Question

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

Answer 1

`Na_2SO_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.

`"32.39 g"/("22.99 g/mol Na")= "1.409 mol Na"`

`"22.53 g"/"32.06 g/mol S" = "0.7027 mol S"`

`"45.07 g"/"16.00 g/mol O" = "2.817 mol O"`

Third, convert the moles into a ratio of atoms.

`"2.817 mol O"/"0.7027 mol S" = "4.009 O"/"1 S"`

`"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

`Na_2 S O_4`