Chapter

1
Chapter 3
Stoichiometry
2
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
mole - the amount of a substance that contains
the same number of entities as there are atoms in
exactly 12g of carbon-12.
This amount is 6.022x1023. The number is called
Avogadros number and is abbrieviated as N.
One mole (1 mol) contains 6.022x1023 entities (to
four significant figures)
4
Counting Objects of Fixed Relative Mass
Figure 3.1
12 red marbles _at_ 7g each 84g 12 yellow marbles
_at_4g each48g
55.85g Fe 6.022 x 1023 atoms Fe 32.07g S
6.022 x 1023 atoms S
5
Oxygen 32.00 g
One Mole of Common Substances
Water 18.02 g
CaCO3 100.09 g
Figure 3.2
Copper 63.55 g
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Table 3.1 Summary of Mass Terminology
Term
Definition
Unit
Isotopic mass
Mass of an isotope of an element
amu
Atomic mass
amu
Average of the masses of the naturally occurring
isotopes of an element weighted according to
their abundance
(also called atomic weight)
Molecular (or formula) mass
amu
Sum of the atomic masses of the atoms (or ions)
in a molecule (or formula unit)
(also called molecular weight)
Molar mass (M)
g/mol
Mass of 1 mole of chemical entities (atoms, ions,
molecules, formula units)
(also called gram-molecular weight)
7
Interconverting Moles, Mass, and Number of
Chemical Entities
8
Sample Problem 3.1
Calculating the Mass and the Number of Atoms in a
Given Number of Moles of an Element
(a) Silver (Ag) is used in jewelry and tableware
but no longer in U.S. coins. How many grams of
Ag are in 0.0342mol of Ag?
(b) Iron (Fe), the main component of steel, is
the most important metal in industrial society.
How many Fe atoms are in 95.8g of Fe?
PLAN
(a) To convert mol of Ag to g we have to use the
g Ag/mol Ag, the molar mass M.
multiply by M of Ag (107.9g/mol)
SOLUTION
3.69g Ag
0.0342mol Ag x
(b) To convert g of Fe to atoms we first have to
find the mols of Fe and then convert mols to
atoms.
PLAN
divide by M of Fe (55.85g/mol)
SOLUTION
95.8g Fe x
1.72mol Fe
multiply by 6.022x1023 atoms/mol
1.04x1024 atoms Fe
1.72mol Fe x
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Sample Problem 3.2
Calculating the Moles and Number of Formula Units
in a Given Mass of a Compound
PROBLEM
Ammonium carbonate is white solid that decomposes
with warming. Among its many uses, it is a
component of baking powder, first extinguishers,
and smelling salts. How many formula unit are in
41.6g of ammonium carbonate?
PLAN
After writing the formula for the
compound, we find its M by adding the masses of
the elements. Convert the given mass, 41.6g to
mols using M and then the mols to formula units
with Avogadros number.
divide by M
multiply by 6.022x1023 formula units/mol
SOLUTION
The formula is (NH4)2CO3.
M (2 x 14.01g/mol N)(8 x 1.008g/mol
H) (12.01g/mol C)(3 x 16.00g/mol O)
96.09g/mol
41.6g (NH4)2CO3 x
x

2.61x1023 formula units (NH4)2CO3
10
Mass of element X
atoms of X in formula x atomic mass of X (amu)
x 100
molecular (or formula) mass of compound(amu)
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Sample Problem 3.3
Calculating the Mass Percents and Masses of
Elements in a Sample of Compound
(a) What is the mass percent of each element in
glucose?
(b) How many grams of carbon are in 16.55g of
glucose?
PLAN
We have to find the total mass of glucose and the
masses of the constituent elements in order to
relate them.
amount(mol) of element X in 1mol compound
multiply by M(g/mol) of X
SOLUTION
mass(g) of X in 1mol of compound
There are 6 moles of C per mole glucose 12
moles H 6 moles O.
divide by mass(g) of 1mol of compound
6 mol C
72 g C
mass fraction of X
multiply by 100
mass X in compound
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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.
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Sample Problem 3.4
Determining the Empirical Formula from Masses of
Elements
PROBLEM
Elemental analysis of a sample of an ionic
compound gave the following results 2.82g of Na,
4.35g of Cl, and 7.83g of O. What are the
empirical formula and name of the compound?
PLAN
Once we find the relative number of moles of each
element, we can divide by the lowest mol amount
to find the relative mol ratios (empirical
formula).
SOLUTION
2.82g Na
0.123 mol Na
mass(g) of each element
divide by M(g/mol)
4.35g Cl
0.123 mol Cl
amount(mol) of each element
7.83g O
0.489 mol O
use of moles as subscripts
preliminary formula
Na1 Cl1 O3.98
NaClO4
Na1 Cl1 O3.98
NaClO4
change to integer subscripts
empirical formula
NaClO4 is sodium perchlorate.
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Sample Problem 3.5
Determining a Molecular Formula from Elemental
Analysis and Molar Mass
PROBLEM
During physical activity. lactic acid
(M90.08g/mol) forms in muscle tissue and is
responsible for muscle soreness. Elemental
anaylsis shows that it contains 40.0 mass C,
6.71 mass H, and 53.3 mass O.
(a) Determine the empirical formula of lactic
acid.
(b) Determine the molecular formula.
PLAN
assume 100g lactic acid and find the mass of each
element
divide each mass by mol mass(M)
amount(mol) of each element
molecular formula
use mols as subscripts
divide mol mass by mass of empirical formula to
get a multiplier
preliminary formula
convert to integer subscripts
empirical formula
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Sample Problem 3.5
Determining a Molecular Formula from Elemental
Analysis and Molar Mass
continued
SOLUTION
Assuming there are 100.g of lactic acid, the
constituents are
40.0g C
6.71g H
53.3g O
3.33mol C
6.66mol H
3.33mol O
C3.33
H6.66
O3.33
CH2O
empirical formula
C3H6O3 is the molecular formula
3
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Information Contained in the Chemical Formula of
Glucose C6H12O6 ( M 180.16 g/mol)
Table 3.2
Oxygen (O)
Carbon (C)
Hydrogen (H)
Atoms/molecule of compound
6 atoms
6 atoms
12 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/moleculeof compound
6(12.01 amu) 72.06 amu
12(1.008 amu) 12.10 amu
6(16.00 amu) 96.00 amu
Mass/mole of compound
96.00 g
72.06 g
12.10 g
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Sample Problem 3.6
Determining a Molecular Formula from Combustion
Analysis
difference (after-before) mass of oxidized
element
PLAN
find the mass of each element in its combustion
product
molecular formula
preliminary formula
empirical formula
find the mols
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Figure 3.4
Combustion Train for the Determination of the
Chemical Composition of Organic Compounds.
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Flow Chart of Mass Percentage Calculation
M (g/mol) of X
Divide by mass(g) of one mol of
compound
Multiply by 100
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Sample Problem 3.6
Determining a Molecular Formula from Combustion
Analysis
continued
SOLUTION
There are 12.01g C per mol CO2 .
1.50g CO2
0.409g C
There are 2.016g H per mol H2O .
0.41g H2O
0.046g H
O must be the difference 1.000g - (0.409
0.049) 0.545
0.0341mol C
0.0461mol H
0.0341mol O
C1H1.3O1
C3H4O3
2.000
C6H8O6