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

$N a_{2} S O_{4}$ $Na_2SO_4$

First, convert percentages into grams.

$32.39 % = 32.39 g$ $32.39 % = 32.39 g$

$22.53 % = 22.53 g$ $22.53% = 22.53 g$

$45.07 % = 45.07 g$ $45.07% = 45.07 g$

Second, convert the grams into moles.

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

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

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

Third, convert the moles into a ratio of atoms.

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

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

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

$N a_{2} S O_{4}$ $Na_2 S O_4$