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

1 Answer
anor277
Dec 19, 2016

#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.

Related questions
See all questions in Empirical and Molecular Formulas
Impact of this question
12739 views around the world
You can reuse this answer
Creative Commons License