CHY203 - Gas Chromatography lab report

Department of Chemical Engineering Gas Chromatography Student name: Student number: Group 2 CHY203, Section 02 March 20, 2013 Instructor: David Naranjit TA: Saif Report Appendix Form for Gas Chromatography Calculations: Unknown Identity Sample calculations of the log (tR/min) from the Run Temperature Programmed Graph of C3 solution for peaks #4: Log (tR/min) = log (5.133) = 0.7104 Table1. Calculation of the log retention time for C3 solutions Peaks # tR/min Log (tR/min) 2 3.779 0.5774 3 4.485 0.6518 4 5.133 0.7104 The unknown identity had been determined by comparing the retention times of the compound C3 with the chromatogram of solution B as shown below: The tR of the unknown from graph of solution C3 = 4.485 Log (tR/min) of unknown = log (4.485) = 0.6518 y = 0.0333x + 0.3114 y represents the log of the unknown retention time x= 10.2 The number of carbons for the unknown at 0.6518 in graph 1 below of C3 solution is 10 carbons The unknown identity in solution is decane C10H22 Graph1. The Temperature Programmed Run in Solutions C3 of log (tR/min) vs. number of carbons in n-alkanes Calculation of the dodecane to octane mass and area ratio in solution E and F Table2. Showing calculation of the solutions E and F mass and area ratio Standard solution E Standard Solution F density of dodecane in g/mL 0.715 0.715 density of octane in g/mL 0.704 0.704 Volume of dodecane in mL 25 x 10-8 75 x 10-8 Volume of octane in mL 5 x 10-7 5 x 10-7 Mass of dodecane in g 1.78 x 10-7 5.36 x 10-7 Mass of octane in g 3.52 x 10-7 3.52 x10-7 Area of dodecane in countss 2.67554 x 104 1.19972 x105 Area of octane in countss 3.09127 x 104 7.19178 x 104 Area ratio 0.8655 1.6682 Mass ratio 0.5057 1.5227 For Standard Solution E: The area ratio for the dodecnce to octane was obtained from the temperature programme graph for E solution as shown below: Area of dodecane at 5.714 min = 2.67554e4 counts. s Area of octane at 2.998 min = 3.09127e4 counts .s 78105016192500Area Ratio = E dodecane E octane = (2.67554 x 104 counts. s) / (3.09127 x 104 counts .s) = 0.8655 The density of the octane C8H18 = 0.704 g/mL The density of the dodecane C12H26 = 0.715 g/mL = m/V The volume that was used in determine the separation of the solution was 0.5L Conversion of the volume: 6667503048000180974186690V= 0.5L 0.001mL 1L = 5 x 10-4 mL The standard volume of octane that used in the solution was 1.0 L/mL solution while the volume used for dodecane was 0.5L/ mL solution Therefore, the volume of octane and dodecane was computed as shown below: V C8H18= (1.0L/mL solution) x (5 x 10-4 mL solution injected) 1L= 0.001 mL = 0.001 mL * 5 x 10-4 mL = 5 x 10-7 mL V C12H26 = (0.5 L/mL solution) x (5 x 10-4 mL solution injected) = 0.0005 mL * 5 x 10-4 mL = 25 x 10-8 mL m C8H18= x V = 0.704 g/mL * 5 x 10-7 mL = 0.000000352 g = 3.52 x 10-7 g m C12H26 = x V = 0.715 g/mL * 25 x 10-8 mL = 0.000000178 g = 1.78 x 10-7 g Mass Ratio = E dodecane 10287001905000 E octane = 1.78 x 10-7 g / 3.52 x 10-7 g = 0.5057 For Standard Solution F: The area ratio of dodecane to octane for solution F was obtained from the graph of Temperature programmed run in F. The area of the dodecane at 5.179 min = 1.19972e5 counts .s The area of the octane at 3.014 min = 7.19178e4 counts.s 49530016573500Ratio = F dodecane Area F octane Area = (1.19972 x 105 counts .s) / (7.19178 x 104 counts.s) = 1.6682 V C8H18 = (0.001 mL) * (5 x 10-4 mL solution injected) = 5 x 10-7 mL V C12H26 = (0.0015 mL) * (5 x 10-4 mL solution injected) = 75 x 10-8 mL m C8H18 = x V = (0.704 g/mL) * (5 x 10-7 mL) = 3.52 x 10-7 g m C12H26 = x V = (0.715 g/mL) * (75 x 10-8 mL) = 5.36 x10-7 g Mass Ratio = F dodecane 10287001905000 F octane = 5.36 x10-7 g / 3.52 x 10-7 g = 1.5227 Graph2. The Area Ratio vs. Mass Ratio of Dodecane/Octane for Standard Solution E and F Determining the concentration of Dodecane in Solution D 96202516764000Unknown #: D422 The area ratio of the D+G was obtained from the temperature programmed graph of D422 as shown below: The area of dodecane at retention time 5.119 min = 7.47353e4 counts.s The area of octane at retention time 3.769 min = 7.08564e4 counts.s Area ratio = D+G dodecane area 1104900254000 D+G octane area = (7.47353 x 104 counts.s) / (7.08564 x 104 counts.s) = 1.7547 The standard volume of octane that used in the solution D+ G was 1.0 L/mL solution The volume of D of 100L mixed with G of 100 (2L octane/mL) Therefore, the volume of octane was computed as shown below: In D solution: V C8H18= (1.0L/mL solution) x (100 L solution mixed) = 0.001 mL * 0.1 mL = 10 x 10-5 mL In G solution: V C12H26= (2.0L/mL solution) x (100 L solution mixed) = 0.002 mL * 0.1 mL = 20 x 10-5 mL m C8H18 = x V = 0.704 g/mL *(20 x 10-5 + 10 x 10-5) mL = 0.00021 g In order to get the mass ratio of the solution, the area ratio is substituted in to the equation of line of graph2 as shown below: y = 0.4167x + 1.3075 Where y is the area ratio x is the mass ratio of the solution 1.7547 = 0.4167x + 1.3075 x = 1.073 914400163195Mass ratio = mass of dodecane in g mass of octane in g m C12H26 = mass ratio x mass of C8H18 in g = 1.073 * 0.00021 g = 2.3 x 10-4 g The concentration of dodecane in solution D was computed as shown below: C C12H26 = m/ = 2.3 x 10-4 g * 0.715 g/mL = 1.6 x 10-4 mL = 0.016 L/mL solutions Dilution factor = total volume (D+G) / volume of D = (20 x 10-5 + 10 x 10-5 mL)/ 10 x 10-5 mL = 3 C= 3 x 0.016 L/mL solutions = 0.048 L/mL solutions Discussion Questions: Compare the two chromatograms from part A. Describe the differences in appearance between the two. Isothermal and temperature programmed are the two chromatograms methods that used to identify a compound by separating the chemical components in gas chromatography device. For isothermal methods, the retention times between the octane and dodecane peaks are higher from that in temperature programs. For example, in temperature programs the retention time for peaks 2 and 3 were 5.124 and 3.773 min respectively. Whereas the retention time for the same solution B but using the isothermal methods, it were 5.232 and 2.191 min for peaks 2 and 3 respectively which showed that the separation of the components takes more time in isothermal. Another thing that can be compare is the peak shapes, by looking and comparing the two graphs from the two programs of solution B it showed that in temperature programs the peaks are sharp and have equal distance comparing with that of isothermal temperature which have wide shape peaks and a large distance between the peaks. For instance, the width between octane and dodecane of solution B in isothermal methods were found to be 0.0175 and 0.0477 min respectively. While the width of the octane and dodecane in temperature programmed were 0.0142 and 0.0140 min respectively. Which method, isothermal or temperature programmed, completely elutes all the components first? Why might this be important? Temperature programmed completely elutes all the components first since it runs at a higher temperature for temperature ramp range (40 to 200oC). Thus, the sample kinetic energy increases as the temperature increase which causes the sample analyte species to move fast through the gas column and hence a short retention time the analyte will take to be completely separate. While isothermal methods, the sample elute from the gas column at a constant temperature of 90oC, the analyte component will take a more retention time to be separated and hence obtaining more accurate result of the species concentration .As a result, temperature programs has a good and fast (in a short retention time) resolution for completely separation a complex chemical components comparing to the isothermal method. Component separation in chromatography is quantified by calculating the column resolution, using the formula 27813001822450Rs = 2 x | tR, A – t R, B| WA + WB Where tR,A and tR,B are the same retention times for components A and B, and WA and WB are the widths of the peaks for A and B, measured at the baseline. Which method producers better separation of the nonane and undecane? Show your calculations. For the temperature- programmed of solution B Let’s A represents the peaks for dodecane B represents the peaks for octane tR, A = 5.124 min tR,B =3.773 min WA = 0.0140 min WB = 0.0142 min Rs = [2* / 5.124- 3.773/ min] / (0.0140 +0.0142) min = 95.8 min For Isothermal sample B iso solution tR, A = 5.232 min tR, B =2.191 min WA = 0.0477 min WB = 0.0175 min Rs = [2* / 5.232- 2.191/ min] / (0.0477 +0.0175) min = 93.3 min By looking at the calculations of the retention times of isothermal and temperature programmed method that obtained. One can conclude that the temperature programmed produce has a better separation of the nonane and undecane since it has a higher retention time comparing with the retention time obtained in isothermal method. Determine the change in column efficiency for the two methods, using the formula 312420023558500N = 5.54 [ tR ]2 W1/2 Where N= number of theoretical plates, tR = retention time of the last peak, and W1/2 = the full width at half height for the last peak in the chromatogram. Which method, isothermal or temperature programmed, results in better column efficiency? Show your calculations. For isothermal elution B: tR = 5.232 min W1/2 = 0.0477 min 5238751663700 N= 5.54 [5.232]2 0.0477 = 3179.3 For temperature programmed run of B solutions: tR = 5.124 min W1/2 = 0.0142 min 5238751663700 N= 5.54 [5.124]2 0.0142 = 10243.3 Based on the calculations for the two methods isothermal and temperatures programmed shown above , the temperature programmed result in better column efficiency since it had higher number of theoretical plates comparing with that of isothermal. Graphs of different solutions of n-alkanes Temperature Programmed Graph for B Solutions Graph of Isothermal Method for B Solutions lefttop Temperature Programmed Graph for C Solutions Temperature Programmed Graph for F Solutions Temperature Programmed Graph for E Solutions Temperature Programmed Graph for Unknown Mixture D+G References: Jim Clark (author) Gas-Liquid Chromatography, Academic Press, 2007 http://www.chemguide.co.uk/analysis/chromatography/gas.html The 92st edition of the CRC Handbook of Chemistry and Physics, 2011-2012 http://www.hbcpnetbase.com/ Instrumental Methods of Analysis Laboratory Manual, Gas Chromatography, Ryerson University 2013, page 52-54 Harris, D.J. ( author) Gas Chromatography, Analytical Chemistry, W.H Freeman NJ,2013, http://en.wikipedia.org/wiki/Gas_chromatography G.A, Gas Chromatograph-Mass Spectrometry, Analytical Chemistry,2013 http://en.wikipedia.org/wiki/Gas_chromatography%E2%80%93mass_spectrometry