Question

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?

Answer 1

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.