A) \[{{C}_{2}}{{H}_{6}}{{N}_{2}}\]
B) \[{{C}_{3}}{{H}_{4}}N\]
C) \[{{C}_{6}}{{H}_{8}}{{N}_{2}}\]
D) \[{{C}_{9}}{{H}_{12}}{{N}_{3}}\]
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 \[={{C}_{3}}{{H}_{4}}N\] \[=\frac{108}{54}=2\] Molecular formula \[=({{C}_{3}}{{H}_{4}}{{N}_{2}})={{C}_{6}}{{H}_{8}}{{N}_{2}})\]
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