How do you find the empirical formulas for 65.2% Sc, 34.8% O?

1 Answer
Mar 4, 2017

Answer:

As with all these problems, we ASSUME a #100*g# mass of compound.........and come up with an empirical formula of #Sc_2O_3#.

Explanation:

The empirical formula is the simplest whole number ratio defining constituent atoms in a species. We assume a #100*g# mass of #Sc_2O_3#, and we come up with a molar ratio:

#"Moles of scandium"# #=# #"Mass of scandium"/"Molar mass of scandium"# #=# #(65.2*g)/(44.96*g*mol^-1)=1.45*mol#.

#"Moles of oxygen"# #=# #"Mass of oxygen"/"Molar mass of scandium"# #=# #(34.8*g)/(16.00*g*mol^-1)=2.175*mol#.

In each instance, I divided thru by the ATOMIC MASS of the element. And now if we divide thru by the smallest molar quantity, (that of the metal), I get a formula of #ScO_(1.5)#. Because, by specification, the empirical formula is a WHOLE number ratio, we quote its empirical formula as #Sc_2O_3#. Capisce?