An unknown compound is analyzed and found to be composed of 14.79% nitrogen, 50.68% oxygen, and 34.53% zinc. What is the empirical formula for the compound?

1 Answer
Jan 1, 2016

Answer:

I follow the normal rigmarole: assume 100 g of unknown stuff and come up with the formula #ZnN_2O_6#.

Explanation:

In 100 g of unknown, there are 50.68 g #O#, 34.53 g #Zn#, and 14.79 g of #N#.

We divide thru by the atomic weights of oxygen, zinc, and nitrogen respectively:

#O: (50.68*g)/(15.99*g*mol^-1) = 3.17# #mol# #O#.

#Zn: (34.53*g)/(65.4*g*mol^-1) = 0.53# #mol# #Zn#.

#N: (14.79*g)/(14.01*g*mol^-1) = 1.17# #mol# #O#.

We divide thru (again) by the lowest ratio (that of Zn) to give:

#ZnN_2O_6#, which is the empirical formula of the compound, that is the simplest whole number ratio that defines constituent atoms in a species. The species is LIKELY zinc nitrate, #Zn(NO_3)_2#, but we don't really know that at present.