# 24.2538 grams of an unknown compound was found to contain 50.15% sulfur and 49.85% oxygen. What is the empirical formula?

##### 1 Answer

#### Explanation:

Here's a great example of an empirical formula problem that can be solved without doing any actual calculation. All you really need here is a **periodic table**.

Notice that *atomic oxygen* has a molar mass of *a little under* *sulfur* has a molar mass of *a little over*

This means that in order to have a **percent composition** of **approximately** **twice as many** moles of oxygen than moles of sulfur.

Why *twice as many*? Because a mole of sulfur atoms is **twice as heavy** than a mole of oxygen atoms.

This means that the empirical formula **for any** compound that has a percent composition of **must be**

#color(green)(|bar(ul(color(white)(a/a)"S"_1"O"_2 implies "SO"_2color(white)(a/a)))|)#

Now, let's do some calculations to prove that this is the case. Your sample of the unknown compound will contain

#24.2538color(red)(cancel(color(black)("g compound"))) * overbrace("50.15 g S"/(100color(red)(cancel(color(black)("g compound")))))^(color(purple)("50.15% S")) = "12.2506 g S"#

#24.2538color(red)(cancel(color(black)("g compound"))) * overbrace("49.85 g O"/(100color(red)(cancel(color(black)("g compound")))))^(color(purple)("49.85% O")) = "12.0905 g O"#

Use each element's **molar mass** to find how many moles you have present in the sample

#"For S: " 12.2506 color(red)(cancel(color(black)("g"))) * "1 mole S"/(32.065color(red)(cancel(color(black)("g")))) = "0.38205 moles S"#

#"For O: " 12.0905color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = "0.75568 moles O"#

To get the *mole ratio* that exists between the two elements in the sample, divide both values by the *smallest one*

#"For S: " (0.38205color(red)(cancel(color(black)("moles"))))/(0.38205color(red)(cancel(color(black)("moles")))) = 1#

#"For O: " (0.75568color(red)(cancel(color(black)("moles"))))/(0.38205color(red)(cancel(color(black)("moles")))) = 1.9780 ~~ 2#

Once again, the **empirical formula** of the unknown compound comes out to be

#color(green)(|bar(ul(color(white)(a/a)"S"_1"O"_2 implies "SO"_2color(white)(a/a)))|)#