Asked by Jake
Pre-Lab Questions
Data:
[mass of copper wire]
0.738 g
[standard solutions]
solution 1: 0.738 g of copper wire dissolved in 50.00 mL solution
solution 2: 1:2 dilution of solution 1
solution 3: 1:4 dilution of solution 1
[absorbance data at lambda(max)]
solution 1: 0.740
solution 2: 0.365
solution 3: 0.195
[least squares analysis of these data pieces]
slope: 49.458
y-intercept: 0.0075
A brass sample of mass 0.643 g was dissolved to make 50.00 mL of solution, which was found to have an absorbance of 0.589 at lambda(max).
Questions:
1) Calculate the concentrations of the three standard solutions (in g/mL).
solution 1: 0.738 g / 50.00 mL = 0.0148 g/mL
solution 2: 0.0148 g/mL / 2 = 0.740 g/mL
solution 3: 0.740 g/mL / 2 = 0.370 g/mL
2) If the concentrations are in units of g/mL, what are the units for the slope?
If absorbance (A) is graphed against concentration (C) then, using the least squares method, the best-fit straight line through the data will be:
A = mC + b
m (slope) = delta y / delta x = delta A / delta C
A is dimensionless. C has units of g/mL. Therefore, the slope for the data presented above is 49.458 (g/mL)^(-1).
3) What are the units for the y-intercept?
The y-intercept (b) is found when concentration is 0. This yield only a numerical value for A without any units.
Therefore, the y-intercept for the data presented above is already correct as 0.0075.
4) Calculate the percent copper in the brass sample.
This is where I'm stuck. I have no idea how to use the absorbances of the various solutions to find the percent copper in the brass sample.
The only step I'm fairly certain about is:
% copper in brass = (mass of copper in brass sample) / (mass of brass sample) *100%
% copper in brass = (mass of copper in brass sample) / (0.643 g) *100%
Any feedback regarding my work for the first 3 questions and any help on the 4th question is greatly appreciated. Thanks in advance.
Data:
[mass of copper wire]
0.738 g
[standard solutions]
solution 1: 0.738 g of copper wire dissolved in 50.00 mL solution
solution 2: 1:2 dilution of solution 1
solution 3: 1:4 dilution of solution 1
[absorbance data at lambda(max)]
solution 1: 0.740
solution 2: 0.365
solution 3: 0.195
[least squares analysis of these data pieces]
slope: 49.458
y-intercept: 0.0075
A brass sample of mass 0.643 g was dissolved to make 50.00 mL of solution, which was found to have an absorbance of 0.589 at lambda(max).
Questions:
1) Calculate the concentrations of the three standard solutions (in g/mL).
solution 1: 0.738 g / 50.00 mL = 0.0148 g/mL
solution 2: 0.0148 g/mL / 2 = 0.740 g/mL
solution 3: 0.740 g/mL / 2 = 0.370 g/mL
2) If the concentrations are in units of g/mL, what are the units for the slope?
If absorbance (A) is graphed against concentration (C) then, using the least squares method, the best-fit straight line through the data will be:
A = mC + b
m (slope) = delta y / delta x = delta A / delta C
A is dimensionless. C has units of g/mL. Therefore, the slope for the data presented above is 49.458 (g/mL)^(-1).
3) What are the units for the y-intercept?
The y-intercept (b) is found when concentration is 0. This yield only a numerical value for A without any units.
Therefore, the y-intercept for the data presented above is already correct as 0.0075.
4) Calculate the percent copper in the brass sample.
This is where I'm stuck. I have no idea how to use the absorbances of the various solutions to find the percent copper in the brass sample.
The only step I'm fairly certain about is:
% copper in brass = (mass of copper in brass sample) / (mass of brass sample) *100%
% copper in brass = (mass of copper in brass sample) / (0.643 g) *100%
Any feedback regarding my work for the first 3 questions and any help on the 4th question is greatly appreciated. Thanks in advance.
Answers
Answered by
DrBob222
I assume you have drawn a graph to find the straight line. I also assume you have measured the absorbance of one or more of the brass samples. Using A from the unknown sample, look at the graph and convert A to C. C will be in g/mL. Use that to convert g/mL to grams of the sample, then calculate % Cu in the usual manner. There is insufficient data and/or explanation to comment on the other parts.
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