1
Chapter Three
Stoichiometry: Chemical Calculations
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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.
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Example 3.1
Calculate the molecular mass of glycerol (1,2,3propanetriol).
Example 3.2
Calculate the formula mass of ammonium
sulfate, a fertilizer commonly used by home
gardeners.
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Determining the molecular mass
of Glycerol (1,2,3-propanetriol).
• Determine molecular formula
– CH2OHCHOHCH2OH = C3H8O3
• Multiply # of atoms of each element by the atomic mass of
that element
 C3 = 3 x 12.011u = 36.033u
 H8 = 8 x 1.008u = 8.064u
 O3 = 3 x 15.999u 47.997u
• Add up the contributions of each element to the molecular
mass
 Molecular mass = 36.033 + 8.064 + 47.997 = 92.094u
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Determining the Formula Mass
of Ammonium Sulfate
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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.
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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?
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Example 3.3
Determine (a) the mass of a 0.0750-mol sample
of Na, (b) the number of moles of Na in a 62.5g sample, (c) the mass of a sample of Na
containing 1.00 × 1025 Na atoms, and (d) the
mass of a single Na atom.
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Example 3.3
Determine (a) the mass of a 0.0750-mol sample of Na, (b) the number of moles of Na in
a 62.5-g sample, (c) the mass of a sample of Na containing 1.00 x 1025 Na atoms, and (d)
the mass of a single Na atom.
Strategy
To perform these calculations, we will use a conversion factor derived from molar
mass to relate moles and grams, and a conversion factor derived from Avogadro’s
number to relate moles and number of atoms. These are the relationships embodied
in Equation (3.1).
Solution
(a) To convert from moles to grams, we need to use the first and third terms in Equation (3.1),
written as a conversion factor: 22.99 g Na/1 mol Na:
(b) Again, we need the first and third terms in Equation (3.1), but this time we write the
conversion factor as the inverse—1 mol Na/22.99 g Na—because we are to convert from
grams to moles:
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Example 3.3 continued
11
Solution continued
(c) Here we can use all three terms in Equation (3.1), written as two conversion
factors, to convert first from number of atoms to number of moles, and then
from moles to the mass in grams:
Alternatively, we can use the second and third terms in Equation (3.1), written
as the conversion factor 22.99 g Na/6.022 x 1023 Na atoms, to convert directly
from number of atoms to mass in grams:
(d) The answer must have the unit grams per sodium atom (g/Na atom). Thus, if we
know the mass of a certain number of Na atoms, our answer is simply that mass
divided by the corresponding number of atoms. And we know these quantities
from the molar mass (22.99 g Na/mol Na) and Avogadro’s number (1 mol
Na/6.022 x 1023 atoms). Our answer is the product of these two factors:
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Example 3.3 continued
Assessment
In these examples, note the common practice of expressing molar mass and Avogadro’s number with
at least one significant figure more than the number of significant figures in the least precisely known
quantity. Doing this ensures that the precision of the calculated results is limited only by the least
precisely known quantity. In part (d), this simple fact should deter you from mistakenly multiplying
instead of dividing by Avogadro’s number: Individual atoms are exceedingly small and possess
masses that are many orders of magnitude less than one gram. It is also worth noting that the
calculated mass is the true mass of a sodium atom—23Na has no isotopes. For elements with two or
more isotopes, the mass calculated for an atom is a weighted average.
Exercise 3.3A
Calculate (a) the mass in milligrams of 1.34 x 10–4 mol Ag and (b) the number of oxygen
atoms in 20.5 mol O2.
Exercise 3.3B
Calculate (a) the number of moles of Al in a cube of aluminum metal 5.5 cm on an edge
(d = 2.70 g/cm3) and (b) the volume occupied by 4.06 x 1024 Br atoms present as Br2
molecules in liquid bromine (d = 3.12 g/mL).
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Solutions to Examples 3.3A,B
Exercise 3.3A
Calculate (a) the mass in milligrams of 1.34 x 10–4 mol Ag and (b) the number of oxygen atoms in
20.5 mol O2.
(a) Mass in mg = 1.34 x 10-4 mol Ag x 107.868g x 1000mg = 14.5mg Ag
1 mol
1g
(b) # O atoms = 20.5 mol x 6.022 x 1023 atoms x _2 atoms = 2.48 x 1025 atoms of oxygen
1 mol
O2 molecule
Exercise 3.3B
Calculate (a) the number of moles of Al in a cube of aluminum metal 5.5 cm on an edge
(d = 2.70 g/cm3) and (b) the volume occupied by 4.06 x 1024 Br atoms present as Br2 molecules in
liquid bromine (d = 3.12 g/mL).
(a)
Moles Al = 166 cm3 of Al x 2.70g x 1 mole Al =
cm3
26.982g
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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
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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.
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We can read formulas in terms of
moles of atoms or ions.
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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
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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
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Percentage Composition of Butane
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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?
Example 3.8
An Estimation Example
Without doing detailed calculations, determine which of
these compounds contains the greatest mass of sulfur
per gram of compound: barium sulfate, lithium sulfate,
sodium sulfate, or lead sulfate.
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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.
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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.
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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.
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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?
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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.
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Elemental Analysis (cont’d)
… H2O, which is absorbed
by MgClO4, and …
The sample is
burned in a
stream of oxygen
gas, producing …
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… CO2, which is
absorbed by NaOH.
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Elemental Analysis (cont’d)
If our sample
were CH3OH,
every two
molecules of
CH3OH …
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… would give two
molecules of CO2 …
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… and four
molecules of H2O.
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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.
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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.
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Writing Chemical Equations
• Sometimes
additional
information
about the
reaction is
conveyed in
the equation.
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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 …
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Guidelines for Balancing
Chemical Equations
• If an element is present in just one compound on each
side of the equation, try balancing that element first.
• Balance any reactants or products that exist as the
free element last.
• In some reactions, certain groupings of atoms (such
as polyatomic ions) remain unchanged. In such cases,
treat these groupings as a unit.
• At times, an equation can be balanced by first using a
fractional coefficient(s). The fraction is then cleared
by multiplying each coefficient by a common factor.
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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.
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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
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1 mol CO
–––––––––––––
1 mol CH3OH
2 mol H2
–––––––––––––
1 mol CH3OH
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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).
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Outline of Simple Reaction Stoichiometry
Note: preliminary and/or follow-up calculations may be needed.
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Example 3.17
When 0.105 mol propane is burned in an
excess of oxygen, how many moles of oxygen
are consumed? The reaction is
C3H8 + 5 O2  3 CO2 + 4 H2O
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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?
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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?
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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.
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Limiting Reactant Analogy
If we have 10
sandwiches, 18
cookies, and 12
oranges …
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… how many
packaged meals
can we make?
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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 …
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… 128 g (0.80 mol)
bromine, we get …
… 150 g of 1,2dibromoethane, and
leftover ethylene!
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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.
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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?
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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
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Actual Yield of ZnS Is Less than
the Theoretical Yield
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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
Example 3.22
A Conceptual Example
What is the maximum yield of CO(g) obtainable from 725 g
of C6H14(l), regardless of the reaction(s) used, assuming no
other carbon-containing reactant or product?
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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 …
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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.
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Preparing 0.01000 M KMnO4
Weigh 0.01000
mol (1.580 g)
KMnO4.
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Dissolve in water. How much
water? Doesn’t matter, as long as
we don’t go over a liter.
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Add more water
to reach the
1.000 liter mark.
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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.
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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.
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Visualizing the Dilution of a Solution
We start and
end with the
same amount of
solute.
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Addition of
solvent has
decreased the
concentration.
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Dilution Calculations …
•
•
•
•
… couldn’t be easier.
Moles of solute does not change on dilution.
Moles of solute = M × V
Therefore …
Mconc × Vconc = Mdil × Vdil
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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?
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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.
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Adding to the previous stoichiometry scheme …
57
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.
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… we can go from
moles of substance B
to either volume of B
or molarity of B.
How?
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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?
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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.
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Chapter Three