What is the empirical formula of a compound that is composed of 1.67 g of cerium and 4.54 g of iodine?

1 Answer
Aug 20, 2016

Answer:

#CeI_3#

Explanation:

#"Moles of cerium"# #=# #(1.67*g)/(140.12*g*mol^-1)# #=# #0.0119*mol#

#"Moles of iodine"# #=# #(4.54*g)/(126.90*g*mol^-1)# #=# #0.0358*mol#

We divide thru by the lowest molar quantity, #0.0119*mol#, that of the metal, to get an empirical formula of #CeI_3#