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Chapter 3
Stoichiometry
Chemical Stoichiometry
• Stoichiometry – The study of quantities of
materials consumed and produced in chemical
reactions.
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2
Micro World
atoms & molecules
Macro World
grams
Atomic mass is the mass of an atom in
atomic mass units (amu)
By definition:
1 atom 12C “weighs” 12 amu
12C
is the standard for atomic mass, with a mass of exactly 12 atomic mass
units (u).
The masses of all other atoms are given relative to this standard.
On this scale
1H
= 1.008 amu
16O
= 16.00 amu
3
4
Copper, a metal known since ancient times, is used in
electrical cables and pennies, among other things.
The atomic masses of its two stable isotopes,
(69.09
percent) and (30.91 percent), are 62.93 amu and 64.9278
amu, respectively.
Calculate the average atomic mass of copper. The relative
abundances are given in parentheses.
Each isotope contributes to the average
atomic mass based on its relative
abundance.
(0.6909) (62.93 amu) + (0.3091)
(64.9278 amu) = 63.55 amu
Multiplying the mass of an isotope by its
fractional abundance (not percent) will give
the contribution to the average atomic
mass of that particular isotope.
5
Average Atomic Mass for Carbon
98.89% of 12 u + 1.11% of 13.0034 u =
exact number
(0.9889)(12 u) + (0.0111)(13.0034 u) =
12.01 u
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6
EXERCISE!
An element consists of 62.60% of an isotope with
mass 186.956 u and 37.40% of an isotope with
mass 184.953 u.
• Calculate the average atomic mass and identify
the element.
186.2 u
Rhenium (Re)
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7
Schematic Diagram of a Mass
Spectrometer
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9
10
The Mole (mol): A unit to count numbers of particles
Dozen = 12
Pair = 2
The mole (mol) is the amount of a substance that
contains as many elementary entities as there
are atoms in exactly 12.00 grams of 12C
1 mol = NA = 6.0221415 x 1023
Avogadro’s number (NA)
11
Avogadro’s number (NA)
6.022 x 1023
= NA
= 1mol
12
eggs
Molar mass is the mass of 1 mole of shoes in grams
marbles
atoms
1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g
1 12C atom = 12.00 amu
1 mole 12C atoms = 12.00 g 12C
1 mole lithium atoms = 6.941 g of Li
For any element
atomic mass (amu) = molar mass (grams)
13
One Mole of:
S
C
Hg
Cu
Fe
14
1 12C atom
12.00 g
1.66 x 10-24 g
x
=
23
12
12.00 amu
6.022 x 10
C atoms
1 amu
M = molar mass in g/mol
NA = Avogadro’s number
15
•
•
•
The number equal to the number of carbon atoms
in exactly 12 grams of pure 12C.
1 mole of something consists of 6.022 × 1023 units
of that substance (Avogadro’s number).
1 mole C = 6.022 × 1023 C atoms = 12.01 g C
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16
EXERCISE!
Calculate the number of iron atoms in a 4.48 mole
sample of iron.
2.70×1024 Fe atoms
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17
•
Mass in grams of one mole of the substance:
Molar Mass of N = 14.01 g/mol
Molar Mass of H2O = 18.02 g/mol
(2 × 1.008 g) + 16.00 g
Molar Mass of Ba(NO3)2 = 261.35 g/mol
137.33 g + (2 × 14.01 g) + (6 × 16.00 g)
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18
CONCEPT CHECK!
Calculate the number of copper atoms in a 63.55 g
sample of copper.
6.022×1023 Cu atoms
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19
CONCEPT CHECK!
Which of the following 100.0 g samples contains
the greatest number of atoms?
a) Magnesium
b) Zinc
c) Silver
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20
EXERCISE!
Rank the following according to number of atoms
(greatest to least):
a) 107.9 g of silver
b) 70.0 g of zinc
c) 21.0 g of magnesium
b)
a)
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c)
21
EXERCISE!
Consider separate 100.0 gram samples of each of
the following:
H2O, N2O, C3H6O2, CO2
– Rank them from greatest to least number of
oxygen atoms.
H2O, CO2, C3H6O2, N2O
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Conceptual Problem Solving
•
Where are we going?
– Read the problem and decide on the final goal.
• How do we get there?
– Work backwards from the final goal to decide
where to start.
• Reality check.
– Does my answer make sense? Is it reasonable?
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23
Determining Chemical Formulas
mass of " A" in whole
mass % " A" 
 100%
mass of the whole
•
Mass percent of an element:
mass of element in compound
mass % =
× 100%
mass of compound
•
For iron in iron(III) oxide, (Fe2O3):
mass % Fe =
2(55.85 g)
111.70 g
=
n 100% = 69.94%
2(55.85 g)+3(16.00 g) 159.70 g
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24
Mass Percentages from Formulas
• Let’s calculate the percent composition of
butane, C4H10.
First, we need the molecular mass of C4H10.
4 carbons @ 12.0 amu/atom  48.0 amu
10 hydrogens @ 1.00 amu/atom  10.0 amu
1 molecule of C4 H10  58.0 amu
Now, we can calculate the percents.
amu C
% C  5848.0
.0 amu total  100%  82.8%C
amu H
% H  5810.0
.0 amu total  100%  17.2%H
25
Percent composition of an element in a compound =
n x molar mass of element
x 100%
molar mass of compound
n is the number of moles of the element in 1 mole
of the compound
2 x (12.01 g)
x 100% = 52.14%
46.07 g
6 x (1.008 g)
%H =
x 100% = 13.13%
46.07 g
1 x (16.00 g)
%O =
x 100% = 34.73%
46.07 g
%C =
C2H6O
52.14% + 13.13% + 34.73% = 100.0%
26
EXERCISE!
Consider separate 100.0 gram samples of each of
the following:
H2O, N2O, C3H6O2, CO2
– Rank them from highest to lowest percent
oxygen by mass.
H2O, CO2, C3H6O2, N2O
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27
Determining Chemical Formulas
• Determining the formula of a compound
from the percent composition.
The percent composition of a compound leads
directly to its empirical formula.
An empirical formula (or simplest formula) for
a compound is the formula of the substance
written with the smallest integer (whole
number) subscripts.
28
Formulas (Empirical and Molecular)
•
•
Example: Glucose
Empirical formula = CH
– Simplest whole-number ratio
• Molecular formula = (empirical formula)n
[n = integer]
• Molecular formula = C6H6 = (CH)6
– Actual formula of the compound
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29
Analyzing for Carbon and Hydrogen
•
Device used to determine the mass percent of each
element in a compound.
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30
EXERCISE!
The composition of adipic acid is 49.3% C, 6.9%
H, and 43.8% O (by mass). The molar mass of the
compound is about 146 g/mol.
– What is the empirical formula?
C3H5O2
– What is the molecular formula?
C6H10O4
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31
Percent Composition and Empirical Formulas
Determine the empirical formula of a
compound that has the following
percent composition by mass:
K 24.75, Mn 34.77, O 40.51 percent.
1 mol K
nK = 24.75 g K x
= 0.6330 mol K
39.10 g K
nMn = 34.77 g Mn x
1 mol Mn
= 0.6329 mol Mn
54.94 g Mn
nO = 40.51 g O x
1 mol O
= 2.532 mol O
16.00 g O
32
Percent Composition and Empirical Formulas
nK = 0.6330, nMn = 0.6329, nO = 2.532
0.6330 ~
K:
~ 1.0
0.6329
Mn :
0.6329
= 1.0
0.6329
2.532 ~
O:
~ 4.0
0.6329
KMnO4
33
Determining Chemical Formulas
• Determining the empirical formula from
the percent composition.
Benzoic acid is a white, crystalline powder
used as a food preservative. The
compound contains 68.8% C, 5.0% H, and
26.2% O by mass. What is its empirical
formula?
In other words, give the smallest wholenumber ratio of the subscripts in the
formula
Cx HyOz
34
Determining Chemical Formulas
Determining the empirical formula from
the percent composition.
Our 100.0 grams of benzoic acid would
contain:
1 mol C
68.8 g C 
 5.73 mol C
12.0 g
1 mol H
5.0 g H 
 5.0 mol H
1.0 g
1 mol O
26.2 g O 
 1.63mol O
16.0 g
This isn’t quite a
whole number
ratio, but if we
divide each
number by the
smallest of the
three, a better
ratio might
35
emerge.
Determining Chemical Formulas
• Determining the empirical formula from
the percent composition.
Our 100.0 grams of benzoic acid would
contain:
now it’s not too
5.73 mol C  1.63  3.50
difficult to see that
the smallest whole
5.0 mol H  1.63  3.0
number ratio is
1.63mol O  1.63  1.00 7:6:2.
The empirical
formula is C7H6O2 .
36
Chemical Reactions
•
A representation of a chemical reaction:
C2H5OH + 3O2
reactants
•
2CO2 + 3H2O
products
Reactants are only placed on the left side of
the arrow, products are only placed on the
right side of the arrow.
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37
Balancing Chemical Equations
C2H5OH + 3O2
•
•
•
2CO2 + 3H2O
The equation is balanced.
All atoms present in the reactants are
accounted for in the products.
1 mole of ethanol reacts with 3 moles of
oxygen to produce 2 moles of carbon dioxide
and 3 moles of water.
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•
•
The balanced equation represents an overall ratio
of reactants and products, not what actually
“happens” during a reaction.
Use the coefficients in the balanced equation to
decide the amount of each reactant that is used,
and the amount of each product that is formed.
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39
Writing and Balancing the Equation for a Chemical Reaction
1. Determine what reaction is occurring. What are the
reactants, the products, and the physical states involved?
2. Write the unbalanced equation that summarizes the
reaction described in step 1.
3. Balance the equation by inspection, starting with the most
complicated molecule(s). The 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.
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Chemical Reactions and Chemical Equations
A process in which one or more substances is changed into one
or more new substances is a chemical reaction
A chemical equation uses chemical symbols to show what
happens during a chemical reaction
3 ways of representing the reaction of H2 with O2 to form H2O
reactants
products
42
How to “Read” Chemical Equations
2 Mg + O2
2 MgO
2 atoms Mg + 1 molecule O2 makes 2 formula units MgO
2 moles Mg + 1 mole O2 makes 2 moles MgO
48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO
IS NOT
2 grams Mg + 1 gram O2 makes 2 g MgO
43
Chemical Reactions: Equations
• Writing chemical equations
A chemical equation is the symbolic representation
of a chemical reaction in terms of chemical formulas.
For example, the burning of sodium and chlorine to
produce sodium chloride is written
2Na  Cl 2  2NaCl
The reactants are starting substances in a chemical
reaction. The arrow means “yields.” The formulas on
the right side of the arrow represent the products.
44
Chemical Reactions: Equations
• Writing chemical equations
In many cases, it is useful to indicate the states of
the substances in the equation.
When you use these labels, the previous equation
becomes
2Na(s )  Cl 2 (g )  2NaCl(s )
45
Chemical Reactions: Equations
• Writing chemical equations
The law of conservation of mass dictates that the
total number of atoms of each element on both
sides of a chemical equation must match. The
equation is then said to be balanced.
CH 4 
O 2  CO 2  H 2O
Consider the combustion of methane to produce
carbon dioxide and water.
46
Chemical Reactions: Equations
• Writing chemical equations
For this equation to balance, two molecules of
oxygen must be consumed for each molecule of
methane, producing one molecule of CO2 and two
molecules of water.
CH 4  2 O 2  CO 2  2 H 2O
Now the equation is “balanced.”
47
Balancing Chemical Equations
1. Write the correct formula(s) for the reactants on
the left side and the correct formula(s) for the
product(s) on the right side of the equation.
Ethane reacts with oxygen to form carbon dioxide and water
C2H6 + O2
CO2 + H2O
2. Change the numbers in front of the formulas
(coefficients) to make the number of atoms of
each element the same on both sides of the
equation. Do not change the subscripts.
2C2H6
NOT
C4H12
48
Balancing Chemical Equations
3. Start by balancing those elements that appear in
only one reactant and one product.
C2H6 + O2
2 carbon
on left
C2H6 + O2
6 hydrogen
on left
C2H6 + O2
CO2 + H2O
start with C or H but not O
1 carbon
on right
multiply CO2 by 2
2CO2 + H2O
2 hydrogen
on right
2CO2 + 3H2O
multiply H2O by 3
49
Balancing Chemical Equations
4. Balance those elements that appear in two or
more reactants or products.
C2H6 + O2
2 oxygen
on left
2CO2 + 3H2O
multiply O2 by 7
2
4 oxygen + 3 oxygen = 7 oxygen
(3x1)
on right
(2x2)
C2H6 + 7 O2
2
2CO2 + 3H2O
2C2H6 + 7O2
4CO2 + 6H2O
remove fraction
multiply both sides by 2
50
Balancing Chemical Equations
5. Check to make sure that you have the same
number of each type of atom on both sides of the
equation.
2C2H6 + 7O2
4CO2 + 6H2O
4 C (2 x 2)
4C
12 H (2 x 6)
12 H (6 x 2)
14 O (7 x 2)
14 O (4 x 2 + 6)
Reactants
4C
12 H
14 O
Products
4C
12 H
14 O
51
Chemical Reactions: Equations
• Balance the following equations.
O2 
P4 
PCl 3 
N 2O 
As2S 3 
Ca3 (PO4 )2 
O2 
POCl 3
P4O6 
As2O 3 
H 3 PO 4 
N2
SO 2
Ca(H2 PO 4 )2
52
Chemical Reactions: Equations
• Balance the following equations.
O 2  2 PCl 3  2 POCl 3
P4  6 N 2O 
P4O6  6 N 2
2 As2S 3  9 O 2  2 As2O 3  6 SO 2
Ca3 (PO4 )2  4 H 3 PO 4  3Ca(H2 PO 4 )2
53
EXERCISE!
Which of the following correctly balances the chemical
equation given below? There may be more than one
correct balanced equation. If a balanced equation is
incorrect, explain what is incorrect about it.
CaO + C
CaC2 + CO2
I.
II.
III.
IV.
CaO2 + 3C
2CaO + 5C
CaO + (2.5)C
4CaO + 10C
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CaC2 + CO2
2CaC2 + CO2
CaC2 + (0.5)CO2
4CaC2 + 2CO2
Notice
•
•
•
•
The number of atoms of each type of element must
be the same on both sides of a balanced equation.
Subscripts must not be changed to balance an
equation.
A balanced equation tells us the ratio of the
number of molecules which react and are produced
in a chemical reaction.
Coefficients can be fractions, although they are
usually given as lowest integer multiples.
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55
Stoichiometric Calculations
•
Chemical equations can be used to relate the
masses of reacting chemicals.
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56
EXERCISE!
Consider the following reaction:
P4(s) + 5 O2(g) ® 2 P2O5(s)
If 6.25 g of phosphorus is burned, what mass of oxygen
does it combine with?
8.07 g O2
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57
EXERCISE!
(Part I)
Methane (CH4) reacts with the oxygen in the air
to produce carbon dioxide and water.
Ammonia (NH3) reacts with the oxygen in the air
to produce nitrogen monoxide and water.
– Write balanced equations for each of these
reactions.
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58
EXERCISE!
(Part II)
Methane (CH4) reacts with the oxygen in the air
to produce carbon dioxide and water.
Ammonia (NH3) reacts with the oxygen in the air
to produce nitrogen monoxide and water.
– What mass of ammonia would produce the
same amount of water as 1.00 g of methane
reacting with excess oxygen?
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59
Stoichiometry: Quantitative Relations in
Chemical Reactions
• Stoichiometry is the calculation of the quantities
of reactants and products involved in a chemical
reaction.
It is based on the balanced chemical equation
and on the relationship between mass and
moles.
Such calculations are fundamental to most
quantitative work in chemistry.
Let’s Think About It
•
Where are we going?
– To find the mass of ammonia that would produce
the same amount of water as 1.00 g of methane
reacting with excess oxygen.
• How do we get there?
– We need to know:
•
•
How much water is produced from 1.00 g of methane
and excess oxygen.
How much ammonia is needed to produce the
amount of water calculated above.
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61
Calculating Masses of Reactants and Products in
Reactions
1. Balance the equation for the reaction.
2. Convert the known mass of the reactant or product
to moles of that substance.
3. Use the balanced equation to set up the
appropriate mole ratios.
4. Use the appropriate mole ratios to calculate the
number of moles of the desired reactant or
product.
5. Convert from moles back to grams if required by
the problem.
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62
Calculating Masses of Reactants and Products in
Reactions
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63
Limiting Reactants
•
•
Limiting reactant – the reactant that runs out first
and thus limits the amounts of products that can be
formed.
Determine which reactant is limiting to calculate
correctly the amounts of products that will be
formed.
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64
Limiting Reactants
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A. The Concept of Limiting Reactants
 Stoichiometric mixture
– N2(g) + 3H2(g)
2NH3(g)
A. The Concept of Limiting Reactants
 Limiting reactant mixture

N2(g) + 3H2(g)  2NH3(g)
A. The Concept of Limiting Reactants
• Limiting reactant mixture
• N2(g) + 3H2(g)
2NH3(g)
• Limiting reactant is the reactant that runs out first.
– H2
Limiting Reactants
•
•
•
The amount of products that can form is limited by
the methane.
Methane is the limiting reactant.
Water is in excess.
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69
CONCEPT CHECK!
Which of the following reaction mixtures could
produce the greatest amount of product? Each
involves the reaction symbolized by the equation:
2H2 + O2
a)
b)
c)
d)
e)
2H2O
2 moles of H2 and 2 moles of O2
2 moles of H2 and 3 moles of O2
2 moles of H2 and 1 mole of O2
3 moles of H2 and 1 mole of O2
Each produce the same amount of product.
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70
Notice
•
We cannot simply add the total moles of all the
reactants to decide which reactant mixture makes
the most product. We must always think about
how much product can be formed by using what we
are given, and the ratio in the balanced equation.
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71
CONCEPT CHECK!
• You know that chemical A reacts with chemical
B. You react 10.0 g of A with 10.0 g of B.
– What information do you need to know in
order to determine the mass of product
that will be produced?
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72
Let’s Think About It
•
Where are we going?
– To determine the mass of product that will be
produced when you react 10.0 g of A with 10.0 g
of B.
• How do we get there?
– We need to know:
•
•
The mole ratio between A, B, and the product they
form. In other words, we need to know the balanced
reaction equation.
The molar masses of A, B, and the product they form.
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73
EXERCISE!
You react 10.0 g of A with 10.0 g of B. What
mass of product will be produced given that the
molar mass of A is 10.0 g/mol, B is
20.0 g/mol, and C is 25.0 g/mol? They react
according to the equation:
A + 3B
2C
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74
Limiting Reagent
• Zinc metal reacts with hydrochloric acid by
the following reaction.
Zn(s)  2 HCl(aq)  ZnCl 2 (aq)  H 2 (g )
If 0.30 mol Zn is added to hydrochloric acid
containing 0.52 mol HCl, how many moles of
H2 are produced?
75
Limiting Reagent
Take each reactant in turn and ask how much product
would be obtained if each were totally consumed. The
reactant that gives the smaller amount is the limiting
reagent.
1 mol H 2
0.30 mol Zn 
 0.30 mol H 2
1 mol Zn
1 mol H 2
0.52 mol HCl 
 0.26 mol H 2
2 mol HCl
Since HCl is the limiting reagent, the amount of H2
produced must be 0.26 mol (0.26*2 = 0.52 g)
76
Reaction Yield
Theoretical and Percent Yield
• The theoretical yield of product is the maximum amount
of product that can be obtained from given amounts of
reactants (if all the limiting reagent reacted).
• Actual Yield is the amount of product actually obtained
from a reaction (experimentally determined).
The percentage yield is the actual yield
(experimentally determined) expressed as a
percentage of the theoretical yield (calculated).
actual yield
%Yield 
 100%
theoretical yield
77
Theoretical and Percent Yield
• To illustrate the calculation of percentage yield,
recall that the theoretical yield of H2 in the
previous example was 0.26 mol (or 0.52 g) H2.
If the actual yield of the reaction had been 0.22 g
H2, then
0.22 g H 2
%Yield 
 100%  42%
0.52 g H 2
•
% Yield is an important indicator of the efficiency of
a particular laboratory or industrial reaction.
78
EXERCISE!
Consider the following reaction:
P4(s) + 6F2(g)
4PF3(g)
– What mass of P4 is needed to produce 85.0 g
of PF3 if the reaction has a 64.9% yield?
46.1 g P4
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79