Standards for Measurement
Preparation for College Chemistry
Columbia University
Department of Chemistry
The Scientific Method
http://antoine.fsu.umd.edu/chem/senese/101/intro/scimethod-quiz.shtml
Observations
(analysis)
Laws
Hypothesis
(explanation)
Experiment
(measurement)
(analysis)
Theory
(Model)
Measurement and Interpretations
Direct Measurement
Interpretation Step
Diameter = 2.5 cm
Radius = Diameter/2
Area =  x r2 = 8.04 cm2
1
2
3
4
5
6
7
8
Art of Chemical Measurement:
• Recognize what can be measured directly
• Devise a way to obtain the desired information from
measurement data
Resting Potential (Squid Experiment)
mV
40
-65
ms
y
x
+ + + + + +
- - - - - -
Na+ K+ Na+ Na+ K+ Na+ K+
- - - - - + + + + + +
Na+ K+ Na+ Na+ K+ Na+ K+
Na+ K+ Na+ Na+ K+ Na+ K+
Na+ K+ Na+ Na+ K+ Na+ K+
Resting Potential
Experimentation
Measured Data
Basic
UNIT
Resting Potential = -65 mV
Derived
Affected by Uncertainty
• Accuracy
NUMERICAL VALUE
• Precision
• Resolution
• Noise
Significant Figures (Sig. Figs.)
The mass of an object weighed on a triple beam balance
(precision + 0.1g) is found to be 23.6 g.
This quantity contains 3 significant figures, i.e., three
experimentally meaningful digits.
If the same measurement is made with an analytical balance
(precision + 0.0001g) , the mass might be 23.5820 g (6 sig. fig.)
Evaluating Zero
Zero is SIGNIFICANT when:
Is between nonzero digits: 61.09 has four sig Figs.
Appears at the end of a number that includes a decimal point 0.500 has
three sig. Figs.; 1000. has four sig. Figs.
Zero is NON SIGNIFICANT when:
Appears before the first nonzero digit. 0.0025 has two sig. Figs. Leading
Zeros are non significant
Appears at the end of a number without a decimal point. 1,000 has
one sig. Fig.; 590 has two sig. Figs.
Exact Numbers
Defined numbers, like 12 inches in a foot, 60 minutes in an hour,
1,000mL in one liter.
Numbers that occur in counting operations.
Exact numbers have an infinite number of sig. figs.
Exact numbers do not limit the number of sig. figs. in a calculation.
Scientific Notation
Number written as a factor between 1 and 10 multiplied by
10 raised to a power.
0.0468  4.68  10 2
0.00003  3.0 10 5 (two digits)
or 3 10 5 (one digit)
Useful to unequivocally designate the significant figures.
1200  1.200  10 3 (four digits)
6,600, 000  6.6 10 6 (two digits)
Multiplication or Division
The answer must contain as many significant figures as in the
least precise quantity (measurement with least precision).
What is the density of a piece of metal weighing 36.123 g
with a volume of 13.4 mL?
Drop these three digits
mass
36.123g
d

 2.69575g / mL
volume 13.4mL
Round off to 7
ANSWER:
2.70g / mL
Addition or Subtraction
Keep only as many digits after the decimal point as there are
in the least precise quantity
Ex. Add 1.223 g of sugar to 154.5 g of coffee:
Total mass = 1.2 g + 154.5 g = 155.7 g
Addition or Subtraction
Note that the rule for addition and subtraction does not relate to
significant figures.
The number of significant figures often decreases upon subtraction.
Mass beaker + sample = 52.169 g
(5 sig. figs.)
- Mass empty beaker = 52.120 g
(5 sig. figs.)
Mass sample = 0.049 g
(2 sig figs)
SI Units Prefixes (Multiples)
Prefix
Symbol
Value
Power
exa
E
1,000,000,000,000,000,000 1018
peta
P
1,000,000,000,000,000
1015
tera
T
1,000,000,000,000,
1012
giga
G
1,000,000,000
109
mega
M
1,000,000
106
kilo
k
1,000
103
hecto
h
100
102
deka
da
10
101
SI Units Prefixes (Submultiples)
Prefix
Symbol
Value
Power
atto
a
0.000000000000000001
10-18
femto
f
0.000000000000001
10-15
pico
p
0.000000000001
10-12
nano
n
0.000000001
10-9
micro
µ
0.000001
10-6
milli
m
0.001
10-3
centi
c
0.01
10-2
deci
d
0.1
10-1
Système International (SI)
The Metric System
Physical Quantity
Name
Symbol
Length
Mass
Time
Electric current
Temperature
Amount of substance
Luminous intensity
meter
kilogram
second
ampere
kelvin
mole
candela
m
kg
s
A
K
mol
cd
Based on seven DIMENSIONALLY INDEPENDENT quantities
Length
Base unit is the meter (m)
1790s: 10-millionth of the distance from the equator to
the North Pole along a meridian.
1889: Distance between two engraved lines on a PlatinumIridium alloy bar maintained at 0°C in Sevres-France.
1960-1983: 1,650,763.73 wavelengths of the orange-red
emision of 18Kr at standard conditions
Since 1983: 1/299,792,458 of the distance traveled by
light in 1 second through vacuum.
Length
Engineering dimensions
1 km = 103 m
1 in. = 2.54cm
1 cm = 10-2 m
1 mile = 1.61km
1 mm = 10-3 m
1 µm = 10-6 m
Atomic dimensions
1 nm = 10-9 m
1 Å = 10-10 m
Mass
Base unit is the kilogram (kg)
International prototype: a platinum-Iridium cylinder
maintained in Sevres-France.
1 kg = 103 g; 1 mg = 10-3 g
A mass of 1 kg has a terrestrial weight of 9.8 newtons (2.2 lbs)
Depending on the precision required and the amount of
material, different balances are used:

The Quadruple Beam Balance (+ 10 mg)

The Top Loading Balance (+ 1 mg).

The Analytical Balance (+ 0.1 mg).
Comparison of Temperature Scales
°F = (1.8 x °C ) + 32
Fahrenheit
K = °C + 273.15
°C = (°F - 32) / 1.8
Kelvin
Celsius
212
100
Freezing point of water
32
0
373.15
100
Boiling point of water
273.15
Temperature Conversion
K = °C + 273.15
°F = (1.8 x °C ) + 32
°C = (°F - 32) / 1.8
Ex. 2.20 Convert 110°F to °C and K
°C = (68° – 32°) / 1.8 = 20°C
K = 20 + 273 = 293 K
Derived Units
Physical quantity
Name
Symbol
Definition
Frequency
Force
Pressure, stress
Energy work, heat
Electric charge
Electromotive Force
Electric Resistance
Hertz
newton
pascal
joule
coulomb
volt
ohm
Hz
N
Pa
J
C
V

s-1
m.kg.s-2
N.m-2 = m-1.kg.s-2
N.m = m2.kg.s-2
A.s
J.C-1 = m2.kg.s-3. A-1
V.A-1 = m2.kg.s-3. A-2
Measurement of Volume
Conversion Factors
2.54cm
1 in
1 in  2,54cm 
or
1 in
2.54cm
Two conversion factors
Unit needed
 Unit needed
Unit given 
Unit given
Dimensional Analysis (pp. 23 - 24)
•Read Problem. What needs to be solved for? Write it down
•Tabulate data given. Label all factors with proper units
•Determine principles involved and unit relationships
•Set up the problem deciding for the proper conversion factor
•Perform mathematical operations
•Check if the answer is reasonable
Simple, One Step Conversions
CBS News reported the barometric pressure to be 99.6 kPa.
Express this in mm Hg.
Conversion factor : 101.3 kPa = 760 mm Hg
Unit needed
pressure (mmHg) = 99.6kPa x 760 mm Hg = 747mmHg
101.3 kPa
Unit given
Unit given
Simple, One Step Conversions
A rainbow trout is measured to be 16.2 in. long. What is the
length in cm?
length in cm = 16.2 in x
2.54 cm
= 41.1 cm
1 in
Note the cancellation of units. To convert from centimeters
to inches, the conversion factor would be 1 in / 2.54 cm.
Multiple Conversion Factors
A baseball is thrown at 89.6 miles per hour. What is the speed
in meters per second?
Mile/hour
m/hour
m/s
1 mile = 1.609 km = 1.609 x 10-3 m; 1 h = 3600 s
1h
miles x 1.609 x 10–3m
= 40.0 m / s
speed = 89.6
x
1 mile
3600 s
hour
Three sig. figs.
Properties of Substances
Intensive vs. Extensive: density vs. mass
Chemical vs. Physical
Density and Solubility
Extensive Properties
Vary with the amount of material
• Mass
• Volume
• Internal Energy
• Enthalpy
• Entropy
Intensive Properties
Independent of the amount of material
Density
(mass per unit volume)
Temperature
(average energy per particle)
Chemical vs. Physical Properties
Chemical Properties
Molecules or ions undergo a change in
structure or composition
Physical Properties
Can be studied without a change in
structure or composition
Density: Conversion factor
mass
volume
An empty flask weighs 22.138 g. You pipet 5.00 mL of octane into
the flask. The total mass is 25.598 g. What is the density?
Octane amount in g:
25.598
-22.138
3.460g
3.460g
d=
= 0.692g / mL
5.00 mL
What is the volume occupied by ten grams of octane?
V = 10.00g x
1 mL
0.692g
= 14.5 mL
Solubility
Expressed as grams of solute per 100 g of solvent in the
CRC (Chemical Rubber Company) Chemistry and Physics
Handbook.
For lead nitrate in aqueous solution:
Solubility (g/100g water)
T (°C)
50
10
140
100
Solubility
How much water is required to dissolve 80 g of lead nitrate
at 100°C?
100g water
Mass water = 80g lead nitrate x
= 57g water
140g lead nitrate
Conversion factor
(from table)
Cool the solution to 10°C. How much lead nitrate remains in
solution?
50g lead nitrate
Mass of lead nitrate = 57g water x
100g water
= 28g lead nitrate
28g lead nitrate remain in solution
80g lead nitrate were initially in solution
80g – 28g = 52g lead nitrate crystallizes out of solution
Classification of Matter
Elements
Compounds
Mixtures
States of Matter
Solid
Liquid
Gas
Plasma
Matter Classification
Matter
Mixtures
Pure Substances
Homogeneous
One Phase
(Solutions)
Elements
Heterogeneous
More than one phase
Compounds
Elements’ Distribution
(earth, sea, atmosphere)
Element
Mass %
Element
Mass %
Oxygen
49.20
Chlorine
0.19
Silicon
25.67
Phosphorus
0.11
Aluminum
7.50
Manganese
0.09
Iron
4.71
Carbon
0.08
Calcium
3.39
Sulfur
0.06
Sodium
2.63
Barium
0.04
Potassium
2.40
Nitrogen
0.03
Magnesium
1.93
Fluorine
0.03
Hydrogen
0.87
Titanium
0.58
All others
0.47
Average Elemental Composition of
Human Body
Element
Mass %
Oxygen
65.0
Carbon
18.0
Hydrogen
10.0
Nitrogen
3.0
Calcium
2.0
Phosphorus
1.0
Traces of other
1.0
elements
Metals, Nonmetals, Metalloids
Main-Group Elements
Main-Group Elements
H
1
Li
3 Be
4
He
2
Transition Metals
Na
11 Mg
12
B
5 C
6 N
7 O
8
F
9 Ne
10
Al 14
13
Si 15
P 16
S 17
Cl Ar
18
19
K Ca
20 21
Sc 22
Ti 23
V 24
Cr Mn
25 26
Fe Co
27 28
Ni Cu
29 Zn
30 Ga
31 Ge
32 As
33 34
Se 35
Br Kr
36
Rb
37 38
Sr 39
Y 40
Zr Nb
41 Mo
42 43
Tc Ru
44 Rh
45 Pd
46 Ag
47 Cd
48 49
In Sn
50 Sb
51 52
Te 53
I Xe
54
Cs
55 Ba
56 La
57 Hf
72 73
Ta 74
W Re
75 Os
76 77
Ir 78
Pt Au
79 Hg
80 Tl
8l Pb
82 83
Bi Po
84 85
At Rn
86
87
Fr Ra
88 Ac
89 104
Rf 105
Ha 106
Sg 107
Bh 108
Hs 109
Mt
Inner-Transition Metals
Ce
58 59
Pr Nd
60 Pm
61 Sm
62 Eu
63 Gd
64 Tb
65 Dy
66 Ho
69 Yb
71
67 68
Er Tmi
70 Lu
Lantanides
Th
90 91
Pa 92
U Np
93 Pu
94 Am
95 Cm
96 Bk
97 98
Cf 99
Md 102
Lr
Es 100
Fm 101
No 103
Actinides
Elements that Exist as Diatomic Molecules
Element
Symbol Molecular
Normal State
Formula
Hydrogen
H
H2
Colorless gas
Nitrogen
N
N2
Colorless gas
Oxygen
O
O2
Colorless gas
Fluorine
F
F2
Pale yellow gas
Chlorine
Cl
Cl2
Yellow-green gas
Bromine
Br
Br2
Reddish-brown liquid
Iodine
I
I2
Bluish-black solid
Depending upon Bonding type
Compounds
Molecular
(Covalent bonds)
Ionic
(Coulombic forces)
Molecules
Cations
Anions
Compounds
Contain two or more elements with fixed mass percents
Covalent:
Glucose:
40.00% C
6.71% H
53.29% O
Ionic:
Sodium chloride:
39.34% Na
60.66% Cl
Information in a Chemical Formula
Ca(NO3) 2
Calcium atom
Nitrate group: two nitrate groups
Per each calcium atom
Total elements:
1 Ca
2N
6O
Allotropic Forms (Allotropes)
Graphite
Carbon
Diamond
Buckyballs (C60 )
Nanotubes
Energy
Heat: Quantitative Measurement
Energy in Chemical Changes
Is the energy available but
not being used or is it in use?
Types of Energy
Forms of Energy
Kinetic Energy (Motion Energy)
Radiant (light)
Thermal (heat)
Chemical
Energy
1 2
Ek  mv
2
(Capacity to do work)
Electrical
Mechanical
Potential Energy (Stored Energy)
o Position,
o Composition
o Condition
Law of Conservation of Energy:
In any chemical or physical change, energy can be converted
from one form to another, but it is neither created nor destroyed
Heat Energy and Specific Heat
Units of Energy:
Joule :
Amount of kinetic energy possessed by a 2kg object moving at a speed of
1m/s. Substituting these values in the equation that defines kinetic energy:
1 2
Ek  mv
2
kgm2
Joule 
s2
Equivalent to the amount of energy you will
feel if you drop 4.4 lb from about 4 in. onto your foot.
calorie (cal) :
Amount of heat energy needed to raise the temperature of one gram of
water by one degree Celsius measured between 14.5 and 15.5°C.
Units of Energy
1 cal = 4.3184 J
The joule and calorie are rather small units.
The large calorie (Cal, C) is used to express the energy
content of foods.
1C = 1kcal = 103 cal
Sprite™ contains 140 C:
1 BTU (British Thermal Unit):
1kcal = 4.3184kJ
140,000 cal of energy is released when the
soft drink is metabolized within the body.
Amount of heat needed to raise the
temperature of a lb of water one °F
1BTU =.818 kcal
Heat Capacity and Specific Heat
Joseph Black (~1750):
“Amount of heat needed to raise the temperature of a
substance by the same amount depends on the substance”
Heat Capacity
Amount of heat needed to raise the temperature of a
given quantity of substance in a specific physical state.
Specific Heat
Amount of heat needed to raise the temperature of 1 g of
a substance in a specific physical state by 1°C
Units: cal /g °C
or J/g °C
The specific heat of a substance changes when the
physical state of the substance changes
Ex.
Water (ice)
Water (steam)
2 .1 J / g °C
2 . 0 J / g °C
Water (liquid)
4 . 18 J / g °C
The higher the specific heat of a substance, the less its
temperature will change when it absorbs a given amount of heat.
metals heat up quickly, but cool quickly
At the beach, sand has a lower specific heat than
water, so it heats up while water stays cool.
Solving problems
Heat transferred = mass x Specific heat x ∆T
q = m x Cs x ∆T
1.
Amount of heat energy needed to cause a fixed amount of a substance
to undergo a specific temperature change without causing a change of
state.
2.
Transfer of heat from one body to another.
I.
Heat always flows from the warmer body to
the colder body.
II.
The heat loss by the warmer body is equal to
the heat gained by the colder body.
Generalizations:
Heat in Chemical Change
Electrolysis
Direct synthesis
H2 + O 2
Potential energy
Potential energy
H2 + O 2
H2 O
H2 O
time
Potential Energy Diagrams
time