A student found that 1.19 g of chromium (#Cr#) formed 1.74 g f chromium oxide. The molar mass of #Cr# is 52.00 g/mol. What is the empirical formula of chromium oxide?

1 Answer
Nov 28, 2016

Answer:

#Cr_2O_3#

Explanation:

As with all these problems, we calculate the molar quantities of each substituent, and normalize them according to the masses of the atomic constituents:

#"Moles of metal"# #=# #(1.19*g)/(52.00*g*mol^-1)=0.0229*mol#

#"Moles of oxygen"# #=# #(1.74*g-1.19*g)/(16.00*g*mol^-1)=0.0343*mol#

And we divide thru by the SMALLEST molar quantity, that of chromium, to give:

#Cr, (0.0229*mol)/(0.0229*mol)=1; O, (0.0343*mol)/(0.0229*mol)=1.50#.

But by definition, the empirical formula is the smallest WHOLE number ratio that defines constituent atoms in a species. To get a whole number ratio, clearly we mulitply the empirical ratio by #2# to #Cr_2O_3# as required. Capisce?

How did I know there were #0.55*g# of oxygen present?