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
anor277
Jan 1, 2016

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.

Related questions
See all questions in Empirical and Molecular Formulas
Impact of this question
3307 views around the world
You can reuse this answer
Creative Commons License