Atomic Emission and Molecular Absorption Spectra v062513_6pm Objective: The student will observe the atomic emission spectra of hydrogen using a spectroscope, determine the identity of an unknown metal ion by conducting a flame test, and plot the absorption spectrum of crystal violet. In the eighteenth century, a Scottish physicist Thomas Melvill reported that when he placed different substances in a flame, they emitted light of different colors. For example, he noted that when table salt was passed through a flame, the resulting emitted light was bright yellow which was a result of sodium (Figure 1). This concept was something that had been known for several centuries before Melvill, but what was important about his Figure 1 discovery is that he found that when the emitted light was passed through a prism and a small slit, there was a band of colors, or a spectrum, that was created that was unique to the specific element in the flame. Figure 2 shows the atomic emission spectrum of sodium. Figure 2 Throughout the nineteenth century, this idea of different elements producing a unique spectrum was utilized as a way of identifying what elements were in substances and was even used to discover new elements. J.J. Balmer, a teacher from Switzerland, made an important discovery in 1885. Balmer was working with hydrogen and its atomic spectrum and he knew that hydrogen emitted light at 4 different wavelengths. Through trial-and-error, he discovered that there was a pattern that the emissions followed that could be represented by the Balmer-Rydberg equation, 1 1 1 = R 2 2 λ n f n i (1) where n f = 2 for lines appearing in the visible light region, n i is a whole number greater than 2, R is the Rydberg constant (1.097 x 10-2 nm -1 ), and λ is the wavelength of the emission line in nanometers. Note that the Rydberg constant necessarily has units of nm -1 because n s are unitless, wavelength in this model is expressed in nm, and its reciprocal is used. Over the next thirty years, the picture of the atom would begin to take shape. Work by J.J. Thompson, E. Rutherford and other scientists showed that the atom was made of posi tively charged protons and neutral neutrons found within the nucleus, and that the nucleus was surro unded by a negatively charged electron cloud. There were several models that tried to describe atomic structure, but one of the most important models was proposed by Niels Bohr in 1913. Observing the atomic spectrum for hydrogen and the work done by Rydberg and Balmer, Bohr suggested that the hydrogen atom was made up of a nucleus with an electron orbiting Spectroscopy 1 Figure 3
that nucleus. He also indicated that the electron could only occupy specific energy levels and that it would not be found between energy levels. The Bohr model of the atom can be seen in Figure 3. Bohr proposed that the electron in the hydrogen atom would exist in a stable, lowest energy or ground state orbital shown on Figure 4 as n = 1. However, when the atom is irradiated by light, the electron absorbs that energy (where the energy of the incoming light matches the difference in the energy level between the orbitals) which results in the electron being excited to a higher energy orbital. The electron would then relax to a lower, more stable energy level. For a more in depth discussion of atomic spectroscopy and the Bohr model, see Chapter 7 Section 3 of Tro 2 nd Edition. The energy difference between the energy levels can be determined using the equation: ΔE = hν (2) Rearranging the speed of light equation (c = νλ) to solve for frequency, the speed of light divided by the wavelength can be substituted in equation 2 for the frequency to give the equation: ΔE = hc! (3)!!!!!!!!!!!!!!!!!λ# In equations 2 and 3, h is Planck s constant (6.626 x 10-34 J s), ν stands for the frequency, c is the speed of light (2.998 x 10 8 m/s), and λ is the wavelength of the light in meters. Bohr found that using his model he could accurately predict the wavelengths of the lines on the hydrogen emission spectrum that Balmer had observed and described using the Balmer-Rydberg equation. He also determined that if an electron fell from an orbital higher than n=2 to the level n=2, the energy emission would be observed in the visible region of the spectrum. The problem with Bohr s model of the atom and the Rydberg-Balmer equation was that they could not accurately predict the wavelengths of spectral lines for atoms with more than one electron due to the fact that this model and equation do not take into account electron-electron repulsions that exist in elements with more than one electron. Though the Bohr model has its fallacies, the idea of different energy states is still applicable to all atoms. - Molecular Absorbance Figure 4 We ve seen that atoms are capable of absorbing and emitting light at specific wavelengths, but absorption of light also occurs at the molecular level. It s this light absorption by molecules that allows us to sense many of the colors we see on an everyday basis. For example, the reason we perceive grass as being green is due to the molecules of chlorophyll within the grass. As seen in Figure 5, chlorophyll absorbs light in the red and violet regions of the visible electromagnetic spectrum. The primary wavelength of light that is not Spectroscopy 2 Figure 5. Absorption spectrum of chlorophyll
absorbed is around 500 nm which corresponds with the green region of the spectrum. This particular wavelength of light is reflected back to the eye and we see that the grass is green. An effective qualitative method of determining the color of a substance that will be observed is by looking at an artist s color wheel (Figure 6). The color absorbed and the color observed tend to be complementary. For chlorophyll the color absorbed is red and the color observed is green. As you can see from the color wheel these two colors are complementary. So how do molecules absorb light? Similar to atomic absorption, molecules have internal energy levels that are responsible for the specific wavelengths of light absorbed. When molecules form, molecular orbitals are formed in which the electrons are found. Similar to atomic orbitals, molecular orbitals have different energy levels in which electrons can be found. Molecular orbitals and molecular orbital theory is discussed in chapter 10 Section 8 of Tro 2 nd Edition. Figure 6. Color wheel relating color with wavelength of light. The best way to discuss light absorption by molecules is to look at the structure of a couple of molecules that absorb light. The molecules in Figure 7 are beta-carotene, the molecule responsible for the orange color of carrots, and the second structure is crystal violet. Figure 7. Stucture of beta-carotene (left) and crystal violet (right). What do these two molecules have in common? The answer is that both molecules have conjugated bond systems. A conjugated bond system means that the molecules have alternating single (as illustrated by a single line) and double (as illustrated by two parallel lines, =) bonds. The presence of the conjugated bond system results in a delocalization of electrons. Just as in atoms, high-energy states exist for molecules. When electrons in the molecule are exposed to energy that matches the difference in energy between two molecular orbital levels, the electrons will transition to a higher energy molecular orbital, leading to a molecular excited state. In the case of molecules with conjugated bonds, the energy difference in the molecular orbital levels is less due to the delocalization of the electrons. As a result, the wavelength of energy falls within the visible region of the electromagnetic spectrum. This allows the molecule to absorb visible light and produces the colors that we see. Spectroscopy 3
Procedure: All measurements should be recorded using the correct number of significant figures appropriate to the equipment being used. All calculated values should be recorded to the correct number of significant digits. Carefully read Chapter 1 Section 7 of Tro 2 nd Edition to re-familiarize yourself with significant digits and being consistent with them in mathematical operations. IF THE SPEC 20 IS NOT ALREADY ON, TURN IT ON NOW TO ALLOW THE INSTRUMENT TO WARM UP BEFORE USE. Flame Test of an Unknown Salt 1. Choose an unknown salt and record its # on your data sheet. 2. Make a loop at the end of a paperclip. 3. Set up and light the Bunsen burner. 4. Scoop a small amount of the unknown salt with the loop end of the paper clip and move the tip of the paper clip into the flame of the Bunsen burner. 5. Record the color of the flame on the data sheet. Atomic Spectrum of Hydrogen 1. Plug in and turn on the power supply with the lamp in place. 2. Set up the spectroscope by moving the slit toward the center of the discharge tube while being careful not to touch the tube. 3. Record the wavelength of the hydrogen emission lines on the data sheet. 4. Turn off the power supply and allow the discharge tube to cool. Absorption Spectrum of Crystal Violet Spec 20 Calibration 1. With no cuvette in the sample compartment, set the wavelength of the Spec 20 to 440 nm using the wavelength selector knob. Be sure that the wavelength range selector on the front of the instrument is set to the correct wavelength range. 2. Set the Spec 20 to show transmittance using the absorbance/transmittance selector. 3. Turn the zero adjust knob until the transmittance reading shows a value of 0. 4. Fill a cuvette approximately 2/3 full with distilled water, wipe the outside of the cuvette with a KimWipe, then place the cuvette into the sample compartment and close the lid. 5. Turn the 100% adjust knob until the Spec 20 shows a transmittance reading of 100. 6. Remove the cuvette containing the distilled 595 0.0 transmittance absorbance concentration factor water and set it aside. You will need the cuvette containing the distilled water when you begin taking readings of the crystal violet. ON/ OFF Wavelength Filter Range lever Absorbance/Transmittance Mode Selector 0% T Adjust empty cavity Wavelength selector knob 100% T Adjust blank- filled cuvette Absorption Spectrum of Crystal Violet Spectroscopy 4
1. Change the absorbance/transmittance knob to read absorbance. 2. Fill a second cuvette approximately 2/3 full of crystal violet solution, wipe the outside of the cuvette with a KimWipe, then place the cuvette into the calibrated instrument and close the lid. 3. Record the absorbance value on the table provided on the data sheet. 4. Remove the crystal violet cuvette and set it aside. 5. Change the Spec 20 wavelength to the next wavelength shown on the data sheet table using the wavelength selector. 6. Change the absorbance/transmittance knob to read transmittance. 7. Turn the zero adjust knob to set the transmittance equal to 0. 8. Place the cuvette containing the distilled water from the calibration step into the instrument. Using the 100% adjust knob set the transmittance equal to 100. Remove the cuvette containing the distilled water and set it aside. 9. Change the absorbance/transmittance knob to read absorbance. 10. Place the cuvette containing the crystal violet solution into the instrument and record the absorbance on the data sheet. 11. Repeat steps 4-10 for all wavelengths shown on the data sheet. Spectroscopy 5
Plotting Data in Excel The exact instructions to use will depend on your version of Excel. The following set of instructions and pictures are similar for all versions of Excel. 1. First, enter the values of your independent variable (x-axis values) into column A, and the values of the dependent variables (y-axis values) into column B. For the crystal violet plot, the wavelength values are the independent variables and the absorbance values are the dependent variables. 2. Using the mouse, highlight the values in columns A and B. Spectroscopy 6
3. Select Insert tab, select the Scatter tab and then select the plot type. 0.6 0.5 0.4 0.3 0.2 Series1 0.1 0-0.1 0 100 200 300 400 500 600 700 4. Click on layout under chart tools a. Chart Title Wavelength vs Absorbance for (compound name) 5. Format the axis. a. Axis titles i. X-axis Wavelength ii. Y-axis Absorbance b. Right click on the X-axis, click on format axis, and change the minimum and maximums to include your data without much excess. Make sure to fix the minimum and maximum. Do the same for the Y-axis. Spectroscopy 7
Wavelength vs Absorbance 0.6 0.5 Absorbance 0.4 0.3 0.2 0.1 0 350 400 450 500 550 600 Wavelength Spectroscopy 8
Pre-Laboratory Assignment 1. An electron in a lithium atom moves from the 2p orbital to the 2s orbital with a ΔE of 2.96 x 10-19 J. When the transition occurs, energy equal to ΔE is released in the form of a photon. What is the wavelength of the light that is emitted? 2. Does the emission line determined question1 fall in the visible region of the electromagnetic spectrum? If so, what color is the light that is emitted? 3. Which of the following molecules, based on its structure, would most likely absorb light? Circle all that apply and explain your reasoning. Spectroscopy 9
Data Sheet Flame Test Unknown # Color Atomic Spectrum - Hydrogen λ 1 = nm color = λ 2 = nm color = λ 3 = nm color = λ 4 = nm color = Molecular Absorbance Spectrum Crystal Violet Wavelength (nm) Absorbance Wavelength (nm) Absorbance 440 570 460 580 480 590 500 600* adjust filter lever 520 610 540 620 560 640 Spectroscopy 10
Results 1. Using the chart provided, what is the identity of the metal in the unknown from the flame test? Ion K Na Li Cu Sr Color Lilac Yellow Red Green Red/Orange Unknown? 2. What is the approximate wavelength of the light emitted from the unknown? From the wavelength, determine ΔE for the transition. Color Wavelength(nm) Red 701 Red/Orange 622 Orange 609 Orange/Yellow 597 Yellow 587 Yellow/Green 577 Green 535 Green/Blue 492 Blue 474 Blue/Violet 455 Violet 423 Wavelength? ΔE (J/photon)? ΔE (kj/mol)? Spectroscopy 11
3. Calculate the energy of each photon associated with each of the wavelengths of the hydrogen lines. Lower energy Purple line Higher energy Purple line Teal-blue line Red line Spectroscopy 12
4. Using the Rydberg equation, calculate the energy level (n initial ) from which the electron fell to produce the emission. Remember: when visible light is produce the electron falls to the 2 nd energy level. See p 301 of Tro 2 nd ed. Lower energy Purple line Higher energy Purple line Teal-blue line Red line Spectroscopy 13
5. Indicate the electron transition (with arrows) and associated energy for each of the four wavelengths of light observed for hydrogen on the diagram below. 6. Plot the absorbance vs. wavelength data of crystal violet in Excel. At what wavelength does the maximum absorbance of light occur? 7. Does the absorbance obtained explain the color of crystal violet? Explain. Spectroscopy 14
Post-Laboratory 1. The Balmer-Rydberg equation fails to accurately predict the quantized energy levels for atoms with more than one electron. a. What physical force is absent in the hydrogen atom, but present in all other atoms, that allows the Rydberg equation to work for hydrogen? b. Can multi-electron atoms still give line spectra? Explain. 2. Use a resource to identify two elements that were discovered because of a unique, previously unobserved, line in their emission spectrum using a Bunsen flame. 3. The line spectrum of a certain system is shown in the figure at right. These lines result from transitions among the four levels shown. a. Which of the photons is the most energetic? b. Add an arrow to the level diagram representing the transition associated with the longest wavelength photon. c. Give the levels associated with each of the following photons: Photon B: Photon C: n"="4" n"="3" n"="2" n"="1" energy"levels"of"system" A" B" C" D" E" F"!"""""shorter"wavelengths" observed"line"spectrum"of"system" Spectroscopy 15
4. Phenolphthalein is an indicator that is colorless in the presence of an acid and pink in the presence of a base as you discovered during the acid / base titration lab. Given knowledge obtained in this laboratory, explain why phenolphthalein behaves as it does in acid/base solutions. (Note: the CO 2 -, called a carboxyl group, contains a charge which indicates extra electrons on the oxygen.) O" O" O" O" O" O"H" phenolphthalein" colorless"in#acid# O"H" O" phenolphthalein" pink"in#base# Spectroscopy 16