What is the empirical formula for a compound that is 31.9% potassium, 28.9% chlorine, and 39.2% oxygen?

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Answer 1 · 2015-12-28T17:18:30

The empirical formula is $KClO_3$ , potassium chlorate.

In 100 g of stuff, there are 31.9 g potassium, 28.9 g of chlorine, and 39.2 g oxygen.

We divide these gram quantities by the respective atomic masses of each of the atoms to get the atomic ratio of constituents.

$K$ : $(31.9*g)/(39.1*g*mol^-1)$ $=$ $0.816*mol$ .

$Cl$ : $(28.9*g)/(35.45*g*mol^-1)$ $=$ $0.816*mol$ .

$O$ : $(39.2*g)/(15.99*g*mol^-1)$ $=$ $2.45*mol$ .

If we divide thru by the least molar quantity (that of potassium or oxygen), we get an empirical formula of $KClO_3$ , potassium chlorate.

Note that because this is an ionic, and not a molecular species, we can quote the empirical as the actual formula.