It's all very mixed up, so in order. First, SCM measurements of the solution from the digestion of soil samples. The results of measuring the concentrations of these elements are known because you measured them. Now calculate what are the concentrations of "samples" in these solutions, as if these samples were substances going completely into solution. First, it is 15 g in 100 mL, which is 150 g in 1 L. Then you dilute these solutions 2 times, so the "sample" concentrations are 75 g/L. This is the value of the denominator of this fraction whose numerator is equal to the result of measuring the concentration of a given element (mg/L). And the value of this fraction is the quotient of the numerator and denominator. For example, the concentration value of a given element measured in a solution of this sample = 10 mg/L, and the concentration value of the "sample" calculated = 75 g/L, i.e. the fraction is 10 mg/75 g, i.e. approx. 0.133 mg/g, i.e. approx. 133 mg/kg.

As for the aqueous samples, you concentrated them 5 times (from 500 mL to 100 mL) and then diluted them 2 times, so your measurements, in mg/L, should be divided by 2.5.

As for the results of SAM measurements, the results of the measurements should be analogously related to the concentrations of "samples" in the measured solutions.

In addition to the conversion of measurement results into concentrations or contents in samples, it would be worth referring to your sample preparation and spectrometric measurements, but maybe at a later date.

ZJ

Sample details: SCM

1. 15g soil digested to 100mL (acid = aqua regia), 500 mL of water samples digested to 100 mL

2. We took 50mL alqt. each, then diluted it to 100 mL

3. Direct measurement of sample was taken using respective lamp, then recorded.

4. ppm in mg/L of metal in sample was determined using SCM curve with x=(Absorbance-b)/m

5. conversion to ppm in mg/kg for soil: im unsure how to go about this and where to include the dilution factor of 2

Fictional example for soil sample

Say there is 1 mg of Ca per 15g of soil

This is 1 mg per (15g/1000g)*Kg of soil

=0.015 mg per Kg soil

=0.015 ppm

At step 1 this 1 mg Ca is diluted by adding 100 mL digestion solution. All of the 1 mg Ca in the 15 g soil will now be in the 100 mL digestion solution.

This is 1 mg per 100 mL, or 10 mg per 1000 mL

=10 ppm

At step 2 this is further diluted 2 x

So the concentration of Ca per 100 mL solution at step 2

=5 ppm

At Step 3 you measured this using AAS and of course, you will get 5 ppm.

At step 4 you need to back calculate per 15 g of soil.

Backtracking

Based on the steps, what you need to do is to times the obtained results (AAS or ICP) by 2

=5 ppm *2

And you will get

=10 ppm or 10 mg/L back

Since you begin with 100 mL sample

Then you get back

=1 mg per 100 mL digested sample

Since you started with 15 g soil

=1 mg Ca is in 15 g soil

=0.015 mg per Kg soil

Once you have reached this step you need to prepare a conversion factor for all samples using the same step above (must also be 15 g of soil)

If the reading after ICP or AAS gave 10 mg/L or 10 ppm

To know the mg of Ca per Kg soil of the new sample just divide the value obtained by (10/0.015) or 666.667

For example

In a new soil sample of 15 g, the readings after ICP gave 21.4 mg/L of Ca

Then the Ca conc in mg/Kg in the soil sample is

21.4/666.667

=0.0321 mg/Kg

Do check the answers with another chemist! I might be wrong.

I only have time for the soil sample.

Thank you for your guidance Dr. Yunus Shukor and Sir Zbigniew JońcaZbigniew Jońca