Isotopes and Atomic Mass PDF

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Isotopes and Atomic Mass gizmo lab...

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SCH3U1 - Privitera

Name: ____________________________

Exploring Isotopes and Atomic Mass - PhET Simulation [

/60]

PART A - Making Isotopes: Open the Isotopes screen, then explore the simulator for a few minutes to develop your own ideas about “isotopes”.

1. What particles determine the mass number? [1] The neutrons and protons are mainly the particles that determine the mass number and are found in the nucleus of the atom. 2. Why is mass number always a whole number? [1] Mass number is always a whole number because it's the number of particles in the nucleus, there can't be fractions. 3. What is the approximate mass of one proton? [1] The approximate mass of one proton is 1.00783 amu 4. What is the approximate mass of one neutron? [1] The approximate mass of one neutron is 1.00627 amu 5. One isotope of carbon (C) has exactly the same mass number and atomic mass since it was used as the definition of the atomic mass unit (amu). Which isotope is it and what is its atomic mass? [2] The isotope is Carbon-12 and its atomic mass is 12 6. Based on what you learned from the simulator, what is the definition of an isotope? In your definition, include the terms: mass number, atomic number, number of protons (p+), neutrons (n0) and electrons (e-). [3]

An isotope is an element that has the same number of protons but the number of neutrons are different, this results in elements having different atomic masses(mass number) and different names like Carbon-12 and Carbon-13. The atomic number will end up being the same amount. PART B - Mixtures and Abundances of Isotopes: Explore the Mixtures screen to develop your ideas about how isotope mixtures are related to average atomic mass. Make sure to explore both My Mix and Nature’s Mix.

1. What is the difference between “My Mix” and “Nature’s Mix”? [1] In my mix I can add as much Hydrogen-1 and -2 as i want, whereas Nature’s Mix has a fixed amount of Hydrogen-1 and Hydrogen-2. Nature’s Mix mainly consists of Hydrogen-1. 2. Do you agree or disagree with the statement “In nature, the chance of finding one isotope of an element is the same for all elements”? Explain your reasoning by referencing what you have learned from the simulator. [2] I disagree with the statement because certain elements are found in different environments in nature and those environments can have the right conditions in order for the isotopes to be in the area. Some elements are rarer than others so it would be harder to find them and their isotopes. Finding oxygen isotopes would be easier to find because there is more oxygen in the environment compared to one of the more harder to find elements. In the sim the elements had different percentages and ratios for each of the isotopes for the selected element. 3. Beryllium (Be) and Fluorine (F) have only one stable isotope. Use the sim and the periodic table to complete the following table: [1] Element

Mass of 1 atom

Average mass of 2 atoms (sim)

Average mass of 3 atoms (sim)

Atomic mass (periodic table)

Be

9.01218 amu

9.01218 amu

9.01218 amu

9.01218 amu

F

18.99840 amu

18.99840 amu

18.99840 amu

18.99840 amu

4. Why are all the values in each row of the table above the same? [1] All the values in each row of the table above are the same because beryllium and fluorine do not have isotopes, there is Fluorine-19 and Beryllium-9. 5. Lithium has only two stable isotopes. Use the sim to determine the following [5]: a. Atomic mass of lithium-6 (in amu) = 6.01512 amu b. Atomic mass of lithium-7 (in amu) = 7.01600 amu c. Average atomic mass of a sample containing three lithium-6 atoms and two lithium-7 atoms (in amu) = 6.41548 amu d. Is the average atomic mass you just determined closer to the mass of lithium-6 or lithium-7? Explain The average atomic mass is close to the mass of lithium-6 because the difference (0.40036 amu) between the average is closer to lithium-6. 6. Describe a method to calculate the average atomic mass of the sample in the previous question using only the atomic masses of lithium-6 and lithium-7, and without using the simulation [3]. First you would multiply the amu of lithium-6 times 3 and add the product of the amu of lithium-7 times 2 to get the average atomic mass of the sample or you could replace the 3 times and 2 times with a percentage 0.6 (60%) and 0.4 (40%). The average atomic mass is 6.41548 amu.

7. You and your friend, Peter, are given a rock that you know has some Silicon. You just learned that there are 3 common isotopes of silicon - Silicon-28, Silicon-29, and Silicon-30. Peter suggests that the rock might have equal parts of each isotope. a. What would be the average mass of Silicon in the rock? Capture your screen from the sim to show a solution that would support Peter’s idea. [2] The average mass of silicon in the rock would be 28.97573 amu

b. How could you check to see if Peter’s ideas are correct? [1] I would check Nature's Mix average atomic mass or compare the amu to the amu of silicon in the periodic table, if the number is off by a lot then Peter’s idea is most likely wrong. 8. Iron has many isotopes but only 4 are found in significant amounts in naturally found mixtures. The amounts by mass percent are: 5.845% of 54Fe (53.9396 amu) 91.754% of 56 Fe (55.9349 amu), 2.119% of 57Fe (56.9354 amu) and 0.282% of 58Fe (57.9333 amu). a. What would you determine the average mass of iron to be? Show your work. [2]

b. How do your results compare to the information on the periodic table? [1] The average mass of iron is heavier in the periodic table compared to the data found in question 8. 9. Answer the following questions in the table. Choose the letter answer that you think is correct and explain why you made that choice [10]: Question:

Your Answer and Explanation:

B) It is carbon-14 because there are 2 extra neutrons which causes the atomic mass to be 14 instead of 12. This gives the element the name carbon-14 instead of carbon-12.

A) 1&2 are an isotope of oxygen and 1 would be oxygen-8 and the other 2 would be oxygen-7. The second one has one less neutron than oxygen-7.

C) I added up all of the atomic masses of the hydrogen atoms and divided it by the number of hydrogen atoms. The approximate average mass of Hydrogen was 1.5 amu. 9/6 = 1.5

C) I added up all the masses (150) and divided by the amount of atoms (4) in order to get the average mass of Argon (37.5 amu).

I would calculate the average mass of nitrogen in the sample and compare it to nature's mix in the simulation. 200(total mass)/14(amount of atoms) = 14.2857 amu.

PART C - Stable and Unstable Isotopes: Instructions: Open the Isotopes screen and click the Green Box (+) symbols next to the words “Symbol” and “Abundance in Nature” to open the tabs. Click on the Atomic Mass (amu) feature on the scale. Data Collection: As you go through the steps, be sure to record all data in the data table below. 1. Use the mouse to select the element helium, He, on the Periodic Table located on the right-hand corner of the screen 2. Create a new isotope of helium by using the mouse to move a neutron out of the nucleus. If the new isotope has an atomic mass, record its data in the Data Table. 3. Select He on the periodic table to reset the isotope. Create a new isotope of helium by using the mouse to move a neutron into the nucleus. Because there is no atomic mass, you may not record this isotope. Make and record as many isotopes as possible for helium by moving neutrons. 4. Once all the isotopes of helium have been found and recorded, repeat steps 2, 3 and 4 for the elements hydrogen, carbon, and oxygen and record the data for all possible isotopes. 5. One number in each ratio must be a “1”. If needed, calculate a whole number ratio of neutrons to protons in the nucleus of all isotopes by dividing the number of neutrons in the atom by the number of protons in the isotope. Table [8]: Isotope Symbol

# of p+

# of n0

# of e-

Atomic mass (amu)

Abundance in nature

Stable or unstable?

Ratio n0:p+

He-3 (Helium-3)

2

1

2

3.01603

0.0001%

Stable

1:2

He-4 (Helium-4)

2

2

2

4.00260

99.9999%

Stable

1:1

H-2(Hydrog en-2)

1

1

1

2.01410

0.0115%

Stable

1:1

H-2 (Hydrogen1)

1

0

1

1.00783

99.9885%

Stable

0:1

H-3 (Hydrogen3)

1

2

1

3.01605

trace

Unstable

2:1

C-13 (Carbon-13)

6

7

6

13.00335

1.07%

Stable

1.16:1

C-14 (Carbon-14)

6

8

6

14.00324

trace

Unstable

1.33:1

O-17 (Oxygen-17 )

8

9

8

16.99913

0.038%

Stable

1.125:1

O-18 (Oxygen-18 )

8

10

8

17.99916

0.205%

Stable

1.25:1

Questions: 1. What do the words stable and unstable mean? [1] Stable means that the isotope will not decay and an unstable isotope will decay. Stable means it is secure and is fixed and will not change. 2. Under what conditions did the nucleus become unstable in the isotope simulation? [1] The nucleus becomes unstable when a neutron is taken away or if three or more neutrons were added. In some cases this resulted in having now atomic mass. 3. What is the relationship between the stability of the nucleus and the ratio of neutrons to protons in the nucleus? Reference your data in your answer. [2] For the unstable isotopes there are more neutrons than protons. Hydrogen-3 had a ratio of 2:1 and was unstable. 1:1 between 1.25:1 seem to be the most stable ratios for the isotopes. 4. Unstable isotopes are also known as radioisotopes, and will decay over time. Based on your data, which isotopes are radioisotopes? How do you know? [2] Hydrogen-3 are radioisotopes because the ratio is 2:1 and the simulation states that the isotope is unstable. Carbon-14 is also a radioisotope and both of the isotopes have a trace chance of appearing in nature.

PART D - Final Practice: 1. Titanium has five common isotopes: 46 Ti (8.00%), mass = 45.953 amu 47 Ti (7.80%), mass = 46.952 amu 48 Ti (73.40%), mass = 47.947 amu 49 Ti (5.50%), mass = 48.948 amu 50 Ti (5.30%), mass = 49.945 amu Calculate the average atomic mass of titanium. [2]

2. The atomic mass of boron is 10.81 amu. Boron has two isotopes: Boron-10 has a mass of 10.01 amu. Boron-11 has a mass of 11.01 amu. What is the percentage of each isotope in boron? (you can check your answer with the simulation) [3]

3. A certain sample of rubidium has just two isotopes, 85Rb (mass = 84.911amu) and 87Rb (mass = 86.909amu). The atomic mass of this sample is 86.231 amu. What are the percentages of the isotopes in this sample? [3]

For the above questions, you may want to do them on a separate piece of paper and add that photo to your doc....