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

Question

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 $XO_2$

Answer 1

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, $X_2O_5$