How to find the empirical formula from mass percentages of oxides [closed]








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Say you've given an element X and it makes two oxides. The first oxide is $X\_\mathrm{4}O\_\mathrm{6}$ and the mass percentage of oxygen is 43.7%. In the second oxide, the mass percentage of X is 43.7%.


Can someone say if my calculations are correct to find the empirical formula of the second oxide.
  

  

  

$$m\_{X\_\mathrm{4}O\_\mathrm{6}} = \frac{16\*6\*100}{43.7} = 219.7g$$
$$m\_{X\_\mathrm{4}} = 219.7 - 96 = 123.7g$$
$$m\_X = \frac{123.7g}{4} = 30.99g$$


$$n\_X = \frac{43.7\%}{30.99} = 1.443\ \ \ \ \ \ \ \ \ \ \ \ n\_O = \frac{56.3\%}{16} = 3.519$$
$$\frac{1.443}{1.443}\ \ \ \ \ \ \ \ \ \ \ \ \ \frac{3.519}{1.443}$$
$$1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2$$
Therefore the oxide $XO\_2$




Here's the answer if anyone is wondering.


$$m\_{X\_\mathrm{4}O\_\mathrm{6}} = \frac{16\*6\*100}{43.7} = 219.7g$$
$$m\_{X\_\mathrm{4}} = 219.7 - 96 = 123.7g$$
$$m\_X = \frac{123.7g}{4} = 30.99g$$


$$n\_X = \frac{43.7\%}{30.99} = 1.443\ \ \ \ \ \ \ \ \ \ \ \ n\_O = \frac{56.3\%}{16} = 3.519$$
$$\frac{1.443}{1.443}\ \ \ \ \ \ \ \ \ \ \ \ \ \frac{3.519}{1.443}$$
$$1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2.439$$
Then you have to round it by multiplying the number by 2
$$2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4.878$$
$$2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5$$
Therefore,
$X\_2O\_5$


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