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?


