Question

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Answer 1

Answer:

$KClO_3$

Explanation:

Hello!

In this case, as we know the mass of the total sample, we can first compute the mass of oxygen:

$m_O=22.9g-7.33g-6.65g=8.92g$

Next, we compute the moles of each element:

$n_K=\frac{7.33g}{39.9g/mol}= 0.184mol\\\\n_{Cl}=\frac{6.65g}{35.45g/mol}=0.188mol \\\\n_O=\frac{8.92g}{16.00g/mol} =0.5575mol$

Now, we divide the moles by 0.184 moles, the fewest ones, to obtain:

$K=\frac{0.184}{0.184}=1.0 \\\\Cl=\frac{0.187}{0.184}=1.0\\\\O=\frac{0.5575}{0.184} =3.0$

Therefore, the empirical formula is:

$KClO_3$

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