Chapter 1: Stoichiometry of Formulas

and Equations
The Mole
Determining the Formula of an Unknown Compound

Writing and Balancing Chemical Equations

Calculating Quantities of Reactant and Product

Fundamentals of Solution Stoichiometry

The Role of Water as a Solvent

Writing Equations for Aqueous Ionic Reactions

3-0
 The Mole

The mole (mol) is the amount of a substance that

contains the same number of entities as there are
atoms in exactly 12 g of carbon-12.

The term “entities” refers to atoms, ions, molecules,

formula units, or electrons – in fact, any type of particle.

One mole (1 mol) contains 6.022x1023 entities (to

four significant figures).
This number is called Avogadro’s number and is
abbreviated as N.

3-1
 For molecular elements and for compounds, the
formula is needed to determine the molar mass.

The molar mass of O2 = 2 x M of O

= 2 x 16.00
= 32.00 g/mol

The molar mass of SO2 = 1 x M of S + 2 x M of O

= 32.07 + 2(16.00)
= 64.07 g/mol

3-2
 Converting Between Amount, Mass, and
Number of Chemical Entities

Mass (g) = amount (mol) x molar mass (g/mol)

1 mol
Amount(mol ) = mass (g) x
Molar mass(g)

6.022x1023 entities
No. of entities = Amount(mol ) x
1 mol
1 mol
Amount(mol ) = no. of entities x
6.022x1023 entities

3-3
 Mass-mole-number relationships for elements.

3-4
 Calculating the Number of Entities in a
Given Amount of an Element

PROBLEM: Gallium (Ga) is a key element in solar panels, calculators and

other light-sensitive electronic devices. How many Ga atoms
are in 2.85 x 10-3 mol of gallium?

PLAN: To convert mol of Ga to number of Ga atoms we need to

use Avogadro’s number.

mol of Ga

multiply by 6.022x1023 atoms/mol

atoms of Ga

3-5
 Sample Problem 3.2

SOLUTION:

2.85 x 10-3 mol Ga x 6.022x1023 Ga atoms

1 mol Ga

= 1.72 x 1021 Ga atoms

3-6
 Calculating the Number of Entities in a
Given Mass of an Element

PROBLEM: Iron (Fe) is the main component of steel and is therefore the
most important metal in society; it is also essential in the body.
How many Fe atoms are in 95.8 g of Fe?

PLAN: The number of atoms cannot be calculated directly from the

mass. We must first determine the number of moles of Fe
atoms in the sample and then use Avogadro’s number.

mass (g) of Fe
divide by M of Fe (55.85 g/mol)

amount (mol) of Fe
multiply by 6.022x1023 atoms/mol

atoms of Fe

3-7
 SOLUTION:

95.8 g Fe x 1 mol Fe
= 1.72 mol Fe
55.85 g Fe

1.72 mol Fe x 6.022x10 23 atoms Fe

1 mol Fe

= 1.04 x 1024 atoms Fe

3-8
 Amount-mass-number relationships for compounds.

3-9
 Calculating the Number of Chemical Entities in
a Given Mass of a Compound I
PROBLEM: Nitrogen dioxide is a component of urban smog that forms
from the gases in car exhausts. How many molecules are
in 8.92 g of nitrogen dioxide?

PLAN: Write the formula for the compound and calculate its molar
mass. Use the given mass to calculate first the number of
moles and then the number of molecules.

mass (g) of NO2

divide by M

amount (mol) of NO2

multiply by NA (molecules/mol)

number of NO2 molecules

3-10
 SOLUTION: NO2 is the formula for nitrogen dioxide.

M = (1 x M of N) + (2 x M of O)
= 14.01 g/mol + 2(16.00 g/mol)
= 46.01 g/mol

8.92 g NO2 x 1 mol NO2

= 0.194 mol NO2
46.01 g NO2

0.194 mol NO2 x 6.022x10 23 molecules NO

2

1 mol NO2

= 1.17 x 1023 molecules NO2

3-11
 Calculating the Number of Chemical Entities in
a Given Mass of a Compound II

PROBLEM: Ammonium carbonate, a white solid that decomposes on

warming, is an component of baking powder, fire
extinguishers and smelling salts.
a) How many formula units are in 41.6 g of ammonium
carbonate?
b) How many O atoms are in this sample?
PLAN:
Write the formula for the compound and calculate its molar mass. Use
the given mass to calculate first the number of moles and then the
number of formula units.
The number of O atoms can be determined using the formula and the
number of formula units.

3-12
 mass (g) of (NH4)2CO3
divide by M

amount (mol) of (NH4)2CO3

multiply by 6.022 x 1023 formula units/mol

number of (NH4)2CO3 formula units

3 O atoms per formula unit of (NH4)2CO3

number of O atoms

SOLUTION: (NH4)2CO3 is the formula for ammonium carbonate.

M = (2 x M of N) + (8 x M of H) + (1 x M of C) + (3 x M of O)
= (2 x 14.01 g/mol) + (8 x 1.008 g/mol)
+ (12.01 g/mol) + (3 x 16.00 g/mol)

= 96.09 g/mol

3-13
 41.6 g (NH4)2CO3 x 1 mol (NH4)2CO3
= 0.433 mol (NH4)2CO3
96.09 g (NH4)2CO3

0.433 mol (NH4)2CO3 x 6.022x10 formula units (NH4)2CO3

23

1 mol (NH4)2CO3

= 2.61x1023 formula units (NH4)2CO3

2.61x1023 formula units (NH4)2CO3 x 3 O atoms

1 formula unit of (NH4)2CO3

= 7.83 x 1023 O atoms

3-14
 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 (g/mol)

x 100
mass (g) of 1 mol of compound

3-15
 Calculating the Mass Percent of Each
Element in a Compound from the Formula

PROBLEM: In mammals, lactose(milk sugar) is metabolized to glucose

(C6H12O6) ,the key nutrient for generating chemical potential
energy. What is the mass percent of each element in
glucose?
PLAN: Find the molar mass of glucose, which is the mass of 1 mole
of glucose. Find the mass of each element in 1 mole of
glucose, using the molecular formula.
The mass % for each element is calculated by dividing the
mass of that element in 1 mole of glucose by the total mass
of 1 mole of glucose, multiplied by 100.

3-16
 PLAN:
amount (mol) of element X in 1 mol glucose
multiply by M (g/mol) of X

mass (g) of X in 1 mol of glucose

divide by mass (g) of 1 mol of glucose

mass fraction of X in glucose

multiply by 100

mass % of X in glucose

3-17
 SOLUTION:
In 1 mole of glucose there are 6 moles of C, 12 moles H, and 6 moles O.

12.01 g C 1.008 g H
6 mol C x = 72.06 g C 12 mol H x = 12.096 g H
1 mol C 1 mol H

16.00 g O
6 mol O x = 96.00 g O M = 180.16 g/mol
1 mol O

72.06 g C
mass percent of C = = 0.4000 x 100 = 40.00 mass %C
180.16 g glucose

12.096 g H
mass percent of H = = 0.06714 x 100 = 6.714 mass %H
180.16 g glucose

3-18
 Mass Percent and the Mass of an Element

Mass percent can also be used to calculate the

mass of a particular element in any mass of a
compound.

Mass of element =

mass of element in 1 mol of compound

mass of compound x
mass of 1 mol of compound

3-19
 Calculating the Mass of an Element in a
Compound

PROBLEM: Use the information from Sample Problem before to

determine the mass (g) of carbon in 16.55 g of glucose.

PLAN: The mass percent of carbon in glucose gives us the relative

mass of carbon in 1 mole of glucose. We can use this
information to find the mass of carbon in any sample of
glucose.

mass of glucose sample

multiply by mass percent of C in glucose

mass of C in sample

3-20
 SOLUTION:
Each mol of glucose contains 6 mol of C, or 72.06 g of C.

Mass (g) of C = mass (g) of glucose x 6 mol x M of C (g/mol)

mass (g) of 1 mol of glucose

= 16.55 g glucose x 72.06 g C

= 6.620 g C
180.16 g glucose

3-21
 Empirical and Molecular Formulas
The empirical formula is the simplest formula for a
compound that agrees with the elemental analysis. It
shows the lowest whole number of moles and gives the
relative number of atoms of each element present.
The empirical formula for hydrogen peroxide is HO.

The molecular formula shows the actual number of

atoms of each element in a molecule of the compound.
The molecular formula for hydrogen peroxide is H2O2.

3-22
 Determining an Empirical Formula from
Amounts of Elements

PROBLEM: A sample of an unknown compound contains 0.21

mol of zinc, 0.14 mol of phosphorus, and 0.56 mol of
oxygen. What is its empirical formula?

PLAN: Find the relative number of moles of each element.

Divide by the lowest mol amount to find the relative mol
ratios (empirical formula).

amount (mol) of each element

use # of moles as subscripts

preliminary formula
change to integer subscripts

empirical formula

3-23
 SOLUTION: Using the numbers of moles of each element given, we
write the preliminary formula Zn0.21P0.14O0.56

Next we divide each fraction by the smallest one; in this case 0.14:
0.21 0.14 0.56
= 1.5 = 1.0 = 4.0
0.14 0.14 0.14
This gives Zn1.5P1.0O4.0

We convert to whole numbers by multiplying by the smallest integer

that gives whole numbers; in this case 2:

1.5 x 2 = 3 1.0 x 2 = 2 4.0 x 2 = 8

This gives us the empirical formula Zn3P2O8

3-24
 Determining an Empirical Formula from
Masses of Elements

PROBLEM: Analysis of a sample of an ionic compound yields 2.82 g

of Na, 4.35 g of Cl, and 7.83 g of O. What is the empirical
formula and the name of the compound?
PLAN: Find the relative number of moles of each element. Divide
by the lowest mol amount to find the relative mol ratios
(empirical formula).
mass (g) of each element
divide by M (g/mol)

amount (mol) of each element

use # of moles as subscripts

preliminary formula
change to integer subscripts
empirical formula

3-25
 1 mol Na
SOLUTION: 2.82 g Na x = 0.123 mol Na
22.99 g Na

4.35 g Cl x 1 mol Cl = 0.123 mol Cl

35.45 g Cl

7.83 g O x 1 mol O = 0.489 mol O

16.00 g O
0.123 0.489
Na and Cl = = 1 and O = = 3.98
0.123 0.123

The empirical formula is Na1Cl1O3.98 or NaClO4;

this compound is named sodium perchlorate.

3-26
 Determining the Molecular Formula

The molecular formula gives the actual numbers of

moles of each element present in 1 mol of compound.
The molecular formula is a whole-number multiple
of the empirical formula.

molar mass (g/mol)

= whole-number multiple
empirical formula mass (g/mol)

3-27
 Determining a Molecular Formula from
Elemental Analysis and Molar Mass
PROBLEM: During excessive physical activity, lactic acid (M = 90.08
g/mol) forms in muscle tissue and causes muscle soreness.
Elemental analysis shows it contains 40.0 mass % C, 6.71
mass % H, and 53.3 mass % O. Determine the empirical
formula and the molecular formula for lactic acid.
PLAN:
assume 100 g lactic acid; then mass % = mass in grams
divide each mass by M

amount (mol) of each element

use # mols as subscripts; convert to integers

empirical formula
divide M by the molar mass for the empirical
formula; multiply empirical formula by this number

molecular formula
3-28
 SOLUTION: Assuming there are 100. g of lactic acid;

40.0 g C x 1 mol C 6.71 g H x 1 mol H 53.3 g O x 1 mol O

12.01 g C 1.008 g H 16.00 g O
= 3.33 mol C = 6.66 mol H = 3.33 mol O

C3.33 H6.66 O3.33

CH2O empirical formula
3.33 3.33 3.33

molar mass of lactate 90.08 g/mol

=3
mass of CH2O 30.03 g/mol

(CH2O)× C3H6O3 is the

3 molecular formula

3-29
 Balancing a Chemical Equation

translate the statement magnesium and oxygen gas react to give

magnesium oxide:
Mg + O2 → MgO

balance the atoms using coefficients;

formulas cannot be changed

2Mg + O2 → 2MgO

adjust coefficients if necessary

check that all atoms balance specify states of matter

2Mg (s) + O2 (g) → 2MgO (s)

3-30
 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

25
C8H18 + O2 8 CO2 + 9 H2O
balance the atoms 2

adjust the coefficients 2C8H18 + 25O2 16CO2 + 18H2O

check the atoms balance 2C8H18 + 25O2 16CO2 + 18H2O

specify states of matter 2C8H18(l) + 25O2 (g) 16CO2 (g) + 18H2O (g)

3-31
 Stoichiometric Calculations

• The coefficients in a balanced chemical equation

– represent the relative number of reactant and product particles
– and the relative number of moles of each.
• Since moles are related to mass
– the equation can be used to calculate masses of reactants
and/or products for a given reaction.
• The mole ratios from the balanced equation are used as
conversion factors.

3-32
Viewed in Reactants Products
Terms of C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)

Molecules 1 molecule C3H8 + 5 molecules O2 3 molecules CO2 + 4 molecules H2O

Amount (mol) 1 mol C3H8 + 5 mol O2 3 mol CO2 + 4 mol H2O

Mass (amu) 44.09 amu C3H8 + 160.00 amu O2 132.03 amu CO2 + 72.06 amu H2O

Mass (g) 44.09 g C3H8 + 160.00 g O2 132.03 g CO2 + 72.06 g H2O

Total Mass (g) 204.09 g 204.09 g

3-33
 Summary of amount-mass-number relationships
in a chemical equation.

3-34
 Calculating Quantities of Reactants and
Products: Amount (mol) to Amount (mol)

PROBLEM: In a lifetime, the average canadian will use about 680 kg of copper
in coins, plumbing and wiring .Copper is obtained from sulfide ores
such an copper(I) sulfide(chalcocite) in a multistep process.It is
“roasted” (heated strongly with oxygen gas) to form powdered
copper(I) oxide and gaseous sulfur dioxide.
How many moles of oxygen are required to roast 10.0 mol of
copper(I) sulfide?

PLAN: write and balance the equation

use the mole ratio as a conversion factor

moles of oxygen

SOLUTION: 2 Cu2S (s) + 3 O2 (g) → 2 Cu2O (s) + 2 SO2 (g)

3 mol O2
10.0 mol Cu2S x = 15.0 mol O2
2 mol Cu2S

3-35
 Calculating Quantities of Reactants and
Products: Amount (mol) to Mass (g)

PROBLEM: During the process of roasting copper(I) sulfide, how

many grams of sulfur dioxide form when 10.0 mol of
copper(I) sulfide reacts?
PLAN: Using the balanced equation from the previous problem,
we again use the mole ratio as a conversion factor.

2 Cu2S (s) + 3 O2 (g) → 2 Cu2O (s) + 2 SO2 (g)

mol of copper(I) sulfide

use the mole ratio as a conversion factor

mol of sulfur dioxide

multiply by M of sulfur dioxide

mass of sulfur dioxide

3-36
 SOLUTION: 2 Cu2S (s) + 3 O2 (g) → 2 Cu2O (s) + 2 SO2 (g)

10.0 mol Cu2S x 2 mol SO2 x 64.07 g SO2

= 641 g SO2
2 mol Cu2S 1 mol SO2

3-37
 Calculating Quantities of Reactants and
Products: Mass to Mass

PROBLEM: During the roasting of copper(I) sulfide, how many

kilograms of oxygen are required to form 2.86 kg of
copper(I) oxide?
PLAN:

3-38
 SOLUTION: 2 Cu2S (s) + 3 O2 (g) → 2 Cu2O (s) + 2 SO2 (g)

3
2.86 kg Cu2O x 10 g x 1 mol Cu2O = 20.0 mol Cu2O
1 kg 143.10 g Cu2O

3 mol O2 32.00 g O2
20.0 mol Cu2O x x x 1 kg = 0.960 kg O2
2 mol Cu2O 1 mol O2 103 g

3-39
 Reactions in Sequence

• Reactions often occur in sequence.

• The product of one reaction becomes a reactant in the
next.
• An overall reaction is written by combining the reactions;
– any substance that forms in one reaction and reacts in the next
can be eliminated.

3-40
 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.

PLAN: Write individual balanced equations for each step.

Adjust the coefficients so that any common substances can
be canceled.
Add the adjusted equations together to obtain the overall
equation.

3-41
 SOLUTION: Write individual balanced equations for each step:
2Cu2S (s) + 3O2 (g) → 2Cu2O (s) + 2SO2 (g)
Cu2O (s) + C (s) → 2Cu (s) + CO (g)

Adjust the coefficients so that the 2 moles of Cu2O formed in reaction

1 are used up in reaction 2:

2Cu2S (s) + 3O2 (g) → 2Cu2O (s) + 2SO2 (g)

2Cu2O (s) + 2C (s) → 4Cu (s) + 2CO (g)

Add the equations together:

2Cu2S (s) + 3O2 (g) + 2C (s) → 2SO2 (g) + 4Cu (s) + 2CO (g)

3-42
 Limiting Reactants

• So far we have assumed that reactants are present in

the correct amounts to react completely.
• In reality, one reactant may limit the amount of product
that can form.
• The limiting reactant will be completely used up in the
reaction.
• The reactant that is not limiting is in excess – some of
this reactant will be left over.

3-43
 Using Molecular Depictions in a Limiting-
Reactant Problem
PROBLEM: Nuclear engineers use chlorine trifluoride to prepare
uranium fuel for power plants. The compound is formed
as a gas by the reaction of elemental chlorine and
fluorine. The circle in the margin shows a representative
portion of the reaction mixture before the reaction starts.
(Chlorine is green, and fluorine is yellow.)

(a) Find the limiting reactant.

(b) Write a reaction table for the process.
(c) Draw a representative portion of the mixture after the reaction
is complete. (Hint: The ClF3 molecule has 1 Cl atom bonded to
3 individual F atoms).

3-44
 PLAN: Write a balanced chemical equation. To determine the limiting
reactant, find the number of molecules of product that would
form from the given numbers of molecules of each reactant.
Use these numbers to write a reaction table and use the
reaction table to draw the final reaction scene.

SOLUTION: The balanced equation is Cl2 (g) + 3F2 (g) → 2ClF3 (g)

There are 3 molecules of Cl2 and 6 molecules of F2 depicted:

3 molecules Cl2 x 2 molecules ClF3 = 6 molecules ClF3

1 molecule Cl2

6 molecules F2 x 2 molecules ClF3

= 4 molecules ClF3
3 molecule Cl2

Since the given amount of F2 can form less product, it

is the limiting reactant.

3-45
 We use the amount of F2 to determine the “change” in the
reaction table, since F2 is the limiting reactant:

Molecules Cl2 (g) + 3F2 (g) → 2ClF3 (g)

Initial 3 6 0
Change -2 -6 +4
Final 1 0 4

The final reaction scene shows that all the F2 has reacted
and that there is Cl2 left over. 4 molecules of ClF2 have
formed:

3-46
 Calculating Quantities in a Limiting-
Reactant Problem: Amount to Amount
PROBLEM: In another preparation of ClF3, 0.750 mol of Cl2 reacts with
3.00 mol of F2.
(a) Find the limiting reactant.
(b) Write a reaction table.
PLAN: Find the limiting reactant by calculating the amount (mol) of
ClF3 that can be formed from each given amount of reactant.
Use this information to construct a reaction table.
SOLUTION: The balanced equation is Cl2 (g) + 3F2 (g) → 2ClF3 (g)

0.750 mol Cl2 x 2 mol ClF3

= 1.50 mol ClF3
1 mol Cl2 Cl2 is limiting, because
it yields less ClF3.
3.00 mol F2 x 2 mol ClF3
= 2.00 mol ClF3
3 mol F2

3-47
 All the Cl2 reacts since this is the limiting reactant. For every 1 Cl2 that
reacts, 3 F2 will react, so 3(0.750) or 2.25 moles of F2 reacts.

Moles Cl2 (g) + 3F2 (g) → 2ClF3 (g)

Initial 0.750 3.00 0
Change -0.750 - 2.25 +1.50
Final 0 0.75 1.50

3-48
 Calculating Quantities in a Limiting-
Reactant Problem: Mass to Mass

PROBLEM: A fuel mixture used in the early days of rocketry consisted of

two liquids, hydrazine (N2H4) and dinitrogen tetraoxide (N2O4),
which ignite on contact to form nitrogen gas and water vapor.
(a) How many grams of nitrogen gas form when 1.00 x 102 g
of N2H4 and 2.00 x 102 g of N2O4 are mixed?
(b) Write a reaction table for this process.

PLAN: Find the limiting reactant by calculating the amount

(mol) of N2 that can be formed from each given mass of
reactant. Use this information to construct a reaction
table.

2N2H4 (l) + N2O4 (l) → 3N2 (g) + 4H2O (g)

3-49
 mass (g) of N2H4 mass (g) of N2O4
divide by M divide by M (g/mol)
(g/mol)
mol of N2H4 mol of N2O4
mole ratio mole ratio

mol of N2 mol of N2

Choose lower number of moles of

N2
multiply by M

mass of N2

3-50
 SOLUTION: 2N2H4 (l) + N2O4 (l) → 3N2 (g) + 4H2O (g)

For N2H4: 1.00x 102 g N2H4 x 1 mol N2H4 = 3.12 mol N2H4
32.05 g N2H4

3.12 mol N2H4 x 3 mol N2 = 4.68 mol N2

2 mol N2H4

For N2O4: 2.00x 102 g N2O4 x 1 mol N2O4

= 2.17 mol N2O4
92.02 g N2O4

3 mol N2
2.17 mol N2O4 x = 6.51 mol N2
1 mol N2O4

N2H4 is limiting and only 4.68 mol of N2 can be produced:

4.68 mol N2 x 28.02 g N2 = 131 g N2

1 mol N2

3-51
 All the N2H4 reacts since it is the limiting reactant. For every 2 moles
of N2H4 that react 1 mol of N2O4 reacts and 3 mol of N2 form:

3.12 mol N2H4 x 1 mol N2O4

= 1.56 mol N2O4 reacts
2 mol N2H4

Moles 2N2H4 (l) + N2O4 (l) → 3N2 (g) + 4H2O (g)

Initial 3.12 2.17 0 0
Change -3.12 - 1.56 +4.68 +6.24
Final 0 0.61 4.68 6.24

3-52
 Reaction Yields

The theoretical yield is the amount of product calculated

using the molar ratios from the balanced equation.

The actual yield is the amount of product actually

obtained.
The actual yield is usually less than the theoretical yield.

% yield = actual yield x 100

theoretical yield

3-53
 Calculating Percent Yield

PROBLEM: Silicon carbide (SiC) is made by reacting sand(silicon

dioxide, SiO2) with powdered carbon at high temperature.
Carbon monoxide is also formed. What is the percent yield
if 51.4 kg of SiC is recovered from processing 100.0 kg of
sand?
PLAN: write balanced equation

find mass of SiO2

g to kg, then divide by M
(g/mol)
find mol SiO2
molar ratio

find mol SiC

Multiply by M(g/mol), Convert g to kg

find kg of SiC percent yield

3-54
 SOLUTION: SiO2(s) + 3C(s) → SiC(s) + 2CO(g)

100.0 kg SiO2 x 10 g x 1 mol SiO2 = 1664 mol SiO

3
2
1 kg 60.09 g SiO2

mol SiO2 = mol SiC = 1664 mol SiC

1664 mol SiC x 40.10 g SiC x 1 kg = 66.73 kg

1 mol SiC 103g

51.4 kg
x 100 = 77.0%
66.73 kg

3-55
 Solution Stoichiometry

• Many reactions occur in solution.

• A solution consists of one or more solutes dissolved in a
solvent.
• The concentration of a solution is given by the quantity
of solute present in a given quantity of solution.
• Molarity (M) is often used to express concentration.
( not to be confused with M,molar molar mass)

C or M=Concentration(mol/L) Moles solute(mol)

volume of solution(L)

3-56
 Calculating the Molarity of a Solution

PROBLEM: Glycine has the simplest structure of the 20 amino acids that
make up proteins .What is the concentration of an aqueous
solution that contains 0.715 mol of glycine (H2NCH2COOH) in
495 mL?
PLAN:
Molarity is the number of moles of SOLUTION:
solute per liter of solution.
0.715 mol glycine x 1000 mL
mol of glycine
495 mL soln 1L
divide by volume

concentration in mol/mL = 1.44 M glycine

103 mL = 1 L

molarity of glycine

3-57
 Calculating Mass of Solute in a Given
Volume of Solution

PROBLEM: Biochemists often study reactions in solutions that contain

phosphate ions.These solutions are commonly found in
cells.What mass of solute is in 1.75 L of 0.460 M sodium
monohydrogen phosphate buffer solution?

PLAN: Calculate the moles of solute volume of solution

using the given concentration and multiply by M
volume. Convert moles to mass moles of solute
using the molar mass of the
multiply by M (g/mol)
solute.
SOLUTION:
grams of solute
1.75 L x 0.460 moles
= 0.805 mol Na2HPO4
1L

0.805 mol Na2HPO4 x 141.96 g Na2HPO4 = 114 g Na2HPO4

1 mol Na2HPO4

3-58
 Preparing a Dilute Solution from a
Concentrated Solution
PROBLEM: Isotonic saline is 0.15 M aqueous NaCl. It stimulates the total
concentration of ions in many cellular fluids and its usess range
from cleaning contact lenses to washing red blood cells. How
would you prepare 0.80 L of isotonic saline from a 6.0 M stock
solution?
PLAN: To dilute a concentrated solution, we add only solvent, so the
moles of solute are the same in both solutions. The volume and
concentration of the dilute solution gives us the moles of solute.
Then we calculate the volume of concentrated solution that
contains the same number of moles.

volume of dilute solution

multiply by M of dilute solution

amount(mol) of NaCl in dilute solution =

amount(mol) NaCl in concentrated solution
divide by M of concentrated solution
Volume(L) of concentrated solution
3-59
 Mdil x Vdil = # mol solute = Mconc x Vconc

SOLUTION:
Using the volume and concentration for the dilute solution:

0.80 L soln x 0.15 mol NaCl = 0.12 mol NaCl

1 L soln

Using the amount of solute and concentration for the concentrated solution:

0.12 mol NaCl x 1 L soln

= 0.020 L soln
6.0 mol NaCl

A 0.020 L portion of the concentrated solution must be

diluted to a final volume of 0.80 L.

3-60
 Calculating Quantities of Reactants and
Products for a Reaction in Solution
PROBLEM: You use 0.10 mol/L HCl solution to simulate the acid concentration
of the stomach. How many liters of “stomach acid” react with a
tablet containing 0.10 g of magnesium hydroxide?

PLAN: Write a balanced equation and convert the mass of

Mg(OH)2 to moles. Use the mole ratio to determine the
moles of HCl, then convert to volume using concentration.

3-61
 SOLUTION:

Mg(OH)2 (s) + 2HCl (aq) → MgCl2 (aq) + 2H2O (l)

0.10 g Mg(OH)2 x 1 mol Mg(OH)2 = 1.7x10-3 mol Mg(OH)2

58.33 g Mg(OH)2

= 1.7x10-3 mol Mg(OH)2 x 2 mol HCl

= 3.4x10-3 mol HCl
1 mol Mg(OH)2

3.4x10-3 mol HCl x 1L HCl soln = 3.4x10-2 L HCl

0.10 mol HCl

3-62
 Solving Limiting-Reactant Problems for
Reactions in Solution
PROBLEM: Mercury and its compounds have many uses, from fillings for
teeth (as a mixture containing silver, copper and tin) in the
past to the current production of chlorine. Because of their
toxicity, however, soluble mercury compounds, such as
mercury(II) nitrate must be removed from industrial
wastewater. One removal method reacts the wastewater
with a sodium sulfide solution to produce solid mercury(II)
sulfide and sodium nitrate solution. In a laboratory
simulation, 0.050 L of 0.010 mol/L mercury(II) nitrate reacts
with 0.020 L of 0.10 mol/L sodium sulfide.
(a) What mass of mercury(II) sulfide forms?

Hg(NO3)2 (aq) + Na2S (aq) → HgS (s) + 2NaNO3 (aq)

3-63
 PLAN: Write a balanced chemical reaction. Determine limiting reactant.
Calculate the grams of mercury(II) sulfide product.

3-64
 SOLUTION: Hg(NO3)2 (aq) + Na2S (aq) → HgS (s) + 2NaNO3 (aq)

0.050 L Hg(NO3)2 x 0.010 mol Hg(NO3)2 x 1 mol HgS

1 L Hg(NO3)2 1 mol Hg(NO3)2
= 5.0x10-4 mol HgS

0.020 L Na2S x 0. 10 mol Na2S x 1 mol HgS = 2.0x10-3 mol HgS

1 L Na2S 1 mol Na2S

Hg(NO3)2 is the limiting reactant because it yields less HgS.

5.0 x 10-4 mol HgS x 232.7 g HgS = 0.12 g HgS

1 mol HgS

3-65
 Water as a Solvent

• Water is a polar molecule

– since it has uneven electron distribution
– and a bent molecular shape.
• Water readily dissolves a variety of
substances.
• Water interacts strongly with its solutes and
often plays an active role in aqueous
reactions.
3-66
 Electron distribution in molecules of H2 and H2O.

A. Electron charge distribution

in H2 is symmetrical. B. Electron charge distribution
in H2O is asymmetrical.

C. Each bond in H2O is polar. D. The whole H2O molecule is polar.

3-67
 An ionic compound dissolving in water.

3-68
 Determining Amount (mol) of Ions in Solution

PROBLEM: What amount (mol) of each ion is in each solution?

(a) 5.0 mol of ammonium sulfate dissolved in water

(b) 78.5 g of cesium bromide dissolved in water
(c) 7.42x1022 formula units of copper(II) nitrate dissolved in water
(d) 35 mL of 0.84 mol/L zinc chloride

PLAN: Write an equation for the dissociation of 1 mol of each

compound. Use this information to calculate the actual number
of moles represented by the given quantity of substance in
each case.

3-69
 SOLUTION:
(a) The formula is (NH4)2SO4 so the equation for dissociation is:
(NH4)2SO4 (s) → 2NH4+ (aq) + SO42- (aq)

5.0 mol (NH4)2SO4 x 2 mol NH4+

= 10. mol NH4+
1 mol (NH4)2SO4

5.0 mol (NH4)2SO4 x 1 mol SO4C = 5.0 mol SO42-

1 mol (NH4)2SO4

3-70
 SOLUTION:

(b) The formula is CsBr so the equation for dissociation is:

CsBr (s) → Cs+ (aq) + Br- (aq)

78.5 g CsBr x 1 mol CsBr x 1 mol Cs +

= 0.369 mol Cs+
212.8 g CsBr 1 mol CsBr

There is one Cs+ ion for every Br- ion, so the number of
moles of Br- is also equation to 0.369 mol.

3-71
 SOLUTION:
(c) The formula is Cu(NO3)2 so the formula for dissociation is:
Cu(NO3)2 (s) → Cu2+ (aq) + 2NO3- (aq)

7.42x1022 formula units Cu(NO3)2 x 1 mol

6.022x1023 formula units

= 0.123 mol Cu(NO3)2

0.123 mol Cu(NO3)2 x 1 mol Cu2+

= 0.123 mol Cu2+ ions
1 mol Cu(NO3)2

There are 2 NO3- ions for every 1 Cu2+ ion, so there are
0.246 mol NO3- ions.

3-72
 SOLUTION:
(d) The formula is ZnCl2 so the formula for dissociation is:
ZnCl2 (s) → Zn2+ (aq) + 2Cl- (aq)

35 mL soln x 1 L x 0.84 mol ZnCl2 = 2.9x10-2 mol ZnCl2

103 mL 1 L soln

2.9x10-2 mol ZnCl2 x 2 mol Cl-

= 5.8x10-2 mol Cl-
1 mol ZnCl2

There is 1 mol of Zn2+ ions for every 1 mol of ZnCl2, so

there are 2.9 x 10-2 mol Zn2+ ions.

3-73
 Ch 17
Redox reactions involve movement of electrons from one
reactant to another ( from the reactant which has less
attraction for electrons to the one with more attraction
for electrons)

Ionic Compounds : involves transfer of electrons,

2Mg(s) + O2 (g) ®2 MgO(s)
Figure 19.1 A

19-74
 Covalent Compounds: involves shift of electrons

H2(g) + Cl2(g) ®2 HCl(g)

Figure 19.1B

19-75
 Overview of Redox Reactions
(ch 17)
Oxidation is the loss of electrons and reduction is the
gain of electrons. These processes occur simultaneously.
In the formation of MgO, Mg loses electrons and O gains
electrons. Mg got oxidized and O got reduced.
Oxidation Mg ® Mg2+ + 2e- ( Mg is oxidized)
Reduction ½O2 + 2e- ® O2- ( O2 is reduced )
The oxidizing agent (O2 , here) takes electrons from
the substance being oxidized. The oxidizing agent is
therefore reduced.
The reducing agent (Mg, here) adds electrons to the
substance being reduced. The reducing agent is
therefore oxidized.
19-76
 Using Oxidation Numbers
Oxidation Number or Oxidation Sate is the charge that the
atom would have if the electrons were transferred completely,
not shared.

A set of rules is used to assign Oxidation Numbers to

an atom.

An Oxidation Number has a sign before the number

(eg +2,-1 etc) where as ionic charge has a sign after
the number ( eg. 2+,1+ etc)

19-77
19-78
 Determine the Oxidation Number of each element in a) zinc chloride,
b) Sulfur trioxide, c) nitric acid and d) dichromate ion.

Plan: From the rules from table 19.1, we can use an equation,
Net Charge on species =∑( Ox:Number of atom X)(number of X atoms in the species)

Solution: a) ZnCl2 We know that Cl has Ox:Number -1.Let

Ox:Number of Zn be y. Net charge on ZnCl2 is zero.
0 = (y)(1) + (-1)(2) ,
therefore y=2 ie, Ox:Number of Zn is +2

b) SO3 Knowing the Ox:Number of O to be -2,

0 = (y)(1) + (-2)(3) ,
y=6 , Ox:Number of S is +6

c) HNO3 Knowing the Ox:Number of O to be -2,and H to be +1,

0 = (+1)(1) +(y)(1) + (-2)(3) ,
y=5 , Ox:Number of N is +5

19-79
 d) Cr2O7 2- Knowing the Ox:Number of O to be -2 and the net charge
on the species is 2-,
-2 = (y)(2) + (-2)(7) ,

Therefore y=6 , Ox:Number of Cr is +6

19-80
 Use oxidation numbers to decide whether each of the following equations
represents a redox reaction:
(a) CaO (s) + CO2 (g) ® CaCO3(s)
(b) 4KNO3(s) ® 2K2O (s) + 2N2(g) + 5O2(g)
(c) H2SO 4(aq) + 2NaOH (aq) ® Na2SO4(aq) + 2H2O (l)

Solution:

Since oxidation numbers are not changing here,

this is NOT a redox reaction

19-81
 This IS a redox reaction

Since oxidation numbers are not changing here,

this is NOT a redox reaction

19-82
 Identify the oxidizing agent and the reducing agent in each of the
following reactions:
(a) 2Al (s) + 3H2SO4(aq) à Al2(SO4)3(aq) + 3H2(g)
(b) PbO(s) + CO (g) ®Pb (s) + CO2(g)
(c) 2H2(g) + O2(g)®2H2O(g)

Solution:
(a)

Al is losing electrons(oxidized) and H+ is gaining electrons

(reduced) So Al is the Reducing agent an H+ is the Oxidizing agent.

19-83
 (b)

C is losing electrons(oxidized) and Pb is gaining electrons

(reduced) so CO is the Reducing agent an PbO is the Oxidizing agent.

(c)

H is losing electrons(oxidized) and O is gaining electrons

(reduced) So H2 is the Reducing agent an O2 is the Oxidizing agent.

19-84
 A Summary of terminology for redox reactions

19-85
 Balancing Redox Equations

Oxidation Number Method

Step 1: assign oxidation numbers to all atoms

Step 2:Identify the oxidized and reduced species

Step 3: Draw tie-lines between the reactant and product species and
write the # of electrons lost(oxidation) and electrons gained
(reduction) on the line.

Step 4: Multiply the # of electrons by factors so that

the # of electrons lost= the # of electrons gained and use the factor
as balancing coefficients

Step 5: Complete the balancing by inspection and add states of matter

19-86
 Problem :Use the oxidation number method to balance the following
equations:
(a) Cu (s) + HNO3 (aq) ® Cu(NO3) 2(aq) + NO 2(g) + H2O(l)
(b) PbS (s) + O2 (g) ® PbO (s) + SO 2(g)
Copper in nitric acid

Solution: (a) Step:1 Assign Oxidation Numbers

Step:2
Identify the oxidized and reduced species.
Cu is oxidized and HNO3 is reduced here

19-87
 Compute electrons lost and gained and draw tie-lines
Step:3

Step:4 Make electrons lost and gained to same number

Step:5 Complete balancing by inspection

2
and add states,

19-88
 Solution: (b)

19-89
 Balancing Redox Reactions in Acidic Solution

Cr2O72-(aq) + I-(aq) → Cr3+(aq) + I2(s)

Step 1: Divide the reaction into half-reactions.

Cr2O72- → Cr3+
I- → I2
Step 2: Balance the atoms and charges in each half-reaction.
For the Cr2O72-/Cr3+ half-reaction:
Balance atoms other than O and H:
Cr2O72- → 2Cr3+
Balance O atoms by adding H2O molecules:

Cr2O72- → 2Cr3+ + 7H2O

19-90
 Balance H atoms by adding H+ ions:
14H+ + Cr2O72- → 2Cr3+ + 7H2O

Balance charges by adding electrons:

6e- + 14H+ + Cr2O72- → 2Cr3+ + 7H2O
This is the reduction half-reaction. Cr2O72- is reduced, and is the
oxidizing agent. The O.N. of Cr decreases from +6 to +3.

For the I-/I2 half-reaction:

Balance atoms other than O and H:
2I- → I2
There are no O or H atoms, so we balance charges by adding electrons:
2I- → I2 + 2e-
This is the oxidation half-reaction. I- is oxidized, and is the reducing
agent. The O.N. of I increases from -1 to 0.

19-91
 Step 3: Multiply each half-reaction, if necessary, by an integer so that
the number of e- lost in the oxidation equals the number of e- gained
in the reduction.
The reduction half-reaction shows that 6e- are gained; the oxidation
half-reaction shows only 2e- being lost and must be multiplied by 3:
3(2I- → I2 + 2e-)
6I- → 3I2 + 6e-

Step 4: Add the half-reactions, canceling substances that appear on

both sides, and include states of matter. Electrons must always cancel.
6e- + 14H+ + Cr2O72- → 2Cr3+ + 7H2O
6I- → 3I2 + 6e-
6I-(aq) + 14H+(aq) + Cr2O72-(aq) → 3I2(s) + 7H2O(l) + 2Cr3+(aq)

19-92
 Balancing Redox Reactions in Basic Solution
An acidic solution contains H+ ions and H2O. We use H+
ions to balance H atoms.
A basic solution contains OH- ions and H2O. To balance
H atoms, we proceed as if in acidic solution, and then
add one OH- ion to both sides of the equation.

For every OH- ion and H+ ion that appear on the same
side of the equation we form an H2O molecule.

Excess H2O molecules are canceled in the final step,

when we cancel electrons and other common species.

19-93
 Balancing a Redox Reaction in Basic
Solution

PROBLEM: Permanganate ion reacts in basic solution with oxalate

ion to form carbonate ion and solid manganese dioxide.
Balance the skeleton ionic equation for the reaction
between NaMnO4 and Na2C2O4 in basic solution:
MnO4-(aq) + C2O42-(aq) → MnO2(s) + CO32-(aq) [basic solution]
PLAN: We follow the numbered steps as described in the text, and
proceed through step 4 as if this reaction occurs in acidic
solution. Then we add the appropriate number of OH- ions
and cancel excess H2O molecules.
SOLUTION:
Step 1: Divide the reaction into half-reactions.
MnO4- → MnO2 C2O42- → CO32-

19-94
 Step 2: Balance the atoms and charges in each half-reaction.
Balance atoms other than O and H:
MnO4- → MnO2 C2O42- → 2CO32-

Balance O atoms by adding H2O molecules:

MnO4- → MnO2 + 2H2O 2H2O + C2O42- → 2CO32-

Balance H atoms by adding H+ ions:

4H+ + MnO4- → MnO2 + 2H2O 2H2O + C2O42- → 2CO32- + 4H+

Balance charges by adding electrons:

3e- + 4H+ + MnO4- → MnO2 + 2H2O 2H2O + C2O42- → 2CO32- + 4H+ + 2e-
[reduction] [oxidation]

19-95
 Step 3: Multiply each half-reaction, if necessary, by an integer so that
the number of e- lost in the oxidation equals the number of e- gained
in the reduction.
x2 6e- + 8H+ + 2MnO4- → 2MnO2 + 4H2O
x3 6H2O + 3C2O42- → 6CO32- + 12H+ + 6e-

Step 4: Add the half-reactions, canceling substances that appear on

both sides.
6e- + 8H+ + 2MnO4- → 2MnO2 + 4H2O
2 6H2O + 3C2O42- → 6CO32- +412H+ + 6e-
2MnO4- + 2H2O + 3C2O42- → 2MnO2 + 6CO32- + 4H+

19-96
 Basic. Add OH- to both sides of the equation to neutralize H+, and
cancel H2O.

2MnO4- + 2H2O + 3C2O42- + 4OH- → 2MnO2 + 6CO32- + [4H+ + 4OH-]

2MnO4- + 2H2O + 3C2O42- + 4OH- → 2MnO2 + 6CO32- + 2 4H2O

Including states of matter gives the final balanced equation:

2MnO4-(aq) + 3C2O42-(aq) + 4OH-(aq) → 2MnO2(s) + 6CO32-(aq) + 2H2O(l)

19-97
 Quantifying Redox Reactions by Titration
In a Redox titration, known concentration of an oxidizing agent is
used to find an unknown concentration of a reducing agent.
Figure 19.3 The redox titration of C2O42- with MnO4 -

19-98
 Calcium ion (Ca2+ ) is necessary for blood clotting and many other
physiological processes. To measure the Ca2+ concentration in
1.00 mL of human blood, Na2C2O4 solution is added, causing Ca2+ to
precipitate as solid CaC2O4. This solid is dissolved in dilute H2SO4 to
release C2O42- and 2.05 mL of 4.88 ×10-4 mol/L KMnO4 is required to
reach the end point. The balanced equation is given below:
2KMnO4(aq) +5CaC2O4(s) + 8H2SO4(aq)®
2MnSO4(aq) + K2SO4(aq) + 5CaSO4(s) +10CO2(g) + 8H2O(l)
Calculate the amount (mol) of Ca2+ in 1.00 mL of blood.

Solution:
1 --L 4.88 × 10-4 mol
Mol of KMnO4 ---
=2.05 mL×---------- ×----------------------- = 1.00× 10-6 mol
1000 mL --- 1 --L

19-99
 Using molar ratio, converting this to mols of CaC2O4 ,

5 mol of CaC2O4
-----------4
mol of CaC2O4 = 1.00× 10-6 mol KMnO ×---------------------
-------------
2 mol KMnO4

= 2.50× 10-6 mol of CaC2O4

--------------- 1 mol of Ca 2+
mol of Ca2+ = 2.50× 10-6 mol of CaC2O4 × -------------------
1 ----------------
mol of CaC2O4

= 2.50× 10-6 mol of Ca2+

19-100