Tasks in Chemistry - Assignment Example

Section A - 35 marks in total Use the data you obtained in the Gel Permeation practical in Term to produce a properly labelled chromatogram for the separation of Blue Dextran from Cobalt Chloride (5 marks) The y-axis represents the absorbance, while the x-axis represents the concentration. 2. Using the HPLC simulation software currently available in the practical folder on Blackboard find the optimum parameters to effect separation of the following mixture of compounds using the Agilent Zorbax column as the stationary phase:acetophenone, p-chlorophenol, benzonitrile, p-xlylene, p-dichlorobenzene, ethylbenzene, bromobenzene, toluene and n-butylbenzene. Once you have identified the correct parameters for separation of this mixture, take a print screen shot of the chromatogram you have produced using the simulation software and paste it to your answer sheet. As part of your answer you also need to list the parameters that gave rise to the separation and also clearly map each of the peaks to the compound it represents from the list above. Factors that lead to separation Mobility of the samples in the stationary phase Solubility of mobile phase Affinity for stationary phase (Note – you can hand write the names of the compounds for each peak, or you can copy paste the print screen to powerpoint and type the names in that way if you prefer) (10 marks) 3. From the right hand side the peaks have been arranged in the following order; acetophenone, p-chlorophenol, benzonitrile, p-xlylene, p-dichlorobenzene, ethylbenzene, bromobenzene, toluene and n-butylbenzene. 4. Discuss each of the chromatograms you have produce in parts 1 and 2 above. Your discussions should make reference/attempt to define by application all of the following: resolution of peaks, band broadening, mechanism of separation for each technique, order of elution, mobile phase properties, stationary phase properties, parameters that gave rise to the chromatograms presented and the effect they had on separation with particular reference to the challenges that presented themselves when trying to establish the optimum parameters. (A maximum word limit of 400 words for part 3 of Section A) (20 marks) PART 1 The chromatogram represents the absorption of different compounds within the mobile phase. Therefore because, they have different absorption co-efficients they will be absorbed at different concentrations. PART 2 The first two peaks of acetophenone and p-chlorophenol have high resolution because they have very short elution time. This is because their mobile phase have low adsorption coefficient as compared to the other compounds. While the rest of the compounds, have high elution time. Some of the peaks have broad bands because of the path, they took to be eluded. It has been found out that some compounds accidentally takes shorter routes in the column, therefore they have fine thin bands. Sample separation depends on the affinity and solubility of the mobile phase on the stationary phase. The samples have been eluded in the order shown above because, acetophenone and p-chlorophenol have low solubility in the stationary phase, therefore they pass the stationary phase very fast. On the other hand bromobenzene, toluene and n-butylbenzene have long retention time because they have high affinity and solubility. Section B – 40 marks in total 1. The following three tables contain data collected over two days from a HPLC analysis of caffeine in blood serum samples. From the data it is quite clear that something has gone wrong during the analysis on one of those days. Use the data presented in the following three tables to answer the questions below: Table 1 - Calibration Data for Day 1 Table 2 - Calibration Data for Day 2 Conc Caffeine Caffeine Internal Standard Conc Caffeine Caffeine Internal Standard g/ml Peak Area Peak Area g/ml Peak Area Peak Area 1 16.1 14.6 1 15.9 14.5 1 16.2 14.7 1 15.9 14.4 1 16.2 14.6 1 15.4 13.9 1 16.2 14.6 1 12.0 10.8 1 16.2 14.8 1 4.1 3.7 2 32.3 14.6 2 32.0 14.5 2 32.4 14.5 2 32.7 14.6 2 32.3 14.5 2 27.8 12.5 2 32.3 14.5 2 33.3 14.9 2 32 14.5 2 31.4 14.2 4 64.1 14.7 4 57.7 13.2 4 64 14.5 4 54.4 12.3 4 63.9 14.5 4 47.9 10.9 4 63.9 14.4 4 45.4 10.2 4 63.9 14.7 4 60.7 14.0 6 94.9 14.6 6 56.9 8.8 6 94.9 14.6 6 52.2 8.0 6 94.7 14.4 6 66.3 10.1 6 95.1 14.6 6 42.8 6.6 6 95 14.5 6 47.5 7.3 8 129 14.5 8 38.7 4.4 8 128.6 14.6 8 36.0 4.1 8 128.7 14.4 8 51.5 5.8 8 128.3 14.7 8 48.8 5.6 8 127.7 14.5 8 44.7 5.1 10 161.3 14.6 10 161.3 14.6 10 160.9 14.6 10 159.3 14.5 10 161.1 14.8 10 157.9 14.5 10 160.8 14.6 10 159.2 14.5 10 159.9 14.7 10 161.5 14.8 Table 3 - Sample Data for Day 1   Internal Standard Peak Area Caffeine Peak Area Vol of Sample Sample Run 1 Run 2 Run 3 Run 1 Run 2 Run 3 ml 1 14.9 14.1 13.6 90.7 90.1 89.8 1.3 2 14 13.6 14.1 27.7 28 28.3 1.1 3 13.9 14.3 14.1 53.7 53.1 54.3 1.3 (a) Calculate, from the data in Tables 1 and 2, the average peak areas for caffeine and the internal standard at each concentration point for each day. Present your results in separate tables. Table 1 Conc. Caffeine g/ml Caffeine average peak area Internal standard Average peak area 1 16.8 14.66 2 32.26 14.66 4 63.96 14.62 6 94.92 14.6 8 128.46 14.58 10 160.8 14.52 Table 2 Conc. Caffeine g/ml Caffeine average peak area Internal standard Average peak area 1 12.66 11.46 2 31.44 14.14 4 53.22 12.12 6 40.92 8.16 8 52.62 5 10 159.84 14.58 (b) Using the calculated average values from (a) above, produce a Normal Calibration Curve for each days data and comment on any difference between the graphs. Day 1 Day 2 Comment Day two samples seems to have been contaminated, or there was a problem with the machine. (c) Using the calculated average values from (a) above, produce an Internal Standard Calibration Curve for each days data and comment on any difference between the graphs. Day 1 Day 2 Comment On day 2, it seems the internal standards were contaminated. (d) The error in the analysis on Day 2 can be clearly seen in one of the Calibration Curves for that day. Explain why we do not see this same effect in the other Calibration Curve for that day. This is because standards are very sensitive, therefore needs much precaution when handling. (e) Calculate the absolute and relative error for each of the data points collected on Day 2, using the corresponding data pointsfrom Day 1 as the accepted true value. Present your answers in a table. Day 1 The slope for caffeine average peak is 16.05 while for internal standard is 0.74. Day 2 The slope for caffeine average is 12.19 while for internal standard is 0.25 Absolute error for caffeine average is 3.86, while for internal standard is 0.49 Relative errors Caffeine = 24%, while for 66% (f) State, giving a reason, whether you believe the error to be systematic or random in nature. The error is random because the deviation of day 2 values are not systematic. (g) Use the Internal Standard Calibration Method to calculate the concentration of caffeine in each of the three samples. Data for the samplesare presented in Table 3 and note that each sample has been analysed in triplicate. Use average values for this data in your calculations. (all sections of parts 1a-1g carry 5 marks each) Table 3 Internal standard peak Caffeine area Vol. (ml) 15.6 89.97 1.3 13.9 28 1.1 14.1 53.7 1.3 From the graph, the first sample concentration will be 8 g/ml, the second sample will be 5.3 g/ml and finally the last sample will be 2.4 g/ml 2. The following graph shows a Standard Addition Calibration Curve for the analysis of an sample known to contain iron using the method of Flame Atomic Absorption Spectroscopy. Using the information provided in this graph calculate the effective concentration of iron in the sample that was analysed. (5 marks) The effective concentration is √0.9997= 0.002 x +0.1008 X= 449.5 ppm Section C – 25 marks in total 1. Nuclear Magnetic Resonance Spectroscopy (NMR), Infrared Spectroscopy (IR) and Mass Spectrometry (MS) are methods of analysis that provide useful information to allow an analyst to determine the identity of a given compound. Briefly discuss the interpretation of the spectra produced by all three methods to enable the identification of compounds. (A maximum of 200 words for the discussion of each method is allowed) (15 marks) NMR technique uses chemical shifts of carbon 13 and 1 hydrogen to identify compounds. Therefore in interpreting the spectra it is necessary that the analyst have the knowledge of chemical shifts of different compounds Mass spectrometry uses fragments for identification of the compounds. It is therefore recommended that an analyst should have a table that has different weights of different elements. Infra-red uses absorption of different elements for qualitative analysis. This is because every element has different absorption co-efficient. 2. Identify the compound in question from the spectra given below. You must identify and explain one piece of evidence obtained from each spectrum that enabled you to identify the compound. MW for the compound is 46 g/mole (10 marks) IR Spectrum The compound is CH3CH2CH2CH2CH2OH. This is because peaks in the spectrum corresponds the compound. 1H NMR Spectrum (high resolution spectrum with additional information from the low resolution spectrum) The compound is CH3CH2OH. This is because the first peak is a quartet. Therefore it means that the hydrogen atom is surrounded by four atoms. Therefore, the hydrogen atom is (CH2) While the rest of the other peaks corresponds to CH3 and OH. 13C NMR The compound is CH3C0CH2 CH3, this is because the first peak corresponds to CH3, while the second peak corresponds to CH2. Mass Spectrum The compound is CH2=CHCH2CH2COCH2NOCH2CHCH3. This is because, the weight fragments for the elements correspond to m/z values of the spectra. Read More