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SCH3U - Unit 1 - Practice Questions

  • Find the number of neutrons, given the atomic mass and atomic number

    • # of neutrons = atomic mass - atomic number

    • Example - Neon-20 has an atomic number of 10

      • # of neutrons = 20 - 10

        = 10 neutrons

  • Find the average atomic mass, given the abundances of isotopes

    • Multiply each mass by the abundance as a decimal

    • Add each product together

    • Use the formula:

      • Avg. amu = m1p1 + m2p2 +… mn-1pn-1 + mnpn , where n is the number of isotopes, m is the mass, and p is the percent of abundance

    • Example - Copper’s abundance is made of Copper-63 (62.9296 amu, 69.17%) and Copper-65 (64.9278 amu, 30.83%). What is the average atomic mass?

      • Avg. amu = (62.9296 amu)(0.6917) + (64.9278 amu)(0.3083)

      • = (43.5284 amu) + (20.0172 amu)

      • = 63.5456

  • Find the abundance of each isotope, given the masses and average atomic mass

    • Use the formulae:

      • Avg. amu = m1p1 + m2p2 and p1 + p2 = 1

      • All of the abundances should equal 1 (or 100%

      • Substitute the known masses and average amu

      • Use substitution to isolate for one abundance, and then solve for the other

LC

SCH3U - Unit 1 - Practice Questions

  • Find the number of neutrons, given the atomic mass and atomic number

    • # of neutrons = atomic mass - atomic number

    • Example - Neon-20 has an atomic number of 10

      • # of neutrons = 20 - 10

        = 10 neutrons

  • Find the average atomic mass, given the abundances of isotopes

    • Multiply each mass by the abundance as a decimal

    • Add each product together

    • Use the formula:

      • Avg. amu = m1p1 + m2p2 +… mn-1pn-1 + mnpn , where n is the number of isotopes, m is the mass, and p is the percent of abundance

    • Example - Copper’s abundance is made of Copper-63 (62.9296 amu, 69.17%) and Copper-65 (64.9278 amu, 30.83%). What is the average atomic mass?

      • Avg. amu = (62.9296 amu)(0.6917) + (64.9278 amu)(0.3083)

      • = (43.5284 amu) + (20.0172 amu)

      • = 63.5456

  • Find the abundance of each isotope, given the masses and average atomic mass

    • Use the formulae:

      • Avg. amu = m1p1 + m2p2 and p1 + p2 = 1

      • All of the abundances should equal 1 (or 100%

      • Substitute the known masses and average amu

      • Use substitution to isolate for one abundance, and then solve for the other

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