Determine the empirical formula of an oxide of nitrogen containing 70% of oxygen if the relative molecular mass of the oxide is 92 deduce its formula

Dear student, 
Elements % by mass atomic mass % by mass / atomic mass simple ratio rouding off simple ratio 
Oxygen 70 16 70/16
= 4.375
 4.375/2.14
=  2.044
2
Nitrogen 30 14 30/14
= 2.14
 2.14/2.14
= 1
1
So, the emperical formula of compound = NO2
Emperical mass = N+2*O = 14+16*2 =46
Now molecular mass = n * emperical formula mass
Where, n = vapour density
n = molecular mass / emperical mass
n = 92/46 = 2
Now molecular formula= emperical formula *n
    = (NO2)2
Therefore, molecular formula = N2O4

Regards

  • 1
What are you looking for?