Chromatography Forum

Hi,
Ive to develop an extraction procedure based on the following example but I can’t out how the original author came up with the concentrations:

Method 1:
20 ul of compound x (1000ng/20ul or 50,000ng/ml) is added to milk (1g). 4 mls of methanol is added. After shaken and centrifugation the methanol layer is removed and evaporated under nitrogen. The extract is resuspended in 1 ml of ethyl acetate. 100 ul of this is diluted in 900 ul of buffer before analysis.
This results in a final concentration 50 ng/ml in the extract.

Method 2:
20 ul of compound x (1000ng/20ul or 50,000ng/ml) is added to milk (1g). 4 mls of methanol is added. After shaken an centrifugation the methanol layer is removed and evaporated under nitrogen. The extract is resuspended in 1 ml of hexane and 250 ul of water/solvent mix. After shaken and centrifugation 200 ul of water/solvent mix is removed and evaporated to dryness. The extract is resuspended in 1 ml of ethyl acetate. 100 ul of this is diluted in 900 ul of buffer before analysis.
This resulted in a final concentration of 40 ng/ml in the extract.

I can’t understand the dilution factor. It looks like compound x was diluted by 1000 in method 1 and 1250 in method 2 but how?

Method 2:
20 ul of compound x (1000ng/20ul or 50,000ng/ml) is added to milk (1g). 4 mls of methanol is added. After shaken an centrifugation the methanol layer is removed and evaporated under nitrogen. The extract is resuspended in 1 ml of hexane and 250 ul of water/solvent mix. After shaken and centrifugation 200 ul of water/solvent mix is removed and evaporated to dryness. The extract is resuspended in 1 ml of ethyl acetate. 100 ul of this is diluted in 900 ul of buffer before analysis.
This resulted in a final concentration of 40 ng/ml in the extract.

I can’t understand the dilution factor. It looks like compound x was diluted by 1000 in method 1 and 1250 in method 2 but how?

You should lose 20% of target analyte compared to method if I see right, did not check calculations though.

200/250 = 40/50

Peter

The assumption here is that the compound X is completely partitioned into the methanol in method 1.

The whole methanolic phase is evaporated (assuming 100% recovery of 'X') and then reconstituted into the 1mL of EtoAc.

Likewise, the assumption is held that the compound X is completely partitioned into the 250 microliters of water/solvent mix phase in method 2 after it has been partitioned into the methanol and then evaporated as was done in method 1, but then redissolved into the hexane and water/solvent mix.

But instead of the entire phase reconstituted into EtoAc as in method 1, only 80% of the water/solvent mix (200 mcL) is taken for reconstitution into EtoAc. This 50 mcL remains behind containing 20% of compound X of the 50,000 ng that was added initially to the milk.

Peter and Krickos are both quite correct.

Do you see where the loss of 10,000 ng of compound X occurs in method 2?

best wishes,

Rod

This sounds like a residue method. It helps if you think of the dilution as happening to the milk, and not the "compound." The 20 uL of "compound" is just a way of getting 1000 ng of the analyte X of interest into 1 g of milk.

So, you've got 1 g of milk with concentration 1000 ng X per g milk.

In method 1, all of the extract ends up in 1 mL EtOAc--1 g milk is pretty close to 1 mL, so no dilution--and in turn this is diluted 1:10 (100 to 100 + 900). So the net dilution is 1:10. If you had 100% recovery, your extract concentration would be 1000/10 or 100 ng/mL. If you only got 50 ng/mL, your recovery was only 50%.

Why do you only have 50% recovery?

For method 1:

1ug (amount of your analyte)/20ul -> added to 1g of milk:
1ug/1g (of milk)
1ug/1g (of milk) + 4 ml MeOH
1ug/x ml of MeOH layer (if 100% recovery) possible place of analyte loose
1ug/evaporated (to dryness ?) possible place of analyte loose
1ug/1ml (ethyl acetate)

You take 100 ul of it so...

0.1ug/100ul

0.1ug/100ul+900ul (buffer)
0.1ug/1ml

Fortification was 1ug/1g
Concetration in vial is 0.1ug/1ml if you have 100% recovery
Dilution factor is 10 for method "1"

For method 2:
1ug (amount of your analyte)/20ul -> added to 1g of milk
1ug/1g (of milk)
1ug/1g (of milk + 4ml MeOH
1ug/x ml of MeOH layer (if 100% recovery) possible place of analyte loose
1ug/evaporated (to dryness ?) possible place of analyte loose
1ug/1ml (hexane) + 250ul (water+solvent) possible place of analyte loose
1ug/250ul (water+solvent) if all analyte move to this layer

Now you take 200ul of water solvent layer so:
0.8ug/200ul
0.8ug/ evaporated to dryness
0.8ug/1ml (ethyl acetate)
0.08ug/100ul
0.08ug.100ul+900ul of buffer
0.08ug/1ml

Fortification was 1ug/1g
Concentration in vial is 0.08ug/ml if you have 100% recovery
Dilution factor is 12.5 for method "2"

Hi,
Ive to develop an extraction procedure based on the following example but I can’t out how the original author came up with the concentrations:

Method 1:
20 ul of compound x (1000ng/20ul or 50,000ng/ml) is added to milk (1g). 4 mls of methanol is added. After shaken and centrifugation the methanol layer is removed and evaporated under nitrogen. The extract is resuspended in 1 ml of ethyl acetate. 100 ul of this is diluted in 900 ul of buffer before analysis.
This results in a final concentration 50 ng/ml in the extract.

Method 2:
20 ul of compound x (1000ng/20ul or 50,000ng/ml) is added to milk (1g). 4 mls of methanol is added. After shaken an centrifugation the methanol layer is removed and evaporated under nitrogen. The extract is resuspended in 1 ml of hexane and 250 ul of water/solvent mix. After shaken and centrifugation 200 ul of water/solvent mix is removed and evaporated to dryness. The extract is resuspended in 1 ml of ethyl acetate. 100 ul of this is diluted in 900 ul of buffer before analysis.
This resulted in a final concentration of 40 ng/ml in the extract.

I can’t understand the dilution factor. It looks like compound x was diluted by 1000 in method 1 and 1250 in method 2 but how?

Method 2 looks a nightmare ! How will you ever get a stable mixture of 2 immiscilble solvents