Which of these three methods for calculating the concentration of a metal sample is true?

Bayu Adi Samodro

asked question --

Which of these three methods for calculating the concentration of a metal sample is true?

I am kind of confused calculating the concentrations of some metal samples in my work. I digested a sample with acid then diluted it in a 250 ml volumetric flask. When I measured with AAS the absorbance is very high; higher than the the absorbance of the greatest concentration of standard solution. So I diluted 25 ml of sample in a 100 ml volumetric flask, which makes the dilution factor (DF) 4. If the absorbance of diluted sample in 100 ml has absorbance X and the absorbance of sample blank is Y, which is true:

A. take the absorbance of sample (X) minus blank absorbance (Y) then multiply with the dilution factor (DF) and to get the concentration using the calibration curve.

(X-Y)*DF

B. the absorbance of sample (X) multiplied by the DF then minus blank absorbance to get the concentration using the calibration curve.

(X*DF)-Y

C. find the concentration with calibration curve first then multiply.

answer

I suggest you should,

1. Find concentration of (sample) X and (blank) Y first from the calibration curve.

2. Minus the concentration of the blank from the sample ( X - Y)

3. Multiply concentration of sample X by D.F

Please note that sometimes high results can mean there is contamination during analysis. If you suspect this, do a repeat.

Also, I advise you to run a known concentration (reference) sample throughout the whole analysis even during dilution steps; a leaf or soil reference would suffice.

Australian Nuclear Science and Technology Organisation

Hello there,

The answer is simple. If the concentration of an analyte in the diluted sample is X (found from the calibration graph), then the concentration of analyte in the original sample is as follows:

concentration of analyte in sample= ((X- blank) * (100/25) * (250 / weight of sample)).

I hope this is clear, if not contact me again.