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,
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
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 |
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