copper-penny-report
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Copper Penny report
General Chemistry (Cornell University)
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Copper Penny report
General Chemistry (Cornell University)
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Copper Penny Report
Natalia Jordan
TA: Sewon Oh
December 12th, 2020
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Ⅰ
. Abstract
This lab was designed to determine the amount of copper in a post-1982 U.S penny through the process of
spectrophotometry. By using the same cuvette, absorbance is directly proportional to concentration, and
the graph of those variables demonstrates Beer’s Law. The penny was first dissolved using nitric acid
followed by ammonium hydroxide. This newly made copper (II) tetraamine complex is referred to as the
“Penny solution” and its absorbance at 620 nm was measured. The absorbance was 0.639. The second
part of the experiment was conducted to measure the absorbance of 5 different concentrations of
Cu(NH3)4+. A standard curve for the concentration of Cu(NH3)4+ was generated
from these values, and
the concentration of Cu(NH3)4+ in the “penny solution” was determined to be 0.012978 M using the
equation of the trendline from the standard curve graph. The mass of copper was then determined to be
Cu from the known molarity and volume of the Cu(NH3)4+ . The percent of error was
.08247
g
0
determined by comparing the measured mass of copper in the penny to the actual mass of copper in a
post-1982 penny, 0.0625g Cu (found from usmint.gov). The percent error was found to be 31.95%.
Ⅱ
. Experimental Section
Part One: First, a post-1982 penny was obtained and weighed to the nearest milligram. A 25 mL
graduated cylinder was used to measure out 15 mL of concentrated nitric acid in the hood. The penny was
then laid flat on the bottom of a 200 mL beaker and the 15 mL of nitric acid were slowly added, in the
fume hood, on top of the penny. Because all of the copper solutions did not completely dissolve, the
contents of the beaker were carefully swirled to ensure that all copper dissolved. The solution initially
appeared to be a green/teal color, then became lighter, and then emitted an orange/brown gas. To the
beaker, 15 mL of distilled water were added. The contents of the beaker turned from a green color to an
aqua blue color. Another beaker was obtained and 25 mL of concentrated ammonium hydroxide was
added to the nitric acid solution in portions. The beaker contents were then swiftly swirled after each
addition of the ammonium hydroxide. After the first addition of ammonium hydroxide, the solution
turned a dark blue and became cloudy after the contents were swirled. In the second addition, the beaker
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contents became warm to the touch and a precipitate was visible in the beaker. In the third addition, the
solution became a darker blue color. Lastly, in the final addition of ammonium hydroxide, a precipitate no
longer remained and the beaker appeared a clear cobalt blue color. The solution was then left alone to
cool to room temperature and 100 mL of copper (II) tetra ammonia complex were measured in a
volumetric flask and diluted with distilled water to the mark. The solution in the volumetric flask was
mixed by inversion several times to ensure it was properly mixed. This solution was labeled the “Penny
Solution''. The absorbance of this solution was measured at 620 nm.
Part two: Five beakers were thoroughly cleaned and dried. The beakers were labeled 1-5. The first beaker
was used to obtain 15 mL of .040 M Cu(NH
3
)
4
+
. A series of 5 serial dilutions was conducted. A
volumetric pipette was used to dispense exactly 10.00 mL of distilled water into each of the four beakers
labeled 2,3,4, and 5. From beaker 1, a volumetric pipette was used to measure exactly 10.00 mL of the
Cu(NH
3
)
4
+
stock solution and it was added into beaker 2. The solution in beaker 2 was swirled to ensure
proper mixing. A volumetric pipette was then used to move 10.00 mL of the solution in beaker 2 into
beaker 3. The solution in beaker 3 was swirled to ensure proper mixing. A volumetric pipette was then
used to move 10.00 mL of the solution in beaker 3 into beaker 4. The solution in beaker 4 was swirled to
ensure proper mixing. A volumetric pipette was then used to move 10.00 mL of the solution in beaker 4
into beaker 5. The solution in beaker 5 was swirled to ensure proper mixing. After the 5 serial dilutions
were performed, the absorbance of the solution in each of the 5 beakers was determined and recorded at
620 nm.
Ⅲ
. Results and Discussion:
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Table 1: Absorbance of Different Solutions measured at 620nm
Table 2: Concentration of Cu(NH3)42+ (Part B) in beaker solutions
Solution
Absorbance (measured at 620nm)
Penny Solution
0.639
Beaker 1
2.323
Beaker 2
1.312
Beaker 3
0.592
Beaker 4
0.275
Beaker 5
0.133
Solution
Concentration (M)
Beaker 1
0.04
Beaker 2
0.02
Beaker 3
0.01
Beaker 4
0.005
Beaker 5
0.0025
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Table 3: Absorbance and Concentration of Cu(NH
3
)
4
2+
(Part B) in beaker solutions
Solution
Absorbance
Concentration (M)
Beaker 5
0.133
0.0025
Beaker 4
0.275
0.005
Beaker 3
0.592
0.01
Beaker 2
1.312
0.02
Beaker 1
2.323
0.04
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Graph 1: Absorbance vs. Concentration for Five Beaker Solutions
with an equation of
A= 53.3x-0.0527 (Table 3 data)
During the experiment, wavelength is held constant and the same cuvette is used. Using the same
cuvette ensures the path length stays the same, thus Graph 1 illustrates Beer’s Law of A= εdc.
Concentration, c, usually has units of M. The distance light travels through the sample, d, usually
has units of cm. The molar absorptivity or extinction coefficient, ε, has units of
.
cm
M
−1
−1
Therefore, absorbance is unitless. If the same d is used for measurements, absorbance becomes
directly proportional to concentration. By using the line of best fit for this graph, it is possible to
determine the concentration of
in the Penny Solution. In this experiment,
u
(
NH
)
C
3 4
2+
pathlength, d, is 1 and the molar absorptivity, ε, is the slope of the best fit line. Ideally, the
y-intercept should be 0, though it appears as -0.0527. Since the Penny Solution’s absorbance is
0.639, its concentration of
is as follows:
u
(
NH
)
C
3 4
2+
3.3
x
.0527
A
= 5
− 0
.639
3.3
x
.0527
0
= 5
− 0
C
=
53.3
(0.639+0.0527)
.012978
M
c
= 0
Concentration of
in “penny solution” =
u
(
NH
)
C
3 4
2+
.012978
M
0
Where c and x are interchangeable
Using this information the grams of copper in the original penny can me calculated:
0.012978 M
in 100mL of solution
u
(
NH
)
C
3 4
2+
00
mL
0012978
mol
Cu
(
NH
)
1
L
0.012978
mol
Cu
(
NH
)
3 4
2+
× 1
×
1
L
1000
mL
= .
3 4
2+
Since Cu and
are in a 1:1 ratio,
u
(
NH
)
C
3 4
2+
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=
0012978
mol
Cu
(
NH
)
.
3 4
2+
.0012978
mol
Cu
0
Using molar mass of Cu (63.546g):
0012978
mol
Cu
.08247
g
Cu
in
the
original
penny
.
×
1
mol
Cu
63.546
Cu
= 0
Assuming a penny is a cylinder with a diameter of 19.0mm, height of 1.5mm, and a density of
8.93g/mL (which is equivalent to 8.93g/cm3), the thickness of copper on the penny can be
calculated:
.08247
g
Cu
.2351
mm
Cu
0
×
1
cm
3
8.93
g
Cu
×
(1
cm
)
3
(10
mm
)
3
= 9
3
surface
Area
)
thickness
)
V
Cu
= (
penny
× (
dh
thickness
)
V
Cu
= π
× (
.2351
mm
(19.0
mm
)(1.5
mm
)
thickness
)
9
3
= π
× (
hickness
of
copper
penny
.03
mm
T
= 1
Assuming that the diameter of a copper atom is
, the thickness of copper can be
.28
0
mm
2
× 1
−7
calculated to be:
.52
0
atoms
thick
(1.03
mm
)
2.28×10
mm
per
atom
−7
= 4
× 1
5
According to the U.S. Mint, post-1982 pennies are composed of 2.5% Cu by mass. Using the
original mass of the penny and this information, the actual mass of copper on a penny can be
determined:
ctual
mass
of
Cu
.500
g
(0.025)
.0625
g
Cu
a
= 2
= 0
As calculated before, the measured mass of Cu from the experiment is as follows:
easured
mass
of
Cu
.08247
g
Cu
M
= 0
Thus the percent error is calculated to be:
ercent
error
00
p
=
actual
value
measured
value
−
actual
value
|
|
× 1
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ercent
error
00
1.95 %
p
=
0.0625
0.08247−0.0625
|
|
× 1
= 3
Our percent error in determining the mass percent of copper in the penny was 31.10%. We
determined that the mass of copper in a penny was 0.08247g when it is really 0.0625 g copper.
The sources of error could have possibly arose from some discrepancies in measuring the
absorbity for the dilution. The most diluted solution appeared slightly cloudy, and its absorbance
was a bit difficult to measure accurately. Likewise, the spectrophotometer may not have been
calibrated correctly and the serial dilutions may have been incorrect. These mistakes would cause
the standard curve of Cu(NH3)42+ to be incorrect, and lead to a miscalculation of the
concentration of copper in the “penny solution,” which would affect the calculation of the copper
cladding’s thickness. For example if the spectrophotometer recorded a lower absorbance than it
should have, this would lead to a flatter slope of the curver and in return a lower estimated
concentration of copper derived from the trendline equation of Cu(NH3)42+. However, since the
R2 value, 0.996, is so close to 1, the line of best fit is a good representation of the data.
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- pnc.com/women O Dashboa x O Course E x 6 The Wavx E CP grou X M Docume X E 3- Qual x G If you ha X purse.html?courseld3D169856748OpenVellumHMAC=624ef941a8d10e711a39b3488025ef9a#10001 Complete the sentences to explain your choice. Match the words in the left column to the appropriate blanks in the sentences on the right. Reset Help nonmetal Lithium is a and sodium is a Lithium is in' group and main group metal sodium is in group transition metal Germanium is a and arsenic is a Germanium is in group metalloid and arsenic is in group 1A (1) Titanium is a and gallium is a 2A (2) Argon is a and bromine is a Argon is in group ,and 3A (13) bromine is in group 4A (14) 5A (15) Nitrogen is a and oxygen is a Nitrogen is in group and oxygen is in group 6A (16) 7A (17) 8A (18) P Pearson Copyright © 2021 Pearson Education Inc. All rights reserved. | Terms of Use | Privacy Policy I Permissions | Contact Us | 88°F Mostly cloudy へ G0 トト "prt sc delete home 144 トII 10 num & 7 backspace lock 8 9.arrow_forwardSafari File Edit View History Bookmarks Develop Window Help Not Secure - a5.chem.binghamton.edu o E LON-CAPA t30p04a O [엘소드/KR Elswrod] 범… MC Kimetsu no Yaiba - Di. As you work on the questions, have your lab manual and text book handy. If you have questions go to one of the office hours and remember, do not leave it for the last minute! In the titration of 25.00 mL of a water sample, it took 19.090 mL of 3.165x 10-3 M EDTA solution to reach the endpoint. Calculate the number of moles of EDTA required to titrate the water sample. (enter your answer with 3 significant figures) 6.04x10-5 You are correct. Your receipt no. is 151-9969 ? Previous Tries The total hardness is due to one or a combination of Ca2+, Mg2+, and Fe2+ in your sample. It is convenient to express this hardness as though it was entirely due to Ca2+. Making this assumption, determine the number of moles of Ca2+ present in the bottled water sample titrated. (enter your answer with 3 significant figures) 6.04x10-5 You are…arrow_forwardRank the following diatomic species of fluorine in order of bond length and bond strength. A. F2 B. F₂+ C. F₂ Bond Length Bond Strength Longest Next Shortest Strongest Next Weakestarrow_forward
- ter 9- CHEM A054 520 2 x Question 12 - Chapter 9 part 2 X с Chapter 9 part 2 Homework i 12 1 points ezto.mheducation.com/ext/map/index.html?_con=con&external_browser=0&launchUrl=https%253A%252F%25 8 01:46:13 eBook Hint Print References Mc Graw Hill CH6_Chem103 - Kenai Peninsu X + °C Determine the freezing point of an aqueous solution that is 0.584 m sucrose. Saved DUarrow_forwardHomework AX /6920146#/6920146/16/2 Macmillan Learning Web 2.0 scientific calculator Williston Forge Jero... Assignment Score: Question 17 of 21 > mass: Home Page - myAl... OO TOOLS ×10% 91.5% X + HUAWEI P30 LEICA TRIPLE CAMERA T Synonyms and Ant... Home Page - myAl.. Resources e Ex Give Up? SOAP2DAY Essayaibot: Free Ess... Hint Hydrogen gas holds promise as an alternative fuel to gasoline. Calculate the mass of H₂ gas in a 38.3 L fuel tank at a pressu 2.42 x 104 mmHg and a temperature of 17.4 °C. Paraphrasing Toof Check Answer Attempt 4 Carrow_forwardAssessment Chemistry x Copy of spread of ista X E The Spread of Islam- x oncourseconnect.com/assessment/1651879/5287e2a3-0d0b-e2c0-c15c-3e68a37149b4 N hbox (004) aarolamex Clase D TPSS Bookmarks CHEMISTRY BENCHMARK TEST 01 CHEMSTRY1IARON JAMES ID 12390724) Aluminum (Al) reacts with chlorine (Cl,) gas to form aluminum chloride. Given the position of aluminum in the periodic table, what charge should the aluminum ion have in this compound? -5 +1 +3 + (P O A O B O D Save/Exit 1 4 5 6 8 9 10 US ASUSarrow_forward
- Provided are spectra for a compound with molecular formula CgH100 Part A) Give the degrees of unsaturation Part B) Identify the peaks that are associated with specific bonds of the structure and list the corresponding functional groups that the specific bonds belong to. Re- create the example table below that organizes this information. Use the wavenumbers that are already labeled on the spectra. Example table: wavenumber (cm-1) Corresponding Bond Type Corresponding Functional Group C(sp3)-H N-H 2978 3300 Part C) Draw the structure and clearly indicate which hydrogens correspond to which signals in the HNMR spectrum ONLY. The signals are already labeled A-D on the HNMR spectrum. Match the hydrogen labeling (HA, HB, etc.) on your drawn structure to the signal labeling provided. 4300 3320-2872 2951 3006 3003 2001 alkyl amine 1613 1510 1000arrow_forward2. MF= C12H160 QE-300 20 200 180 160 140 120 100 80 60 40 20 2 2 3. 2 11 10 9 8 7 6 5 4 3 1arrow_forwardCUTE ASS Gwinn X B Homepage-2021S x A Classwork for 2021 x E Danesha Patterson x Commonlit | The S docs.google.com/document/d/1TNETJ72LOKTHX-Ab-VC2FSAnLbFjT2a5pa79lj8 CommonLit | The Scramble for Africa |Free Reading Passages and Literac. izzes - 2021S1 H. commonlit.org anesha Patterson - Copy of Stoichiometry DLD Review 2020 (2).DOCX @ e Edit View Insert Format Tools Add-ons Help Last edit was 2 days ago E A, P Normal text Arlal 100% +BIUA E = E E tE 1E E - E - 4. 11 w .. gi - 2 3. 4 7 **Draw your T-charts on a piece of paper and take a pic to upload*** Use the equation given below for questions 1-3 , - 1co, + 2 H,O How many moles of methane (CH,) are needed to produce 3.7 moles of water (H,O)? 1CH, + 20, 1. 2. How many grams of oxygen (0,) react if the reaction produces 0.9 moles of carbon dioxide (CO,)? How many grams of water (H,O) are produced if 136 grams of methane (CH,) react completely in this reaction? If you start with 136 grams of methane (CH4), what is the theoretical…arrow_forward
- Mg3(PO4)2 (s) Mg2*(aq) + _ PO43- (aq) Mg2* (aq) PO43 (aq) initial change eqm Write Ksn in terms of x. 36 x5 12 x3 3x4 108 x5arrow_forwardComplete the table with at least 3 Major NMR shifts. Normal NMR Analysis BIU X₂ X²- Chemical shift, ppm Integration Multiplicity fxe Part of structure Txarrow_forwardExample 2arrow_forward
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