# Question #87503

##### 1 Answer

#### Explanation:

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

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