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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.

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 (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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