Empirical formula?

Question

A compound contains 3.24 grams of sodium, 2.26 grams of sulfur, and 4.51 grams of oxygen. What is the empirical formula for this compound?

Answer 1

So, the goal of this is to find the mole fractions between sodium, sulfur and oxygen.

If we take 3.24g of Sodium, we can multiply it by the fraction 1mol/22.9g to get 3.24g Na = .141mol Na

Similarly, we can do the same for Sulfur and Oxygen.

2.26g S * 1mol/32.1g = .07 mol S

4.51g O * 1mol/16g = .28 mol O

From this, we divide all of these decimals by the lowest quantified element, in this case, we would divide the number of moles of Na and O by S because we have the least amounts of moles of S.

If we do that, we see that

.141 mol Na / .07 mol S = 2 mol Na/ 1 mol S

.28 mol O / .07 mol S = 4 mol O / 1 mol S.

We can then assemble the empirical formula of the compound. For every 1 mol (and therefore atom) of S, we have 2 moles of Na and 4 moles of O.

Therefore, our empirical element is Na2SO4