How to find the empirical formula from mass percentage

Elemental analysis of a compound showed that it consisted of: $52.17\%C$, $13.05\%H$, $34.78\%O$. Find the empirical formula of the compound.

SOLUTION:
We can assume that 100gr of the substance contain 52.17gr of carbon, 13.05gr of hydrogen, and 34.78gr of oxygen.
Remembering the atomic weight, we can calculate the number of mole (n):

$$n_C=\frac{52.17gr}{12gr/mol}=4.35mol$$

$$n_H=\frac{13.05gr}{1.01gr/mol}=12.9mol$$

$$n_O=\frac{34.78gr}{16gr/mol}=2.17mol$$

The empirical formula is $C_xH_yO_z$; so we have:
$x:y:z=4.35:12.9:2.17$

We divide now by the smallest number (oxygen), to obtain:
$$x:y:z=\frac{4.35}{2.17}:\frac{12.9}{2.17}:\frac{2.17}{2.17}=2:6:1$$

So, the empirical formula is $C_2H_6O$.

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PROBLEM 5:
Determine the empirical formula of a compound that showed the following analytical results: $26.6\%K$, $35.4\%Cr$, $38.0\%O$.