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
1/17/2015
1
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)
1/17/2015
2
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
1/17/2015
3
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
1/17/2015
4
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
1/17/2015

5
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
1/17/2015
6
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
7
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

1/17/2015
The Unknowns have from 2 to 4 of the compounds
in the standard mixture
8
Gas Chromatography – Acetates

The Report

The Gas Chromatograph instrument settings and the
processing of the samples to get the chromatograms for
both the standard equimolar and unknown mixtures is
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
9
Gas Chromatography – Acetates



1/17/2015
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

Record Instrument readings (Place in GC procedure “Results”)
 Injection Port Temp
For the instruments in the 409 lab,
all three temperatures are read from
 Column Temp
the single dial on the front of cabinet
 Detector Temp
 Gas Flow Rate
– 60 mL / min
 Chart Speed
– Generally 5 cm /min
 Moving Liquid Phase – (DC-200)
1/17/2015
11
Gas Chromatography – Acetates

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 as a short vertical line

Insert the needle into the sample vial and pull back on the
plunger pulling sample into syringe

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 “B” Port through the rubber septum

Coordinating with the chart recorder operator, 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 chart recorder operator
simultaneously pushes the button to start the recorder
1/17/2015
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
1/17/2015
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
1/17/2015
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 17-20
1/17/2015
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

Adjust the peak areas for non-linear thermal response
using the algorithm described in slides 17-20

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 next slide
1/17/2015
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
1/17/2015
Baseline
Baseline
17
Gas Chromatography – Acetates

Thermal Response Factor
1/17/2015

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 (nonlinearity) 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 nonlinear area instrument response is constructed as follows:

Determine the area of each peak in an equimolar
mixture

Compute the ratio (TR) 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 =
1/17/2015
Area1
Area1
TR 2 =
Area1
Area 2
TR 3 =
Area1
Area 3
TR 4 =
Area1
Area4
19
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 × TR 1
areab  adj = areab × TR 2
areac  adj = areac × TR 3
aread  adj = aread × TR 4
1/17/2015

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
20
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)
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
1/17/2015
21
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
1/17/2015
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
1/17/2015
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
areai
moles i TR i
=
•
areas
moless TR s

(1)
If the equation is rearranged to indicate an adjustment to
the measured areas
moles i
area i TR s
=
•
(2)
moless
areas TR i
Note the subscripts relative to the TR factor
1/17/2015
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:
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
1/17/2015
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
1/17/2015
26
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
1/17/2015
=
=
=
=
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%
27
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
1/17/2015
/ 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
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
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
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= 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%
29