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

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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

$Na_2 S O_4$