User:Allison K. Alix/Notebook/Thesis Research/2013/04/09: Difference between revisions
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==Procedures== | ==Procedures== | ||
Part 1: Preparation of 2.98μM ethane thiol | Part 1: Preparation of 1mL of 2.98μM ethane thiol | ||
* original concentrationof AuNP = 19.4nM | |||
* desired concentration of ethanethiol = 19.4nM x(15393) = 298624.2nM = 298μM | |||
* density of ethanethiol = 0.839g/mL | |||
1) Calculating concentration of ethanethiol | |||
0.839g/mL x 0.025mL = 0.020975g | |||
0.020975g/(62.13 g/mol) = 3.376 x 10<sup>-4</sup> mol | |||
3.376 x 10<sup>-4</sup> mol/1mL x (1000mL/1L) = 0.34M = 340mM | |||
340mM (x) = (30mM)(1mL) | |||
x = 0.088mL = 88μL in 912μL H<sub>2</sub>O | |||
30mM (x) = 3mM (1mL) | |||
x = 100μL in 900μL | |||
3mM(x) = 298μM (1mL) | |||
x = 99μL in 901μL to obtain 298μM ethanethiol | |||
2) Reacting AuNP with ethanethiol | |||
* react 200μL 298μM ethane thiol with 200μL 19.4nM AuNP (synthesized by Dr. Miller) | |||
Revision as of 07:47, 9 April 2013
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Objectives
ProceduresPart 1: Preparation of 1mL of 2.98μM ethane thiol
1) Calculating concentration of ethanethiol 0.839g/mL x 0.025mL = 0.020975g 0.020975g/(62.13 g/mol) = 3.376 x 10-4 mol 3.376 x 10-4 mol/1mL x (1000mL/1L) = 0.34M = 340mM 340mM (x) = (30mM)(1mL) x = 0.088mL = 88μL in 912μL H2O 30mM (x) = 3mM (1mL) x = 100μL in 900μL 3mM(x) = 298μM (1mL) x = 99μL in 901μL to obtain 298μM ethanethiol 2) Reacting AuNP with ethanethiol
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