User:Dhea Patel/Notebook/Phosphorylation/2012/02/02: Difference between revisions
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==Entry title== | ==Entry title== | ||
This entry is a continuation of Jordans' 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 H<sub>2</sub>O. | |||
==Calculations== | |||
2 tetrabutyl ammonia, [CH<sub>3</sub>(CH<sub>2</sub>)]<sub>4</sub>N has a molar mass of 242.53 g/mol | |||
ATP, C<sub>10</sub>H<sub>16</sub>N<sub>5</sub>O<sub>13</sub>P<sub>3</sub> has a molar mass of 507.22 g/mol | |||
Na<sub>2</sub>ATP 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 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) | |||
= '''1.792 g''' | |||
*"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)<sub>2</sub>(phen-IA)](PF<sub>6</sub>)<sub>2</sub>) = 1.489225 g [Ru(bpy)<sub>2</sub>(phen-IA)](PF<sub>6</sub>)<sub>2</sub> | |||
4 mmol × (32.04 g/1000 mmol MeOH) = 0.12816 g MeOH | |||
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Revision as of 10:35, 2 February 2012
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Entry titleThis entry is a continuation of Jordans' Notebook.
Liquid vial: 2.3803 g Solid vial: 0.1294 g Aqueous solution vial: 1.6780 g Total mass: 4.1877 g
Calculations2 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.
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 g
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
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 | |
