A) \[{{C}_{2}}{{H}_{5}}O\]
B) \[{{C}_{3}}{{H}_{6}}{{O}_{3}}\]
C) \[{{C}_{4}}{{H}_{10}}{{O}_{2}}\]
D) \[{{C}_{5}}{{H}_{14}}O\]
Correct Answer: C
Solution :
Empirical formula mass =\[{{C}_{2}}{{H}_{5}}O\] = 24+ 5 +16= 45. \[n=\frac{\text{Mol}\text{. mass}}{\text{Emp}\text{. mass}}=\frac{90}{45}=2\] Mol. formula = \[{{({{C}_{2}}{{H}_{5}}O)}_{2}}={{C}_{4}}{{H}_{10}}{{O}_{2}}\].
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