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Chapter 3
Mass Relationships in
Chemical Reactions
Chemical Formulas, Composition, &
Stoichiometry
Atomic Mass (Atomic Weight)

Atomic mass is expressed in amu (atomic mass
units)
◦ An amu is 1/12 the mass of 12C atom.
◦ 1 amu = 1.660 x 10-24 g
◦ This yields a proton and neutron mass of approximately 1
Proton mass:
Neutron mass:
Electron mass:

1.00728 amu
1.00867 amu
0.00054 amu
Atomic mas is approximately the sum of the
masses of the number of protons, neutrons, and
electrons
Atomic Mass
Weighted Average
In nature, chlorine (17 protons) has 2 main
isotopes:
35 18 neutrons (75.53%) – AW = 34.97
37 20 neutrons
t
(24
(24.47%)
47%) – AW = 36.97
36 97
Average atomic mass:
(0.7553 x 34.97 ) + (0.2447 x 36.97 )
= 35.46
1
Atoms and Molecules
Atoms and Molecules are very, very,
very, very, very, very, very, very
….small
 Too small for us to measure in
practical manner
 Therefore, we invented moles

Mole

Defined as the number of atoms in 12 g of
= 6.022 x 1023 items of anything
12C
 Avogadro’s number
 One mole of 12C weighs exactly 12 g by definition
definition.
 One mole of carbon occurring in its natural isotopic
ratio weighs 12.011 g
 because a few carbon atoms contain more than 12
neutrons
Moles
 A mole
is a simply a count – like a
dozen
◦ 1 dozen eggs = 12 eggs
◦ 1 mole of eggs = 6
6.022
022 x 1023 eggs
2
Molar Mass




Molar mass (M) is the mass in grams of one
mole of atoms
Equal to the mass in grams of one mole of
the atoms or g/mol
Cl Chlorine
35.453 g/mol of atoms
The periodic table list molar mass for any
particular element – weighted averages of
naturally occurring isotopes
Sometimes called atomic mass or incorrectly
atomic weight
For Elements
Divide by
6.02 x 1023
Multiply
by
Molar Mass
GRAMS

MOLES
Divide by
Molar Mass

NUMBER
OF
ATOMS
Multiply by
6.02 x 1023
Grams to Moles
How many moles of Fe are there in 136.9 g
of iron?
3
Moles to Atoms

How many atoms of C are there in 10.0
moles of carbon?
Atoms to Grams

What is the mass in grams of one
hydrogen atom?
How many atoms in 1.00 g of C?
First, ask: how
many moles of
C?
4
Moles

How many stars in a mole of stars?
Molecular Formula
The molecular formula is the actual number of
atoms present in a single molecule of a
substance.
CH4
C
CO2
O2
HCl
H2S
NaCl
Methane
et a e
Carbon Dioxide
Oxygen (Dioxygen)
Hydrogen Chloride
Hydrogen Sulfide
Sodium Chloride
Subscripts
indicate the
number of
atoms of a
particular
element in
a molecule
Molar Mass

The Molar Mass of a compound or molecule is sum of the
atomic mass or molar masses of the elements in the compound
each taken the number of times the element occurs

Equal to the mass in amu of one molecule
◦ Sometimes called molecular weight or formula weight
H2O
Water
(2 x 1.01) + (1 x 16.00) = 18.02 amu/molecule

Equal to the mass in grams of one mole of the molecules
CH4
Methane
(1 x 12.01) + (4 x 1.01) = 16.05 g/mole
5
Molar Mass

What is the Molar Mass for:
◦ Na2CO3
Sodium Carbonate
◦ CH3COOH
Acetic Acid
◦ Ca3(PO4)2
Calcium Phosphate
◦ CuSO4• 5H2O
Copper(II) Sulfate Pentahydrate
For Compounds
Divide by
6.02 x 1023
Multiply
by
FW
GRAMS

MOLES
Divide by
FW

NUMBER
OF
Molecules
Multiply by
6.02 x 1023
Mole Concept

How moles in 32.084 g of methane
(CH4)? How many molecules?
◦ MM of CH4= 1 x 12.01 g/mole + 4 x 1.008 g/mole
= 16.04 g/mole
6
Mole Concept

A billion (1.00 x 109) molecules of SO2 would
have what mass?
Percent Composition

What percent of acetone (C3H6O) is
carbon, hydrogen, oxygen?
Percent Composition

What is the percent composition of
Ni(CO)4?
7
Simplest (Empirical) Formula

Compound A: Cu 30.03 %, C 22.70 %, H
1.91 %, O 45.37%? What is the simplest
formula of compound A?
 Assume you have 100 g of A. Then each %
represents the mass (g) of the particular
element.
Simplest (Empirical) Formula

Divide the number of moles of each
element by the smallest number of moles of
all the elements (0.473 moles of Cu in this
case).




0.473 moles Cu /0.473 moles = 1 mole of Cu = 1
1.892 moles of C/0.473 moles = 4.0 moles of C = 4
1.91 moles of H/0.473 moles = 4.04 moles of H = 4
2.836 moles of O/ 0.473 moles = 5.99 moles of O = 6
Percent Composition & MW

A compound with a MW = 46.1 g/mol is composed
of C (52.1 %), H (13.1 %) and O (34.7 %).
What is the molecular formula?
8
Percent Purity
Percent Purity
= Mass (g) of pure substance x 100 %
Mass (g) of impure substance
Percent Purity
18 karat gold is 75 % gold
 If an 18 karat gold ring weighs 12 grams,
how much gold (g) is in the ring?

Chemical Reactions
Chemical reactions always involve
changing one or more substances into
one or more different substances
 Chemical reactions rearrange atoms
or ions to form other substances
CH4 + 2 O2  CO2 + 2 H2O

Fe

X
Au
9
Chemical Reaction Equations




Chemists use a shorthand notation to
represent a chemical reaction: EQUATION.
Reactants  Products
Where the reactants and products are
represented by symbols for atoms and for
formulas of molecules.
CH4 + O2  CO2 + H2O
◦ The order of reactants is not important
◦ The arrow reads “yields”
Chemical Reaction Equations
In order for the equation to accurately
represent the chemical reaction, it must
be BALANCED
 Balanced means that for any particular
element, the number of atoms on the
right side of the equation must be equal
to the number of atoms on the left

◦ Conservation of mass in chemical
reactions
Balancing Chemical Equations



Balance chemical equations by placing simple,
WHOLE number coefficients in front each
molecule to make the number of atoms on both
sides of the equation equal
Subsc
pts represent
ep ese the
e number
u be o
o s in a
Subscripts
of a
atoms
compound (molecule) and can NOT be changed to
balance the chemical equation
Balance methane burning in oxygen.
CH4 + O2 
CH4 + 2 O2 
CO2 + 2 H2O
CO2 + 2 H2O
10
Balancing Chemical Equations
CH4 + 2 O2 
CO2 + 2 H2O
Think it is balanced, always double check

C:

1
C 1
H 4

2x2=4
O 2x2 = 4

2 +2 = 4
This is called balancing by inspection
◦ Trial and error process (with some guidelines)
Balancing Chemical Equations
1.
2.
3.
4.
Look for elements in only one
reactant and one product, balance
these first.
If free, uncombined elements appear
on either
ith side
id b
balance
l
th
these llast.
t
If you end up with ½ a molecule in
the equation, multiply the entire
equation by 2.
Balance the equations with the
lowest (smallest) numbers possible.
Balancing Chemical Equations
N2 + H2  NH3
11
Balancing Chemical Equations
C8H18 + O2  H2O + CO2
Balancing Chemical Equations
UO2 + HF  UF4 + H2O
Balancing Chemical Equations
Fe2O3 + CO  Fe + CO2
12
Balancing Chemical Equations
NH3 + O2  NO + H2O
Balancing Chemical Equations
NH3 + O2  NO + H2O
2 NH3 + O2  NO + 3 H2O
2 NH3 + O2 2 NO + 3 H2O
(2 NH3 + 5/2O2 2 NO + 3 H2O)x2
4 NH3 + 5 O2  4 NO + 6 H2O
N
H
O
4
4x3
5x2



4
4x1
6x2
6x1
What do you get from a balance
chemical equation?

Molar or Molecular Reaction Ratios
CH4 +
2 O2 
CO2 +
2H2O

If you burn 1 molecule of CH4, you get 1
molecule
l
l off CO2 and
d 2 molecules
l
l off H2O

If you burn 47 molecules of CH4, you get 94
molecules of H2O

If you burn 3 moles of CH4, you get 3 moles
of CO2 and 6 moles of H2O
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Conservation of Matter
 CO2 +
CH4 +
2 O2
1 mol
2 mol
1 mol
2 mol
16.0 g
2(32.0 g)
44.0 g
2(18.0 g)
16.0 g
64. 0 g
44.0 g
36.0 g
80.0 g reactants
2 H2O
80.0 g of products
Mass of Reactant Required or Formed
CH4 +
32.0g
16.0g/mole
2 O2 
excess
CO2
+
2H2O
?g
18.0g/mole
Mass of Reactant Required or Formed
CH4 +
32.0g
16.0g/mole
32.0 g CH4
2 O2 
excess
CO2
+
2H2O
?g
18.0g/mole
x 1 mole CH4
x 2 mole H20
x 18.0 g H20
16.0 g
1 mole CH4
1 mole
= 72.0 g H20
Mole ratio from the
balanced chemical equation
14
Mass of Reactant Required or Formed
heat
Fe3O4 +
4 H2
?g
excess

3Fe
+
4H2O
18.75 g
231.5 g/mole
18.02 g/mole
Mass of Reactant Required or Formed

Calculate the mass of water required to
exactly react with 100.0 g of potassium
2 K((s)) +
2 H2O ((l))
100.0 g
?g

2 KOH ((aq)
q) + H2 ((l))
39.10 g/mole 18.02 g/mole
Limiting Reagent Concept

Sometimes one reactant is used up and the
other is in excess
◦ This is called a limiting reaction
M:

CH4 +
32.0 g
2 O2 
96.0 g
CO2 +
16.0 g/mol
32.0 g/mol
44.0 g/mol
2H2O
How much (mass, ie. g) CO2 is formed?
15
Limiting Reaction Concept
CH4
+
2 O2
32.0 g

32 g CH4 x 1 mole/16 g = 2 mol CH4
Required Moles:
1.5 mol CH4
Subtract:
CO2
+
2H2O
96.0 g
96 g O2 x 1mol/32 g = 3 mol O2
4 mol O2
Use the molar ratio to
determine required moles
+0.5 mol CH4
In Excess (+)
-1 mol O2
Limiting Reagent (-)
Limiting Reaction Concept

Methane is in excess, oxygen is the limiting
reagent. The number of CO2 moles will be
determined by the number of moles of O2.
96 g O2 x 1 mole O2
32.0 g
x 1 mole CO2
2 mole O2
x 44.0 g CO2
= 66.0 g
1 mole
CO2
Limiting Reagent (the easy way)
How many grams of Ni(OH)2 could be prepared
from reacting 25.9 g of NiCl2 and 10.0 g NaOH?
NiCl2 + 2NaOH  Ni(OH)2 + 2NaCl
25 9 g 10.0
25.9
10 0 g
?g
Work the problem twice – use the answer with lowest result
16
Limiting Reagent (the easy way)
How many grams of Ni(OH)2 could be prepared
from reacting 25.9 g of NiCl2 and 10.0 g NaOH?
NiCl2 + 2NaOH  Ni(OH)2 + 2NaCl
25 9 g
25.9
10
10.0
0g
?g
25.9 g NiCl2
10.0 g NaOH
1 mol NiCl2
1 mol Ni(OH)2
92.7 g Ni(OH)2
129.6g
1 mol NiCl2
1 mol
1 mol NaOH
1 mol Ni(OH)2
92.7 g Ni(OH)2
40.0g
2 mol NaOH
1 mol
=
18.5 g Ni(OH)2
=
11.6 g Ni(OH)2
Thus, we get 11.6 g of Ni(OH)2. There is not enough
NaOH to make more. NaOH is the limiting reactant.
Mass of Reactant Required or Formed

Given 100.0 g of potassium and 100.0 g of
water, which is the limiting reagent and what
mass KOH is produced by the reaction?

2 K(s) +
2 H2O (l)
100 0 g
100.0
100 0 g
100.0
39.10 g/mol
18.02 g/mol
2 KOH (aq) + H2 (l)
? g
56.11 g/mol
Percent Yield Calculations
Yield is the amount of a product produced.
% Yield = actual amount obtained x100 %
theoretical amount
CH4 +
32.0 g
2 O2 
excess
CO2 +
2H2O
66.0 g
72.0 g comes from calculating the mass of product formed
assuming a perfect reaction (a few slides before this one)
17