Question

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?

Answer 1

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

Explanation:

`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