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Lesson 2.03

Lesson 2.03. Isotopes and Weighted Averages. Introduction. How can more than one form of atom make up an element? What can be different and what is always the same among the different atoms of a given element?. Objectives. After completing this lesson, you will be able to :

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Lesson 2.03

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  1. Lesson 2.03 Isotopes and Weighted Averages

  2. Introduction • How can more than one form of atom make up an element? • What can be different and what is always the same among the different atoms of a given element?

  3. Objectives After completing this lesson, you will be able to: • Identify the structural differences between the isotopes of elements. • Determine the weighted average mass of an element when given the appropriate data.

  4. What do you know? • The number of protons in an atom’s nucleus identifies the atom as a specific element. • Proton’s never change for a particular element EXAMPLE: All oxygen atoms have eight protons. We know this because Oxygen’s atomic number = 8

  5. What is an Isotope? • Isotopes are atoms of the same element that have different masses due to their varying numbers of neutrons

  6. Ways to represent an Isotope Method One: Add the mass number of the isotope, with a hyphen, to the end of the element’s name. EXAMPLES: an isotope of a carbon (C) atom that has six protons, six neutrons, and six electrons you can represent the isotope as Carbon-12. The 12 is the mass number of the isotope (six protons + six neutrons). Uranium-235would be an isotope of uranium with a mass of 235 (92 protons + 143 neutrons)

  7. Ways to represent an Isotope Method Two: The mass number is given as a superscript to the upper left side of the element’s symbol, with the atomic number as a subscript to the lower left side of the symbol. EXAMPLE: If you have an isotope of a carbon (C) atom that has six protons, eight neutrons, and six electrons you can represent the isotope by writing C with 14 a superscript on the upper left side and six as a subscript in the lower left side of the symbol. The 14 is the mass number and the six is the atomic number. Mass Number 14 C 6 Atomic Number

  8. Method Two Examples EXAMPLE: An isotope of hydrogen (H) that has a mass of two would be written as H with two as a superscript on the upper left side and one as a subscript in the lower left side of the symbol. The two is the mass number and the one is the atomic number. 2 H 1

  9. Practice • Refer to the periodic table and each element’s atomic number to answer the following questions: 1- How many protons, neutrons, and electrons are in an atom of chlorine-37? Chlorine has an atomic number of 17. 2-How many protons, neutrons, and electrons are in an atom of Br with an atomic number of 35 and a mass number of 80?

  10. More Practice 3-How many protons, neutrons, and electrons in an atom of beryllium-9? Beryllium has an atomic number of 4. 4- How many protons, neutrons, and electrons are in an atom of Ar with an atomic number of 18 and a mass number of 38?

  11. Check your Answers 1-How many protons, neutrons, and electrons are in an atom of chlorine-37? Chlorine has an atomic number of 17. Protons = 17 (atomic number) Electron = 17 (atomic number) Neutrons = 20 (neutrons = mass number (37)—atomic number (17) 2-How many protons, neutrons, and electrons are in an atom of Br with an atomic number of 35 and a mass number of 80? Protons = 35 (atomic number) Electron = 35 (atomic number) Neutrons = 45 (neutrons = mass number (80)—atomic number (35)

  12. Check Your Answers 3- How many protons, neutrons, and electrons in an atom of beryllium-9? Beryllium has an atomic number of 4. Protons = 4 (atomic number) Electron = 4 (atomic number) Neutrons = 5 (neutrons = mass number (9)—atomic number (4) 4- How many protons, neutrons, and electrons are in an atom of Ar with an atomic number of 18 and a mass number of 38? Protons = 18 (atomic number) Electron = 18 (atomic number) Neutrons = 20 (neutrons = mass number (38)—atomic number (18)

  13. Atomic Mass • Have you noticed that the atomic masses on the periodic table have decimal values even though the mass numbers are whole numbers?

  14. Average Atomic Mass • The average mass of an element calculated by averaging the mass values for all of that element’s isotopes. • Because pure elements are made up of mixtures of isotopes, it is important to take into account the percentage and mass of each isotope, especially when some isotopes make up a greater percentage of the mixture than other isotopes. • the average atomic mass of each element is calculated using a weighted average.

  15. What is a weighted average? • A weighted average accounts for the percent abundance and mass of each isotope in an element. • Percent abundance is a measure of how common or rare that isotope is. • The average atomic mass of an element can be calculated by multiplying the mass of each isotope by its relative abundance (the percentage represented in decimal form) and adding the results • Isotope A [percent abundance × mass] + Isotope B [percent abundance × mass] + continue for each isotope = average atomic mass

  16. Example Step 1: Convert the percentages to decimals by dividing by 100

  17. Example continued Step 2- multiply each isotopes mass by the decimal of the abundance Step 3- add the individual totals together (15.96192 + 0.00646 + 0.036) = 16.0u

  18. Practice 1 Copper has two naturally occurring isotopes. Copper-63, with an atomic mass of 62.94 u, makes up 69.17 percent of the sample, and Copper-65, with an atomic mass of 64.93 u, makes up the other 30.83 percent. Calculate the weighted average atomic mass of copper based on the given information. Please give your answer in a total of four significant figures.

  19. Practice 2 Three isotopes of argon (Ar) occur in nature. Argon-36 (35.968 u) makes up 0.337 percent of the sample, Argon-38 (37.963 u) makes up 0.063 percent of the sample, and Argon-40 (39.962 u) makes up the other 99.6 percent of the sample. Calculate the weighted average atomic mass of argon, giving your answer in a total of five significant figures.

  20. Answers to Practice Question 1- [.6917 × 62.94 u] + [.3083 × 64.93 u] = 63.55 u (average atomic mass) Question 2- [0.00337 × 35.968 u] + [0.00063 × 37.963 u] + [0.99600 × 39.962 u] = 39.947 u (average atomic mass)

  21. Review

  22. Congrats! You should now be able to: • Identify the structural differences between the isotopes of elements. • Determine the weighted average mass of an element when given the appropriate data. Please review this presentation again if you need additional practice

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