Hi i have a few questions to ask you people out there. I have just started a science course and we just started chromatography and some of my assignment questions are fairly hard and i cant seem to find any of the information:

1.Why is it important to stop development time before the mobile phase reaches the top of the paper??

2. If two chromatograms beign developed with exactly the same stationary, mobile and analyte phases, except one was developed for twenty minutes and the other for two hours. Would there be a change in Rf values for the same components(explain your answer)?

3. Why do some components exhibit excessive tailing?

4. How could we use column chromatography to remove chloride ions from contaminated water?

Question 1: When the mobile phase reaches the top of the paper it should stop flowing, except for the effects of evaporation which will keep it flowing upwards (although at a lower rate). As a result, sample components keep moving although the solvent front has stopped; therefore, any measurements of Rf are meaningless.
Question 2: In theory there should be no difference, but again there are the possible effects of evaporation which would cause the mobile phase to move faster than the position of the solvent front would indicate. That is why it is important to use a well-sealed chamber for paper chromatography, especially when the solvents are volatile or the development times are long.
Question 3: One possibility is that these components are interacting with the paper by two or more mechanisms. One is a stronger interaction which causes the components to move more slowly. A molecule coming off the paper into the mobile phase can reinteract by either of the two mechanisms. The result would be a strong spot or band with a streak of more strongly retained material following behind, like the tail of a comet. Near the front of the streak, the molecules have interacted in the weak way most of the time; near the back, the molecules have interacted in the strong way most of the time.
Question 4: One way to do it is by anion-exchange chromatography (look it up). An anion-exchange column has immobile cations; if the counterion (which is in the solution) is hydroxide, when you pass the contaminated water through the column the chloride exchanges with the hydroxide and forms a salt with the immobile cations. In the water, the chloride has been replaced by hydroxide.
I notice you have not had any answers before this one and hope it arrives soon enough to be useful. This is my first visit to the forum so I couldn't have answered earlier. Good luck, and I hope you enjoy the science course.

hi im doing an asignment also, please could u asner the following questions. it is do with gas liquid chrmatography.

what is the problem with the chromatogram produced at 60 degrees?
In which order do the n- alkanes separate on the chromatogram?
Will the retention time of a peak change if the ovewn temperature is changed?
What other factors affect the retention time of a compound?

I would be very grateful if u could help me as im a complete novice when it comes to chromatography

Thankyou- Michael

Hi Michael
1. What problem are you seeing and what system are you running.
2. This may be dependant on GC column but for example on an econo cap EC5 30M x 0.53mm ID x 1.2um Alltech GC (I am not from Alltech by the way and do not sell the columns, Helium carrier at 8ml/min, 40 to 310 degree ramp over 12 min and FID set to 310 and injector at 280, the retention will be nDecane (C10), nDoDecane (C12), nTetraDecane (C14), nHexaDecane(C16), nOctaDecane(C18), nEicosane(C20), nDocosane(C22), nTetracosane (C24). These are the constituents of Diesel for example.
3. Yes the retention eill change. The higher the temp or the quicker the speed to ramp the temperature will result in a shorter retention and sharper, more sensitive chromatography. The down side is that you may get peaks merging into each other or they may just come through with the injection front and therefore not be quantifiable.
4. a)the type of column, length of column, diameter of column, packing material, injector temperature, carrier gas flow, split ratio (only if overloading is occurring without split.
Hope this helps. Contact me for more detail

Hi,

I would like to know if itīs normal that the area dicreases when the flow rate is increasing ( inverse proportionality)?

Thanks

Damien

This is already answered in another conversation.Have a look at:

Should Flow Rate Effect Detector Response Like This?

And the answer is:
By Uwe Neue on Tuesday, March 13, 2001 - 04:41 pm:

Your detector sees concentration. Your integrator creates concentration x time as peak area. The idea of using peak area is to get mass. To get mass, you would need to multiply the peak area of your integrator with the flow rate ( mass = (mass/volume)* time * (volume/time) ).
If I do this for your three flow rates, I get 1332, 1338, and 1344, the same number within 0.5%. The remainder are either integration errors or flow rate imprecision.

wich is better for playing with temperatures
below 0 nitrogen or CO2?

Hi guys i have a problem with my Thin Layer Chromotography Assignment, could you please help me to start a Conclusion and discussino about TLC? there are 3 spots on the plate, the first one is a small amount of benzoic acid in 5 mL of methanol + 1 drop of concentrated HCL. The Second spot is a small amount of benzoic acid dissolved in 2mL of methanol. The Third Spot is a small amount of methyl benzonate dissolved in 2 mL of methanol. The mobile phase is a solution of cyclohexane, thyl acetate, dichloromethane and formic acid (conc.) om the ratio 11:6:4:1. Volumes of 5,5; 3; 2; and 0,5 mL. acetate could be replaced by butyl acetate.
The spots are located as follow: 1st spot=(22mm Rf= 0.37) and (34 mm Rf= 0.57) the distance moved by mobole phase is (60 mm) the 2nd spot ( 21 mm Rf = 0.35 mm) the 3rd spot (36 mm Rf= 0.60).

The questions are as follow: 1) Write a balanced equation for the reaction in part 1 above.
calculate the average Rf value of each spot on the plate. (this is answered)

2) Identify the probale composition of each spot, explaining your reasoning.

3) carefully explain why there is a seperation of components of spot 1. Look at the molecular structure of the components of the spots and the mobile and stationary phases.


4) If the same procedure is repeated but the cyclohexane is ommited from the moblie phase, all spot components are carried to the top of the solvent front and no seperation occurs. Explain why this would be so.

THANKS a LOT Simon

Hi guys i have a problem with my Thin Layer Chromotography Assignment, could you please help me to start a Conclusion and discussino about TLC? there are 3 spots on the plate, the first one is a small amount of benzoic acid in 5 mL of methanol + 1 drop of concentrated HCL. The Second spot is a small amount of benzoic acid dissolved in 2mL of methanol. The Third Spot is a small amount of methyl benzonate dissolved in 2 mL of methanol. The mobile phase is a solution of cyclohexane, thyl acetate, dichloromethane and formic acid (conc.) om the ratio 11:6:4:1. Volumes of 5,5; 3; 2; and 0,5 mL. acetate could be replaced by butyl acetate.
The spots are located as follow: 1st spot=(22mm Rf= 0.37) and (34 mm Rf= 0.57) the distance moved by mobole phase is (60 mm) the 2nd spot ( 21 mm Rf = 0.35 mm) the 3rd spot (36 mm Rf= 0.60).

The questions are as follow: 1) Write a balanced equation for the reaction in part 1 above.
calculate the average Rf value of each spot on the plate. (this is answered)

2) Identify the probale composition of each spot, explaining your reasoning.

3) carefully explain why there is a seperation of components of spot 1. Look at the molecular structure of the components of the spots and the mobile and stationary phases.


4) If the same procedure is repeated but the cyclohexane is ommited from the moblie phase, all spot components are carried to the top of the solvent front and no seperation occurs. Explain why this would be so.

THANKS a lot Simon
simoncity@web.de