Question

A compound containing sodium, chlorine, and oxygen is 25.42% sodium by mass. A 3.25 g sample gives 4.33×1022 atoms of oxygen. What is the empirical formula?

Answer 1

step  one 
calculate  the  %  of  oxygen
from  avogadro  constant
1moles =  6.02  x  10  ^23  atoms
what  about    4.33  x10^22  atoms
= ( 4.33  x  10^ 22 x 1 mole )  /  6.02  10^23=   0.0719 moles
mass=  0.0719  x16=  1.1504   g
% composition   is therefore= ( 1.1504/3.25)  x100 = 35.40%
 step  two
calculate the  %  composition  of  chrorine
100-  (25.42  +  35.40)=39.18%

step  3
calculate the  moles   of  each  element
that   is  
Na  =  25.42  /23=1.1052  moles
Cl=  39.18  /35.5=1.1037moles
O=  35.40/16=  2.2125   moles
step  4
find  the  mole  ratio  by  dividing  each  mole  by  1.1037  moles
that  is
Na  =  1.1052/1.1037=1.001
Cl= 1.1037/1.1037=  1
0=2.2125 = 2
therefore  the  empirical  formula= NaClO2