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=(mass)/(Ar)=\\\\mol=(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 (84.1)/(126.9045)=0.663`

mol ratio of Cobalt and Iodine in the compound :

`\tt 0.221/ 0.663=1/ 3`