Question

Please help make sure its correct thanks

Answer 1

The empirical formula : FeBr₃

Further explanation

A 8.310 g sample of Iron, so mol of Iron(Ar=55.845 g/mol) :

`\tt mol=(mass)/(Ar)\\\\mol=(8.310)/(55.845)\\\\mol=0.149`

Mass of metal bromide formed : 43.98 g, so mass of Bromine :

`\tt =mass~metal~bromide-mass~Iron\\\\=43.98-8.310\\\\=35.67~g`

then mol Bromine (Ar=79,904 g/mol) :

`\tt (35.67)/(79,904)=0.446`

mol ratio of Iron and Bromine in the compound :

`\tt 0.149/ 0.446=1/ 3`