Analytical Chem Midterm
What is sensitivity? (Measure of a method to show significant response difference):
Slope of a calibration curve (ie. molar absorptivity for absorbance versus concentration.
What is a plasma and why is it generally more desirable to use for atomic emission than just an acetylene torch?
A plasma when used for atomic emission is partially ionized argon gas that contains both argon ions and electrons. In general, a plasma is just ionized gas that contains both cations and anions of the gas that is ionized. Due to its ability to reach higher temperature than an acetylene torch or flame it is able to provide better atomization efficiency and produce more excited states.
Show a likely absorbance as a function of organic solvent composition curve on the plot below and justify your answer.
A slight increase in absorbance as the % organic solvent increases due to the fact that the surface tension of the solution is lowered, allowing smaller droplets to form which can more easily reach the flame.
Flame emission (like we did with the airbrush) can be an effective way to determine alkali metals but not transition metals. Why is this, considering N*/N = (g*/g)e^(-E/kT)
Alkali metals can be easily excited therefore the change of E, the excitation energy, is low. Flame emission has a high enough temperature to do this so the N*/N ratio is quite high. Transition metals have a much higher excitation energy so a hotter source like the ICP plasma is needed to provide a good N*/N ratio.
How does the ICP torch work?
An inductively couples plasm (ICP) torch is used to deliver hot plasma for atomic emission, as it provides a more efficient atomization and an increase in the number of excited states. It is structured by three quartz tubes and a radio-frequency induction coil. An ICP torch also contains a nebulizer that is connected to an internal plasma tube. As a sample is introduced into the torch, it interacts with Argon molecules within the nebulizer. Next, the argon / sample mixture is carried through the plasma tube, and plasma is formed by the spark of a Tesla coil. Here, the induction coil creates a magnetic field inside the ICP torch causing the mixture to move in circular motions, allowing heat to generate in the plasma. These high temperatures are what allows the plasma torch to work more efficiently than other atomization tools; however, a constant flow of argon allows the quartz tubes to remain cool, preventing the destruction of the instrument.
Raman has this (these) significant advantages(s) over IR spectroscopy.
Aqueous solvents and glass cuvettes IR can do solid sampling; photochemistry caused by Raman laser is undesirable.
What does this equation represent: M* or M+* -> M or M+ + hv?
Atomic emission This equation represents the excited metal species dropping back to the ground state emitting light. This is the atomic emission signal. M* is the excited metal atom. M and M+ are the ground state metal atom and metal ion. E = hv.
Why are atomic emission spectra a series of lines while UV spectra of molecules are broad bands?
Atomic emission spectra are a series of lines while UV spectra are broad bands due to the difference intermediate energy levels that a molecule can go through when absorbing UV radiation that it wont experience in atomic absorption. When an analyte absorbs UV radiation, it can not only transition between electronic energy states but all vibrational states which appears as closely speared bands that have merged together to create one broad absorption band. In an atomic emission spectra there is a series of lines because vibrational and rotational energy levels are not present. This means that there is a difference between excited and ground energy levels that doesn't vary.
What is the difference between atomization and excitation and doe we want to minimize or maximize N*/No of the Boltzmann equation for atomic emission?
Atomization is when molecules in a solid or liquid are heated up and converted in a free atom in the gaseous state. Excitation is when these free atoms absorb energy and move up to a higher energy level state. To increase the atomic emission intensity, the N*/No part of the Boltzmann equation needs to be maximized. The more molecules in the excited state, N*, the more emission seen.
Why do UV-VIS spectra show broad bands while the bands in IR spectra are sharper?
Because IR spectra are only based on transitions between vibration energy levels, this gives ride to sharper fundamental absorption lines and UV-VIS spectra bands are broader because not only due they include that transition between vibrational energy levels.
One limitation of Beer's law is it is generally considered valid (linear) at analyze concentrations of 0.01M or less. Provide a specific explanation.
Beer's law is generally considered valid at analyze concentration of 0.01M or less for a couple reasons. The main reason is that at concentrations higher than 0.01M, the individual particles of the analyze are not independent which causes the particles of the analyze to interact in such a way that analyte's absorptivity is changed. Another reason is due to a change in the solution's refractive index that occurs at higher concentrations. The analyte's absorptivity is dependent of this refractive index, so a change in it will cause a change to the absorptivity.
What is the Limit of Linearity (LOL)?
Concentration range over which instrument response is proportional to concentration. Correlation coefficient R2 > 0.995 defines paper limit
Describe two advantages of Raman over IR spectroscopy
Despite the work for Raman and IR spectroscopy to be complimentary techniques, there are advantages of Raman spectroscopy over IR. The first being that since signals in the Raman spectra typically in the visible or near-IR region, then glass or quartz cuvettes/cells can be utilized. Glass cannot be utilized in IR because glass absorbs IR radiation. Using these types of cells allows the experimenter to avoid using other atmospherically unstable materials for their cells such as sodium chloride. Another advantage of Raman spectroscopy over IR spectroscopy is that water can be used as a solvent in Raman. This is because water doesn't provide a strong scatter in Raman as it does in IR with its broad signal. Due to this water can be used as a solvent and aqueous solutions can be tested using Raman.
The possible energy transitions when a metal is introduced into a flame are...
Electronic; metals can only undergo electronic transitions, no molecular bonds.
What is electro thermal atomization and indicate one advantage and one disadvantage of this approach?
Electrothermal atomization uses an electrically heated graphite furnace instead of a flame to cause atomization of the metal (M* to M0). One advantage is that it eliminates the sample matrix first before the atomization step and another advantage is that it has better detection limits than flame AA. A disadvantage is it is difficult to automate and it is more expensive as compared to flame AA.
External conversion is one likely alternative pathway for fluorescence. What is external conversion and how can it be minimized using temperature?
External conversion is the interaction of solvent molecules with the fluorescent compound of interest, causing the compound in the excited state to come back to the ground state without emitting light. This quenching pathways can be minimized by the cooling the solution, decreasing the solvent collisions.
Using the 532 nm laser instrument, what happened to the Raman signal when fluorescein was added to the hydrogen peroxide solution and why?
Fluorescein can undergo excitation at this laser wavelength and generate fluorescent light which gives a large background signal obscuring the Raman signal.
Why is the use of high-power laser as the excitation source advantageous for fluorescence by not absorption? F=kC and A=log(Po/P)
Fluorescence is proportional to the power of the excitation source Po which is part of the constant k in F=kC where F= fluorescence signal and c= concentration. Since absorbance depends on the ratio of Po/P, increasing the source power will not change the relative ratio of Po/P. As Po increases, P will also increase.
Summarize the derivation of Beer's law
In order to derive the Beer's law equation of A=bC, we must note that electromagnetic radiation experiences a decrease in power which can be written as dP when it passes through the thin later of a sample of thickness, which can be expressed as dx. The sample's thickness along with the concentration of the analyze, C, is proportional to the decrease in power which gives us the following equation where P is just the proportionality constant : -dPP = Cdx Finding ourselves with a derivative function, we must integrate both -dPP and dx, we find ourselves with the equation: ln(P0/PT)=bC This equation is the natural log of the power of the monochromatic electromagnetic radiation before entering the thin layer of sample over the power after it has exited the layer which is proportional to the analyze concentration, path length, and the proportionality constant. After conversion the natural log to log, the equation is: A=abC where a is the absorptivity. This commonly replaced with the molar absorptivity, when using molar it's concentration, so that equation would be A=bC.
What is the Limit of Quantification (LOQ)
Lower limit of linear range Instrument: [Blank signal + 10(signal noise)]/sensitivity Method based on calibration curve: 10(std. deviation of intercept)/slope
Fluorescence sensitivity can be increased by monitoring the light emission from a larger volume of the Yvette. Explain how this could be done by changing either the slit dimensions or the slit orientation.
Increasing the dimensions of the slit will cause an increase in sampling volume. This directly increases the amount of emission monitored, resulting in higher fluorescent sensitivity. Similarly, orienting the slit in order to get an optimal sample volume will increase the amount of espionage that is monitored, therefore increasing the sensitivity of fluorescence. The effects of these alterations can be seen in this figure, turning the slit from a vertical orientation to a horizontal orientation increased the sampling volume from 0.03 cm cubed to 0.09 cm cubed.
Internal conversion and external conversion are likely alternative pathways for fluorescence. What are these conversion processes and why might they be favored over fluorescence?
Internal conversion occurs when a molecule this is in a ground vibrational level of an excited state passes straight through to a higher vibrations; energy level of a lower energy electronic state. When this occurs with vibrational relaxation, the molecule in an excited electronic state can return to the ground state without the emission of a photon. External conversion is a little different as it requires an external source in which excess energy can be transferred to such a solvent. Both of these alternative pathways might be favored over fluorescence because they act much faster than emitting a photon which means that they are more efficient.
A significant limitation to Beer's Law can be noted if a light emitting diode (LED) source is used with no monochromator which is a common colorimeter design. What is this limitation and why is it rarely a problem with a photodiode array (PDA) instrument?
LED source does not generate monochromatic light (narrow wavelength range of 1 nm precision bandwidth) but has a broader wavelength range of 10-20 nm bandwidth source. A PDA instrument has multiple detectors to measure light with 0.5 nm precision and therefore monochromatic light is being used as the basis of the Beer's Law linearity.
Explain the figure indicating how the energy of the stokes and anti-stokes transitions compare to that of the Rayleigh transition. Which is the most common?
Looking at the figure, the energy of both types of Raman scattering, the stoke and anti-stokes, transitions differ from that of the Rayleigh transitions. The energy of the stoke transitions is weaker than the energy of the Rayleigh transition while the energy of the anti-stoke transitions are greater than the Rayleigh transition. Rayleigh scattering is more common that the Raman scatters because in Rayleigh scattering energy is transferred to molecules in the ground state and then these molecules are returned to the ground state by remission. This process has the highest probability of occurring out of the different types of transitions which is why it is the most commonly occurring. In Raman scattering, the transitions occur between molecules that are not already in the ground state or first energy level state which is more uncommon.
Why is the best range to take absorbance measurements from 0.3-0.8 absorbance units?
Low and high absorbance have large amounts of uncertainty and error below .3 and above 0.8 absorbance units. Placement of the sample in the cuvette is a contribution to uncertainty for low absorbance values. The sample is hit by the wave differently based on how it is positioned because the optics in the sample are not the same at every position. Radiation from photon detectors is an indeterminate error that causes higher absorbance values to be affected and have more uncertainty. The proportion of the wave that exits the sample to the wave that enters, readout resolution, is a cause of uncertainty but can be manipulated with the sample we are using to help the precision. When the absorbance is low, redefining the 0% readout resolution by using higher concentration of the analyze and if the absorbance is too high, redefining the 100% readout resolution using a lower concentration of the analyte.
what is selectivity?
Measure of how specific a method is for an analyze Want ratio of sensitivity for interfering/sensitivity for analyze to be a small number
What is robustness?
Method good for many different sample matrices
In the two figures, the signal intensity can be considered to be proportional to the probability of such an energy transition. Explain why the peaks for the anti-stokes peaks are smaller than the stoke peaks.
On a Raman spectra, the stokes peaks represent lower-frequency emissions, while anti-stokes peaks represent higher-frequency emissions. Because electrons more often populate the ground vibrational level, there is a higher probability that ground level transition will occur; this directly correlates to the stokes peaks being larger and more intense than the anti-stoke peaks. The smallness of the anti-stokes peaks usually result in their omission when analyzing the Raman spectra.
How are the photodiode array spectrophotometer and multichannel atomic emission spectrometer different but also similar in design?
One way in which the photodiode array spectrophotometer and multichannel atomic emission spectrometer differ is in the type of radiation they measure. A PDA spectrophotometer measures the absorption or transmission of light through a sample while a multichannel atomic emission spectrometer measures the emission of light from excited atoms in the sample. They both use a detector array to capture the signal. In a photodiode array spectrophotometer, the detector array measures the intensity of the light after it has passed through the sample. In a multichannel atomic emission spectrometer, the detector array measures the intensity of the light emitted by the excited atoms in the sample. Secondly, both instruments are capable of analyzing multiple samples simultaneously. In and PDA spectrophotometer, this is achieved by using a multi-cell holder, which allows several samples to be analyzed at once. In a multichannel atomic emission spectrometer, a multi-sample probe is used to introduce multiple samples to the instrument at the same time.
When Orange Crush (colored soda) is added to quinine solution, would the fluorescence be expected to increase or decrease? Considering quinine fluoresces at 450 nm, justify your answer.
Orange Crush is complementary in color to blue based on the color wheel and can absorb light in the 450 nm range. Therefore, the fluorescence produced by quinine will be absorbed causing a decrease in the fluorescence signal that reaches the detector.
The atomic absorption signal for Ca can be lowered significantly by the presence of..
Phosphate and Aluminum cause calcium molecular formation
Using the color wheel, what color LED should be used to measure the absorbance of a green solution and explain why.
Red LED because it is complementary in color to the solution. The LED complementary in color to the solution should be used for absorbance measurements.
What is surface enhanced Raman (SERS) and indicate the main analytical advantage of this technique.
SERS involves the determination of an analyze absorbed to an AG or At surface or mixed with a colloidal suspension (nano particles) of either of these metals. The Raman signal using SERS is substantially enhanced compound to standard Raman, allowing ppl detection limits.
What is self-absorption and why does it often control the limit of linearity for a calibration curve in ICP atomic emission?
Self-absorption is a type of chemical interface that can occur when using flame emission. Self-absorption occurs when an excited state atom of the analyte at the center of the flame emits a photon and an atom at a cooler ground state absorbs that photon. This chemical interference happens because the concentration of the excited state atoms of the analyte is greatest at the center of the flame due to the higher temperature at this location. Self-absorption limits linearity of the calibration curve in ICP atomic emission at high analyte concentrations because the linearity of the calibration curve is the relationship between emission intensity and the amount of excited state atoms of the analyte and when self-absorption occurs it prevents the emission to be seen since it is just being simultaneously;y absorbed. The measured population of excited state atoms would not be proportional to the emission intensity because the emission intensity would be observed as less since the photons are absorbed before being seen.
What is the Limit of Detection (LOD)?
Smallest amount of analyze that can be determined with confidence Instrument: [blank signal + 3(signal noise)]/sensitivity
Why can't trinitrotoluene (TNT) be determined by fluorescence?
TNT is a single aromatic ring with three NO2 substituents and a methyl group. It is due to its structure that TNT cannot be determined by fluorescence because a single aromatic ring structure does not produce any type of fluorescent emission. Along with that, the NO2 side groups are electron-withdrawing groups that also decrease the fluorescent quantum yield.
What is the purpose of the RF coil current in the ICP torch below?
The RF coil generates a magnetic field and upon sparking the argon gas, electrons are accelerated forming a hot Ar+ plasma.
The Raman signal is not due to the absorbance of light or emission of light. What is the Raman signal due to, and describe the light source and its wavelength that generates the Raman signal?
The Raman signal is due to shifts in vibrational frequency. The photons are inelastically scattered by the sample molecule and either gains or losses energy in the process. Because this type of spectroscopy is based on inelastic scattering, it can detect molecular separations that are IR inactive. This difference in energy relates to the vibrational energy and can therefore be used to measure the vibrational energy of the molecule. The light source is always a high-power laser, preferably one which is monochromatic and stable. The five most commonly used lasers include wavelengths ranging from 405-1064 nm.
Explain why the source and detector for the fluorescence instrument at a 90 degree angle.
The detector is at a right angle to the source light so only fluorescence from the cuvette, not the source light, reaches the detector.
What is the advantage of signal averaging and explain how the following equation is derived. (S/N)n = sir root (n)(Sx/Nx). Define the variables in this equation.
The diode array detector is advantageous because it is able to collect a massive amount of data over a short period of time. Because of its great speed, many data points are collected and averaged to obtain a "final spectrum." This method actually increases the signal to noise ratio which is what we want. With signal averaging, the signal increases as nSx, while the noise increase at a rate of the square root of n. This increases the signal to noise ratio. An is the signal to noise ratio, and (SN)n is the ratio after n amount of scans. As the number of spectra increases (n increases), the sum of the signals increases. This is shown as nSx. The noise is a random event, so it increases as nNx. This gives the equation (SN)n=nSxnNx which can be simplified to (SN)n = n(SxNx) as seen above. The variables are: (SN)n : signal to noise ratio after n scans N : number of spectra Sx ; the signal Nx : noise at any point Therefore, if you have the number of spectra run and the frequency of signal and noise, you can see how the amplification affected the overall spectra produced.
Why must the excitation wavelength in Raman spectroscopy be carefully chosen? Give three reasons
The excitation wavelength in Raman spectroscopy must be chosen carefully for multiple reasons. The first reason being because at certain shorter wavelengths photo decomposition of the sample can occur. Another reason is due to the production of high fluorescence at shorter wavelength sources. The final reason that the excitation wavelength in this type of spectroscopy must be chosen carefully is because samples that are colored or certain solvent samples will absorb the Raman-scattering radiation which will affect the outcome of the spectroscopy. Source that emit near-IR radiation which have longer wavelengths are used commonly as excitation sources as they do not cause photodecomposition and typically there is either no fluorescence or the fluorescence is less intense when these lasers are used.
For the valence shell energy level diagram for Na, which wavelength transition is most intense and why?
The excited states with the lowest energy are most probable to be formed. Therefore, 3s to 3p is the most likely excitation transition and similarly 3p to 3s is the most intense emission.
Aromatic compounds such as benzene or naphthalene are easily detected by UV spectrophotometry at 260 nm using a mercury lamp which has strong emission at 254 nm. What is the molecular functional group involved and the type of electronic transition at 260 nm.
The molecular functional group that is involved when aromatic compounds are detected by UV spectrophotometry at 260 nm is a carbon to carbon double bond, C=C. The transition at this point is pi to pi*
What is the quantum yield of a molecule and how is it affected by temperature and why?
The quantum yield of a molecule is essentially a measurement of the fluorescent efficiency. It is the fraction of molecules in the excited state that become related down to the ground state via fluorescence that ranges between 0 to 1. Quantum yield is affected by temperature and will generally decrease with an increase in temperature. This occurs because as temperature increases, more collisions occur between molecules which results in an increase in external conversion. This means that more excited state molecules are returning to the ground state via a pathway that is not fluorescence.
Why are the IR and Raman spectra in the figure both quite similar but also show differences? Explain what must happen to the molecule for an IR transition to be observed and what must happen to the molecule to see a Raman transition peak.
The similarities are due to similar energy changes within the spectrums. The differences in IR and Raman come down to charge or electron distribution to get that energy to change to the excited state. For IR, there is a change in the charge distribution during the vibration to excite the molecule caused by radiation interacting with the molecule. In Raman, there is just a split second where the electrons around the bonds are distorted to cause a dipole then this disappears once it goes back to the remission state. There is also a difference in probabilities for transitions that will likely happen within both IR and Raman. This will cause the intensities of the bands in the spectra to be different. IR: asymmetric stretching vibration Raman: symmetric stretching vibration
Why is the symmetric stretch for CO2 not IR active but it is Raman active?
The symmetry of the C-O stretch means there is no permanent dipole which is necessary for an IR transition. However, this symmetric stretch does cause a change in electron distribution temporality (polarizability) which permits a Raman peak transition.
Explain what happens when no stray-light is allowed in the colorimeter and when stray-light is allowed in?
With the colorimeter lid closed, there is no stray light and Beer's Law should be followed with a linear relationship of absorbance and concentration. With the colorimeter lid open, the potential prescence of stray light will cause flattening of the absorbance-concentration curve and nonlinearity.
