User:Michael F. Nagle/Notebook/Chem 571/2012/08/29: Difference between revisions

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==Procedure==
==Procedure==
#25mL of stock solution of 10mM [[AU_Biomaterials_Design_Lab:Materials/HAuCl4|HAuCl<sub>4</sub>]] was prepared.
#25mL of stock solution of 7.721mM [[AU_Biomaterials_Design_Lab:Materials/HAuCl4|HAuCl<sub>4</sub>]] was prepared.
## (moles [[AU_Biomaterials_Design_Lab:Materials/HAuCl4|HAuCl<sub>4</sub>]])/.025L H<sub>2</sub>O = .01M
## (moles [[AU_Biomaterials_Design_Lab:Materials/HAuCl4|HAuCl<sub>4</sub>]])/.025L H<sub>2</sub>O = .01M
## 2.5*10<sup>-4</sup> mol [[AU_Biomaterials_Design_Lab:Materials/HAuCl4|HAuCl<sub>4</sub>]] is the amount of gold needed to make this concentration with .025L
## 2.5*10<sup>-4</sup> mol [[AU_Biomaterials_Design_Lab:Materials/HAuCl4|HAuCl<sub>4</sub>]] is the amount of gold needed to make this concentration with .025L

Revision as of 18:58, 6 December 2012

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Objective

  • Prepare stock solutions of HAuCl4 and BSA and use them to make solutions of HAuCl4 and BSA at varying mole ratios.
    • These are to be analyzed by UV/Vis next week to find the optimal mole ratio for nucleation of gold nanoparticles.

Procedure

  1. 25mL of stock solution of 7.721mM HAuCl4 was prepared.
    1. (moles HAuCl4)/.025L H2O = .01M
    2. 2.5*10-4 mol HAuCl4 is the amount of gold needed to make this concentration with .025L
    3. 2.5*10-4 mol * 339.785mol/g HAuCl4 = .0849g HAuCl4 is needed
    4. .0197g of the .0849g stuck to the weigh paper, leaving .0656gHAuCl4.
    5. .0656g HAuCl4 * (1molHAuCl4/339.79g/molHAuCl4)= xM HAuCl4
    6. .007721M HAuCl4 was the molarity of stock solution made
    7. .0849g HAuCl4 was put in 25mL to make a 10mM solution
  2. 25mL stock solution of 15μM BSA was prepared
    1. (moles BSA)/.025L = .000015
    2. 3.75*10^-7mol BSA is the amount of BSA needded to make a 15μM solution with .025L.
    3. 3.75*10^-7mol BSA * 66,463g/mol BSA = .0249g BSA needed
    4. .0249g BSA was put in 25mL to make a 15μM solution
  3. 6mL solutions were prepared with the mole ratios of (HAuCl4/BSA) 60, 80, 100, 120, 128, 130, 132, 133, 134, 136, 138, 140, 160, and 170. .6mL of HAuCl4 stock was added to each tube, so each was 1.5mM HAuCl4
  • The formula m1v1=m2v2 was used to determine how much BSA was needed for mole ratios (BSA/Au) of 60-170
    • 15μM*(volume BSA stock solution) = (1.5*6)
      • 0.6mL BSA stock solution in each tube
    • 7720μM*x mL=6mL*(1.5*60μM)
      • 0.069mL HAuCl4 stock solution
      • 5.331 mL H2O
    • 7720μM*x mL=6mL*(1.5*80μM)
      • 0.093mL HAuCl4 stock solution
      • 5.307mL H2O
    • 7720μM*x mL=6mL*(1.5*100μM)
      • 0.117mL HAuCl4 stock solution
      • 5.283mL H2O
    • 7720μM*x mL=6mL*(1.5*120μM)
      • 0.134mL HAuCl4 stock solution
      • 5.266mL H2O
    • 7720μM*x mL=6mL*(1.5*128μM)
      • 0.149mL HAuCl4 stock solution
      • 5.251mL H2O
    • 7720μM*x mL=6mL*(1.5*130μM)
      • .152mL HAuCl4 stock solution
      • 5.248mL H2O
    • 7720μM*x mL=6mL*(1.5*132μM)
      • 0.154mL HAuCl4 stock solution
      • 5.246mL H2O
    • 7720μM*x mL=6mL*(1.5*133μM)
      • 0.155mL HAuCl4 stock solution
      • 5.245mL H2O
    • 7720μM*x mL=6mL*(1.5*134μM)
      • 0.156mL HAuCl4 stock solution
      • 5.254mL H2O
    • 7720μM*x mL=6mL*(1.5*136μM)
      • 0.159mL HAuCl4 stock solution
      • 5.244mL H2O
    • 7720μM*x mL=6mL*(1.5*138μM)
      • 0.161mL HAuCl4 stock solution
      • 5.242mL H2O
    • 7720μM*x mL=6mL*(1.5*140μM)
      • 0.163mL HAuCl4 stock solution
      • 5.237mL H2O
    • 7720μM*x mL=6mL*(1.5*160μM)
      • 0.187mL HAuCl4 stock solution
      • 5.213mL H2O
    • 7720μM*x mL=6mL*(1.5*170μM)
      • 0.198mL HAuCl4 stock solution
      • 5.202mL H2O
    • The tubes were capped with tin foil and heated at 85°C for 4 hours.

Results

  • The BSA solution created was 15μM
  • The actual molarity for the HAuCl4 stock solution created turned out to be 7.721mM rather than 10mM. This is because unlike BSA, which is easy to handle, HAuCl4 comes in large chunks which don't break easily and is hydroscopic, leaving residue on the weigh paper.


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