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:

#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#