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

1 Answer
Mar 4, 2017

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?