Introduction:
Unless you’re a coin collector, you probably think all United States pennies are pretty much the same. To the casual observer, all the pennies in circulation do seem to be identical in size, thickness, and composition. But just as elements have one or more isotopes with different masses, the pennies in circulation have different masses.
In this investigation, you are going to use pennies with different masses to represent different “isotopes” of an imaginary element called pennium, or Pe. Remember that chemical isotopes are atoms that have the same number of protons, but different numbers of neutrons. Thus, chemical isotopes have nearly identical chemical properties, but some different physical properties.
In this investigation, you will determine the relative abundance of the isotopes of pennium and the masses of each isotope. You will then use this information to determine the atomic mass of pennium. Recall that the atomic mass of an element is the weighted average of the masses of the isotopes of the element. This average is based on both the mass and the relative abundance of each isotope as it occurs in nature.
Purpose
To determine whether the total mass changes during a chemical reaction.
Materials:
Laboratory balance
25 pennies in a resealable bag
Procedure
Remove the pennies from the resealable bag and count them to make sure that there are 25. Determine and record the combined mass of your 25 pennies. (66.2 g)
Find the mass of each penny separately. In the Data Table, record the year the penny was minted and its mass to the nearest 0.01 gram. (NOT 0.1 gram – in other words, round off one decimal).
Place the 25 pennies in the resealable bag and return the pennies and the balance to the area designated. Clean up and wash your hands, pennies are dirty!
Table 1: Mass of Each Penny by Year
| Penny # | Year | Mass (g) |
| 1 | 1962 | 3.00 g |
| 2 | 1969 | 3.05 g |
| 3 | 1973 | 3.01 g |
| 4 | 1973 | 3.03 g |
| 5 | 1977 | 3.11 g |
| 6 | 1981 | 3.05 g |
| 7 | 1982 | 3.13 g |
| 8 | 1985 | 2.50 g |
| 9 | 1988 | 2.47 g |
| 10 | 1988 | 2.50 g |
| 11 | 1989 | 2.47 g |
| 12 | 1989 | 2.48 g |
| 13 | 1993 | 2.49 g |
| 14 | 1995 | 2.45 g |
| 15 | 1995 | 2.45 g |
| 16 | 1996 | 2.51 g |
| 17 | 1999 | 2.45 g |
| 18 | 1999 | 2.54 g |
| 19 | 2000 | 2.50 g |
| 20 | 2001 | 2.53 g |
| 21 | 2004 | 2.56 g |
| 22 | 2005 | 2.48 g |
| 23 | 2007 | 2.50 g |
| 24 | 2013 | 2.49 g |
| 25 | 2016 | 2.48 g |
Calculations
Inspect your data carefully. Determine the number of isotopes of Pe that are present (How many different masses of Pe are there?).
Two isotopes of Pe are present: Pennies made up until 1982 and pennies made after 1982.
Calculate the fractional abundance of each isotope in your sample. To check your math, add up your fractional abundances, they should total to 1.
Fractional Abundance (F.A) =(# of pennies for each isotope)/(Total # of pennies)
F.A of Pe-Pre 1982 = 7/25 = 0.28 (28% abundance)
F.A of Pe-Post 1982 = 18/25 = 0.72 (72% abundance)
Calculate the average atomic mass of each isotope.
Average atomic mass =(total mass of pennies of each isotope)/(number of pennies of that isotope)
Avg atomic mass of Pe-Pre 1982 = 21.38/7 = 3.05 g
Avg atomic mass of Pe-Post 1982 = 44.85/18 = 2.49 g
Using the fractional abundance and the average atomic mass of each isotope, calculate the average atomic mass of Pe.
Average Atomic Mass = (avg mass isotope #1) x (F.A. #1) + (avg mass isotope 2) x (F.A #2)
Average atomic mass of Pe = (3.05 x 0.28) + (2.49 x 0.72) = 2.65 g
Questions
Was the mass of 25 pennies equal to 25 times the mass of one penny? Explain, using your data.
The mass of the 25 pennies was not equal to 25 times the mass of one penny. This is because there are different isotopes of Pe. As observed in the lab, the two different isotopes of Pe had their fractional abundances. The different fractional abundances relate to the proportion of each isotope, meaning that the masses would not be constant between the two isotopes. Therefore, an average atomic mass for each isotope is required to produce an accurate number for the mass of the pennies.
In what year(s) did the mass of Pe change? How could you tell?
The mass of Pe changed after the year 1982. When recording the mass of each penny by year in Table 1, it was observed that the masses of the pennies shifted after 1982.
How can you explain the fact that there are different “isotopes” of Pennium? Use the US Mint website to research.
The phenomenon of there being different “isotopes” of Pennium can be explained because of the difference in the material used to construct the pennies. Pennies used to be made out of pure copper. However, in 1982, pennies started to be made with primarily zinc and only 2.5% copper, reducing the mass of the pennies in the process.
Why are the atomic masses for most elements, not whole numbers?
The atomic masses for most elements are not whole numbers because most of these elements have many different isotopes. The atomic mass of each element is the average mass of all naturally occurring isotopes of the element.
How are the three isotopes of hydrogen (hydrogen-1, hydrogen-2, and hydrogen-3) alike? How are they different?
The three isotopes of hydrogen are alike because they are all naturally occurring isotopes of the same element. However, they differ in the number of neutrons.
Copper has 2 isotopes, copper-63 and copper-65. The relative abundance of copper-63 is 69.1% and copper65, 30.9%. Calculate the average atomic mass of copper.
Average atomic mass of Cu = (0.691 x 62.93) + (0.309 x 64.93)
Average atomic mass of Cu = (43.48) + (20.06)
Average atomic mass of Cu = 63.5 amu
Discussion of Error
Clearly explain why we only recorded two decimals when recording the mass of each penny. If we had recorded one decimal, how might our estimate of the number of isotopes of Pennium have changed?
The decision to record the mass of the pennies with two decimal places was made to provide reasonable accuracy while keeping the data manageable. If one were to record the masses with one decimal place instead, the mass of the pennies would have to be rounded to the nearest tenth of a decimal. This would cause a change in the data, potentially affecting the number of isotopes found. However, in this case, with Pennium, the masses of the pennies were separated into two distinct groups. As a result, rounding the masses would not significantly impact the results, as there would still be two distinct isotopes of Pennium.
Based on the results of the experiment, it appears that there were little to no errors in the execution of the experiment. However, it is worth noting that during the measurement of one of the pennies, the battery for the laboratory scale died. Fortunately, the batteries were replaced and the same scale was used to continue the measurements. It is important to acknowledge that if a different scale had been used at that point, it could have introduced a slight variation in the data, potentially affecting the results to some degree.
Conclusion
The primary objective of this experiment was to unravel the masses and relative abundances of isotopes of Pennium (Pe) and to calculate its average atomic mass. Two distinct isotopes of Pe were uncovered by measuring the masses of various pennies. The fractional abundances of these isotopes were calculated to be 0.28 and 0.72, revealing the proportion of each isotope within the sample. Utilizing fundamental chemistry formulas, the average atomic mass of Pennium was calculated to be 2.65 grams. This result confirms the success of the lab objective and emphasizes the vital role of isotopic analysis in comprehending element composition, showcasing the diverse nature of scientific exploration and discovery.
