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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 |
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Answer» 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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