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Organic Compound's Mass from Vapor Pressure

  1. Dec 4, 2006 #1
    Is this even correct?


    A natural product (MW = 150 g/mol) distills with steam at a boiling temperature of 99 C at atmospheric pressure. The vapor pressure of water at 99 C is 733 mmHg.

    a. Calculate the weight of the natural product that codistills with each gram of water at 99 C.


    (P*_a)/(P*_b) = moles A / moles B
    P*_a = 760 torr
    P*_b = 733 torr

    ratio 760 moles/733 moles water = 1.0368 mol/1 mol water
    1.0368 mol (150 g/mol) = 155.52 g organic compound / (18.0 g water) = 8.64 g/ 1 g water ?



    b. How much H2O must be removed by steam distillation to recover this natural product from 0.5 g of spice that contains 10 % of the desired substance?

    5 g*(.10) = .05 g desired substance

    0.05 g (1 g water/8.64 g organic) = 0.00579 g water

    Thanks.
     
  2. jcsd
  3. Dec 4, 2006 #2

    GCT

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    Science Advisor
    Homework Helper

    No, the amount of the organic compound should be much smaller, with respect to the mole value. Assuming the effects
    of boiling point elevation is not important here, you can simply account for the mole ratio of the solute....

    (760 mmHg-733 mmHg)/733 mmHg=mole ratio of solute/moles of water

    The method here seems appropriate.
     
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