chemistry chapter 1 (foundation)

8/13/2019 Chemistry chapter 1 (foundation)

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Chapter 1

Lecture Presentation

2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition

Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward

Matter and

Measurement


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Chemistry

In this science we studymatter, its properties,

and its behavior.




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Matter

We define matter as anything that has mass and takes

up space.




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Matter


Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc.

Atoms are the building blocks of matter.

Each element is made of the same kind of atom.

A compound is made of two or more different kinds of

elements.


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States of Matter




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Classification of Matter




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2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward

2012 Pearson Education, Inc.


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Matter

And

Measurement


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Matter

And

Measurement


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Sample Exercise 1.1 Distinguishing among Elements, Compounds,

and Mixtures

Solution

White gold contains gold and a white metal, such as palladium. Two samples of white gold differ in the relative

amounts of gold and palladium they contain. Both samples are uniform in composition throughout. Use Figure 1.9

to classify white gold.


Aspirin is composed of 60.0% carbon, 4.5%hydrogen, and 35.5% oxygen by mass, regardless

of its source. Use Figure 1.9 to classify aspirin.

Answer:


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Sample Exercise 1.1 Distinguishing among Elements, Compounds,and Mixtures

Because the material is uniform throughout, it ishomogeneous. Because its composition differs for

the two samples, it cannot be a compound.

Instead, it must be a homogeneous mixture.

Solution







Answer:


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Sample Exercise 1.1 Distinguishing among Elements, Compounds,and Mixtures

Because the material is uniform throughout, it ishomogeneous. Because its composition differs for

the two samples, it cannot be a compound.

Instead, it must be a homogeneous mixture.

Solution







Answer: It is a compound because it has constant

composition and can be separated into several

elements.


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Properties and


Changes of Matter


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Types of Properties Physical Properties

Can be observed without changing a substanceinto another substance.

Boiling point, density, mass, volume, etc.


Chemical Properties Can only be observed when a substance is

changed into another substance.

Flammability, corrosiveness, reactivity with acid,etc.


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Types of Properties

Intensive Properties

Are independent of the amount of the substance

that is present.

Densit boilin oint color etc.


Extensive Properties

Depend upon the amount of the substance

present.

Mass, volume, energy, etc.


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Types of Changes

Physical Changes

These are changes in matter that do not change thecomposition of a substance.

Chan es of state tem erature volume etc.


Chemical Changes

Chemical changes result in new substances.

Combustion, oxidation, decomposition, etc.


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Chemical Reactions



In the course of a chemical reaction, the reacting

substances are converted to new substances.


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Separation of


Mixtures


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Filtration

In filtration, solid

substances are separated

from liquids and solutions.



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Distillation

Distillation uses

differences in the

boiling points of

substances to se arate a


homogeneous mixtureinto its components.


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ChromatographyThis technique separates substances on the basis of

differences in solubility in a solvent.



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Units of


Measurement


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SI Units


Systme International dUnits

A different base unit is used for each quantity.


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Metric SystemPrefixes convert the base units into units that are

appropriate for the item being measured.



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Sample Exercise 1.2 Using SI Prefixes

Solution

What is the name of the unit that equals (a) 109 gram, (b) 106 second, (c) 103 meter?

(a) How many picometers are there in one meter?

(b) Express 6.0 103 m using a prefix to replace the

power of ten. (c) Use exponential notation to

Practice Exercise


express 4.22 mg in grams. (d) Use decimal notation

to express 4.22 mg in grams.

Answers:


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Sample Exercise 1.2 Using SI Prefixes

We can find the prefix related to each power of ten in

Table 1.5: (a) nanogram, ng, (b) microsecond, s,

(c) millimeter, mm.

Solution

What is the name of the unit that equals (a) 109 gram, (b) 106 second, (c) 103 meter?

(a) How many picometers are there in one meter?

(b) Express 6.0 103 m using a prefix to replace the

power of ten. (c) Use exponential notation to

Practice Exercise


express 4.22 mg in grams. (d) Use decimal notation

to express 4.22 mg in grams.

Answers:


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Volume

The most commonly usedmetric units for volume are

the liter (L) and the milliliter


.

A liter is a cube

1 decimeter (dm) long on

each side.

A milliliter is a cube1 centimeter (cm) long on

each side.


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Temperature

By definitiontemperature is a

measure of the average


particles in a sample.


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Temperature In scientific measurements,

the Celsius and Kelvin

scales are most often used. The Celsius scale is based

on the properties of water.


0C is the freezing pointof water.

100 C is the boiling point

of water.


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Temperature

The kelvin is the SI unit

of temperature.

It is based on the


.

There are no negative

Kelvin temperatures.

K = C + 273.15


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Temperature

The Fahrenheit scale is

not used in scientificmeasurements.

=


C = 5/9(F 32)


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Sample Exercise 1.3 Converting Units of Temperature

(a)

(b)

Solution

A weather forecaster predicts the temperature will reach 31 C. What is this temperature (a) in K, (b) in F?


Ethylene glycol, the major ingredient in antifreeze, freezes at -11.5 C. What is the freezing point in (a) K,

(b) F?

Answers:

Practice Exercise


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(a) Using Equation 1.1, we have K = 31 + 273 = 304 K.

(b) Using Equation 1.2, we have

Solution




(b) F?

Answers:

Practice Exercise


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(a) Using Equation 1.1, we have K = 31 + 273 = 304 K.

(b) Using Equation 1.2, we have

Solution




(b) F?

Answers: (a) 261.7 K, (b) 11.3 F

Practice Exercise


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Derived Units

Density is a physical property of asubstance.


,

derived from the units for mass and volume.

d =

m

V


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Sample Exercise 1.4 Determining Density and Using Density toDetermine Volume or Mass

(a) We are given mass and volume, so Equation

1.3 yields

Solution

(a) Calculate the density of mercury if 1.00 102 g occupies a volume of 7.36 cm3.(b) Calculate the volume of 65.0 g of liquid methanol (wood alcohol) if its density is 0.791 g/mL.

(c) What is the mass in grams of a cube of gold (density = 19.32 g/cm3) if the length of the cube is 2.00 cm?


(b) So v ng Equat on 1.3 or vo ume an then

using the given mass and density gives

(c) We can calculate the mass from the volume of

the cube and its density. The volume of a cube

is given by its length cubed:

Solving Equation 1.3 for mass and

substituting the volume and density of the

cube, we have


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(a) We are given mass and volume, so Equation

1.3 yields

Solution

(a) Calculate the density of mercury if 1.00 102 g occupies a volume of 7.36 cm3.(b) Calculate the volume of 65.0 g of liquid methanol (wood alcohol) if its density is 0.791 g/mL.

(c) What is the mass in grams of a cube of gold (density = 19.32 g/cm3) if the length of the cube is 2.00 cm?


(b) So v ng Equat on 1.3 or vo ume an then

using the given mass and density gives

(c) We can calculate the mass from the volume of

the cube and its density. The volume of a cube

is given by its length cubed:

Solving Equation 1.3 for mass and

substituting the volume and density of the

cube, we have

Volume = (2.00 cm)3 = (2.00)3 cm3 = 8.00 cm3

Mass = volume density = (8.00 cm3)(19.32 g/cm3) = 155 g


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Continued

(a) Calculate the density of a 374.5-g sample of copper if it has a volume of 41.8 cm3. (b) A student needs

15.0 g of ethanol for an experiment. If the density of ethanol is 0.789 g/mL, how many milliliters of ethanol

are needed? (c) What is the mass, in grams, of 25.0 mL of mercury (density = 13.6 g/mL)?

Answers:

Practice Exercise



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Uncertainty in



Measurement

U i i M


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Uncertainty in Measurements

Different measuring devices have different uses and

different degrees of accuracy.




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Significant Figures

The term significant figures refers to digitsthat were measured.



,

attention to significant figures so we do notoverstate the accuracy of our answers.


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Significant Figures When addition or subtraction is performed,

answers are rounded to the least significantdecimal place.



performed, answers are rounded to thenumber of digits that corresponds to the

leastnumber of significant figures in any

of the numbers used in the calculation.


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Sample Exercise 1.5 Relating Significant Figures to the Uncertainty


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Sample Exercise 1.5 Relating Significant Figures to the Uncertaintyof a Measurement

The value 4.0 has two significant figures, whereas 4.00 has three. This difference implies that the 4.0 has

more uncertainty. A mass reported as 4.0 g indicates that the uncertainty is in the first decimal place. Thus,

the mass might be anything between 3.9 and 4.1 g, which we can represent as 4.0 0.1 g. A mass reported

as 4.00 g indicates that the uncertainty is in the second decimal place. Thus, the mass might be anythingbetween 3.99 and 4.01 g, which we can represent as 4.00 0.01 g. (Without further information, we cannot

be sure whether the difference in uncertainties of the two measurements reflects the precision or the

accuracy of the measurement.)

Solution

What difference exists between the measured values 4.0 g and 4.00 g?



A sample that has a mass of about 25 g is placed on a balance that has a precision of 0.001 g. How manysignificant figures should be reported for this measurement?

Answer: five, as in the measurement 24.995 g, the uncertainty being in the third decimal place

Sample Exercise 1.6 Determining the Number of Significant Figures


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p g g gin a Measurement

Solution

How many significant figures are in each of the following numbers (assume that each number is a measuredquantity): (a) 4.003, (b) 6.023 1023, (c) 5000?

How many significant figures are in each of the following measurements: (a) 3.549 g, (b) 2.3 104 cm,3

Practice Exercise



.

Answers:



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p g g gin a Measurement

(a) Four; the zeros are significant figures. (b) Four; the exponential term does not add to the number of

significant figures. (c) One; we assume that the zeros are not significant when there is no decimal point

shown. If the number has more significant figures, a decimal point should be employed or the number

written in exponential notation. Thus, 5000. has four significant figures, whereas 5.00103 has three.

Solution

How many significant figures are in each of the following numbers (assume that each number is a measuredquantity): (a) 4.003, (b) 6.023 1023, (c) 5000?

How many significant figures are in each of the following measurements: (a) 3.549 g, (b) 2.3 104 cm,3

Practice Exercise



.

Answers: (a) four, (b) two, (c) three



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p g g gin a Calculated Quantity

In reporting the volume, we can show only as many significant figures as given in the dimension with the

fewest significant figures, that for the height (two significant figures):

Solution

The width, length, and height of a small box are 15.5 cm, 27.3 cm, and 5.4 cm, respectively. Calculate the volumeof the box, using the correct number of significant figures in your answer.



It takes 10.5 s for a sprinter to run 100.00 m. Calculate her average speed in meters per second, and express

the result to the correct number of significant figures.

Answer:

Practice Exercise



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p g g gin a Calculated Quantity

In reporting the volume, we can show only as many significant figures as given in the dimension with the

fewest significant figures, that for the height (two significant figures):

A calculator used for this calculation shows 2285.01, which we must round off to two significant figures.

Because the resulting number is 2300, it is best reported in exponential notation, 2.3 103, to clearly

Solution

The width, length, and height of a small box are 15.5 cm, 27.3 cm, and 5.4 cm, respectively. Calculate the volumeof the box, using the correct number of significant figures in your answer.



n cate two s gn cant gures.

It takes 10.5 s for a sprinter to run 100.00 m. Calculate her average speed in meters per second, and express

the result to the correct number of significant figures.

Answer: 9.52 m/s (three significant figures)

Practice Exercise



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in a Calculated Quantity

To calculate the density, we must know both the mass and the volume of the gas. The mass of the gas is just

the difference in the masses of the full and empty container:

(837.6 836.2) g = 1.4 gIn subtracting numbers, we determine the number of significant figures in our result by counting decimal

places in each quantity. In this case each quantity has one decimal place. Thus, the mass of the gas, 1.4 g,

has one decimal place.

Solution

A gas at 25 C fills a container whose volume is 1.05 103 cm3. The container plus gas has a mass of 837.6 g. The

container, when emptied of all gas, has a mass of 836.2 g. What is the density of the gas at 25 C?



s ng t e vo ume g ven n t e quest on, . cm , an t e e n t on o ens ty, we ave



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in a Calculated Quantity

To calculate the density, we must know both the mass and the volume of the gas. The mass of the gas is just

the difference in the masses of the full and empty container:

(837.6 836.2) g = 1.4 gIn subtracting numbers, we determine the number of significant figures in our result by counting decimal

places in each quantity. In this case each quantity has one decimal place. Thus, the mass of the gas, 1.4 g,

has one decimal place.

Solution

A gas at 25 C fills a container whose volume is 1.05 103 cm3. The container plus gas has a mass of 837.6 g. The

container, when emptied of all gas, has a mass of 836.2 g. What is the density of the gas at 25 C?



s ng t e vo ume g ven n t e quest on, . cm , an t e e n t on o ens ty, we ave

In dividing numbers, we determine the number of significant figures in our result by counting the number of

significant figures in each quantity. There are two significant figures in our answer, corresponding to the

smaller number of significant figures in the two numbers that form the ratio. Notice that in this example,following the rules for determining significant figures gives an answer containing only two significant

figures, even though each of the measured quantities contained at least three significant figures.

Sample Exercise 1.8 Determining the Number of Significant Figuresi C l l d Q i


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in a Calculated QuantityContinued

To how many significant figures should the mass of the container be measured (with and without the gas) in

Sample Exercise 1.8 for the density to be calculated to three significant figures?

Answer:

Practice Exercise




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Accuracy versus Precision


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Accuracy versus Precision

Accuracy refers to the proximity of a

measurement to the true value of a

quantity.



rec s on re ers o e prox m y o

several measurements to each other.

Dimensional Analysis


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Dimensional Analysis

We use dimensional analysis to convert one

quantity to another.

Most commonly, dimensional analysis utilizesconversion factors (e.g., 1 in. = 2.54 cm)

1 in. 2.54 cm



2.54 cm

1 in.

or


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Dimensional AnalysisUse the form of the conversion factor that

puts the sought-for unit in the numerator:




Given unit = desired unitdesired unit

given unit

Conversion factor


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Dimensional Analysis For example, to convert 8.00 m to inches,

convert m to cm

convert cm to in.




8.00 m 100 cm

1 m

1 in.

2.54 cm= 315 in.

Sample Exercise 1.9 Converting Units


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Because we want to change from pounds to grams, we look for a relationship between

these units of mass. From the back inside cover we have 1 lb = 453.6 g. To cancel pounds

and leave grams, we write the conversion factor with grams in the numerator and pounds

in the denominator:

Solution

If a woman has a mass of 115 lb, what is her mass in grams? (Use the relationships between units given on the back

inside cover of the text.)



By using a conversion factor from the back inside cover, determine the length in kilometers of a 500.0-mi

automobile race. 1 km = 0.62137 mi

Answer:

Practice Exercise

Sample Exercise 1.9 Converting Units


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Because we want to change from pounds to grams, we look for a relationship between

these units of mass. From the back inside cover we have 1 lb = 453.6 g. To cancel pounds

and leave grams, we write the conversion factor with grams in the numerator and pounds

in the denominator:

Solution

If a woman has a mass of 115 lb, what is her mass in grams? (Use the relationships between units given on the back

inside cover of the text.)



The answer can be given to only three significant figures, the number of significant

figures in 115 lb. The process we have used is diagrammed in the margin.

By using a conversion factor from the back inside cover, determine the length in kilometers of a 500.0-mi

automobile race. 1 km = 0.62137 mi

Answer: 804.7 km

Practice Exercise

Sample Exercise 1.10 Converting Units Using Two or MoreConversion Factors


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Conversion Factors

To go from the given units, m/s, to the desired units, mi/hr, we must convert meters to miles and seconds to

hours. From our knowledge of SI prefixes we know that 1 km = 103 m. From the relationships given on the

back inside cover of the book, we find that 1 mi = 1.6093 km. Thus, we can convert m to km and then

convert km to mi. From our knowledge of time we know that 60s = 1 min and 60 min = 1 hr. Thus, we can

convert s to min and then convert min to hr. The overall process is

SolutionThe average speed of a nitrogen molecule in air at 25

C is 515 m/s. Convert this speed to miles per hour.



Applying first the conversions for distance and then those for time, we can set up one long equation inwhich unwanted units are canceled:

Sample Exercise 1.10 Converting Units Using Two or MoreConversion Factors


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To go from the given units, m/s, to the desired units, mi/hr, we must convert meters to miles and seconds to

hours. From our knowledge of SI prefixes we know that 1 km = 103 m. From the relationships given on the

back inside cover of the book, we find that 1 mi = 1.6093 km. Thus, we can convert m to km and then

convert km to mi. From our knowledge of time we know that 60s = 1 min and 60 min = 1 hr. Thus, we can

convert s to min and then convert min to hr. The overall process is

SolutionThe average speed of a nitrogen molecule in air at 25

C is 515 m/s. Convert this speed to miles per hour.



Applying first the conversions for distance and then those for time, we can set up one long equation inwhich unwanted units are canceled:

Sample Exercise 1.11 Converting Volume UnitsEarths oceans contain approximately 1 36 109 km3 of water Calculate the volume in liters


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From the back inside cover, we find 1 L = 103 m3, but there is no relationship listed involving km3. From

our knowledge of SI prefixes, however, we know 1 km = 103 m and we can use this relationship between

lengths to write the desired conversion factor between volumes:

Thus, converting from km3 to m3 to L, we have

Solution

Earth s oceans contain approximately 1.36 109 km3 of water. Calculate the volume in liters.



If the volume of an object is reported as 5.0 ft3, what is the volume in cubic meters?

Answer:

Practice Exercise

Sample Exercise 1.11 Converting Volume UnitsEarths oceans contain approximately 1 36 109 km3 of water Calculate the volume in liters


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From the back inside cover, we find 1 L = 103 m3, but there is no relationship listed involving km3. From

our knowledge of SI prefixes, however, we know 1 km = 103 m and we can use this relationship between

lengths to write the desired conversion factor between volumes:

Thus, converting from km3 to m3 to L, we have

Solution

Earth s oceans contain approximately 1.36 109 km3 of water. Calculate the volume in liters.



If the volume of an object is reported as 5.0 ft3, what is the volume in cubic meters?

Answer: 0.14 m3

Practice Exercise

Sample Exercise 1.12 Conversions Involving DensityWhat is the mass in grams of 1 00 gal of water? The density of water is 1 00 g/mL


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Before we begin solving this exercise, we note the following:

1. We are given 1.00 gal of water (the known, or given, quantity) and asked to calculate its mass in

grams (the unknown).

2. We have the following conversion factors either given, commonly known, or available on the back

inside cover of the text:

Solution

What is the mass in grams of 1.00 gal of water? The density of water is 1.00 g/mL.



desired result, whereas the last conversion factor must be inverted in order to cancel gallons:

Sample Exercise 1.12 Conversions Involving DensityWhat is the mass in grams of 1 00 gal of water? The density of water is 1 00 g/mL


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Before we begin solving this exercise, we note the following:

1. We are given 1.00 gal of water (the known, or given, quantity) and asked to calculate its mass in

grams (the unknown).

2. We have the following conversion factors either given, commonly known, or available on the back

inside cover of the text:

Solution

What is the mass in grams of 1.00 gal of water? The density of water is 1.00 g/mL.



desired result, whereas the last conversion factor must be inverted in order to cancel gallons:

The unit of our final answer is appropriate, and weve taken care of our significant figures. We can further

check our calculation by estimating. We can round 1.057 off to 1. Then focusing on the numbers that do not

equal 1 gives 4 1000 = 4000 g, in agreement with the detailed calculation.

chemistry chapter 1 (foundation)

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