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

##### 1 Answer

May 19, 2016

#### Answer:

#### Explanation:

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

#(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:

- 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