terms.
A categorical variable provides qualitative information that we can use to describe the samples relative to each other, or that we can use to place the samples into groups. For the data in Table 1, “bag id,” “type,” and “rank” are categorical variables.
A numerical variable provides quantitative information on which we can perform a meaningful calculation; for example, we can use “# yellow M&Ms” and “total M&Ms” to calculate the new variable “% yellow M&Ms.” For the data in Table 1, “year,” “weight (oz),” “# yellow M&Ms,” “% red M&Ms,” and “total M&Ms” are numerical variables.
Some students will include “year” as a categorical variable, which is not an unreasonable choice as it might serve as a useful way to group samples; however, it is listed here as a numerical variable because it can serve as a useful predictive variable in a regression analysis. Some students will include “rank” as a numerical variable, essentially rewriting the entries as numerals; however, there are no meaningful calculations that we can complete using this variable.
Investigation 2.
Suppose we decide to code the type of M&M using 1 for plain and 2 for peanut. Does this change your answer to Investigation 1? Why or why not?
No. Although it is tempting to assume that a number must imply a numerical variable, we need to remember that we can convert any descriptive phrase into a number even if the number does not convey quantitative information. For example, although we might choose to code samples of plain M&Ms using the integer 1 and code samples of peanut M&Ms using the integer 2, we would never report that the average sample is of type {(4)(2) + (2)(2)}/6 = 1.33 as this does not have any meaningful interpretation.
Not all students are familiar with databases or with coding, and may ask why we might choose to code a variable if replacing a descriptive phrase with an integer provides us with no advantage and if it comes at the cost of making it more difficult for others to read our table. When this question arises, it is helpful to note that there are several reasons we might choose to replace a descriptive phrase with an integer when creating a computerized database, particularly if the database has many records: storage space (it takes less space to store an integer than it does to store a character string); search speed (it takes less time to search for an integer than it does to search for a character string); and fewer errors when entering data (consider how easy it is to type penut for peanut).
Investigation 3.
Categorical variables are described as nominal or ordinal. Define the terms nominal and ordinal and assign each of the categorical variables in Table 1 to one of these terms.
A nominal categorical variable does not carry with it any implied order; an ordinal categorical variable, on the other hand, coveys a meaningful sense of order. For the categorical variables in Table 1, “bag id” and “type” are nominal variables, and “rank” is an ordinal variable.
Some students may interpret the use of consecutive alphabetical letters for “bag id” as implying order, but there is nothing to suggest that this order is meaningful.
Investigation 4.
A numerical variable is described as either ratio or interval depending on whether it has (ratio) or does not have (interval) an absolute reference. Explain what it means for a variable to have an absolute reference and assign each of the numerical variables in Table 1 as either a ratio variable or an interval variable. Why might this difference be important?
A numerical variable has an absolute reference if it has a meaningful zero—that is, a zero that means a measured quantity of none—against which we reference all other measurements of that variable. For the numerical variables in Table 1, “year” is an interval variable because our scale for time is referenced to an arbitrary point in time, 1 CE, and not to the beginning of time; “weight (oz),” “# yellow M&Ms,” “% red M&Ms,” and “total M&Ms” are ratio variables because each has a meaningful zero.
For a ratio variable, we can make meaningful absolute and relative comparisons between two results, but only meaningful absolute comparisons for an interval variable. For example, consider sample e, which was collected in 1994 and which has 331 M&Ms, and sample d, which was collected in 2000 and which has 24 M&Ms. We can report a meaningful absolute comparison for both variables: sample e is six years older than sample d and sample e has 307 more M&Ms than sample d. We also can report a meaningful relative comparison for the total number of M&Ms—there are 331 ÷ 24 = 13.8 times as many M&Ms in sample e as in sample d—but we cannot report a meaningful relative comparison for year because a sample collected in 2000 is not 2000 ÷ 1994 = 1.003 times older than a sample collected in 1994.
Investigation 5.
Numerical variables also are described as discrete or continuous. Define the terms discrete and continuous and assign each of the numerical variables in Table 1 to one of these terms.
A numerical variable is discrete if it can take on only specific values—typically, but not always, an integer value—between its limits; a continuous variable can take on any possible value within its limits. For the numerical data in Table 1, “year,” “# yellow M&Ms,” and “total M&Ms” are discrete in that each is limited to integer values. The numerical variables “weight (oz)” and “% red M&Ms,” on the other hand, are continuous variables.
Students will sometime ask why weight is not a discrete variable given that a balance records the weight to a set number of decimal points. Here it is helpful to remind students that what makes a variable discrete is not our ability to measure it, but a property inherent in the variable itself. In the context of this data, each M&M is an indivisible unit and the number of units is discrete; however, two M&Ms with masses of 0.8561 g and 0.8559 g have different weights even if our balance reads, and we report, both as 0.856 g. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Introduction_to_Data_and_the_Analysis_of_Data/Instructors_Guide/01_Part_I%3A_Ways_to_Describe_Data.txt |
Investigation 6.
Use the dot plot in Figure 1 to deduce the general structure of a box and whisker plot, giving particular attention to the position along the x-axis of the three vertical lines that make up the yellow box and the two vertical lines that make up the whiskers on either side of the yellow box. You might begin by tabulating the number of samples that fall to the left of the box, that fall within the box, including its boundaries, and that fall to the right of the box, and the number of samples that lie to the left and to the right of line inside the box.
Of the 30 samples, seven are on the left side of the box, 17 are within the box, and six are on the right side of the box; relative to the box’s middle line, 14 lie to the left and 13 lie to the right. One reasonable interpretation of these observations is that the box contains approximately the middle 50% of the data (17 of 30 samples, or 57%) and that the line inside the box divides the data approximately in half (14 of 30 samples, or 47%, are left of the line and 13 of 30 samples, or 43%, are right of the line).
The two whiskers extend to encompass all but one of the 30 samples. Clearly the whiskers convey information about the overall variability of the data, but there is insufficient information in this one example to suggest exactly how the length of the whiskers are determined (although, at least for this example, the whiskers do not include the one sample that lies at a distance of more than 1.5 × w, where w is the width of the box.
For students who have difficulty accepting 57%, 47%, and 43% as being suggestive of 50%, it helps to have them consider the effect on the percentages of the limited number of samples (30) and the fact that multiple samples have the same result. For further details on box and whisker plots, see https://en.Wikipedia.org/wiki/Box_plot.
There are a variety of ways to define the whiskers and to handle points that fall outside of a whisker. The method used here is to draw a whisker to the data point whose value is no greater than +1.5 × w of the box’s largest value (in this case 17+1.5 × 4 = 23), and to draw a whisker to the data point whose value is no less than -1.5 × w of the box’s smallest value (in this case 13 - 1.5 × 4 = 7). Results that fall outside of the whiskers are flagged using a dot (•), even when individual results are not shown using a dot plot.
Investigation 7.
The box and whisker plot in Figure 1 is perfectly symmetrical in that each side of the box is two units from the box’s middle line, and each whisker is six units from the box’s nearest edge. What does this symmetry suggest about how the results are distributed? Is the actual distribution of the 30 results perfectly symmetrical? If no, is this a problem?
The symmetry of the box and the whiskers suggests that there is a symmetrical distribution of the data set’s individual results around its middle. The data itself is not perfectly symmetrical—for example, there are five samples within ±2 of the left whisker, but just three samples within ±2 of the right whisker. This difference between the symmetry of the data and the symmetry of the box and whisker plot is not a problem as we use a box and whisker plot simply to develop a general understanding of our data’s structure.
Investigation 8.
In Figure 1 we see that the result for sample 22 falls outside the range of values included within the whiskers. Why might a result that falls outside the whiskers concern us? Does the presence of this particular point suggest a problem? How might your response change if this sample’s reported value is 0 yellow M&Ms? How might your response change if this sample’s reported value is 45 yellow M&Ms?
If we assume that the box and the whiskers should include all samples for which the results are not subject to an error—then we might wish to look more closely at a sample that falls outside of the whiskers as it may suggest a problem with our data, either in the counting of M&Ms, in the recording of that count, or in the manufacturing process. In this case, the result for sample 22 does not bother us as it is not that different from the next lowest value and, more important, an error in counting M&Ms does seem not likely when the bag contains just 55 M&Ms (a counting error is more likely if a bag has 550 M&Ms). For the same reason, we are not likely to question a result of 0. A result of 45 yellow M&Ms, however, seems unreasonable as it is almost twice as many as the next highest value; in this case we might suspect that an error was made when entering the result into the data table.
Investigation 15 introduces the difference between samples and populations, so this language is not used here; if you wish to discuss this difference here, you may wish to begin the case study with a discussion of samples and populations.
Investigation 9.
Figure 2 shows box and whisker plots and dot plots for all six colors of M&Ms included in Table 2 (note: even with jittering, you will not be able to see all 30 samples in these dot plots). Based on these plots, where do you see similarities and where do you see differences in the distribution of M&Ms? What do these similarities and differences suggest to you? For those distributions that do not appear symmetrical, suggest one or more reasons for the lack of symmetry. What do the relative positions of the data for brown and for green M&Ms suggest about their relative abundance in 1.69-oz packages of plain M&Ms?
There are many observations we can make using this data, a few of which are gathered here. One observation is that finding a sample outside of the whiskers is a rare event as it happens just once in 180 measurements (sample 22, yellow). Another observation is that the boxes for brown M&Ms and for yellow M&Ms overlap each other but do not overlap with the other four colors of M&Ms (although the upper edge of the box for red abuts the lower edge of the box for yellow); this suggests that yellow M&Ms and brown M&Ms are much more common than the other four colors. Another interesting difference is that the lower whiskers for blue, green, and orange M&Ms are much shorter than their respective upper whiskers; this suggests that their distributions are not symmetrical, a result that is not surprising given that the we cannot have fewer than zero M&Ms with any particular color. Finally, the relative positions of the box and whisker plots for green M&Ms and for brown M&Ms suggests that it is a rare bag that has more green M&Ms than brown M&Ms, which places a hard limit on the data’s lower boundary; indeed, this happens just once (sample 30, which has 9 green M&Ms and 8 brown M&Ms).
Investigation 10.
Figure 3 shows box and whisker plots and dot plots for yellow M&Ms grouped by the store where the packages of M&Ms were purchased. Based on these plots, where do you see similarities and where do you see differences in the distribution of yellow M&Ms? What do these similarities and differences suggest to you? In what ways might this data be reassuring to us? Give an example of a result that might suggest we look more closely at our data.
Although the box and whisker plots are quite different in terms of the relative sizes of the boxes and the relative length of the whiskers, the dot plots suggest that the distribution of the underlying data is relatively similar in that most values are in the range of 12–18 yellow M&Ms with a maximum of 22 or 23 yellow M&Ms and a minimum of eight yellow M&Ms (setting aside sample 22, which, as noted in the response to Investigation 9, is the only result in 180 measurements that does not fall within the span of its whiskers). These observations are reassuring because we do not expect the source of the bags of M&Ms to affect the composition of their contents. If we saw evidence that the source did affect our results, then we would need to look more closely at the bags themselves for evidence of a poorly controlled variable, such as type (Did we accidently purchase bags of peanut butter M&Ms from one store?) or the product’s lot number (Did the manufacturer change the composition of colors between lots?).
As a reminder, the division of the 30 samples among these three sources is artificial and is done solely to illustrate the concept of grouping and the analysis of a common variable (yellow M&Ms) between different groups.
Investigation 11.
Draw a box and whisker plot and an accompanying dot plot for the total number of M&Ms. Compare your plots to those in Figure 2 and discuss any similarities and differences.
The total number of M&Ms in the 30 samples are, in order 57, 56, 59, 56, 57, 54, 57, 57, 56, 55, 59, 58, 55, 56, 55, 58, 56, 56, 56, 60, 58, 55, 57, 56, 55, 59, 59, 57, 54, and 56. A box & whisker plot and a dot plot are shown below.
The most interesting observation for this data is that the box does not appear to have a middle line. Of course, it actually does have a middle line, but it simply is the same as the box’s lower limit. We already know from Investigation 7 that the box’s middle line divides the data in half, so we know that half of the bags have 56 or fewer M&Ms and that half have 56 or more M&Ms. We also know that we have a greater number of unique results above the middle value (57, 58, 59, and 60 M&Ms) than below the middle value (54 and 55 M&Ms). Although the box and the whisker plot looks symmetrical, the results are skewed somewhat toward larger numbers of M&Ms.
Students will benefit from drawing this plot by hand. Although Excel does not include a command for drawing a box and whisker plot, an on-line search will yield methods for creating a plot that will mimic the traditional box and whisker plot. The statistical program R has a built in boxplot command.
Investigation 12.
For the histograms in Figure 4, where do you see similarities and where do you see differences in the distribution of M&Ms? How do the results seen here compare with your interpretation of the box and whisker plots and the dot plots in Figure 2?
The information here is very similar to what we saw in the box and whisker plots. In particular, the colors of M&Ms with the least symmetrical whiskers—blue, green, and orange—have histograms that are not symmetrical and that tend to decrease in value more slowly when moving from the bin that contains the greatest number of samples to bins that contain samples with greater numbers of M&M. The lack of symmetry for yellow M&Ms, which decreases in value more slowly as we move from the bin that contains the greatest number of samples toward bins that contains samples with a smaller number of M&Ms, is more evident here than in the box and whisker plots.
Investigation 13.
The histograms in Figure 5, from left-to-right, use bins widths of 1, 2, and 3 units, respectively. Note that the x-axis shows the specific results gathered into each bin. How does the choice of bin size affect your understanding of this data? Which of these histograms provides the best representation of the data? As part of your answer, identify what you see as the limitations of the other two histograms.
Using a bin size of 1 unit makes it easy to see that there were no bags with 9, 12, or 14 M&Ms; this information is not available when the bins have sizes of 2 units or of 3 units. The histograms using bins of 1 unit and of 2 units are similar in shape: if we draw a smooth curve through the data—ignoring the noise due to our limited number of samples—both histograms suggest that the frequency of an outcome decreases as we move from the smallest number of M&Ms (four) to the largest number of M&Ms (15); a smooth curve through the histogram using a bin of 3 units, however, suggests that the frequency increases and then decreases as we move from the smallest number of M&Ms (four) to the largest number of M&Ms (15).
Of the three options, the best representation of the data is the one with a bin size of 2 units. Although the histogram using a bin size of 1 unit does show us possible outcomes that did not occur, the resulting histogram is much noisier. The histogram using a bin size of 3 units is the least noisy, but its first bin includes as possible outcomes results that are not in our data (two and three orange M&Ms), which is somewhat misleading.
Investigation 14.
Draw a histogram for the total number of M&Ms and explain the reason(s) for your choice of bin size. Compare your plots to those in Figure 4 and discuss any similarities and any differences.
The total number of M&Ms in the 30 samples are, in order 57, 56, 59, 56, 57, 54, 57, 57, 56, 55, 59, 58, 55, 56, 55, 58, 56, 56, 56, 60, 58, 55, 57, 56, 55, 59, 59, 57, 54, and 56. Gathering these into a frequency table
number of M&Ms
54
55
56
57
58
59
60
frequency
2
5
9
6
3
4
1
suggests that a bin size of 1 unit is a good option as a bin size of 2 units has just four total bins, one of which must include a result either less than 54 or greater than 60; the resulting histogram is shown below.
This histogram is consistent with our observations from the box and whisker plot in Investigation 12, but it presents us with a much clearer picture of the data.
Students will benefit from drawing this plot by hand. Excel’s Data Analysis tools provides a method for creating histograms, as does R using the hist command. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Introduction_to_Data_and_the_Analysis_of_Data/Instructors_Guide/02_Part_II%3A_Ways_to_Visualize_Data.txt |
Investigation 15.
Before we consider ways to summarize our data, we need to draw a distinction between a sample and a population. We collect and analyze samples with the hope that we can deduce something about the properties of the population. Using our data for M&Ms as an example, define the terms sample and population.
For our data, the structure of which is in Table 2, each row is single sample of plain M&Ms in the form of individual 1.69-oz packages. These samples are drawn from the much larger population of all plain M&Ms (or, at least all plain M&Ms manufactured at the time the samples were packaged).
Investigation 16.
Using the data for yellow M&Ms, calculate the mean and the median for each store and discuss your results. If the mean and the median are equal to each other, what might you reasonably conclude about your data? If the mean is larger than the median, or if the mean is smaller than the median, what might you reasonably conclude about your data? A measure of central tendency is considered robust when it is not changed by one or more results that differ substantially from the remaining results. Which measure of central tendency is more robust? Why?
To help us understand how we arrive at each value, we will use the data in Figure 3 for yellow M&Ms in bags purchased at CVS. To begin, let’s construct a frequency table, which shows the eight unique results, ordered from smallest-to-largest, and the number of bags with each unique result.
number ($N$)
5
8
13
15
16
17
19
23
frequency ($f$)
1
1
2
2
1
1
1
1
To calculate the mean using a frequency table, we multiply each unique result by its frequency, sum up the values, and divide by the number of samples; thus
number ($N$)
5
8
13
15
16
17
19
23
frequency ($f$)
1
1
2
2
1
1
1
1
$N×f$
5
8
26
30
16
17
19
23
The sum of the values in the last row is 144, which gives the mean as
$\bar{x} =\dfrac{144}{10} =\textrm{14.4 yellow M&Ms}$
For the 10 samples from CVS, the median is the average of the 5th and the 6th values when ordered by rank. Using the frequency data, the 5th value is 15 and the 6th value is 15, which gives the median as 15 yellow M&Ms. The following table summarizes the means and the medians for yellow M&Ms by store.
store
mean
median
CVS
14.4
15.0
Kroger
14.2
15.0
Target
14.9
14.5
For each store, we see that the mean and the median are similar in value; we also see that the means and the medians between the three stores are similar. Both are reasonable results as, discussed in the response to Investigation 10.
If the mean and the median are equal to each other, then the distribution of the individual values must be perfectly symmetrical about the mean and median. If the mean is larger than the median, then the data likely is skewed toward the right, and if the mean is smaller than the median, then the data likely is skewed toward the left.
The median is more robust than the mean because the median uses the rank, not the value, of each data point, which makes it relatively insensitive to an unusually large or small result. For example, if the sample from CVS with 19 yellow M&Ms has, instead, 29 yellow M&Ms, then the mean increases from 14.4 to 15.4, but the median remains unchanged.
Students should, of course, calculate the mean (and other statistics) using a calculator, a spreadsheet, such as Excel, or a statistical program, such as R. There is benefit, however, in seeing how the sample’s data comes together to give the mean, which is the reason for detailing the calculation using a frequency table; the same approach is used in the next investigation.
Although generally it is true that data is skewed to the right when the mean is greater than the median and skewed to the left when the mean is less than the median, this ‘rule’ does not hold true in all cases. In particular, it may not hold for a discrete distribution when the areas to the left and to the right of the median are not equal (because many samples share the median’s value). It also fails with multimodal distributions and in distributions where there is a long tail in the direction of the skew, but a heavy tail in the other direction. See von Hippel, P. T. “Mean, Median, and Skew,” J. Statistics Education, 2005, 13(2) (www.amstat.org/publications/jse/v13n2/vonhippel.html) for additional details.
Investigation 17.
Using the data for yellow M&Ms, calculate the variance, the standard deviation, the range, and the IQR for each store and discuss your results. Is there a relationship between the standard deviation, the range, or the IQR? A result is considered robust when its value is not changed by one or more values that differ substantially from the remaining values. Which measure of spread—the variance, the standard deviation, the range, or the IQR—is the most robust? Why? Which is the least robust? Why?
To help us understand how we arrive at each value, we will use the data in Figure 3 for yellow M&Ms in bags purchased at CVS. To begin, let’s use the same frequency table from Investigation 16, which shows the eight unique results, ordered from smallest-to-largest, and the number of bags with each unique result.
number ($N$)
5
8
13
15
16
17
19
23
frequency ($f$)
1
1
2
2
1
1
1
1
To calculate the variance, we first calculate each unique difference relative to the mean $(x_i-\bar{x})$, square these unique differences, multiply each unique squared difference by its frequency, sum up the values, and divide by $n-1$; thus
number ($N$)
5
8
13
15
16
17
19
23
frequency ($f$)
1
1
2
2
1
1
1
1
$(x_i-\bar{x})$
–9.4
–6.4
–1.4
0.6
1.6
2.6
4.6
5.6
$(x_i-\bar{x})^2$
88.36
40.96
1.96
0.36
2.56
6.76
21.16
73.96
$f×(x_i-\bar{x})^2$
88.36
40.96
3.92
0.72
2.56
6.76
21.16
73.96
The sum of the values in the last row is 238.40, which gives the variance as
$s^2 =\dfrac{238.40}{10-1}=26.49$
and the standard deviation as 5.15 yellow M&Ms. To find the range, we subtract the smallest value (5 yellow M&Ms) from the largest value (23 yellow M&Ms), which makes the range 18 yellow M&Ms. To find the IQR, we use the median to divide the 10 samples into a lower half with values of 5, 8, 13, 13, and 15 yellow M&Ms, and an upper half of 15, 16, 17, 19, and 23 yellow M&Ms; the median of the upper half is 17 yellow M&Ms and the median of the lower half is 13 yellow M&Ms, which makes the IQR 4 yellow M&Ms. The following table summarizes the variance, the standard deviation, the range, and the IQR for yellow M&Ms by store.
store
variance
standard deviation
range
IQR
CVS
26.49
5.15
18
4
Kroger
21.96
4.69
15
3
Target
15.43
3.93
15
7
These results are consistent with our observations from Investigation 10.
This is a nice set of data to show that there is no general relationship between the variance, the standard deviation, the range, and the IQR as measures of spread. For example, the store with the smallest standard deviation (Target) is the store with the largest IQR.
For the reasons outline in the response to Investigation 16, the IQR is the most robust measure of spread as it uses the rank, not the value, of each data point. The least robust measure of spread is the range. For example, if the sample from CVS with 19 yellow M&Ms has, instead, 29 yellow M&Ms, then the range increases from 18 to 24, but the IQR remains unchanged.
Students often ask why we divide by n - 1 instead of by n. Although a rigorous explanation is beyond the scope of this case study, here is an intuitive way for them to think about this. In the numerator of the equation for variance we sum up the squared differences between the result for each sample, $x_i$, and the mean of these samples, $\bar{x}$. Because the sample’s mean is calculated from the individual samples, we reasonably might expect that this sum is smaller than the result if we used the population’s mean (which is unknown to us and which might be quite different from sample’s mean); dividing by n - 1 instead of by n compensates for this difference. For further details on what is called Bessel’s correction, see https://en.Wikipedia.org/wiki/Bessel's_correction. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Introduction_to_Data_and_the_Analysis_of_Data/Instructors_Guide/03_Part_III%3A_Ways_to_Summarize_Data.txt |
Investigation 18.
So, what does it mean to build a model? Consider the histograms in Figure 4. What property of the population are we attempting to model? What do your responses imply about the model’s general mathematical form? What does it mean to test a model and how might we accomplish this?
For the histograms in Figure 4, we wish to model the number of each color of M&M in a 1.69-oz bag of plain M&Ms. A suitable mathematical model will need to predict the probability, p, of drawing X M&Ms of a particular color when we select a sample of N M&Ms from the population of all M&Ms; thus, we expect the equation to be a function of the form P = f(X,N).
To test a model, we need to compare the results predicted by our model to the results of our experiments. For example, Table 2 contains results for the distribution of colors and the net weight of plain M&Ms in 1.69-oz bags. If we develop a model that predicts successfully the average number of yellow M&Ms in a 1.69-oz bag, then we have some confidence that the model is reasonable. Of course, we need to define how we decide if a model’s predictions agree with our data, which we will explore in greater detail in Part V.
Investigation 19.
The box and whisker plot in Figure 1 includes data from the analysis of 30 samples of 1.69-oz bags of plain M&Ms. Collectively, the samples have 1699 M&Ms, of which 435 are yellow. If you pick one M&M at random from these 1699 M&Ms, what is the probability, p, that it is yellow? Suppose that this probability applies to the population of all plain M&Ms. If we draw a sample of five M&Ms from this population, what is the probability that the sample contains no yellow M&Ms? Repeat for each of 1–5 yellow M&Ms. Construct a histogram of your results and report the mean and the variance. Repeat this analysis for green M&Ms. Compare your two histograms and discuss their similarities and their differences. Using the data in Table 2, comment on the suitability of the binomial distribution for modeling the number of yellow M&Ms in samples of five M&Ms.
Given the data from our samples, the probability of selecting a single yellow M&M is
$p=\dfrac{435}{1699} =0.256$
If we assume that this probability applies to the population of plain M&Ms and if we assume a binomial distribution, then the probability of drawing zero yellow M&Ms in a sample of five M&Ms is
$P(0,5)= \dfrac{5!}{0! (5 -0 )!} × (0.256)^0 ×(1 -0.256)^{5 -0} =0.228$
The remaining probabilities are 0.392, 0.270, 0.093, 0.016, and 0.001 for P(1, 5) to P(5, 5), respectively; the resulting distribution is shown below on the left. The mean and the variance are
$μ=Np=5×0.256=1.28$
$σ^2 =Np(1-p)=5×0.256×(1-0.256)=0.95$
For green M&Ms, there are 110 in the combined sample of 1699 plain M&Ms, or a probability of 0.0647. The probabilities for drawing zero to five green M&Ms in a sample of five M&Ms are, respectively, 0.716, 0.248, 0.034, 0.002, 0.000, and 0.000; the resulting histogram is shown above to the right. The mean and the variance are 0.32 and 0.30, respectively.
In terms of similarities, both histograms encompass a net probability of 1.00 as they span the six possible outcomes when drawing a sample of five M&Ms, and neither histogram is symmetrical around its most probable outcome. The most importance difference between the two histograms is the relative frequencies of the possible outcomes; in particular, a sample of five M&Ms is more than 3× as likely to have no green M&Ms than to have no yellow M&Ms, and is more than 3× as likely to have three or more yellow M&Ms than three or more green M&Ms. Given their relative abundances—25.6% of the 1699 total M&Ms are yellow versus just 6.47% for green—these differences make sense.
From Table 2, we know that actual distribution of results for yellow M&Ms in the first five sampled is seven with no yellow M&Ms, 13 with one yellow M&M, eight with two yellow M&Ms, two with three yellow M&Ms, and zero with four and with five yellow M&Ms. The following table compares the results of our experiment and the predicted results from our model.
P(X,N)
experiment
model
absolute error
P(0,5)
0.233
0.228
0.005
P(1,5)
0.433
0.392
0.041
P(2,5)
0.267
0.270
–0.003
P(3,5)
0.067
0.093
–0.026
P(4,5)
0.000
0.016
–0.016
P(5,5)
0.000
0.001
–0.001
As one sample is 1/30th, or 0.033, of the 30 samples, the absolute errors represent an oversampling of approximately one for P(1,5) and an undersampling of approximately one for P(3,5), an experimental uncertainty that seems reasonable given the relatively small number of samples and the relatively small value for N.
Investigation 20.
Explain why we cannot use the binomial distribution to model the distribution of yellow M&Ms in 1.69-oz bags of plain M&Ms.
The binomial distribution predicts the probability of a particular event, X, such as drawing five yellow M&Ms, in samples of fixed size, N, where XN. Because the number of M&Ms varies from bag-to-bag, the value of N varies from bag-to-bag and we cannot model the distribution of M&Ms in 1.69-oz bags of plain M&Ms; we could, however, model the distribution of yellow M&Ms in all 1.69-oz bags that contain the same total number of M&Ms.
Investigation 21.
The histograms in Figure 4 include data from the analysis of 30 samples of 1.69-oz bags of plain M&Ms. Collectively, the samples have an average of 14.5 yellow M&Ms per bag. Suppose this rate applies to the population of all 1.69-oz bags of plain M&Ms. If you pick a 1.69-oz bag of plain M&Ms at random, what is the probability that it contains exactly 11 yellow M&Ms? Repeat for each of 0–29 yellow M&Ms. Construct a histogram that shows the actual distribution of bags of M&Ms for each of 0–29 yellow M&Ms, using a bin size of 1 unit, and overlay a line plot that shows the predicted distribution of bags; be sure to you use the same scale for each plot’s y-axis. Comment on your results.
If we assume that the average rate of 14.5 yellow M&Ms per bag applies to the population of plain M&Ms, and assume a Poisson distribution, then the probability of finding 11 yellow M&Ms in a 1.69-oz package of plain M&Ms is
$P(11,14.5)=\dfrac{e^{-14.5}14.5^0}{11!}=0.0753$
or 2.3 out of 30 bags of M&Ms. The actual number of bags of M&Ms that contain each of 0–29 yellow M&Ms and the predicted probabilities are gathered in the following table; the predicated probabilities from the Poisson equation are multiplied by 30 so that the two results are on the same scale.
X
Actual
P(X,14.5)
0
0
0.000
1
0
0.000
2
0
0.002
3
0
0.008
4
0
0.028
5
1
0.081
6
0
0.195
7
1
0.405
8
3
0.733
9
0
1.181
10
1
1.713
11
0
2.258
12
1
2.729
13
3
3.043
14
4
3.152
15
3
3.047
16
5
2.761
17
2
2.355
18
2
1.897
19
1
1.448
20
0
1.050
21
0
0.725
22
1
0.478
23
2
0.301
24
0
0.182
25
0
0.106
26
0
0.059
27
0
0.032
28
0
0.016
29
0
0.008
The resulting histogram for the actual distribution of yellow M&Ms in the 30 samples and the predicted distribution are shown here
Two factors make difficult any comparison of the actual counts to the predicted counts: the actual counts are discrete (we can have 0 or 1 bag with five yellow M&Ms, but we cannot have 0.08 bags with five yellow M&Ms), and the number of samples, at 30, is too small to allow for predicted counts of at least one bag for outcomes with a small probability (we would need to sample 370 bags of M&Ms to have a predicted count of one bag with five yellow M&Ms). Still, the overlap of the actual and the predicted values, and the general shape of the actual distribution suggests that the Poisson distribution provides a reasonable model of our data.
Investigation 22.
Explain why we cannot use the binomial distribution or the Poisson distribution to model data for the net weight of M&Ms in Table 2.
The binomial distribution and the Poisson distribution are useful for modeling discrete events, such as the number of yellow M&Ms in a sample of fixed size, or the number of green M&Ms in bags of a particular size. The net weight of a sample of M&Ms, however, is a continuous variable, which requires a different type of mathematical model.
Investigation 23.
Using the curves in Figure 6 as an example, discuss the general features of a normal distribution, giving particular attention to the importance of variance. How do you think the areas under the three curves from -∞ to +∞ are related to each other? Why might this be important?
Here are three observations based on these three examples of normal distribution curves: (a) a normal distribution is symmetric about μ, with half of its outcomes on either side of μ; (b) the most likely outcome in a normal distribution is when x = μ; and (c) as the variance increases, the normal distribution’s maximum value decreases and the spread of its distribution on either side of μ becomes wider.
The area under the curve must equal the total probability of obtaining the outcome x; thus, the area under all three curves is the same and is equal to 1. This is important because it means that the area between any two limits defined in terms of μ and σ is the same for any value of μ and σ.
Investigation 24.
Assuming that the mean, $\bar{x}$, and the standard deviation, s, for the net weight of our samples of M&Ms are good estimates for the population’s mean, μ, and standard deviation, σ, what is the probability that the contents of a 1.69-oz bag of plain M&Ms selected at random will weigh less than the stated net weight of 1.69 oz? Suppose the manufacturer wants to reduce this probability to no more than 5%: How might they accomplish this?
For our 30 samples, the mean net weight is 48.98 g with a standard deviation of 1.433 g. The stated net weight of 1.69 oz is equivalent to 47.9 g. To find the probability that the M&Ms in a randomly selected 1.69-oz bag have a mass less than 47.9 g, we first calculate the deviation, z, taking the mean and the standard deviation for our samples as estimates for μ and σ
$z=\dfrac{x-μ}{σ}=\dfrac{47.9 -48.98}{1.433} =-0.747$
and then use the table in Appendix 3 to find the area under the normal curve to the left of 47.9, finding that it is 0.228, or 22.8%; the figure below shows this area highlighted in blue
To decrease this probability to 5%, or 0.050, we use Appendix 3 to find that this corresponds to a z of –1.645. Substituting this into the equation for z gives
$z= \dfrac{x-μ}{σ}=\dfrac{47.9 -μ}{σ}=-1.645$
With one equation and two unknowns, there are many possible combinations of mu and sigma that will work. We can place an upper limit on each by maintaining σ as 1.433 and calculating μ
$\dfrac{47.9 -μ}{1.433} =-1.645$
and calculating μ as 50.27 g, or by maintaining μ as 48.98
$\dfrac{47.9 -48.98}{σ}=-1.645$
and calculating σ as 0.650 g. Given that the average bag has a mean net weight of 48.98 g and a mean number of M&Ms of 56.63, the average plain M&M has a mass of 0.865 g. To increase the mean net weight from 48.98 g to 50.27 g, we need to increase the mean number of M&Ms per bag to 57.92, or we need to decrease the standard deviation by the equivalent of ±0.91 M&Ms.
Investigation 25.
Suppose we arrange to collect samples of plain M&Ms such that each sample contains 330 M&Ms—an amount roughly equivalent to a 10-oz bag of plain M&Ms—drawn from the same population as the data in Table 2. Can we model this data using a normal distribution in place of the binomial distribution or the Poisson distribution? What advantages are there in being able to use the normal distribution? How might you apply this to more practical analytical problems, such as determining the concentration of Pb2+ in soil?
When N is equal to 5—as is the case in Investigation 22—it is impossible to use a normal distribution to approximate a binomial distribution because there is no probability, p, where both N × p ≥ 5 and N × (1 - p) ≥ 5 are true. If we increase N to 330, then any value of p that is greater than 0.0152 or that is smaller than 0.984 will allow us to approximate the data using a normal distribution; for the data in Table 2, the smallest value of p is for green M&Ms (0.0647, or 6.47%) and the largest value of p is for brown M&Ms (0.258, or 25.8%); thus, we expect that it is possible to model the data using a normal distribution.
The average count per 1.69-oz bag, λ, of each color of M&M ranges from a minimum of 3.67 for green M&Ms to a maximum of 14.8 for brown M&Ms, neither of which meets the criterion of λ ≥ 5 needed to use a normal distribution to approximate the Poisson distribution. If we increase N from an average of 56.63, for the data in Table 2, to 330, then the average count for any color will increase by 5.83×; thus, the smallest value of λ for any color is 21.3 for green M&Ms, which suggests that it is possible to model the data using a normal distribution.
The primary advantage to us in using the normal distribution is being able to use a single distribution to model diverse types of data, which often simplifies our analysis of data.
A sample of soil consists of many different types of materials—some organic and some inorganic—each of which has a different μ and σ for its concentration of Pb2+. Because individual particles of these materials are small, in any reasonable sample the value of N for each particle is sufficiently large that the concentration of Pb2+ likely follows a normal distribution even if the underlying distribution of particles follows a binomial or a Poisson distribution. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Introduction_to_Data_and_the_Analysis_of_Data/Instructors_Guide/04_Part_IV%3A_Ways_to_Model_Data.txt |
Investigation 26.
A z of 1.96 corresponds to a 95% confidence interval. Using Appendix 2, show that this is correct. What value of z corresponds to a 90% confidence inteval, and what value of z corresponds to a 99% confidence inteval? Report the 90%, the 95% and the 99% confidence intervals for the net weight of a single 1.69-oz bag of plain M&Ms drawn from a population for which μ is 48.98 g and σ is 1.433 g. For the data in Table 2, how many of the 30 samples have net weights that fall outside of the 90% confidence interval? Does this result make sense given your understanding of a confidence interval?
From the table in Appendix 2, we know that the area to the right of a z of 1.96 is 2.5% (a probability of 0.025) of the total area. Because the normal distribution is symmetrical, we know that the area to the left of a z of –1.96 also is 2.5% of the total area. The combined area of 5% is the percentage of samples excluded from the confidence interval; thus, the term ± encompasses 95% of all samples, which is the meaning of a 95% confidence interval.
For a 90% confidence interval, we look for the value of z that corresponds to an area of 5%, which is a z of 1.645. For a 99% confidence interval, we look for the value of z that corresponds to an area of 0.5%, which is a z of 2.576. In both cases, the value of z is interpolated using the two neighboring values from the table. For example, from the table we see that z of 2.56 corresponds an area of 0.523% and a z of 2.58 corresponds to an area of 0.494%; to find the value z that corresponds to an area of 0.500% we set up the following equation
$\dfrac{2.56 -z}{2.56 -2.58} =\dfrac{0.523 -0.500}{0.523 -0.494}$
and solve for z, obtaining a result of 2.576.
The 90%, 95%, and 99% confidence intervals for a single 1.69-oz bag of plain M&Ms are
90% confidence interval: $\textrm{48.98 g ± (1.645)(1.433 g) = 48.98 g ± 2.36 g}$
95% confidence interval: $\textrm{48.98 g ± (1.96)(1.433 g) = 48.98 g ± 2.81 g}$
99% confidence interval: $\textrm{48.98 g ± (2.576)(1.433 g) = 48.98 g ± 3.69 g}$
For the data in Table 2, four of the samples have a net weight that falls outside of our 90% confidence interval, which extends from 46.62 g to 51.34 g; these samples are number 6 (46.405 g), number 29 (46.577 g), number 26 (51.682 g), and number 20 (51.730 g). With 30 samples and a 90% confidence interval, we expect, on average, to find that three samples have net weights that fall outside the confidence interval; finding four such samples is not an unreasonable outcome given that each sample represents just 3.3% of our pool of 30 samples.
Investigation 27.
Suppose we draw four 1.69-oz bags of M&Ms from a population for which μ is 48.98 g and σ is 1.433 g. What are the 90%, the 95% and the 99% confidence intervals for the mean, $\bar{x}$, of these samples? Prepare a plot that shows how n affects the width of the 95% confidence interval, expressed as $±zσ/\sqrt{n}$, and discuss the significance of your plot. Suppose we wish to decrease the confidence interval by a factor of 3× solely by increasing the number of samples taken. If the original confidence interval is based on the mean of four samples, how many additional samples must we acquire?
The 90%, 95%, and 99% confidence intervals for the mean of four 1.69-oz bag of plain M&Ms are
90% confidence interval: $\mathrm{48.98\: g ± (1.645)(1.433\: g) ⁄\sqrt{4} = 48.98\: g ± 1.18\: g}$
95% confidence interval: $\mathrm{48.98\: g ± (1.96)(1.433\: g) ⁄ \sqrt{4} = 48.98\: g ± 1.40\: g}$
99% confidence interval: $\mathrm{48.98\: g ± (12.576)(1.433\: g) ⁄ \sqrt{4} = 48.98\: g ± 1.85\: g}$
Note that each confidence interval is half of that for the analysis of a single sample because the square root of the number of samples is two.
The figure below shows how the confidence interval changes as we increase the value of n from 1 to 25. The most important feature of this plot is to note how the rate at which the confidence interval’s width becomes smaller slows down as we increase the number of samples. For example, increasing the number of samples from one to four, decreases the confidence interval by a factor of 2× from ±2.81 to ±1.40. A further two-fold decrease in the confidence interval to ±0.70 requires 16 total samples, or an additional 12 samples.
To achieve a three-fold improvement in the width of the confidence interval—that is, a decrease in the confidence interval’s width by a factor of three—requires that we increase the number of samples from n1 to n2, where
$\dfrac{\dfrac{zσ}{\sqrt{n_1}}}{\dfrac{zσ}{\sqrt{n_2}}}=\dfrac{\sqrt{n_2}} {\sqrt{n_1}}=3$
Solving shows that the ratio n2n1 is 32 or 9; thus, if the original confidence interval is based on four samples, then to achieve the desired smaller confidence interval we need a total of 9 × 4 = 36 samples, or an additional 32 samples.
Investigation 28.
Our data for 1.69-oz bags of plain M&Ms includes 30 measurements of the net weight. What are the 90%, the 95% and the 99% confidence intervals for the mean, $\bar{x}$, of these samples? Using the 99% confidence interval as an example, explain the meaning of this confidence interval. Is the stated net weight of 1.69 oz a reasonable estimate of the true mean for the population of 1.69-oz bags of plain M&Ms?
The 90%, 95%, and 99% confidence intervals for the mean of 1.69-oz bag of plain M&Ms are
90% confidence interval: $\mathrm{48.98\: g ± (1.699)(1.433\: g) ⁄\sqrt{30} = 48.98\: g ± 0.45\: g}$
95% confidence interval: $\mathrm{48.98\: g ± (2.045)(1.433\: g) ⁄ \sqrt{30} = 48.98\: g ± 0.54\: g}$
99% confidence interval: $\mathrm{48.98\: g ± (2.756)(1.433\: g) ⁄ \sqrt{30} = 48.98\: g ± 0.72\: g}$
Using the 99% confidence interval as an example, and assuming that there are no errors in our measurements, there is a 99% probability that the confidence interval’s range of net weights—from a low of 48.26 g to a high of 49.70 g—includes the true mean for the population of all 1.69-oz bags of plain M&Ms; there is a 1% probability that population’s mean falls outside of this range. Given that the 99% confidence interval does not include the stated net weight of 1.69 oz (47.9 g), we can safely conclude that 1.69 oz is not a good estimate for the population’s mean net weight.
Investigation 29.
In 1996, Mars, the manufacturer of M&Ms, reported the following distribution for the colors of plain M&Ms: 30% brown, 20% red, 20% yellow, 10% blue, 10% green, and 10% orange. Pick any one color of M&Ms and, using the data in Table 2, calculate the percentage of that color in each of the 30 samples. Report the mean and the standard deviation for your color and use a t-test to determine whether your sample’s mean is consistent with the result reported by Mars. Gather results for the remaining five colors from other students and discuss your pooled results. Assuming that the distribution of colors reported by Mars is correct, what can you conclude about the manufacturing process.
For this problem the null hypothesis is $H_0\textrm{: }\bar{x}=μ$ and the alternative hypothesis is $H_A\textrm{: }\bar{x}≠μ$, where $\bar{x}$ is the mean of the 30 samples for the color of interest and μ is the mean reported by Mars for the color of interest. The following table summarizes the results of the t-test, by color
color
$\bar{x}$ (%)
s (%)
μ (%)
t
reject H0 at
brown
25.81
5.004
30
4.581
$α$ < 0.01
red
17.64
6.600
20
1.962
0.05 < $α$ <0.10
yellow
25.52
7.594
20
3.984
$α$ < 0.01
blue
11.75
6.667
10
1.440
$α$ > 0.10
green
6.52
4.743
10
4.018
$α$ < 0.01
orange
12.75
4.983
10
3.024
$α$ < 0.01
where s is the standard deviation for the 30 samples, t is the calculated experimental value of t based on $\bar{x}$, s, and μ, and the last column defines the value of $α$ for which we can reject the null hypothesis and accept the alternative hypothesis.
With the exception of the color blue, we have good evidence that the distribution of colors for these 30 samples is not in agreement with the manufacturer’s stated distribution. For brown, yellow, green, and orange, the 99% confidence interval does not include μ; for red, the 90% confidence interval does not include μ. These results are not particularly suprising. Although the distribution of colors in a production batch presumably matches the percentages provided by Mars, the mixing of the M&Ms at the level at which individual bags are filled likely is far from homogeneous.
In a response to a query regarding the proportions of colors in bags of M&Ms, the manufacturer noted that “[e]ach large production batch is blended to [this] ratio and mixed thoroughly. However, since the individual packages are filled by weight on high-speed equipment, and not by count, it is possible to have an unusual color distribution.” The full response can be seen here. It seems more likely that the color distribution is a function of sampling uncertainty associated with the population’s homogeneity at the level of sampling.
06 Part VI: Now Its Your Turn
Answers, of course, will vary.
In addition to duplicating some of the questions considered in this case study using the additional data sets or newly collected data, students should be able to extend the concept of a t-test to include comparisons of different colors of M&Ms or off different types of M&Ms, and a consideration of paired vs. unpaired data. Students also can explore their ability to model data using the binomial distribution to model the distribution of samples of fixed size, or using the Poisson distribution to model the frequency of a rare event, such as the presence of damaged M&Ms in bags of a fixed size.
Other types of analyses that go beyond what is presented in this case study include linear regression (for example, mass as a function of diameter using Data Set 2), analysis of variance, and quality control charts.
07 Appendix
The following R code, with comments, was used to generate the net weights for the 30 samples. The file data.csv was identical in structure to Table 2, but did not include the final column of net weights. This code reads in the original data, calculates the total number of M&Ms in each sample, draws the appropriate number of M&Ms for each sample and calculates the average weight of an M&M in the sample, calculates the net weight of M&Ms in each sample, and adds the net weights to the original data and saves the data as a new file.
# Create data frame ‘rawdata’ to store data for the samples
rawdata = read.csv(“data.csv”)
# Create vector ‘total’ to store number of M&Ms in each sample, calculated
# by summing, by row (1), the number of M&Ms in columns 3–8 of ‘rawdata’
total = apply(rawdata[ , 3:8], 1, sum)
# Create vector ‘avg.weight’ to store average weight of M&Ms in each sample
avg.weight = seq(1:30)
# Create separate vectors for the population mean and the standard deviation
# with values determined using the masses of 462 plain M&Ms available at
# the Puget Sound Data Hoard (http://stat.pugetsound.edu/hoard/Default.aspx)
mu = 0.86483
sig = 0.046199
# For each sample, calculate the average mass for its M&Ms, with the number
# of M&Ms defined by the vector ‘total,’ using a random draw from a normal
# distribution defined by the population’s mean standard deviation
for (i in 1:30) avg.weight[i] = mean(rnorm(total[i], mu, sig))
# Find the net weight for each sample by multiplying the number of M&Ms in
# the sample by the average weight of an M&M in the sample
net.weight = total * avg.weight
# Create final data set by adding net.weight to the original data and save
mmdata = data.frame(rawdata, net.weight)
write.csv = (mmdata, file = “mmdata.csv”) | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Introduction_to_Data_and_the_Analysis_of_Data/Instructors_Guide/05_Part_V%3A_Ways_to_Draw_Conclusions.txt |
This module introduces students to the relationship between chromatograms and mass spectra using hands-on group work with molecules built from LEGO™ bricks and environmental case studies involving the analysis of pesticides by GC-MS and oil spill dispersant chemicals by LC-MS/MS.
The first of three in-class exercises guides students through the process of creating total and extracted ion chromatograms and mass spectra using molecules built from LEGO™ bricks. The goals of this hands-on exercise are to build a solid conceptual understanding of mass spectral peak origin, interpretation, and connection to the chromatogram. Two environmental case studies, involving pesticides (GC-MS) and oil spill dispersants (LC-MS/MS), reinforce the relationships between chromatography and mass spectrometry, explore the mass spectrum of an isotopically-labeled internal standard, require recall of the factors influencing peak broadening, and introduce the idea of tandem mass spectrometry. The module assumes students have had prior exposure to chromatographic separations, the basic block diagram of a mass spectrometer, and common ionization mechanisms (e.g., EI and ESI).
Resources
• In-Class Exercises (Word, PDF)
• Data Sheet (Excel, PDF)
• Mass Spectra and Total Ion Chromatogram Simulation (Excel)
• Mass Spectra Simulation (PDF)
• Total Ion Chromatogram Simulation (PDF)
• Key (Word, PDF)
• Instructor’s Guide (Word, PDF)
• Assessment Questions: For Assessment questions that accompany this module, please contact David Griffith ([email protected])
Mass Spectrometry: LEGO Exercise
Learning Objectives
After completing these exercises, you will be able to…
(1) Explain the relationship between a total ion chromatogram (TIC), an extracted (or selected) ion chromatogram (XIC), and a tandem (MS/MS) mass spectrum.
(2) Interpret GC-MS and LC-MS/MS data and predict/explain simple fragmentation patterns
Exercise 1
Within your group of three, assign and distribute a different color (RED, BLUE, YELLOW) “Duplo” molecule to each person.
In the table below, record the number of individual blocks that make up each color. These will represent the molecular masses of your group’s three molecules.
Molecular Mass
RED
BLUE
YELLOW
1. Predict the elution order of the three molecules following GC chromatographic separation. Assume that RED has the lowest boiling point and YELLOW has the highest.
Elution order: 1st_________________ 2nd________________ 3rd_______________
1. What logic could you use to predict how each Duplo molecule would break apart (“fragment”) if dropped on the ground?
2. How would this same logic apply to the fragmentation of actual molecules during GC-MS? (In other words, why do molecules break apart in certain places and not in others?)
3. Fill in the table below with your predictions for the “masses” (# of blocks) of the likely fragments of each molecule. [Note: each molecule does not necessarily need to form five different mass fragments.]
Fragment 1
Fragment 2
Fragment 3
Fragment 4
Fragment 5
Molecular Mass
Mass
Mass
Mass
Mass
Mass
RED
BLUE
YELLOW
Once your group has agreed on an elution order and predicted fragment sizes, bring your intact molecules up to the front of the class and place them in the correct order. If another group has a different elution order than you, discuss your reasoning and come to a consensus.
Now let’s look at how each molecule fragments in the mass spectrometer.
1. Will every fragment be detected?
2. “Each fragment is formed by the breakdown of the next largest fragment.”
1. Describe how you would plot the total ion chromatogram (TIC) for this example and explain which ions contribute to the TIC signal.
2. Describe how you would plot an extracted ion chromatogram (XIC) for this example.
Exercise 2
(Adapted from Bullen, Fitch, Kelly, and Larive, 2013, Environmental Analysis – Lake Nakuru Flamingos: Pesticides, Analytical Sciences Digital Library, Online.)
Mass spectrometry is widely used in analytical chemistry, forensics, bioanalysis, and environmental analysis. Some advantages of mass spectrometry include its sensitivity and low detection limits. Both qualitative and quantitative results can be obtained using a mass spectrometer. However, mass spectrometers are generally more costly than electrochemical and spectroscopic instruments.
A total ion chromatogram (TIC) was collected by GC-MS for a standard solution containing a mixture of 20 organochlorine pesticides:
Consider the chromatographic peak at 21.876 min corresponding to the insecticide, DDT.
1. List three factors responsible for the broadness of the peak.
2. What does the y-axis label “Abundance” refer to?
3. What has to happen for a DDT molecule to create a detector response?
4. Explain why neutral molecules are not detected in MS.
To help confirm the identity of the peak at 21.876 min as DDT, it would be common to compare a mass spectrum (m/z vs. abundance) from the middle portion of the peak to a reference spectrum for DDT, taken under similar ionization conditions. A useful source for reference EI mass spectra is the NIST Chemistry WebBook.
The NIST mass spectrum (GC-MS; EI; 70eV) of DDT (MW = 354.5 g/mol) is given below.
1. Predict how the mass spectrum of DDT-d8 would differ from the mass spectrum above. (DDT-d8 is an isotopically labeled DDT in which 8 hydrogen atoms have been replaced by 8 deuterium atoms.)
We often use internal standards to correct for instrument variability and losses during sample processing. Ideally, the internal standard is an isotopically labeled version of the analyte of interest. For example, if you were analyzing a water sample for DDT concentration, you might inject a known amount of DDT-d8 at the time of collection, then prepare your sample (e.g., extract, concentrate, clean up) for injection on the GC-MS.
1. Consider the TIC for a water sample extract that contained DDT as well as the internal standard DDT-d8 that you added. Is there a problem? If so, how would you solve it?
Exercise 3
In 2010, the Deep Water Horizon oil spill released 5 million barrels of crude oil into the bottom waters of the Gulf of Mexico. One strategy to minimize surface slicks was to release chemical dispersants (i.e., surfactants) into the water by plane or into the deep water near the wellhead, 5000 feet below the surface. This was the first time that dispersants were released into deep water on a large scale (2.1 million gallons of dispersants), so there was considerable uncertainty about what would happen to the dispersant chemicals over time. The following figure was taken from a paper titled “Fate of Dispersants Associated with the Deepwater Horizon Oil Spill” by Elizabeth Kujawinski et al. (2011).
1. Have each person in your group pick a panel in the figure, take a couple minutes to study and annotate your panel (referring to other panels as needed), then take turns describing your panel to the other members of your group. In your explanations be sure to include a full description of the type of data plotted, the axes, and the meaning of the information on the right side of your panel. [Bonus: can you find the error in the figure caption?]
1. The table above (modified from Glover et al., 2014, Chemosphere 111: 596-602) lists some of the components of Corexit, a dispersant mixture used during the Deep Water Horizon oil spill. Assuming the LC/MS analysis by Kujawinski et al. utilized electrospray ionization in negative mode, which component did they detect in their field sample? Explain your reasoning.
2. How do you think Kujawinski et al. quantified the amount of DOSS in their field sample?
In fact, the data from Figure S2 were collected using a liquid chromatograph coupled to an ion trap mass spectrometer operated in MS/MS mode. In this mode, the mass spectrometer uses the mass analyzer (an ion trap) to isolate ions of one particular m/z, fragment them, then sort/detect the resulting fragment m/z values. In this way Kujawinski et al. were able to create an MS/MS spectrum (not shown), which plots the m/z values for the fragments of the selected ion. The bottom panel in Figure S2 shows the XIC for two characteristic fragments (m/z 227 and m/z 291) of the molecular ion m/z 421.
1. Describe a situation in which it would be advantageous to use an MS/MS (“tandem”) instrument? Explain your reasoning. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Mass_Spectrometry%3A_LEGO_Exercise/In-Class_Exercises__Chromatograms_and_Mass_Spectra.txt |
Exercise = black text; Key = blue text; Instructor’s Guide = bracketed italicized text with yellow highlight
[Before starting this exercise students should understand the basics of GC and LC separations and have had a brief introduction into mass spectrometry, including instrument design (basic block diagram), ionization sources (EI and ESI), and simple quadrupole or magnetic sector mass analyzers.]
[Build molecules out of large LEGO™ (“Duplo”) bricks or an equivalent snap together building brick (see configurations below). Add one “charge” (round drafting sticker) to a single exterior brick of each molecule; vary the charge placement on each molecule.]
[Print three data sheets (attached) - one for each molecule type (color).]
[Use duct tape to create large axes (retention time vs. mass) on the floor of the space where you will “fragment” the Duplo molecules. The space should ideally have enough room for all the students in your class to view and walk around the plot they will create on the floor. Label the axes with a permanent marker. The time axis can simply indicate the direction of increasing retention time. The mass axis should include numbers from 1 to 16.]
Learning Objectives
After completing these exercises, you will be able to…
(1) Explain the relationship between a total ion chromatogram (TIC), an extracted (or selected) ion chromatogram (XIC), and a tandem (MS/MS) mass spectrum.
(2) Interpret GC-MS and LC-MS/MS data and predict/explain simple fragmentation patterns
Exercise 1
[Set aside ~45 minutes of class time. Exercise 1 works well as an in-class activity; alternatively, it may be used as an exercise directly preceding a laboratory period focused on mass spectrometry.]
Within your group of three, assign and distribute a different color (RED, BLUE, YELLOW) “Duplo” molecule to each person. [If extra molecules are available, it may be helpful to give each group one extra molecule of each color to use as reference when figuring out how each might fragment.]
In the table below, record the number of individual blocks that make up each color. These will represent the molecular masses of your group’s three molecules.
Molecular Mass
RED
12 (compact)
BLUE
12 (branched)
YELLOW
16
1. Predict the elution order of the three molecules following GC chromatographic separation. Assume that RED has the lowest boiling point and YELLOW has the highest.
Elution order: 1st_____ RED ____________ 2nd_____ BLUE ___________ 3rd_____YELLOW__________
[Groups may sometimes predict the reverse order. If they do, you might ask the group to think about evaluating the three molecules in terms of volatility. What does it mean that RED has the lowest boiling point of the three? It suggests that it is the most volatile and thus most likely to partition into the mobile (gas) phase, move through the GC column quickly, and be detected first.]
1. Figure out where the weak points are within the molecule and evaluate which parts appear to be more exposed
2. Should be related to bond strength and stability of molecules and fragments
3. Fill in the table below with your predictions for the “masses” (# of blocks) of the likely fragments of each molecule. [Note: each molecule does not necessarily need to form five different mass fragments.]
Fragment 1
Fragment 2
Fragment 3
Fragment 4
Fragment 5
Molecular Mass
Mass
Mass
Mass
Mass
Mass
RED
12 (compact)
BLUE
12 (branched)
YELLOW
16
[Draw a time axis on the board ahead of time or while students are discussing the questions above. It works well if the axis is drawn increasing to the right as is typical for chromatograms.]
Once your group has agreed on an elution order and predicted fragment sizes, bring your intact molecules up to the front of the class and place them in the correct order. If another group has a different elution order than you, discuss your reasoning and come to a consensus.
[At this point all of the RED Duplo molecules should be on stage right, BLUE in the center, and YELLOW on stage left. Each color will have it’s own clear plastic tub. You could discuss the idea that each molecule will have a characteristic “retention” time, but for our purposes here only the order matters.]
[If necessary, this would be a good time to move the class to the space where you’ve taped the time vs. mass axes on the floor and will be fragmenting the molecules. At this point, it is useful to have the students group up according to the color of the molecule they handled in their original group of three (e.g., all those who worked with RED get together; same for BLUE and YELLOW).]
Now let’s look at how each molecule fragments in the mass spectrometer.
[Select one representative for each color to come to the front, pick up their tub and get positioned at the appropriate location on the time axis taped to the floor (Elution order: RED, BLUE, YELLOW). Prompt the class to describe what happens after each color band elutes from the GC column (ionization/fragmentation via EI; a good chance to review how fragmentation happens in EI), then let the representative for the RED group simulate fragmentation by dumping the tub containing all the RED molecules onto the floor from approximately chest height. Have the RED group arrange all of the fragments on the floor in order of mass and perpendicular to the retention time axis. Repeat with BLUE (middle of the time axis) then YELLOW (near the end of the time axis), both from the same approximate height as RED.]
[Note here the importance of the height (a proxy for energy; 70eV for EI) from which the molecules were dropped, both in terms of fragmentation pattern and consistency. You might also ask the students why most GC-MS EI sources are set to 70eV and what will happen if we drop the Duplo molecules from the height of the roof instead (higher energy yields more smaller fragments). Usually the class intuits that fragmentation patterns should be related to the energy of the electron beam in EI, that mass spectra comparison/database matching is only possible if fragmentation patterns are consistent from instrument to instrument, and that higher energies will result in a shift towards smaller fragments.]
1. Will every fragment be detected? [Have the students discuss this question with all other students that handled the same color as they did. After students realize that only charged fragments will be detected, have them find the fragments with a round drafting sticker attached (representing a “charge”), and remove all other fragments. Note that based on the location of the charge and the way that each molecule fragments, some fragments will be more likely to be observed by the detector than others. Record the number and mass of each charged fragment for each color molecule (using the data sheet; one sheet per color).]
No, because only the ones with charge will get through the mass analyzer to reach the detector.
2. “Each fragment is formed by the breakdown of the next largest fragment.”
[A common misconception among students is that mass spectra reflect the orderly breakdown of the molecular ion into the largest fragment followed by serial fragmentation into increasingly smaller masses. It is worth mentioning that fragmentation patterns are better described in probabilistic terms. Asking students to envision how the Duplo molecules fragmented when dumped on the floor can help them appreciate that some hit the floor first and hardest, while others were “cushioned” or “shielded”; the orientation and timing of each Duplo molecule hitting the floor was different; the actual height from which each was dropped varied slightly due to the starting position of the Duplo molecule in the tub; even the strength of each linkage within two of the same color Duplo molecules is likely to be slightly different. However, each of these differences are layered on top of the inherent defining structural features of each Duplo molecule to determine which fragments dominate.]
1. To get the TIC you would sum up all of the charged fragments at every time point along the chromatogram. Ions contribute to the signal if they are within the range of m/z values selected by the mass analyzer.
2. An “extracted ion chromatogram” (XIC) displays the intensity of only a particular ion (or ions) of interest over time.
[After groups have some ideas and report out, review the difference between a chromatogram (TIC, XIC) and a mass spectrum. Now use the fragment data sheet to fill in the Excel MS tab, which will show a mass spectrum for each colored molecule. Discuss the convention that spectra are often normalized so that the most abundant peak is set to 100, and point out that each total mass spectrum has a RT associated with it. Does each type of molecule contain a characteristic/unique mass? Are any masses common to more than one type of molecule? The plots below (also included as a pptx file) may be useful as a way to connect the Duplo molecule activity to the “look” of real data from a GC-MS.]
Exercise 2
(Adapted from Bullen, Fitch, Kelly, and Larive, 2013, Environmental Analysis – Lake Nakuru Flamingos: Pesticides, Analytical Sciences Digital Library, Online.) [Set aside ~25 minutes of class time]
Mass spectrometry is widely used in analytical chemistry, forensics, bioanalysis, and environmental analysis. Some advantages of mass spectrometry include its sensitivity and low detection limits. Both qualitative and quantitative results can be obtained using a mass spectrometer. However, mass spectrometers are generally more costly than electrochemical and spectroscopic instruments.
A total ion chromatogram (TIC) was collected by GC-MS for a standard solution containing a mixture of 20 organochlorine pesticides:
Consider the chromatographic peak at 21.876 min corresponding to the insecticide, DDT.
1. List three factors responsible for the broadness of the peak. ]
Longitudinal diffusion, mass transport lags, multiple overlapping compounds
[Students will often come up with the first two answers if they have recently completed a unit on separations and the factors that influence peak broadening. One approach to having students discover the last answer (multiple overlapping compounds) is to ask them to consider the chromatogram at the base of the peak at 21.876 min. They should notice the small peak that overlaps with the peak of interest. Can they know from the TIC alone that other compounds are not co-eluting with the peak of interest? Once students agree that the answer is no, a nice way to reinforce connections to previous material is to ask them to think of ways to determine if the peak was “pure”. Their answers may include techniques to improve chromatographic resolution or utilize MS data to plot an XIC or evaluate mass spectral consistency through the width of the peak.]
1. Number of ions that reach the detector (creating electrical signals that can be quantified or “counted”)
2. It has to be injected into the instrument, survive the GC column and inlet intact, create ions (molecular ions or fragment ions) in the source, make it through the mass analyzer (quadrupole, TOF, etc.), and reach the detector.
3. Neutral molecules are unaffected by the electromagnetic fields used in most mass analyzers.
To help confirm the identity of the peak at 21.876 min as DDT, it would be common to compare a mass spectrum (m/z vs. abundance) from the middle portion of the peak to a reference spectrum for DDT, taken under similar ionization conditions. A useful source for reference EI mass spectra is the NIST Chemistry WebBook.
The NIST mass spectrum (GC-MS; EI; 70eV) of DDT (MW = 354.5 g/mol) is given below.
[If time allows, the DDT mass spectrum shown above provides a good opportunity to ask students to explain the y-axis label (“Rel. Abundance”), how it differs from other mass spectra y-axis labels (“Abundance”; “Detector Response”), and the nature of the raw signal used to construct this particular mass spectrum. These questions may also be useful as a way to transition into a subsequent discussion about mass analyzers and detectors. Some students might also notice that the molecular ion peak (and others) appear as a groups of peaks, a feature that can be used to talk about characteristic isotope patterns common for halogen-containing compounds.]
1. The masses of the molecular ion cluster would be 8 units higher. Depending on where the deuterium labels were placed, you would also expect increases (of 0 to 8 units) in fragment m/z values.
[Students sometimes assume that all fragments of DDT-d8 will be 8 units higher. To correct this misunderstanding, ask students to imagine where the deuterium atoms might be located on the DDT structure then predict where the deuterium atoms will be when the structure fragments. Students should recognize that fragments can’t all increase by 8 and some fragments (those without a deuterium label) may not increase in mass at all compared to the unlabeled DDT.]
We often use internal standards to correct for instrument variability and losses during sample processing. Ideally, the internal standard is an isotopically labeled version of the analyte of interest. For example, if you were analyzing a water sample for DDT concentration, you might inject a known amount of DDT-d8 at the time of collection, then prepare your sample (e.g., extract, concentrate, clean up) for injection on the GC-MS.
1. Yes, the problem is that DDT and DDT-d8 will probably co-elute (you might see only one peak in the TIC at 21.876 min), so you wouldn’t be able to quantify either one. You could solve the problem by picking one or two fragments that are specific to DDT, and then have the computer sum them up at each time along the chromatogram, ignoring all the other fragment m/z values. This new (simplified subset) chromatogram is called an extracted (or selected) ion chromatogram (XIC). You could then do the same thing using fragment m/z values specific to DDT-d8. This would allow you to separate the DDT peak (XIC) from the DDT-d8 peak (XIC) and integrate (quantify) each.
Exercise 3
[Set aside ~25 minutes of class time]
In 2010, the Deep Water Horizon oil spill released 5 million barrels of crude oil into the bottom waters of the Gulf of Mexico. One strategy to minimize surface slicks was to release chemical dispersants (i.e., surfactants) into the water by plane or into the deep water near the wellhead, 5000 feet below the surface. This was the first time that dispersants were released into deep water on a large scale (2.1 million gallons of dispersants), so there was considerable uncertainty about what would happen to the dispersant chemicals over time. The following figure was taken from a paper titled “Fate of Dispersants Associated with the Deepwater Horizon Oil Spill” by Elizabeth Kujawinski et al. (2011).
1. Bottom plot: Shows the XIC for two fragments (m/z 227 and m/z 291) of the m/z 421 molecular ion
y-axis: relative abundance (signal intensity adjusted so that the highest value is 100)
NL: normalization level: the actual signal intensity (e.g., counts per second) for the highest value
ESI: electrospray ionization (negative mode)
Bonus: The figure shows chromatograms, not spectra.
1. DOSS (dioctyl sodium sulfosuccinate): The observed m/z value for the molecular ion at 6 min is consistent with ionization via the removal of a hydrogen atom from DOSS, which we would write as [DOSS – H]. The fragment ions’ (m/z 227 and 291) retention time aligns with that of the molecular ion and both are plausible fragments based on the structure of DOSS. Kujawinski confirmed their assignment by injecting a pure standard solution of DOSS and observing the same fragments and retention times.
[In the table, DOSS is shown as the sodium salt (MW 444 g/mol). If students are confused about why this produces a molecular ion with m/z 421, ask them to describe what will happen to the sodium salt under ESI(–) conditions. They typically realize that the sodium salt of DOSS will form the [DOSS – H] molecular ion, and the positive sodium ion (MW 23) will not be observed or attached.]
1. They determined the peak area in the XIC for m/z 421 then used a DOSS calibration curve to calculate the concentration. Why not use the fragment ions? Because they give less signal, thus have lower signal to noise ratio, and likely a higher LOD.
In fact, the data from Figure S2 were collected using a liquid chromatograph coupled to an ion trap mass spectrometer operated in MS/MS mode. In this mode, the mass spectrometer uses the mass analyzer (an ion trap) to isolate ions of one particular m/z, fragment them, then sort/detect the resulting fragment m/z values. In this way Kujawinski et al. were able to create an MS/MS spectrum (not shown), which plots the m/z values for the fragments of the selected ion. The bottom panel in Figure S2 shows the XIC for two characteristic fragments (m/z 227 and m/z 291) of the molecular ion m/z 421.
1. -To determine the structure of an unknown organic compound by fragmenting the molecular ion, collecting those fragments, fragmenting the fragment ions, collecting those fragments, and so on. This process can continue n times and is thus sometimes referred to as MSn. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Mass_Spectrometry%3A_LEGO_Exercise/Instructors_Guide__Chromatograms_and_Mass_Spectra.txt |
Thumbnail: White light is dispersed by a prism into the colors of the visible spectrum. (CC BY-SA 3.0; D-Kuru).
01 In Class Problems: Molecular Spectroscopy
Electromagnetic Radiation
1. What is the relationship between the energy (E) and frequency ($\nu$) of electromagnetic radiation?
2. What is the relationship between the energy and wavelength ($\lambda$) of electromagnetic radiation?
3. Write the types of radiation observed in the electromagnetic spectrum going from high to low energy. Also include what types of processes occur in atoms or molecules for each type of radiation.
Beer’s Law
1. What factors influence the absorbance that you would measure for a sample? Is each factor directly or inversely proportional to the absorbance?
2. If you wanted to measure the concentration of a particular species in a sample, describe the procedure you would use to do so.
3. Suppose a small amount of stray radiation (PS) always leaked into your instrument and made it to your detector. This stray radiation would add to your measurements of Po and P. Would this cause any deviations to Beer's law? Explain.
4. The derivation of Beer's Law assumes that the molecules absorbing radiation don't interact with each other (remember that these molecules are dissolved in a solvent). If the analyte molecules interact with each other, they can alter their ability to absorb the radiation. Where would this assumption break down? Guess what this does to Beer's law?
5. Beer's law also assumes purely monochromatic radiation. Describe an instrumental set up that would allow you to shine monochromatic radiation on your sample. Is it possible to get purely monochromatic radiation using your set up? Guess what this does to Beer's law.
6. Is there a disadvantage to reducing the slit width?
7. Consider the relative error that would be observed for a sample as a function of the transmittance or absorbance. Is there a preferable region in which to measure the absorbance? What do you think about measuring absorbance values above 1?
8. What are some examples of matrix effects and what undesirable effect could each have that would compromise the absorbance measurement for a sample with an unknown concentration?
02 Out-of-class Ques
Sources
1. Describe the desirable features of a radiation source for a spectrophotometer.
2. Plot the relative intensity of light emitted from an incandescent light bulb (y-axis) as a function of wavelength (x-axis). This plot is a classic observation known as blackbody radiation. On the same graph, show the output from a radiation source that operated at a hotter temperature.
3. Examining the plots above, what does this suggest about the power that exists in radiation sources for the infrared portion of the spectrum?
4. Explain the advantages of a dual- versus single-beam spectrophotometer.
Lasers
1. Why is it impossible to create a 2-level laser?
2. Using your understanding of a 2-level system, explain what is meant by a 3-level and 4-level system. 3- and 4-level systems can function as a laser. How is it possible to achieve a population inversion in a 3- and 4-level system?
3. Which of the two (3- or 4-level system) is generally preferred in a laser and why?
Monochromators
1. Explain in general terms the mechanism in a prism and grating that leads to the attainment of monochromatic radiation. Compare the advantages and disadvantages of each type of device. What is meant by second order radiation in a grating? Describe the difference between a grating that would be useful for the infrared region of the spectrum and one that would be useful for the ultraviolet region of the spectrum.
2. Explain the significance of the slit width of a monochromator. What is the advantage(s) of making the slit width smaller? What is the disadvantage(s) of making the slit width smaller?
Detectors
1. Explain how a photomultiplier tube works. What are any advantages or disadvantages of a photomultiplier tube?
2. Describe a photodiode array detector. What advantages does it offer over other detection devices?
03 In-class Question
General aspects of UV/VIS absorption spectra
1. Compare and contrast the absorption of ultraviolet (UV) and visible (VIS) radiation by an atomic substance (something like helium) with that of a molecular substance (something like ethylene).
2. Do you expect different absorption peaks or bands from an atomic or molecular substance to have different intensities? If so, what does this say about the transitions?
3. Compare a molecular absorption spectrum of a dilute species dissolved in a solvent at room temperature versus the same sample at 10K.
4. Are there any other general processes that contribute to broadening in an absorption spectrum?
5. Compare the UV absorption spectrum of 1-butene to 1,3-butadiene.
1. Rank these from high to low energy.
2. Compare the UV absorption spectrum of benzene and pyridine.
1. The peaks in the 320-380 nm portion of the UV absorption spectrum of pyridine shift noticeably toward the blue (high energy) portion of the spectrum on changing the solvent from hexane (C6H14) to methanol (CH3OH). Account for this change.
2. The peaks in the UV spectrum of benzene shift slightly toward the red (low energy) portion of the spectrum on changing the solvent from hexane (C6H14) to methanol (CH3OH). Account for this change.
UV/VIS spectroscopy as a qualitative and quantitative tool
1. Is UV/VIS spectroscopy useful as a qualitative tool?
2. Is UV/VIS spectroscopy useful as a quantitative tool?
3. If you were using UV spectroscopy for quantitative analysis, what criteria would you use in selecting a wavelength for the analysis?
4. What variables influence the recording of UV/VIS absorption spectra and need to be accounted for when performing qualitative and quantitative analyses?
5. Provided the UV/VIS absorption spectra of HA and A differ from each other, describe a method that you could use to measure the pKa of the acid. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Molecular_and_Atomic_Spectroscopy/01_In_Class_Problems%3A_Molecular_Spectroscopy/01_In-class_Question.txt |
Energy level diagrams for organic molecules
1. Draw an energy level diagram for a typical organic compound with $\pi$ and $\pi$ * orbitals and indicate which orbitals are filled and which are empty.
2. Now consider the electron spin possibilities for the ground and excited state. Are there different possible ways to orient the spins (if so, these represent different spin states).
3. Do you think these different spin states have different energies?
4. Which one do you expect to be lower in energy?
5. If the spin state is defined as (2S + 1) where S represents the total electronic spin for the system, try to come up with names for the ground and possible excited states for the system that are based on their spin state.
6. Draw a diagram of the energy levels for such a molecule. Draw arrows for the possible transitions that could occur for the molecule.
7. What do you expect for the lifetime of an electron in the T1 state?
8. Why is phosphorescence emission weak in most substances?
9. Which transition ($\pi$*-$\pi$ or $\pi$*-n) would have a higher fluorescent intensity? Justify your answer.
Instrumental considerations for luminescence measurements
1. What would constitute the basic instrumental design of a fluorescence spectrophotometer?
2. What would be the difference between an excitation and emission spectrum in fluorescence spectroscopy?
3. Draw representative examples of the excitation and emission spectrum for a molecule.
4. Describe a way to measure the phosphorescence spectrum of a species that is not compromised by the presence of any fluorescence emission.
5. If performing quantitative analysis in fluorescence spectroscopy, which wavelengths would you select from the spectra you drew in the problem above?
6. Which method is more sensitive, absorption or fluorescence spectroscopy?
Variables that influence fluorescence measurements
1. What variables influence fluorescence measurements? For each variable, describe its relationship to the intensity of fluorescence emission.
2. Consider the reaction shown below for the dissociation of 2-naphthol. This reaction may be either slow (slow exchange) or fast (fast exchange) on the time scale of fluorescence spectroscopy. Draw the series of spectra that would result for an initial concentration of 2-naphthol of 10-6 M if the pH was adjusted to 2, 8.5, 9.5, 10.5, and 13 and slow exchange occurred. Draw the spectra at the same pH when the exchange rate is fast.
1. Devise a procedure that might allow you to determine the pKa of a weak acid such as 2-naphthol.
2. Which compound will have a higher quantum yield: anthracene or diphenylmethane?
05 In-class Question
Background information
1. Can infrared spectra be recorded in air? If so, what does this say about the major constituents of air?
2. Why don’t the major constituents of air absorb infrared radiation? It might be worth noting that a molecule such as hydrogen chloride (HCl) does absorb infrared light.
3. Describe the vibrations of carbon dioxide (CO2) and determine which ones absorb infrared radiation.
Specialized techniques
1. One technique is called non-dispersive infrared (NDIR) spectroscopy. NDIR is usually used to measure a single constituent of an air sample. Think what the name implies and consider how such an instrument might be designed.
Fourier-transform Infrared Spectroscopy
1. Consider the light path for a Michelson interferometer and plot the intensity of radiation at the sample versus the position of the moveable mirror for monochromatic radiation of wavelength x, 2x or 4x.
1. What are the advantages of FT-IR spectrophotometers over conventional IR spectrophotometers that use a monochromator?
In-class Questions%3
1. Consider the molecular vibrations of carbon dioxide and determine whether or not they are Raman active.
2. Which set of lines, Stokes or anti-Stokes, is weaker?
3. What effect would raising the temperature have on the intensity of Stokes and anti-Stokes lines?
4. What would be the ideal source to use for measuring Raman spectra?
5. The molecule carbon tetrachloride (CCl4) has three Raman-active absorptions that occur at 218, 314 and 459 cm-1 away from the laser line. Draw a representation of the Raman spectrum of CCl4 that includes both the Stokes and anti-Stokes lines.
6. Why do the anti-Stokes lines of carbon tetrachloride have the following order of intensity: 219 > 314 > 459 cm-1?
02 In Class Problems: Atomic Spectroscopy
Atomization Sources
1. What are the relative advantages and disadvantages of using a flame or furnace as an atomization source?
Instrumental Design Features
1. If you were to run an analysis using an atomic absorption spectrophotometer, you would note that a separate source lamp called a hollow cathode lamp is needed for each individual element that you wish to measure. For example, a lead lamp emits the specific lines of light that are absorbed by lead. Why is the cathode designed with a hollow configuration?
2. Needing a different lamp for each element is expensive and not as simple as using a continuum source with a monochromator. Why is it apparently not feasible to use a broadband continuum source with a monochromator when performing atomic absorption spectroscopy?
3. One thing you might consider is whether continuum lamps have enough power in the part of the electromagnetic spectrum absorbed by elements. In what part of the electromagnetic spectrum do most atoms absorb (or emit) light?
4. Do powerful enough continuum sources exist in this region of the electromagnetic spectrum?
5. A more helpful thing to consider is the width of an atomic line. What are the two major contributions to the broadening of atomic lines? (Hint: We went over these earlier in the course).
6. When these contributions to line broadening are considered, the width of an atomic line is observed to be in the range of 0.002-0.005 nm. Using information about the width of an atomic line, explain why a continuum source will not be suitable for measuring atomic absorption.
7. Why does the hollow cathode lamp have a low pressure instead of a high pressure of argon filler gas?
8. Flame noise (either emission from the flame or changes in the flame background as a sample is introduced) presents a significant interference in atomic methods. Can you design a feature that could be incorporated into an atomic absorption spectrophotometer to account for flame noise?
9. Particulate matter in a flame will scatter light from the hollow cathode lamp. This is a problem since a detector cannot distinguish the difference between light that is scattered and light that is absorbed.
Molecular species in a flame exhibit broadband absorption of light. Again, a detector cannot distinguish broadband absorption from molecular species from absorption by atomic species.
Can you design to feature that could be incorporated into an atomic absorption spectrophotometer than can be used to account for both scattered light and light absorbed by molecular species?
Chemical Interferences
1. Metal complexes with low volatility are often difficult to analyze when performing atomic absorption measurements because the atomization efficiency is reduced to unacceptably low levels. Can you devise a strategy or strategies for eliminating the problem of a non-volatile metal complex?
2. Can you devise a strategy to overcome unwanted ionization of the analyte?
Matrix Effects
1. Devise a general method that can be used to account for the presence of unknown matrix effects.
Thumbnail: White light is dispersed by a prism into the colors of the visible spectrum. (CC BY-SA 3.0; D-Kuru). | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Molecular_and_Atomic_Spectroscopy/01_In_Class_Problems%3A_Molecular_Spectroscopy/04_In-class_Question.txt |
Thumbnail: White light is dispersed by a prism into the colors of the visible spectrum. (CC BY-SA 3.0; D-Kuru).
04 Learning Objectives
After completing this unit the student will be able to:
1. Explain what it means to use spectroscopic methods for qualitative and quantitative analysis.
2. Identify the terms in and describe deviations to Beer’s Law.
3. Describe the effect of changing the slit width and the impact it will have on qualitative and quantitative analyses.
4. Qualitatively determine the relative error in absorbance measurements and determine the optimal range for measurement purposes.
5. Describe the desirable features of a radiation source.
6. Explain the advantages of a dual versus single-beam spectrophotometer.
7. Explain the difference between a 3- and 4-level laser and why it is not possible to have a 2-level laser.
8. Compare the output of and advantages of prisms and gratings as dispersing elements.
9. Explain how a photomultiplier tube works.
10. Explain how an array detector works and describe the advantages of using an array detector.
02 Ultraviolet Visible Spectroscopy
After completing this unit the student will be able to:
1. Compare and contrast atomic and molecular spectra.
2. Explain why atomic spectra consist of lines whereas molecular spectra at room temperature are broad and continuous.
3. Justify the difference in molecular spectra at room temperature and 10K.
4. Describe the cause of Doppler broadening.
5. Determine the effect of conjugation on a UV/Vis absorption spectrum.
6. Determine the effect of non-bonding electrons on a UV/Vis absorption spectrum.
7. Determine the effect of solvent on the energy of n-$\pi$* and $\pi$-$\pi$* transitions.
8. Evaluate the utility of UV/Vis spectroscopy as a qualitative and quantitative method.
9. Describe a procedure by which UV/Vis spectroscopy can be used to determine the pKa of a weak acid.
03 Molecular Luminescence
After completing this unit the student will be able to:
1. Describe the difference between a singlet and triplet state.
2. Draw an energy level diagram and identify the transitions that correspond to absorption, fluorescence, internal conversion, radiationless decay, intersystem crossing and phosphorescence.
3. Explain why phosphorescence emission is weak in most substances.
4. Draw a diagram that shows the layout of the components of a fluorescence spectrophotometer.
5. Describe the difference between a fluorescence excitation and emission spectrum.
6. Draw representative examples of fluorescence excitation and emission spectra.
7. Describe a procedure for measuring phosphorescence free of any interference from fluorescence.
8. Justify why fluorescence measurements are often more sensitive than absorption measurements.
9. Describe the meaning and consequences of self-absorption.
10. Identify variables including the effect of pH that can influence the intensity of fluorescence.
11. Identify the features that occur in organic molecules that are likely to have high fluorescent quantum yields.
12. Compare two molecules and determine which one will undergo more collisional deactivation.
04 Infrared Spectroscopy
After completing this unit the student will be able to:
1. Describe the selection rule for infrared-active transitions.
2. Determine the vibrations for a triatomic molecule and identify whether they are infrared-active.
3. Draw the design of a non-dispersive infrared spectrophotometer and describe how it functions.
4. Describe the difference between time and frequency domain spectra.
5. Explain how a Michelson Interferometer can be used to obtain a time domain spectrum.
6. Explain the advantages of Fourier Transform infrared spectroscopy over conventional infrared spectroscopy.
05 Raman Spectroscopy
After completing this unit the student will be able to:
1. Determine whether the molecular vibrations of a triatomic molecule are Raman active.
2. Explain the difference between Stokes and anti-Stokes lines in a Raman spectrum.
3. Justify the difference in intensity between Stokes and anti-Stokes lines.
4. Draw the Stokes and anti-Stokes lines in a Raman spectrum of a compound when given the energies of the different transitions.
06 Atomic Spectroscopy
After completing this unit the student will be able to:
1. Compare and contrast the advantages of flame, furnace and inductively coupled plasma atomization sources.
2. Justify why continuum radiation sources are usually not practical to use for atomic absorption spectroscopy.
3. Describe the design of a hollow cathode lamp and justify the reasons for a hollow cathode configuration and low pressure of argon filler gas.
4. Devise an instrumental procedure to account for flame noise in atomic absorption spectroscopy.
5. Devise an instrumental procedure to account for molecular absorption and scatter from particulate matter in atomic absorption spectroscopy.
6. Describe three possible strategies that can be used to overcome the problem of non-volatile metal complexes.
7. Devise a procedure to overcome excessive ionization of an analyte.
8. Devise a procedure to account for matrix effects. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Molecular_and_Atomic_Spectroscopy/04_Learning_Objectives/01_Background_Material.txt |
The problem sets on spectroscopy can be used in at least two different manners. The primary intent is to use these as in-class, collaborative learning exercises. Groups of 3-4 students work together in discussing and working through the problems. When using the problem sets in this manner, the instructor must actively facilitate and guide students through the material. This manual will guide instructors through each of the problem sets, identifying possible student responses to the questions and the response and activities of the instructor during the progression of the problem.
An alternative to the use of the problems in class is to assign them as out-of-class activities, preferably done as a group activity among students or as a peer-led team-learning activity. The accompanying text provides a detailed discussion of each step of the question, such that students could work back and forth between the problem and text on an iterative basis to gain an understanding of the material.
There is no perfect way to assemble groups for such collaborative learning activities. I gather information on the first day of class (year in college, major, prior chemistry courses) and then use this to set groups of 3-4 students that start on the second day of class. I try to make the groups as heterogeneous as possible and they work together for the entire semester. Another strategy is to assign groups for a shorter period of time that might encompass completion of a specific topic or unit (e.g. fluorescence spectroscopy), and to then create new groups for the next unit. One other possibility is to have different groups every day of class. Since it is important for groups to work well together, having new groups every day may be less successful than allowing groups to work together for more extended periods of time. I would recommend that the instructor assign groups rather than allowing the students to pick their own. This avoids the potential problem of friends who want to be in the same group but who then do not work well together or stay focused on the assigned task. It also avoids the problem of the student who is left without a group at the end of the selection process, something that can be especially problematic if it is a member of a minority group. When using collaborative groups, it is also important for the instructor to monitor the functioning of the groups and to step in to address either dysfunctional groups or the recalcitrant individual within a group. Peer-evaluation processes are often used by instructors who employ group activities as a way of assessing how well groups are working.
I also expect the groups to meet outside of class for any homework assignments, something that is aided because I am at a residential college. An alternative to this is to schedule a room on the evening before a homework assignment is due and encourage them to come to this place and work in any arrangement they wish on the homework. I have run such sessions for several years now and attend them as a facilitator (one result is that it has cut down considerably the individual traffic to my office seeking help on the homework problems) and it has been an excellent way to promote collaboration among the students.
The instructor has an especially important role to fulfill during the group activities. I have observed that the more engaged that I am in the process in helping to guide the students through the material, the more effective the learning that occurs. In many instances, it seems that the students are initially stumped by the question, that they begin to explore things that they do know that might apply to answering the question, and that help from the instructor either by letting them know that they are on the right track or by suggesting another direction in which to take their thinking is necessary. As they begin a question, I roam around the room listening in on conversations and looking over their shoulders at what might be written in their notebook. If I hear something interesting, I indicate that to the group. If I see that someone has written something interesting and relevant in their notebook, I tell other group members that they ought to talk with this individual about what they have written, and that the individual should explain to the other group members why they wrote that down. If I hear a group going entirely in the wrong direction, I probe them on why they are heading in that way and then offer suggestions about things to consider that will set them off in the right direction. When all groups have realized an important point, I call time out and summarize the concept at the board. Then I send them back to continue with the next part of the problem. Most of the problems are handled in such an iterative manner where the students work through some important part of the problem, I summarize it at the board when they have developed the concept, and then they return to the next part of the problem. Occasionally a group will just not see something, whereas every other group has gotten the point, and it may require a direct intervention from the instructor with that group to explain the concept. Similarly, there are times when I call their attention to the board to summarize a point when one of the groups still has not gotten the concept but waiting would slow down the remainder of the class to an unacceptable level.
When using these materials, I want the students to discuss and discover the concepts inherent in the problems, so they do not have the text when working on the problems. After they have completed a particular problem, I then give them access to a copy of that portion of the text. The text thoroughly explains each problem or concept and I encourage the students to read it over that evening to reinforce the concepts developed in class that day. I also give homework problems designed to reinforce the concepts developed in class.
After a very brief introduction to the general concept of spectroscopy (probing chemical species with electromagnetic radiation; that different species and different processes absorb different components of the electromagnetic spectrum; that this can be used for purposes of identification and quantification) I give the students the first set of in-class questions that cover general background information on spectroscopy.
Italicized items throughout this instructor’s manual are questions or prompts I often give the students to help them solve the problem.
05 Instructors Manual Molecular and Atomic Spectroscopy
Electromagnetic Radiation
1. What is the relationship between the energy (E) and frequency ($\nu$) of electromagnetic radiation?
Students usually remember the equation that E = h$\nu$ without any prompting on my part and determine that there is a direct proportionality between the two.
1. What is the relationship between the energy and wavelength ($\lambda$) of electromagnetic radiation?
Students usually remember that c = $\lambda\nu$ without any prompting on my part and determine that there the energy and wavelength of radiation are inversely proportional.
1. Write the types of radiation observed in the electromagnetic spectrum going from high to low energy. Also include what types of processes occur in atoms or molecules for each type of radiation.
The students’ individual ability to identify all the different types of electromagnetic radiation and rank them in energy usually varies widely. Within a group most are able to generate a complete or close to complete list and rank those that they are most familiar with. One of the most perplexing to most students is where to put microwave radiation in the energy ranking.
Identifying the types of processes that occur in atoms or molecules for each type of radiation presents more difficulties.
What type of process do you already know about in molecules and what radiation produces them?
Within groups they can determine that UV/VIS involves transitions of valence electrons and IR corresponds to molecular vibrations. Many are familiar with the idea of a nuclear spin flip from their organic chemistry course, although they may or may not remember that RF radiation is used to excite nuclear spin flips. Some know that it is possible to rotationally excite molecules, although they often do not know that rotational excitation occurs in the microwave region of the spectrum. It is uncommon for them to know what processes occur with $\gamma$-rays and X-rays. Many are not familiar with the idea of an electron spin flip in paramagnetic substances and that it occurs in the microwave region of the spectrum.
At this point, I briefly discuss the difference between absorbance and emission. I also discuss how different spectroscopic methods are of different utility for compound identification and compound quantification. Some techniques (e.g., NMR spectroscopy) are useful for interpretation and identification, whereas others (e.g., IR spectroscopy) are useful for identification but not that amenable to interpretation and instead require use of a computer library to determine the best match.
I then go over the basic design of an absorption spectrophotometer and present them with the following series of questions on Beer’s Law.
Beer’s Law
1. What factors influence the absorbance that you would measure for a sample? Is each factor directly or inversely proportional to the absorbance?
Students quickly realize that the absorbance relates to the concentration and that it is a direct proportion. They often do not think of path length as a variable, I suspect because they are given specific cuvettes to use in any prior measurement they have performed and then don’t think the path length is something they could adjust.
What would be the effect of increasing the path length?
This is usually sufficient for them to see that there ought to be a direct relationship between path length and absorbance. Some students are familiar with the concept of an extinction coefficient from other courses, but rarely do they have an exact understanding of the meaning of the extinction coefficient. I indicate that molar absorptivity is another term for the extinction coefficient and at this point we can write Beer’s Law on the board.
How would you measure a spectrum and draw an example of a UV/VIS absorbance spectrum of a chemical species?
Someone in each group usually has enough prior experience and knows that recording a spectrum involves measuring the absorbance as the wavelength is scanned. If so, they can draw a spectrum where the absorbance varies with wavelength so that there are regions of high absorbance and regions of low absorbance. Some may think of an atomic (line) spectrum whereas others think of a molecular absorbance spectrum, and I clarify that they are different but that we can consider the nature of the extinction coefficient using either of them. I also prompt them to consider the concentration and path length when someone records a spectrum, and they realize that both are fixed.
Explain why the absorbance is high in some regions and low in others (or that lines in an atomic emission spectrum have different intensities).
They can usually rationalize that chemical species have the ability to absorb some wavelengths of light and not others, and when pushed on the differences in intensities of lines in an atomic emission spectrum, that some transitions must have a higher likelihood of occurrence than others. They sometimes wonder whether the difference in intensities reflects differences in the detector response, so it is important to point out that it is a fundamental process taking place in the chemical species. At this point, we can now discuss how the extinction coefficient or molar absorptivity is a measure of the probability that a particular wavelength of light can be absorbed. We discuss the aspect of energy transitions and that different transitions within a chemical species have different probabilities of occurrence. I introduce the idea of selection rules and that it is appropriate to talk about the degree to which a transition is allowed. I also introduce the idea that there are some transitions that are not allowed or forbidden.
1. If you wanted to measure the concentration of a particular species in a sample, describe the procedure you would use to do so.
The groups’ first response to this is often rather superficial. They tend to think more of putting the sample into a cuvette, measuring the absorbance and somehow equating that with concentration without explicitly stating that you first need to select a wavelength to use and prepare a standard curve.
Referring back to the spectra from the problem above that are still on the board, I point out that the analyst must set a wavelength.
Which wavelength would you choose?
They usually see right away that $\lambda$max is the preferable one and that with the highest molar absorptivity would provide the largest response. I also push them to examine how $\lambda$max would provide the lowest possible detection limits of any of the wavelengths.
Can you imagine a situation where you would not use $\lambda$max for the analysis?
Most groups quickly realize that you would need a different wavelength if the sample had another substance in it that absorbed at $\lambda$max.
Having selected the proper wavelength, how would you relate the absorbance of the sample with an unknown concentration to the actual concentration?
At this point, they realize the need to examine a sample with a known concentration and some students realize that they will need a standard curve with several concentrations whereas others may think only one known concentration is acceptable.
We can then discuss the concept of a blank solution and examine how a standard curve ought to be a linear plot that goes through the origin. We also examine how the slope of the standard curve can be used to determine the molar absorptivity.
1. Suppose a small amount of stray radiation (PS) always leaked into your instrument and made it to your detector. This stray radiation would add to your measurements of Po and P. Would this cause any deviations to Beer's law? Explain.
It is helpful to draw a picture on the board that shows a basic design of the spectrophotometer and indicates Po, P and Ps.
Consider the situation of a sample with a high concentration and another sample with a low concentration of analyte, and think about the relative magnitudes of the different terms at these different conditions.
It can also be useful to indicate on the board the way in which the stray radiation gets incorporated into the expression for the absorbance [A = log(Po + Ps)/(P + Ps)], and to examine these terms at the extremes of high and low concentrations. At this point, the groups can usually rationalize that the (P + Ps) term will approach Ps or a constant as the concentration of analyte is increased. When asked to draw the standard curve that would be observed, they can draw one that shows a negative deviation at higher concentrations.
1. The derivation of Beer's Law assumes that the molecules absorbing radiation don't interact with each other (remember that these molecules are dissolved in a solvent). If the analyte molecules interact with each other, they can alter their ability to absorb the radiation. Where would this assumption break down? Guess what this does to Beer's law?
Groups usually realize that the molecules are more likely to interact with each other at high concentration.
1. Beer's law also assumes purely monochromatic radiation. Describe an instrumental set up that would allow you to shine monochromatic radiation on your sample. Is it possible to get purely monochromatic radiation using your set up? Guess what this does to Beer's law.
What is meant by “purely monochromatic radiation”?
We have not yet discussed the specific details of a monochromator, but based on earlier discussion related to selecting a $\lambda$max value, they already know that some form of wavelength selection device is necessary. They are familiar with the ability of a prism to disperse radiation. I point out that gratings are more commonly used and that we will discuss gratings in more detail later in the course. With a drawing of a prism on the board, and prompted as to how they would direct only one wavelength on a sample, they realize that it will be necessary to use a slit that blocks out the unwanted wavelengths, but that the radiation passing through the device will never be purely monochromatic. At this point, without explaining it further, I indicate that polychromatic light will lead to negative deviations from Beer’s Law, especially at higher concentrations.
1. Is there a disadvantage to reducing the slit width?
What varies as one goes from a wide to a narrow slit width?
The groups realize that a wide slit width gives more power (and I point out how we will equate the number of photons with power) and a wider range of wavelengths, whereas a narrow slit width gives fewer photons and a smaller range of wavelengths.
Do you want high or low source power?
Based on our prior discussion of the effect of stray radiation they usually realize that higher power is desirable. This is also a useful time to further introduce the presence of noise and discuss how the signal-to-noise ratio is an important consideration in spectroscopic measurements. They realize that this argues for the use of wide slits.
Is there any situation where you would want to use a small slit width?
The groups can usually figure out that the ability to distinguish two nearby peaks is improved with smaller slit widths. This allows us to discuss what is meant by “resolution” in spectroscopic measurements.
Finally, we can examine Figure 1.5 to look at the effect of polychromatic radiation on the deviation from Beer’s law. I also draw Figure 1.6 on the board and ask what wavelength they would use for the analysis and to justify why. They readily appreciate that the broader region is better for use because of the prior discussion about deviations to Beer’s Law and because slight changes in the setting of the monochromator will have less significant effects on the measured absorbance.
1. Consider the relative error that would be observed for a sample as a function of the transmittance or absorbance. Is there a preferable region in which to measure the absorbance? What do you think about measuring absorbance values above 1?
Examine separately the extent of error that would occur at low and high concentration.
Groups can usually determine that the error at the extremes of concentration is more pronounced and there must be some mid-range absorbance measurements where the error is minimized.
Determine the percent transmittance that gives an absorbance value of 1 and consider the likelihood that negative deviations to Beer’s law occur in this region?
They can figure out that an absorbance of 1 equals only 10% transmittance and realize that negative deviations are likely to occur, enhancing the error of the measurement and reducing the number of significant figures that could be measured. I also discuss with them whether it would ever be acceptable to use a non-linear standard curve.
It is ever acceptable to extrapolate a standard curve to higher concentrations?
They have enough understanding to determine that the possible onset of negative deviations to Beer’s Law means that one cannot reliably extrapolate to higher concentrations.
1. What are some examples of matrix effects and what undesirable effect could each have that would compromise the absorbance measurement for a sample with an unknown concentration?
After describing to the class what is meant by a matrix effect, groups are given a few minutes to discuss this question and I have them report out on what they identified. In the aggregate the class is usually to arrive at important variables such as pH, the possibility that other species might absorb at $\lambda$max, and the possibility that another species interacting with the analyte could alter the value of $\lambda$max. Prompting may be required to have them consider scatter from suspended particulate matter and that the solvent may have an effect as well.
Are there any special constraints that must be considered in selecting a buffer?
They can usually identify that the buffer can’t absorb at $\lambda$max. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Molecular_and_Atomic_Spectroscopy/05_Instructors_Manual__Molecular_and_Atomic_Spectroscopy/01_In-clas.txt |
The following set of questions is given as an out-of-class exercise. There are two ways to consider having the students answer them. One is to provide the text that accompanies this unit and have the students read the text and answer the questions. The other is to send them to a resource such as the Analytical Sciences Digital Library or more broadly the internet, to see what they can come up with as answers to the questions. Having worked on the assignment, we then spend a class period developing the different topics covered by the questions in the assignment.
Notes are provided below for questions I ask in class to prompt their consideration and our discussion of the topics.
Sources
1. Describe the desirable features of a radiation source for a spectrophotometer.
2. Plot the relative intensity of light emitted from an incandescent light bulb (y-axis) as a function of wavelength (x-axis). This plot is a classic observation known as blackbody radiation. On the same graph, show the output from a radiation source that operated at a hotter temperature.
I find it desirable to explain something about the origin of blackbody radiation and Planck’s explanation of blackbody radiation.
1. Examining the plots above, what does this suggest about the power that exists in radiation sources for the infrared portion of the spectrum?
2. Explain the advantages of a dual- versus single-beam spectrophotometer.
Lasers
1. Why is it impossible to create a 2-level laser?
Whether or not the students have read the textbook or found other sources, the concept of the population of energy states, a saturated transition and a population inversion are usually quite new and foreign to them. I find that it is necessary to take time to explain each of these terms, as well as the process of stimulated emission. Understanding absorption and stimulated emission, the students can then examine and explain why it is impossible to create a 2-level laser.
1. Using your understanding of a 2-level system, explain what is meant by a 3-level and 4-level system. 3- and 4-level systems can function as a laser. How is it possible to achieve a population inversion in a 3- and 4-level system?
It is helpful to discuss the concept of excited state lifetimes and ask the students to consider the relative lifetimes needed to create a 3- and 4-level laser.
1. Which of the two (3- or 4-level system) is generally preferred in a laser and why?
Having considered the 2- and 3-level systems and the corresponding populations, the students can usually explain why a 4-level laser is preferable.
Monochromators
1. Explain in general terms the mechanism in a prism and grating that leads to the attainment of monochromatic radiation. Compare the advantages and disadvantages of each type of device. What is meant by second order radiation in a grating? Describe the difference between a grating that would be useful for the infrared region of the spectrum and one that would be useful for the ultraviolet region of the spectrum.
2. Explain the significance of the slit width of a monochromator. What is the advantage(s) of making the slit width smaller? What is the disadvantage(s) of making the slit width smaller?
This has already been discussed in class.
Detectors
1. Explain how a photomultiplier tube works. What are any advantages or disadvantages of a photomultiplier tube?
2. Describe a photodiode array detector. What advantages does it offer over other detection devices? | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Molecular_and_Atomic_Spectroscopy/05_Instructors_Manual__Molecular_and_Atomic_Spectroscopy/02_Out-of-.txt |
Given the background we have already developed on spectroscopy, I give them this set of in-class questions without providing any further background on UV/VIS absorption spectroscopy.
General aspects of UV/VIS absorption spectra
1. Compare and contrast the absorption of ultraviolet (UV) and visible (VIS) radiation by an atomic substance (something like helium) with that of a molecular substance (something like ethylene).
2. Do you expect different absorption peaks or bands from an atomic or molecular substance to have different intensities? If so, what does this say about the transitions?
The background for answering these two questions has usually come up earlier in the course when the students were asked to draw a spectrum because some students think of atomic spectra and others think of molecular spectra. Also, the discussion of molar absorptivity introduced the concept of the probability of transitions explaining the difference in intensities.
Why do atomic spectra consist of discrete lines whereas molecular spectra are broadened and continuous in nature? Are there processes that can occur in molecules that cannot occur in atoms?
With these two prompts, the groups can usually figure out that atoms only undergo electronic excitation whereas molecules can also be vibrationally and rotationally excited. They also know that vibrations are excited in the IR and, from our discussion very early in the class, that rotations are excited in the microwave region. In addition they know that the IR and microwave region of the spectrum is lower in energy than the UV/Vis region of the spectrum. This allows us to draw a diagram similar to Figure 2.4 from the accompanying text showing how the molecule will have many more closely spaced transitions than an atom that lead to the continuous nature of the absorbance spectrum.
1. Compare a molecular absorption spectrum of a dilute species dissolved in a solvent at room temperature versus the same sample at 10K.
What would happen to the sample at a temperature of 10K?
The students quickly realize that it will freeze to a solid. It will also help to point out that the freezing of many solvents will produce a transparent glassy material that UV/VIS radiation readily passes through so that it is possible to obtain an absorption spectrum on a frozen sample. I briefly describe the use of molecular beam apparatus as another way to obtain species at ultracold temperatures.
What takes place in a liquid sample that does not occur in a solid sample?
The groups come up with the realization that the molecules in the liquid sample are moving around and colliding with each other.
What happens when molecules collide with each other?
It may help to remind them that molecules consist of nuclei surrounded by electron “clouds”. They can also reason out that the electron clouds become distorted because of the collisions.
What effect would the distortion of the electron clouds have on the energy of the electrons? Do different molecules collide with different degrees of force?
With these prompts, they come to realize that the collisions of molecules leads to a broadening of the spectrum because the energies of a specific transition is slightly different in different molecules. That allows them to understand the rationale for the spectra observed in Figure 2.5.
1. Are there any other general processes that contribute to broadening in an absorption spectrum?
They are usually stumped by this question.
Have you ever heard of the Doppler effect and, if so, what do you know about it?
Some of them have heard of the Doppler Effect in physics or astronomy courses, and we discuss its significance to broadening in spectroscopic measurements.
1. Compare the UV absorption spectrum of 1-butene to 1,3-butadiene.
Consider the molecular orbitals involved in 1-butene and to draw an energy level diagram that shows the relative energies of the orbitals.
From organic chemistry, they know that $\sigma$- and $\pi$-orbitals are involved, and they usually know the relative order in which to put them. This enables a brief discussion of the HOMO and LUMO and the lowest energy transition ($\pi$-$\pi$*) for the molecule, which is the important one to consider in answering the question. They also know from organic chemistry that 1,3-butadiene is a conjugated compound. At this point, I direct them to the next question.
1. Rank these from high to low energy.
They have some experience with this from organic chemistry and are able to draw pictures like those in Figures 2.8, 2.12 and 2.13. They also can usually rank them from low to high energy, and determine which ones are filled with electrons and which ones are empty.
Compare the energy of the HOMO to LUMO.
• They can usually see that it will be lower in 1,3-butadiene than in 1-butene (Figures 2.9 versus 2.14). I then show them the spectra in Figures 2.10 and 2.15.
What would happen to the HOMO to LUMO transition in the absorbance spectra for the series of fused ring polycyclic aromatic compounds benzene, naphthalene, anthracene and pentacene?
They reason that the energy of the HOMO to LUMO transition ought to move toward the red as more rings are added to the compound. I then show them the spectra in Figure 2.16.
1. Compare the UV absorption spectrum of benzene and pyridine.
What distinguishes pyridine from benzene?
They quickly realize that the nitrogen atom in pyridine has a lone pair of electrons, and I introduce the idea that from a spectroscopic and energy level standpoint, we think of these as non-bonding electrons.
Draw representative energy level diagrams for benzene and pyridine, show which orbitals are filled, and to compare the energy of the HOMO to LUMO transition for the two compounds. Most are able to reason that the non-bonding electrons ought to go between the $\pi$ and $\pi$* orbitals and that the lowest energy transition of pyridine ought to be red-shifted relative to that of benzene. I then show them the two spectra in Figures 2.18 and 2.20 to confirm their conclusion.
1. The peaks in the 320-380 nm portion of the UV absorption spectrum of pyridine shifts noticeably toward the blue (high energy) portion of the spectrum on changing the solvent from hexane (C6H14) to methanol (CH3OH). Account for this change.
What would occur with pyridine that is different in the two solvents?
Groups have the background to know that methanol will form a hydrogen bond with the nitrogen atom of the pyridine.
What would this do to the energy of the non-bonding electrons?
Usually they are able to determine that it would stabilize them and lower their energy.
Would potential hydrogen bonding by methanol have any effect on electrons in the $\pi$ or $\pi$* orbitals of the pyridine (pointing out that when a molecule of pyridine absorbs radiation, it promotes an electron to the $\pi$* orbital)?
They often seem to think that association of the slightly positive hydrogen atom of methanol would have more of a stabilizing effect on the $\pi$-electrons rather than the $\pi$* electrons.
Draw the locations of these orbitals in the molecule and which is more exterior and exposed to the solvent?
With this prompt, they realize that an electron in the $\pi$*-orbital extends out further and shows more stabilization than the $\pi$-electrons. They also realize that the stabilization of the non-bonding electrons will be much greater than the other orbitals, thereby allowing them to explain the blue shift.
1. The peaks in the UV spectrum of benzene shift slightly toward the red (low energy) portion of the spectrum on changing the solvent from hexane (C6H14) to methanol (CH3OH). Account for this change.
Based on the information developed in answering the previous question, it is relatively straightforward for the students to explain the observation in this situation.
UV/VIS spectroscopy as a qualitative and quantitative tool
I first remind them what we mean by a qualitative and quantitative tool before giving them the following question.
1. Is UV/VIS spectroscopy useful as a qualitative tool?
Consider the general features of a UV/VIS absorption spectrum for organic compounds. Is this sufficient to identify an unknown compound?
We also explore whether a match in the UV/VIS spectrum might be used to confirm the identity of an unknown if the person has an idea what its identity might be. We also explore that many transition metal species have more distinct absorption spectra that can be used to confirm the identity of a species.
1. Is UV/VIS spectroscopy useful as a quantitative tool?
Consider things like the power of UV/VIS sources, the sensitivity of detectors, the magnitude of extinction coefficients, and given a certain extinction coefficient (e.g., 5,000) with a certain minimal reading for absorbance (e.g., 0.01), what would be the minimal concentration that could be measured?
With that, they see that UV/VIS spectroscopy is a useful quantitative tool.
1. If you were using UV spectroscopy for quantitative analysis, what criteria would you use in selecting a wavelength for the analysis?
This was covered earlier but serves as a review that $\lambda$max is preferable provided there are no interferences in the sample.
1. What variables influence the recording of UV/VIS absorption spectra and need to be accounted for when performing qualitative and quantitative analyses?
Some of these have been covered as well during the discussion of solvent effects. The students can usually think of things like solvent effects and pH as possible variables may alter the value of $\lambda$max.
Are there inorganic species that might be in the sample? If so, what effect would they have?
These questions are sufficient to get them to think about the potential impact of metal ions in a sample.
1. Provided the UV/VIS absorption spectra of HA and A differ from each other, describe a method that you could use to measure the pKa of the acid.
It will be necessary to set this up by writing the appropriate reaction and discussing the time scale of the reaction relative to the time scale of absorption of a photon. With the understanding that the reaction is slow relative to the process of absorption, students can appreciate that a species is either in the HA or A form during absorption. It also helps to examine the possible resonance forms for a carboxylate ion (RCO2-) versus its corresponding carboxylic acid (RCO2H) to have the students appreciate that the absorption spectra of the two species are likely to be different. At this point, there are usually some students in the class who remember the Henderson-Hasselbalch equation. If not, I ask them:
Do you remember an equation in acid-base chemistry that had an HA and A- species in it?
With the equation in hand, they sometimes can figure out that they would need to examine a solution that was strongly acidic and another that was strongly basic to determine the extinction coefficients for HA and A- respectively. If not, I ask them:
How could you prepare a solution with essentially all HA and then one with essentially all A-? By this point they usually recognize that buffered solutions at intermediate pH will give some of both and with the known pH and measured concentrations, they can use the Henderson-Hasselbalch equation to calculate the pKa.
I save a discussion of the relative value of $\lambda$max for the HA and A form for our unit on fluorescence, but it could be done at this time as well. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Molecular_and_Atomic_Spectroscopy/05_Instructors_Manual__Molecular_and_Atomic_Spectroscopy/03_In-clas.txt |
I provide a very brief introduction on the meaning of luminescence and indicate that the process of fluorescence – a subcategory of luminescence – involves emission of radiation from a species that has first been excited by light, the details of which we will develop through a series of questions. I also describe how a fluorescent light works.
Energy level diagrams for organic molecules
1. Draw an energy level diagram for a typical organic compound with $\pi$ and $\pi$ * orbitals and indicate which orbitals are filled and which are empty.
Given our prior discussion of UV/VIS absorption, groups can immediately write the answer to this question.
1. Now consider the electron spin possibilities for the ground and excited state. Are there different possible ways to orient the spins (if so, these represent different spin states).
The groups can usually see that there is only one way to write the ground state. They often see that in the excited state it is possible to have the spins of the electrons paired or parallel. It is worth indicating that for the situation where the excited state has paired spins, it does not matter which one is spin-up and spin-down as these are identical so long as the spins are paired.
1. Do you think these different spin states have different energies?
Many students just make the intuitive guess that they do have different energies, although they really may not yet understand why this is.
1. Which one do you expect to be lower in energy?
I generally find that students get this wrong and think that the excited state with the spins paired would have the lower energy. Even if they think the state with the parallel spins is lower, they rarely if ever provide the proper justification for this answer. I point out that they actually learned something in general chemistry while doing atomic structure that would allow them to answer this question and justify their answer. If this doesn’t lead them to the proper answer, I direct them to consider the following:
Think back to atomic structure and remember what happened when two electrons were put into a set of p-orbitals.
They do remember that the electrons go into separate orbitals with parallel spins. We can then discuss why this was the case – it takes energy to pair electrons, therefore parallel spins in degenerate orbitals will be the lower energy of the two possibilities.
Do you think the energy state with parallel spins in non-degenerate $\pi$ and $\pi$* orbitals will be lower in energy than paired spins?
At this point they decide that the state with the parallel spins will be lower in energy.
1. If the spin state is defined as (2S + 1) where S represents the total electronic spin for the system, try to come up with names for the ground and possible excited states for the system that are based on their spin state.
They may need to be reminded about the values of electron spin quantum numbers. Recognizing that the ground state and excited state with pair spins have an S of zero and a spin state of one is obvious to them. I point out that they only need to use +1/2 for the electron spins for the case where both have parallel spins and they arrive at the number three for the spin state. With those numbers, they usually can come up with the name singlet and triplet or something close to that to distinguish the two states.
1. Draw a diagram of the energy levels for such a molecule. Draw arrows for the possible transitions that could occur for the molecule.
With the knowledge from the previous section of the course on UV-Vis absorption spectroscopy, the students are usually able to draw a reasonable first approximation for the energy level diagram needed to discuss fluorescence. They know about the first and second excited states, and that there are vibrational and rotational levels superimposed on each. They can usually locate the first excited triplet state, and I usually point out that we often write this off to the right side of the first excited singlet state to make the diagram a bit clearer.
As they start to put transitions onto the diagram, they are usually quick to realize there will be absorption into the excited singlet states because that was discussed earlier in the unit on UV-Vis spectroscopy. If not, I may prompt them to show possible absorption transitions on their diagram. They also usually draw an absorption transition to the T1 state, and I point out to them that absorption transitions that involve a spin flip are forbidden by selection rules (if the idea of selection rules has not been discussed yet in the course, this is a good time to discuss that or remind them how selection rules were important in determining something like the molar absorptivity).
Now that the electron is excited, what are its various options for getting back to the ground state?
They can usually determine that the excited species can either lose the extra energy as heat (I indicate that this is something we can denote on the figure with a squiggly line) or lose it as radiation (I indicate that this is something we can denote on the figure with a solid line). We can then illustrate both transitions for a system in the S1 state and I indicate that the loss of energy as heat is referred to as radiationless decay and the lost of energy as radiation is referred to as fluorescence. I remind them that fluorescence specifically refers to a system that has been excited by absorbing light (as distinct from being excited by absorbing heat), and that the emission of light must be a transition that goes from one singlet state to another.
What process would promote radiationless decay?
They can usually reason out that collisions of the excited molecule with surrounding compounds in the matrix will promote radiationless decay.
What can happen to an electron excited into the S2 state (or higher vibrational level within the S2 state)?
They usually think it is possible to have both radiationless decay and fluorescence, and I indicate that only on a few rare occasions have people discovered molecules that fluorescence from the S2 state.
At this point, it is useful to mention that it is important to consider the lifetimes of the different states.
Can you guess the lifetime of an electron is a first excited singlet stage, second excited singlet state, and vibrational state?
They usually propose that the lifetime of the second excited singlet state is probably less than that of the first excited singlet state. I then have them take guesses and play a game of warmer and colder as we hone on in the values. With the lifetimes now at their disposal, they can appreciate why systems excited to higher energy states than the S1 state will quickly undergo radiationless decay.
It is now necessary to discuss the process of internal conversion that allows systems in the S2 state to convert into the S1 state and to indicate these on the diagram.
Is there only one possible fluorescence transition from S1?
Since they are already familiar with the fact that molecules can be excited into higher vibrational levels of electronic states, they usually reason out that molecules can perhaps undergo fluorescent transitions into higher energy vibrational states of S0. We then discuss how fluorescence from a molecule can have a variety of different wavelengths. Also that there will be different probabilities for each of the fluorescent transitions and that the S1-S0 may or may not be the most probable one.
At this point we have Figure 3.4 from the text complete except for a consideration of the T1 level.
Would it be possible for a molecule to undergo a transition from S1 to T1?
They remember that we just said transitions that involved spin flips are forbidden, but they usually suspect an S1 to T1 transition can happen because we have just spent so much time developing the idea of the triplet state. We then discuss what may account for the intersystem crossing process to occur. We also discuss how the transition can occur into an upper vibrational level of T1 and then undergoes radiationless decay into the T1 state.
1. What do you expect for the lifetime of an electron in the T1 state?
They usually realize that the only way for the electron to get out of the T1 state is to first undergo a spin-flip, and this will take some time. We play another guessing game of warmer-colder to get the range of lifetimes of systems in the T1 state.
When prompted they usually think the only possible route to lose the extra energy is through an intersystem crossing back into a higher vibrational level of S0 followed by radiationless decay. But I also indicate that even though emission of a photon from T1 to S0 is “forbidden”, in some systems it actually happens, and we call this phosphorescence. I point out examples of where they have seen phosphorescence (glow in the dark substances, their TV screen for a few seconds after they turn it off).
1. Why is phosphorescence emission weak in most substances?
They can usually rationalize out that, in most compounds, intersystem crossing is weak to begin with (I use this as an occasion to point out the fact that processes like radiationless decay, fluorescence, and intersystem crossing are all in competition with each other), which is one reason for the weak intensity of phosphorescence. Similarly, they can usually reason out that the long time the system spends in the T1 state means more collisions will occur.
What could you do to a sample to enhance the likelihood that phosphorescence would occur over radiationless decay?
Usually someone in each group can come up with the idea of freezing the sample into a solid to reduce the number of collisions.
1. Which transition ($\pi$*-$\pi$ or $\pi$*-n) would have a higher fluorescent intensity? Justify your answer.
Think back to our unit on UV-Vis spectroscopy – we discussed at least one important reason that will help answer this question.
Some of the students usually remember that the molar absorptivity for n-$\pi$* transitions is lower than that for $\pi$-$\pi$* transitions so that there will be fewer excited molecules and fewer to undergo fluorescence.
I then tell them that the lifetime of the n-$\pi$* excited state is longer than that of the $\pi$-$\pi$* excited state, and ask how this will influence fluorescence intensity.
With that information, they realize that there will be more collisional decay for the molecule in the n-$\pi$* state and therefore lower fluorescence.
Instrumental considerations for luminescence measurements
1. What would constitute the basic instrumental design of a fluorescence spectrophotometer?
The students think of things like a light source, monochromator, sample holder and detector. They rarely seem to be able to think it through to the point of realizing that fluorescence spectrophotometers will need two monochromators, one for the excitation beam and one for the emitted light. I almost always have to remind them of the process – radiation is used to excite the molecule, light is emitted by the molecule – before someone realizes that there are two sources of light to consider and they will need a monochromator for each.
Where would you position the emission monochromator relative to the excitation monochromator?
Some of them realize that the molecules will emit the light in all directions such that the emission monochromator can be placed somewhere else besides at 180o to the excitation monochromator.
Would a 90o placement have any advantage over a 180o placement?
The groups can usually determine that the 180o placement is less favorable since the source light will strike the detector and that at 90o only the fluorescence will be measured.
1. What would be the difference between an excitation and emission spectrum in fluorescence spectroscopy?
They can usually figure out that one involves scanning the excitation and the other the emission monochromator. However, I often need to point out that when scanning the emission monochromator, the excitation monochromator needs to be set to a known excitation wavelength, and when running an excitation spectrum, the emission monochromator needs to be set to a known emitting wavelength.
1. Draw representative examples of the excitation and emission spectrum for a molecule.
The students are usually stumped by this question and I need to provide some prompts to get them to think it through. The first is to redraw the energy level diagram on the board showing the possible absorption and fluorescent transitions. The second is to draw two sets of axes on the board, one on top of the other, where the y-axis is fluorescent intensity and the x-axis is wavelength. I indicate that the wavelength scale on the two axes is identical and ask them to draw this diagram in their notes.
Draw the excitation spectrum on the top.
After all groups have a satisfactory representation of an excitation spectrum, I then instruct them to:
Draw the emission spectrum on the bottom set of axes.
Some of the students finally realize that the two spectra will be mostly offset from each other except for the S0-S1 transition. If not, I may prompt them by asking:
Are there any transitions that have exactly the same energy in the excitation and emission spectrum?
Once both are on the board and understood, I ask them which one most resembles an absorption spectrum and they immediately realize it is the excitation spectrum.
1. Describe a way to measure the phosphorescence spectrum of a species that is not compromised by the presence of any fluorescence emission.
They are usually perplexed by this problem, often remembering back that phosphorescence occurs from solids, but now knowing whether that is useful. Also, they want to think of a way to promote intersystem crossing to get all of the excited-state molecules into the T1 state so that no fluorescence occurs.
Can they remember any significant differences that exist between the S1 and T1 energy states?
Someone usually brings up the difference in lifetimes of the excited states, which leads them to ask whether there is some way to excite the compound, turn off or block the excitation source, and then build a delay into the measurement of the emission. I then discuss the idea of using a pulsed source, and the concepts of having a delay and gate time.
1. If performing quantitative analysis in fluorescence spectroscopy, which wavelengths would you select from the spectra you drew in the problem above?
Because we have discussed wavelength selection in UV-Vis spectroscopy, they usually know to select the $\lambda$max for both the excitation and emission spectrum.
Can they think of any problem with setting both the excitation and emission monochromator to the S0-S1 transition?
Some are able to think of the possibility of scattering taking place. We then discuss how scatter will occur in all directions and be indistinguishable from fluorescence.
1. Which method is more sensitive, absorption or fluorescence spectroscopy?
Before giving them this question, I do a brief lecture on the meaning of a quantum yield. I define the meaning of the term (ratio of the number that fluoresce relative to the number that were excited).
Identify all of all the processes that compete against fluorescence from the S1 state.
With this decided, we then write the expression for the quantum yield using the rate constants for the different processes. We then examine how the fluorescent quantum yield will range from 0-1 and that it will be rare for a molecule to have a quantum yield of 1. I indicate that molecules with fluorescent quantum yields of 0.01 or higher are useful for analysis by fluorescence spectroscopy.
I then provide them with the question about relative sensitivity of absorption and fluorescence spectroscopy and provide the following additional prompt:
Think about the instrumental setup that is used for absorption and fluorescence measurements and consider exactly what is measured or compared. Also think about each technique when a very low concentration sample is being measured.
At this point, they are able to reason that the values of P0 and P in absorption spectrophotometry will be quite close in magnitude whereas the fluorescence measurement is just a small signal.
I now ask them to consider an observation with their own eyes where they would examine the difference in intensity between a 99- and 100-Watt light bulb (comparable to absorption spectrophotometry) compared to the difference between a 1-Watt light bulb and darkness.
Which difference would be easier to detect?
They immediately say it would be easier to distinguish the 1-Watt bulb over darkness than the difference between a 99- and 100-Watt bulb. We then discuss how electronic devices can also distinguish this better and fluorescence (or any emission method) has an inherent sensitivity advantage over an absorption method.
Variables that influence fluorescence measurements
1. What variables influence fluorescence measurements? For each variable, describe its relationship to the intensity of fluorescence emission.
I often preface this question by suggesting that the students think back to things we already considered in our unit on UV/Vis absorption spectroscopy. Students come up with the obvious one of concentration, and usually realize that the path length and molar absorptivity will influence this as well. Some even realize that the source power is important in fluorescence as well. This then allows us to write the equation for fluorescence intensity.
What would the curve look like at high concentration?
Recognizing the negative deviation that occurred in Beer’s law, they usually realize that a negative deviation would be expected as well in fluorescence spectroscopy. I then introduce the idea of self-absorption that can occur at high concentrations in emission methods.
What would you do with a sample to insure that it was not at a concentration range where self-absorption was occurring?
They almost immediately recommend diluting the sample.
Are there other variables that influence the magnitude of the quenching?
The importance of collisions and its contribution to radiationless decay has been discussed earlier so some realize that this is likely an issue. Most then realize that temperature is an important variable and can determine that higher temperatures should reduce fluorescent intensity.
Many also suggest that solvent is a variable, and we revisit some of the discussion we had in UV/Vis spectroscopy on the effect of solvent on $\lambda$max for $\lambda$-$\lambda$* and n-$\lambda$* transitions. Many also suggest that pH is a variable.
I then explain how paramagnetic substances promote intersystem crossing.
Can you think of any important paramagnetic substances that might be present in a sample?
Usually each group has someone who identifies paramagnetic metal ions and dissolved oxygen as possibilities.
Can you think of a way to eliminate dissolved oxygen from a sample?
Some are familiar with using a ultrasonicator to degas a solution and sometimes a student even comes up with the idea of purging the solution with a gas that is not paramagnetic.
We discuss how removing paramagnetic metal ions from a sample is a more difficult problem.
1. Consider the reaction shown below for the dissociation of 2-naphthol. This reaction may be either slow (slow exchange) or fast (fast exchange) on the time scale of fluorescence spectroscopy. Draw the series of spectra that would result for an initial concentration of 2-naphthol of 10-6 M if the pH was adjusted to 2, 8.5, 9.5, 10.5, and 13 and slow exchange occurred. Draw the spectra at the same pH when the exchange rate is fast.
This is a complex question with several parts to it, although we did cover some background for this in the unit on UV/Vis spectroscopy. If they have forgotten this, I prompt them to focus on the two extremes and ask them first to describe where the reaction lies at a pH of 2. The groups realize this will be virtually all of the protonated form. I then ask them to describe where the reaction lies at a pH of 13 and they realize it will be virtually all of the deprotonated form.
Do you expect any difference in the emission spectrum of the two species in these two forms, and if so, what might the difference be?
Intuitively, they expect the two forms to have a different emission spectrum – if not, why are we even talking about this problem! After giving them a bit of time to talk about it in their groups and hearing some of the things they are thinking about – rarely do they come up with something that is the actual explanation for why the deprotonated form will fluoresce at a longer wavelength than the protonated form.
Draw the resonance forms for the two species.
Some are rusty at this but others quickly realize and explain to their group members that the naphtholate ion will have more resonance forms than the naphthol species.
What does the presence of more resonance forms do to the energy of the excited state?
Some propose that it may lower the energy of the excited state and therefore the energy gap such that the transition shifts toward the red.
At this point, we can draw spectra for the two extremes.
Consider the point where the pH equals the pKa, and think about the difference that would be observed between fast and slow exchange.
Usually the groups can reason out what would occur for the two different situations and we can fully draw and discuss the spectra represented in Figure 3.11 of the text.
Which to you think actually happens – fast or slow exchange?
I prompt them to think back about UV/Vis spectroscopy and lifetimes of states as well. Groups arrive at the conclusion that slow exchange is likely to occur on the time scale of fluorescence spectroscopy.
Questions always arise about whether the two species would have the same intensity or not and we have some discussion about that question.
1. Devise a procedure that might allow you to determine the pKa of a weak acid such as 2-naphthol.
We have examined a similar question in UV/Vis spectroscopy so the students usually hone in on the procedure that would be done.
I then use this as an opportunity to talk about the complication that arises if this experiment is actually performed using fluorescence spectroscopy – namely that the calculated pKa values at each pH will be different because the pKa value of the excited state of the compound is different than the pKa value of the ground state.
1. Which compound will have a higher quantum yield: anthracene or diphenylmethane?
Students usually want to answer this question by considering the relative extent of conjugation of the system rather than the role of collisional deactivation for the two compounds.
Would collisional deactivation be different for the two molecules?
Not all the students will immediately realize that anthracene will exhibit less collisional deactivation than diphenylmethane (they might propose that since it’s bigger, they will collide into each other with more force – if they give this response, it is important to remind them that most of the collisions actually occur with solvent).
Which would come out worse in a collision – a Greyhound bus or a car towing a boat on a trailer?
They usually realize that the collisions with diphenylmethane will lead to more radiationless decay than those with anthracene.
I then discuss how relatively few compounds exhibit intense fluorescence – that fluorescence spectroscopy is far more selective than UV/Vis absorption spectroscopy – and that derivatization of compounds with fluorescent chromophores is sometimes done for the analysis because of the enhanced sensitivity of fluorescence.
Finally, I present to them a brief discussion of topics like chemiluminescence, bioluminescence and triboluminescence. We also talk briefly about the process that takes place in glow sticks. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Molecular_and_Atomic_Spectroscopy/05_Instructors_Manual__Molecular_and_Atomic_Spectroscopy/04_In-clas.txt |
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