IGEM:American University/2009/Notebook/Advanced Experimental Chemistry: Fun Times in Beeghly/2013/02/15
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(Autocreate 2013/02/15 Entry for IGEM:American_University/2009/Notebook/Advanced_Experimental_Chemistry:_Fun_Times_in_Beeghly) 
Current revision (16:28, 15 February 2013) (view source) (→Welcome to Advanced Experimental Chemistry) 

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==Welcome to Advanced Experimental Chemistry==  ==Welcome to Advanced Experimental Chemistry==  
  to  +  
+  =Today's ToDo List=  
+  
+  1) Crosslink Monday's Films: This was done with the same procedures as always: Placed in 0.7M HCl (70 degrees) for 1hr then into 1.68g/100mL Sodium BiCarbonate 15 minute then soaked in water for at least an hour. This was done in the hood.  
+  
+  2) Remove and dry the films crosslinked on monday on paper towels in the hood.  
+  
+  3) Preparing PVOH and 3Chloro2Hydroxypropyltrimethylammonium chloride reacted crystals. Procedure and Calculations below:  
+  
+  From previous calculations, there are 1.369E22 OH groups in 1g PVOH MW22000. We want to prepare 2g, so it's actually 2.738E22 OH groups.  
+  We want to prepare a solution of the 3Chloro2Hydroxypropyltrimethylammonium chloride such that it reacts w 10% of OH groups in 2g of PVOH in a 1:1 reaction.  
+  2.738E22 OH groups x 0.1 = 2.738E21 molecules of 3Chloro2Hydroxypropyltrimethylammonium chloride react w 10% of OH groups in 2g PVOH.  
+  2.738E21 molecules of 3Chloro2Hydroxypropyltrimethylammonium chloride / 6.022 molecules per mole = 0.0045 moles 3Chloro2Hydroxypropyltrimethylammonium chloride.  
+  0.0045 moles 3Chloro2Hydroxypropyltrimethylammonium chloride x 188.1g/mole 3Chloro2Hydroxypropyltrimethylammonium chloride = 0.855g of 3Chloro2Hydroxypropyltrimethylammonium chloride would react w 10% of the OH groups in PVOH.  
+  
+  4 sets of 2g PVOH were left to dissolve (as much as possible on a stir plate) in 5mL 0.2M NaOH for 30 minutes.  
+  0.855g of 3Chloro2Hydroxypropyltrimethylammonium chloride was left to dissolve in 5mL 0.2M NaOH for 30 minutes.  
+  
+  So... we want solutions of PVOH and 3Chloro2Hydroxypropyltrimethylammonium chloride in 10 mL of 0.2M NaOH such that 3Chloro2Hydroxypropyltrimethylammonium chloride reacts with 0.01%, 0.1%, 1% and approximately 10% of the OH groups in 2g PVOH. Thus, since our "stock" solution of 3Chloro2Hydroxypropyltrimethylammonium chloride is 10% and 0.885g in 5mL NaOH (0.2M). 0.855g/5mL=0.171g/mL 3Chloro2Hydroxypropyltrimethylammonium chloride "stock" solution...  
+  
+  0.01%: x/0.855g=0.01%/10%, x=0.000855g. 0.000855g/0.171(g/mL)= 0.005mL of the "stock" 3Chloro2Hydroxypropyltrimethylammonium chloride solution. Thus to make our 0.01% 3Chloro2Hydroxypropyltrimethylammonium chloride:PVOH, would require the 2gPVOH dissolved in 5mL NaOH 0.2M + 0.005mL of the "stock" 3Chloro2Hydroxypropyltrimethylammonium chloride filled to 10mL NaOH 0.2M [4.995mL extra NaOH 0.2M]. Similarly...  
+  
+  0.1%: Required the 2gPVOH dissolved in 5mL NaOH 0.2M + 0.05mL of the "stock" 3Chloro2Hydroxypropyltrimethylammonium chloride filled to 10mL NaOH 0.2M [4.95mL extra NaOH].  
+  
+  1%: Require the 2gPVOH dissolved in 5mL NaOH 0.2M + 0.5mL of the "stock" 3Chloro2Hydroxypropyltrimethylammonium chloride filled to 10mL NaOH 0.2M [4.5mL extra NaOH].  
+  
+  To make the "10%" solution, we used the 4.445mL left over "10% stock" 3Chloro2Hydroxypropyltrimethylammonium chloride solution. To solve for the actual percentage this makes... (4.445mL)(0.171g/mL 3Chloro2Hydroxypropyltrimethylammonium chloride) = 0.760095g 3Chloro2Hydroxypropyltrimethylammonium chloride  
+  (0.760095g)/(0.855g)=x%/10% x=8.89% so, the solution is actually '''8.89%''' of PVOH OH groups reacting w 3Chloro2Hydroxypropyltrimethylammonium chloride.  
+  
+  The PVOH and 3Chloro2Hydroxypropyltrimethylammonium chloride wasn't dissolving properly after they were combined (the PVOH never really dissolved in the first place) so the solutions were moved to the hood to dissolve on the hot plate.  
Current revision
iGEM Project name 1  Main project page Previous entry Next entry 
Welcome to Advanced Experimental ChemistryToday's ToDo List1) Crosslink Monday's Films: This was done with the same procedures as always: Placed in 0.7M HCl (70 degrees) for 1hr then into 1.68g/100mL Sodium BiCarbonate 15 minute then soaked in water for at least an hour. This was done in the hood. 2) Remove and dry the films crosslinked on monday on paper towels in the hood. 3) Preparing PVOH and 3Chloro2Hydroxypropyltrimethylammonium chloride reacted crystals. Procedure and Calculations below: From previous calculations, there are 1.369E22 OH groups in 1g PVOH MW22000. We want to prepare 2g, so it's actually 2.738E22 OH groups. We want to prepare a solution of the 3Chloro2Hydroxypropyltrimethylammonium chloride such that it reacts w 10% of OH groups in 2g of PVOH in a 1:1 reaction. 2.738E22 OH groups x 0.1 = 2.738E21 molecules of 3Chloro2Hydroxypropyltrimethylammonium chloride react w 10% of OH groups in 2g PVOH. 2.738E21 molecules of 3Chloro2Hydroxypropyltrimethylammonium chloride / 6.022 molecules per mole = 0.0045 moles 3Chloro2Hydroxypropyltrimethylammonium chloride. 0.0045 moles 3Chloro2Hydroxypropyltrimethylammonium chloride x 188.1g/mole 3Chloro2Hydroxypropyltrimethylammonium chloride = 0.855g of 3Chloro2Hydroxypropyltrimethylammonium chloride would react w 10% of the OH groups in PVOH. 4 sets of 2g PVOH were left to dissolve (as much as possible on a stir plate) in 5mL 0.2M NaOH for 30 minutes. 0.855g of 3Chloro2Hydroxypropyltrimethylammonium chloride was left to dissolve in 5mL 0.2M NaOH for 30 minutes. So... we want solutions of PVOH and 3Chloro2Hydroxypropyltrimethylammonium chloride in 10 mL of 0.2M NaOH such that 3Chloro2Hydroxypropyltrimethylammonium chloride reacts with 0.01%, 0.1%, 1% and approximately 10% of the OH groups in 2g PVOH. Thus, since our "stock" solution of 3Chloro2Hydroxypropyltrimethylammonium chloride is 10% and 0.885g in 5mL NaOH (0.2M). 0.855g/5mL=0.171g/mL 3Chloro2Hydroxypropyltrimethylammonium chloride "stock" solution... 0.01%: x/0.855g=0.01%/10%, x=0.000855g. 0.000855g/0.171(g/mL)= 0.005mL of the "stock" 3Chloro2Hydroxypropyltrimethylammonium chloride solution. Thus to make our 0.01% 3Chloro2Hydroxypropyltrimethylammonium chloride:PVOH, would require the 2gPVOH dissolved in 5mL NaOH 0.2M + 0.005mL of the "stock" 3Chloro2Hydroxypropyltrimethylammonium chloride filled to 10mL NaOH 0.2M [4.995mL extra NaOH 0.2M]. Similarly... 0.1%: Required the 2gPVOH dissolved in 5mL NaOH 0.2M + 0.05mL of the "stock" 3Chloro2Hydroxypropyltrimethylammonium chloride filled to 10mL NaOH 0.2M [4.95mL extra NaOH]. 1%: Require the 2gPVOH dissolved in 5mL NaOH 0.2M + 0.5mL of the "stock" 3Chloro2Hydroxypropyltrimethylammonium chloride filled to 10mL NaOH 0.2M [4.5mL extra NaOH]. To make the "10%" solution, we used the 4.445mL left over "10% stock" 3Chloro2Hydroxypropyltrimethylammonium chloride solution. To solve for the actual percentage this makes... (4.445mL)(0.171g/mL 3Chloro2Hydroxypropyltrimethylammonium chloride) = 0.760095g 3Chloro2Hydroxypropyltrimethylammonium chloride (0.760095g)/(0.855g)=x%/10% x=8.89% so, the solution is actually 8.89% of PVOH OH groups reacting w 3Chloro2Hydroxypropyltrimethylammonium chloride. The PVOH and 3Chloro2Hydroxypropyltrimethylammonium chloride wasn't dissolving properly after they were combined (the PVOH never really dissolved in the first place) so the solutions were moved to the hood to dissolve on the hot plate.
