User:Mary Mendoza/Notebook/CHEM 571 Experimental Biological Chemistry I/2012/08/29: Difference between revisions
Mary Mendoza (talk | contribs) |
Mary Mendoza (talk | contribs) |
||
Line 57: | Line 57: | ||
X<sub>Au</sub>= 1.5 μM × 140 (mole ratio) = 210 μM of Au | X<sub>Au</sub>= 1.5 μM × 140 (mole ratio) = 210 μM of Au | ||
2. After obtaining the mole concentrations of Au, these mole concentrations were applied into the dilution equation used earlier to obtain the volume of Au needed. The | |||
Revision as of 13:07, 14 September 2012
Project name | <html><img src="/images/9/94/Report.png" border="0" /></html> Main project page Next entry<html><img src="/images/5/5c/Resultset_next.png" border="0" /></html> |
Preparation of Gold and Bovine Serum Albumin Stock Solutions
.025 L of water × (15×10-6 mM of BSA ÷ 1 L of water) = 3.75×10-7 mM of BSA 3.75×10-7 mM of BSA × 66430 g (formula weight of BSA) = .0249 g of BSA
.025 L of water × (10×10-3 mM of HAuCl4 ÷ 1L of water)= .00025 mM of HAuCl4 .00025 mM of HAuCl4 × 339. 79 g (formula weight of HAuCl4)= .085 g of HAuCl4 Preparation of different mole ratios of Au/BSA
M1 = 15 μM V1 = Volume needed from 15 μM BSA stock solution to complete ratio M2 = 1.5 μM (concentration of BSA for mole ratio 60) V2 = 6 mL (total volume of final solution)
1. First obtained the mole concentration needed for Au in correspondence to its mole ratio using the following factors: X= mole concentration needed for Au 1.5 μM is the mole concentration of BSA used since this is held constant XAu= 1.5 μM × 133 (mole ratio) = 199.50 μM of Au XAu= 1.5 μM × 134 (mole ratio) = 201 μM of Au XAu= 1.5 μM × 136 (mole ratio) = 204 μM of Au XAu= 1.5 μM × 138 (mole ratio) = 207 μM of Au XAu= 1.5 μM × 140 (mole ratio) = 210 μM of Au 2. After obtaining the mole concentrations of Au, these mole concentrations were applied into the dilution equation used earlier to obtain the volume of Au needed. The
|