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
Mar 6, 2016



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 #"16.0 g mol"^(-1)# and the sulfur has a molar mass of a little over #"32.0 g mol"^(-1)#.

This means that in order to have a percent composition of approximately #50%# for both elements, a given compound (that only contains sulfur and oxygen, of course) needs to have 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 #~~50%# for both oxygen and sulfur 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)))|)#