User:Dhea Patel/Notebook/Phosphorylation/2012/02/02: Difference between revisions
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*"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" | * "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 |
Revision as of 13:36, 5 February 2012
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Preparing for the General Synthesis of Modified ATPThis entry is a continuation of Jordan's 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 |