CHAPTER 1:
CHEMISTRY AND MEASUREMENT
Vanessa Prasad-Permaul
Valencia College
CHM 1045
1
Properties of Matter
 Chemistry: The study of composition,
properties, and transformations of matter
 Matter: Anything that has both mass &
volume
 Hypothesis: Interpretation of results
 Theory: Consistent explanation of
observations
2
Conservation of Mass
 Law of Mass Conservation: Mass is neither
created nor destroyed in chemical reactions.
3
Example 1: Conservation of Mass
C(s) + O2(g)  CO2(g)
a)
12.3g C reacts with 32.8g O2, ?g CO2
12.3g + 32.8g = 45.1g
a)
0.238g C reacts with ?g O2 to make .873g CO2
0.238g + x = 0.873g = 0.873g-0.238g = 0.635g of O2
a)
?g C reacts with 1.63g O2 to make 2.24g CO2
x + 1.63g = 2.24g = 2.24g - 1.63g = 0.61g C
4
Example 1: Conservation of Mass
Exercise 1.1
1.85g of wood is placed with 9.45g of air in a sealed vessel. It is
heated and the wood burns to produce ash and gases. The ash
is weighed to yield 0.28g. What is the mass of the gases in the
vessel?
1.85g Wood + 9.45g Air
heat
0.28g Ash + ? g gases
1.85 + 9.45 - 0.28 = 11.02g of gases
What is the mass of wood that is converted to gas by the end of
the experiment?
1.85g of Wood – 0.28g of ash = 1.57g
5
Matter
 Matter is any substance that has mass and
occupies volume.
 Matter exists in one of three physical states:
 solid
 liquid
 gas
6
Solid
 In a solid, the particles of matter are tightly
packed together.
 Solids have a definite, fixed shape.
 Solids cannot be compressed and have a
definite volume.
 Solids have the least energy of the three states
of matter.
7
Liquid
 In a liquid, the particles of matter are loosely
packed and are free to move past one another.
 Liquids have an indefinite shape and assume the
shape of their container.
 Liquids cannot be compressed and have a
definite volume.
 Liquids have less energy than gases but more
energy than solids.
8
Gases
 In a gas, the particles of matter are far apart
and uniformly distributed throughout the
container.
 Gases have an indefinite shape and assume the
shape of their container.
 Gases can be compressed and have an
indefinite volume.
 Gases have the most energy of the three states
of matter.
9
Phases
10
Properties of Matter
 A physical change is a change in the form of matter
but not in its chemical identity
 A chemical change or a chemical reaction is a change
in which one of more kinds of matter are transformed
into a new kind of matter or several new kinds of
matter
11
Properties of Matter
 Physical Properties can be determined without
changing the chemical makeup of the sample.
 Some typical physical properties are:
 Melting Point, Boiling Point, Density, Mass, Touch,
Taste, Temperature, Size, Color, Hardness,
Conductivity.
 Some typical physical changes are:
 Melting, Freezing, Boiling, Condensation, Evaporation,
Dissolving, Stretching, Bending, Breaking.
12
Properties of Matter
 Chemical Properties are those that do change
the chemical makeup of the sample.
 Some typical chemical properties are:
 Burning, Cooking, Rusting, Color change, Souring of
milk, Ripening of fruit, Browning of apples, Taking a
photograph, Digesting food.
 Note: Chemical properties are actually chemical
changes
13
Properties of Matter
Exercise 1.2
Potassium (K) is a soft, silvery-colored metal that melts @ 64oC.
It reacts vigorously with water (H2O), Oxygen (O2) and Chlorine
(Cl2).
Identify all physical properties:
• Soft
• Silvery-colored
• Melting point of 64oC
Identify all chemical properties:
• Metal (its chemical identity)
• K reacts vigorously with H2O
• K reacts vigorously with O2
• K reacts vigorously with Cl2
14
Classifications of Matter
 Matter can be divided into two classes:
 mixtures
 pure substances
 Mixtures are composed of more than one
substance and can be physically separated into
its component substances.
 Pure substances are composed of only one
substance and cannot be physically separated.
15
Pure Substances
 There are two types of pure substances:
 Compounds
 Elements
 A compound is a substance composed of two or
more elements chemically combined
 Compounds can be chemically separated into
individual elements.
 Water is a compound that can be separated into
hydrogen and oxygen.
 An element cannot be broken down further by
chemical reactions.
16
Dalton’s Atomic Theory
 Law of Definite Proportions: Different
samples of a pure chemical substance always
contain the same proportion of elements by
mass.
 Any sample of H2O contains 2 hydrogen atoms for
every oxygen atom
17
Mixtures
 There are two types of mixtures:
 homogeneous mixtures
 heterogeneous mixtures
 Homogeneous mixtures have uniform
properties throughout.
 Salt water is a homogeneous mixture.
 Heterogeneous mixtures do not have uniform
properties throughout.
 Sand and water is a heterogeneous mixture.
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Example 2: Matter
Which of the following represents a mixture?
19
20
Accuracy, Precision, and Significant Figures in
Measurement
 Accuracy is how close
to the true value a
given measurement is.
 Precision is how well a
number of independent
measurements agree
with one another.
21
Example 8: Accuracy & Precision
 Which of the following is precise but not
accurate?
22
Accuracy, Precision, and Significant Figures in
Measurement
 Significant Figures are the total number of
digits in the measurement.
 The results of calculations are only as
reliable as the least precise measurement!!
 Rules exist to govern the use of significant
figures after the measurements have been
made.
23
Accuracy, Precision, and Significant Figures in
Measurement
 Rules for Significant Figures:
 Zeros in the middle of a number are significant
 Zeros at the beginning of a number are not
significant
 Zeros at the end of a number and following a
period are significant
 Zeros at the end of a number and before a period
may or may not be significant.
24
Example 4: Significant Figures
How many Significant Figures ?
a) 0.000459 = 3
b) 12.36 = 4
c) 36,450 = 4
d) 8.005 = 4
e) 28.050 = 5
25
Accuracy, Precision, and Significant Figures in
Measurement
 Rules for Calculating Numbers:
 During multiplication or division, the answer can’t
have more significant figures than any of the
original numbers.
26
Example 5: Significant Figures
a) 218.2 x 79 = 17237.8 = 1.7 x 104
a) 12.5 / 0.1272 = 94.33962264150943 = 94.3
b) 0.2895 x 0.29 = 0.083955 = 0.084
c) 32.567 / 22.98 = 1.417188859878155 = 1.417
27
Accuracy, Precision, and Significant Figures in Measurement
-During addition or subtraction, the answer
can’t have more digits to the right of the
decimal point than any of the original numbers.
28
Example 6: Significant Figures
a) 218.2 + 79 = 297.2 = 297
b) 12.5 - 0.1272 = 12.3728 = 12.4
c) 0.2895 + 0.29 = 0.5795 = 0.58
d) 32.567 - 22.98 = 55.547 = 55.55
e) 185.5+2.224 = 187.724 = 187.7
29
Accuracy, Precision, and Significant Figures in
Measurement
 Rules for Rounding Numbers:
 If the first digit removed is less than 5
 round down (leave # same)
 If the first digit removed is 5 or greater
 round up
 Only final answers are rounded off, do not round
intermediate calculations
30
Example 7: Rounding and Significant Figures
Round off each of the following measurements
a) 3.774499 L to 4 sig. figs. = 3.774L
b) 255.0974 K to 3 sig. figs. = 255K
c) 55.265 kg to 4 sig. figs. = 55.27kg
d) 1.2151ml to 3 sig. figs. = 1.22ml
e) 1.2143g to 3 sig. figs. = 1.21g
31
SIGNIFICANT FIGURES
Exercise 1.3
Give answers to the following arithmetic setups. Round to the
correct number of significant figures:
a)
5.61 x 7.891
9.1
= 4.864671
b)
8.91 - 6.435 =
c)
6.81 – 6.730
d)
38.91 x (6.81-6.730) = 38.91 x 0.08 = 3.1128 = 3
2.475
= 0.08
= 4.9
= 2.48
= 0.08
32
Scientific Notation
 Changing numbers into scientific notation
 Large # to small #
 Moving decimal place to left, positive exponent
123,987 = 1.23987 x 105
 Small # to large #
 Moving decimal place to right, negative exponent
0.000239 = 2.39 x 10-4
How to put into calculator
33
Example 3: Scientific Notation
Put into or take out of scientific notation
a)
1973 = 1.973 x 103
b) 5.5423 x 10-4 = 0.00055423
c)
0.775 = 7.75 x 10-1
d) 3.55 x 107 = 35,500,000
e) 8500 = 8.5 x 103
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Measurement and Units
SI Units
Physical Quantity
Mass
Length
Temperature
Amount of substance
Time
Electric current
Luminous intensity
Name of Unit
kilogram
meter
kelvin
mole
second
ampere
candela
Abbreviation
kg
m
K
mol
s
A
cd
35
Measurement and Units
Some prefixes for multiples of SI units
*
*
*
*
*
*
*
Factor
1,000,000,000 = 109
1,000,000 = 106
1,000 = 103
100 = 102
10 = 101
0.1 = 10-1
0.01 = 10-2
0.001 = 10-3
0.000,001 = 10-6
0.000,000,001 = 10-9
0.000,000,000,001 = 10-12
Prefix
giga
mega
kilo
hecto
deka
deci
centi
milli
micro
nano
pico
Symbol
G
M
k
h
da
d
c
m
µ
n
p
* Important
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Exercise 1.4
Express the following quantities using an SI prefix and a
base unit. For instance, 1.6 x 10-6m = 1.6mm. A quantity such
as0.000168g could be written 0.168mg or 168mg.
1.84 x 10-9 m = 1.84 nm (nanometer)
5.67 x 10-12 s = 5.67 ps (picosecond)
7.85 x 10-3 g = 7.85 mg (milligram)
9.7 x 103 m = 9.7 km (kilometer)
0.000732 s = 0.732 ms (millisecond)
= 732us (microsecond)
f) 0.000000000154 m = 0.154nm (nanometer)
= 154pm (picometer)
a)
b)
c)
d)
e)
37
Changes in Physical State
 Most substances can exist as either a solid,
liquid, or gas.
 Water exists as a solid below 0 °C; as a liquid
between 0 °C and 100 °C; and as a gas above
100°C.
 A substance can change physical states as the
temperature changes.
38
Solid  Liquid
 When a solid changes to a liquid, the phase
change is called melting.
 A substance melts as the temperature increases.
 When a liquid changes to a solid, the phase
change is called freezing.
 A substance freezes as the temperature decreases.
39
Liquid  Gas
 When a liquid changes to a
gas, the phase change is
called vaporization.
 A substance vaporizes as
the temperature increases.
 When a gas changes to a
liquid, the phase change is
called condensation.
 A substance condenses as
the temperature decreases.
40
Solid  Gas
When a solid changes directly to a gas,
the phase change is called sublimation.
A substance sublimes as the temperature
increases.
When a gas changes
directly to a solid, the phase
change is called deposition.
A substance undergoes
deposition as the
temperature decreases.
41
Temperature
Diagram of the various phases of temperature change
42
Temperature
 Temperature
Conversions:
The Kelvin and Celsius
scales have equal size
units (a
change of 1oC is
180oF
100 oC
100 K
equivalent to a change
of 1K)
43
 Temperature Conversions:
 Celsius (°C) — Kelvin (K) temperature conversion:
Kelvin (K) = t°C x 1K + 273.15K
1oC
 Fahrenheit (°F) — Celsius (C) temperature conversions:
there are exactly 9oF for every 5oC. Knowing that 0oC =
32oF
tF = tC x 9oF + 32
5o C
tC = 5oC x (toF – 32)
9oF
44
Example 9: Temp. Conversions
Carry out the indicated temperature conversions:
a) –78°C = ? K = (-78oC x 1K/1oC) +273.15K = 195.15 = 195K
b) 158°C = ? °F = (158oC x 9oF/5oC)+32oF = 316.4 = 316oF
c) 373.15 K = ? °C = (373.15K x 1oC/1K)– 273.15K = 100K
d) 98.6°F = ? °C = 5oC/9oF x (98.6oF – 32oF) = 37oC
e) 98.6°F = ? K = (37oC x 1K/1oC) +273.15K = 310.15 = 310K
45
Exercise 1.5
A person with a fever has temperature of 102.5oF.
What is this temperature in oC? A cooling mixture
of dry ice and isopropyl alcohol has a temperature
of -78 oC. What is the temperature in kelvins?
a) oC = 5oC x (oF – 32 ) = 0.555 x (102.5 – 32) = 39.2oC
9 oF
b) K = oC + 273.15 = -78 + 273.15 = 195 K
46
Volume
 Volume: how much three-dimensional space a
substance (solid, liquid, gas) or shape occupies or
contains often quantified numerically using the SI
derived unit (m3) the cubic meter.
 The volume of a container is generally understood to be
the capacity of the container, i. e. the amount of fluid
(gas or liquid) that the container could hold, rather than
the amount of space the container itself displaces.
47
Volume
 units of Volume:
 m3 or cm3 (cc)
 Traditionally chemists use liter (L)
 1cm3 = 1cc = 1mL
48
Measurement and Units
 Density: relates the mass of an object to its
volume.
Density = mass / Volume
D = m /V
V=m/D
m =V  D
 Density decreases as a substance is heated
because the substance’s volume increases.
49
Density
What is the density of glass (in mL) if a sample
weighing 26.43 g has a volume of 12.40 cm3?
d =?
m = 26.43 g
V = 12.40 cm3 = 12.40 mL
d = m = 26.43 g = 2.13145 = 2.131 g/mL
V
12.40 mL
50
Density
What is the volume of an unknown solution if the
mass is 12.567 g and the density is 14.621 g/mL ?
d = m/V
Vxd=m
V = m/d
V = 12.567 g / 14.621 g/mL = 0.85952 mL
12.567g x
1mL
14.621g
= 0.85952 mL
51
Density
What is the mass of an unknown solution if the
volume is 20.2 mL and the density is 2.613 g/mL?
d = m/V
m = d xV
m = 2.613g x 20.2 mL = 52.7826 = 52.8 g
mL
52
Exercise 1.6
A piece of metal wire has a volume of 20.2 cm3 and a
mass of 159 g. What is the density of the metal?
D = m = 159 g = 7.87128712 = 7.87 g /cm3
V
20.2cm3
We know that the following metals have the following
densities. Which metal is the wire made of?
Mn = 7.21 g/cm3
Fe = 7.87 g/cm3
Ni = 8.90 g/cm3
53
Exercise 1.7
Ethanol (grain alcohol) has a density of 0.789 g/cm3.
What volume (mL) of ethanol must be poured into
a graduated cylinder to equal 30.3 g?
d = m/V
Vxd=m
V=m/d
V = 30.3 g x 1 cm3 = 38.4cm3
0.789 g
54
Dimensional Analysis & Units
 Dimensional-Analysis method uses a conversion
factor to express the relationship between units.
Original quantity x conversion factor = equivalent
quantity
Example: express 2.50 kg  lb.
Conversion factor: 1.00 kg = 2.205 lb
2.50 kg x 2.205 lb = 6.00 lb
1.00 kg
55
56
Exercise 1.8
The oxygen molecule (O2) consists of two oxygen
atoms a distance of 121 pm apart. How many
millimeters (mm) is this distance?
121 pm x 10-12 m x 1mm = 1.21 x 10-7 mm
1 pm
10-3
57
Exercise 1.9
A large crystal is constructed by stacking small identical
pieces of crystal. A unit cell is the smallest piece from
which a crystal can be made. A unit cell of a crystal of
gold metal has a volume of 67.6 A3. What is the volume
in dm3?
3
3
67.6 A3 x 10-10 m x 10 dm = 6.76 x 10 -26 dm3
1 A3
1m
58
Exercise 1.10
Using the following definitions, obtain the
conversion factor for yards to meters. How many
meters are there in 3.54 yd?
1 in = 2.54cm (exactly)
1 yd = 36in (exactly)
1 yd x 1 in x 1 cm = 1.093613298 = 1.094 yd/m
36 in 2.54 cm 10-2 m
3.54 yd x 1 m
= 3.24 m
1.094 yd
59
Conversions
a) 1.267 km  m  cm
1.267km x 1000m x 100cm = 126700cm = 1.267 x 105
1km
1m
b) 0.784 L  mL
0.784L x 1000mL = 784L
1L
c) 3.67 x 105 cm  in
3.67 x 105cm x 1in = 144488.1889in = 1.44 x 105in
2.54cm
60
Conversions
d) 79 oz  g
79oz x 28.35g = 2239.65g = 2.2 x 103g
1oz
e) 9.63 x 10-3 yd  ft
9.63 x 10-3yd x 1m x 1km x 0.62137mile x 5280ft
1.0936yd 1000m
1km
1mile
= 0.0289ft
f) 23.5 cm2  m2
23.5cm2 x 1m2 = 0.235m2
100cm2
61
Conversions
g) 1.34 x 1012 pm  m
1.34 x 1012pm x 1m
= 1.34 x 1024m
10-12pm
h) 4.67 x 10-7 nm  pm
4.67 x10-7nm x 1m x 10-12pm = 4.67 x 10-12pm
10-9nm
1m
62