Ex 5 Atomic Spectra - Lab report 5 PDF

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RP-1 Chemistry 134L Laboratory Report

5. Atomic spectra, Molecular spectra, and Mass spectra Purposes of the experiment and brief overview of the experimental methods

There are three purposes to this study. The first is to study the characteristic absorption and emission behavior of hydrogen gas. We do this by measuring the diffraction of hydrogen emission behavior of hydrogen gas, which was accomplished by measuring the diffraction of hydrogen through observation of a hydrogen discharge lamp's diffraction grating onto a meter stick. We also observed the peaks of the spectrum by looking at a Vernier graph set up by the TA. From these values, we are able to calculate the value of R. The second purpose is to observe the visible absorption spectrum of dyes in M&M's using a Vernier spectrophotometer. We extracted the dyes by dissolving the color coating from M&M's and mixing the resulting dye solution with water-holding cuvettes, which were then analyzed by the spectrophotometer. The wavelengths absorbed the most were then identified and recorded for each color. The last purpose was to analyze two mass spectrum graphs of chloromethane and bromomethane and determine which molecules were present, and in which intensities, after mass spectrometry was carried out.

Results and observations (Be careful with significant figures; show work and units wherever appropriate). 1. Atomic Spectrum of Hydrogen Grating ruling: # of lines per inch = 13400

# of lines cm-1 = 5276

Distance d between two successive grating lines (show work below) = 0.0001895 d=1/(number of lines cm^-1) d=1/5276 =0.0001895

Distance b from grating to center T along meter stick B = 40.5 cm Table 1. Values of a in cm on meter stick A

RP-2 Caution: a refers to the actual distance in cm; not the reading on the meter stick. Line # Left Right Mean Color 1

50-32.1=17.9

67.8-50=17.8

17.9

red

2

12.8

12.7

12.8

Green-blue

3

11.3

11.3

11.3

purple

Calculations: Table 2 Using the average value of a (), calculate sin  and . Line #

Color of line

b

sin 

 ( = d sin ) / nm

1 2 3

red Green-blue

17.9 12.8

40.5 40.5

0.404 0.301

766 571

purple

11.3

40.5

0.269

509

In the space below, show one sample calculation for sin  and (in nm). sin  a/⎷(a2+b2)= 17.9/⎷(17.9^2+40.5^2)=0.404cm = 0.0001895 * 0.404 = 0.0000766 = 766 cm

Wavelengths (in nm) of color lights from the ruler arrangement Red: 766 nm

Green-blue: 571 nm

Purple: 509 nm

Wavelengths (in nm) of color lights from the Vernier spectrometer Red: 658 nm

Green-blue: 488 nm

Purple: 436 nm

Violet: 412 nm

Wavelengths (in cm) of color lights from the Vernier spectrometer Red:0.0000658

Green-blue:0.0000488

Purple: 0.0000546

Violet: 0.0000412

Show work below for converting from the unit nm to the unit cm: 658nm *(1*10^-7cm/ 1 nm) = 0.0000658 cm Now prepare the table given below using the wavelengths obtained from the spectrometer. Then, prepare plots of (1 /) vs. (1 / n2initial) for various nfinal values. You need to plot l/ on the vertical axis, and l/n2initial on the horizontal axis. All three plots should be on the same graph.

Table 3 Color of light  / cm

1/

Red

15200

0.0000

n2initial

n final 1

4

n2initial

n final 2

9

n final 3

n2initial 16

RP-3 Green-blue

Purple

Violet

658 . 000048 8 . 000043 4 . 000041 2

20500

1

9

2

16

3

25

23000

1

16

2

25

3

36

24400

1

25

2

36

3

49

Start with nfinal = 1; then ninitial can be 2, 3, 4 etc. With n final = 1, you need to associate 1/ of the red line with ninitial = 2 (why not ninitial = 3?), and so on. Use the same label for these points with n final = 1; draw the graph using a computer. Draw the ‘best’ straight line as well. Print the equation of the best straight line on the graph. If you need help, ask Dr. G. Then calculate R from the slope and intercept. Show work for the calculation of R below. R= -slope = 42837 cm^-1 = 42800 cm^-1 Intercept= R/nfinal2 R=Intercept * nfinal2= 25740*1 = 25700 cm^-1 Now, let nfinal = 2; then ninitial can be 3, 4, 5 etc. With nfinal = 2, you need to plot, l/ of the red line with ninitial =3, and so on. Use the same label for these points with n final = 2. Draw the ‘best’ straight line through this second set of points. Then calculate R from the slope and intercept. Similarly, let nfinal = 3, and plot (1 /) vs. (1 / n2initial).

RP-4 n=2: R= -slope= 110181 cm^-1 = 1.10*10^5 cm^-1 Intercept= R/nfinal2 R=Intercept * nfinal2= 27424*4 = 109696 cm^-1 = 1.10*10^5 cm^-1 N=3: R= -slope= 219744 cm^-1 = 2.20*10^5 cm^-1 Intercept= R/nfinal2 R=Intercept * nfinal2= 29053*9 = 261477 cm^-1 = 2.61*10^5 cm^-1

Of the three plots, which one do you choose as the correct one? See if you can reach a decision by the linearity of the plot, and/or by comparing the values of R from each of the plots with its theoretical value. State clearly which straight line you chose, and explain why you made this choice. We chose the red line as the correct line because its calculated R value is the most accurate to the theoretical value of R. The R value calculated from the slope of this line (by taking its negative) was 1.10*10^5 cm^-1, which is equivalent to the theoretical value of R calculated below, of 1.10*10^7 m^-1

Experimental value of Rydberg’s constant R from the slope for nf = 1 (work shown above): 4.28*10^4cm-1

Experimental value of Rydberg’s constant R from the slope for nf = 2: 1.10*10^5 cm-1 Experimental value of Rydberg’s constant R from the slope for nf = 2: 2.20*10^5 cm-1

Experimental value of Rydberg’s constant R from the intercept for nf = 1 (show work below): 2.57*10^4 cm1

Experimental value of Rydberg’s constant R from the intercept for nf = 2: 2.20* 10^5cm-1 Experimental value of Rydberg’s constant R from the intercept for nf = 3: 2.61*10^5 cm-1

RP-5 m e4 8 ε 2 c h3

0 Calculate the theoretical value of R (in cm-1) using RTheory = Show work below and or on the other side of this sheet, and units. If you have not already done so, refer to the lab manual for information on this equation. In calculating the value of R Theory, treat the mantissas together, and the exponents together. Examples: 2.3 x 10 -34 x 5.2 x 10-47 = 2.3 x 5.2 x 10(34+47) = 1.2 x 10-80; (6.6 x l0-34)3 = (6.6)3 x 10-(3x34) = 287. x 10-102 = 2.9 x 10-100

m e4 8 ε 2 c h3

0 = RTheory = = (9.1044*10^-31 kg)(1.6022 * 10^-19 C)^4/ (8(9.9542*10^-12) J^-1 C^2 m^-1)^2(2.9979 *10^8 m s^-1)(6.626 * 10^-34)^3 = 1.10*10^7 m^-1 =1.10*10^5 cm^-1

RP-6 2. Visible Absorption spectrum of m&m dyes

Plot absorbance on the vertical axis versus wavelength (from 400 to 700 nm) for the five dyes, all five in one graph. Label each spectrum appropriately, mark and label max for each. Calculate the energy of a photon of wavelength corresponding to each max, and tabulate them below. Show one sample calculation. Color of dye max Energy of one photon / J Energy per mol / kJ mol-1 red 508 3.91 x 10-19 235 orange 484 4.10 x 10-19 247 -19 yellow 418 4.75 x 10 286 green 628 3.16 x 10-19 190 blue 630 3.15 x 10-19 190 Sample calculation (be careful with sigfigs; show units): Energy of photon with max = 508 nm = h c /  x 10-34 J s x 2.998 x 108 m s-1 / 5.08*10^7m = 3.91 x 10-19 J Energy per mole = Energy photon-1 x 6.022 x 1023 photons mol-1= 3.91 x 10-22 kJ x 6.022x1023 photons mol-1

RP-7 = 235 kJ/mol

3. Mass Spectrometry Using the mass spectrum of chloromethane (CH3Cl) provided in the lab (also in Canvas), fill in the following table. Pick the top five prominent peaks (in terms of relative intensity / relative height), list each one in the table (see example below), in the order of decreasing prominence, and label each peak in the spectrum by the number you assigned in the table. Include a picture of the appropriately labeled spectra in your lab report. Peak # 1 2 3 4 5

Mass of Chemical species particle / amu 12 50 CH335Cl+ (Is there any other possibility?) 12 52 CH337Cl 12 15 CH3 35 35 Cl 37 37 Cl

Relative peak height 100 31 72 6 2

1

3

2 5 4

RP-8 Using the mass spectrum of bromomethane (CH3Br) provided in the lab (also in Canvas), fill in the following table. Pick the top five prominent peaks (in terms of relative intensity / relative height), and list each one in the table, in the order of decreasing prominence; see example below. Pick the top five prominent peaks (in terms of relative intensity / relative height), list each one in the table (see example below), in the order of decreasing prominence, and label each peak in the spectrum by the number you assigned in the table. Include a picture of the appropriately labeled spectra in your lab report.

Peak # 1 2 3 4 5

Mass of Chemical species particle / amu 12 94 CH379Br 12 95 CH381Br 12 15 CH3 79 79 Br 81 81 Br

Relative peak height 100 94 47 6 6

1 2

3

4

5

RP-9 Questions For each question, first list the question in italics, and then list the answer right below it in regular style. Be brief and to the point. Brevity and relevance are more important than volume. 1. Was your experimental value of Rydberg’s constant consistent with its theoretical value? Include percent deviation(s). If not, what feature(s) of the experiment / procedure was responsible for the inconsistency? The value of Rydberg's constant that was calculated from the slope of the n=2 line was most consistent with the experimental value of R with a 0% deviation: both values were calculated to be 1.10 x 10^5. The values calculated from the slopes of the n=1 and n=3 lines wvaried greatly from the true value of R, with values of 4.28 x 10^4 (deviation of 61.1%) and 2.20x10^5 (deviation of 100%), respectively. Similar deviations were seeni nthe R values calculated from the intercept of the n=1, n=2, and n=3 lines, with values of 2.57 x 10^4 (deviation of 76.6%), 2.20 x 10^5 cm^-1 (Deviation of 100%), and 2.61 x 10^5 cm^-2 (deviation of 137%) respectively. This inconsistency may be due to experimental error. The spectral lines are present because as electrons in the hydrogen atom drop down energy levels after being excited, they release energy which is manifested as color.

2. What is the origin of the spectral lines in the H spectrum according to Bohr’s theory? Besides, the four wavelengths we saw, are there others? If yes, how can they be ‘seen’? There are more wavelengths than the four that we saw and these would be the ones in the UV and infrared ranges that are not visible to the human eye. They can be "seen" with UV and infrared cameras. 3. Did the visible absorption spectra of the different dyes support the idea of complimentary colors? Explain why or why not. Be succinct. Is the green color from the green m&m a pure color or a composite? What about the orange color in orange m&ms? Explain succinctly. Yes. For each dye observed, the highest absorbance occured at the wavelength of its complimentary color (Ex: red with green, blue with orange, etc.). The only exception is the green colored dye, which had peaks at orange and purple wavelengths, which occurs because green is a composite color. Although orange should be a composite color, it absorbs the most wavelgths of light in its complement of the blue range as expected. 4. What features of the mass spectrum show the presence of isotopes? How could we determine the abundances of isotopes of an element? Choose an(y) atom of your choice represented in the mass spectra of interest in today’s experiment, and use that to answer this question. The presence of isotopes can be seen in chloromethane’s mass spectrum – the presence of relatively high intensities at masses that were not immediately expected indicate that isotopes are present. Chlorine has isotopes with molecular weights of 35 and 37, and with C-12 the molecular weight of these atoms would be 50 amu and 52 amu respectively, and this is seen in the lines at each of these molecular weights. The height of the line indicates intensity, otherwise known as how heavily present that particular molecule was in the sample, and the molecules with the Cl-35 isotope are more heavily present with an intensity of 100, while the molecule with the Cl-37 isotope has a relative intensity of 31, suggesting that the Cl-35 isotope may appear more frequently in nature. This reasoning can

RP-10 be used to determine the abundance of isotopes of an element. NOTES: Notes were not taken because I directly recorded all information onto this word document on account of not being able to write with a broken arm....