The empirical formula of metal iodide : CoI₃(Cobalt(III) Iodide)
Further explanation
13.02 g sample of Cobalt , then mol Co(MW=58.933 g/mol) :
[tex]\tt mol=\dfrac{mass}{Ar}=\\\\mol=\dfrac{13.02~g}{58.933}\\\\mol=0.221[/tex]
Mass of metal iodide formed : 97.12 g, so mass of Iodine :
[tex]\tt =mass~metal~iodide-mass~Cobalt\\\\=97.12-13.02\\\\=84.1~g[/tex]
Then mol iodine (MW=126.9045 g/mol) :
[tex]\tt \dfrac{84.1}{126.9045}=0.663[/tex]
mol ratio of Cobalt and Iodine in the compound :
[tex]\tt 0.221\div 0.663=1\div 3[/tex]
