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