chem lab report

  

I need to write a long lab report. I attached the instructions of the lab, and my post lab data, and ACS style writing guide. I need all the parts that are listed in ACS style writing guide, including abstract, intro, methods and materials, results, discussion, references, and appendix. If you don’t have chemistry background, please don’t do this.
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Chem 305 L – Spectra of Conjugated Dyes
Post Lab
Calculations of Nb values
1,1- diethyl-2,2-cyanine iodide
Nb = P+3
Nb = 3+3 =6
1,1- diethyl-2,2-carbocyanine iodide
Nb = P+3
Nb = 5+3 =8
1,1- diethyl-2,2-dicarbocyanine iodide
Nb = P+3
Nb = 7+3 =10
L.. (not sure about the equation.. lost prelab)
From pre-lab part 4
L2=3.03×10-4×(N+1)λ
i)1,1- diethyl-2,2-cyanine iodide
λ = 524 nm and N = 6
L2=3.03×10-4×(6+1)(524)
L = 1.0542 nm
ii)1,1- diethyl-2,2-carbocyanine iodide
λ = 604 nm and N = 8
L2=3.03×10-4×(8+1)(604)
L = 1.2834 nm
iii)1,1- diethyl-2,2-dicarbocyanine iodide
λ = 692 nm and N = 10
L2=3.03×10-4×(10+1)(692)
L = 1.5187 nm
Effective length
Effective length = L/N
i)1,1- diethyl-2,2-cyanine iodide
Effective length = L/N
= 1.0542/6
= 0.1757 nm
ii)1,1- diethyl-2,2-carbocyanine iodide
Effective length = L/N
= 1.2834/8
= 0.1604 nm
iii)1,1- diethyl-2,2-dicarbocyanine iodide
Effective length = L/N
= 1.4480/10
= 0.1448 nm
graph
Slope = I = 0.1161 nm
Y intercept = y* = 0.5887 nm
alternative way
L=0.139Nb
i)1,1- diethyl-2,2-cyanine iodide
L=0.139(6) = 0.834nm
ii)1,1- diethyl-2,2-carbocyanine iodide
L=0.139(8) = 1.112nm
iii)1,1- diethyl-2,2-dicarbocyanine iodide
L=0.139(10) = 1.39nm
Sample calculations
Calculations of Nb values
1,1′-diethyl-4,4′-cyanine iodide
Nb = P+3
Nb = 7+3 =10
1,1′-diethyl-4,4′-carbocyanine iodide
Nb = P+3
Nb = 9+3 =12
1,1′-diethyl-4,4′-dicarbocyanine iodide
Nb = P+3
Nb = 11+3 =14
L
L2=3.03×10-4×(N+1)λ
i) 1,1′-diethyl-4,4′-cyanine iodide
λ = 592 nm and N = 10
L2=3.03×10-4×(10+1)(592)
L = 1.4047 nm
ii) 1,1′-diethyl-4,4′-carbocyanine iodide
λ = 708 nm and N = 12
L2=3.03×10-4×(12+1)(708)
L = 1.6045 nm
iii) 1,1- diethyl-2,2-dicarbocyanine iodide
λ = 813 nm and N = 14
L2=3.03×10-4×(14+1)(813)
L = 1.8571 nm
Effective length
Effective length = L/N
i)1,1- diethyl-2,2-cyanine iodide
Effective length = L/N
= 1.4047/10
= 0.140470 nm
ii)1,1- diethyl-2,2-carbocyanine iodide
Effective length = L/N
= 1.6045/12
= 0.133708 nm
iii)1,1- diethyl-2,2-dicarbocyanine iodide
Effective length = L/N
= 1.8571/14
= 0.132650 nm
graph
From the graph
Slope = I = 0.1131 nm
Y intercept = y* = 0.4911 nm
alternative way
L=0.139Nb
i) 1,1′-diethyl-4,4′-cyanine iodide
L=0.139(10) = 1.39nm
ii) 1,1′-diethyl-4,4′-carbocyanine iodide
L=0.139(12) = 1.668nm
iii) 1,1′-diethyl-4,4′-dicarbocyanine iodide
L=0.139(14) = 1.946nm
Data set 1 (experiment data)
Molecule
b
1,1- diethyl-2,2-cyanine
iodide
1,1- diethyl-2,2carbocyanine iodide
Nb
λ (nm)
L(N,λ)
C-C bond
4
6
524
1.0542
0.1757
L=
0.139Nb
0.834
6
8
604
1.2834
0.1604
1.112
I (nm)
Y*
0.1161
0.5887
1,1- diethyl-2,2dicarbocyanine iodide
Data set 2
Molecule
1,1′-diethyl-4,4′-cyanine
iodide
1,1′-diethyl-4,4’carbocyanine iodide
1,1′-diethyl-4,4’dicarbocyanine iodide
8
10
692
1.5187
0.1148
1.390
b
Nb
λ (nm)
L(N,λ)
C-C bond
4
6
592
1.4047
0.1405
L=
0.139Nb
1.390
6
8
708
1.6045
0.1337
1.668
8
10
813
1.8571
0.1327
1.940
I (nm)
Y*
0.1131
0.4911
L values are not the same. Therefore, they don’t agree well
Percentage error = {(Experiment value-estimated value)/Experiment value}100%
= (1.0542-0.834)/ (1.0542) *100%
= 20.9%
But the data set 2, values of L (N, λ) and L = 0.139Nb are nearly equal. I think there could be
some experimental errors.
Chem 305 L – Spectra of Conjugated Dyes
UV-vis spectroscopy will be used to explore the electronic structure of a family of three
conjugated dyes:
1,1- diethyl-2,2-cyanine iodide
1,1- diethyl-2,2-carbocyanine iodide
1,1- diethyl-2,2-dicarbocyanine iodide
A conjugated π-electron system is well represented by the particle-in-a-box (PIB) model. In this
model an electron is free to move within a box of fixed length where the potential inside the box
is zero. At the walls of the box, the potential is infinite. The potential difference causes the
electron to stay within the box.
As you have learned previously, the addition of energy can excite an electron from the highest
occupied molecular orbital (HOMO) to the lowest unoccupied molecular orbital (LUMO). The
HOMO energy level is found by taking the number of π -electrons in the conjugated chain, N,
and dividing by 2, a reflection of the Aufbau and Pauli Exclusion Principles. Each energy level in
our system can contain only two electrons. The next highest state is the LUMO. The assumption
we make in today’s experiment is that the wavelength of maximum absorbance (λ max)
represents the π -electron transition from the HOMO (N/2) to the LUMO (N/2 +1). The difference
in energy is given by
ΔE= h2Δn2/8mL2
(eq. 1)
where Δn2 is [(N/2 +1)2 – (N/2)2], m is the mass of the electron, and L is the length of the box.
Since E =hv=hc/λ, we can obtain
L2= (N+1)hλ/8mc
(eq. 2)
Upon substituting in values for h (Planck’s constant), m (mass of electron), and c (speed of light)
and converting L and λ to nanometers, you can obtain an equation for L in terms of N and λ.
This derivation is part of your prelab.
It has been shown experimentally that for symmetric molecules, the PIB model needs to be
considered more carefully due to the effect of electron donating or withdrawing groups on either
side of the conjugated system. The potential at the walls is no longer infinite. As a result, the
length of the box appears to be shorter with electron donating groups and longer with electron
withdrawing groups. The correction parameter is α. If -1< α <1, then the PIB model is a good representation. The value of α allows us to quantify the extent to which the length of the box is extended or retracted. The parameter α should be consistent within a family of dyes, but has often times proven to be not. Calculation of another parameter will yield more consistent results while also allowing us to calculate the average bond length. L= (b x l) + γ (eq 3) In equation 3, b represents the number of carbon bonds in the chain (between nitrogens in our molecules), l is the average bond length along the chain in a series of dyes, and γ is a parameter that describes the change in the chain length due to electron wtihdrawing/donating groups. γ yields a consistent result within a series of dyes while a previous model that calculates α often did not. A plot of L (calculated from equation 2) vs b will give values of l from the slope and γ from the y-intercept. Since we have three dyes, we will have a graph with three points. Experiment: We will make stock solutions of 0.1mM dye with methanol. Instead of measuring out the tiny masses, we will simply use a few grains of dye in about 5-10mL of solvent. When we take the UV-vis, if the absorbance is beyond 2,we will dilute our samples until the absorbances are below 2. Scan the blank and samples from 350-750 nm. Pre lab: 1. Determine the structure of the 3 dyes. 2. Count the number of π -electrons in each molecule (along the carbon chain only). 3. What should we use for the blank? 4. Substitute values for h, m, and c and convert L and ƛ to nanometers in equation 2 (L2= (N+1)hλ,/8mc, to obtain an equation for L in terms of N and ƛ. Post-lab: The conjugated system in between nitrogen atoms of the cyanine dyes was analyzed. One nitrogen atom has a + charge and the other has a lone pair. The lone pair contributes to the extended π –orbital system. Consequently, the total number of π -electrons, designated the letter Nb is equal to the number of carbons in the conjugated system + 3 (or Nb= p+3). What is Nb for each of the molecules? Use the Nb value to calculate the L, length of the box, for each molecule. Use the equation derived in the pre-lab. Also, calculate the effective length of each carbon-carbon bond. Graph L vs. b to determine l and γ for the dyes. An alternate way to estimate the box length is to make the approximation that the conjugated bonds are essentially the same as 1.5 bonds in benzene, which are 0.139 nm. Estimate L by multiplying 0.139xNb for each of the molecules. Given the λmax’s for another family of dyes, do the same calculations as above for the following: 592 nm for 1,1'-diethyl-4,4'-cyanine iodide 708 nm for 1,1'-diethyl-4,4'-carbocyanine iodide 813 nm for 1,1'-diethyl-4,4'-dicarbocyanine iodide Show sample calculations, include your plots (properly labeled), and produce a table of your results. An example, which you should include in your plot, is given below for the molecule belonging to the 2nd set of dyes, 1,1'-diethyl-4,4'-tricarbocyanine iodide: L= 0.139Nb molecule b Nb λ (nm) L (N, λ ) c-c bond 1,1'-diethyl-4,4'tricarbocyanine iodide 14 16 929 2.19 0.136875 2.22 l (nm)* γ* * one value for series of dyes Compare the values of L (N, λ) and L = 0.139Nb. Comment on the agreement/pattern. Do they agree well or not? Integrated Writing Guide Abstract The rate of hydrolysis of sucrose by beta-fructofuranosidase from Saccharomyces cerevisiae was measured by polarimetry. Experiments were performed at 300 K, 310 K, and 320 K. The enzyme catalyzed activation energy was 29.2 ± 0.2 kJ/mol. The rate constant at 300 K was found to be 0.065 ± 0.001 min-1. The activation energy was comparable to the activation energy for betafructofuranosidases isolated from plant sources. The abstract should be a concise (short) and specific summary of the report that allows readers to decide whether they want to read the report. It should include purpose, methods, scope, results, and conclusions. A technical document is not a mystery novel. Give a very brief version of your conclusions right away and support them later. For more information, see the ACS Style Guide1, pp. 21-22. • purpose • method • numerical results with error limits and correct units • conclusion • clear • concise • no major conceptual errors • no grammar errors 1 The ACS Style Guide: Effective communication of scientific information, 3rd ed.; Coghill, A.M, Garson, L.R., Eds.; Oxford University Press: New York, 2006. 1 Introduction Plants and yeasts use -Dfructofuranosidases (EC 3.2.1.26) to catalyze the hydrolysis of the disaccharide sucrose into the two monosaccharides, fructose and glucose. Because sucrose rotates planepolarized light to the right, while an equimolar solution of glucose and fructose rotates planepolarized light to the left, -fructofuranosidase is commonly called invertase. Similarly, the mixture of glucose and fructose is called invert sugar. Invertase is used in the confectionary industry, because invert sugar is less prone than sucrose to form grainy crystals.1 Invertases can be isolated from plants2 and from yeasts.3 The kinetics of these enzymes under various conditions are of interest because they can be so used.4 Invertase activity in fruits and vegetables contributes to spoilage in stored food.4 The rate of reaction for the invertasecatalyzed hydrolysis of sucrose can be measured by following changes in optical rotation of an aqueous solution of sucrose and invertase. When sucrose is hydrolyzed to a mixture of glucose and fructose, the optical rotation changes from clockwise (+) to counter clockwise (-). The introduction must accomplish two objectives: it must give the purpose of the report and acquaint the reader with the experiment. This should set the background and the context of your experiment. Primarily, you will be trying to explain what you were trying to do and why it is significant. Describe what others have done in this area and cite the relevant references. Ask your instructor if you need help searching for literature articles. The introduction should also provide whatever background theory or formulas the reader needs to know to understand your paper. You may have to define the terms used in stating the subject and provide background such as theory or history of the subject. Much of this information will be in the lab manual, but the instructor will usually expect you to show your own comprehension of the problem by describing it in your own words. For more information, see the ACS Style Guide, pp. 22-23. • background • at least 4 literature references to prior work • equations • significance of topic • purpose • clear • no major conceptual errors • no grammar errors 2 The optical rotation is monitored with an optical polarimeter and then used to calculate the amount of sucrose left unhydrolyzed. The angle of rotation is determined at HO HO O O HO OH O HO HO OH OH invertase HO O HO OH HO OH O HO + OH HO OH Figure 1. Invertase catalyzes the hydrolysis of sucrose into glucose and fructose. Sucrose has a positive specific rotation ([ ]D = 66.5°), while the equimolar mixture of glucose and fructose that results from hydrolysis has a negative specific rotation ([ ]D = -22.0°) . Any figures used should be labeled with a reference number and a caption. The captions should consist of complete sentences, and give the reader enough information to understand the figure without going back to the main body of the text. Figures should be neatly drawn and centered on the page. If you need to draw chemical structures your lab instructor will help you find the necessary software. For more information, see the ACS Style Guide, p. 365 and pp. 375-383. • Captions in complete sentences • Numbered sequentially • Neat, centered the beginning of the experiment (a0) and at equilibrium (aeq). The algebraic difference (a0 - aeq ) is a measure of the original sucrose concentration. The concentration of water during the reaction remains essentially constant, since it is present in large excess. The reaction is known to be first order in All equations should be given a reference sucrose, so the concentration of sucrose as a number, so that you may refer to them later function of time follows the relationship in the text. Center equations, and format them neatly. Introduce all variables in the text the first time you use them in an C (1) ln t = kt equation, so that the reader can interpret the C0 equation correctly. If you need help editing equations, ask your lab instructor for help. where C0 is the initial concentration of sucrose, For more information, see the ACS Style Ct is the sucrose concentration at time t after Guide, pp. 218-222. the addition of invertase, and k is the first order • Numbered sequentially rate constant for the reaction. Since Ct is • Properly formatted proportional to (at - aeq) and C0 is proportional • All variables introduced in text to (a0 -aeq), where at is defined as the rotation • Grammar angle at time t, equation 1 can be re-written as follows: 3 at aeq = kt ln a a 0 eq (2) This implies that the rate constant k can be determined from a plot of ln (at aeq ) versus time. The rate constant k depends on the absolute temperature T according to the Arrhenius Equation: k = Ae Ea RT (3) where Ea is the activation energy. Thus if the rate constant is determined at several different temperatures, the activation energy can be determined via a modification of equation (3) by plotting ln(k) versus 1/T. ln k = ln A Ea RT (4) Once the plot is constructed the slope can be used to calculate the activation energy as follows. Ea = -R•slope (5) The data treatment consists of determining the rate constants for the invertase-catalyzed hydrolysis of sucrose at three different temperatures followed by determining the activation energy from an Arrhenius plot (ln(k) versus 1/T). 4 Methods and Materials Sucrose (Sigma-Aldrich, #47289) and invertase (from baker’s yeast, Sigma-Aldrich, #I9274) were used as received and all solutions were made with distilled water. Optical rotation was measured at a wavelength of 589 nm using a digital polarimeter (Jasco, DIP-360) and a quartz cell with a 10 cm path length. The concentration of the sucrose solution used in each experiment was 50 g/L. The invertase solution was prepared in an acetate buffer with a pH of 5.0 at a concentration of 0.04 g/L. Temperature was controlled with a refrigerated circulating water bath (VWR, 1140) connected to the polarimeter cell. All solutions were placed in the temperature bath prior to mixing for thermal equilibration. Experiments were conducted at 300K, 310K, and 320K using the same procedure. First the polarimeter cell was filled with sucrose solution (~10 mL) and the optical rotation was measured producing a0. Next, 5 mL of sucrose solution was removed and 5 mL of the invertase solution was added. After the addition of invertase the optical rotation was measured every 5 minutes until the measured value approached the equilibrium value (aeq). The equilibrium optical rotation (aeq) was measured at t = 60 minutes for each temperature. Start with a description of the chemicals you used. Include the source, grade, and method you used to purify them. If you used chemicals as supplied without further purification, say so. A diagram of instrumentation used in the experiment can be helpful. Be complete, accurate, and precise. Do not give details that are common knowledge in the field, but do provide information of particular interest, such as the brand name and model or a complicated apparatus or unusual equipment (for example, Oscilloscope – Tectronix Model 561B-CRO158, Serial #123456789). For the experimental procedure, you should use clear paragraph organization to list all steps in the correct order. Be complete. You should provide enough information so that another researcher in your field could use your description to replicate the experiment. State what you really did and what actually happened, not what was supposed to happen or what the textbook said. If you deviated from the given procedure, describe the changes you made and explain how the affected your outcome. For more information, see the ACS Style Guide, pp. 22-23. • source and grade of chemicals • chemical purification/sample prep • make and model of instrument(s) • diagram of apparatus if needed • complete • clear • no grammar errors • past tense narrative format (not a list) 5 Results Figures 2 through 4 show the change in ln(at - aeq) as a function of time for 300K, 310K, and 320K respectively. As described above by Equation 2, the slope of the graph of ln(at -aeq) versus time will be equal to the negative of the rate constant. The rate constants for each temperature are indicated in each figure and will be used to determine the activation energy for the catalyzed reaction via an Arrhenius Plot. Errors in the rate constants were determined by a linear regression using Excel. The linearity of each graph is quite good, R2 values greater than 0.99 in each case, which confirms the reaction is first order with respect to the sucrose concentration. Further, as expected the rate constants increase with increasing temperature. State your actual, not expected, results. Although results are usually presented quantitatively, you should always introduce each block of information with simple, clear language. Do not rely upon figures, graphs and tables exclusively to convey essential information. Merely supplying the equations or diagrams and expecting the reader to interpret them without guidance from you is not sufficient. You should describe all significant results in words, clearly and simply. Refer to the raw data only to point out trends and identify special features. For more information, see the ACS Style Guide, pp. 23. • • All results presented Results explained On all graphs, label both axes and include the units of any physical quantity. If you have fit a curve to the data, state the equation and the values for the parameters in the caption. Also, scale the axes so the data occupies the entire graph; don’t crowd the data into one corner or side of the graph. For more information, see the ACS Style Guide, pp. 344-360. • • • axis labeled and units regression equation(s) Scaling Figure 2: The optical rotation of sucrose is graphed versus time for hydrolysis at 300K. The data were f ... Purchase answer to see full attachment