Organic Semiconductor Solution
Preparation
Procedure No.:
Procedure Title:
Revision:
Date:
Physics CR-012
Organic Semiconductor Solution Preparation
1
20th June 2009
Author:
Checked By:
Colin Belton
Xuhua Wang
Organic Semiconductor Solution Preparation
1. Decide what solution concentration and volume are needed for your requirements
(See attached Notes On Solution Concentration Units). Ensure that you know what
mass of solute and volume of solvent necessary to make your solution.
2. Place a piece of cleanroom paper beside the scales to work on.
3. In the fumehood, clean a spatula using a cleanroom wipe, which has been damped
with acetone. Always be sure not to contaminate stock containers of material, as
some of these are extremely difficult/costly to replace. It is good practice to take your
own smaller stock of material and use this to make your solutions rather than
repeatedly using the main stock bottle. Note it is best to use a clean vial, which has
been covered in tin foil to store material; this prevents excessive exposure to UV light,
which can cause photo-degradation.
4. Measure the mass of solute (material) into a clean sample vial as follows:
a. Place the vial onto the scales and allow the reading to stabilise.
b. Press the “Zero” button.
c. Remove the vial from the scales and place it on the cleanroom paper.
d. Using the spatula, decant a small amount of material into the vial
e. Replace the vial on the scales and allow the reading to stabilise; the mass of
solute will be displayed on the scales.
f. Repeat this steps d-e until you have the desired mass in the sample vial. Do
not decant material into the vial when it is on the scales. Stray material on the
scales can give rise to inaccurate readings as well as get stuck under the pan
and cause damage to the scales.
5. The solvent may now be added to the vial, using a graduated pipettor in the
fumehood. Decant the required volume of solvent into the vial as follows:
a. Set the required volume on the graduated pipettor using the adjustment
wheel at the top of the pipettor (units set in L). Allow the pipettor to “rest” for
at least a minute, after changing the volume; this results in a more accurate
dispensed volume.
b. Pour a small amount of solvent from one of the Winchester stock bottles into
a clean beaker.
c. Place a pipette tip onto the graduated pipettor.
d. In air, depress the plunger until the first stop is encountered. Place the tip of
the pipette into the solvent and slowly release the plunger, to draw the liquid
into the pipette tip. Dispense the solvent back into the beaker. This is known
as pre-wetting the pipette and increases the accuracy of the process.
e. Now depress the plunger to the first stop again, in air. Once again place the
tip into the solvent and draw up the full volume allowed by the first stop
graduation.
f. Withdraw the pipette from the solvent and insert the tip into the vial. Hold the
pipette at 45 degrees to the horizontal and touch the tip onto the side of the
vial. Slowly depress the plunger to decant the solvent, until the first stop is
encountered. Then depress the plunger to the second stop. The pipette is
calibrated to deliver the required amount of solvent, only if this procedure is
followed.
g. Dispose of the pipette tip into the sharps bin by pointing the pipette tip at the
entrance of the bin and pressing the tip removal button. NOTE. Do not lay the
pipettor flat while a wet tip is in place, as the solvent will damage it; Always
keep the pipettor upright when using it to prevent contamination by organic
solvents/samples.
6. Place a clean magnetic stirrer bar into the solution.
7. Place a cap onto the sample vial and seal with parafilm; the vial may now be handled
outside the fumehood.
8. To ensure complete salvation of the material, the sample should be either put into the
ultrasonic bath for 15 mins or placed on a magnetic stirrer for certain time at an
appropriate temperature. This step would depend on what type of the materials used.
Notes On Solution Concentration Units
Concentration of Polymer Solutions
Due to the distribution of their molar mass it is best to use the mg/mL unit for measuring the
concentration of polymer solutions. This is a simple expression of the concentration of a
solution and only requires the mass of solute and the volume of the solvent. The
concentration is then given by the ratio:
Mass of Solute
Volume of Solvent
Example: Prepare a 4 mL solution of 75 mg/mL PMMA in acetone.
Concentration
Mass of PMMA
=
= Concentration  Volume of Solvent
= 75 mg/ml  4ml
= 300 mg
Dissolve 300mg of PMMA in 4 mL of acetone.
Molarity
The most common unit of solution concentration is molarity (M). The molarity of a solution is
defined as the number of moles of solute per one litre of solution. Note that the unit of volume
for molarity is litres, not millilitres or some other unit. Also note that one litre of solution
contains both the solute and the solvent. Molarity, therefore, is a ratio between moles of
solute and litres of solution. To prepare laboratory solutions, usually a given volume and
molarity are required. To determine molarity, the formula weight or molar mass of the solute is
needed. The following examples illustrate the calculations for preparing solutions.
If starting with a solid, use the following procedure:
 Determine the mass in grams of one mole of solute, the molar mass, MM.
 Decide volume of solution required, in litres, V.
 Decide molarity of solution required, M.
 Calculate mass of solute (m) required using
mi
=
MMi  M  V
Example: Prepare an 800 mL solution of 2 M sodium chloride in water.
MMNaCl
=
58.45 g/mol
mNaCl
=
58.45 g/mol  2 mol/L  0.8 L
=
93.52 g NaCl
Dissolve 93.52 g of NaCl in about 400 mL of distilled water; then add more water until final
volume is 800 mL.
If starting with a solution or liquid reagent: use the following procedure:
 When diluting more concentrated solutions, decide what volume (V2) and molarity
(M2) the final solution should be. Volume can be expressed in litres or millilitres.
 Determine molarity (M1) of starting, more concentrated solution.
 Calculate volume of starting solution (V1) required using
M1V1
=
M2V2
Note: V1 must be in the same units as V2.
Example: Prepare 100 mL of 1.0 M hydrochloric acid from concentrated (12.1 M) hydrochloric
acid.
M1V1
=
M2V2
(12.1 M)V1
V1
=
=
(1.0 M)(100 mL)
8.26 mL conc. HCl
Add 8.26 mL of concentrated HCl to about 50 mL of distilled water, stir, then add water up to
100 mL.
Percent Solutions
Mass percent solutions are defined based on the grams of solute per 100 grams of solution.
Example: 20 g of sodium chloride in 100 g of solution is a 20% by mass solution.
Volume percent solutions are defined as millilitres of solute per 100 mL of solution.
Example: 10 mL of ethyl alcohol plus 90 mL of H2O (making approx. 100 mL of solution) is a
10% by volume solution.
Mass-volume percent solutions are also very common. These solutions are indicated by
w/v% and are defined as the grams of solute per 100 millilitres of solution. Example: 1 g of
phenolphthalein in 100 mL of 95% ethyl alcohol is a 1 w/v% solution.
Conversion Between Percent Solutions
You may wish to convert mass percent to volume percent or vice versa. If so, follow this
procedure:
A 10% by mass solution of ethyl alcohol in water contains 10 g of ethyl alcohol and 90 g of
water.
1. The formula for determining the volume of the component (ethyl alcohol in our example) is:
Volume
=
Mass of ethyl alcohol
Density of ethyl alcohol
2. Determine the volume of the total solution by dividing the mass of the solution by the
density of the solution.
3. Determine the percent by volume by dividing the volume of the component by the volume
of the solution.
Let’s solve 1, 2, and 3 above as follows:
1. Mass of ethyl alcohol = 10 g (given)
Density of ethyl alcohol = 0.794 g/mL (from handbook)
Volume
=
Mass of ethyl alcohol
Density of ethyl alcohol
Volume
=
10 g
0.794 g/mL
=
2.
12.6 mL
Mass of solution = 100 g (given)
Density of solution (10% ethyl alcohol) = 0.983 g/mL (from CRC handbook)
Volume
=
100 g
0.983 g/mL
=
101.8 mL
=
Volume of ethyl alcohol
Total volume of solution
3. Volume percent of solution
Volume
=
12.6 mL
101.8 mL
=
12.4%
Reverse the procedure to convert volume percent to mass percent.
Calculating Molarity From Percent Solutions
To determine the molarity of a mass percent solution, the density of the solution is required.
Use the following procedure:
1. Determine the mass of solution by multiplying the volume of the solution by the density of
the solution.
Mass = Volume  Density
2. Determine concentration in percent by mass of the solute in solution. Change to the
decimal equivalent.
3. Calculate the molar mass of the compound, MM.
4. Multiply mass (step 1) by mass % (step 2) and divide by molecular mass (step 3) to find the
number of moles present in the whole solution.
5. Divide the number of moles (step 4) by the volume in litres of the solution to find the
molarity of the solution.
Example: Determine molarity of 37.2% hydrochloric acid (density 1.19 g/mL).
 Mass of solution = 1,000 mL x 1.19 g/mL = 1,190 g
 Mass % = 37.2 % = 0.372
 Molar mass of hydrochloric acid = 36.4 g/mol
Mass  Mass%
MMHCl
=
=
1.190 g  0.372
36.4 g/mol
12.1 moles
Molarity = moles/liters = 12.1 moles/1 liter = 12.1 M