An oxide of chromium is found to have the following composition: 28.2% Cr and 71.8% O. How would you determine this compound's empirical formula?

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Answer 1 · 2017-07-01T02:32:04

Change the % to grams, change grams to moles, and then find a mole ratio.

$28.2 % = 28.2 g$ Cr.

$71.8 % = 71.8 g$ O

If there were 100 grams of the compound 28.2 grams would be Cr and 71.8 grams would be oxygen.

$(28.2 g)/(52.0 g//mol) = 0.542 mol Cr$

$(71.8 g)/(16.0 g//mol) = 4.49 mol O$

To set up a ratio divide the moles Cr into the moles O

$4.49/0.542 = 8.28$ round this off to a 8:1 ratio.

The empirical formula based on the data given would be

$CrO_8$ which does not make sense chemically