Question

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

Answer 1

`CrO_3`

Step-by-step 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:

`n_(Cr)=52g*(1mol)/(52g)=1mol\\ \\n_O=48g*(1mol)/(16g)=3mol`

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

`Cr=(1)/(1) =1\\\\O=(3)/(1) =3`

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

`CrO_3`

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

Regards.