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CHAPTER 1: THE FOUNDATIONS OF CHEMISTRY
CLASS NOTES FOR CHEMISTRY 1411
Mr. Peter Yang-Ying Lin
CHEMISTRY
The class note is based on WHITTEN, DAVIS, PECK AND STANLEY
9TH EDITION. It may be slightly different from McGraw Hills textbook.
1-1 MATTER AND ENERGY
Definitions:
(1) Matter: Anything that has mass and occupies space. (Mass has to do with quantity of
matter and does not vary with location; weight is a measure of gravitational attraction and
does vary with location.)
(2) Energy: Capacity for doing work or transferring heat (many forms such as
mechanical, light, electrical or heat)
Kinetic energy: Energy of motion
Potential energy: Energy of position, condition or composition
(3) Chemistry: The study of matter and the changes matter undergoes (organic,
inorganic, analytical, physical, and biochemistry)
(4) Physics:
The study of energy and the changes that energy undergoes.
(5) Law of Conservation of Matter: There is no observable change in the amount of
matter during a chemical reaction or in a physical change.
(6) Law of Conservation of Energy: Energy cannot be created or destroyed in a chemical
reaction or in a physical change. It can only be converted from one form to another.
(7) Law of Conservation of Matter and Energy: The combined amount of matter and
energy in the universe is fixed.
1-2 CHEMISTRY-A MOLECULAR VIEW OF MATTER
Macroscale: sample large enough to see and handle
Nanoscale: exceedingly small examples are atoms and molecules
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Elements: Basic building blocks of all other matter—can not be broken down to simpler
substances by chemical changes—periodic table is a list of the elements—each element
has a name and a symbol (abbreviation) which is usually 1 or 2 letters, sometimes from
Latin names. Be familiar with the elements in Table 1-3 in the text, page 18.
Compounds: Two or more elements combined in a definite ratio and can’t be separated
by physical means. Examples: H2O, water; H2O2, hydrogen peroxide
Atom: The smallest particle of an element that has the properties of that element through
all chemical and physical changes.
Subatomic particles(fundamental particles): The major particles that make up the atom—
proton, neutron and electron
Particle abbreviation
proton
p
neutron
n
electron
e-
charge
1+
0
1-
mass
1 amu
1 amu
0 amu
Both charge and mass are relative numbers, not absolute. Amu = atomic mass units
If the carbon atom is assigned the mass of 12 amu, then the proton is one-twelth as much.
Atomic number: The atomic number, Z, of an element is equal to the number of protons
in the nucleus of the atom. It is also equal to the number of electrons, if the atom is
neutral. Hydrogen has one proton; helium has two protons, lithium has three protons, etc.
Atomic mass number (atomic weight): The atomic weight of an element is equal to the
sum of the protons and neutrons. Hydrogen has one proton and no neutrons. Helium has
two protons and two neutrons. Lithium has three protons and four neutrons.
Molecule: A molecule is the smallest particle of an element or a compound that can
exist independently.
Substance
Silver
Oxygen
Chlorine
Water
Ammonia
Atom
Ag
O
Cl
-
Molecule
Ag (monatomic molecule)
O2 (diatomic molecule)
Cl2
H2O
NH3
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1-3 STATES OF MATTER
Solids: definite shape and definite volume; ~not compressible
Liquids: no definite shape, but definite volume; ~not compressible
Gases: no definite shape and no definite volume; highly compressible
1-4 CHEMICAL AND PHYSICAL PROPERTIES
Physical properties (characteristics): Properties that we can observe and the
composition of the substance is not changed. (non-destructive) Examples: color,
density, hardness, melting point, boiling point, thermal and electrical conductivities.
(Doesn’t depend upon amount for intensive properties.)
Chemical properties: Properties that we can observe and the composition of the
substance is changed. (destructive) Examples: Will the substance burn? Will the
substance tarnish? Will the substance rust? (Doesn’t depend upon amount.)
1-5 CHEMICAL AND PHYSICAL CHANGES
Physical changes: Changes that we observe and there is no change in composition of
the substance (non-destructive), but energy is released or absorbed. Examples: All
phase changes are physical changes. The processes of dissolving or precipitating out
are also physical.
Chemical changes: Changes that we can observe and there is a change in
composition of the substance. New substances are formed and energy is released or
absorbed. Examples: Burning, tarnishing, rusting, digesting
1-6 MIXTURES, SUBSTANCES, COMPOUNDS, AND ELEMENTS
Matter
Pure Substances
Elements
Compounds
Mixtures
Heterogeneous
Homogeneous
Elements: Basis building blocks of all other matter—cannot be broken down to simpler
substances by chemical changes—periodic table is a list of the elements—each element
has a name and a symbol (abbreviation) which is usually 1 or 2 letters, sometimes from
Latin names. Be familiar with the elements in Table 1-3 in the text, page 18.
Compounds: Two or more elements combined in a definite ratio and can’t be separated
by physical means. Examples: H2O, water; H2O2, hydrogen peroxide
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Mixtures: Two or more substances mixed together and each substance retains its own
properties—no definite ratio and usually easy to separate by physical means such as
filtering, decanting, evaporation of a liquid, etc.
Heterogeneous mixtures: Mixtures that to the eye do not appear to be the same
throughout. Example: sand in water
Homogeneous mixtures: Mixtures that to the eye appear the same throughout.
Example: salt dissolved in water.
1-7 MEASUREMENTS IN CHEMISTRY
In science use the metric system or the SI system.
1-8 UNITS OF MEASUREMENT
Prefixes that we use a lot in the metric system:
Kilo or K = 1000 or 1 x 103 (text book uses a small k)
Centi or c =
Milli or m =
1
100
or 0.01 or 1 x 10-2
1
1000
or 0.001 or 1 x 10-3
Mass: Basic unit is the gram (g) in metric system. A nickel is about 5 grams.
kilogram (Kg)
milligram (mg)
1Kg = 1000 g
1000 mg = 1 g
Can make conversion factors out of these:
1 𝐾𝑔
1000 𝑔
1000 𝑔
or
1 𝐾𝑔
1000 𝑚𝑔
1𝑔
or
1𝑔
1000 𝑚𝑔
Length: Basic unit is the meter, (m). A meter is about 1 yard.
kilometer (Km)
centimeter (cm)
millimeter (mm)
1 Km = 1000 m
1 m = 100 cm
1 m = 1000 mm
Can make conversion factors out of these:
1 𝐾𝑚
1000 𝑚
1𝑚
100 𝑐𝑚
or
or
1000 𝑚
1 𝐾𝑚
100 𝑐𝑚
1𝑚
5
1𝑚
1000 𝑚𝑚
or
1000 𝑚𝑚
1𝑚
Volume: Basic unit is the liter, (L). A liter is about 1 quart.
Milliliter (mL)
1 L = 1000 mL
Can make a conversion factor:
1 cc = 1 cm3 = 1 mL
1𝐿
1000 𝑚𝐿
One other unit to know: 1 Angstrom = 1 x 10-8 cm
Or 1 Angstrom = 1 x 10 -10 m
or
1000 𝑚𝐿
1𝐿
or 1 A = 1.0 x 10-8 cm
1-9 USE OF NUMBERS
Scientific notation:
4300000 = 4.3 x 106 To put this number in your calculator:
Press 4.3; press EE or EXP (Do not press times 10); press 6
It will probably look like 4.3 E 06 on your calculator.
0.000348 = 3.48 x 10-4 To put this number in your calculator:
Press 3.48; press EE or EXP; press +/- ; press 4
It will probably look like 3.49 E -04 on your calculator
Significant Figures (sig figs)
RULES:
1. Nonzero digits are always significant. 38.57 has 4 sig figs.
2. Zeroes are sometimes significant, and sometimes they are not.
(a) Zeroes at the beginning of a number are never significant. 0.052 has 2 sig figs.
(b) Zeroes between nonzero digits are always significant. 2007 has 4 sig figs.
(c) Zeroes at the end of the number that contains a decimal place are significant.
440.0 has 4 sig figs.
(d) Zeroes at the end of the number that does not contain a decimal point may or may
not be significant. 23000 may have 2, 3, 4 or 5 sig figs depending on how it was
measured. (OWL electronic homework will say two sig figs for this number.)
3. Exact numbers can be considered as having an unlimited number of significant
figures. Counting number of people in class would give an exact number; there should be
no uncertainty in the number.
4. In addition or subtraction, the answer contains no more decimal places than the least
number of decimal places used in the operation.
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Example 1-3
37.24 mL
+10.3 mL
47.54 mL but round to 47.5 mL
First number has two decimal places and the second number has one decimal place, thus,
the answer should have one decimal place. Put units on answer.
5. In multiplication and division, an answer contains no more significant figures than
the least number of significant figures used in the operation.
Example 1-4
(12.34 cm) × (1.23 cm) = 15.1782 cm 2 but round to 15. 2 cm2
The first number has four sig figs and the second number has three sig figs,
thus, the answer should have three sig figs. Put units on answer.
Rounding hints:
1. When the number to be dropped is less than 5, the preceding number is left
unchanged.
2. When the number to be dropped is greater than 5, the preceding number is increased
by 1.
3. When the number to be dropped is exactly 5, the preceding number is set to the
nearest even number.
4. Recommendation: If the problem you are working has several steps, wait to round
off until the final step.
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1-10 THE UNIT FACTOR METHOD (DIMENSIONAL ANALYSIS)
Example 1-5
English-English conversion
1.47 miles = ________ inches
Facts given: 1 mile = 5280 ft
1 ft = 12 inches
Therefore the plan is: change miles to feet to inches
1.47 𝑚𝑖𝑙𝑒𝑠
1
×
5280 𝑓𝑡
1 𝑚𝑖𝑙𝑒
×
12 𝑖𝑛𝑐ℎ𝑒𝑠
1 𝑓𝑡
= 93139.2 inches
But round to 93100 inches to have 3 sig figs. Or 9.31 x 104 inches.
Put units on all your answers and give the correct number of sig figs. Notice that units all
cancel, but for the inches in Example 1-5. This is a good safety check on your work; if
the units do not cancel, probably one of the conversion factors is inverted.
The conversion factors themselves should be treated as absolute numbers and therefore,
do not dictate the number of sig figs in the answer. The part that makes the problem
unique should dictate sig figs.
Example 1-6
1.10 A = _______cm
3 sig figs in 1.10 A
Fact: 1 Angstrom or 1 A = 1.0 x 10-8 cm; make a conversion factor from this fact
1.10 𝐴
1
×
1.0 𝑥 10−8 𝑐𝑚
1𝐴
= 1.10 x 10-8 cm
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Example 1-8
Metric-metric conversion
0.234 mg = __________ g = ____________Kg
0.234 𝑚𝑔
1
0.234 𝑚𝑔
1
×
×
1𝑔
1000 𝑚𝑔
1𝑔
1000 𝑚𝑔
3 sig figs in number
= 0.000234 g or 2.34 x 10-4 g
×
1 𝐾𝑔
1000 𝑔
= 0.000000234 Kg or 2.34 x 10-7
An easier way is to write:
KHD unit dcm To help you remember: (Ken’s Horse Died. Unit Diane’s Cat
Meowed.)
First letter in each word in the two sentences supplies the hints.
Where K = kilo
H = hecto
D = deka
Unit
d = deci
c = centi
m = milli
1000
Each is different from the next one by a power of 10.
100
10
1
0.1
0.01
0.001
To use this system for Example 1-8: We start at m (or mg) and want to go to the basic
unit. On the line KHD unit dcm, this would be three moves to the left. Take the decimal
point and move it three places in the same direction in the number. Thus, 0.234 mg
becomes 0.000234g We can always add zeroes.
0.234 mg = ____________Kg
On the line KHD unit dcm, this would be six moves to
the left; move the decimal point in the number, six places to the left. The answer is
0.000000234 Kg
With this system it is possible to move left or right and also across the basic unit. The
basic unit in mass would be gram, in volume it would be liter and in length it would be
meter. Most people find this is an easier way to do metric-metric conversions, rather than
the unit analysis method. The unit analysis should be used for English-English
conversions, metric-English or English-metric conversions.
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For any English-metric or metric-English, you should know the following:
Mass 1 lb = 454 g
Length 1 in = 2.54 cm
Volume 1 qt = 0.946 L
Example 1-11
English-metric conversion
1.0 gal = ________ mL There are 2 sig figs. We should expect a large answer.
Plan: gallons to quarts to liters to mL’s
Fact: 1 gal = 4 qt
1.0 𝑔𝑎𝑙
1
×
4 𝑞𝑡
1 𝑔𝑎𝑙
You will be given any English-English conversion factors needed.
×
0.946 𝐿
1 𝑞𝑡
×
1000 𝑚𝐿
1𝐿
= 3.8 x 103 mL Answer to 2 sig figs.
Is this reasonable? 1 gallon is 4 quarts and a quart is about 1 liter, thus we have about 4
liters or 4000 mL and we got 3800 mL. Thus, our answer is reasonable.
1-11 PERCENTAGE
Example 1-12
U.S. pennies since 1982 are 97.6 % zinc and 2.4 % copper.
If the mass of a penny was found to be 1.494 g, how many grams of zinc does it
contain?
97.6 𝑔 𝑧𝑖𝑛𝑐
Per cent means parts per hundred. Thus, 97.6 % is 100 𝑔 𝑜𝑓 𝑠𝑎𝑚𝑝𝑙𝑒
1.494 𝑔 𝑠𝑎𝑚𝑝𝑙𝑒
1
×
97.6 𝑔 𝑍𝑛
100 𝑔 𝑠𝑎𝑚𝑝𝑙𝑒
= 1.46 g of zinc
Answer has 3 sig figs. and the units are g of zinc.
1-12 DENSITY AND SPECIFIC GRAVITY
Density = mass/volume
Units are usually g/mL or g/cm3
density.
or D =
𝑚
𝑣
Since 1 mL = 1 cm3 , these give identical numbers for
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Example 1-13
47.3 mL of ethyl alcohol (ethanol) has a mass of 37.32 g. What is its density?
D=
𝑚
D=
This equation has three variables; two must be given in any density problem.
𝑣
37.32 𝑔
47.3 𝑚𝐿
The answer should have 3 sig figs.
D = 0.789 g/mL
Example 1-14
116 g of ethanol is needed for a chemical reaction, what volume of liquid would you measure out,
if you knew that the density was 0.789 g/mL?
D=
𝑚
This time we know D and m, but not V.
𝑣
Multiply both sides by V and we get:
Dv = m
Divide both sides by D, so that v stands alone.
v =
𝑚
𝐷
116 𝑔
0.789 𝑔/𝑚𝐿
= 147 mL
We want 3 sig figs. in answer.
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Notice that g and g cancel and that leaves 1/𝑚𝐿 a complex fraction for the units. mL comes around
on top to give us mL as the final unit.
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Example 1-16
Specific gravity is defined:
specific gravity =
𝐷 𝑜𝑓 𝑆𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒
𝐷 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
It’s a comparison to the density of
water.
(The density of water is 1.00 g/mL at 20 oC)
𝑚
D=
𝑣
If the density of sodium chloride is 2.16 g/mL, what is its specific gravity?
S. G. =
2.16 𝑔/𝑚𝐿
1.00 𝑔/𝑚𝐿
= 2.16 with no units.
Example 1-17
Battery acid is 40.0% sulfuric acid and 60.0% water by mass. Its specific gravity is 1.31.
Calculate the mass of pure sulfuric acid in 100.0 mL of the acid solution.
First change S. G. to density. D = 1.31 g/mL (Just add the units back).
100.0 mL of solution would be how many g of solution?
D=
𝑚
𝑣
D = Multiply both sides by v.
Dv = m
(1.31 g/mL)(100.0 mL) = 131 g of solution
It is 40.0 % sulfuric acid, so we take 40.0% of the 131 g of solution.
131 𝑔 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
1
×
40.0 𝑔 𝑜𝑓 𝑠𝑢𝑙𝑓𝑢𝑟𝑖𝑐 𝑎𝑐𝑖𝑑
100 𝑔 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
Notice three sig figs. in answer.
= 52.4 g of sulfuric acid
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.
1-13 HEAT AND TEMPERATURE
Heat is a form of energy. Temperature is a measure of the intensity of the heat—the
hotness or coldness of the object.
In this country, we use Fahrenheit, oF. Water freezes at 32 oF and boils at 212 oF at 1 atm
pressure.
In chemistry, we usually use the Celsius scale, oC. On this scale at 1 atm pressure, water
freezes at 0.0 oC and boils at 100.0 oC.
The equation to convert is:
solve for the other.
o
F = 1.8 oC + 32
Either oF or oC will be given and then
Example 1-18
o
100
F= ________oC
o
F = 1.8 oC + 32
Rearrange by subtracting 32 from each side and then divide both sides by 1.8,
so that C stands alone.
𝐹−32 o
= C
1.8
100−32
1.8
= 38oC
Another temperature scale is the Kelvin, K, or absolute scale.
If –273oC is the lowest temperature possible, then to avoid negative numbers, on the
Kelvin scale, this was set at 0. The size of the degree is the same on Kelvin and Celsius,
thus, K = oC + 273
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Example 1-19
400 K = ________oF
Need two equations:
K = oC + 273
or rearrange K – 273 = oC
o
400 – 273 = 127 C
o
F = 1.8oC + 32
o
F = 1.8(127) + 32 = 261 oF
1-14 HEAT TRANSFER AND THE MEASUREMENT OF HEAT
Specific heat of a substance: Amount of heat required to raise the temperature of one
gram of the substance by one degree C.
Heat can be measured in calories or in Joules.
Calorie: Amount of heat needed to raise the temperature of one gram of water from
14.5oC to 15.5oC.
One calorie = 4.18 Joules (J) to 3 sig figs.
The specific heat of water in the liquid state is: 4.18 J/g . oC
Example 1-20
How much heat is required to raise the temperature of 205 g of water from 21.2oC to
91.4oC?
Heat = (mass of substance)(specific heat of substance)(change in temperature)
Heat = (205 g)×(4.18 J/g . oC)×(70.2oC) = 60200 J to 3 sig figs. Or 6.02 x 104 J
Or the answer could be expressed as 60.2 KJ.
All units cancel, but J in the equation.
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Example 1-21
A 588 g piece of iron is heated to 97.5 oC. It is then dropped into 247 g of water originally
at 20.7 oC. When the system has equilibrated, the water and iron are both at 36.2 oC.
Calculate the specific heat of iron. Specific heat of water is 4.18 J/g . oC
We must assume that heat lost by the iron is exactly equal to the heat gained by the water.
Heat lost = heat gained
Heat lost =(mass iron)×(spec. heat of iron)×(change of temp. for the iron)
=(588)×(x)×(61.3)
Heat gained = (mass of water)(spec. heat of water)(change of temp. for the water)
=(247)×(4.18)×(15.5)
Set these two equal to each other and solve for x. x = 0.444 J/g . oC
Example 1-22
We add the same amount of heat to 10.0 g of each of the following substances starting at
20.0 o C liquid water, liquid mercury, liquid benzene and solid aluminum. Rank the
samples from lowest to highest final temperature. Use the following values for specific
heats:
H2O(l) Specific heat = 4.18 J/g C
Hg(l)
Specific heat = 0.138 J/g C
C6H6(l) Specific heat = 1.74 J/g C
Al(s) Specific heat = 0.900 J/g C.
Water < benzene < aluminum < mercury
END OF CHAPTER 1 NOTES
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