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Understanding Isotopes, Atomic Mass, and Moles in Chemistry

Dive into the core concepts of isotopes, atomic mass, and the mole in this comprehensive overview. Learn how to determine the number of neutrons in an element by subtracting atomic number from mass number with examples including Chlorine-37. Gain insights into calculating average atomic mass through natural abundance, and master the mole concept with Avogadro's number, including practical problems involving copper and aluminum. This resource offers a blend of theoretical and practical approaches to enhance your understanding of these essential chemistry topics.

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Understanding Isotopes, Atomic Mass, and Moles in Chemistry

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  1. Chemistry chapter 3 Sample Problems

  2. Isotopes • The number of neutrons is found by subtracting the atomic number from the mass number. • Mass # (235)---atomic #(92) =# of nuetrons (143)

  3. Sample Problem 3-1 • How many protons, electrons, and neutrons are there in an atom of Chlorine-37? • Given: name and mass # of chlorine-37 • Unknown: #’s of protons , electron and neutrons (cont)

  4. Sample Problem 3-1 • Plan: • Atomic # =# of protons = # of neutrons • Mass # =# of neutrons + # of Protons • Compute: • Mass # of element - atomic number of element = # of neutrons in element • (cont)

  5. Sample Problem 3-1 • Mass # -atomic # = 37-17 = 20 neutrons • An atom of chlorine-37 contains 17 electons 17 protons and 20 neutrons

  6. Calculating Average Atomic mass • Background- the average atomic mass of an element depends on both the mass amd the relative abundance of each of the element’s isotopes. • Copper • .6917 (% natural abundance) X 62.929 599 amu + 0.3083 X 64.927 793 amu = 63.55 amu

  7. The Mole • A mole is the amount of substance that contains as many particles as there are atoms in exactly 12 g of Carbon 12

  8. Avogadro’s # • Is the number of particles in exactly one mole of a pure substance. = 6.022 X 1023 • / means divide

  9. Relationships • Amount of element in moles X 6.022 X 1023 = Number of atoms of element • Number of atoms of element X 1/ 6.022 X 10 23 X molar mass of element = Mass of element in grams • Mass of element in grams X 1/molar mass of element = Amount 0f element in moles

  10. Relationships • Mass of element in grams X 1/ molar mass of element X 6.022 X 1023 = Number of atoms of element. • Amount of element in moles X molar mass of element = Mass of element in grams • Number of atoms of element = 6.022 X 1023 X Amount of element in moles

  11. Sample Problem 3-2 • What is the mass in grams of 3.50 mol of the element copper, Cu? • Given: 3.50 mol Cu Unknown : mass of Cu in grams • Amount of Cu in moles ------- mass of Cu in grams • Moles of Cu X grams of Cu/moles of Cu = grams of Cu

  12. Sample Problem 3-2 • Compute: • 3.50 moles Cu X 63.5 g Cu/mol Cu = 222g Cu

  13. Sample Problem 3-3 • A chemist produced 11.0 g of Al. How many moles of Al were produced? • Given: 11.9 g Al Unknown: amount of Al in moles • Grams Al X moles Al/grams of Al = moles of Al • 11.9 grams Al X mol Al/26.98 g Al =0.441 mol Al

  14. Sample Problem 3-4 • How many moles of silver, Ag are in 3.01 X 1023 atoms of Ag? • Given 3.01 X 1023 atoms of Ag Unknown: Amount of Ag in moles • Ag atoms X moles Ag/Avogadro’s # of Ag atoms = moles of Ag • 3.01 X 1023 Ag atoms X mol Ag/6.022 X 1023 Ag atoms = 0.50 mol Ag

  15. Sample Problem 3-5 • What is the mass in grams of 1.20 X 108 atoms of Cu? • Given: 1.20 X 108 atoms of Cu • Unknown: mass of Cu in grams • Plan: number of atoms of Cu  amount of Cu in Moles  mass of Cu in Grams

  16. Sample 3-5 • Plan: • Cu atoms X moles Cu/Avogadro’s # X grams Cu/moles Cu = grams Cu • Compute: • 1.20 X 108 Cu atoms X 1 mol Cu/ 6.022 X 1023 Cu atoms X 63.55 g Cu/ mol Cu = 1.27 X 10-14 g Cu

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