Transcript significant digit

Chapter 2
Scientific
Measurements
Vanessa N. Prasad-Permaul
CHM 1025
Valencia Community College
1
Uncertainty in Measurements
 A measurement is a number with a unit
attached.
 It is not possible to make exact measurements,
thus all measurements have uncertainty.
 We will generally use metric system units. These
include:
 The meter, m, for length measurements
 The gram, g, for mass measurements
 The liter, L, for volume measurements
2
Length Measurements
 Let’s measure the length of a candy cane.
 Ruler A has 1 cm divisions, so we can estimate
the length to ± 0.1 cm. The length is 4.2 ± 0.1
cm.
 Ruler B has 0.1 cm divisions, so we can
estimate the length to ± 0.05 cm. The length
is 4.25 ± 0.05 cm.
3
Uncertainty in Length
 Ruler A: 4.2 ± 0.1 cm; Ruler B: 4.25 ± 0.05 cm.
 Ruler A has more uncertainty than Ruler B.
 Ruler B gives a more precise measurement.
4
EXAMPLE 2.1 Uncertainty in Measurement
Which measurements are consistent with the metric rulers shown in Figure 2.2?
(a) Ruler A: 2 cm, 2.0 cm, 2.05 cm, 2.5 cm, 2.50 cm
(b) Ruler B: 3.0 cm, 3.3 cm, 3.33 cm, 3.35 cm, 3.50 cm
Figure 2.2 Metric Rulers for Measuring Length On Ruler A, each division is 1 cm. On
Ruler B, each division is 1 cm and each subdivision is 0.1 cm.
Solution
Ruler A has an uncertainty of ±0.1 cm, and Ruler B has an uncertainty of ±0.05
cm. Thus,
(a) Ruler A can give the measurements 2.0 cm and 2.5 cm.
(b) Ruler B can give the measurements 3.35 cm and 3.50 cm.
EXERCISE 2.1 Uncertainty in Measurement
Figure 2.2 Metric Rulers for Measuring Length On Ruler A, each division is 1 cm.
On Ruler B, each division is 1 cm and each subdivision is 0.1 cm.
Practice Exercise
Which measurements are consistent with the metric rulers shown
in Figure 2.2?
(a) Ruler A: 1.5 cm, 1.50 cm, 1.55 cm, 1.6 cm, 2.00 cm
(b) Ruler B: 0.5 cm, 0.50 cm, 0.055 cm, 0.75 cm, 0.100 cm
Mass Measurements
 The mass of an
object is a measure
of the amount of
matter it possesses.
 Mass is measured
with a balance and is
not affected by
gravity.
 Mass and weight are
not interchangeable.
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Mass Versus Weight
 Mass and weight are not the same.
 Weight is the force exerted by gravity on an object.
8
Volume Measurements
 Volume is the amount of space occupied by a
solid, a liquid, or a gas.
 There are several instruments for measuring
volume, including:
 Graduated cylinder
 Syringe
 Buret
 Pipet
 Volumetric flask
9
Significant Digits
 Each number in a properly recorded
measurement is a significant digit (or
significant figure).
 Significant digits express the uncertainty in
the measurement.
 When you count significant digits, start
counting with the first nonzero number.
 Let’s look at a reaction measured by three
stopwatches.
10
Significant Digits,
Continued
 Stopwatch A is calibrated to seconds (±1 s);
Stopwatch B to tenths of a second (±0.1 s);
and Stopwatch C to hundredths of a second
(±0.01 s).
• Stopwatch A reads 35 s; B reads 35.1 s; and C reads 35.08 s.
– 35 s has two significant figure.
– 35.1 s has three significant figures.
– 35.08 has four significant figures.
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EXAMPLE 2.2 Significant Digits
State the number of significant digits in the following
measurements:
(a) 12,345 cm (b) 0.123 g
(c) 0.5 mL (d) 102.0 s
Solution
In each example, we simply count the number of
digits. Thus,
(a) 5
(b) 3
(c) 1
(d) 4
Notice that the leading zero in (b) and (c) is not part
of the measurement but is inserted to call attention to
the decimal point that follows.
EXERCISE 2.2 Significant Digits
Practice Exercise
State the number of significant digits in the following
measurements:
(a) 2005 cm (b) 25.000 g
(c) 25.0 mL (d) 0.25 s
Significant Digits and
Placeholders
 If a number is less than 1, a placeholder zero
is never significant.
 Therefore, 0.5 cm, 0.05 cm, and 0.005 cm all
have one significant digit.
 If a number is greater than 1, a placeholder
zero is usually not significant.
 Therefore, 50 cm, 500 cm, and 5000 cm all
have one significant digit.
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EXAMPLE 2.3 Significant Digits
State the number of significant digits in the following
measurements:
(a) 0.025 cm (b) 0.2050 g
(c) 25.0 mL
(d) 2500 s
Solution
In each example, we count the number of significant
digits and disregard placeholder zeros. Thus,
(a) 2
(b) 4
(c) 3
(d) 2
EXERCISE 2.3 Significant Digits
Practice Exercise
State the number of significant digits in the following
measurements:
(a)
(c)
0.050 cm
50.00 mL
(b)
(d)
0.0250 g
1000 s
Exact Numbers
 When we count something, it is an exact number.
 Significant digit rules do not apply to exact
numbers.
 An example of an exact
There are seven
on this slide.
number:
coins
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Rounding Off Nonsignificant
Digits
 All numbers from a measurement are
significant. However, we often generate
nonsignificant digits when performing
calculations.
 We get rid of nonsignificant digits by
rounding off numbers.
 There are three rules for rounding off
numbers.
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Rules for Rounding Numbers
1. If the first nonsignificant digit is less than 5, drop all
nonsignificant digits.
2. If the first nonsignificant digit is greater than or
equal to 5, increase the last significant digit by 1
and drop all nonsignificant digits.
3. If a calculation has two or more operations, retain
all the nonsignificant digits until the final operation
and then round off the answer.
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Rounding Examples
 A calculator displays 12.846239 and 3 significant
digits are justified.
 The first nonsignificant digit is a 4, so we drop all
nonsignificant digits and get 12.8 as the answer.
 A calculator displays 12.856239 and 3 significant
digits are justified.
 The first nonsignificant digit is a 5, so the last
significant digit is increased by one to 9. All the
nonsignificant digits are dropped, and we get 12.9
as the answer.
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Rounding Off and Placeholder
Zeros
 Round the measurement 151 mL to two
significant digits.
 If we keep two digits, we have 15 mL, which is only
about 10% of the original measurement.
 Therefore, we must use a placeholder zero: 150 mL
 Recall that placeholder zeros are not significant.
 Round the measurement 2788 g to two
significant digits.
 We get 2800 g.
• Remember, the placeholder zeros are not significant, and 28
grams is significantly less than 2800 grams.
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EXAMPLE 2.4 Rounding Off
Round off the following numbers to three significant
digits:
(a) 22.250
(b) 0.34548
(c) 0.072038 (d) 12,267
Solution
To locate the first nonsignificant digit, count three
digits from left to right. If the first nonsignificant digit
is less than 5, drop all nonsignificant digits. If the first
nonsignificant digit is 5 or greater, add 1 to the last
significant digit.
(a) 22.3 (Rule 2)
(b) 0.345 (Rule 1)
(c) 0.0720 (Rule 1)
(d) 12,300 (Rule 2)
In (d), notice that two placeholder zeros must be
added to 123 to obtain the correct decimal place.
EXERCISE 2.4 Rounding Off
Practice Exercise
Round off the following numbers to three significant
digits:
(a) 12.514748 (b) 0.6015261
(c) 192.49032 (d) 14652.832
Adding and Subtracting
Measurements
 When adding or subtracting measurements, the
answer is limited by the value with the most
uncertainty.
• Let’s add three mass
measurements.
• The measurement 106.7 g has the
greatest uncertainty (± 0.1 g).
• The correct answer is 107.1 g.
106.7
0.25
+ 0.195
107.145
g
g
g
g
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EXAMPLE 2.5 Addition/Subtraction and Rounding Off
Add or subtract the following measurements and round
off your answer:
35.45 mL – 30.5 mL
In addition or subtraction operations, the answer is limited
by the measurement with the most uncertainty.
Solution
Since 30.5 mL has the most uncertainty (±0.1 mL), we
round off to one decimal place. The answer is
5.0 mL and is read “five point zero milliliters.”
EXERCISE 2.5 Addition/Subtraction and Rounding
Off
Practice Exercise
Add or subtract the following measurements and
round off your answer:
(a) 8.6 cm + 50.05 cm (b) 34.1 s – 0.55 s
Multiplying and Dividing
Measurements
 When multiplying or dividing measurements,
the answer is limited by the value with the
fewest significant figures.
 Let’s multiply two length measurements:
(5.15 cm)(2.3 cm) = 11.845 cm2
 The measurement 2.3 cm has the fewest
significant digits—two.
 The correct answer is 12 cm2.
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EXAMPLE 2.6 Multiplication/Division and
Rounding Off
Multiply or divide the following measurements and round off
your answer:
(a) 50.5 cm  12 cm (b) 103.37 g/20.5 mL
Solution
In multiplication and division operations, the answer is limited by
the measurement with the least number of significant digits.
(a) In this example, 50.5 cm has three significant digits and 12 cm
has two.
(50.5 cm) (12 cm) = 606 cm2
The answer is limited to two significant digits and rounds off to 610
cm2 after inserting a placeholder zero. The answer is read “six
hundred and ten square centimeters.”
EXAMPLE 2.6 Multiplication/Division and
Rounding Off
(b) In this example, 103.37 g has five significant digits
and 20.5 mL has three.
The answer is limited to three significant digits and
rounds off to 5.04 g/mL. Notice that the unit is a
ratio; the answer is read as “five point zero four
grams per milliliter.”
EXERCISE 2.6 Multiplication/Division and
Rounding Off
Practice Exercise
Multiply or divide the following measurements and
round off your answer.
(a) (359 cm) (0.20 cm)
(b) 73.950 g/25.5 mL
Exponential Numbers
 Exponents are used to indicate that a number
has been multiplied by itself.
 Exponents are written using a superscript;
thus, (2)(2)(2) = 23.
 The number 3 is an exponent and indicates
that the number 2 is multiplied by itself 3
times. It is read “2 to the third power” or “2
cubed”.
 (2)(2)(2) = 23 = 8
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Powers of 10
 A power of 10 is a number that results when 10 is raised
to an exponential power.
 The power can be positive (number greater than 1) or
negative (number less than 1).
32
EXAMPLE 2.7 Converting to Powers of 10
Express each of the following ordinary numbers as a
power of 10:
(a) 100,000
(b) 0.000 000 01
Solution
The power of 10 indicates the number of places the
decimal point has been moved.
(a) We must move the decimal five places to the left;
thus, 1  105.
(b) We must move the decimal eight places to the
right; thus, 1  10–8.
EXERCISE 2.7 Converting to Powers of 10
Practice Exercise
Express each of the following ordinary numbers as a
power of 10:
(a) 10,000,000
(b) 0.000 000 000 001
Concept Exercise
Which of the following lengths is less: 1  103 cm or 1  10–3
cm?
EXAMPLE EXERCISE 2.8 Converting to
Ordinary Numbers
Express each of the following powers of 10 as an
ordinary number:
(a) 1  104 (b) 1  10–9s
Solution
The power of 10 indicates the number of places the
decimal point has been moved.
(a) The exponent in 1  104 is positive 4, and so we
must move the decimal point four places to the right
of 1, thus, 10,000.
(b) The exponent in 1  10–9 is negative 9, and so we
must move the decimal point nine places to the left of
1, thus, 0.000 000 001.
EXAMPLE EXERCISE 2.8 Converting to
Ordinary Numbers
Practice Exercise
Express each of the following powers of 10 as an
ordinary number:
(a) 1  1010
(b) 1  10–5
Concept Exercise
Which of the following masses is less: 0.000 001 g or
0.000 01 g?
Scientific Notation
 Numbers in science are often very large or very
small. To avoid confusion, we use scientific
notation.
 Scientific notation utilizes the significant digits in a
measurement followed by a power of 10. The
significant digits are expressed as a number
between 1 and 10.
power of 10
D.DD
n
x 10
significant digits
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Applying Scientific Notation
 To use scientific notation, first place a
decimal after the first nonzero digit in the
number followed by the remaining significant
digits.
 Indicate how many places the decimal is
moved by the power of 10.
 A positive power of 10 indicates that the decimal
moves to the left.
 A negative power of 10 indicates that the decimal
moves to the right.
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Scientific Notation,
Continued
There are 26,800,000,000,000,000,000,000 helium
atoms in 1.00 L of helium gas. Express the
number in scientific notation.
 Place the decimal after the 2, followed by the
other significant digits.
 Count the number of places the decimal has
moved to the left (22). Add the power of 10 to
complete the scientific notation.
2.68 x 1022 atoms
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Another Example
The typical length between two carbon atoms in a
molecule of benzene is 0.000000140 m. What is
the length expressed in scientific notation?
 Place the decimal after the 1, followed by the
other significant digits.
 Count the number of places the decimal has
moved to the right (7). Add the power of 10 to
complete the scientific notation.
1.40 x 10-7 m
40
EXAMPLE EXERCISE 2.9 Scientific Notation
Express each of the following values in scientific notation:
(a) There are 26,800,000,000,000,000,000,000 helium atoms in a
one liter balloon filled with helium gas.
(b) The mass of one helium atom is 0.000 000 000 000 000 000
000 006 65 g.
Solution
We can write each value in scientific notation as follows:
(a) Place the decimal after the 2, followed by the other significant digits
(2.68). Next, count the number of places the decimal has moved. The
decimal is moved to the left 22 places, so the exponent is +22. Finally, we
have the number of helium atoms in 1.00 L of gas: 2.68  1022 atoms.
(b) Place the decimal after the 6, followed by the other significant digits
(6.65). Next, count the number of places the decimal has shifted. The
decimal has shifted 24 places to the right, so the exponent is –24. Finally,
we have the mass of a helium atom: 6.65  10–24 g.
EXAMPLE EXERCISE 2.9 Scientific Notation
Practice Exercise
Express each of the following values as ordinary numbers:
(a) The mass of one mercury atom is 3.33  10–22 g.
(b) The number of atoms in 1 mL of liquid mercury is 4.08 
1022.
Concept Exercise
Which of the following masses is greater: 1  10–6 g or 0.000 01 g?
Scientific Calculators
 A scientific calculator has an exponent key (often
“EXP” or “EE”) for expressing powers of 10.
 If your calculator reads 7.45 E-17,
the proper way to write the answer
in scientific notation is 7.45 x 10-17.
 To enter the number in your
calculator, type 7.45, then press
the exponent button (“EXP” or
“EE”), and type in the exponent
(17 followed by the +/– key).
43
Unit Equations
 A unit equation is a simple statement of two
equivalent quantities.
 For example:
 1 hour = 60 minutes
 1 minute = 60 seconds
 Also, we can write:
 1 minute = 1/60 of an hour
 1 second = 1/60 of a minute
44
Unit Factors
 A unit conversion factor, or unit factor, is a
ratio of two equivalent quantities.
 For the unit equation 1 hour = 60 minutes, we
can write two unit factors:
1 hour
or
60 minutes
60 minutes
1 hour
45
EXAMPLE EXERCISE 2.10 Unit Conversion Factors
Write the unit equation and the two corresponding unit factors for each of the following:
(a) pounds and ounces
(b) quarts and gallons
Solution
We first write the unit equation and then the corresponding unit factors.
(a) There are 16 ounces in a pound, and so the unit equation is 1 pound = 16
ounces. The two associated unit factors are
(b) There are 4 quarts in a gallon, and so the unit equation is 1 gallon = 4 quarts.
The two unit factors are
Practice Exercise
Write the unit equation and the two corresponding unit factors for each of the
following:
(a) hours and days (b) hours and minutes
EXAMPLE EXERCISE 2.10 Unit Conversion Factors
Continued
Concept Exercise
How many significant digits are in the following unit equation?
1 hour = 3600 seconds
Unit Analysis Problem
Solving
 An effective method for solving problems in
science is the unit analysis method.
 It is also often called dimensional analysis or
the factor-label method.
 There are three steps to solving problems
using the unit analysis method.
48
Steps in the Unit Analysis
Method
1. Write down the unit asked for in the answer.
2. Write down the given value related to the answer.
3. Apply a unit factor to convert the unit in the given
value to the unit in the answer.
49
Unit Analysis Problem
How many days are in 2.5 years?
 Step 1: We want days.
 Step 2: We write down the given: 2.5 years.
 Step 3: We apply a unit factor (1 year = 365
days) and round to two significant figures.
50
Another Unit Analysis
Problem
A can of soda contains 12 fluid ounces. What is the
volume in quarts (1 qt = 32 fl oz)?
 Step 1: We want quarts.
 Step 2: We write down the given:
12 fl oz.
 Step 3: We apply a unit factor
(1 qt = 12 fl oz) and round to two
significant figures.
51
EXAMPLE EXERCISE 2.11 Unit Analysis Problem Solving
A can of soda contains 12 fluid ounces (fl oz). What is the
volume in quarts (given that 1 qt = 32 fl oz)?
Dr. Pepper A 12 fl oz can of soda contains 355 mL
EXAMPLE EXERCISE 2.11 Unit Analysis Problem Solving
Continued
Strategy Plan
Step 1: What unit is asked for in the answer?
Step 2: What given value is related to the answer?
Step 3: What unit factor should we apply? Since the unit
equation is 1 qt = 32 fl oz, the two unit factors are 1 qt/32 fl oz,
and its reciprocal 32 fl oz/1 qt.
Unit Analysis Map
EXAMPLE EXERCISE 2.11 Unit Analysis Problem Solving
Continued
Solution
We should apply the unit factor 1 qt/32 fl oz to cancel fluid ounces
, which appears in the denominator.
The given value, 12 fl oz, limits the answer to two significant digits. Since the unit factor 1 qt/32 fl oz is
derived from an exact equivalent, 1 qt = 32 fl oz, it does not affect the significant digits in the answer.
Practice Exercise
A can of soda contains 355 mL. What is the volume in liters (given that 1 L
= 1000 mL)?
Concept Exercise
How many significant digits are in the following unit equation?
1 L = 1000 mL
Another Unit Analysis Problem,
Continued
A marathon is 26.2 miles. What is the distance in
kilometers (1 km = 0.62 mi)?
 Step 1: We want km.
 Step 2: We write down the given:
26.2 mi.
 Step 3: We apply a unit factor
(1 km = 0.62 mi) and round to
two significant figures.
55
EXAMPLE EXERCISE 2.12 Uncertainty in Measurement
A marathon covers a distance of 26.2 miles (mi). If 1 mile is exactly equal to 1760 yards, what is the distance of the
race in yards?
Strategy Plan
Step 1: What unit is asked for in the answer?
Step 2: What given value is related to the answer?
Step 3: What unit factor should we apply? Since the
unit equation is 1 mi = 1760 yd, the two unit
factors are 1 mi/1760 yd, and its reciprocal
1760 yd/1 mi.
Boston Marathon Marathon
athletes run a distance of
26.2 miles.
EXAMPLE EXERCISE 2.12 Uncertainty in Measurement
Continued
Unit Analysis Map
Solution
We should apply the unit factor 1760 yd/1 mi to cancel miles
, which appears in the denominator.
The given value, 26.2 mi, limits the answer to three significant digits. Since the unit factor 1760 yd/1 mi is
derived from an exact equivalent, 1 mi = 1760 yd, it does not affect the significant digits in the answer.
EXAMPLE EXERCISE 2.12 Uncertainty in Measurement
Continued
Practice Exercise
Given that a marathon is 26.2 miles, what is the distance in kilometers (given that 1 km = 0.62 mi)?
Answer: 42 km
Concept Exercise
How many significant digits are in the following unit equation?
1 km = 0.62 mi
Critical Thinking: Units
 When discussing measurements, it is critical
that we use the proper units.
 NASA engineers mixed metric and English
units when designing software for the Mars
Climate Orbiter.
 The engineers used kilometers rather than miles.
 1 kilometer is 0.62 mile.
 The spacecraft approached too close to the
Martian surface and burned up in the atmosphere.
59
The Percent Concept
 A percent, %, expresses the amount of a
single quantity compared to an entire
sample.
 A percent is a ratio of parts per 100 parts.
 The formula for calculating percent is shown
below:
quantity of interest
% 
x 100%
total sample
60
Calculating Percentages
 Sterling silver contains silver and copper. If a
sterling silver chain contains 18.5 g of silver
and 1.5 g of copper, what is the percent of
silver in sterling silver?
18.5 g silver
x 100 %  92.5% silver
(18.5  1.5) g
61
Percent Unit Factors
 A percent can be expressed as parts per 100
parts.
 25% can be expressed as 25/100 and 10% can
be expressed as 10/100.
 We can use a percent expressed as a ratio as a
unit factor.
 A rock is 4.70% iron, so
4.70 g iron
100 g of sample
62
EXAMPLE EXERCISE 2.13 The Percent Concept
Sterling silver contains silver and copper metals. If a sterling silver chain contains 18.5 g of silver and 1.5 g of
copper, what is the percent of silver?
Sterling Silver Sterling silver has a high luster and is found in fine utensils and jewelry.
EXAMPLE EXERCISE 2.13 The Percent Concept
Continued
Strategy Plan
Step 1: What is asked for in the answer?
Step 2: What given value is related to the answer?
Step 3: What unit factor should we apply?
No unit factor is required.
Solution
To find percent, we compare the mass of silver metal to the total mass of the silver and copper in the chain,
and multiply by 100%.
Genuine sterling silver is cast from 92.5% silver and 7.5% copper. If you carefully examine a piece of
sterling silver, you may see the jeweler’s notation .925, which indicates the item is genuine sterling silver.
EXAMPLE EXERCISE 2.13 The Percent Concept
Continued
Practice Exercise
A 14-karat gold ring contains 7.45 g of gold, 2.66 g of silver,
and 2.66 g of copper. Calculate the percent of gold in the 14karat ring.
Concept Exercise
If a gold alloy contains 20% silver and 5% copper, what is the
percent of gold in the alloy?
Percent Unit Factor
Calculation
The Earth and Moon have a similar composition;
each contains 4.70% iron. What is the mass of
iron in a lunar sample that weighs 235 g?
 Step 1: We want g iron.
 Step 2: We write down the given: 235 g sample.
 Step 3: We apply a unit factor (4.70 g iron = 100 g
sample) and round to three significant figures.
4.70 g iron
235 g sample x
 11.0 g iron
100 g sample
66
EXAMPLE EXERCISE 2.14 Percent as a Unit Factor
The Moon and Earth have a similar composition and each contains 4.70% iron, which supports the theory that the
Moon and Earth were originally a single planet. What is the mass of iron in a lunar sample that weighs 235 g?
Strategy Plan
Step 1: What unit is asked for in the answer?
Step 2: What given value is related to the answer?
Step 3: What unit factor should we apply?
From the definition of percent, 4.70 g iron = 100 g
sample; the two unit factors are 4.70 g iron/100 g
sample, and its reciprocal 100 g sample/4.70 g iron.
Unit Analysis Map
EXAMPLE EXERCISE 2.14 Percent as a Unit Factor
Continued
Solution
We should apply the unit factor 4.70 g iron/100 g sample to cancel grams sample
the denominator.
, which appears in
The given value and unit factor each limits the answer to three significant digits.
Practice Exercise
A Moon sample is found to contain 7.50% aluminum. What is the mass of
the lunar sample if the amount of aluminum is 5.25 g?
Concept Exercise
Water is 11.2% hydrogen by mass. What two unit factors express the
percent hydrogen in water?
Chemistry Connection: Coins
 A nickel coin contains 75.0 % copper metal and 25.0
% nickel metal, and has a mass of 5.00 grams.
 What is the mass of nickel metal in a nickel coin?
25.0 g nickel
50.0 g coin x
 12.5 g nickel
100 g coin
69
Chapter Summary
 A measurement is a number with an attached
unit.
 All measurements have uncertainty.
 The uncertainty in a measurement is dictated
by the calibration of the instrument used to
make the measurement.
 Every number in a recorded measurement is a
significant digit.
70
Chapter Summary, Continued
 Placeholding zeros are not significant digits.
 If a number does not have a decimal point, all
nonzero numbers and all zeros between
nonzero numbers are significant.
 If a number has a decimal place, significant
digits start with the first nonzero number and
all digits to the right are also significant.
71
Chapter Summary, Continued
 When adding and subtracting numbers, the
answer is limited by the value with the most
uncertainty.
 When multiplying and dividing numbers, the
answer is limited by the number with the
fewest significant figures.
 When rounding numbers, if the first
nonsignificant digit is less than 5, drop the
nonsignificant figures. If the number is 5 or
more, raise the first significant number by 1,
and drop all of the nonsignificant digits.
72
Chapter Summary, Continued
 Exponents are used to indicate that a number
is multiplied by itself n times.
 Scientific notation is used to express very
large or very small numbers in a more
convenient fashion.
 Scientific notation has the form D.DD x 10n,
where D.DD are the significant figures (and is
between 1 and 10) and n is the power of ten.
73
Chapter Summary, Continued
 A unit equation is a statement of two
equivalent quantities.
 A unit factor is a ratio of two equivalent
quantities.
 Unit factors can be used to convert
measurements between different units.
 A percent is the ratio of parts per 100 parts.
74
State the number of significant
digits in the measured quantity
0.8020 mL.
a.
b.
c.
d.
two
three
four
five
© 2011 Pearson Education, Inc.
State the number of significant
digits in the measured quantity
0.8020 mL.
a.
b.
c.
d.
two
three
four
five
© 2011 Pearson Education, Inc.
To the correct number of
significant digits, what is the
sum of 22.9898 g + 79.904 g?
a.
b.
c.
d.
102.89 g
102.893 g
102.894 g
102.8938 g
© 2011 Pearson Education, Inc.
To the correct number of
significant digits, what is the
sum of 22.9898 g + 79.904 g?
a.
b.
c.
d.
102.89 g
102.893 g
102.894 g
102.8938 g
© 2011 Pearson Education, Inc.
To the correct number of
significant digits, what is the
product of
2.1
cm × 1.45 cm ×
3
a. 11 cm
3.654 cm?
b. 11.1 cm3
c. 11.12 cm3
d. 11.13 cm3
© 2011 Pearson Education, Inc.
To the correct number of
significant digits, what is the
product of 2.1 cm × 1.45 cm ×
3.654 cm?
a.
b.
c.
d.
11 cm3
11.1 cm3
11.12 cm3
11.13 cm3
© 2011 Pearson Education, Inc.
Which is the largest mass?
a.
b.
c.
d.
0.000392 × 102 g
0.0453 g
4.73 × 10–2 g
5.27 × 10–3 g
© 2011 Pearson Education, Inc.
Which is the largest mass?
a.
b.
c.
d.
0.000392 × 102 g
0.0453 g
4.73 × 10–2 g
5.27 × 10–3 g
© 2011 Pearson Education, Inc.
Water is 11.9% hydrogen and 88.1%
oxygen. What is the mass of
hydrogen and oxygen in 100.0 g of
water?
a. 0.119 g hydrogen and 0.881 g oxygen
b. 1.19 g hydrogen and 8.81 g oxygen
c. 11.9 g hydrogen and 88.1 g oxygen
d. 119 g hydrogen and 881 g oxygen
© 2011 Pearson Education, Inc.
Water is 11.9% hydrogen and 88.1%
oxygen. What is the mass of
hydrogen and oxygen in 100.0 g of
water?
a. 0.119 g hydrogen and 0.881 g oxygen
b. 1.19 g hydrogen and 8.81 g oxygen
c. 11.9 g hydrogen and 88.1 g oxygen
d. 119 g hydrogen and 881 g oxygen
© 2011 Pearson Education, Inc.