An organic compound contains 53.1% Carbon, 6.2% Hydrogen, 12.4% Nitrogen, and 28.3% Oxygen by mass. What is the molecular formula of the compound if its vapour density is 56.5? [ C =12, H = 1, N = 14, O = 16].
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C\(_5\)H\(_7\)NO\(_2\)
- C\(_5\)H\(_6\)NO\(_2\)
- C\(_3\)H\(_6\)NO\(_2\)
- C\(_3\)H\(_7\)NO\(_2\)
The correct answer is: A
Explanation
C H N O
\(\frac{53.1}{12}\) \(\frac{6.2}{1}\) \(\frac{12.4}{14}\) \(\frac{28.3}{16}\)
\(\frac{4.425}{0.886}\) \(\frac{6.2}{0.886}\) \(\frac{0.886}{0.886}\) \(\frac{1.769}{0.886}\)
5 : 7 : 1 : 2
C\(_5\) H\(_7\) N O\(_2\) - Empirical formula
(Empirical formula) n = Molecular formula
(C\(_5\)H\(_7\)NO\(_2\))n = 2 x vapour density
([12x5] + [1x7] + 14 + [16x2] ) n = 2 x 56.5
113n = 113
n = \(\frac{113}{113}\)
n = 1
Therefore, the molecular formula is C\(_5\)H\(_7\)NO\(_2\)
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