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The empirical formula of the compound is KClO₃.
This is the simplest structural formula a compound can have.
The given mass of the elements;
potassium, K = 5.99 g
chlorine, Cl = 5.43 g
oxygen, O = 18.77 g - (5.99 + 5.43)g = 7.35 g
The empirical formula of the compound is calculated as follows;
[tex]K: \ \ \frac{5.99}{39} = 0.15 \\\\Cl: \ \ \frac{5.43}{35.5} = 0.15\\\\O: \ \ \frac{7.35}{16} = 0.46[/tex]
Divide through by the smallest fraction;
[tex]K: \ \frac{0.15}{0.15} = 1\\\\Cl: \ \frac{0.15}{0.15} =1 \\\\O: \ \frac{0.46}{0.15} = 3[/tex]
empirical formula = KClO₃
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