User:Dhea Patel/Notebook/Phosphorylation/2012/02/02: Difference between revisions

From OpenWetWare
Jump to navigationJump to search
(5 intermediate revisions by 2 users not shown)
Line 6: Line 6:
| colspan="2"|
| colspan="2"|
<!-- ##### DO NOT edit above this line unless you know what you are doing. ##### -->
<!-- ##### DO NOT edit above this line unless you know what you are doing. ##### -->
==Entry title==
==Preparing for the General Synthesis of Modified ATP==




This entry is a continuation of Jordans' Notebook.
This entry is a continuation of [[User:Jordan_L._Metsky/Notebook/Phosphorylation/2012/01/27|Jordan's Notebook.]]


*t-butyl ammonium ATP was extracted from a round-bottom flask into 3 vials. Most of it was a gel-like liquid, though a crude, wax-like material was at the bottom of the flask.  
*t-butyl ammonium ATP was extracted from a round-bottom flask into 3 vials. Most of it was a gel-like liquid, though a crude, wax-like material was at the bottom of the flask.  
Line 33: Line 33:




*"Calculating Theoretical Yield"
* '''Calculating Theoretical Yield'''


0.5 grams Na<sub>2</sub>(ATP) / (553.2 g/mol) Na<sub>2</sub>(ATP) = 9.038E-4 mol
0.5 grams Na<sub>2</sub>(ATP) / (553.2 g/mol) Na<sub>2</sub>(ATP) = 9.038E-4 mol


9.038E-4 mol Na<sub>2</sub>(ATP) × (2 mol (t-butyl)ammonia/1 mol ATP) × (992.28 g [(t-butyl) ammonium]<sub>2</sub>(ATP)
9.038E-4 mol Na<sub>2</sub>(ATP) × (2 mol (t-butyl)ammonia/1 mol ATP) × (992.28 g [(t-butyl) ammonium]<sub>2</sub>(ATP)
= '''1.792 g'''


*"Assume, knowing that the assumption is wrong, that there is 1.2 g (t-butyl) ammonium ATP"
= '''1.792 grams'''
 
* '''Assume, knowing that the assumption is wrong, that there is 1.2 g (t-butyl) ammonium ATP'''


1.2 g / (507.22g ATP) = 0.002366 mol ATP or 2.3658 mmol ATP in 1.2 g (t-butyl) ammonium ATP
1.2 g / (507.22g ATP) = 0.002366 mol ATP or 2.3658 mmol ATP in 1.2 g (t-butyl) ammonium ATP
Line 46: Line 48:
2.3658 mmol ATP/2.3803 g product = 0.99391 mmol ATP/g product
2.3658 mmol ATP/2.3803 g product = 0.99391 mmol ATP/g product


*"Calculating mass of reactants for General synthesis of modified ATP"
 
* '''Calculating mass of reactants for General synthesis of modified ATP'''


0.5 mmol ATP / (0.99391 mmol/g ATP) = 0.50306 g product
0.5 mmol ATP / (0.99391 mmol/g ATP) = 0.50306 g product

Revision as of 13:38, 5 February 2012

Project name <html><img src="/images/9/94/Report.png" border="0" /></html> Main project page
<html><img src="/images/c/c3/Resultset_previous.png" border="0" /></html>Previous entry<html>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</html>Next entry<html><img src="/images/5/5c/Resultset_next.png" border="0" /></html>

Preparing for the General Synthesis of Modified ATP

This entry is a continuation of Jordan's Notebook.

  • t-butyl ammonium ATP was extracted from a round-bottom flask into 3 vials. Most of it was a gel-like liquid, though a crude, wax-like material was at the bottom of the flask.

Liquid vial: 2.3803 g Solid vial: 0.1294 g Aqueous solution vial: 1.6780 g Total mass: 4.1877 g

  • Calculations were run to determine the mmol ATP/gram of product was in the liquid vial.
  • 0.5121 g of t-butyl ammonium ATP product was vacuumed overnight to get rid of any excess H2O.

Calculations

2 tetrabutyl ammonia, [CH3(CH2)]4N has a molar mass of 242.53 g/mol

ATP, C10H16N5O13P3 has a molar mass of 507.22 g/mol

Na2ATP has a molar mass of 552.2 g/mol

0.5 grams of ATP was used in the reaction.


  • Calculating Theoretical Yield

0.5 grams Na2(ATP) / (553.2 g/mol) Na2(ATP) = 9.038E-4 mol

9.038E-4 mol Na2(ATP) × (2 mol (t-butyl)ammonia/1 mol ATP) × (992.28 g [(t-butyl) ammonium]2(ATP)

= 1.792 grams


  • Assume, knowing that the assumption is wrong, that there is 1.2 g (t-butyl) ammonium ATP

1.2 g / (507.22g ATP) = 0.002366 mol ATP or 2.3658 mmol ATP in 1.2 g (t-butyl) ammonium ATP

2.3658 mmol ATP/2.3803 g product = 0.99391 mmol ATP/g product


  • Calculating mass of reactants for General synthesis of modified ATP

0.5 mmol ATP / (0.99391 mmol/g ATP) = 0.50306 g product

2.5 mmol × (162.15 g/1000 mmol carbon diimidazole) = 0.405375 g carbon diimidazole

2.5 mmol × (73.09 g/1000 mmol DMF) = 0.182725 g DMF

2.5 mmol × (595.69g/1000 mmol [Ru(bpy)2(phen-IA)](PF6)2) = 1.489225 g [Ru(bpy)2(phen-IA)](PF6)2

4 mmol × (32.04 g/1000 mmol MeOH) = 0.12816 g MeOH


Retrieved from "https://openwetware.org/mediawiki/index.php?title=User:Dhea_Patel/Notebook/Phosphorylation/2012/02/02&oldid=580755"