User:Carly M. Montanero/Notebook/CHEM-571/2013/09/11: Difference between revisions
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==Calculating the Number of Gold Atoms in Citrate-AuNP== | ==Calculating the Number of Gold Atoms in Citrate-AuNP== | ||
# The equation of the best fit line from the class data was y = 3x10<sup>-5</sup>x<sup>2</sup> + 0.0209x. | |||
# The average of our citrate-AuNP concentration was 0.1287 μg/mL. | |||
# Setting 0.1287 equal to y yields the equation 0.1287 = 3x10<sup>-5</sup>x<sup>2</sup> + 0.0209x. | |||
# Solving the quadratic equation gives two answers: x = -702.771 and x = 6.104. | |||
# Disregarding the negative value, the 6.014 μg/mL corresponds to the concentration of gold. | |||
# The molecular weight of gold is 196.97 g/mol. | |||
# Converting the concentration of gold to molarity yields a concentration of 3.099x10<sup>-5</sup> M gold. | |||
# Dividing the concentration of gold (3.099x10<sup>-5</sup>) M by the concentration of citrate-AuNP (([[User:Carly_M._Montanero/Notebook/CHEM-571/2013/08/28 | 9.17×10<sup>-9</sup>M]]) gives the final number of gold atoms per citrate-AuNP molecule. | |||
# Though these calculations, there are 3379 gold atoms per citrate-AuNP molecule. | |||
==Notes== | ==Notes== |
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Objective
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Calculating the Number of Gold Atoms in Citrate-AuNP
NotesThe UV-Vis spectra of trial 4 Bradford Assay of horseradish peroxidase had some serious errors. We discarded the data. |