Question

What is the empirical formula of a compound that contains 25.92 percent nitrogen and 74.07 percent oxygen?

Answer 1

`N_2O_5`

Explanation:

As with all these problems we assume a `100*g` mass of compound. And then we divide the individual elemental masses by the ATOMIC masses of each constituent.

`"Moles of nitrogen"` `=` `(25.92*g)/(14.01*g*mol)=1.85*mol`.

`"Moles of oxygen"` `=` `(74.07*g)/(15.999*g*mol)=4.62*mol`.

And we divide thru the individual molar quantities by the smallest such quantity, that of nitrogen.

`N:(1.85*mol)/(1.85*mol)=1; O:(4.62*mol)/(1.85*mol)=2.50`

Because, by definition, empirical formulae are the simplest WHOLE number ratio definining constituent atoms in a species, we double this ratio to get:

`N_2O_5`, `"dinitrogen pentoxide"`. Of course, this is an empirical formula; we would need a molecular weight determination before we declared that `N_2O_5` was the mollykule.