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Isotope notes. Dalton was proved incorrect and his theory was modified. Protons define the element. Atoms that have the same number of protons are always atoms of a specific element . Example: Carbon. Neutrons can vary. BUT

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Isotope notes


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    1. Isotope notes Dalton was proved incorrect and his theory was modified

    2. Protons define the element • Atoms that have the same number of protonsare always atoms of a specific element. • Example: Carbon

    3. Neutrons can vary BUT • atoms can have different numbers of neutrons and still be an atom of a specific element.

    4. Isotopes • This is because elements can have isotopes (basically atoms of the same element with a different number of neutrons in their nuclei).

    5. Dalton was wrong • When Dalton stated his atomic theory in the early 1800’s, he assumed that all of the atoms of a given element were identical.

    6. James Chadwick • Over 100 years after Dalton, James Chadwick discovered that the nuclei of most atoms contains neutrons as well as protons.

    7. Dalton’s Theory Changes • Dalton’s theory now states: • All atoms of the same element contain the same number of protons and electrons, but atoms of a given element may have different numbers of neutrons.

    8. The Isotopes of Hydrogen • Hydrogen – 1 • Also written H-1 • Also known as protium • Hydrogen has an atomic number of 1, so it has 1 proton • The hyphen notation above tells us that the mass number of H-1 is 1 • Number of neutrons = mass number – atomic number • So H-1 must have 0 neutrons

    9. The Isotopes of Hydrogen • Hydrogen – 2 • Also written H-2 • Also known as Deuterium • Hydrogen has an atomic number of 1, so it has 1 proton • The hyphen notation above tells us that the mass number of H-2 is 2 • Number of neutrons = mass number – atomic number • So H-2 must have 1 neutron

    10. The Isotopes of Hydrogen • Hydrogen – 3 • Also written H-3 • Also known as Tritium • Hydrogen has an atomic number of 1, so it has 1 proton • The hyphen notation above tells us that the mass number of H-3 is 3 • Number of neutrons = mass number – atomic number • So H-3 must have 2 neutron

    11. Calculate the number of neutrons • For chlorine found on the periodic table (the most common form of chlorine that is found in nature) Chlorine-35 • For Chlorine-37 • Chlorine-35 = 18 neutrons, Chlorine-37 = 20 neutrons • For Cobalt found on the periodic table Cobalt-59 • For Cobalt-60 • Cobalt-59 = 32 neutrons, Cobalt-60 = 33 neutrons

    12. Calculating Average Atomic Mass Average atomic mass is the atomic mass that appears on the periodic table. For example – Copper has an average atomic mass of 63.55 amu.

    13. Calculating Average Atomic Mass Yet, in nature, most elements are found as mixtures of two or more isotopes. For example, copper consists of • 69.17% copper-63 which has a relative atomic mass of 62.94 amu AND • 30.83% copper-65which has a relative atomic mass of 64.93 amu

    14. Calculating Average Atomic Mass To find the averageatomic mass, multiply the decimal equivalent of the percent (for example 69.17% = 0.6917) of each isotope by the respective relative atomic mass and add the results. (0.6917 X 62.94 amu) + (0.3083 X 64.93 amu) = 63.55 amu

    15. Practice Calculating Average Atomic Mass • Boron – 10 is found 19.9% of the time in nature and has a relative atomic mass of 10.013 amu • Boron – 11 is found 80.1% of the time in nature and has a relative atomic mass of 11.009 amu • Calculate the average atomic mass of Boron

    16. Practice Calculating Average Atomic Mass Boron (0.199 X 10.013) + (0.801 X 11.009) = 10.81 amu

    17. Practice Calculating Average Atomic Mass • Magnesium – 24 is found 78.99% of the time in nature and has a relative atomic mass of 23.985042 amu • Magnesium – 25 is found 10.00% of the time in nature and has a relative atomic mass of 24.985837 amu • Magnesium – 26 is found 11.01% of the time in nature and has a relative atomic mass of 25.982593 amu • Calculate the average atomic mass of Magnesium

    18. Practice Calculating Average Atomic Mass Magnesium (0.7899 X 23.985042) + (0.1000 X 24.985837) + (0.1101 X 25.982593) = 24.306 amu