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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
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
slide6
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

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.

slide32
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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