Isotope

The average atomic mass of each element uses the masses of the various isotopes of an element

An isotope of an element is the same element with the same # of p+, but a varying number of neutrons and thus a varying atomic mass

NOT all elements contain isotopes

Isotope Determinationex. H

There are 3 isotopes of Hydrogen

atomic mass atomic # #p+ #n

1.001H 1.00 1 1 0

2.001H 2.00 1 1 1

3.001H 3.00 1 1 2

The average atomic mass of H = 1.0079

Isotope Determination

There are three isotopes of Carbon

12C Carbon 12

13C Carbon 13

14C Carbon 14

Groups - # valence e- Periods - Size of atom

Noble gasesGroup VIII

HalogensGroup VII

AlkalliMetalsGroup I

MetalsNonmetals

Metalloids

Alkali EarthMetals

Group II

Formula WeightsH2

N2 O2 F2

Cl2

Br2

I2

Chemical EquationsSodium + Diatomic Chlorine ------> Sodium Chloride

Reactant(s) -----yields-----> Product(s)

Na + Cl2 ----------> NaCl

Equations must be balanced due to the

Law of Conservation of Matter

Chemical EquationsBookkeep the Atoms

Count the individual atoms of reactants and products line up under each atom in separate rows:

Na + Cl2 NaCl

1 Na 1 Na

2 Cl 1 Cl

Use whole number multipliers to equalize the atoms of reactants and products.

Place these whole number prefixes before the element or the compound

Chemical EquationsBookkeep the Atoms

Na + Cl2 NaCl

1 Na 1 Na

2 Cl 1 Cl

Chemical Equations

2Na + Cl2 2NaCl

2 Sodium atoms react with 1 chlorine molecule to form 2 “Molecules” of Sodium Chloride

Chemical Equations

1. (NH4)SO4 H2SO4 + NH3

5. Fe + H2O Fe3O4 + H2

TRY in groups

#6 and #9

% of an Element in a Compound

(Elemental Analysis) Mass of Element x 100

FW of Compound

% H in H2O 2H x 100 = 2.00 amu x 100 = 11.1%

H2O 18.00 amu

% O in H2O O x 100 = 16.00 amu x 100 =

88.89% H2O 18.00 amu

Group Work

Determine the Elemental Analysis of:

» C6H12O6

» NH3

» CO2

Definition

Mole (mol or n)» 1 mole contain the same number of entities

(atoms, molecules, ions, particles) as there are in exactly 12.01 grams of Carbon and a mole contains 6.022 x 1023

number of entities. This number is called Avogadro’s Number.

1.00 gH

12.01 gC

18.00gH2O

35.45 gCl-

180.06gC6H12O6

6.022 x1023

atoms

6.022 x1023

atoms6.022 x1023

molecules

6.022 x1023

ions6.022 x1023

molecules

Analogy

1 pair = 2 1 trio = 3 1 dozen = 12 1 mol = 6.022 x 1023

Molar Mass (MM)

To Determine the Molar Mass (MM):

For an element - it’s the atomic weight

in grams

For a compound - it’s the formula weight

in grams

A mole contains Avogadro’s Number

(6.022 x 1023) particles

(atoms, ions, molecules).

Working with Moles

Formula given grams of a compound or element, determine the number of moles

Moles = grams MM

OR Dimensional Analysis

Working with Moles

Formula given moles of a compound or element, determine the number of grams

moles (MM) = grams

OR Dimensional Analysis

Multiply by Avogadro’s #, Divide by Molar Mass

Multiply by Molar Mass, Divide by Avogadro’s #

Number of Moles

Number of atoms, ions,

molecules

Mass in

grams