# What is the empirical formula of a compound that contains 32.39 percent sodium, 22.53 percent sulfur, and 45.07 percent oxygen?

Jun 23, 2017

$N {a}_{2} S {O}_{4}$

#### Explanation:

First, convert percentages into grams.

 32.39 % = 32.39 g

22.53% = 22.53 g

45.07% = 45.07 g

Second, convert the grams into moles.

$\text{32.39 g"/("22.99 g/mol Na")= "1.409 mol Na}$

$\text{22.53 g"/"32.06 g/mol S" = "0.7027 mol S}$

$\text{45.07 g"/"16.00 g/mol O" = "2.817 mol O}$

Third, convert the moles into a ratio of atoms.

$\text{2.817 mol O"/"0.7027 mol S" = "4.009 O"/"1 S}$

$\text{1.409 mol Na"/"0.7027 mol S" = "2.005 Na"/"1 S}$

Rounding to integers, the ratio is 1 S: 2 Na : 4 O

$N {a}_{2} S {O}_{4}$