Mass fraction of nitrogen in the of oxide nitrogen equal to 36.84 %. Output the simplest formula this of oxide .

given:
ω(N) = 36,84  %
find :
NхОу  -?
Solution:
m(N)=(m(N_х О_у )•ω(N))/(100 %)= (100 г•36,84 %)/(100 %)=36,84 g;
m(О) = m(NхОу) -  m(N) = 100 g -36,84 g = 61,16 g
М(N) = 14 g/mol,
ν(N)=(36,84 g )/(14 g/mol)   =2,63 mol ;
М(О) = 16 g/mol,
ν(О)=(61,16 g )/(16 g/mol)   =3,95 mol ;
ν(N)  :  ν(О)  =  х  :  у =  2,63 : 3,95
х∶у= (2,63)/(2,63) ∶ (3,95)/(2,63)=1∶1,5=2∶3 .
simplest formula oxide N2О3.
Answer: N2О3.