# What is the empirical formula of a compound where the mass ratio of sulphur and oxygen is 2:3?

This is how I tried to solve it:

$$\frac{2}{3} = \frac{0.67}{1}$$ $$\frac{32.06}{16} = \frac{2.00}{1}$$ $$\frac{0.66}{2.00} = \frac{0.33}{1}$$

I assumed by the result that if for every $$\pu{1g}$$ of $$\ce{O}$$ there are $$\pu{0.33g}$$ of $$\ce{S}$$ then if we have roughly $$\pu{1g}$$ of $$\ce{S}$$ there must be $$\ce{3 O}$$ atoms. Is this result acceptable or do you know easier, clearer methods for this kind of example?

• This is pretty much how I would do it. And indeed the compound is $\ce{SO3}$. Jan 24 '20 at 0:26