Hi Elmira

It is simple.

solution 1: you should add 0.15(g) poly A to 5 ml DMF 

solution 2: you should add 0.1(g) poly B to 5 ml DMF

then mix them together. the final solution consist of 1.5% poly A and 1% poly B.

thanks but the final composition consists following composition:

0.15 gr A

0.1 gr B

DMF : 10 ml (density= 0.95 gr/cm3) or 10*0.95=9.5 gr

total weight= 0.15+0.1+9.5= 9.75

% poly A= 0.15/9.75=0.0153*100=1.53 %NOT 1.5 % ?!!!!!!! Why are there differences?

% poly B=0.1/9.75=0.0102*100= 1.02 % NOT 1.00 % ?!!!!!! Why are there differences?

% DMF= 9.5/9.75= 0.9743*100= 97.34 % NOT 97.5 %?!!!!!! Why are there differences?

MY PROBLEM is these differences, I don't know why?! 

Hello dear, You have to employ these proportions to prepare 10 mL of the final solution:

Weight 0.15 g of component 1 and 0.1 g of component 2, then add DMF (9.75 g), this if the concentration is wt% (W/W), but if it is (V/V) you can dissolve the components in separate form by using 5 ml of DMF in both cases.

If you want to obtain this proportion: (Poly A )/ (Poly A+Poly B) = 60 %

You can use 3 grams of Poly A and 2 grams of Poly B.  The errors in your calculation is that the proportion of DMF is 9.75 grams. 

Find out a common solvent for both of the polymer.Make the equimolar solutions of both of the polymer. Mix them in lukewarm condition.

Hi If these are in wt-% then I think your mistake is looking at the density (I have used a density of DMF of 0.944 g/cm^3) rather than just using the weights 

If your values are wt-% then:

0.15 g of Poly A + 4.85g DMF = 3 wt-% Poly A (vol of DMF = 5.14 mL)

0.1g of Poly B + 4.9g DMF = 2 wt-% Poly B (vol of DMF = 5.19 mL)

Total weight of Poly A + Poly B + DMF = 10 g

Mix the two together and you have 

0.15 g Poly A in 10 g = 1.5 wt-% Poly A

0.1g Poly B in 10g = 1 wt-% Poly B

9.75g of DMF in 10 g = 97.5 wt-% (vol of DMF = 10.32 mL)

thanks all

It never ceases to amaze me how the education system fails to teach simple logical thinking and how to do these types of calculations.

First of all, let's make some abbreviations. PA = poly A; PB = poly B; PAS = poly A solution; PBS = poly B solution; FINAL = final solution. According to the "total composition" section, it seems we need to work in wt/wt. (If this is found in a published article, it is shameful that it was published without the units stated explicitly.} However, you desire a VOLUME quantity for your final use. (Are you certain that it said to use 10 mLs and not 10 grams of FINAL?) So that is why your calculations seem to come out a bit strange.   So because of this, if you REALLY want to use 10 mL of final solution (say by pipetting), then it will NOT contain the weights of PA and PB that you desired! (but they will be very close.)

PAS: 3 g PA / 97 g DMF

PAB:  2g PB / 98 g DMF

FINAL: (1.5 g PA + 1 g PB ) / 97.5 g DMF

So, how to obtain 1.5 g PA from the PAS, and 1 g PB from the PBS?

1.5 g PA * (100 g PAS /3 g PA) = 50 g PAS needed.

1 g PB * (100 g PBS /2 g PBS) = 50 g PAB needed.

So, by mixing together 50 g PAS with 50 g PBS, you get 100 g FINAL.

To verify this meets the "final composition" requirements:

50 g PAS * (97 g DMF / 100 g PAS) = 48.5 g DMF present.

50 g PBS * (98 g DMF / 100 g PBS) = 49 g DMF present.

So the total amount of DMF in FINAL is 97.5 g. And we already did the calculations that showed there are 1.5 g PA and 1 g PB in FINAL.

Now, you probably want to scale down because you only desired 10 mLs of FINAL.

So, make 1/10 the size.

PAS: 300 mg PA dissolved in 9.7 g DMF.

PBS: 200 mg PB dissolved in 9.8 g DMF.

Mix the two together to give 20 grams FINAL.

Now you can take a 10 mL aliquot for your desired solution amount.

Note that you COULD instead dissolve 150 mg PA in 4.85 g DMF and 100 mg PB in 4.9 g DMF and mix together to give 10 g FINAL, but this would be a bit tricky to take from this a 10 mL aliquot.

(I am interested in reading this publication, if you can post a copy or the URL  or DOI for it.)

Elmira, I see that you made a mistake.

First, you gave this composition.

poly A: 1.5 %

poly B: 1 %

DMF: 97.5 %

But later, you do the following for composition

0.15 gr A

0.1 gr B

DMF: 1ml (density = 0.95 gr/cm3) or 10*0.95=9.5 gr

total weight= 0.15+0.1+9.5=9.75

The last two lines are the mistakes.

We must assume from the first section that the concentration is wt/wt. But in your calculation of DMF, you changed the DMF from wt to volume. You must leave the DMF as weight as well, to give the following.

0.15 g A

0.1 g B

9.75 g DMF

total weight = 0.15 + 0.1 + 9.75 = 10.00.

Now after you prepare these 10 g of final solution, you remove your 10 mL aliquot, as I explained in the previous posting. (This is why I questioned if the amount of the final material you will use is 10 grams or 10 mL.)

Simply add 5 ml of solution of polymer A into 5 ml of solution of polymer B to make the total volume 10 ml and the required % concentration of 1.5% PA and 1% PB in DMF.