A $22.35g$ mass of an unknown compound consisting of potassium, chlorine, and oxygen, contained $7.13g$ of potassium, $6.47*g$ of chlorine. What is its empirical formula?

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Answer 1 · 2016-09-19T18:31:31

$KClO_3$

As always, we divide each constitiuent mass thru by the ATOMIC mass of each constituent atom.

$"Moles of potassium"$ $=$ $(7.13*g)/(39.10*g*mol^-1)$ $=$ $0.182*mol$ .

$"Moles of chlorine"$ $=$ $(6.47*g)/(35.45*g*mol^-1)$ $=$ $0.182*mol$ .

$"Moles of oxygen"$ $=$ $((22.35-7.13-6.47)*g)/(15.999*g*mol^-1)$ $=$ $0.546*mol$ .

And now we divide thru by the LOWEST molar quantity, that of potassium/chlorine, to give an empirical formula of $KClO_3$ .

How did I know that that there were $(22.35-7.13-6.47)*g$ of $O$ ? The mass of oxygen was not given in the problem.

A 22.35*g22.35*g mass of an unknown compound consisting of potassium, chlorine, and oxygen, contained 7.13*g7.13*g of potassium, 6.47*g6.47*g of chlorine. What is its empirical formula?