Dalton s atomic theory
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Dalton’s Atomic Theory. Elements - made up of atoms Same elements, same atoms. Different elements, different atoms. Chemical reactions involve bonding of atoms. The Atom. Made up of: Protons – (+) charged Electrons – (-) charged neutrons. Periodic Table. Alkaline Metals – Grps. I & II

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Dalton’s Atomic Theory

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Dalton s atomic theory

Dalton’s Atomic Theory

  • Elements - made up of atoms

  • Same elements, same atoms.

  • Different elements, different atoms.

  • Chemical reactions involve bonding of atoms


The atom

The Atom

  • Made up of:

    • Protons – (+) charged

    • Electrons – (-) charged

    • neutrons


Periodic table

Periodic Table

  • Alkaline Metals – Grps. I & II

  • Transition Metals

  • Non-metals

  • Halogens – Group VII

  • Noble Gases –Group VIII - little chemical activity


Periodic table1

Periodic Table

  • Atomic Mass - # at bottom

    • how much element weighs

  • Atomic Number - # on top

    • gives # protons = # electrons


  • Periodic table2

    Periodic Table

    • Atomic Mass

      • number below the element

      • not whole numbers because the masses are averages of the masses of the different isotopes of the elements


    Dalton s atomic theory

    Ions

    • Are charged species

    • Result when elements gain electrons or lose electrons


    2 types of ions

    2 Types of Ions

    • Anions – (-) charged

      • Example:F-

  • Cations – (+) charged

    • Example:Na+


  • Highly important

    Highly Important!

    • Gain of electrons makes element (-)=anion

    • Loss of electrons makes element (+)=cation


    Isotopes

    Isotopes

    • Are atoms of a given element that differ in the number of neutrons and consequently in atomic mass.


    Example

    Example

    Isotopes% Abundance

    12C 98.89 %

    13C 1.11 %

    14C

    11C


    Dalton s atomic theory

    • For example, the mass of C = 12.01 a.m.u is the average of the masses of 12C, 13C and 14C.


    Determination of aver mass

    Determination of Aver. Mass

    • Ave. Mass = [(% Abund./100) (atomic mass)] + [(% Abund./100) (atomic mass)]


    Take note

    Take Note:

    • If there are more than 2 isotopes, then formula has to be re-adjusted


    Sample problem 1

    Sample Problem 1

    • Assume that element Uus is synthesized and that it has the following stable isotopes:

      • 284Uus (283.4 a.m.u.)34.6 %

      • 285Uus (284.7 a.m.u.)21.2 %

      • 288Uus (287.8 a.m.u.)44.20 %


    Solution

    Solution

    • Ave. Mass of Uus =

    • [284Uus](283.4 a.m.u.)(0.346)

    • [285Uus] +(284.7 a.m.u.)(0.212)

    • [288Uus] +(287.8 a.m.u.)(0.4420)

    • = 97.92 + 60.36 + 127.21

    • = 285.49 a.m.u (FINAL ANS.)


    Oxidation numbers

    Oxidation Numbers

    • Is the charge of the ions (elements in their ion form)

    • Is a form of electron accounting

    • Compounds have total charge of zero (positive charge equals negative charge)


    Oxidation states

    Oxidation States

    • Are the partial charges of the ions. Some ions have more than one oxidation states.


    Oxidation states1

    Oxidation States

    • - generally depend upon the how the element follows the octet rule

    • Octet Rule – rule allowing elements to follow the noble gas configuration


    Nomenclature

    Nomenclature

    • - naming of compounds


    Periodic table3

    Periodic Table

    • Rows (Left to Right) - periods

    • Columns (top to bottom) - groups


    Determination of aver mass1

    Determination of Aver. Mass

    • Ave. Mass = [(% Abund./100) (atomic mass)] + [(% Abund./100) (atomic mass)]


    Sample problem 11

    Sample Problem 1

    • Assume that element Uus is synthesized and that it has the following stable isotopes:

      • 284Uus (283.4 a.m.u.)34.6 %

      • 285Uus (284.7 a.m.u.)21.2 %

      • 288Uus (287.8 a.m.u.)44.20 %


    Solution1

    Solution

    • Ave. Mass of Uus =

    • [284Uus](283.4 a.m.u.)(0.346)

    • [285Uus] +(284.7 a.m.u.)(0.212)

    • [288Uus] +(287.8 a.m.u.)(0.4420)

    • = 97.92 + 60.36 + 127.21

    • = 285.49 a.m.u (FINAL ANS.)


    Stoichiometry

    STOICHIOMETRY

    • For example, the mass of C = 12.01 a.m.u is the average of the masses of 12C, 13C and 14C.


    Formula weight molecular weight

    Formula Weight & Molecular Weight

    • The FORMULA WEIGHT of a compound equals the SUM of the atomic masses of the atoms in a formula.

    • If the formula is a molecular formula, the formula weight is also called the MOLECULAR WEIGHT.


    The mole

    The MOLE

    • Amount of substance that contains an Avogadro’s number (6.02 x 10 23)of formula units.


    The mole1

    The MOLE

    • The mass of 1 mole of atoms, molecules or ions = the formula weight of that element or compound in grams. Ex. Mass of 1 mole of water is 18 grams so molar mass of water is 18 grams/mole.


    Formula for mole

    Formula for Mole

    Mole = mass of element formula weight of element


    Sample mole calculations

    Sample Mole Calculations

    1 mole of C = 12.011 grams

    • 12.011 gm/mol

  • 0.5 mole of C = 6.055 grams

    • 12.011 gm/mol


  • Avogadro s number

    Avogadro’s Number

    • Way of counting atoms

    • Avogadro’s number = 6.02 x 1023


    Point to remember

    Point to Remember

    One mole of anything is 6.02 x 1023 units of that substance.


    Dalton s atomic theory

    And……..

    • 1 mole of C has the same number of atoms as one mole of any element


    Formula weight molecular weight1

    Formula Weight & Molecular Weight

    • The FORMULA WEIGHT of a compound equals the SUM of the atomic masses of the atoms in a formula.

    • If the formula is a molecular formula, the formula weight is also called the MOLECULAR WEIGHT.


    Summary

    Summary

    • Avogadro’s Number gives the number of particles or atoms in a given number of moles

    • 1 mole of anything = 6.02 x 10 23 atoms or particles


    Sample problem 2

    Sample Problem 2

    • Compute the number of atoms and moles of atoms in a 10.0 gram sample of aluminum.


    Solution2

    Solution

    • PART I:

    • Formula for Mole:

      • Mole = mass of element

        atomic mass of element


    Solution cont

    Solution (cont.)

    • Part II:To determine # of atoms

    • # atoms = moles x Avogadro’s number


    Problem 2

    Problem # 2

    • A diamond contains 5.0 x 1021 atoms of carbon. How many moles of carbon and how many grams of carbon are in this diamond?


    Molar mass

    Molar Mass

    • Often referred to as molecular mass

      • Unit = gm/mole

    • Definition:

      • mass in grams of 1 mole of the compound


    Example problem

    Example Problem

    • Determine the Molar Mass of C6H12O6


    Solution3

    Solution

    • Mass of 6 mole C = 6 x 12.01 = 72.06 g

    • Mass of 12 mole H = 12 x 1.008 = 12.096 g

    • Mass of 6 mole O = 6 x 16 = 96 g

    • Mass of 1 mole C6H12O6 = 180.156 g


    Problem 3

    Problem #3

    • What is the molar mass of (NH4)3(PO4)?


    Molar mass1

    Molar Mass

    • Often referred to as molecular mass

      • Unit = gm/mole

    • Definition:

      • mass in grams of 1 mole of the compound


    Sample problem

    Sample Problem

    • Given 75.99 grams of (NH4)3(PO4), determine the ff:

    • 1. Molar mass of the compound

    • 2. # of moles of the compound

    • 3. # of molecules of the compound

    • 4. # of moles of N

    • 5. # of moles of H

    • 6. # of moles of O

    • 7. # of atoms of N

    • 8. # of atoms of H

    • 9. # of atoms of O


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