What is the empirical formula of an oxide of chromium that is 48% oxygen

Explanation:

Hello,

In this case, considering that the compound is 48% oxygen, we infer it is 52% chromium, therefore, by using their atomic masses we compute the moles by assuming those percentages as masses:

Next, by dividing by the smallest moles, we find the subscripts in the empirical formula:

Which are also expressed in the smallest whole numbers, therefore the empirical formula is:

Which corresponds to the chromic oxide or chromium (IV) oxide.

Regards.

i think its 100

i hope this !