Beanium
Description
Beanium is an imaginary element whose atoms are macroscopic. The Beanium model is based upon an analogy to isotopes.
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Set
- Imagine this scenario:
- Recently discovered as residues on plates in the school cafeteria is a substance thought to be a new element and named Beanium.
- Students have been asked to examine this material in the chemistry laboratory and to determine some of its properties.
- Since Beanium atoms are visible to the eye, this research will be fairly easy. (No other known element has atoms visible to the naked eye, or even to the most powerful optical microscopes.)
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Procedure
The margarine containers hold Beanium atoms. The differences between the isotopes of Beanium are very distinct.
- Find the total mass of Beanium in the sample.
- Sort the Beanium atoms into groups, each group representing a different isotope.
- Record the total number of each isotope (beans) in your sample.
- On the data sheet, sketch a picture of each isotope emphasizing differences between them.
- Atomic masses are determined by comparing the mass of an atom to a standard mass, the mass of a carbon-12 atom. In this experiment, however, the standard we select is the mass of one milliliter of water at 4 °C, or 1.0000g. In this way, the balance reading is automatically in the desired units.
- Determine the atomic mass of each isotope in the following manner.
- Find the total mass of each isotope; record.
- Divide the total mass of each isotope by the number of atoms in that particular sample. This will give the mass of an isotope, and, since we have chosen the particular mass standard that we have, the actual mass.
- Determine the percent abundance of each isotope in the entire sample by dividing the number of atoms of the isotope by the total number of atoms held in the container.
- Determine the atomic mass for Beanium based on the percent abundances of each isotope and their atomic masses. This is the method of weighted averages used to determine atomic masses of the elements.
- atomic mass =
- [(% isotope 1 x mass of one atom of isotope 1)
- +(% isotope 2 x mass of one atom of isotope 2)
- + ...
- +(% isotopen x mass of one atom of isotopen)] ÷100%
- Compute the total mass of the Beanium based upon the number of atoms, and the average mass.
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Handout
Name _____________________________ Class _______
Teacher______________________________
DoChem 026 Beanium
- State the atomic difference between isotopes of the same element?
- There are 125 navy beans, 56 pinto beans and 144 blackeye peas in a container. Find the percent by number of navy beans in the container.
- Suppose your chemistry grade is broken down so that 60% of it is based on exams, 20% on lab reports and 20% on homework. Find your average score based upon these individual scores:
- exams 78
- labs 85
- homework 91
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- Total number of beans in container =
- Total mass of Beanium =
| |
Type |
# of beans |
Mass of sample |
| a. |
|
|
|
| b. |
|
|
|
| c. |
|
|
|
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- Calculate the following:
- % abundance of each bean
- The isotopic mass of each isotope
- The average atomic mass of Beanium for this sample.
- There is a distinction between percent abundance by number and percent abundances by mass. Compute the percent of the mass of the bean sample contributed by navy beans.
- Neon-20 has a mass of 19.9924 amu, Neon-21 has a mass of 20.9940 amu, and Neon-22 has a mass of 21.9914 amu. The relative abundance of 20Ne is 90.92%, 21Ne is 0.257% and 22Ne is 8.82%. Calculate the average atomic mass for neon, expected to be found on the periodic table.
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Handout Makeup
Name _____________________________ Class _______
Teacher______________________________
DoChem 026 Beanium
- State the atomic difference between isotopes of the same element?
- There are 125 navy beans, 56 pinto beans and 144 blackeye peas in a container. Find the percent by number of navy beans in the container.
- Suppose your chemistry grade is broken down so that 60% of it is based on exams, 20% on lab reports and 20% on homework. Find your average score based upon these individual scores:
- exams 78
- labs 85
- homework 91
Watch the movies.
In terms of this analogy, what do you think is the most important difference between the beans in one of the cups and the atoms in a pure isotope sample?
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- A student had the following data:
- Total sample mass = 207.97 g
- Mass of 220 navy beans = 85.80 g
- Mass of 194 pinto beans = 46.95 g
- Mass of 402 blackeye peas = 75.13 g
- Calculate the following:
- % abundance of each bean
- The isotopic mass of each isotope
- The average atomic mass of Beanium for this sample.
- There is a distinction between percent abundance by number and percent abundances by mass. Compute the percent of the mass of the bean sample contributed by navy beans.
- Neon-20 has a mass of 19.9924 amu, Neon-21 has a mass of 20.9940 amu, and Neon-22 has a mass of 21.9914 amu. The relative abundance of 20Ne is 90.92%, 21Ne is 0.257% and 22Ne is 8.82%. Calculate the average atomic mass for neon, expected to be found on the periodic table.
- State the atomic difference between isotopes of the same element?
- There are 125 navy beans, 56 pinto beans and 144 blackeye peas in a container. Find the percent by number of navy beans in the container.
- Suppose your chemistry grade is broken down so that 60% of it is based on exams, 20% on lab reports and 20% on homework. Find your average score based upon these individual scores:
- exams 78
- labs 85
- homework 91
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Teachers Guide
Purpose
To illustrate the concept of isotopes through a "bean" analogy.
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Materials
(per 10 students)
- 5 margarine containers and covers
- 2500 beans (pinto, navy, blackeye)
- 5 triple beam balances
- 20 paper cups
- Lab Hints:
- Students can bring in the materials, particularly containers.
- Coins, ball bearings, plastic beads, copper BBs, and other suitably uniform objects may replace the beans.
- Vary the numbers of each type of bean from student to student; many different combinations of the three work well.
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Time
Teacher initial preparation time is variable; time is needed to buy beans and collect margarine containers.
Teacher preparation (when materials available): 5 minutes.
Class Time: 30 to 40 minutes
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Hazards
There are no unusual hazards in this experiment.
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Precautions
No special precautions are required in this experiment. Follow routine laboratory precautions.
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Disposal
Save the materials for subsequent classes.
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Set?
Set Questions: (now in handout)
- State the atomic difference between isotopes of the same element?
- There are 125 navy beans, 56 pinto beans and 144 blackeye peas in a container. Find the percent by number of navy beans in the container.
- Suppose your chemistry grade is broken down so that 60% of it is based on exams, 20% on lab reports and 20% on homework. Find your average score based upon these individual scores:
- exams 78
- labs 85
- homework 91
Answers to Set Questions:
- Isotopes of an element differ in the number of neutrons present in the nucleus.
- {(125 navy)/( 125 + 56 + 144 beans)} x 100% = 38.5% navy beans.
- Chemistry average score is:
- { [ (60)(78) + (20)(85) + (20)(91)]/100 }x 100%= 82%.
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Presentation?
Presentation Question:
- In terms of this analogy, what do you think is the most important difference between the beans in one of the cups and the atoms in a pure isotope sample?
- In order for the analogy to hold, each and every bean for a given type of bean must have exactly the same mass. That is, all of the navy beans must have exactly the same mass, and so forth. This is not the case; there are measurable differences between beans of a given type. In a pure sample of one isotope, all of the atoms do have the same mass.
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Sample Data
Total number of beans in container = 518
Total mass of Beanium = 136.32 g.
| Type |
# of beans |
Mass of sample |
| navy |
165 |
64.85g |
| pinto |
143 |
32.89g |
| blackeye |
210 |
38.64g |
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Closure?
Closure Questions:
- A student had the following data:
- Total sample mass = 207.97 g
- Mass of 220 navy beans = 85.80 g
- Mass of 194 pinto beans = 46.95 g
- Mass of 402 blackeye peas = 75.13 g
- Calculate the following:
- % abundance of each bean
- The isotopic mass of each isotope
- The average atomic mass of Beanium for this sample.
- There is a distinction between percent abundance by number and percent abundances by mass. Compute the percent of the mass of the bean sample contributed by navy beans.
- Neon-20 has a mass of 19.9924 amu, Neon-21 has a mass of 20.9940 amu, and Neon-22 has a mass of 21.9914 amu. The relative abundance of 20Ne is 90.92%, 21Ne is 0.257% and 22Ne is 8.82%. Calculate the average atomic mass for neon, expected to be found on the periodic table.
Answers to Closure Questions:
- a. 27.0%, 23.8%, 49.3%; b. 0.390 g/atom, 0.242 g/atom, 0.187 g/atom; c. 0.255 g/atom.
- 41.26% by mass navy beans.
- [((90.92%) x (19.9924))
- + ((0.257%)x(20.9940))
- + ((8.82%)x(21.9914))] ÷ 100% = 20.171
- (actual value: 20.183)
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Background
- Any atom can be selected as the measuring standard, and the ratios calculated for that atom. An atom of oxygen-16 is 1.33291 times heavier than carbon-12, for example. Carbon-12 is selected as the standard by which masses of atoms are measured. By assigning 12C a mass of exactly 12.00000... atomic mass units, it turns out that protons and neutrons each have about 1 amu of mass, and that all isotopes have nearly integral mass values.
- There are several roundoff errors possible that you might point out. Not all percents add to 100%. Also, the observed total sample masses are usually different from the sum of the masses of the samples, and the calculated mass is different from the observed total mass in nearly every case. These represent roundoff errors and experimental errors.
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Data Analysis
Individual atomic masses:
- 64.85 g/165 navy isotope= 0.393 g/navy isotope
- 32.89 g/143 pinto isotope = 0.230 g/pinto isotope
- 38.64 g/210 blackeye isotope = 0.184 g/blackeye isotope
% abundance:
- 165/518 x 100 = 31.9%
- 143/518 x 100 = 27.6%
- 210/518 x 100 = 40.5%
Average atomic mass :
- [(31.9)x(0.393 g) + (27.6)x(0.230 g) + (40.5)x(0.184 g)] ÷ 100% = 0.263 g/atom
- Total mass = 518 atoms x 0.263 g/atom = 136 g.
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Key Words
- charge
- positive charge
- negative charge
- Coulomb's Law
- opposite charge
- like charge
- force
- repel
- attract
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