A) \[C{{H}_{2}}O\]
B) CHO
C) \[C{{H}_{4}}O\]
D) \[C{{H}_{3}}O\]
Correct Answer: D
Solution :
\[C=\frac{38.7}{12}=3.22=\frac{3.22}{3.22}=1\] \[H=\frac{9.67}{1}=9.67=\frac{9.67}{3.22}=3\] \[O=\frac{51.63}{16}=3.22=\frac{3.22}{3.22}=1\] \[\therefore \]Empirical formula is \[C{{H}_{3}}O\]You need to login to perform this action.
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