Linear Combination

We will make use of the numbers used in the file Summation: 5 molecules of D.P.=10, 3 molecules of D.P.=30, and 2 molecules of D.P.=60.

We could calculate the average D.P. by:

Now, the total mass of:

5- 1000 amu molecules is 5000;
3- 3000 amu molecules is 9000;
2- 6000 amu molecules is 12,000;
and the total mass of the system is 5000+9000+12,000= 26,000

The weight average fractions are thus as follows:

  • 5000/26,000 = .192
  • 9000/26,000 = .346
  • 12,000/26,000 = .461

  • .192 + .346 + .461 = .999 The D.P.=10 polymers comprise 19.2% of the total mass
    The D.P.=30 polymers comprise 34.6% of the total mass
    The D.P.=60 polymers comprise 46.1% of the total mass

    .192(1000) + .346(3000) + .461(6000) =

    192 + 1038 + 2766 = 3996 [which makes sense because .999(4000) = 3996]



    We can try the above equation, using the right-hand side to get 
    
5000(1000) + 9000(3000) + 12000(6000)   104,000,000   
------------------------------------- = ----------- = 4000 
     5000 + 9000 + 12000                  26,000
    Why bother with weight average fraction?

     POSSIBLE MISTAKE -The following assumes a GPC hooked up to a thermal detector. It has been suggested that this isn't done. I am looking into this. (7/23/95- wld)

    Detection may be through a continuous-flow sample that measures a concentration dependent phenomenon such as conductivity, radioactivity, light absorbance, or refractive index. (Elias- 130)

    It comes down to this: I have two peaks coming through the detector at different times.
    1. This peak consists of 10 molecules of D.P.=100
    2. This peak consists of 5 molecules of D.P.=200
    If the detector is analyzing mass, both peaks will register the same peak area,
    But, if the detector is analyzing concentration, then the first peak will have twice the area of the second.

    Follow-up is continuing.

    The following is true if your     above.

    It is left to the reader to prove that s/he can take weight average fraction data and molecule weights (0.192, 5000) (0.346, 3000) and (0.462, 6000) and correctly calculate number average molecular weight.



    It doesn't matter in terms of the calculation, but stricting speaking, the values, 0.192, 0.346, and 0.462 are not really the w(i) values that the above equation refers to: the w(i) values are actually 5000, 9000, and 12,000. In each case you are "off by a constant" but since you do this in both the numerator and the denominator, a cancellation of the error occurs, and you get the correct answer.

    Last Update- July 1, 1995- wld