I prepared samples for ICP-MS analysis. During digestion I mixed about 0.5 g of soil samples and final solution has a volume of 50 ml. Now, the solution was still concentrated according to laboratory technician. She took 1 ml and mixed it with 10ml pure water. During sample counting the mass and dilution factor were not entered into the windows menu to be calculated into the results. ICP-MS results came out in ppb. Example, 431.31 ppd was recorded.
Work backwards.
Your ICP-MS results are reflective (assuming it is properly calibrated) of the sample that you placed in it. Because your technician mixed 1 mL with 10 mL of pure water, you have a solution that is 11-fold diluted (i.e. has a dilution factor of 11). Confirm that this is the case and she didn't actually mix with 9 mL pure water, which would produce a 10-fold dilution (dilution factor of 10) - which is a round number that is easier to work with.
When interpreting your ICP-MS result, multiply it by the dilution factor to receive the concentration in your undiluted solution. Then, multiply by the volume of the undiluted solution (50 mL) to determine the mass of the analyte in the solution. Since your 50 mL solution is reflective of the 0.5 g of soil sample you used to create it, you can compare the mass of the analyte (converted to mg) to the mass of soil (converted to kg) to determine your results in mg/kg
431.31ppb=0.43131mg/Kg=0.00043131mg/g
0.00043131mg/g x 10 x 50 = 0.2156 mg/g = 215.6 mg/kg
0.5 g in 50 mL, final volume is 50 mL, dilution is (0+50)/0.5= 100
1 mL +10 mL, final volume is 11 mL, dilution factor is (1+10)/1=11
Total dilution factor is 100x11 = 1,100 x
GC-MS gave 431.31 ppd (I assume this is ppb)
Equal to 431.31 ug/Kg
Equal to 431.31/1000 mg/Kg =0.43131 mg/Kg
0.43131 mg/Kg x 1,100 dilution factor
Answer=474.441 mg/Kg sample
Formula
(GC-MS value/1000)*total dilution factor
Thanks you very much everybody for clarifying on this challenging approach of ICP-MS samples analysis. You all contributed equally and this has recharged my knowledge and interest in mass spectroscopic techniques. The document from the ICP-MS is attached here.
Same answer as someone else got, but this is the way I think of it and work through in my spreadsheets:
-Step 1 - Take your non-dilution corrected value (ppb) and multiply by the dilution factor (11)
-Step 2 - Take your dilution corrected ppb (ug/L) and multiply by the volume of liquid (in liters) you used to pickup your solid (0.05L) to get ug of mass in the liquid
-Step 3 - Divide that number by your original mass in grams of soil that you digested. You then have ug/g, which is the same as mg/kg
1. 431.31 ppb (concentration non-dilution corrected) x 11 = 4744.41 ppb (concentration dilution-corrected)
2. ppb is the same as micrograms/L, so 4744.41 ug/L x 0.050 L = 237.22 ug of mass of your element was in that solution
3. Divide the 237.22 ug by 0.5 g original mass = 474.44 ug of the element per g of media (or mgX/kg media)
Umer Bakali Vaino Indongo
Thanks for Vaino and all the good answers. I would like to ask a similar question:
I had quite similar experiment steps to yours:
0.1 gram of sample was digested and filled up to 50 ml. then 1 ml thereof was diluted to 10 ml (1:10) for ICP measurement.
The result was 0.91 mg/l.
Therefore, I am confused about dilutions,
1. did I dilute the sample 2 times?? (DF: 50ml/0.1g * 10 =5000??)
2. OR I just care about the second dilution 1:10?
(factorsanswers provided different dilution factor)
Thank you so much and looking forward to any help!!