Correct Answer: Option D

Explanation

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

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