Experiment 6 Absorbance and Fluorescence Spectroscopy PDF

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Absorbance and Fluorescence Spectroscopy: Determination of Common Compounds in Energy and Sports Drinks Research Question(s) What are the concentrations of dyes and riboflavin (vitamin B2) in common beverages? What methods can you use to determine the concentration of one or more unknown compounds in a solution? How can you ensure that your experiment and calculations give reliable data? Learning Goals It is often a goal of a chemist to identify and quantify compounds that are present in an unknown. What compounds are present? How much of each is there? Think of a common analysis, like trying to determine if a person has enough iron in her blood. The blood sample is a complicated mixture, and data must be acquired carefully and interpreted consistently to give the patient an accurate diagnosis. In this experiment you will quantify the food dyes and riboflavin found commonly in some of the most popular beverages today. In this experiment you will learn two different techniques for determining the concentration of species in solution. One technique, ultraviolet-visible (UV/Vis) spectroscopy, relies on the ideas that many molecules absorb light, and the amount absorbed is proportional to the concentration of the molecules in solution. The second technique, fluorescence spectroscopy, relies on similar principles but is more selective of the compounds that can be analyzed. Certain compounds, once they have absorbed light energy, release that energy at a characteristic wavelength in a process called emission. Like absorbance, emission can be correlated to compound concentration. In this lab you’ll be using absorbance spectroscopy to determine the concentrations of two food dyes in a sports drink, and you’ll use fluorescence spectroscopy to determine the concentration of riboflavin in an energy drink. You need to understand how the instrument responds to the species present. By generating a calibration curve, which is a plot of absorbance or emission measurements for samples of a given compound at known concentrations, you can account for any instrument-based variations that might be present in your data. You can also use these curves to calculate the concentration of compounds in unknown solutions. You will then use that calibration curve to determine the concentration of an unknown solution provided to you.

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Absorbance Spectroscopy When an atom or molecule absorbs a single photon, an electron makes a transition to a higher energy orbital. Thus, the number of photons a sample of material absorbs is directly related to the number of particles in the path of the beam light. This important concept means that by measuring the change in intensity or number of photons (I) of a light beam before (I 0) and after (It) it passes through a solution of molecules, we can measure the number of molecules in the solution. The intensity change is related to the number of molecules the beam encounters. More technically, the log of the ratio of I 0 to It is called the absorbance (A). 𝐼

𝐴 = log ( 0 ) 𝐼𝑡

Red Orange Yellow Green Blue Indigo Violet

I0

It

Red Orange Yellow Green Blue Indigo Violet

It

Red Orange Yellow Green Blue Indigo Violet

Blank 0.35

I0 Sample

Light Source

0.30

Absorbance (a.u.)

Red Orange Yellow Green Blue Indigo Violet

0.25 0.20 0.15 0.10 0.05 0.00

Detector Detector

400

500

600

700

800

Wavelength (nm)

Figure 1: For the Red Dye #3 sample above, all of the red and some of the blue light pass through the sample, whereas the yellow, orange, and green light is absorbed. Thus, the absorbance values of yellow, orange, and green light are high and the absorbance values of red and blue are lower. Absorbance specifically depends on certain properties of the sample: the wavelength of the measurement , the molar absorptivity at that wavelength,  , the path length of the sample cell, l, and the concentration of the sample, c. The absorbance, A, of the solution containing a given species is equal to the molar absorptivity 𝜀𝜆 (in L mol−1 cm-1) times the path length l (in cm) times the concentration c (in mol L-1). If the concentration of the sample is expressed in units other than molarity,  is referred to as the extinction coefficient. 𝐴𝜆 = 𝜀𝜆 𝑙𝑐 𝐴𝜆 𝑡𝑜𝑡𝑎𝑙 = 𝜀1 𝑏𝑐1 + 𝜀2 𝑏𝑐2 +…

(Beer’s Law) (for mixtures)

Notice that there is a linear relationship between absorbance and concentration. The slope of the line corresponds to the molar absorptivity times the path length of the sample,  ·b.

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Special sample cuvettes are used which have a path length of 1.000 cm. The molar absorptivity depends on the wavelength of the light and is determined by measuring the absorbance of a series of standard solutions of known concentration. Once  and b are known, the concentration of an unknown solution can be determined by measuring the absorbance of the unknown and using the Beer-Lambert Law given above. Alternatively, we can also explain this phenomenon by quantifying the transmission of a sample, or the number of photons that the sample does not absorb. Transmission, It ⁄I0 , is related to absorbance: I 10-A = ( t ) = T I 0

In the example spectrum shown above, the transmission values of red and blue wavelengths of light would be high but those of orange, green, and yellow would be low. Scientists use both absorbance and transmission data to report the optical properties of a sample depending on what information they want to convey. Fluorescence Spectroscopy In absorbance spectroscopy, the measurement is focused on the energies absorbed by the sample. In emission spectroscopy, the detector focuses on the light emitted by the molecules. Electrons in molecules are associated with various types of energies. You may be familiar with electronic energy states, but electrons also occupy rotational and vibrational energy states (see Figure 2). When a molecule absorbs light, the electrons can be promoted to many different types of excited states. For example, an electron can be excited from a ground electronic state to a higher electronic state, i.e. from S 0 to S1 (represented by the h1abs transition in Figure 2). It could also be excited to a different vibrational state, i.e. from the ground vibrational state of S 0 to an excited vibrational state in S 2 (represented by h2abs in Figure 2). The molecule can relax back to its ground state through several different processes of energy emission, one of which is fluorescence.

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Figure 2: Physical processes that occur after a molecule absorbs a photon. When a molecule absorbs a photon, an electron is promoted to higher energy excited state (e.g. S 0 → S1). When a molecule emits a photon, an electron moves to a lower energy state (e.g. S2 → S0). If the excited molecule immediately relaxes back to the ground state, light will be emitted at the same wavelength as the light that was absorbed in a process called resonance fluorescence. In the figure above, this is represented by h1abs and h1fluor, which are equivalent. But if the molecule loses some energy to collisions with solvent or loss of heat (a process called internal conversion) then relaxes back down to the ground state, the energy emitted will be lower than that absorbed originally. This is true for the absorption/emission pair of h2 abs and h2fluor in the figure, where h2abs > h2fluor. It is common that excitation occurs with light in the UV range (higher energy) while emitted light is in the visible range (lower energy), which is part of what makes fluorescent molecules so fun to work with. The emission spectrum is often roughly the mirror image of the absorption spectrum. In absorption, the transition from the ground vibration level to the lowest excited vibrational energy level is λ0. All subsequent transitions are to higher energy vibrational levels (λ+1, λ+2, etc.). Thus, λ0 is the lowest energy transition and λ+2 is the highest energy transition. After absorption, the molecules relax back to the lowest energy excited vibrational state through internal conversion. This means that for emission, λ0 (the transition from the lowest vibration energy level to the ground vibrational energy level) is the highest energy transitional. All subsequent transitions are of lower energy. Figure 3 demonstrates the symmetry of absorption and emission transitions.

Figure 3: Energy-level diagram showing why absorption and emission spectra are roughly mirror images of each other.

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As with absorption spectroscopy, the intensity of emitted light is proportional to the concentration of the molecules in the sample. 𝐼 = 𝑘𝐼0 𝑐 The intensity of light emitted depends on the intensity of light supplied for excitation, I0, and the concentration, c, of the sample. The constant k is a term that includes the molar absorptivity at the excitation wavelength and the path length of excitation through the center of the cell. However, this relationship is only linear at lower concentrations because if the sample gets too concentrated, the chance of self-absorption of the emitted light is greater. In that case one molecule might emit light that a neighboring molecule absorbs, and that photon would not be detected. This is sometimes referred to as fluorescence quenching. For more information about fluorescence spectroscopy, please consult your Harris textbook. Absorbance and Fluorescence Measurements Spectrophotometric determinations based on the Beer-Lambert Law are among the most widely used analytical procedures. These methods involve the measurement of the fraction of incident electromagnetic radiation that is absorbed by a sample (Figur e 4). To determine the concentration of a colored species in solution, a cuvette containing the solution is placed in the spectrometer. Spectrometers consist of a built-in light source, which produces a narrow beam of light that passes through a sample cuvette (I0). The light that passes through the sample (It) is then split into its component wavelengths by a diffraction grating. A specialized (charge-coupled device or CCD) detector then measures the amount of light that reaches it at each wavelength. A computer then converts this into an absorbance reading using the relationship 𝐴 = log(𝐼0 ⁄𝐼𝑡 ).

Figure 4: Schematic of an absorbance measurement using a spectrometer (top view).

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Fluorescence measurements are very similar to absorbance. However, the molecules in the sample are excited by a specific wavelength of light (which can be done for absorbance spectroscopy but is not in this lab), and the emission of light is detected at 90˚ relative from the excitation (Figure 5).

Figure 5: Schematic of a fluorescence measurement using a spectrometer (top view) Calibrations No instrument will automatically determine the concentration of a sample. Instead you must calibrate the instrument using a known solution and compare the response of an unknown. By measuring the absorbance of a set of standard solutions of known concentrations, you can create a calibration curve that allows you to determine the concentration of an unknown of the same species. This calibration curve correlates to how the spectrometer responds to various concentrations of standard solutions. Beer’s Law may fail if a solution is too concentrated or too dilute. This means that you can only accurately determine the concentration of an unknown if its absorbance falls within the linear range of the dataset. An example for determining the linear range of a dataset using Red Dye #3 is shown below. Figure 1 (previous page) shows that the maximum absorbance for Red Dye #3 is at 526 nm. Figure 6 shows the absorbance measurements for a series of eight standard solutions of Red Dye #3 at 526 nm. Notice that at high concentrations the data no longer appears linear and

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the R2 value is only 0.9691. For a perfectly linear response, the R 2 value is 1. The closer the R2 value is to 1, the more reliable your data is. A linear regression of the five most dilute samples gives a much more linear response, R2 = 0.9995. The linear response range for a compound must be experimentally determined at the wavelength of interest. The extinction coefficient for Red Dye #3 at 526 nm can be determined from the slope of the calibration curve using the Beer’s Law data. 𝐴𝜆 = 𝜀𝜆 𝑙 ∙ 𝑐 y

m x

A plot of absorbance versus concentration will have a slope of 𝜀𝜆 𝑙. For Red Dye #3 the equation for the calibration curve is y = 0.0805x − 0.0791 as shown in Figure 2. Most sample holders (l) are 1 cm. 𝑚 0.0805 𝐿/𝑚𝑔 𝑚 = 𝜀𝜆 𝑙 ⇒ 𝜀𝜆 = = = 0.0805 𝐿 ∙ 𝑚𝑔−1 ∙ 𝑐𝑚−1 1 𝑐𝑚 𝑙

2.500

Absorbance at 526 nm

Absorbance at 526 nm

An unknown solution with absorbance of 0.9054 would correspond to a concentration of 12.23 mg/L of Red Dye #3.

2.000 1.500 1.000 y = 0.0644x + 0.088 R² = 0.9691

0.500 0.000 0.00

10.00 20.00 30.00 Concentration of Red#3 (mg/L)

40.00

2.000 1.500 1.000 0.500

y = 0.0805x - 0.0791 R² = 0.9995

0.000 0.00

5.00

10.00

15.00

20.00

25.00

Concentration of Red#3 (mg/L)

Figure 6: Beer’s law plot for Red Dye #3 using full and linear portion data set at 526 nm Rather than trying to analyze really concentrated solutions, it is considered good lab practice to quantitatively dilute solutions to concentrations that give absorbance values below 1.5 absorbance units. For more information on how to perform these dilutions, look at question 1 in the pre-lab.

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Simultaneous Determination of Two Dyes

M+N N

Absorbance

You will determine the concentrations of two dyes simultaneously by spectrophotometric analysis. Since the total absorbance of the solution at a given wavelength is equal to the sum of the absorbances of the individual components, it is possible to analyze the individual components of a mixture even when their spectra overlap. Consider the spectra of M and N given in Figure 7.

M λ1

λ2 Wavelength (nm)

There is obviously no wavelength Figure 7: Absorbance spectra of components M, at which the absorbance of this N and the spectra of both components together. mixture is due simply to one of the components; thus, an analysis for either M or N by a single measurement is not possible. However, the absorbances of the mixture at the wavelengths 1 and 2 may be expressed as follows: 1 1 1 At 1 : A =  M bc M +  N bc N At 2 : A 2 =  2Mbc M +  2Nbc N

The four molar absorptivities,  1M ,  2M ,  1N , and  N2 can be evaluated from standard solutions containing only M or N. Then, if the absorbances of the mixture are measured at 1 and 2, the concentrations of the individual components can be calculated by solving the two equations given above simultaneously. The best accuracy in an analysis of this sort is obtained by choosing wavelengths at which the differences in molar absorptivities between the two species is large. For your unknown, you will use the absorption spectra of yellow and red dyes to choose two wavelengths, 1, and 2, for analysis. For each dye what is the wavelength that corresponds to the highest absorption measurement? This value is called max and is the wavelength at which the instrument is most sensitive to the species being analyzed. Additional Reading Quantitative Chemical Analysis (Harris, 8th ed.): Chapter 2.4-2.6 (Burets, Volumetric Flasks, Pipets and Syringes) Chapter 2.10-2.11 (Graphing with Microsoft Excel) Chapter 17 (Fundamentals of Spectrophotometry) Chapter 18.1 (Analysis of a Mixture)

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Pre-Lab Exercises 1. Imagine that you and your lab partner made solutions according to the scheme below using a buret to dispense dye into a volumetric flask. Calculate the exact concentration of the final solutions in the last column. Show one example calculation.

Standard

Initial concentration of dye (M)

Target volume of dye (mL)

#1 #2 #3 #4

5.00 × 10-5 5.00 × 10-5 5.00 × 10-5 5.00 × 10-5

0.5 1.0 1.5 2.0

Actual volume of dye dispensed (mL) 0.48 1.10 1.51 2.05

Total solution volume* (mL)

Final concentration of dye (M)

5.00 5.00 5.00 5.00

2. For the solutions shown above, you and your partner did not dispense exactly the target volume from the buret. The target volume of dye was 3.5 mL, and you dispensed 3.48 mL. Does this mean the solutions have to be made over again? Why or why not? 3. Go to the PhET interactive simulation site: http://phet.colorado.edu/sims/html/beerslaw-lab/latest/beers-law-lab_en.html. Double click on the Beer’s Law icon to begin the simulation. a. Adjust the concentration of the red drink to 50mM and record %T and absorbance. b. Now double the concentration to 100mM and record %T and absorbance. Repeat for 200mM concentration. What happens to the ability to detect the solution at high concentrations? c. Set the wavelength of the light to variable. What colors does the red drink absorb well? What colors are transmitted? How does the absorbance and transmittance correspond to the color of the solution? 4. In fluorescence spectroscopy, the light detected is 90˚ from the incident light. What is the reason for this? What would happen if the detector was 180˚ (directly across) the incident light like it is in absorbance spectroscopy? Experimental Work in groups of three to complete this experiment. Student #1 should prepare the Red #40 calibration curve, and Student #2 should prepare the Yellow #5 calibration curve (Part 1). Student #3 should prepare the riboflavin calibration curve (Part 2). All three students

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should work together to measure the absorbance of a mixture of dyes in a sports drink and the fluorescence of riboflavin an energy drink (Rockstar). Part 1: Absorbance of food dyes Goal Divide work amongst your group to create calibration curves for Red #40 and Yellow #5 dye. Use these calibration curves for the simultaneous determination of red and yellow food dye in a sports drink. Procedure Group work: Student #1 and #2 work together to create dye stock solutions. Obtain approximately 2-3 mL of Yellow #5 and Red #40 dyes (approximately 2.5 × 10-3 M). Record the exact concentration of the stock solution for both dyes. You will need to create dilutions in order to measure the molar absorptivity. What is the difference between molar absorptivity and the extinction coefficient? For each dye, dilute your solution to a final concentration of 5.00 × 10-5 M using volumetric glassware. You only need to make 50-mL of each solution from a few mL or less of the stock solution . Your GSI will guide you through this process. How does your final concentration compare to your target concentration of 5.00 × 10-5 M? This will be the stock solution you use for your red and yellow dye calibration curves. You will need to create further dilutions of each of your 5.00 × 10-5 M solutions in order to measure the constant 𝜀𝜆 in 𝐴𝜆 = 𝜀𝜆 𝑙 ∙ 𝑐 for each dye.

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Buret and volumetric flask Group work: Student #1 - Red #40, Student #2 - Yellow #5. Work together to choose two wavelengths that will produce the greatest accuracy in determining the concentration of mixture of red and yellow dye. Use the same spectrometer for both dyes (share between Students #1 and #2). Using the buret and flask method (Appendix A), prepare one standard solution at a time, transferring each solution to a cuvette for analysis. Follow the scheme shown in the table below. Dilute to the 10.00 mL mark with distilled water. When reusing the same glassware or cuvettes, it is good analytical practice to proceed from low to high concentration. Why? Record the exact volumes you actually dispense and use these to calc...