I have followed the modified BCR method to study the speciation of trace metals in the sediment samples. As per the procedure I have taken 1g of sediment sample and extracted it with different reagents to find the metals bound to carbonates & acid extractable fraction, Fe/Mn oxides, organic matter & sulphides and Residual fractions. For the first two steps I added 40ml of prepared reagents and for the third step I have added 50ml of reagent. For the final fraction I have digested the remaining residue from all the steps with aqua-regia and make upto 100ml. Then I have analyzed all the four extracted fraction in the AAS.
For Example [if the values directly recorded from AAS]
Element Cu (AAS reading) Fraction 1(11.44), Fraction 2(1.734), Fraction 3(0.202), Residual Fraction(0.086) means whether I have to do the
Dilution factor Calculation like this 11.44*40; 1.734*40; 0.202*50; 0.086*100;
My question is "whether for calculating the dilution factor, I have to multiple the reading value which I got form AAS with amount of acid which I added to the sediment in each separate fractions or I have to use any other different calculation factor"?
Dear Kumar
For dilution factor you should divide the volume of your final solution by the weight of sediment used. For example 50mL/1g=50. That means you should multiply the AAS values by 50. Also we should know whether you have use sequential or non-sequential method for digestion.
karbassi
Dear Prof. Karbassi,
Thank you for your kind reply. I have used the sequential method for digestion.
Regards,
Anand.
Dear Kumar,
Your calculation way is true, no problem. Because, your sample weight is 1.00 g and in the first BCR step you used 40 mL of 0.11 M acetic acid solution. First, after the measurement of this solution in AAS, the absorbance value should be converted to the concentration unit by using the calibration curve, let's say you found 11.44 mg/L (or µg/mL) Cu. Next step is in the folowing:
[(11.44 µg/mL) x 40 mL] / (1.00 g) = 11.44 µg/g Cu (or 11.44 ppm Cu). That is in this calculation your dilution factor is 40. The only condition is the volumes must be measured truly at first and in the end of the experiments.
Regards,
Senol Kartal
I have a question; I have a dilution factor of 500 (Final volume=50ml, initial sample=0.1 g), while the readings of AAS for minerals is for example, 32 ppm. should I like multiply my readings with dilution factor.? it goes like 16000, is it correct?
I have a confusion, whether I am doing right or wrong, please tell me.
I have taken 0.5059g of fortified rice powder and which is digested with 10ml of mixed sulphuric acid and nitric acid(3:1). Then after digestion, milipore water added and made up the volume to 100ml. AAS reading was 0.2887ppm or mg/L.
I converted or calculated following this formula
=(AAS reading x volume make up x dilution factor)/Sample weight
so dilution factor is = volume make up/weight of sample=100/0.5059=197.66 (is it right?)
so= 0.2887 x 100 x 197.66)/0.5059
=11279.78microgram/kg
=11.27978mg/kg
=0.00127978mg/gm
=0.127978mg/100g
If wrong correct me. I need your help.
Hi Elizabeth,
Sorry for the delay in reply. The final calculation has to be like this
=(AAS reading x volume make up)/Sample weight
0.2887*100/0.5059
=57.06 mg/kg.
If you have any queries let me know.
Regards,
Anand.
The formulae for this calculation is:
Conc. (mg/kg) = AAS reading x Dilution factor (for water analysis)
where the dilution factor = Volume of digest/Volume made up to
But for solid samples
Conc. (mg/kg) = AAS reading x Dilution factor/Mas of sample digested
Sorry, i have a question with regard to Elizabeth question. How to calculate dilution factor in solid sample. As i know in solutions we just deal with final volume/initial volume but in solid we have weight of sample which dissolved in certain liquid(reagent) and diluted again to a certain amount.
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