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A compound on analysis gave the followin...

A compound on analysis gave the following percentage composition by weight: hydrogen = 9.09, oxygen = 36.36 carbon = 54.55
Its `VD` is 44. Find the molecular formula of the compound.

Text Solution

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The correct Answer is:
B, C, D
`{:("Element", "Moles", "Least ratio"),(C, (54.55)/(12) = 4.54, (4.54)/(2.27) = 2),(H, (9.09)/(1) = 9.09, (9.09)/(0.27) = 4),(O, (36.36)/(16) = 2.27, (2.27)/(2.27) = 1):}`
Empirical formula `= C_(2) H_(4) O`.
Empirical formula weight `= 12 xx 2 + 1 xx 4 + 16 = 44`
`VD = 44`,
Molecular Weight `= 2 xx VD = 2 xx 44 = 88`
`n = ("Molecular Weight")/("Empirical formula weight") = (88)/(44) = 2`
`= 2 xx (C_(2) H_(4) O) = C_(4) H_(8) O_(2)`
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