Chapter 3
Stoichiometry of Formulas and Equations
3-1
Mole - Mass Relationships in Chemical Systems
3.1 The Mole
3.2 Determining the Formula of an Unknown Compound
3.3 Writing and Balancing Chemical Equations
3.4 Calculating the Amounts of Reactant and Product
3.5 Fundamentals of Solution Stoichiometry
3-2
mole(mol) - the amount of a substance that contains the
same number of entities as there are atoms in exactly
12 g of carbon-12.
This amount is 6.022x1023. The number is called
Avogadro’s number and is abbreviated as N.
One mole (1 mol) contains 6.022x1023 entities (to four
significant figures)
3-3
Figure 3.1
Counting objects of fixed relative mass.
12 red marbles @ 7g each = 84g
12 yellow marbles @4g each=48g
3-4
55.85g Fe = 6.022 x 1023 atoms Fe
32.07g S = 6.022 x 1023 atoms S
Figure 3.2
Oxygen
32.00 g
One mole of
common
substances.
CaCO3
100.09 g
Water
18.02 g
Copper
63.55 g
3-5
Table 3.1
Summary of Mass Terminology
Term
Definition
Unit
Isotopic mass
Mass of an isotope of an element
amu
Atomic mass
Average of the masses of the naturally
occurring isotopes of an element
weighted according to their abundance
amu
Sum of the atomic masses of the atoms
(or ions) in a molecule (or formula unit)
amu
(also called
atomic weight)
Molecular
(or formula) mass
(also called
molecular weight)
Molar mass (M)
Mass of 1 mole of chemical entities
(also called
(atoms, ions, molecules, formula units)
gram-molecular weight)
3-6
g/mol
Information Contained in the Chemical Formula of Glucose
C6H12O6 ( M = 180.16 g/mol)
Table 3.2
Carbon (C)
Hydrogen (H)
Oxygen (O)
Atoms/molecule
of compound
6 atoms
12 atoms
6 atoms
Moles of atoms/
mole of compound
6 moles
of atoms
12 moles
of atoms
6 moles
of atoms
Atoms/mole of
compound
6(6.022 x 1023)
atoms
12(6.022 x 1023)
atoms
6(6.022 x 1023)
atoms
Mass/molecule
of compound
6(12.01 amu)
=72.06 amu
12(1.008 amu)
=12.10 amu
6(16.00 amu)
=96.00 amu
12.10 g
96.00 g
Mass/mole of
compound
3-7
72.06 g
Interconverting Moles, Mass, and Number of Chemical Entities
Mass (g) = no. of moles x
No. of moles = mass (g) x
no. of grams
1 mol
1 mol
no. of grams
No. of entities = no. of moles x
6.022x1023 entities
1 mol
No. of moles = no. of entities x
3-8
1 mol
6.022x1023 entities
g
M
Figure 3.3
MASS(g)
MASS(g)
of
ofelement
element
Summary of the mass-molenumber relationships for
elements.
M (g/mol)
AMOUNT(mol)
AMOUNT(mol)
of
ofelement
element
Avogadro’s
number
(atoms/mol)
ATOMS
ATOMS
of
ofelement
element
3-9
Sample Problem 3.1
PROBLEM:
Calculating the Mass and the Number of Atoms in a
Given Number of Moles of an Element
(a) A graduated cylinder contains 32.0 cm3 of mercury. If the density
of mercury at 25 ºC is 13.534 g/cm3, how many moles of mercury are
in the cylinder?
(b) How many atoms of mercury are there?
3-10
Figure 3.3
MASS(g)
MASS(g)
of
ofcompound
compound
Summary of the mass-molenumber relationships for
compounds.
M (g/mol)
AMOUNT(mol)
AMOUNT(mol)
of
ofcompound
compound
chemical
formula
Avogadro’s
number
(molecules/mol)
MOLECULES
MOLECULES
(or
(orformula
formulaunits)
units)
of
ofcompound
compound
3-11
AMOUNT(mol)
AMOUNT(mol)
of
ofelements
elementsinin
compound
compound
Sample Problem 3.2
Calculating the Moles and Number of Formula Units
in a Given Mass of a Compound
PROBLEM: You have 16.5 g of the common compound oxalic acid, C2H2O4.
(a) How many moles are represented by this mass?
(b) How many molecules of oxalic acid are in 16.5 g?
(c) How many atoms of carbon are in 16.5 g oxalic acid?
(d) Calculate the mass of one molecule of oxalic acid.
3-12
Mass percent from the chemical formula
Mass % of element X =
atoms of X in formula x atomic mass of X (amu)
x 100
molecular (or formula) mass of compound(amu)
Mass % of element X =
moles of X in formula x molar mass of X (amu)
molecular (or formula) mass of compound (amu)
3-13
x 100
Sample Problem 3.3
Calculating the Mass Percents of Elements in a
Sample of Compound
PROBLEM: Calculate the percentage composition of C12H22O11
3-14
Empirical and Molecular Formulas
Empirical Formula The simplest formula for a compound that agrees with
the elemental analysis and gives rise to the smallest set
of whole numbers of atoms.
Molecular Formula The formula of the compound as it exists, it may be a
multiple of the empirical formula.
3-15
Sample Problem 3.4
Determining the Empirical Formula from Masses
of Elements
PROBLEM: Mercury forms a compound with chlorine that is 73.9 % mercury
and 26.1 % chlorine by mass. What is the empirical formula of this
chloride of mercury?
3-16
Sample Problem 3.5
Determining a Molecular Formula from Elemental
Analysis and Molar Mass
PROBLEM: Eugenol is the active compound of oil of cloves. It is composed
of carbon, hydrogen and oxygen. Given the following information
for eugenol:
C 73.14 %, H 7.37 %, molecular mass = 164.2 g/mol
(a) Determine the empirical formula of eugenol.
(b) Determine the molecular formula.
3-17
Figure 3.4
Combustion train for the determination of the
chemical composition of organic compounds.
m
m
CnHm + (n+ ) O2 = n CO(g) +
H O(g)
2
2 2
3-18
Sample Problem 3.6
PROBLEM:
3-19
Determining a Molecular Formula from Combustion
Analysis
Isopropyl alcohol (M = 60.09 g/mol), a substance sold as
rubbing alcohol, is composed of C, H, and O. Combustion of
0.255 g of isopropyl alcohol produces 0.561 g CO2 and 0.306 g
H2O. Determine the empirical and molecular formulas of
isopropyl alcohol.
Table 3.3 Some Compounds with Empirical Formula CH2O
(Composition by Mass: 40.0% C, 6.71% H, 53.3% O)
Molecular
Formula
Whole-Number
Multiple
formaldehyde
CH2O
1
30.03
disinfectant; biological preservative
acetic acid
C2H4O2
2
60.05
acetate polymers; vinegar(5% soln)
lactic acid
C3H6O3
3
90.09
sour milk; forms in exercising muscle
erythrose
C4H8O4
4
120.10
part of sugar metabolism
ribose
C5H10O5
5
150.13
component of nucleic acids and B2
glucose
C6H12O6
6
180.16
major energy source of the cell
Name
CH2O
3-20
C2H4O2
C3H6O3
M
(g/mol)
C4H8O4
Use or Function
C5H10O5
C6H12O6
Table 3.4 Two Pairs of Constitutional Isomers
C4H10
Property
Butane
C2H6O
2-Methylpropane
Ethanol
Dimethyl Ether
M(g/mol)
58.12
58.12
46.07
46.07
Boiling Point
-0.50C
-11.060C
78.50C
-250C
Density at 200C 0.579 g/mL 0.549 g/mL
(gas)
(gas)
Structural
formulas
Space-filling
models
3-21
H
H
H
H
C
C
C
C
H
H
H
H
H
H
H
H
H
H
C
C
C
C
H
H
H
H
H
0.789 g/mL 0.00195 g/mL
(liquid)
(gas)
H
H
H
H
C
C
H
H
H
H
OH H
C
H
O
C
H
H
Figure 3.6
The formation of HF gas on the macroscopic and molecular levels.
3-22
translate the statement
balance the atoms
adjust the coefficients
check the atom balance
specify states of matter
3-23
Sample Problem 3.7
PROBLEM:
Balancing Chemical Equations
Write balanced equations for the following reactions:
(a) P4(s) + Cl2(g) → PCl3(l)
(b) Combustion of propane, C3H8
(c) Combustion of butane, C4H10
3-24
Figure 3.8
Summary of the mass-mole-number relationships
in a chemical reaction.
MASS(g)
MASS(g)
of
ofcompound
compoundAA
MASS(g)
MASS(g)
of
ofcompound
compoundBB
M (g/mol) of
compound A
AMOUNT(mol)
AMOUNT(mol)
of
ofcompound
compoundAA
molar ratio from
balanced equation
Avogadro’s number
(molecules/mol)
MOLECULES
MOLECULES
(or
(orformula
formulaunits)
units)
of
ofcompound
compoundAA
3-25
M (g/mol) of
compound B
AMOUNT(mol)
AMOUNT(mol)
of
ofcompound
compoundBB
Avogadro’s number
(molecules/mol)
MOLECULES
MOLECULES
(or
(orformula
formulaunits)
units)
of
ofcompound
compoundBB
Sample Problem 3.8
PROBLEM:
Calculating Amounts of Reactants and Products
In the balanced reaction between P4(s) and Cl2(g) to give
PCl3(l):
(a) calculate the mass of Cl2 required if all of 1.45 g of P4 is to
react.
(b) What mass of PCl3(l) is formed?
3-26
Sample Problem 3.9
PROBLEM:
3-27
Calculating Amounts of Reactants and Products
An alloy has the composition: 93.7 % Al and 6.3 % Cu by mass
and density = 2.85 g/cm3. A piece of alloy of volume 0.691
cm3 reacts with excess HCl(g) to afford aluminum chloride
and hydrogen gas only. Assuming all of the Al, but no Cu,
reacts, calculate the mass of H2 obtained.
Sample Problem 3.10 Writing an Overall Equation for a Reaction Sequence
PROBLEM:
Pure TiO2 is the pigment commonly used in white paint. It can
be purified by reacting crude TiO2 with elemental carbon in
the presence of chlorine. The resulting TiCl4 is a liquid,
which can be purified by distillation and then reacted with
oxygen to reproduce pure TiO2.
2TiO2(s) + 3C(s) + 4Cl2(g) → 2TiCl4(l) + CO2(g) + 2CO(g)
TiCl4(l) + O2(g) → TiO2(s) + 2Cl2(g)
(a) What is the overall balanced equation for this reaction?
(b) How much carbon is required to produce 1.00 kg of pure
TiO2(s)?
3-28
Figure 3.10
3-29
An ice cream sundae analogy for limiting reactions.
Table 3.5 Information Contained in a Balanced Equation
Viewed in
Terms of
molecules
amount (mol)
mass (amu)
mass (g)
total mass (g)
3-30
Reactants
C3H8(g) + 5O2(g)
Products
3CO2(g) + 4H2O(g)
1 molecule C3H8 + 5 molecules O2
3 molecules CO2 + 4 molecules H2O
1 mol C3H8 + 5 mol O2
3 mol CO2 + 4 mol H2O
44.09 amu C3H8 + 160.00 amu O2
132.03 amu CO2 + 72.06 amu H2O
44.09 g C3H8 + 160.00 g O2
132.03 g CO2 + 72.06 g H2O
204.09 g
204.09 g
Sample Problem 3.11 Calculating Amounts of Reactant and Product in
Reactions Involving a Limiting Reactant
PROBLEM:
3-31
In the balanced reaction between P4(s) (125 g) and Cl2(g) (323
g) to give PCl3(l), calculate the mass of PCl3(l) that can be
formed.
Figure 3.11
The effect of side reactions on yield.
A +B
C
(reactants)
(main product)
D
(side products)
3-32
Sample Problem 3.12 Calculating Percent Yield
PROBLEM:
3-33
In the balanced reaction between P4(s) (125 g) and Cl2(g) (323
g) to give PCl3(l), the mass of PCl3(l) that is actually isolated is
372 g. Calculate the percentage yield for the reaction.
Sample Problem 3.13 Calculating the Molarity of a Solution
PROBLEM:
3-34
Calculate the molarity of a solution made by dissolving 23.4 g of
sodium sulfate (Na2SO4) in enough water to form 125 mL of
solution.
Figure 3.12
Summary of
mass-mole-number-volume
relationships in solution.
MASS (g)
of compound
in solution
M (g/mol)
AMOUNT (mol)
of compound
in solution
Avogadro’s number
(molecules/mol)
MOLECULES
(or formula units)
of compound
in solution
3-35
M (g/mol)
VOLUME (L)
of solution
Sample Problem 3.14 Calculating Mass of Solute in a Given Volume of
Solution
PROBLEM:
3-36
How many grams of Na2SO4 are required to make 0.350 L of
0.500 M Na2SO4?
Figure 3.12
Laboratory preparation of molar solutions.
A
•Weigh the solid needed.
•Transfer the solid to a
volumetric flask that contains
about half the final volume of
solvent.
3-37
C Add solvent until the solution
reaches its final volume.
B Dissolve the solid
thoroughly by swirling.
Figure 3.13
Converting a concentrated solution to a dilute solution.
3-38
Sample Problem 3.15 Preparing a Dilute Solution from a Concentrated
Solution
PROBLEM: How many milliliters of 3.0 M H2SO4 are needed to make 450
mL of 0.1 M H2SO4?
MdilxVdil = #mol solute = MconcxVconc
3-39
Sample Problem 3.16 Calculating Amounts of Reactants and Products for
a Reaction in Solution
PROBLEM:
The following is an example of an acid-base neutralization
reaction that we will meet in the next chapter:
Ca(OH)2 (s) + 2HNO3 (aq) → Mg(NO3)2 (aq) + 2H2O (l)
How many grams of Ca(OH)2 are needed to neutralize (react
with completely) 25.0 mL of 0.100 M HNO3?
3-40
Sample Problem 3.17 Solving Limiting-Reactant Problems for Reactions
in Solution
PROBLEM:
A sample of 70.5 mg of potassium phosphate is added to 15.0
mL of 0.050 M silver nitrate, resulting in the formation of a
precipitate of silver phosphate:
K3PO4 (aq) + 3AgNO3 (aq) → Ag3PO4 (s) + 2KNO3 (aq)
(a) Calculate the theoretical yield, in grams, of the precipitate
that forms.
3-41
Figure 3.15
3-42
An overview of the key mass-mole-number
stoichiometry relationships.