Gas Chromatography – Acetates

1. Gas Chromatography – Acetates • Gas Chromatography, Refractive Index & Distillation • The next two (2) experiments introduce Gas Chromatography and Simple & Fractional Distillation. • They are then tied together along with the Refractive Index technique in a third experiment. • This Week • Gas Chromatography – Acetates • Pavia – p. 817 – 836 • Slayden – p. 39 - 31 • 2nd Week • Distillation of a Mixture • Slayden – p. 43 - 46 • 3rd Week • Gas Chromatography and Refractive Index of Distillates from “Distillation of Mixture” Experiment • Slayden –p. 47

2. Gas Chromatography – Acetates • Gas Chromatography • Uses • Separation and analysis of organic compounds • Testing purity of compounds • Determine relative amounts of components in mixture • Compound identification • Isolation of pure compounds (microscale work) • Similar to column chromatography, but differs in 3 ways: • Partitioning process carried out between Moving Gas Phase and Stationary Liquid Phase • Temperature of gas can be controlled • Concentration of compound in gas phase is a function of the vapor pressure only. • GC also known as Vapor-Phase Chromatography (VPC) and Gas-Liquid Partition Chromatography (GLPC)

3. Gas Chromatography – Acetates • Gas Chromatograph • Microliter Syringe • Heated injection port with rubber septum for inserting sample • Heating chamber with carrier gas injection port • Oven containing copper, stainless steel, or glass column • Column packed with the Stationary Liquid Phase, a non-volatile liquid, wax, or low melting solid-high boiling hydrocarbons, silicone oils, waxes or polymeric esters, ethers, and amides. We use DC200 from Dow Chemical • Liquid phase is coated onto a support material, generally crushed firebrick

4. Gas Chromatography – Acetates • Principals of Separation • Column is selected, packed with Liquid Phase, and installed • Sample injected with microliter syringe into the injection port where it is vaporized and mixed into the Carrier Gas stream (helium, nitrogen, argon) • Sample vapor becomes partitioned between Moving Gas Phase and Stationary Liquid Phase • The time the different compounds in the sample spend in the Vapor Phase is a function of their Vapor Pressure • The more volatile (Low Boiling Point / Higher Vapor Pressure) compounds arrive at the end of the column first and pass into the detector

5. Gas Chromatography – Acetates • Principals of Detection • Two Detector Types • Thermal Conductivity Detector (TCD) (we use this) • Flame Ionization • TCD is electrically heated “Hot Wire” placed in carrier gas stream • Thermal conductivity of carrier gas (helium in our case) is higher than most organic substances • Presence of sample compounds in gas stream reduces thermal conductivity of stream • Wire heats up and resistance decreases • Two detectors used: one exposed to sample gas and the other exposed to reference flow of carrier gas • Detectors form arms of Wheatstone Bridge, which becomes unbalanced by sample gas • Unbalanced bridge generates electrical signal, which is amplified and sent to recorder

6. Gas Chromatography – Acetates • Factors Affecting Separation • Boiling Points of Components in Sample • Low boiling point compounds have higher vapor pressures • High boiling point compounds have lower vapor pressures requiring more energy to reach equilibrium vapor pressure, i.e., atmospheric pressure • Boiling point increases as molecular weight increases • Flow Rate of Carrier Gas • Choice of Liquid Phase • Molecular weights, functional groups, and polarities of component molecules are factors in selecting liquid phase • Length of Column • Similar compounds require longer columns than dissimilar compounds. Isomeric mixtures often require quite long columns

7. Gas Chromatography – Acetates • The Experiment • Purpose – Introduce the theory and technique of gas chromatography Identify a compound by it retention time From the relationship between peak area and mole content calculate the mole fraction and mole percent of a compound in a mixture • Approach • Obtain chromatograph of a known equimolar mixture of four (4) esters - Ethyl, Propyl, Butyl, Hexyl Acetate • Obtain chromatograph of unknown mixture (one or more compounds in the known mixture) • Determine Retention Times • Calculate Peak Areas • Adjust Peak Areas for Thermal Response • Calculate Total Area from Adjusted Areas • Calculate Mole Fraction • Calculate Mole Percentage

8. Gas Chromatography – Acetates • The Experiment (Con’t) • Groups – Work in groups of three (2) • Each group will obtain 2 copies of the chromatogram for the standard (equimolar) mixture • Each Student will run their own unknown • Samples • The Standard (Equimolar) Mixture has 4 esters: Ethyl Acetate, Propyl Acetate, Butyl Acetate, Hexyl Acetate • The Unknowns have from 2 to 4 of the compounds in the standard mixture

9. Gas Chromatography – Acetates • The Report • The Gas Chromatograph instrument settings and the processing of the samples to get the chromatograms are considered one (1) procedure • When multiple samples or sub-samples are processed with the same procedure, it is not necessary to set up a separate procedure for each sample • Setup a suitable template in “Results” section to report all of the results obtained • Thus, the process to obtain Gas Chromatograms of the “Known” mixture of 4 acetates and the “Unknown” mixture utilize the same procedure • The computation of the Peak Area, Adjusted Peak Areas, Total Peak area, Mole Fraction, and Mole % are considered “separate” procedures

10. Gas Chromatography – Acetates • Data Summary Procedure – Using complete sentences summarize, in paragraph form, all of the results obtained in the experiment • Analysis & Conclusion Section • Develop a set of arguments to prove the identity of the unknown compounds in the unknown mixture • Comment on the equivalency of the peak areas and equimolar content of the known mixture • Why was it necessary to apply the Thermal Response Correction Factor to the measured peak areas? • Chromatograms • Copied chromatogram sets for each team member must be copied at the same scale, otherwise retention time computations will be wrong • Tape the trimmed chromatograms to a blank sheet of paper and attach to end of report

11. Gas Chromatography – Acetates • Record Instrument readings (Place in GC procedure “Results”) • Injection Port Temp • Column Temp • Detector Temp • Gas Flow Rate – 60 mL / min • Chart Speed – Generally 5 cm /min • Moving Liquid Phase – (DC-200) • Injecting the Sample • Sample is injected into the “B” port with the Microsyringe • The Microsyringe is fragile and expensive – BE CAREFUL • Mark “Starting Point” on chart – short vertical line • Insert needle fully into “B” Port through the rubber septum • Coordinating with chart recorder operator, inject the sample into the heated chamber, while simultaneously starting the chart recorder For the instrument in the 407 lab, all three temperatures are read from the single dial on the front of cabinet

12. Gas Chromatography – Acetates • Determine the Retention Time • The period following injection that is required for a compound to pass through the column to the point where the detector current is maximum, i.e. maximum pen deflection or maximum peak height • For a given set of constant conditions (carrier gas, flow rate of carrier gas, column temperature, column length, liquid phase, injection port temperature), the retention time of any compound is always constant • Retention Time is similar to the “Retardation Factor, Rf” in Thin Layer Chromatography • Compute Retention Time from the Chart Speed (5 or 10 cm/min) and the distance on the chart from the time of injection to the point on the chart where the perpendicular line drawn from the peak height intersects the base line

13. Gas Chromatography – Acetates • Determination of Retention Time • Since Velocity (v) = Distance / Time = d / t • Ret Time (t) = Distance(cm) / Velocity(cm/min) = d / v Note: Disregard “Air Peaks” in all calculations • Retention Time Distances • Mark Starting Point On Chart (t= 0) • Draw vertical Line from Peak Top to Base Line • Measure Distance from Starting Point to Base of Peak Distance

14. Gas Chromatography – Acetates • Quantitative Analysis • The area under a gas chromatograph peak is proportional to the amount (moles) of the compounds eluted • The molar percentage composition of a mixture can be approximated by comparing the relative areas of the peaks in the chromatogram • This approach assumes that the detector is equally sensitive to all compounds and its response is linear • This assumption is usually not valid and will be addressed by adjusting the peak areas using the Thermal Response algorithm described on slides 17-24

15. Gas Chromatography – Acetates • Triangulation Method of Determining Area Under Peak • Multiply the height of peak (in mm) above the baseline* by the width of the peak at half the height. • Baseline is a straight line connecting side arms of the peak. Best if peaks are symmetrical. • Add the individual areas to get the total area • Divide each area by total area to get mole fraction • Multiply mole fraction by 100 to get adjusted mole % • See algorithm development on next slide • Adjust the peak areas for non-linear thermal response using the algorithm described in slides 17-28

16. Gas Chromatography – Acetates • Draw Baseline connecting peak bottoms • Peak Area by the Triangulation Method Peak Area = h * w½ Where h = Peak Height from baseline w½ = width of peak at ½ the peak height • Adjust Peak Area for thermal response See discussion on following slides • Total Adjusted Peak Area (TA) = A + B • Mole Fraction (MF) A/TA B/TA • Mole Percent = MF x 100 Baseline Baseline

17. Gas Chromatography – Acetates • Thermal Response Factor • The areas of gas chromatogram peaks are proportional to the molar content of the mixture • Compounds with different functional groups or widely varying molecular weights do not all have the same thermal conductivity. This can cause the instrument to produce response variations, which cause deviations (non-linearity) in the relationship between peak area and molar content • A correction factor called “The Thermal Response Factor” for a given compound can be established from the relative peak areas of an equimolar solution • Equimolar mixtures contain compounds with the same molar content, i.e., the same number of moles • Thus, equimolar mixtures should produce peaks of equal area, if the instrument response is linear

18. Gas Chromatography – Acetates Thermal Response Ratios GC Peak Area Correction Factor (approach 1) • The ratio of one peak area to another in a GC chromatogram should be proportional to the molar ratio of the components in the mixture • The expression for modifying the Peak Areas for a non-linear area instrument response is constructed as follows: • Determine the area of each peak in an equimolar mixture • Compute the ratio of one of the peaks selected as the “basis for computation” relative to each peak area

19. Gas Chromatography – Acetates • Thermal Response Correction Factor (con’t) • Multiply the area of each peak by the respective Thermal Response Factor (TRx) • Compute the Total Adjusted Area • Compute the Adjusted Mole Fraction • Compute the Adjusted Mole Percent

20. Gas Chromatography – Acetates Thermal Response Ratios Example – Ethyl Acetate (S=2) is used as basis for calculations Total Adjusted Area  2.14 + 2.88 + 2.63 + 2.26 = 9.91 Mole Fraction EtAc — 2.14 / 9.91 = 0.216 Mole Fraction ProAc — 2.88 / 9.91 = 0.291 Mole Fraction Bu Ac — 2.63 / 9.91 = 0.265 Mole Fraction HexAc — 2.26 / 9.91 = 0.228

21. Gas Chromatography – Acetates Thermal Response Ratios GC Peak Area Correction Factor (alternate approach) • The ratio of one peak area to another in a GC chromatogram should be proportional to the molar ratio of the components in the mixture • If the peaks of an equimolar mixture do not have the same area, the relationship between the area of a peak and the mole fraction of the compound in the mixture is incorrect and would have to be adjusted by some factor • The Thermal Response Factor (TR) is determined from an “Equimolar” Mixture

22. Gas Chromatography – Acetates • The derivation that follows utilizes ratios between any two compounds in a mixture, one of which will be designated as the “basis for computation” • Assuming an equimolar mixture of 4 acetates: Ethyl Acetate, Propyl Acetate, Butyl Acetate, Hexyl Acetate • In the equation development below, the subscript “i” will be used to designate the compounds in a mixture: i(1,2,3,4) = Ethyl(1), Propyl(2), Butyl(3), Hexyl(4) • In the derivation and examples that follow, Ethyl Acetate will be used as the basis for the calculations (designated by subscript (s), but any of the other compounds could also be used, such as in the case where the unknown mixture does not contain any Ethyl Acetate

23. Gas Chromatography – Acetates Thermal Response Ratios (Con’t) • The following expression equates corrected area ratios to an adjustment of the molar ratios • The area ratio (mole ratio) of each component (i) is shown relative to the selected base of computation compound (s) in the mixture • If the equation is rearranged to indicate an adjustment to the measured areas Note the subscripts relative to the TR factor

24. Gas Chromatography – Acetates • Compute the TRs/TRi ratios from the measured peak areas from the standard equimolar mixture: For an equimolar mixture: molei/moles = 1 Thus, substitution in equation 2 gives: Again: note the relative position of the subscripts • From equation (3), each individual TRs/TRi ratio is calculated from the peak areas of the standard equimolar mixture

25. Gas Chromatography – Acetates Thermal Response Ratios (Con’t) • Adjusting the Peak Areas of the Unknown Mixture • Using each TRs/TRi ratio, the mole ratio of each component in the unknown mixture, relative to the base compound, is calculated from equation (2) • The Molei/Moles values from equation 2 now represent adjusted peak areas, and thus are proportional to the molar content of the unknown mixture • The adjusted Molei/Moles values are summed • The new Mole Fractions are computed by dividing each Molei/Moles value by the total • The new Mole % is computed by multiplying the mole fraction by 100 .

26. Gas Chromatography – Acetates Thermal Response Ratios (Con’t) Example – Ethyl Acetate (S=2) is used as basis for calculations Apply TRs/Tri correction factor to measured area ratios using equation #2 EtAc / EtAc = mol2 / mol2 = area2 / area2  TR2 / TR2 = 2.14 / 2.14 1.00 = 1.00 ProAc / EtAc = mol3 / mol2 = area3 / area2  TR2 / TR3 = 2.18 / 2.14 1.32 = 1.34 BuAc / EtAc = mol4 / mol2 = area4 / area2 TR2 / TR4 = 2.12 / 2.14  1.24 = 1.23 HexAc / EtAc = mol6 / mol2 = area6 / area2 TR2 / TR6 = 1.54 / 2.14  1.47 = 1.06  moli/mol2 = 1.00 + 1.34 + 1.23 + 1.06 = 4.63  mole % EtAc = 1.00 / 4.63 * 100 = 21.6%  mole % ProAc = 1.34 / 4.63 * 100 = 28.9%  mole % BuAc = 1.23 / 4.63 * 100 = 26.6%  mole % HexAc = 1.06 / 4.63 * 100 = 22.9%

27. Gas Chromatography – Acetates Thermal Response Ratios (Con’t) Example # 2 – Ethyl Acetate (S=2) is used as basis for calculations Apply TRs/Tri correction factor to measured area ratios using equation #2 EtAc / EtAc = mol2 / mol2 = area2 / area2  TR2 / TR2 = 2.14 / 2.14 1.00 = 1.00 ProAc / EtAc = mol3 / mol2 = area3 / area2  TR2 / TR3 = 2.18 / 2.14 0.69 = 0.70 BuAc / EtAc = mol4 / mol2 = area4 / area2  TR2 / TR4 = 2.12 / 2.14 0.62 = 0.61 HexAc / EtAc = mol6 / mol2 = area6 / area2  TR2 / TR6 = 1.54 / 2.14 0.61 = 0.44  moli/mol2 = 1.00 + 0.70 + 0.61 + 0.44 = 2.75  mole% EtAc = 1.00 / 2.75 * 100 = 36.4%  mole% ProAc = 0.70 / 2.75 * 100 = 25.4%  mole% BuAc = 0.61 / 2.75 * 100 = 22.2%  mole% HexAc = 0.44 / 2.75 * 100 = 16.0%

28. Gas Chromatography – Acetates Thermal Response Ratios (Con’t) Ex. 3 - Assumes the unknown is missing Ethyl Acetate and Propyl Acetate (S=3) is used as basis for calculations Apply TRs/Tri correction factor to measured area ratios using equation #2 • ProAc / EtAc = mol3 / mol3 = area3 / area3  TR3 / TR3 = 2.18 / 2.18 1.00 = 1.00 • BuAc / EtAc = mol4 / mol3 = area4 / area3  TR3 / TR4 = 2.12 / 2.18 0.89 = 0.87 • HexAc / EtAc = mol6 / mol3 = area6 / area3  TR3 / TR6 = 1.54 / 2.18 0.89 = 0.63 • moli/mol3 = 1.00 + 0.87 + 0.63 = 2.50 •  mole% ProAc = 1.00 / 2.50 * 100 = 40.0% •  mole% BuAc = 0.87 / 2.50 * 100 = 34.8% •  mole% HexAc = 0.63 / 2.50 * 100 = 25.2%