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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 $XO_2$

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    $\begingroup$ Good practice is starting with symbolic algebraic formulas and keeping it this way until all formulas are ready to finally put there literal numbers. It helps focusing on principles and improves orientation, spotting mistakes, the text reusability and its permanent value. $\endgroup$
    – Poutnik
    Jan 5 at 10:47
  • $\begingroup$ What keeps you from checking it yourself, e.g. verifying what is X percentage context in XO2 ? $\endgroup$
    – Poutnik
    Jan 5 at 10:50
  • $\begingroup$ Yup I was so dump I wasn't rounding the number properly thanks found the answer X2O5. Thanks $\endgroup$ Jan 5 at 13:09
  • $\begingroup$ The first oxide is $\ce{P4O6}$ and not $\ce{XO2}$ as Dulsara stated. The second oxide is $\ce{P4O10}$ as Dulsara said. My final question is : Why publish a question where its author knows the answer ? $\endgroup$
    – Maurice
    Jan 5 at 18:11
  • $\begingroup$ Sorry about that. First when I got XO2 I did in fact check the mass percentages and didn't get the correct answers and thought my approach was wrong for the entire question and later found out that it was just a rounding problem. Because I round 2.5 as 2 at first. $\endgroup$ Jan 5 at 18:28
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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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