Question

A compound is found be 36.5% `Na`, 25.3% `S`, and 38.0% `O`. What is its empirical formula?

Answer 1

`Na_2SO_3`

Explanation:

To solve for an empirical formula, when given the percent compositions, first, divide each percent composition by the atomic mass of THAT element.

1. `(36.5% Na)/(22.99 g/(mol) Na` `(25.3% S)/(32.07 g/(mol) S` `(38.0% O)/(16 g/(mol) O`
   (considering that this is not oxygen gas). For these atomic masses, I rounded to the nearest hundredth.

You get: `1.587` `0.788` `2.375` (in corresponding order to the above equations- these values don't have units).

1. Then, you divide each of these values by the SMALLEST of these values.
   `1.587/0.788` `0.788/0.788` `2.375/0.788`
   Then, still corresponding to Na, S, and O respectively, we get:
   `2.013` mol Na= `2` mol Na
   `1` mol S
   `3.013` mol O= `3` mol O
   (after rounding the answers)

Therefore, the empirical formula is `Na_2SO_3`