klt

Thin Layer Chromatograph Page 1

I. EXPERIMENT TITLE : Thin Layer Chromatography

II. EXPERIMENT DATE : May 18th , 2015 at 7.30 p.m

III. END OF THE EXPERIMENT : May 18th , 2015 at 10.00p.m

IV. EXPERIMENT PURPOSE :

To determine the proper composition of eluen using concentrated ring method.

To determine the value of Rf from plant dye using thin layer chromatography plate.

V. BASIC THEORY

A. Chromatography

Analytical separations may be classified in three ways: by the physical state of the

mobile phase and stationary phase; by the method of contact between the mobilephase and

stationary phase; or by the chemical or physical mechanism responsiblefor separating the

samples constituents. The mobile phase is usually a liquid or agas, and the stationary phase,

when present, is a solid or a liquid film coated on asolid surface. Chromatographic techniques

are often named by listing the type ofmobile phase, followed by the type of stationary phase.

(David Harvey, 2000: 546)

B. Thin Layer Chromatography

Programming of mobile-phase properties in thin layer chromatography (TLC)

naturally, in many cases, influences substantially the separation of the analyzed mixtures. For

example, if the programming of the mobile-phase composition (and of, consequently, its

properties) is implemented before a chromatographic plate, there are methods and devices

developed for their implementation. In the given review, the new variant of TLC with the

programming of composition and properties of a mobile phase (MP) is described. The new

method of programming is the result of dissolving the active volatile reagent being in the

chamber atmosphere and dissolving in the MP migrating on the plate adsorption layer. The

active volatile compounds (the modifiers) as a result of dissolving in the MP change the

properties of the latter in a direction necessary for an experimenter. It is interesting to note

that according to the generally opinion, presence of a gas phase in contact with a TLC plate is

considered as a negative factor of TLC by most of the researches. However, a gas phasehas

been shown by us to play a positive role as well, improving separation of analyzed mixtures.


The given review is devoted to the description of the bases of the new TLC method

with a controlled gas phase (TLC-CGP). The authors hope that the method described will

arouse the readers interest and wish to use this method in ones work and to propose new

variants of its further development and practical application.

( Eli Grushka and Nelu Grinberg, 2014: 283)

Chromatography means a method of analysis in which a mobile phase passes over a

phase in such a way that a mixture of substances is separated into its components. The term

thin-layer chromatography, introduced by E. Stahl in 1956, means achromatographic

separation process in which the stationary phase consists of a thinlayer applied to a solid

substrate or support. For some years, TLC has alsobeen referred to as planar

chromatography. However, apart from the fact that paper.

An essential precondition is that the substances or mixtures of substances to be

analyzed should be soluble in a solvent or mixture of solvents.

TLC is used if:

the substances are nonvolatile or of low volatility

the substances are strongly polar, of medium polarity, nonpolar or ionic

a large number of samples must be analyzed simultaneously, cost-effectively, andwithin

a limited period of time

the samples to be analyzed would damage or destroy the columns of LC

(liquidchromatography) or GC (gas chromatography)

the solvents used would attack the sorbents in LC column packings

the substances in the material being analyzed cannot be detected by the methods ofLC or

GC or only with great difficulty

after the chromatography, all the components of the sample have to be detectable(remain

at the start or migrate with the front)

the components of a mixture of substances after separation have to be detected

individually or have to be subjected to various detection methods one after the other(e.g.

in drug screening)

no source of electricity is available

(Elke Hahn-Deinstrop, 2007:1-2)


Preparation of the plate

In thin-layer chromatography a variety of coating materials is available, but silica

gel is most frequently used. A slurry of the adsorbent (silica gel, cellulose powder, etc.) is

spread uniformly over the plate by means of one of the commercial forms of spreader, the

recommended thickness of adsorbent layer being 150-250 pm. After air-drying overnight, or

oven-drying at 80-90 OC for about 30 minutes, it is ready for use. Ready to use thin-layers

(i.e. pre-coated plates or plastic sheets) are commercially available; the chief advantage of

plastic sheets is that they can be cut to any size or shape required, but they have the

disadvantage that they bend in the chromatographic tank unless supported. Two points of

practical importance may be noted here:

1. care should be exercised in handling the plate to avoid placing fingers on the active

adsorbent surface and so introducing extraneous substances;

2. pre-washing of the plate is advisable in order to remove extraneous material contained

in the layer, and this may be done by running the development solvent to the top of

the plate

Development of plates

The chromatogram is usually developed by the ascending technique in which the

plate is immersed in the developing solvent (redistilled or chromatographic grade solvent

should be used) to a depth of 0.5 cm. The tank or chamber used is preferably lined with

sheets of filter paper which dip into the solvent in the base of the chamber; this ensures that

the chamber is saturated with solvent vapour (Fig. 8.6). Development is allowed to proceed

until the solvent front has travelled the required distance (usually 10-15 cm), the plate is then

removed from the chamber and the solvent front immediately marked with a pointed object.

The plate is allowed to dry in a fume cupboard or in an oven; the drying conditions

should take into account the heat- and light-sensitivity of the separated compounds. The

positions of the separated solutes can be located by various methods. Coloured substances

can be seen directly when viewed against the stationary phase whilst colourless species may

usually be detected by spraying the plate with an appropriate reagent which produces

coloured areas in the regions which they occupy.


(G.H. Jeffery, et al., 1989: 230-231)

C. TLC Solvent Choice

When you need to determine the best solvent or mixture of solvents (a "solvent

system") to develop a TLC plate or chromatography column loaded with an unknown

mixture, vary the polarity of the solvent in several trial runs: a process of trial and error.

Carefully observe and record the results of the chromatography in each solvent system. You

will find that as you increase the polarity of the solvent system, all the components of the

mixture move faster (and vice versa with lowering the polarity). The ideal solvent system is

simply the system that gives the best separation.

TLC elution patterns usually carry over to column chromatography elution patterns.

Since TLC is a much faster procedure than column chromatography, TLC is often used to

determine the best solvent system for column chromatography. For instance, in determining

the solvent system for a flash chromatography procedure, the ideal system is the one that

moves the desired component of the mixture to a TLC Rf of 0.25-0.35 and will separate this

component from its nearest neighbor by difference in TLC Rf values of at least 0.20.

Therefore a mixture is analyzed by TLC to determine the ideal solvent(s) for a flash

chromatography procedure.

Beginners often do not know where to start: What solvents should they pull off the

shelf to use to elute a TLC plate? Because of toxicity, cost, and flammability concerns, the

common solvents are hexanes (or petroleum ethers/ligroin) and ethyl acetate (an ester).

Diethyl ether can be used, but it is very flammable and volatile. Alcohols (methanol, ethanol)

can be used. Acetic acid (a carboxylic acid) can be used, usually as a small percentage

component of the system, since it is corrosive, non-volatile, very polar, and has irritating


vapors. Acetone (a ketone) can be used. Methylene chloride or and chloroform (halogenated

hydrocarbons) are good solvents, but are toxic and should be avoided whenever possible. If

two solvents are equal in performance and toxicity, the more volatile solvent is preferred in

chromatography because it will be easier to remove from the desired compound after

isolation from a column chromatography procedure.

(Departement of Chemistry and Biochemistry University of Colorado, 2015:

http://orgchem.colorado.edu//Technique/Procedures/TLC/TLC.html)

The easiest way to choose the proper eluen that used ring concentrated method. The

result that obtained is compared with this picture:

(Pirim Setiarso, et al., 2015: 8)

Stain samples Solvent front

Less Polar Quite Polar Too Polar


D. Rf Factor

TheRf factor is used for the qualitative evaluation of a TLC separation. It is the

quotient of the distance of the substance zone from the sample origin to the front ofthe mobile

phase (zf). Historically, Goppelsrder was the first person to use thisRfvalue (relation to front

expression) to characterize planar separations.

where

zs distance of the substance zone from the sample origin (mm)

solvent from migration distance (mm)

distance between the immersion line and sample origin (mm)

By definition, theRf value cannot exceed 1. To avoid the decimal point, theRfvalue is

sometimes multiplied by 100 and then described as the hRf value. The value of the retardation

factor in a given separation system at constant temperaturedepends entirely on the

characteristic properties of the separated substances. It is important for identification

purposes that Rfvalues are accurate and reproducible,but this is difficult to achieve, since it is

almost impossible to adequately control all the experimental conditions that influence the

separation process.

This problem is avoided by defining a retardation factor for a standard

substance(Rst) that has been already separated in the system:

Here,

distance of the substance zone from the sample origin (mm)

distance of the standard substance zone from the sample origin (mm)


(Bend Spangenberg, et al., 2011: 23)

E. Ring oven technique

This technique relates to the testing of a single droplet of solution on a filter paper to

determine the components of the solution. The solution was dropped once on the center of the

filter paper (circular), then one or more components of the sample will be bound to the paper

by precipitation. Components dissolved moved with a suitable solvent so that the shift from

the center to the edges of the filter paper. For the solution reaches the edge, solvent

evaporation occurs and solute will be left behind in an area in the form of rings showed that

the separation of droplets spotting early.

(S.M. Khopkar, 1990: 130-131)

F. The content of suji leaf

Prangdimurti as listed in Back to Biology Site showed that the fresh leave that

contain water 73,25% of pandan suji contain 3.773,9 ppm chlorophyll that consist of 2.542,6

ppm chlorophyll a dan 1250,3 ppm chlorophyll b.

The structure of chlorophyll A


The structure of chlorophyll B

(Putri M. Wahyuni, 2014 http://backtobiologysite.blogspot.com)


VI. TOOLS AND MATERIALS

A. Tools

Aluminum foil

Arloji glass

Beaker glass

Capillary pipe

Chamber

Clamp

Measuring glass

Mechanical pencil

Metal Sheet

Mortal

Separated Funnel

Stative

B. Materials

Eluen with comparison:

Eluen Hexane Chloroform Ethanol

A 1 44,5 4,5

B 3 4 3

C 3 3 4

D 4 3 3

E 4,5 4,5 1

F 4,5 1 4,5

Pandan Suji leaves

Turmeric

Ethanol

Methanol

Chloroform


VII. FLOW CHART

A. Preparation of Sample

-

B. Preparation of Plate

Upper layer Bottom layer

Form two layer

filtrate residue

Pandan Suji Leaves

The color of solution dark enough

-Blend

-Weight 25 grams

- Add 15 mL of methanol

-Wait

Filtered

-entered with separated funnel

-Add with 25 mL chloroform

-Shake while open to release

-wait

Using as sample pigment

Take


Form two layer

filtrate residue

Turmeric


-Blend

-Weight 25 grams

- Add 15 mL of ethanol

-Wait

Filtered




-wait


Take

Plate (3x5)cm

Plate for concentrate

ring method

-take into oven 10 minutes

-sign 6 dots with pencil

distance between dots is 1cm

Plate (3x5)cm

Plate KLT for

determine Rf

-take into oven 10

minutes

-sign bottom border 1 cm


C. Preparation of Eluen

D. Dropping Sample Stage

a. For Concentrated Ring

b.

c. For Rf Determined

Filter Paper

Preparation of Rf done

-put into glass until close all the glass

but does not close the upper glass

-Put 5 mL the best eluen

-close with glass plate until filter paper soaked

Plate which is dotting (A-F)

with turmeric extract

The best ratio of eluen

-Put A-F eluen based on the

Spot that written

-Compare the ring that formed


with Pandan Suji extract

The best ratio of eluen


Spot that written


Plate which is dotting (A-B)


Rf

Immersed into chamber

Which is filled with

the best eluen



Rf



the best eluen


VIII. DATA OF EXPERIMENT

No. Procedure Data of Experiment Hypothesis

/Reactions Conclusion

Before After

A.Preparation of Sample

Pandan suji leaf =

green (+++)

Chloroform =

colorless

Pandan suji +

methanol = green

(+++)

-Residue=

green(+++)

-Filtrate=

green(+++) solution

-Filtrate + CHCl3=

two layer

Upper layer=

yelllow turbid

solution

Bottom layer=

green(++) solution

Pandan

suji leave

is non

polar and

turmeric is

non polar

too. Both

of them

can

dissolve in

CHCl3

which is

semi polar

solution

The color

of pandan

suji extract

is green

solution

(++)

The color

of turmeric

extract is

orange

solution

(++)

The color of pandan

suji extract is green

solution (++)


Form two layer

filtrate residue

Pandan Suji Leaves


-Blend

-Weight 25 grams

- Add 15 mL of methanol

-Wait

Filtered



-Shake while open to release, wait


Take


Turmeric = orange

(++)

Turmeric +

chloroform =

orangish yellow

(+++)

Filtrate = yellow

solution (++)

Residue = orangish

yellow (+++)

Filtrate +

chloroform = two

layer

-Upper layer =

yellow turbid

solution

-Bottom layer

yellow solution

(++)

The color of turmeric

extract is yellow

solution (++)


Form two layer

filtrate residue

Turmeric


-Blend

-Weight 25 grams

- Add 15 mL of ethanol

-Wait

Filtered




-wait


Take


B.Preparation of Plate

The color of plate

= white block =

silver

The color of plate =

white block = silver

The plate

is heat the

stationary

in plate to

decrease

the percent

water.

Plate after heat in

oven is ready to used.

C.Preparation of Eluen

Mixture

A=Colorless

B=Colorless

C=Colorless

D=Colorless

E=Colorless

F=Colorless

Hexane:

Chloroform:

ethanol

A 1 : 4,5 : 4,5

B 3 : 4 : 3

C 3 : 3 : 4

D 4 : 3 : 3

E 4,5 : 4,5 : 1

F 4,5 : 1 : 4,5

Hexane : non polar

Chloroform : semi

polar

Ethanol : polar

-Eluen is

ready to be

used

-The

function of

filter paper

is for

knowing

the

saturated

of eluen in

chamber.

Eluen is ready to be

used

Plate (3x5)cm

Plate for concentrate

ring method


-sign 6 dots with pencil

distance between dots is 1cm

Plate (3x5)cm

Plate KLT for

determine Rf


-sign bottom border 1 cm

and upper border 0,5 cm

With pencil

Filter Paper

Preparation of Rf done

-put into glass until close all the glass

but does not close the upper glass

-Put 5 mL the best eluen

-close with glass plate until filter paper soaked


D.Dotting of Sample Stage

-For Concentrated ring

A.Turmeric :

Orange ++

solution eluen

A,B,C,D,E,F :

colorless solution

pandan suji leaves

: green ++ solution

eluen

A,B,C,D,E,F :

colorless solution

The composition of

eluen that proper to

turmeric and pandan

suji leave is mixture

of E solution, with the

comparison

Hexane : chloroform :

ethanol. 4,5 : 4,5 : 1.

That show by E spot

is not spread to for

from the spot

-For Rf Determined

Turmeric = orange

++ solution

Pandan suji leave

= green ++

solution

Eluen E =

colorless solution

There are two spot

in pandan suji

leaves extract for

thin layer

chromatography the

distance is long

enough

There are three spot

in turmeric for TLC

the distance is near

each other.

The Rf of turmeric

are

Rf1 = 0,727, Rf2=

0,818, Rf3=0,864

The Rf of pandan

suji leave are

Rf1 = 0,773

Rf2 = 0,982

Tumeric

Extract

separated

in 3

component

.

Pandan

extract

separated

in 5

component

.

The Rf of turmeric are

Rf1 = 0,727, Rf2=

0,818, Rf3=0,864

The Rf of pandan suji

leave are

Rf1 = 0,773

Rf2 = 0,982



The best ratio


Spot that written




The best ratio of


Spot that written


Plate which is dotting (A-

B) with turmeric extract

Rf

Immersed into

chamber


the best eluen



Rf



the best eluen


IX. EXPLANATION

1. Preparation of sample

Thin layer cromatography experiment has purpose todetermine the proper

composition of eluen using concentrated ring method and determine the value of Rf from

plant dye using thin layer chromatography plate. First we had to prepare the sample. We used

ripe tumerics and pandan suji leafs as sample. We cleaned tumerics and pandan suji leafs.

Than we grinded the tumerics, orange (++) and pandan suji leafs (+++) until smooth. After

that we weighed the tumerics and pandan suji leafs, the tumerics was ----gram and pandan

suji leafs was -----gram. Then we dissolved the pandan suji leafs with 10 mL of methanol,

colourless. And we dissolved the tumerics with 10 mL of ethanol, colourless solution. We did

it to dissolve the content of panadan suji and tumeric.

Pandan suji has a substance called chlorophyll, chlorophyll had green dye in its

leafs, while there was a substance in turmeric called carotenoids which gave a strong yellow

color of turmeric that was carotene. Both of these campound was easy to be studied its

chromatographic process.

The structure of chlorophyll A

The structure of chlorophyll B


And the structure of carotenoids

Chlorophyll was organic compound that had a properties of hydrocarbon like which

is non polar hydrocarbon compound with the polar part of carbonyl group. Chlorophyll B had

aldehyde group which was more polar than alkene group of chlorophyll A. In suji leafs,

chlorophyll A was more dominant. We used methanol as solvent because methanol was the

most polar alcohol, and chlorophyll easy to dissolve in alcohol.

Tumeric contain-carotene and -carotene, both of them were non polar and soluble

to less polar organic solvent like alcohol. Ethanol was less polar organic solvent. So tumeric

was soluble in ethanol.

We set aside the mixture of pandan suji leaf with methanol and the mixture of

tumeric with ethanol for 20 minutes, to give the time for the chlorophyll of pandan suji

dissolved in methanol and the carotene in tumeric dissolved in ethanol, until the colour of the

solution became dark enough, it was a sign that the content of tumeric dissolved in ethanol

and pandan suji leave dissolved in methanol. The colour of tumerics and ethanol became

orangish yellow (+++) and the colour of pandan suji leafs with methanol was green (+++).

After that we did decantation to the mixture of tumeric with ethanol using spatula

and we put the filtrate into separated funnel and the residue still in beaker glass. The residue

was orangish yellow (+++) and the filtrate was orangish yellow (+++). And we did

decantation to the mixture of pandan suji leafs with methanol and using spatula and we put

the filtrate into separated funnel and the residue still in beaker glass. The colour of residue

was green (+++) and the colour of filtrate was green (+++). Decantation process was

-carotene

-carotene


preferred over maintaining the filtration due to chlorophyll pigments in the filtrate. If using a

filtration process, chlorophyll possible to be filtered on filter paper. Similarly, the carotene

pigment in turmeric, decantation process used was for maintaining the carotene pigment.If

using a filtration process, carotene possible to be filtered on filter paper.

Then both of the filtrate was extracted in separated funnel with the addition of

semipolar solvent of chloroform, colourless solution. We shaked the separating funnel and we

opened the faucet to exit the gas result from chloroform. The addition of chloroform due to

get pure chlorophyll and carotene from other contaminants in water phase. The organic phase

and the water phase of pandan suji extract and tumeric extract separated. Both of them form

two layer. In pandan suji leafs, the upper layer was yellow turbid solution and the bottom

layer was green (++) solution. In tuemric, the upper layer was yellow turbid solution and the

bottom layer was yellow (++) solution. The bottom layer was used for analysis, so we got the

colour of tumeric extract was yellow solution (++). The colour of pandan suji extract was

green (++) solution.

2. Preparation of plate

We prepared plate 4 plates, two plates size 3 cm x 5 cm and two plates size 2 cm x 5

cm. We sign 6 dots in in plate size 3 cm x 5 cm, the distance between dots was 1 cm. We

gave name A, B, C, D, E, and F. The purpose of the distance of each dots was for gave place

for the sample development. And the dots was given to make the development of sample and

its spread clearer, the pencil dots not react with sample and eluen. We signed the plates with

size 2 cm x 5 cm, 1 cm from bottom border and 0,5 cm from upper border with pencil. We

gave two dots and gave name A and B. Then we put the plates in oven for 10 minutes. The

purpose we put the plates in oven was for activate the stationary in plate and removed the

water content in plates. The content of water in plates would affected the sample

development, if there was water content the sample development process would be slow and

affected the value of Rf.

3. Preaparation of eluen and concentrated ring

Composition of 6 eluens which were supplied in analytics laboratory:

Eluen A composed by comparison of hexane : chloroform : ethanol = 1 : 4,5 : 4,5

Eluen B composed by comparison of hexane : chloroform : ethanol = 3 : 4 : 3

Eluen C composed by comparison of hexane : chloroform : ethanol = 3 : 3 : 4

Eluen D composed by comparison of hexane : chloroform : ethanol = 4 : 3 : 3

Eluen E composed by comparison of hexane : chloroform : ethanol = 4,5 : 4,5 : 1


Eluen F composed by comparison of hexane : chloroform : ethanol = 4,5 : 1 : 4,5.

All the eluen above was colourless solution, hexane was non polar solution,

chloroform was semipolar solution, and ethanol was polar solution. After the plate was ready

to be used. We dropped each dots of plate 3 cm x 5 cm with tumeric extract and the other

plate 3 cm x 5 cm with pandan suji extract used capillary pipe. We used capillary pipe

because capillary pipe was small enough to use for dropped the sample and eluen. After that

we dropped the eluen A, B, C, D, E, and F in each dots A, B, C, D, E, and F.

Dot A, B, C, D and F showed the spot was spread far from the initial spot, so eluen

A, B, C, D, and F was too polar. The eluen E was polar enough for tumeric sample and

pandan suji leafs. The result was based on the spread of spot in concentrated ring was not too

far and not too small, it was said the spread was enough.

4. Dropping in concentrated ring and determination of Rf

For determine Rf we put filter paper into glass, we put of 5 mL of right eluen based

on concentrated ring, eluen E was the right eluen based on our experiment that we would

explain more in dropping in concentrated ring. And we closed the glass chamber until filter

paper all soaked. The function of filter paper was for knowing the saturated of eluen in

chamber. The wet of filter paper showed that the chamber was saturated with vapor of

solvent. We closed the glass to make the chamber was saturated with the eluen.

The plates with size 2 cm x 5 cm, 1 cm from bottom border and 0,5 cm from upper

border with pencil. We gave two dots and gave name A and B. Each dots we dropped with

tumeric extract, and the other plate we dropped with pandan suji extract. We opened the

chamber and we put the plate used tweezers, the bottom part of plates must touched the

bottom of chamber and the plates must be upright so the spot would not be sloping. We set

aside until the eluen reached the upper border of plate. We took the plate, and we marked the

spot with pencil while we dried the plates then we covered with cellophane tape. We got

Concentrated ring of Pandan Suji Concentrated ring of Tumeric


three spot for tumeric extract and two spot for panadan suji extract. We should get 6 spot of

pandan suji extract but we just got two, we woul discuss in discussion.

The next step was calculated the Rf of tumeric extract and pandan suji extract. We

used formula

Rf determination of Pandan Suji Rf determination of Tumeric


From the calculation that we showed in attachment we got the Rf of tumeric extract was

= 0,727; = 0,818; = 0,864. And Rf of pandan suji extract was = 0,773; =

0,982.

X. DISCUSSION

In the experiment to determine the Rf of pandan suji we just got two spot because we

did not do the procedure well. We did not mark and cover the plate directly. The eluen

and spot had been evaporated, and got two spot only.

XI. CONCLUSION

From the experiment that we did, we could conclude that:

1. Based on explanation that we did about concentrated ring in pandan suji and tumeric

sample the suitable eluen was eluen E which composed by the comparison of hexane :

chloroform : ethanol = 4,5 : 4,5 : 1; because the spot of eluen E was polar enough, the

spread of sample from the initial spot was not too far.

2. The value of Rf in tumeric and pandna betawi dye pigment as follows:

A. Rf of tumeric: = 0,727; = 0,818; = 0,864

B. Rf of pandan suji: = 0,773; = 0,982.

XII. QUESTION AND ANSWER

1. What will happen if eluen that used as solvent in Thin Layer Chromatography too

polar or less polar? Why?

If the eluen that used in TLC was too polar, the spot that we dropped in plate

would up to upper border without any sepration because there were no interaction

between eluen and the stationary phase in plate which was non polar.

If the plate that used as TLC was less polar, the spot that dropped in plate would

not be move because non polar eluen would get interaction with the stationary

phase in plate which was non polar.

2. What is the function of filter paper in the experiment determination of Rf?

The function of filter paper in the experiment determination of Rf was for knowing the

saturated of eluen in the chamber, it is marked by the all part of filter paper is wet.

3. Why the surface of Thin Layer Chromatography can not be broken?


The surface of TLC can not be broken because can affect the spot separation result in

development of spot. If the stationary phase is broken, the line of spot is broken or

disconnect and affect the value of Rf.

4. Why the TLC plate that used must be heat in oven?

Thin layer that we used must be dried first in oven in temperature 1100 before we used

to remove the water molecule that binded and activate the stationary phase of plate

(absorbent).

5. Why the upper border and bottom boder of plate must be marked with pencil?

We used pencil as border to make it clear the upper and bottom border, and pencil not

react with solvent and sample.If we used bolpoint, the spot of bolpoint would eluted

and develop.


REFERENCES

Departement of Chemistry and Biochemistry University of Colorado. 2015. Procedure of

TLC.(Online),(http://orgchem.colorado.edu//Technique/

Procedures/TLC/TLC.html, accessed on May 6th

2015).

Grushka, Eli and Nelu Greinberg. 2014. Advance in Chromatography. Massachusettes:

Taylor & Francis Group.

Hahn-Deinstrop, Elke. 2007. Applied Thin-Layer Chromatography. Weinheim: WILEY-

VCH Verlag GmbH & Co. KgaA.

Harvey, David. 2000. Modern Analytical Chemistry. USA: Mc Graw Hill Company.

Jeffery, G.H. et al. 1989. Vogels Textbook of Quantitative Analytical Chemical Analysis 5 th

Edition. London: Longman Group.

Pirim Setiarso, et al. 2015. Petunjuk Praktikum Kimia Analitik II (DDPK). Surabaya:

FMIPA.

Putri M. Wahyuni. 2014. Pigmen Klorofil DaunSuji (Pleomele Angustifolia) Sebagai

Alternatif Pewarna Makanan Alami, (Online)(http://backtobiologysite.

blogspot.com/2014/04/pigmen-klorofil-daun-suji-pleomele.html, accessed on

May 9th

08.00).

S.M. Khopkar. 1990. Konsep Dasar Kimia Analitik. Jakarta: Universitas Indonesia.

Spangenberg, Bernd et al. 2011. Quantitative Thin Layer Chromatography A Practical

Survey. Germany: Springer-Verlag Berlin Heidelberg.

Surabaya, May 18th

, 2015

Mengetahui,

Dosen/Asisten/Pembimbing, Praktikan,

(.) ()


ATTACHMENT

A. Preparation of Sample

Pandan Suji leaves 25 gram Pandan Suji that

had been pounded

10 mL of methanol,

colorless solution

Waiting for extract of

pandan suji+methanol

Extract of pandan suji

has green color

Green ++ color

solution

Light green color

solution


Shaking the separated funnel to

separated it well

Separated the solution to

get the extract

25 gram of Turmeric that had

been pounded has orange

color

10 mL of ethanol has

colorless solution

Waiting for extract of

turmeric+ethanol

The extract of turmeric was

poured into separated funnel

Extract of Pandan

Suji

Mix ponded turmeric

and 10 mL ethanol

Shaking the separated

funnel to separated it well


B. Dotting of Sample Stage

Extract of turmeric

which is formed 2 layers

Turmeric extract

Dotting the metal sheet and give the

label

The metal sheet that had been dotting


The metal sheet for

Rf determined

A Eluen, 1:4,5:4,5

(hexane:chloroform:ethanol)

B Eluen 3:4:3

(hexane:chloroform:etha

nol)

Darker color of

turmeric extract

Light color of turmeric

extract


C Eluen 3:3:4


F Eluen 4,5:1:4,5


D Eluen 4:3:3


Metal sheet that had been dotting with turmeric extract

and give eluen

E Eluen 4,5:4,5:1


Dotting the metal sheet


Metal sheet that had been dotting with Pandan Suji

extract and eluen


Chamber that will used

Metal sheet that used to determined

Rf was immersed with E eluen

Metal sheet of turmeric extract

that had been immersed

Metal sheet of Pandan Suji

extract that had been immersed


CALCULATION

Turmeric

1. R3A = 4,7 cm R3B = 4,8 cm Reluen = 5,5 cm

Rf3A =

=

= 0,855

Rf3B =

=

= 0,872

Rf3eluen =

=

= 0,864


Rf2A =

=

= 0,818

Rf2B =

=

= 0,818

Rf2eluen =

=

= 0,818

3. R1A = 4 cm R1B = 4 cm Reluen = 5,5 cm

Rf1A =

=

= 0,727

Rf1B =

=

= 0,727

Rf1eluen =

=

= 0,727

Pandan suji


Rf1A =

=

= 0,764

Rf1B =

=

= 0,782

Rf1eluen =

=

= 0,773


Rf2A =

=

= 0,982

Rf2B =

=

= 0,982

Rf2eluen =

=

= 0,982

klt

Documents

phase properties

mobilephase composition

mobile phase mp

layer chromatography

mobile phase passes

type of stationary phase

layer chromatography

new tlc method