1/17/2015 1 gas chromatography, refractive index & distillation the next two (2) experiments...

$: 1/17/2015 1 Gas Chromatography, Refractive Index & Distillation The next two (2) experiments introduce Gas Chromatography and Simple & Fractional Distillation$
1/17/2015 1

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

Gas Chromatography – Acetates

1/17/2015 2

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)


1/17/2015 3

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


1/17/2015 4

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


1/17/2015 5

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


1/17/2015 6

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


1/17/2015 7

The Experiment Purpose – Introduce the theory and technique of gas

chromatographyIdentify a compound by it retention timeFrom 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


1/17/2015 8

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


1/17/2015 9

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


1/17/2015 10

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


1/17/2015 11

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

1/17/2015 12

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


1/17/2015 13

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 (t = 0)

Draw vertical Line from Peak Top to Base Line

Measure Distance from Starting Point to Base of Peak Distance

Note: Disregard “Air Peaks” in all calculations

1/17/2015 14

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


1/17/2015 15

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


1/17/2015 16

Draw Baseline connecting peak bottoms Peak Area by the Triangulation Method

Peak Area = h * w½

Whereh = Peak Height from baseline

w½ = width of peak at ½ the peak height

Adjust Peak Area for thermal responseSee 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

1/17/2015 17

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


1/17/2015 18

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


1 1 1 11 2 3 4

1 2 3 4

Area Area Area AreaTR = TR = TR = TR =

Area Area Area Area

1/17/2015 19

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


a a 1

b b 2

c c 3

d d 4

area adj = area ×TR




1/17/2015 20

Thermal Response RatiosExample – Ethyl Acetate (S=2) is used as basis for calculations

Gas Chromatography – AcetatesEtAc (2)EtAc (2) ProAc (3)ProAc (3) BuAc (4)BuAc (4) HexAc (6)HexAc (6)

StandardStandard

EquimolarEquimolar

MixtureMixture

MeasuredMeasured

Peak AreaPeak Area1.441.44 1.091.09 1.161.16 0.980.98

TRTRss/TR/TRii = A = Ass/A/Aii

(s=2)

1.441.44 = 1.00= 1.001.441.44

1.441.44 = 1.32 = 1.321.091.09

1.441.44 = 1.24= 1.241.161.16

1.44 1.44 = 1.47 = 1.470.980.98

UnknownUnknown

MixtureMixture

MeasuredMeasured


Adjusted AreasAdjusted Areas 2.14 * 1.00 = 2.142.14 * 1.00 = 2.14 2.18 * 1.32 = 2.882.18 * 1.32 = 2.88 2.12 * 1.24 = 2.632.12 * 1.24 = 2.63 1.54 * 1.47 = 2.261.54 * 1.47 = 2.26

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

1/17/2015 21

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


1/17/2015 22

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


1/17/2015 23

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


i i i

s s s

area moles TR = • (1)

area moles TR

si i

s s i

TRmoles area = • (2)

moles area TR

1/17/2015 24

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


s s si

s i i i

TR area TRarea1 = • or = (3)

area TR area TR

1/17/2015 25

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


.

1/17/2015 26


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%

EtAc (2)EtAc (2) ProAc (3)ProAc (3) BuAc (4)BuAc (4) HexAc (6)HexAc (6)

StandardStandard

EquimolarEquimolar

MixtureMixture

MeasuredMeasured



(s=2)

1.441.44 = 1.00= 1.001.441.44

1.441.44 = 1.32 = 1.321.091.09

1.441.44 = 1.24= 1.241.161.16

1.44 1.44 = 1.47 = 1.470.980.98

UnknownUnknown

MixtureMixture

MeasuredMeasured


areaareaii/area/areass

(s=2)(s=2)

2.142.14 = 1.00= 1.002.142.14

2.182.18 = 1.02= 1.022.142.14

2.122.12 = 0.99= 0.992.142.14

1.541.54 = 0.72= 0.722.142.14

Apply TRs/Tri correction factor to measured area ratios using equation #2

Thermal Response Ratios (Con’t)Example – Ethyl Acetate (S=2) is used as basis for calculations

1/17/2015 27


EtAc (2)EtAc (2) ProAc (3)ProAc (3) BuAc (4)BuAc (4) HexAc (6)HexAc (6)

StandardStandard

EquimolarEquimolar

MixtureMixture

MeasuredMeasuredPeak AreaPeak Area 128128 186186 208208 210210


(s=2) 128128 = 1.00= 1.00 128128

128128 = 0.69= 0.69 186186

128128 = 0.62= 0.62 208208

128128 = 0.61= 0.61 210210

UnknownUnknown

MixtureMixture

MeasuredMeasuredPeak AreaPeak Area 2.142.14 2.182.18 2.122.12 1.541.54

areaareaii/area/areas s

(s=2)(s=2)2.142.14 = 1.00= 1.002.142.14

2.182.18 = 1.01= 1.012.142.14

2.122.12 = 0.99= 0.992.142.14

1.541.54 = 0.72= 0.722.142.14

Thermal Response Ratios (Con’t)Example # 2 – Ethyl Acetate (S=2) is used as basis for calculations

EtAc / EtAc = mol2 / mol2 = area2 / area2 TR2 / TR2 = 2.14 / 2.14 1.00 = 1.00ProAc / EtAc = mol3 / mol2 = area3 / area2 TR2 / TR3 = 2.18 / 2.14 0.69 = 0.70BuAc / EtAc = mol4 / mol2 = area4 / area2 TR2 / TR4 = 2.12 / 2.14 0.62 = 0.61HexAc / 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%


1/17/2015 28

Gas Chromatography – AcetatesThermal Response Ratios (Con’t)Ex. 3 - Assumes the unknown is missing Ethyl Acetate and Propyl Acetate (S=3) is used as basis for calculations

ProAc (3)ProAc (3) BuAc (4)BuAc (4) HexAc (6)HexAc (6)

Standard

Equimolar

Mixture

MeasuredMeasuredPeak AreaPeak Area

186186 208208 210210


(s=3) 186 186 = 1.0= 1.0 186186

186186 = 0.89= 0.89 208208

186186 = 0.89= 0.89 210210

Unknown

Mixture

MeasuredMeasuredPeak AreaPeak Area

2.182.18 2.122.12 1.541.54

areaareaii/area/areass

(s=3) 2.182.18 = 1.0 = 1.0 2.182.18

2.122.12 = 0.97= 0.97 2.182.18

1.541.54 = 0.71= 0.71 2.182.18

ProAc / EtAc = mol3 / mol3 = area3 / area3 TR3 / TR3 = 2.18 / 2.18 1.00 = 1.00BuAc / EtAc = mol4 / mol3 = area4 / area3 TR3 / TR4 = 2.12 / 2.18 0.89 = 0.87HexAc / 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%


1/17/2015 1 gas chromatography, refractive index & distillation the next two (2) experiments...

Documents

moving gas phase

rd week gas chromatography

sample gas unbalanced

vapor phase

column chromatography

carrier gas stream helium

vaporphase chromatography

argon sample vapor