Determine the formula of the substance which has: 43.39%Na, 11.32%C, 45.28%O.

First of all, let's establish the formula for the substance as being:NaxCyOz, where

x=the number of Na atoms;

y= the number of C atoms;

z= the number of O atoms; 

Then, we'll establish the atomic mass for each element, the notation used for the atomic mass being A:

A(Na)=23

A(C)=12

A(O)=16

Now, we'll calculate the number of gram-atoms of each element from molecule, by dividing the quantitative percentage to the atomic mass of the element. So:

x=43.39/23=1.89 gram-atoms of Na

y=11.32/12=0.94 gram-atoms of C

z=45.28/16=2.83 gram-atoms of O

The mixing up proportion of atoms:

x:y:z=1.89:0.94:2.83

We'll divide all obtained numbers to the smaller obtained, which is 0.94.

x:y:z=1.89/0.94:0.94/0.94:2.83/0.94

x:y:z= 2:1:3

The chemical formula of the substance is: Na2CO3