# Calculation of Atomic weight percentage in AAS

I am currently doing doping studies in a system similar to this ($\ce{A_{x}B_{1-x}}$) and I have to find atomic weight percentage of $\ce{B}$ element using atomic absorption spectroscopy(AAS). I have prepared different PPM of solutions according to the weight percentage of $\ce{B}$ element.

Example to prepare 10 atomic weight percentage of $\ce{B}$ in $5~\mathrm{ppm}$ solution, I have used this equation to calculate molecular weight: $$\text{Actual mol weight}=\text{mol.wt}(A)\cdot x + \text{mol.wt}(B) \cdot (1-x).$$

In this case $x = 0.9$ and $1-x = 0.1$.

From this, I have prepared $5~\mathrm{ppm}$ of a solution of the doped material assuming that $10\%$ of $\ce{B}$ is doped completely (but in reality only very low percentage is doped).

I got the below data from AAS:

B (At wt %) Conc of sol(ppm)    AAS Conc(µg/ml)
10          5                 1.5
20          5                 0.9


From the above data, I have to calculate actual atomic weight percentage of $\ce{B}$. My colleague has suggested this method in order to find $\text{Actual atomic weight percentage}$

$$\text{act.At.Wt.%} =\frac{\text{Obs.AAS.Conc(µg/ml)}}{\text{Sol.Conc(ppm)}}\cdot \text{At.Wt.percentage.of.}\ce{B}$$

Using this I got the results below.

B (At wt %) Conc of sol(ppm)    AAS Conc(µg/ml)   Actual B (At wt %)
10                5                  1.25            2.5
20                5                  0.9             3.6


But I have doubt in this method as you can see in the table the observed AAS concentration for 10 and 20 At wt % $\ce{B}$ is 1.25 and 0.9 (µg/ml) respectively. You can see that the AAS concentration of 20 At.wt % is lower than 10 A.wt %. But when we calculate actual $\ce{B}$ At.wt.% we can see that the 10 A.wt % is lower comparing 20 A.wt %. Can you tell me whether the equation suggested by my colleague is correct or not? If not can you suggest any other method in order to find the actual atomic weight percentage?