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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2005

  • question_answer
    A compound \[(60\,\,g)\] on analysis produce carbon, hydrogen and oxygen\[24\,\,g\], \[4\,\,g\] and \[32\,\,g\] respectively. The empirical formula is:

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

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

    C) \[C{{H}_{2}}O\]

    D)        \[{{C}_{2}}{{H}_{4}}{{O}_{6}}\]

    Correct Answer: C

    Solution :

    \[\because \]  \[60\,\,g\]of compound contains \[24\,\,g\] of carbon \[\therefore \]\[%\]of carbon\[=\frac{24\times 100}{60}=40%\] Similarly, percentage of hydrogen                 \[=\frac{4\times 100}{60}=6.66%\] \[\therefore \]  percentage of oxygen                 \[=100-(40+6.66)\]                 \[=53.34%\]
    Element Percentage Atomic mass Relative number of atoms Ratio
    C \[40%\] 12 \[40/12\] \[=3.33\] \[3.33/3.3\] \[3=1\]
    H \[6.66%\] 1 \[6.66/1\] \[=6.66\] \[6.66/3.3\] \[3=2\]
    O \[53.34%\] 16 \[53.34/16\] \[=3.33\] \[3.33/3.3\] \[3=1\]
    Therefore, empirical formula\[=C{{H}_{2}}O\]


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