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## Lab: PhET Isotopes and Atomic Mass

This lab is a computer-simulation-based activity. The simulator is available for immediate use in any browser at https://phet.colorado.edu/sims/html/isotopes-and-atomic-mass/latest/isotopes-and-atomic-mass_en.html. Or, perhaps more easily accessed by performing a search for “PhET Isotopes and Atomic Mass”.

### Background

Atoms of an element come in varieties called isotopes. Isotopes are atoms of the same element which have different mass numbers. Remember, the mass number of a specific atom is the sum of the protons and neutrons in its nucleus. In order to be atoms of the same element, isotopes must have the same number of protons. They have different mass number due to their having different numbers of neutrons.

Some isotopes of an element are stable, that is, they are not radioactive. Others may exist in nature or only in the lab when they are made artificially. Still others are simply impossible to make. For example, helium has two naturally occurring isotopes, helium-3 (32He ), and helium-4 (42He ). Both of these are stable but one of them is a lot easier to find in the environment than the other. Only one out of a million helium atoms is a helium-3 atom. There are absolutely no helium-2 atoms in the world and none can ever be made. A helium-2 atom would have two protons but zero neutrons and with no neutrons an atomic nucleus can’t remain whole.

The naturally occurring isotopes of an element each have a natural abundance. The natural abundance of an isotope is the percentage of all atoms of that element which have the mass number of that isotope. For example, 80.1% of all boron atoms in nature have the mass number 11. So we say boron-11 (115B) has an abundance of 80.1%. The only other naturally occurring isotope of boron is boron-10 (105B), which has an abundance of 19.9%. How can we understand the idea of abundance? Imagine we have 1,000 atoms of boron. On average, 801 of them will be boron-11 and 199 of them will be boron-10.

Isotopes are identified by their mass numbers. Natural abundances are measured by scientists using an instrument called a mass spectrometer. The same instrument can be used to measure the third important characteristic of isotopes which is their atomic mass. The atomic mass of an isotope is the actual mass of that isotope in amu. For example, boron-10 atoms have an atomic mass of 10.01294 amu and boron-11 atoms have a mass of 11.00931 amu.

Chemists need to know the mass of atoms because we use that information to be able to count atoms by weighing them. That’s what the atomic mass that’s shown on the periodic table is for. If we take a look at the box for boron on the periodic table we can see that the atomic mass of boron is 10.81 amu. This is neither 10.01294 amu nor is it 11.00931 amu. All atoms of boron have either a mass of 10.01294 amu or 11.00931 amu and no atoms of boron have a mass of 10.81 amu. So what does the 10.81 amu represent? It is the average atomic mass of boron. The average atomic mass of an element is the average mass of its isotopes. This average must take the natural abundances of the isotopes into account. To find the average it is not enough to add up the isotopes’ masses and divide by the number of isotopes. For boron, this would give:

```(10.01294 amu) + (11.00931 amu)
------------------------------- = 10.51 amu
2
```

The figure 10.51 amu is wrong. The correct average mass must account for the fact that 801 out of 1,000 boron atoms have a mass of 11.00931 amu and 199 out of 1,000 have a mass of 10.01294 amu. Here is how that calculation works:

```
(0.199)x(10.01294 amu) + (0.801)x(11.00931 amu) = 10.81 amu

```

You convert each percentage for the natural abundance to a decimal, multiply each one by its matching isotope mass, and add them together. Another way to do it is to imagine you have 1,000 atoms of boron. In that case you multiply each isotope mass by the number of atoms of that isotope, add it all together and divide by 1,000:

```
(199)x(10.01294 amu) + (801)x(11.00931 amu)
-------------------------------------------- = 10.81 amu
1,000
```

Because the natural abundance of an isotope is included in the calculation of the average atomic mass of all of the atoms of an element it has an effect on that average. The average atomic mass will be closest to the isotope with the larger natural abundance. For example, the average atomic mass of boron is closer to the atomic mass of boron-11 than it is to atomic mass of boron-10. This makes sense because boron-11 makes up 80.1% of all boron atoms.

page break

### Part I

When you open the simulator you will be on the Home tab. Click on the Isotopes tab at the bottom of the screen. When you get to that tab, click the buttons that will make the “Symbol” and “Abundance in Nature” displays open up.

Fill in the following table by creating all of the stable isotopes of each element in the table below with the simulator and recording what you see on your screen. To switch elements you use the segment of the periodic table in the upper-right part of the screen. You will have to switch back and forth between “Mass Number” and “Atomic Mass (amu)” on the display under the atomic model. If you are doing this on a piece of paper, please take care to set up a neat table like the one you see on your screen.

 Isotope Name Complete Isotope Symbol Number of Protons Number of Neutrons Mass Number Atomic Mass (amu) Abundance in Nature Hydrogen Hydrogen Boron Boron Carbon Carbon Fluorine Neon Neon Neon

### Part II

Next, click on the Mixtures tab at the bottom of the simulation screen. On that screen select Hydrogen (H) from the periodic table and open the “Percent Composition” and the “Average Atomic Mass” displays. At bottom-right click the button that says “My Mix”. Just under the main display, which is currently a black rectangle, there is a button with an icon that looks like a bucket and another that looks like a slider. Click on the slider button.

1. Set up a 50-50 mixture of hydrogen-1 and hydrogen-2. On the right there is a display called “Average Atomic Mass”. What is the average atomic mass for the mixture you have made?
2. Use the following formula to calculate the average mass yourself. The atomic masses are the numbers with many decimal places in the table you filled in Part I. You have to hit Enter on your calculator after adding but before dividing by 2.
```   [(Mass of Hydrogen-1) + (Mass of Hydrogen-2)]
------------------------------------------------ =
2
```
3. Now use a different formula to calculate the average mass. The atomic masses are the numbers with many decimal places in the table you filled in Part I.
```   (Mass of Hydrogen-1) × 50 + (Mass of Hydrogen-2) × 50
-------------------------------------------------------- =
100
```

This should give you the same exact answer that you got for the previous calculation and which should be displayed on the simulator.
4. The average atomic mass depends on the masses of the atoms but also on how many of each kind there are. Change the mixture of hydrogen atoms so there are 10 hydrogen-1 atoms and 90 hydrogen-2 atoms. Use the following formula to calculate the average atomic mass:
```   (Mass of Hydrogen-1) × 10 + (Mass of Hydrogen-2) × 90
-------------------------------------------------------- =
100
```

This is a different average because the relative number of each isotope of hydrogen has been changed.
5. A simpler way to calculate average atomic mass is to take the abundances of each isotope as percentages (like 90 out of 100 is 90%) and rewriting them as decimals (like 90% is 0.90). Use the same mixture of hydrogen atoms (10 hydrogen-1 atoms and 90 hydrogen-2 atoms). Use the following formula to calculate the average atomic mass:
```   (Mass of Hydrogen-1) × (0.10) + (Mass of Hydrogen-2) × (0.90) =
```

Notice you get the same result.

page break

1. Look at the true “Percent Composition” for hydrogen using “Nature’s Mix”. Use the percentages to calculate how many atoms of each isotope there are in a sample of 10,000 hydrogen atoms. (For example, if 7.5% of 10,000 people have brown hair then that would be 0.075 × 10,000 = 750.)

Hydrogen-1: ______________ out of 10,000 atoms

Hydrogen-2: ______________ out of 10,000 atoms
2. The true average atomic mass of hydrogen is 1.00794. Use the percentages given for hydrogen-1 and hydrogen-2 on the “Percent Composition” display to calculate this average mass using the exact atomic masses of each isotope:
```   (Mass of Hydrogen-1) × (0.999885) + (Mass of Hydrogen-2) × (0.000115) =
```

This should give you precisely the true average atomic mass of hydrogen.
3. Why is the true average atomic mass of hydrogen closer to the mass of the hydrogen-1 isotope than to the mass of the hydrogen-2 isotope?
4. Switch the simulator to show the element carbon (C). Look at the true “Percent Composition” for carbon (C) using “Nature’s Mix”. Use the percentages to calculate how many atoms of each isotope there are in a sample of 10,000 carbon atoms.

Carbon-12: ______________ out of 10,000 atoms

Carbon-13: ______________ out of 10,000 atoms
5. The true average atomic mass of carbon is 12.01070 amu. Why is the average mass closer to the mass of carbon-12 (12.000000 amu) than it is to the mass of carbon-13 (13.00335 amu)?
6. Switch the simulator to show the element fluorine (F). Use the simulator to find out, based on “Nature’s Mix”, why the average atomic mass of fluorine is the same as the exact atomic mass of the isotope fluorine-19. Why is that?
7. Hide the “Average Atomic Mass” display. Switch the simulator by selecting boron (B) from the periodic table. Make your own mixture of boron isotopes by selecting “My Mix”. Select 30 boron-10 atoms and 70 boron-11 atoms. Using the method with the percentages, calculate the average atomic mass of your mixture. Show you work:
8. Why was your answer closer to the mass of boron-11 than to the mass of boron-10?
9. Show the “Average Atomic Mass” display again. Make your own mixture of boron isotopes by selecting “My Mix” and adjusting the numbers of boron-10 atoms and boron-11 atoms until the average mass is 10.16239 amu, which is closer to the mass of boron-10 than boron-11. Hint: Make sure that the total number of atoms is always 100. What are the percentages of boron-10 and boron-11?
10. Show your work for a calculation of the average atomic mass of boron, using data you can find in the simulator. Average atomic masses are calculated using this general formula:
```(Mass of    (Percent        (Mass of    (Percent
Isotope ×   Composition  +  Isotope ×   Composition  + … = Average Atomic Mass
One)        One)            Two)        Two)

If you have more than two isotopes, you just insert the mass and percent composition of each additional isotope.
```
11. Look at the true “Percent Composition” for magnesium (Mg). Use the percentages to calculate how many atoms of each isotope there are in a sample of 10,000 magnesium atoms.

Magnesium-24: ______________ out of 10,000 atoms

Magnesium-25: ______________ out of 10,000 atoms

Magnesium-26: ______________ out of 10,000 atoms
12. Show your work for a calculation of the average atomic mass of magnesium, using data you can find in the simulator.

Last updated: Nov 27, 2020       Home