This problem is describing the percent composition of a compound formed by chlorine and oxygen, whereas the former has a percent composition of 52.6 % and asks for the empirical formula.
In this case, according to the given information, it turns out firstly possible to realize, that the percent composition of oxygen will be 47.4 % as both must sum 100 %.
Next, for the determination of the empirical formula, we can assume this percents as masses, to calculate the moles of each atom in the formula:
[tex]n_{Cl}=52.6gCl*\frac{1molCl}{35.45gCl}=1.484molCl\\\\n_{O} =47.4gO*\frac{1molO}{16.0gO} =2.963molO[/tex]
After that, we divide the moles of both Cl and O by 1.484 (moles of Cl), in order to calculate the mole ratios which also provide the subscripts in the empirical formula:
[tex]Cl:\frac{1.484}{1.484}=1\\\\O:\frac{2.963}{1.484}=2[/tex]
Hence, the formula is:
[tex]ClO_2[/tex]
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