User:Mary Mendoza/Notebook/CHEM 571 Experimental Biological Chemistry I/2012/08/29: Difference between revisions
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* Calculation of the required Au concentration for all other mole ratios was divided amongst other group members, Michael Nagle and Puja Moody. Calculations for mole ratios 133, 134, 136, 138, and 140 are provided below: | * Calculation of the required Au concentration for all other mole ratios was divided amongst other group members, Michael Nagle and Puja Moody. Calculations for mole ratios 133, 134, 136, 138, and 140 are provided below: | ||
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 | |||
X<sub>Au</sub>= 1.5 μM × 133 (mole ratio) = 199.50 μM of Au | |||
X<sub>Au</sub>= 1.5 μM × 134 (mole ratio) = 201 μM of Au | |||
X<sub>Au</sub>= 1.5 μM × 136 (mole ratio) = 204 μM of Au | |||
X<sub>Au</sub>= 1.5 μM × 138 (mole ratio) = 207 μM of Au | |||
X<sub>Au</sub>= 1.5 μM × 140 (mole ratio) = 210 μM of Au | |||
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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
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