Empirical formula and rounding

I am very confused with the procedure to find the empirical formula. In the part where we have to multiply by subscripts to make them integers, are there clear rules?


For example, if I have the formula (I don't know if it exists in reality)
$$C\_{2}H\_{3.75}O\_{1}$$


In this case, it seems that I have to multiply by 4 and it would be:


$$C\_{8}H\_{15}O\_{4}$$


But, if I work with more or fewer digits, I could end up with something like this:
$$C\_{2}H\_{3.60}O\_{1}$$


and I should multiply by 5:
$$C\_{10}H\_{18}O\_{5}$$


So which one is correct?
And another question is when can I just approximate and not multiply by an integer?
Are the empirical formulas we find only probable?
Is there a way to verify the empirical formula?





> 
> In the part where we have to multiply by subscripts to make them integers, are there clear rules?
> 
> 
> 


No. The number of significant figures in the given data can give some guidance. If there is a large uncertainty in the underlying data, you should round more.



> 
> Are the empirical formulas we find only probable?
> 
> 
> 


Anything based on measurement is only probable. In this case, you would go for the simplest empirical formula (lowest numbers) that is a reasonably good fit to the data.



> 
> Is there a way to verify the empirical formula?
> 
> 
> 


Yes. You could either repeat the experiment or, even better, do a different experiment that gives you supplemental data. If you somehow measure the molar mass, this will put a limit of the magnitude of the subscripts in the empirical formula. For example, if the molar mass of hydrogen peroxide is measured as 35 g/mol, an empirical formula of $\ce{HO}$ would be a better fit than an empirical formula of $\ce{H134O135}$.



> 
> So which one is correct?
> 
> 
> 


Without knowing the underlying data, there is no way to give an answer. $\ce{C8H15O4}$ would be suspicious to someone familiar with organic chemistry (it looks like a radical), but without knowing the pedagogical situation in which a question was asked, I would not reject $\ce{C8H15O4}$ on those grounds.


 7