CHEMISTRY
The Molecular Nature of Matter and Change
Third Edition
Chapter 3
Lecture Outlines*
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Chapter 3
Stoichiometry
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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
3-3
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 Avogadro’s number and is abbrieviated
as N.
One mole (1 mol) contains 6.022x1023 entities
(to four significant figures)
3-4
Counting Objects of Fixed Relative Mass
Figure 3.1
12 red marbles @ 7g each = 84g
12 yellow marbles @4g each=48g
3-5
55.85g Fe = 6.022 x 1023 atoms Fe
32.07g S = 6.022 x 1023 atoms S
One Mole of
Common Substances
CaCO3
100.09 g
Oxygen
32.00 g
Water
18.02 g
Figure 3.2
Copper
63.55 g
3-6
Molecular Mass vs. Molar Mass ( M )
The Molecular mass of a compound expressed in amu is
numerically the same as the mass of one mole of the
compound expressed in grams.
For water:
H 2O
Molecular mass = (2 x atomic mass of H ) + atomic mass of O
= 2 ( 1.008 amu) + 16.00 amu = 18.02 amu
Mass of one molecule of water = 18.02 amu
Molar mass = ( 2 x atomic mass of H ) + atomic mass of O
= 2 ( 1.008 g ) + 16.00 g = 18.02 g
18.02 g H2O = 6.022 x 1023 molecules of water = 1 mole H2O
3-7
Sample Problem 3.1
PROBLEM:
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 the molar mass of Ag, M
SOLUTION: 0.0342mol Ag x 107.9g Ag = 3.69g Ag
mol Ag
PLAN: (b) To convert g of Fe to atoms we first
have to find the #mols of Fe and then
convert mols to atoms.
SOLUTION: 95.8g Fe x mol Fe
= 1.72mol Fe
55.85g Fe
6.022x1023atoms Fe = 1.04x1024 atoms
1.72mol Fe x
Fe
mol Fe
3-8
amount(mol) of Ag
multiply by M of Ag
(107.9g/mol)
mass(g) of Ag
mass(g) of Fe
divide by M of Fe
(55.85g/mol)
amount(mol) of Fe
multiply by 6.022x1023
atoms/mol
atoms of Fe
Calculating the Number of Moles and Atoms
in a Given Mass of Element
Problem: Tungsten (W) is the element used as the filament in light
bulbs, and has the highest melting point of any element
3680oC. How many moles of tungsten, and atoms of the
element are contained in a 35.0 mg sample of the metal?
Plan: ??
Solution: ??
Answer:
1.90 x 10 - 4 mol W
1.15 x 1020 atoms of Tungsten
3-9
Table 3.1 Summary of Mass Terminology
Term
Definition
Unit
Isotopic mass
Mass of an isotope of an element
amu
Atomic mass
(atomic weight)
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
Mass of 1 mole of chemical entities
(atoms, ions, molecules, formula units)
g/mol
Molecular mass
(formula mass or
molecular weight)
Molar mass (M)
(gram-molecular
weight)
3-10
Interconverting Moles, Mass, and Number of Chemical Entities
no. of grams
Mass (g) = no. of moles x
1 mol
No. of moles = mass (g) x
1 mol
no. of grams
No. of entities = no. of moles x
6.022x1023 entities
1 mol
1 mol
No. of moles = no. of entities x
6.022x1023 entities
3-11
Sample Problem 3.2
PROBLEM:
Calculating the Moles and Number of Formula Units
in a Given Mass of a Compound
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 Avogadro’s
number.
SOLUTION:
The formula is (NH4)2CO3.
mass(g) of (NH4)2CO3
divide by M
amount(mol) of (NH4)2CO3
multiply by 6.022x1023
formula units/mol
number of (NH4)2CO3 formula units
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
mol (NH4)2CO3
96.09g (NH4)2CO3
x
6.022x1023 formula units (NH4)2CO3
mol (NH4)2CO3
2.61x1023 formula units (NH4)2CO3
3-12
=
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 and Masses of
Elements in a Sample of Compound
PROBLEM: Glucose (C6H12O6) is the most important nutrient in the
living cell for generating chemical potential energy.
(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.
SOLUTION:
There are 6 moles of C per mole
glucose; 12 moles H; 6 moles O.
6 mol C 12 g C
mol C
= 72 g C
amount(mol) of element
X in 1mol compound
multiply by M(g/mol) of X
mass(g) of X in 1mol of
compound
divide by mass(g) of
1mol of compound
mass fraction of X
multiply by 100
mass % X in compound
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
Information Contained in the Chemical Formula of Glucose
C6H12O6 ( M = 180.16 g/mol)
Table 3.2
Carbon (C)
Hydrogen (H)
Oxygen (O)
Atoms per molecule
of compound
6 atoms
12 atoms
6 atoms
Moles of atoms per
mole of compound
6 moles
of atoms
12 moles
of atoms
6 moles
of atoms
Atoms per mole of
compound
6(6.022 x 1023)
atoms
12(6.022 x 1023)
atoms
6(6.022 x 1023)
atoms
Mass per 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 per mole of
compound
3-16
72.06 g
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 mol Na
= 0.123 mol Na
22.99g Na
mass(g) of each element
mol Cl
divide by M(g/mol)
4.35g Cl
= 0.123 mol Cl
35.45g Cl
amount(mol) of each element
mol O
7.83g O
= 0.489 mol O
use # of moles as subscripts
16.00 O
preliminary formula
Na1 Cl1 O3.98
NaClO4
change to integer subscripts
empirical formula
3-17
NaClO4 is sodium perchlorate.
Sample Problem 3.5
Determining a Molecular Formula from Elemental
Analysis and Molar Mass
PROBLEM: During physical activity. lactic acid (M=90.08g/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
preliminary formula
convert to integer subscripts
empirical formula
3-18
divide mol mass by
mass of empirical
formula to get a
multiplier
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 mol C
12.01g C
3.33mol C
C3.33
3.33
6.71g H
mol H
53.3g O
1.008g H
16.00g O
6.66mol H
3.33mol O
H6.66 O3.33
3.33 3.33
molar mass of lactate
CH2O
empirical formula
90.08g
3
mass of CH2O
3-19
mol O
30.03g
C3H6O3 is the
molecular formula
Sample Problem 3.6
PROBLEM:
Determining a Molecular Formula from Combustion
Analysis
Vitamin C (M=176.12g/mol) is a compound of C,H, and O
found in many natural sources especially citrus fruits. When a
1.000-g sample of vitamin C is placed in a combustion chamber
and burned, the following data are obtained:
mass of CO2 absorber after combustion
=85.35g
mass of CO2 absorber before combustion
=83.85g
mass of H2O absorber after combustion
=37.96g
mass of H2O absorber before combustion
=37.55g
What is the molecular formula of vitamin C?
PLAN:
difference (after-before) = mass of oxidized element
find the mass of each element in its combustion product
find the mols
3-20
preliminary
formula
empirical
formula
molecular
formula
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
3-21
Flow Chart of Mass Percentage Calculation
moles of X in one
mol of compound
M = (g/mol) of X
mass (g) of X in one
mol of compound
Divide by mass(g) of one mol
of compound
Mass fraction of X
Multiply by 100
Mass % of X
3-22
Sample Problem 3.6
continued
Determining a Molecular Formula from Combustion
Analysis
SOLUTION:
CO2
85.35g-83.85g = 1.50g
There are 12.01g C per mol CO2 .
37.96g-37.55g = 0.41g
H2O
12.01g CO2
1.50g CO2
= 0.409g C
44.01g CO2
There are 2.016g H per mol H2O .
0.41g H2O
2.016g H2O
= 0.046g H
18.02g H2O
O must be the difference:
0.409g C
= 0.0341mol C
12.01g C
C1H1.3O1
1.000g - (0.409 + 0.049) = 0.545
0.046g H
= 0.0461mol H
1.008g H
C3H4O3
= 0.0341mol O
16.00g O
176.12g/mol
88.06g
3-23
0.545g O
= 2.000
C6H8O6