**'a sample of a compound is 67.6% mercury (Hg), 10.8% sulfur (S), and 21.6% oxygen (O). what is the empirical formula for this compound? use the molar masses Hg= 200.6 g/mol, 2= 32.1 g/mol, O= 16.0 g/mol.**

**Answer:**

Okay, so the trick to approaching this is you have to assume some given amount of the sample to start. In this case, and in most cases, assuming 100 grams of the sample is best as it makes the math easiest. Doing so we see that if we have a 100 grams of the sample, then we have 67.6g of Hg, 10.8g of S, and 21.6g of O.

Now, we divide each of these amounts by their molecular masses to get the moles of each constitutent:

67.6g Hg / 200.6 g/mol Hg = 0.3369890329 moles Hg

10.8g S / 32.1 g/mol S = 0.33644859813 moles S

21.6g O / 16.0 g/mol O = 1.35 moles O

Now, what we do is we divide each of the moles by the LOWEST number of moles- in this case the moles of sulfur.

Moles Hg/ moles S = 1.001606292

Moles S/ moles S = 1

Moles O/ moles S = 4.0125

So, this tells us that for every sulfur atom, there’s 4 oxygen atoms and 1 mercury atom. This means our compound will have an empirical formula of HgSO4

And that’s it!