ta có :
%Al=22,13%
%P=25,40%
% O = 100- 22,13 + 25,40= 52,47%
=> $n_{Al}$=$\frac{%m x M_A}{M_Al}$ =$\frac{22,13% x 122}{27}$ ≈≈ 1 (mol)
=> $n_{P}$= $\frac{25,40% x 122}{31}$≈ 1 ( mol)
=> $n_{O}$= $\frac{52,47% x 122}{16}$≈ 4 ( mol)
Vậy CTHH là $AlPO_{4}$