Download Chapter Three

1
Chapter Three
Stoichiometry: Chemical Calculations
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
2
Molecular Masses and
Formula Masses
• Molecular mass: sum of the masses of the atoms
represented in a molecular formula.
• Simply put: the mass of a molecule.
• Molecular mass is specifically for molecules.
• Ionic compounds don’t exist as molecules; for
them we use …
• Formula mass: sum of the masses of the atoms or
ions present in a formula unit.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
3
Molecular Mass
Molecular mass is the sum of the masses of the
atoms represented in a molecular formula.
1 Oxygen atom
Example:
water - H2O
2(1.0079 u) + 15.9994 u
= 18.0152 u
2 Hydrogen atoms
EOS
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
4
Formula Mass
Formula mass is the sum of the masses of the
atoms or ions present in a formula unit.
ClNa+
Cl-
Na+
Cl-
One Na+ and one Cl– make a
formula unit for sodium chloride
Na+
Cl-
Na+
The mass of one formula unit is:
Crystal of
= 22.9898 u + 35.4527 u
sodium chloride
Prentice Hall © 2005
= 58.4425 u
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
EOS
Chapter Three
5
Example 3.1
Calculate the molecular mass of glycerol
(C3H8O3).
Example 3.2
Calculate the formula mass of ammonium
sulfate, a fertilizer commonly used by home
gardeners.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
6
Determining the Formula Mass
of Ammonium Sulfate
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
7
The Mole & Avogadro’s Number
• Mole (mol): amount of substance that contains as
many elementary entities as there are atoms in
exactly 12 g of the carbon-12 isotope.
• Atoms are small, so this is a BIG number …
• Avogadro’s number (NA) = 6.022 × 1023 mol–1
• 1 mol = 6.022 × 1023 “things” (atoms, molecules,
ions, formula units, oranges, etc.)
– A mole of oranges would weigh about as much as the
earth!
• Mole is NOT abbreviated as either M or m.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
8
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
9
One Mole of
Four Elements
One mole each of helium,
sulfur, copper, and mercury.
How many atoms of helium
are present? Of sulfur? Of
copper? Of mercury?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
10
The Mole and Molar Mass
• Molar mass is the mass of one mole of a
substance.
• Molar mass is numerically equal to atomic mass,
molecular mass, or formula mass. However …
• … the units of molar mass are grams (g/mol).
• Examples:
1 atom Na = 22.99 u
1 mol Na = 22.99 g
1 molecule CO2 = 44.01 u
1 mol CO2 = 44.01 g
1 formula unit KCl = 74.56 u
1 mol KCl = 74.56 g
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
11
Conversions involving Mass, Moles,
and Number of Atoms/Molecules
1 mol Na = 6.022 × 1023 Na atoms = 22.99 g Na
We can use these equalities to construct conversion factors,
such as:
1 mol Na
–––––––––
22.99 g Na
22.99 g Na
–––––––––
1 mol Na
1 mol Na
––––––––––––––––––
6.022 × 1023 Na atoms
Note: preliminary and follow-up calculations may be needed.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
12
We can read formulas in terms of
moles of atoms or ions.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
13
Hints for Doing Mass/Mole/Atom
Problems
• HINT #1: If you see, ANYWHERE AT ALL
IN THE PROBLEM, “grams”, you will NEED
to make a SAMT chart!
• HINT #2: If you see, ANYWHERE AT ALL
IN THE PROBLEM, “atoms/molecules/etc”,
you will NEED to use Avogadro’s #.
• HINT #3: If you see, MOLE in the given or
find, you will only do a 2-step problem.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
14
Examples
• Given: 3.50 mol Cu; Find: # g of Cu.
• Given: 11.5 g of H2O; Find: # mol of H2O.
• Given: 0.55 mol of Al; Find: # atoms of Al.
• Given: 3.73 x 10 23 molecules of HCl; Find: #
mol of HCl.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
15
Examples – Part 2
• These are more difficult, and are 3-steps:
• Given = 11.5 g of Ca; Find = # atoms of Ca.
• Given = 1.22 x 10 23 molecules of NO; Find = #
g of NO.
• Given = 13.5 g of HCl; Find = # particles of
HCl.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
16
Examples
• Calculate the mass in milligrams of
1.34 x 10-4 mol Ag.
• Calculate the number of oxygen atoms in
20.5 mol O2.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
17
Examples
• Calculate both the (a) # of moles and (b) the
# of atoms of Al in a cube of aluminum
metal 5.5 cm on an edge. The density of
aluminum is known to be 2.70 g/cm3.
• Calculate the volume occupied by 4.06 x 1024
Br atoms present as Br2 molecules in liquid
bromine. (d = 3.12 g/mL)
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
18
Example 3.4
Determine (a) the number of NH4+ ions in a 145-g
sample of (NH4)2SO4 and (b) the volume of 1,2,3propanetriol (glycerol, d = 1.261 g/mL) that contains
1.00 mol O atoms.
Example 3.5 An Estimation Example
Which of the following is a reasonable value for the
number of atoms in 1.00 g of helium?
(a) 4.1 × 10–23
(c) 1.5 × 1023
(b) 4.0
(d) 1.5 × 1024
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
19
Mass Percent Composition
from Chemical Formulas
The mass percent composition of a compound refers
to the proportion of the constituent elements,
expressed as the number of grams of each element
per 100 grams of the compound. In other words …
X g element
X % element = –––––––––––––– OR …
100 g compound
g element
% element = ––––––––––– × 100
g compound
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
20
Percentage Composition of Butane
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
21
Example 3.6
Calculate, to four significant figures, the mass percent
of each element in ammonium nitrate.
Example 3.7
How many grams of nitrogen are present in 46.34 g
ammonium nitrate?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
22
Chemical Formulas from Mass
Percent Composition
• We can “reverse” the process of finding
percentage composition.
• First we use the percentage or mass of each
element to find moles of each element.
• Then we can obtain the empirical formula by
finding the smallest whole-number ratio of moles.
– Find the whole-number ratio by dividing each number
of moles by the smallest number of moles.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
23
Examples
• Analysis shows that a compound contains 32.31%
sodium, 22.67% sulfur, and 45.02% oxygen. Find
the empirical formula for this compound.
• HINT = Whenever you are given %s, we assume
we ALWAYS have a 100-gram sample, so
immediately convert you %s to grams.
• ANS = Na2SO4
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
24
Examples (cont.)
• Find the empirical formula of a compound found to
contain 26.56% K, 35.41% Cr, and the remainder O.
• ANS = K2Cr2O7
• Analysis of a 10.150-g sample of a compound is
known to contain only phosphorus and oxygen.
There is a phosphorus content of 4.433-g. What is
the empirical formula of this compound?
• ANS = P2O5
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
25
Example 3.9
Phenol, a general disinfectant, has the composition
76.57% C, 6.43% H, and 17.00% O by mass.
Determine its empirical formula.
Example 3.10
Diethylene glycol, used in antifreeze, as a softening
agent for textile fibers and some leathers, and as a
moistening agent for glues and paper, has the
composition 45.27% C, 9.50% H, and 45.23% O by
mass. Determine its empirical formula.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
26
Molecular Formulas
• Molecular formulas are the actual formulas for the
compound.
• Example: In our last example, we determined that the
empirical formula was P2O5. Experimentation shows
that the molar mass of this compound is 283.89 g/mol.
What is the compound’s molecular formula?
• Hint = The molar mass given in the problem statement
is the numerator!!!!
• ANS = P4O10
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
27
Relating Molecular Formulas
to Empirical Formulas
• A molecular formula is a simple integer multiple
of the empirical formula.
• That is, an empirical formula of CH2 means that
the molecular formula is CH2, or C2H4, or C3H6,
or C4H8, etc.
• So: we find the molecular formula by:
molecular formula mass
= integer (nearly)
empirical formula mass
We then multiply each subscript in the empirical formula by the
integer.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
28
Examples
• Determine the molecular formula of the
compound with an empirical formula of CH
and a formula mass of 78.110 amu (g/mol).
• ANSWER = C6H6
• A sample of a compound with a formula mass
of 34.00 amu (g/mol) is found to consist of
0.44 gH and 6.92 gO. Determine (a) empirical
and (b) molecular formulas for this compound.
• ANSWER = (a) HO; (b) H2O2
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
29
Example 3.11
The empirical formula of hydroquinone, a
chemical used in photography, is C3H3O, and
its molecular mass is 110 u. What is its
molecular formula?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
30
Elemental Analysis …
• … is one method of determining empirical
formulas in the laboratory.
• This method is used primarily for simple organic
compounds (that contain carbon, hydrogen,
oxygen).
– The organic compound is burned in oxygen.
– The products of combustion (usually CO2 and H2O)
are weighed.
– The amount of each element is determined from the
mass of products.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
31
Elemental Analysis (cont’d)
… H2O, which is absorbed
by MgClO4, and …
The sample is
burned in a
stream of oxygen
gas, producing …
Prentice Hall © 2005
… CO2, which is
absorbed by NaOH.
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
32
Elemental Analysis (cont’d)
If our sample
were CH3OH,
every two
molecules of
CH3OH …
Prentice Hall © 2005
… would give two
molecules of CO2 …
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
… and four
molecules of H2O.
Chapter Three
33
Example 3.12
Burning a 0.1000-g sample of a carbon–
hydrogen–oxygen compound in oxygen yields
0.1953 g CO2 and 0.1000 g H2O. A separate
experiment shows that the molecular mass of
the compound is 90 u. Determine (a) the mass
percent composition, (b) the empirical formula,
and (c) the molecular formula of the compound.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
34
Writing Chemical Equations
• A chemical equation is a shorthand description of a
chemical reaction, using symbols and formulas to represent
the elements and compounds involved.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
35
Writing Chemical Equations
• Sometimes
additional
information
about the
reaction is
conveyed in
the equation.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
36
Balancing Equations Illustrated
How can we tell
that the
equation is not
balanced?
… not by changing
the equation …
… and not by
changing the
formulas.
The equation is
balanced by changing
the coefficients …
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
37
Chemical Reactions
• What is a chemical reaction?
• A chemical reaction is any process by which
1 or more substances are changed into one
or more different substances.
• A +
B

C
+
D
• A and B are referred to as REACTANTS; C
and D are referred to as PRODUCTS.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
38
Chemical Equation
• Chemical reactions are described by
chemical equations.
• Chemical equation = an equation that
represents, with symbols and formulas,
the identities and relative molecular
amounts of the reactants and products in
a chemical reaction.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
39
Indications of a Chemical Reaction
• The following methods are ways we tell a chemical
reaction occurs:
• #1. Evolution of energy as heat and/or light.
• #2. Production of a gas.
• #3. Formation of a precipitate.
• Precipitate = a solid that is produced as a result of a
chemical reaction of 2 or more solutions.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
40
Characteristics of a Chemical
Equation
• 1. The equation must represent known facts. All
reactants and products MUST be identified.
• 2. The equation must contain the correct formulas for
the reactants and products.
• 3. The law of conservation of mass must be satisfied.
• NOTE = we are ONLY allowed to change
coefficients, NOT subscripts in a chemical equation!
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
41
Coefficients vs. Subscripts
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
42
Word and Formula Equations
• __H2O 
__H2 +
__O2
• What are the reactant(s)?
• What are the product(s)?
• What are the coefficients for each material?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
43
Word Equations
• A word equation is an equation in which the
reactants products in a chemical reaction are
represented by words.
• Remember – what are our diatomic elements?
• Ex#1: Solid aluminum reacts with gaseous
oxygen to produce solid aluminum oxide.
• Ex#2: Methane, in the presence of oxygen,
combusts into water and carbon dioxide.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
44
A Few More Examples…
• Example #3:
• Solid sodium oxide is added to water at room
temperature and forms sodium hydroxide.
• Example #4:
• Hydrazine, N2H4, is used as rocket fuel.
Hydrazine reacts violently with oxygen to
produce gaseous nitrogen and water.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
45
A Few Hints…
• When balancing chemical equations, here are a few
hints:
• #1. Balance most elements OTHER THAN hydrogen
and oxygen 1st, then go with hydrogen, and save
oxygen for LAST.
• #2. Sometimes, it is easier to call water HOH.
• #3. Look for polyatomic ions on both sides of the
equation. If they are there, you can use a polyatomic
as a “pseudo” element.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
46
What Do I Mean by Hint #3?
• Example:
• Magnesium phosphate reacts with lithium chloride
to form magnesium chloride and lithium
phosphate.
• Calcium hydroxide and sodium fluoride react to
form calcium fluoride and sodium hydroxide.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
47
Pop Quiz….
• Try the following….think about what you would
get:
• #1. __Li
+
__O2

???
• #2. __NaCl

???
• #3. __H2
+
__F2
• #4.
Prentice Hall © 2005
__MgCO3


???
???
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
48
5 Basic Types of Chemical
Reactions
• There are several ways to classify chemical
reactions.
•
•
•
•
•
•
We will look at the 5 basic types of reactions:
1. synthesis (composition)
2. decomposition
3. single displacement (replacement)
4. double displacement (replacement)
5. combustion
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
49
Synthesis Reactions
• Synthesis reactions are AKA composition
reactions.
• Synthesis reactions = a reaction in which 2
or more substances combine to form a new
compound.
• A +
Prentice Hall © 2005
X

AX
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
50
Types of Synthesis Reactions
• Type #1: Reactions of elements with
oxygen and sulfur.
• Remember, oxygen is diatomic (O2). Sulfur
is odd – it can be S8.
• __Rb
+
__S8 
__Rb2S
• __Fe
+
__O2

Prentice Hall © 2005
__Fe2O3
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
51
Types of Synthesis Rxns
(cont.)
• Type #2: Reactions of metals with
halogens.
• __Na
+
__Br2

__NaBr
• __F2
+
__Mg

__MgF2
• Remember – ALL halogens are diatomic!
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
52
Types of Synthesis Rxns.
(cont.)
• Type #3: Synthesis reactions with oxides.
• __CaO + __H2O
• __SO2
+

__H2O
__Ca(OH)2

__H2SO4
• Remember, when balancing, call water HOH.
• Usually, you get a metal hydroxide.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
53
Decomposition Reactions
• In a decomposition reaction, a single compound
undergoes a reaction that produces 2 or more
simpler substances.
• Decomposition reactions are usually the opposite of
synthesis reactions.
• AX 
A
+
X
• Decomposition reactions usually require heat or
electricity in order to occur.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
54
Types of Decomposition
Reactions
• Type #1: Decomposition of a Binary
Compound
• 2H2O

2 H2 +
O2 (electrolysis)
• 2HgO 
2Hg +
O2 (under heat)
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
55
Types (cont.)
• Type #2: Decomposition of Metal Carbonates
• What IS a metal carbonate?
• CaCO3

CaO +
CO2
• Breaks down into a metal oxide and carbon
dioxide gas.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
56
Types (cont.)
• Type #3: Decomposition of a Metal
Hydroxide
• What IS a metal hydroxide?
• Ca(OH)2 
CaO +
HOH
• Breaks down into a metal oxide and water.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
57
Types (cont.)
• Type #4: Decomposition of a Metal Chlorate
• What IS a metal chlorate?
• 2KClO3 
2KCl +
3O2
• Breaks down into a metal chloride and
oxygen.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
58
Types (cont.)
• Type #5: Decomposition of Acids
• Think acids that are NOT binary for these!
• H2CO3

CO2 +
H2 O
• H2SO4

SO3 +
H2 O
• Breaks down into water and “what’s left”.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
59
Single Displacement Reactions
• Single displacement reactions are also
known as single replacement reactions.
• We will look at this type of reaction more in
the next section…
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
60
Double Displacement
Reactions
• In double displacement (replacement)
reactions, the ions of 2 compounds exchange
places in an aqueous solution to form 2 new
compounds.

• AX +
BY
• __KI
+__Pb(NO3)2__PbI2 +__KNO3
Prentice Hall © 2005
AY
+
BX
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
61
Combustion Reactions
• In a combustion reaction, a hydrocarbon
reacts with oxygen , releasing a LARGE
amount of energy, as well as water and
carbon dioxide.
• __CH4 +
__O2  __CO2 + __H2O
• __C3H8 +
__O2 __CO2 + __H2O
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
62
Example 3.13
Balance the equation
Fe + O2  Fe2O3
(not balanced)
Example 3.14
Balance the equation
C2H6 + O2  CO2 + H2O
Example 3.15
Balance the equation
H3PO4 + NaCN  HCN + Na3PO4
Example 3.16
A Conceptual Example
Write a plausible chemical equation for the reaction between
water and a liquid molecular chloride of phosphorus to form an
aqueous solution of hydrochloric acid and phosphorus acid. The
phosphorus-chlorine compound is 77.45% Cl by mass.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
63
Stoichiometric Equivalence
and Reaction Stoichiometry
• A stoichiometric factor or mole ratio is a
conversion factor obtained from the stoichiometric
coefficients in a chemical equation.
• In the equation: CO(g) + 2 H2(g)  CH3OH(l)
– 1 mol CO is chemically equivalent to 2 mol H2
– 1 mol CO is chemically equivalent to 1 mol CH3OH
– 2 mol H2 is chemically equivalent to 1 mol CH3OH
1 mol CO
–––––––––
2 mol H2
Prentice Hall © 2005
1 mol CO
–––––––––––––
1 mol CH3OH
2 mol H2
–––––––––––––
1 mol CH3OH
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
64
Concept of Stoichiometric Equivalence
One car may be equivalent
to either 25 feet or 10 feet,
depending on the method of
parking.
One mole of CO may be
equivalent to one mole of
CH3OH, or to one mole of CO2,
or to two moles of CH3OH,
depending on the reaction(s).
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
65
Outline of Simple Reaction Stoichiometry
Note: preliminary and/or follow-up calculations may be needed.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
66
Example 3.17
When 0.105 mol propane is burned in an
excess of oxygen, how many moles of oxygen
are consumed?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
67
Outline of Stoichiometry Involving Mass
… we’ve added a
conversion from
mass at the
beginning …
To our simple
stoichiometry
scheme …
Substances A and B
may be two reactants,
two products, or
reactant and product.
… and a
conversion to
mass at the end.
Think: If we are given
moles of substance A
initially, do we need
to convert A to
grams?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
68
Examples
• Magnesium reacts with oxygen to
produce magnesium oxide.
• Given = 2.00 mol Mg
• Find = # grams of MgO
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
69
Examples
• Tin reacts with hydrofluoric acid to
produce tin (II) fluoride and hydrogen
gas.
• Given = 30.00 grams of HF
• Find = # grams of SnF2
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
70
Example 3.18
The final step in the production of nitric acid involves
the reaction of nitrogen dioxide with water; nitrogen
monoxide is also produced. How many grams of nitric
acid are produced for every 100.0 g of nitrogen dioxide
that reacts?
Example 3.19
Ammonium sulfate, a common fertilizer used by
gardeners, is produced commercially by passing
gaseous ammonia into an aqueous solution that is 65%
H2SO4 by mass and has a density of 1.55 g/mL. How
many milliliters of this sulfuric acid solution are required
to convert 1.00 kg NH3 to (NH4)2SO4?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
71
Limiting Reactants
• Many reactions are carried out with a limited
amount of one reactant and a plentiful amount of
the other(s).
• The reactant that is completely consumed in the
reaction limits the amounts of products and is
called the limiting reactant, or limiting reagent.
• The limiting reactant is not necessarily the one
present in smallest amount.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
72
Limiting Reactant Analogy
If we have 10
sandwiches, 18
cookies, and 12
oranges …
Prentice Hall © 2005
… how many
packaged meals
can we make?
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
73
Limiting vs. Excess
• When looking at any chemical reaction, the reaction
is rarely carried out as per the balanced equation.
• We will be looking at reactions with 2 reactants.
• (A) limiting reactant = the reactant in the
chemical reaction that is completely “used up” –
none will remain when the reaction is complete.
• (B) excess reactant = the reactant that is NOT
completely used up in a chemical reaction.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
74
What to Do
• Silicon dioxide, aka quartz, is usually quite
unreactive but reacts readily with hydrogen
fluoride to produce silicon tetrafluoride and water.
• If 6.0g of HF is added to 4.5g of SiO2, which is
the limiting reactant?
• Step #1: BALANCE THE EQUATION!
• __SiO2 + __HF 
Prentice Hall © 2005
__SiF4
+
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
__H2O
Chapter Three
75
What to Do (cont.)
• SiO2 +
4HF

SiF4
+
2H2O
• Step #2: Pick a product, any product (if the reactant is
limiting for 1 product, it will be limiting for the other) = I
pick SiF4.
• Step #3: Convert BOTH reactants to grams of product.
• Step #4: The smaller value gives us the reactant that is
LIMITING!
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
76
Example
• Fe2O3 is a substance that can be made in the chem lab via
the following, UNBALANCED, equation:
• __Fe
+__H2O  __Fe2O3 +__H2
• (a) Balance the equation
• (b) If 36.0 g water is mixed with 67.0 g Fe, determine the
limiting reactant.
• (c) What mass of iron oxide is produced, based on the
limiting reactant?
• (d) What mass of excess reactant remains when the
reaction is completed?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
77
Molecular View of the Limiting Reactant Concept
1. Why is ethylene left
over, when we started
with more bromine
than ethylene? (Hint:
count the molecules.)
2. What mass of ethylene
is left over after
reaction is complete?
(Hint: it’s an easy
calculation; why?)
When 28 g (1.0
mol) ethylene
reacts with …
Prentice Hall © 2005
… 128 g (0.80 mol)
bromine, we get …
… 150 g of 1,2dibromoethane, and
leftover ethylene!
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
78
Recognizing and Solving
Limiting Reactant Problems
•
•
•
We recognize limiting reactant problems by the
fact that amounts of two (or more) reactants are
given.
One way to solve them is to perform a normal
stoichiometric calculation of the amount of
product obtained, starting with each reactant.
The reactant that produces the smallest amount
of product is the limiting reactant.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
79
Example 3.20
Magnesium nitride can be formed by the
reaction of magnesium metal with nitrogen gas.
(a) How many grams of magnesium nitride can
be made in the reaction of 35.00 g of
magnesium and 15.00 g of nitrogen? (b) How
many grams of the excess reactant remain after
the reaction?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
80
Yields of Chemical Reactions
• The theoretical yield of a chemical reaction is the
calculated quantity of product in the reaction.
• The actual yield is the amount you actually get
when you carry out the reaction.
• Actual yield will be less than the theoretical yield,
for many reasons … can you name some?
actual yield
Percent yield = ––––––––––––– × 100
theoretical yield
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
81
Actual Yield of ZnS Is Less than
the Theoretical Yield
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
82
Example 3.21
Ethyl acetate is a solvent used as fingernail polish remover.
What mass of acetic acid is needed to prepare 252 g ethyl
acetate if the expected percent yield is 85.0%? Assume that
the other reactant, ethanol, is present in excess. The
equation for the reaction, carried out in the presence of
H2SO4, is
CH3COOH + HOCH2CH3  CH3COOCH2CH3 + H2O
Acetic acid
Ethanol
Ethyl acetate
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
83
Solutions and
Solution Stoichiometry
• Solute: the substance being dissolved.
• Solvent: the substance doing the dissolving.
• Concentration of a solution: the quantity of a
solute in a given quantity of solution (or solvent).
– A concentrated solution contains a relatively large
amount of solute vs. the solvent (or solution).
– A dilute solution contains a relatively small
concentration of solute vs. the solvent (or solution).
– “Concentrated” and “dilute” aren’t very quantitative …
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
84
Molar Concentration
Molarity (M), or molar concentration, is the amount
of solute, in moles, per liter of solution:
moles of solute
Molarity = ––––––––––––––
liters of solution
• A solution that is 0.35 M sucrose contains 0.35
moles of sucrose in each liter of solution.
• Keep in mind that molarity signifies moles of
solute per liter of solution, not liters of solvent.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
85
Preparing 0.01000 M KMnO4
Weigh 0.01000
mol (1.580 g)
KMnO4.
Prentice Hall © 2005
Dissolve in water. How much
water? Doesn’t matter, as long as
we don’t go over a liter.
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Add more water
to reach the
1.000 liter mark.
Chapter Three
86
Example 3.23
What is the molarity of a solution in which 333 g potassium
hydrogen carbonate is dissolved in enough water to make
10.0 L of solution?
Example 3.24
We want to prepare a 6.68 molar solution of NaOH (6.68 M
NaOH).
(a) How many moles of NaOH are required to prepare 0.500
L of 6.68 M NaOH?
(b) How many liters of 6.68 M NaOH can we prepare with
2.35 kg NaOH?
Example 3.25
The label of a stock bottle of aqueous ammonia indicates
that the solution is 28.0% NH3 by mass and has a density of
0.898 g/mL. Calculate the molarity of the solution.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
87
Dilution of Solutions
• Dilution is the process of preparing a more dilute
solution by adding solvent to a more concentrated
one.
• Addition of solvent does not change the amount of
solute in a solution but does change the solution
concentration.
• It is very common to prepare a concentrated stock
solution of a solute, then dilute it to other
concentrations as needed.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
Visualizing the Dilution of a Solution
We start and
end with the
same amount of
solute.
Prentice Hall © 2005
88
Addition of
solvent has
decreased the
concentration.
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
89
Dilution Calculations …
•
•
•
•
… couldn’t be easier.
Moles of solute does not change on dilution.
Moles of solute = M × V
Therefore …
Mconc × Vconc = Mdil × Vdil
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
90
Example 3.26
How many milliliters of a 2.00 M CuSO4 stock
solution are needed to prepare 0.250 L of 0.400 M
CuSO4?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
91
Solutions in Chemical
Reactions
• Molarity provides an additional tool in
stoichiometric calculations based on
chemical equations.
• Molarity provides factors for converting
between moles of solute (either reactant or
product) and liters of solution.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
Adding to the previous stoichiometry scheme …
92
If substance A is a
solution of known
concentration …
If substance B is in
solution, then …
… we can start with
molarity of A times
volume (liters) of the
solution of A to get
here.
Prentice Hall © 2005
… we can go from
moles of substance B
to either volume of B
or molarity of B.
How?
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
93
Example 3.27
A chemical reaction familiar to geologists is that
used to identify limestone. The reaction of
hydrochloric acid with limestone, which is largely
calcium carbonate, is seen through an
effervescence—a bubbling due to the liberation of
gaseous carbon dioxide:
CaCO3(s) + 2 HCl(aq)  CaCl2(aq) + H2O(l) + CO2(g)
How many grams of CaCO3(s) are consumed in a
reaction with 225 mL of 3.25 M HCl?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three
94
Cumulative Example
The combustion in oxygen of 1.5250 g of an
alkane-derived compound composed of
carbon, hydrogen, and oxygen yields 3.047 g
CO2 and 1.247 g H2O. The molecular mass of
this compound is 88.1 u. Draw a plausible
structural formula for this compound. Is there
more than one possibility? Explain.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Three