# What is the empirical formula of a compound containing 60.0 g sulfur and 90.0 g oxygen?

Apr 24, 2016

${\text{SO}}_{3}$

#### Explanation:

Your strategy here will be to use molar masses of the two elements to determine how many moles of each you have present in this sample.

Once you know that, use the number of moles of sulfur and oxygen to determine the smallest whole number ratio that exists between the two elements in the compound.

So, sulfur has a molar mass of ${\text{32.065 g mol}}^{- 1}$, which means that one mole of sulfur has a mass of $\text{32.065 g}$.

60.0 color(red)(cancel(color(black)("g"))) * "1 mole S"/(32.065color(red)(cancel(color(black)("g")))) = "1.8712 moles S"

Oxygen has molar mass of ${\text{15.9994 g mol}}^{- 1}$, which means that one mole of oxygen will have a mass of $\text{15.9994 g}$.

90.0 color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = "5.6525 moles O"

Now, a compound's empirical formula tells you the smallest whole number ratio that exists between the compound's constituent elements.

To get the mole ratio that exists between sulfur and oxygen in this compound, divide both values by the smallest one

"For S: " (1.8712 color(red)(cancel(color(black)("moles"))))/(1.8712color(red)(cancel(color(black)("moles")))) = 1

"For O: " (5.6525color(red)(cancel(color(black)("moles"))))/(1.8712color(red)(cancel(color(black)("moles")))) = 3.021 ~~ 3

Since $1 : 3$ is the smallest whole number ratio that can exist between the two elements, it follows that the compound's empirical formula will be

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {\text{S"_1"O"_3 implies "SO}}_{3} \textcolor{w h i t e}{\frac{a}{a}} |}}}$