PowerPoint Lecture Presentation by
J. David Robertson
University of Missouri
Copyright
© The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Micro World atoms & molecules
Macro World grams
By definition:
1 atom 12 C “weighs” 12 amu
On this scale
1 H = 1.008 amu
16 O = 16.00 amu
3.1

Natural lithium is:
7.42% 6 Li (6.015 amu)
92.58% 7 Li (7.016 amu)
7.42 x 6.015 + 92.58 x 7.016
100
= 6.941 amu
3.1

Average atomic mass (6.941)

The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of 12 C
1 mol = N
A
= 6.0221367 x 10 23
Avogadro’s number ( N
A
)
3.2

eggs
Molar mass is the mass of 1 mole of in grams atoms
1 mole 12 C atoms = 6.022 x 10 23 atoms = 12.00 g
1 12 C atom = 12.00 amu
1 mole 12 C atoms = 12.00 g 12 C
1 mole lithium atoms = 6.941 g of Li
For any element atomic mass (amu) = molar mass (grams)
3.2

“ What is a mole ?”
1 dozen = 12, An Egg = 70 g = 0.070 kg
1 mole = 6.022x 10 23 , A Textbook = 4.9 lbs = 2.2kg
Earth = 6.1 x 10 24 kg
• 1 dozen Egg = 12 Egg (70g/Egg) = 840 g ~ 1 kg
• 1 mole Egg = 6.022x10
23 Egg(70g/Egg) = 4.2x10
22 kg
• Mass Earth / Mass of One Egg ~ 8.7x10
25 ~ 150 mols
• Mass Earth / Mass of Textbook ~ 3 x10 24 ~ 5 mols
• Namely, 150 moles of eggs or 5 moles of the textbook is equivalent to the mass of Earth.

C
One Mole of:
S
Hg
Cu
Fe
3.2

1 12 C atom
12.00 amu x
12.00 g
6.022 x 10 23 12 C atoms
=
1.66 x 10 -24 g
1 amu
1 amu = 1.66 x 10 -24 g or 1 g = 6.022 x 10 23 amu
M
= molar mass in g/mol
N
A
= Avogadro’s number
3.2

How many atoms are in 0.551 g of potassium (K) ?
1 mol K = 39.10 g K
1 mol K = 6.022 x 10 23 atoms K
0.551 g K x
1 mol K
39.10 g K x
6.022 x 10 23 atoms K
=
1 mol K
8.49 x 10 21 atoms K
3.2

Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule.
SO
2
1S 32.07 amu
2O + 2 x 16.00 amu
SO
2
64.07 amu
For any molecule molecular mass (amu) = molar mass (grams)
1 molecule SO
2
1 mole SO
2
= 64.07 amu
= 64.07 g SO
2
3.3

How many H atoms are in 72.5 g of C
3
H
8
O ?
1 mol C
3
H
8
O = (3 x 12) + (8 x 1) + 16 = 60 g C
3
H
8
O
1 mol C
3
H
8
O molecules = 8 mol H atoms
1 mol H = 6.022 x 10 23 atoms H
72.5 g C
3
H
8
O x
1 mol C
3
H
8
O
60 g C
3
H
8
O x
8 mol H atoms
1 mol C
3
H
8
O x
6.022 x 10 23 H atoms
1 mol H atoms
=
5.82 x 10 24 atoms H
3.3

(1) Ion Generator w/ Electron gun
(2) Ion Accelerator
(3) Electrostatic Analyzer:
Select ions w/ same KE: KE = (1/2) m v 2 v = (2 x KE/m) 1/2
(4) Magnetic Analyzer: F = q x v x B
3.4

Percent composition of an element in a compound = n x molar mass of element molar mass of compound x 100% n is the number of moles of the element in 1 mole of the compound
%C =
2 x (12.01 g)
46.07 g
%H =
6 x (1.008 g)
46.07 g
%O =
1 x (16.00 g)
46.07 g x 100% = 52.14% x 100% = 13.13% x 100% = 34.73%
C
2
H
6
O
52.14% + 13.13% + 34.73% = 100.0%
3.5

Combust 11.5 g ethanol
Collect 22.0 g CO
2 and 13.5 g H
2
O g CO
2 mol CO
2 mol C g C 6.0 g C = 0.5 mol C g H
2
O mol H
2
O mol H g of O = g of sample – (g of C + g of H) g H 1.5 g H = 1.5 mol H
4.0 g O = 0.25 mol O
Empirical formula C
0.5
H
1.5
O
0.25
Divide by smallest subscript (0.25)
Empirical formula C
2
H
6
O
3.6

A process in which one or more substances is changed into one or more new substances is a chemical reaction
A chemical equation uses chemical symbols to show what happens during a chemical reaction
3 ways of representing the reaction of H
2 with O
2 to form H
2
O reactants products
3.7

2 Mg + O
2
2 MgO
Photo Flash Reaction
2 atoms Mg + 1 molecule O
2 makes 2 formula units MgO
2 moles Mg + 1 mole O
2 makes 2 moles MgO
48.6 grams Mg + 32.0 grams O
2 makes 80.6 g MgO
2 grams Mg + 1 gram O
2 makes 2 g MgO
3.7

1. Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation.
Ethane reacts with oxygen to form carbon dioxide and water
C
2
H
6
+ O
2
CO
2
+ H
2
O
2. Change the numbers in front of the formulas
( coefficients ) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts.
2C
2
H
6
NOT C
4
H
12
3.7

3. Start by balancing those elements that appear in only one reactant and one product.
C
2
H
6
+ O
2
CO
2
+ H
2
O start with C or H but not O
2 carbon on left
C
2
H
6
+ O
2
1 carbon on right
2 CO
2
+ H
2
O multiply CO
2 by 2
6 hydrogen on left
C
2
H
6
+ O
2
2 hydrogen on right
2 CO
2
+ 3 H
2
O multiply H
2
O by 3
3.7

4. Balance those elements that appear in two or more reactants or products.
C
2
H
6
+ O
2
2 CO
2
+ 3 H
2
O multiply O
2 by
7
2
2 oxygen on left
C
2
H
6
7
2 2
2 C
2
H
6
+ 7 O
2
4 oxygen + 3 oxygen = 7 oxygen
(2x2) (3x1) on right
2 CO
2
+ 3 H
2
O remove fraction multiply both sides by 2
4 CO
2
+ 6 H
2
O
3.7

5. Check to make sure that you have the same number of each type of atom on both sides of the equation.
2 C
2
H
6
+ 7 O
2
14 O ( 7 x 2)
4 CO
2
+ 6 H
2
O
14 O ( 4 x 2 + 6 )
Reactants
4 C
12 H
14 O
Products
4 C
12 H
14 O
3.7

1. Write balanced chemical equation
2. Convert quantities of known substances into moles
3. Use coefficients in balanced equation to calculate the number of moles of the sought quantity
4. Convert moles of sought quantity into desired units
3.8

Methanol burns in air according to the equation
2CH
3
OH + 3O
2
2CO
2
+ 4H
2
O
If 209 g of methanol are used up in the combustion, what mass of water is produced?
grams CH
3
OH moles CH
3
OH molar mass
CH
3
OH moles H coefficients chemical equation
2
O grams H molar mass
H
2
O
2
O
209 g CH
3
OH
1 mol CH
3
OH x
32.0 g CH
3
OH x
4 mol H
2
O
2 mol CH
3
OH x
18.0 g H
2
O
1 mol H
2
O
=
235 g H
2
O
3.8

3.9

• M + W > MC Fr + __ W > Bk
10 10 ( ) 10 10 ( )
20 10 ( ) 20 10 ( )
15 40 ( ) 15 40 ( )
15mol 40mol ( )mol
Limiting Reactant vs. Reactant in Excess

g Al g Fe
2
Do You Understand Limiting Reagents?
In one process, 124 g of Al are reacted with 601 g of Fe
2
O
3
Thermite reaction: used in welding of steel and iron
2Al + Fe
2
O
3
Al
2
O
3
+ 2Fe
Calculate the mass of Al
2
O
3 formed.
O
3 mol Al mol Fe
2
O
3 mol Fe
2
O
3 needed
OR mol Al needed g Fe
2
O
3 needed g Al needed
124 g Al x
1 mol Al
27.0 g Al x
1 mol Fe
2
O
3
2 mol Al x
160. g Fe
2
O
3
1 mol Fe
2
O
3
= 367 g Fe
2
O
3
Start with 124 g Al need 367 g Fe
2
O
3
Have more Fe
2
O
3
(601 g) so Al is limiting reagent
3.9

Use limiting reagent (Al) to calculate amount of product that can be formed.
g Al mol Al
2Al + Fe
2
O
3 mol Al
2
O
3
Al
2
O
3
+ 2Fe g Al
2
O
3
124 g Al x
1 mol Al
27.0 g Al x
1 mol Al
2
O
3
2 mol Al x
102. g Al
2
O
3
1 mol Al
2
O
3
= 234 g Al
2
O
3
3.9

Theoretical Yield is the amount of product that would result if all the limiting reagent reacted.
Actual Yield is the amount of product actually obtained from a reaction.
% Yield =
Actual Yield
Theoretical Yield x 100
3.10

Plants need: N, P, K, Ca, S, & Mg
3H
2
( g ) + N
2
( g ) 2NH
3
( g )
NH
3
( aq ) + HNO
3
( aq ) NH
4
NO
3
( aq ) fluorapatite
2Ca
5
(PO
4
)
3
F ( s ) + 7H
2
SO
4
( aq )
3Ca(H
2
PO
4
)
2
( aq ) + 7CaSO
4
( aq ) + 2HF ( g )