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You do this by balancing the ionic charges.
In your question you have not named the oxide completely. You are correct in saying chromium has multiple charges. These are referred to as oxidation states, which, in this case refers to the charge on the ion. To show this the oxidation state is written after the element in Roman numerals surrounded by brackets.
The most stable oxide of chromium is chromium (III) oxide. This tells us that it contains ##Cr^(3+}## ions.
Since oxide ions are ##O^(2-)## you can see that we need 2 ##Cr^(3+)## for every 3 ##O^(2-)## since this gives 6+ charges and 6- charges.
So the formula is ##Cr_2O_3##
The other, less stable oxide, is chromium(VI)oxide.
See if you can work out its formula by the same method.
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