Analysis By Gas Chromatography Biology Essay

oven temperature on the separation of methyl and ethyl esters. Besides the optimal separation status for the analyte ( methyl, ethyl and unknown ester ) is to be identified. Finally for this portion of the experiment, the consequence of ‘split injection ‘ and ‘splitless injection ‘ is determined. ‘Split injection ‘ is when the split valve is kept unfastened when sample is injected while ‘splitless injection is when the split valve is closed before the sample is introduced1.

Methods

Apparatus

Gas chromatography machine. Agilent technologies 7802A GC system, consecutive figure CN10022002 ITL9002 GC6.

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Column: Agilent J & A ; W GC column, Part No 19091J-413 HP-5, Serial No US9035232L, Length 30m, I.D 0.320mm, movie thickness 0.25 Aµm

Autosampler: G4513A Serial No: CN95303257

Instrumental status

Nitrogen bearer gas flow: 2ml/min ( split ratio 1:50 ) .

Oven temperature =100-150oC ( isothermal or temperature gradient programmed )

Injector temperature = 200oC

Detector temperature = 250oC

Air and H flows preset

Detector: FID

Further Instrumental status

Condition 1: Initial oven temperature: 100 oC ; Concluding temperature: 100oC tally clip: 6min.

Condition 2: Initial oven temperature: 150 oC ; Concluding temperature: 150oC ; hold clip: 6min ; ramp rate: 0oC/min ; entire tally clip: 6min.

Condition 3: Initial temperature: 100oC ; concluding temperature: 150 oC ; hold clip: 0min ; ramp rate: 10A°C/min ; entire tally clip: 5min.

Condition 4: Initial temperature: 100oC ; concluding temperature: 150 oC ; hold clip: 0min ; ramp rate: 15A°C/min ; entire tally clip: 3.33min.

Condition 5: Initial temperature: 100oC ; concluding temperature: 150 oC ; hold clip: 0min ; ramp rate: 20A°C/min ; entire tally clip: 2.5min.

Condition 6: Initial temperature: 55oC ; initial clip: 1min ; optimum incline rate to 150A°C ; purging valve: unfastened 0.7min after the start of the tally.

Reagents

Sample A: Methyl ester mixture in hexane: methyl pentanoate 2.0 % ; methyl hexanoate 2.5 % ; methyl heptanoate 3.0 % ; methyl octanoate: 3.5 %

Sample B: Methane.

Sample C: Ethyl ester mixture in hexane: ethyl butanoate 2 % ; ethyl pentanoate 2.25 % ; ethyl hexanoate 2.75 % ; ethyl heptanoate 3 %

Sample D: Unknown Eastern time 1212 esters.

Sample Tocopherol: Diluted Ethyl ester mixture.

Stairss

From the Lab manual

Measure 1: Inject 0.1I?l Sample A utilizing a split injection into GC utilizing Condition 1.

Measure 2: Inject 100Aµl Sample B utilizing a split injection into GC utilizing Condition 1

Measure 3: Inject 0.1I?l Sample A utilizing a split injection into GC utilizing Condition 2.

Measure 4: Repeat the injection ( 0.1I?l ) of Sample A utilizing Condition 3, 4 and 5.

Step5: Inject Sample C into GC utilizing Condition 5.

Step6: Run Sample D the mixture incorporating three unknown esters ( EST1212 ) utilizing Condition 5.

Step7: Repeat the analysis of the Sample C utilizing Condition 5 and choosing the splitless option in the injector parametric quantities window ( split/splitless )

Step8: Dilute the Sample C five hundred crease by reassigning 20 Aµl into a 10ml volumetric flask and thining to the with hexane and inject 1I?l of the diluted sample into GC utilizing Condition 5.

Measure 9: Repeat the analysis, but this clip uses the “ sandwich ” injection technique to shoot 1I?l of the diluted Sample C.

Set the injector parametric quantities to the followers:

1.0I?l hexane, 0.2I?l air, 1.0I?l sample, 0.2I?l air, 0.2I?l hexane, 1.0I?l air. Inject the whole 3.6I?l sample.

Remark

Condition 5 resulted in optimal separation of Sample A ( methyl ester ) as shown in page 21. This status gave a short keeping and a good chromatogram compared to other conditions

In the splitless injection ( step 7 ) for the ester extremums are looking as shown in page 24. This is because the bearer gas continuously mixes with the vapor in the injector, doing it more and more dilute but ne’er wholly blushing the sample from the injector2. Splitless injection is non well-suited for volatile compounds3.

Consequences and Discussion

Summary of consequence

Table 1: Consequence of the separation of methyl esters utilizing status 1 ( as shown in the chromatogram in page 16 and some belongingss of the mixture of methyl esters

Compound

thulium ( min )

tR ( min )

t’R ( min )

log t’R ( min )

Boiling point ( oC )

Carbon figure

Methyl pentanoate

1.500

1.974

0.474

-0.324

126.0

6

Methyl hexanoate

2.447

0.947

-0.024

151.0

7

Methyl heptanoate

3.356

1.856

0.269

172.5

8

Methyl octanoate

5.100

3.600

0.556

193.0

9

Where

thulium is the keeping clip of methane.

tR is the keeping clip of methyl esters.

t’R is the adjusted keeping clip gotten by utilizing t’R = tR – thulium

Boiling points2,3

Example for the computation of adjusted keeping clip.

Using methyl pentanoate values

From the formula2 ( t’R = tR – thulium )

t’R = 1.974 – 1.500 = 0.474

Figure 1: The secret plan of log t’R ( adjusted keeping clip ) against figure of C atoms

The relationship shows a additive dependance between the log t’R and the figure of C atom

Figure2: The secret plan of log t’R ( adjusted keeping clip ) against boiling point

The graph shows a additive dependance of the log t’R ( min ) on the boiling point ( oC ) .

Table 2: Consequence of the separation of methyl esters utilizing status 1 ( as shown in the chromatogram on page 16a and the efficiency

Compound

thulium ( min )

tR ( min )

t’R ( min )

W1/2 ( min )

Nitrogen

Methyl pentanoate

1.500

1.974

0.474

0.017

74698

Methyl hexanoate

1.500

2.447

0.947

0.033

30461

Methyl heptanoate

1.500

3.356

1.856

0.046

29488

Methyl octanoate

1.500

5.100

3.600

0.083

20917

Where

thulium ( min ) is the keeping clip for methane ( chromatogram on page 17 )

W1/2 ( min ) is the breadth at half tallness and was measured straight from extremums on page 16b

N ( dimensionless ) is the efficiency, calculated utilizing the formula4

Remark

The efficiency of a column is determined by two factors2:

The difference in the elution times between extremums: the farther apart, the better their separation.

The other factor is how wide the extremums are: the wider the extremums, the poorer their separation.

Therefore, the efficiency of the column for each of the methyl ester extremums utilizing 100oC isothermal analysis is just because the elution clip is non farther apart and some of the extremums are excessively wide.

Table 3: Consequence of the separation of the Methyl esters, ethyl esters and the unknown ester sample ( EST1212 ) utilizing Condition 5

Compound

thulium ( min )

tR ( min )

t’R ( min )

Page

Methyl pentanoate

1.500

1.756

0.256

16a

Methyl

Hexanoate

1.500

1.974

0.474

16a

Methyl heptanoate

1.500

2.299

0.799

16a

Methyl

Octanoate

1.500

2.738

1.238

16a

Ethyl

Butanoate

1.500

1.715

0.215

22

Ethyl

Pentanoate

1.500

1.908

0.408

22

Ethyl

Hexanoate

1.500

2.199

0.699

22

Ethyl

Heptanoate

1.500

2.600

1.100

22

Unknown ester 1

1.500

1.908

0.408

23

Unknown ester 2

1.500

1.974

0.474

23

Unknown ester 3

1.500

2.602

1.102

23

Where

thulium is the keeping clip of methane. ( chromatogram on page 17 )

tR is the keeping clip of methyl esters, ethyl esters and unknown esters.

t’R is the adjusted keeping clip gotten by utilizing t’R = tR – thulium

From the tabular array above

t’R ( min ) for ethyl pentanoate = 0.408= t’R ( min ) for unknown ester 1

t’R ( min ) for methyl hexanoate = 0.474 = t’R ( min ) for unknown ester 2

t’R ( min ) for ethyl heptanoate ( 1.100 ) is tantamount to t’R ( min ) for unknown ester 3 ( 1.102 ) .

Therefore, the unknown esters in the unknown ester sample ( EST1212 ) are

Ethyl pentanoate

Methyl hexanoate

Ethyl heptanoate

Answers to inquiries

Comparison of split and splitless chromatograms

The initial oven temperature is lowered to 50oC for the splitless injection because the sample dissolver hexane has a boiling point of 69oC and any initial oven temperature that is above the temperature of hexane will take to the solvent extremum shadowing, and the early eluting of compounds have wide extremum forms and are ill resolved from one another2. Therefore, the initial oven temperature is lowered to enable the constituents condense, organizing a narrow “ bullet ” of mixture to be injected onto the column, therefore minimise peak broadening4.

The duplicability of the split and splitless analysis of ethyl esters can be affected by mutual opposition and the induced vapor force per unit area volume1.

Mentions

K. Grob Classical split and splitless injection in capillary gas chromatography: with some comments on PTV injection. Heidelberg ; New York: A. Huethig ; 1986. pp. 97,155,248-250.

D. C. Harris, 1948- . Quantitative chemical analysis. 6th erectile dysfunction. New York: W. H. Freeman ; 2002. pp. 556-557,588-599.

H. M. McNair, 1933- , J. M. Miller, 1933- . Basic gas chromatography. New York ; Chichester: John Wiley ; 1998. p. 99.

K. Grob Split and splitless injection or quantitative gas chromatography: constructs, procedures, practical guidelines, beginnings of mistake. 4th erectile dysfunction. Weinheim ; Cambridge: Wiley-VCH ; 2001. pp. 64-70

Name: Ime Cletus Usen Registration Number: B212859 Partner ‘s Name: Lufeng Zhao Date: 26-10-2012

Part 2: Quantitative Analysis of Ethanol in Alcoholic Beverages by Internal Standards

Abstraction

A method is given for the quantitative analysis of ethyl alcohol in alcoholic drinks by gas chromatography. This method uses an internal criterion and fire ionisation sensor for the accurate and precise finding of ethyl alcohol in alcoholic drinks ( Quantitative analysis ) compared to other methods of analysis normally used. For this experiment, propan-1-ol is used as an internal criterion to find the comparative responds factor for ethyl alcohol, which is so usage to determine the concentration of ethyl alcohol in the alcoholic drink. This experiment has explored and seen the effectivity of utilizing an internal criterion ( propan-1-ol ) for the finding of the concentration of ethyl alcohol in alcoholic drinks.

Introduction

The purpose of this experiment is to detect the consequence of the internal criterion ( propan-1-ol ) for the finding of ethyl alcohol in alcoholic drinks by gas chromatography ( with fire ionisation sensor ) . The highest preciseness for quantitative GC is obtained utilizing internal criterions because the uncertainnesss introduced by sample injection, flow rate, and fluctuation in column status are minimised.1,4 An internal criterion is a known sum of a compound, different from the analyte, that is added to the unknown2. The internal criterion should hold the undermentioned features

It should elute near the extremums of involvement but must be good resolved from them3.

It should be chemically similar to the analytes of involvement and non respond with any sample components3.

Like any criterion, it must be available in high purity3.

Be similar in functional group type to the constituent ( s ) of involvement. If such a compound is non readily available, an appropriate hydrocarbon should be substituted5.

Be sufficiently non-volatile to let for storage of standard solutions for important periods of time5.

A standardization curve is so plotted for the ratio of the analyte peak country to the internal criterion peak country as a map of the analyte concentration of the standard1,4.

Safety

I ensured that the injection syringe was carefully injected into the injection valve.

Method

Summary of instrument and instrumental status

Instrument

Gas Chromatography machine. Varian CP-3380, Serial Number 05469 ITL1724, GC 2.

Pipette 20-200Aµl and 500-5000 Aµl. eppendorf research, US Patent No 5531131.

Unknown Sample: Andrew Peace Chardonnay, South Eastern Australia 75CLe 12.5 % vol. Bottled by W1507 at NR104BG, UK for Bottle Green Ltd LS184BH South Eastern Australia, Andrew Peace vinos, Murray valley main road, Piangil, Victoria 3597. www.apwines.com

Column: Agilent J & A ; W GC column, Part No 19091J-413 HP-5, Serial No US9035232L, Length 30m, I.D 0.320mm, movie thickness 0.25 Aµm.

SGE Syringe IBR-7 Cat # 2477L

Instrumental status

Nitrogen 2ml/min

Oven temperature = 45oC

Injector temperature = 150oC

Detector temperature = 200oC

Air and H flows preset

Fire ignited – allow to brace for 10mins

Checked sensor signal is less than 20 and stable.

Preparation of criterion

100Aµl 10 % v/v aqueous ethyl alcohol was pipetted into a sample phial.

700Aµl of distilled H2O and 200Aµl 15 % v/v aqueous propan-1-ol was added to try vial incorporating aqueous ethyl alcohol and was cap instantly to forestall doomed of volatiles.

These was repeated with changing sum of ethyl alcohol and distilled H2O but with 200Aµl 15 % v/v aqueous propan-1-ol in separate phials as shown in Table a below

Table a. Description of the readying of ethanol/propan-1-ol criterions.

Vial No

Volume 10 % aq Ethanol ( Aµl )

Volume 15 % aq Propanol ( Aµl )

Volume

Water ( Aµl )

1

100

200

700

2

200

200

600

3

300

200

500

4

400

200

400

5

500

200

300

Each of the criterions in the sample phial was injected into the GC utilizing the GC syringe. ( 0.1 Aµl of the solution in each sample vial were injected at about 2 proceedingss between injections ) .

Besides triplicate solution of the unknown sample ( beverage sample ) was prepared by pipetting 200 Aµl of the unknown sample into a sample phial and adding 600 Aµl of distilled H2O and 200 Aµl of 15 % v/v aqueous propan-1-ol as shown in tabular array B below.

Table B. Description of the readying of the unknown sample ( beverage sample )

Vial No

Volume Unknown sample ( Aµl )

Volume 15 % aq Propanol ( Aµl )

Volume

Water ( Aµl )

6

200

200

600

7

200

200

600

8

200

200

600

0.1Aµl of the solution in each sample phial ( 6, 7, 8 ) were injected into the GC at about 2 proceedingss between injection.

Result and Discussion

Table c. Summary of informations

Vial No

Page

Ce ( Aµl )

Cp ( Aµl )

Ae ( mV.s )

Ap ( mV.s )

Ce/Cp

Ae/Ap

1

10

30

377.497

1593.899

0.333

0.237

2

20

30

752.516

1507.398

0.667

0.499

3

30

30

1229.607

1567.497

1.000

0.784

4

40

30

1878.929

1880.070

1.333

0.999

5

50

30

474.427

495.571

1.667

0.957

6

Unknown

30

729.010

1294.222

n/a

0.563

7

Unknown

30

813.806

1286.530

n/a

0.633

8

Unknown

30

1127.405

1749.031

n/a

0.645

Where

Ce is the concentration of ethyl alcohol

Cp is the concentration of propan-1-ol

Ae is the peak country of ethyl alcohol

Ap is the peak country of propan-1-ol.

Example for the computation of Ae/Ap, Ce, Cp, and Ce/Cp.

Using Vial 1 of table degree Celsius, where Ae= 377.497mV.s and Ap= 1593.899mV.s

Ae/Ap= 377.497mV.s / 1593.899mV.s = 0.236839

Using Vial 1 of tabular array a,

Ce= 10 % v/v *100 Aµl

Ce= 10/100* 100 Aµl

Ce= 10 Aµl.

Cp= 15 % v/v*200 Aµl

Cp= 15/100*200 Aµl

Cp= 30 Aµl

Using the consequence of Ce and Cp above

Ce/Cp= 10 Aµl / 30 Aµl

Ce/Cp= 0.333333

Figure. A secret plan of Ae/Ap versus Ce/Cp

Y = 0.5824x + 0.1131

RA? = 0.9021

Slope of the graph= 0.5824

The fluctuation in the last point is due to some experimental mistake.

Answers to Questions

The comparative response factor for ethyl alcohol is the incline of the graph ( ) . Therefore RF= 0.5824 from figure 1. This is important because it has to be usage to find the concentration of ethyl alcohol in the unknown sample.

Table vitamin D: Showing peak country of the unknown sample

Vial

Ae ( mV.s )

Ap ( mV.s )

Ae/Ap

6

729.010

1294.222

0.563

7

813.806

1286.530

0.633

8

1127.405

1749.031

0.645

Mean of Ae/Ap = 0.563+0.633+0.645 = 1.841 = 0.614

3 3

The relationship between Ce/Cp=x and Ae/Ap=y is y=0.5824x+0.1131 from the graph ( Figure 1 )

Therefore, the average = y= Ae/Ap=0.614

And ten = y-0.1131 = 0.614 – 0.1131 = 0.8600

0.5824 0.5824

Therefore x= Ce/Cp=0.8600

And Cp is known as 15 % v/v

So Ce= CpA-0.8600=15 % v/vA-0.8600= 12.9 % v/v

The associated uncertainness is the standard divergence of the mean6,7.

hypertext transfer protocol: //standard-deviation.appspot.com/images/standard-deviation-1.png

Where

I? is the standard divergence

ten is each value of Ae/Ap

ten is the mean of the values of Ae/Ap

N is the figure of values.

Table vitamin E: Valuess for the computation of standard divergence

ten

ten

x – ten

( ten -x ) 2

0.563

0.614

– 0.051

2.601 x 10-3

0.633

0.614

0.019

3.610 x 10-4

0.645

0.614

0.031

9.610 x 10-4

a?‘ ( ten – ten ) 2 = 2.601 x10-3 + 3.610 ten 10-4 + 9.610 ten 10-4 = 3.923 x 10-3

Therefore

I? = a?s 3.923 x 10-3 = 0.036 =3.6 %

3

Therefore, the concentration of ethyl alcohol in the unknown drink and the associated uncertainness is

12.9 % v/v A±3.6 % = 25.8Aµl A± 7.2 is the concentration of ethyl alcohol in the unknown sample.

A good internal criterion must hold a close extremum to the analyte and must be good from the analyte and must be chemically similar to the analytes3,4. Therefore propan-1-ol is a suited internal criterion for the finding of ethyl alcohol because it gives a close extremum to ethanol and it is good separated and besides has similar chemical belongingss.

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