# 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?

Jul 1, 2017

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.

$\frac{28.2 g}{52.0 g / m o l} = 0.542 m o l C r$

$\frac{71.8 g}{16.0 g / m o l} = 4.49 m o l O$

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

$\frac{4.49}{0.542} = 8.28$ round this off to a 8:1 ratio.

The empirical formula based on the data given would be

$C r {O}_{8}$ which does not make sense chemically