Chemical Quantities: The Mole Concept and Molar Mass
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CHAPTER OUTLINE
3
THE MOLE
CONCEPT
Chemists find it more convenient to use mass
relationships in the laboratory, while chemical
reactions depend on the number of atoms
present.
In order to relate the mass and
number of atoms, chemists use
the SI unit
mole
(abbreviated
mol).
4
THE MOLE
CONCEPT
The number of particles in a mole is called
Avogadro’s number and is 6.02x10
23
.
1 mole
equals to
6.02 x 10
23
Avogadro’s
number (N
A
)
5
THE MOLE
CONCEPT
A mole is a very large quantity
6.02x10
23
If 10,000 people started to count
Avogadro’s number and counted at the
rate of 100 numbers per minute each
minute of the day, it would take over 1
trillion years to count the total number.
MOLE
6
THE MOLE
CONCEPT
1 mole H
2
molecules
= 6.02x10
23
H
2
molecules
= 2 x (6.02x10
23
) H atoms
1 mole Na
+
ions
1 mole H
2
O molecules
= 6.02x10
23
H
2
O molecules
= 2 x (6.02x10
23
) H atoms
= 6.02x10
23
O atoms
= 6.02x10
23
Na
+
ions
1 mole H atoms
= 6.02x10
23
H atoms
7
THE MOLE
CONCEPT
The atomic mass of one atom expressed in amu
is numerically the same as the mass of 1 mole of
atoms of the element expressed in grams.
Mass of 1 H atom = 1.008 amu
Mass of 1 mol H atoms = 1.008 grams
Mass of 1 Mg atom = 24.31 amu
Mass of 1 mol Mg atoms = 24.31 g
Mass of 1 Cl atom = 35.45 amu
Mass of 1 mol Cl atoms = 35.45 g
8
MOLAR MASS
1 mol O atom =
2 mol H atoms =
Mass of one mole of H
2
O
2 (1.01 g) = 2.02 g
1 (16.00 g) = 16.00 g
18.02 g
Molar mass
The mass of one mole of a substance is called
molar mass, and is measured in grams.
9
MOLAR MASS
2 mol O atoms =
2 mol H atoms =
Mass of one mole of Ca(OH)
2
2 (1.01 g) = 2.02 g
2 (16.00 g) = 32.00 g
74.10 g
Molar mass
1 mol Ca atom =
1 (40.08 g) = 40.08 g
10
Calculate the molar mass of each compound shown
below:
Example 1:
Lithium carbonate
Li
2
CO
3
Li
2 x 6.94
= 13.88 g
C
1 x 12.01
= 12.01 g
O
3 x 16.00
= 48.00 g
Molar mass
= 73.89 g/mol
11
Calculate the molar mass of each compound shown
below:
Example 2:
Salicylic acid
C
7
H
6
O
3
C
7 x 12.01
= 84.07 g
H
6 x 1.01
= 6.06 g
O
3 x 16.00
= 48.00 g
Molar mass
= 138.13 g/mol
12
CALCULATIONS
USING THE MOLE
Conversions between mass, mole and particles
can be done using molar mass and Avogadro’s
number.
Mass of a
substance
Moles of a
substance
MM
Particles of
a substance
N
A
Molar
mass
Avogadro’s
number
13
How many moles of iron are present in 25.0 g of iron?
Example 1:
3 significant figures
55.85
1
Molar mass
0.448
14
What is the mass of 5.00 mol of water?
Example 2:
3 significant figures
1
18.02
90.1
Molar mass
15
How many Mg atoms are present in 5.00 g of Mg?
Example 3:
1
24.30
mass
mol
atoms
Avogadro’s
number
1
6.02 x 10
23
1.24x10
23
atoms Mg
Molar mass
3 significant figures
16
How many molecules of HCl are present in 25.0 g
of HCl?
Example 4:
1
36.46
mass
mol
molecules
1
6.02 x 10
23
4.13 x 10
23
molecules HCl
3 significant figures
17
MOLES OF ELEMENTS
IN A FORMULA
The subscripts in a chemical formula of a
compound indicate the number of atoms of
each type of element.
For example, 1 molecule of aspirin contains:
C
9
H
8
O
4
18
MOLES OF ELEMENTS
IN A FORMULA
The subscripts also indicate the number of
moles of each element in one mole of the
compound.
For example, 1 mole of aspirin contains:
C
9
H
8
O
4
19
MOLES OF ELEMENTS
IN A FORMULA
Using the subscripts from the aspirin formula,
one can write the following conversion factors
for each of the elements in 1 mole of aspirin.
C
9
H
8
O
4
20
Determine the number of moles of C atoms in 1
mole of each of the following substances:
Example 1:
a) Acetoaminophen used in Tylenol
C
8
H
9
NO
2
b) Zinc dietary supplement
Zn(C
2
H
3
O
2
)
2
21
How many carbon atoms are present in 1.50 moles
of aspirin, C
9
H
8
O
4
?
Example 2:
1.50 mol C
9
H
8
O
4
x
x
= 8.13x10
24
atoms
22
SUMMARY OF
MASS-MOLE CALCULATIONS
23
STOICHIOMETRY
Stoichiometry is the quantitative relationship
between the reactants and products in a
balanced chemical equation.
A balanced chemical equation provides several
important information about the reactants and
products in a chemical reaction.
24
MOLAR
RATIOS
For example:
1 N
2
(g) + 3 H
2
(g)
2 NH
3
(g)
1 molecule
3 molecules
2 molecules
100 molecules
300 molecules
200 molecules
10
6
molecules
3x10
6
molecules
2x10
6
molecules
1 mole
3 moles
2 moles
This is the molar
ratios between
the reactants and
products
25
Example 1:
Determine each mole ratio below based on the
reaction shown:
2 C
4
H
10
+ 13 O
2
8 CO
2
+ 10 H
2
O
26
STOICHIOMETRIC
CALCULATIONS
Stoichiometric calculations can be classified as
one of the following:
MOLES of
compound A
MOLES of
compound B
molar ratio
Mole-mole
calculations
MASS of
compound A
MM
Mass-mole
calculations
MASS of
compound B
MM
Mass-mass
calculations
27
MOLE-MOLE
CALCULATIONS
Relates moles of reactants and products in a
balanced chemical equation
MOLES of
compound A
MOLES of
compound B
molar ratio
28
How many moles of nitrogen will react with 2.4
moles of hydrogen to produce ammonia as shown in
the reaction below?
Example 1:
1 N
2
(g) + 3 H
2
(g)
2 NH
3
(g)
2.4 mol H
2
1
3
= 0.80 mol N
2
Mole ratio
29
How many moles of ammonia can be produced from
32 moles of hydrogen? (Assume excess N
2
present)
Example 2:
1 N
2
(g) + 3 H
2
(g)
2 NH
3
(g)
32 mol H
2
2
3
= 21 mol NH
3
Mole ratio
30
In one experiment, 6.80 mol of ammonia are
prepared. How many moles of hydrogen were used
up in this experiment?
Example 3:
1 N
2
(g) + 3 H
2
(g)
2 NH
3
(g)
6.80 mol NH
3
3
2
= 10.2 mol H
2
Mole ratio
31
MASS-MOLE
CALCULATIONS
Relates moles and mass of reactants or products
in a balanced chemical equation
MOLES of
compound A
MOLES of
compound B
molar ratio
MASS of
compound A
MM
32
How many grams of ammonia can be produced
from the reaction of 1.8 moles of nitrogen with
excess hydrogen as shown below?
Example 1:
1 N
2
(g) + 3 H
2
(g)
2 NH
3
(g)
1.8 mol N
2
2
1
= 61 g NH
3
Mole ratio
1
17.04
Molar mass
33
How many moles of hydrogen gas are required to
produce 75.0 g of ammonia?
Example 2:
1 N
2
(g) + 3 H
2
(g)
2 NH
3
(g)
75.0 g NH
3
1
17.04
= 6.60 mol H
2
2
3
Molar mass
Mole ratio
34
How many moles of ammonia can be produced from
the reaction of 125 g of nitrogen?
Example 3:
1 N
2
(g) + 3 H
2
(g)
2 NH
3
(g)
125 g N
2
1
28.02
= 8.92 mol NH
3
1
2
Molar mass
Mole ratio
35
MASS -MASS
CALCULATIONS
Relates mass of
reactants and products
in a balanced chemical
equation
36
What mass of oxygen will be required to react
completely with 96.1 g of propane, C
3
H
8
, according
to the equation below?
Example 1:
1 C
3
H
8
(g) + 5 O
2
(g)
3 CO
2
(g) + 4 H
2
O (g)
Mass of
propane
Moles of
propane
Moles of
oxygen
Mass of
oxygen
37
Example 1:
1 C
3
H
8
(g) + 5 O
2
(g)
3 CO
2
(g) + 4 H
2
O (g)
96.1 g C
3
H
8
= 349 g O
2
Molar mass
Mole ratio
Molar mass
38
What mass of carbon dioxide will be produced from
the reaction of 175 g of propane, as shown?
Example 2:
1 C
3
H
8
(g) + 5 O
2
(g)
3 CO
2
(g) + 4 H
2
O (g)
Mass of
propane
Moles of
propane
Moles of
carbon
dioxide
Mass of
carbon dioxide
39
Example 2:
1 C
3
H
8
(g) + 5 O
2
(g)
3 CO
2
(g) + 4 H
2
O (g)
175 g C
3
H
8
= 524 g CO
2
Molar mass
Mole ratio
Molar mass
40
LIMITING
REACTANT
When 2 or more reactants are combined in non-
stoichiometric ratios, the amount of product
produced is limited by the reactant that is not in
excess.
This reactant is referred to as limiting reactant.
When doing stoichiometric problems of this
type, the limiting reactant must be determined
first before proceeding with the calculations.
41
LIMITING REACTANT
ANALOGY
Consider the following recipe for a sundae:
42
LIMITING REACTANT
ANALOGY
How many sundaes can be prepared from the
following ingredients:
The number of sundaes possible is limited by the
amount of syrup, the limiting reactant.
43
GUIDE TO
LIMITING REACTANT CALC.
Assume reactant 1
is limiting
Assume reactant 2
is limiting
C
ompare
44
How many moles of H
2
O can be produced by
reacting 4.0 mol of hydrogen and 3.0 mol of oxygen
gases as shown below:
Example 1:
2 H
2
(g) + 1 O
2
(g)
2 H
2
O (g)
45
Example 1:
4.0 mol H
2
2
2
= 4.0 mol H
2
O
2 H
2
(g) + 1 O
2
(g)
2 H
2
O (g)
Assume H
2
is LR
Assume O
2
is LR
3.0 mol O
2
2
1
= 6.0 mol H
2
O
Correct
answer
46
A fuel mixture used in the early days of rocketry
was a mixture of N
2
H
4
and N
2
O
4
, as shown below.
How many grams of N
2
gas is produced when 100. g
of N
2
H
4
and 200. g of N
2
O
4
are mixed?
Example 2:
2 N
2
H
4
(l) + 1 N
2
O
4
(l)
3 N
2
(g) + 4 H
2
O (g)
47
Example 2:
Assumes
N
2
H
4
is LR
Assumes
N
2
O
4
is LR
48
Example 2:
100. g N
2
H
4
32.06
4.68 mol N
2
2
3
2 N
2
H
4
(l) + 1 N
2
O
4
(l)
3 N
2
(g) + 4 H
2
O (g)
Assume N
2
H
4
is LR
1
49
Example 2:
200. g N
2
O
4
92.02
6.52 mol N
2
1
3
2 N
2
H
4
(l) + 1 N
2
O
4
(l)
3 N
2
(g) + 4 H
2
O (g)
Assume N
2
O
4
is LR
1
50
Example 2:
6.52 mol N
2
2 N
2
H
4
(l) + 1 N
2
O
4
(l)
3 N
2
(g) + 4 H
2
O (g)
Assume N
2
O
4
is LR
Assume N
2
H
4
is LR
4.68 mol N
2
Correct
amount
N
2
H
4
is
LR
51
Example 2:
4.68 mol N
2
28.02
= 131 g N
2
2 N
2
H
4
(l) + 1 N
2
O
4
(l)
3 N
2
(g) + 4 H
2
O (g)
Calculate mass of N
2
1
52
How many moles of Fe
3
O
4
can be produced by
reacting 16.8 g of Fe with 10.0 g of H
2
O as shown
below:
Example 3:
3 Fe (s) + 4 H
2
O (l)
Fe
3
O
4
(s) + 4 H
2
(g)
53
Example 3:
16.8 g Fe
0.100 mol Fe
3
O
4
Assume Fe is LR
3 Fe (s) + 4 H
2
O (l)
Fe
3
O
4
(s) + 4 H
2
(g)
54
Example 3:
10.0 g H
2
O
0.139 mol Fe
3
O
4
Assume H
2
O is LR
3 Fe (s) + 4 H
2
O (l)
Fe
3
O
4
(s) + 4 H
2
(g)
55
Example 3:
0.139 mol Fe
3
O
4
Assume H
2
O is LR
Assume Fe is LR
0.100 mol Fe
3
O
4
Correct
amount
Fe is
LR
3 Fe (s) + 4 H
2
O (l)
Fe
3
O
4
(s) + 4 H
2
(g)
56
How many grams of AgBr can be produced when
50.0 g of MgBr
2
is mixed with 100.0 g of AgNO
3
, as
shown below:
Example 4:
MgBr
2
+ 2 AgNO
3
2 AgBr + Mg(NO
3
)
2
57
Example 4:
50.0 g MgBr
2
184.10
1
2
Assume MgBr
2
is LR
1
MgBr
2
+ 2 AgNO
3
2 AgBr + Mg(NO
3
)
2
187.77
1
102 g AgBr
58
Example 4:
100.0 g AgNO
3
169.88
2
2
Assume AgNO
3
is LR
1
MgBr
2
+ 2 AgNO
3
2 AgBr + Mg(NO
3
)
2
187.77
1
111 g AgBr
59
Example 4:
111 g AgBr
Assume AgNO
3
is LR
Assume MgBr
2
is LR
102 g AgBr
Correct
amount
MgBr
2
is LR
MgBr
2
+ 2 AgNO
3
2 AgBr + Mg(NO
3
)
2
60
PERCENT YIELD
The amount of product calculated through
stoichiometric ratios are the maximum amount
product that can be produced during the
reaction, and is thus called theoretical yield.
The actual yield of a product in a chemical
reaction is the actual amount obtained from the
reaction.
61
PERCENT YIELD
The percent yield of a reaction is obtained as
follows:
62
In an experiment forming ethanol, the theoretical
yield is 50.5 g and the actual yield is 46.8 g. What is
the percent yield for this reaction?
Example 1:
92.7 %
63
Silicon carbide can be formed from the reaction of
sand (SiO
2
) with carbon as shown below:
Example 2:
1 SiO
2
(s) + 3 C (s)
1 SiC (s) + 2 CO (g)
When 125 g of sand are processed, 68.4 of SiC is
produced. What is the percent yield of SiC in this
reaction?
64
Example 2:
125 g SiO
2
60.09
83.4 g SiC
1
1
Calculate theoretical yield
1
1 SiO
2
(s) + 3 C (s)
1 SiC (s) + 2 CO (g)
1
40.10
65
Example 2:
82.0 %
Calculate percent yield
66
In an experiment to prepare aspirin, the theoretical
yield is 153.7 g and the actual yield is 124.0 g. What
is the percent yield of this reaction?
Example 3:
80.68 %
67
Carbon tetrachloride (CCl
4
) was prepared by
reacting 100.0 g of Cl
2
with excess carbon disulfide
(CS
2
), as shown below. If 65.0 g was prepared in
this reaction, calculate the percent yield.
Example 4:
CS
2
+ 3 Cl
2
CCl
4
+ S
2
Cl
2
68
Example 4:
100.0 g Cl
2
70.90
3
1
Calculate theoretical yield
1
1
153.81
CS
2
+ 3 Cl
2
CCl
4
+ S
2
Cl
2
72.31 g
69
Example 4:
89.9 %
Calculate percent yield
70
ENERGY CHANGES
IN CHEMICAL REACTIONS
Heat is energy change that is lost or gained when a
chemical reaction takes place.
The system is the reactants and products that we
are observing. The surroundings are all the things
that contain and interact with the system, such as
the reaction flask, the laboratory room and the air
in the room.
The direction of heat flow depends whether the
products in a reaction have more or less energy
than the reactants.
71
ENERGY CHANGES
IN CHEMICAL REACTIONS
For a chemical reaction to occur, the molecules of the
reactants must collide with each other with the proper
energy and orientation.
The minimum amount of energy
required for a chemical reaction
to occur is called the activation
energy.
The heat of reaction is the
amount of heat absorbed or
released during a reaction and is
designated by the symbol
H.
72
ENERGY CHANGES
IN CHEMICAL REACTIONS
When heat is released
during a chemical reaction,
it is said to be exothermic.
For exothermic reactions,
H is negative and is
included on the right side of
the equation.
H=
22.0 kcal
73
ENERGY CHANGES
IN CHEMICAL REACTIONS
When heat is gained during
a chemical reaction, it is
said to be endothermic.
For endothermic reactions,
H is positive and is
included on the left side of
the equation.
H= +137 kcal
74
CALCULATING HEAT
IN A REACTION
The value of
H in a chemical reaction refers to the
heat lost or gained for the number of moles of
reactants and products in a balanced chemical
equation.
For example, based on the chemical equation shown
below:
H= +137 kcal
or
or
75
Given the reaction shown below, how much heat, in kJ,
is released when 50.0 g of NH
3
form?
Example 1:
H=
92.2 kJ
50.0 g NH
3
x
= 135 kJ
76
How many kJ of heat are needed to react 25.0 g of HgO
according to the reaction shown below:
Example 2:
H= 182 kJ
25.0 g HgO
x
= 10.5 kJ
77
THE END
Chemists use the mole concept to relate mass and the number of atoms in chemical reactions. Avogadro's number, molar mass, stoichiometry, and energy changes in reactions are key concepts explored in this chapter. The mole is a vital unit in chemistry, enabling scientists to quantify substances and make accurate calculations in various chemical processes.
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Presentation Transcript
Chapter 6B Chemical Quantities 1
CHAPTER OUTLINE The Mole Concept Molar Mass Calculations Using the Mole Stoichiometry & Molar Ratios Mole-Mole & Mass-Mole Calculations Mass-Mass Calculations Limiting Reactant Percent Yield Energy Changes in Chemical Reactions 2
THE MOLE CONCEPT Chemists find it more convenient to use mass relationships in the laboratory, while chemical reactions depend on the number of atoms present. In order to relate the mass and number of atoms, chemists use the SI unit mole (abbreviated mol). 3
THE MOLE CONCEPT The number of particles in a mole is called Avogadro s number and is 6.02x1023. 6.02 x 1023 equals to 1 mole Avogadro s number (NA) 4
THE MOLE CONCEPT A mole is a very large quantity 6.02x1023 If 10,000 people started to count Avogadro s number and counted at the rate of 100 numbers per minute each minute of the day, it would take over 1 trillion years to count the total number. 5
THE MOLE CONCEPT 1 mole H atoms = 6.02x1023 H atoms 1 mole H2 molecules = 6.02x1023 H2 molecules = 2 x (6.02x1023) H atoms 1 mole H2O molecules = 6.02x1023 H2O molecules = 2 x (6.02x1023) H atoms = 6.02x1023 O atoms 1 mole Na+ ions = 6.02x1023 Na+ ions 6
THE MOLE CONCEPT The atomic mass of one atom expressed in amu is numerically the same as the mass of 1 mole of atoms of the element expressed in grams. Mass of 1 H atom = 1.008 amu Mass of 1 Mg atom = 24.31 amu Mass of 1 Cl atom = 35.45 amu Mass of 1 mol H atoms = 1.008 grams Mass of 1 mol Mg atoms = 24.31 g Mass of 1 mol Cl atoms = 35.45 g 7
MOLAR MASS The mass of one mole of a substance is called molar mass, and is measured in grams. Mass of one mole of H2O 2 mol H atoms = 2 (1.01 g) = 2.02 g 1 mol O atom = 1 (16.00 g) = 16.00 g Molar mass 18.02 g 8
MOLAR MASS Mass of one mole of Ca(OH)2 1 mol Ca atom = 1 (40.08 g) = 40.08 g 2 mol O atoms = 2 mol H atoms = 2 (16.00 g) = 32.00 g 2 (1.01 g) = 2.02 g Molar mass 74.10 g 9
Example 1: Calculate the molar mass of each compound shown below: Lithium carbonate Li2CO3 Li C O 2 x 6.94 = 13.88 g 1 x 12.01 = 12.01 g 3 x 16.00 = 48.00 g Molar mass = 73.89 g/mol 10
Example 2: Calculate the molar mass of each compound shown below: Salicylic acid C7 H6O3 C H O 7 x 12.01 = 84.07 g 6 x 1.01 3 x 16.00 = 48.00 g = 6.06 g Molar mass = 138.13 g/mol 11
CALCULATIONS USING THE MOLE Conversions between mass, mole and particles can be done using molar mass and Avogadro s number. Avogadro s number Molar mass Mass of a substance Moles of a substance Particles of a substance MM NA 12
Example 1: How many moles of iron are present in 25.0 g of iron? mole g 55.85 1 25.0 g Fe x mol Fe 0.448 = Molar mass 3 significant figures 13
Example 2: What is the mass of 5.00 mol of water? g mol= 1 18.02 5.00 mol H O x g H O 90.1 2 2 3 significant figures Molar mass 14
Example 3: How many Mg atoms are present in 5.00 g of Mg? mass mol atoms atoms mol 1 mol g 24.30 1 6.02 x 1023 5.00 g Mg x x = 1.24x1023 atoms Mg Avogadro s number Molar mass 3 significant figures 15
Example 4: How many molecules of HCl are present in 25.0 g of HCl? mass mol molecules molecules mol 1 mol g 36.46 1 6.02 x 1023 25.0 g HCl x x = 4.13 x 1023 molecules HCl 3 significant figures 16
MOLES OF ELEMENTS IN A FORMULA The subscripts in a chemical formula of a compound indicate the number of atoms of each type of element. For example, 1 molecule of aspirin contains: C9H8O4 8 H atoms 9 C atoms 4 O atoms 17
MOLES OF ELEMENTS IN A FORMULA The subscripts also indicate the number of moles of each element in one mole of the compound. For example, 1 mole of aspirin contains: C9H8O4 9 mole C atoms 8 mole H atoms 4 mole O atoms 18
MOLES OF ELEMENTS IN A FORMULA Using the subscripts from the aspirin formula, one can write the following conversion factors for each of the elements in 1 mole of aspirin. C9H8O4 9 moles C 1 mole C9H8O4 8 moles H 1 mole C9H8O4 4 moles O 1 mole C9H8O4 19
Example 1: Determine the number of moles of C atoms in 1 mole of each of the following substances: a) Acetoaminophen used in Tylenol C8 H9NO2 8 moles C 1 mole C8H9 NO2 b) Zinc dietary supplement Zn(C2 H3O2)2 4 moles C 1 mole Zn(C2 H3O2)2 20
Example 2: How many carbon atoms are present in 1.50 moles of aspirin, C9H8O4? 9 moles C 1 mole C9H8 O4 x6.02x1023 atoms 1 mole C 1.50 mol C9H8O4x = 8.13x1024 atoms 21
SUMMARY OF MASS-MOLE CALCULATIONS 22
STOICHIOMETRY Stoichiometry is the quantitative relationship between the reactants and products in a balanced chemical equation. A balanced chemical equation provides several important information about the reactants and products in a chemical reaction. 23
MOLAR RATIOS For example: 1 N2 (g) + 3 H2 (g) 2 NH3 (g) 3 molecules This is the molar ratios between the reactants and products 1 molecule 2 molecules 100 molecules 300 molecules 200 molecules 106 molecules 3x106 molecules 2x106 molecules 1 mole 3 moles 2 moles 24
Example 1: Determine each mole ratio below based on the reaction shown: 8 CO2 + 10 H2O 2 C4H10 + 13 O2 mol O mol CO 13 8 = 2 2 mol C H mol H O 2 10 = 4 10 2 25
STOICHIOMETRIC CALCULATIONS Stoichiometric calculations can be classified as one of the following: calculations Mass-mass Mass-mole calculations MASS of compound B MASS of compound A Mole-mole calculations MM MM MOLES of compound A MOLES of compound B molar ratio 26
MOLE-MOLE CALCULATIONS Relates moles of reactants and products in a balanced chemical equation MOLES of compound A MOLES of compound B molar ratio 27
Example 1: How many moles of nitrogen will react with 2.4 moles of hydrogen to produce ammonia as shown in the reaction below? 1 N2 (g) + 3 H2 (g) 2 NH3 (g) 1 mol N x mol H 3 2 2.4 mol H2 = 0.80 mol N2 2 Mole ratio 28
Example 2: How many moles of ammonia can be produced from 32 moles of hydrogen? (Assume excess N2 present) 1 N2 (g) + 3 H2 (g) 2 NH3 (g) mol NH mol H 3 2 x 3 32 mol H2 = 21 mol NH3 2 Mole ratio 29
Example 3: In one experiment, 6.80 mol of ammonia are prepared. How many moles of hydrogen were used up in this experiment? 1 N2 (g) + 3 H2 (g) 2 NH3 (g) mol H x mol NH 2 3 2 6.80 mol NH3 = 10.2 mol H2 3 Mole ratio 30
MASS-MOLE CALCULATIONS Relates moles and mass of reactants or products in a balanced chemical equation MASS of compound A MM MOLES of compound A MOLES of compound B molar ratio 31
Example 1: How many grams of ammonia can be produced from the reaction of 1.8 moles of nitrogen with excess hydrogen as shown below? 1 N2 (g) + 3 H2 (g) 2 NH3 (g) mol NH mol N 1 g NH x mol NH 1 17.04 2 1.8 mol N2 x 3 3 = 61 g NH3 2 3 Mole ratio Molar mass 32
Example 2: How many moles of hydrogen gas are required to produce 75.0 g of ammonia? 1 N2 (g) + 3 H2 (g) 2 NH3 (g) 1 3 mol H x mol NH 2 mol NH g NH 17.04 75.0 g NH3 x 2 3 = 6.60 mol H2 3 3 Molar mass Mole ratio 33
Example 3: How many moles of ammonia can be produced from the reaction of 125 g of nitrogen? 1 N2 (g) + 3 H2 (g) 2 NH3 (g) 1 2 mol NH mol N 1 mol N g N 28.02 125 g N2 x x 3 2 = 8.92 mol NH3 2 2 Molar mass Mole ratio 34
MASS -MASS CALCULATIONS Relates mass of reactants and products in a balanced chemical equation 35
Example 1: What mass of oxygen will be required to react completely with 96.1 g of propane, C3H8, according to the equation below? 1 C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g) Moles of oxygen Mass of propane Moles of propane Mass of oxygen 36
Example 1: 1 C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g) 1 5 mol C H x g C H 44.11 mol O x mol C H 1 g O x mol O 1 3 8 2 96.1 g C3H8 3 8 3 8 32.00 2 = 349 g O2 2 Molar mass Mole ratio Molar mass 37
Example 2: What mass of carbon dioxide will be produced from the reaction of 175 g of propane, as shown? 1 C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g) Moles of carbon dioxide Mass of propane Moles of propane Mass of carbon dioxide 38
Example 2: 1 C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g) 1 3 mol C H x g C H 44.11 mol CO x mol C H 1 3 8 2 175 g C3H8 3 8 3 8 g CO x mol CO 1 44.01 2 = 524 g CO2 2 Molar mass Molar mass Mole ratio 39
LIMITING REACTANT When 2 or more reactants are combined in non- stoichiometric ratios, the amount of product produced is limited by the reactant that is not in excess. This reactant is referred to as limiting reactant. When doing stoichiometric problems of this type, the limiting reactant must be determined first before proceeding with the calculations. 40
LIMITING REACTANT ANALOGY Consider the following recipe for a sundae: 41
LIMITING REACTANT ANALOGY How many sundaes can be prepared from the following ingredients: amount of syrup, the limiting reactant. The number of sundaes possible is limited by the Limiting reactant Excess reactants 42
GUIDE TO LIMITING REACTANT CALC. Assume reactant 1 is limiting Assume reactant 2 is limiting Compare 43
Example 1: How many moles of H2O can be produced by reacting 4.0 mol of hydrogen and 3.0 mol of oxygen gases as shown below: 2 H2 (g) + 1 O2 (g) 2 H2O (g) Mole-mole calculations Limiting reactant 44
Example 1: Correct answer 2 H2 (g) + 1 O2 (g) 2 H2O (g) Assume H2 is LR 2 mol H O mol H 2 4.0 mol H2 x 2 = 4.0 mol H2O 2 Correct assumption Assume O2 is LR 2 mol H O mol O 1 3.0 mol O2 = 6.0 mol H2O x 2 2 45
Example 2: A fuel mixture used in the early days of rocketry was a mixture of N2H4 and N2O4, as shown below. How many grams of N2 gas is produced when 100. g of N2H4 and 200. g of N2O4 are mixed? 2 N2H4 (l) + 1 N2O4 (l) 3 N2 (g) + 4 H2O (g) Limiting reactant Mass-mass calculations 46
Example 2: Assumes N2O4 is LR Assumes N2H4 is LR 47
Example 2: 2 N2H4 (l) + 1 N2O4 (l) 3 N2 (g) + 4 H2O (g) Assume N2H4 is LR 1 3 mol N H x g N H 32.06 mol N mol N H 2 x = 2 100. g N2H4 2 4 2 4 2 4 4.68 mol N2 48
Example 2: 2 N2H4 (l) + 1 N2O4 (l) 3 N2 (g) + 4 H2O (g) Assume N2O4 is LR 1 3 mol N mol N O 1 mol N O x g N O 92.02 x = 2 200. g N2O4 2 4 2 4 2 4 6.52 mol N2 49
Example 2: 2 N2H4 (l) + 1 N2O4 (l) 3 N2 (g) + 4 H2O (g) Assume N2H4 is LR 4.68 mol N2 Assume N2O4 is LR 6.52 mol N2 Correct amount N2H4 is LR 50
