1.

Determine the molecular formula of the compound which contains `H = 6.67 %, C = 40 %` and the rest is oxygen. 0.6 g of the compound occupy 224 cc at N.T.P

Answer» Correct Answer - `C_(2)H_(4)O_(2)`
Step I. Percentage of oxygen `= 100 - (6.67 + 40) = 100-46.67 = 53.33`
Step II. Empirical formula of the compound
`{:("Element","Percentage","Atomic mass","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("C",40.00,12,(40.0)/(12)=3.33,(3.33)/(3.33)=1.0,1),("H",6.67,1,(6.67)/(1)=6.67,(6.67)/(3.33)=2.0,2),("O",53.33,16,(53.33)/(16)=3.33,(3.33)/(3.33)=1.0,1):}`
Empirical formula of the compound `= CH_(2)O`
Step III. Molecular formula of the compound
224 cc of the vapours of the compound at N.T.P occupy mass = 0.6 g
22400 cc of the vapours of the compound at N.T.P occupy mass `= 0.6//224 xx 22400 = 60.0 g`
`:.` Molecular mass of the compound = 60 g = 60 u
Empirical formula mass `= 12 + 2 xx 1 + 16 = 30 u , n = ("Molecular mass")/("Empirical formula mass")=((60u))/((30u))=2`
`:.` Molecular formula `= 2 xx CH_(2)O=C_(2)H_(4)O_(2)`.


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