1.

An organic compound on analysis gave the following percentage composition : `C = 57.8 %, H= 3.6 %` and the rest is oxygen. The vapour density of the compound was found to be 83. Find the molecular formula of the compound.

Answer» Correct Answer - `C_(8)H_(6)O_(4)`
Step I. Percentage of oxygen `100 - (57.8 + 3.6) = 100 - 61.4 = 38.6`
Step II. Empirical formula of organic compound
`{:("Element","Percentage","Atomic mass","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("C",57.8,12,(57.8)/(1)=4.82,(4.82)/(2.41)=2.0,4),("H",3.6,1,(3.6)/(1)=3.6,(3.60)/(2.41)=1.5,3),("O",38.6,16,(38.6)/(16)=2.41,(2.41)/(2.41)=1.0,2):}`
Empirical formula of the compound `= C_(4)H_(3)O_(2)`
Step II. Molecular formula of the compound
Empirical formula mass `= 4 xx 12 + 3xx 1+ 2 xx 16 = 83 u`
Molecular mass `= 2 xx V.D = 2 xx 83 = 166u`
`n=("Molecular mass")/("Empirical formula mass")=((166u))/((83u))=2`
`:.` Molecular formula of compound `= 2 xx C_(4)H_(3)O_(2) = C_(8)H_(6)O_(4)`.


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