Chapter 3
Atomic Mass
 amu = Average Atomic Mass Unit
 Based on 12C as the standard.
 12C = exactly 12 amu
 The average atomic mass (weight) of
an element is equal to the sum of the
products of each isotope’s mass (in amu)
multiplied by it’s relative abundance.
EXAMPLE OF AVERAGE
ATOMIC MASS PROBLEM
 Naturally occurring chlorine is 75.53% 35Cl
which has an atomic mass of 34.969 amu,
and 24.47% 37Cl, which has an atomic mass of
36.966 amu.
 Calculate the average atomic mass of
chlorine.
EXAMPLE OF AVERAGE ATOMIC MASS
PROBLEM (CONT)
 Average atomic Mass = [ (%/100)
(Atomic Mass) ]
 Average atomic mass =
(0.7553) (34.969 amuCl35) + (0.2447) (36.966
amuCl37)
= 26.41 amu + 9.045 amu
= 35.46 amu
Formula Weight
 AKA: molar mass (g/mol)
 Sum of all atomic weights of each atom in
its chemical formula
Ex: H2S04
1(2H) + 32(1S) + 16(4O) = 98 amu
MOLE
Avogadro’s Number
 1 mole of any
substance is
6.02 x 1023 particles
Question
 How many nitrogen
atoms are in 0.25
mol of Ca(NO3)2
Answer
6.02 x 1023 molec Ca(NO3)2 x_2N___
X .25
1 mol
molec Ca(NO3)2
=
atoms
Law of Conservation of Mass
 Mass is never created
or destroyed
Lavosier Says :
2 Na +
 Reason for balancing
chemical equations
Cl 2
46.0 g 70.9 g

2NaCl
116.9 g
Balancing Equations
1. Write the correct formulas for the reactants
and the products
2.Chose the compound that has the greatest
number of atoms, then look to the element in
that compound that has the greatest number of
atoms.
3.Balance this element first by placing a
coefficient in front of the corresponding
compound on the other side of the equation.
Balancing Equations Cont.
4. Balance H then O
5. Check all coefficients to see that they
are in the lowest possible ratio.
Examples:
C2H6 + O2 -> CO2 + H2O
C2H6O + O2 -> CO2 + H2O
CaCO3 + H3PO4 -> Ca3(PO4)2 + CO2 + H2O
ANSWER
2C2H6 + 7O2 -> 4CO2 + 6H2O
C2H6O + 3O2 -> 2CO2 + 3H2O
3CaH6O3 + 2H3PO4 -> Ca3(PO4)2 + 3CO2 + 3H2O
Percent Composition
We can describe composition in two ways
1. number of atoms
2. % (by mass) of its elements.
We can find % mass from formula mass, by
comparing each element present in 1
mole of compound to the total mass of 1
mole of compound
Example of % Comp
 Calculate the percentage of nitrogen in
Ca(NO3)2
Answer
% N = # N atoms(m.w N)
m. w Ca(NO3)2
% N = 2(14.02 N amu)
164.12 Ca(NO3)2amu
= 17%
X 100%
X 100%
Inter-converting
Grams -> moles -> molec -> atoms
How many oxygen atoms are present in
4.20 grams NaHCO3?
Answer
4.20 g NaHCO3
atoms
(1mole NaHCO3)
84 g NaHCO3
(6.02e 23molec)
1 mol
3 Oxygen
1 molec NaHCO3
= 9.033 x 10 22 atoms of Oxygen in 4.20 grams NaHCO3
Question
 Determine the mass in grams of
3.00 x 1020 N2 molecules
Answer
3.00 x 1020 molec N2 (1 mol)
(28g N2)
6.02e23 molec 1 mol N2
= 0.0140 g
Determining empirical
formula from mass percent
 Recall: Empirical formula: simplest whole
# ratio of atoms in a compound.
Example: Vitamin C is composed of
40.92% C, 4.58% H, and 54.50% O by
mass. What is the empirical formula?
Answer:
1.
2.
3.
CHO
Convert mass % into grams (assume 100g)
Convert grams to moles
Divide each mol by the smallest number of moles
present. You may round to nearest whole #
40.92 g C 1mol C
12 g C
= 3.4 moles C /3.4 = 1 C
54.40 g O 1mol O
16g O
= 3.4 moles O / 3.4 = 1 O
4.48 g H
= 4.48 moles H/ 3.4 = 1 H
1mol H
1gH
Determine Molecular formula
from Empirical Formula
Recall: Molecular formula: the exact formula of a molec, giving
types of atoms and the number of each type.
1.
2.
3.
Using mass % and molar mass, determine mass of each
element in 1 mole of compound
Determine number of moles of each element in 1 mole of
compound
The integers from the previous step represent the
subscripts in the molecular formula
Let’s look back at our work
40.92 g C 1mol C
12 g C
= 3.4 moles C /3.4 = 1 C
54.40 g O 1mol O
16g O
= 3.4 moles O / 3.4 = 1 O
4.48 g H
= 4.48 moles H/ 3.4 = 1.3 H
1mol H
1gH
C3H4O3 = molecular formula
Shortcut
n = Molecular Weight
empirical Formula Weight
The molecular weight of butyric acid is
88 amu. If the empirical formula is C2H4O.
What is the molecular formula?
C2H4O = 12 + 12+ 1+1+1+1+16 = 44 amu
n = 88 amu = 2
44
Molecular formula = (empirical)n
(C2H4O)2 = Molecular Formula = C4H8O2
Stoichiometry: mixing exactly
enough chemical so that all is used
Mass-Mass problems
g given  mol given  mol required  g required
(grams to moles to moles to grams)
Silicon carbide is made by heating silicon dioxide
to high temperatures.
SiO2 (s) + 3C (s)  SiC(s) + 2CO (g)
How many grams of CO are formed by complete
rxn of 5.00 g SiO2?
AP EXAM HINT: always make sure your equation
is balanced first or mole ratios will be wrong.
Given: 5.00 g SiO2
Find : CO g
SiO2 (s) + 3C (s)  SiC(s) + 2CO (g)
grams to moles to moles to grams
5.00 g SiO2 1mol SiO2 2 mol CO
60 g SiO2 1 mol SiO2
28 g CO = 4.6 g CO
1 mol CO
Mole ratio
How many moles of sulfuric acid would be
needed to produce 4.80 moles of molecular
iodine (I2) according to the following
balanced equation.
10HI + 2KMnO4 + 3H2SO4  5I2 + 2MnSO4 + K2SO4 + 8H2O
4.80 mol I2 3 mol H2SO4
5 mol I2
= 2.88 mol H2SO4
Limiting reagent
This will be at least 1 AP question!
The number of products that can form is
limited by the amount of reactant present.
The limiting reactant is the one that gives
the least amount of product.
Reactants  Products
When a mixture of silver and sulfur is heated,
silver sulfide is formed:
16 Ag (s) + S8 (s)  8 Ag2S (s)
What mass of Ag2S is produced from a mixture
of 2.0 g of Ag and 2.0 g of S?
2.0 g Ag 1mol Ag
8 mol Ag2S
107.9 g Ag 16 mol Ag
247.9 g Ag2S
1 mol Ag2S
= 2.3 g Ag2S
2.0 g S8
1mol S8
256.8 S8
8 mol Ag2S
1 mol S8
= 15 g Ag2S
247.9 g Ag2S
1 mol Ag2S
Theoretical / Percent yield
 The amount of product that is calculated
based on the limiting reactant.
% Yield = Actual yield
X 100%
theoretical yield
\
HW
 7, 19, 21, 31,33
 35, 56,57, 58, 62,63, 67, 69. 71,74
81,82,83, 95, 98, 100a-b,105
Types of reactions
Chemical Reactivity
Combination/Synthesis Reaction:
2 or more substances react to form one new
product
A+B C
+

solid magnesium and oxygen gas react to
produce solid magnesium oxide
2Mg (s) +
Metal
O2(g) 
nonmetal
Diatomic
2MgO (s)
ionic compound
2+ 2-
Decomposition Rxn
 One substance undergoes a reaction to
produce two or more substances.
 Typically occurs when things are heated.
AX  A + X

+
Solid calcium carbonate reacts to
produce solid calcium oxide and
carbon dioxide gas
CaCO3 (s)  CaO (s)
2+ (2-)
2+ 2-
+ CO2 (g)
4+ 2(2-)
Single displacement
 One element replaces a similar element in
a compound
A + BX  AX + B
BX + Y  BY + X
+

+
Solid copper is dissolved in aqueous silver
nitrate to produce solid silver and aqueous
copper nitrate.
Cu(s) + AgNO3 (aq) Ag(s) + Cu(NO3)2 (aq)
Write the sentence for this reaction:
Fe (s) + Cu(NO3)2 (aq)  Fe(NO3)2 (aq)+ Cu
(s)
Double Replacement Rxn/
Metathesis
 The ions of two compounds exchange places
in an aqueous solution to form two new
compounds.
AX + BY  AY + BX
 One of the compounds formed is usually a
precipitate, an insoluble gas that bubbles out
of solution, or a molecular compound,
usually water.
Double Replacement Rxn/
Metathesis
AX + BY  AY + BX
+

+
Write the sentence for these double replacement
reactions
KOH (aq) + H2SO4 (aq)  K2SO4 (aq) + H2O (l)
FeS (aq) + HCl (aq)  FeCl2 (aq) + H2S (aq)
Combustion Reaction
A substance combines with oxygen, releasing a
large amount of energy in the form of light and
heat.
C3H8 (g)+ 5O2 (g)  3CO2 (g) + H2O (g)
Usually CO2 (carbon dioxide) / CO (carbon monoxide)
and water are produced.
 Reactive elements combine with
oxygen
P4(s) + 5O2(g)  P4O10 (s)
(This is also a synthesis reaction)
 The burning of natural gas, wood,
gasoline
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)