Calculation of formula from chemical analysis

The percentage composition of an unknown compound is: $27.05\%Na$, $16.48\%N$, $56.47\%O$. Find the empirical formula of the compound.

SOLUTION:
First, remember the atomic weight of Na, N, O:
$$\begin{cases}Na=23\\ N=14\\ O=16\end{cases}$$
We can say that if we have 100gr of the compound, 27.05gr is Na, 16.48gr is N and 56.47gr is O. Now we can found the number of mole (n) of sodium, nitrogen and oxygen in 100gr of the compound:

$$n_{Na}=\frac{27.05gr}{23gr/mol}=1.18mol$$

$$n_N=\frac{16.48gr}{14gr/mol}=1.18mol$$

$$n_O=\frac{56.47gr}{16gr/mol}=3.54mol$$

The empirical formula is $Na_xN_yO_z$;
$x:y:z=1.18:1.18:3.54$
Dividing by the smallest number, we find:
$$x:y:z=\frac{1.18}{1.18}:\frac{1.18}{1.18}:\frac{3.54}{1.18}=1:1:3$$

So, the empirical formula is $NaNO_3$.

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PROBLEM 4:
From the following analytical results (percentage by weight), determine the empirical formula fro the compound analyzed: $21.8\%Mg$, $27.9\%P$, $50.3\%O$.