The empirical formula : FeBr₃
Further explanation
A 8.310 g sample of Iron, so mol of Iron(Ar=55.845 g/mol) :
[tex]\tt mol=\dfrac{mass}{Ar}\\\\mol=\dfrac{8.310}{55.845}\\\\mol=0.149[/tex]
Mass of metal bromide formed : 43.98 g, so mass of Bromine :
[tex]\tt =mass~metal~bromide-mass~Iron\\\\=43.98-8.310\\\\=35.67~g[/tex]
then mol Bromine (Ar=79,904 g/mol) :
[tex]\tt \dfrac{35.67}{79,904}=0.446[/tex]
mol ratio of Iron and Bromine in the compound :
[tex]\tt 0.149\div 0.446=1\div 3[/tex]
