Gas Chromatography Acetates

Gas Chromatography – Acetates     Gas Chromatography, Simple & Fractional Distillation  The next two (2) experiments introduce Gas Chromatogr...
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Gas Chromatography – Acetates 







Gas Chromatography, Simple & Fractional Distillation  The next two (2) experiments introduce Gas Chromatography and Simple & Fractional Distillation This Week  Gas Chromatography – Acetates  Pavia – p. 817 - 836  Slayden – p. 39 - 31 2nd Week  Simple & Fractinal Distillation of a Mixture  Slayden – p. 43 - 46 3rd Week  Gas Chromatography of the Distillates from the Distillation of a Mixture Experiment  Slayden – p. 47

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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)

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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

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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

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Gas Chromatography – Acetates Principals of Detection  Two Detector Types  Thermal Conductivity Detector (TCD) (we use this)  Flame Ionization  TCD is an electrically heated “Hot Wire” placed in the 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 the heats up and the 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 1/17/2015 

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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

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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 1/17/2015  Calculate Mole Percentage

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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 

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The Unknowns have from 2 to 4 of the compounds in the standard mixture

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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

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Gas Chromatography – Acetates 





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Data Summary Procedure – Using complete sentences summarize, in paragraph form, all of the results obtained in the experiment Analysis & Conclusion Section  Using results, 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 10

Gas Chromatography – Acetates 

Procedure 

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Preparing the Microsyringe 

The Microsyringe is fragile and expensive – BE CAREFUL



Insert the needle into the sample vial and pull back on the plunger pulling sample into syringe



Expel the sample onto a Chemwipe



Repeat 3-4 times



This process removes any contaminants from the syringe barrel



This process is repeated for each new sample

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Gas Chromatography – Acetates SRI 310C GC Instructions 

The mouse operator clicks on the “Blank” page icon on the left side of the menu line



The mouse operator then clicks on the “0” box between the + and – boxes on the left side of the screen



The mouse operator clicks on the “Acquisition” icon



The syringe operator inserts the needle into the sample vial and pulls back on the plunger pulling sample into the syringe (2.0 uL



Secure the plunger with thumb and finger so that no liquid escapes syringe



Grasping syringe barrel with other hand, carefully insert the needle fully into the injection port through the rubber septum. Note: Some resistance may be encountered

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Gas Chromatography – Acetates SRI 310C GC Instructions (con’t) 

Coordinating with the student handling the computer mouse, push the plunger all the way in to inject the sample into the heated chamber, pause ½ second and retract the syringe quickly



Upon injection of the sample, the mouse operator simultaneously clicks on “Run” under the “Aquisition” menu item to begin timing of the sample passing through the column



When the last peak is recorded on the chart (3-4 minutes), the mouse operator clicks on “STOP” under the “Acquisition” menu to stop the sample run



Instrument Settings are printed on the Gas Chromatogram and should be recorded in the “Results” box of the procedure

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Gas Chromatography – Acetates SRI 310C GC Instructions (con’t) 

Print Chromatogram Note: Instructor normally sets print parameters at beginning of class as follows 1. Click on “Print” a. Click on “Format” b. Click on “On Peaks” under “Labels”

c. Under “Available” click on “Time” (retention time) d. Click “>>” to move selection to “Selected” e. Repeat for “Area” (Peak Area)

2. Click OK twice 

Label chromatogram with sample ID and student name



Where applicable print multiple copies of chromatogram

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Select Blank Page

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Acquisition Run Stop

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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

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Gas Chromatography – Acetates 

Determination of Retention Time  Since Velocity (v) = Distance / Time = d / t  Ret Time (t) = Distance(cm) / Velocity(cm/min) = d / v

Retention Time Distances Mark Starting Point On Chart, i.e., start time = 0.0 min Draw a base line from left base of first peak to right base of right-most peak Draw vertical Lines from Peak tops to Base Line Measure Distance from starting point (t = 0) to Base of each vertical line Calculate Retention Time 1/17/2015

Note: Disregard “Air Peaks” in all calculations

Distance 17

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 22-25

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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.



Adjust the peak areas for non-linear thermal response using the algorithm described in slides 22-25



Add the individual adjusted areas to get the total area



Divide each area by total area to get mole fraction



Multiply mole fraction by 100 to get mole %



See algorithm development on the following slides

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Gas Chromatography – Acetates 

Draw Baseline connecting peak bottoms



Peak Area by the Triangulation Method Baseline

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



Baseline

B/TA

Mole Percent = MF x 100

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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

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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 nonlinear 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



For example, if the first peak in the equimolar mixture is selected as the basis, then: TR1 =

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Area1 Area1

TR 2 =

Area1 Area2

TR 3 =

Area1 Area3

TR 4 =

Area1 Area4 22

Gas Chromatography – Acetates 

Thermal Response Correction Factor (con’t) 

Multiply the area of each peak by the respective Thermal Response Factor (TRx)

areaa  adj = areaa × TR1 areab  adj = areab × TR 2 areac  adj = areac × TR 3 aread  adj = aread × TR 4



Note: The correction factors determined for the equimolar mixture are also used for the unknown mixture



Compute the Total Adjusted Area



Compute the Adjusted Mole Fraction



Compute the Adjusted Mole Percent

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Gas Chromatography – Acetates EtAc (2)

ProAc (3)

BuAc (4)

HexAc (6)

1.44

1.09

1.16

0.98

Measured Standard Equimolar Mixture

Peak Area TRs/TRi = As/Ai

1.44

(s=2)

1.44

Measured Unknown Mixture

Peak Area Adjusted Areas

= 1.00

1.44 1.09

= 1.32

1.44 1.16

= 1.24

1.44

= 1.47

0.98

2.14

2.18

2.12

1.54

2.14 * 1.00 = 2.14

2.18 * 1.32 = 2.88

2.12 * 1.24 = 2.63

1.54 * 1.47 = 2.26

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 1.000 (mole fraction must sum to 1.000

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Gas Chromatography – Acetates Thermal Response Ratios GC Peak Area Correction Factor alternate approach – do not use unless instructed to do so 

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

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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

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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

areai molesi TR i = • areas moless TR s 

(1)

If the equation is rearranged to indicate an adjustment to the measured areas molesi areai TR s = • (2) moless areas TR i Note the subscripts relative to the TR factor

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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:

TR s areai 1 = • areas TR i

areas TR s or = areai TR i

(3)

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

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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

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Gas Chromatography – Acetates Measured Standard Equimolar Mixture

Peak Area

ProAc (3)

BuAc (4)

HexAc (6)

1.44

1.09

1.16

0.98

TRs/TRi = As/Ai

1.44

(s=2)

1.44

Measured Unknown Mixture

EtAc (2)

= 1.00

2.14

Peak Area areai/areas

2.14

(s=2)

2.14

= 1.00

1.44 1.09

1.44

= 1.32

1.16

2.18

2.18 2.14

= 1.24

1.44

2.12

2.12

= 1.02

2.14

= 1.47

0.98

1.54

= 0.99

1.54 2.14

= 0.72

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 ProAc BuAc HexAc

/ EtAc = / EtAc = / EtAc = / EtAc =

mol2 / mol2 mol3 / mol2 mol4 / mol2 mol6 / mol2

 moli/mol2

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= = = =

area2 / area2 area3 / area2 area4 / area2 area6 / area2

 TR2 / TR2  TR2 / TR3  TR2 / TR4  TR2 / TR6

= 1.00 + 1.34 + 1.23 + 1.06

 mole % EtAc =  mole % ProAc =  mole % BuAc =  mole % HexAc =

1.00 / 4.63 * 100 1.34 / 4.63 * 100 1.23 / 4.63 * 100 1.06 / 4.63 * 100

= = = =

2.14 2.18 2.12 1.54

/ 2.14 / 2.14 / 2.14 / 2.14

 1.00 = 1.00  1.32 = 1.34  1.24 = 1.23  1.47 = 1.06

= 4.63 = = = =

21.6% 28.9% 26.6% 22.9%

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Gas Chromatography – Acetates Thermal Response Ratios (Con’t) Example # 2 – Ethyl Acetate (S=2) is used as basis for calculations Standard Equimolar Mixture

Unknown Mixture

Measured Peak Area TRs/TRi = As/Ai (s=2)

EtAc (2)

ProAc (3)

BuAc (4)

HexAc (6)

128

186

208

210

128 = 1.00 128

128 = 0.69 186

128 = 0.62 208

128 = 0.61 210

Measured Peak Area

2.14

2.18

2.12

1.54

areai/areas (s=2)

2.14 = 1.00 2.14

2.18 = 1.01 2.14

2.12 = 0.99 2.14

1.54 = 0.72 2.14

Apply TRs/Tri correction factor to measured area ratios using equation #2 EtAc ProAc BuAc HexAc

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/ EtAc = / EtAc = / EtAc = / EtAc =

mol2 / mol2 mol3 / mol2 mol4 / mol2 mol6 / mol2

= = = =

area2 / area2  TR2 / TR2 area3 / area2  TR2 / TR3 area4 / area2  TR2 / TR4 area6 / area2  TR2 / TR6

= = = =

2.14 2.18 2.12 1.54

 moli/mol2

= 1.00 + 0.70 + 0.61 + 0.44 = 2.75

 mole% EtAc  mole% ProAc  mole% BuAc  mole% HexAc

= = = =

1.00 / 2.75 * 100 0.70 / 2.75 * 100 0.61 / 2.75 * 100 0.44 / 2.75 * 100

= = = =

/ 2.14 / 2.14 / 2.14 / 2.14

36.4% 25.4% 22.2% 16.0%

   

1.00 0.69 0.62 0.61

= = = =

1.00 0.70 0.61 0.44

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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 Standard Equimolar Mixture Unknown Mixture

Measured Peak Area TRs/TRi = As/Ai (s=3) Measured Peak Area areai/areas (s=3)

ProAc (3)

BuAc (4)

HexAc (6)

186

208

210

186 = 1.0 186 2.18 2.18 = 1.0 2.18

186 = 0.89 208

186 = 0.89 210

2.12

1.54

2.12 = 0.97 2.18

1.54 = 0.71 2.18

Apply TRs/Tri correction factor to measured area ratios using equation #2 ProAc / EtAc = mol3 / mol3 BuAc / EtAc = mol4 / mol3 HexAc / EtAc = mol6 / mol3

= area3 / area3  TR3 / TR3 = 2.18 / 2.18  1.00 = area4 / area3  TR3 / TR4 = 2.12 / 2.18  0.89 = area6 / area3  TR3 / TR6 = 1.54 / 2.18  0.89

moli/mol3  mole% ProAc  mole% BuAc  mole% HexAc 1/17/2015

= 1.00 + 0.87 + 0.63

=

= 1.00 = 0.87 = 0.63

2.50

= 1.00 / 2.50 * 100 = 40.0% = 0.87 / 2.50 * 100 = 34.8% = 0.63 / 2.50 * 100 = 25.2% 32

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