They represent elements by using symbols
Answer:
![n_{Mg}=3.50molMg\\\\ n_{Cl}=7.00molCl\\\\n_O=28.0molO](https://tex.z-dn.net/?f=n_%7BMg%7D%3D3.50molMg%5C%5C%5C%5C%20n_%7BCl%7D%3D7.00molCl%5C%5C%5C%5Cn_O%3D28.0molO)
Explanation:
Hello there!
In this case, according to the given information it turns out possible for us to realize that one mole of the given compound, Mg(ClO₄)₂, has one mole of Mg, two moles of Cl and eight moles of O; thus, we proceed as follows:
![n_{Mg}=3.50molMg(ClO_4)_2*\frac{1molMg}{1molMg(ClO_4)_2}=3.50molMg\\\\ n_{Cl}=3.50molMg(ClO_4)_2*\frac{2molCl}{1molMg(ClO_4)_2}=7.00molCl\\\\n_O=3.50molMg(ClO_4)_2*\frac{8molO}{1molMg(ClO_4)_2}=28.0molO](https://tex.z-dn.net/?f=n_%7BMg%7D%3D3.50molMg%28ClO_4%29_2%2A%5Cfrac%7B1molMg%7D%7B1molMg%28ClO_4%29_2%7D%3D3.50molMg%5C%5C%5C%5C%20n_%7BCl%7D%3D3.50molMg%28ClO_4%29_2%2A%5Cfrac%7B2molCl%7D%7B1molMg%28ClO_4%29_2%7D%3D7.00molCl%5C%5C%5C%5Cn_O%3D3.50molMg%28ClO_4%29_2%2A%5Cfrac%7B8molO%7D%7B1molMg%28ClO_4%29_2%7D%3D28.0molO)
Best regards!
The empirical formula is XeO₃.
<u>Explanation:</u>
Assume 100 g of the compound is present. This changes the percents to grams:
Given mass in g:
Xenon = 73.23 g
Oxygen = 26.77 g
We have to convert it to moles.
Xe = 73.23/
131.293 = 0.56 moles
O = 26.77/ 16 = 1.67 moles
Divide by the lowest value, seeking the smallest whole-number ratio:
Xe = 0.56/ 0.56 = 1
O = 1.67/ 0.56 = 2.9 ≈3
So the empirical formula is XeO₃.