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Extraction efficiency problem PLease help

  1. Oct 15, 2007 #1
    Extraction efficiency problem PLease help....

    1. The problem statement, all variables and given/known data

    The distribution coefficient, k = (conc. in CH2Cl2/conc in H2O), between dichloromethane and water for solute A is 7.3.

    What weight of A would be removed from a solution of 10 g of A in 100 mL of water by a single extraction with 100 mL of dichloromethane?

    What weight of A would be removed by three successive extractions with 33 mL portions of dichloromethane?

    How much dichloromethane would be required to remove 98.5% of A in a single extraction?
     
  2. jcsd
  3. Oct 15, 2007 #2
    is this method right>>?>??

    Basic numbers :

    10 g of A total
    100 ml of H2O, 100 ml of dichloromethan ( Iwilll abbreviate as dich.)
    10 g of A in 100 ml of h2o by a single extraction w/ 100 mL of dich.?
    if (x) g = the amount that stays in the water, 7.3(x)g of A must go into dich. phase.

    Wt. of A in Dich. = 100 ml * 7.3(x)g/ml = 730 (x)g
    Wt. of A in H2O = 100ml * (x) g/ml = 100(x)g

    Total weight of A = 10 g (now partitioned between two layers)

    add 100 (x) g + 730(x)g = 830 (x)g

    830 (x)g = 10g
    (x) = .0120481928

    now, plugging x back into the equations:

    wt. of A in Dich. = 730 (.0120481928)g = 8.7965 g

    wt. of A in H20 = 100 (.0120481928)g = 1.205 g
    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    Pt. 2: what weight og A owuld be removed by 3 successive extractions with 33 ml portions of Dich.?

    Everything is the same, except instead of one extraction there will be 3 extractions with 33 mL portion s of dich. instead of 100 mL.

    Wt. of A in dich. = (33mL)(7.3)(x)g/mL = 240.9 (x)g
    Wt. of A in H20 = 100mL (x)g/ml = 100 (x) g

    Total weight = 10 g

    after adding 100 x + 240.9 x = 340.9 x

    340.9(x)g = 10 g

    x= 0.0293341156

    after substituting back into equations:

    Wt. of A in dich. = (33mL)(7.3)(.0293341156)g/mL = 6.824697g
    Wt. of A in H20 = 100mL (.0293341156)g/ml = 2.83341156 g

    DIch. later is removed. and second round begins

    Wt. of A in dich. = 240.9g
    wt. of A in H20 = 100g

    total wieght = 2.833 g

    340.9 (x)g = 2.833g

    (x) = .0083103549

    Weight of A in Dich. = 240.9 (.0083103549) = 1.99947g
    weight of A in H2O = 100 (.0083103549) = .86103549g

    Now layer is removed and third round begins

    weight of A in Dich. = 240.9
    Wt. of A in water = 100

    total weight = 340.9

    there fore,

    340.9 x = .83g

    x= .00243g

    wt. of A in dich = 240.9 * .00243g = .5855387g
    wt. of A in H2o = 100 * .00243g = .243g

    total % weight = 6.825 + 1.999+ .5855 = 9.04095 from 10 g

    (9.04095 / 10) * (100%) = 94%
     
  4. Oct 15, 2007 #3
    continuation


    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    Pt 3:

    How much dich. would be required to remove from 98.5% of A in a single extraction?

    98.5% = 9.85g of total weight of A in dichloro

    wt. of A in Dich = (x)(y) = .985g

    x= x
    y = orig wt. of dich.
    z = x*y

    z= 10g

    wt. of A in Dich. = y * 7.3x = y7.3 x
    wt. of A in water = 100mL * 7.3 x = 730x


    to be continued....

     
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