# What is the empirical formula of a compound which has a percent composition of 40.04% S and 59.96% O?

Dec 13, 2015

The empirical formula is $\text{SO"_3}$.

#### Explanation:

An empirical formula represents the lowest whole number ratio of elements in a compound.

Since the percentages of the elements equal 100%, we can assume we have a 100 g sample, so that the percentage of each element becomes grams.

40.04%rArr40.04 "g S"
59.96%rArr59.96 "g O"

First determine the moles of each element by dividing its given mass by its molar mass. The molar mass of an element is its atomic weight (relative atomic mass) on the periodic table in g/mol.

Molar mass of S$=$$\text{32.06 g/mol}$
Molar mass of O$=$$\text{15.999 g/mol}$

$\text{S} :$$40.04 \cancel{\text{g S"xx(1"mol S")/(32.06cancel"g S")="1.249 mol S}}$

$\text{O} :$$59.96 \cancel{\text{g O"xx(1"mol O")/(15.999cancel"g O")="3.748 mol O}}$

Next determine the mole ratios of each element by dividing the moles of each each element by the smallest number of moles.

$\text{S} :$(1.249 "mol")/(1.249"mol")=1.000"

$\text{O} :$(3.748"mol")/(1.249"mol")=3.000"

Since the mole ratios are the lowest whole number ratio of the elements, the empirical formula is $\text{SO"_3}$.