Chapter 3
Stoichiometry of Formulas and Equations
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
1
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)
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)
g/mol
2
Information Contained in the Chemical Formula of Glucose
C6H12O6 ( M = 180.16 g/mol)
Table
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
72.06 g
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
g
M
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
Figure 3.3
MASS(g)
MASS(g)
ofofelement
element
Summary of the mass-molenumber relationships for
elements.
M (g/mol)
AMOUNT(mol)
AMOUNT(mol)
ofofelement
element
Avogadro’s
number
(atoms/mol)
ATOMS
ATOMS
ofofelement
element
Figure 3.3
MASS(g)
MASS(g)
ofofcompound
compound
Summary of the mass-molenumber relationships for
compounds.
M (g/mol)
AMOUNT(mol)
AMOUNT(mol)
ofofcompound
compound
chemical
formula
AMOUNT(mol)
AMOUNT(mol)
ofofelements
elementsinin
compound
compound
Avogadro’s
number
(molecules/mol)
MOLECULES
MOLECULES
(or
(orformula
formulaunits)
units)
ofofcompound
compound
4
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 41.6 g 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, 41.6 g 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 14.01 g/mol N)+(8 x 1.008 g/mol H)
+(12.01 g/mol C)+(3 x 16.00 g/mol O) = 96.09 g/mol
mol (NH4)2CO3 6.022x1023 formula units (NH4)2CO3
41.6 g (NH4)2CO3 x
x
=
96.09 g (NH4)2CO3
mol (NH4)2CO3
2.61x1023 formula units (NH4)2CO3
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)
x 100
molecular (or formula) mass of compound (amu)
5
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.
amount(mol) of element X
in 1mol compound
multiply by M(g/mol) of X
mass(g) of X in 1mol of
compound
SOLUTION:
(a)
divide by mass (g) of
1mol of compound
Per mole glucose there are
6 moles of C, 12 moles H, 6 moles O
mass fraction of X
Molar mass of glucose :
multiply by 100
M = 180.16 g/mol
mass % X in compound
continued
Calculating the Mass Percents and Masses of Elements in a Sample
of Compound
6 mol C ×
12.01 g C
6 mol O ×
16.00 g O
(b)
= 72.06 g C
12 mol H ×
mol C
= 96.00 g O
1.008 g H
= 12.096 g H
mol H
M = 180.16 g/mol
mol O
mass percent of C =
72.06 g C
180.16 g glucose
12.096 g H
mass percent of H =
mass percent of O =
180.16 g glucose
96.00 g O
180.16 g glucose
= 0.3999 × 100% = 39.99 mass %C
= 0.06714 × 100% = 6.714 mass %H
= 0.5329 × 100% = 53.29 mass %O
6
Empirical and Molecular Formulas
Empirical Formula An empirical formula indicates the relative number of atoms of each
element in the compound. It is the simplest type of 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 A molecular formula shows the actual number of atoms of each element
in a molecule of the compound.
The formula of the compound as it exists, it may be a multiple of the
empirical formula.
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)
x 100
molecular (or formula) mass of compound (amu)
7
Determining the Empirical Formula from Masses of Elements
PROBLEM: Elemental analysis of a sample of an ionic compound showed 2.82 g of
Na, 4.35 g of Cl, and 7.83 g 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: 2.82 g Na
mass(g) of each element
divide by M (g/mol)
4.35 g Cl
mol Na
22.99 g Na
mol Cl
35.45 g Cl
amount(mol) of each element
use # of moles as subscripts
7.83 g O
mol O
16.00 g O
= 0.123 mol Na
= 0.123 mol Cl
= 0.489 mol O
preliminary formula
Na1 Cl1 O3.98
NaClO4
change to integer subscripts
NaClO4 is sodium perchlorate.
empirical formula
Determining a Molecular Formula from Elemental Analysis and Molar Mass
PROBLEM: During physical activity. lactic acid (M=90.08 g/mol) forms in muscle tissue
and is responsible for muscle soreness. Elemental anaylsis shows that this
compound 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
use # mols as subscripts
preliminary formula
convert to integer subscripts
empirical formula
divide mol mass by mass of empirical formula to get a multiplier
molecular formula
8
continued
Determining a Molecular Formula from Elemental Analysis and Molar Mass
Assuming there are 100. g of lactic acid, the constituents are
SOLUTION:
40.0 mass % C, 6.71 mass % H, and 53.3 mass % O.
40.0 g C
mol C
6.71 g H
12.01g C
mol H
1.008 g H
3.33 mol C
H6.66 O3.33
3.33
3.33 3.33
CH2O
molar mass of lactate
90.08 g
mass of CH2O
30.03 g
Chemical Equations
CO2 + H2O
C8H18 + 25/2 O2
2C8H18 + 25O2
2C8H18(l) + 25O2 (g)
8 CO2 + 9 H2O
16CO2 + 18 H2O
16 CO2 (g) + 18 H2O (g)
mol O
16.00 g O
6.66 mol H
C3.33
C8H18 + O2
53.3 g O
3.33 mol O
empirical formula
3
C3H6O3 is the
molecular formula
Reactants is denoted as the
substance involved in a chemical
reaction. (write on the LEFT side)
Chemical reaction yields one or
more products which are, in
general, different from the
reactants. (write on the RIGHT
side)
The coefficients – the relative
amount of the reactants and
products. (write before formula)
• Chemical reaction is defined as a process that results in the
interconversion of chemical substances.
• Chemical reaction is characterized by a chemical change.
• Chemical reaction encompasses changes that strictly involve the
motion of electrons in the forming and breaking of chemical bonds.
• Chemical reaction is also called in particular the notion of a chemical
equation.
9
Law of conservation of mass/matter
Law of mass/matter conservation
translate the statement
Lomonosov Lavoisier law
balance the atoms
Law of mass/matter conservation states that
the mass of a closed system of substances will
remain constant,
constant regardless of the processes
acting inside the system.
Law of mass/matter conservation indicated
that matter cannot be neither created nor
destroyed,
destroyed although it may change form.
Las of mass/matter conservation implies that
for any chemical process in a closed system,
the mass of the reactants must equal the
mass of the products.
products
adjust the coefficients
check the atom balance
specify states of matter
Balancing Chemical Equations
PROBLEM: Within the cylinders of a car’s engine, the hydrocarbon octane (C8H18), one
of many components of gasoline, mixes with oxygen from the air and burns
to form carbon dioxide and water vapor. Write a balanced equation for this
reaction.
PLAN:
SOLUTION:
translate the statement
C8H18 + O2
CO2 + H2O
balance the atoms
C8H18 + 25/2 O2
adjust the coefficients
2C8H18 + 25O2
16CO2 + 18H2O
check the atom balance
2C8H18 + 25O2
16CO2 + 18H2O
specify states of matter
2C8H18(l) + 25O2 (g)
8 CO2 + 9 H2O
16CO2 (g) + 18H2O (g)
10
Stoichiometry : mass-mole-number relationships
Summary of the mass-mole-number relationships in a chemical reaction.
MASS
MASS(g)
(g)
ofofcompound
compoundAA
MASS(g)
MASS(g)
ofofcompound
compoundBB
M (g/mol) of
compound A
M (g/mol) of
compound B
molar ratio from
AMOUNT
AMOUNT(mol)
(mol)
ofofcompound
compoundAA
AMOUNT(mol)
AMOUNT(mol)
ofofcompound
compoundBB
balanced equation
Avogadro’s number
(molecules/mol)
Avogadro’s number
(molecules/mol)
MOLECULES
MOLECULES
(or
(orformula
formulaunits)
units)
ofofcompound
compoundAA
MOLECULES
MOLECULES
(or
(orformula
formulaunits)
units)
ofofcompound
compoundBB
Calculating Amounts of Reactants and Products
PROBLEM:
In a lifetime, the average American uses 1750 lb (794 g) of copper in
coins, plumbing, and wiring. Copper is obtained from sulfide ores,
such as chalcocite, or copper(I) sulfide, by a multistep process. After
an initial grinding, the first step is to “roast” the ore (heat it strongly
with oxygen gas) to form powdered copper(I) oxide and gaseous sulfur
dioxide.
(a) How many moles of oxygen are required to roast 10.0 mol of
copper(I) sulfide?
(b) How many grams of sulfur dioxide are formed when 10.0 mol of
copper(I) sulfide is roasted?
(c) How many kilograms of oxygen are required to form 2.86 kg of
copper(I) oxide?
PLAN:
write and balance equation
find mols O2
find mols SO2
find g SO2
find mols Cu2O
find mols O2
find kg O2
11
Calculating Amounts of Reactants and Products
SOLUTION:
2Cu2S(s) + 3O2(g)
2Cu2O(s) + 2SO2(g)
(a) How many moles of oxygen are required to roast 10.0 mol of copper(I) sulfide?
(a)
3 mol O2
10.0 mol Cu2S ×
= 15.0 mol O2
2 mol Cu2S
(b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide is roasted?
(b) 10.0 mol Cu S ×
2
2 mol SO2
×
2 mol Cu2S
64.07g SO2
= 641g SO2
mol SO2
(c) How many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide?
3
(c) 2.86 kg Cu O × 10 g Cu2O × mol Cu2O
2
143.10 g Cu2O
kg Cu2O
20.0 mol Cu2O ×
3 mol O2
×
2 mol Cu2O
32.00 g O2
mol O2
×
= 20.0 mol Cu2O
kg O2
103 g O2
= 0.959 kg O2
Take-homoe message
Writing an Overall Equation for a Reaction Sequence
PROBLEM:
Roasting is the first step in extracting copper from chalcocite, the ore
used in the previous problem. In the next step, copper(I) oxide reacts
with powdered carbon to yield copper metal and carbon monoxide gas.
Write a balanced overall equation for the two-step process.
SOLUTION:
PLAN:
write balanced equations for each step
cancel reactants and products common to
both sides of the equations
sum the equations
2Cu2S(s) + 3O2(g)
2Cu2O(s) + 2C(s)
2Cu2S(s)+3O2(g)+2C(s)
2Cu2O(s) + 2SO2(g)
4Cu(s) + 2CO(g)
4Cu(s)+2SO2(g)+2CO(g)
12
One reactant is present in limited supply
An ice cream sundae analogy for limiting reactions.
Calculating Amounts of Reactant and Product in Reactions
Involving a Limiting Reactant
PROBLEM:
A fuel mixture used in the early days of rocketry is composed of two
liquids, hydrazine(N2H4) and dinitrogen tetraoxide(N2O4), which ignite
on contact to form nitrogen gas and water vapor. How many grams of
nitrogen gas form when 1.00×102 g of N2H4 and 2.00×102 g of N2O4
are mixed?
PLAN: We always start with a balanced chemical equation and find the number of mols
of reactants and products which have been given.
In this case one of the reactants is in molar excess and the other will limit the
extent of the reaction.
mass of N2H4
mass of N2O4
limiting mol N2
divide by M
multiply by M
mol of N2H4
mol of N2O4
g N2
molar ratio
mol of N2
mol of N2
13
Calculating Amounts of Reactant and Product in Reactions Involving a
Limiting Reactant
2 N2H4(l) + N2O4(l)
SOLUTION:
mol N2H4
1.00×102g N2H4
= 3.12 mol N2H4
32.05g N2H4
3 mol N2
3.12 mol N2H4 ×
3 N2(g) + 4 H2O(l)
N2H4 is the limiting
reactant because it
produces less
product, N2, than
does N2O4.
= 4.68 mol N2
2mol N2H4
2.00 × 102g N2O4 ×
2.17mol N2O4 ×
mol N2O4
92.02 g N2O4
3 mol N2
= 2.17 mol N2O4
= 6.51 mol N2
mol N2O4
The percent yield
The effect of side reactions on yield.
A +B
C
(main product)
(reactants)
D
(side products)
write balanced equation
find mol reactant & product
find g product predicted
actual yield / theoretical yield × 100%
percent yield
14
Calculating Percent Yield
PROBLEM:
Silicon carbide (SiC) is an important ceramic material that is made by
allowing sand(silicon dioxide, SiO2) to react with powdered carbon at
high temperature. Carbon monoxide is also formed. When 100.0 kg
of sand is processed, 51.4 kg of SiC is recovered. What is the percent
yield of SiC from this process?
PLAN:
SOLUTION:
SiO2(s) + 3C(s)
write balanced equation
find mol reactant & product
100.0 kg SiO2
SiC(s) + 2CO(g)
103 g SiO2
mol SiO2
kg SiO2
60.09 g SiO2
= 1664 mol SiO2
mol SiO2 = mol SiC = 1664
find g product predicted
1664 mol SiC
actual yield/theoretical yield x 100
103g
mol SiC
51.4 kg
percent yield
40.10 g SiC kg
= 66.73 kg
x100 =77.0%
66.73 kg
Calculating the Molarity of a Solution
PROBLEM:
PLAN:
Glycine (H2NCH2COOH) is the simplest amino acid. What is the
molarity of an aqueous solution that contains 0.715 mol of glycine in
495 mL?
Molarity is the number of moles of solute per liter of solution.
mol of glycine
divide by volume
concentration (mol/mL) glycine
SOLUTION:
0.715 mol glycine
1000mL
495 mL soln
1L
103mL = 1L
molarity(mol/L) glycine
= 1.44 M glycine
15
Determining a Molecular Formula from Combustion Analysis
PROBLEM:
Vitamin C (M = 176.12 g/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
= 85.35 g
mass of CO2 absorber before combustion
= 83.85 g
mass of H2O absorber after combustion
= 37.96 g
mass of H2O absorber before combustion
= 37.55 g
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
preliminary
formula
find the mols
empirical
formula
molecular
formula
continued
Determining a Molecular Formula from Combustion Analysis
SOLUTION:
CO2
85.35 g - 83.85 g = 1.50 g
There are 12.01 g C per mol CO2
H2O
1.50 g CO2
37.96 g - 37.55 g = 0.41 g
12.01 g CO2
= 0.409 g C
44.01 g CO2
There are 2.016 g H per mol H2O.
0.41 g H2O
2.016 g H2O
= 0.046 g H
18.02 g H2O
O must be the difference: 1.000 g - (0.409 + 0.046) = 0.545
0.409 g C
= 0.0341 mol C
12.01 g C
= 0.0456 mol H
1.008 g H
C0.0341H0.0456O0.0341
C1H1.3O1
0.046 g H
Empirical formula
Assume : (C3H4O3)n
176.12 g / mol
88.06 g / mol
= 0.0341 mol O
16.00 g O
Preliminary formula
C3H4O3
0.545 g O
= 2.000
C6H8O6
Molecular formula
16
Fundamental premise of hydrated compounds
1.
Hydrate – water molecules that are chemically bound to the ions of salt as
part of the structure of the compound.
2.
Waters of crystallization - chemically bound to the ions of the salts in its
crystalline structure.
3.
Anhydrous salt – the salts has water molecules loosely bound to the ions
and water can be removed after heat-treatment.
4.
Efflorescent – hydrated salts spontaneously lose water molecules to
atmosphere.
5.
Deliquescent – the salts absorb water. (In our lab, CaCl2 is a deliquescent
salt which can be used as a desiccant in desiccators since CaCl2 can absorb
water)
6.
Name salt as before and add hydrate (H2O) with Greek prefix:
CoBr3Ÿ6H2O = Cobalt (III) bromide hexahydrate
Hydrated salts and calculation of water percent
Hydrate: water molecules are chemically bound to the ions of the salt as part of its
crystalline structure. These waters are called waters of crystallization
MgSO4 • 7H2O (epsom salt),
CuSO4 • 5H2O,
FeCl3 • 6H2O,
CaCl2 • 2H2O
∆ ( > 200 °C)
MgSO4 • 7H2O
MgSO4
+
Mass of water
The mass percent of water =
7H2O
× 100 %
Mass of water + mass of MgSO4
Assume there is 1 mol of Epsom salt,
mass of water = 126.1 g,
mass of MgSO4 = 120.4 g
Percent by mass of water in the salt is 51.16 %
17