2. Sample Papers
  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main - Mock Test - 25

  • question_answer
    An organic compound contains \[49.3%\] carbon, \[6.84 %\] hydrogen and its vapour density is 73. Molecular formula of the compound is:

    A) \[{{C}_{3}}{{H}_{5}}{{O}_{2}}\]                  

    B)        \[{{C}_{4}}{{H}_{10}}{{O}_{2}}\]

    C)        \[{{C}_{6}}{{H}_{10}}{{O}_{4}}\]                

    D)        \[{{\operatorname{C}}_{3}}{{H}_{10}}{{O}_{2}}\]

    Correct Answer: C

    Solution :

    Element % Relative no. of atoms Simplest ratio of atoms
    C 49.3 \[49.3/12=4.1\] \[1.5\times 2=3\] \[4.1/2.74=1.5\]
    H 6.84 \[6.84/1=6.84\] \[6.84/2.74=2.5\] \[=\,\,\,2.5\times 2=5\]
    O 43.86 \[43.86/16=2.74\] \[2.74/2.74=1\] \[1\times 2=2\]
    \[\therefore \,\,\,Empirical\,\,formula=\,\,{{C}_{3}}{{H}_{5}}{{O}_{2}}\] Empirical formula mass \[=\left( 3 \times  12 \right)+\left( 5 \times  1 \right)+\left( 2\times  16 \right)=36+5+32=73\] \[\operatorname{Molecular} mass = 2 \times  Vapour density\] \[=\,\,\,2\,\,\times 73=146\] \[\operatorname{n}=\,\,\frac{molecular\,\,mass}{empirical\,\,formula\,mass}=146/73=2\,\] \[\operatorname{Molecular} formula = Empirical formula \times  2\] \[=\,\,\,({{C}_{3}}{{H}_{5}}{{O}_{2}})\times 2={{C}_{6}}{{H}_{10}}{{O}_{4}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner