Zumdahl10e_PPT_Ch03.pptx

Chapter 3
Stoichiometry
Copyright ©2018 Cengage Learning. All Rights Reserved.
Cover Page

Section 3.1
Counting by Weighing
Chemical Stoichiometry
 Study of the quantities of materials consumed
and produced in chemical reactions
 Requires understanding the concept of relative
atomic masses
Copyright © Cengage Learning. All rights reserved 2

Section 3.1
Counting by Weighing
Counting Objects by Their Mass
 Average mass of objects is required to count the
objects by weighing
 For purposes of counting, objects behave as though
they are all identical
 Samples of matter can contain huge numbers of
atoms
 Number of atoms in a sample can be determined by
finding its mass

Section 3.2
Atomic Masses
Modern System of Atomic Masses
 Instituted in 1961
 Standard - 12C
 12C is assigned a mass of exactly 12 atomic mass units
(u)
 Masses of all other atoms are given relative to this standard

Section 3.2
Atomic Masses
Mass Spectrometer
 Helps to accurately compare the masses of atoms

Section 3.2
Atomic Masses
Average Atomic Mass
 Known as atomic weight (as per IUPAC’s
declaration), which is dimensionless by custom
 IUPAC - International Union of Pure and Applied
Chemistry
 Since elements occur in nature as mixtures of
isotopes, atomic masses are usually average
values

Section 3.2
Atomic Masses
Average Atomic Mass of Carbon
 Natural carbon is a mixture of 12C, 13C, and 14C
 Atomic mass of carbon is an average value of these
three isotopes
 Composition of natural carbon:
 12C atoms (mass = 12 u) - 98.89%
 13C atoms (mass = 13.003355 u) - 1.11%

Section 3.2
Atomic Masses
Average Atomic Mass of Carbon (continued)
 Calculation of average atomic mass for natural
carbon
 For stoichiometric purposes, assume that carbon is
composed of only one type of atom with a mass of
12.01
     
98.89% of 12 u + 1.11% of 13.0034 u
= 0.9889 12 u + 0.0111 13.0034 u
= 12.01 u

Section 3.2
Atomic Masses
Exercise
 An element consists of:
 1.40% of an isotope with mass 203.973 u
 Calculate the average atomic mass, and identify the element
Mass = 207.2 u
The element is lead (Pb)

Section 3.2
Atomic Masses
Example 3.1 - Average Mass of an Element
 When a sample of natural copper is vaporized
and injected into a mass spectrometer, the
results shown in the graph are obtained
 Use these data to compute the
average mass of natural copper
 Mass values for 63Cu and 65Cu are
62.93 u and 64.93 u, respectively

Section 3.2
Atomic Masses
Example 3.1 - Solution
 Where are we going?
 To calculate the average mass of natural copper
 What do we know?
 63Cu mass = 62.93 u
 65Cu mass = 64.93 u
 How do we get there?
 As shown by the graph, of every 100 atoms of natural
copper, 69.09 are 63Cu and 30.91 are 65Cu

Section 3.2
Atomic Masses
Example 3.1 - Solution (continued 1)
 Thus, the mass of 100 atoms of natural copper is
 Average mass of a copper atom is
 This mass value is used in doing calculations involving the
reactions of copper
69.09 atoms
  u
62.93
atom
+ 30.91 atoms
 
 
 
  u
64.93
atom
= 6355 u
 
 
 
6355 u
= 63.55 u/atom
100 atoms

Section 3.3
The Mole
Mole (mol)
 Unit of measure established for use in counting
atoms
 Number equal to the number of carbon atoms in
exactly 12 grams of pure 12C
 Techniques such as mass spectrometry determine this
number to be 6.02214 ×1023
 Avogadro’s number: One mole of something consists of
6.022× 1023units of that substance

Section 3.3
The Mole
Figure 3.4 - One-Mole Samples of Several Elements

Section 3.3
The Mole
Table 3.1 - Comparison of 1 Mole Samples of Various
Elements

Section 3.3
The Mole
Using the Mole in Chemical Calculations
 12 g of 12C contains 6.022×1023 atoms, and
12.01-g sample of natural carbon also contains
6.022×1023 atoms
 Ratio of the masses of the samples (12 g/12.01 g)
is the same as the ratio of masses of individual
components (12 u/12.01 u)
 Both samples contain the same number of atoms
(6.022×1023 atoms)

Section 3.3
The Mole
Using the Mole in Chemical Calculations (continued 1)
 Mole is defined such that a sample of a natural
element with a mass equal to the element’s
atomic mass expressed in grams contains 1 mole
of atoms
 Since 6.022×1023 atoms of carbon (each with a
mass of 12 u) have a mass of 12 g, then
23
6.022 × 10 atoms
  12 u
atom
23
= 12 g
and 6.022 × 10 u = 1 g
 
 
 
Exact number

Section 3.3
The Mole
Using the Mole in Chemical Calculations (continued 2)
 Relationship can be used to derive the unit factor
required to convert between atomic mass units
and grams

Section 3.3
The Mole
Interactive Example 3.5 - Calculating the Number of
Moles and Mass
 Cobalt (Co) is a metal that is added to steel to
improve its resistance to corrosion
 Calculate both the number of moles in a sample of
cobalt containing 5.00 × 1020 atoms and the mass of
the sample

Section 3.3
The Mole
Interactive Example 3.5 - Solution
 To calculate the number of moles and the mass of a
sample of Co
 Sample contains 5.00 × 1020 atoms of Co

Section 3.3
The Mole
Interactive Example 3.5 - Solution (continued 1)
 Note that the sample of 5.00 × 1020 atoms of cobalt is
less than 1 mole (6.022 × 1023 atoms) of cobalt
 What fraction of a mole it represents can be determined as
follows:
20
5.00 × 10 atoms Co 23
1 mol Co
×
6.022 × 10 atoms Co
4
= 8.30 ×10 mol Co


Section 3.3
The Mole
 Since the mass of 1 mole of cobalt atoms is 58.93 g,
the mass of 5.00 × 1020 atoms can be determined as
follows:
 Reality check - Sample contains 5 × 1020 atoms,
which is approximately 1/1000 of a mole
 Sample should have a mass of about (1/1000)(58.93)
≅ 0.06, and the answer of ~ 0.05 makes sense
4
8.30 × 10 mol Co
 58.93 g Co
×
1 mol Co
2
= 4.89 × 10 g Co


Section 3.3
The Mole
Join In (4)
 Which of the following is closest to the average
mass of one atom of copper?
a. 63.55 g
b. 52.00 g
c. 58.93 g
d. 65.38 g
e. 1.055 × 10–22 g

Section 3.3
The Mole
Molecular Mass
 Molecular mass (or molecular weight) is the sum of the
atomic masses (in amu) in a molecule.
 SO2
2
1S 32.07amu
2O +2×16.00amu
SO 64.07amu
For any molecule
molecular mass (amu) − molar mass (grams)
1 molecule SO2 = 64.07 amu
1 mole SO2 = 64.07 g SO2
24

Section 3.3
The Mole
Example 3.6 1
 Methane (CH4) is the principal
component of natural gas.
 How many moles of CH4 are present
in 6.07 g of CH4?
 CH4
 Steve Allen/Stockbyte/Getty Images 25

Section 3.3
The Mole
Example 3.6 2
 Strategy
 We are given grams of CH4 and asked to solve for moles of
CH4.
 What conversion factor do we need to convert between grams
and moles?
 Arrange the appropriate conversion factor so that grams cancel
and the unit moles are obtained for your answer.
26

Section 3.3
The Mole
Example 3.6 3
 Solution
 The conversion factor needed to convert between grams and moles is the
molar mass. First we need to calculate the molar mass of CH4, following
the procedure in Example 3.5:
 molar mass of CH4 = 12.01 g + 4(1.008
g)
=16.04 g
Because
1 mol CH4 = 16.04 g CH4
the conversion factor we need should have grams in the denominator so that
the unit g will cancel, leaving the unit mol in the numerator:
4
4
1mol
16.04 g
CH
CH
27

Section 3.3
The Mole
Example 3.6 4
 We now write,
6.07 g CH 4
4
1mol
16.04 g
CH

CH
4
4
0.378 mol CH

 Thus, there is 0.378 mole of CH4 in 6.07 g of CH4.
Check
Should 6.07 of CH4 equal less than 1 mole of CH4?
What is the mass of 1 mole of CH4?
28

Section 3.3
The Mole
Example 3.7 1
 How many hydrogen atoms
are present in 25.6 g of urea
[(NH2)2CO], which is used
as a fertilizer, in animal feed,
and in the manufacture of
polymers?
 The molar mass of urea is
60.06 g.
urea
 Ken Karp/McGraw-Hill 29

Section 3.3
The Mole
Example 3.7 2
 Strategy
 We are asked to solve for atoms of hydrogen in 25.6 g of urea.
We cannot convert directly from grams of urea to atoms of
hydrogen.
 How should molar mass and Avogadro’s number be used in
this calculation?
 How many moles of H are in 1 mole of urea?
30

Section 3.3
The Mole
Example 3.7 3
Solution
To calculate the number of H atoms, we first must convert grams
of urea to moles of urea using the molar mass of urea. This part is
similar to Example 3.2.
The molecular formula of urea shows there are four moles of H
atoms in one mole of urea molecule, so the mole ratio is 4:1.
Finally, knowing the number of moles of H atoms, we can
calculate the number of H atoms using Avogadro’s number. We
need two conversion factors: molar mass and Avogadro’s number.
31

Section 3.3
The Mole
Example 3.7 4
 We can combine these conversions
 grams of urea → moles of area → moles of H → atoms of
H
into one step:
 
 
   
23
2 2
2 2
2 2
2 2
1mol 4 mol 6.022 10 atoms
25.6 g
60.06 g 1mol 1mol
NH CO H H
NH CO
NH CO NH CO H

   
25.6 g
  
2 2
NH CO
 
2 2
1mol NH

CO
60.06 g  
2 2
NH CO
4 mol

H
 
2 2
1mol NH CO
23
6.022 10 atoms
1 mol
H


H
24
1.03 10 atoms
H
 
Check
Does the answer look reasonable?
How many atoms of H would 60.06 g of urea contain?
32

Section 3.3
The Mole
Formula Mass
 Formula mass is the sum of the atomic masses (in amu) in a
formula unit of an ionic compound.
1 Na 22.99 amu
NaCl 1Cl 35.45 amu
NaCl 58.44 amu

For any ionic compound
formula mass (amu) = molar mass (grams)
1 formula unit NaCl = 58.44 amu
1 mole NaCl = 58.44 g NaCl
33

Section 3.3
The Mole
Percent Composition
Percent composition of an element in a compound
molar mass of element
100%
molar mass of compound
n
 
n is the number of moles of the element in 1 mole of the
compound
C2H6O
 
2 12.01g
% 100% 52.14%
46.07 g
C

  
 
6 1.008 g
% 100% 13.13%
46.07 g
H

  
 
1 16.00 g
%0 100% 34.73%
46.07 g

  
52.14% 13.13% 34.73% 100.0%
  
34

Section 3.3
The Mole
Example 3.8 1
 Phosphoric acid (H3PO4) is a
colorless, syrupy liquid used in
detergents, fertilizers, toothpastes,
and in carbonated beverages for a
“tangy” flavor.
 Calculate the percent composition
by mass of H, P, and O in this
compound.  H3PO
4
35

Section 3.3
The Mole
Example 3.8 2
 Strategy
 Recall the procedure for calculating a percentage. Assume that
we have 1 mole of H3PO4.
 The percent by mass of each element (H, P, and O) is given by
the combined molar mass of the atoms.
 Of the element in 1 mole of H3PO4 divided by the molar mass
of H3PO4, then multiplied by 100 percent.
36

Section 3.3
The Mole
Example 3.8 3
 Solution
 The molar mass of H3PO4 is 97.99 g. The percent by mass of each
of the elements in H3PO4 is calculated as follows:
 
3 4
3 1.008 g
% 100% 3.086
97.99 g
H
H
H PO
  
3 4
30.97 g
% 100% 31.61%
97.99 g
P
P
H PO
  
 
3 4
4 16.00 g
% 100% 65.31%
97.99 g
O
O
H PO
  
Check
Do the percentages add to 100 percent? The sum of the percentages is
(3.086% + 31.61% + 65.31%) = 100.01%. The small discrepancy from
100 percent is due to the way we rounded off.
37

Section 3.7
Determining the Formula of a Compound
Determining the Formula of a Compound - Process
 Weigh the sample of the compound by:
 Decomposing the sample into its constituent elements
 Reacting it with oxygen
 Resulting substances are then collected and
weighed
 This type of analysis is done by a combustion device
 Results of such analyses provide the mass of each type
of element in the compound, which can be used to
determine the mass percent of each element

Section 3.7
Combustion Device
 Used to analyze substances for hydrogen and
carbon
 Helps determine the mass percent of each
element in a compound

Section 3.7
Empirical Formula
 Any molecule that can be represented as (CH5N)n
has the empirical formula CH5N
 n - Integer
 Molecular formula: Exact formula of the molecules
present in a substance
 Molecular formula = (empirical formula)n
 Requires the knowledge of the molar mass

Section 3.3
The Mole
Percent Composition and Empirical
Formulas
 Access the text alternative for slide
images.
41

Section 3.3
The Mole
Example 3.9 1
 Ascorbic acid (vitamin C)
cures scurvy.
 It is composed of 40.92
percent carbon (C), 4.58
percent hydrogen (H), and
54.50 percent oxygen (O) by
mass.
 Determine its empirical
formula.
42

Section 3.3
The Mole
Example 3.9 2
 Strategy
 In a chemical formula, the subscripts represent the ratio of the
number of moles of each element that combine to form one
mole of the compound.
 How can we convert from mass percent to moles?
 If we assume an exactly 100-g sample of the compound, do we
know the mass of each element in the compound?
 How do we then convert from grams to moles?
43

Section 3.3
The Mole
Example 3.9 3
Solution
If we have 100 g of ascorbic acid, then each percentage can be converted
directly to grams. In this sample, there will be 40.92 g of C, 4.58 g of H, and
54.50 g of O. Because the subscripts in the formula represent a mole ratio, we
need to convert the grams of each element to moles. The conversion factor
needed is the molar mass of each element. Let n represent the number of moles
of each element so that.
40.92 g
C
n  C
1mol
12.01 g
C

C
3.407 mol C

40.92 g
H
n  H
1mol
1.008 g
H

H
4.54 mol H

54.50 g
O
n  O
1mol O

16.00 g O
3.406 mol O

Thus, we arrive at the formula C3.407H4.54O3.406,
44

Section 3.3
The Mole
Example 3.9 4
 Thus we arrive at the formula C3.407H4.54O3.406,
 which gives the identity and the mole ratios of atoms present.
However, chemical formulas are written with whole numbers.
Try to convert to whole numbers by dividing all the subscripts by
the smallest subscript (3.406):
3.407 4.54 3.406
C: 1H : 1.33 O : 1
3.406 3.406 3.406
  
 where
the
  sign means “approximately equal to.”
 This gives CH1.33O as the formula for ascorbic acid. Next, we need to
convert 1.33, the subscript for H, into an integer.
45

Section 3.3
The Mole
 This can e done by a trial-and-error procedure:
46

Section 3.7
Problem-Solving Strategy - Empirical Formula
Determination
 Base the calculation on 100 g of compound as
mass percentage gives the number of grams of a
particular element per 100 g of compound
 Each percent will then represent the mass in grams of
that element
 Determine the number of moles of each element
present in 100 g of compound using the atomic
masses of the elements present

Section 3.7
Problem-Solving Strategy - Empirical Formula
Determination (continued)
 Divide each value of the number of moles by the
smallest of the values
 If each resulting number is a whole number (after
appropriate rounding), these numbers represent the
subscripts of the elements in the empirical formula
 If the numbers obtained are not whole numbers,
multiply each number by an integer so that the results
are all whole numbers

Section 3.7
Critical Thinking
 One part of the problem-solving strategy for
empirical formula determination is to base the
calculation on 100 g of compound
 What if you chose a mass other than 100 g?
 Would this work?
 What if you chose to base the calculation on 100
moles of compound?
 Would this work?

Section 3.7
Problem-Solving Strategy - Determining Molecular
Formula from Empirical Formula
 Obtain the empirical formula
 Compute the mass corresponding to the empirical
formula
 Calculate the ratio:
 Integer from the previous step represents the
number of empirical formula units in one
molecule
Molar mass
Empirical formula mass

Section 3.7
Formula from Empirical Formula (continued)
 When the empirical formula subscripts are
multiplied by this integer, the molecular formula
results
 This procedure is summarized by the equation:
molar mass
Molecular formula empirical formula
empirical formula mass
 

Section 3.7
Interactive Example 3.11 - Determining Empirical and
Molecular Formulas II
 A white powder is analyzed and found to contain
43.64% phosphorus and 56.36% oxygen by mass
 Compound has a molar mass of 283.88 g/mol
 What are the compound’s empirical and molecular
formulas?

Section 3.7
 To find the empirical and molecular formulas for the
given compound
 Percent of each element
 Molar mass of the compound is 283.88 g/mol

Section 3.7
 What information do we need to find the empirical
formula?
 Mass of each element in 100.00 g of compound
 Moles of each element
 What is the mass of each element in 100.00 g of
compound?
 Mass of P = 43.64 g
 Mass of O = 56.36 g

Section 3.7
 What are the moles of each element in 100.00 g of
compound?
 What is the empirical formula for the compound?
 Dividing each mole value by the smaller one gives
43.64 g P
1 mol P
×
30.97 g P
= 1.409 mol P
56.36 g O
1 mol O
×
16.00 g O
= 3.523 mol O
1.409 3.523
= 1 P and = 2.5 O
1.409 1.409

Section 3.7
 This yields the formula PO2.5
 Since compounds must contain whole numbers of atoms,
the empirical formula should contain only whole numbers
 To obtain the simplest set of whole numbers, we multiply
both numbers by 2 to give the empirical formula P2O5

Section 3.7
 What is the molecular formula for the compound?
 Compare the empirical formula mass to the molar mass
 Molecular formula is (P2O5)2, or P4O10
Empirical formula mass = 141.94 g/mol
Given molar mass = 283.88 g/mol
Molar mass 283.88
= = 2
Empirical formula mass 141.94

Section 3.7
 Note
 Structural formula for this interesting compound is
given below

Section 3.7
Exercise
 A compound contains 47.08% carbon, 6.59%
hydrogen, and 46.33% chlorine by mass
 Molar mass of the compound is 153 g/mol
 What are the empirical and molecular formulas of the
compound?
Empirical formula - C3H5Cl
Molecular formula - C6H10Cl2

Section 3.7
Formula from Mass Percent and Molar Mass
 Using the mass percentages and the molar mass,
determine the mass of each element present in 1
mole of compound
 Determine the number of moles of each element
present in 1 mole of compound
 Integers in this step represent the subscripts in the
molecular formula

Section 3.7
Interactive Example 3.12 - Determining a Molecular
Formula
 Caffeine, a stimulant found in coffee, tea, and
chocolate, contains 49.48% carbon, 5.15%
hydrogen, 28.87% nitrogen, and 16.49% oxygen
by mass and has a molar mass of 194.2 g/mol
 Determine the molecular formula of caffeine

Section 3.7
 To find the molecular formula for caffeine
 Percent of each element - 49.48% C, 28.87% N, 5.15% H, and
16.49% O
 Molar mass of caffeine is 194.2 g/mol
 What information do we need to find the molecular
formula?
 Mass of each element (in 1 mole of caffeine)
 Mole of each element (in 1 mole of caffeine)

Section 3.7
 What is the mass of each element in 1 mole (194.2 g)
of caffeine?
49.48 g C
100.0 g
194.2 g
×
caffeine
96.09 g C
=
mol mol caffeine
5.15 g H
100.0 g
194.2 g
×
caffeine
10.0 g H
=
mol mol caffeine

Section 3.7
28.87 g N
100.0 g
194.2 g
×
caffeine
56.07 g N
=
mol mol caffeine
16.49 g O
100.0 g
194.2 g
×
caffeine
32.02 g O
=
mol mol caffeine
 What are the moles of each element in 1 mole of
caffeine?
96.09 g C
Carbon:
1 mol C
×
mol caffeine 12.01 g C
8.001 mol C
=
mol caffeine
10.0 g H
Hydrogen:
1 mol H
×
mol caffeine 1.008 g H
9.92 mol H
=
mol caffeine

Section 3.7
56.07 g N
Nitrogen:
1 mol N
×
mol caffeine 14.01 g N
4.002 mol N
=
mol caffeine
32.02 g O
Oxygen:
1 mol O
×
mol caffeine 16.00 g O
2.001 mol O
=
mol caffeine
 Rounding the numbers to integers gives the molecular
formula for caffeine: C8H10N4O2

Section 3.7
Join In (10)
 Empirical formula of styrene is CH; its molar mass
is 104.1 g/mol
 What is the molecular formula of styrene?
a. C2H4
b. C8H8
c. C10H10
d. C6H6

Section 3.8
Chemical Equations
Chemical Change
 Involves the reorganization of atoms in one or
more substances
 Atoms are neither created nor destroyed
 Represented by a chemical equation
 Reactants: Presented on the left side of an arrow
 Products: Presented on the right side of the arrow
4 2 2 2
CH + O CO + H O


Reactants Products

Section 3.8
Chemical Equations
Balancing a Chemical Equation
 All atoms present in the reactants must be
accounted for among the products
 Same number of each type of atom should be present
on the products side and on the reactants side of the
arrow

Section 3.8
Chemical Equations
Balancing a Chemical Equation (continued)
 Example
4 2 2 2
Unbalanced equation: CH + O CO + H O


4 2 2 2
Balanced equation: CH + 2O CO + 2H O



Section 3.8
Chemical Equations
Information Provided by Chemical Equations
 Nature of the reactants and products
 Compounds involved in the reaction
 Physical states of the reactants and products

Section 3.8
Chemical Equations
Information Provided by Chemical Equations (continued)
 Relative numbers of reactants and products
 Indicated by coefficients in a balanced equation
 Mass remains constant
 Atoms are conserved in a chemical reaction

Section 3.9
Balancing Chemical Equations
Things to Remember
 An unbalanced equation is of limited use
 Identities of the reactants and products of a
reaction must be recognized by experimental
observation
 Formulas of compounds must never be changed
while balancing a chemical equation
 Subscripts in a formula cannot be changed, nor can
atoms be added or subtracted from a formula

Section 3.9
Critical Thinking
 What if a friend was balancing chemical equations
by changing the values of the subscripts instead
of using the coefficients?
 How would you explain to your friend that this was
the wrong thing to do?

Section 3.9
Problem-Solving Strategy - Writing and Balancing the
Equation for a Chemical Reaction
1. Determine what reaction is occurring
 Determine the reactants, the products, and the
physical states involved
2. Write the unbalanced equation that summarizes
the reaction

Section 3.9
Problem-Solving Strategy - Writing and Balancing the
Equation for a Chemical Reaction (continued)
3. Balance the equation by inspection, starting with
the most complicated molecule(s)
 Determine what coefficients are necessary
 Same number of each type of atom needs to appear on both
reactant and product sides
 Do not change the formulas of any of the reactants or
products

Section 3.9
Critical Thinking
 One part of the problem-solving strategy for
balancing chemical equations is starting with the
most complicated molecule
 What if you started with a different molecule?
 Could you still eventually balance the chemical equation?
 How would this approach be different from the suggested
technique?

Section 3.9
Interactive Example 3.14 - Balancing a Chemical
Equation II
 At 1000°C, ammonia gas, NH3(g), reacts with
oxygen gas to form gaseous nitric oxide, NO(g),
and water vapor
 This reaction is the first step in the commercial
production of nitric acid by the Ostwald process
 Balance the equation for this reaction

Section 3.9
 Steps 1 and 2
 Unbalanced equation for the reaction is
 Step 3
 Because all the molecules in this equation are of about
equal complexity, where we start in balancing it is
rather arbitrary
       
3 2 2
NH + O NO + H O
g g g g


Section 3.9
 Let’s begin by balancing the hydrogen
 A coefficient of 2 for NH3 and a coefficient of 3 for H2O give
six atoms of hydrogen on both sides
 Nitrogen can be balanced with a coefficient of 2 for
NO
 Finally, note that there are two atoms of oxygen on
the left and five on the right
       
3 2 2
2NH + O NO + 3H O
g g g g

       
3 2 2
2NH + O 2NO + 3H O
g g g g


Section 3.9
 Oxygen can be balanced with a coefficient of for O2
 Usual custom is to have whole-number coefficients
 We simply multiply the entire equation by 2
 Reality check - There are 4 N, 12 H, and 10 O on
both sides, so the equation is balanced
5
2
       
3 2 2
5
2NH + O 2NO + 3H O
2
g g g g

       
3 2 2
4NH + 5O 4NO + 6H O
g g g g


Section 3.9
 Visual representation of the balanced reaction

Section 3.9
Exercise
 Balance the following equations:
a.
b.
       
2 3 4 2 3 4 2
Ca(OH) + H PO H O + Ca (PO )
aq aq l s

       
2 3 4 2 3 4 2
3Ca(OH) + 2H PO 6H O + Ca (PO )

aq aq l s
       
3 2 4 2 4 3
AgNO + H SO Ag SO + HNO

aq aq s aq
       
3 2 4 2 4 3
2AgNO + H SO Ag SO + 2HNO

aq aq s aq

Section 3.9
Join In (11)
 When the equation NH3 + O2  NO + H2O is
balanced with the smallest set of integers, the
sum of the coefficients is
a. 4
b. 12
c. 14
d. 19
e. 24

Section 3.9
Join In (12)
 How many of the following correctly balance this
chemical equation:
CaO + C  CaC2 + CO2
i. CaO2 + 3C  CaC2 + CO2
ii. 2CaO + 5C  2CaC2 + CO2
iii. CaO + (5/2)C  CaC2 + (1/2)CO2
iv. 4CaO + 10C  4CaC2 + 2CO2

Section 3.9
Join In (12) (continued)
a. 0
b. 1
c. 2
d. 3
e. 4

Section 3.9
Join In (13)
 How many of the following statements are true
concerning balanced chemical equations?
I. Number of molecules is conserved
II. Coefficients indicate mass ratios of the substances
involved
III. Atoms are neither created nor destroyed
IV. Sum of the coefficients on the left side equals the
sum of the coefficients on the right side

Section 3.9
a. 0
b. 1
c. 2
d. 3
e. 4

Section 3.10
Stoichiometric Calculations: Amounts of
Reactants and Products
Problem-Solving Strategy - Calculating Masses of
Reactants and Products in Reactions
 Balance the equation for the reaction
 Convert the known mass of the reactant or
product to moles of that substance
 Use the balanced equation to set up the
appropriate mole ratios
 Use the appropriate mole ratios to calculate the
number of moles of the desired reactant or
product

Section 3.10
Problem-Solving Strategy - Calculating Masses of
Reactants and Products in Reactions (continued)
 Convert from moles back to grams if required by
the problem

Section 3.10
Critical Thinking
 Your lab partner observed that you always take
the mass of chemicals in lab, but then you use
mole ratios to balance the equation
 “Why not use the masses in the equation?” your
partner asks
 What if your lab partner decided to balance equations
by using masses as coefficients?
 Is this even possible?
 Why or why not?

Section 3.10
Interactive Example 3.15 - Chemical Stoichiometry I
 Solid lithium hydroxide is used in space vehicles
to remove exhaled carbon dioxide from the living
environment by forming solid lithium carbonate
and liquid water
 What mass of gaseous carbon dioxide can be
absorbed by 1.00 kg of lithium hydroxide?

Section 3.10
 To find the mass of CO2 absorbed by 1.00 kg LiOH
 Chemical reaction -
 1.00 kg LiOH
 What information is needed to find the mass of CO2?
 Balanced equation for the reaction
       
2 2 3 2
LiOH + CO Li CO + H O
s g s l


Section 3.10
1. What is the balanced equation?
2. What are the moles of LiOH?
 To find the moles of LiOH, we need to know the molar
mass
       
2 2 3 2
2LiOH + CO Li CO + H O
s g s l

Molar mass of LiOH = 6.941 + 16.00 + 1.008
= 23.95 g/mol

Section 3.10
 Now we use the molar mass to find the moles of LiOH
3. What is the mole ratio between CO2 and LiOH in the
balanced equation?
1.00 kg LiOH
1000 g LiOH
×
1 kg LiOH
1 mol LiOH
×
23.95 g LiOH
= 41.8 mol LiOH
2
1 mol CO
2 mol LiOH

Section 3.10
4. What are the moles of CO2?
5. What is the mass of CO2 formed from 1.00 kg LiOH?
Thus, 920 g of CO2(g) will be absorbed by 1.00 kg of LiOH(s)
41.8 mol LiOH 2
1 mol CO
×
2 mol LiOH
2
= 20.9 mol CO
2
20.9 mol CO 2
2
44.0 g CO
×
1mol CO
2
2
= 9.20 10 g CO


Section 3.10
Exercise
 Over the years, the thermite reaction has been
used for welding railroad rails, in incendiary
bombs, and to ignite solid-fuel rocket motors
 Reaction is as follows:
       
2 3 2 3
Fe O + 2Al 2Fe + Al O
s s l s


Section 3.10
Exercise (continued)
 What masses of iron(III) oxide and aluminum must be used
to produce 15.0 g iron?
 What is the maximum mass of aluminum oxide that could be
produced?
Mass of iron (III) oxide = 21.5 g
Mass of Aluminum = 7.26 g
13.7 g Al2O3

Section 3.11
The Concept of Limiting Reactant
Stoichiometric Mixture
 A type of mixture that contains relative amounts
of reactants that match the numbers in the
balanced equation
     
2 2 3
3H + N 2NH
g g g


Section 3.11
When Hydrogen Is the Limiting Reactant

Section 3.11
Limiting Reactant
 One that runs out first and thus limits the amount
of product that can form
 To determine how much product can be formed from
a given mixture of reactants, one must look for the
reactant that is limiting
 Some mixtures can be stoichiometric
 All reactants run out at the same time
 Requires determining which reactant is limiting

Section 3.11
Determination of the Limiting Reactant Using
Reactant Quantities
 Compare the moles of reactants to ascertain
which runs out first
 Use moles of molecules instead of individual
molecules
 Use the balanced equation to determine the limiting
reactant
 Determine the amount of limiting reactant formed
 Use the amount of limiting reactant formed to
compute the quantity of the product

Section 3.11
Determination of Limiting Reactant Using Quantities of
Products Formed
 Consider the amounts of products that can be
formed by completely consuming each reactant
 Reactant that produces the smallest amount of
product must run out first and thus be the limiting
reactant

Section 3.11
Interactive Example 3.17 - Stoichiometry: Limiting
Reactant
 Nitrogen gas can be prepared by passing gaseous
ammonia over solid copper(II) oxide at high
temperatures
 Other products of the reaction are solid copper and
water vapor
 If a sample containing 18.1 g of NH3 is reacted with
90.4 g of CuO, which is the limiting reactant?
 How many grams of N2 will be formed?

Section 3.11
 To find the limiting reactant
 To find the mass of N2 produced
 Chemical reaction
 18.1 g NH3
 90.4 g CuO
         
3 2 2
NH + CuO N + Cu + H O
g s g s g


Section 3.11
 What information do we need?
 Moles of NH3
 Moles of CuO
 To find the limiting reactant, determine the balanced
equation
         
3 2 2
2NH + 3CuO N + 3Cu + 3H O
g s g s g


Section 3.11
 What are the moles of NH3 and CuO?
 To find the moles, we need to know the molar masses
NH3 17.03 g/mol
CuO 79.55 g/mol
3
18.1 g NH 3
3
1 mol NH
×
17.03 g NH
3
= 1.06 mol NH
90.4 g CuO
1 mol CuO
×
79.55 g CuO
= 1.14 mol CuO

Section 3.11
A. First we will determine the limiting reactant by
comparing the moles of reactants to see which one is
consumed first
 What is the mole ratio between NH3 and CuO in the
balanced equation?
3
3 mol CuO
2 mol NH

Section 3.11
 How many moles of CuO are required to react with 1.06
moles of NH3?
 Thus 1.59 moles of CuO are required to react with 1.06
moles of NH3
 Since only 1.14 moles of CuO are actually present, the
amount of CuO is limiting; CuO will run out before NH3 does
3
3
3 mol CuO
1.06 mol NH × = 1.59 mol CuO
2 mol NH

Section 3.11
2NH3 + 3CuO → Products
Before 1.06 1.14
Change –1.06 –1.59
 There is not enough CuO to react with all of the NH3, so
CuO is the limiting reactant

Section 3.11
B. Alternatively we can determine the limiting reactant
by computing the moles of N2 that would be formed
by complete consumption of NH3 and CuO
3
1.06 mol NH 2
3
1 mol N
×
2 mol NH
2
= 0.530 mol N
1.14 mol CuO 2
1 mol N
×
3mol CuO
2
= 0.380 mol N

Section 3.11
 As before, we see that the CuO is limiting since it
produces the smaller amount of N2
 To find the mass of N2 produced, determine the moles
of N2 formed
 Because CuO is the limiting reactant, we must use the
amount of CuO to calculate the amount of N2 formed

Section 3.11
 What is the mole ratio between N2 and CuO in the
balanced equation?
 What are the moles of N2?
2
1 mol N
3 mol CuO
1.14 mol CuO 2
1 mol N
×
3 mol CuO
2
= 0.380 mol N

Section 3.11
 What mass of N2 is produced?
 Using the molar mass of N2 (28.02 g/mol), we can calculate
the mass of N2 produced
2
0.380 mol N 2
2
28.02 g N
×
1 mol N
2
= 10.6 g N

Section 3.11
Concept of Yield
 Theoretical yield: Amount of product formed
when the limiting reactant is entirely consumed
 Amount of product predicted is rarely obtained
because of side reactions and other complications
 Percent yield: Actual yield of product
 Often provided as a percentage of the theoretical yield
Actual yield
× 100% = percent yield
Theoretical yield

Section 3.11
Interactive Example 3.18 - Calculating Percent Yield
 Methanol (CH3OH), also called methyl alcohol, is
the simplest alcohol
 It is used as a fuel in race cars and is a potential
replacement for gasoline
 Methanol can be manufactured by combining gaseous
carbon monoxide and hydrogen

Section 3.11
Interactive Example 3.18 - Calculating Percent Yield
(continued)
 Suppose 68.5 kg CO(g) is reacted with 8.60 kg
H2(g)
 Calculate the theoretical yield of methanol
 If 3.57 ×104 g CH3OH is actually produced, what is the
percent yield of methanol?

Section 3.11
 To calculate the theoretical yield of methanol
 To calculate the percent yield of methanol
 Chemical reaction
 68.5 kg CO(g) and 8.60 kg H2 (g)
 3.57×104 g CH3OH is produced
     
2 3
H + CO CH OH
g g l


Section 3.11
 What information do we need?
 Moles of H2
 Moles of CO
 Which reactant is limiting
 Amount of CH3OH produced

Section 3.11
 To find the limiting reactant, balance the chemical
equation
 What are the moles of H2 and CO?
     
2 3
2H + CO CH OH
g g l


Section 3.11
 To find the moles, we need to know the molar masses
H2 2.016 g/mol
CO 28.02 g/mol
68.5 kg CO
1000 g CO
×
1 kg CO
1 mol CO
×
28.02 g CO
3
= 2.44 × 10 mol CO
2
8.60 kg H
2
1000 g H
×
2
1 kg H
2
2
1 mol H
×
2.016 g H
3
2
= 4.27 ×10 mol H

Section 3.11
 We can determine the limiting reactant by calculating
the amounts of CH3OH formed by complete
consumption of CO(g) and H2(g):
 Since complete consumption of the H2 produces the smaller
amount of CH3OH, the H2 is the limiting reactant as we
determined above
3
2.44 ×10 mol CO 3
1 mol CH OH
×
1 mol CO
3
3
= 2.44 × 10 mol CH OH
3
2
4.27 ×10 mol H 3
2
1 mol CH OH
×
2 mol H
3
3
= 2.14 × 10 mol CH OH

Section 3.11
 To calculate the theoretical yield of methanol,
determine the moles of methanol formed
 We must use the amount of H2 and the mole ratio between
H2 and CH3OH to determine the maximum amount of
methanol that can be produced:
 Determine the theoretical yield of CH3OH in grams
3
2
4.27 ×10 mol H 3
2
1 mol CH OH
×
2 mol H
3
3
= 2.14 × 10 mol CH OH
3
3
2.14 × 10 mol CH OH 3
3
32.04 g CH OH
×
1 mol CH OH
4
3
= 6.86 10 g CH OH


Section 3.11
 Thus, from the amount of reactants given, the
maximum amount of CH3OH that can be formed is
6.86 ×104 g
 This is the theoretical yield
 Percent yield of CH3OH
4
3
4
3
Actual yield (grams) 3.57 × 10 g CH OH
× 100 = × 100% = 52.0%
Theoretical yield (grams) 6.86 × 10 g CH OH

Section 3.11
Problem-Solving Strategy
 Solving a stoichiometry problem involving masses
of reactants and products
1. Write and balance the equation for the reaction
2. Convert the known masses of substances to moles
3. Determine which reactant is limiting

Section 3.11
Problem-Solving Strategy (continued)
4. Using the amount of the limiting reactant and the
appropriate mole ratios, compute the number of
moles of the desired product
5. Convert from moles to grams, using the molar mass

Section 3.11
Join In (14)
 In the reaction 2A + 3B  C, 4.0 moles of A react
with 4.0 moles of B
 Which of the following choices best answers the
question, “which reactant is limiting?”

Section 3.11
a. Neither is limiting because equal amounts (4.0 mol) of
each reactant are used
b. A is limiting because 2 is smaller than 3 (the coefficients in
the balanced equation)
c. A is limiting because 2 mol is available but 4.0 mol is
needed
d. B is limiting because 3 is larger than 2 (the coefficients in
the balanced equation)
e. B is limiting because 4.0 mol is available but 6.0 mol is
needed

Section 3.11
Join In (15)
 Limiting reactant in a reaction:
a. Has the smallest coefficient in a balanced equation
b. Is the reactant for which you have the fewest
number of moles
c. Has the lowest ratio of [moles available/coefficient
in the balanced equation]
d. Has the lowest ratio of [coefficient in the balanced
equation/moles available]
e. None of these

Section 3.11
Join In (16)
 Consider the following balanced equation:
A + 5B  3C + 4D
 Which of the following choices best answers the
question, “when equal masses of A and B are reacted,
which is limiting?”

Section 3.11
a. If the molar mass of A is greater than the molar mass of B,
then A must be limiting
b. If the molar mass of A is less than the molar mass of B,
then A must be limiting
c. If the molar mass of A is greater than the molar mass of B,
then B must be limiting
d. If the molar mass of A is less than the molar mass of B,
then B must be limiting

Section 3.11
Join In (17)
 Which of the following reaction mixtures would
produce the greatest amount of product
according to the following chemical equation:
2H2 + O2  2H2O

Section 3.11
a. 2 mol H2 and 2 mol O2
b. 2 mol H2 and 3 mol O2
c. 2 mol H2 and 1 mol O2
d. 3 mol H2 and 1 mol O2
e. All of these choices would produce the same
amount of product

Zumdahl10e_PPT_Ch03.pptx

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