Question #87503

You know the percent composition of the gas to be 21.9% sulfur and 78.1% fluorine.

What that means is that for any sample of gas you have, sulfur will always account for 21.9% of the total mass and and fluorine for 78.1% of the total mass.

To make the calculations easier, let's assume that you're dealing with a 100-g sample of gas. According to the aforementioned percent composition, you know that this sample must contain

#100color(red)(cancel(color(black)("g gas"))) * "21.9 g S"/(100color(red)(cancel(color(black)("g gas")))) = "21.9 g S"#

and of course

#100color(red)(cancel(color(black)("g gas"))) * "78.1 g F"/(100color(red)(cancel(color(black)("g gas")))) = "78.1 g F"#

Now you need to use the molar masses of the two elements to figure out how many moles of each you'd get in that sample

#21.9color(red)(cancel(color(black)("g S"))) * "1 mole S"/(32.07color(red)(cancel(color(black)("S")))) = "0.683 moles S"#

and

#78.1color(red)(cancel(color(black)("g F"))) * "1 mole F"/(19.0color(red)(cancel(color(black)("g F")))) = "4.11 moles F"#

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

#"For S": (0.683color(red)(cancel(color(black)("moles"))))/(0.683color(red)(cancel(color(black)("moles")))) = 1#

#"For F": (4.11color(red)(cancel(color(black)("moles"))))/(0.683color(red)(cancel(color(black)("moles")))) = 6.02 ~=6#

This means that the empirical formula of the gas is #"SF"_6#. To get its molecular formula you need to know its molar mass.

As practice, try to start with samples of different mass to see that the empirical formula comes out the same every time.