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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• Transition Metals• Non-metals• Halogens – Group VII• Noble Gases –Group VIII - little
chemical activity
Periodic Table
• Atomic Mass - # at bottom•how much element weighs
• Atomic Number - # on top•gives # protons = # electrons
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
Ions
• Are charged species
• Result when elements gain electrons or lose electrons
2 Types of Ions
• Anions – (-) charged•Example: F-
• Cations – (+) charged•Example: Na+
Highly Important!
• Gain of electrons makes element (-) = anion
• Loss of electrons makes element (+) = cation
Isotopes
• Are atoms of a given element that differ in the number of neutrons and consequently in atomic mass.
Example
Isotopes % Abundance12C 98.89 %13C 1.11 %14C 11C
–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
• Ave. Mass = [(% Abund./100) (atomic mass)] + [(% Abund./100) (atomic mass)]
Take Note:• If there are more than 2
isotopes, then formula has to be re-adjusted
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
• 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
• 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
• Are the partial charges of the ions. Some ions have more than one oxidation states.
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
• - naming of compounds
Periodic Table
• Rows (Left to Right) - periods
• Columns (top to bottom) - groups
Determination of Aver. Mass
• Ave. Mass = [(% Abund./100) (atomic mass)] + [(% Abund./100) (atomic mass)]
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
• 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
–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
• 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
• Amount of substance that contains an Avogadro’s number (6.02 x 10 23)of formula units.
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
Mole = mass of element formula weight of
element
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
•Way of counting atoms
• Avogadro’s number = 6.02 x 1023
Point to Remember
One mole of anything is 6.02 x 1023 units of that substance.
And……..
• 1 mole of C has the same number of atoms as one mole of any element
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
• 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
• Compute the number of atoms and moles of atoms in a 10.0 gram sample of aluminum.
Solution
• PART I:• Formula for Mole:
–Mole = mass of elementatomic mass of element
Solution (cont.)
• Part II: To determine # of atoms
• # atoms = moles x Avogadro’s number
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• Often referred to as molecular mass–Unit = gm/mole
• Definition: –mass in grams of 1 mole of the
compound
Example Problem
• Determine the Molar Mass of C6H12O6
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
•What is the molar mass of (NH4)3(PO4)?
Molar Mass• Often referred to as molecular mass–Unit = gm/mole
• Definition: –mass in grams of 1 mole of the
compound
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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