A) \[{{C}_{2}}{{H}_{6}}{{N}_{2}}\]
B) \[{{C}_{3}}{{H}_{4}}N\]
C) \[{{C}_{6}}{{H}_{8}}{{N}_{2}}\]
D) \[{{C}_{9}}{{H}_{12}}N\]
Correct Answer: C
Solution :
C H N \[9\] \[1\] \[3.5\] \[9/12=0.75\] \[1/1=1\] \[3.5/14=0.25\] \[\frac{0.75}{0.25}=3\] \[\frac{1}{0.25}=4\] \[\frac{0.25}{0.25}=1\] So, empirical formula\[={{C}_{3}}{{H}_{4}}N\] \[n=\frac{108}{54}=2\] Molecular formula\[={{({{C}_{3}}{{H}_{4}}N)}_{2}}={{C}_{6}}{{H}_{8}}{{N}_{2}}\]
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