• Atomic mass provides a means to count atoms by measuring the mass of a sample
• The periodic table on the inside cover of the text gives atomic masses of the elements
• The mass of an atom is called its atomic mass
• When using atomic masses, retain a sufficient number of significant figures so the atomic mass data contributes only slightly to the uncertainty of the result

• The molar mass allows counting of molecules by mass
• The molar mass is the sum of atomic masses of the atoms in the compounds formula
– For example the molar mass of water, H
2
O, is twice the mass of hydrogen (1.008) plus the mass of oxygen (15.999) = 18.015
• Strictly speaking, ionic compounds do not have a “molar mass” because they don’t contain molecules

• The mass of the formula unit is called the formula mass
• Formula masses are calculated the same way as molecular masses
– For example the formula mass of calcium oxide, CaO, is the mass of calcium (40.08) plus the mass of oxygen (15.999) = 56.08
• One mole of a substance contains the same number of formula units as the number of atoms in exactly 12 g of carbon-12

• One mole of a substance has a mass in grams numerically equal to its formula mass
• The mass of one mole of a substance is also called its molar mass
• One mole of any substance contains the same number of formula units
• This number is called Avogadro’s number or constant
1 mol formula units = 6.02 x 10 23 formula units

• Counting formula units by moles is no different than counting eggs by the dozen
(12 eggs) or pens by the gross (144 pens)
• Avogadro’s number is huge because atoms and molecules are so small: a huge number of them are needed to make a lab-sized sample
• Avogadro’s number links moles and atoms, or moles and molecules and provides an easy way to link mass and atoms or molecules

•
Atoms – elements
•
Molecules – molecular compounds
•
Ions – charged particles and
•
Formula units – Ionic compounds
•
Photons – electromagnetic radiation

• Using water (molar mass 18.015) as an example:
1 mole H
2
O

6.022 x 10 23 molecules H
2
O
1 mole H
2
O

18.015 g H
2
O
18.015 g H
2
O

6.022 x 10 23 molecules H
2
O
• Within chemical compounds, moles of atoms always combine in the same ratio as the individual atoms themselves so:
1 mole H
2
O

2 mole H
1 mole H
2
O

1 mole O

•
Stoichiometry is the study of the mass relationships in chemical compounds and reactions
• A common use for stoichiometry is to relate the masses of reactants needed to make a compound
• These calculations can be solved using the factor-label method and equivalence relations relating molecular masses and/or formula masses

• Example: How many grams of iron are in a
15.0 g sample of iron(III) oxide?
ANALYSIS: 15.0 g Fe
2
O
3
LINKS: 1 mol Fe
2
O
3

? g Fe

2 mol Fe
1 mol Fe
2
O
3

159.7 g Fe
2
O
3
1 mol Fe

55.85 g Fe
SOLUTION:
15 .
0 g Fe
2
O
3
 1 mol Fe
2
O
3
159 .
9 g Fe
2
O
3

2 mol Fe
1 mol Fe
2
O
3
 55 .
85 g Fe
1 mol Fe

10 .
5 g Fe

• The usual form for describing the relative masses of the elements in a compound is a list of percentages by mass
• This is called the percentage composition or percentage composition by mass
• The percentage by mass is the number of grams of the element in 100 g of the compound and can be calculated using:
 mass of element mass of whole sample


• Example: A sample was analyzed and found to contain 0.1417 g nitrogen and 0.4045 g oxygen. What is the percentage composition of this compound?
ANALYSIS: Find sample mass and calculate %
LINKS: whole sample = 0.5462 g
SOLUTION:
% N
 0.1417
g N
0.5462
g sample

100%

25.94
%
% O
 0.4045
g O
0.5462
g sample

100%

74.06
%

• Hydrogen peroxide consists of molecules with the formula H
2
O
2
• This is called the molecular formula
• The simplest formula for hydrogen peroxide is HO and is called the empirical formula
• It is possible to calculate the empirical formula for a compound from mass data
• The goal is to produce the simplest whole number mole ratio between atoms

• Example: A 2.012 g sample of a compound contains 0.522 g of nitrogen and 1.490 g of oxygen. Calculate its empirical formula
ANALYSIS: We need the simplest whole number mole ratio between nitrogen and oxygen
SOLUTION:
0 .
522 g N

1 mol N
14.01
g N

0.0373
mol N
1 .490
g O

1 mol O
15.999
g O

0.0931
mol
N
0 .0 3 7 3
O
0 .0 3 7 3
0 .0 9 3 1

0 .0 3 7 3
N
1.00
O
2.50
N
1.00

2
O
2.50

2

N
2.00
O
5.00

O
N
2
O
5

• Empirical formulas may also be calculated indirectly
• When a compound made only from carbon, hydrogen, and oxygen burns completely in pure oxygen only carbon dioxide and water are produced
• This is called combustion
• Empirical formulas may be calculated from the analysis of combustion information

• Example: The combustion of a 5.217 g sample of a compound of C, H, and O gave
7.406 g CO
2 and 4.512 g of H
2
O. Calculate the empirical formula of the compound.
ANALYSIS: This is a multi-step problem. The mass of oxygen is obtained by difference: g O = 5.217 g sample – ( g C + g H )
The masses of the elements may then by used to calculate the empirical formula of the compound
SOLUTION:

7.406
g CO
2
 12 .
011 g C
44.009
g CO
2

2.022
g C
4.512
g H
2
O
 2.0158
g H
18.015
g H
2
O

0.5049
g H total mass of C and H

2.473
g mass O

5.217
g 2.527g

2.690
g
C : 2.022
g C

1 mol C
12.011
g C

0.1683
mol C
H : 0.5049
g H

1 mol H
1.008
g H

0.5009
mol H
O : 2.690
g O

1 mol O
15.999
g O

0.1681
mol O
C
0 .1 6 8 3
H
0 .1 6 8 1
0 .5 0 0 9
O
0 .1 6 8 1
0 .1 6 8 1
0 .1 6 8 1

C
1.001
H
2.980
O
1.000

CH
3
O

• The formula for ionic compounds is the same as the empirical formula
• For molecules, the molecular formula and empirical are usually different
• If the experimental molecular mass is available, the empirical formula can be converted into the molecular
• The molecular formula will be a common multiplier times all the coefficients in the empirical formula

• Example: The empirical formula of hydrazine is NH
2
, and its molecular mass is
32.0. What is its molecular formula?
ANALYSIS: The molecular mass (32.0) is some simple multiple of the mass calculated from the empirical formula (16.03)
SOLUTION:

32.0
16.03

1

2.00
2

2.00

2
4

• The coefficients of a balanced chemical equation provide the mole-to-mole ratios for the substances involved in the reaction
• Whenever a problem asks you to convert between different substances, the calculation usually involves a mole-to-mole relationship
• How to detect unbalanced equations will be covered shortly

• Example: If 0.575 mole of CO
2 is produced by the combustion of propane, C
3
H
8
, how many moles of oxygen are consumed? The balanced equation is:
C
3
H
8
+ 5 O
2

3 CO
2
+ 4 H
2
O
ANALYSIS: Relating two compounds usually requires a mole-to-mole ratio
SOLUTION:
0 .
575 mol C
3
H
8
 5 mol O
2
1 mol C
3
H
8

2.88
mol O
2

• In many problems you will also need to perform one or more mole-to-mass conversions
• These types of stoichiometry problems are summarized with the flowchart:

• Example: How many grams of Al
2
O
3 produced when 41.5 g Al react?
are
2Al( s ) + Fe
2
O
3
( s )

Al
2
O
3
( s ) + 2 Fe( l )
ANALYSIS:
41.5 g Al

? g Al
2
O
3
SOLUTION: grams Al
 moles Al
 moles Al
2
O
3
 grams Al
2
O
3
41.5
g Al

1 mol Al
26.98
g Al
 1 mol Al
2
O
2 mol Al
3
 102.0
g Al
2
1 mol Al
2
O
O
3
3

78.45
g Al
2
O
3

•
Chemical equations provide quantitative descriptions of chemical reactions
•
Conservation of mass is the basis for balancing equations
•
To balance an equation:
1) Write the unbalanced equation
2) Adjust the coefficients to get equal numbers of each kind of atom on both sides of the arrow

•
Guidelines for Balancing Equations:
1) Balance elements other than H and O first
2) Balance as a group any polyatomic ions that appears unchanged on both sides of the arrow
3) Balance separately those elements that appear somewhere by themselves
•
As a general rule you should use the smallest whole-number coefficients when writing balanced chemical equations

• All reactions eventually use up a reactant and stop
• The reactant that is consumed first is called the limiting reactant because it limits the amount of product that can form
• Any reagent that is not completely consumed during the reactions is said to be in excess and is called an excess reactant
• The computed amount of product is always based on the limiting reagent

• Example: How many grams of NO can form when 30.0 g NH
3 according to: and 40.0 g O
2 react
4 NH
3
+ 5 O
2

4 NO + 6 H
2
O
ANALYSIS: This is a limiting reactant problem
SOLUTION:
30 .
0 g NH
3
 1 mol NH
3
17 .
03 g NH
3

4 mol NO
4 mol NH
3
 30 .
01 g NO
1 mol NO

52 .
9 g NO
40.0
g O
2
 1 mol O
2
32 .
00 g O
2

4 mol NO
5 mol O
2
 30 .
01 g NO
1 mol NO

30.01
g NO
O
2 is the limiting reagent and 30.01
g NO form

• The amount of product isolated from a chemical reactions is almost always less than the calculated, or maximum, amount
• The actual yield is the amount of the desired product isolated
• The theoretical yield is the amount that would be recovered if no loss occurred (the calculated, maximum amount)
• The percentage yield is the actual yield as a percentage of the theoretical yield

 actual yield theoretica l yield

• When working with percentage yield:
– Remember they involve a measured (actual yield) and calculated (theoretical yield) quantity
– The calculation may be done in either grams or moles
– The result can never be a number larger than
100%