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An organic compound (A) contains C and H only (C=90%). (A) produces another compound (B) on treatment with HBr. (B) contains 79.2% of Br. If the molecular weight of (A) is 40, find the formulae of (A) and (B). (Br= 80) |
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Answer» Solution :MOLES of `C= (90)/(12)= 7.5` Moles of `H= (10)/(1)=10` `therefore` moles of `C:H= 7.5: 10` =`3:4` `therefore` empirical formula is `C_(3)H_(4)` As molecular weight is 40, the molecular formula, will be the same as the empirical formula i.e., (A) is `CH_(3)- C -= CH or, CH_(2)=C= CH_(2)` SINCE with HBr, (A) will produce `CH_(3)-C(Br_(2))- CH_(3)` in which Br % `=(160)/(202)XX 100=79.2%` Which is the same as the GIVEN value, (B) will be `CH_(3)-C(Br_(2))-CH_(3)` |
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