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

Question

19580 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-23T00:14:57

#Na_2SO_4#

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#