Elements :  C N  O H

% By mass 53.1 12.4  28.3  6.2

Divide through by atomic mass \(\frac{53.1}{12}\) \(\frac{12.4}{14}\) \(\frac{28.3}{16}\)  \(\frac{6.2}{1}\)

 4.425  0.8857 1.7688  6.2

Divide through by the least amount \(\frac{4.425}{0.8857}\) \(\frac{0.8857}{0.8857}\) \(\frac{1.7688}{0.8857}\) \(\frac{6.2}{0.8857}\)

 5  : 1  : 2 : 7

Empirical formula : C\(_5\)N\(_1\)O\(_2\)H\(_7\) ⇒ C\(_5\)H\(_7\)NO\(_2\)

Recall that (Empirical formula)n = Relative molecular mass 

But Relative molecular mass =  2 x Vapour density

So, (Empirical formula)n = 2 x Vapour density

 [C\(_5\)H\(_7\)NO\(_2\) ] n = 2 x 56.5

 [ (12 x 5) + (1 x 7) + 14 + (16 x 2) ] n = 113

  [60 + 7 + 14 + 32] n = 113

 113n = 113

 n  = \(\frac{113}{113}\) = 1

 [C\(_5\)H\(_7\)NO\(_2\) ] n = [C\(_5\)H\(_7\)NO\(_2\) ] 1 ⇒ C\(_5\)H\(_7\)NO\(_2\) - Option D