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 :
\[\begin{align} & \text{C}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{H}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{N} \\ & 9\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3.5 \\ & 9/12=0.75\,\,\,\,\,\,\,\,\,\,1/1=1\,\,\,\,\,\,\,\,\,\,\,\,\,\,3.5/14=0.25 \\ & \frac{0.75}{0.75}=3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{0.25}=4\,\,\,\,\,\,\,\,\frac{0.25}{0.25}=1 \\ \end{align}\] 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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