Question

Please help make sure its correct thanks

Answer 1

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) :

$\tt mol=\dfrac{mass}{Ar}=\\\\mol=\dfrac{13.02~g}{58.933}\\\\mol=0.221$

Mass of metal iodide formed : 97.12 g, so mass of Iodine :

$\tt =mass~metal~iodide-mass~Cobalt\\\\=97.12-13.02\\\\=84.1~g$

Then mol iodine (MW=126.9045 g/mol) :

$\tt \dfrac{84.1}{126.9045}=0.663$

mol ratio of Cobalt and Iodine in the compound :

$\tt 0.221\div 0.663=1\div 3$