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CHEM 1405:
Introductory Chemistry
Houston Community College
Dr. Laura Jakubowski
Chapter 2 – The Metric System
Textbook “Introductory Chemistry: Concepts and Critical Thinking” Seventh Edition by Charles H. Corwin
© 2014 Pearson Education, Inc.
Basic Units and Symbols
• Up until the 1800s, the English system was most commonly used for
measurements
• The metric system was later adopted, as it offers simplicity and basic
units – the meter (m) for length, the gram (g) for mass, the liter (L) for
volume, and the second (s) for time
• Original metric references:
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Metric Prefixes
• The metric system is a decimal system, it uses a prefix to express a
multiple or a fraction of a basic unit
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The International System of Units (SI)
• The metric system is sometimes referred to as the International
System of measurement (symbol SI)
• SI is similar to the metric system, but much more comprehensive
length
mass
time
temperature
electric current
light intensity
amount of substance
meter
kilogram
second
Kelvin
ampere
candela
mole
m
kg
s
K
A
cd
mol
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Metric Conversion Factors
• A unit equation relates two quantities that are equal
• For example, 1000 meters = 1 kilometer (1000 m = 1 km), also known as an
exact equivalent
• Further examples:
• 1 kg = 1000 g
• 1 s = 1,000,000,000 ns
• 1 L = 1 × 106 µL
• A unit factor is the ratio of two equivalent quantities, the numerator and
denominator are equivalent – ratio can be inverted to reciprocal
1 m = 100 cm
unit factors are:
1m
100 cm
and
100 cm
1m
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Metric-Metric Conversions
• The unit analysis method is very effective for the problems found in
introductory chemistry
• Step 1: Write down the unit asked for in the answer
• Step 2: Write down the given value related to the answer
• Step 3: Apply unit factor(s) to convert the unit in the given value to the
unit in the answer
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Metric-Metric Conversions
• Problem: find the mass in grams of a 325 mg aspirin tablet
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Metric-Metric Conversions
• Problem: find the mass in grams of a 325 mg aspirin tablet
325 mg
unit
factor
×
=
g
1 g = 1000 mg
1g
1000 mg
325 mg
×
or
1g
1000 mg
1000 mg
1g
= 0.325 g
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Metric-Metric Conversions
• The mass of Earth is 5.98 × 1024 kg. What is the mass in megagrams?
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Metric-Metric Conversions
• The mass of Earth is 5.98 × 1024 kg. What is the mass in megagrams?
given value
5.98 × 1024 kg
unit factor 1
×
5.98 ×
1 kg
1000 g
or
1000 g
1kg
1024
×
unit factor 2
unit in answer
1 Mg
1,000,000 g
or
1,000,000 g
1 Mg
= ? Mg
1000 g
1 Mg
kg ×
×
= 5.98 × 1021 Mg
1kg
1,000,000 g
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Metric-English Conversions
• The United States is the last major world nation to formally adopt the
metric system – and progress is slow to achieve full compliance
• Therefore, the following conversions are useful to know:
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Metric-English Conversions
• A can of soda contains 12.0 fl oz (1 quart = 32 fluid ounces). What is the
volume of soda in milliliters?
• If a tennis ball weighs 2.0 oz (16 oz = 1 lb), what is the mass of the
tennis ball in grams?
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Metric-English Conversions
• A can of soda contains 12.0 fl oz (1 quart = 32 fluid ounces). What is the
volume of soda in milliliters?
12.0 fl oz ×
1 qt
946 mL
×
= 355 mL
32 fl oz
1 qt
• If a tennis ball weighs 2.0 oz (16 oz = 1 lb), what is the mass of the
tennis ball in grams?
454 g
1 lb
2.0 oz ×
×
= 57 g
1 lb
16 oz
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Set Up the Following Conversion Equations
a) An elephant weighs 5,100 kg. What is its mass in pounds?
b) A television screen measures 42 in. What is the size of the screen in
meters?
c) Light travels through the universe at a velocity of 3.00 × 1010 cm/s.
How many gigameters does light travel per second?
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Set Up the Following Conversion Equations
a) An elephant weighs 5,100 kg. What is its mass in pounds?
1000 g
1 lb
5,100 kg ×
×
= 11,000 lb
1 kg
454 g
b) A television screen measures 42 in. What is the size of the screen in
meters?
1m
2.54 cm
42 in. ×
1 in.
×
100 cm
= 1.1 m
c) Light travels through the universe at a velocity of 3.00 × 1010 cm/s.
How many gigameters does light travel per second?
3.00 × 1010 cm/s ×
1m
1 Gm
×
= 0.300 Gm/s
9
100 cm
1 × 10 m
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The Percent Concept
• A percent (symbol %) expresses the amount of a single quantity
compared to an entire sample
• A dime is 10% of a dollar, and a quarter is 25% of a dollar
one quantity
× 100% = N%
total sample
• After 1971, 5 cent coins were composed of both nickel and copper. If a
coin contains 3.80 g copper and 1.27 g nickel, what is the percent of
copper in the coin?
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The Percent Concept
• A percent (symbol %) expresses the amount of a single quantity
compared to an entire sample
• A dime is 10% of a dollar, and a quarter is 25% of a dollar
one quantity
× 100% = N%
total sample
• After 1971, 5 cent coins were composed of both nickel and copper. If a
coin contains 3.80 g copper and 1.27 g nickel, what is the percent of
copper in the coin?
3.80 g
= 0.750 × 100% = 75.0%
(3.80 + 1.27) g
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Percent Unit Factors
• Percent can be expressed as parts per 100 parts, 10% and 25%
correspond to 10 and 25 parts per 100 parts, respectively
10
100
25
• 25% corresponds to
100
• 10% corresponds to
• The Moon contains 4.70% iron. What is the mass of iron in a lunar
sample that weighs 235 g?
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Percent Unit Factors
• Percent can be expressed as parts per 100 parts, 10% and 25%
correspond to 10 and 25 parts per 100 parts, respectively
10
100
25
• 25% corresponds to
100
• 10% corresponds to
• The Moon contains 4.70% iron. What is the mass of iron in a lunar
sample that weighs 235 g?
4.70 g iron
235 g sample ×
= 11.0 g iron
100 g sample
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Volume by Calculation
• Volume of a rectangular solid can be calculated if length (l), width (w),
and thickness (t) are known
t
w
l
l × w × t = volume
• If solid measures 3 cm by 2 cm by 1 cm:
3 cm × 2 cm × 1 cm = 6 cm3
• The unit cm3 is referred to as a cubic centimeter
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Thickness Calculation
• A sheet of aluminum foil measures 25.0 mm by 10.0 mm and the volume
is 3.75 mm3. What is the thickness of the foil in mm?
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Thickness Calculation
• A sheet of aluminum foil measures 25.0 mm by 10.0 mm and the volume
is 3.75 mm3. What is the thickness of the foil in mm?
l × w × t = volume
25.0 mm × 10.0 mm × t = 3.75 mm3
(solve for t):
250. mm2 × t = 3.75 mm3
3.75 mm3
t=
250 mm2
t = 0.0150 mm
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Volumes of Solids, Liquids, and Gases
• A liter (L) is equivalent to the volume
occupied by a cube, 10 cm per side
• Calculated volume of a liter:
1 L = 10 cm × 10 cm × 10 cm = 1000 cm3
• Recall that 1 L = 1000 mL
1000 cm3 = 1 L = 1000 mL
1000 cm3 = 1000 mL
• Therefore:
1 cm3 = 1 mL
• Note – in medicine, a cm3 is often
abbreviated as cc (cubic centimeter)
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Metric-English Volume Conversion
• An automobile engine has a volume displacement of 498 cm 3 in each
cylinder, express the volume in cubic inches (in.3).
• An SUV has a 244 in3 engine, express the engine volume in liters.
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Metric-English Volume Conversion
• An automobile engine has a volume displacement of 498 cm 3 in each
cylinder, express the volume in cubic inches (in.3).
498 cm3 ×
1 in.
1 in.
1 in.
×
×
= 30.4 in.3
2.54 cm
2.54 cm
2.54 cm
• An SUV has a 244 in3 engine, express the engine volume in liters.
244 in.3 ×
1L
2.54 cm
2.54 cm
2.54 cm
×
×
×
= 4.00 L
3
1000 cm
1 in.
1 in.
1 in.
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Set Up the Following Conversion Equations
a) A patient receives an 850 cc (cm3) infusion of fluid. How many liters of
fluid did the patient receive?
b) A moon sample is found to contain 7.50% aluminum. What is the total
mass of the lunar sample if the amount of aluminum is 5.25 g?
c) A windowpane is 0.53 cm thick and 95 cm wide. How tall is the
windowpane, assuming its volume is 12,600 cm3?
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Set Up the Following Conversion Equations
a) A patient receives an 850 cc (cm3) infusion of fluid. How many liters of
fluid did the patient receive?
850 cm3 ×
1L
= 0.85 L
3
1000 cm
b) A moon sample is found to contain 7.50% aluminum. What is the total
mass of the lunar sample if the amount of aluminum is 5.25 g?
100 g sample
5.25 g aluminum ×
= 70.0 g sample
7.50 g aluminum
c) A windowpane is 0.53 cm thick and 95 cm wide. How tall is the
windowpane, assuming its volume is 12,600 cm3?
l × 95 cm × 0.53 cm = 12,600 cm3
l = 250 cm
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Volume of a Solid by Displacement
• The volume of an irregular solid
cannot be determined directly by
multiplying its measurements
• The volume can be determined
indirectly by measuring the amount of
water it displaces – called volume by
displacement
• The difference between the final water level and the initial water level
in the graduated cylinder (10.5 mL) is equal to the volume of the solid
piece of jade
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Volume of a Gas by Displacement
• The volume displacement method
can also be used to determine the
volume of a gas
• Using a setup as seen, a sample is
heated, produces a gas, and that
gas displaces a certain volume of
water
• The volume of water displaced is
equal to the volume of gas
produced
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The Density Concept
• Which weighs more – a ton of feathers or a ton of bricks?
They have the same mass, but very different volume!
cnx.org
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The Density Concept
• The density (symbol d) expresses the concentration of its mass
• Density is defined as the amount of mass per unit volume
mass
= density
volume
• Different units can be used to express density – for solids and liquids
usually g/mL or g/cm3, and for gases usually g/L
Substance:
Density: 0.917 g/cm3
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1.00 g/cm3
11.3 g/cm3
22.5 g/cm3
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The Density Concept
• Identify the liquids and solids
Choose from:
• water
• ethyl ether
• chloroform
• rubber
• ice
• aluminum
L2
L3
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The Density Concept
• Identify the liquids and solids
Choose from:
• water
ethyl ether
• ethyl ether
• chloroform
water
• rubber
L
• ice
• aluminum chloroform
L
2
3
ice
rubber
aluminum
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Density Calculations
• If a platinum nugget has a mass of 214.50 g and a volume of 10.0 cm3,
what is the density of the metal?
• Carbon tetrachloride is a solvent used for degreasing electronic parts. If
25.0 mL of carbon tetrachloride has a mass of 39.75 g, what is the
density of the liquid?
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Density Calculations
• If a platinum nugget has a mass of 214.50 g and a volume of 10.0 cm3,
what is the density of the metal?
mass
= density
volume
214.50 g
= 21.5 g/cm3
3
10.0 cm
• Carbon tetrachloride is a solvent used for degreasing electronic parts. If
25.0 mL of carbon tetrachloride has a mass of 39.75 g, what is the
density of the liquid?
39.75 g
= 1.59 g/mL
25.0 mL
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Applying Density as a Unit Factor
• The unit analysis method can be used to solve density problems
• For example, the density of mercury is 13.6 g/mL, this can be expressed as:
13.6 g
1 mL
or
1 mL
13.6 g
• What is the volume in mL of 75.5 g of liquid mercury?
• A 1.00-in cube of copper measures 2.54 cm on a side. What is the mass
in grams of the copper cube (d of copper = 8.96 g/cm3)?
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Applying Density as a Unit Factor
• The unit analysis method can be used to solve density problems
• For example, the density of mercury is 13.6 g/mL, this can be expressed as:
13.6 g
1 mL
or
1 mL
13.6 g
• What is the volume in mL of 75.5 g of liquid mercury?
75.5 g ×
1 mL
= 5.55 mL
13.6 g
• A 1.00-in cube of copper measures 2.54 cm on a side. What is the mass
in grams of the copper cube (d of copper = 8.96 g/cm3)?
(2.54 cm)(2.54 cm)(2.54 cm) = 16.4
cm3
8.96 g
×
= 147 g
1 cm3
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Specific Gravity
• The ratio of the density of a liquid to the density of water (at 4 °C) is
called specific gravity (symbol sp gr)
• Specific gravity is a unitless quantity
density of object (g/mL)
specific gravity =
density of water (g/mL)
• The density of water is 1.00 g/mL, therefore the specific gravity is 1.00
• Diagnostic medical testing often includes the specific gravity of body
fluids
• The specific gravity of urine may be 1.02 and of blood may be 1.06, meaning
both are more dense than water
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Temperature
• Temperature is the average kinetic energy of individual particles in
motion, and it is measured with a thermometer
• Three temperature scales:
Water Freezes
Water Boils
• Fahrenheit degree (symbol °F)
• Celsius degree (symbol °C)
• Kelvin unit (symbol K)
• Lowest possible temperature is
-273.15 °C, 0 K
• No highest temperature, but the
sun’s interior is ~10,000,000K
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Temperature Conversions
• 180 Fahrenheit units are equivalent to 100 Celsius units (212 °F - 32 °F)
• The difference between the freezing point from °F to °C is 32
100 °C
(°F - 32 °F) ×
= °C
180 °F
180 °F
°C ×
100 °C
+ 32 °F = °F
• Celsius units are equivalent to Kelvin units, but the Kelvin scale is 273
units above the Celsius scale
°C + 273 = K
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Temperature Conversions
• Normal body temperature is 98.6 °F. What is normal body temperature in
degrees Celsius?
• The temperature in Celsius of liquid nitrogen is -196 °C, what is the
Kelvin temperature?
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Temperature Conversions
• Normal body temperature is 98.6 °F. What is normal body temperature in
degrees Celsius?
100 °C
(98.6 °F - 32 °F) ×
= °C
180 °F
66.6 °F ×
100 °C
= 37.0 °C
180 °F
• The temperature in Celsius of liquid nitrogen is -196 °C, what is the
Kelvin temperature?
-196 °C + 273 = 77 K
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The Heat Concept
• Heat is the flow of energy from an object at higher temperature to an
object at lower temperature – heat is a measure of total energy and
temperature is a measure of average energy
• Which feels hotter – quickly drinking a cup of hot tea, or a spoon of hot tea?
The cup of tea has more heat than the spoon of tea, even though both are 100 °C
Lesser total heat energy
(lesser volume)
Greater total heat energy
(greater volume)
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The Heat Concept
• Heat energy is often expressed in units of calories
• A calorie (symbol cal) is the amount of heat necessary to raise 1 gram
of water 1 degree on the Celsius scale
• A kilocalorie (kcal) is the amount of heat necessary to raise 1000 grams
of water 1 degree on the Celsius scale
• Nutritional Calorie (Cal) is equal to 1 kilocalorie, that is 1000 calories
• The SI unit of energy is the joule (symbol J), where 1 cal = 4.184 J
• The heat produced by chemical reactions is often expressed in
kilocalories, as well as in kilojoules (kJ), where 1 kcal = 4.184 kJ
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Energy Conversion
• Burning one liter of natural gas produces 9.46 kcal of heat energy.
Express the energy in kilojoules (given that 1 kcal = 4.184 kJ).
• Burning one gram of gasoline produces 47.9 kJ of energy. Express the
heat energy in kilocalories.
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Energy Conversion
• Burning one liter of natural gas produces 9.46 kcal of heat energy.
Express the energy in kilojoules (given that 1 kcal = 4.184 kJ).
4.184 kJ
9.46 kcal ×
= 39.6 kJ
1 kcal
• Burning one gram of gasoline produces 47.9 kJ of energy. Express the
heat energy in kilocalories.
1 kcal
47.9 kJ ×
= 11.4 kcal
4.184 kJ
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Set Up the Following Conversion Equations
a) The average surface temperature of Mars is -55 °C. What is the
average temperature in degrees Fahrenheit?
b) A diamond displaces 3.4 mL of water. What is the mass of the diamond
in grams, assuming its density is 3.52 g/mL?
c) Burning one gram of gasoline produces 47.9 kJ of energy. How many
kcal of energy is in 1 kg gasoline?
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Set Up the Following Conversion Equations
a) The average surface temperature of Mars is -55 °C. What is the
average temperature in degrees Fahrenheit?
180 °F
-55 °C ×
100 °C
+ 32 °F = -67 °F
b) A diamond displaces 3.4 mL of water. What is the mass of the diamond
in grams, assuming its density is 3.52 g/mL?
3.52 g
3.4 mL ×
= 12.0 g
1 mL
c) Burning one gram of gasoline produces 47.9 kJ of energy. How many
kcal of energy is in 1 kg gasoline?
1000 g
47.9 kJ
1 kcal
1 kg ×
×
×
= 11,400 kcal
1 kg
1g
4.184 kJ
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Specific Heat
• Specific heat is the amount of heat required to raise the temperature of
1 gram of substance 1 degree Celsius – units are cal / (g × °C)
• Water has a relatively high specific heat – 1 cal / (g × °C), this helps to
regulate climates and maintain moderate temperatures on Earth,
because oceans resist wide swings in temperature
Ice
Water
Iron
1g
1g
1g
Temperature after receiving 1 cal of heat:
Specific heat of substance (in cal / (g × °C)):
1.0 °C
1.00
Silver
2.0 °C
0.492
9.3 °C
0.108
1g
17.7 °C
0.0566
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Specific Heat Calculations
• If it takes 3.35 cal of heat energy to raise the temperature of 10.0 g of
mercury by 10.0 °C, what is the specific heat of mercury?
• The specific heat of ethanol is 2.46 J / (g × °C). How much heat energy
(in J) is required to raise the temperature of ethanol by 5 °C?
50
Specific Heat Calculations
• If it takes 3.35 cal of heat energy to raise the temperature of 10.0 g of
mercury by 10.0 °C, what is the specific heat of mercury?
3.35 cal
= 0.0335 cal / (g × °C)
10.0 g × 10 °C
• The specific heat of ethanol is 2.46 J / (g × °C). How much heat energy
(in J) is required to raise the temperature of 1 g of ethanol by 5 °C?
1 g × 5 °C ×
2.46 J
= 12.3 J
g × °C
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Summary
• The metric system uses meters (m) for distance, grams (g) for mass, liters (L) for
volume, and seconds (s) for time
• The metric system base units are modified using prefixes to reduce or enlarge the
base units by factors of ten
• The unit analysis method is a tool to convert between units of a given value to units
asked for in the answer
• A percent expresses the amount of a single quantity compared to an entire sample
• Volume of a solid = l × w × t; it is often reported in cm3
• Volume by displacement is used for irregular solids and gases
• Density = mass/volume; specific gravity is the ratio of given density to that of water
• Temperature is the average energy of molecules in motion; units are degrees
Fahrenheit (°F), degrees Celsius (°C) and the Kelvin unit (K)
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Summary
• Heat is the total energy of molecules in motion, expressed in calories (cal) or joules
(J), where 1 cal = 4.184 J; specific heat is the amount of heat necessary to raise 1 g
of substance 1 °C, for water the value is 1.00 cal / (g × °C)
Key terms: English system, metric system, meter (m), gram (g), liter
(L), second (s), International System (SI), unit equation, exact
equivalent, unit factor, reciprocal, unit analysis method, percent (%),
cubic centimeter (cm3), volume by displacement, density, specific
gravity (sp gr), temperature, Fahrenheit degree, Celsius degree,
Kelvin unit, heat, calorie (cal), joule (J), specific heat
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