PART I: INORGANIC CHEMISTRY
UNIT I: Introduction to Chemistry
SCIENCE is a systematized knowledge based on facts and principles taken or derived from hypothesis
and experimentation which help man understand the world around him.
Fields of Science
 Social Science deals with institutions or organizations functioning in a society; and of human
behaviors and relationships
 Politics
 Economics
 History
 Natural Science deals with the study of nature (measurable phenomena)
 Biological sciences – living things
o Zoology
o Botany
o Microbiology
 Physical sciences – nonliving things
o Physics
o Chemistry
o Geology
o Astronomy
o Meteorology
 Applied Science deals with the application of the theoretical sciences (social and natural sciences)
 Medicine
 Engineering
 Architecture
CHEMISTRY describes matter, its properties, the changes it undergoes, and the energy that accompany
those processes.
CHEMICAL TECHNOLOGY deals with the application of the principles of chemistry for practical use.
1
History of Chemistry
 The earliest practical knowledge of chemistry was concerned with metallurgy, pottery and dyes.
These crafts were developed with considerable skill, but with no understanding of the principles
involved, as early as 3500 B.C in Egypt and Mesopotamia.
 Basic ideas of element and compound were first formulated by the Greek philosophers in 500 BC –
300 BC. Generally there were 4 elements (fire, air, water, earth) combined in varying
proportions to form all things.
 Aristotle’s definition of a simple body as ‘one into which other bodies can be decomposed and
which itself is not capable of being divided’ is close to the modern definition of element.
 Ancient Egyptian industrial arts and Greek philosophical speculations were fused into a new
science at the beginning of the Christian era in Alexandria.
This became the beginning of alchemy (medieval form of chemistry) which used occultism and
magic.
Interest of the period:
 transmutation of base metals into gold
 imitation of precious gems
 search for the elixir of life ( thought to grant immortality)
 Muslim conquest in 7th century A.D. diffused the remains of Hellenistic civilization to the
Arab world.
 The first chemical treatises to become well-known in Europe were Latin translations of Arabic
works; hence, it is often erroneously supposed that chemistry originated among the Arabs.
 Alchemy developed extensively during the middle Ages, cultivated largely by itinerant scholars
who wandered over Europe looking for patrons.
Evolution of Modern Chemistry
Scientist
Robert Boyle (1627-1691)
Contribution
He performed experiments under reduced pressure using an air
pump and discovered that volume and pressure are inversely
related in gases.
Robert Hooke (1635-1703)
He gave the first rational explanation of combustion as
combination with air.
John Mayow (1641-1679)
He postulated that the air needed in lighting a candle is the
same air needed by animals to survive.
Johann Joachim Becher (1635-1682)
George Ernst Stahl (1660-1734)
They introduced the phlogiston theory of combustion which held
that the substance phlogiston is contained in all combustible
bodies and escapes when the bodies burn.
2
Jan Baptist van Helmont (1589-1644)
He was the first to describe carbon dioxide as “gas sylvestre”
given off by burning charcoal and this gas was the same as that
produced by fermenting grapes which sometimes renders the air
of caves unbreathable.
Joseph Black (1728-1799)
He was the first person to isolate carbon dioxide in a perfectly
pure state using carbonates.
Joseph Priestly (1733-1804)
He discovered the “dephogisticated air” by heating a
nonmetallic red power (mercuric oxide) in a jar inverted in a
bowl of mercury. Antoine Lavoisier named the dephogisticated
air as oxygen.
Henry Cavendish (1731-1810)
He isolated a very light, highly flammable gas by reacting metals
with acids; when this gas is burned, pure water was produced.
Antoine Lavoisier named the gas hydrogen.
Branches of Chemistry
Branch
Inorganic Chemistry
Description
It is the study of almost all the elements and their compounds except
carbon and its compounds. Some simple carbon compounds are traditionally
classified as inorganic since they are derived from mineral sources.
Organic Chemistry
It is the study of carbon and its compounds, considering at least carboncarbon or carbon-hydrogen chain except urea.
Analytic Chemistry
It is the identification of the composition, both quantitative (how much is
present) and qualitative (what is present) of a substance.
Biochemistry
It is the study of the composition and processes that occur in living
organisms.
Physical Chemistry
It is the study of theories, principles and laws that govern the structure
and changes in matter.
Environmental Chemistry
It is the study of the composition, nature and concentration of pollutants
and other substances affecting the environment.
Nuclear Chemistry
It is the study of atomic nucleus, its reactions and the products of such
reactions.
Industrial Chemistry
It is the study of the physical and chemical processes applied in the
manufacture of substances
3
The Scientific Method is a systematic approach in problem solving.
 stating the problem
 clear statement
 manageable scope
 gathering information on the problem
 observation or research
 formulation of hypothesis
 proposed solution to a problem
 tentative answer-models
 performing the experiment
 variable – factor that is being tested (experimental set-up)
 recording and analyzing data
 data – recorded observations and measurements
 data table – is a chart that organizes information in rows and columns
 graph – is data shown in the form of a picture
 stating a conclusion
 conclusion – final answer to the problem
 theory – logical explanation of events that occur in nature supported by results obtained from
repeated experiments where further experiments, new phenomena could be discovered
 law – statement that summarizes the results of observations and experiments; a precise
statement which describes a mode of behavior or pattern of action about a large number of
related facts
Scientific Attitudes
1. curiosity
2. determination
3. open-mindedness
4. acceptance of failure
5. objectivity
6. humility
7. skepticism
8. patience
Importance of Chemistry
1. In the field of _____, it enables scientists to investigate and discover the properties of raw
materials in the purpose of processing them into new products as well as improve existing ones.
2. In agriculture, chemistry is used in soil analysis and in the manufacture of fertilizers and
insecticides increasing production.
3. In medicine, chemistry is employed to discover new minerals and vitamins to improve health and to
understand human ailments and to invent drugs to cure diseases.
4. In industry, chemistry makes possible the manufacture of new and better economic goods, the
invention of better appliances and the development of new technologies for processing goods for
our consumption
Concerns of Chemistry
1. worldwide food shortage
2. development of pharmaceutical chemicals
3. dwindling natural resources and pollution
4
FUNDAMENTALS OF MEASUREMENT
Measurement is a process of comparing a thing with a standard. It is very important in giving more
precise quantitative and qualitative description of an object.
Methods of measurements
 Direct measurement
 Measuring things with the use of an instrument
 Indirect measurement
 Measuring things using mathematical calculation
Kinds of measurable quantities
 Fundamental quantities
 Measured by direct method
 Derived quantities
 Formulated based on the fundamental quantities
BASE UNITS
Fundamental units of the international system
Measurement
Length
Mass
Time
Thermodynamic
temperature
Electric current
Amount of substance
Luminary intensity
Unit
Symbol
Meter
Kilogram
Second
Kelvin
m
kg
s
K
Ampere
Mole
Candela
A
mol
cd
DERIVED UNITS
AREA
VOLUME
DENSITY
m x m = m2
cm x cm x cm = cm3
ρ = kg/cm3 = kg/cm3
Multiply two lengths
Multiply 3 lengths together
Mass per unit volume (D= M/V)
5
System of measurement
 Metric system/SI/international system of units
Variations:
MKS – meter-kilogram-second
CGS – centimeter-gram-second
 English or British system
FPS system - foot-pound-second
The following are prefixes used to express bigger and smaller units:
Symbol
Prefix
Exponential
yotta
Y
1024
zetta
Z
1021
exa
E
1018
peta
P
1015
tera
T
1012
giga
G
109
mega
M
106
kilo
k
103
hecto
h
102
deca
da
101
deci
d
10¯1
centi
c
10¯2
milli
m
10¯3
micro
µ
10¯6
nano
n
10¯9
pico
p
10¯12
femto
f
10¯15
atto
a
10¯18
zepto
z
10¯21
yocto
y
10¯24
6
SIGNIFICANT FIGURES
 They comprise all digits that are known with certainty plus the first digit that is uncertain. The
position of the decimal point is relevant.
Rules in counting significant figures
Rule
Measurement
Expression
Significant
Figures
1. All nonzero digits are significant.
14567.3
34.128
3.986
6
5
4
2. All zeros between two nonzero digits are
significant.
34.0045
1.00006017
20037089
6
9
8
3. Zeros to the right of a nonzero digit, but to the
left of an understood decimal point, are not
significant unless specifically indicated as
significant by a bar placed above the rightmost,
such zero becomes significant.
108,000000
108,0000 _
108,000000
3
4
9
4. All zeros to the right of a decimal point but to
the left of a nonzero digit are not significant.*
0.000509
0.02876
0.0000036
3
4
2
5. All zeros to the right of a decimal point and to
the right of a nonzero digit are significant.
0.4087
0.030670890
700.00000000
4
8
11
*The single zero conventionally placed to the left of a decimal point in such an expression is never
significant. It is just used to locate the decimal point.
OPERATIONS WITH SIGNIFICANT FIGURES
 Addition and Subtraction
The answer must contain the same number of decimal places as the term with the least number of
significant figures.
Examples:
a. 20.63 + 6.6 cm + 3. 786 cm = 31.016 cm
b. 387.876 L – 197.23L = 190.646 L
ans. 31.0 cm
ans. 190.65 L
7
 Multiplication and Division
The answer must contain the same number of significant figures as the term with the least number
of significant figures.
Examples:
a. 9.25 m x 0.52 m x 11. 35m = 54.5935 m3
b. 69.48 m by 3.62 s = 19.19337017 m/s
ans. 55 m3
ans. 19.2 m/s
SCIENTIFIC NOTATION
 Method of writing or expressing very large or very small numbers into its exponential form.
Form: M x 10n
Where: M - number lower than one and not greater than ten
N - number of times the decimal point is moved. It can be
a positive or negative integer.
Rules:
1.
Determine M by moving the decimal point in the original number to the left or right so that
the only one nonzero digit is to the left of it.
2. Determine n by counting the number of places the decimal point has been moved.
If moved to the left, n is positive.
If moved to the right, n is negative.
Examples:
Diameter of the earth
Diameter of a hydrogen atom
Speed of light
Positional form
Exponential form
= 1 300 000 000
ans. 1.3 x 109 cm
= 0.000 000 01 cm
ans. 1 x 10- 8 cm
= 30 000 000 000 cm/s ans. 3 x 1010cm/s
ROUNDING OFF NUMBERS
1.
If the figure to be dropped is five, or greater than five, increase by one the value of the last
figure to be retained.
Examples:
a) 78.567 rounded off to the nearest hundredths place is 78.57
b) 123.345 rounded off to the nearest hundredths place is 123.35
c) 457.9679 rounded off to the nearest thousandths place is 457.968
2. If the figure to be dropped is less than five, the last figure to be retained should not be changed.
Examples:
a) 5.052
rounded off to the nearest hundredths place is 5.05.
b) 136.324
rounded off to the nearest ones place is 136
c) 98230.478 rounded off to the nearest tens place is 98230
8
CONVERSION OF UNITS
 A unit conversion factor is used to covert a quantity in one system of units to corresponding quantity
in another system of units.
METRIC to ENGLISH
1cm = 0.3937 in = 0.03281 ft
1m = 39.37 in 3.281 ft = 1.094 yd
ENGLISH to METRIC
1 in = 2.54 cm = 0.0254 m
1ft = 30.5 cm = 0.305 m
1 km = 3281 ft = 0.6214 mi
1 yd = 91.4 cm = 0.914 m
1cm3 = 0.0610 in3 0.0000 353 ft3
1 mi = 1609 m =1.609 km
1L = 1.06 qt = 0.265 gal = 0.0353 ft3
1 qt = 946 ml = 0.946 L
1g = 0.0353 oz = 0.00220 lb
1 oz = 28350 mg = 28.35 g
1kg = 2.20 lb = 0.00110 tn
1lb = 453.6 g = 0.4536 kg
1 metric tn (103kg)= 2200 lb = 1.10 tn
1 tn =907 kg = 0.907 metric tn
UNCERTAINTIES IN MEASUREMENT
The experimental process is an essential part of chemistry, and reliable measurement
information is an essential part of the experimental process. Unfortunately, the measurement of any
physical quantity is subject to some uncertainty. Consequently, the complete expression of a measured
quantity must include the number value, the measurement unit used, and some indications of how reliable
the number is.
Contributions to the uncertainty in measurements of physical quantities accrue from limitations in
accuracy and limitations in precision inherent in all measurement process.
PRECISION AND ACCURACY
 Dependent on measuring device used and the skill of the person measuring.
PRECISION refers to the degree to which successive measurements agree with each other.
ACCURACY is how close or the nearness of a measurement to its accepted value.
PERCENTAGE ERROR
% = Difference between the two values / true value x 100 %
9
SAMPLE PROBLEMS:
DENSITY
ρ = m/v
1.
2.
3.
4.
5.
6.
7.
8.
9.
Calculate the density of a material that has a mass of 52.457 g and a volume of 13.5 cm 3. (3.89
g/cm3)
A student finds a rock on the way to school. In the laboratory he determines that the volume of the
rock is 22.7 cm3, and the mass in 39.943 g. What is the density of the rock? (1.76 g/cm3)
If 30.943 g of a liquid occupy a space of 35.0 ml, what is the density of the liquid in g/cm 3? (0.884
g/cm3)
The density of silver is 10.49 g/cm3. If a sample of pure silver has a volume of 12.993 cm3, what
would the mass? (136.3g)
How many grams of tin would occupy 5.5 L, if it has a density of 7.265 g/cm 3? (4.0 x 104 g)
What is the mass of a 350 cm3 sample of pure silicon with a density of 2.336 g/cm3? (820 g)
Pure gold has a density of 19.32 g/cm3. How large would a piece of gold be if it had a mass of 318.97
g? (16.51 cm3)
How many cm3 would a 55.932 g sample of copper occupy if it has a density of 8.92 g/cm 3? (6.270
cm3)
The density of lead is 11.342 g/cm3. What would be the volume of a 200.0 g sample of this metal?
(17.63 cm3)
PERCENT ERROR
% error =
accepted value  exp erimental value
x 100
accepted value
10. Working in the laboratory, a student finds the density of a piece of pure aluminum to be 2.85 g/cm 3.
The accepted value for the density of aluminum is 2.699 g/cm3. What is the student's percent error?
(5.59%)
11. A student experimentally determines the specific heat of water to be 4.29 J/g x C o. He then looks up
the specific heat of water on a reference table and finds that is is 4.18 J/g x C o. What is his
percent error? (2.63%)
12. A student takes an object with an accepted mass of 200.00 grams and masses it on his own balance.
He records the mass of the object as 196.5 g. What is his percent error? (1.75%)
TEMPERATURE
K = C + 273
13.
14.
15.
16.
17.
18.
250 Kelvin to Celsius
339 Kelvin to Celsius
17 Celsius to Kelvin
55 Celsius to Kelvin
89.5 Fahrenheit to Celsius
383 Kelvin to Fahrenheit
C = K – 273
C = (F - 32) x 5/9
-23 oC
66 oC
290 K
328 K
31.9 oC
230. F
10
F = (C x 9/5) + 32
SPECIFIC HEAT CAPACITY
Q = mc(T)
where ΔQ is the heat energy put into or taken out of the substance
m is the mass of the substance
c is the specific heat capacity
ΔT is the temperature differential.
19. How many calories of heat are required to raise the temperature of 550 g of water from 12.0 oC to
18.0 oC? (remember the specific heat of water is 1.00 cal/g x oC) = 3300 cal
20. How much heat is lost when a 640 g piece of copper cools from 375 oC, to 26 oC? (The specific heat
of copper is 0.38452 J/g x oC) = 86 000 J
21. The specific heat of iron is 0.4494 J/g x oC. How much heat is transferred when a 24.7 kg iron ingot
is cooled from 880 oC to 13 oC? = 9 600 000 J
22. How many grams of water would require 2.20 x 104 calories of heat to raise its temperature from 34.0
o
C to 100.0 oC? (Remember the specific heat of water is 1.00 cal/g x oC)= 333 g
23. 8750 J of heat are applied to a piece of aluminum, causing a 56 oC increase in its temperature. The
specific heat of aluminum is 0.9025 J/g x oC. What is the mass of the aluminum? = 170 g
24. Find the mass of a sample of water if its temperature dropped 24.8 oC when it lost 870 J of heat.
(remember that 4.18 J = 1.00 cal) = 8.4 g
25. How many degrees would the temperature of a 450 g ingot of iron increase if 7600 J of energy are
applied to it? (The specific heat of iron is0.4494 J/g x oC) = 38 oC
26. A 250 g sample of water with an initial temperature of 98.8 oC loses 7500 joules of heat. What is the
final temperature of the water? (Remember, final temp = initial temp - change in temp) = 92 oC
27. Copper has a specific heat of 0.38452 J/g x oC. How much change in temperature would the addition
of 35 000 Joules of heat have on a 538.0 gram sample of copper? = 170 oC
28. Determine the specific heat of a certain metal if a 450 gram sample of it loses 34 500 Joules of heat
as its temperature drops by 97 oC. = 0.79 J/g x oC
29. 4786 Joules of heat are transferred to a 89.0 gram sample of an unknown material, with an initial
temperature of 23.0 oC. What is the specific heat of the material if the final temperature is 89.5
o
C? = 0.809 J/g x oC
30. The temperature of a 55 gram sample of a certain metal drops by 113 oC as it loses 3500 Joules of
heat. What is the specific heat of the metal? = 0.56 J/g x oC
11
NAME:_________________________________
COURSE YEAR & SCHED:________________
SCORE:_________
DATE:___________
ACTIVITY 1.1
1. Given the following data, solve for the required values.
TRIAL 1
METRIC
ENGLISH
SYSTEM(cm)
SYSTEM
(in)
Length
Width
Thickness
Length
1
13.3
6.2
6.5
5.3
2
13.4
6.4
6.3
5.2
Average
Width
2.3
2.1
Thickness
2.4
2.3
Answer
a)____________________
b)____________________
c)____________________
d)____________________
e)____________________
f)____________________
a) Area of the wooden block
b) Experimental value
c) Percentage error if the true value is 86.45 cm2
d) Volume of the wooden block
e) Experimental value
f) Percentage error if the true value is 538.39 cm3
2. Group the class into dyads and present a figure for each group to measure.
Problem:
What is the distance between points A and B following the arrow?
Materials:
Any measuring device or material found inside the classroom.
Procedure:
1. Discuss within your group the possible ways of solving the problem.
2. Write down your proposed procedure.
3. Perform the activity and record the data gathered.
4. Show the procedure you followed to your instructor.
5. Fill in the table with appropriate data and do the necessary conversion of units stated below.
Table 1
Name of Student
1.
2.
3.
Average
Table 2
Distance
A to B
mm
Measurement
cm
M
Note: Score will be based on percentage error.
12
km
NAME:__________________________________________
COURSE YEAR AND SCHED.______________________
SCORE:_________
DATE: _________
ACTIVITY 1.2
1. Express each measurement to three significant figures in scientific notation.
a)
b)
c)
d)
e)
f)
8,567, 800,000,000 km
45.4236 g
0.0056231 L/mole
0.00000000000428 m/s
98.6700000 kg
99.98000 g/cm
________________________
________________________
________________________
________________________
________________________
________________________
2. Express each measurement in positional notation or exponential form.
a)
b)
c)
d)
e)
2.35 x 105 g
33 000 000 mg
5.56 x 10 -8 mole
670 000 000 000 cg
100 000 000 cm3
________________________
________________________
________________________
________________________
________________________
3. Give the correct answer to the following using the rules in the operations with significant figures.
1. 286.05 m + 11.7 m + 3.66 m
________________________
2. 35.77 L - 4.6 L
________________________
3. 235 cm x 35 cm x 0.7cm
________________________
4. 98.568 sec / 35 sec / 13.4 sec ________________________
5. 395 km / 0.003 hr
________________________
4. A basketball player is undergoing therapy after a mild surgery from his right leg due to bone
translocation. During his exercise and physical examination, the following data were gathered. He stands 6
feet 3 inches tall. He weighs 194 pounds. He plays in a game for 35 minutes. He throws the ball with the
speed of 100 miles per hour during fast breaks and he gains a body temperature of 102 degrees
Fahrenheit during workout.
a) How tall is the patient in
(1) ______meter
b) How much does he weigh in (1) ______kg
(2) ______cm
(2) ______g
and (3) ______mm
and
(3) ______mg
c) How long does he play in seconds? ______
d) How fast does he throw in (1) ______km/hr
e) How hot is his body in
(1) ______0C
(2) ______ft/sec
(2) ______K
13
(3) ______m/s
NAME:__________________________________________
COURSE YEAR AND SCHED.______________________
SCORE:_________
DATE: _________
ACTIVITY 1.3
Solve the following problems correctly. Show all solutions.
1.
A patient has a body temperature of 1040F. What is his temperature in Celsius?
2. Which patient has a higher fever, one with a temperature of 102.20F or one with a temperature of
39.00C?
3. Normal urine density is 1.025g/ml. What is the weight of a 100ml sample of urine?
4. An order for a medication reads: Give 0.25ml/kg body weight. How much medication should be
given to a patient weighing 175 lbs?
5. For the following problems, refer to the given food label:
a. In each serving, what is the amount of
total fats in kilograms present?
b. Determine the amount, in grams of cholesterol.
c. How many μg of sodium is present?
6. Consider the following label on a vial of medication:
Cefuroxime
Pediatric Dosage
200mg/kg/day
14
a.
How much of the antibiotic should be administered per day to an infant who weighs 14.7 lbs?
b. If the medication is to be administered three times a day, how much medication should be
given per dose?
c.
Compute for the total dosage in grams if the antibiotic is to be given for 7 days.
7. The doctor orders a 500 μg dose of a drug. Only 1mg tablet is on hand. How many would you give?
8. Consider the following medication order:
Name: Pascual, Piolo
Height: 5’8”
Weight: 160lbs
Medication Order:
Meperidine IV q8h
Dosage Instruction: 50mg/m2
a.
Room-Bed: 100-A
What is the patient’s height in inches?
b. What is the patient’s weight in kilograms?
c.
What is the patient’s body surface area?
d. Compute the amount of the drug to be administered to the patient.
9. A patient has been prescribed 15 mg of Diazepam. The tablet is available only at 0.005g dose. How
many tablets should be given to the patient?
10. 800 ml of sodium chloride 0.9% (normal saline) is dripping into Mrs. Fernandez’ arm at 25
drops/min. The IV set delivers 20 drops/ml. How long will it take her to receive the infusion?
15
UNIT II: Matter and Energy
Matter is anything that occupies space and has mass.
Matter
Pure
Substances
Element
Mixtures
Compound
metal
acid
non-metal
base
metalloids
salt
Homogeneous
solution
Heterogeneous
colloid
suspension
oxide
PURE SUBTANCES
Pure substances which consist of one particular kind of atom are homogeneous materials with definite
chemical properties.
ELEMENTS
Element is a type of matter composed of atoms which all have exactly the same positive charge on their
nuclei. All atoms of an element have the same atomic number.
Classifications of elements:
Metals
Physical properties:
Chemical properties:
Usually solid at room temperature
Ductile (can be drawn into sheets or wire without breaking)
Malleable (can be hammered, pounded or pressed into different shapes
without breaking)
Good conductor of heat and electricity
Show metallic luster when polish
Combine with other metals to form an alloy
Ex: bronze (Cu and Sn) and brass (Cu & Zn)
React with non0metal to form ionic compound or salts
Ex: NaCl, CaF2, K2O
Lose electron forming positive ion (for base forming elements)
Ex: Al, Mg, Ca, K
16
Non-metals
Physical properties:
Chemical properties:
Metalloids or Semi-metals
Properties:
Seldom have metallic luster
Tend to be colorless or brilliantly colored
Often gases at room temperature
Generally have low melting point and densities
Serve as poor conductor of heat and electricity
Are Brittle
Combine with non-metals to form covalent compounds
Ex: CO2, H2O, CH4
Gain electrons forming a negative ion (for acid forming elements)
Ex: C, H, O, Br, S
Have properties that lie between metals and non-metals
Often look like metal but are brittle like non-metal
Are neither conductor nor insulator but make excellent semi-conductor
Ex: B, Si, As, Ge
COMPOUNDS
 Are Pure substances
 Are made up of atoms of two or more elements combined in fixed ratios
 Are chemically combined elements
 May be broken or decomposed into simpler substances
Classification of Compounds
Acid comes from the Latin word acidus which means ‘sharp’
 hydrogen compound whose hydrogen can be replaced by a metal
 the negative portion of the acid is a non-metal or an acid radical
 properties of acids:
o have sour taste
o turn blue litmus paper to red
o react with certain metals to produce hydrogen
o react with bases to produce salts and water
 classification of acids:
o monoprotic acid
ex: HNO3, HCl, HBr
o diprotic acid
ex: H2SO4, H2C2O4
o triprotic acid
ex: H3PO4, H3AsO3
Base





hydroxides of metals
alkalis – soluble bases
compound which forms hydroxyl group (OH) as the only negative group in the solution
proton acceptor
OH-1 radical
17

Salt



properties of bases:
o have bitter taste
o have soapy or slippery feeling
o turn red litmus paper to blue
o react with an acid to produce salt and water
o most metallic hydroxides are insoluble in water
substance that consisst of a metal or a metallic radical
combined with a non-metal or an acid radical
an ionic compound consisting of a positive ion except hydrogen and a negative ion except hydroxide
or oxide ion
types of salts:
o normal salt
 contains only a metal or a metallic radical, combined with a non-metal or an acid radical
 is neutral to litmus paper
 ex: NaCl
o basic salt
 contains one or more OH radical
 reacts to litmus paper changing red to blue
 ex: Bi(OH)2 NO3
o acid salt
 compound in which a part of the hydrogen of the acid has been replaced by a metal
 reacts to litmus paper changing blue to red
 ex: NaHSO3
o double salt
 contains two metals combined with one radical
 ex: (NaAl(SO4))2 · 12 H2O
Oxide
 compound which contains oxygen and only one other element
 kinds of oxides
 metallic oxide
 is a combination of a metal and oxygen
 considered as a basic anhydride
 base = water + basic anhydride
 ex: Na2O, CaO
Na2O + H2O
NaOH
 non-metallic oxide
 the element other than oxygen is a non-metal
 also called acid anhydride
 acid = non-metallic oxide + water
 ex: CO2 + H2O
H2CO3
18
MIXTURE
 contains two or more substances that intermingle without being chemically combined
 has components that could be separated by either physical or chemical means
 has variable composition
 has constituents that retain their original identity since simple mixing has not changed their
physical and chemical properties
 types of mixtures:
o homogeneous mixture
 appears to be made up of only one phase to the naked eye
 ex: sugar solution, alloy, starch and flour
o heterogeneous mixture
 components are easily identified or distinguished
 ex: rice and corn, sugar and sand, oil and water
 ways in separating mixtures
o filtration
o evaporation
o decantation
o magnetism
o distillation
o centrifugation
o flotation
o chromatography
PROPERTIES OF MATTER
Extensive property
 changes when the size of the sample changes
 ex: mass, volume, length, total charge
Intensive property
 doesn’t change when some of the samples are taken away
 serves as clues in identifying unknown substances
 ex: temperature, color, hardness, taste, melting point, boiling point, pressure, molecular weight,
density, malleability, ductility
types of intensive property
Physical property
 physical appearance
o color, taste, odor, volume, mass
 physical changes
o boiling point, melting point, freezing point
Chemical property
 describes how a material reacts or fails to react with other materials
 reactivity
o ability of a substance to combine readily with other substance
 combustibility
o ability of a substance to combine with oxygen
19




stability
o ability of a substance to resist change or inability to decompose under ordinary
conditions
deliquescence
o ability of a substance to become wet when exposed to the air because it attracts and
absorbs moisture from air. (ex. NaCl, ZnCl2)
efflorescence
o ability of a substance to become very dry, crisp and powdery when exposed to air
because it loses its water of crystallization. (ex: washing soda)
effervescence
o ability of a substance to form bubbles or foams as it gives off gas. (ex: hydrogen
peroxide)
PHASES OF MATTER
 solids
 substances having definite volume and definite shape
 generally rigid materials and are in compact form
types of solids
crystalline substances
o have definite arrangement and the atoms have definite relative positions
o assume definite geometric shape or figure
o have definite melting points
o examples: table sugar, salt, diamonds
amorphous substances
o do not assume any definite form
o have no definite melting points
o are sometimes called “super cooled liquids
o examples: wax, paraffin, glass, plastic
 liquids
 are substances having definite volume but no definite shape
 assume the shape of their containers
 are practically incompressible
 gases
 are substances having no definite shape and no definite volume
 are compressible
 have low densities
 plasma
 is an ionized gas which may result from the breaking of an atom or molecule of a gas when
sufficient energy is supplied to it.
 contains free electrons and free positive ions, which are equal in number moving independently
20
CHANGES IN MATTER
Physical change
 rearranges molecules but doesn’t affect their internal structures
 affects the state and appearance of matter
 no new substance is formed
o boiling water
o dissolving sugar in water
o dicing potatoes
 processes
o evaporation – liquid to gas
o condensation – gas to liquid
o freezing – liquid to solid
o melting – solid to liquid
o sublimation – solid to gas
Chemical change
 results in the formation of a new chemical substance
o iron rusting (iron oxide forms)
o gasoline burning (water vapor and carbon dioxide form)
 processes
o oxidation – process where oxygen is added, producing an oxide
 2Na + O2
Na2O
o Reduction – process where oxygen is removed from the substance
 2HgO
2Hg + O2
o Neutralization – reaction of acid and base producing salt and water
 HCl + NaOH
H2O + NaCl
o Hydrolysis – reaction of salt and water producing acid and base
 KCl + H2O
HCl + KOH
o Fermentation – reaction caused by microorganisms producing alcohol and CO 2
 Used in beer industries
o Saponification – reaction between an alkali or base and fats and oils producing soap and
glycerol

Evidences of chemical change
o Taking and giving off energy
o Color/odor change
o Evolution of gas
o Formation of precipitate
21
Nuclear change
 is a change in the composition of nuclei of atoms
 is the fusion of hydrogen atoms to produce helium and energy
 nuclear fission
o heavy nucleus splits into two much lighter nuclei, emitting several small particles at the
same time
o atomic bomb – uncontrolled fission
 exceeds the critical mass - (chain reaction)
 1 kg of fissionable uranium – used bomb dropped on Hiroshima, Japan
o Nuclear energy reactors – controlled fission
 Nuclear power plant generates heat to produce steam, which turns a turbine
attached to an electric generator
 Heat is produced by splitting uranium

nuclear fusion
o two lighter nuclei combine to form a heavier one
o ultimate source of nearly all the energy on Earth
 nuclear fusion in the sun
 deuterium + tritium  helium + energy
 1.7 x 109 kJ/mol of energy with no radioactive by-products
**
Lise Meitner (1878-1968)
Until very recently, this extraordinary physicist received little of the acclaim she deserved.
Meitner worked in the laboratory of the chemist Otto Hahn, and she was responsible for the discovery of
protactinium (Pa) and numerous radioisotopes. After leaving Germany in advance of the Nazi domination,
Meitner proposed correct explanation of nuclear fission. In 1944 Hahn received the Nobel Prize in
chemistry, but he did not even acknowledge Meitner in his acceptance speech. Today, most physicists
believe Meitner should have received the prize. Despite controversy over names for elements 104 to 109,
it was widely agreed that element 109 should be named meitnerium (Mt).
22
ENERGY
 every change in matter involves energy
 ability to do work.
Kinetic Energy
Energy possessed by a body in motion by virtue of
its motion
Potential Energy
Energy possessed by a body at rest on the basis of
its position, condition or composition
Forms of Kinetic Energy:
Forms of Potential Energy:
Electrical Energy is the movement of electrical
charges. It may operate electrical devices
Chemical Energy is energy stored in the bonds of
atoms and molecules. It is the energy that holds
these particles together.
Radiant Energy is electromagnetic energy that
travels in transverse waves. Radiant energy includes
visible light, x-rays, gamma rays and radio waves.
Thermal Energy, or heat, is the internal energy in
substances––the vibration and movement of the
atoms and molecules within substances.
Nuclear Energy is energy stored in the nucleus of
an atom––the energy that holds the nucleus
together.
Gravitational Energy is the energy of position or
place.
Sound Energy is produced when a force causes an
object or substance to vibrate––the energy is
transferred through the substance in a wave.
CHANGES IN ENERGY
1. Exothermic change
 energy is given off
 it involves heat flow from the system to the surroundings
 example: combustion
2. Endothermic change
 energy is absorbed
 it involves heat flow from the surroundings to the system
 example: decomposition of water; melting of ice
Enthalpy - a thermodynamic quantity that is the sum of the internal energy plus the product of the
pressure and volume
 exothermic - decrease in the enthalpy of the system (H final < Hinitial)
 endothermic - increase in the enthalpy of the system (H final > Hinitial)
 melting of ice (heat flows into the ice from the surroundings)
Entropy – a thermodynamic quantity related to the number of ways the energy of a system can be
dispersed through the motions of its particles
 exothermic – heat lost by the system is gained by the surroundings
 endothermic – heat gained by the system is lost by the surroundings
23
LAWS OF MATTER AND ENERGY
Law
Proponent
Law of Conservation Antoine Lavoisier
of Matter
Law of Conservation
of Energy
Gottfried
Wilhelm Leibniz
Law
of
Definite
Proportion or
Law of Constant
Proportion
Joseph Proust
Law
of
Proportion
John Dalton
Multiple
Theory
Matter is neither created
nor destroyed in any
chemical reaction.
Energy is neither created
nor destroyed but can be
transformed from one
form to another.
A pure compound is
always
made
up
or
consists of the same
element combined in the
same proportions by mass.
When two elements A and
B form one compound, the
amount of A that is
combined
in
this
compound with a fixed
amount of B stand in
small, whole-number ratio.
Example
180 g glucose + 192 g oxygen gas
 264 g carbon dioxide + 108 g
water (372 g = 372 g)
Calcium carbonate (CaCO3)
Ca = 40 g = 40%
C = 12 g = 12%
O = 48 g = 48%
100 g 100%
Carbon oxide I
57.1% oxygen
42.9% carbon
Ratio: (57.1/42.9) = 1.33
Carbon oxide II
72.7% oxygen
27.3% carbon
Ratio: (72.7/27.3) = 2.66
2.66 2
=
(whole number ratio)
1.33 1
Mass fraction
 fraction by mass
 part of the compound’s mass contributed by the element
 dividing the mass of each element by the total mass of the compound
Mass percent
 percent by mass
 fraction by mass expressed as a percentage
24
GAS LAWS
Some Gases and their Uses
GASES
1. carbon dioxide
2. dry ice
3. krypton
4. xenon
4. liquid helium
5. Argon
6. Argon-nitrogen mixture
7. steam or gaseous water
8. ozone
USES
Used as a refrigerant
Insufflation gas for minimal invasive surgery to enlarge and
stabilize body cavities for better visibility of the surgical field
For cryotherapy
Used in preserving, freezing, and transporting food
Used to preserve and ship biological samples.
Used by cosmetic surgeons to freeze warts for easy removal.
Used in lasers and photography
Used in coagulation of retina (krypton laser)
Used in anesthetics
Used in magnetic resonance imaging (MRI)
Used as a semi-conductor
Used for treatment of benign eyelid tumor
Used in electric light bulbs
Used to power electric generators
Used to protect man from too much exposure to the sun’s
ultraviolet rays
Manifests bactericidal, virucidal and fungicidal actions
Has the capacity to stimulate blood circulation and the immune
response
The Kinetic Molecular Theory of Gases
1. Particles of gases are constant, random, and straight-line motion.
2. Molecules of a gas are separated by great distances which makes the gas mostly empty space.
3. Molecules collide with one another and with the walls of their container.
4. Each molecule acts independently of all the others.
5. Individual molecules upon collision may or may not gain energy.
Atmospheric Pressure
 it is the force exerted by air upon a given volume of matter
 measured by a barometer (Evangelista Torricelli)
 measured by a manometer
 1 atmosphere (1 atm) = 101.3 kPa
 14.7 lb/in2
 76 cmHg
 760 mmHg
 760 torr
 1.01325 bar
 1013.25 mb
Standard Temperature and Pressure (STP)
 0 °C (273 K)
 1 atm
25
Physical Properties
 indefinite shape and volume
 gases exert pressure
 low density
 diffusion
Gas Laws
Boyle’s Law
 Relation of volume to pressure at constant temperature
 Robert Boyle
 “At constant temperature, the volume of a confined gas is inversely proportional to pressure”.
 P α 1/V
P = k/V
PV = k
 P1V1 = P2V2
Example:
a. If the volume of a gas is 8.0 L under a pressure of 4.2 atm, what would be its volume if the
pressure were increased to 7.4 atm?
i. Given:
Required: V2
1. P1 =4.2 atm
2. V1 = 8.0 L
3. P2 = 7.4 atm
ii. Solution:
1. P1V1 = P2V2
P1V 1
P2
4.2 x8.0
 4.54 L
3. V2 =
7.4
2. V2 =
b. A sample of oxygen occupies 18.5 L under a pressure of 800 torr. At what pressure would it
occupy 23.4 L if the temperature did not change?
c. A gas at 1 atm expands to a final volume of 6.45 L having a final pressure of 600 atm. What is
the initial volume of the gas?
Charles’ Law
 Relation of volume to temperature at constant pressure
 Jacques Alexandre Cesar Charles
 “The volume of a confined gas is directly proportional to temperature when pressure is held
constant”.
 V α T
V = kT
V/T = k

V1 V 2
=
T1 T 2
Example
a. A sample of nitrogen occupies 230 mL at 290 K. At what temperature in K would it occupy 340
mL if the pressure is held constant?
i. Given:
Required: T2
1. T1 = 290 K
2. V1 = 230 mL
3. V2 = 340 mL
26
ii. Solution:
V1 V 2
=
T1 T 2
T 1V 2
2. T2 =
V1
290 x340
3. T2 =
 428.70 L
230
1.
b. A tank contains 600 L of oxygen at 34 °C. What will be the final volume if the temperature is
increased to 40 °C?
c. What is the initial volume of a 45 mL in an inflated balloon at 25 °C if the final temperature is
39.6 °C?
Combined Gas Laws
 Boyle’s Law and Charles’ Law
 “The volume occupied by a given amount of gas is proportional to the absolute temperature divided
by the pressure”.


T
T
V =
P
P
P1V 1 P 2V 2

T1
T2
V α
Examples:
a. A sample of neon occupies 120 L at 34 °C under a pressure of 900 torr. What is its volume at
standard temperature and pressure (STP)?
iii. Given:
Required: V2
1. T1 = 34 °C + 273 = 307 K
2. V1 = 120 L
3. P1 = 900 torr
4. T2 = 273 K
5. P2 = 760 torr
iv. Solution:
P1V 1 P 2V 2

T1
T2
P1V 1T 2
2. V2 =
P 2T 1
900 x120 x 273
 126.37 L
3. V2 =
760 x307
1.
b. What is the volume of methane at 273 K and 760 torr, if its original volume at 300 K and 769
torrs is 40 cc?
27
Gay Lussac’s Law of Combining Volumes
 Joseph Louis Gay Lussac
 “Gases combine in simple and definite proportions by volumes or gases react with one another in
volume ratios of small whole numbers”
o 2 H2
+
O2
2 H 2O
2L
1L
2L
Avogadro’s Law
 Amadeo Avogrado
 “At the same temperature and pressure, equal volumes of all gases contain the same number of
molecules”.
 1 mol gas = 22.4 L at STP
 V α n
V = kn
V/n = k

V1 V 2

n1 n 2
Dalton’s Law of Partial Pressures
 John Dalton
 “The total pressure exerted by a mixture of ideal gases is the sum of the partial pressures of
those gases”.
 Ptotal = PA + PB + PC + …..
The Ideal Gas Law
 Boyle’s Law, Charles’ Law and Avogadro’s Law
 PV = nRT

R (universal gas constant) = 0.082057
L atm
mol K
Examples:
a. What pressure in atm is exerted by 7.8 mol of Xe in a 1.2 L flask at 287 K?
v. Given:
Required: P
1. n = 7.8 mol
2. V = 1.2 L
3. T = 287 K
4. R = 0.082057
L atm
mol K
vi. Solution:
1. PV = nRT
nRT
V
7.8 x0.082057 x 287
 153.08 atm
3. P =
1.2
2. P =
b. What is the volume of a gas balloon filled with 3 moles of He when the atmospheric pressure
is 790 torr and the temperature is 30 °C?
c. Nitric acid, a very important industrial chemical, is made by dissolving the gas nitrogen dioxide
NO2 in water. Calculate the number of moles needed at 2 atm and 340 K.
28
Graham’s Law of Diffusion
 The rates of effusion or diffusion of two different gases are inversely proportional to the square
roots of their molar masses
 Diffusion is the movement of a fluid from an area of higher concentration to an area of
lower concentration.
 Effusion is the escape of gas in its container

RgasA
=
RgasB
MgasB
MgasA
Where: R is the rate of diffusion of effusion of gas A or B
M is the molar mass of either gas A or gas B


The rates of diffusion of two gases is inversely proportional to the square roots of their densities
RgasA
=
RgasB
DgasB
DgasA
Where: R is the rate of diffusion of effusion of gas A or B
D is the density of either gas A or gas B
Example:
Complete the relative rates of H2 and CO2 through a fine pinhole.
Solution:
Molecular weight of H2 = 2 amu
Molecular weight of CO2 = 44 amu
RgasA
=
RgasB
MgasB
MgasA
RH 2
=
RCO 2
MCO2
MH 2
RH 2
=
RCO 2
44
22
Answer:
RH2 = 4.69 times the rate of RCO2
NAME:__________________________________________
29
SCORE:_________
COURSE YEAR AND SCHED.______________________
DATE: _________
ACTIVITY 2.1
MATCHING TYPE
D
O
O
O
O
O
O
O
O
O
O
E
O
O
O
O
O
O
O
O
O
O
Match Column I with Column II. Shade the circle that best corresponds to your
answer (one per number). Erasures and superimpositions will invalidate correct
answers.
A
O
O
O
O
O
O
O
O
O
O
B
O
O
O
O
O
O
O
O
O
O
C
O
O
O
O
O
O
O
O
O
O
Column I
1. Ag
2. aluminum
3. baking soda
4. distilled water
5. ice cream
6. jelly
7. mineral water
8. peanut butter
9. styrofoam
10. table sugar
Column II
A. colloid
B. compound
C. element
D. solution
E. suspension
A
O
O
O
O
O
B
O
O
O
O
O
C
O
O
O
O
O
Column I
11. burning of a candle
12. condensation of water vapor
13. radioactive decay of uranium
14. cutting of diamonds
15. fireworks (pyrotechnics)
Column II
A. nuclear change
B. physical change
C. chemical change
A
O
O
O
O
O
B
O
O
O
O
O
C
O
O
O
O
O
D
O
O
O
O
O
Column I
16. electronegativity
17. flammability
18. temperature
19. volume
20. acidity
Column II
A. chemical property
B. both
C. physical property
D. neither A nor C
A
O
O
O
O
O
B
O
O
O
O
O
C
O
O
O
O
O
D
O
O
O
O
O
E
O
O
O
O
O
Column I
Column II
21. thermal equilibrium
A. Law of Conservation of Mass
22. conservation of energy
B. Second Law of Thermodynamics
23. zero heat, maximum entropy (disorder)
C. E = mc2
24. matter can’t be created
D. Zeroth Law of Thermodynamics
25. mass is converted to an amount of energy E. First Law of Thermodynamics
A
O
O
O
O
O
B
O
O
O
O
O
C
O
O
O
O
O
D
O
O
O
O
O
E
O
O
O
O
O
Column I
26. bird sitting on a tree branch
27. battery
28. mp3 file playing
29. nuclear explosion
30. friction in a roller coaster
Column II
A. kinetic to electrical energy
B. potential energy
C. mechanical to thermal energy
D. electrical to sound energy
E. kinetic to radiant energy
30
PROBLEM SOLVING Show all derivations and solutions in detail for full credit of points.
Enclose/Emphasize all final answers at the end of every solution. Round off
final answers to 2 SD.
___________________1. A sample of a gas occupies 360 ml under a pressure of 0.75 atm. If the
volume is decreased to 56 ml, what is the final pressure, the temperature
remaining constant?
___________________2. A sample of a gas has a volume of 79.5 ml at -45 °C at 1 atm. What
volume will the sample occupy at -25 °C at 1 atm?
___________________3. A 20-liter container is filled with a gas to a pressure of 10 atm at 0°C. At
what temperature (in °C) will the pressure in the container be 2.5 atm?
___________________4. P1 = 3.48 atm, V1 = 1.68 L, T1 = 994.2 K, P2 = 4,142 torr, V2 = 6,060 mL,
T2 = ____ in °C
Solve for the unknown values in the table below:
P
V
5.
2.00 atm
_____li
6.
456 torr
1.00 li
7.
3382 mm Hg
50.00 ml
8.
_____Pa
1250 cc
n
1.5 mol
_____mol
0.0105 mol
2.6 mol
31
T
100 °C
100 °K
_____°K
75 °C
NAME:__________________________________________
COURSE YEAR AND SCHED.______________________
SCORE:_________
DATE: _________
ACTIVITY 2.2
1.
Use the general gas law to find the volume of air exhaled under the following conditions:
P1 = 98 kPa
P2 = 105 kPa
0
T1 = 25 C
T2 = 350C
V1 = 500 ml
V2 = ?
Focus Questions:
1. Define ventilation. Explain how air moves in and out of the lungs, using the concept of
pressure.
2. What is meant by negative pressure in the lungs? How does intrapleural pressure prevent the
lungs from collapsing?
3. What is diffusion? Use your definition to account for the movement of gases to and from the
blood through the walls of the alveoli and capillaries? What factors will affect this process?
4. Why is the presence of a surfactant in the fluids of the alveoli important?
5. Explain the process by which oxygen is transported in the blood and released to the cells in
the body?
32
UNIT III: Atomic Structure
Atom
 comes from the Greek word “atomos” which means “uncut” or “indivisible”
Atomic Theory
Scientists
Democritus
Aristotle
Daniel Bernoulli
John Dalton
Contribution
He proposed that ‘matter is made up of tiny, indivisible atoms’.
He did not accept the theory of Democritus.
He proposed the 4-element theory making up matter.
He attempted to explain the behavior of matter quantitatively;
Dalton’s Atomic Theory may be summed up as follows:
 An element is composed of extremely small, indivisible particles called atoms.
 All atoms of a given element have identical properties that differ from those
of other elements.
 Atoms cannot be created, destroyed, or transformed into atoms of another
element.
 Compounds are formed when atoms of different elements combine with one
another in small whole-number ratios.
 The relative numbers and kinds of atoms are constant in a given compound.
Atomic Models:
Dalton’s Model
1808
Thomson’s Model
1890
Rutherford’s Model
1911
Electron Cloud Model
33
Lord Kelvin’s Model
1900
Bohr’s Model
1913
Men who contributed in the Development of Modern Atomic Concept
Scientists
Joseph Jon Thomson
Wilhelm Konrad Roentgen
Robert Millikan
Eugen Goldstein
Sir James Chadwick
Henri Becquerel
Ernest Rutherford
Max Planck
Neils Bohr
Henry Moseley
Sommerfield
Louis de Broglie
Erwin Schrödinger
Weiner Heisenberg
Plucker
Contribution
He discovered the electron using the cathode ray tube.
He named his atomic model as ‘raisin or plum pudding model’.
He discovered the x-rays (penetrating rays without mass or charge)
using the cathode ray tube.
He discovered the charge of an electron which is 1.60 x 10-19 coulomb
using x-rays.
He discovered the protons using the canal ray tube.
He studied the alpha particles in beryllium, which lead to the discovery
of neutron.
He pioneered the study on radioactive rays together with Marie
Sklodowska Curie and Pierre Curie.
He perceived the atom as a miniature solar system.
He used the alpha scattering apparatus leading to the discovery of the
nucleus.
 alpha (α)
o +2, stopped by a thin sheet of paper, injure normal cells in
the body
 Beta (β)
o -1, stopped by a 1 cm thick of aluminum foil, harmful to the
body
 Gamma (γ)
o No charge, stopped by lead, cause mutations
He described the radiant energy as consisting of quanta.
He developed a theory that an electron in the atom gains or loses
energy by absorbing or emitting energy in orbits.
He noted that the number of positive charges increases from atom to
atom by single electronic unit.
He introduced sub-levels and assuming electrons travel in an elliptical as
well as circular orbits.
He extended the particle-wave duality concept that matter must be
both as particle and a wave (matter waves).
He developed the equation that relates the wavelength of an electron to
its energy, which described the probability that an electron will be at a
certain point in space
He stated that it is impossible to pinpoint or to determine accurately
both the momentum and the position of the electron simultaneously.
(referred to as the Uncertainty principle)
He experimented the cathode by a steam of high speed of electron.
34
Modern Atomic Theory
 The modern atomic theory involves the small nucleus and the three elementary particles called the
protons, electrons, and neutrons.
 The nucleus contains the protons and the neutrons surrounded by an electron cloud.
Atomic particle
Electron
Proton
Neutron
actual mass (g)
9.110 x 10-29
1.673 x 10-24
1.675 x 10-24
Mass Number (A) - “massenzahl”
A
Z
Electrical charge (C) relative mass (amu)
-1
1/1837
(- 1.602 x 10-19)
+1
1
(+ 1.602 x 10-19)
0
1
X
symbol of element (X)
Atomic Number (Z) – “atomzahl”
A = p + n
e = p = Z
Element
A
B
C
A
15
Z
8
p
e
6
42
18
24
35
n
Nucleons
 Name of the protons and neutrons inside the nucleus
Ions
 Electrically charged atoms (positive and negative ions)
Isotopes
 Atoms having different atomic weights or mass but of the same atomic number
 Frederick Soddy
Element
Hydrogen
Symbol
1
1H
2
1H
Mass (amu)
Percent
abundance
1.0078
99.985
2.0141
0.015
3.0160
negligible
12.00
98.89
Average atomic
mass
1.0079
3
1H
Carbon
12
6C
12.011
13
6C
Oxygen
13.003
1.11
15.995
99.759
16.995
0.037
34.969
75.77
16
8O
15.999
17
8O
Chlorine
35
17Cl
35.453
37
17Cl
36.966
Separation of Isotopes
 Centrifuge
 Thermal diffusion
 Electrolysis
 Fractional distillation
36
24.23
Radioisotopes
 Are artificial radioactive isotopes
 Discovered by Irene and Frederic Joliot Curie
 Are prepared by the bombardment of naturally occurring atoms
Radioisotopes
1. Iodine – 131
Iodine – 123
2. Strontium – 90
3. Cobalt – 59
Cobalt - 60
4. Thallium – 201
5. Chromium - 151
6. Boron - 10
7. Fluorine – 18
8. Strontium-89
Samarium 153
Rhenium-186
9. Phosphorus - 32
Uses
Diagnostic testing of thyroid gland functions
Treatment of small lesions
Radiation therapy of most tumors/cancerous cells
Studying defects in vitamin B12 absorption
Myocardial imaging
Detection and prognosis of coronary artery disease
Determining the flow of blood through the heart
Important in the diagnoses of anemia
Detection of malignant tumor cells in the brain
Cardiac and brain imaging
Relief of cancer-induced bone pain
Controlling the excess Polycythemia vera, an excess of red blood
cells is produced in the bone marrow
10. Technetium - 99
Imaging the skeleton and heart muscle in particular, but also for
brain, thyroid, lungs (perfusion and ventilation), liver, spleen,
kidney (structure and filtration rate), gall bladder, bone marrow,
salivary and lacrimal glands, heart blood pool, infection and
numerous specialized medical studies
11. Bismuth - 213
Targeted alpha therapy (TAT), especially cancers
12. Copper – 64
Studying genetic diseases affecting copper metabolism, such as
Wilson's and Menke's diseases
13. Dysprosium - 165
Aggregated hydroxide for synovectomy treatment of arthritis
14. Holmium - 166
Diagnosis and treatment of liver tumors
15. Iron - 59
Studying of iron metabolism in the spleen
16. Selenium – 75
Inn the form of seleno-methionine for studying the production of
digestive enzymes
17. Sodium – 24
Studying of electrolytes within the body
18. Xenon – 133
Pulmonary (lung) ventilation studies
19. Ytterbium – 169
Cerebrospinal fluid studies in the brain
20. Yttrium – 90
Cancer brachytherapy and as silicate colloid for the relieving the
pain of arthritis in larger synovial joints
Source: James R. Fromm (jfromm@3rd1000.com); http://www.bookrags.com;
Radioisotopes in Medicine. World Nuclear Association (http://www.world-nuclear.org)
37
Quantum Theory
 Study of the discrete nature of phenomena at the atomic and subatomic levels
 Quanta – indivisible units of energy
Plank’s



Theory
Max Plank (German)
Explained why spectra of radiation changes with the temperature
Photon – a quantum or packet of electromagnetic radiation
o Photoelectric effect – Einstein
o Compton effect – Arthur Compton
Dualistic Nature of Light
 Louis de Broglie (French)
 Light is both a wave and a particle
 Matter waves – matter behaves like a wave and that light waves behave like particles of matter
The Uncertainty Principle
 Werner Heisenberg
 The actual position and velocity of the particle cannot both be determined at the same time
The Present Day Atom
 Erwin Schrödinger
 Mathematic theory of wave mechanics
 Wave function or orbitals
 Electron cloud
Energy



level
Energy shell
Corresponds to the period of an element in the periodic table
A group of atomic orbitals that have the same value for (n), the principal quantum number
Sub-level
 A division of an electron shell
 An electron shell may hold two or more orbitals
1. s (sharp)
 spherical shape
 1 orbital (2 electrons)
2. p (principal)
 dumbbell shaped
 px, py, pz (3 orbitals) (6 electrons)
3. d (diffuse)
 four-leaf clover, an hour glass and a ring
 5 orbitals (10 electrons)
4. f (fundamental)
 complex shapes
 7 orbitals (14 electrons)
38
Valence electron
 Electron found in the outermost shell in the ground state
Valence shell
 The outermost energy level
Orbital
 A region around the nucleus of an atom where an electron with a given energy is likely to be found
Principal Energy
n=1
n=2
n=3
n=4
Sublevels
1s
2s, 2p
3s, 3p, 3d
4s, 4p, 4d, 4f
Orbitals
1
8
27
64
Max Electrons
2
8
18
32
Quantum Numbers
1. principal quantum number (n)
o refers to the main energy levels
o K, L, M, N, O, P, Q
2. azimuthal quantum number (l)
o describes the way the electron moves around the nucleus
o describes the orbital shape
o s, l = 0; p, l = 1; d, l = 2; f, l = 3
3. magnetic quantum number (m)
o describes the possible orientation in space of the electrons in a magnetic field
o s, m = 0 (no effect on a magnetic field)
o p, m = -1, 0, +1 (for px, py, and pz)
o d, m = -2, -1, 0, +1, +2 (5 d orbitals)
o f, m = -3, -2, -1, 0, +1, +2, +3 (7 f orbitals)
4. spin quantum number (s)
o describes the way the electron spins or rotates on its own axis as it moves about the
nucleus
o +1/2 spin (clockwise direction)
o -1/2 spin (counterclockwise direction)
 Parallel spins – both are cw or cc
 Opposite spins – one cw and the other cc
 Paramagnetism – property of a substance to be attracted by a magnet (unpaired
electrons)
 Diamagnetic – repelled by a magnet (paired electrons)
39
Example
1. 2p4
n = 2; l = 1; m = -1; s = +1/2
2. 3p3
n = 3; l = 1; m = +1; s = -1/2
3. 3d6
n = 3; l = 2; m = -2; s = +1/2
Electron Configuration
 Orderly distribution of electrons among the orbitals of an atom
Mnemonic Device
Source: http://www.mpcfaculty.net/mark_bishop/memory_aid_e_config.jpg
Ground State
 Lowest, most stable energy state of the atom
Aufbau Principle
 German word “aufbaue” – to build up
 Electrons are added one at a time to the lowest energy orbitals
Pauli Exclusion Principle
 No two electrons can occupy the same quantum state simultaneously in an atom
o An orbital can hold a maximum of 2 electrons
o The electrons can occupy the same orbital if they spin in opposite direction
 clockwise (upward arrow)
 counterclockwise (downward arrow)
Hund’s Rule
 When electrons occupy orbitals of equal energy, one electron enters each orbital until all the
orbitals contain one electron with parallel spins
 Second electrons then add to each orbital pairing the spins of the first electrons
40
Name: _________________________
Schedule: ______________________
Date: __________________
Rating: _________________
Activity 3.1
Given the atomic number of the following elements:
A. Write the orbital diagram of the outermost configuration.
B. Write the electronic configuration.
C. Determine the family and period where the element is located.
1. A12
Orbital diagram: ____________________________________________________________
Electronic configuration: ______________________________________________________
Family: __________
Period: __________
2. B23
Orbital diagram: ____________________________________________________________
Electronic configuration: ______________________________________________________
Family: __________
Period: __________
3. C30
Orbital diagram: ____________________________________________________________
Electronic configuration: ______________________________________________________
Family: __________
Period: __________
3. C67
Orbital diagram: ____________________________________________________________
Electronic configuration: ______________________________________________________
Family: __________
Period: __________
3. C80
Orbital diagram: ____________________________________________________________
Electronic configuration: ______________________________________________________
Family: __________
Period: __________
41
Valence Electron
 Electrons in the outermost energy level
 Determines the chemical properties of elements
Lewis Dot Structure
 Gilbert Lewis
 Noble gas configuration
o Rule of Two or Duet Rule
o Octet Rule
 Consists of the element’s symbol surrounding dots to represent the number of valence electrons
Examples:
Phosphorus
Chlorine
Arsenic
Chemical Bonds
1. Ionic Bond
 Transferring of electrons
 Metal donates the electron to a non-metal
 Metal + non-metal
 Properties of Ionic Compounds
o Are solid with high melting point ( › 400°C)
o Are soluble in polar solvents
o Conduct electricity when in aqueous solution or in molten state
Examples:
Source:
www.hcc.mnscu.edu/.../V.12/LiF_ionic_bond.jpg
2. Covalent Bond
 Sharing of electrons
 Non-metal + non-metal
o Single bond (H2O, CH4)
o Double bond (O2, CO2)
o Triple bond (N2)
 Properties of Covalent Compounds
o Are usually gas, liquid or solid with low melting points ( < 300°C)
o Mostly are soluble in non-polar solvents
42
Examples:
a. Cl2 and H2
b. O2
c. N2
3. Metallic Bond
 Is the force of attraction that holds metals together
 Consists of the attraction of the free-floating valence electrons for the positively charged
metal ions
 Explains many physical properties of metals
o Metals are good conductors of electricity
o Metals are ductile
o Metals are malleable
43
4. Intermolecular Forces of Attraction
 Are the forces of attractions that exist between molecules in a compound
 Cause the compound to exist in solid, liquid, or gas; and affect the melting and boiling points of
compounds as well as the solubility of one substance in another
 Also called Van der Waals forces (Johannes van der Waals 1837-1923)
Types of Intermolecular Forces of Attraction
 Dispersion forces
 Weakest of all molecular interactions among non-polar molecules (H2, Cl2)
 The more electrons that are present in the molecule, the stronger the dispersion
forces
 Dipole to dipole interaction
 Occurs between polar molecules
 The partial positive charge on one molecule is electrostatically attracted to the partial
negative charge on a neighboring molecule
 Hydrogen bonds
 Are the strongest of the intermolecular forces
 Important in determining the properties of water and biological molecules like protein
 Occur between molecules from hydrogen being covalently bonded to either fluorine,
oxygen or nitrogen
44
Name: _________________________
Schedule: ______________________
Date: __________________
Rating: _________________
Activity 3.2
I.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Fill-in the Blanks: Write the word or group of words on the blank to complete each
sentence.
Atom comes from the ________________ word ‘atomos’ which means uncut or indivisible.
Daniel Bernoulli attempted to explain the behavior of matter _______________.
_______________ is the study of the attraction of materials to charged bodies.
JJ Thomson discovered the electron using _______________.
Goldstein used the _______________ in discovering proton.
Cations are ______________ charged ions.
The element hydrogen has _____________ isotopes.
The Swedish chemist, _______________, used the first one or two letters of the name of the
element for its chemical symbol.
_______________ is the Latin word for mercury.
_______________ proposed the particle-wave nature of matter.
II.
Fill-in the table with the correct information:
Element
Mass Number Atomic Number
1. B
12
7
2. L
3. Es
36
4. S
10
5. U
45
P
e
9
n
12
24
9
15
III.
Provide the (a) electronic configuration of the following elements. Illustrate the (b) orbital
diagram of the outermost configuration and determine its (c) set of quantum numbers. (30
pts)
1. A6
a) ______________________________________________________________________
b) ______________________________________________________________________
c) n = _____ l = _____
m = _____
s = _____
2. B10
a) ______________________________________________________________________
b) ______________________________________________________________________
c) n = _____ l = _____
m = _____
s = _____
3. C17
a) ______________________________________________________________________
b) ______________________________________________________________________
c) n = _____ l = _____
m = _____
s = _____
45
Name: _________________________
Schedule: ______________________
Date: __________________
Rating: _________________
Activity 3.3
Illustrate the Lewis Dot and Structural Formulas (Cationic kernels are underlined). Then
identify the type (and subtype, if applicable) of the chemical bond. Minimize erasures.
KCN
carbon tetrachloride
O2
water molecule
nitrogen gas
chlorine gas
oxygen gas
CH4
Structural
Formula
Lewis Dot
Formula
Structural
Formula
Lewis Dot
Formula
Structural
Formula
Lewis Dot
Formula
Ca(OH)2
46
UNIT IV: The Periodic Table
Chemical Symbols
 Swedish chemist J.J. Berzellius (1779 – 1848) used the first one or two letters of the name
of the element
o C, H, O, N, S, I, Al, Br, Ca, Co, Si, Ti
 Derived from the name of the discoverer
o Lawrencium (Lr)
o Einsteinium (Es)
o Fermium (Fm)
o Mendelevium (Md)
 Name of gods
o Thorium (Th)
o Vanadium (V)
 From place of the founder
o Francium (Fr)
o Americium (Am)
o Californium (Cf)
o Europium (Eu)
o Gallium (Ga)
o Germanium (Ge)
o Polonium (Po)
 From Latin words
o Antimony
o Copper
o Gold
o Iron
o Lead
o Mercury
o Potassium
o Silver
o Sodium
o Tin
Stibium
Cuprum
Aurum
Ferrum
Plumbum
Hydrargyrum
Kalium
Argentum
Natrium
Stannum
Sb
Cu
Au
Fe
Pb
Hg
K
Ag
Na
Sn
47
Periodic Table
 table of the chemical elements arranged to show patterns of recurring chemical and physical
properties
 one way of arranging known elements and provides a possible way of determining elements yet to
be discovered
Development of the Periodic Table
Scientists
Contribution
Johannes Dobereiner
He arranged the elements by ‘triad’ according to order of physical
properties.
Examples: Ca, Ba, Sr / Cl, I, Br / S, Te, Se / Fe, Co, Ni
John Newlands
He arranged the elements by ‘law of octaves’ according to order of
increasing atomic weights.
Examples: H, F, Cl, Co&Ni, Br, Pd, I, Pt&Ir
Li, Na, K, Cu, Rb, Ag, Cs, Tl
Lothar Meyer
He arranged the elements according to periodic trends.
 Melting points, boiling points, and chemical activities
Dmitri Mendeleev
He arranged the elements in order of increasing atomic mass.
Henry Moseley
He arranged the elements in order of atomic number giving rise to the
‘modern periodic table’.
Modern Periodic Table
 Periods – seven horizontal rows
 Groups or Families – eighteen vertical columns
o Representative elements – Group 1A through Group O
o Transition elements – Group B elements
o Lanthanides
o Actinides
Periodic Law
 The physical and chemical properties of the elements are periodic functions of their atomic number
Uses of the Periodic Table
1. The periodic table is a helpful tool used by scientists and students to find the structured sequence in
terms of physical and chemical properties which identify an element.
2. If a person knows the main properties of each of the groups in the periodic table and how chemical
properties vary within a group, he or she can predict the properties of any particular element with a
reasonable degree of confidence.
48
Name: _________________________
Schedule: ______________________
Date: __________________
Rating: _________________
Activity 4.1
Puzzle: Encircle the names of the element symbols in the puzzle.
F
R
A
N
C
I
U
M
U
I
N
A
R
U
C
R
A
R
I
O
R
K
Y
Y
N
T
A
O
O
A
E
D
G
T
P
O
L
O
N
I
U
M
B
X
R
B
I
O
R
O
N
W
L
R
R
C
A
O
Y
B
O
U
N
A
T
I
N
D
R
A
L
M
R
G
O
P
M
T
Y
Q
Y
I
R
T
T
A
I
O
E
N
O
M
A
G
N
E
S
I
U
M
P
L
N
N
A
E
Y
U
E
X
Q
I
N
N
A
S
O
Y
M
A
Y
Z
I
N
C
C
L
G
G
N
E
V
T
A
M
R
A
P
I
U
B
V
S
S
I
E
E
O
B
U
U
P
L
C
I
N
E
O
T
L
N
U
S
A
I
C
P
K
K
E
A
R
S
E
N
I
C
M
I
R
R
H
I
E
W
J
A
F
N
A
D
A
I
T
T
E
E
M
L
M
D
A
A
W
R
O
N
U
K
T
M
R
P
F
P
O
T
A
S
S
I
U
M
A
Y
1. Fr
2. Hg
3. W
4. Mg
5. C
6. Co
7. Ar
8. Zn
9. Po
10. K
11. I
12. Ni
13. B
14. Ag
15. As
49
16. O
17. Os
18. Sn
19. Y
20. U
Name: _________________________
Schedule: ______________________
Date: __________________
Rating: _________________
Activity 4.2
Case Study:
Mrs. Anne Curtis is in her late 50’s. On admission she had a weight of 175Kg. She has a
history of adult-onset diabetes mellitus. She is in hospital recovering from a septic ulcer on her
left foot. A glucose tolerance test administered on admission indicated a level of glucose in the
blood of over 150mg/mL indicating loss of insulin control. One reason for this seems to be her
poor attention to diet. On her chart are the following information, placed there for reference
by student nurses:
Clinical note for Anne Curtis: Nursing care for this patient would involve continuous
monitoring of such chemistry-related values of serum osmolality, electrolytes, glucose,
hemoglobin, hematocrit, and urine ketones. Also, there may be a need to administer prescribed
regular insulin when serum glucose falls within the range of 250-800mg/dL.
1. Explain how Anne Curtis’ dietary treatment is helping to deal with her diabetes and her
obesity. Be specific in your discussion, referring to insulin, glucose, and the organic
molecules in her diet.
2. Anne Curtis is ordered 36 units of ‘Actfast’ insulin each morning and 18units each night.
What are the two volumes you would give if the 10mL vial contains 100U/mL.
50
Periodic Trends
Property
Atomic Radius
Description
Distance of the electron cloud from the nucleus
Trend
Family – increases
Period - decreases
Family – decreases
Period – increases
Ionization Energy
Energy required to remove an electron from an
atom
 Nuclear charge – the larger the nuclear charge,
the greater is the ionization energy
 Shielding effect – the greater the shielding
effect, the less is the ionization energy
 Atomic size – the bigger the atom, the lesser is
its ionization energy
 Sub-energy level – an electron from a full or
half full sublevel requires additional energy to
be removed
Electron Affinity
Energy required to attract an additional electron
 Metals – low electron affinity
 Non-metals – high electron affinity
Family – decreases
Period – increases
Electronegativity
Ability of an atom to attract a shared pair of
electrons in a chemical bond
Family – decreases
Period – increases
Ionic radius
Distance between the nuclei of two like atoms
after gaining or losing electrons
Family – increases
Period - decreases
Metallicity
Tendency of an element to lose electrons forming
positive ion
Family – increases
Period – decreases
Non-metallicity
Tendency of an element to gain electrons forming
negative ion
Family – increases
Period - increases
The Beneficial and Harmful Elements to the Environment
Question: What elements are harmful to living organisms and how do they get into our environment?
Trace essential elements such as fluorine, copper, selenium, molybdenum, and others can be hazardous to
living organisms if present at high levels. Nonessential heavy metals such as arsenic, lead, mercury,
cadmium, chromium are usually toxic to organisms as much lower levels than trace essential elements.
Depending on the association that these nonessential elements may form with natural geologic materials
such as organic matter, other elements or minerals, and adsorbers (such as clays), these elements can
range from being safe to being extremely toxic.
(http://geology.er.usgs.gov/eastern/environment/environ.html)
51
What harmful elements are contained in auto exhaust gas?
(01/17/2004)
Auto expel large amount of carbon monoxide, hydrocarbon, nitro-oxide, fine articles and sulfide, etc. All
these primary pollutants will form such secondary pollutants as actinological fume and acid disposition,
etc. Out of the pollutants, carbon monoxide can put people in a state of oxygen deficiency, and result in
chronicle intoxication with long time contact; hydrocarbon is the substance incurring cancer; and nitrooxide will do harm to respiratory system and immune system of human body. Lead pollution will be
brought about after petroleum-containing lead is burned and 85% of lead are let out into air. In the
past half century, amount of lead released into air after auto exhaustion has reached millions of tons,
which has become a pollution source globally. Lead will do harm to a lot of human organs and systems
such as deterioration of intelligence, kidney damnification, infertility and high blood pressure. Lead
pollution in air does the most harm to children. If amount of lead increases in children, it can result in
deterioration of intelligence, and impose negative impact on growth of children, even bring about some
diseases. According to some research, amount of lead in blood increases by 10gramma per deciliter, and
average IQ of children will decrease 2.5.
(Copyright 2007 Shanghai Municipal Government All Rights Reserved .)
52
UNIT V: Chemical Nomenclature of Inorganic Compounds and Chemical Equations
Chemical Nomenclature
Steps in writing formulas:
1. Write the symbol of the positive particle or cation first before the symbol of the negative
particle or anion.
Example:
Na+
Al3+
O2F2. Criss-cross the valence numbers, disregarding the sign, so that the charge of the cation becomes
the subscript of the anion and the charge of the anion becomes the subscript of the cation.
Example:
Na+ + FK- + O2Na1 F1
K2O1
a. If the subscript is 1, it is no longer written.
Example:
NaF
K 2O
b. If the subscript of both cation and anion are the same, omit writing the subscripts.
Example:
Mg2+ + S2MgS
c. Reduce the subscript to the simplest whole number ratio.
Example:
W6+ + O2W2O6 ------- WO3
d. When the polyatomic ion needs a subscript, enclose it first with a parenthesis. If the
original polyatomic ion already contains a parenthesis, use a bracket.
Example:
Al3+ + Cr2O72B3+ + Fe(CN)64Al2(Cr2O7)3
B4[Fe(CN)6]3
Naming Chemical Formulas
Rules for naming inorganic compounds
1. Binary compounds containing a metal and a non-metal
 Give the name of the cation first followed by the anion
 Transition metals
a.
Old/traditional system
i. FeBr2 – ferrous bromide
ii. FeBr3 – ferric bromide
b. Stock system
i. FeBr2 – iron (II) bromide
ii. FeBr3 – iron (III) bromide
53
Naming Ions Using the Stock System and Old System
Ion
Stock System
2+
Fe
Iron (II)
Fe3+
Iron (III)
Cu+
Copper (I)
2+
Cu
Copper (II)
Sn2+
Tin (II)
4+
Sn
Tin (IV)
2+
Pb
Lead (II)
Pb4+
Lead (IV)
Hg+
Mercury (!)
Hg2+
Mercury (II)
Au+
Gold (I)
2+
Au
Gold (III)
Old System
Ferrous
Ferric
Cuprous
Cupric
Stannous
Stannic
Plumbous
Plumbic
Mercurous
Mercuric
Aurous
Auric
2. Binary compounds containing two non-metals
 Use prefixes to indicate the numbers of atoms present
 Mono is never used in the first element
o NO2 – nitrogen dioxide
o P2O5 – diphosphorus pentaoxide
3. Binary acids
 Non-oxygen acids
 Use the prefix hydro followed by the anion and the word acid added at the end
o HCl – hydrochloric acid
o H2S – hydrosulfuric acid
4. Ternary compounds
 Compounds containing more than two elements
 Give first the name of the cation followed by the anion
o K2SO4 – potassium sulfate
o NH4NO3 – ammonium nitrate
5. Ternary acids
 Oxyacids
 -ite (ous ending)
 -ate (ic ending)
o HNO2 – nitrous acid (hydrogen + nitrite)
o HNO3 – nitric acid (hydrogen + nitrate)
o HClO2 – chlorous acid (hydrogen + chlorite)
o HClO3 – chloric acid (hydrogen + chlorate)
54
Name: _________________________
Schedule: ______________________
Date: __________________
Rating: _________________
Activity 5.1
I. Name the following compounds. (10 pts)
Traditional System
1. CuBr
________________________
Stock System
_____________________
2. CuCl2
________________________
_____________________
3. PbSO4
________________________
_____________________
4. Fe(NO3)3
________________________
_____________________
5. HgO
________________________
_____________________
II. Write the formula of the following compounds. (10 pts)
1. calcium oxide
_______________ 11. diphosphorus trioxide
______________
2. ammonium acetate
_______________
12. aluminum carbonate
______________
3. cesium iodide
_______________
13. lithium sulfide
______________
4. phosphoric acid
_______________
14. stannic nitrate
______________
5. hydrobromic acid
_______________
15. mercurous phosphate
______________
6. magnesium oxide
_______________
16. aurous oxide
______________
7. ammonium cyanide
_______________
17. boron carbonate
______________
8. cesium bromide
_______________
18. potassium sulfide
______________
9. sulfuric acid
_______________
19. stannous nitrate
______________
10. hydroiodic acid
_______________
20. mercuric phosphate
______________
55
Chemical Reactions
Symbols
Meaning
Yield
Reversible reaction
Evolution of gas
Formation of precipitate
Δ
(s)
(l)
(aq)
(+)
(g)
Steps
1.
2.
3.
Heat
Solid state
Liquid state
Aqueous solution
Added to
Gaseous state
in writing a chemical equation
Formulate a word equation.
Make the skeleton equation.
Balance the equation.
Balancing Chemical Equations
Tips;
1. Always check the equation by counting atoms on each side of the equation. For each element, the
number of atoms on the left has to be equal to the number of atoms on the right.
2. Start by putting a “1” in front of the most complex compound (the one containing the greatest
number and variety of atoms)
3. Balance elements that appear in multiple reactants or multiple products last. In combustion
equations, this means that the oxygen is always balanced last.
4. Whole number coefficients are preferred. But use fractions during the balancing process. The
equation can be scaled to remove them later.
5. Be certain that the total charge is the same on each side after you have mass-balanced equation.
6. Don’t touch the subscripts. Changing the subscripts in the formula results in an entirely different
chemical compound. Remember that the coefficients show how much compound is involved in the
reaction, while the subscripts show how much element is in the compound.
7. When you notice that the same polyatomic ions appear on both side of the equation, it’s much
easier to balance the equation as though the polyatomic ions were ‘atoms’.
8. A missing coefficient is understood to be a ‘1’
9. Write the equation so that the coefficients are the smallest set of integers possible. For
example, if all of the coefficients are divisible by 2, divide them all by 2.
Example: When silver oxide is heated, silver metal and oxygen gas are produced.
1. word equation:
silver oxide
silver metal + oxygen gas

2. skeleton equation: Ag2O
Ag
+
O2

3. balanced equation: 2Ag2O
4Ag
+
O2

56
Types of Chemical Reactions
1. Combination or synthesis
a. Metal + non-metal

Mg + S

b. Non-metal + non-metal

C + 2H2

c. Non-metal + oxygen

C + O2

d. Metal oxide + water
K 2O + H 2O 
e. Non-metal oxide + water 
SO2 + H2O 
f. Metal oxide + non-metal oxide
Na2O + SO3 

ionic compound
MgS
covalent compound
CH4
non-metal oxide
CO2
metal hydroxide (base)
2KOH
oxyacid
H2SO4
salt
Na2SO4
2. Decomposition reaction
a. Hydrates yield water and anhydrous salt
i. CuSO4 · 5H2O 
CuSO4 + 5H2O
b. Chlorates when heated form metal chloride and oxygen gas
i. 2KClO3 
KCl + H2O
c. Metal oxide when heated form the metal and oxygen gas
i. 2HgO
 2Hg + O2
d. Carbonates when heated yield metal oxide and carbon dioxide
i. MgCO3 
MgO + CO2
e. Bicarbonates when heated form metal oxide, water and carbon dioxide
i. 2NaHCO3  Na2O + 2CO2 + H2O
f. Electricity
i. 2H2O 
2H2 + O2
3. Simple/single displacement
a. A + BC 
AC + B
b. AB + C 
AC + B
i. Mg + Zn (NO3)2
ii. Mg + LiNO3
4. Double displacement
a. AB + CD
i. AgNO3 + HCl
Mg(NO3)2 + Zn
no reaction
AD + CB
AgCl + HNO3
57
Activity Series of Elements
Source:
www.avon-chemistry.com/ch9_chart11.jpg
OXIDATION – REDUCTION REACTIONS
 redox reactions
 reactions that are concerned with the transfer of electrons between and among reactants
 oxidation – loss of electrons
 reduction – gain of electrons
 oxidizing agent – substance that is reduced during chemical reaction; it gains electrons
 reducing agent – substance that is oxidized during chemical reaction; it loses electrons
58
Oxidation Number
 The charge that atom would carry if the compound were composed of ions
Rules:
1. The oxidation number of a free and uncombined element whether monoatomic or diatomic is zero.
Thus, the atoms O2, N2, Cl2 have an oxidation number of 0.
2. The oxidation number of elements belonging to group IA, IIA, and IIIA is the same as their
valence. Thus, Na in NaCl has an oxidation number of +1, the oxidation number of Ba in BaO is +2
and oxidation number of Al is AlCl3 is +3.
3. The oxidation of hydrogen is +1 when it is combined with a non-metal. Hydrogen is therefore in
the +1 oxidation state in CH4, NH3, H2O, and HCl.
4. The oxidation number of hydrogen is -1 when it is combined with a metal. Hydrogen is therefore
in the -1 oxidation state in LiH, NaH, CaH2.
5. The oxidation number of oxygen is -2 when it is combined oxygen. Thus the oxidation number of
oxygen in H2O, NaNO3 is -2.
6. The oxidation number of oxygen in peroxide is -1. Peroxides have a general formula R2O2 (for
group IA) and R1O2 (for group IIA). Thus in H2O2 and CaO2 the oxidation number of oxygen is -1.
7. The sum of the oxidation number in a neutral compound is zero and equals the overall charge for
an ionic species.
Balancing Oxidation-Reduction Equations
Methods:
1. Oxidation State Method or Valence-Change Method
(http://preparatorychemistry.com/Bishop_Balancing_Redox.htm)
a. Write the correct formula for all the reactants and products in the equation.
HNO3 + H3AsO3  NO + H3AsO4 + H2O
b. Assign the oxidation numbers to the atoms in the equation.
H+N5+O32- +
H3+ As3+O32-

N2+O2-
+
c.
H3+As5+O42-
+
H2+O2-
Identify which atoms are reduced and oxidized.
As
+3 to +5
N
+5 to +2
d. Compute the total change in oxidation number both for the reduced and oxidized atom.
As
Net Change = +2
N
Net Change = −3
e. Make the number of electrons lost equal to the number of electrons gained by using
appreciation factors.
2HNO3 + 3H3AsO3  NO + H3AsO4 + H2O
f.
Balance the rest of the equation by inspection.
2HNO3 + 3H3AsO3  2NO + 3H3AsO4 + H2O
Balance the following reactions:
Cu + HNO3  Cu(NO3)2 + NO + H2O
2. NO2 + H2  NH3 + H2O
1.
59
2. Half-reaction Method or Ion-electron Method
(http://preparatorychemistry.com/Bishop_Balancing_Redox.htm)
Acidic Solutions: Cr2O72−(aq) + HNO2(aq)  Cr3+(aq) + NO3−(aq)
a. Write the skeleton equation of the oxidation and reduction half-reactions.
Cr2O7 2−
Cr3+

HNO2
NO3 −

b. Balance all elements except hydrogen (H) and oxygen (O).
Cr2O72−
2Cr3+

HNO2
NO3 −

c. Balance the oxygen atoms by adding H2O molecules on either side of the equation where O
atoms are needed.
Cr2O72−
2Cr3+ + 7H2O

HNO2 + H2O 
NO3 −
d. Then, balance the hydrogen atoms by adding H+ ions on either side of the equation where H
atoms are needed.
Cr2O72− + 14H+  2Cr3+ + 7H2O
HNO2 + H2O
 NO3− + 3H+
e. Balance the charge by adding electrons.
6e− + Cr2O72− + 14H+
2Cr3+ + 7H2O

HNO2 + H2O
NO3− + 3H+ + 2e−

f. Make the number of electrons lost in the oxidation half-reaction equal to the number of
electrons gained in the reduction half-reaction.
6e− + Cr2O72− + 14H+
2Cr3+ + 7H2O

3(HNO2 + H2O
NO3− + 3H+ + 2e−)

6e− + Cr2O72− + 14H+
2Cr3+ + 7H2O

3HNO2 + 3H2O
3NO3− + 9H+ + 6e−

g. Add the 2 half-reactions algebraically.
Cr2O72− + 3HNO2 + 5H+
2Cr3+ + 3NO3− + 4H2O

h. Re-check to make sure that the number of atoms and the charge balance.
Basic Solutions: Cr(OH)3 + ClO3−  CrO42− + Cl−
a. Write the skeleton equation of the oxidation and reduction half-reactions.
Cr(OH)3
CrO42−

ClO3−
Cl−

b. Balance all elements except hydrogen (H) and oxygen (O).
Cr(OH)3
CrO42−

ClO3−
Cl−

c. Balance the oxygen atoms by adding H2O molecules on either side of the equation where O
atoms are needed.
Cr(OH)3 + H2O
CrO42
−
ClO3
Cl− + 3H2O

d. Then, balance the hydrogen atoms by adding H+ ions on either side of the equation where H
atoms are needed.
Cr(OH)3 + H2O
CrO42− + 5H+

−
+

ClO3 + 6H
Cl− + 3H2O
60
e.
Balance the charge on each side of the equation by adding electrons.
Cr(OH)3 + H2O
CrO42− + 5H+ + 3e
ClO3− + 6H+ + 6e−
Cl− + 3H2O

f. Make the number of electrons lost in the oxidation half-reaction equal to the number of
electrons gained in the reduction half-reaction.

2(Cr(OH)3 + H2O
CrO42− + 5H+ + 3e−)

ClO3− + 6H+ + 6e−
Cl− + 3H2O
2Cr(OH)3 + 2H2O
2CrO42− + 10H+ + 6e
ClO3− + 6H+ + 6e−
Cl− + 3H2O

g. Add the 2 half-reactions algebraically.
2Cr(OH)3 + ClO32CrO42− + Cl− + H2O + 4H+

−
h. Add enough OH ions to each side of the equation to cancel the H+ ions.

2Cr(OH)3 + ClO3− + 4OH2CrO42− + Cl− + H2O + 4H+ + 4OH
+
−
i. Combine the H ions and OH ions that are on the same side of the equation to form water
molecule/s.

2Cr(OH)3 + ClO3− + 4OH
2CrO42− + Cl− + H2O + 4H2O
j. Cancel or combine the H2O molecules.
2Cr(OH)3 + ClO3− + 4OH− 
2CrO42− + Cl− + 5H2O
k. Re-check to make sure that the number of atoms and the total charge on both side of the
equation are balanced.
61
Name: _________________________
Schedule: ______________________
Date: __________________
Rating: _________________
Activity 5.2
MATCHING TYPE
A B C D E
O O O
O O O
O O O
O O O
O O O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
Match Column I with Column II. Shade the circle that best corresponds to
your answer (one per number). Erasures, superimpositions and alterations
will invalidate correct answers.
Column I
Column II
A. O2 + PbO
B. O2 + Pb
C. Pb + O
1. PbO + heat →
A. NaO + H2O B. NaClO + H2
C. NaCl + H2O
2. HCl +NaOH →
+Cl
A. Al3O2
B. Al2O
C. Al2O3
3. Al + O2 →
A. H2 + ZnCl2
B. H2 + ZnCl
C. H + ZnCl
4. Zn + HCl →
AgCl
+ B.
AgCl
+ C. AgCl2 + Ba(NO3)2
5. BaCl2 + AgNO3 A.
→
Ba(NO3)2
Ba2(NO3)
6. Na + H2O →
7. NaNO3 + heat →
8. Na2SO4 + BaCl2 →
9. Na2O + H2O →
10. Xhalogen + YZ → Z + YX
F. Metathesis
G. Analysis
H. single displacement
I. Synthesis
J. None
Balance by change-in-oxidation-number method.
__ H+1 + __ Cr2O7 + __ H2S → __ Cr+3 + __ S + __ H2O
Reduction:
Oxidation:
Balance the following reactions by ion-electron method.
The following reaction occurs in acidic solution:
__ ____ + __ Cr2O7-2 + __ Cl-1 → __ Cr+3 + __ Cl2 + __ ____
Reduction:
Oxidation:
The following reaction occurs in alkaline solution:
__ ____ + __ Br2 → __ BrO3-1 + __ Br-1 + __ ____
Reduction:
Oxidation:
62
Name: _________________________
Schedule: ______________________
I.
Date: __________________
Rating: _________________
Activity 5.3
Indicate the oxidation number of the underlined elements in the following compounds. (10 pts)
1. Cu2O
__________
6. Hg2O
__________
2. KCl
__________
7. NaF
__________
3. HClO3
__________
8. H3BO3
__________
4. (PO4)-3
__________
9. (SO4)-2
__________
5. HBrO3
__________
10. HNO3
__________
BALANCING
AB
O O
O O
O O
AB
O O
O O
O O
AB
O O
O O
O O
O O
AB
O O
O O
O O
O O
AB
O O
O O
O O
O O
O O
O O
C
O
O
O
C
O
O
O
C
O
O
O
O
C
O
O
O
O
C
O
O
O
O
O
O
D E
O O
O O
O O
D E
O O
O O
O O
D E
O O
O O
O O
O O
D E
O O
O O
O O
O O
D E
O O
O O
O O
O O
O O
O O
F
O
O
O
F
O
O
O
F
O
O
O
O
F
O
O
O
O
F
O
O
O
O
O
O
G
O
O
O
G
O
O
O
G
O
O
O
O
G
O
O
O
O
G
O
O
O
O
O
O
H
O
O
O
H
O
O
O
H
O
O
O
O
H
O
O
O
O
H
O
O
O
O
O
O
Using trial-and-error method, balance the following equations. Then choose
from the letters provided below and shade the circle that corresponds to your
answer (one per number). Erasures and superimpositions will invalidate correct
answers.
A=1
D=4
G=7
J = 10 or more
B=2
E=5
H=8
C=3
F=6
I=9
I J
O O 1.
O O 2.
_1_S8 + _2_F2 → _3_SF6
O O3
I J
O O4
O O5
_4_S8 + _5_O3 → _6_SO2
O O 6
I J
O O 7
O O 8
_7_CaCl2 + _8_AgNO3 → _9_AgCl + _10_Ca(NO3)2
O O 9
O O 10
I J
O O 11
O O 12
_11_C2H6 + _12_O2 → _13_CO2 + _14_H2O
O O 13
O O 14
I J
O O 15
O O 16
O O 17
_15_FeTiO3 + _16_C + _17_Cl2 → _18_FeCl3 + _19_TiCl4 + _20_CO2
O O 18
O O 19
O O 20
63
UNIT VI: Stoichiometry
Atomic mass
o Mass of an atom
o Total mass of protons, electrons and neutrons in an atom
Atomic weight/relative atomic mass
o The average of the atomic masses of all the chemical element's isotopes
o H = 1 g/mole
o O = 16 g/mole
Molecular weight (MW)
o Also called formula weight
o Sum of all the atomic weight of all the atoms in a molecule
o NaCl:
Na: 1 x 23 = 23 g/mole
Cl: 1 x 35 = 35 g/mole
58 g/mole
MOLE




Refers to the measure of the amount of particles in matter
Called the “chemist’s dozen”
SI unit for the amount of substance
1 mole of particles = 6.02 x 1023 particles (Avogadro’s number)
o 1 mole of O = 6.02 x 1023 atoms
o 1 mole of NaCl = 6.02 x 1023 molecules
number of atoms
Avogadro' s Number
Number Of Molecules
Number of moles =
Avogadro' s Number
Weight In Grams Of A Subs tan ce
Number of moles =
Molecular Weight Of A Subs tan ce
Number of moles =
Sample Problems:
1. How many moles are 3.40 x 1024 atoms of oxygen?
No. of moles =
number of atoms
Avogadro' s Number
= 3.40 x 1024 atoms
6.02 x 1023 atoms/mole
No. of moles = 5.65 moles of O
2. How many moles are 6.71 x 1026 molecules of salt?
No. of moles =
Number Of Molecules
Avogadro' s Number
No. of moles = 6.71 x 1026 molecules
6.02 x 1023 molecules/mole
No. of moles = 1.11 x 103 moles NaCl
64
3. Calculate the number of moles in 65 grams of H 2O?
No. of moles =
Weight In Grams Of A Subs tan ce
Molecular Weight Of A Subs tan ce
MWH2O : H
O
2 x 1 g/mole = 2 g/mole
1 x 16 g/mole = 16 g/mole
18 g/mole
No. of moles = 65 grams
18 g/mole
No. of moles = 3.61 moles of H2O
Percentage Composition
 This is the relative measure of the mass of each element found in the compound
% mass of element =
Mass In Grams Of Element
x 100
Mass In Grams Of Compound
Sample Problems:
1. Compute for the percentage composition of the compound formed from the reaction of 45 grams of Na
and 34 grams of Cl.
Mass of the Compound = 45 grams Na + 34 grams Cl
Mass of the Compound = 79 grams
Na =
45 grams
x 100
79 grams
Cl =
= 56.96%
34 grams
x 100
79 grams
= 43.04%
2. Calculate the percentage composition of potassium permanganate.
Potassium permanganate = KMnO4
MW of KMNO4 :
K
1 x 39
39
Mn
1 x 55
55
O
4 x 16
64
158 g/mole
K=
39 grams / mole
x 100
158 grams / mole
= 24.68%
Mn =
55 grams / mole
x 100
158 grams / mole
= 34.81%
O=
64 grams / mole
158 grams / mole
= 40.51%
65
Empirical Formula
 Gives the lowest whole number ratio of elements in a compound
 Example: benzene (C6H6) has a ratio of 6:6 or 1:1; therefore the empirical formula is CH
Sample Problems:
1. Find the empirical formula of a compound that is 48.38% carbon, 8.12% hydrogen, and 43.5% oxygen
by mass.
a. Get the mass of each element by assuming a certain overall mass for the sample (100 g).
C = 48.38%
= 48.38 g
H = 8.12%
= 8.12 g
O = 43.50%
= 43.50 g
100.00 g
b. Convert the mass of each element to moles of each element using their atomic masses.
C = (48.38 g) (1 mol/12 g) = 4.03 mol C
H = (8.12 g) (1 mol/1 g) = 8.12 mol H
O = (43.50 g) (1 mol/16 g) = 2.72 mol O
c.
Find the ratio of each element by dividing the number of moles of each by the smallest
number of moles.
C = (4.03/2.72) = 1.48 ≈ 1.50
H = (8.12/2.72) = 2.99 ≈ 3
O = (2.72 /2.72) = 1
d. Use the mole ratio to write the empirical formula. (because not all the answers are whole
numbers, multiply the ratios by a number to make them whole).
C = 1.50 mol x 2 = 3
H = 3 mol x 2 = 6
O = 1 mol x 2 = 2
Empirical Formula: C3H6O2
2. Calculate the empirical formula of a compound composed of 4.5 g of carbon and 1.5 g of hydrogen.
a. Compute for the number of moles using the atomic mass of each element.
C = (4.5 g) (1 mol/12g) = .0375 mol C
H = (1.5 g) (1 mol/1g) = 1.5 mol H
b. Find the ratio of each element by dividing the number of moles of each by the smallest
number of moles.
C = (0.375/.0375) = 1
H = (1.5/0.375) = 4
c. Use the mole ratio to write the empirical formula.
Empirical formula = CH4
66
Molecular Formula
 Gives the actual formula of a compound
 Could either be the same as the empirical formula or be a whole-number multiple of the empirical
formula
 Example: C6H6 is the molecular formula of the empirical formula CH
Sample Problems:
1. The empirical formula for vitamin C is C3H4O3. Based from an experimental data, the molecular mass
of vitamin C is about 180 g/mole. What is the molecular formula of vitamin C?
a. Divide the molecular weight by the empirical formula weight to find a multiple or factor.
Molecular weight = 180 g/mole
EF weight of C3H4O3 = (3 x 12 g/mole) + (4 x 1 g/mole) + (3 x 16 g/mole)
= 88 g/mole
Factor/Multiple =
Molecular Weight
Empirical Formula Weight
Factor/Multiple = 180 g/mole
88 g/mole
Factor/Multiple = 2
b. Multiply the subscript of the empirical formula with the factor or multiple.
(C3H4O3)2
= C3x2 H4x2 O3x2
MF:
C6H8O6
2. A compound is composed of 40.00% C, 6.72% H, and 53.29% O; and has a molecular weight of 180
g/mol. Compute its molecular formula.
a. Find the empirical formula of the compound.
C = 40.00%
= 40.00 g
H = 6.72%
= 6.72 g
O = 53.28%
= 53.28 g
100.00 g
C = (40.00 g) (1 mol/12 g) = 3.33 mol C
H = (6.72 g) (1 mol/1 g) = 6.72 mol H
O = (53.28 g) (1 mol/16 g) = 3.33 mol O
C = (3.33/3.33) = 1
H = (6.72/3.33) = 2
O = (3.33 /3.33) = 1
EF: CH2O
b. Divide the molecular weight by the empirical formula weight to find a multiple or factor.
Molecular weight = 180 g/mole
EF weight of CH2O = 12 g/mole + 2 g/mole + 16 g/mole
= 30 g/mole
Factor/Multiple =
c.
Molecular Weight
Empirical Formula Weight
Factor/Multiple = 180 g/mole
30 g/mole
Factor/Multiple = 6
Multiply the subscript of the empirical formula with the factor or multiple.
(CH2O)6 = C1x6 H2x6 O1x6
MF:
C6H12O6
67
Stoichiometry
 the quantitative description of the relationships between reactants and between reactants and
products in a chemical reaction
1. Mole to Mole Relationship
How many moles of NH3 could be produced if 4 moles of N2 were reacted with Hydrogen?
a. Write the balanced equation.
N2 + 3 H2 ---> 2 NH3
b. Determine the mole relationship between the given component and the required component.
Relative moles (from the equation):
1 mole N2 = 2 moles NH3
Given moles (from the problem):
4 moles N2 = X moles NH3
c. Form the proportion of the given and required components. Then solve for the unknown.
Given mole N2
=
Given mole NH3
Rel. mole N2
Rel. mole NH3
4 moles N2
=
1 mole N2
X = 8 moles NH3
X moles NH3
2 moles NH3
2. Mole to Mass Relationship
From the equation: N2 + 3 H2 ---> 2 NH3, determine the mass in grams of H2 that will react with the 4
moles of N2.
a. Write the balanced equation
N2 + 3 H2 ---> 2 NH3
b. Convert the moles of the given component to moles of the requested component.
Relative moles (from the equation):
1 mole N2 = 3 moles H2
Given moles (from the problem):
4 moles N2 = X moles H2
Given mole N2
=
Given mole NH3
Rel. mole N2
Rel. mole NH3
4 moles N2
=
1 mole N2
X = 12 moles H2
c.
X moles H2
3 moles H2
Convert the moles of the requested component to mass in grams.
MW H2: 2 x 1 g/mole = 2 g/mole H2
12 moles H2 x 2 g H2 = 24 grams H2
1 mole H2
3. Mole to Volume Relationship
How many liters of NH3 can be produced at 27 °C and 760 torr if 20 moles of N2 are utilized? (Use
the equation: N2 + 3 H2 ---> 2 NH3)
a. Convert the moles of the given component to the moles of the requested component using the
coefficients of the balanced equation.
1 mole N2 = 2 moles NH3
20 moles N2 x 2 moles NH3
= 40 moles NH3
1 mole N2
68
b. Convert the moles of the requested component to its volume using the Ideal Gas Law Equation.
PV = nRT
T = 27 + 273 = 300 K
P = 760 torr x 1 atm = 1 atm
760 torr
V=
nRT
P
= (40 moles) (0.0821 L-atm/mol-K) (300 K) = 985.20 L NH3
1 atm
4. Mass to Mass Relationship
From the equation: N2 + 3 H2 ---> 2 NH3, compute the grams of NH3 that will be produced if 10 grams
of H2 reacted with N2.
a. Balance the equation.
N2 + 3 H2 ---> 2 NH3
b. Compute for the moles of the given component using its mass.
MW H2: 2 x 1 g/mole = 2 g/mole H2
10 g H2 x 1 mol H2
= 5 moles H2
2 g H2
c.
Convert the moles of the given component to moles of the requested component using the
coefficients of the balanced equation.
3 moles H2 = 2 moles NH3
5 moles H2 x 2 moles NH3
= 3.33 moles NH3
3 moles H2
d. Convert moles of the requested component to its mass.
MW NH3: N = 1 x 14
= 14 g/mole
H=3x1
= 3 g/mole
17 g/mole
3.33 moles NH3 x 17 grams NH3 = 56.61 grams NH3
1 mole NH3
5. Mass to Volume Relationship
How many liters of H2 will be required at 300 K and 3 atm to cosume 56 grams of N 2? (Use the
reaction: N2 + 3 H2 ---> 2 NH3)
a. Balance the equation.
N2 + 3 H2 ---> 2 NH3
b. Compute the moles of the given component using its mass.
MW N2: 2 x 14 g/mole = 28 g/mole N2
56 g N2 x 1 mol N2
= 2 moles N2
28 g N2
69
c.
Convert the moles of the given component to moles of the requested component using the
balanced equation.
1 moles N2 = 3 moles H2
2 moles N2 x 3 moles H2
= 6 moles H2
1 mole N2
d. Convert moles of the requested component to its volume using the Ideal Gas Law Equation.
PV = nRT
V=
nRT
P
= (6 moles H2) (0.0821 L-atm/mol-K) (300 K) = 49.26 L H2
3 atm
6. Volume to Volume Relationship
What is the volume of NH3 that will be produced if 10 liter of N2 are used? (Use the reaction: N2 + 3
H2 ---> 2 NH3)
a. Balance the equation.
N2 + 3 H2 ---> 2 NH3
b. Convert the volume of the given component to liters of the required component using the
coefficients of the balanced equation.
1 liter N2 = 2 liters NH3
10 liters N2 x 2 liters NH3 = 20 liters NH3
1 liter N2
7. Theoretical and Actual Yield
Theoretical yield
 maximum amount of product that could be computed given the amounts of the reactants.
Actual yield
 amount of product produced when the reaction is carried out in the laboratory
Percent Yield =
Acutal Yield
x 100
Theoretical Yield
Example:
1. What is the percent yield for a reaction if you predict the formation of 21 grams of C6H12 and actually
recovered only 3.8 grams?
Percent Yield =
Acutal Yield
x 100
Theoretical Yield
= 3.8 grams x 100
21 grams
Percent Yield = 18.10%
70
2. When 10 grams of methane in an excess of oxygen is burned, 19.8 grams of water is produced. What is
the percent yield of water?
a. Balance the equation.
CH4 + 2O2 -> CO2 + 2H2O
b. Convert the mass of the given component to moles using the balanced equation.
MWCH4: C = 1 x 12 g/mole = 12 g/mole
H = 4 x 1 g/mole = 4 g/mole
16 g/mole
10 grams CH4 x 1 mole CH4 = 0.625 mole CH4
16 g CH4
c.
Form the proportion of the given and required components. Then solve for the unknown.
1 mole CH4 = 2 moles H2O
0.625 mole CH4 =
1 mole CH4
X = 1.25 moles H2O
X moles H2O
2 moles H2O
d. Convert the computed number of moles to mass of the required component to get the
theoretical yield.
MW H2O: H = 2 x 1 g/mole = 2 g/mole
O = 1 x 16 g/mole = 16 g/mole
18 g/mole
1.25 moles H2O x 18 g H2O
= 22.5 grams H2O
1 mole H2O
e.
Solve for the percent yield.
Percent Yield =
Acutal Yield
x 100
Theoretical Yield
= 19.8 g x 100
22.5 g
Percent Yield = 88 %
71
Name: _________________________
Schedule: ______________________
Date: __________________
Rating: _________________
Activity 6.1
Compute for the unknown. Show all solutions.
___________________11. How many moles are there in 50 g CCl4 ?
___________________12. How many molecules are there in 25 g H2SO4?
___________________13. How many moles are there in 1.269 X 1024 molecules of NaNO3?
___________________14. What is the mass, in grams, of 0.75 mol Ca3(PO4)2 ?
___________________15. What is the mass, in grams, of 1.35 X 1023 molecules C6H12O6 ?
___________________16. How many particles are there in 1.38 mol CuSO4 · NH4NO3 · 12 H2O ?
III. Identify / Solve the following items in this hydrate, FeSO4 · 6 H2O. (Show derivation of %
composition of each component at the back of this sheet.)
Components
No. of Atoms
Atomic Mass
Relative Mass
% Composition
% composition of O = ___________
Solution:
72
Name: _________________________
Schedule: ______________________
Date: __________________
Rating: _________________
Activity 6.2
Compute for the unknown. Show all solutions.
1. The MW of cholesterol is 386.73 g/mol and the compound contains 83.9 % C, 12.0 % H and the
remaining is oxygen.
___________________a. Compute for the mole ratio of C.
___________________b. Compute for the mole ratio of H.
___________________c. Compute for the mole ratio of O.
___________________d. What is the empirical formula of the compound?
___________________e. What is the molecular formula of the compound?
2. When 6.65 g of the hydrate NiSO4 · XH2O was heated in a vacuum, the water was driven off and
3.67 g of anhydrous NiSO4 remained.
___________________a. Calculate the number of moles of NiSO4.
___________________b. Calculate the number of moles of H2O.
___________________c. What is the empirical formula ( the value of X) ?
3. In a convenient laboratory preparation of pure nitrogen gas, NaN3 decomposes to Na and N2 upon
heating.
___________________a. Write a balanced equation.
___________________b. What number of moles of NaN3 is required to prepare 1.00 mol of N2 ?
___________________c. What mass of N2 is produced by the decomposition of 2.5 g of NaN3 ?
___________________d. What volume of Na is produced when 11.75 li of N2 is prepared ?
73
Name: _________________________
Schedule: ______________________
Date: __________________
Rating: _________________
Activity 6.3
1.
The acid secreted by the cells of the stomach lining is a hydrochloric acid solution that typically
contains 0.282 g of HCl per 50.00mL of solution. What is the concentration of this acid?
2. A sample of tin having a mass of 3.996 grams was oxidized and found to have combined with 1.077
grams of oxygen. Calculate the empirical formula of this oxide of tin.
3. Dibutyl succinate is an insect repellant used against household ants and roaches. Its composition is
62.58% C, 9.63% H and 27.79% O. Its composition experimentally determined molecular mass is
230amu. What are the empirical and molecular formulas of dibutyl succinate?
4. Fructose is a very sweet naturally occurring sugar present in honey, fruits, and fruit juices. It has
a molar mass of 180.1 g/mol and a composition of 40.0% C, 6.7% H, and 53.3% O. Calculate the
molecular formula of fructose.
5. Calculate the number of moles corresponding to:
a. 0.10g of the secondary extracellular cation, Ca++
b. 4.0g of the major intracellular cation, K+
6. Isopropyl alcohol, commonly known as rubbing alcohol, is used as an antiseptic. It is composed of C,
H, and O. Combustion of isopropyl alcohol produces 0.561g CO 2 and 0.306 g H2O. Determine the
empirical formula of isopropyl alcohol.
74
UNIT VII: Solutions
Solution
 A homogenous mixture of two or more substances in which settling does not occur
 A homogenous mixture in which one substance is dissolved in another substance
o Solvent – the substance that does the dissolving
o Solute – the substance that is dissolved
Types of Solutions
Type
Solvent
Solute
Example
Gas
Gas
Air
Gas
Liquid
Moist air
Gas
Solid
Soot in air
Liquid
Liquid
Solid
Brine, syrup
Liquid
Liquid
Alcohol in water, vinegar
Liquid
Gas
Carbonated beverages
Solid
Solid
Solid
Copper in nickel, brass, steel
Solid
Liquid
Mercury in water
Solid
Gas
Hydrogen dissolved in platinum metal
Source: Interactive Chemistry by Aurora Franco, et. al. 2000: 208)
Gaseous
The Solution Process
 The solute particles are separated from the surface of the solid solute. Dissolving is aided by the
attraction between the solute and the solvent molecules.
This needs heat or energy
(endothermic)
 Solvent particles are moved apart to allow solute molecules to enter the liquid. This needs energy
(endothermic).
 The solute molecules are attracted to solvent molecules. This gives up energy (exothermic).
Rate of Solution
 agitating or stirring the solution
 heating the solute
 powdering the solid solute
Solubility
 measures of how much of that solute can be dissolved in a given amount of solvent under certain
conditions
Factors affecting the solubility of substance:
 Nature of solute and solvent
 Non-polar solvents – benzene and CCl4
 Polar solvents - water
 Effect of Temperature
 Effect of Pressure
75
Concentration of Solution
 Refers to the amount of solute dissolved in a certain amount of solvent
 Concentrated solution – much of solute is dissolved in a solvent



saturated solution
o solution that has dissolved the maximum amount of solute for a given amount of
solvent at a given temperature
unsaturated solution
o solution that contains less solute than it can hold at a certain temperature and
pressure
supersaturated solution
o solution that can hold more solute than that which is normal for that temperature
Expressing Concentrations
1. Percent by mass
mass
mass
of
of
solute
x 100%
solution

% by mass =
a.
A bottle of a certain ceramics tile cleanser which is a solution of hydrogen chloride, contains 130
g HCl and 750 g of H2O. What is the % by mass of HCl in this cleanser?
Given: mass of solute = 130 g HCl
Mass of solvent = 750 g H2O
Required: % by mass of HCl
Answer: 14.77 %
b. A sample of 0.892 g of potassium chloride is dissolved in 54.6 g of water. What is the percent by
mass of KCl in this solution?
2. Percent by volume
volume
volume
of
of
solute
x 100%
solution

% by volume =
a.
What is the percent by volume of ethanol in the final solution when 75 mL of ethanol is diluted to
a volume of 250 mL of water?
Given: volume of solute = 75 mL ethanol
Mass of solvent = 250 mL H2O
Required: % by volume of ethanol
Answer: 23.08 %
b. What is the percent by volume of acetic acid in a fiscal volume when 75 mL of acetic acid is
diluted to a volume of 250 mL of water?
76
3. Molarity (M)
moles
of
solute
liter
of
solution
mass
of
solute
 Molarity (M) =
MWsolute
X
Vsolution

Molarity (M) =
a.
Calculate the molar concentration of a solution that contains 15 g of KOH in 225 mL of
solution.
Given:
mass of solute = 15 g KOH
Volume of solution = 225 mL (0.225 L)
MW of KOH = 56 g/mole
Number of moles = 0.27 moles KOH
Answer:
1.19 M
b. Calculate the molarity of the solution containing 10 grams of sulfuric acid in 500 mL of
solution.
Given:
mass of solute = 10 g H2SO4
Volume of solution = 500 mL (0.500 L)
MW of H2SO4 = 98 g/mole
Answer:
0.204 M
4. Molality (m)
moles
ki log ram
mass
 Molality =
MWsolute
of
of
of
X
solute
solvent
solute
Ksolvent

Molality =
a.
2.3 grams of ethanol is added to 500 grams of water. Determine the molality of the resulting
solution.
Given: mass of solute = 2.3 grams
Mass of solvent = 500 grams (0.5 kg)
MW of solute = 46 g/mole
Required: molality of the solution
Answer: 0.1 molal
b. Determine the mass of water to which 293 grams of NaCl is added to obtain a 0.25 molal solution.
Given: mass of solute = 293 grams
Molality of the solution = 0.25 molal
Required: mass of water
Answer: 20 kg of water
77
5. Normality (N)
mass
Eq
of
solute
x
Lsolution
MW
of
solute
 Equivalent weight =
Factor(F )
Eq
 N = molarity x
mole

N=
a.
F for Acid is the number of replaceable hydrogen ion
 HCl ; F = 1
 H2SO4 ; F = 2
b. F for Base is the number of replaceable hydroxyl ion
 NaOH; F = 1
 Ba (OH)2; F = 2
c.
F for Salt is the valence of the positive ion
 NaNO3; F = 1
 K2CO3; F = 2
**Calculate the normality of a solution containing 2.45 grams of H 2SO4 in 2 liters of solution?
Given: mass of H2SO4 = 2.45 grams
Volume of solution = 2 liters
MW of the solution = 98 g/mol
Required: Normality of the solution
Solution:
Equivalent weight of H2SO4 = (98/2) = 49 g/mol
N = [2.45/(49 x 2)] = 0.025 normal
**How many grams of sulfuric acid are contained in 3 liters of 0.500 N solution?
Given: Volume of solution = 2 liters
MW of the solution = 98 g/mol
N = 0.500 normal
Required: mass of solute
Solution:
Equivalent weight of H2SO4 = (98/2) = 49 g/mol
Mass of solute = (0.5 x 49 x 3)] = 73.5 g
78
Name: _________________________
Schedule: ______________________
Date: __________________
Rating: _________________
Activity 7.1
A Ca(OH)2 solution has a density of 0.55 g/ml and weighs 50 g. It is prepared from 23 grams of
Ca(OH)2. Solve for:
____________________a. molarity
____________________b. normality
____________________c. molality
____________________d. mole fraction of solute
____________________e. mole fraction of solvent
____________________f. % by volume of solvent if the volume and weight of the solvent are equal in
value
5. If the solute weighs 183.72 g, what volume, in liters, of the following compounds should be used to
prepare:
____________________a. 12.65 M of NaOH
____________________b. 9.05 N of KMnO40 → KMnO4-2
79
Name: _________________________
Schedule: ______________________
Date: __________________
Rating: _________________
Activity 7.2
1. Sam is a nursing student. He has been asked to intravenously inject his patient with 10cc
of an isotonic preparation. Mistakenly, he injected his patient with pure water. What is
likely to happen to the red blood cells near the site of injection?
2. Anne is Sam’s patient. She was carefully watching him fill an IV bottle with a mixture
specially ordered by the physician. She noticed that he accidentally filled the bottle with
concentrated (10%) NaCl solution instead of the mixture from the pharmacy. Why should
she refuse to allow Sam to attach the IV bottle to her system?
3. Sam is now assigned to the surgical unit. He is given a piece of a living tissue and asked to
put it in a fluid before taking the sample to the pathology laboratory. What essential
characteristics must such fluid have in order not to damage the tissue?
80
pH
 The rough measure of the acidity of a solution. The p stands for “potenz” which means the
potential to be and the H stands for Hydrogen.
 Denotes negative logarithm of the hydrogen ion concentration.
 is divided into fourteen units
 7 is neutral
 Basic solution has a ph> 7
 Acidic solution has pH < 7
0
1
2
3
4
more acidic
5
6
7
8
neutral
9
10
11
12
13
14
more basic
 pH = - log [H+]
 pOH = - log [OH-]
 Kw = 1.0 x 10-14 = [H+][OH-]
 If we take the negative logarithm of both sides of the equilibrium expression, it becomes,
 -log Kw = -log ([H+][OH-])
 -log Kw = -log [H+] + –log [OH-]
 -log 1.0 x 10-14 = -log [H+] + – log [OH-]
 14 = pH + pOH
***The autoionization of water
H 2O + H 2O
H3O + OH
*** Kw (equilibrium constant)
 Known as the ion product for water
 Pure water at 25 °C (H3O+ = OH- = 1.0 x 10-7 mol/L)
Calculation of pH
1. Calculate the pH of a solution in which the H+ concentration is 0.070 mol/L.
a. [H+] = 0.070 M = 7.0 x 10-2 M
b. pH = -log [H+] = -log [7.0 x 10-2] = 1.15
2. Compute for the pH of a solution that contains 3.65 g of HCl in 2.00 L of solution.
a. Molarity = number of moles of solute
number of liters of solution
1molHCl
? molHCl
3.65 gHCl
=
x
= 0.0500 mol HCl/L or 0.0500 M
Lsolution 2.00 Lsolution 36.5 gHCl
b. [H+] = 0.0500 M = 5.0 x 10 -2 M
c. pH = -log [H+] = -log [5.0 x 10-2] = 1.30
3. The pH of a solution if 3.301. What is the concentration of H+ in this solution?
a. pH = -log [H+]
b. 3.301 = -log [H+]
c. -3.301 = log [H+]
d. antilog -3.301 = H+
e. H+ = 5.00 x 10-4 M
81
Calculation of pOH
1. Calculate [H+], pH, [OH-1], and pOH for a 0.015 M HNO3 solution.
a. [H+] = 0.015 M (because it is an acid)
b. pH = - log [H+] = -log [0.015] = 1.82
c. pH + pOH = 14.00
d. pOH = 14.00 – pH = 14.00 – 1.82 = 12.18
e. [H+] [OH-] = 1.0 x 10-14
f. [OH-] = (1.0 x 10-14) ÷ [H+] = (1.0 x 10-14) ÷ 0.015 = 6.7 x 10-13 M
2. Calculate [H+], pH, [OH-1], and pOH for a 0.015 M Ca(OH)2 solution.
a. [OH-] = 2 x 0.015 M = 0.030 M
b. pOH = -log [OH-] = -log (0.030) = 1.52
c. pH + pOH = 14.00
d. pH = 14.00 – pOH = 14.00 – 1.52 = 12.48
e. [H+] [OH-] = 1.0 x 10-14
f. [H+] = (1.0 x 10-14) ÷ [OH-] = (1.0 x 10-14) ÷ 0.030 = 3.3 x 10-13 M
82
Name: _________________________
Schedule: ______________________
Date: __________________
Rating: _________________
Activity 7.3
A. Solve for the H+, OH-, pH, and pOH values. (20 pts)
H+
Solution
a. 0.045 M LiOH
OH-
pH
pOH
b. 0.185 M HBr
c. 0.230 M Mg(OH)2
d. 0.0099 M HI
e. 2.30 x 10
-4
M HF
B. Calculate the following values for each solution. (24 pts)
H+
OH-
pH
2.36
pOH
11.32
10.78
7.21
3.58
4.97
12.43
4.65
C. The normal pH of Human Blood ranges from 7.35 to 7.45. Calculate the concentrations of H3O+
and OH- ions in human blood that has a pH of 7.45.
83
Colligative Properties of Solutions
 are properties that depend upon the concentration or number of solute molecules or ions, but not upon
the identity of the solute
1. vapor pressure lowering
2. boiling point elevation
3. freezing point depression
4. osmotic pressure
Vapor Pressure
 the pressure of the vapor resulting from evaporation above a sample in a closed container
 the rate of condensation of the gas becomes equal to the rate of evaporation of the liquid or solid
 Raoult’s Law
o The vapor pressure of a volatile or of an ideal solution is dependent on the vapor pressure of
each component and the mole faction of the component present in the solution.
o Vapor Pressure of Solution = (Mole Fraction of solvent) (Vapor Pressure of Pure Solvent)
o
Mole Fraction of a solvent =
number of moles of solvent
total number of moles of solution and of solvent
Example:
Determine the vapor pressure of a solution at 25 °C that has 45 grams of glucose dissolved in 72 grams of
water. The vapor pressure of pure water at 25 °C is 23.8 torr.
a. Compute the number of moles of the substance/solute by using its mass.
MWglucose: C = 6 x 12 = 72 g/mole
H = 12 x 1 = 12 g/mole
O = 6 x 16 = 96 g/mole
180 g/mole
45 grams x 1mole C6H12O6
= 0.25 mole
180 grams C6H12O6
b. Compute the number of moles of the solvent by using its mass.
MWH2O: 18 g/mole
72 grams x 1 mole H2O
= 4 moles
18 grams H2O
c. Determine the mole fraction of the solute
Mole fraction of solvent=
number of moles of solvent
total number of moles of solution and of solvent
= ________4 moles H2O
0.25 moles C6H12O6 + 4 moles H2O
Mole fraction of solvent = 0.94
d. Determine the vapor pressure of the solution
Vapor Pressure of Solution = (Mole Fraction of solvent) (Vapor Pressure of Pure Solvent)
= (0.94) (23.8 torr)
V.P. of Solution = 22.38 torr
84
Freezing Point Depression
 Change that occurs in the temperature at which a liquid freezes when a solute is dissolved in it
 The solution has a lower freezing point than a pure solvent
 Freezing point depression
o ΔTf = Kfm
where ΔTf = freezing point depression
Kf = molal freezing point constant of the solvent
m = molal concentration
o Freezing point of solution = FPsolvent - ΔTf
Example:
Compute the freezing point of a solution of 90 grams of glucose dissolved in 750 grams of water. The K f
of water is -1.86 °C/m. The freezing point of pure water is 0 °C.
a. Compute the number of moles of the substance/solute by using its mass.
MWglucose: C = 6 x 12 = 72 g/mole
H = 12 x 1 = 12 g/mole
O = 6 x 16 = 96 g/mole
180 g/mole
90 grams x 1mole C6H12O6
= 0.5 mole
180 grams C6H12O6
b. Convert the mass of solvent to kilograms.
750 grams H2O x 1 kg H2O
= 0.75 kg H2O
1000 g H2O
c.
Calculate the molal concentration.
m = moles of solute
kilogram of solvent
= 0.5 mole C6H12O6
0.75 kg H2O
m = 0.67 m
d. Compute the freezing point of solution.
ΔTf = Kfm
= (-1.86 °C/m) (0.67 m)
ΔTf = -1.25 °C
Freezing point of solution = FPsolvent - ΔTf
= 0 °C – (-1.25 °C)
Freezing point of solution = 1.25 °C
Boiling Point Elevation
 Change that occurs in the temperature at which a liquid boils when a solute is dissolved in it
 The solution has a higher boiling point than a pure solvent
 Boiling Point Elevation
o ΔTb = Kbm
where ΔTb = boiling point elevation
Kb = molal boiling point constant of the solvent
m = molal concentration
o Boiling point of solution = BPsolvent + ΔTb
85
Example:
The normal boiling point of benzene is 80.1 oC. It has a boiling point elevation constant of 2.53 oC/m. If we
make up a 0.5 molal solution of Br2 in benzene, what is the boiling point of the mixture?
ΔTb = Kbm
= (2.53 °C/m) (0.5 m)
ΔTb = 1.27 °C
Boiling point of solution = BPsolvent + ΔTb
= 80.1 °C + 1.27 °C
Boiling point of solution = 81.37 °C
Osmotic Pressure
 Osmosis is the diffusion of small particles through a semi-permeable membrane
 Osmotic pressure is a hydrostatic pressure produced by the difference of solute concentration on
the opposite sides of the semi-permeable membrane
 π = MRT
where: π = osmotic pressure
M = molarity of the solution
R = gas constant (0.08205 L-atm/mole-K)
T = temperature in Kelvin
Example:
What is the osmotic pressure of a sugar solution that contains 100 grams of sucrose C 12H22O11 which is
dissolved in water to make 1 liter of solution at 25 °C?
a. Compute the number of moles of the substance/solute by using its mass.
MWsucrose: C = 12 x 12 = 144 g/mole
H = 22 x 1 = 22 g/mole
O = 11 x 16 = 176 g/mole
342 g/mole
100 grams x 1mole C12H22O11
342 grams C12H22O11
= 0.29 mole
b. Determine the molarity concentration.
Molarity = moles of solute
Liters of solution
= 0.29 mole C12H22O11
1 L solution
Molarity = 0.29 M
c.
Calculate the osmotic pressure.
T = 25 + 273 = 298 K
π = MRT
= (0.29) (0.08205 L-atm/mole-K) (298 K)
π = 7.09 atm
86
Name: _________________________
Schedule: ______________________
Date: __________________
Rating: _________________
Activity 7.4
1.
What is the osmotic pressure at 25 °C of an aqueous solution that is 0.0010 M C12H22O11
(Sucrose)?
2. What mass of urea [CO (NH2)2] would you dissolve in 225 mL of solution to obtain an osmotic
pressure of 0.015 atm at 25 °C?
3. What is the a) Molality, b) Freezing point, and c) Boiling point of a solution containing 2.68 g of
Naphthalene (C10H8) in 18.4 of Benzene (C6H6)?
4. The freezing point of a solution of 8.00 g of an unknown compound dissolved in 60.0 g of acetic
acid if 13.2 0C. Calculate the molar mass of the compound.
5. A solution is prepared by dissolving 6.85 g of ordinary sugar (sucrose, C12H22O11, 34 g/mol) in 34.0
g of water. Calculate the boiling point of the solution.
6. Sucrose is a non volatile, non-ionizing solute in water. Determine the vapour pressure lowering at
25 °C, of the 1.25 molal sucrose solution. Assume that the solution behaves ideally. The vapour
pressure of pure water at 25 °C is 23.8 torr.
7. Calculate the (a) lowering of the vapour pressure and (b) the vapour pressure of a solution
prepared by dissolving 25.5 g of Naphthalene, C 10H8 ( a non-volatile electrolyte) in 150.0 g of
Benzene C6H6 at 20 °C. Assume that the solution is ideal. The vapour pressure of pure benzene is
74.6 torr at 20 °C.
87