Question

A compound is found to contain 25.24 % sulfur and 74.76 % fluorine by mass.

What is the empirical formula for this compound?

Answer 1

`S_(2)`
`F_(3\\)`, the empical formula

Step-by-step explanation:

Let's assume we have 100 grams of the compound:

The 100 grams will consist of:

25.24 grams of S

74.76 grams of F

100.0 grams

We need to find the numbers of atoms in each of these masses. Divide each by that element's molar mass to find atoms of each:

Sulfur: (25.24 g)/(32 g/mole S) = 0.789 moles S

Flourine: (74.76 g)/(19.0 g/mole F) = 5.93 moles F

There are 0.8 atom of S for every 6 atoms of F

We can;t have 0.8 atom of anything, so let's multiply it by a factor to make it a whole number. 5 is the best I can find. Both need to be increased by the same factor:

That makes 4 atoms of S and 30 atoms of fluorine.

`S_(4)`
`F_(6\\)`

We can reduce that to
`S_(2)`
`F_(3\\)`, the empical formula