Bài giảng hóa học đại cương i (đại học thủy lợi)

28-Aug-16

WHAT IS THE TEXT BOOK?

GENERAL CHEMISTRY 1

•

Chemistry - The Molecular
Nature of Matter and
Change (5th edition)

(CHEMW2014)

•

Syllabus for 57NKN

•

A copy of the text book is on
reserve at the TLU Library.

Thanh M. Le

CHEMW2014 will be studied
from Chapter 1 to Chapter 12
in the text book.

1

28-Aug-16

28-Aug-16

2

WHAT DO STUDENTS NEED TO
COMPLETE?

THE COURSE GRADING

• Homework = 10 recitation assignments
• Exam = 4 chapter exams + 1 final exam

Recitations = 10*10 points

100 points

You will not be allowed to attend to the final exam, IF:

Chapter exams = 4*100 points

400 points

• you miss out any chapter exam without the

Final exam = 1*200 points

200 points

Total possible points:

700 points

lecturer’s permission.
• you are present in the CHEMW2014 class less than
80% of the prescribed time
3

28-Aug-16

4

28-Aug-16

www.facebook.com/groups/generalchemistry1

5

28-Aug-16

28-Aug-16

28-Aug-16

6

6

28-Aug-16

Lecture PowerPoint

Chapter 1

Chemistry
The Molecular Nature of
Matter and Change

Keys to the Study of
Chemistry

Martin S. Silberberg

28-Aug-16

2

• Physical Properties: properties a substance shows by itself

1.1 Some Fundamental Definitions

without interacting with another substance - color, melting point,
• Chemistry: is the study of matter, its properties, the

boiling point, density.

changes that matter undergoes, and the energy associated

• Chemical Properties: properties a substance shows as it

with these changes.

interacts with, or transforms into, other substances -

• Matter: anything that has both mass and volume - the “stuff”

flammability, corrosiveness.

of the universe: books, planets, trees, professors, students

• A physical change: occurs when a substance alters its

• Composition: the types and amounts of simpler substances

physical form, not its composition.

that make up a sample of matter

• A chemical change, also called a chemical reaction: occurs

• Properties: the characteristics that give each substance a

when a substance (or substances) is converted into a different

unique identity

substance (or substances).

28-Aug-16

3

28-Aug-16

4

Example 1: The scenes below represent an atomic-scale view of
substance A undergoing two different changes (B or C). Decide whether

Exercise 1: Decide whether each of the following processes is

each scene shows a physical or a chemical change?

primarily a physical or a chemical change, and explain briefly:
(a) Frost forms as the temperature drops on a humid winter night.
(b) A cornstalk grows from a seed that is watered and fertilized.
Solution:

(c) A match ignites to form ash and a mixture of gases.

a) A  B: each particle of substance A is composed of one blue and two red spheres.
Sample B is composed of two different types of particles, some have two red spheres

(d) Perspiration evaporates when you relax after jogging.

while some have one red and one blue. As A changes to B, the chemical composition
has changed. A  B is a chemical change.

(e) A silver fork tarnishes slowly in air.

b) A  C: each particle of C is still composed of one blue and two red spheres, but the
particles are closer together and are more organized. The composition remains
unchanged, but the physical form is different. A  C is a physical change.

28-Aug-16

28-Aug-16
5

6


28-Aug-16
The States of Matter
Energy in Chemistry

• A solid has a fixed shape and volume.

• Energy is the ability to do work.

Solids may be hard or soft, rigid or flexible.

• Potential energy is energy due to the position of an object.

• A liquid has a varying shape that conforms

• Kinetic energy is energy due to the movement of an object.

to the shape of the container, but a fixed

Total Energy = Potential Energy + Kinetic Energy
• Lower energy states are more stable, and are favored over

volume. A liquid has an upper surface.

higher energy states.

• A gas has no fixed shape or volume and

• Energy is neither created nor destroyed, it is conserved, and

therefore does not have a surface.

can be converted from one form to another.

28-Aug-16

7

28-Aug-16

8

Example 2: A lower energy state is more stable.

1.4 Chemical Problem Solving
• All measured quantities consist of a number and a unit.
• Units are used like numbers, in other words, units can be

multiplied, divided, and canceled.
• Try to make a habit of including units in all calculations
whenever you practice.
ã Example 3:
Area = 3 cm ì 4 cm = (3 × 4) (cm × cm) = 12 cm2

9

28-Aug-16

28-Aug-16

10

Conversion Factors
Exercise 2: To wire your stereo equipment, you need 325
• A conversion factor is a ratio of equivalent quantities used to
express a quantity in different units.

centimeters of speaker wire that sells for $0.15/ft. What is the
price of the wire? The conversion factor:

• Example 4: some conversions factors:

1 foot (ft) = 30.48 centimeters (cm).

 Between mile and feet is: 1 mi = 5280 ft
 Between mile and kilometer is: 1 mi = 1.609 km

Exercise 3: A furniture factory needs 31.5 ft2 of fabric to

 Between joule and calorie is: 1 J = 0.000239 cal

upholster one chair. Its Vietnamese supplier sends the fabric in
bolts of exactly 200 m2. What is the maximum number of chairs
that can be upholstered by 3 bolts of fabric (1 m = 3.281 ft)?

28-Aug-16

11

28-Aug-16

12


28-Aug-16
SI Base Units (International System of Units)

1.5 Measurement in Scientific Study

Physical Quantity

Unit

Unit

(Dimension)

Name

Abbreviation

kilogram

kg

meter

m

second

s

• Our current system of measurement began in 1790, when the
Mass

newly formed National Assembly of France.

Length

• In 1960, another international committee met in France to
Time

establish the International System of Units. The units of this

Temperature

system are called SI units, from the French Système

Electric Current
Amount of substance

International d’Unités.

Luminous intensity

28-Aug-16

13

kelvin

K

ampere

A

mole

mol

candela

cd
14

28-Aug-16

Common Decimal Prefixes
used with SI Units

Common SI-English
equivalent quantities
Quantity

SI to English Equivalent

English to SI Equivalent

Length

1 km = 0.6214 mile
1 m = 1.094 yard
1 m = 39.37 inches
1 cm = 0.3937 inch

1 mi = 1.609 km
1 yd = 0.9144 m
1 ft = 0.3048 m
1 in = 2.54 cm

Volume

1 cubic meter (m3) = 35.31 ft3
1 dm3 = 0.2642 gal
1 dm3 = 1.057 qt
1 cm3 = 0.03381 fluid ounce

1 ft3 = 0.02832 m3
1 gal = 3.785 dm3
1 qt = 0.9464 dm3
1 qt = 946.4 cm3
1 fluid ounce = 29.57 cm3

Mass

1 kg = 2.205 pounds (lb)
1 g = 0.03527 ounce (oz)

1 lb = 0.4536 kg
1 oz = 28.35 g

28-Aug-16

16

Exercise 4: A graduated cylinder contains 19.9
mL of water. When a small piece of galena, an

Some volume equivalents in SI

ore of lead, is added, it sinks and the volume

1

m3

= 1000

dm3 =

1000 L

increases to 24.5 mL. What is the volume of the
piece of galena in cm3 and in L?

1 dm3 = 1000 cm3 = 1 L = 1000 mL

Exercise 5: Many international computer communications are

1 cm3 = 1000 mm3 = 1 mL = 100= μL

carried out by optical fibers in cables laid along the ocean floor.
If one strand of optical fiber weighs 1.19 x 10-3 lb/m, what is the

1 mm3 = 1 μL

mass (in kg) of a cable made of six strands of optical fiber,
each long enough to link New York and Paris (8.94 x 103 km)?
28-Aug-16

17

28-Aug-16

18

28-Aug-16
Density

Densities of Some Common Substances

• At a given temperature and pressure, the density of a
substance is a characteristic physical property and has a
specific value.

*At

28-Aug-16

19

Substance

Physical State

Density (g/cm3)

Hydrogen

gas

0.0000899

Oxygen

gas

0.00133

Grain alcohol

liquid

0.789

Water

liquid

0.998

Table salt

solid

2.16

Aluminum

solid

2.70

Lead

solid

11.3

Gold

solid

19.3

room temperature (20°C) and normal atmospheric pressure (1atm).

28-Aug-16

20

Temperature
• Distinguish between temperature and heat:
Exercise 6: Lithium, a soft, gray solid with the lowest density of

 Temperature (T) is a measure of how hot or cold a substance is relative to another

any metal, is a key component of advanced batteries. A slab of

 Heat (Q) is the energy that flows between objects that are at different temperatures.

substance.

lithium weighs 1.49x103 mg and has sides that are 20.9 mm by
11.1 mm by 11.9 mm. Find the density of lithium in

• Temperature Scales
 Kelvin (K): the “absolute temperature scale” begins at absolute zero and has only

g/cm3.

positive values. Note that the kelvin is not used with the degree sign (°).
 Celsius (oC): the Celsius scale is based on the freezing and boiling points of water.
This is the temperature scale used most commonly around the world. The Celsius and

Exercise 7: A cube of gold 3 cm on a side weighs 18.3 ounces.

Kelvin scales use the same size degree although their starting points differ.

What is the density of gold in g/cm3? The conversion factor:

 Fahrenheit (oF): the Fahrenheit scale is commonly used in the US. The Fahrenheit
scale has a different degree size and different zero points than both the Celsius and

1 ounce (oz) = 28.3495 grams (g)

Kelvin scales.

T (K) = T (oC) + 273.15;

28-Aug-16

T (oF) = T (oC)*9/5 - 32

21

Example 5: the number of significant figures in a measurement.

1.6 Uncertainty in Measurement:
Significant Figures
• Every measurement includes some uncertainty. The
rightmost digit of any quantity is always estimated.
• The recorded digits, both certain and uncertain, are called
significant figures.
• The greater the number of significant figures in a quantity, the
greater its certainty.

28-Aug-16
28-Aug-16

23

24


28-Aug-16
Example 6: Determining which digits are significant.

Determining Which Digits are Significant
• All digits are significant, except zeros that are used only to
position the decimal point.
1. Make sure the measured quantity has a decimal point.


Start at the left and move right until you reach the first nonzero

digit.



Count that digit and every digit to its right as significant.

2. If no decimal point is present, zeros at the end of the number
are not significant
28-Aug-16

25

28-Aug-16

26

Example 7: How many digits in these numbers are significant.
a) 1.030 mL

b) 5300. L

c) 5300 L

d) 0.00004715 m

e) 0.0000007160 cm3

Exercise 8: For each of the following quantities, determine the
number of significant figures in each quantity.

Solution:
a) 1.030 mL has 4 significant figures.

(a) 0.0030 L

(b) 0.1044 g

(c) 53,069 mL

(d) 0.00204010 m

(e) 57,600. s

(f) 0.0007001 cm3

b) 5300. L has 4 significant figures.
c) 5300 L has only 2 significant figures.

Exercise 9: Express the number in 8c), 8d), 8e), 8f) in

d) 0.00004715 m has 4 significant figures, = 4.715x10-5 m

exponential notation.

e) 0.0000007160 cm3 has 4 significant figures, = 7.160x10-7 cm3
28-Aug-16

27

28-Aug-16

28

Rules for Significant Figures
in Calculations
Example 8: Significant Figures in Calculations
1. For multiplication and division. The answer contains the

a) Multiplication:
9.21 cm x 6.8 cm x 0.3744 cm = 23.4225 cm3 = 23 cm3

same number of significant figures as there are in the

b) Addition:

measurement with the fewest significant figures.

83.5 mL + 23.28 mL = 106.78 mL = 106.8 mL
2. For addition and subtraction. The answer has the same

c) Subtraction:
865.9 mL - 2.8121 mL = 863.0879 mL = 863.1 mL

number of decimal places as there are in the
measurement with the fewest decimal places.
28-Aug-16

29

28-Aug-16

30


28-Aug-16
Rules for Rounding Off Numbers
Example 9:

• If the digit removed is more than 5, the preceding number

a) 5.379 rounds to 5.38 if 3 significant figures are retained.

increases by 1.

b) 0.2413 rounds to 0.241 if 3 significant figures are retained.

• If the digit removed is less than 5, the preceding number is

c) 17.75 rounds to 17.8; but 17.65 rounds to 17.6.

unchanged.

d) 17.6500 rounds to 17.6; but 17.6513 rounds to 17.7

• If the digit removed is 5, followed by zeros, or with no
following digits, the preceding number increases by 1 if it is

•Be sure to carry two or more additional significant figures

odd and remains unchanged if it is even.

through a multistep calculation and round off the final answer

• If the 5 is followed by other nonzero digits, rule 1 is

only.

followed.
28-Aug-16

31

28-Aug-16

32

Example 10: Significant figures of measuring devices.
Example 11: Perform the following calculations and round
each answer to the correct number of significant figures.

(a)

16.3521 cm2 - 1.448 cm2

4.80x104 mg

(b)

1g
1000 mg

11.55 cm3

7.085 cm

Solution:
(a)

16.3521 cm2 - 1.448 cm2
7.085 cm

(b)

4.80x104 mg

1g
1000 mg

11.55 cm3

The mass (6.8605 g)

The volume (68.2 mL)

=

14.904 cm2
7.085 cm

=

48.0 g
11.55 cm3

= 2.104 cm

= 4.16 g/ cm3

34

28-Aug-16

Precision, Accuracy, and Error

Example 12a: Precision and accuracy in a laboratory calibration.

• Precision refers to how close the measurements in a series
precise and accurate

are to each other.
• Accuracy refers to how close each measurement is to the
actual value.
• Systematic error produces values that are either all higher
or all lower than the actual value. This error is part of the
experimental system.

precise but not accurate

• Random error produces values that are both higher and
lower than the actual value.

28-Aug-16

35

28-Aug-16

36


28-Aug-16
Example 12b: Precision and accuracy in a laboratory calibration.
random error

Chapter 1: Suggested Practice Problems
1, 4, 8, 26-42(e), 44, 45, 52-68(e), 72, 73
systematic error

37

28-Aug-16

28-Aug-16

38

28-Aug-16

40

39

28-Aug-16
28-Aug-16

39


30-Aug-16
2.1 Elements, Compounds, and Mixtures

Chapter 2

• Element - the simplest type of substance with unique physical and
chemical properties. An element consists of only one type of atom.

The
Components
of Matter

• Atom - the smallest constituent unit of ordinary matter that has the
properties of a chemical element.
• Molecule - a structure that consists of two or more atoms which are
chemically bound together and thus behaves as an independent unit.
• Compound - a substance composed of two or more elements which
are chemically combined.
• Mixture - a group of two or more elements and/or compounds that

Thanh M. Le
30-Aug-16

are physically intermingled.
1

Example 1:
• Elements,
• Atoms,
• Molecules,
• Compounds
• Mixtures
on the atomic
scale

30-Aug-16

2

Exercise 1: The scenes below represent an atomic-scale view
of 7 samples of matter. Describe each sample as an element,
compound, or mixture.

30-Aug-16

3

30-Aug-16

4

• Mass fractions (or mass percents) of any element in a

2.2 Mass Conservation & Mass fractions

compound AxBy:
• Law of Conservation of Mass: “the total mass of substances
does not change during a chemical reaction”.
• Mass of element in sample:
=
• Example 3: a) Calculate %mN in NH4NO3.

• Example 2:
CaO

+

56.08g
(calcium oxide)

30-Aug-16

+

CO2

→

CaCO3

44.00g

=

100.08g

(carbon dioxide)

b) Calculate mN in 140 kg of NH4NO3.
Solution:

(calcium carbonate)

5

30-Aug-16

6


30-Aug-16
Mass Number, Atomic Number,
Atomic Symbol, and Isotopes,

2.5 The Atomic Theory today

Nucleus
Shell of
atom

Relative Mass
(amu)

Relative
charge

Proton (p+)

1.00727

+1

Neutron (n0)

1.00866

0

Electron (e-)

0.00054858

-1

• Isotope = atoms of an element with the same number of
protons, but a different number of neutrons
• Number of neutrons = N
• Number of protons = P
• We have:
Z=P
A=Z+N
• Each element normally
has many isotopes:

• Example 4: isotopes of
carbon

ã The atomic mass unit (amu)
= 1.66054ì10-24 g;

30-Aug-16

7

• The atomic mass (also called atomic weight) of an element is
the average mass of the masses of isotopes which are
weighted according to their abundances.
• Atomic mass is calculated from its isotopes mass:
Isotope
Mass (amu)
Percent Abundance (%)

A1X

A2X

A3X

M1
y1

M2
y2

M3
y3

30-Aug-16

Exercise 2: Silicon(Si) is essential to the computer industry as
a major component of semiconductor chips. It has three
naturally occurring isoltopes:

28Si, 29Si,

and 30Si. Determine

the number of protons, neutrons, and electrons in each silicon
isotope, if atomic number of silicon is 14.

Exercise 3: Boron (5B) has two naturally occurring isotopes.

• Example 5: Silver (47Ag) has two occur naturally, 107Ag and
109Ag, with mass 106.90509 (amu) and 108.90476 (amu), with
abundance 51.84% and 48.16%, repectively. So, the atomic
mass of silver is:

Find the percent abundances of 10B and 11B given the atomic
mass of B = 10.81 amu, the isotopic mass of 10B = 10.0129
amu, and the isotopic mass of

11B

= 11.0093 amu.

30-Aug-16

10

Mendeleev's 1871 periodic table

2.6 Periodic Table
• At the end of the 18th century,
Lavoisier compiled a list of the 23
elements known at that time.
• By 1870, 65 were known.
• In 1871, the Russian chemist
Dmitri Mendeleev published the
most successful of these
Dmitri Mendeleev (1836–1907)
organizing schemes as a table of the elements.
• By 1925, 88 were known. Today, there are 116 and still counting!

30-Aug-16

11

30-Aug-16

12


30-Aug-16
Organization of the Periodic Table

• Each element has a box that contains its atomic number, atomic
The Modern Periodic Table

symbol, and atomic mass.
• The boxes lie in order of increasing atomic number (number of
protons) as you move from left to right.
• The boxes are arranged into a grid of periods (horizontal rows)
and groups (vertical columns).
• The elements are classified as metals, nonmetals, or metalloids
(semimetals).

13

30-Aug-16

14

2.8 Formulas, Names of Compounds

Names of common Ions (1)

• Types of Chemical Formulas:
 Empirical formula: shows the relative number of atoms
of each element in the compound.
 Molecular formula: shows the actual number of atoms of
each element in a molecule of the compound.
 Structural formula: shows the number of atoms and the

H+ hydrogen
Li+ lithium

Na+ sodium

Mg2+ magnesium
Ca2+ calcium
Sr2+ strontium

HFCl-

hydride
fluoride
chloride

K+ potassium
Cs+ cesium
Ag+ silver

Ba2+
Zn2+
Cd2+
Al3+

BrIO2S2-

bromide
iodide
oxide
Sulfide

N3-

nitride

barium
zinc
Cadmium
aluminum

bonds between them.
Cu+
Cu2+
Fe2+
Fe3+
30-Aug-16

15

30-Aug-16

Names of common Ions (2)
NH4+
O+

H3
Co2+
Co3+
Mn2+
Mn3+
Sn2+
Sn4+

30-Aug-16

ammonium

OH-

hydroxide

hydronium
cobalt(II)
cobalt (III)
manganese(II)

CH3COOCNClO3NO2-

acetate
cyanide
chlorate
nitrite

manganese(III)
tin(II)/ stannous
tin(IV)/ stannic

-

NO3
MnO4CO32SO42PO43-

nitrate

permanganate
carbonate
sulfate
phosphate

CrO42Cr2O72-

chromate
dichromate

copper(I)/ cuprous
copper(II)/ cupric
iron(II)/ ferrous
iron(III)/ ferric
16

Names of compounds (3)
• The cation name = the metal name.
• The anion name = the nonmetal name + the suffix “ide”.
• Name of ionic compound = the cation name + the anion name.
• Example 6: Name the ionic compound formed from the
following pairs of elements:

a) Mg and N;

b) I and Cd;

Solution:
a) Mg = magnesium → Mg2+ = magnesium; N = nitrogen →
N3- = nitride;  Mg3N2 = magnesium nitride.

b) I = iodine → I- = iodide; Cd = cadmium → Cd2+ = cadmium
 CdI2 = cadmium iodide.
17

30-Aug-16

18


30-Aug-16
Exercise 4: Name the ionic compound formed from the following

Names of multi cations of a family (4)

pairs of elements:
• With metal have more than one cation:
 The cation name with lower charge = the Latin metal root
+ the suffix -ous.

a) Zn & O;

b) Ag & Br;

c) Li & Cl;

d) Al & S;

e) Sr and F;

f) S and Cs;

 The cation name with higher charge = the Latin metal root
Exercise 5: Give the formulas for the following compounds:

+ the suffix -ic.

a) tin(II) fluoride;

• Example 7: Name the following ions:
a) Fe2+ and Fe3+

b) Cu+ and Cu2+

c) Sn2+ and Sn4+

a) Fe = iron = ferrum (Latin name) →

Fe2+

= ferrous;

Fe3+

c) lead(IV) oxide;

d) mercuric chloride; e) ferrous chloride; f) aluminum sulfide;
g) calcium bromide

Solution:

b) ferric oxide;

h) chromium(III) iodide

i) stannous fluoride

= ferric

b) Cu = copper = cuprum (Latin name) → Cu+ = cuprous; Cu2+ = cupric.
c) Sn = tin = stannum (stannum) → Sn2+ = stannous; Sn4+ = stannic
30-Aug-16

19

30-Aug-16

20

Names of multi anions of a family (5)

Names of acids (6)

• With a family has 2 oxoanions (anions including oxygen atoms):

• Name of binary acid (HX) = Prefix “hydro” + nonmetal root

 The anion name with more O = the nonmetal root + suffix “ate”.
 The anion name with fewer O = the nonmetal root + suffix “ite”.

name + suffix “ic” + “acid”.

• Name of oxoacid:

• With a family has 4 oxoanions:
 the anion name with most O atoms = the prefix “per” + the
nonmetal root + the suffix “ate”;
 the anion name with least (three fewer) O atoms = the prefix
“hypo” + the root + the suffix "ite”.

 The acid name with more O atoms = the nonmetal root name +
suffix “ic” + “acid”.
 The acid name with fewer O atoms = the nonmetal root name +
suffix “ous” + “acid”.

• Example 8: Name the following ions:
a) SO42- and SO32b) NO3- and NO2Solution:

 The prefixes “hypo” and “per” are kept with the families
including 4 oxoacids.

a) S = sulfur (Latin name) → SO42- more O atoms than SO32→ SO42- = sulfate; SO32- = sulfite.
b) N = nitrum (Latin name) → NO3- more O atoms than NO2→ NO3- = nitrate; NO2- = nitrite.
30-Aug-16

• Example 9: HCl = hydrochloric acid

21

30-Aug-16

HNO2 = nitrous acid

HBrO4 = perbromic acid

H2SO4 = sulfuric acid

HIO2 = iodous acid

HIO3 = iodic acid
22

Exercise 6: Give the names of the following formulas :

Names of Hydrates and Binary Covalent Compounds (7)

a) Cs2S;

b) CdI2;

c) CrI3;

• Name of hydrates = name of compound + a Greek numerical

d) CoS;

e) Fe2O3;

f) Fe(ClO4)2;

g) Ba(CH3COO)2

h) Na2SO3

i) Fe2(SO4)3

• Name of binary covalent compound (compound of 2 non-metal

k) NaOH

l) KOH

m) Cu(OH)2

elements) = a Greek numerical prefix + name of element 1 + a

prefix + “hydrate”

Greek numerical prefix + name of element 2 + suffix “ide”.
• The Greek numerical prefix = mono (1); di (2); tri (3); tetra (4);

Exercise 7: Give the names of the following acids :

penta (5); hexa (6); hepta (7); octa (8); nona (9); deca (10).

a) HBr;

b) HNO3;

c) CH3COOH;

d) HCN;

e) H2SO3;

f) HClO4;

Ba(OH)2.8H2O = barium hydroxide octahydrate

g) HClO3;

h) HClO2;

i) HClO;

CS2 = carbon disulfide

k) H2CO3

l) H2S

m) HF

PCl5 = phosphorus pentachloride

30-Aug-16

• Example 10: CuSO4.5H2O = copper(II) sulfate pentahydrat.

23

30-Aug-16

N2O4 = dinitrogen tetraoxide

24


30-Aug-16
Exercise 8: Explain what is wrong with the following name or

2.9 Mixtures: Classification and Separation

formula. If they are wrong, please correct it:
• There are two broad classes of mixtures:

a) Ba(CH3COO)2 is called barium diacetate.

 A heterogeneous mixture: has one or more visible

b) Sodium sulfide has the formula (Na)2SO3.

boundaries between the components.

c) Iron(II) sulfate has the formula Fe2(SO4)3.

 A homogeneous mixture has no visible boundaries

d) Cesium carbonate has the formula Cs2(CO3).

because the components are mixed as individual atoms,

e) SF4 is monosulfur pentafluoride.

ions, and molecules. A homogeneous mixture is also called

f) Dichlorine heptaoxide is Cl2O6.

a solution.

g) N2O3 is dinitrotrioxide.

 Solutions in water, called aqueous solutions.

30-Aug-16

25

30-Aug-16

26

heterogeneous mixture
homogeneous mixture

Chapter 2: Suggested Practice Problems
36, 39, 41-49(odd), 52, 55, 56, 58
84-110(even)

30-Aug-16

27

30-Aug-16

28

The Mass Spectrometer and Its Data

29

30-Aug-16
30-Aug-16

29

30-Aug-16

30


09-Oct-16
3.1 The Mole

Chapter 3

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

Stoichiometry of

contains the same number of entities as there are atoms in
exactly 0.012 kg of carbon-12.

Formulas and

ã One mole (1 mol) contains 6.022ì1023 entities (to four
significant figures). This number is called Avogadro’s number

Equations

(abbreviated NA)
• The term “entities” = any type of particle (atoms, ions,

Chemistry is a practical science
09-Oct-16

molecules, or electrons).
1

Molar Mass
• The molar mass (M) of a substance is the mass per mole of its
entites (atoms, molecules).
• Molar mass has units of grams per mole (g/mol)
• Elements:

09-Oct-16

2

Interconverting Moles (n), Mass (m), &
Number of Chemical Entities (N)

 For monatomic elements (Ne, Fe, Ba...), the molar mass is the
same as the atomic mass in gram per mole.. For example: the molar
mass of Ne = 20.18 g/mol.
 For molecular elements (O2, H2, S8...): you must know the molecular
formula to determine the molar mass. For example: the molar mass of
O2 = 2ìMO = 2ì16.00 g/mol = 32.00 g/mol.

ã Compounds. The molar mass of a compound is the sum of the
molar masses of the atoms of the elements in the formula.
 For example: the molar mass of K2S = 2×MK + MS = (2ì39.10 g/mol)+
32.07 g/mol = 110.27 g/mol
09-Oct-16

3

09-Oct-16

4

ã Example 1:

Mass Percent & Mass of an Element

a) How many grams of Ag are in 0.0342 mol of Ag?
b) How many Ga atoms are in 2.85 x 10-3 mol of gallium?
c) How many Fe atoms are in 95.8 g of Fe?

• Assume with a compound AxBy

Solution

a)
b)

c)

09-Oct-16

5

09-Oct-16

6


09-Oct-16
• Example 2: Glucose (C6H12O6) is a key nutrient for generating
chemical potential energy in biological systems.

• Exercise 1: Nitrogen dioxide is a component of urban smog

a) What is the mass percent of each element in glucose?

that forms from the gases in car exhausts. Assume we have a

b) How many grams of carbon are in 16.55 g of glucose?

sample of 8.92 g of nitrogen dioxide.

Solution

a) How many molecules are in there?

a)

b) How many mol of N, mol of O are in the sample?

• Exercise 2: Ammonium carbonate, a white solid that
decomposes on warming, is an component of baking powder.
b)

a) How many molecules are in 41.6 g of ammonium carbonate?
b) How many O atoms are in this sample?
09-Oct-16

7

09-Oct-16

8

3.2 Determining the Formula of an
Unknown Compound

• Exercise 3: Tetraphosphorus decaoxide reacts with water to
formphosphoric acid, a major industrial acid. In the laboratory,
the oxide is used as a drying agent.

• The empirical formula is the simplest formula for a compound

a) What is the mass (in g) of 4.65×1022 molecules of tetra-

that agrees with the elemental analysis. It shows the lowest

phosphorus decaoxide?

whole number of moles and gives the relative number of atoms

b) How many P atoms are present in this sample?

of each element present. For example, the empirical formula for
glucose (C6H12O6) is CH2O.

• Exercise 4: Ammonium nitrate is a common fertilizer.

• The molecular formula shows the actual number of atoms of

Agronomists base the effectiveness of fertilizers on their
nitrogen content.

each element in a molecule of the compound. For example, the

a) Calculate the mass percent of N in ammonium nitrate.

molecular formula for glucose is C6H12O6.

b) How many grams of N are in 35.8 kg of ammonium nitrate?
09-Oct-16

9

Steps to determine
empirical formula & molecular formula

09-Oct-16

10

• Example 3: A sample of an unknown compound contains 0.21
mol of zinc, 0.14 mol of phosphorus, and 0.56 mol of oxygen.

• Step 1: Find the number of moles of each element and show
the preliminary formula.
• Step 2: Divide by the lowest mol amount.
• Step 3: If the results is not whole numbers, multiplying by the
smallest integer that gives whole numbers to get empirical
formula.
• Step 4: Divide Mmolecule for Mempirical to find a conversion factor.
• Step 5: Multiply the empirical formula to the factor to get
molecular formula.

What is its empirical formula?
Solution
• The preliminary formula = Zn0.21P0.14O0.56. Divide each
fraction by the smallest mol, in this case 0.14

• The result = Zn1.5P1.0O4.0. We convert to whole numbers by
multiplying by the smallest integer that gives whole numbers; in
this case 2, then we get empirical formula = Zn3P2O8.

09-Oct-16

11

09-Oct-16

12


09-Oct-16
• Example 4: Lactic acid (M = 90.08 g/mol) 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?
Solution
• Assuming there are 100. g of lactic acid: CxHyOz.

Combustion Analysis of Organic Compounds
• In the combustion analysis of Organic Compounds (CxHyOz).
o

t
CxHyOz + O2 →
CO2 + H2O

 All the H in the Organic Compounds is converted to H2O.
 All the C in the Organic Compounds is converted to CO2.
 The mass of O in the Organic Compounds = the mass of

• The result give us the preliminary formula = C3.33P6.66O3.33.
• Divide each fraction by the smallest mol, in this case 0.33, we

get empirical formula = CH2O.
• Find the conversion factor:

• Multiply to the factor, the molecular formula = C3H6O3.
09-Oct-16

organic compounds minus the sum of the C & H masses

13

• Example 5: When a 1.000 g sample of vitamin C
(M= 176.12 g/mol) 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?
Solution
• mCO2 = 85.35 g - 83.85 g = 1.50 g → mC = 0.409 g C
• mH2O= 37.96 g - 37.55 g = 0.41 g → mC = 0.046 g H
• mO = mvitamin C – mC – mH = 1.000-(0.409 + 0.046) = 0.545 g O
• Convert mC, mH, mO to moles, we get the preliminary formula =
C0.0341P0.0456O0.0341.
• Divide each fraction by the smallest mol, in this case 0.0341, then
multiply by the smallest integer, in this case 3, we get empirical
formula = C3H4O3.
• Find the conversion factor, in this case 2, multiply to the factor, the
molecular formula = C6H8O6.
15

14

Molecular Formula = CH2O = Formaldehyde
Molecular Formula = C2H4O2 = acetic acid
Empirical
Formula
CH2O

Molecular Formula = C3H6O3 = lactic acid
Molecular Formula = C4H8O4 = erythrose
Molecular Formula = C5H10O5 = ribose
Molecular Formula = C6H12O6 = glucose

09-Oct-16

16

Isomers

Chemical Formulas

• Isomers are molecules that have the same molecular formula,

• The types of Chemical Formulas:

but have a different arrangement of the atoms in space.
 General formula. For example: CxHyOz.

 Isomer compounds are the same molecular formula but

 Preliminary formula. For example: (C1.5H3O1.5)

different properties.

 Empirical formula. For example: (CH2O) or (CH2O)n
 Molecular formula. For example: C6H12O6.
 Condensed formula (semi-structural formula).
For example: CH2OH[CH(OH)]4CHO.
 Structural formula. For example:
09-Oct-16

17

09-Oct-16

18


09-Oct-16
• Exercise 7: A dry-cleaning solvent (M=146.99 g/mol) that
• Exercise 5: Analysis of a sample of an ionic compound yields

contains C, H, and Cl is suspected to be a cancer-causing

2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What is the

agent. When a 0.250-g sample was studied by combustion

empirical formula and the name of the compound?

analysis, 0.451 g of CO2 and 0.0617 g of H2O formed. Find the
molecular formula?

• Exercise 6: One of the most widespread environmental
carcinogens (cancer-causing agents) is benzo pyrene

• Exercise 8: Menthol (M=156.3 g/mol), a strong-smelling

(M=252.30 g/mol). It is found in coal dust, in cigarette smoke,
and even in charcoal-grilled meat. Analysis of this hydrocarbon

substance used in cough drops, is a compound of carbon,
hydrogen, and oxygen. When 0.1595 g of menthol was

shows 95.21 mass % C and 4.79 mass % H. What is the

subjected to combustion analysis, it produced 0.449 g of CO2

molecular formula of benzo pyrene?

and 0.184 g of H2O. What is menthol’s molecular formula?
09-Oct-16

19

3.3 Writing and Balancing
Chemical Equations

09-Oct-16

20

Steps to Balance a Chemical Equation
• Step 1: Translate the chemical statements to its chemical

• A chemical equation uses formulas to express the identities
and quantities of substances involved in a physical or chemical

signs. Reactants are written on the left. Products are written on

change.

the right.

• The equation must be balanced; the same number and type of

• Step 2: Put coefficients to balance the 2 sides of the equation;

each atom must appear on both sides.

do not change the formulas.

A yield arrow points from
reactants to products.

• Step 3: Check that all atoms balance & adjust the coefficients if
Coefficients

2Mg

+

O2

necessary.

2MgO

• Step 4: Write physical states of reactants and products.
Reactants are written on the left.
Products are written on the right.
09-Oct-16

21

09-Oct-16

22

• Example 7: The following molecular scenes depict an reaction. The
blue spheres represent nitrogen while the red spheres represent
oxygen. Write a balanced equation for this reaction.

• Example 6: 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?
Solution
• C8H18 +

O2 → CO2 +

• C8H18 +

25/2O2 → 8CO2 +

• 2C8H18 +

H2O

25O2 → 16CO2 +

• 2C8H18(g) +

9H2O
18H2O

25O2(g) → 16CO2(g) +

18H2O(l)
23

Solution
• The reactant circle shows only one type of molecule, composed of 2
N and 5 O atoms. The formula is thus N2O5. There are 4 N2O5
molecules depicted.
• The product circle shows two types of molecule; one has 1 N and 2 O
atoms while the other has 2 O atoms. The products are NO2 and O2.
There are 8 NO2 molecules and 2 O2 molecules shown.

4 N2O5 → 8 NO2 + 2 O2.
2N2O5 → 4NO2 + O2.
2N2O5 (g) → 4NO2 (g) + O2 (g)
24


09-Oct-16
• Exercise 9: Write a balanced equation for each of the following
chemical statements:
a) A characteristic reaction of Group 1A elements: chunks of
sodium react violently with water to form hydrogen gas and
sodium hydroxide solution.
b) The destruction of marble statuary by acid rain: aqueous nitric
acid reacts with calcium carbonate to form carbon dioxide,
water, and aqueous calcium nitrate.
c) Halogen compounds exchanging bonding partners:
phosphorus trifluoride is prepared by the reaction of
phosphorus trichloride and hydrogen fluoride; hydrogen
chloride is the other product. The reaction involves gases only.
d) Explosive decomposition of dynamite: liquid nitroglycerine
(C3H5N3O9) explodes to produce a mixture of gases-carbon
dioxide, water vapor, nitrogen, and oxygen.
09-Oct-16

25

• Example 8: Copper is obtained from copper(I) sulfide by roasting it in
the presence of 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 form when 10.0 mol of copper(I)
sulfide reacts?
c) How many kilograms of oxygen are required to form 2.86 kg of
copper(I) oxide?
Solution
2Cu2S(s) + 3O2(g) → 2Cu2O(s) + 2SO2(g)
a)

3.4 Calculating Amounts of
Reactant and Product
• 1. Write a balanced equation for the reaction.
• 2. Convert the given mass (or number of entities) of the first
substance to amount (mol).
• 3. Use the appropriate molar ratio from the balanced equation
to calculate the amount (mol) of the second substance.
• 4. Convert the amount of the second substance to the desired
mass (or number of entities).

09-Oct-16

26

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.
• For example:

2 reactions
in sequence

b)

c)

nCu2O = mCu2O/MCu2O = 2.86×103/143.1 = 20.0 mol → nO2 = ...

27

Limiting Reactants

09-Oct-16

28

• Example 9: Chlorine reacts with fluorine to form chlorine trifluoride.

• Limiting reagent (limiting reactants) = is the substance that

The reaction is run with 0.750 mol of Cl2 and 3.00 mol of F2. What

is totally consumed when the chemical reaction is complete.

mass of chlorine trifluoride will be produced?

The amount of product formed is limited by this reagent,

Solution

since the reaction cannot continue without it.

Cl2(g) + 3F2(g) → 2ClF3(g)

If one or more other reagents are present in excess of the
• Compare:

quantities required to react with the limiting reagent, they are
described as excess reagents or excess reactants.

, the ratio is smaller than the other, that

reactant will be limiting reactant:

• There are 2 methods to find out which reactant is limiting reagent

, so Cl2 will be the

limiting reactant.

Method 1: Comparison of reactant amounts.
• Calculate the product always have to use the limiting reactant.

Method 2: Comparison of product amounts which can be
formed from each reactant.
09-Oct-16

29

30


09-Oct-16
• Example 10: Silicon carbide (SiC) is made by reacting sand

Percent Yields

(silicon dioxide, SiO2) with powdered carbon (C) at high to.
• The theoretical yield is the amount of product calculated using
the molar ratios from the balanced equation.

Carbon monoxide is also formed. What is the % yield if 51.4 kg
of SiC is formed from processing 100.0 kg of sand?

• The actual yield is the amount of product actually obtained.

Solution

The actual yield is usually less than the theoretical yield.

SiO2(s) + 3C(s) → SiC(s) + 2CO(g)

• Yields in Multistep Syntheses:
 Overall % yield = multiply every yields of all the steps
Overall % yield = %yield1 ì %yield2 ì %yield3 ì... ì %yieldn
09-Oct-16

31

32

ã Exercise 12: Marble (calcium carbonate) reacts with
hydrochloric acid solution to form calcium chloride solution,

• Exercise 10: A fuel mixture used in the early days of rocketry
consisted of two liquids, hydrazine (N2H4) and dinitrogen

water, and carbon dioxide. What is the percent yield of carbon

tetraoxide (N2O4), which ignite on contact to form nitrogen gas

dioxide if 3.65 g of the gas is collected when 10.0 g of marble

and water vapor:

reacts?

a) Write the balanced chemical equation for the reaction?
b) How many grams of nitrogen gas form when 1.00 x 102 g of

• Exercise 13: When 56.6 g of calcium and 30.5 g of nitrogen

N2H4 and 2.00 x 102 g of N2O4 are mixed?

gas undergo a reaction that has a 93.0% yield, what mass of
calcium nitride forms?

• Exercise 11: How many grams of solid aluminum sulfide can
be prepared by the reaction of 10.0 g of aluminum and 15.0 g of

sulfur? How much of the non-limiting reactant is in excess?
09-Oct-16

33

09-Oct-16

34

• Molarity = a concentration unit, defined to be the number of

3.5 Fundamentals of
Solution Stoichiometry

moles of solute divided by the number of liters of solution.

• A solution = a homogeneous mixture of two or more substances.
A solution may exist in any phase.

• Preparing a Dilute Solution from a Concentrated Solution: the
moles of solvent is constant when the solution is diluted.

• A solvent = the component of a solution that is present in the

nsolute = CM1×V1 = CM2×V2

greatest amount. It is the substance in which the solute is
dissolved. For example: the solvent for seawater is water.
• A solute = the substance that is dissolved in a solution.
• Concentration = a measurement of the amount of solute present

in a chemical solution, with respect to the amount of solvent.
09-Oct-16

35

09-Oct-16

36


09-Oct-16
• Exercise 14: What mass of solute is in 1.75 L of 0.460 M
sodium monohydrogen phosphate buffer solution?

• Example 11: What is the molarity of an aqueous solution that
contains 0.715 mol of glycine (H2NCH2COOH) in 495 mL?
Solution

• Exercise 15: A 0.10 mol/L HCl solution is used to simulate the
acid concentration of the stomach. What is the volume (L) of
“stomach acid” react with a tablet containing 0.10 g of
magnesium hydroxide as the following reaction?
Mg(OH)2 (s) + 2HCl (aq) → MgCl2 (aq) + 2H2O (l)

• Example 12: “Isotonic saline” is a 0.15 M aqueous solution of
NaCl. How would you prepare 0.80 L of isotonic saline from a
6.0 M stock solution?
Solution
• nsolute = CM1×V1 = CM2×V2 → 6.0 (M) × V1 = 0.15 (M) × 0.8 (L)
→ V1 = 0.020 (L) 6.0 M stock solution

• A 0.020 L portion of the concentrated solution must be diluted to
a final09-Oct-16
volume of 0.80 L.

37

• Exercise 16: In a simulation mercury removal from industrial
wastewater, 0.050 L of 0.010 mol/L mercury(II) nitrate reacts
with 0.020 L of 0.10 mol/L sodium sulfide. What is the mass of
mercury(II) sulfide that would form from the following reaction?
Hg(NO3)2 (aq) + Na2S (aq) → HgS (s) + 2NaNO3 (aq)
09-Oct-16

38

Chapter 03: Suggested Practice Problems
2,7,8-20(even),23,25,27,28,31-41(odd),44,47,5155(o),63-71(odd), 59,63-87(odd),88,92106(even),124,130,137,145,149

09-Oct-16
09-Oct-16

09-Oct-16

39

40

40

23-Oct-16

Types of Chemical Reactions
Chapter 4:

• There are some types of chemical reactions:
 Precipitation Reactions.

Three Major Types of

 Acid-Base Reactions (Neutralization Reactions).
 Oxidation-Reduction Reactions (Redox Reactions).

Chemical Reactions

 Metathesis Reactions (Displacement Reactions)
AB + CD → AC + BD, or X + YZ → XZ + Y.
 Combination Reactions (X + Y → Z)
 Decomposition Reactions (X → Y + Z)
 Combustion Reactions.

23-Oct-16

1

23-Oct-16

2

Ionic Compounds in Water

4.1 The Role of Water as a Solvent
• Water is a polar molecule, since:
 it has uneven electron distribution
 it has 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.

Each bond in H2O is polar.
23-Oct-16

The whole H2O molecule is polar.

3

• Many ionic compounds dissolve in water (NaCl, KNO3...) →

23-Oct-16

4

• When an ionic compound dissolves, its aqueous solution

“soluble” compound. However, many others do not (AgCl,

conduct an electric current → the ionic compound = an

BaSO4,...) → “insoluble” compound.

electrolyte. For example, NaCl in water:

 In the latter cases, the electrostatic attraction among ions in
the compound remains greater than the attraction between
ions and water molecules, so the solid stays largely intact.
• Example 1: Solubilities of NaCl & AgCl in water:
 Solubility of NaCl in H2O at 20oC = 365 g/L
→ NaCl = a “soluble” compound.
 Solubility of AgCl in H2O at 20oC = 0.009 g/L
→ AgCl = a “insoluble” compound.
23-Oct-16

5

23-Oct-16

6


23-Oct-16
• Example 2: The beakers shown below (A, B, C, D) contain
aqueous solutions of the strong electrolyte potassium sulfate.
Which beaker best represents the compound in solution? (H2O
molecules are not shown).

Covalent Compounds in Water
• Water can dissolves many covalent compounds, however,
many other covalent substances do not dissolve appreciably in
water.
 In the former cases, even though these substances

dissolve, they do not dissociate into ions but remain as
intact molecules. As a result, their aqueous solutions do not
conduct an electric current, so these substances are called

Solution
• The formula of potassium sulfate is K2SO4, so the equation for
dissociation is:
K2SO4(s) → 2K+(aq) + SO42-(aq)
• There should be 2 cations for every 1 anion; beaker C
represents this correctly.

nonelectrolytes.

23-Oct-16

7

• Example 3: How many 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 molecules of copper(II) nitrate dissolved in water.
Solution
a) The formula of ammonium sulfate is (NH4)2SO4 so the equation
for dissociation is:
(NH4)2SO4(s) → 2NH4+(aq) + SO42-(aq)
5.0 mol
2×5.0 mol 5.0 mol
b) The formula of cesium bromide is CsBr, and 78.5 g CsBr =
0.369 mol CsBr, so the equation for dissociation is:
CsBr(s) → Cs+(aq) + Br-(aq)

0.369mol 0.369mol 0.369mol
c) The formula of copper(II) nitrate is Cu(NO3)2, and 7.42x1022
molucules Cu(NO3)2 = 0.123 mol Cu(NO3)2, so the equationis:
Cu(NO3)2(s) Cu2+(aq) + 2NO3-(aq)
0.123 mol
0.123 mol 2ì0.123 mol
23-Oct-16
9

23-Oct-16

8

ã Exercise 1: How many moles of each ion are in each solution?
a) 2 mol of potassium perchlorate dissolved in water
b) 354 g of magnesium acetate dissolved in water
c) 1.88×1024 molecules of ammonium chromate dissolved in water
d) 1.32 L of 0.55 M sodium bisulfate.

• Exercise 2: Calculate the molarity of sodium sulfate & each ion
in a 23.55-mL solution which contains 28.24 mg of sodium
sulfate (used in dyeing and printing textiles), M = 139.04 g/mol.

23-Oct-16

10

4.2 Equations for Aqueous Ionic Reactions
• Example 4: Write 3 types of equations to represent aqueous
• Three types of equations to represent aqueous ionic reactions:

ionic reaction between silver nitrate & sodium chromate.

 Molecular equation: shows all reactants and products as
if they were intact, undissociated compounds

Solution
• Molecular equation:

 Total ionic equation: shows all soluble ionic substances

2AgNO3(aq) + Na2CrO4(aq) → Ag2CrO4(s) + 2NaNO3(aq)

dissociated into ions.

• Total ionic equation:

 Net ionic equation: eliminates the spectator ions and

2Ag+(aq)+2NO3-(aq)+2Na+(aq)+CrO42-(aq)→Ag2CrO4(s)+2Na+(aq)+2NO3-(aq)

shows only the actual chemical change. Spectator ions

• Net ionic equation:

are ions that are not involved in the actual chemical

2Ag+(aq) + CrO42-(aq) → Ag2CrO4(s)

change. Spectator ions appear unchanged on both sides

• Spectator ions = NO3-, Na+.

of the total ionic equation.
23-Oct-16

11

23-Oct-16

12


23-Oct-16
• Example 5: Write 3 types of equations to represent aqueous
ionic reaction between barium hydroxide + sulfuric acid.

• Exercise 3: Write 3 types of equations to represent aqueous

Solution

ionic reactions between:

• Molecular equation:

a) Sodium carbonate & Calcium nitrate

Ba(OH)2(aq) + H2SO4(aq) → BaSO4(s) + 2H2O(l)

b) Ammonium sulfate & Barium iodide

• Total ionic equation:

c) Aluminum nitrate & Sodium phosphate

Ba2+(aq) + 2OH-(aq) + 2H+(aq) + SO42-(aq) → BaSO4(s) + 2H2O(l)
• Net ionic equation:
Ba2+(aq) + 2OH-(aq) + 2H+(aq) + SO42-(aq) → BaSO4(s) + 2H2O(l)

• Exercise 4: If 38.5 mL of lead(II) nitrate solution reacts

• The net ionic equation is the same as the total ionic equation

completely with excess sodium iodide solution to yield 0.628 g

since there are no spectator ions.

of precipitate, what is the molarity of lead(II) ion in the original

• This reaction is both a neutralization reaction and a

solution?.

precipitation reaction.
23-Oct-16

13

23-Oct-16

14

4.3 Precipitation Reactions

Predicting whether a precipitate will form

• In a precipitation reaction: two soluble ionic compounds react

1) Attention to the ions present in the reactants.
2) Consider all possible cation-anion combinations.

to give an insoluble products, called a precipitate.

3) Use the solubility rules to decide whether any of the ion

• The key event in a precipitation reaction is: the formation of an

combinations is insoluble.

insoluble product through the net removal of solvated ions from

4) An insoluble combination identifies the precipitate that will

solution.

form.

• It is possible for more than one precipitate to form in such a

• For example: Will a precipitate form from these reactions?

reaction.

a) Ammonium sulfate + barium chloride → ?

• Most of the precipitation reactions are metathesis reactions

b) FeSO4(aq) + Sr(OH)2(aq) → ?

(or double displacement reactions).
23-Oct-16

15

Solubility Rules
Soluble Ionic Compounds
1. All common compounds of Group 1A ions (Li+, Na+, K+, etc.) and
ammonium ion (NH4+) are soluble.
2. All common nitrates (NO3-), acetates (CH3COO- or C2H3O2-) and most
perchlorates (ClO4-) are soluble.
3. All common chlorides (Cl-), bromides (Br-) and iodides (I-) are soluble,
except those of Ag+, Pb2+, Cu+, and Hg22+. All common fluorides (F-) are
soluble except those of Pb2+ and Group 2A.
4. All common sulfates (SO22-) are soluble, except those of Ca2+, Sr2+, Ba2+,
Ag+, and Pb2+.
Insoluble Ionic Compounds
1. All common metal hydroxides are insoluble, except those of Group 1A
and the larger members of Group 2A(beginning with Ca2+).
2. All common carbonates (CO32-) and phosphates (PO43-) are insoluble,
except those of Group 1A and NH4+.

3. All common sulfides are insoluble except those of Group 1A, Group 2A
and NH4+.
23-Oct-16
17

23-Oct-16

16

• Example 6: The precipitation of Calcium fluoride.
Solution
2NaF(aq) + CaCl2(aq) → CaF2(s) + 2NaCl(aq)
2Na+

(aq)

+ 2F-(aq) + Ca2+(aq) + 2Cl-(aq) → CaF2(s) + 2Na+(aq) + 2Cl-(aq)
2F-(aq) + Ca2+(aq) → CaF2(s)

• Example 7: The precipitation of Lead(II) iodide.
Solution
2NaI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2NaNO3(aq)
2Na+

(aq)+

2I-(aq)+ Pb2+(aq) + 2NO3-(aq) → PbI2(s) + 2Na+(aq) + 2NO3-(aq)
Pb2+(aq) + 2I-(aq) → PbI2(s)

23-Oct-16

18


23-Oct-16
• Exercise 6: The following molecular views show reactant
solutions for a precipitation reaction (with H2O molecules
omitted for clarity).

• Exercise 5: Predict whether or not a reaction occurs when
each of the following pairs of solutions are mixed. If a reaction
does occur, write balanced molecular, total ionic, and net ionic
equations, and identify the spectator ions.
a) Potassium fluoride (aq) + strontium nitrate (aq) →
b) Ammonium perchlorate (aq) + sodium bromide (aq) →
c) Iron(III) chloride(aq) + cesium phosphate(aq) →

a) Which compound is dissolved in beaker A: KCl, Na2SO4,
MgBr2, or Ag2SO4? Which compound is dissolved in beaker B:
NH4NO3, MgSO4, Ba(NO3)2, or CaF2?
b) Name the precipitate and spectator ions when solutions A and
B are mixed, and write balanced molecular, total ionic, and net
ionic equations for this process.

d) Sodium hydroxide(aq) + cadmium nitrate(aq) →
e) Magnesium bromide(aq) + potassium acetate(aq) →
f) Silver nitrate(aq) + barium chloride(aq) →
23-Oct-16

19

4.4 Acid-Base Reactions

23-Oct-16

20

• The Bronsted-Lowry Acid-Base Theory.
 Acids are substances that can donate H+ ions to bases.

• The Arrhenius Acid-Base Theory.

 An acid is a "proton donor“.

 An acid = a substance that produces H+ ions when
dissolved in H2O.

 A base is a "proton acceptor".
• Since a hydrogen atom is a proton and one electron, technically

 A base = a substance that produces OH- ions when

an H+ ion is just a proton.

dissolved in H2O.

• The acid-base reaction is essentially a proton transfer.

• An acid-base reaction = a neutralization reaction.
H+(aq) + OH-(aq) → H2O(l)

•

H+

interacts strongly with H2O to form H3O+ in aqueous solution.
H+(aq) + H2O(l) → H3O+(aq)

23-Oct-16

21

Strong Bases

• hydrochloric acid, HCl

• sodium hydroxide, NaOH

• hydrobromic acid, HBr

• potassium hydroxide, KOH

• hydriodic acid, HI

• calcium hydroxide, Ca(OH)2

• nitric acid, HNO3

• strontium hydroxide, Sr(OH)2

• sulfuric acid, H2SO4

• barium hydroxide, Ba(OH)2

• Weak acids and weak bases dissociate very little into ions
in aqueous solution. They are weak electrolytes and conduct
poorly in solution. For example:
• Weak acids and weak bases are kept unchanged in total
ionic equations. For example:
 Molecular equation:
CH3CH2COOH(aq) + NH3(aq) → CH3CH2COONH4(aq)
 Net ionic equation:
CH3CH2COOH(aq) + NH3(aq) → CH3CH2COO-(aq) + NH4+(aq)

• perchloric acid, HClO4
23-Oct-16

22

• Strong acids and strong bases dissociate completely into
ions in aqueous solution. They are strong electrolytes and
conduct well in solution. For example:

Strong Acid & Base
Strong Acids

23-Oct-16

23

23-Oct-16

24


23-Oct-16
Acid-Base Titrations
•

Titration (titrimetry) = a common laboratory method of quantitative
chemical analysis, a known concentration of one reactant is used to
determine the concentration of the other.

•

In a typical acid-base titration, a standardized solution (titrant) of base,
one whose concentration is known, is added slowly to an acid solution
(analyte) of unknown concentration.

•

An acid-base indicator has different colors in acid and base, and is used to
monitor the reaction progress.

•

At the equivalence point, the amount of H+ from the acid equals the amount
of OH- ion produced by the base.

•

The end point occurs when there is a slight excess of base and the
indicator changes color permanently.

23-Oct-16

25

23-Oct-16

26

• Example 8: A 50.00 mL sample of HCl is titrated with 0.1524 M
• Exercise 7: How many moles of H+(aq) are present in 1.5 L of

NaOH. The buret reads 0.55 mL at the start and 33.87 mL at

mixture solution 1.4 M nitric acid & 0.7 M sulfuric acid?

the end point. Find the concentration of the HCl solution.
Solution

• Exercise 8: Write balanced molecular, total ionic, and net ionic

• The balanced equation for the reaction

equations for the following acid-base reactions & identify the

NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

spectator ions.

• Volume of base NaOH = 33.87 mL – 0.55 mL = 33.32 mL
ã Moles of NaOH = 0.3332ì0.1524 =

5.078ì10-3

mol NaOH

a) hydrochloric acid (aq) + potassium hydroxide (aq) →

• Since 1 mol of HCl reacts with 1 mol NaOH, the amount of HCl
= 5.078x10-3 mol.

b) strontium hydroxide (aq) + perchloric acid (aq) →
c) hydriodic acid (aq) + calcium hydroxide (aq) →

• The concentration of the HCl = (5.078×10-3 mol):(0.0500 L) =

d) potassium hydroxide (aq) + propionic acid (aq) →

0.1016 M HCl.
23-Oct-16

27

23-Oct-16

28

4.5 Oxidation-Reduction Reactions

• Exercise 9: What volume of 0.1292 M Ba(OH)2 would

• Oxidation is the loss of electrons. Reduction is the gain of

neutralize 50.00 mL of the HCl solution standardized above in

electrons. Oxidation and reduction occur together.

the sample problem?

• The reducing agent loses electrons and is oxidized. The
oxidizing agent gains electrons and is reduced.

• Exercise 10: In a titration 42.0 mL of 0.210 M H2SO4 was

• A redox reaction involves electron transfer.

needed to neutralize 50.0 mL of a NaOH solution. Determine

• For example:

the molarity of the sodium hydroxide solution.

23-Oct-16

29

23-Oct-16

30