Derived from Brighthub
This activity will simulate atomic isotopes and atomic mass calculations. We will use plastic eggs to represent atoms. In each atom will be some dark beans to represent protons and light-colored beans to represent neutrons.
- Each table gets an Eggium atom.
- Determine the atomic number (protons = dark-colored beans), the number of neutrons (light-colored beans), the number of electrons (assume a balanced atomic charge), and the mass number for your atom. Record this information in your notes using the table format shown below.
- The symbol for Eggium is Eg. Put the information you've gathered for your Eggium atom in the proper isotope notation. Example: 126C
Number of protons Number of neutrons Mass number Number of electrons Information written in isotope notation
- Write your atom information in the isotope notation on the whiteboard. Spreadsheet
- Record the class data for number of atoms of a particular isotope:
Number of atoms Atomic mass number
- Are all of the atoms in the class the same element? How do you know?
- Are all of the atoms of Eggium the same mass?
- Why do some atoms have more mass than others?
- What is consistent from atom to atom?
- What do you call atoms that have the same number of protons but variable numbers of neutrons?
- Do some of the isotopes occur more often than others?
- What is the average atomic mass for eggium? Use method one (By Counting Atoms) shown below. Show your calculations !
- Which method do you think scientists use to calculate the average atomic mass of an element?
- Why do you think that?
- Use method two (By Natural Abundance) to calculate the percentage occurrence of each eggium isotope. Show your calculations. For example:
- If you have 14 atoms total and 4 of them are a certain isotope
- 100 (4 / 14) ≈ 28%.
- the percentage abundance for that isotope ≈ 28%.
Methods for Finding Average Atomic Mass
- By Counting Atoms:
- Add the masses of each atom and divide by n.
- 144 / 14 = 10.3
- The atomic mass of Eggium is 10.3
- By Natural Abundance:
- Total atoms of Eggium = 14
- % blue = 100 (# of blue / total ) = 100 (3 / 14) = 21%
- % red = 100 (# of red / total ) = 100 (7 / 14) = 50%
- % green = 100 (# of green / total ) = 100 (4 / 14) = 29%
- (21% x 9) + (50% x 10) + (29% x 11) = 10.3