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Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Introductory Chemistry, 3rd Edition
Nivaldo Tro
Chapter 2
Measurement and
Problem Solving
2009, Prentice Hall
Graph of global
Temperature rise in 20th
Century. Cover page
Opposite page 11.
Outline
Tro's "Introductory Chemistry",
Chapter 2
2
2.1 Measuring Global Temperature
2.2 Scientific Notation: Writing Large and Small Numbers
2.3 Significant Figures: Writing Numbers to Reflect Precision
2.4 Significant Figures in Calculations
2.5 The Basic Units of Measure
2.6 Converting One Unit to Another
2.7 Solving Multistep Conversions
2.8 Units Raised to a Power
2.9 Density
2.10 Numerical Problem Solving Strategies and the
Solution Map
2.1 Measuring Global
Temperature
Tro's "Introductory Chemistry",
Chapter 2
3
Tro's "Introductory Chemistry",
Chapter 2
4
Scientists have Measured the Average
Global Temperature Rise over the Past
Century to be 0.6 °C
• 0.6 – a number, quantitative:between 0.5 and 0.7 °C.
• °C – unit of a standardized scale.
• 0.60 °C - indicates a number between 0.59 and 0.61°C
• Number + unit = a measurement
Chapter 2 -Measurement and
Problem Solving Topics in this
Chapter.• Measurements
• Precision
• Basic Units
• Converting Measurements
• Derived Units
Tro's "Introductory Chemistry",
Chapter 2
5
Tro's Introductory Chemistry, Chapter
2
6
Scientific Notation
A way of writing large and small
numbers.
A small number :
Radius of a hydrogen nucleus is about
0.00000000000005 m
(13 zeros between decimal and 5)
A large number:
Distance to nearest star other than the sun is
24,500,000,000,000 miles
24.5 trillion miles
(245 + 11 zeroes)
Tro's "Introductory Chemistry",
Chapter 2
7
Scientific Notation
• Expresses a number as a number between 1
and 10 multiplied by a power of 10.
• Example: 3.89 x 105 = 389,000
• More compact way of writing large or small
numbers
• Easier to compare numbers
• Easier to determine precision
Tro's "Introductory Chemistry",
Chapter 2
8
Exponents or Powers of Ten
• Positive exponent = larger number
• Negative exponent = smaller number
• Write out powers
10 = 101 1/10 = 10-1
100 =102 1/100 = 10-2
1,000= 103 1/1,000 = 10-3
10,000 = 104 1/10,000 = 10-4
1 = 100
2.2 Scientific Notation: Writing
Large and Small Numbers
Tro's "Introductory Chemistry",
Chapter 2
9
Tro's "Introductory Chemistry",
Chapter 2
10
Scientific Notation
• Parts of a number in scientific notation
Figure from the center of page 12 showing the parts
Of a number written in scientific notation
exponent
1.35 x 10-8
exponential part (10)
decimal point of number between 1 and 10
Tro's "Introductory Chemistry",
Chapter 2
11
Writing Numbers in Scientific Notation
for Numbers Greater than 1. 1. Move the decimal to give a number between 1 and 10. Only
one digit to the left of the decimal.
2. Decimal from right to left, the number is now smaller than
the original number so you must multiply by a number
greater than 1. This means increase in the exponent.
3. Example 34,780 move decimal 4 places to left ()
4. 34,780 = 3.4780 x 10,000 = 3.4780 x 104
5. Decimal to left, increase exponent, the number of places you
moved the decimal point.
Our Examples (1)• A large number:
• Distance to nearest star other than the sun is
• 24,500,000,000,000 miles
• 24.5 trillion miles
• (245 + 11 zeroes)
• To get a number between 1 and 10 the decimal is moved --
(left) by 13 places. 2,45 is smaller than the actual
number so we need a positive exponent.
• 2.45. x 10+13 miles
Tro's "Introductory Chemistry",
Chapter 2
12
Tro's "Introductory Chemistry",
Chapter 2
13
Writing Numbers in Scientific Notation
for Numbers Less than 1.
1. Move the decimal to give a number between 1 and 10. Only
one digit will be to the left of the decimal.
2. Decimal from left to right, the number is now larger than the
original number so you must multiply by a number less than
1. This means decrease in the exponent.
3. Example 0.0067 move decimal 3 places to right ()
4. 0.0067 = 6.7 x 1/1,000 = 6.7 x 10-3
5. Decimal to right, decrease exponent, the number of places
you moved the decimal point.
Our Examples (2)
• A small number :
• Radius of a hydrogen nucleus is about
• 0.00000000000005 m
• (13 zeros between decimal and 5)
• To get a number between 1 and 10 the decimal is moved --
(right) by 14 places. 5. is bigger than the actual number
so we need a negative exponent.
• 5. x 10-14 m
Tro's "Introductory Chemistry",
Chapter 2
14
Tro's "Introductory Chemistry",
Chapter 2
15
Writing a Number in Scientific Notation,
Other Examples Continued
• The U.S. population in 2007 was estimated
to be 301,786,000 people. Express this
number in scientific notation.
• 301,786,000 people = 3.01786 x 108 people
Tro's "Introductory Chemistry",
Chapter 2
16
Example with a Number that
Already has a Power of Ten
• 458.98 x 105
• Move decimal 4.5898 This is left () 2 places, so exponent increase by 2 (5 +2 = 7)
• 4.5898 x 107
Tro's "Introductory Chemistry",
Chapter 2
17
Inputting Scientific Notation into a Calculator
• Learn how your calculator handles scientific
notation.
• Usually through a key marked EXP
Tro's "Introductory Chemistry",
Chapter 218
Reporting Measurements
• Measurements depend upon the equipment
being used.
• All the digits except the last are certain.
• The last is an estimate.
24 25 cm24.7? cm ? Is a guess 24.72cm
3 certain one
digits + guess
= four significant digits
Range is 24.71 to 24.73 cm
2.3 Significant Figures: Writing
Numbers to Reflect Precision
Tro's "Introductory Chemistry",
Chapter 2
19
Tro's "Introductory Chemistry",
Chapter 2
20
Significant Figures
Writing numbers to reflect precision.
Why do we need significant figures?
All measures have some degree of
uncertainty.
Precision Increased by Dividing?
Consider traveling 237 miles in 3.7 hours.
237 mi
3.7 h
Tro's "Introductory Chemistry",
Chapter 2
21
= 64.05405405405 mi/h
Significant Figures in a Single
Measurement
• Nonzero numbers are significant (23.89 cm)
• Zeros
Interior zeros – always significant (1.5067 in)
Trailing zeros after a decimal – always significant
(45.900 L)
Leading zeros – never significant (0.0000456 m)
Trailing zeros before a decimal – ambiguous,
avoid by using scientific notation.
\
Tro's "Introductory Chemistry",
Chapter 2
22
Exact Numbers have Unlimited
Significant figures
• Counted numbers (3 trials of an experiment)
• Defined numbers (1 in = 2.54 cm)
• Numbers as part of a formula (2pr = C)
Tro's "Introductory Chemistry",
Chapter 2
23
Tro's "Introductory Chemistry",
Chapter 2
24
Examples of —Determining the Number of
Significant Figures in a Number
2.4 Significant Figures in
Calculations
Tro's "Introductory Chemistry",
Chapter 2
25
Tro's "Introductory Chemistry",
Chapter 2
26
Multiplication and Division with
Significant Figures
• The overall answer is limited by the measurement
with the fewest number of significant figures.
5.02 × 89,665 × 0.10 = 45.0118 = 45
3 sig. figs. 5 sig. figs. 2 sig. figs. 2 sig. figs.
5.892 ÷ 6.10 = 0.96590 = 0.966
4 sig. figs. 3 sig. figs. 3 sig. figs.
Tro's "Introductory Chemistry",
Chapter 2
27
Rounding
• Find the number that is the last significant figure. Look at the number to the right.
1. For 0 to 4, keep the last significant figure the same .
34.8244 to four significant figures 34.82
1. For 5 to 9, add 1 the last significant figure the same.
45.967346 to four significant figures 45.97
Tro's "Introductory Chemistry",
Chapter 2
28
Examples of Multiplication and Division
Tro's "Introductory Chemistry",
Chapter 2
29
Addition and Subtraction with
Significant Figures
• The overall answer is limited by the measurement
with the fewest number of decimal places.
5.74 2 decimal places
+0.823 3 decimal places
+2.651 3 decimal places
9.214 2 decimal places rounds to 9.21
Tro's "Introductory Chemistry",
Chapter 2
30
Both Multiplication/Division and
Addition/Subtraction with
Significant Figures.
• Follow the standard order of operations.
Please Excuse My Dear Aunt Sally.
Keep track of number of significant figures at each step, but round at the end. Example next page.
- n
Tro's "Introductory Chemistry",
Chapter 2
31
Example 1.6—Perform the Following Calculations
to the Correct Number of Significant Figures
4555.30015.45120.010.1 a)
5820.100
1.105
355.0
b)For addition and subtraction, it is useful to draw a line
To help determine the correct decimal place.
Addition or Subtraction and
Multiplication and Division
Together
Tro's "Introductory Chemistry",
Chapter 2
32
Do the steps in parentheses and apply the rules to determine the
Correct number of significant figures or decimal p[laces in the
Intermediate answer.
Keep track of significant figures and only round at the last step.
Both Multiplication/Division and
Addition/Subtraction with
Significant Figures (cont.)
33
12.95 × (12.96 – 11.346)
2 right 3 right
decimal dcecimal
12.95 × (1.614)
2 right decimal (by subt. rules)
12.95 × (1.614) = 20.9013 20.9
four three three
sig. fig sig. fig. sig. fig
(by multip. Rules)
Tro's "Introductory Chemistry",
Chapter 2
34
Example 1.6—Perform the Following Calculations
to the Correct Number of Significant Figures,
Continued
5379904.5233.4526755.45299870.3562.4 c)
1.0142.002.855.084.14 d)
Tro's "Introductory Chemistry",
Chapter 2
35
2.5 The Basic Units of Measure
Tro's "Introductory Chemistry",
Chapter 2
36
Measurement = Number + Unit• Units - agreed upon standardized quantities
• Scientists use the SI system which is a form of the
metric system
• SI = Système International
• Base units are the units for fundamental quantities
Tro's "Introductory Chemistry",
Chapter 2
37
Quantity Unit Symbol
Length meter m
Mass kilogram kg
Time second s
Temperature kelvin K
Base Units in the SI System
38
Length• Measure of a 1-D distance
• SI unit = meter = 39.37 in, 3.37 in longer than a yard
• Commonly used centimeters (cm) = 1/100 of a meter. 1 inch = 2.54 cm (exactly, by definition)
A figure showing a meterstick slightly longer than a yardstick
Tro's "Introductory Chemistry",
Chapter 2
39
Mass• Measure of the amount of matter present
in an object.
• SI unit = kilogram (kg) = 2.2 lb
• A replica of the standard mass is shown
• Commonly measure in the lab mass in grams (g) or milligrams (mg).
• 1 kg = 1,000 g = 1,000,000 mg
• 1 oz = 23.35 g
• 1 lb = 453.6 g
• A nickel (5 cent piece) weighs about 5 g
Standard
Mass
Figure 2.6
Top loading
Balance
Page 23
Tro's "Introductory Chemistry",
Chapter 2
40
Time
• Measure of the duration of an event.
• SI units = second (s)
• 1 s is as a certain number of repeats
of a Cs atom in a so called atomic
clock.
Tro's "Introductory Chemistry",
Chapter 2
41
Temperature• Measure of the average energy of
motion of molecules in a sample.
• Is a result of molecular motions
• Faster molecular motions = higher
temperature
• Heat flows from warmer objects to
cooler objects through molecular
collisions until they reach the same
temperature.
• Unit is Kelvin (K). Water boils at 373.15
K, and freezes at 272.15 K
Three figures
Student in 22 °C
Room temp.
Ice at 0 °C
Death Valley at
40 ° C
Tro's "Introductory Chemistry",
Chapter 2
42
Nonbase Units in the
SI System
• Units in SI:
Prefix (a multiplier) + base unit
Example centimeter = centi + meter = centimeter0.01 x meter = 0.01 meter
Tro's "Introductory Chemistry",
Chapter 2
43
Common Prefixes in the
SI System (see the stuff to memorize on the
web page)
Prefix SymbolDecimal
EquivalentPower of 10
mega- M 1,000,000 Base x 106
kilo- k 1,000 Base x 103
deci- d 0.1 Base x 10-1
centi- c 0.01 Base x 10-2
milli- m 0.001 Base x 10-3
micro- m or mc 0.000 001 Base x 10-6
nano- n 0.000 000 001 Base x 10-9
Tro's "Introductory Chemistry",
Chapter 2
44
Prefixes Indicate What Power of 10 the
Base Unit has been Multipied by
• Kilometer = kilo (1000) + meter = 1000 x meter (km)
• Nanosecond = nano (10-9) + second = 10-9 x second (ns)
• Milligram = milli (10-3) + gram = 10-3 x gram (mg)
• If the multiplier is greater than 1 the new unit will be
larger than the base unit.
• If the multiplier is less than 1 the new unit will be
smaller than the base unit.
Tro's "Introductory Chemistry",
Chapter 2
45
Volume, an Example of
a Derived Unit
• Derived unit. A combination of the base units.
• This may be a multiplication, division, addition, subtraction or other mathimatical combination
• Examples Area = length2 Volume = length3
• SI system = Area = square meter (m2)
Volume = cubic meter (m3)
Graduated
Cylinder
A cube 10 cm
On a side which
Is the volume =
1 Liter
More Convenient
Volume Units
• Liter (L) is defined as a cubic decimeter (dm3).
• A dm = 10 cm so a Liter is a cube with 10 cm sides.
• 10 cm x 10 cm x 10 cm = 1000 cm3
• So a Liter also equals 1000 cm3
• Milliliter = 1/1000 L = 1 cm3
• REMEMBER: 1 mL = 1 cm3
• Solids usually use cm3 and liquids usually use mL
Tro's "Introductory Chemistry",
Chapter 2
46
Tro's "Introductory Chemistry",
Chapter 2
47
Table 2.3 (page 24) Common Units
and Their Equivalents
Length
1 kilometer (km) = 0.6214 mile (mi)
1 meter (m) = 39.37 inches (in.)
1 meter (m) = 1.094 yards (yd)
1 foot (ft) = 30.48 centimeters (cm)
1 inch (in.) = 2.54 centimeters (cm) exactly
Tro's "Introductory Chemistry",
Chapter 2
48
Table 2.3 (cont.) Common Units and
Their Equivalents, Continued
Volume
1 liter (L) = 1000 milliliters (mL)
1 liter (L) = 1000 cubic centimeters (cm3)
1 liter (L) = 1.057 quarts (qt)
1 U.S. gallon (gal) = 3.785 liters (L)
Mass
1 kilogram (km) = 2.205 pounds (lb)
1 pound (lb) = 453.59 grams (g)
1 ounce (oz) = 28.35 (g)
2.6 Converting One Unit to
Another
Tro's "Introductory Chemistry",
Chapter 2
49
•Metric metric (easy, a power of 10)
•Metric English (a little harder, a conversion factor)
•Arithmetic operations apply to numbers and units
•Keep track of units by dimensional analysis
Dimensional Analysis
• Using units to help solve problems is called
dimensional analysis
• Units may be operated on mathematically
like numbers
Tro's "Introductory Chemistry",
Chapter 2
50
SI to SI Conversions
• This involves a change in the unit that
differs by a power of 10. The number will
change by moving the decimal.
• A way to guide you through the problem is
a plan that your text calls solution map.
• A solution map lists the steps one mus go
through to solve the problem
Tro's "Introductory Chemistry",
Chapter 2
51
Example SI to SI
• How many mm are there in 34.9 m?
• Solution map:
m mm
• How do we make the change? Using a
conversion factor
Tro's "Introductory Chemistry",
Chapter 2
52
Conversion Factor
• A conversion factor is derived from an
equation. It is a fraction = 1
• Multiplying by a conversion factor does not
change a quantities size, but does change
the unit used to measure it.
• Example from the table 1 mm = 0.001 m
• 1mm 0.001 m
0.001 m 1mmTro's "Introductory Chemistry",
Chapter 2
53
Conversion Factors (cont)
• Notice the equation gives two conversion
factors.
• The one to choose is determined by which
unit is supposed top cancel out
• Back to our problem 34.9 m is how many
mm?
Tro's "Introductory Chemistry",
Chapter 2
54
Conversions Convert an Old Unit
to a New Unit
new unit
• old unit × = new unit
old unit
• In the solution map
• Solution map: m mm
• m = old unit and mm = new unit
55
Carrying out the Solution
1 mm
• 34.9 m × = 34,900 mm
0.001 m 3.49 x 104 mm
• Writing in scientific notation to give 3 sig. fig.
Tro's "Introductory Chemistry",
Chapter 2
56
Problem to Work in Class
• How many kg are there in 4,392 mm?
• Solution map:
mm m km
• Could be done in one step, but it is easy to
convert to the base unit.
Tro's "Introductory Chemistry",
Chapter 2
57
SI to Other Systems of Measure
• A piece of notebook paper is 8.50 in wide,
what is that length in cm?
• Solution map:
in cm
• Need a conversion factor from the
definition 1 in = 2.54 cm
• 1in or 2.54 cm
2.54 cm 1in
Tro's "Introductory Chemistry",
Chapter 2
58
Solution
• 8.50 in × 2.54 cm = 21.59 cm
1 in 21.6 cm
( 3 sig. fig. )
Tro's "Introductory Chemistry",
Chapter 2
59
2.7 Solving Multistep
Conversions
Tro's "Introductory Chemistry",
Chapter 2
60
Some Conversions Require Two
or More Conversion Factors
• Given that 2 Tablespoons (tbsp) = 1 oz, 8 oz
= 1 cup, 4 cup = 1 qt, and 1 L = 1.097 qt
• How many mL are there in 1.00 tbsp?
• Solution map:
tbsp oz cupqtLmL
• Each step represents a conversion. 5 steps.
(Solution next slide)
Tro's "Introductory Chemistry",
Chapter 2
61
• 1.00 tbsp × 1 oz = 0.500 oz
2 tbsp
• 0.500 oz × 1 cup = 0.0625 cup
8 oz
• 0.0625 cup × 1 qt = 0.015625 qt
4 in
• 0.015625 qt × 1 L = 0.0142339 L
1.097 in
• 0.0142339 L × 1000 mL = 14.2339 mL
1 L 14.2 mL
Tro's "Introductory Chemistry",
Chapter 2
62
2.8 Units Raised to a Power
Tro's "Introductory Chemistry",
Chapter 2
63
Units Raised to a Power
• An injection of a drug has a volume of 4.52
cm3. What is the volume in cubic inches
(in3)?
• Solution map: cm3 in3
• 4.52 cm3 × 1 in = 1.78 in3 (NO!!!!)
2.54 cm
• What does cm3 really mean
Tro's "Introductory Chemistry",
Chapter 2
64
Units Raised to a Power
• 4.52 cm3 = 4.53 cm × cm × cm
• For squared or cubed units, one must also
square or cube the entire conversion factor
• 4.52 cm3 × 1 in × 1 in × 1 in =
2.54 cm 2.54 cm 2.54 cm
or 3
• 4.52 cm3 × 1 in = 4.52 cm3× 13 in3 =
2.54cm (2.54) 3cm3
Tro's "Introductory Chemistry",
Chapter 2
65
Units Raised to a Power
• 4.52 cm3 × 1 in3 = 0.276 in3
16.387 cm3
Tro's "Introductory Chemistry",
Chapter 2
66
Tro's "Introductory Chemistry",
Chapter 2
67
2.9 Density
Tro's "Introductory Chemistry",
Chapter 2
68
Mass and Volume• Two main characteristics of matter.
• Mass or volume alone cannot be used to
identify what type of matter something is.
If you are given a large glass containing 100 g
of a clear, colorless liquid and a small glass
containing 25 g of a clear, colorless liquid, are
both liquids the same stuff?
• Even though mass and volume are
individual properties, for a given type of
matter they are related to each other!
Tro's "Introductory Chemistry",
Chapter 2
69
Mass vs. Volume of BrassMass
grams
Volume
cm3
20 2.4
32 3.8
40 4.8
50 6.0
100 11.9
150 17.9
This slide was a graph of the
mass as y and the volume as x
resulting in a straight line
this indicates that the ratio
of mass to volume is a constant =
the density of a substance
For these brass samples d = 8.3
g/mL
Tro's "Introductory Chemistry",
Chapter 2
70
Tro's "Introductory Chemistry",
Chapter 2
71
Density• Ratio of mass:volume.
• Its value depends on the kind of material, not the amount.
• Solids = g/cm3
1 cm3 = 1 mL
• Liquids = g/mL
• Gases = g/L
• Volume of a solid can be determined by water displacement—Archimedes Principle. Explain
• Density : solids > liquids > gases
Except ice is less dense than liquid water!
Volume
MassDensity
For a Table of Sample Densities
see Table 2.4 on page 33
Tro's "Introductory Chemistry",
Chapter 2
72
Volume by Water Displacement
• Graduated cylinder – water is added = Vinitial
• A solid object is added to the cylinder.
• Volume of water rises = Vfinal
• Vobject = Vfinal - Vinitial
Tro's "Introductory Chemistry",
Chapter 2
73
Tro's "Introductory Chemistry",
Chapter 2
74
Using Density in Calculations
Three Possible Formulas
Volume
MassDensity
Density
Mass Volume
Volume Density Mass
Solution Maps:
m, V D
m, D V
V, D m
Density Problems
Tro's "Introductory Chemistry",
Chapter 2
75
•Think of density as a conversion factor between
mass and volume of a given substance.
•Problems from text for class 2.96, 2.100,
Problem 2.96
Tro's "Introductory Chemistry",
Chapter 2
76
An aluminum engine block has a volume of 4.77 L and a
mass of 12.88 kg. What is the density of the aluminum
in grams per cubic centimeter?
Solution map
Volume, L Volume, mL=cm3
mass
Density =
Mass, kg mass, g volume
Problem 2.96 (cont.)
Tro's "Introductory Chemistry",
Chapter 2
77
Volume to cm3:
4.77 L × 1000 mL = 4,770 mL = 4,770 cm3
1 L
Mass to g:
12.88 kg × 1000 g = 12,880 g
1 kg
Density using the formula:
12,880 g = 2.7002096 g/cm3 2.70 g/cm3
4,770 cm3 3 sig. fig.
Problem 2.100
Tro's "Introductory Chemistry",
Chapter 2
78
Acetone (fingernail polish remover) has a density of 0.7857 g/cm3.
(a) What is the mass in grams of 28.56 mL of acetone?
(b) What is the volume in milliliters of 6.54 g of acetone?
There are two basic ways to work this problem. 1)Plug the
variables into the formula for density and solve for the unknown.
Or
2) Treat the density as a conversion factor between mass and
volume of a substance.
We will use # 2
Solution maps (a) ml g (b) g mL
Problem 2.100 (cont.)
Tro's "Introductory Chemistry",
Chapter 2
79
A density of 0.7857 g/cm3 means that 1 cm3 or mL of the substance
Has a mass of 0.7857 g. (i.e. 1 mL = 0.7857 g of acetone)
From this equation we can derive conversion factors.
(a) The conversion is from mL to g so we want mL to cancel
28.56 mL × 0.7857g = 22.44 g (4 sig. fig.)
1 mL
(a) The conversion is from g to mL so we want g to cancel
6.54 g × 1 mL = 8.32 g (3 sig. fig.)
0.7857g