Answer:
[tex]CrO_3[/tex]
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:
[tex]n_{Cr}=52g*\frac{1mol}{52g}=1mol\\ \\n_O=48g*\frac{1mol}{16g}=3mol[/tex]
Next, by dividing by the smallest moles, we find the subscripts in the empirical formula:
[tex]Cr=\frac{1}{1} =1\\\\O=\frac{3}{1} =3[/tex]
Which are also expressed in the smallest whole numbers, therefore the empirical formula is:
[tex]CrO_3[/tex]
Which corresponds to the chromic oxide or chromium (IV) oxide.
Regards.
