[Sara8]

Hi all  ,
This is my first time here so I hope I can  find some *delete me*
I am doing HPLC-FLD analysis of fumonisins from rice and I am stuck with the dilution factor that I have to consider when I do all the calculations about my samples with respect to the calibration curve.

The process is as it follows:
1. Mix 5 grams of the grain with 50 ml of acetonitrile/water (50/50, v/v) and blend 5 min.
2. Filter the mixture through Whatman No 1 filter paper into 100 ml Erlenmeyer flasks.
3. Take two ml of the filtrate and dilute it in 5 ml of 1% (w/v) KCl aqueous solution.
4. Several steps to condition C18 cartridge.
5. Pass the sample volume (7 ml) through C18 cartridge.
6. Wash the cartridge with 3 ml of 1% (w/v) KCl aqueous solution and 2 ml of ACN/1% KCl (20/80, v/v).
7. Elute the fumonisins with 2 ml of methanol/water (70/30, v/v) into Eppendorf tubes.
8. Allow them to dry
9. Suspend the samples in 200 µl methanol, take 50 µl to sylanized HPLC vials and add 200 µl of derivatization agent. Then inject in chromatograph.

Thanks in advance!

[AWK]

Forget points 2 to 7.
From point 1, 8 and 9 you know that you injected sample contains fumonisins from 50/200 =1/4 of your 5 g rise sample. The rest calculations depends on units of your calibration curve.

[Sara8]

Shouldn't I consider that I am diluting 2 ml of my filtrate with 5 ml of another solution?

[AWK]

No, but my fault, point 3 is important, though unclear. Acetonitrile is quite volatile and during filtration one may expect substantial loss of solvent. You take 2 ml of this solution (filtrate from operation 2) from unknown volume (assuming 50 ml or filtrate should be slightly diluted with solvent to volume 50 ml in volumetric flask).

The following operations, up to evaporation (point theoretically  do not change amount of fumosinins.

So additional factor 2/50 is needed (you analyse (2/50) x (1/4) = 1/100 of you sample content).