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Chemistry chapter 3

Chemistry chapter 3. Sample Problems. Isotopes. The number of neutrons is found by subtracting the atomic number from the mass number. Mass # (235)---atomic #(92) =# of nuetrons (143). Sample Problem 3-1. How many p rotons, electrons, and neutrons are there in an atom of Chlorine-37?

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Chemistry chapter 3

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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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