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?

  • 5
    $\begingroup$ This is pretty much how I would do it. And indeed the compound is $\ce{SO3}$. $\endgroup$ Jan 24 '20 at 0:26

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