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A compound on analysis gave the follwing result C=54.54%,H=9.09% and vapour density of compound = 88.Determine the molecular formula of the compound :-A. `C_(8)H_(16)O_(0)`B. `C_(4)H_(16)O_(8)`C. `C_(2)H_(4)O`D. `CH_(4)O_(2)` |
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Answer» Correct Answer - A `{:("Atom", "atomic mass", %wt, %//"Atomic wt.", "simple ratio"),(C,12,54.54,4.54,2),(H,1,9.09,9.09,4),(O,16,36.37,2.27,1):}` Empirical formula `= C_(2)H_(4)O` Empirical formula weight = 44 Molecular formula weight `= 88xx2=176` `n = (176)/(44)=4` Hence molecular formula = `C_(8)H_(16)O_(4)` |
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