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ChemQuest 1
Name: ____________________________
Date: _______________
Hour: _____
Information: Qualitative vs. Quantitative
The following observations are qualitative.
The building is really tall.
It takes a long time for me to ride my bike to the store.
I live really far away.
The following observations are quantitative.
The river is 31.5 m deep.
The cheese costs \$4.25 per pound.
It is 75o F outside today.
Critical Thinking Questions
1. What is the difference between qualitative and quantitative observations? (Your answers should
reveal an understanding of the definitions for qualitative and quantitative.)
Qualitative refers to the general quality of a characteristic (tall, short, long, wide, etc.) whereas
quantitative includes the quantity of the characteristic (how tall, how short, how long, how wide,
etc.). Therefore, quantitative observations will always contain a number to specify more
precisely than qualitative.
2. Write an example of a quantitative observation that you may make at home or at school.
Any of the following are examples: I like to eat 2 bowls of cereal each day. My shoes are size
10.
3. Why are instruments such as rulers, scales (balances), thermometers, etc. necessary?
Instruments help us to communicate with each other effectively. Without instruments, we could
only speak in generalities. They also make it possible to know exact quantities.
Information: Units
The following tables contain common metric (SI) units and their prefixes.
Table 1: metric base units
Quantity
Unit
Unit
Symbol
Length
meter
m
Mass
kilogram
kg
Time
second
s
Temperature
Kelvin
K
Volume
Liter
L
Amount of substance mole
mol
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Table 2: prefixes for metric base units.
Prefix
Mega
M
kilo
k
deci
d
centi
c
milli
m
micro
µ
nano
n
pico
p
Symbol
Meaning
million
thousand
tenth
hundredth
thousandth
millionth
billionth
trillionth
Note the following examples:
• “milli” means thousandth so a milliliter (symbol: mL) is one thousandth of a Liter and it
takes one thousand mL to make one L.
• “Mega” means million so “Megagram” (Mg) means one million grams NOT one millionth
of a gram. One millionth of a gram would be represented by the microgram (µg). It takes
one million micrograms to equal one gram and it takes one million grams to equal one
Megagram.
• One cm is equal to 0.01 m because one cm is “one hundredth of a meter” and 0.01 m is
the expression for “one hundredth of a meter”
Critical Thinking Questions
4. How many milligrams are there in one kilogram? 1,000,000 mg
5. 21.5 km is how many meters? 21,500 m
6. Is it possible to answer this question: How many mg are in one km? Explain.
No, because milligrams and kilometers are different units of measure.
7. What is the difference between a Mm and a mm? Which is larger one Mm or one mm?
A Mm is a megameter (1,000,000 meters) and a mm is a millimeter (1/1000 of a meter).
Information: Scientific Notation
“Scientific notation” is used to make very large or very small numbers easier to handle. For
example the number 45,000,000,000,000,000 can be written as “4.5 x 1016 ”. The “16” tells you that
there are sixteen decimal places between the right side of the four and the end of the number.
Another example: 2.641 x 1012 = 2,641,000,000,000 the “12” tells you that there are 12
decimal places between the right side of the 2 and the end of the number.
Very small numbers are written with negative exponents. For example, 0.00000000000000378 can
be written as 3.78 x 10-15. The “-15” tells you that there are 15 decimal places between the right side
of the 3 and the end of the number.
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Another example: 7.45 x 10-8 = 0.0000000745 the “-8” tells you that there are 8 decimal
places between the right side of the 7 and the end of the number.
In both very large and very small numbers, the exponent tells you how many decimal points are
between the right side of the first digit and the end of the number. If the exponent is positive, the
decimal places are to the right of the number. If the exponent is negative, the decimal places are to
the left of the number.
Critical Thinking Questions
8. Two of the following six numbers are written incorrectly. Circle the two that are incorrect.
a)
3.57 x 10-8
b)
4.23 x 10-2
c)
75.3 x 102
d)
2.92 x 109
e)
0.000354 x 1014
f)
9.1 x 104
What do you think is wrong about the two numbers you circled?
C should be written 7.53x103 and E should be written 3.54x1010. The correct format for
writing numbers in scientific notation is with only one digit to the left of the decimal place. Also,
the number should not begin with zeros.
9. Write the following numbers in scientific notation:
a) 25,310,000,000,000,000 = 2.531x1016
b) 0.000000003018 = 3.018x10-9
10. Write the following scientific numbers in regular notation:
a) 8.41 x 10-7 = 0.000000841
b) 3.215 x 108 = 321,500,000
Information: Multiplying and Dividing Using Scientific Notation
When you multiply two numbers in scientific notation, you must add their exponents. Here are two
examples. Make sure you understand each step:
(4.5x1012) x (3.2x1036) = (4.5)(3.2) x 1012+36 = 14.4x1048 1.44x1049
(5.9x109) x (6.3x10-5) = (5.9)(6.3) x 109+(-5) = 37.17x104 3.717x105
When you divide two numbers, you must subtract denominator’s exponent from the numerator’s
exponent. Here are two examples. Make sure you understand each step:
2.8 x1014 2.8
=
x1014−7 = 0.875x107 = 8.75 x106
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3.2 x10
3.2
5.7 x1019 5.7
=
x1019−( −9) = 1.84 x1019+9 = 1.84 x10 28
−9
3.1
3.1x10
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Critical Thinking Questions
11. Solve the following problems.
a) (4.6x1034)(7.9x10-21) = (4.6)(7.9) x 1034+(-21) = 36.34 x1013 3.634x1014
b) (1.24x1012)(3.31x1020) = (1.24)(3.31)x1012+20 = 4.1x1032
12. Solve the following problems.
a)
8.4 x10 −5 8.4
=
x10 − 5−17 = 2.05 x10 − 22
4.1x10 17 4.1
b)
5.4 x10 32 5.4
=
x10 32 −14 = .74 x1018 = 7.4 x1017
7.3 x10 14 7.3
Information: Adding and Subtracting Using Scientific Notation
Whenever you add or subtract two numbers in scientific notation, you must make sure that they have
the same exponents. Your answer will them have the same exponent as the numbers you add or
subtract. Here are some examples. Make sure you understand each step:
4.2x106 + 3.1x105 make exponents the same, either a 5 or 6 42x105 + 3.1x105 = 45.1x105 = 4.51x106
7.3x10-7 - 2.0x10-8 make exponents the same, either -7 or -8 73x10-8 – 2.0x10-8 = 71x10-8 = 7.1x10-7
Critical Thinking Questions
13. Solve the following problems.
a) 4.25x1013 + 2.10x1014 = first, change exponents so both are either 13 or 14
4.25x1013 + 21.0x1013 = 25.25x1013 2.525 x 1014
b) 6.4x10-18 – 3x10-19 = first, change exponents so both are either -18 or -19
6.4x10-18 – 0.3x10-18 =
6.1 x 10-18
c) 3.1x10-34 + 2.2x10-33 = first, change exponents so both are either -34 or -33
3.1x10-34 + 22x10-34 = 25.1x10-34 2.51x10-33
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ChemQuest 2
Name: ____________________________
Date: _______________
Hour: _____
Information: Significant Figures
We saw in the last ChemQuest that scientific notation can be a very nice way of getting rid of
unnecessary zeros in a number. For example, consider how convenient it is to write the following
numbers:
32,450,000,000,000,000,000,000,000,000,000,000 = 3.245 x 1034
0.000000000000000127 = 1.27 x 10-16
There are a whole lot of zeros in the above numbers that are not really needed. As another example,
consider the affect of changing units:
21,500 meters = 21.5 kilometers
0.00582 meters = 5.82 millimeters
Notice that the zeros in “21,500 meters” and in “0.00582 meters” are not really needed when the units
change. Taking these examples into account, we can introduce three general rules:
1. Zeros at the beginning of a number are never significant (important).
2. Zeros at the end of a number are not significant unless… (you’ll find out later)
3. Zeros that are between two nonzero numbers are always significant.
Therefore, the number 21,500 has three significant figures: only three of the digits are important—
the two, the one, and the five. The number 10,210 has four significant figures because only the zero
at the end is considered not significant. All of the digits in the number 10,005 are significant because
the zeros are in between two nonzero numbers (Rule #3).
Critical Thinking Questions
1. Verify that each of the following numbers contains four significant figures. Circle the digits
that are significant. Significant digits are in red italics type.
a) 0.00004182
b) 494,100,00
c) 32,010,000,000
d) 0.00003002
2. How many significant figures are in each of the following numbers?
__5__ a) 0.000015045
__2__ b) 4,600,000
__4__ c) 2406
__1__ d) 0.000005
__6__ e) 0.0300001
__2__ f) 12,000
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Information: The Exception to Rule #2
There is one exception to the second rule. Consider the following measured values.
It is 1200 miles from my town to Atlanta.
It is 1200.0 miles from my town to Atlanta.
The quantity “1200.0 miles” is more precise than “1200 miles”. The decimal point in the quantity
“1200.0 miles” means that it was measured very precisely—right down to a tenth of a mile.
Therefore, the complete version of Rule #2 is as follows:
Rule #2: Zeros at the end of a number are not significant unless there is a decimal point in
the number. A decimal point anywhere in the number makes zeros at the end of a number
significant.
Not significant because these are at
the beginning of the number!
0.0000007290
This zero is significant because it is
at the end of the number and there
is a decimal point in the number.
Critical Thinking Questions
3. Verify that each of the following numbers contains five significant figures. Circle the digits
that are significant. Digits in red italics are significant.
a) 0.00030200
b) 200.00
c) 2300.0
d) 0.000032000
4. How many significant figures are there in each of the following numbers?
__6__ a) 0.000201000
__5__ b) 23,001,000
__3__ c) 0.0300
__7__ d) 24,000,410
__7__ e) 2400.100
__2__ f) 0.000021
Information: Rounding Numbers
In numerical problems, it is often necessary to round numbers to the appropriate number of
significant figures. Consider the following examples in which each number is rounded so that each
of them contains 4 significant figures. Study each example and make sure you understand why they
were rounded as they were:
42,008,000 42,010,000
12,562,425,217 12,560,000,000
0.00017837901 0.0001784
120 120.0
Critical Thinking Questions
5. Round the following numbers so that they contain 3 significant figures.
a) 173,792
b) 0.0025021
c) 0.0003192
d) 30
__174,000___
__0.00250___
__0.000319__
__30.0______
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6. Round the following numbers so that they contain 4 significant figures.
a) 249,441
b) 0.00250122
c) 12,049,002
d) 0.00200210
__249,400___
__0.002501__
__12,050,000__
___0.002002__
Information: Multiplying and Dividing
When you divide 456 by 13 you get 35.0769230769… How should we round such a number? The
concept of significant figures has the answer. When multiplying and dividing numbers, you need to
simple steps:
1) Count the number of significant figures in each number.
Here’s an example:
3 significant figures
4560
= 325.714285714 = 330
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2 significant figures
have only 2 significant figures
since 2 is the least number of
significant figures in this
problem.
Here’s another example:
13.1× 1.2039 = 15.77109 = 15.8
3 significant
figures
5 significant
figures
have 3 significant figures since 3
is the least number of significant
figures in this problem.
Critical Thinking Questions
7. Solve the following problems. Make sure your answers are in the correct number of
significant figures.
a) (12.470)(270) = ___3400________
b) 36,000/1245 = ___29_________
c) (310.0)(12) = ____3700_________
d) 129.6/3 = ____40____________
e) (125)(1.4452) = ___181_________
f) 6000/2.53 = ____2000________
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Information: Rounding to a Decimal Place
As you will soon discover, sometimes it is necessary to round to a decimal place. Recall the names
of the decimal places:
The hundred The ten
thousands
thousands
place
place
The
thousands
place
The
hundreds
place
The
tens
place
The
ones
place
The
tenths
place
The
hundredths
place
The
thousandths
place
If we rounded the above number to the hundreds place, that means that there can be no significant
figures to the right of the hundreds place. Thus, “175,400” is the above number rounded to the
hundreds place. If we rounded to the tenths place we would get 175,398.4. If we rounded to the
thousands place we would get 175,000.
Critical Thinking Questions
8. Round the following numbers to the tens place.
a) 134,123,018 = ___134,123,020__
b) 23,190.109 = __23,190__________
c) 439.1931 = ___440___________
d) 2948.2 = ___2950______________
Did you know that 30,000 plus 1 does not always equal 30,001? In fact, usually 30,000 + 1 =
30,000! I know you are finding this hard to believe, but let me explain…
Recall that zeros in a number are not always important, or significant. Knowing this makes a big
difference in how we add and subtract. For example, consider a swimming pool that can hold 30,000
gallons of water. If I fill the pool to the maximum fill line and then go and fill an empty one gallon
milk jug with water and add it to the pool, do I then have exactly 30,001 gallons of water in the pool?
because “30,000 gallons” is not a very precise measurement. So we see that sometimes 30,000 + 1 =
30,000!
Rounding numbers when adding and subtracting is different from multiplying and dividing. In
adding and subtracting you round to the least specific decimal place of any number in the problem.
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The tens place
contains a
significant figure.
350.04
+720
1070.04
1070
The hundredths
place contains a
significant figure.
The answer gets rounded to the
least specific place that has a
significant figure. In this case, the tens
place is less specific than the
hundredths place, so the answer is
rounded to the tens place.
Example #2: Subtracting
The thousands
place contains a
significant figure.
7000
- 1770
5230
The tens place
contains a
significant figure.
The answer gets rounded to the
least specific place that has a
5000
Critical Thinking Questions
significant figure. In this case, the
thousands place is less specific than the
tens place, so the answer gets rounded
to the thousands place.
9. a) 24.28 + 12.5 = ___36.8___________
b) 120,000 + 420 = ____120,000________
c) 140,100 – 1422 = __138,700_______
d) 2.24 – 0.4101 = ___1.83_____________
e) 12,470 + 2200.44 = __14,670______
f) 450 – 12.8 = _____440______________
10. The following are problems involving multiplication, dividing, adding, and subtracting. Be
careful of the different rules you need to follow!
a) 245.4/120 = ___2.0______________
b) 12,310 + 23.5 = ___12,330___________
c) (31,900)(4) = ___100,000_________
d) (320.0)(145,712) = __46,000,000______
e) 1420 – 34 = ___1390____________
f) 4129 + 200 = ___4300_______________
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ChemQuest 3
Name: ____________________________
Date: _______________
Hour: _____
Information: Changes in Matter
Books are made of matter. You are made of matter. “Matter” is a fancy word for the “stuff” of
which all objects are made. Every day, matter is changed in different ways. For example, paper can
be changed in many ways—it can be torn, folded, or burned.
A chemical change is any alteration that changes the identity of matter. For example, by passing
electricity through water it can be broken down into hydrogen and oxygen. Burning paper is a
chemical change because after the change takes place, the paper has been changed into different
substances (like ash, carbon dioxide, etc.).
A physical change is any alteration that does not change the identity of the matter. Shredding paper
does not change the paper into a different substance. Dissolving salt in water is a physical change
because after the change, the salt and water are both still there.
Critical Thinking Questions
1. Explain why each of the following is a physical change.
a) boiling water until no water remains
Boiling the water simply changes the state of matter, but it is still H2O before and after it
is boiled.
b) mixing sugar with coffee
The sugar is still sugar before and after you mix it with coffee. Neither the sugar nor the
coffee changes into something entirely different.
2. Explain why each of the following is a chemical change.
a) a car rusting
When a car rusts, the iron metal in the car changes into something else (iron oxide) which
has completely different properties.
b) food digesting
The chemicals in the food are broken down into other chemicals during digestion.
3. Identify each of the following changes as chemical or physical by placing a C or P in each
blank.
__C__ a) acid rain corroding the statue of liberty
__ P__ d) melting steel
__P__ b) dissolving salt in water
__C__ e) dissolving steel in acid
__ P__ c) boiling salt water until just salt remains
__ P__ f) cracking ice
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Information: Elements, Compounds, Mixtures
Examine the following tables. Following the name of each element or compound is the “chemical
formula” of the element or compound; please see the periodic table for the meaning of some of the
symbols (i.e. Na = sodium). Italics tell you that substance is organic.
Elements
Sodium (Na)
Chlorine (Cl)
Carbon (C)
Oxygen (O)
Hydrogen (H)
Compounds
Water (H2O)
Methane (CH4)
Sodium chloride, salt (NaCl)
Carbon dioxide (CO2)
Hydrogen Peroxide (H2O2)
Pure Substances
Salt (NaCl)
Hydrogen (H)
Carbon dioxide (CO2)
Water (H2O)
Aluminum (Al)
Mixtures
Salt water (NaCl and H2O)
Sand
Hydrogen (H) and Oxygen (O)
Sodium (Na) and Chlorine (Cl)
Kool-aid (sugar, water, etc.)
Critical Thinking Questions
4. How are elements different from compounds?
Elements are composed of only one type of atom, but compounds are composed of more than
one.
5. How are compounds different from mixtures? Compounds are formed by a chemical change
(i.e. two hydrogen and one oxygen atom bonding to form a water molecule), but mixtures are
formed by a physical change (i.e. stirring salt and water together.
6. How are pure substances different from mixtures?
Pure substances are not mixed with anything else, but mixtures are composed of two or more
things physically (not chemically) combined.
7. Can something be both a mixture and a pure substance? Explain using examples from the
tables.
No, there is nothing from the table that is in both categories.
8. Is it always possible to identify something as an element, compound, pure substance or
mixture just by looking at it? Explain using examples from the tables.
No, some things look the same, but are not the same at the microscopic scale. For example,
water (a compound) and salt water (a mixture) look exactly the same.
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9. Formulate a definition for each of the following terms.
a) element:
Matter that is composed of only one kind of atom.
b) compound:
Matter that is composed of two or more kinds of atoms chemically combined together
(that is, it is made by a chemical change occurring between two or more atoms).
c) mixture:
Matter that is composed of two or more pure substances physically combined together.
d) pure substance: Matter that is not mixed—either a pure element or a pure compound.
10. Categorize each of the following as an element, compound, mixture, or pure substance. If
more than one label applies, then include both labels. (You will need more than one label
sometimes.)
a) ____Mixture_________ Popsicle
c) __Element, Pure Substance_ Gold
b) Compound, Pure Substance Sugar
d) ______Mixture___________ Dishwater
11. If you have a container with hydrogen gas and oxygen gas in it do you have water? Why or
why not?
You do not have water because the hydrogen and oxygen atoms are simply mixed together;
they are not bonded together.
12. Give an example of something that is an element. Your example should not already be on this
sheet.
Any substance listed on the periodic table. For example, phosphorus or nitrogen.
13. Give an example of something that is a compound. Your example should not already be on
this sheet.
Octane, hydrochloric acid, carbon monoxide, etc.
14. Give an example of something that is a mixture. Your example should not already be on this
sheet.
15. What do all organic substances have in common?
They all contain carbon.
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Information: Homogeneous and Heterogeneous Mixtures
Examine the following table.
Example of Mixture
Salt water
Oil and water
Sugar and salt (no water)
Sugar and salt in water
Sand and water
Carbon dioxide, water, and ice
14 kt. gold (mixture of silver
and gold)
# of phases in
mixture
1
2
2
1
2
3
1
How many kinds of
states in mixture
2
1
1
2
2
3
1
Homogeneous or
heterogeneous?
Homogeneous
Heterogeneous
Heterogeneous
Homogeneous
Heterogeneous
Heterogeneous
Homogeneous
Critical Thinking Questions
16. What is the difference between a "phase of matter" and a "state of matter"? Define each term
as best you can.
The “state” of matter tells us whether the substance is a solid, liquid, or gas. A “phase” of
matter is a section of matter that is the same throughout. For example, in oil and water, there
is one state—the liquid state—and there are two phases—the oil phase and the water phase.
17. What relationship exists between a homogeneous mixture and the number of phases in the
mixture?
A homogeneous mixture must have only one phase.
18. What is the difference between homogeneous and heterogeneous mixtures?
A homogeneous mixture is the same throughout—one small sample of the mixture is an exact
representation of the whole. A heterogeneous mixture has different parts—one small sample
will not have the same composition as the rest of the mixture.
19. If you had to categorize elements as homogeneous or heterogeneous, what category would
you put them in?
All elements are homogeneous. For example, one small piece of gold should have the same
composition as a larger piece of gold.
20. If you had to categorize compounds as homogeneous or heterogeneous, what category would
you put them in?
All compounds are homogeneous. For example, one drop of water will have the same
composition as another drop—all H2O molecules.
21. Categorize each of the following as homogeneous or heterogeneous.
heterogeneous b) ice water
heterogeneous c) dishwater
homogeneous d) 14 kt. Gold
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ChemQuest 4
Name: ____________________________
Date: _______________
Hour: _____
Information: Phase diagrams
Figure 1: Phase diagram for water. Note that the unit for pressure on the diagram is atmospheres
(atm). Another unit is the kilopascal (kPa). Standard pressure is the pressure at sea level and it is
equal to 1 atm, which is equivalent to 101.325 kPa. Standard temperature is 273 Kelvin (K), which
equals 0oC. The abbreviation STP is used for "standard temperature and pressure" and it denotes a
temperature of 0oC and 1 atm (or 273K and 101.325 kPa).
Triple
point
A phase diagram is a graph that illustrates under what conditions the states of matter exist. For
example, in the phase diagram of water above, it should be noted that at 1 atm (which equals 101.325
kPa) of pressure and 50 oC, H2O exists as a liquid. The dark solid lines represent the boundaries
between different states. The term “vapor” is used instead of “gas” because vapor describes a
substance that is normally a liquid at STP. The line that divides the liquid area from the vapor area
has a special name: the vapor pressure curve. (The vapor pressure of water at any given temperature
can be found looking at the vapor pressure curve.) To the left of the line, liquid exists. To the right of
the line, vapor (gaseous H2O) exists. Right on the line, for example when the pressure is 1 atm and
the temperature is 100oC, both liquid and vapor exist at the same time. When two or more things
coexist at the same time it is called “equilibrium”. During equilibrium, liquid changes to vapor and
vapor changes to liquid all at the same time and rate.
Critical Thinking Questions
1. When a substance melts, what happens to the motion of the molecules of the substance?
The molecules begin to move faster and get farther apart.
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2. What is the freezing point and boiling point of water when the pressure on the water is 1 atm?
(I.e. at what temperature will the water freeze and boil when the pressure is 1 atm?)
Freezing point = 0oC; Boiling point = 100oC
3. How does lowering the pressure affect the boiling point?
The boiling point also gets lower.
4. Estimate the boiling point of water when the pressure on the water is 0.1 atm.
Approximately 73oC.
5. The triple point of a substance is the conditions under which a solid, liquid and gas all exist in
equilibrium. On the phase diagram of water, label the location of the triple point.
See diagram.
6. Dry ice sublimes; that is, it changes directly from a solid to a gas. Is it ever possible for ice to
sublime? If so, describe the conditions necessary for sublimation to occur.
Yes, ice will sublime at pressures below approximately 0.007 atm.
7. What is meant by the “vapor pressure” of water?
The pressure that gaseous water exerts, or, the force that water molecules exert after the
molecules have evaporated.
8. From the phase diagram, estimate the vapor pressure of water at 25 oC.
Approximately 0.01 atm.
9. What is the vapor pressure of water when the temperature is 100 oC?
Approximately 1 atm.
10. When the atmospheric pressure equals 1 atm, what is the boiling point of water?
100oC
11. From your answers to questions 9 and 10, formulate a definition for boiling point in terms of
atmospheric pressure and vapor pressure of the liquid.
The boiling point is the temperature at which the vapor pressure of a liquid equals the
atmospheric pressure. As the atmospheric pressure decreases, so does the liquids vapor pressure
and, thus, the boiling point will also decrease.
12. Is it ever possible for solid H2O to exist at a temperature above 0 oC?
Yes, at lower pressures (pressures below 1 atm).
13. Is it ever possible for solid H2O to exist at a temperature above 25 oC?
No, there is no solid region on the phase diagram at any temperature above roughly 5oC.
14. Describe the region where a solid-liquid equilibrium exists.
A solid liquid equilibrium exists along the line that divides the liquid region from the vapor
region.
15. Why might an equilibrium situation be described as a “reversible change”?
Equilibrium describes a situation where two equal and opposite processes occur at the same rate.
In this case, water evaporates and some vapor condenses, thereby “reversing” the change.
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16
ChemQuest 5
Name: ____________________________
Date: _______________
Hour: _____
Information: Heat and Temperature
When a substance absorbs heat energy, the temperature of the substance increases. There are a
number of factors (such as mass of the substance) that affect how much the temperature of a
substance changes.
Critical Thinking Questions
1. Consider two pots of water. Each pot has the same diameter, but pot A is deeper than pot B and
so there is more water in pot A. If both of the pots are exposed to exactly the same amount of
heat for five minutes on a stove, which pot will contain the hottest water after heating?
The water in Pot B will be hotter.
2. Propose an explanation for the fact that even though both pots were exposed to the same amount
of heat, one got hotter.
Since Pot A has more water in it, it will take more heat to increase the temperature.
3. Fill in the blank: The greater the mass of a substance (like water in questions 1 and 2), the
_______lower_________ the temperature change when heat is applied.
greater or lower
4. Consider the metal hood of a car on a warm sunny day. Next to the car is a large puddle of water.
The puddle of water is so large that it has the same mass as the hood of the car. Assume that the
hood of the car and the puddle are exposed to the same amount of sunlight. Which will be hotter
after two hours—the hood of the car or the puddle?
The hood of the car will be hotter.
5. Propose an explanation for the fact that even though both the hood and the puddle were exposed
to the same amount of heat energy and their masses were the same, one still got much warmer
than the other.
The metal must heat up faster than water does. There must be something about metal that allows
it to heat up faster than water.
6. In general is it harder to change the temperature of metal (like on a car hood) or of water? In
other words, would it require more heat energy to change the temperature of metal or of water?
It would require more heat energy to change the temperature of water.
17
7. In one of the following blanks you will need to write “400” and in the other blank you will need
to write “200”.
If we wanted to change the temperature of water by 4 oC, then _400___ Joules of heat energy
are required, but to change the temperature of metal by 4oC, then __200___ Joules of heat energy
are required.
Information: Specific Heat
In questions 1 and 2 above you probably recognized that the temperature change of a substance
depends on the mass of the substance. You also have probably experienced the fact that different
substances heat at different rates as was discussed in questions 3 and 4 above. Each substance has its
own specific heat capacity. The specific heat capacity is a measure of the amount of energy needed
to change the temperature of the substance. The higher the specific heat, the more energy is required
to change the temperature of the substance.
Critical Thinking Questions
8. Which substance has a higher specific heat—water or a metal like the aluminum in a car hood?
Since it requires more heat energy change the temperature of water and since specific heat is a
measure of how much energy is needed to change the temperature of a substance, water must
have a higher specific heat.
9. Consider 200 Joules (J) of heat energy applied to several objects and fill in the blank: The higher
the specific heat of the object, the __lower___________ the temperature change of the object.
lower or higher
10. Given the following symbols and your answers to the above questions, which of the following
equations is correct. Make sure you have the correct answer before proceeding to the next
questions!
∆T = temperature change of a substance
Cp = specific heat capacity
m = mass of the substance
q = amount of heat energy applied to the substance
(C p )(m)
q
(q)(m)
A) ∆T = qmCp
B) ∆T =
C) ∆T =
D) ∆T =
(m)(C p )
Cp
q
HINT: equation D is not correct because according to that equation, a large mass (m) will lead
to a large temperature change (∆T), but this is not consistent with question 3.
11. If ∆T is measured in oC, m is measured in grams (g) and q is measured in Joules (J), what are the
units for specific heat capacity?
q
q
J
∆T =
Cp =
Cp =
(m)(C p )
(m)(∆T )
(g)( o C)
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18
12. What is the temperature change of a 120 g piece of aluminum whose specific heat is 1.89 J/goC
after 1800 J of heat energy is applied?
1800 J
q
∆T =
∆T =
= 7.936508 o C = 7.9 o C
J
(m)(C p )
(120 g)(1.89 o )
g C
13. A beaker containing 250.0 g of water is heated with 1500.0 J of heat energy. If the temperature
of the water changed from 22.0000oC to 23.4354oC, what is the specific heat of water?
∆T = 23.4354 oC – 22.0000 oC = 1.4354 oC
q
1500 J
J
q
∆T =
Cp =
Cp =
= 4.180 o
o
(m)(C p )
(m)(∆T )
(250 g)(1.4354 C)
g C
14. Heat energy equal to 25,000 J is applied to a 1200 g brick whose specific heat is 2.45 J/goC.
a) What is the change in temperature of the brick?
∆T =
q
25,000
=
= 8.503401 = 8.5 o C
(m)(C p ) (1200)(2.45)
b) If the brick was initially at a temperature of 25.0oC, what is the final temperature of
the brick?
25.0 + 8.5 = 33.5 oC
19
ChemQuest 6
Name: ____________________________
Date: _______________
Hour: _____
Information: Specific Heat Constants Depend on the State of Matter
Recall that specific heat is the energy necessary to raise the temperature of one gram of a substance
by one degree Celsius. The specific heat of liquid water is a constant equal to 4.18 J/(g-oC). This
means that it takes 4.18 Joules (J) of heat energy to raise the temperature of each gram of water by
one degree Celsius. For ice, the specific heat is 2.06 J/goC. Steam’s specific heat is 2.02 J/goC. You
will want to remember the three different specific heat values for H2O.
Critical Thinking Questions
1. How much energy is required to heat 32.5g of water from 34oC to 75oC?
∆T = 75-34 = 41
q = (m)(Cp)(∆T) = (32.5)(4.18)(41) = 5569.9 J
2. How many J of energy are needed to heat 45.0g of steam from 130oC to 245oC? Why don’t
you use 4.18 J/goC in this calculation?
∆T = 245-130 = 115
q = (m)(Cp)(∆T) = (45.0)(2.02)(115) = 10,453.5 J
You don’t use 4.18 because 4.18 is the specific heat for liquid water; 2.02 is for steam.
3. Why is it impossible for you to answer the following question right now: How much energy is
required to heat 35g of H2O from 25oC to 150oC?
The problem involves both liquid water and steam and we don’t know how to combine them
in the same calculation….YET
Information: Energy Involved in Changing State
Obviously, it takes energy to heat a substance such as water. There is also an energy change when a
substance changes state. For example, when water freezes, energy is taken away from the water and
when water boils, energy is being added to the water.
When water is heated up to 100oC, it will NOT automatically boil unless more energy is added. Once
water is at the correct temperature (100oC), it will boil IF you add energy to change it from a liquid to
a gas. The extra energy is needed to separate the molecules from one another. Note: at the boiling
point of a liquid energy is added to the liquid, but the temperature of the liquid does NOT
change. The temperature of H2O will increase above 100oC only after all of the water has been
changed to steam. After the phase change, if more energy is applied, the temperature will go up.
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20
Figure 1: Graph of the temperature of H2O vs. energy added to the water.
100oC
Temperature of
H2O (oC)
0 oC
0
25
50
75 100 125 150 175 200 225 250 275 300
Energy added to water in kilojoules (kJ)
Critical Thinking Questions
4. What is the significance of the horizontal portions of the graph? What is going on during
those times?
The horizontal portions indicate a phase change such as melting or boiling.
5. Label the temperature scale on the graph.
See graph.
6. Does it require more energy to “vaporize” water at the boiling point or to melt water at the
melting point? Explain.
It takes more energy to vaporize than to melt because the horizontal section of the graph is
much longer during the vaporizing, or boiling.
Information: Mathematics of Changes of State
Recall the equation: q = (m)(Cp)(∆T) where q is the heat energy, Cp is the specific heat of the
substance, m is the mass of the substance, and ∆T is the change in the temperature. This equation
works ONLY when there is no phase change. You use it to find how much energy is required to
change the temperature of a certain substance as long as you know the specific heat of the substance
and as long as no change in state occurs.
Now for when there is a change in state: The energy required for a change of state is given a special
name called enthalpy. So the “enthalpy of vaporization” (symbolized by ∆Hvap) of water is the
energy needed to vaporize (or boil) one gram of water when the water is already at its boiling point.
The ∆Hvap of water is 2260 J/g. So for each gram of hot (100oC) water, 2260 J are required to
vaporize it. Similarly, the enthalpy of fusion (∆Hfus) of H2O is 334 J/g. So each gram of ice at 0oC
there are 334 J of energy required to melt it. (Fusion is melting.)
21
Critical Thinking Questions
7. Verify that it takes 4140.7 J of energy to heat 12.7 g of water from room temperature (22oC)
to the boiling point (100oC). Recall that the specific heat (Cp) of water is 4.18 J/(g-oC).
∆T = 100-22 = 78
q = (m)(Cp)(∆T) = (12.7)(4.18)(78) = 4140.7 J
8. Given the fact that the ∆Hvap of water is 2260 J/g we know that 2260 J of energy is required to
vaporize 1 gram of water. To vaporize 2 grams of water it requires 4520 J. For 3 grams,
6780 J are needed. Verify that it takes 28,702 J of energy to vaporize 12.7 g of water, when
the water is already at its boiling point.
(2260 J/g)(12.7g) = 28,702 J; note: a helpful equation for phase changes is q=(m)(∆H)
9. Verify that it takes 1513.6 J of energy to heat 12.7 g of steam from 100oC to 159oC. Note: the
specific heat of steam (Cp) is 2.02 J/(g-oC).
∆T = 159-100 = 59
q = (m)(Cp)(∆T) = (12.7)(2.02)(59) = 1513.6 J
10. Using only your answers to questions 7-9, calculate how many Joules of energy it takes to
change 12.7 g of water at 22oC to steam at 159oC.
Questions 7-9 are each different “steps” for taking 12.7g of water from 22oC to 159oC so you
can simply add the answers: 4140.7 + 28,702 + 1513.6 = 34,356.3 J
11. Calculate how many Joules of energy would be required to change 32.9 g of water at 35oC to
steam at 120oC. You will need to break this problem into four steps.
a) Find the Joules needed to heat the water to the boiling point.
q = (m)(Cp)(∆T) = (32.9)(4.18)(65) = 8938.9 J
b) Find the Joules needed to vaporize the water.
q=(m)(∆H) = (32.9)(2260) = 74,354 J
c) Find the Joules needed to heat the vapor (steam) from the boiling point to 120oC.
q = (m)(Cp)(∆T) = (32.9)(2.02)(20) = 1329.2 J
8938.9 + 74,354, + 1329.2 = 84,622 J
12. How much heat energy would be required to change the temperature of 125g of ice from
-32.9oC to liquid water at 75oC?
q = (m)(Cp)(∆T) = (125)(2.06)(32.9) = 8471.75 J
q=(m)(∆H) = (125)(334) = 41,750 J
q = (m)(Cp)(∆T) = (125)(4.18)(75) = 39,187.5 J
Total: 8471.75 + 41,750 + 39,187.5 = 89,406 J
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22
ChemQuest 7
Name: ____________________________
Date: _______________
Hour: _____
Information: Density
Object
1
2
3
4
5
6
7
Mass of Object (g)
21.50
12.6
41.90
32.90
59.5
0.594
17.23
Volume of Object (mL)
18.40
14.7
31.60
12.85
61.7
0.574
21.67
Density of Object (g/mL)
1.168
0.857
1.326
2.560
0.964
1.035
0.7951
Critical Thinking Questions
1. Consider the data for objects 1, 2 and 3. Which of the following equations correctly show the
relationship(s) between mass (M), volume (V) and density (D)? There may be more than one
A)
D=
V
M
F) V = DM
B)
M=
V
D
G) M = DV
C)
D=
M
V
D)
V=
D
M
E)
V=
M
D
H) D = MV
2. For objects 4, 5, 6 and 7 there are blanks in the table. Using your answers to question 1, fill in the
blanks.
See the table above.
3. What are the units for density if the mass of an object was measured in kilograms and the volume
in liters?
kg/L
4. In your own words, define “density”.
Density: the amount of mass an object has per unit volume. Density includes the idea of how
much mass something has in comparison to its size.
5. Calculate the density of an object that has a mass of 45.0 kg and a volume of 20.0 L. Include
units.
m 45.0kg 2.25 kg/L
D= =
=
V
20.0 L
23
Information: Mass and Weight
The following is a table that relates the mass of an object and the weight of the same object. The pull
of gravity is a constant. As long as you stay in the same place, the pull of gravity (g) does not
change. For example, at sea level g is always equal to 9.8 m/s2.
Object
1
2
3
4
Place
Earth, sea level
Earth, sea level
Earth, Mt. Everest
Moon
Pull of Gravity
9.8 m/s2
9.8 m/s2
9.1 m/s2
1.6 m/s2
Mass
42 kg
29 kg
38 kg
51 kg
Weight
411.6 N
284.2 N
345.8 N
81.6 N
Critical Thinking Questions
6. What is the relationship between mass (M), the pull of gravity (g) and weight (W)?
A) M = gW
B) g = MW
C) W = Mg
7. What is the weight of a 42 kg object (like object 1) if the object was on the moon where g is
always 1.6 m/s2?
W = Mg = (42kg)(1.6 m/s2) = 67.2 N = 67 N
8. Mass measures how much matter an object has. It is an indication of how much “stuff” is in
the object. It does not describe the size of the object or the weight of the object. What is the
difference between mass and weight?
Weight is affected by gravity, but mass is not.
9. In question 8, you calculated the weight of object 1 on the moon. How did that weight compare
to the object’s weight on Earth?
Object 1 and all other objects weigh less on the moon than they do on Earth. The mass,
however, of any object will be the same on Earth or on the moon.
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24
ChemQuest 8
Name: ____________________________
Date: _______________
Hour: _____
Information: Structure of the Atom
Note the following symbols: (they are not to scale)
= proton (positive charge)
= electron (negative charge)
= neutron (no charge)
The following three diagrams are hydrogen atoms:
1
1
2
1
H
H
3
1
H
The following three diagrams are carbon atoms:
12
6
13
6
C
(6 protons, 6 neutrons)
Notice the type of notation used for atoms:
C
(6 protons, 7 neutrons)
A
Z
14
6
C
(6 protons, 8 neutrons)
X
X = chemical symbol of the element
Z = “atomic number”
A = “mass number”
12
6
1
1
C,
13
6
C, and 146 C are notations that represent isotopes of carbon.
3
H , 21 H and 1 H are notations that represent isotopes of hydrogen.
The part of the atom where the protons and neutrons are is called the nucleus.
25
Critical Thinking Questions
1
1. How many protons are found in each of the following: 1 H ? in
Each has one proton, given by the atomic number of one.
1
2
1
H ? in 31 H ?
2
3
2. How many neutrons are found in each of the following:1 H ? in 1 H? in 1 H ?
2
1
3
1 H has one neutron
1 H has zero neutrons
1 H has two neutrons
and subtracting the atomic
Calculate the number of neutrons by taking the mass number
number.
3
1
2
3. How many electrons are found in each of the following: 1 H ? in 1 H ? in 1 H ?
Each has one electron. Every atom has the same number of electrons as protons.
4. What structural characteristics do all hydrogen atoms have in common?
Each hydrogen atom has the same number of protons (1) and the same number of electrons
(1).
5. What structural characteristics do all carbon atoms have in common?
Each carbon atom has the same number of protons (6) and the same number of electrons (6).
6. What does the mass number tell you? Can you find the mass number of an element on the
periodic table? It tells you the total number of protons plus neutrons. The mass number is not
found on the periodic table.
7. What does the atomic number tell you? Can you find the atomic number of an element on the
periodic table? It tells you the number of protons that an atom has. The atomic number is found
on the periodic table.
8. Define the term isotope. The term “isotope” refers to an atom that has the same number of
protons as another atom, but a different number of neutrons.
9. How does one isotope of carbon differ from another isotope of carbon?
The isotopes have different numbers of neutrons (but the same number of protons).
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26
Information: Atoms, Ions, Masses of Subatomic Particles
The atomic mass unit (amu) is a special unit for measuring the mass of very small particles such as
atoms. The relationship between amu and grams is the following: 1.00 amu = 1.66 x 10-24g
Note the following diagrams of a fluorine atom and a fluorine ion.
atom
ion
9 protons
10 neutrons
9 protons
10 neutrons
19
9
F
mass = 18.9980 amu
atom
19
9
F -1
mass = 18.9985 amu
ion
12 protons
12 neutrons
12 protons
12 neutrons
24
12
Mg
mass = 23.9978 amu
24
12
Mg + 2
mass = 23.9968 amu
Critical Thinking Questions
10. What is structurally different between an atom and an ion? Note: This is the ONLY structural
difference between an atom and an ion.
An ion has a different number of electrons than protons. An atom always has the same
number of protons as electrons. For example, the fluoride ion has one more electron than
protons and the magnesium ion has two fewer electrons than protons.
11. In atomic mass units (amu), what is the mass of an electron? Since the fluoride ion has one extra
electron than the fluorine atom, you can subtract their masses to find the mass of one electron.
18.9985 – 18.9980 = 0.0005 amu
12. Is most of the mass of an atom located in the nucleus or outside the nucleus? How do you know?
Most of the mass is inside the nucleus. The electrons only have a mass of 0.0005 amu each.
All of the rest of the mass is in the nucleus, where the protons and neutrons are located.
27
13. If protons and neutrons have the same mass, what is the approximate mass of a proton and
neutron in atomic mass units (amu)? Almost all of the nearly 19 amu of a fluorine atom is located
in the nucleus, where there are a total of 19 protons and neutrons. The 19 protons and neutrons
add up to a mass of approximately 19 amu and thus each proton and each neutron has a mass of
14. The mass of carbon-14 (mass number = 14) is about 14 amu. Does this agree with what you
determined in questions 11 and 13?
Yes, since each proton and each neutron has a mass of about 1 amu (question #13) and the
electrons have almost no mass (question #11) the mass of an atom that has 14 protons and
neutrons should be about 14 amu.
15. The charge (in the upper right hand corner of the element symbol) is –1 for a fluorine ion. Why
isn’t it +1 or some other number?
The fluoride ion has one more electron (negative charge) than protons (positive charge).
16. What is the charge on every atom? Why is this the charge?
Every atom has a charge of zero (neutral) because of equal numbers of protons and electrons.
17. How do you determine the charge on an ion?
Compare the number of protons and electrons. Take the number of protons and subtract the
number of electrons.
18. An oxygen ion has a –2 charge. (Use your periodic table if necessary)
a) How many protons does the oxygen ion have? From the periodic table we see that oxygen
is atomic number 8 and therefore it has 8 protons.
b) How many electrons does the oxygen ion have? 10 electrons because it must have two
more electrons than protons.
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28
ChemQuest 9
Name: ____________________________
Date: _______________
Hour: _____
Information: Weighted Averages
Examine the table of student test scores for five tests they have taken.
Test
1
2
3
4
5
Student A
95
74
82
92
81
84.8
Student B
76
88
90
81
72
81.4
Critical Thinking Questions
1. Calculate the average grade for students A and B and enter the average in the table above.
See table above.
2. If you know a student’s average grade can you tell what the student’s individual test scores were?
Explain.
No, two students with different test scores could still end up with the same average grade.
3. Suppose student C had an average of 83%. On each of his five tests he scored either 65% or
95%. Which score occurred more often? Explain.
The score of 95% occurred more often because 95% is closer to the average of 83%.
4. What if the teacher decided that test five would count for 40% of the final grade and test four
would count for 30% of the final grade and each of the other tests would count for 10%.
Calculate the new average for each student. Note: this is called the weighted average.
Student A’s new average: ____85.1____
Student B’s new average: ____78.5_______
Student A= (0.40)(81) + (0.30)(92) + (0.10)(82) + (0.10)(74) + (0.10)(95) = 85.1
Student B = (0.40)(72) + (0.30)(81) + (0.10)(90) + (0.10)(88) + (0.10)(76) = 78.5
Note: when finding the weighted average you do not divide at the end.
29
Information: Average Atomic Mass
On the periodic table you can find the average atomic mass for an element. This average is a
weighted average and it tells you the average mass of all the isotopes of an element. The periodic
table does not contain mass numbers for individual atoms, instead you can find the average mass of
atoms. The average atomic mass is calculated just how you calculated the weighted average in
question 4 above.
Critical Thinking Questions
5. Neon has three different isotopes. 90.51% of neon atoms have a mass of 19.992 amu. 0.27% of
neon atoms have a mass of 20.994 amu. 9.22% of neon atoms have a mass of 21.991 amu. What
is the average atomic mass of neon?
(0.9051)(19.992) + (0.0027)(20.994) + (0.0922)(21.991) = 20.179 amu
6. Chlorine-35 is one isotope of chlorine. (35 is the mass number.) Chlorine-37 is another isotope
of chlorine. How many protons and how many neutrons are in each isotope of chlorine?
Chlorine has an atomic number of 17, so each isotope has 17 protons. Neutrons are found by
taking the mass number and subtracting the atomic number. Chlorine-35 has 18 neutrons (35-17)
and Chlorine-37 has 20 neutrons (37-17).
7. Of all chlorine atoms, 75.771% are chlorine-35. Chlorine-35 atoms have a mass of 34.96885
amu. All other chlorine atoms are chlorine-37 and these have a mass of 36.96590. Calculate the
average atomic mass of chlorine.
Percent of chlorine atoms that are chlorine-37: 100% - 75.771% = 24.229%
(0.75771)(34.96885) + (0.24229)(36.96590) = 35.4527 amu
8. Do your answers for questions 5 and 7 agree with the average atomic masses for neon and
chlorine on the periodic table?
Yes, both answers should agree with the periodic table.
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30
Skill Practice
9. Complete the following table.
Symbol
# of neutrons
16
# of protons
15
# of electrons
15
Atomic #
15
Mass #
31
15
13
10
13
28
42
38
38
38
80
69
50
50
50
119
Po
126
84
84
84
210
N −3
8
7
10
7
15
31
2815
13
P
Al +3
80
38
119
50
Sr
210
84
15
7
Sn
10. A certain element has two isotopes. One isotope , which has an abundance of 72.15% has a mass
of 84.9118 amu. The other has a mass of 86.9092 amu. Calculate the average atomic mass for
this element.
% of other isotope = 100% - 72.15% = 27.85%
(0.7215)(84.9118) + (0.2785)(86.9092) = 85.468 amu
11. Given the following data, calculate the average atomic mass of magnesium.
Isotope
Magnesium-24
Magnesium-25
Magnesium-26
Mass of Isotope
23.985
24.986
25.983
(0.7870)(23.985) + (0.1013)(24.986) + (0.1117)(25.983) = 24.309 amu
Abundance
78.70%
10.13%
11.17%
31
ChemQuest 10
Name: ____________________________
Date: _______________
Hour: _____
Information: Bohr’s Solar System Model of the Atom
All the planets are attracted to the sun by gravity. The reason that the Earth doesn’t just float out into
space is because it is constantly attracted to the sun. Similarly, the moon is attracted to the Earth by
Earth’s gravitational pull. So all of the planets are attracted to the sun but they never collide with the
sun.
Negative charges are attracted to positive ones. Therefore the negative electrons in an atom are
attracted to the positive protons in the nucleus. In the early 1900’s scientists were looking for an
explanation to a curious problem with their model of the atom. Why don’t atoms collapse? The
negative electrons should collapse into the nucleus due to the attraction between protons and
electrons. Why doesn’t this happen? Scientists were at a loss to explain this until Neils Bohr
proposed his “solar system” model of the atom.
Critical Thinking Questions: Bohr’s reasoning
1. Consider swinging a rock on the end of a string in large circles. Even though you are constantly
pulling on the string, the rock never collides with your hand. Why is this?
The rock must move so quickly that the pull from your hand is not enough to pull the rock into
your hand. If you were to let go, it would fly away from you in a straight line. The pull from
your hand keeps the rock near you, but because of its initial velocity it doesn’t get closer to you.
2. Even though the Earth is attracted to the sun by a very strong gravitational pull, what keeps the
Earth from striking the sun?
The fact that the earth is moving keeps it from striking the sun. The earth would move away from
the sun, but the gravity of the sun attracts it, causing a circular orbit for the earth.
3. Why doesn’t the moon strike the Earth?
The reasoning is the same as that for questions 1 and 2.
4. How could it be possible for electrons to not “collapse” into the nucleus?
The electrons must be moving in orbits so fast that the pull from the nucleus is not enough to
attract them toward the nucleus—this is Bohr’s “solar system model” of the atom.
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32
Information: Light
Recall that light is a wave. White light is composed of all the colors of light in the rainbow. All light
travels at the same speed (c), 3.00 x 108 m/s. (The speed of all light in a vacuum is always equal to
3.0 x 108 m/s.) Different colors of light have different frequencies (f) and wavelengths (λ). The
speed (c), frequency (f) and wavelength (λ) of light can be related by the following equation:
c=f •λ
It is important that the wavelength is always in meters (m), the speed is in m/s and the frequency is in
hertz (Hz). Note: 1 Hz is the inverse of a second so that 1 Hz = 1/s.
Critical Thinking Questions
5. As the frequency of light increases, what happens to the wavelength of the light?
As the frequency increases, the wavelength must be smaller. This is because the frequency times
the wavelength must always equal 3.00x108 m/s.
6. What is the frequency of light that has a wavelength of 4.25 x 10-8 m?
3.00 x10 8 m / s
c = f •λ → f = =
= 7.06 x1015 Hz
−8
λ
4.25 x10 m
7. What is the wavelength of light that has a frequency of 3.85 x 1014 Hz?
c 3.00 x108 m / s
c = f •λ → λ = =
= 7.79 x10 −7 m
14
f
3.85 x10 Hz
c
Information: Energy levels
After Bohr proposed the Solar System Model (that electrons orbit a nucleus just like planets orbit the
sun), he called the orbits “energy levels”.
Consider the following Bohr model of a Boron atom:
= proton (positive charge)
= electron (negative charge)
= neutron (no charge)
1st
energy
level
2nd
energy
level
Higher energy levels are further from the nucleus. For an electron to go into a higher energy level it
must gain more energy. Sometimes the electrons can absorb light energy. (Recall that different
colors of light have different frequencies and wavelengths.) If the right color (and therefore,
frequency) of light is absorbed, then the electron gets enough energy to go to a higher energy level.
The amount of energy (E) and the frequency (f) of light required are related by the following
equation:
E=h•f
E is the energy measured in Joules (J), f is the frequency measured in Hz, and h is Planck’s constant
in units of J/Hz which is the same as J-second. Planck’s constant, h, always has a value of
6.63x10-34J-s.
33
An electron that absorbs energy and goes to a higher energy level is said to be “excited.” If an
excited electron loses energy, it will give off light energy. The frequency and color of light depends
on how much energy is released. Again, the frequency and energy are related by the above equation.
When an excited electron loses energy, we say that it returns to its “ground state”. Since not all
electrons start out in the first energy level, the first energy level isn’t always an electron’s ground
state.
Critical Thinking Questions
8. Does an electron need to absorb energy or give off energy to go from the 2nd to the 1st energy
level?
It needs to give off energy to go to a lower energy level.
9. How is it possible for an electron go from the 3rd to the 4th energy level?
The electron must absorb energy to go to a higher energy level.
10. Red light of frequency 4.37 x 1014 Hz is required to excite a certain electron. What energy did the
electron gain from the light?
E = h • f = (6.63 x10 −34 )(4.37 x1014 ) = 2.90 x10 −19 J
11. The energy difference between the 1st and 2nd energy levels in a certain atom is 5.01 x 10-19 J.
What frequency of light is necessary to excite an electron in the 1st energy level?
E 5.01x10−19
E = h• f → f = =
= 7.56 x1014 Hz
−34
h 6.63x10
12. a) What is the frequency of light given off by an electron that loses 4.05 x 10-19 J of energy as it
moves from the 2nd to the 1st energy level?
E 4.05 x10 −19
E = h• f → f = =
= 6.11x1014 Hz
− 34
h 6.63x10
b) What wavelength of light does this correspond to (hint: use c = f λ)?
c 3.00 x10 8 m / s
c = f •λ → λ = =
= 4.91x10 −7 m
14
f
6.11x10 Hz
13. Do atoms of different elements have different numbers of electrons?
Yes, atoms of different elements have different numbers of electrons, just as they have
different numbers of protons.
14. If all of the electrons in atom “A” get excited and then lose their energy and return to the ground
state the electrons will let off a combination of frequencies and colors of light. Each frequency
and color corresponds to a specific electron making a transition from an excited state to the
ground state. Consider an atom from element “B”. Would you expect the excited electrons to let
off the exact same color of light as atom “A”? Why or why not?
A different color would be let off because atom B and atom A have different numbers of
electrons. Therefore, we would expect at least a slight difference in colors that are let off.
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ChemQuest 11
Name: ____________________________
Date: _______________
Hour: _____
Information: Energy Levels and Sublevels
As you know, in his solar system model Bohr proposed that electrons are located in energy
levels. The current model of the atom isn’t as simple as that, however.
Sublevels are located inside energy levels just like subdivisions are located inside cities.
Each sublevel is given a name. Note the following table:
TABLE 1
Energy Level
1st energy level
2nd energy level
3rd energy level
4th energy level
Names of sublevels that exist in the energy level
s
s and p
s, p, and d
s, p, d, and f
Note that there is no such thing as a “d sublevel” inside of the 2nd energy level because there are only
s and p sublevels inside of the 2nd energy level.
Critical Thinking Questions
1. How many sublevels exist in the 1st energy level?
One: only the s sublevel exists.
2. How many sublevels exist in the 2nd energy level?
Two: the s and the p sublevels exist.
3. How many sublevels exist in the 3rd energy level?
Three: the s, p, and d sublevels exist.
4. How many sublevels would you expect to exist in the 5th energy level?
The number of sublevels equals the energy level number, so in the 5th energy level we should
expect 5 sublevels to exist.
5. Does the 3f sublevel exist? (Note: the “3” stands for the 3rd energy level.)
No, in the 3rd energy level there are only s, p, and d sublevels. The following sublevels exist
in the 3rd energy level: 3s, 3p, and 3d.
35
Information: Orbitals
So far we have learned that inside energy levels there are different sublevels. Now we will look at
orbitals. Orbitals are located inside sublevels just like streets are located inside subdivisions.
Different sublevels have different numbers of orbitals.
TABLE 2
# of Orbitals
Sublevel
Possible
s
1
p
3
d
5
f
7
Here’s an important fact: only two electrons can fit in each orbital. So, in an s orbital you can have a
maximum of 2 electrons; in a d orbital you can have a maximum of 2 electrons; in any orbital there
can only be two electrons.
Since a d sublevel has 5 orbitals (and each orbital can contain up to two electrons) then a d sublevel
can contain 10 electrons (= 5 x 2). Pay attention to the difference between “sublevel” and “orbital”.
Critical Thinking Questions
6. How many orbitals are there in a p sublevel? 3
7. How many orbitals are there in a d sublevel? 5
8. a) How many total sublevels would be found in the entire 2nd energy level? Two, since the 2nd
energy level has s and p sublevels.
b) How many orbitals would be found in the entire 2nd energy level? 4 orbitals: the 2s
sublevel has 1 orbital and the 2p sublevel has 3 orbitals for a total of 4 orbitals.
9. a) How many electrons can fit in an f sublevel?
(2 electrons per orbital) x (7 orbitals in an f sublevel) = 14 electrons
b) How many electrons can fit in an f orbital?
Any orbital can only hold 2 electrons.
10. How many electrons can fit in a d orbital? in a p orbital? in any kind of orbital?
Each orbital can hold a maximum of 2 electrons.
11. In your own words, what is the difference between a sublevel and an orbital?
Orbitals are contained within sublevels. A sublevel is a grouping of orbitals.
12. How many electrons can fit in each of the following energy levels:
1st energy level = 2 because only 2 can fit in an s sublevel and the first energy
level only has an s sublevel.
2nd energy level = 8; 2 in the 2s sublevel and 6 in the 2p sublevel giving a total
of 8.
3rd energy level = 18; 2 in the 3s sublevel, 6 in the 3p sublevel, and 10 in the
3d sublevel giving a total of 18.
4th energy level = 32; 2 in the 4f, 6 in the 4p, 10 in the 4d, and 14 in the 4f.
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Information: Representing the Most Probable Location of an Electron
The following is an “address” for an electron—a sort of shorthand notation. The diagram below
represents an electron located in an orbital inside of the p sublevel in the 3rd energy level.
EXAMPLE #1:
Some important facts about the above diagram:
• The arrow represents an electron.
• The upward direction means that the electron is spinning clockwise.
• “3p” means that the electron is in the p sublevel of the 3rd energy level.
• Each blank represents an orbital. Since there are three orbitals in a p sublevel, there are also
three blanks written beside the p.
• In the diagram, the electron is in the first of the three p orbitals.
Here’s another example:
EXAMPLE #2:
Critical Thinking Questions
13. In example #2, why are there 5 lines drawn next to the d?
In a d sublevel there are 5 orbitals and therefore there needs to be 5 lines drawn.
14. In example #2, what does it mean to have the arrow pointing down?
A downward arrow indicates that the electron is spinning counterclockwise.
15. Write the notation for an electron in a 2s orbital spinning clockwise.
16. Write the notation for an electron in the first energy level spinning clockwise.
Note: in the first energy level, the only sublevel is a 1s.
17. What is wrong with the following notation? You should find two things wrong.
There is no such thing as a d sublevel in the 2nd energy level. Also, a d sublevel needs to have
5 lines drawn for the 5 orbitals in the d sublevel.
18. Write the notation for an electron in the 4th energy level in an f sublevel spinning clockwise.
37
ChemQuest 12
Name: ____________________________
Date: _______________
Hour: _____
Information: Quantum Numbers
Quantum mechanics is a set of complex mathematics that is used to describe the most probable
location of an electron. Shortly after Bohr, a man by the name of Heisenberg proposed an
uncertainty principle, which stated that it is impossible to know both the exact position and the
exact velocity of a small particle at the same time. The location of an electron in an atom is based on
probability—the most likely location for an electron.
To locate the most probable location of a person you need 4 things. If you know 4 things: state, city,
street and house number then you know the most probable location of the person.
You also need 4 things, called “quantum numbers”, to describe the most probable location for an
electron. Each “number” is actually symbolized by a letter:
TABLE 1
Quantum number
Principal quantum number, n
Azimuthal quantum number, l
Magnetic quantum number, ml
Spin quantum number, ms
What the Quantum number tells us
which energy level the electron is in
which sub-level within the energy level the electron is in
which orbital within the sub-level the electron is in
direction of electron spin (clockwise or counterclockwise)
The four quantum numbers—n, l, ml and ms—come from a very complex equation. Together all four
of them (n, l, ml, ms) will describe the most probable location of an electron kind of like how (x, y, z)
describes the location of a point on a graph.
TABLE 2: Rules governing what values quantum numbers are allowed to have
Quantum number
Possible values
n
1, 2, 3, 4, …integer values
l
0, 1, 2, …integers up until n-1
ml
-l, ..., 0, …, +l
ms
+½ or -½
Examples:
• If n = 3, then the electron is in the 3rd energy level and l is allowed to have only a value of 0,
1, or 2. It cannot equal anything higher than 2 because n-1=2.
• If l = 1, then ml is allowed to have a value of -1, 0, 1.
• If l = 2, then ml is allowed to have a value of -2, -1, 0, 1, or 2.
• ms can only be +½ meaning that the electron is spinning clockwise or -½ meaning that the
electron is spinning counterclockwise.
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Critical Thinking Questions
1. Using the quantum numbers (n, l, ml, ms) explain why each of the following is not an allowed
combination of quantum numbers. The first one is done for you.
a) (3, 4, 1, +½) Here n = 3 and so l can only have values of 0, 1, or 2. Here, however, l
has a value of 4, which is impossible.
b) (2, 1, -2, -½) Here l = 1 and so ml can have values of -1, 0, or 1. Therefore an ml
value of -2 is incorrect.
c) (1, 0, 0, -1) ms = -1, but it can only have values of +½ or -½
2. If n=4, what are the possible values of l? 0, 1, 2, or 3
3. If l = 3, what are the possible values for ml? -3, -2, -1, 0, 1, 2, or 3
4. Fill in the blanks in the following table.
Principle Quantum
Number (n)
n=1
n=2
n=3
n=4
Sublevels that are possible
Possible values for l
s
s and p
s, p, and d
s, p, d and f
0
0 or 1
0, 1, or 2
0, 1, 2, or 3
5. Given the above table and remembering that the quantum number l tells us which sublevel (s,
p, d, or f), complete the following statements. The first is done for you.
• For an s sublevel, l equals ____0____.
•
For a p sublevel, l equals ___1____.
•
For a d sublevel, l equals ____2___.
•
For an f sublevel, l equals ___3____.
Information: Correlating quantum numbers and sublevels
TABLE 3
Azimuthal Quantum
Number (l)
Sublevel
0
1
2
3
s
p
d
f
# of Orbitals
Possible in
the sublevel
1
3
5
7
Possible ml values
0
-1, 0, +1
-2, -1, 0, +1, +2
-4, -3, -2, -1, 0, 1, 2, 3, 4
The same as Table 2 from ChemQuest 11
39
Critical Thinking Questions
6. How many ml values are possible for a d sublevel? 5 values are possible. (Since l = 2 for the d
sublevel, the possible ml values are -2, -1, 0, 1, and 2—a total of 5 values possible.)
7. How many orbitals are there for a d sublevel? 5
8. How many ml values are possible for an f sublevel? 7 values are possible. (Since l = 3 for the
f sublevel, the possible ml values are -3, -2, -1, 0, 1, 2, and 3—a total of 7 values possible.)
9. How many orbitals are there for a d sublevel? 7
between the number of orbitals and the number of ml values possible?
The number of possible ml values equals the number of orbitals for a sublevel. Each value of
ml corresponds to an orbital.
Information: Correlating quantum numbers to what you already know
ms = +½ for this electron
ml = +1 for this orbital
n = 3 for the 3rd
energy level.
l = 1 for the p
sublevel.
ml = 0 for this orbital
ml = -1 for this
orbital
The quantum numbers (n, l, ml, ms) for the above diagram are: (3, 1, -1, +½).
• n = 3 indicating the third energy level
• l = 1 indicating the p sublevel
• ml = -1 indicating that the electron is the first of the three orbitals.
• ms = +½ indicating a spin in the clockwise direction.
Critical Thinking Questions
11. Explain why the set of quantum numbers (n, l, ml, ms) is (4, 3, -2, -½) for the following
electron.
ms = -½ when an arrow is pointing down
n=4 for the 4th
energy level
l=3 for the
f sublevel
ml = -2 for
this orbital
12. Draw an orbital diagram for an electron whose quantum numbers are (5, 2, 0, +½).
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40
ChemQuest 13
Name: ____________________________
Date: _______________
Hour: _____
Information: Energy of Sublevels
Each sublevel has a different amount of energy. For example, orbitals in the 3p sublevel have more
energy than orbitals in the 2p sublevel. The following is a list of the sublevels from lowest to highest
energy:
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d…
To help you, here is the above list with the orbitals included. Recall that each blank represents an
orbital:
1s _, 2s _, 2p _ _ _, 3s _, 3p _ _ _, 4s _, 3d _ _ _ _ _ , 4p _ _ _ , 5s _, 4d _ _ _ _ _ , 5p _ _ _ , 6s _, 4f _ _ _ _ _ _ _
Note that d and f sublevels appear to be out of place. This is because they have extra high energies.
For example, the 3d sublevel has a higher energy than a 4s sublevel and the 4f sublevel has a higher
energy than the 6s sublevel.
When electrons occupy orbitals, they try to have the lowest amount of energy possible. (This is
called the Aufbau Principle.) An electron will enter a 2s orbital only after the 1s sublevel is filled up
and an electron will enter a 3d orbital only after the 4s sublevel is filled. Recall that only two
electrons can fit in each orbital. (This is called the Pauli Exclusion Principle.) When two electrons
occupy the same orbital they must spin in opposite directions—one clockwise and the other
counterclockwise.
Critical Thinking Questions
1. a) How many electrons would an atom need to have before it can begin filling the 3s sublevel?
10; Before the 3s can be filled, all these orbitals must be filled with two electrons in each
orbital: 1s _2s _2p _ _ _
b) What is the first element that has enough electrons to have one in the 3s sublevel? (Use your
periodic table.)
The 11th electron goes into the 3s sublevel. The atom with 11 electrons is sodium.
2. a) How many electrons would an atom need to have before it can begin filling the 3d sublevel?
20 electrons because all of these orbitals need to be filled: 1s _2s _2p _ _ _3s _3p _ _ _4s _
b) What is the first element that has enough electrons to begin placing electrons in the 3d
sublevel?
The 21st electron goes into the 3d sublevel. The atom with 21 electrons is scandium.
41
Information: Hund’s Rule
Electrons can be “paired” or “unpaired”. Paired electrons share an orbital with their spins parallel.
Unpaired electrons are by themselves. For example, boron has one unpaired electron. Boron’s
orbital diagram is below:
If we added one more electron to boron’s orbital diagram we will get carbon’s orbital diagram. One
important question is: where does the next electron go? The electron has a choice between two equal
orbitals—which of the 2p orbitals will it go in?
Hund’s rule tells us which of the above choices is correct. Hund’s rule states: when electrons have a
choice of entering two equal orbitals they enter the orbitals so that a maximum number of unpaired
electrons result. Also, the electrons will have parallel spins. Therefore Choice A is carbon’s actual
orbital diagram because the p electrons are in separate orbitals and they have parallel spins!
The following are the electron orbital diagrams for the next elements, nitrogen and oxygen. Notice
that nitrogen’s 2p electrons are all unpaired to obey Hund’s Rule. The 2p electrons are forced to
begin pairing up in oxygen’s configuration.
Critical Thinking Questions
3. How many “unpaired” electrons are in a nitrogen atom? 3
4. Why does carbon’s sixth electron have to go into another p orbital? Why can’t it go into a 2s
orbital? Why can’t it go into a 3s orbital?
The 2s sublevel is already full. It can’t go into a 3s orbital until all of the 2p orbitals are filled.
5. Write the electron orbital diagram for phosphorus.
6. Write the electron orbital diagram for arsenic.
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42
7. Compare the orbital diagrams for nitrogen (given above), phosphorus and arsenic. What is
similar about the electrons in the last sublevel for each of them?
They all have 3 electrons in a p sublevel.
Information: Electron Configurations vs. Orbital Diagrams
The electron orbital diagram of an atom can be abbreviated by using what is called electron
configurations. The following is the electron configuration for carbon: 1s2 2s2 2p2. The following is
the electron configuration for several elements whose orbital diagrams are given above:
Carbon: 1s2 2s2 2p2
nitrogen: 1s2 2s2 2p3
oxygen: 1s2 2s2 2p4
Critical Thinking Questions
8. What are the small superscripts (for example, the 4 in oxygen) representing in an electron
configuration?
The superscripts tell us how many electrons are in the sublevel. The 4 in oxygen tells us that
there are 4 electrons in the 2p sublevel.
9. What information is lost when using electron configurations instead of orbital diagrams? When
might it be more helpful to have an orbital diagram instead of an electron configuration?
We cannot tell as easily if the electrons are unpaired or paired. It would be more helpful to
have an orbital diagram when evaluating paired or unpaired electrons.
10. How many unpaired electrons are in a sulfur atom? What did you need to answer this question—
an orbital diagram or an electron configuration?
There are two unpaired electrons. The orbital diagram was necessary (below).
11. Write the electron configuration for zirconium (atomic # = 40).
1s22s22p63s23p64s23d104p65s24d2
12. Write the configuration for argon (atomic # = 18).
1s22s22p63s23p6
13. Write the electron configuration for calcium (atomic # = 20). Notice that calcium has all of
argon’s electrons plus two additional ones in a 4s orbital.
1s22s22p63s23p64s2
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ChemQuest 14
Name: ____________________________
Date: _______________
Hour: _____
Information: Valence Electrons
The electrons in the highest energy level are called valence electrons. Valence electrons are the
electrons located farthest from the nucleus. Valence electrons are always in the highest energy level.
The valence electrons are the most important electrons in an atom because they are the electrons that
are the most involved in chemical reactions and bonding.
The electron configuration for thallium (#81) is:
1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p1
The outermost energy level (not sublevel) is the 6th energy level. How many total electrons does
thallium have in the sixth level? 3, they are in boldfaced type above. Therefore, thallium has 3
valence electrons.
Critical Thinking Questions
1. Write the electron configurations for
a) oxygen: 1s22s22p4
b) sulfur: 1s22s22p63s23p4
2. How many valence electrons does oxygen have? In its outer energy level (the 2nd level)
oxygen has a total of 6 valence electrons (2 in the 2s and 4 in the 2p).
3. How many valence electrons does sulfur have? In its outer energy level (the 3rd level) sulfur
has a total of 6 valence electrons (2 in the 3s and 4 in the 3p).
4. Verify that selenium (atomic number = 34) has six valence electrons by drawing an electron
configuration and giving a brief explanation.
The electron configuration for selenium = 1s22s22p63s23p64s23d104p4. Selenium’s outermost
level is the 4th energy level. There are a total of 6 electrons in selenium’s 4th energy level: 2
are in the 4s sublevel and 4 are in the 4p sublevel.
Information: Bohr Diagrams
Below are seven “Bohr diagrams” for atoms #3-9.
FIGURE 1:
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44
Critical Thinking Questions
5. In each of the Bohr diagrams in Figure 1, the first energy level only has two electrons drawn
in it. Why is this?
No more than 2 electrons can fit in the first energy level of any atom.
6. What is the maximum number of electrons that the second energy level can have? How many
electrons can the 3rd energy level have?
The 2nd energy level can hold 8 electrons and the 3rd can hold 18. (See ChemQuest 9,
question 12).
7. Draw Bohr diagrams for the following atoms.
a) magnesium
b) phosphorus
c) argon
Information: Electron Dot Diagrams
Below are electron dot diagrams, also known as “Lewis Structures” for atoms #3-10.
FIGURE 2:
The position of the dots is important. For example, another atom that has three dots in its Lewis
structure is aluminum. Aluminum’s three dots must be positioned the same way as boron’s. Thus,
aluminum’s Lewis structure is:
FIGURE 3:
Critical Thinking Questions
8. What relationship exists between an atom’s valence electrons and the number of dots in the
Lewis structure of the atom?
The number of dots in the Lewis structure equals the number of valence electrons.
9. Why does nitrogen’s Lewis Structure has five dots around it while nitrogen’s Bohr diagram
contains 7 dots around it.
The Lewis structure only shows the valence electrons, but the Bohr diagram shows all of the
electrons. Nitrogen has 7 total electrons, 5 of which are valence.
45
10. Recall from questions 1-4 that oxygen, sulfur and selenium all have the same number of
valence electrons (6). They also are in the same column of the periodic table. Predict how
many valence electrons tellurium (Te) will have.
6, since tellurium is also in the same column.
11. Comparing Figure 2 and Figure 3 we see that boron and aluminum have the same number of
dots in their Lewis structures. Notice they are in the same column of the periodic table.
Write the Lewis structure for gallium (Ga).
12. Write the Lewis structure for sulfur and selenium. Compare the structures you write with
oxygen’s Lewis structure from Figure 2.
a) Sulfur
b) Selenium
13. In question seven, you drew the Bohr diagram for magnesium. Now write the Lewis
Structure for magnesium. What similarities exist between the Lewis Structures for
magnesium and beryllium?
Both have the same number of valence electrons and therefore the same number of dots.
14. Complete this statement: If elements are in the same column of the periodic table, they must
have Lewis Structures that are ___similar______.
similar or different
15. Why does sodium have the same Lewis structure as lithium?
Because sodium is in the same column of the periodic table as lithium.
16. Lewis structures are easier to draw than Bohr diagrams, but what information is lost by
drawing a Lewis structure instead of a Bohr Diagram?
In a Lewis structure, you have no information about inner level electrons.
17. Draw the Lewis structure for the following elements.
a) germanium
b) bromine
c) xenon
d) potassium
e) arsenic
18. You should be able to tell how many valence electrons an atom has by which column of the
periodic table the element is in. How many valence electrons are in each of the following
atoms?
a) bromine
b) tin
c) krypton
d) rubidium
7
4
8
1
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46
ChemQuest 15
Name: ____________________________
Date: _______________
Hour: _____
Information: Relating Electron Configurations to the Periodic Table
In this section you will see how the periodic table serves as a road map for writing electron
configurations. Get your periodic table out and get ready. Remember that a row on the periodic table
goes horizontally from left to right. Columns are vertical (up and down).
Critical Thinking Questions
1. Write the electron configurations for Li, Na and K. (Remember for electron configurations,
arrows are not necessary.)
Li = 1s22s1
Na = 1s22s22p63s1
K = 1s22s22p63s23p64s1
2. What is similar about the electron configurations of all of the elements in question 1? Look at
the very end of their configurations.
Each of them ends with s1.
3. Lithium (Li) is in row 2 of the periodic table, sodium (Na) is in row 3, and potassium is in
row 4. How do their row numbers affect how their electron configurations end?
Li in row 2 ends with 2s1. Na in row 3 ends in 3s1. K in row 4 ends in 4s1. The row number
is the same as the ending number of the sublevel.
4. Write the electron configurations for Be, Mg, and Ca.
Be = 1s22s2
Mg = 1s22s22p63s2
Ca = 1s22s22p63s23p64s2
5. What is similar about the ending of the electron configurations for all of the elements in
question 4?
Each of them ends with s2.
6. Beryllium (Be) is in row 2 of the periodic table, magnesium (Mg) is in row 3, and Calcium
(Ca) is in row 4. How do their row numbers affect how their electron configurations end?
Be in row 2 ends with 2s2. Mg in row 3 ends in 3s2. Ca in row 4 ends in 4s2. Again, the row
number is the same as the ending number of the sublevel.
7. Given what you have done in questions 1-6, complete the following statement.
Elements ending in s1 are in column number _1 or IA_ and those ending in s2 are in column
number _2 or IIA_ of the periodic table.
47
8. Name the element that have an electron configuration ending with… (the first is done for
you)
c) 7s1 _Francium (Fr)__
a) 5s1 _Rubidium (Rb)__
b) 6s2 _Barium (Ba)___
9. Write the electron configurations for B, Al, and Ga and note their similarities.
B = 1s22s22p1
Al = 1s22s22p63s23p1
Ga = 1s22s22p63s23p64s23d104p1
10. Write the electron configurations for N, P and As and note their similarities.
N = 1s22s22p3
P = 1s22s22p63s23p3
As = 1s22s22p63s23p64s23d104p3
11. Complete the following statement: Elements ending in p1 are in column number 13 or IIIA of
the periodic table and elements ending in p3 are in column number 15 or VA of the periodic
table. Therefore, elements ending in p2 must be in column 14 or IVA of the periodic table.
12. Name the elements that have an electron configuration ending with…(the first is done for
you)
a) 3p4 __Sulfur (S)__
b) 5p6 __Xenon (Xe)___
c) 6p5 __Astatine (As)_
13. Write the electron configurations for Ti, Zr, and Hf and note their similarities.
Ti = 1s22s22p63s23p64s23d2
Zr = 1s22s22p63s23p64s23d104p65s24d2
Hf = 1s22s22p63s23p64s23d104p65s24d105p66s24f145d2
14. Write the electron configurations for Cr, Mo, and W and note what is similar about them.
Cr = 1s22s22p63s23p64s23d4
Mo = 1s22s22p63s23p64s23d104p65s24d4
W = 1s22s22p63s23p64s23d104p65s24d105p66s24f145d4
15. Complete the following: Elements ending in d2 are in column number 4 or IVB and those
ending in d4 are in column number 6 or VIB. Therefore, elements ending in d3 are in column
number 5 or VB. Elements ending in d7 must be in column number 9 or VIIIB.
16. Notice from question 6 that an element that ends in 3s is in the 3rd row. Also, from questions
9-12 it should be clear that an element that ends in 3p is in the 3rd row. However, notice that
an element that ends in 3d is in the 4th row instead of the 3rd row. Offer an explanation for
this.
The d sublevels have extra high amounts of energy. Remember for example, that is why the
3d sublevel comes after the 4s sublevel. This same explanation accounts for the fact that the d
sublevels are “misplaced” by one row on the periodic table.
17. Name the elements that have an electron configuration ending with…(the first is done for
you)
a) 4d3 _Niobium (Nb)_
b) 5d8 _Platinum (Pt)__
c) 3d6 _Iron (Fe)______
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18. There are three major divisions on the periodic table: the “s block”, the “d block” and the “p
block”. Where are these blocks of elements located? Give the column numbers of their
locations. (Yes, there is also an “f block” but we won’t use that much.)
s block: columns __1-2__
d block: columns __3-12_
p block: columns _13-18_
Information: Abbreviating the Electron Configurations
Electron configurations can be shortened using a special group of elements called the noble gases.
They are found in the column furthest to the right on the periodic table: helium, neon, argon, krypton,
xenon, and radon. These gases are very non-reactive.
All of the noble gases have electron configurations that end in p6. Because of their unique nonreactivity, they are often used to abbreviate long electron configurations.
Take note of krypton’s electron configuration: 1s22s22p63s23p64s23d104p6.
Now notice that strontium’s electron configuration is the same as krypton’s except that strontium has
38 electrons instead of 36. Strontium’s electron configuration is 1s22s22p63s23p64s23d104p65s2.
Strontium has the same electron configuration as krypton and then two additional electrons in the 5s
orbital. Therefore strontium’s electron configuration can be abbreviated as [Kr]5s2. This notation
means that strontium has all of krypton’s electrons plus two more in the 5s sublevel.
As another example, consider iodine. Instead of writing a long electron configuration, we can
1. Going backward from iodine on the periodic table, find the previous noble gas. It is
krypton.
2. Take note of what krypton’s electron configuration ends with. It ends with 4p6 since it is
in the 4th row and in the p6 column.
3. Iodine has 17 more electrons than krypton and so you can begin by writing [Kr] followed
by orbitals for 17 more electrons. After 4p6 comes 5s2…
Iodine = [Kr]5s24d105p5
Any of the noble gases can be used for abbreviations. Here are a few more examples:
iron = [Ar]4s23d6
cesium = [Xe]6s1
phosphorus = [Ne]3s23p3
Study the above examples and make sure you understand why they are written that way.
Critical Thinking Questions
19. Write abbreviated electron configurations for the following elements:
a) Ruthenium (Ru) Ru = [Kr]5s24d6
b) Arsenic (As) As = [Ar]4s23d104p3
c) Tellurium (Te) Te = [Kr]5s24d105p4
49
ChemQuest 16
Name: ____________________________
Date: _______________
Hour: _____
Information: Shielding
FIGURE 1: “Bohr Diagrams” of boron, carbon and nitrogen
nucleus
charge = +5
Boron
nucleus charge = +6
Carbon
nucleus charge = +7
Nitrogen
Because the nucleus is positively charged, it exerts an attractive force on the electrons. However, the
three electrons in boron’s outer energy level do not feel the full +5 attraction from the 5 protons in
boron’s nucleus. Before the +5 attraction gets to the outer energy level it gets partially cancelled (or
“shielded”) by the two electrons in the first energy level. The two electrons in the first energy level
weaken the attractive force by two. Therefore to the outer energy level it only “feels” like a +3
charge rather than a +5 charge from the nucleus.
Consider the diagram of carbon. An electron in the outer energy level only “feels” a charge of +4
coming from the nucleus because the two electrons in the first energy level shield two of the positive
charges from the nucleus.
Critical Thinking Questions
1. How large is the charge that the second energy level of nitrogen “feels” from the nucleus?
Nitrogen’s second energy level feels a charge of +5 from the nucleus. The total charge of +7 gets
weakened by the two electrons in nitrogen’s first energy level.
2. Why does the first energy level in each of the three above diagrams only contain two electrons?
The first energy level cannot hold more than two electrons.
3. How many electrons can fit in the second energy level of any atom?
8
4. How many electrons can fit in the third energy level?
18
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50
5. How many energy levels does aluminum have? How many electrons should be in each energy
level?
3 energy levels; 2 electrons in the first, 8 in the second, and that leaves 3 for the third.
6. Draw a Bohr diagram for aluminum similar to those above.
Nucleus
has a +13
charge
7. How large is the charge that the second energy level of an aluminum atom “feels” from the
nucleus?
It feels a +11 charge. The +13 charge of the nucleus gets weakened to a +11 by the 2 electrons in
the first energy level.
8. How large is the charge that the third energy level of an aluminum atom “feels” from the nucleus?
It feels a +3 charge. The +13 charge of the nucleus gets weakened by the 2 electrons in the first
energy level and the eight electrons in the second energy level.
Information: Charge and Distance
As you know, opposite charges attract. Examine the following diagrams of charged metal spheres.
FIGURE 2:
Diagram A
+3
Diagram B
-3
+4
Diagram C
-4
+5
-5
The attraction between the two charged metal spheres in each diagram is represented by an arrow.
The metal spheres are pulled closer together in diagram C because of the +5 to -5 attraction is
stronger than the +4 to -4 attraction in Diagram B and the +3 to -3 attraction in Diagram A.
Electrons behave the same way as the metal spheres are depicted in Figure 2. Consider the Bohr
diagrams of boron, carbon and nitrogen in Figure 1. Recall that Boron’s outer electrons feel a +3
attraction from the nucleus. Carbon’s outer electrons feel a +4 attraction. In question 1, you found
out that nitrogen’s outer electrons feel a +5 attraction.
This attraction between the nucleus and outer energy level determines the size of the atom. If the
attraction is strong the atom is small; if the attraction is weak, the atom spreads out and is larger.
Critical Thinking Questions
9. Which atom is larger: nitrogen or carbon? Why?
Carbon, because nitrogen’s nucleus pull of the outer electrons harder (+5 attraction) than carbon’s
nucleus (+4 attraction). Therefore nitrogen’s electrons are pulled closer to the nucleus making
nitrogen smaller.
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10. In a silicon atom, the force of attraction between the nucleus and outer energy level is +4 to –4.
Using a Bohr diagram of a silicon atom as an illustration, explain in detail why this is true.
By the time the +14 charge of the nucleus is felt
by the 3rd energy level, it has to pass through a
total of 10 electrons (2 in the 1st energy level and
8 more in the 2nd). This decreases the +14 charge
by 10 to a +4 charge. Thus, the attraction from
the nucleus is +4.
Nucleus
has a +14
charge
11. Draw Bohr diagrams for sulfur and chlorine.
Nucleus
has a +16
charge
Nucleus
has a +17
charge
12. a) Find the size of the charge attraction from the nucleus to the outer energy level for sulfur and
for chlorine.
Sulfur: +6
Chlorine: +7
b) Which atom do you predict to be larger: sulfur or chlorine?
Sulfur
c) Explain, in detail, your reasoning to part b.
Sulfur has a weaker attraction than chlorine. Therefore, sulfur’s electrons won’t be held as
tightly as chlorine’s.
13. Notice and compare the locations of boron, carbon and nitrogen on the periodic table. Now
compare their sizes. Do the same with sulfur and chlorine. There is a general trend in size as you
proceed from left to right across the periodic table. What is this trend? In other words, how do
atoms in the same row of the periodic table compare to each other in size?
As you go from left to right across a row in the periodic table, the size of atoms decreases.
Information: Bohr Diagrams and the Size of Atoms
Examine the following “Bohr Diagrams” of three atoms from the periodic table.
FIGURE 3:
Atom A
Atom B
Atom C
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Critical Thinking Questions
14. Give the name and atomic number of each atom.
Atom A
Name of the element
Sodium
Atomic number
11
Atom B
Atom C
Lithium
Hydrogen
3
1
15. a) Concerning atoms A, B and C, what is similar about their location in the periodic table?
They are in the same column.
b) In atoms A, B, and C compare the force of attraction from the nucleus to the outer level of
electrons.
The force of attraction in all of them is +1.
16. Draw Bohr diagrams for neon and argon.
17. a) What is similar about the location of neon and argon in the periodic table
They are in the same column.
b) Compare the force of attraction between the outer level electrons and the nucleus for neon and
argon.
Each of them has a +8 attraction from the nucleus.
c) Using your answers to 15b and 17b, what can be said about elements in the same column and
the force of attraction between their outer energy level and nucleus?
Elements in the same column have the same attraction between the nucleus and outer level
electrons.
18. Atom A is larger than atom B. The attraction between the nucleus and outer level electrons is
equal in atoms A and B, so what other reason could there be for Atom A’s larger size? Propose
an explanation based on that structure of atoms A and B.
Atom A has an additional level of electrons. Each additional level increases the size of the atom.
19. Which is larger—neon or argon? Why is this atom the largest?
Argon because argon has an additional energy level.
20. In general, there is a trend in the sizes of atoms as you move down a column of the periodic table.
a) What is this trend?
As you move down a column, atoms get larger.
b) Why does this trend exist? (Explain the basis for the trend.)
Going down a column also means that the number of energy levels are increasing.
Increasing the energy levels causes the atoms to be larger.
21. Order the following lists of elements in order from smallest to largest.
b) P, Sb, N
c) S, Ca, Mg, Cl
a) K, As, Br
Br, As, K
N, P, Sb
Cl, S, Mg, Ca
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ChemQuest 17
Name: ____________________________
Date: _______________
Hour: _____
Information: Separating Charges
Examine Figure 1 below where there are three pairs of metal spheres that have different amounts of
charge on them. The spheres in diagram C are closer than the others because they have the strongest
attraction.
FIGURE 1: Attraction between metal spheres.
Diagram A
+3
-3
+4
Diagram B
Diagram C
-4
+7
-7
Critical Thinking Questions
1. Which would be harder to separate: the two spheres in Diagram A or the two spheres in Diagram
B? Why?
The spheres in Diagram B would be harder to separate because the force of attraction is greater
due to the higher charge difference (+4 and -4) and also the spheres are closer together which
makes them harder to separate just like magnets that are very close together.
2. Draw Bohr diagrams of the following atoms: lithium, nitrogen and fluorine.
3. For each of the atoms in question two compare the attraction between the nucleus and the outer
level of electrons. Which atom has the strongest attraction between the nucleus and outer
electrons?
Fluorine has the strongest attraction. A +7 force from the nucleus is attracting the outer level.
(Nitrogen’s outer level feels a +5 attraction and lithium’s outer level feels a +1 attraction from the
nucleus.)
4. Which would be more difficult: if you wanted to remove an outer electron from an atom of
fluorine or from an atom of nitrogen?
For similar reasons as question 1, it would take more energy to remove an outer electron from an
atom of fluorine than from nitrogen. The outer electron from fluorine feels a stronger force of
attraction from the nucleus.
5. Would it be easier to remove an outer electron from lithium or nitrogen?
It would be easier to remove an outer electron from lithium because lithium’s outer electron
experiences a weaker force of attraction from the nucleus.
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Information: Ionization Energy
The amount of energy that it takes to completely remove an electron from an atom is called
ionization energy. The first ionization energy is the energy required to remove one electron from an
atom’s outer energy level. The second ionization energy is the energy needed to remove a second
electron from the energy level. The second ionization energy is always higher than the first
ionization energy. Because noble gases have eight electrons in their outer energy level, they are very
stable and therefore it takes a very high amount of energy to remove an electron from a noble gas.
Critical Thinking Questions
6. Which would have a higher first ionization energy: phosphorus or aluminum? (You may want to
Phosphorus, because it has a higher attraction between (+5) the nucleus and outer electrons than
aluminum (+3 attraction).
7. a) Phosphorus and aluminum are in the same row of the periodic table. Lithium, nitrogen and
fluorine are also in the same row. Using your answers to questions 3, 4 and 6 what do you notice
about the ionization energy of elements proceeding from left to right across a row of the periodic
table? Does the ionization energy increase or decrease as you go across a period?
As you move from left to right across a row (or “period”), the ionization energy increases.
b) Explain why you believe this trend exists.
This trend exists because as you move from left to right across a row, the attraction between
the nucleus and the outer level electrons also increases. For atoms with a stronger attraction
for the outer electrons, it requires more ionization energy to remove the electron.
8. Draw a Bohr diagram of sodium. Do you expect the first ionization energy to be high or low for
sodium?
Sodium has a low first ionization energy because of the low attraction
from the nucleus (+1) for the electron in the third energy level.
9. The second ionization energy of all elements is higher than the first ionization energy, but in
sodium the second ionization energy is extra large. Considering the structure of the sodium atom
explain why this might be. (HINT: after one electron is removed, what is sodium’s electron
arrangement like?).
After the first electron is removed, sodium has the same electron configuration as neon with 10
electrons. It is like a noble gas and therefore to remove the next electron is very difficult.
10. A certain atom in the third period has a third ionization energy that is unusually high. Using the
same reasoning you used for question 9, name this atom.
Magnesium because it is in the 3rd row (period) of the periodic table and it has two more electrons
than neon (a noble gas). After the first two electrons are removed, the 3rd one is extra hard to
remove because it will need to be removed from a noble gas-like configuration.
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Information: Trend in Ionization Energy in Groups
Consider the following figure of diagrams of two magnets that you wish to separate.
FIGURE 2: Separating Magnets
Diagram D
Diagram E
Critical Thinking Questions
11. Would it be easier to separate the magnets in Diagram D or those in Diagram E in Figure 2?
Assume that each magnet is attracted to the other and that the size of the attraction is the same.
Take only the distance between magnets into consideration.
Magnets are easier to separate when they are farther apart, so it will be easier to separate the
magnets in Diagram E.
12. How does the distance between the outer level of electrons and the nucleus in phosphorus
compare to the distance between the outer level of electrons and the nucleus in nitrogen?
Since phosphorus is a larger atom, the distance between the nucleus and the outer level electrons
is greater in it than in nitrogen.
13. Based on your answer to question 12, would it be easier to remove an electron from nitrogen or
from phosphorus?
It would be easier to remove an electron from phosphorus than from nitrogen since the outer
electron is farther away in the phosphorus atom.
14. Nitrogen and phosphorus are in the same column on the periodic table. From your answers to
questions 12 and 13 what happens to the ionization energy as you move down a column in the
periodic table?
As you move down a column of the periodic table, the ionization energy decreases because the
size of the atoms increases.
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56
ChemQuest 18
Name: ____________________________
Date: _______________
Hour: _____
Information: Ions
Figure 1: Below are four Bohr diagrams of atoms and ions. The two diagrams on the left are atoms;
the two on the right are ions.
12 protons
12 protons
Ion of Atom A
Atom A
17 protons
Atom B
17 protons
Ion of Atom B
Critical Thinking Questions
1. Prove that both Atom A and Atom B are neutral (have a charge of zero).
Both atoms have an equal number of protons and electrons.
2. What is the identity of Atom A and of Atom B?
The atom with 12 protons and 12 electrons is magnesium.
57
3. Given the above diagrams, how does an atom become an ion?
An atom can become an ion by either gaining or losing electrons.
4. What is the charge on Ion A? What is the charge on Ion B?
Atom A = 12 protons and 10 electrons = +2 charge
Atom B = 17 protons and 18 electrons = -1 charge
5. Write the electron configuration (ex: 1s22s22p6….) for each ion and atom shown in the Bohr
diagrams.
Atom A: __1s22s22p63s2______________
Ion A: __1s22s22p6____________________
Atom B: __1s22s22p63s23p5____________
Ion B: ___1s22s22p63s23p6______________
6. Consider the electron configuration that you wrote for Ion A. What atom has the same electron
configuration as this ion?
Neon
7. Consider the electron configuration that you wrote for Ion B. What atom has the same electron
configuration as this ion?
Argon
8. Bromine atoms always gain one electron when they become an ion. Which atom has the same
number of electrons as a bromine ion?
Krypton
9. Cesium atoms always lose one electron to become an ion. Which atom has the same number of
electrons as a cesium ion?
Xenon
10. Consider your answers to questions 6-9. What do all of the atoms you named have in common?
They are all noble gases with eight outer-level electrons.
11. Knowing what you know about the atoms that you named in questions 6-9, why do you think
atoms want to form ions the way they do?
The noble gases have very stable electron configurations and so atoms want to gain or lose
electrons so that their electron configuration is the same as a noble gas and more stable.
Information: Ions
As you know, all of the noble gases are very stable. Ions form in such a way so that the ion will have
the same number of electrons as a noble gas. Take oxygen, for example. Oxygen has 8 electrons.
To become like a noble gas it could either gain two to become like neon or it could lose six to
become like helium. So what will oxygen do—gain two or lose six? As a general rule, atoms will
gain or lose the fewest number of electrons possible.
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Critical Thinking Questions
12. What does an oxygen atom do when becoming an ion? (Does it gain or lose electrons and how
many?)
It will gain 2 electrons rather than losing 6.
13. An oxygen atom has an overall neutral charge because it has an even number of protons and
electrons. What is the overall charge on an oxygen ion? After gaining 2 negative electrons the
charge will be -2.
14. Consider an aluminum atom.
a) To become like argon, would aluminum have to gain or lose electrons? How many?
Aluminum would have to gain 5 electrons.
b) To become like neon, would aluminum have to gain or lose electrons? How many?
Aluminum would have to lose 3 electrons.
c) Considering your answers to parts a and b, what does an aluminum atom do to become an
ion?
Since aluminum wants to gain or lose the fewest number of electrons possible, aluminum will
lose 3 electrons.
d) What is the charge on an aluminum ion?
After losing 3 electrons, the charge will be a net +3.
15. When each of the following atoms becomes an ion, what will the charge be?
a) Ca
+2
b) Cl
-1
c) N
-3
d) K
+1
e) S
-2
f) B
+3
g) P
-3
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ChemQuest 19
Name: ____________________________
Date: _______________
Hour: _____
Information: Naming Ions
To write an ion, you write the symbol of the atom and put the charge in the upper right corner.
Consider the following examples: Al3+, O2-, Mg2+. You should verify that each of the charges is
correct.
Positive and negative ions are named differently. Positive ions retain the same name as the
parent atom. For example, Al3+ is called the “aluminum ion” and Mg2+ is called the “magnesium
ion.” Negative ions are named a little differently. For negative ions, you change the ending of the
name to “-ide”. Therefore, O2- is named oxide and P3- is named phosphide.
Critical Thinking Questions
1. Write the symbol (including the charge) and name for each of the ions for each of the following:
a) Ca
b) Cl
c) N
d) K
e) S
f) B
g) P
Ca2+
ClN3K+
S2B3+
P3calcium ion
chloride
nitride potassium ion
sulfide
boron ion phosphide
Information: Ionic Bonding and Formulas
There are two ways in which atoms can “bond” to each other and form a compound. The means of
bonding that we will consider now is called ionic bonding, which occurs between a metal and a
nonmetal. As you know, opposite charges attract. Ionic bonding is when two ions of opposite charge
attract and bond to each other forming an ionic compound. Consider the following examples of
formulas for ionic compounds:
•
•
•
•
One Na+ (sodium ion) and one Cl- (chloride ion) bond to make NaCl, “sodium chloride.”
One Mg2+ (magnesium ion) and two F- (fluoride ion) bond to make MgF2, “magnesium fluoride.”
Three Ca2+ (calcium ion) and two N3- (nitride ion) bond to make Ca3N2, “calcium nitride.”
One Al3+ (aluminum ion) and one N3- (nitride ion) bond to make AlN, “aluminum nitride.”
The small numbers at the bottom right of each symbol in a formula are called “subscripts”.
Subscripts tell us how many of each type of atom are present. For example in the formula Mg3N2
there are three magnesium ions and two nitride ions.
Critical Thinking Questions
2. Consider the formula NaCl in the above example. It tells us that one Na+ ion is bonded to one Clion. What is the overall charge for NaCl? Is it positive, negative, or neutral?
Neutral: one +1 ion (Na+) and one -1 ion (Cl-) together are neutral.
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60
3. Consider MgF2. This formula tells us that one Mg2+ ion bonds with two F- ions. What is the
overall charge on MgF2?
Neutral: one +2 ion (Mg2+) and two -1 ions (F-) together are neutral.
4. What is the overall charge on any ionic compound?
Neutral
5. Why is calcium nitride written like Ca3N2 and not something like CaN2 or Ca2N3? In other words
why do exactly three calcium ions bond with exactly two nitride ions?
Three +2 ions (Ca2+) are required to go along with two -3 ions (N3-) so that the overall charge is
neutral. Mathematically: 3(+2) + 2(-3) = 0
6. The formula Ca3N2 can never be written as N2Ca3. To find out why, take note of each of the four
example formulas given above.
a) In terms of charge, what do the first ions named all have in common?
They are positively charged.
b) In terms of charge, what do the second ions named all have in common?
They are negatively charged.
c) Now, why can’t Ca3N2 ever be written like N2Ca3?
The positive ion (Ca2+) must be written first.
7. There are two rules to follow when writing formulas for ionic compounds. One has to do with
charges (see questions 4 and 5) and the other has to do with which atom to write first and which
one to write second (see question 6). What are these two rules?
1. Ionic compounds need to be written with the ions combined so that the final compound is
neutral.
2. The positive ion must be written first.
8. What is wrong with the following formulas?
b) PNa3
a) Al2S
This compound is not neutral. Two
Al3+ ions will combine with three
S2- ions to give the correctly
written, neutral compound: Al2S3.
c) Mg2S2
Although this compound is neutral,
the positive ion (Na+) is not written
first. The correct way to write it
would be Na3P.
It is generally best to write ionic
compounds in their lowest multiple
so it should be written MgS and not
Mg2S2 or Mg3S3 or Mg4S4, etc.
9. Write the formula and name for the compound that forms when the following atoms form ionic
compounds. The first is done for you.
a) nitrogen and sodium
b) barium and sulfur
c) magnesium and iodine
Na3N
BaS
MgI2
sodium nitride
barium sulfide
magnesium iodide
d) oxygen and aluminum
e) calcium and phosphorus f) sodium and sulfur
Al2O3
Ca3P2
Na2S
aluminum oxide
calcium phosphide
sodium sulfide
10. Given the following compounds, determine the charge on the unknown ion “X”.
c) X3P2
a) X2S
b) MgX
X must be +1 since two +1
X ions would be needed to
balance out the -2 sulfide
ions.
X must be -2 since a -2
X ion would be needed to
balance out the +2
magnesium ion.
X must be +2 since three +2
X ions would be needed to
balance out the two -3
phosphide ions.
61
ChemQuest 20
Name: ____________________________
Date: _______________
Hour: _____
Information: Polyatomic Ions
The word, “polyatomic” means “many atoms”. A polyatomic ion, therefore, is an ion that is made of
more than one atom. An example of a polyatomic ion is the sulfate ion, SO42-. Sulfate is composed
of one sulfur atom and four oxygen atoms and overall sulfate has a negative two charge.
Some polyatomic ions:
Sulfate: SO42Cyanide: CNAcetate: C2H3O2-
Phosphate: PO43Ammonium: NH4+
Hydroxide: OH-
Nitrate: NO3Chlorate: ClO3Carbonate: CO32-
Critical Thinking Questions
1. What do all of the polyatomic ions that have the suffix “-ate” have in common?
They all contain oxygen atoms.
2. Which two atoms do you think compose the polyatomic ion called “silicate”?
silicon and oxygen
3. What is the difference between calcium nitride and calcium nitrate?
The –ate ending indicates that calcium nitrate has oxygen in it, but calcium nitride does not.
Information: Writing Formulas With Polyatomic Ions
First of all, you must remember that you can never change the formula for a polyatomic ion. Sulfate
is always SO42- and never S2O84- or something else. Following are some examples of chemical
formulas that contain polyatomic ions.
Ammonium chloride is formed from one ammonium ion (NH4+) and one chloride ion (Cl-) to give the
formula: NH4Cl. Sodium sulfate requires two sodium ions (Na+) because sulfate (SO42-) has a
negative two charge; the formula is: Na2SO4.
Consider calcium hydroxide. Calcium has a positive two charge (Ca2+) and hydroxide has a negative
one charge (OH-). We need two hydroxide ions to combine with one calcium ion so that the overall
charge ends up being zero. We write calcium hydroxide like Ca(OH)2.
Following are some more examples:
potassium acetate: KC2H3O2
barium phosphate: Ba3(PO4)2
magnesium nitrate: Mg(NO3)2
calcium carbonate: CaCO3
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Critical Thinking Questions
4. As mentioned above, calcium hydroxide is written like Ca(OH)2. Why can’t it be written like
CaOH2?
Ca(OH)2 means that two OH- have bonded with a Ca2+ ion. CaOH2 means that two H+ ions
have bonded with one O2- ion and a Ca2+ ion. CaOH2 is not a neutral compound.
5. As mentioned above, barium phosphate is written as Ba3(PO4)2. Why can’t it be written like
Ba3PO42?
Ba3(PO4)2 means that two PO43- ions have bonded with three Ba2+ ions. Ba3PO42 means that
three Ba3+ ions have bonded with one P3- ion and 42 O2- ions, which would not yield a neutral
compound.
6. Name the following compounds. Each includes at least one polyatomic ion.
a) Na3PO4
b) (NH4)2SO4
c) Mg(C2H3O2)2
sodium phosphate
ammonium sulfate
magnesium acetate
d) (NH4)2S
ammonium sulfide
e) CaCO3
calcium carbonate
f) Ba(NO3)2
barium nitrate
7. Write formulas for the following ionic compounds. Note that each includes a polyatomic ion.
a) lithium phosphate
b) ammonium oxide
c) barium hydroxide
Li3PO4
(NH4)2O
Ba(OH)2
d) calcium cyanide
Ca(CN)2
e) sodium chlorate
NaClO3
f) potassium sulfate
K2SO4
8. In question 3, you were asked the difference between calcium nitride and calcium nitrate.
Now write the formula for each of them.
calcium nitride: Ca3N2
calcium nitrate: Ca(NO3)2
Information: Formulas for Acids
Acids are compounds that contain positive hydrogen ions (H+) bonded to a negative ion. For
example, carbonic acid is formed when the carbonate ion (CO32-) bonds with two hydrogen ions (H+)
to give H2CO3. Other common acids are listed below:
Hydrochloric acid: HCl
Sulfuric acid: H2SO4
Nitric Acid: HNO3
Acetic Acid: HC2H3O2
Critical Thinking Questions
9. Why do carbonic and sulfuric acid require two H+ ions to bond, but HCl and HNO3 only have
one H+? Carbonate and sulfate both have a -2 charge and therefore require two H+ ions (each
having a +1 charge) so that overall the formulas will be neutral.
10. Phosphoric acid is made from the phosphate ion and H+ ions. Write the formula for
phosphoric acid.
H3PO4 Three H+ ions are needed because phosphate has a -3 charge.
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ChemQuest 21
Name: ____________________________
Date: _______________
Hour: _____
Information: Charges of Some Transition Elements
So far you have learned that you can predict the charge that an ion will have based on its location on
the periodic table. However, the transition elements are not easy to predict. A few common
transition elements are listed below. You should memorize their charges.
Silver: Ag+
Zinc: Zn2+
Critical Thinking Questions
1. Write the formulas for the following compounds:
a) silver nitrate
AgNO3
b) zinc phosphate
Zn3(PO4)2
CdCl2
Information: More Than One Possible Charge
Many transition elements can have more than one charge when they become an ion. Copper ions, for
example, can be Cu+ or Cu2+. As another example, iron ions are sometimes Fe2+ and sometimes Fe3+.
Critical Thinking Questions
2. Copper and iron are in the “d block” and so you need to calculate their charge by comparing
what bonds to them. Find the charge on copper and iron in each of the following compounds.
a) CuCl2
+2
b) CuCl
c) FeSO4
d) Fe2(SO4)3
+1
+2
+3
(This is similar to what you did in question 10 for ChemQuest 16)
3. Give your best attempt at naming the compounds from question 2. (They are rewritten
below.)
a) CuCl2
copper chloride
b) CuCl
copper chloride
c) FeSO4
iron sulfate
d) Fe2(SO4)3
iron sulfate
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Information: Formulas Containing Roman Numerals
You probably put the same name for the compounds in question 3a and 3b. You may also have put
the same name for the compounds in 3c and 3d. BUT these are not the same compound! You cannot
have the same name for two different compounds. Here are the correct names for the compounds in
questions 2 and 3:
a) CuCl2
copper(II) chloride
b) CuCl
copper(I) chloride
c) FeSO4
iron(II) sulfate
d) Fe2(SO4)3
iron(III) sulfate
Critical Thinking Questions
4. Compare your answers for questions 2 & 3 with the names of the compounds given in the
information section. What do the Roman numerals stand for?
The Roman numerals stand for the charge on the transition metal ion.
5. Why is MnO2 called manganese(IV) oxide?
Manganese has a charge of +4 indicated by the Roman numeral IV.
6. Name the following compounds. Note: assume that anytime you have a transition element (d
block element) you must use a Roman numeral unless the element is silver, zinc, or cadmium.
(The first one is done for you.)
a) NiNO3
b) Cr2(CO3)3
c) FeNO3
nickel(I) nitrate chromium(III) carbonate
e) Cu3(PO4)2
copper(II) phosphate
f) MnS
iron(I) nitrate
g) ZnCl2
manganese(II) sulfide
zinc chloride
d) CoCl2
cobalt(II) chloride
h) AgNO3
silver nitrate
(No Roman numerals are needed for zinc or silver, but if you included them it is OK.)
7. Write the formulas for the following compounds. (The first one is done for you.)
a) mercury(II) acetate
b) chromium(III) sulfate
Hg(C2H3O2)2
Cr2(SO4)3
c) iron(I) carbonate
Fe2CO3
d) potassium carbonate
e) strontium nitride
f) manganese(IV) chlorate
K2CO3
Sr3N2
Mn(ClO3)4
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ChemQuest 22
Name: ____________________________
Date: _______________
Hour: _____
Information: Terminology
Recall that an ionic bond results from the combination of a metal and a nonmetal. A covalent bond is
the type of bond between two nonmetals. Covalent bonds are formed by neutral atoms that share
electrons rather than by charged ions. When a compound is formed by sharing electrons, the
compound is called a molecule or molecular compound. It is important to note that ionic compounds
are not called molecules. The largest class of molecules are called organic molecules. Carbon is the
distinguishing mark of organic compounds.
Critical Thinking Questions
1. Circle any of the following compounds that would properly be called a “molecule”.
a) H2O
b) CO2
c) NaCl
d) Mg3P2
e) N2O5
Any compound with no metal and no polyatomic ion is considered a “molecule” and covalent.
Information: Naming Covalent Compounds
There are several prefixes used to name molecules. The name “carbon oxide” is not sufficient
because carbon and oxygen sometimes form CO2 and sometimes CO. Prefixes are necessary to
distinguish between them.
Formula
Name
N2 O4
dinitrogen tetraoxide
SF6
sulfur hexafluoride
XeCl5
xenon pentachloride
SO3
sulfur trioxide
CO
carbon monoxide
Critical Thinking Questions
2. Fill in the table to indicate which prefix is used to represent the numbers. The first one is done
for you.
Number
Prefix
1
mono
2
di
3
tri
4
tetra
5
penta
6
Hexa
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66
3. Name each of the following molecules.
b) CF4
c) SCl3
d) SO
a) N2O5
dinitrogen pentoxide carbon tetrafluoride sulfur trichloride
sulfur monoxide
the mono is dropped and we assume there is only one of the atoms.)
4. Which of the above compounds would be classified as “organic”?
Only carbon tetrafluoride (b) because it is the only one containing carbon.
Information: Empirical Formulas
Molecules can be represented by using either a molecular formula or an empirical formula. The
molecular formula tells you exactly how many atoms of each element are in the compound. For
example, in the table below, compound #2 has exactly 4 carbons and 8 hydrogens in each molecule.
Observe the table below that shows four organic molecules along with a molecular and empirical
formula for each one:
Molecule
Molecular Formula Empirical Formula
#1
C 2 H4
CH2
#2
C 4 H8
CH2
#3
C 3 H8
C 3 H8
#4
C8H18
C 4 H9
Critical Thinking Questions
5. What is an empirical formula?
An empirical formula is a way to show the lowest, simplified ratio of elements in a compound. It
tells us the ratio of atoms in a compound and NOT how many atoms are in the compound.
6. How can molecules #1 and #2 have the same empirical formula even though they are different
molecules?
Both molecules #1 and #2 have a 1 to 2 ration of carbon to hydrogen and since an empirical
formula gives the ratio, they both have the same empirical formula.
7. Given the empirical formula for a compound is it possible to determine the molecular formula? If
so, explain how.
No, because the empirical formula does not tell us how many of each atom is present in a
compound.
8. Given the molecular formula for a compound is it possible to determine its empirical formula? If
so, explain how.
Yes, you must divide each subscript by the same number. For example if you are given the
molecular formula C4H6, you could reduce it by dividing each subscript by 2 to obtain C2H3.
9. Give the empirical formula for each of the molecules below:
b) C2H4O2
c) C4H14
a) N2O6
NO3
CH2O
C 2 H7
d) C3H5
C 3 H5
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ChemQuest 23
Name: ____________________________
Date: _______________
Hour: _____
Information: Drawing Covalent Compounds
For covalent bonding, we often want to draw how the atoms share electrons in the molecule. For
example, consider CCl4 and NF3 as drawn below:
Each line represents two electrons
that are being shared
Notice that the atoms share electrons so that they all have 8 electrons. If you count the electrons
around carbon, you will get a total of eight (each line is two electrons). If you count the electrons
around each chlorine atom, you will find that there are eight of them.
Critical Thinking Questions
1. How many valence electrons does a carbon atom have (before it bonds)? Hint: find this based
on carbon’s column on the periodic table.
4; since carbon is in column 14 (or “4A”) on the periodic table it has 4 valence electrons.
2. How many valence electrons does a chlorine atom have (before it bonds)?
7; since chlorine is in column 17 (or “7A”) it has 7 valence electrons.
3. Since CCl4 is made up of one carbon and four chlorine atoms, how many total valence
question 2.
4 + 4(7) = 4 + 28 = 32
4. Verify that there are 32 electrons pictured in the drawing of CCl4.
Count each dot in the picture above: there are 24. There are also 8 more electrons represented
by the 4 lines (each line represents two electrons).
5. Find the sum of all the valence electrons for NF3. (Add how many valence electrons one
nitrogen atom has with the valence electrons for three fluorine atoms.)
26; nitrogen has 5 valence electrons; fluorine has 7. Since there are one nitrogen and 7
fluorine atoms, the total is obtained as follows: 5 + 7(3) = 5 + 21 = 26
6. How many electrons are pictured in the drawing of NF3 above?
26; count each dot in the picture above: there are 20. There are also 6 more electrons
represented by the 3 lines (each line represents two electrons).
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7. In CCl4 carbon is the “central atom”. In NF3 nitrogen is the “central atom”. What is meant by
“central atom”?
The central atom is the atom that is in the middle of other atoms in the Lewis structure. Other
atoms branch off from the central atom.
8. In SF3 sulfur is the central atom. You can tell which atom is the central atom simply by
looking at the formula. How does the formula give away which atom is the central atom?
The central atom is usually the first atom written in the formula. Also, you can recognize the
central atom in the formula because there is only one of the atom. For example, in the above
formula for SF3 there is only one sulfur atom.
9. Identify the central atom in each of the following molecules:
A) CO2
carbon
B) PH3
phosphorus
C) SiO2
silicon
10. For each of the compounds from question 9, add up how many valence electrons should be in
the bonding picture. A is done for you.
A) CO2
4 + 2(6) = 16
B) PH3
5 + 3(1) = 8
C) SiO2
4 + 2(6) = 16
11. The number of electrons that should appear in the bonding picture for CO3 is 22. The number
of electrons that appear in the picture for CO32- is 24. Offer an explanation for why CO32- has
24 electrons instead of 22. (Where did the extra two electrons come from?)
The -2 charge on CO32- means that it has two extra electrons.
12. The number of electrons that should be included in the picture of NH4 is 9. The number of
electrons in the picture for NH4+ is 8. Offer an explanation for why NH4+ has 8 electrons
The +1 charge on NH4+ means that it has one fewer electron.
13. Considering questions 11 and 12, we can formulate a rule: For each negative charge on a
polyatomic ion, we must ____add______ an electron and for each positive charge we must
___subtract____ an electron.
14. For each of the polyatomic ions or molecules below, determine the total number of valence
electrons.
a) NO3b) SCl4
c) H3O+
d) PO435 + 3(6) + 1 = 24
6 + 4(7) = 34
3(1) + 6 -1 = 8
5 + 4(6) +3 = 32
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Information: Steps for Drawing Lewis Structures for Covalent Compounds
Study the two examples in the table of how to write structures for CO32- and NH3. Make sure you
understand each of the five steps.
Step #1: Add up the number of
valence electrons that should be
included in the Lewis Structure.
CO32-
NH3
4 + 3(6) + 2 = 24
(carbon has four and each
oxygen has six; add two for the
-2 charge)
5 + 3(1) = 8
(nitrogen has five; each
hydrogen has one)
Step #2: Draw the “skeleton
structure” with the central atoms
and the other atoms, each
connected with a single bond.
Step #3: Add six more electron
dots to each atom except the
central atom. Also, never add dots
to hydrogen.
(no change)
Step #4: Any “leftover” electrons
are placed on the central atom.
Find the number of leftovers by
taking the total from Step #1 and
subtracting the number of
electrons pictured in Step #3.
24 – 24 = 0 leftover electrons
Step #5: If the central atom has 8,
then you are done. If not, then
move two electrons from a
different atom to make a multiple
bond. Keep making multiple
bonds until the central atom has 8
electrons.
a total of 4 electrons are
shared here
2 electrons were
moved to form a
“double bond”
8 – 6 = 2 leftover
electrons; placed around
nitrogen
(no change)
(no change)
Critical Thinking Questions
15. Write the Lewis Structure for nitrate, NO3-1. Hint: when you are done it should look very
similar to CO32- in the table above.
16. Draw the Lewis Structure for SO2.
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70
ChemQuest 21
Name: ____________________________
Date: _______________
Hour: _____
Introduction Questions
1. Draw the Lewis structures for the following molecules. Part c is done for you.
a) H2O (oxygen is central)
b) SO2
c) N2
2. In question 1 above you should see three different kinds of bonds. Define each of these kinds
of bonds as best you can.
a) single bond: two electrons being shared between atoms; represented by one line
b) double bond: four electrons being shared between atoms; represented by two lines
together.
c) triple bond: six electrons being shared between atoms; represented by three lines
together.
3. In general, which kind of bond do you think would be harder to break—a single, double or
triple bond? Why?
Triple bonds are harder to break because six electrons being shared means that there is more
“glue” than between atoms that share fewer electrons.
Information: Bond Order, Bond Length and Bond Energy
Study the following table relating bond order, bond length, and bond energy. Note that bond length
is the distance between two bonding atoms in picometers and bond energy is the amount of energy
(in kilojoules) required to break one mole of bonds.
Carbon—Carbon Bonds
(note: these are drawings of bonds, not entire
molecules)
Bond Order
for C—C
bonds
C—C Bond
Length (pm)
C—C Bond
Energy (kJ/mole)
1
150
350
2
130
600
3
120
830
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Critical Thinking Questions
4. Is your answer to question 3 consistent with the data in the table?
Hopefully you see that triple bonds require more energy to break—they are harder to break
than other bonds.
5. What is the relationship between bond order and single, double, and triple bonds?
A single bond has a bond order of 1, a double bond has bond order of 2, and a triple bond has
a bond order of 3.
6. What relationship exists between bond length and bond energy?
Shorter bonds are also harder to break.
7. Which bonds are shortest—single, double or triple?
Triple bonds are the shortest.
8. A structure for benzene is given below.
H
H
All of these single bonds are
1st order and have lengths of
approximately 150 pm (as
estimated from the table
above)
C
C
C
H
H
C
C
C
All of these double bonds
are 2nd order and have
lengths of approximately
130 pm (as estimated from
the table above)
H
H
a) Based on the Lewis structure, what is the bond order for each C—C bond? Label your
predictions on the structure above.
See labels above.
b) What would be the approximate lengths of the carbon-carbon bonds? Label these
predictions on the structure also.
See the labels where the double bonds have lengths of approximately 130 pm and
single bonds have lengths of about 150 pm.
Information: Resonance Structures
Experimentally, all of the carbon-carbon bonds in benzene are exactly the same length. All of the
bonds also require the same energy to break. This would indicate that all of the carbon-carbon bonds
are identical. The bond order of each of the C—C bonds is approximately 1.5. The bonds have
characteristics in between those of first order and second order bonds. Therefore, the structure of
benzene is best represented by what is called a resonance structure. The structure is given at the top
of the next page.
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72
The resonance hybrid representation of benzene:
H
H
H
H
C
H
C
H
H
C
C
C
C
C
C
C
C
C
H
H
H
C
H
H
Critical Thinking Questions
9. a) The nitrate ion, NO3-, is best represented by three resonance structures. Draw the three
structures below.
b) Why were three structures required instead of two?
The double bond can be in three different, but equivalent places and thus, three different
resonance structures are needed.
c) Describe in general terms what you think nitrate’s nitrogen-oxygen bonds are like in terms
of bond order, bond length, and bond energies.
Each bond has a bond order of 1.333 (or 1 1/3). Each bond will be equal in length,
somewhere between 130 pm and 150 pm . Each bond will have the same bond energy,
between 350 kJ and 600 kJ. (These values are given on the table above.)
As an explanation for the bond order you could mentally picture it like this: The bond
order is 1 and 1/3 because the double bond is at each of the positions 1/3 of the time. This
means that each bond is a single bond and 1/3 of another bond.
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ChemQuest 25
Name: ____________________________
Date: _______________
Hour: _____
Information: More than one possible Lewis Structure.
Sometimes when you draw a Lewis structure you discover that there is more than one possible way to
draw it. For example, consider the following Lewis Structures for sand, which is silicon dioxide:
Diagram #1:
Structure A
Structure B
Critical Thinking Questions
1. What is the total number of electrons allowed in the Lewis structure for SiO2? Do both of the
above Lewis structures have the correct number of electrons pictured?
16; you find this by adding up the total valence electrons of each atom involved. Oxygen has
6 valence electrons since it is in column 16. Each silicon has 4 valence electrons. Adding it
all up looks like this: 6(2) + 4 = 16.
2. Is each of the proposed structures above a legitimate Lewis structure for SiO2? (In other
words, in each of the proposed structures are all of the “rules” that you know followed?)
Explain.
Both Lewis Structures look good according to the rules we know of so far. There are the
correct number of electrons in the picture (16) and every atom is sharing enough electrons so
that each has 8.
3. According to a large number of experiments, both of the silicon-oxygen bonds in SiO2 are
identical. Given this information, which Lewis structure, A or B, is a better description of the
bonding in SiO2. Explain.
Given this experimental evidence, Structure A is better because both of the bonds in Structure
A are identical.
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Information: Formal Charges
When there are two structures that are possible for a compound, then scientists in a lab can determine
which one is the correct one. Other than laboratory work, there is another way to determine which
Lewis structure is correct—using “formal charges”.
Let’s examine Structure B for SiO2. In the structure that was drawn, each atom has the eight
electrons that they need. Let’s look at how many formal electrons each atom has. To understand
“formal electrons” study the diagram of Structure B below:
Diagram #2:
Of the two electrons being shared, one is
a formal electron for oxygen and one is a
formal electron for silicon.
Electrons that
aren’t shared
belong to the atom
that they are
around, so these six
electrons are six of
oxygen’s formal
electrons.
Of the six electrons being shared, three
are formal electrons for oxygen and three
are formal electrons for silicon.
Oxygen Atom #2
Oxygen Atom #1
Electrons that aren’t
shared belong to the
atom that they are
around, so these two
electrons are two of
oxygen’s formal
electrons.
You should see from the above diagram that the silicon atom has a total of four formal electrons,
oxygen atom #1 has 7 formal electrons, and oxygen atom #2 has 5 formal electrons. Now we can
calculate the formal charge of each atom.
Found from the column of the periodic table that the atom is in
Formal charge = (# of valence electrons) – (# of formal electrons)
Here are the formal charges for each of the atoms in structure B for SiO2.
Oxygen atom #1: 6 – 7 = -1
Oxygen atom #2: 6 – 5 = 1
Si = 4 – 4 = 0
Critical Thinking Questions
4. Verify that the formal charge for each of the atoms is zero in Structure A from Diagram #1.
Each oxygen has six valence electrons and six formal electrons formal charge is zero.
Silicon has four valence electrons and four formal elcectrons formal charge is zero.
5. Consider questions 3 and 4. If two different structures are possible, does the best structure
have the fewest or the most formal charges?
The best structure will have the fewest formal charges.
6. Draw a Lewis structure for the carbonate ion, CO32-. On your drawing label the formal charge
of each atom.
This carbon’s formal charge is zero because C has 4
These oxygens have a
formal charge of -1, having
6 valence electrons and 7
formal electrons in the
picture: 6 – 7 = – 1
valence electrons and there are 4 formal electrons
pictured here: 4 – 4 = 0
This oxygen’s formal charge is zero because
oxygen has 6 valence electrons and here there
are 6 formal electrons; 6 – 6 = 0
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7. Add up all of the formal charges for all the atoms from question 6.
(-1) + (-1) + 0 + 0 = – 2
8. Draw the Lewis structure for the ammonium ion, NH4+ and label the formal charge of each
atom.
Each hydrogen has a
formal charge of
zero since hydrogen
has one valence
electron and each
has one formal
electron in the
picture.
Nitrogen’s formal charge here is +1
since nitrogen has 5 valence electrons
and in the picture it has 4 formal
electrons. 5 – 4 = +1
9. Add up all of the formal charges for the atoms from question 8.
+1 + 0 + 0 + 0+ 0 = +1
10. Draw the Lewis structure for SO3 and label the formal charge of each atom.
This sulfur has a formal charge of +2, having 6
valence electrons and only 4 formal electrons in
the picture: 6 – 4 = + 2
These oxygens have a
formal charge of -1, having
6 valence electrons and 7
formal electrons in the
picture: 6 – 7 = – 1
This oxygen’s formal charge is zero because
oxygen has 6 valence electrons and here there
are 6 formal electrons; 6 – 6 = 0
11. Add up all of the formal charges for the atoms from question 10.
+2 + (-1) + (-1) + 0 = 0
12. Consider questions 6-11 and then fill in the blanks: The SUM of the formal charges for all the
atoms in a structure always equals the _charge_ of the molecule or ion. For example, the
charge on CO32- is ___-2___ and the sum of the formal charges (question 6) is ___-2___.
13. Which of the following is the best structure for CH2O2? Explain your reasoning.
Structure #1
Structure #2
Structure #1 is better because all of the atoms have a formal charge of zero. In Structure #2,
one of the oxygen atoms has a formal charge of -1 and the other oxygen has a formal charge
of +1. The best structure always has the least number of formal charges.
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76
ChemQuest 26
Name: ____________________________
Date: _______________
Hour: _____
Information: Definition of Electronegativity
Electronegativity is a measure of how much an atom attracts an electron. The higher the
electronegativity, the greater the atom’s attraction for electrons. Atoms that become negative ions
have a much greater electronegativity than atoms that become positive ions. Below is a table of the
electronegativities of many elements from the periodic table.
1
H
2.20
3
4
Li
Be
0.97 1.47
11
12
Na Mg
1.01 1.23
19
20
K
Ca
0.91 1.04
5
6
7
8
9
B
C
N
O
F
2.01 2.50 3.07 3.50 4.10
13
14
15
16
17
Al
Si
P
S
Cl
1.47 1.74 2.06 2.44 2.83
31
32
33
34
35
Ga Ge
As
Se
Br
1.82 2.02 2.20 2.48 2.74
2
He
—
10
Ne
—
18
Ar
—
36
Kr
—
Critical Thinking Questions
1. Why do you think there are no values for the noble gases?
Noble gases are so stable that they do not attract electrons.
2. In terms of electrons, what is the difference between a covalent bond and an ionic bond?
Electrons are shared in a covalent bond, but atoms lose or gain them in ionic bonds.
3. What type of bond (covalent or ionic) would you expect to form between an atom with a high
electronegativity and an atom of low electronegativity. Explain and give an example.
Ionic bond because the atom with high electronegativity will attract an electron away from the
atom with low electronegativity. Example: NaF (between a metal and a nonmetal)
4. What type of bond (covalent or ionic) would you expect to form between two atoms of somewhat
high electronegativity? Explain and give an example.
Covalent bond because both atoms will be trying to attract an electron and so they will end up
sharing it. Example: CO2 (both atoms have high electronegativity compared to the metals)
5. Consider the ionic compound, sodium chloride (NaCl). Which atom has a greater attraction for
electrons—sodium or chlorine? Which atom forms the negative ion?
Chlorine has the highest electronegativity and therefore the strongest attraction. It also forms the
negative ion and sodium forms a positive ion.
77
6. In an ionic bond, the atom with the highest electronegativity will always form a
_______negative_________ ion.
(positive or negative)
7. Consider the covalent compound, carbon monoxide (CO).
a) Draw the Lewis dot structure for carbon monoxide.
b) In the Lewis structure you drew, you should see that there is a triple bond between carbon and
oxygen. The carbon and oxygen share 6 electrons. All 6 electrons are not shared equally,
however, because carbon and oxygen don’t have equal attraction for electrons. The 6
electrons spend a little more time near one of the atoms—predict which one and explain.
The electrons will spend more time near the oxygen because oxygen has a greater
electronegativity than carbon and therefore attracts the electrons more.
8. Why are the electrons in a nitrogen-phosphorus covalent bond NOT shared equally? Which atom
do the electrons spend more time around? Explain.
They are not shared equally because nitrogen and phosphorus do not have the same
electronegativity. The atoms spend more time around the most electronegative atom—
nitrogen.
9. True or false: In an ionic bond, the difference in electronegativities between the two bonding
atoms is greater than the difference in a covalent bond.
True; there is a greater difference between a metal and nonmetal than between 2 nonmetals.
10. In terms of electronegativity, explain why this statement is true: “Carbon monoxide is more
‘ionic’ than carbon monosulfide”.
The electronegativity difference is greater between carbon and oxygen (like in carbon monoxide)
than between carbon and sulfur (like in carbon monosulfide)
11. Which bond is more like an ionic bond—a nitrogen-oxygen bond or a carbon-oxygen bond?
Explain.
A carbon oxygen bond is more like an ionic bond because there is a greater difference in
electronegativity between carbon and oxygen than between nitrogen and oxygen.
12. Which compound is more like an ionic compound—NH3 or PH3? Explain.
NH3, because of the greater electronegativity difference between N and H than between P and H.
Information: “Polar” Bonds
In critical thinking questions 10, 11, and 12 we used the term “ionic” when describing covalent
bonds. This can get confusing and so instead of the term “ionic” we will use the term “polar”. A
polar covalent bond then is a covalent bond in which the electrons are not shared equally by the two
atoms involved in the bond. For example, in carbon monoxide the electrons spend more time near
oxygen than carbon because oxygen has a greater electronegativity. Because of oxygen’s greater
electronegativity, it attracts the electrons more than carbon.
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Critical Thinking Questions
13. Which bond is more polar—a phosphorus-chlorine bond or a phosphorus-fluorine bond? Explain
why.
P—F bond is more polar than the P—Cl bond because there is a greater electronegativity
difference between P and F than between P and Cl.
14. Consider a carbon-fluorine bond. One atom in the bond is “partially negative” and the other atom
is “partially positive”. Which atom is which? Explain how you know.
The fluorine is “partially negative” because it has a higher electronegativity than carbon and
therefore the electrons that are shared between carbon and fluorine spend more time near fluorine.
15. In your own words explain what you think the term “partially negative” means and explain why
an atom might be partially negative when it bonds covalently. (I.e. What makes it partially
negative?)
If an atom is “partially negative” it means that the electrons are more attracted to that atom than
to the other atom covalently bonded to it.
16. In your own words explain what you think the term “partially positive” means and explain why
an atom might be partially positive when it bonds covalently. (What makes it partially positive?)
If an atom is partially positive, then the electrons it shares with another atom will spend more
time around that other atom.
Information: Polar vs. Nonpolar
Not all covalent compounds are categorized as polar. For example, a carbon-hydrogen bond is made
up of two atoms that have very similar electronegativities. Because of this, carbon and hydrogen
share the electrons just about equally and we say that a carbon-hydrogen bond is “nonpolar”. As a
general measure, if the difference in electronegativity between two bonding atoms is less than about
0.5, then the bond is nonpolar.
Critical Thinking Questions
17. For the following bonds, indicate whether they are ionic (I), polar covalent (P), or nonpolar
covalent (NP).
__NP_ a) C—P
__P__ b) S—O
__I__ c) Fe—O
18. What are the three most electronegative atoms on the periodic table?
Nitrogen, oxygen, and fluorine
__P__ d) N—H
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ChemQuest 27
Name: ____________________________
Date: _______________
Hour: _____
Information: Shapes of Molecules
Name
Lewis
Structure
Methane, CH4
H
Ammonia, NH3
H
H C H
H
Tetrahedral shape
Water, H2O
H
N
O
H
H
Trigonal pyramidal
shape
H
Bent shape
3-D
Shape
Bond angle =109.5o
Bond angle =106.5o
Bond angle =104.5o
Total # of electron regions
# of Bonding electron regions
# of lone pair electron regions
4
4
0
4
3
1
4
2
2
Name
Carbonate, CO32-
Ozone, O3
Carbon dioxide, CO2
Lewis
Structure
O
Trigonal planar shape
3-D
Shape
Bond angle =120o
Total # of electron regions
# of bonding electron regions
# of lone pair electron regions
3
3
0
Bent shape
Bond angle =118.6o
C
O
Linear shape
Bond angle =180o
3
2
1
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2
2
0
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Critical Thinking Questions
1. What is an electron region?
A place around the central atom where electrons (either bonding or lone pair) can be found.
2. What is a "lone pair electron region"?
A place around the central atom where electrons not bonding with another atom can be found.
3. What is a "bonding electron region"?
A place around the central atom where electrons are shared with another atom.
4. The number of electron regions determines the bond angle. With this in mind, complete the
following sentence: "Any molecule that has bond angles of approximately 105-109o will have
____4_______ total electron regions; any molecule that has bond angles of approximately 120o
how many?
will have ____3______ total electron regions; and any molecule with bond angles of
how many?
approximately 180o will have ____2______ total electron regions."
how many?
5. The molecules in the above table are representative of many other molecules. Therefore, it can be
said that any molecule with 3 bonding electron regions and 1 lone pair electron region has a
geometrical shape called "trigonal pyramidal". Draw Lewis dot structures for the following
structures and name the geometrical shape.
A) NO3B) NF3
C) CF4
trigonal planar
trigonal pyramidal
tetrahedral
6. A certain molecule has a bent shape with bond angles of about 119o. Is the molecule SO2 or SH2?
Explain. (Hint: draw the Lewis structures for SO2 and SH2.)
The molecule is SO2 because SO2 has 3 electron domains which corresponds to bond angles near
120o. SH2 has 4 electron domains which would correspond to bond angles near 109o.
Information: VSEPR
The geometry of molecules is based on a theory called "Valence Shell Electron Pair Repulsion"
(VSEPR) theory. The word "repulsion" is the key word because this theory states that all the electron
pairs repel each other and so they want to get as far away from each other as possible. The atoms in a
tetrahedral molecule are as far apart as geometrically possible at bond angles of 109.5o. There is no
way that the atoms can get farther apart.
Critical Thinking Questions
7. In the tables above, there are 3 molecules that have a total of 4 electron regions. The bond angles
are slightly different because of lone pair electrons. What takes up more room--a lone pair of
electrons or a bonding pair of electrons? Offer proof from the table above.
A lone pair takes up more room, which “squeezes” atoms a little closer together and causes the
smaller bond angles.
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8. If you know how many bonding regions and lone pair regions surround an atom you can predict
the bond angles around the atom, even in complex situations. Examine the following "big"
molecules. By each arrow that points to an atom, write the bond angle for that atom; you should
write 109o, 120o, or 180o to represent the approximate bond angle. One of them is done for you.
109o
H
C
H
H
H
H
H
C
C
C
C
H
C
C
H
H
C
H
N
H
109o
120o because of 3 bonding regions
and no lone pair regions
H
H
H
H
O
120o
109o
120o
H
H
C
H
H
C
C
H
H
H
H
H
H
C
H
N
C
H
H
O
H
O
C
C
C
N
O
180o
109o
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ChemQuest 28
Name: ____________________________
Date: _______________
Hour: _____
Information: Introduction to Reactions
During a chemical reaction, new substances are formed. Reactants are transformed into different
products. Atoms are never created or destroyed, but they are rearranged. A chemical equation
represents what happens during a reaction. The following is an example of a chemical equation:
Example Equation: Ca + HNO3 Ca(NO3)2 + H2
This equation describes the reaction of calcium (Ca) with nitric acid (HNO3) to produce calcium
nitrate (Ca(NO3)2) and hydrogen gas (H2). You may notice that there are more total atoms on the
right side than there are on the left side of the equation. If this seems strange to you, don’t worry
about it now; we’ll fix this later.
Note in the above equation that hydrogen gas is written as H2 and not simply as H. There are a few
elements that exist as diatomic molecules. If a substance is diatomic then the substance must always
be bonded to something. A hydrogen atom is diatomic and so it must be bonded to something else
like in HCl or HNO3. If nothing is available for it to bond to, it will bond to itself by forming H2. All
of the diatomic substances are listed below:
Br
I
N
Cl
H
O
F
When by themselves these elements exist as Br2, I2, N2, Cl2, H2, O2, and F2. By the way, you can
remember these by recalling a made-up name: Mr. Brinclhof.
Critical Thinking Questions
1. Consider the bromine atoms in this reaction: LiBr + P Li3P + Br2.
a) Why is bromine written as Br2 on the right side?
Bromine is a diatomic molecule and always needs to be bonded to something; even
bonding to itself works.
b) Why is it not necessary for LiBr to be written as LiBr2?
The Br is already bonded to something else (the Li). If it were written LiBr2 it would not
be a neutral compound.
2. What are the reactants in the example equation in the above information section?
Reactants are written on the left side of the equation and so the reactants are Ca and HNO3.
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3. Answer the questions that follow based on this chemical equation: Na + MgCl2 NaCl + Mg.
a) Why can’t NaMg be produced?
Na+ cannot bond with Mg2+ because they are both positive.
b) Why can’t NaCl2 be produced?
Na+ only requires one Cl-. NaCl2 is not a neutral compound.
c) Are NaCl and Mg the only products that can be produced?
Yes.
4. Given the following equation: Li + Ca3(PO4)2 Li3PO4 + Ca.
a) Why can’t CaLi2 be produced?
Ca2+ and Li+ won’t bond because they are both positive.
b) Why can’t Li3P be produced?
Although Li3P is neutral, there is no P3- in the equation. There is only PO43- and we will
almost never be breaking up polyatomic ions like PO43-. Don’t mess with polyatomic ions!
c) Are Li3PO4 and Ca the only substances that can be produced?
Yes.
5. Write chemical equations for the following reactions.
a) Aluminum sulfate reacts with barium to produce barium sulfate and aluminum.
Al2(SO4)3 + Ba BaSO4 + Al
b) Magnesium reacts with copper(I) nitrate to produce magnesium nitrate and copper.
Mg + CuNO3 Mg(NO3)2 + Cu
c) Sodium reacts with calcium phosphide to produce sodium phosphide and calcium.
Na + Ca3P2 Na3P + Ca
d) Phosphorus reacts with sodium chloride to produce sodium phosphide and chlorine.
(Note that chlorine is diatomic!)
P + NaCl Na3P + Cl2
6. Each of the reactions you wrote in question 5 follows a similar pattern. The same pattern is
followed by the equations in questions 3 and 4. Describe this pattern.
A single atom reacts with a compound and replaces one of the atoms in that compound.
7. a) How are reactions 5c and 5d different?
Na forms a positive ion (Na+) and replaces another positive ion, but P forms a negative ion (P3-)
and replaces another negative ion.
b) How are reactions 5c and 5d similar?
In both reactions a single atom replaces another atom from a compound that it is reacting with.
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8. Complete the following reactions:
a) NaCl + Ag AgCl + Na
(note: order is not important, so you could also write
Na + AgCl)
b) Li + Ca3(PO4)2 Li3PO4 + Ca
(again, order is not important)
Information: Single and Double Replacement Reactions
Each of the equations that you looked at in the above section is called a single replacement reaction.
Notice that in each of them, a single atom replaces an ion from another reactant. Study what happens
in the following reactions. They are called double replacement reactions.
MgCl2 + NaF NaCl + MgF2
Al2S3 + Na2O Al2O3 + Na2S
Ca(NO3)2 + Na3P NaNO3 + Ca3P2
Critical Thinking Questions
9. What is the difference between single replacement reactions and double replacement reactions?
In single replacement reactions, one of the reactants is made of only one kind of atom. In double
replacement reactions, both reactants contain more than one kind of atom.
10. Complete the following reactions by providing the formulas for the missing compound(s).
a) Cu(NO3)2 + _____NaCl________ NaNO3 + CuCl2
b) ZnI2 + ______Al2(SO4)3________ ZnSO4 + AlI3
c) K2O + MgBr2 ____MgO_______ + ______KBr_______
11. Name the two products in the reaction between calcium phosphate and sodium iodide.
Calcium iodide and sodium phosphate
(The reaction is Ca3(PO4)2 + NaI CaI2 + Na3PO4)
12. Explain why when you mix the following reactants, no reaction occurs: Na2SO4 + NaCl Both reactants contain sodium. Therefore, if the sodium atoms replaced each other, the same
products would be formed—no change would occur.
Information: Combustion, Synthesis, and Decomposition Reactions
Another type of reaction is a combustion reaction. During combustion, a hydrocarbon reacts with
oxygen. The products for complete combustion are always the same—water and carbon dioxide and
energy. The following equation is an example of the combustion of a hydrocarbon.
C3H8 + O2 CO2 + H2O
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Two other types of reactions are synthesis and decomposition. During a synthesis reaction, several
reactants combine to make a single product. During a decomposition, one reactant decomposes into
two or more products. The following table shows some examples of these types of reactions.
Synthesis
H2 + O2 H2 O
Na + Cl2 NaCl
Decomposition
H2 O H 2 + O 2
NaCl Na + Cl2
Critical Thinking Questions
13. Categorize each of the following reactions as single replacement (SR), double replacement (DR),
synthesis (S), decomposition (D) or combustion (C).
__S__ a) Ca + O2 CaO
__SR_ b) Mg(NO3)2 + Cu CuNO3 + Mg
__C__ c) C4H10 + O2 CO2 + H2O
__D__ d) Al2O3 Al + O2
__SR_ e) SrCl2 + F2 SrF2 + Cl2
__DR_ f) BaF2 + Na2O BaO + NaF
14. Write an equation for the combustion of C3H6.
C3H6 + O2 CO2 + H2O
(Combustion reactions always involve combining with O2 to form CO2 and H2O.)
15. Write an equation for the decomposition of calcium oxide.
CaO Ca + O2
(Oxygen is diatomic, so it is written as O2 when by itself.)
Practice Problems
1. Complete the following reactions.
a) Na2CO3 + AlN Na3N + Al2(CO3)3
(Remember that the order is never important.)
b) BaCl2 + F2 BaF2 + Cl2
c) CuNO3 + Ag AgNO3 + Cu
2. Fill in the blanks for the missing reactant or product and then in the blank to the left of each
equation indicate whether the reaction is a single replacement (SR), double replacement (DR),
synthesis (S), decomposition (D) or combustion (C).
__DR_ a) LiCl + ___Zn(NO3)2____ ZnCl2 + LiNO3
__SR_ b) ____Na______ + CaBr2 NaBr + Ca
__S__ c) K + Cl2 ____KCl_______
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ChemQuest 29
Name: ____________________________
Date: _______________
Hour: _____
Introduction Questions
1. Write the equation for the reaction of sodium and chlorine to form sodium chloride.
Na + Cl2 NaCl
2. Write the equation for the reaction of calcium nitride and sodium chloride to produce calcium
chloride and sodium nitride.
Ca3N2 + NaCl CaCl2 + Na3N
Information: Subscripts and Coefficients
A subscript is a small number that tells you how many atoms are in a compound. For example, in
CaCl2 the two is the subscript and it tells us that there are two chloride ions bonded to one calcium.
A coefficient tells also tells us how many atoms or compounds there are, but in a different way. For
example in the expression “3 H2O” the three is the coefficient. The three tells us that there are three
molecules of water present. In the expression “3 H2O” there are a total of 6 hydrogen atoms and 3
oxygen atoms.
Critical Thinking Questions
3. Verify that in 4Ca3(PO4)2 there are 32 oxygen atoms present.
Multiply the subscripts by each other and then by the coefficient: 4 x 2 x 4 = 32
4. How many oxygen atoms are in each of the following:
__3__ a) Al2O3
__3__ b) 3 Na2O
__16_ c) 4 Na2SO4
__30_ d) 5 Mg(NO3)2
Information: How To Balance Equations
Consider the reaction of sodium and chlorine to produce sodium chloride from question one:
Na + Cl2 NaCl
Remember that when chlorine is by itself it is always written as Cl2. On the reactant side of the
reaction (left side) there are a total of two atoms of chlorine, but on the product side there is only one
atom of chlorine. Atoms cannot simply disappear. In order for the equation to make sense, we need
to balance the equation. This can be done, first, by adding a “2” to the product side:
Na + Cl2 2 NaCl
Now the equation reads that one atom of Na reacts with one molecule of Cl2 to produce two units of
NaCl. However, now the Na atoms are not balanced because there is one atom of the reactant side,
but two atoms of Na on the product side. This can be fixed by adding another two:
2 Na + Cl2 2 NaCl
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Let us consider another example, the equation you wrote in question two above:
Ca3N2 + NaCl CaCl2 + Na3N
Notice that none of the atoms are “balanced”. There are three calcium atoms on one side and only
one on the other. There are two nitrogen atoms on one side and one on the other. How can we fix
this? Begin by “balancing” one atom at a time:
1. First, let’s balance the calcium atoms by placing a three in front of CaCl2:
Ca3N2 + NaCl 3 CaCl2 + Na3N
2. Next, let’s balance the nitrogen atoms, by placing a 2 in front of Na3N:
Ca3N2 + NaCl 3 CaCl2 + 2 Na3N
3. Now we will balance the sodium atoms by placing a 6 in front of NaCl.
Ca3N2 + 6 NaCl 3 CaCl2 + 2 Na3N
4. Finally, examine the chlorine atoms and notice that they are already balanced.
5. Double check each atom to make sure there are equal numbers of each on both
sides of the equation.
When balancing equations you NEVER change subscripts. Only change the coefficients.
Critical Thinking Questions
5. Which of the following equations are properly and completely balanced? (Circle as many as
apply.)
A) 2 NaCl + Cu CuCl2 + Na
B) C3H8 + 3 O2 3 CO2 + H2O
C) MgCl2 + F2 MgF2 + Cl2
D) FeCl2 + O2 2 FeO + Cl2
6. Balance each of the following equations by inserting the correct coefficient in each blank.
Remember to balance one atom at a time. If the number “one” belongs in the blank you may
either leave it blank or insert the number one.
A) _____ CaCl2 + _____ Li2O _____ CaO + __2__ LiCl
B) _____ Al2S3 + __3__ Cu __3__ CuS + __2__ Al
C) __2__ Na3P + __3__ MgBr2 _____ Mg3P2 + __6__ NaBr
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A) CaCl2 + Li2O CaO + 2 LiCl
B) Al2S3 + 3 Cu 3 CuS + 2 Al
C) 2 Na3P + 3 MgBr2 Mg3P2 + 6 NaBr
Information: Balancing Equations Containing Polyatomic Ions
When an equation contains polyatomic ions, it may look a little more difficult to balance. But
balancing an equation containing polyatomic ions is really not much different from the ones you just
did. In the equations above, you kept in mind that you must balance only one atom at a time. With
polyatomic ions, keep in mind that you balance one polyatomic ion at a time. For example, consider
the following unbalanced equation.
Na2SO4 + Ca3(PO4)2 Na3PO4 + CaSO4
The first thing you should do is take note of which atoms/ions are not balanced. Keep the polyatomic
ions together. In other words, do not balance the phosphorus and oxygen atoms separately; instead,
balance the phosphate ions on each side of the reaction. Follow these steps:
1. Balance the sodium so that there are 6 sodium atoms on each side of the reaction:
3 Na2SO4 + Ca3(PO4)2 2 Na3PO4 + CaSO4
2. Now there are three sulfate ions on the left and one on the right. Add a three to the
right.
3 Na2SO4 + Ca3(PO4)2 2 Na3PO4 + 3 CaSO4
3. Take inventory of all the atoms and ions to make sure they are balanced. The
equation is now balanced.
Critical Thinking Questions
8. Which of the following equations are properly balanced?
A) Ba(NO3)2 + Fe2SO4 2 FeNO3 + BaSO4
B) 3 SrCl2 + Al(NO3)3 3 Sr(NO3)2 + AlCl3
C) Ca3(PO4)2 + 3 Na2CO3 2 Na3PO4 + 3 CaCO3
D) Fe(NO3)2 + Mg Mg(NO3)2 + Fe
9. Balance each of the following equations by inserting the correct coefficient in each blank.
Remember to balance one atom/ion at a time. If the number “one” belongs in the blank you may
either leave it blank or insert the number one.
A) _____ Ca(NO3)2 + __2__ Na __2__ NaNO3 + _____ Ca
B) _____ Al2(CO3)3 + __3__ MgCl2 __2__ AlCl3 + __3__ MgCO3
C) _____ Ba3(PO4)2 + __3__ Na2CO3 __2__ Na3 PO4 + __3__ BaCO3
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10. Make sure you got the following answers for question 9:
A) Ca(NO3)2 + 2 Na 2 NaNO3 + Ca
B) Al2(CO3)3 + 3 MgCl2 2 AlCl3 + 3 MgCO3
C) Ba3(PO4)2 + 3 Na2CO3 2 Na3 PO4 + 3 BaCO3
11. Complete and balance the following equations.
A)
2 HCl +
Mg MgCl2 + H2
B)
Na2S +
Be(OH)2 2 NaOH + BeS
C)
BaF2 +
D)
Ca3(PO4)2 +
E)
Zn(NO3)2 +
F)
3 Ca(NO3)2 +
2 NaNO3 Ba(NO3)2 + 2 NaF
3 K2SO4 2 K3PO4 + 3 CaSO4
Mg Mg(NO3)2 + Zn
Al2(CO3)3 3 CaCO3 + 2 Al(NO3)3
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ChemQuest 30
Name: ____________________________
Date: _______________
Hour: _____
Information: Atomic Mass Using Atomic Mass Units (amu) and Grams (g)
One atomic mass unit (amu) is equal to 1.6611x10-24 grams.
Critical Thinking Questions
1. According to the periodic table, a single carbon atom has a mass of 12.011 amu. What is the
mass of a single carbon atom in grams?
(12.011 amu) x (1.6611x10-24 g/amu) = 1.9951x10-23
2. How many carbon atoms does it take to equal 12.011 grams?
(12.011g) / (1.9951x10-23 g/atom) = 6.0201x1023 atoms
3. According to the periodic table, a single phosphorus atom has a mass of 30.973 amu. What is the
mass of a single phosphorus atom in grams?
(30.973 amu) x (1.6611x10-24 g/amu) = 5.1449x10-23
4. How many phosphorus atoms does it take to equal 30.973 grams?
(30.973g) / (5.1449x10-23 g/atom) = 6.0201x1023 atoms
They are both approximately the same. Therefore, it takes about 6.02x1023 atoms of carbon to
equal the atomic mass in grams and it takes 6.02x1023 atoms of phosphorus to equal its atomic
mass in grams.
Information: What is a Mole?
should be about 6.02x1023. The quantity, 6.02x1023 is Avogadro’s constant and we call it the “mole”.
Just like the quantity “12” is called a “dozen”, so the quantity 6.02x1023 is called a “mole”.
By definition, a mole is the quantity of atoms necessary to equal the element’s atomic mass in grams.
So, according to the periodic table, one atom of sodium has an atomic mass of about 22.99 amu. If
you weighed out 22.99 grams on a balance, you would have 6.02x1023 atoms of sodium present.
Look at your periodic table and find gold (atomic number = 79). What is the mass of one gold atom?
You should note that one gold atom has a mass of 196.97 amu. How many gold atoms would you
need to get 196.97 grams? You would need to put 6.02 x 1023 atoms of gold on the balance before
you would have 196.97 grams of gold. One mole of an atom will always equal the atom’s atomic
mass in grams.
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Critical Thinking Questions
6. What is the mass of one atom of aluminum? (include units)
26.98 amu (from the periodic table)
7. If you had 6.02 x 1023 atoms of aluminum, what mass of aluminum would you have? (include
units)
Since 6.02x1023 is a “mole” there will be 26.98 grams of aluminum.
8. What is atomic mass?
It is the mass of a single atom, usually measured in units of amu (atomic mass units).
9. If you had one mole of pennies, how many pennies would you have?
You would have 6.02x1023 pennies.
10. If you had 3 moles of sand, how many grains of sand would you have?
(6.02x1023)(3) = 1.81x1024 grains of sand.
Information: Molecular Mass (also known as Formula Mass) and Molar Mass
Just as atomic mass is the mass of an atom, molecular mass is that mass of a molecule. It is found by
adding up all of the masses of the atoms in the molecule. Because ionic compounds are not properly
called molecules, the term formula mass is used in place of molecular mass for ionic compounds.
Consider the following examples:
1. The atomic mass of hydrogen 1.0 amu and the atomic mass of oxygen is 16.0 amu. One
molecule of water (H2O) has a molecular mass of 18.0 amu. This number is obtained by
adding the masses of two hydrogens (each at 1.0 amu) and the mass of one oxygen (16.0
amu).
2. Aluminum chloride (AlCl3) has a formula mass of about 133.5 amu. This is found by
adding the mass of one aluminum atom (27.0 amu) to the mass of three chlorine atoms (3
x 35.5 amu). Verify this on your calculator.
Just as one mole of atoms equals the atomic mass of an atom in grams, so also one mole of molecules
equals the molar mass of the molecule in grams. Molar mass is the mass (in grams per mole) of one
mole of a substance. Therefore we expect that 6.02 x 1023 molecules of water will have a mass of
18.0 g and 6.02 x 1023 formula units of AlCl3 will have a mass of 133.5 g. We say, then, that the
molar mass of water is 18.0 g/mol and the molar mass of AlCl3 is 133.5 g/mol. (g/mol is read grams
per mole where mol is the abbreviation for mole.)
Critical Thinking Questions
11. Verify using a periodic table and calculator that the molecular mass of N2O5 is approximately 108
amu.
2(14.0) + 5(16.0) = 108 amu; note: 14.0 and 16.0 are found on the periodic table for
nitrogen and oxygen, respectively.
12. How many molecules of N2O5 are required to equal 108 grams? 6.02x1023
Since 108 amu is the molecular mass, it will take 6.02x1023 molecules to equal 108 grams.
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13. Why is the term “molecular mass” applied to water, but the term “formula mass” applied to
aluminum chloride? Although molecular and formula masses are calculated the exact same way,
the term “molecular” cannot apply to any substance that has a metal in it (such as aluminum).
14. Find the molecular or formula mass for each of the following (include units):
a) magnesium phosphide
b) sodium sulfate
Mg3P2
Na2SO4
3(24.31) + 2(30.97) = 134.87 amu
2(23.0) + (32.06) + 4(16.0) = 142.06 amu
(NOTE: all numbers in parentheses are found on the periodic table.)
c) Ca(NO3)2
(40.1) + 2(14.0) + 6(16.0) = 164.1 amu
d) C4H8
4(12.0) + 8(1.01) = 56.1 amu
15. Find the molar mass of each of the following (include units):
a) CaCl2
b) barium nitrate
40.1 + 2(35.5) = 111.1 g/mol
Ba(NO3)2
137.3 + 2(14.0) + 6(16.0) = 261.3 g/mol
16. What is the difference between the terms molecular mass and molar mass.
Molecular refers to the mass of a single molecule, but molar refers to the mass of a mole of
something (or the mass of 6.02x1023 atoms or molecules) .
Information: Beginning Mole Conversions
The mole (often abbreviated as mol) is the link between the microscopic (atoms and molecules) and
the macroscopic (things measured in grams). If you know how many grams of a substance you have
and you know the molar mass, you can find out how many molecules you have. Two examples of
how to do this using a method similar to converting units is shown below:
1. How many atoms of carbon does it take to equal 23.5 g?
23.5 g •
1mol 6.02x1023 atoms
•
= 1.18x1024 atoms
12 .0 g
mol
molar mass of carbon, from the periodic table
2. How many molecules of carbon monoxide gas does it take to equal 50.0 g?
50.0g •
mol 6.02 x10 23 molecules
•
= 1.08x10 24 molecules
28.0g
mol
molar mass of carbon monoxide found by adding molar mass of C and of O
93
Critical Thinking Questions
17. Using your calculator, verify that if you have 125 g of gold, you have about 3.82x1023 atoms of
gold. Show the calculations below.
mol 6.02x10 23 atoms
125g •
•
=
197.0 g
mol
3.82x10 23 atoms
18. Consider 210 g of N2O5. How many molecules are present?
mol 6.02x10 23 molecules
210g •
•
= 1.17 x10 24 molecules
108g
mol
19. If you exhale 7.25 x 1024 molecules of CO2…
a) How many moles of CO2 do you exhale? (Hint: use the conversion factor that one mole =
6.02x1023 molecules)
7.25 x10 24 molecules •
mol
= 12.04 mol
6.02 x10 23 molecules
b) How many grams of CO2 do you exhale? (Hint: find how many grams are in a mole by
finding the molar mass of CO2. Use this as a conversion factor.)
12.04mol •
44.0 g
= 529.9 g
mol
20. In a bag full of pennies, you may have 2.15 moles of copper. How many grams do you have?
2.15mol •
63.55 g
= 136.6 g
mol
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94
ChemQuest 31
Name: ____________________________
Date: _______________
Hour: _____
Information: Percent Composition
Sometimes it is needful to know the composition of a compound. For example, 39.3% of the mass of
sodium chloride is due to sodium. The other 60.7% of the mass is from chlorine. So, in a 100 g
sample of sodium chloride, there are 39.3 g of sodium and 60.7 g of chlorine. This type of data is
known as percent composition. The percent composition tells you the percentage by mass of an
element in a compound. There is a convenient formula for finding the percent composition of an
element in a compound:
(obtained from periodic table)
percent composition of element " X" =
mass of x in one mole of the compound
• 100
mass of one mole of the compound
(obtained from periodic table)
Let us look at how the percent composition of calcium (Ca) in calcium chloride (CaCl2) was
determined.
percent composition of Ca =
mass of Ca in one mole of CaCl 2
• 100
mass of one mole of CaCl 2
from periodic table for calcium
40.1 g
percent composition of Ca =
• 100 = 36.1%
111.1 g
from periodic table for calcium + 2 chlorines;
40.1 + 2(35.5) = 111.1
As another example, consider calculating the percent composition of nitrogen in Ca3N2:
From periodic table for 2 nitrogen atoms: 2(14.0)=28.0
percent composition of N =
28.0 g
• 100 = 18.9%
148.3 g
from periodic table for 3 calcium + 2 nitrogen atoms
3(40.1) + 2(14.0) = 148.3
Critical Thinking Questions
1. Verify that in C4H10 the percent composition of carbon is approximately 82.6%.
4(12.0)
•100 = 82.6%
[4(12.0) + 10(1.01)]
Note: 12.0 and 1.01 are the molar masses for carbon and hydrogen on the periodic table.
95
2. Calculate the percent composition of sodium in Na2S.
2(23.0)
•100 =
[2(23.0) + (32.1)]
58.9%
Information: Formulas and Percent Composition
Table 1: Percent composition and formulas of some compounds.
Name
Structural Formula
Hexene
H2C
Propene
CH
CH2
H2C
CH2
CH
CH2
CH3
CH3
Molecular
Formula
% Comp.
of H
% Comp.
of C
C6H12
14.4
85.6
C 3 H6
14.4
85.6
C 6 H6
7.8
92.2
C 4 H4
7.8
92.2
C 6 H6
7.8
92.2
CH
Benzene
HC
CH
HC
CH
CH
HC
CH
HC
CH
CH2
CH2
HC
C
C
CH
Critical Thinking Questions
3. Verify that the percent composition of C and H given for hexene in Table 1 are correct.
For carbon:
For hydrogen: 100-85.6 = 14.4%
6(12.0)
• 100 = 85.6%
[6(12.0) + 12(1.01)]
4. Fill in the blanks in Table 1 by determining the percent composition and the molecular
formulas of each compound.
See table above.
5. Can you determine a compound’s structural formula if you are given the molecular formula?
Explain.
No, for example, hexene and cyclohexane each have the same molecular formula, but
different structural formulas.
6. What is true about the percent composition of two different compounds that each have the
same molecular formula?
If two compounds have the same molecular formula they will have the same percent
composition as well.
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96
7. Can you determine a molecule’s molecular formula solely from the percent composition?
Explain.
No. As can be seen in the Table 1, compounds with different molecular formulas may still
have the same percent composition.
8. Using only the information in Table 1 (no calculators or periodic tables), predict the percent
composition of each of the following compounds:
Molecular Formula
% Composition of H
% Composition of C
C 8 H8
7.8
92.2
C10H20
14.4
85.6
Note: any compound with a 1:1 ratio of carbon to hydrogen will have 92.2% carbon and 7.8%
hydrogen. Any compound with a 1:2 ratio will have 85.6% carbon and 14.4% hydrogen.
Information: Empirical Formulas
An empirical formula is a formula that describes the lowest whole-number ratio of elements in a
compound. An example of an empirical formula is CH, which is the empirical formula for benzene
whose molecular formula was given in Table 1.
Critical Thinking Questions
9. What is the empirical formula of a compound whose percent composition is 92.2% carbon
and 7.8% hydrogen? (See question 8 and Table 1.)
The empirical formula must be CH as demonstrated in question 8 and Table 1.
10. If you know the percent composition of each atom in a molecule, can you determine the
empirical formula for the compound?
Yes you can, although right now you may not be sure exactly how to do this.
Information: Calculating the Empirical Formula
When you know the percent composition of each element in a compound, you can calculate the
empirical formula of that compound. The following example will illustrate how to do this.
Example 1: A certain compound is 30.4% nitrogen and 69.6% oxygen by mass. What is the
empirical formula of the compound?
Step #1: Divide each percentage by the molar mass from the periodic table:
For Nitrogen: 30.4 = 2.17
For Oxygen: 69.6 = 4.35
14.0
16.0
Step #2: Find the ratio of nitrogen to oxygen. To do this, find the smallest number
obtained in step #1. In this example, the smallest answer is 2.17. Now divide each of
your answers to step #1 by this smallest number. In this example, you should divide
2.17
4.35
= 1.00
= 2.00
For Nitrogen:
For Oxygen:
2
.
17
2
.
17
Step #3: Write the formula. The answers from step #2 are the subscripts in the
formula! The formula therefore is NO2.
97
If in step #2 you get something like Nitrogen = 1.00 and Oxygen = 2.50 then the formula you write in
step #3 would be NO2. 5. This doesn’t make sense because all subscripts must be whole numbers.
You would need to double each subscript. The formula would be N1x2O2.5x2 = N2O5.
Critical Thinking Questions
11. Find the empirical formula for a compound that contains 82.4% nitrogen and 17.6%
hydrogen.
5.89 is smaller than 17.4
82
.
4
For hydrogen: 17.6
For nitrogen:
= 5.89
= 17.4
14.0
1.01
Divide by the smallest: 5.89/5.89 = 1
17.4/5.89 = 2.95 3.0
The formula therefore is N1H3, or just NH3.
Information: Calculating the Molecular Formula from the Empirical Formula
Remember that the empirical formula is just a simplification of the molecular formula. For example,
consider the empirical formula NO. There are several possible molecular formulas including: N2O2,
N3O3, N4O4, etc. Which one is it? Notice that the possible formula N2O2 is made up of two of the
empirical formulas, NO. Similarly, N3O3 is made up of three of the empirical formulas, NO. How do
we know which empirical formula is correct? We need is the molar mass of the molecular formula.
Critical Thinking Questions
12. The empirical formula for a certain compound is NO. The molar mass of the compound is
60.0 g/mol.
a) What is the molar mass of the empirical formula? (Use the periodic table.)
14.0 + 16.0 = 30.0
b) Divide the molar mass of the compound (given in the question) by the molar mass of the
empirical formula found in part a.
60.0/30.0 = 2.0
c) Your answer to part b tells you how many empirical formulas are in the molecular
formula. You now should be able to write the correct molecular formula, which is N2O2.
Verify that the correct molecular formula is N2O2.
We found in part b that there are 2 empirical formulas in the molecular formula, therefore
the molecular formula must be N1x2O1x2 = N2O2.
13. a) What is the empirical formula of a compound whose percent composition by mass is 85.7%
carbon and 14.3% hydrogen?
14.3
85.7
= 7.14
= 14.2
For carbon:
For hydrogen:
12.0
1.01
Divide by smallest: 7.14/7.14 = 1
14.2/7.14 = 1.99 2
The empirical formula is therefore: CH2
b) If the compound has a molar mass of 56 g/mol, what is the molecular formula?
Molar mass of empirical formula from part a = 12.0 + 2(1.01) = 14.02
56/14.02 = 3.99 4; Molecular formula = C1x4H2x4 = C4H8
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ChemQuest 32
Name: ____________________________
Date: _______________
Hour: _____
Information: Mole Ratios in Equations
Propane is burned in many rural homes for heat in the winter. Below is the balanced equation for the
combustion of propane (C3H8).
C3H8 + 5 O2 3 CO2 + 4 H2O
For each molecule of propane that is burned, there needs to be five molecules of oxygen present.
Likewise, if there were a dozen molecules of propane, five dozen molecules of oxygen would be
required. Similarly, for each mole of propane, five moles of oxygen are needed. Also, for each mole
of propane burned three moles of carbon dioxide and 4 moles of water are produced. The numbers
of moles of each substance in a chemical equation are related by the ratio of the coefficients of
each substance.
Critical Thinking Questions
Note: For questions 1-6, refer to the balanced equation for the combustion of propane.
1. a) How many moles of water are produced when 1.45 moles of propane are combusted?
It is a 4:1 ration so the moles of water = (1.45)(4) = 5.8 moles
b) How many molecules of water is this? (Remember each mole has 6.02x1023 molecules.)
(5.8)(6.02x1023) = 3.49x1024
2. If 2.35 moles of CO2 are produced in a reaction, how many moles of H2O would be produced?
It is a 4:3 ration so the moles of water = 2.35(4/3) = 3.13 moles
3. Why is this statement false: “If 10 grams of propane burn, you need 50 grams of oxygen.”
The coefficients in the balanced equation relate mole ratios and not gram ratios. The above
statement would be true if it read: “If 10 moles of propane burn, you need 50 moles of oxygen.”
4. a) If 27.3 moles of carbon dioxide are produced during the combustion of a certain amount of
propane, how many moles of propane were combusted?
Since it is a 1:3 ratio, the moles of propane = 27.3(1/3) = 9.1 moles
We need the molar mass of propane using the periodic table: 3(12.0) + 8(1.01) = 44.08 g/mol
Now, using our answer from part a: (9.1 moles)(44.08 g/mol) = 401.1 g
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5. If you have 410 grams of propane and want to know how many grams of oxygen are required to
burn it, you can follow these steps…
a) Find the number of moles of propane that you have. Convert grams to moles!
410g/(44.08g/mol) = 9.30 mol; (note: 44.08 is the molar mass found using the periodic table)
b) The moles of propane are related to the moles of oxygen by the ratio of coefficients in the
balanced chemical equation. Find the number of moles of oxygen you need given the moles
of propane from part a.
Since it is a 5:1 ratio, the moles of oxygen are: 9.30(5) = 46.5 mol
c) Find the grams of oxygen from the moles of oxygen. Convert the moles of oxygen (answer to
part b) to grams of oxygen (O2)! (Note: use the molar mass for O2, not just O. You should
get approximately 1490 g of oxygen.)
(46.5 mol)(32.0 g/mol) = 1488 g
6. Verify that this statement is correct: If 315 grams of propane combusts, then approximately 515
grams of water are produced.
(315 g) / (44.08g/mol) = 7.15 mol propane
4:1 ratio so the mol of water = (7.15)(4) = 28.6 mol
(28.6 mol)(18 g/mol) = 514.8 g
7. Consider the decomposition of ammonia: 2 NH3 3 H2 + N2. If you start with 425 g of NH3,
how many grams of H2 and N2 can be produced?
(425 g) / (17.0 g/mol) = 25 mol NH3
(25 mol NH3)(3/2) = 37.5 mol H2
(37.5 mol H2)(2.0 g/mol) = 75 g H2
(25 mol NH3)(½) = 12.5 mol N2
(12.5 mol N2)(28.0 g/mol) = 350 g N2
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100
ChemQuest 33
Name: ____________________________
Date: _______________
Hour: _____
Information: Limiting Reactant
Again consider the combustion of propane: C3H8 + 5 O2 3 CO2 + 4 H2O. If you had 10 moles
of propane to burn, you would need 50 moles of oxygen according to the ratio in the balanced
equation. If you only had 20 moles of oxygen you could not combust all 10 moles of propane. The
reaction has been limited by the amount of oxygen you have—you don’t have enough oxygen to
burn all of the propane. In this case, oxygen is called the “limiting reactant” because it limits how
much propane can react. Notice that the limiting reactant isn’t always the substance that is present in
the fewest number of moles. In this example, propane (C3H8) is the “excess reactant” because after
the reaction there will be some of it left over. It is important to remember that everything in a
chemical equation is related by mole ratios. If you only know the mass (grams) of the substances,
you need to convert to moles.
Critical Thinking Questions
8. a) In the above discussion, it was evident that 20 moles of oxygen was not sufficient to combust
10 moles of propane. How many moles of the propane can be combusted with 20 moles of
oxygen?
4 moles. For every 5 moles of oxygen, 1 mole of propane can be combusted, so the number of
moles of propane that can be combusted with 20 moles of oxygen is 20/5 = 4 moles
b) How many moles of carbon dioxide will be produced? (Base the answer to this question on the
number of moles of propane that actually get combusted—which is your answer to part a.)
12 moles. For every mole of propane that combusts 3 moles of CO2 are produced, so the number
of moles of CO2 that can be produced when 4 moles of propane combusts = 4(3) = 12.
c) Verify that if 12.5 moles of propane and 63.2 moles of oxygen were present, then propane is
the “limiting reactant” and oxygen is the excess reactant.
For each mole of C3H8 five moles of O2 are required, so for 12.5 moles of C3H8, the number
of moles of O2 needed are (12.5)(5) = 62.5 moles. Since we have more than 62.5 moles
(according to the question we have 63.2 moles), then we have excess O2 and therefore C3H8 is
the limiting reactant.
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9. Consider the following chemical reaction: 3 MgCl2 + 2 Na3PO4 6 NaCl + Mg3(PO4)2.
Assume that 0.75 mol of MgCl2 and 0.65 mol of Na3PO4 are placed in a reaction vessel.
a) Verify that Na3PO4 is the excess reactant and MgCl2 is the limiting reactant.
Using the 3:2 ratio, we find that for 0.75 mol of MgCl2, we need (0.75)(2/3) = 0.5 moles of
Na3PO4. We have more than that and therefore Na3PO4 is the excess reactant and MgCl2 the
limiting.
b) How many moles of the excess reactant are left over after the reaction stops?
We have 0.65 mol and we use up 0.5 mol (calculated in part a), so the number of moles left
over is 0.65 – 0.5 = 0.15 mol
c) How many moles of NaCl will be produced in this reaction? (Remember—you must base this
answer on how many moles of the limiting reactant that reacted.)
1.5 mol. 0.75 mol of MgCl2 is the limiting reactant, so we use the 6:3 ratio of NaCl to MgCl2
and multiply (0.75)(6/3) = 1.5 mol.
10. Consider the double replacement reaction between calcium sulfate (CaSO4) and sodium iodide
(NaI). If 34.7 g of calcium sulfate and 58.3 g of sodium iodide are placed in a reaction vessel,
how many grams of each product are produced? (Hint: Do this problem in the steps outlined
below.)
a) Write the balanced chemical equation for the reaction.
CaSO4 + 2 NaI CaI2 + Na2SO4
b) Find the limiting reactant. First, convert 34.7g and 58.3g from grams to moles using the
molar masses from the periodic table. Next, compare the number of moles of each reactant.
Ask yourself: Do I have enough NaI to use up all of the CaSO4? Do I have enough CaSO4 to
use up all of the NaI? Whichever one will get used up is the limiting reactant.
(34.7g) ÷ (136.2g/mol) = 0.255 mol CaSO4; (58.3g) ÷ (149.9g/mol) = 0.389 mol NaI
To use up all of the NaI, I need (0.389)(½) = 0.1945 mol CaSO4 I have more than this
much, so NaI is limiting reactant.
c) Use the number of moles of the limiting reactant to calculate the number of moles of each
product produced using the coefficients from the balanced chemical equation in part a.
mol of CaI2 = mol of Na2SO4 because the coefficients are both “1” in the balanced equation.
Using the mol of limiting reactant and the mole ratio of 1:2, you can calculate the mol of CaI2
and Na2SO4. (0.389)(½) = 0.1945 mol.
d) In part c you found the moles of each product produced. Now convert moles to grams using
the molar mass from the periodic table. You have now answered the question.
g of CaI2 = (0.1945 mol)(293.9g/mol) = 57.2g
g of Na2SO4 = (0.1945 mol)(142.1g/mol) = 27.6g
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102
11. If 181.1g of Al(NO3)3 react with 102.1g of CaO in a double replacement reaction, how many
grams of each product will be produced? (Note: this is just like the last question, but parts a-d are
not spelled out for you.)
First, write the balanced equation:
2 Al(NO3)3 + 3 CaO 3 Ca(NO3)2 + Al2O3
Next, find the limiting reactant:
(181.1g) ÷ (213 g/mol) = 0.850 mol Al(NO3)3
(102.1g) ÷ (56.1g/mol) = 1.82 mol CaO
To use up all 0.850 mol of Al(NO3)3, I need (0.850)(3/2) = 1.275 mol CaO. Since you have
more than this amount, CaO is present in excess and Al(NO3)3 is the limiting reactant.
Use the moles of limiting reactant to calculate the moles of each product produced:
mol Ca(NO3)2 = (0.850)(3/2) = 1.275 mol
mol Al2O3 = (0.850)(½) = 0.425 mol
Convert the mol of each product to grams of each product:
Grams Ca(NO3)2 = (1.275 mol)(164.1g/mol) = 209.2 g
Grams Al2O3 = (0.425mol)(102g/mol) = 43.4 g
103
ChemQuest 34
Name: ____________________________
Date: _______________
Hour: _____
Information: Sources of Error in the Real World
When we perform experiments in the laboratory we never get all of the product we theoretically
could get. Reactions in the laboratory are not ideal. For example, consider the reaction of sodium
metal and chlorine gas to produce salt:
2 Na + Cl2 2 NaCl
If we react 23.0g of sodium metal with plenty of chlorine gas, calculations tell us that we should get
about 58.5g of sodium chloride. However, if you tried this in the lab and then took the mass of your
final product on a balance, you would usually get somewhat less than 58.5g. The more careful you
are, the closer you can get to the mark of 58.5g, but you will never be able to make 58.5g of sodium
chloride.
Critical Thinking Questions
1. What are some practical things—some “sources of error”—that can happen in the lab and in a
reaction that would cause us to get less than 100% of the product that we are trying to
produce?
Spilling some of the product before weighing; evaporation of some product that are
liquids
2. The information section above made it sound like we always would get less than 58.5 g of
salt. However, it is possible that after the reaction we could take the mass of the product and
find that it is greater than 58.5g. According to our calculations this should be impossible, but
what “sources of error” may cause this?
Contamination of product with a substance that would add weight.
3. When 40.5 g of sodium metal reacts with plenty of chlorine gas, how many grams of sodium
chloride should be produced? (Note: when questions are worded this way, you automatically
know that sodium is the limiting reactant.)
40.5g ÷ 23g/mol = 1.76 mol Na = mol NaCl because of a 1:1 mol ratio
(1.76 mol NaCl)(58.5 g/mol) = 103g NaCl
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104
Information: Calculating Percent Yield
Percent yield is a measure of how much of a product you actually obtained in a reaction compared to
the amount that you could have obtained according to calculations. There is a handy formula for
calculating percent yield: percent yield = actual yield •100
theoretical yield
“Theoretical yield” is the amount of a product that we should be able to make. You calculated the
say that you actually attempted this experiment in a lab with 40.5g of sodium and when you were
finished you collected 84.0g of sodium chloride—this amount is your “actual yield”, the amount of
product you actually collected in the lab.
Using the formula given above, we can calculate our percent yield:
84.0g
percent yield =
•100 = 81.6%
103g
Critical Thinking Questions
4. Consider the combustion of 215.0g of butane, C4H10 with plenty of oxygen. If you are able to
collect 295.3g of water in the laboratory, what was your percent yield. Hint: first calculate
your theoretical yield using the balanced equation and mole ratios!
215 g
333.6g theoretical yield
divide 2 C4H10 + 13 O2 8 CO2 + 10 H2O
58 g/mol
18 g/mol
multiply
3.71 mol
18.53 mol
x 10/2
Percent yield = (295.3g) ÷ (333.6) · 100 = 88.5%
5. Consider the double replacement reaction of excess sodium chloride with 203.9g of silver
nitrate. If you are able to produce 150.4g of silver chloride what was your percent yield?
203.9 g
171.98g theoretical yield
Divide
NaCl + AgNO3 NaNO3 + AgCl
169.87g/mol
143.32 g/mol Multiply
1.20 mol
1.20 mol
Percent yield = (150.4) ÷ (171.98) · 100 = 87.45%
6. If you make 46.8g of calcium carbonate by reacting 87.5g of calcium nitrate with plenty of
sodium carbonate, what is your percent yield?
87.5 g
53.35g theoretical yield
Divide
Ca(NO3)2 + Na2CO3 CaCO3 + 2 NaNO3
164.1g/mol
100.1g/mol
multiply
0.533 mol
0.533 mol
Percent yield = (46.8) ÷ (53.35) · 100 = 87.7%
105
ChemQuest 35
Name: ____________________________
Date: _______________
Hour: _____
Information: Gas Pressure
Figure 1: Two containers of gas molecules
Container 1
Container 2
Gas molecules move randomly in their containers, colliding with the walls of their containers causing
“gas pressure.” Pressure can be defined as the force pushing on an area. It can be described with the
equation:
F
2
P=
A where P is pressure (in kPa), F is force (in N) and A is area (in m ).
The more molecules collide with the wall and the faster the molecules are going when they strike the
wall, the greater the force on the wall and therefore, the higher the pressure.
Critical Thinking Questions
1. Use the pressure equation to explain why it would be more likely for an ice skater to fall through
the ice on a lake than it would be for someone walking across the lake with regular shoes on.
The blade of an ice skate has a much smaller surface area than the sole of a shoe. With a smaller
area (A), the pressure (P) on the ice is larger.
2. Which container in Figure 1 has the highest pressure? Explain.
Container 2, because there are more gas molecules just like a tire filled with more air.
3. If I heated container 1 and did not heat container 2, could I get the pressure in container 1 to equal
container 2? Explain.
Yes, increasing the temperature increases pressure by speeding up gas molecules.
4. If container 2 was made of an elastic material and if I expanded container 2, could I make the two
containers have equal gas pressures? Explain.
Yes, increasing volume decreases pressure giving both containers an equal number of molecules
per volume.
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106
Information: Gas Laws
Observe the following experimental data concerning a container of gas. The pressure, volume and
temperature of a gas are all related. The table of data was obtained by making measurements of the
pressure, volume and temperature of a sample of a gas. Several different kinds of gases were used
and all had identical results. The variable that was not changed was the amount of gas present. The
number of moles of gas always remained the same during these trials.
Table 1: Experimental Data for Gases
Trial
A
B
C
D
E
F
G
H
Pressure (P)
Units: kPa
120
135
195
150
135
100
225
262
Volume (V)
Units: L
3.2
2.5
2.3
2.0
4.2
3.0
3.2
2.8
Temperature (T)
Units: K
324.3
285.0
378.7
254.4
480.8
254.4
608.0
620.4
Critical Thinking Questions
P1 P2
=
5. Verify that this equation is true when the volume is unchanged:T1 T2
(Hint: You must use two sets of data where the volume does not change like in trials A and G.
Note: the subscript 1 refers to the pressure and temperature for the first trial you select and the
subscript 2 refers to the pressure and temperature for the second trial you select.)
PA PG
120
225
=
=
= 0.370
TA TG
324.3 608.0
6. Scientists often look for relationships between variables. If you wanted to see how the volume
and pressure are related you would need to compare data from different trials when the
temperature does not change. Why?
Temperature would affect volume and pressure.
7. Find two sets of data in the table that have constant temperature. Which of the following
mathematical relationships is true (there may be more than one) when the temperature remains
unchanged? This relationship is known as “Boyle’s Law”, named after the person who first
discovered it.
a) P1 P2
b) (P1)(V1) = (P2)(V2)
c) P1 + V1 = P2 + V2
d) P1 V1
=
=
V1 V2
P2 V2
B is the correct choice. Using data from trials D and F, we see:
(PD)(VD) = (PF)(VF) (150)(2.0) = (100)(3.0) = 300
107
8. At constant pressure, which of the following equations is/are true? This relationship is known as
V1 T1
“Charles’ Law”, named after the person who first discovered it.
=
a) V1 V2
b) (T1)(V1) = (T2)(V2)
c) P1 + V1 = P2 - V2
d) V2 T2
=
T1 T2
Substituting data from trials B and E you will see that a is correct.
9. Complete the following. You may want to consider the equations from questions 5, 7 and 8.
a) At constant volume, as the temperature increases, the pressure always ___increases______.
b) At constant temperature, as the volume increases, the pressure always ___decreases______.
c) At constant pressure, as the temperature increases, the volume always ___increases______.
10. If the volume is not constant, is the statement you completed in 9a always true? Justify your
answer by citing experimental data from the data table.
No, in trial A, the temperature is greater, but the pressure is lower than in trial B.
11. If the temperature or pressure is not constant are your statements in 9b and 9c correct? Justify
No. In trial C, the volume is greater, but the pressure is higher than in trial D. In trial C, the
temperature is greater, but the volume is lower than in trial B.
12. Would the equations you discovered still be true if the temperature was measured in degrees
Celsius (oC) instead of Kelvin (K)? Recall that K = oC + 273 or oC = K – 273.
No, if you changed all the temperatures to oC, none of the equations would work.
13. Which of the following quantities is a constant?
a) PV
b) PV + T
c) PT
T
V
For each trial, if you multiply the pressure and volume and then divide by the temperature you
get approximately 1.18.
P1 V1 P2 V2
=
14. Based on your answer to question 13, verify that this equation is true:
This
T
T
1
2
equation is called the “Combined Gas Law”. Notice that it contains all of the equations combined
into one!
Again, the pressure times the volume divided by the Kelvin temperature equals 1.18 for every
trial in the table.
15. What needs to remain constant in order for equation 14 to be true? (You may need to refer to the
information section.)
The number of moles of gas in the container must be the same. If the moles changes we
probably would not get (P)(V)/T to equal 1.18.
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108
16. Prove that when the temperature remains constant, the combined gas law becomes Boyle’s Law.
If the temperature is constant, then T2 = T1 and we can write:
P1V1 P2V2
multiplying both sides by T1 P1V1 = P2V2
=
T1
T1
17. Prove that when the pressure remains constant, the combined gas law becomes Charles’s Law.
If pressure is constant, then P2 = P1 and we can write:
P1V1 P1V2
=
dividing both sides by P1 T1
T2
V1 V2
=
T1 T2
18. You have discovered several new mathematical relationships among gases. Now is your chance
to practice using these equations!
a) At constant temperature, the volume of a gas expands from 4.0 L to 8.0 L. If the initial
pressure was 120 kPa, what is the final pressure?
P1V1 = P2V2 (120)(4.0) = P2 (8.0) P2 = 60 kPa
b) At constant pressure, a gas is heated from 250 K to 500 K. After heating, the volume of the
gas was 12.0 L. What was the initial volume of the gas? Notice: as the temperature doubled,
what happened to the volume?
V1 V2
=
T1 T2
V1 12.0
=
V1 = 6.0 L
250 500
c) The volume of a gas was originally 2.5 L; its pressure was 104 kPa and its temperature was
270K. The volume of the gas expanded to 5.3 L and its pressure decreased to 95 kPa. What
is the temperature of the gas?
P1V1 P2V2
(104)(2.5) = (95)(5.3)
=
T1
T2
270
T2
T2 = 522.9 kPa
19. At constant temperature, if you increase the volume by a factor of two (doubling the volume), the
pressure ____decreases_____ by a factor of ______2_______. (Refer to 18a for a hint.)
increases or decreases
what number
109
ChemQuest 36
Name: ____________________________
Date: _______________
Hour: _____
Information: The Gas Constant
We previously examined the fact that PV is a constant as long as the number of moles of a gas
T
remains unchanged.
This means that any time you measure a gas, the pressure times the volume divided by the
temperature (in Kelvin) will always equal the same number unless the amount of gas in the container
changes. We are now going to explore how the pressure, volume and temperature depend on the
number of moles of gas that are present.
Table 1: Experimental Data for Three Different Samples of Gas That Contain the Same Amount of
Gas (# of moles of gas is constant)
Gas
A
B
C
# of moles of Gas in
container (mol)
0.3008
0.3008
0.3008
Volume of
Container (L)
14.0
16.0
10.0
Temperature
of Gas (K)
672.0
1280
340.0
Pressure of Gas
(kPa)
120.0
200.0
85.0
Table 2: Experimental Data for Three Different Samples of Gases That Contain Different Amounts
of Gas (# of moles of gas is not constant)
Gas
D
E
F
# of moles of Gas in
container
3.3485
0.7306
0.85950
Volume of
Container (L)
40.0
22.5
16.0
Temperature
of Gas (K)
230.0
315
280
Pressure of Gas
(kPa)
160.0
85
125
Critical Thinking Questions
1. Prove that
PV
is a constant for the data in Table 1, but not a constant for the data in Table 2.
T
PAVA/TA = PBVB/TB = 2.5; but in Table 2, PDVD/TD does not equal PEVE/TE
PV
2. Why is T a constant in Table 1, but not in Table 2?
The number of moles of gas is not constant in Table 2.
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110
3. Which of the following is a constant for Table 1 and Table 2? (Note that “n” means number of
moles of gas in the container.)
A) PT
B) PTn
C) (n2)PV+T
D) PV
nV
V
nT
4. Consider your answer to question 3. Compute the numerical value of the constant by putting
values into the equation indicated by your answer to question 3. This constant is given a special
name, “The Ideal Gas Constant” and it has the symbol “R”.
PV
= 8.31 = R
nT
5. Which of the following equations is true? The equation is known as the “Ideal Gas Law”.
A)
PT = nRV
B) PV = nRT
C) TV = nRP
PV
= R PV = nRT
nT
6. At 215 kPa of pressure and a temperature of 318 K, 2.35 mol of a gas has what volume? (Use the
ideal gas law and the ideal gas constant, R, to calculate V. Your answer should be 28.9 L.)
PV = nRT (215)V = (2.35)(8.31)(318) V = 28.88 L
7. What is the temperature of a gas if 4.11 mol of a gas has a pressure of 315 kPa and a volume of
12.5 L?
PV = nRT (315)(12.5) = (4.11)(8.31)T T = 115.3 K
8. 23.5 g of CO2 gas is at a temperature of 45oC and a pressure of 125 kPa. What is the volume of
the container? (Hint: change grams to moles and 45oC to Kelvin.)
23.5g ÷ 44.0g/mol = 0.534 mol; T = 45oC + 273 = 318K
PV = nRT (125)V = (0.534)(8.31)(318) V = 11.3 L
Information: Molar Volume
Molar volume: the volume of one mole of a gas. Depending on the temperature and pressure a mole
of gas may have different volumes. Whenever we are speaking of molar volume, we can assume that
we are talking about one mole of the gas. Thus, whenever we need to calculate the molar volume,
n=1.
111
Critical Thinking Questions
9. What is the molar volume of oxygen gas if the gas has a pressure of 120 kPa and a temperature of
38oC? Use the ideal gas law (from question 5) with n=1 for one mole of gas and solve for V;
remember to use Kelvin temperature and that R always equals 8.31.
PV = nRT (120)V = (1)(8.31)(311) V = 21.5 L
10. Prove that if two different gases are at the same pressure and temperature, each gas has the same
molar volume.
In question 9 we looked at oxygen gas at 120 kPa and 38oC (311 K). Now, I’ll pick another
gas—nitrogen—at 120 kPa and 38oC.
PV = nRT (120)V = (1)(8.31)(311) V = 21.5 L
11. What is the molar volume of any gas at STP? Recall that STP is standard temperature (273 K)
and standard pressure (101.325 kPa).
PV = nRT (101.325)V = (1)(8.31)(273) V = 22.4 L
12. Why do you think all gases have the same molar volume at STP?
As we found in question 10, the molar volume of a gas depends on the temperature and pressure
and not the identity of the gas.
Information: Using the Ideal Gas Law to Calculate Molar Mass
You should know that to find the number of moles (n) of a substance, you take the mass (m) of the
substance and divide it by its molar mass (M) according to the following equation:
# of moles =
mass
Molar mass
⇒ n=
m
M
Now consider the Ideal Gas Law, PV=nRT. Since n=m/M we can replace n with m/M in the ideal
gas law to get:
m
PV = RT
M
By rearranging it a bit, we can get a more useful form of the equation: MPV = mRT.
This equation can then be used when you are dealing with grams and molar mass instead of moles.
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112
Critical Thinking Questions
13. If 128g of a certain gas in a container with a volume of 21.5 L has a pressure of 132 kPa and a
temperature of 45oC, what is the molar mass of the gas?
MPV = mRT M(132)(21.5) = (128)(8.31)(318) M = 119.2 g/mol
14. How many grams of CO2 are in a 5.0 L container if the pressure in the container is 101 kPa and
the temperature is 280 K?
MPV = mRT (44.0)(101)(5.0) = m(8.31)(280) 9.55 g
15. If 4.0 L of a gas were produced at STP and the mass of the gas was found to be 12.8g, then what
is the molar mass of the gas?
MPV = mRT M(101.325)(4.0) = (12.8)(8.31)(273) 71.6 g/mol
113
ChemQuest 37
Name: ____________________________
Date: _______________
Hour: _____
Critical Thinking Questions
1. How many moles and how many grams of hydrogen gas can be produced by reacting 32.8 g of
magnesium according to the following balanced equation?
Mg (s) + 2 HCl (aq) MgCl2 (aq) + H2 (g)
2.72 g
divide
32.8g
2.72 g
Mg (s) + 2 HCl (aq) MgCl2 (aq) + H2 (g)
24.3 g/mol
2.02 g/mol
1.35 mol
1.35 mol
multiply
2. Considering your answer to question 1, what volume of hydrogen gas was produced if this
reaction was carried out at STP?
n = 1.35, so using PV=nRT (101.325)V = (1.35)(8.31)(273) V = 30.2 L
3. Nitrogen gas can be produced by reacting Na3N with chlorine gas. If the reaction was carried out
at STP with 34.8 g of Na3N, what volume of nitrogen gas can be produced? (Hint: first use
stoichiometry to calculate the number of moles of nitrogen produced and then use the ideal gas
law.)
34.8g
2 Na3N + 3 Cl2 N2 + 6 NaCl
83 g/mol
0.419 mol x ½ = 0.2095 mol N2 = n
PV=nRT (101.325)V = (0.2095)(8.31)(273) V = 4.7 L
Information: Mole Ratios in Chemical Equations and Volume
As you know, the coefficients in balanced equations relate the number of moles of reactants to each
other. For gaseous reactions, the coefficients relate not only the moles but also the number of liters
of reactants and products to each other. For example, consider the following reaction, noting that
“(g)” means that the substances are gaseous:
N2 (g) + 2 O2 (g) 2 NO2 (g)
The coefficients tell you that for every mole of N2, two moles of O2 react and two moles of NO2 are
produced. It also tells you that for each liter of N2, two liters of O2 react and two liters of NO2 are
produced.
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114
Critical Thinking Questions
4. Verify that for the following reaction to be completed, 4 liters of hydrogen would require at least
2 liters of oxygen gas at the same pressure and temperature.
2 H2 (g) + O2 (g) 2 H2O (g)
It is a 2:1 ratio of hydrogen to oxygen, so for every two liters of hydrogen one liter of oxygen
is required.
5. In a certain combustion reaction involving propane, 34.4 L of CO2 was produced. How many
liters of propane (C3H8) were combusted? Assume the reaction took place at STP.
C3H8 + 5 O2 3 CO2 + 4 H2O
There is a 3:1 ratio of CO2 to C3H8, so there needs to be three times as much CO2 as there is
C3H8. Therefore, the volume of C3H8 required is 34.4÷3 = 11.5 L
6. How many grams of sodium are required to produce 31.9 L of H2 at 97 kPa and 290 K according
to the following balanced equation? 2 HNO3 + 2 Na 2 NaNO3 + H2
PV=nRT (97)(31.9) = n(8.31)(290) n = 1.28 mol H2
(1.28 mol H2)(2) = 2.56 mol Na (2:1 mol ratio of Na to H2)
(2.56 mol Na)(23.0g/mol) = 58.9 g
7. At room temperature (25oC) and pressure (100 kPa), 381g of octane (C8H18) was burned in an
atmosphere of plenty of oxygen gas. How many liters of each product were produced?
2 C8H18 + 25 O2 16 CO2 + 18 H2O
Liters of C8H18: MPV = mRT (114.18)(100)V = 381(8.31)(298) V = 82.6 L
Liters are related by the ratios of coefficients in the balanced equation:
Liters CO2 = 82.6(16/2) = 660.8 L
Liters H2O = 82.6(18/2) = 743.4 L
115
ChemQuest 38
Name: ____________________________
Date: _______________
Hour: _____
Information: Collecting Gas Over Water
When gas is collected in a container, it is often collected using a technique called, “water
displacement.” In water displacement, a container is filled with water and then gas is bubbled into
the container. In this way, the container can be filled with relatively pure gas without air in it.
Figure 1A: A beaker is filled with water and then turned upside down in a bucket of water. Rubber
tubing is attached to a source of gas so that the gas is bubbling up through the water in the beaker.
These gas bubbles force some of the water out of the beaker.
Beaker upside down in water
Rubber tubing carrying gas
Gas bubbles
Gas Generator
Water
Bucket
Figure 1B: The same setup after 5 minutes.
Rubber tubing carrying gas
Beaker upside down in water
Collected Gas
Gas Generator
Water
Bucket
Critical Thinking Questions
1. To collect the gas, why is the beaker first filled completely with water? What purpose does
the water serve?
By filling the beaker with the water you are making sure that there is no air in the beaker.
2. Examine the beaker in Figure 1A and 1B. What causes the water level to go down after 5
minutes?
The gas has replaced the water and the water has gone into the bucket.
3. In Figure 1B, is the gas in the beaker pure? Explain.
It is quite pure; however, there is some water vapor left behind.
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116
Information: Dry Gas
The gas that is collected is not “dry” because there is some water vapor left behind in the beaker. If
you attempt to collect pure oxygen, you will actually get mostly pure oxygen with a little water
vapor. You can get an idea of how much water is left behind by examining the water vapor pressure
at various temperatures.
Table 1: Vapor pressure of water at various temperatures.
Temperature (oC)
0
5
10
15
20
25
30
35
40
45
Vapor Pressure (kPa)
0.6
0.9
1.2
1.6
2.3
3.2
4.2
5.6
7.4
9.8
Critical Thinking Questions
4. Why is it logical to expect that the vapor pressure of water would increase as the temperature
increases?
As the temperature increases, water is more likely to evaporate and the evaporated molecules
move faster, colliding with more force (or pressure).
5. Examine the two containers below. Both contain gases that were collected over water.
Which one was collected at the higher temperature—gas A or gas B? Explain your answer.
Gas A =
Gas B =
Water vapor =
Water vapor =
Gas B was collected at higher temperatures because there is more water vapor mixed with the
gas and the water vapor pressure increases as the temperature increases.
6. Consider Gas A from question 5. The total pressure in the container is 104 kPa. If the
temperature of the container is 20oC, calculate the pressure of Gas A when it is “dry”. Hint:
find the vapor pressure of water at 20oC from Table 1 and then subtract it from the total
pressure in the container.
104 – 2.3 = 101.7 kPa
7. Consider Gas B from question 5. The total pressure in the container is 110 kPa. If the
temperature of the container is 35oC, calculate the pressure of Gas B when it is “dry”.
110 – 5.6 kPa = 104.4 kPa
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Information: Dalton’s Law of Partial Pressures
John Dalton was one of the first scientists to quantitatively state a mathematical relationship
involving the total gas pressure in a container and the individual gases in the container. Dalton’s law
states that the total pressure of any mixture of gases in a container is the sum of all the individual
gas’s partial pressures. In equation form, Dalton’s law can be written as:
PTotal = PGas A + PGas B + PGas C +…
Critical Thinking Questions
8. A container of gas with a pressure of 450 kPa contained three different gases—hydrogen,
oxygen and nitrogen. If the partial pressure of hydrogen was 210 kPa and the partial pressure
of oxygen was 125 kPa, what was the partial pressure of nitrogen?
450 – (210 + 125) = 115 kPa
9. A tank held neon gas at a pressure of 350 kPa, helium at a pressure of 275 kPa and argon at a
pressure of 410 kPa. What was the pressure in the tank?
350 + 275 + 410 = 1035 kPa
10. A certain gas was collected over water. The total pressure of the container was 100.0 kPa.
The pressure of the dry gas was 94.4 kPa. At what temperature was the gas collected?
100.0 – 94.4 = 5.6 kPa (the water vapor pressure)This corresponds to a temperature of 35oC
11. Hydrogen gas was collected over water at a pressure of 102.5 kPa and a temperature of 25oC.
After sealing the container, it was heated to 210oC. What is the final pressure of the dry
hydrogen gas? (Hints: You must first find P1 for the gas by subtracting out the water vapor
from the given pressure; solve for P2 using the combined gas law with constant volume.)
P1 = 102.5 kPa – 3.2 kPa = 99.3 kPa; T1 = 25oC + 273 = 298K; T2 = 210oC+273= 483 K
P1 P2
99 .3 P2
=
→
=
→ P2 = 160 .9 kPa
T1 T2
298 483
12. Oxygen gas was collected over water at 30oC. The pressure in the 2.5 L container was 110
kPa. If the container was allowed to expand to 7.0 L and if the temperature was decreased
to -30oC, what is the final pressure of the dry oxygen gas?
P1=110-4.2 = 105.8 kPa; T1 = 30oC+273 = 303 K; T2 = -30oC + 273 = 243 K
P1 V1 P2 V2
105.8( 2.5) P2 (7.0)
=
→
=
→ 30.3 kPa
T1
T2
303
243
13. A quantity of oxygen gas was collected over water at 30oC in a 475 mL container. The
pressure in the container was measured to be 105 kPa. How many grams of oxygen gas were
collected?
P = 105 – 4.2 = 100.8 kPa; molar mass of O2 = 32.0g/mol; V = 0.475 L (not mL); T = 303 K
MPV = mRT (32.0)(100.8 kPa)(0.475) = m(8.31)(303) m = 0.608 g
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118
ChemQuest 39
Name: ____________________________
Date: _______________
Hour: _____
Information: Molarity
Concentration is a term that describes the amount of solute that is dissolved in a solution.
Concentrated solutions contain a lot of dissolved solute, but dilute solutions contain only a little.
Critical Thinking Questions
1. Consider the terms "concentrated" and "dilute". Are these qualitative or quantitative terms?
These are qualitative terms: very general and do not include the magnitude or quantity.
2. One way of quantitatively measuring solution concentration is with units of molarity, symbolized
by M. You see 1.7 liters (L) of a sodium chloride and water solution. The label on the bottle
reads "1.5 M NaCl". You don't know what molarity is, but you decide to find out. After
evaporating the water out of the solution you discover that there are about 149 grams of salt.
Using this information, which of the following formulas is/are correct for finding molarity?
grams of solute
moles of solute
A) Molarity =
B) Molarity =
moles of solute
liters of solution
149g ÷ 58.5g/mol = 2.547 mol
1.5 = 2.547 mol ÷ 1.7 L
3. Using the equation you discovered in question two, calculate the molarity of each of the
following solutions.
A) A solution is prepared by dissolving 24.9 g of CaCl2 in 210 mL (which 0.210 L) of solution.
1.07 M
Change g to mol: 24.9g ÷ 111.1g/mol = 0.224 mol
Molarity = mol ÷ L = 0.224 ÷ 0.210 = 1.07 M
B) A solution contains 12.9 g of Na2SO4 in 325 mL of solution.
0.279 M
Change g to mol: 12.9 ÷ 142.1g/mol = 0.09078 mol
Molarity = mol ÷ L = 0.09078mol ÷ 0.325L = 0.279 M
4. Verify that I need 2.15 moles of Ca(NO3)2 to make 358 mL of a 6.00 molar solution.
mol = (Molarity)(Liters) = (6.00M)(0.358L) = 2.15 mol
5. Verify that it takes 80.8 g of sodium chloride to make 425 mL of a 3.25 M solution.
mol = (Molarity)(Liters) = (3.25M)(0.425L) = 1.38 mol
(1.38 mol)(58.5g/mol) = 80.7 g
119
6. Consider 670 mL of a 4.10 M solution of Mg(NO3)2 setting in a beaker. If you evaporate all 670
mL of the solution, how many grams of solute would be left in the beaker?
407.8g
mol = (Molarity)(Liters) = (4.10M)(0.670L) = 2.747 mol
convert mol to g: (2.747mol)(148.3g/mol) = 407.8g
Information: Molality
Molality is another way of expressing solution concentration. The symbol for molality is m.
Whereas molarity (M) represents the ratio of moles solute to liters of solution, the molality (m) is the
ratio of moles solute to kilograms of solvent. It can be expressed using the following formula:
moles of solute
molality =
kg solvent
Critical Thinking Questions
7. Consider a solution that is prepared by adding 1.34 moles of sodium nitrate to 2.5 kg of water.
What is the molality of the solution?
0.536 m
m = 1.34mol ÷ 2.5kg = 0.536 m
8. Considering the data given in question 7, is this enough data to find the molarity? If so, calculate
the molarity. If not, explain why not.
No, because we don’t know the total volume of the solution.
9. What is the molality of a solution that is made by dissolving 32.6 g of Na2SO4 in 475 g of water?
0.483 m
Convert g to mol: 32.6g ÷ 142.1g/mol = 0.229 mol
molality = mol/kg = 0.229mol ÷ 0.475kg = 0.483 m
10. Consider 2.35 moles of sodium chloride are dissolved in 1.21 kg of solution to make 1.29 liters.
Calculate and compare the molarity and molality.
Molarity: 1.82 M
M = mol ÷ L = 2.35mol ÷ 1.29 L = 1.82 M
molality: 1.94 m
m = mol ÷ kg = 2.35mol ÷ 1.21 kg = 1.94 m
11. If 26.45g of Na2SO4 are dissolved in 1.10 kg of solution to make 1.24 L, calculate both the
molarity and the molality of the resulting solution.
Convert g to mol: 26.45g ÷ 142.1g/mol = 0.186mol
Molarity: 0.150 M
M = mol ÷ L = 0.186mol ÷ 1.24 L = 0.150 M
molality: 0.169 m
m = mol ÷ kg = 0.186mol ÷ 1.10kg = 0.169 m
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Information: Mole Fraction
Another way of expressing solution concentration is called “mole fraction”. The mole fraction
(symbolized by X) of the solute or of the solvent can be calculated using the following equations:
mol solute
mol solvent
X solute =
X solvent =
(mol solute + mol solvent )
(mol solute + mol solvent )
Note: both the solute and the solvent must be converted to moles when finding the mole fraction!
Critical Thinking Questions
12. Prove that the mole fraction of salt (XNaCl) equals 0.049 when 14.25 g of NaCl is dissolved in
85.0 g of H2O.
85.0g ÷ 18.0g/mol = 4.722 mol H2O
14.25g ÷ 58.5g/mol = 0.2436 mol NaCl
mol NaCl
0.2436
X NaCl =
=
= 0.049
(mol NaCl + mol H 2O ) (0.2436 + 4.722)
13. Find the mole fraction of water (Xwater) for the solution described in question 12.
mol H 2O
4.722
X H 2O =
=
= 0.951
(mol H 2O + mol NaCl ) (0.2436 + 4.722)
14. Prove that Xsolute + Xsolvent = 1.
The answer from question 12 + answer to question 13 = 1
15. In a certain salt water solution, the mole fraction of salt is 0.18. Find the mole fraction of water.
1.00 – 0.18 = 0.82
Information: Mass Percent Composition
Mass percent composition is similar to the mole fraction except the amounts of solute and solvent are
in grams instead of moles. Here is the formula for finding the mass percent of a solute:
mass solute
mass% solute =
• 100
(mass solute + mass solvent )
Critical Thinking Questions
16. Prove that the mass percent of salt is 14.36% in the solution described in question 12.
mass NaCl
14.25
mass% NaCl =
• 100 =
• 100 = 14.36%
(mass NaCl + mass H 2 O )
14.25 + 85.0
17. Calculate the mass percent of sodium phosphate if 12.5g of it are dissolved in 250 mL of water.
(Note: 1 mL of water has a mass of 1 g.)
Because 1 mL = 1g, the mass of H2O = 250 g
mass Na 3 PO 4
12.5
mass% Na 3 PO 4 =
• 100 =
• 100 = 4.76%
(mass Na 3PO 4 + mass H 2 O )
12.55 + 250
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ChemQuest 40
Name: ____________________________
Date: _______________
Hour: _____
Introduction Question: Melting Ice
1. In colder climates during the winter, people put salt on the roads and walkways to melt ice.
Why do people do this? Why does salt melt the ice?
Salt actually lowers the freezing point of water so that water will not freeze until much
lower temperatures. Salt water remains a liquid at 0oC
Information: Dissociation and Total Molality of Particles
When you dissolve a solute in a solvent, the resulting solution has slightly different properties than
the original solvent. For example, salt water has a different freezing point and boiling point than pure
water. The salt interferes with water’s ability to freeze and boil.
When ionic compounds dissolve, they dissociate. When an ionic compound dissociates that means
that it breaks up into ions. For example, salt (sodium chloride) breaks up into sodium ions and
chloride ions. This process is represented in the following balanced equation:
NaCl Na+ + ClNote for the above equation that Cl- does not need to be written as Cl2 because Cl- is a chloride ion
and not a lone chlorine atom.
Since calcium nitrate is an ionic compound it also dissociates as shown below:
Ca(NO3)2 Ca2+ + 2 NO3Covalent molecules do not dissociate. Although they may dissolve, they do not break up into ions.
Critical Thinking Questions
2. Write the balanced equation for the dissociation of magnesium chloride.
MgCl2 Mg2+ + 2 Cl3. Write the balanced equation for the dissociation of ammonium sulfate.
(NH4)2SO4 2 NH4+ + SO42-
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4. Consider calcium nitrate. Each calcium nitrate breaks up into one calcium ion and two nitrate
ions according to the balanced equation given in the information section. If you take one
mole of calcium nitrate and put it in water, it will dissociate.
a) How many moles of calcium ions and how many moles of nitrate ions will there be in the
solution?
One mole of calcium ions and two moles of nitrate ions.
b) What is the total number of moles of all ions in the solution?
3; one mole of calcium ions plus two moles of nitrate ions equal a total of three moles.
5. A solution is made so that it is 2.5 M Ca(NO3)2. Therefore the concentration of Ca2+ is 2.5 M
and the concentration of NO3- is 5.0 M. The total concentration of all particles is 7.5 M.
Explain.
Since each mole of Ca(NO3)2 breaks up into 1 mole of Ca2+, the concentration of Ca2+ would
be the same as the concentration of Ca(NO3)2. Since each mole of Ca(NO3)2 breaks up into 2
moles of NO3-, the concentration of NO3- is twice the concentration of Ca(NO3)2.
6. A solution is made so that it the concentration is 3.0 m MgCl2. What is the molality of Mg2+
and Cl- ions? What is the total molality of all particles in the solution?
Mg2+ ___3.0m___
Cl- ___6.0m___
Total molality of all particles: ___9.0m_____
7. A solution is prepared by dissolving 45.7 g of sodium carbonate in 200 g of water.
a) What is the molality of the sodium carbonate?
45.7 g
0.431mol
= 0.431mol
= 2.16m
106g / mol
0.200kg
b) What is the total molality of all particles in the solution?
6.48m; Na2CO3 2 Na+ + CO32- ; Since Na2CO3 breaks into 3 total ions, the total
molality is 3(2.16) = 6.48m
8. Consider sugar (C6H12O6) , a covalent molecule. If a solution is made so that the
concentration is 3.5 m in sugar, then what is the total molality of particles?
3.5m; Covalent compounds don’t break up into ions so the total molality is still 3.5m.
Information: Total Molality of Particles and Changes in Boiling/Freezing Points
You may be wondering how all of this ties together. We have seen that adding a solute changes the
boiling and freezing points of solvents. The amount of the change depends on how much solute is
added. Equations relating the change in boiling or freezing point and the molality is shown below:
∆Tbp = (mT)(Kbp) for boiling point
Note: mT is the total molality of particles. Kbp and Kfp are constants called the molal boiling point
elevation constant and the molal freezing point depression constant respectively. Kbp for water is
0.515 oC/m and Kfp for water is 1.853 oC/m.
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Critical Thinking Questions
9. What is the freezing point of a 2.5 m solution of salt water. Hints: first find ∆Tfp and then
subtract the change from the original freezing point (0oC for water). Also, remember mT is
not 2.5 m in this problem.
mT = 2.5(2) = 5.0 (recall that NaCl breaks into Na+ and Cl-)
∆Tfp = (mT)(Kfp) = (5.0)(1.853) = 9.3oC
Tfp = 0oC – 9.3oC = – 9.3oC
10. Find the boiling point of a 3.7 m solution of calcium chloride.
mT = (3.7)(3) = 11.1 (recall that CaCl2 breaks into Ca2+ and two Cl-)
∆Tbp = (mT)(Kbp) = (11.1)(0.515) = 5.72oC
Tbp = 100oC + 5.72oC = 105.72oC
11. What is the freezing point of a sugar solution in which the concentration of sugar is 2.25m?
Note: sugar is covalent so it dissolves but it does not dissociate.
∆Tfp = (mT)(Kfp) = (2.25)(1.853) = 4.17oC
Tfp = 0oC – 4.17oC = – 4.17oC
Information: Raoult’s Law
A solution will almost always have a lower vapor pressure than the pure solvent. For example, salt
water will have a lower vapor pressure than pure water. The vapor pressure of a solution (Psolution) is
related to the vapor pressure of the pure solvent (Psolvent) by the mole fraction of the solvent (Xsolvent)
in an equation known as Raoult’s Law:
Psolution = (Xsolvent)(Psolvent)
Critical Thinking Questions
12. What is the vapor pressure of water at 20oC is 2.3 kPa. What is the vapor pressure of a
solution formed by dissolving 21.5g of LiCl in 84.3g of H2O?
21.5 g
= 0.507mol
42.44 g / mol
x solvent =
84.3g
= 4.68mol
18.02 g / mol
4.68
= 0.902 Psolution = (Xsolvent)(Psolvent) = (0.902)(2.3) = 2.08 kPa
(4.68 + 0.507)
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ChemQuest 41
Name: ____________________________
Date: _______________
Hour: _____
Information: Molar Mass
As you know, the molar mass of any substance is how much mass (measured in grams) one mole of a
substance has. The units for molar mass are g/mol. If you know that 2.0 moles of water have a mass
of 36.0 g, you can find the molar mass of water by dividing 36.0 g by 2.0 moles to get 18.0 g/mol.
Critical Thinking Questions
1. If 2.75 moles of a certain compound has a mass of 125.9 g, what is the molar mass of the
compound?
125.9 g
= 45.8 g/mol
2.75mol
2. If you put 2.53x1024 molecules of an unknown compound on a balance you will discover that the
mass is 757.8 g. What is the identity of the unknown compound? (Hint: find the molar mass and
then calculate the molar masses of the following compounds.)
A) C10H12O
2.53 x10 24
= 4.20 mol
6.02 x10 23
B) C6H12O6
C) C8H18
D) C10H8NO
757.8 g
= 180.3 g/mol B has a molar mass of 180 g/mol also
4.20mol
Information: Relating Molar Mass and Colligative Properties
One of the ways to determine the molar mass of a compound is by experiments involving colligative
properties. By measuring the temperature changes of solutions, it is possible to calculate the molality
of the solution and from the molality you can determine the molar mass. The following calculations
walk you through the process. Please note that the process used here is valid only for covalent
compounds.
125
Critical Thinking Questions
3. When 135 g of an unknown covalent compound is dissolved in 450 g of water, the freezing point
of the solution is –3.21oC. Find the molar mass of the covalent compound. Follow these steps…
a) What is the molality of the solution? (Use ∆Tfp = mT Kfp and solve for mT. Note that since
this is a covalent compound mT equals the molality of the solute because covalent compounds
don’t dissociate.)
∆Tfp = mT Kfp mT = ∆Tfp ÷ Kfp = (3.21) ÷ 1.853 = 1.73 m
b) How many moles of solute were dissolved? (Multiply your answer to part a by the kilograms
of solvent.)
(1.73m)(0.450kg) = 0.779 mol
c) Calculate the molar mass of the compound. (Take the mass of the solute given in the problem
135g ÷ 0.779mol = 173 g/mol
4. Find the molar mass of a covalent compound if 210 g of the substance is dissolved in 810 g of
water changes the boiling point of the solution to 101.3 oC.
∆Tbp = mT Kbp mT = ∆Tbp ÷ Kbp = (1.3) ÷ 0.515 = 2.52 m
(2.52m)(0.810kg) = 2.04 mol
210 g ÷ 2.04 mol = 103 g/mol
5. Find the molar mass of a covalent compound if 38.5 g dissolves in 520 g of water to give a
freezing point of –2.15oC.
∆Tfp = mT Kfp mT = ∆Tfp ÷ Kfp = 2.15 ÷ 1.853 = 1.16 m
(1.16m)(0.520 kg) = 0.603 mol
38.5g ÷ 0.603 mol = 63.8 g/mol
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126
6. When 282.5 g of a covalent compound is dissolved in 630 g of water, the vapor pressure lowers
from 3.2 kPa to 2.1 kPa. What is the molar mass of the compound? Use the following steps…
a) Use Raoult’s Law to find the mole fraction of the water. (Solve for xsolvent.)
Psolution = (Xsolvent)(Psolvent) Xsolvent = Psolution ÷ Psolvent = 2.1 ÷ 3.2 = 0.656
b) Convert grams of water to moles of water.
630g ÷ 18.02 g/mol = 34.96 mol
c) Use the mole fraction equation (below) and your answer to part b to solve for molsolute.
mol solvent
x solvent =
(mol solvent + mol solute )
mol solute =
mol solvent
34.96
− mol solvent =
− 34.96 = 18.33 mol
0.656
x solvent
d) Divide the mass of the compound given in the question by your answer to part c to find the
molar mass.
282.5g ÷ 18.33 mol = 15.4 g/mol
7. Find the molar mass of a covalent solute if when 371 g of the solute are added to 650 g of water,
the vapor pressure changes from 4.2 kPa to 2.8 kPa.
Psolution = (Xsolvent)(Psolvent) Xsolvent = Psolution ÷ Psolvent = 2.8 ÷ 4.2 = 0.667
650 g ÷ 18.02g/mol = 36.07 mol of water
mol solvent
36.07
mol solute =
− mol solvent =
− 36.07 = 18.0 mol
0.667
x solvent
371g ÷ 18.0 mol = 20.6 g/mol
8. 230 g of a covalent solute are dissolved in 520 g of water. The vapor pressure changes from 3.5
kPa to 2.9 kPa. What is the freezing point of the solution? Hint: first find the moles of solute
using Raoult’s Law and then find the molality and use ∆Tfp = mT Kfp.
Psolution = (Xsolvent)(Psolvent) Xsolvent = Psolution ÷ Psolvent = 2.9 ÷ 3.5 = 0.829
molsolvent = 520g ÷ 18.02g/mol = 28.86 mol
mol solvent
28.86
− mol solvent =
− 28.86 =
x solvent
0.829
molality = mT = 5.95mol ÷ 0.520kg = 11.45 m
∆Tfp = mT Kfp = (11.45)(1.853) = 21.2
Tf = 0oC – 21.2oC = –21.2oC
mol solute =
5.95 mol
127
ChemQuest 42
Name: ____________________________
Date: _______________
Hour: _____
Information: Average Rate of Reaction
Recall that during a chemical reaction reactants are transformed into products: A + B C + D. A
very important question is: how fast do such processes happen? For example, we need to know how
long it will take for a medicine to work in our bodies and how long will it take to produce chemicals
in industry.
How fast do reactants A and B disappear? How fast do products C and D form? We can express
such questions in the form of an equation as shown below.
rate of disappearance
change in molarity of reactant A
∆[reactant A]
=−
of reactant A = −
change in time
∆time
Keep in mind that all rates are written as positive numbers; thus, the function for the negative sign in
the above equation is to yield a positive result for the rate.
Critical Thinking Questions
1. What is molarity?
Molarity is a measure of the moles of solute per liter of solution.
2. What do the symbols ∆ and [ ] mean in the above equation?
∆ means “change” and [ ] means concentration in molarity.
3. What units would you expect for the rate of disappearance of a reactant? (Assume time is
measured in seconds.)
Molarity per second
4. How is the change in molarity of a reactant calculated? Would this be a positive or a negative
number?
Final molarity minus initial molarity. It would be a negative number because the molarity of
a reactant decreases as the reaction proceeds.
5. How is the change in molarity of a product calculated? Would this be a positive or a negative
number?
Final molarity minus initial molarity. It would be a positive number because the molarity of a
product increases as the reaction proceeds.
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128
6. When writing the equation for the rate of formation of a product, the negative sign in the above
equation is not needed. Explain why.
The only purpose of the negative sign is to give a positive number and the negative sign is
only needed for calculating the change in concentration for a reactant because the change will
be a negative number, as mentioned in our answer to question 4.
Information: Stoichiometry and Average Rate
Consider the decomposition of N2O5 and the experimental data for the reaction taking place inside a
container that has a volume of 3.0 L.
2 N2O5 (g) 4 NO2 (g) + O2 (g)
Time (s)
0
600
1200
1800
Moles N2O5
0.4320
0.4194
0.4104
0.4032
Moles NO2
0
0.0252
0.0432
0.0576
Moles O2
0
0.0063
0.0108
0.0144
Critical Thinking Questions
7. Calculate the change in moles of each reactant and product between 600 and 1200 seconds.
change in moles of N2O5:
0.4194 – 0.4104 = 0.0090
change in moles of NO2:
0.0432 – 0.0252 = 0.0180
change in moles of O2:
0.0108 – 0.0063 = 0.0045
8. Compare your answers in question 7. What is the relationship between the coefficients in the
balanced chemical equation and the number of moles used up or produced in a reaction?
The change in N2O5 is twice the change in O2 which corresponds to the 2:1 ratio of the
balanced equation. Also, the change in NO2 is twice the change of N2O5, which also
corresponds to the 4:2 ratio of the balanced equation.
9. Fill in the missing blanks in the table above.
∆[O2] = 0.0144-0.0108 = 0.0036; ∆[NO2] = 4(0.0036) = 0.0144; ∆[N2O5]=2(0.0036) = 0.0072
10. Calculate the following values. Be sure to use molarity in your calculations.
a) Rate of disappearance of N2O5 between time 0 and 600 s.
0.4194 mol 0.4320 mol
−
0.1398 − 0.144
3.0 L
3.0 L
=
= −7.0 x10 −6 M/s
600 s - 0 s
600
b) Rate of appearance of NO2 between time 0 and 600 s.
0.0252 mol 0 mol
−
3.0 L
3.0 L = 0.0084 = 1.4 x10 − 5 M/s
600 s - 0 s
600
c) Rate of appearance of O2 between time 0 and 600 s.
0.0063 mol 0 mol
−
3.0 L
3.0 L = 0.0021 = 3.5 x10 − 6 M/s
600 s - 0 s
600
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11. Compare each of the rates. What relationship exists between the rates and the coefficients in the
balanced chemical equation?
They are related by the coefficients in the balanced equation so that
Rate O2 = 2(Rate N2O5) = 4(Rate NO2)
12. Calculate the following values. Be sure to use molarity in your calculations.
a) Rate of disappearance of N2O5 between time 600 and 1200 s.
0.4104 mol 0.4194 mol
−
0.1368 − 0.1398
3.0 L
3.0 L
=
= −5.0 x10 − 6 M/s
600 s - 0 s
600
b) Rate of appearance of NO2 between time 600 and 1200 s.
Rate NO2 = 2(RateN2O5) = 2(5.0x10-6) = 1.0x10-5 M/s
c) Rate of appearance of O2 between time 600 and 1200 s.
Rate O2 = 1/2(Rate N2O5) = 2.5x10-6 M/s
13. As the reaction proceeds are the rates constant? What affects the rate?
No, they are not constant. The rates are gradually slowing. For example, between 0 and 600
seconds N2O5 was disappearing at a rate of 7.0x10-6 M/s, but between 600 and 1200 seconds
N2O5 was disappearing at a rate of 5.0x10-6 M/s. The other reactants also slowed. It seems that
rate is dependent on concentration.
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130
ChemQuest 43
Name: ____________________________
Date: _______________
Hour: _____
Information: Reaction Order
Below is a table of data corresponding to the following balanced equation:
2 ClO2 (aq) + 2 OH- (aq) ClO3- (aq) + ClO2- (aq) + H2O (l) .
Six experiments were carried out with differing concentrations of ClO2 and OH- in each experiment.
The measurement of how quickly ClO2 disappears is given in the table for each of the six
experiments. How quickly a reactant disappears (or how quickly a product forms) is a good
measurement of how fast a reaction takes place.
Table 1: Experimental data for the reaction of ClO2
Experiment
Initial [ClO2]
Initial [OH-]
Initial Rate of disappearance
of ClO2 (M/s)
1
0.020
0.030
0.00276
2
0.040
0.030
0.01104
3
0.020
0.060
0.00552
4
0.040
0.060
0.02208
5
0.040
0.090
0.03312
6
0.120
0.030
0.09936
Critical Thinking Questions
1. What happens to the rate of a reaction as the concentrations of the reactants increases? Justify
The higher the concentration of reactants, the faster the rate. For example, the rate
corresponding to Experiment 4 is greater than the rate in Experiment 1 and Experiment 4 also
has a greater concentration of reactants.
2. Consider the molecular level of what is happening when ClO2 reacts with OH- to form products.
Offer an explanation for why changing the concentration of reactants changes the rate of a
reaction.
Changing the concentration changes how many molecules collide and react with each other.
3. Does the reaction depend on the concentration of ClO2 and the concentration of OH- equally? In
other words, is the rate more dependent on ClO2, on OH-, or is it equally dependent on the
The rate depends more on the concentration of ClO2 than on OH-. For example, comparing
experiments 1 and 2 we see that when [ClO2] is doubled and [OH-] held constant the rate
increases by a factor of 4. When we compare experiments 2 and 4 when [OH-] is doubled, we
see that the rate only increases by a factor of 2.
131
4. If you wanted to know how the rate of reaction depends on the concentration of ClO2 you could
compare experiments 1 and 2. But if you compared experiments 1 and 4, you would not be able
to accurately see how the reaction depends on the concentration of ClO2. Why?
You always need a control--one of the concentrations needs to remain constant. In
experiments 1 and 4 both the [ClO2] and [OH-] change.
5. Which two experiments would you want to compare to determine how much the rate of reaction
depends upon the concentration of OH-?
A) 1 and 4
B) 5 and 6
C) 1 and 6
D) 1 and 3
6. Considering [ClO2] in experiments 1 and 2, complete the following sentence.
When [ClO2] increases by a factor of 2, the rate of reaction increases by a factor of 2 to the _2nd_
power.
7. Considering [OH-] in the two experiments you identified in question 5, complete the following
sentence.
When [OH-] increases by a factor of 2, the rate of reaction increases by a factor of 2 to the _1st_
power.
8. Considering [ClO2] in experiments 2 and 6, complete the following sentence.
When [ClO2] increases by a factor of 3, the rate of reaction increases by a factor of 3 to the _2nd_
power.
9. Considering [OH-] in experiments 2 and 5, complete the following sentence.
When [OH-] increases by a factor of 3, the rate of reaction increases by a factor of 3 to the _1st_
power.
10. The rate dependence with respect to [ClO2] is said to be second order. Given your answers to
questions 6 and 8, explain what “second order” means.
Second order means that if the concentration increases by a factor of X, then the rate increases
by a factor of X to the 2nd power.
11. The rate dependence with respect to [OH-] is said to be first order. Given your answers to
questions 7 and 9, explain what “first order” means.
First order means that if the concentration increases by a factor of X, then the rate increases
by a factor of X to the 1st power.
12. Can you find the order for a reactant just by looking at the balanced equation?
No, you must have experimental data. The orders of reactants are not the same as the
coefficients in the balanced equation as can be seen in the balanced equation given at the
beginning of the information section.
13. The overall “order” of the reaction for this reaction is third order. Explain how the overall order
of a reaction is found.
Overall order is found by adding the orders for each of the reactants.
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Information: Rate Law
Once you know the order with respect to each reactant, you can determine the “rate law” for the
reaction. Each reaction has a different rate law. The rate law is a convenient way of expressing how
the rate of a reaction depends upon concentration. The rate law for the reaction we have been
considering so far is
Rate = k[ClO2]2[OH-]1
Each reaction has a rate constant, given the symbol k. As you will soon see, the rate constant can be
determined from experiments in a similar fashion to how you determined the orders for reactants.
Critical Thinking Questions
14. What is the relationship between the order of the reactant and the exponent for the reactant in the
rate law?
The order for the reactant equals the exponent in the rate law.
15. Using the data from any of the six experiments in Table 1, verify that the rate constant is 230
1/(M2s). For example, let’s pick experiment 3 to use. The rate is given and so are the
concentrations of ClO2 and OH-. Plug these given data into the rate law and solve for k. Use
units in your calculation to verify that the units for k are 1/(M2s).
Rate = k[ClO2]2[OH-]1 0.00552M/s = k (0.020M)2(0.060M) k = 230 1/M2s
16. Now that you know the rate constant, you can calculate the rate for any concentration of
reactants. For example, calculate the rate of reaction when the concentration of ClO2 is 0.32 and
the concentration of OH- is 0.42.
Rate = k[ClO2]2[OH-]1 Rate = (230)(0.32)2(0.42) = 9.89 M/s
Skill Practice
17. In a certain reaction, it is discovered that if the concentration of a reactant is tripled, then the rate
of the reaction increases from 0.0670 M/s to 1.809 M/s. What is the order with respect to this
reactant?
1.809 ÷ 0.0670 = 27; 27 = 33; since the exponent is a 3, the order is 3rd order.
18. Given the following data, write the rate law for the reaction. Then find the rate constant (include
units).
H2O2 + 2 HI 2 H2O2 + I2
Experiment
[H2O2]
[HI]
Rate (M/s)
1
0.1
0.1
0.0076
2
0.1
0.2
0.0608
3
0.2
0.2
0.2432
Comparing experiments 1 and 2 we see that the rate increases by a factor of 23 when [HI] doubles.
Comparing experiments 2 and 3 we see that the rate increases by a factor of 22 when [H2O2] doubles.
Therefore we can write: Rate = k[H2O2]2[HI]3.
Rate = k[H2O2]2[HI]3 (plug in data from expt. #1 0.0076 = k(0.1)2(0.1)3 k = 760 1/M4s
133
ChemQuest 44
Name: ____________________________
Date: _______________
Hour: _____
Information: First Order Reactions with One Reactant
As we consider what affects the rate of reactions it is desirable to examine how the concentration of a
reactant changes with time. Let us consider reactions that have only one reactant and we will further
restrict our considerations to first order reactants. As an example, consider the following reaction:
SO2Cl2 SO2 + Cl2
Table 1: Experimental data for the decomposition of SO2Cl2
Time (s)
[SO2Cl2]
0
0.0250
60
0.0228
120
0.0208
180
0.0190
It can be shown that the natural log of the concentration of a first order reactant varies directly with
the time. So in this case ln[SO2Cl2] varies in direct proportion to the time.
1. Using the above data, prove on the graph below that ln[SO2Cl2] = - kt + ln[SO2Cl2]0 is a straight
line when graphed. Note the expression [SO2Cl2]0 is the concentration of SO2Cl2 at a time of
zero seconds. Label the axes.
-3.680
Calculate ln[SO2Cl2] values and plot:
Time
ln[SO2Cl2]
0
-3.689
60
-3.781
120
-3.873
180
-3.963
-3.730
-3.780
ln[SO2Cl2]
-3.830
-3.880
-3.930
-3.980
0
30
60 90 120 150 180 210 240
Time (s)
2. Given this relationship between concentration and time (ln[SO2Cl2] = - kt + ln[SO2Cl2]0), find the
rate constant k.
Choose values to plug into equation; I'll use at time 120:
ln(0.0208) = -k (120) + ln(0.0250) k = 0.00153 1/s
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134
3. Find the half-life for this reaction. What this means is that you need to find the time it takes for
half of the reactant to get used up.
Find the time (t) when [SO2Cl2] = 1/2(0.0250) = 0.0125
ln(0.0125) = -(0.00153)(t) + ln(0.0250) t = 453 seconds
Information: Second Order Reactions with One Reactant
Consider the following reaction: 2 NO2 2 NO + O2. The following experimental data was
gathered for this reaction:
Table 2: Experimental data for the decomposition of NO2
Time (s)
[NO2]
0
0.0370
45
0.0338
90
0.0311
135
0.0288
If you attempted a plot of ln vs. t as you did in question 2 above you would not get a straight line.
Instead, for second order reactants, the inverse of the concentration varies directly with time.
Critical Thinking Questions
4. Using the following graph and the above data, prove that the following equation yields a straight
line.
1
1
= kt +
[ NO 2 ]
[ NO 2 ]0
37
35
Calculate 1/[NO2] and plot this data:
Time (s)
1/[NO2]
0
27.03
45
29.59
90
32.15
135
34.72
33
1/[NO2]
31
29
27
25
0
20
40
60
80
100 120 140 160
Time (s)
5. What is the value of the rate constant, k, for this reaction?
1/0.0311 = k (90) + 1/0.0370 k = 0.0570 1/Ms
135
6. Find the half-life for this reaction.
Need to find the time, t, when [NO2] = 1/2(0.0370) = 0.0185 M
1/0.0185 = (0.0570) (t) + 1/0.0370 t = 474 s
Skill Practice Question
7. Given the following reaction and table of experimental data, answer the following questions.
2 N2O5 4NO2 + O2
Time (s)
[N2O5]
0
0.0200
100
0.0169
200
0.0142
300
0.0120
400
0.0101
500
0.0086
600
0.0072
700
0.0061
a) Is the reaction 1st order or second order with respect to N2O5? How do you know?
It is 1st order because the data fits the equation ln[N2O5] = -kt + ln[N2O5]0 much more
closely than it fits the 2nd order equation.
b) What is the value for the rate constant?
Using ln[N2O5] = -kt + ln[N2O5]0 and plugging in data…
ln(0.0086) = -k(500) + ln(0.0200) k = 0.0017 1/s
c) What is the half life for this reaction?
ln(0.0100) = -(0.0017)(t1/2) + ln(0.0200) t = 408 s
d) How many seconds are required for the concentration of N2O5 to reach a level of 0.0025
M?
ln(0.0025) = -(0.0017)(t) + ln(0.0200) 1223 s
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136
ChemQuest 45
Name: ____________________________
Date: _______________
Hour: _____
Information: Forward and Reverse Processes
When most people think of chemical reactions, they think of reactants being transformed into
products. For example, take the reaction of carbon monoxide with hydrogen gas to form methane gas
and water vapor:
CO (g) + 3 H2 (g) CH4 (g) + H2O (g)
If you were watching the molecules of this reaction, you would see that at the very beginning, there is
no methane and no water in the container. Soon methane and water would begin to form and then the
reaction would appear to stop. However, at this point there would still be some carbon monoxide
and hydrogen present.
What happens at the molecular level is this: as the products (methane and water) begin to form, they
react with each other and begin forming the reactants (carbon monoxide and hydrogen) again. There
is a forward reaction and a reverse reaction. The forward reaction is written above. The reverse
reaction is below:
CH4 (g) + H2O (g) CO (g) + 3 H2 (g)
The best way to represent the reaction, then, is as follows:
CO (g) + 3 H2 (g)  CH4 (g) + H2O (g)
The reaction appears to stop when the rate of the forward reaction equals the rate of the reverse
reaction. At this point we say that the reaction has reached equilibrium. The reaction is still
occurring, but the products and reactants are being formed at the same time so there is no net change
in their amounts.
Critical Thinking Questions
1. Consider the reaction above involving carbon monoxide and hydrogen. At the beginning of the
reaction, there are 2.50 moles of carbon monoxide and 5.10 moles of hydrogen gas placed in a 5.0
L container. Of course, at the very beginning there is no methane or water in the container.
When equilibrium is reached, there are 1.02 moles of water in the container. Calculate the
number of moles of methane, hydrogen and carbon monoxide in the container.
Hint: Calculate the change in the number of moles of water; by how many moles did the
water increase? The number of moles of methane increased by this same amount. The
number of moles of carbon monoxide decreased by this same amount. The number of moles
of hydrogen decreased by three times this amount. We know this because of the coefficients
in the balanced chemical equation.
Change in moles of H2O, CO and of CH4 = 1.02 because coefficients in balanced equation are equal.
Change in moles of H2 = 3(1.02) = 3.06. Therefore the final moles of each are as follows…
CH4 = 1.02 mol; H2 = 5.10-3.06 = 2.04 mol; CO = 2.50 – 1.02 = 1.48 mol
137
2. Consider the following reaction: N2 (g) + 3 H2 (g)  2 NH3 (g). Initially, 4.25 moles of
nitrogen gas and 6.33 moles of hydrogen gas are placed in a 3.35 L container. At equilibrium,
2.15 moles of NH3 (ammonia) was present. Calculate the number of moles of nitrogen and
hydrogen at equilibrium.
There were initially 0 moles of NH3, but at equilibrium there were 2.15 moles. From this we
can calculate the change in moles of N2 and H2 and then the amount at equilibrium
N2 (g) + 3 H2 (g)  2 NH3 (g)
Initial moles: 4.25
6.33
0
Change in moles: -1.075
-3.225
+2.15
Final moles: 3.175
3.105
2.15
3. As mentioned already, equilibrium occurs when the rate of the forward reaction equals the rate of
the reverse reaction. Assume that all reactions discussed so far are elementary reactions. This
means that the exponents in the rate law are the same as the coefficients in the balanced equation.
a) Write the rate law for the forward reaction of carbon monoxide and hydrogen. Use kf to
symbolize the rate constant for the forward reaction.
Rate = kf [CO][H2]3
b) Write the rate law for the reverse reaction of carbon monoxide and hydrogen. Use kr to
symbolize the rate constant for the reverse reaction.
Rate = kr [H2O][CH4]
4. Now divide the reverse rate law by the forward rate law and complete the following equation.
rate reverse
=
rate forward
kr [H2O][CH4]
kf [CO][H2]3
5. At equilibrium, the reverse rate equals the forward rate. What does the left side of the equation in
question 4 equal?
Since ratereverse = rateforward, then ratereverse ÷ rateforward = 1
6. Taking into account your answer to question 5, rearrange the equation that you wrote in number
four and get kf and kr on the left side of the reaction and everything else on the right side.
kf
kr
=
[H2O][CH4]
[CO][H2]3
Information The Equilibrium Constant
The equilibrium constant is a constant that allows us to compare the concentrations of products and
reactants in a chemical reaction. The equilibrium constant (K) is defined as kf/kr.
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Critical Thinking Questions
7. Given your answer to question 6 and also the information above, write the expression for the
equilibrium constant for the reaction of carbon monoxide with hydrogen.
[H2O][CH4]
K=
[CO][H2]3
8. Calculate the equilibrium constant for the reaction using your answers to question number 1. You
will first need to find the molarity of each reactant and product. You should get a value of 2.07.
We need to divide each of the final moles from Q 1 by 5.0L to obtain molarity.
[CH4] = [H2O] = 1.02 ÷5= 0.204 M; [H2]=2.04÷5= 0.408 M; [CO]=1.48÷0.296 M
[H2O][CH4]
(0.204)(0.204)
=
[CO][H2]3
= 2.07
(0.296)(0.408)3
9. Verify that the equilibrium constant expression for the reaction described in question two can be
written as
K=
[NH 3 ] 2
[ N 2 ][H 2 ] 3
Yes, in general, equilibrium constants are written as the concentration of the products divided by
the concentration of the reactants, with each substance raised to the power of its coefficient.
10. Calculate the numeric value of the equilibrium constant from question 9 using your answers and
the data in question 2.
[NH3] = 2.15mol÷3.35L = 0.642M; [N2] = 3.175mol÷3.35L = 0.948M; [H2]=3.105÷3.35L= 0.927M
[NH3]2
(0.642)2
[N2][H2]3
=
(0.948)(0.927)3
= 0.546
11. Considering questions 7 and 9, what relationship exists between the coefficients in the balanced
chemical equation and the expression for the equilibrium constant?
In general, equilibrium constants are written as the concentration of the products divided by
the concentration of the reactants, with each substance raised to the power of its coefficient.
12. Write the equilibrium constant expression for each of the following reactions.
a) 2 HI  H2 + I2
b) 2 CO2  2 CO + O2
[H2][I2]
[CO]2[O2]
[HI]2
[CO2]2
139
Information: Calculating Equilibrium Constants
Consider a 100 L container that holds 80 moles of hydrogen iodide. Over time, the hydrogen iodide
decomposes into hydrogen and iodine. At equilibrium, there are 8.84 moles of iodine. The balanced
equation for this process is: 2 HI  H2 + I2
Critical Thinking Questions
13. Find the initial concentration of HI and the equilibrium concentration of I2.
[HI] = 80 ÷ 100 = 0.80 M
[I2] = 8.84 ÷ 100 = 0.0884 M
14. Consider the balanced equation for a moment. How will the change in iodine concentration
compare to the change in hydrogen iodide? (Hint: it depends on the coefficients in the balanced
equation.)
The change in HI will be twice the change in I2.
15. What was the initial concentration of I2? Note: “initial” means before any reaction takes place.
Zero
16. What was the initial concentration of H2?
Zero
17. What was the change in concentration for I2? (Remember that change in concentration is simply
the final minus the initial concentration.)
0.0884 M
18. Calculate the change in H2 and the change in HI concentration.
∆ in [H2] = ___0.0884 M__
∆ in [HI] = (0.0884)(2) = 0.177 M
19. What is the equilibrium concentration (i.e. concentration at equilibrium) of HI? Note the
equilibrium concentration of HI is equal to the initial concentration of HI minus the change in
concentration.
0.80 – 0.177 = 0.623 M
20. What is the equilibrium concentration of H2? Note the equilibrium concentration of H2 is equal
to the initial concentration of H2 plus the change in concentration of H2.
0 + 0.0884 = 0.0884 M
21. In question 19, you subtracted the change in concentration, but in question 20, you added it.
Why?
Question 19 referred to a reactant and the concentration of a reactant is decreasing with time.
Question 20 referred to a product and the concentration of product increases with time.
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140
22. Write the equilibrium constant expression for this reaction.
[H2][I2]
[HI]2
23. Calculate the equilibrium constant for this reaction.
(0.0884)(0.0884)
= 0.0201
(0.623)2
24. This problem combines all of the previous eleven questions and asks you to find the equilibrium
constant for a reaction. Consider the following reaction: 2 NO + O2 2 NO2. 5.25 moles of
NO and 3.15 moles of O2 are combined in a 12.0 L container. At equilibrium 3.20 moles of NO2
are in the container. Verify that the equilibrium constant for this reaction is 18.88. Don’t forget
to use molarity instead of moles!
Initial concentrations:
[NO] = 5.25 ÷ 12.0 = 0.4375 M
[O2] = 3.15 ÷ 12.0 = 0.2625 M
Equilibrium (final) concentrations:
[NO2] = 3.20 ÷ 12.0 = 0.2667 M
Changes in concentration:
∆[NO2] = 0.2667 M (because there was zero to start with and at equilibrium there was 0.2667 M)
∆[NO] = ∆[NO2] = 0.2667 M (because coefficients are same in balanced equation)
∆[O2] = ½ (∆[NO2]) = 0.1333 M (because there is a 1:2 ratio in balanced equation.)
Equilibrium (final) concentrations:
[NO2] = 0.2667 M
[NO] = 0.4375 – 0.2667 = 0.1708 M
[O2] = 0.2625 – 0.1333 = 0.1292 M
[NO2]2
K=
[NO]2[O2]
(0.2667)2
=
(0.1708)2(0.1292)
= 18.87
141
ChemQuest 46
Name: ____________________________
Date: _______________
Hour: _____
Information: Equilibrium Constant for Gas Pressure
So far we have looked at equations involving gases in terms of the molarity of the gas. For example,
if 3 moles of a gas was in a 1.5 L container we used the value 2.0 M in calculating the equilibrium
constant. The equilibrium constant, K, when dealing strictly with molarity actually has the symbol
Kc. It has been unnecessary to write Kc until now.
Molarity is not the only way to express the concentration of a gas in a container. Pressure, for
example, may also be used. We know that for a 1.5 L container, the higher the pressure the greater
the concentration. When only pressure information about a gaseous reaction we can still calculate an
equilibrium constant, Kp. It is calculated in a very similar way as Kc is calculated. Consider the
following reaction.
N2 (g) + 3 H2 (g)  2 NH3 (g)
The equilibrium constant expression, Kp for this reaction is:
2
p NH 3
Kp =
3
pN 2 pH 2
Note: p stands for the partial pressure of each gaseous reactant or product.
Critical Thinking Questions
1. Consider the following reaction: 2 NO (g) + Br2 (g)  2 NOBr (g). Write the expression for
Kp .
[NOBr]2
Kp =
[NO]2[Br2]
Information: Relating Kc and Kp
The values Kc and Kp are related by the following equation:
Kp = Kc(RT)∆n
R = 0.0821 (L-atm)/(mol-K) or 8.31 (L-kPa)/(mol-K). It is customary to use
atmospheres (atm) to measure pressure, so we will use the value 0.0821 for R.
T = Kelvin temperature
∆n = the change in the number of moles of gas as the reaction proceeds.
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Critical Thinking Questions
2. Verify that ∆n = -1 for the reaction in question 1. (Show how you can obtain -1.)
There are three moles of gaseous reactants and only two moles of gaseous products.
Therefore, the number of moles of gas has decreased by one; thus, ∆n = -1.
3. Verify that ∆n = -2 for the reaction N2 (g) + 3 H2 (g)  2 NH3 (g).
The four total moles of gaseous reactants decreases by two moles to produce two moles of
gaseous products.
4. For the reaction referred to in questions 1 and 2 Kc equals 0.45 at 650oC. What is Kp at this
temperature? You should get a value of about 0.00594.
Kp = Kc(RT)∆n = (0.45)[(0.0821)(923)]-1 = 0.00594
5. For the reaction in question 3, it was found that Kp equals 1.25x10-4 at 425oC. What is Kc at this
temperature?
Kp = Kc(RT)∆n Kc = Kp ÷ (RT)∆n = 1.25x10-4 ÷ [(0.0821)(698)]-2 = 0.410
Information: Equilibrium Constants from the Sum of Chemical Equations
Consider the following two reactions for which the equilibrium constants are known:
Reaction 1: CO + 3 H2  CH4 + H2O; K1 = 3.92
Reaction 2: CH4 + 2 H2S  CS2 + 4 H2; K2 = 3.3x104
Now consider the following reaction for which the equilibrium constant is unknown:
Reaction 3: CO + 2 H2S  CS2 + H2O + H2; K3 = ?
It is possible to obtain the equilibrium constant K3 from K1 and K2. The following questions will
show you how.
Critical Thinking Questions
6. Which of the following is the relationship between reaction 3 and reactions 1 and 2?
A) Reaction 3 = Reaction 1 + Reaction 2
B) Reaction 3 = Reaction 2 – Reaction 1
7. Write the equilibrium constant expressions for K1 and K2.
[CS2][H2]4
[CH4][H2O]
K1 =
[CO][H2]3
8. Now write the equilibrium constant expression for K3.
[CS2][H2O]
K3 =
[CO][H2O]2
K2 =
[CH4][H2S]2
143
9. Consider your answers to questions 6 through 8. Which of the following equations is true:
A) K3 = K2/K1
B) K3 = K1 + K2
C) K3 = K1K2
D) K3 = K1/K2
10. Calculate the equilibrium constant K3.
K3 = K1K2 = (3.92)(3.3x104) = 1.29x105
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144
ChemQuest 47
Name: ____________________________
Date: _______________
Hour: _____
Critical Thinking Questions
1. If Kc for a given reaction is very large would there be a large amount of products or reactants in
the mixture?
Large amounts of products.
2. If Kc for a given reaction is very small would there be a large amount of products or reactants in
the mixture?
Large amounts of reactants.
3. Offer a mathematical explanation for your answers to questions 1 and 2.
Since Kc equals the concentration of products divided by reactants, a large Kc means a large
numerator (products) whereas a small Kc indicates a small numerator.
Information: The Reaction Quotient
The reaction quotient, Qc, is calculated in the same way as you would calculate the equilibrium
constant. For the reaction aA + bB  cC + dD, the reaction quotient is:
[C] c [ D ] d
Q=
[A] a [ B] b
It is important to keep in mind that the reaction quotient does not involve equilibrium concentrations.
The concentrations used to calculate Qc are at any time, not just at equilibrium.
Critical Thinking Questions
4. Consider the following reaction: CO + 3H2  CH4 + H2O. While carrying out a reaction
between carbon monoxide and hydrogen, a scientist analyzed the mixture and found that in the
3.5 L container there were 0.35 moles of CO, 0.42 moles of H2, 0.29 moles of CH4, and 0.38
moles of H2O. What is the reaction quotient for this mixture?
[CO] = 0.35÷3.5 = 0.10 M; [H2] = 0.42÷3.5 = 0.12 M; [CH4] = 0.29÷3.5 = 0.0829 M;
[H2O] = 0.38÷3.5 = 0.109 M
[CH4][H2O]
(0.0829)(0.109)
Qc =
=
= 52.3
3
3
[CO][H2]
(0.10)(0.12)
Information: What Qc Tells Us
As a reaction proceeds it will always tend to go toward equilibrium. For example, the equilibrium
constant for the reaction described in question 4 is 3.92. The concentration of products and reactants
will adjust themselves so that as the reaction progresses until the products divided by reactants
(raised to the appropriate power) will equal 3.92.
145
Critical Thinking Questions
5. Given your answer to question 4 and the fact that Kc equals 3.92 for the reaction, what must
happen for the reaction to reach equilibrium?
A) more products must form
B) more reactants must form
Qc is 0.75 and it must increase to 3.92 as the reaction proceeds. To increase the value of Q,
the concentration of the products must increase.
6. At a certain time during a reaction whose equilibrium constant was 12.5, it was found that the
reaction quotient was 4.2. Predict what will happen to the concentration of reactants and products
as the reaction progresses.
4.2 must increase to 12.5 by increasing the concentration of products and decreasing the
concentration of reactants.
7. At a certain time during a reaction whose equilibrium constant was 0.45, it was found that the
reaction quotient was 2.1. Predict what will happen to the concentration of products and reactants
as the reaction progresses.
2.1 must decrease to 0.45 by decreasing the concentration of products and increasing the
concentration of reactants.
8. Given your answers to questions 6 and 7, complete the following sentences.
If Qc is greater than Kc, then the concentration of products needs to __decrease______.
If Qc is less than Kc, then the concentration of products needs to ____increase________.
9. Consider the equilibrium reaction of hydrogen gas reacting with nitrogen gas to produce
ammonia, NH3. Kc for the reaction is 0.500. A 50.0 L reaction vessel contains 1.00 mol N2, 3.00
mol H2, and 0.500 mol of NH3. Will more NH3 be formed or will more N2 and H2 form as the
reaction proceeds?
3 H2 + N2  2 NH3
[H2] = 3.00÷50.0 = 0.0600 M; [N2] = 1.00 mol÷50.0 = 0.0200 M; [NH3] = 0.500÷50.0 = 0.0100 M
[NH3]2
Q=
3
[H2] [N2]
(0.0100)2
=
= 23.1
(0.0600) (0.0200)
3
Since Q is greater than Kc, the amount of product (NH3) needs to decrease and more reactants
(H2 and N2) will form.
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146
ChemQuest 48
Name: ____________________________
Date: _______________
Hour: _____
Information: Definitions of Acids and Bases
Arrhenius definitions
1) acid: substance that when dissolved in water increases [H+]; (note: H+ exists bonded to
water as the hydronium ion, H3O+, so [H+] and [H3O+] are equivalent expressions)
2) base: substance that when dissolved in water increases [OH-]
Bronsted-Lowry definitions
1) acid: substance that donates a proton, H+, in a reaction
2) base: substance that accepts a proton, H+, in a reaction
Table 1: Equilibrium constants (at 25oC) for some acid-base equilibrium reactions.
Reaction
Kc
1.07 x 10-11
1. C2H3O2- + H2O  HC2H3O2 + OH4.9 x 10-11
2. HCN + SO42- + H2O  HSO4- + CN- + H2O
3. HC2H3O2 + H2O  H3O+ + C2H3O23.09 x 10-7
+
4. H2CO3 + H2O  H3O + HCO3
7.82 x 10-9
5. HCl + H2O  H3O+ + Cl2.0 x 104
Critical Thinking Questions
1. List all of the reactants in Table 1 that are Arrhenius acids.
HC2H3O2, H2CO3, and HCl
2. List all of the reactants in Table 1 that are Arrhenius bases.
C2H3O23. List all of the reactants in Table 1 that are Bronsted-Lowry acids.
H2O, HCN, HC2H3O2, H2CO3, and HCl
4. List all of the reactants in Table 1 that are Bronsted-Lowry bases.
C2H3O2-, SO42-, H2O
5. Is it possible for an ion to act as a base? Explain.
Yes, C2H3O2- and SO42- act like bases in the above reactions.
6. Can a substance act as both an acid and a base under different conditions? Explain.
Yes, for example in the above table H2O acts like a base and an acid in different reactions.
7. Is this statement true: “All substances that are Arrhenius acids are also Bronsted-Lowry acids”?
Explain.
Yes, all of the substances in our answer to question 1 also are found in our answer to question 3.
147
8. If a substance is a Bronsted-Lowry acid, can we conclude that the substance is also an Arrhenius
acid? Explain.
No because not all of the substances from question 3 are listed in question 1.
9. a) What is the strongest acid among the reactants in reactions 3-5 inTable 1? Explain.
HCl because the large Kc indicates that it forms H3O+ in large concentrations.
b) What is the weakest acid among the reactants in reactions 3-5 inTable 1? Explain.
HCN, because of its very small Kc.
10. Consider Reaction 3.
a) What substance is formed (by the acid) after the acid loses a proton?
C2H3O2b) Is this substance an acid or a base? (Hint: look at reaction 1.)
It is a base since it produces OH- in reaction 1.
11. Drawing a conclusion from question 10, what can be said about a substance after it loses a
proton? Is the substance formed acidic or basic?
After an acid loses an H+, the substance formed is a base.
Information: Conjugate Acid-Base Pairs
After an acid loses a proton in a reaction, the substance formed behaves like a base. Verify this by
examining Reactions 3 and 1 in Table 1. Notice from reaction 3 that HC2H3O2 is an acid. After it
loses a proton it becomes the acetate ion, C2H3O2-. The acetate ion is a base, as seen in reaction 1;
there is a special name for this base: it is a conjugate base. So, C2H3O2- is the conjugate base of
HC2H3O2. Similarly, HSO4- is the conjugate acid of SO42-. Verify this by examining Reaction 2.
Critical Thinking Questions
12. Describe how a conjugate base is formed.
It is formed by an acid losing a H+.
13. How is a conjugate acid formed?
It is formed by a base gaining a H+.
14. For each of the acids below, write the reaction of the acid with water and circle the formula of the
a) H2SO4
H2SO4 + H2O  H3O+ + HSO4b) HCO3-
HCO3- + H2O  H3O+ + CO32-
c) HF
HF + H2O  H3O+ + F-
15. For each of the bases below, write the reaction of the base with water and circle the formula of
the conjugate acid in your reaction.
a) NH3
NH3 + H2O  NH4+ + OHb) OH-
OH- + H2O  H2O + OH-
c) NO3-
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148
ChemQuest 49
Name: ____________________________
Date: _______________
Hour: _____
Information: Measuring Acidity
Many people have heard of “pH”, and many people know that it probably refers to acidity, but not
very many understand what exactly it measures. The pH is a measure of the concentration of
hydrogen ions (H+) in a solution. Recall that in water H+ ions bond to water to form H3O+
(hydronium ions). Keep this in mind because hydrogen (H+) and hydronium (H3O+) are often used
interchangeably.
The pH of a solution can range from 0-14. The equation for calculating pH is:
pH = -log[H+]
Critical Thinking Questions
1. A certain solution has a concentration of hydrogen ions of 2.5 x 10-3 M. What is the pH?
pH = -log[H+] = -log(2.5x10-3) = 2.60
2. The same solution as question 1 has a concentration of hydroxide ions of 4.0 x 10-12. What is the
pOH?
pOH = -log[OH-] = -log(4.0x10-12) = 11.40
3. A certain solution has a pH of 5.2. What is the concentration of hydrogen ions?
[H+] = 10(-pH) = 10(-5.2) = 6.3x10-6 M
4. The same solution as question 3 has a pOH of 8.8. What is the concentration of hydroxide ions?
[OH-] = 10(-pOH) = 10(-8.8) = 1.6x10-9 M
5. Use the information given and your answers to questions 1 and 2 to find the following:
a) Divide the pH by the pOH of the solution.
2.60 ÷ 11.40 = 0.228
b) Multiply the pH by the pOH of the solution.
2.60 x 11.40 = 29.64
c) Add the pH and the pOH together.
2.60 + 11.40 = 14.00
d) Multiply [H+] and [OH-] together.
2.5 x 10-3 x 4.0 x 10-12 = 1.0x10-14
e) Divide [H+] by [OH-].
2.5x10-3 ÷ 4.0x10-12 = 6.25x108
149
6. Use the information given and your answers to questions 3 and 4 to find the following:
a) Divide the pH by the pOH of the solution.
5.2 ÷ 8.8 = 0.591
b) Multiply the pH by the pOH of the solution.
5.2 x 8.8 = 45.76
c) Add the pH and the pOH together.
5.2 + 8.8 = 14.0
d) Multiply [H+] and [OH-] together.
6.3x10-6 x 1.6x10-9 = 1.0x10-14
e) Divide [H+] by [OH-].
6.3x10-6 ÷ 1.6x10-9 = 3937.5
7. Consider your answers to questions 5 and 6. Were there any that gave the same answer (or just
5c and 6c: pH + pOH = 14
5d and 6d: [H+]x[OH-] = 1.0x10-14
a) Find the pH of a solution that has a pOH of 4.6.
9.4; pH + pOH = 14 pH = 14 – pOH = 9.4
b) The [H3O+] of a solution is 2.55x10-3 M. What is the [OH-]?
3.92x10-12 M; [H+] x [OH-] = 1.00x10-14 [H+] = 1.00x10-14 ÷ 2.55x10-3 = 3.92x10-12 M
9. What is the pOH of a solution that has a [H+] = 4.22 x 10-3 M.
11.625; pH = -log(4.22x10-3) = 2.375 pOH = 14 – pH = 14 – 2.375 = 11.625
10. A certain solution has a pH of 3.25.
a) Find the pOH of the solution.
pOH = 14 – 3.25 = 10.75
b) Find the [H+] of the solution.
[H+] = 10(-pH) = 10(-3.25) = 5.623x10-4 M
c) Find the [OH-] of the solution.
[OH-] = 1x10-14 ÷ [H+] = 1x10-14 ÷ 5.623 x 10-4 = 1.778x10-11 M
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11. Find the pH of a 0.035 M solution of HNO3. (Hint: First find the [H+] by realizing that nitric acid
dissociates according to this balanced equation: HNO3 H+ + NO3-. If the concentration of
HNO3 is 0.035, then [H+] is also 0.035.)
pH = -log(0.035) = 1.46
12. Find the pH of a 0.16 M solution of HCl.
pH = -log(0.16) = 0.80
13. Find the pH of a 0.045 M solution of NaOH. (NaOH Na+ + OH-)
12.65; pOH = -log[OH-] = -log(0.045) = 1.35; pH = 14 – pOH = 14-1.35 = 12.65
14. Find the pH of a 0.0045 M solution of LiOH.
11.65; pOH = -log[OH-] = -log(0.0045) = 2.35; pH = 14 – pOH = 14-2.35 = 11.65
15. If 14.5 g of NaOH is dissolved in 720 mL of water, what is the pH?
13.702
14.5g ÷ 40 g/mol = 0.363 mol; 0.363 mol ÷ 0.720 L = 0.503 M = [NaOH] = [OH-]
pOH = -log[OH-] = -log(0.503) = 0.298
pH = 14 – pH = 14 – 0.298 = 13.702
16. A certain solution contains 0.35 mol of HCl in 1200 mL of water.
a) Find the pH and pOH of the solution.
pH = 0.535 pOH = 13.465
[HCl] = [H+] = 0.35mol ÷ 1.200L = 0.292 M
pH = -log(0.292) = 0.535; pOH = 14 – pH = 14 – 0.535 = 13.465
b) Find the [H+] and the [OH-] in the solution.
[H+] = 10(-pH) = 10(-0.535) = 0.292 M
[OH-] = 10(-pOH) = 10(-13.465) = 3.428x10-14 M
151
ChemQuest 50
Name: ____________________________
Date: _______________
Hour: _____
Information: Constants and Equations
You have discovered two fundamental constants represented by the following equations:
1. pH + pOH = 14
2. [OH-][H+] = 1x10-14
A third equation is related to equation 2 above:
3. (Ka)(Kb) = 1x10-14
This equation holds true for any conjugate acid-base pair. The Ka for the acid multiplied by the Kb
for the conjugate base will always equal 1x10-14.
The following questions will give you some practice using the above three relationships combined
with the equations for pH and pOH that you should already be familiar with.
Critical Thinking Questions
1. The Ka for a certain weak acid is 1.8x10-9. What is the Kb for the conjugate base of this acid?
Kb = 1x10-14 ÷ Ka = 1x10-14 ÷ 1.8x10-9 = 5.6x10-6
2. The Kb for NH3 is 1.8x10-5. Write the formula for and calculate the Ka for the conjugate acid of
NH3.
The formula is NH4+ since conjugate acids are formed when a base gains a H+.
Ka = 1x10-14 ÷ 1.8x10-5 = 5.6x10-10
3. The pH of a certain solution was 1.45. What was the pOH?
pOH = 14 – 1.45 = 12.55
4. If the hydroxide ion concentration of a certain solution was 2.6x10-4 M, what is the pH?
pOH = -log(2.6x10-4) = 3.59 pH = 14 – 3.59 = 10.41
Information: Strong Acids and Bases
Acids and bases are called “strong” if they dissociate completely (or almost completely) in water.
For example, hydrochloric acid (HCl) exists almost completely as H+ ions and Cl- ions in water.
Therefore HCl is a strong acid. If you put 2.0 moles of HCl in water, the HCl will dissociate and you
will end up having about 2.0 moles of H+ and 2.0 moles of Cl- in the water. Each HCl molecule
breaks up into two ions when dissolved: HCl H+ + Cl-. For every mole of HCl added to water,
you will get one mole of H+.
Critical Thinking Questions
5. Consider a strong acid like HCl. If the concentration of the acid is 0.025 M, what is the
concentration of hydrogen ions? Explain your reasoning.
0.025 M because all of the HCl breaks up into H+ and Cl- ions.
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6. Consider a strong base such as NaOH. If the concentration of the base is 0.75 M, what is the
concentration of the hydroxide ions? Explain.
For a strong base, [OH-] = [base]. Therefore, here the [OH-] = 0.75 M
7. What is the pH of a solution that has a concentration of HCl equal to 0.0049 M?
[H+] = [HCl] = [0.0049]; pH = -log(0.0049) = 2.31
8. If 2.5 moles of HCl were dissolved in 42 L of water, what would the pH be? What would the
pOH be?
[H+] = [HCl] = 2.5mol ÷ 42L = 0.595 M
pH = -log(0.595) = 0.23
pOH = 14 – 0.23 = 13.77
9. The strong base, NaOH, was dissolved in water. If 4.2 moles of NaOH was dissolved in 245 L of
water, what is the pH of the solution?
[OH-] = [NaOH] = 4.2mol ÷ 245L = 0.017 M; pOH = -log(0.017) = 1.77
pH = 14 – 1.77 = 12.23
Information: Weak Acids and Bases
Weak acids and bases do not dissociate completely in water. For example, consider acetic acid
(HC2H3O2):
HC2H3O2  H+ + C2H3O2- ; Ka = 1.7x10-5
For HCl, the concentration of HCl equaled the concentration of H+ so finding the pH was easy.
However, for acetic acid, [HC2H3O2] does not equal [H+] because not all of the HC2H3O2 dissociates
into H+ ions. To find the pH of a solution of acetic acid, you first must calculate the [H+] using the
Ka. This will be an equilibrium problem! Once you know [H+], you calculate pH using pH = log[H+].
Critical Thinking Questions
10. What is the pH of a solution that was 0.75 M acetic acid? (Note: 0.75 M is the initial
concentration of acetic acid.)
a) Find the equilibrium concentration of H+ using the equilibrium constant, Ka, as shown
below. Recall that [C2H3O2-] and [H+] will equal “x” and [HC2H3O2] will equal 0.75x.
[C H O - ][H + ]
Ka = 2 3 2
[HC 2 H 3O 2 ]
1.7 x10
−5
[x][ x]
x2
−5
=
→ 1.7 x10 =
→ x = 0.036 = [ H + ]
[0.75 - x ]
0.75
b) Now that you know the equilibrium concentration of H+, calculate the pH.
pH = -log(0.036) = 2.45
153
11. Find the pH of a 0.85 M solution of hydrocyanic acid, HCN, whose Ka = 4.9x10-10.
pH = 4.69
[CN-][H+]
Ka =
(x)(x)
=
[HCN]
4.9x10
-10
(0.85-x)
x2
=
x = 2.04x10-5 = [H+]
0.85
pH = -log(2.04x10-5) = 4.69
12. Find the pH of a 0.5 M solution of ammonia, NH3, whose Kb is 1.8x10-5.
pH = 11.48
[NH4+][OH-]
Kb =
1.8x10-5 =
[NH3]
(x)(x)
x = 0.003 M = [OH-]
(0.5 – x)
pOH = -log(0.003) = 2.52
pH = 14 – 2.52 = 11.48
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154
ChemQuest 51
Name: ____________________________
Date: _______________
Hour: _____
Information: Dilutions
When water is added to a solution, the concentration decreases. It is often desirable to be able to
calculate the concentration of solutions that have been diluted. For doing this, keep in mind that
molarity is equal to the moles of solute divided by the total liters of solution. Therefore, the
following equations are valid where M is molarity, Lsolution is liters of solution and molsolute is the
moles of solute:
Equation #1:
Equation #2:
molsolute
M=
molsolute = (M)(Lsolution)
Lsolution
Critical Thinking Questions
For the following questions, assume that liquid volumes are additive.
1. A certain solution is prepared by dissolving 4.0 moles of salt (NaCl) in enough water to make 400
mL of solution. Later, the solution was diluted with enough water so that the volume of the
solution was 650 mL. Calculate the molarity of the solution before and after dilution.
Mbefore = 4.0mol ÷ 0.400L = 10.0 M
Mafter = 4.0mol ÷ 0.650L = 6.15 M
2. A 6.0 M solution of salt has a volume of 500 mL. Later, 275 mL of water is added. Confirm that
the molarity of the resulting is approximately 3.87 M. (Hint: first find the moles of salt present
before the additional 275 mL of water was added by using equation #2 and then find the new
molarity using equation #1.)
moles of salt = 6.0M x 0.5L = 3.0mol
M = 3.0mol ÷ (0.500 + 0.275)L = 3.87 M
3. Calculate the molarity of the solution formed by taking 350 mL of 2.25 M HCl and adding 420
mL of water.
mol solute = 2.25M x 0.350L = 0.788 mol
M = 0.788mol ÷ (0.350 + 0.420) = 1.02 M
4. Imagine that you have 300 mL of a stock solution of 2.8 M HCl solution. Describe how I could
prepare 50 mL of 1.2 M HCl solution by using some of stock solution and diluting it with water.
Be specific.
mol HCl needed = 1.2M x 0.050L = 0.060 mol
(2.8M)(xL) = 0.060 x = 0.0214 L = 21.4 mL
Take 21.4 mL of the stock solution and dilute it to 50 mL.
155
Information: Mixing Strong Acids and Strong Bases
Usually when we speak of “salt” we mean table salt, which is sodium chloride (NaCl). A salt is a
general term for an ionic compound formed when an acid and a base mix. Whenever an acid and a
base react, water and a salt are formed. For example consider the following reactions in which nitric
acid (HNO3) and hydrochloric acid (HCl) react with the base sodium hydroxide (NaOH):
HCl + NaOH NaCl + H2O
HNO3 + NaOH NaNO3 + H2O
Notice that in each reaction water and a salt (sodium chloride one reaction and sodium nitrate in
another) were formed.
If equal moles of strong acid and strong base react, then they neutralize each other and form a
solution of salt water. If there are more moles of acid than base then the resulting solution will be
acidic. If there are more moles of base than acid, then the resulting solution will be basic.
Critical Thinking Questions
5. Consider the reaction of 2.5 moles of hydrochloric acid with 1.9 moles of sodium hydroxide.
a) If this reaction took place in 2.0 L of solution, what is the concentration of leftover
hydrochloric acid after the reaction?
mol leftover = 2.5 – 1.9 = 0.6 mol ÷ 2.0L = 0.3 M
b) From your answer to part a, verify that the pH of the solution after the reaction is
approximately 0.52.
pH = -log(0.3) = 0.52
6. Question 5 could be rewritten like this: Consider the reaction of 1.0 L of _2.5_ M hydrochloric
acid with 1.0 L of _1.9_ M sodium hydroxide. Fill in the blanks with the appropriate numbers
indicating the molarity.
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156
7. 320 mL of 3.1 M HCl is mixed with 240 mL of 4.1 M NaOH. Use the following steps to find the
pH of the resulting solution.
a) Calculate the moles of HCl and the moles of NaOH that are reacting using Equation #2.
mol HCl = (3.1M)(0.320L) = 0.992 mol
mol NaOH = (4.1M)(0.240L) = 0.984 mol
b) Find out which substance is left over and find out how many moles of this substance is left
over.
mol HCl leftover = 0.992 – 0.984 = 0.008 mol
c) Divide the moles left over by the total volume in liters to get the concentration of the left over
substance.
0.008 mol ÷ (0.320+0.240) = 0.014 M
d) Your answer to part c is also the concentration of H+ (if the acid is left over) or the
concentration of OH- (if the base is left over). From this information calculate the pH of the
[H+] = [HCl] = 0.014M
pH = -log(0.014) = 1.85
8. Calculate the pH of a solution formed by mixing 450 mL of 0.79 M HCl with 430 mL of 1.2 M
NaOH. Hint: this is very similar to question 7.
13.26
mol HCl = (0.79M)(0.450L) = 0.356mol; mol NaOH = (1.2M)(0.430L) = 0.516mol
mol NaOH leftover = 0.516 – 0.356 = 0.16 mol ÷ (0.45+0.43) = 0.182 M = [OH-]
pOH = -log(0.182) = 0.74 pH = 14 – 0.74 = 13.26
9. Calculate the pH of a solution formed by mixing 820 mL of 1.2 M HNO3 with 700 mL of 0.9 M
NaOH.
0.633
mol HNO3 = (1.2)(0.820) = 0.984 mol; mol NaOH = (0.9)(0.700) = 0.630 mol
mol HNO3 leftover = 0.984 – 0.630 = 0.354 mol ÷ (0.700+0.820) = 0.233 M = [H+]
pH = -log(.233) = 0.633
10. Consider 400 mL of a 2.5 M HCl solution. How many milliliters of 1.25 M NaOH will be needed
to neutralize the HCl?
800 mL
mol NaOH must equal mol HCl = (2.5M)(0.400L) = 1.00 mol
L = mol ÷ M = 1.00 ÷ 1.25 = 0.800 L 800 mL
157
ChemQuest 52
Name: ____________________________
Date: _______________
Hour: _____
Information: Review Definitions of Acids and Bases
Recall the definitions of acids and bases according to Bronsted-Lowry:
1) acid: substance that donates a proton, H+, in a reaction
2) base: substance that accepts a proton, H+, in a reaction
Recall also, that we have seen that it is possible for ions to act as acids or bases. We will now
examine the acidity or basicity of ions.
Critical Thinking Questions
1. If the chloride ion (Cl-) or the cyanide ion (CN-) were dissolved in water, would you expect the
ions act as acids or bases? Ask yourself: could it accept a H+? Does it have a H+ to give away?
Explain.
Since they don’t have a H+ to give away, we can guess that they would act like bases and
2. If the ammonium ion (NH4+) were dissolved in water, would you expect it to act as an acid or
base? Explain.
We can expect NH4+ to act like an acid since it appears to have at least one H+ to give away.
3. Although not all ions act as acids or bases, we can generalize from the above two questions that
IF an ion acts as an acid or a base the following rules will be observed:
a) When in water, positive ions may act like ____acids_______.
b) When in water, negative ions may act like ____bases_______.
Information: Determining Whether an Ion Affects the pH
It was mentioned above that not all ions will act as an acid or a base. Some, in fact, will not affect
the pH of a solution at all. We are now going to look at what determines whether an ion affects the
pH or not.
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Consider the chloride ion. If it affects the pH, it will react with water in the following way:
Equation 1: Cl- + H2O  HCl + OHNow consider the cyanide ion. If it affects the pH, it will react with water in the following way:
Equation 2: CN- + H2O  HCN + OHHopefully, these equations confirm what you wrote for the questions above: negative ions, if they
affect the pH, will act like bases because they produce OH- ions. In fact, the cyanide ion (CN-) is the
conjugate base of the weak acid HCN. The base ionization constants (Kb) for equations 1 and 2 are
shown below:
K b1 =
[ HCl ][OH − ]
[Cl − ]
K b2 =
[ HCN ][OH − ]
[CN − ]
Critical Thinking Questions
4. The expression for Kb1 assumes that HCl exists in the solution without breaking into ions
completely. Similarly, the expression for Kb2 assumes that HCN exists in solution without
breaking into ions completely. However, one of these—HCl or HCN—completely breaks up into
ions when dissolved. Which substance breaks up completely into ions? Explain.
HCl completely breaks up into ions and exists as H+ and Cl- in water. We know this because
HCl is a strong acid.
5. One of the ions—Cl- or CN-—does not affect the pH because it will not react as depicted in
equation 1 or 2 above. Given your answer to question 4, which ion do you think will not affect
the pH of a solution?
Cl- will not affect the pH because it will not accept a H+ to form HCl.
6. Complete the following sentence: The negative ions formed from a _____strong________ acid
strong/weak
will not affect the pH, but the negative ion formed from a _____weak_______ acid will raise the
strong/weak
pH and act like a(n) _____base________.
acid/base
7. Similar reasoning applies to the positive ions formed from strong and weak bases. Using
question six as a pattern, complete the following sentence:
The positive ions formed from a ______strong______ base will not affect the pH, but the
strong/weak
positive ion formed from a ____weak______ base will lower the pH and act like a(n)
strong/weak
_____acid______.
acid/base
159
8. Determine whether each of the following will act like an acid (A) or a base (B). If the ion will
not affect the pH, place an X in the blank.
__X__ a) Na+ (from the strong base NaOH)
__ X__ b) NO3- (from the strong acid HNO3)
__A__ c) NH4+ (from the weak base NH3)
__ X__ d) K+ (from the strong base KOH)
__B__ e) NO2- (from the weak acid HNO2)
__ B_ f) C2H3O2- (from the weak acid HC2H3O2)
9. Write the chemical equation for acetic acid dissociating.
HC2H3O2  H+ + C2H3O2- or HC2H3O2 + H2O  H3O+ + C2H3O210. From the Ka of acetic acid (1.7x10-5), calculate the Kb for the acetate ion.
Kb = 1x10-14÷1.7x10-5 = 5.88x10-10
11. When ammonia (NH3) reacts with water, the ammonium ion is formed (NH4+). Write the
chemical equation for this process.
NH3 + H2O  NH4+ + OH12. Given the Kb for ammonia (1.8x10-5), calculate the Ka for the ammonium ion.
Ka = 1x10-14÷1.8x10-5 = 5.56x10-10
Information: Salts
Salts are ionic compounds. They are similar to strong acids and bases because they dissociate
completely in water. However, not all salts affect the pH. Consider the salt sodium acetate
(NaC2H3O2). If you place 2.0 moles of sodium acetate in 1.0 L of water, it will dissociate completely
as follows:
NaC2H3O2 Na+ + C2H3O2Each mole of NaC2H3O2 breaks up into a mole of Na+ and a mole of C2H3O2-. Therefore, because
you started out with 2.0 moles of NaC2H3O2 in 1.0 L of water, the concentration of Na+ is 2.0 M and
the concentration of C2H3O2- is 2.0 M. We have seen already that Na+ will not affect the pH since it
is derived from the strong base NaOH. C2H3O2- (derived from the weak acid HC2H3O2) will affect
the pH by producing OH- ions: C2H3O2- + H2O  HC2H3O2 + OH-.
Critical Thinking Questions
13. Determine whether each of the following salts will act like an acid (A) or a base (B). If the salt
will not affect the pH, place an X in the blank.
__X__ a) NaCl
__A__ b) NH4Cl
__B__ c) NaCN
__X__ d) KNO3
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14. What is the pH of a 0.45 M solution of NaC2H3O2? (Follow the following steps.)
a) Will Na+ or will C2H3O2- affect the pH?
C2H3O2b) Hopefully your answer to part a is C2H3O2-. Find the concentration of C2H3O2- in the
solution. (Note: remember that salts dissociate completely.)
[C2H3O2-] = [NaC2H3O2] = 0.45 M
c) Now that you know [C2H3O2-]initial complete the necessary equilibrium calculations
using the Kb for C2H3O2- (see question 10) to calculate [OH-].
[HC2H3O2][OH-]
(x)(x)
x2
-10
Kb =5.88x10 =
=
=
x = 1.63x10-5 M
C2H3O2(0.45 – x)
0.45
d) Calculate the pOH of the solution.
pOH = -log(1.63x10-5) = 4.79
e) From the pOH, verify that the pH is 9.2.
pH = 14 – 4.79 = 9.21 ~ 9.2
15. Find the pH of a 0.32 M solution of NH4NO3.
4.88
NH4+ affects pH: NH4+  H+ + NH3 ; Ka = 5.56x10-10 (see question 12)
[H+][NH3]
(x)(x)
x2
-10
Ka = 5.56x10 =
=
=
x = 1.33x10-5 = [H+]
[NH4+]
0.32-x
0.32
pH = -log(1.33x10-5) = 4.88
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ChemQuest 53
Name: ____________________________
Date: _______________
Hour: _____
Information: Internal Energy
Thermodynamics involves the study of the energy and disorder of a system. Every system has
“internal energy”. Internal energy is the total amount of all the potential and kinetic energy that a
system has. For an example of a “system” let’s take a car. A car has kinetic energy (the energy of
motion) if it is moving and it has potential energy (chemical energy of the gasoline) if it has gas in
the tank. A car always has a certain amount of energy associated with it. A car has more total energy
(chemical potential energy) with a full tank of gas than with a half full tank. Therefore, after driving
a car for a while the total amount of energy (called the internal energy) that the car has decreases.
Critical Thinking Questions
1. If we use KE to symbolize kinetic energy and PE to symbolize potential energy, and U to
symbolize internal energy, then write an equation for internal energy.
U = KE + PE
2. Consider a car as described in the table below. Fill in the blanks in the table with the hypothetical
values for kinetic energy, potential energy, and internal energy. (Assume that the overall mass of
the car stays the same even though the amount of gasoline in the tank decreases.)
Kinetic
Energy
Potential
Energy
Internal
Energy
A car is setting in
the driveway
trip. The gas tank
is full.
0
The car has set out on
the trip and has been
driving for a while.
Cruise control is set
for a constant speed.
20
100
100
The car is still
driving at the
same speed.
The car ran out
of gas and is
parked along
the highway.
20
0
70
30
0
90
50
0
3. The Law of the Conservation of Energy states that energy isn’t created or destroyed. The car in
question 2, lost energy. If the energy was not destroyed, hypothesize what happened to it.
The energy changed forms into sound energy, friction, heat, etc. and was lost to the
surroundings.
4. What would be the internal energy of the car in question 2 if someone brought enough gas to fill
the tank back up completely while the car was stranded at the side of the highway?
100
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162
Information: First Law of Thermodynamics
As you saw in question 2 above, the internal energy of a system can change. What happens to the
energy? The energy merely changes form. In the car example, the internal energy was changed into
heat energy as the gasoline burned. Also, some of the internal energy was used to do work by
moving the car. The First Law of Thermodynamics states that the change in internal energy of a
system equals the sum of the heat and the work. Internal energy of a system can increase or decrease.
If a system is heated, it gains internal energy, but if the system loses heat then it decreases in internal
energy. If a system does work it loses internal energy, but if work is done on the system then it gains
internal energy.
Critical Thinking Questions
5. For each of the following, state what the “system” is and also state whether the internal energy of
the system would increase or decrease in each situation.
a) Lifting a ball from the ground and holding it six feet off the ground.
System: ball; internal energy increases
b) Heating up a piece of pizza in the microwave.
System: pizza; internal energy increases
c) The logs in the fireplace are burning.
System: logs; internal energy decreases
d) An ice cube is melting while it sets on the kitchen counter.
System: ice cube; internal energy increases
e) Two molecules bond together and heat energy is released (exothermic).
System: molecules; internal energy decreases
Information: Enthalpy
Heat energy is a large factor in how much the internal energy of a system changes. We will
now turn our attention to chemical reactions. In your answer to question 5e above you should have
identified the two molecules as the “system” and you should have noted that their overall internal
energy decreased since they lost heat. Verify that this is correct. The heat energy involved in a
chemical reaction is given the name “enthalpy”. The change in heat energy for a reaction is called
the change in enthalpy and it is given the symbol ∆H and it has units of kilojoules (kJ). An example
of 5e taking place is when one mole of hydrogen molecules reacts with ½ mole of oxygen molecules:
Example Reaction 1: H2 (g) + ½ O2 (g) H2O (g) ; ∆Hf = -241.8 kJ
In the following reaction, energy is absorbed by the system (endothermic) instead of being released.
Example Reaction 2: C (s) + 2 S (s) CS2 (g) ; ∆Hf = 117 kJ
Critical Thinking Questions
6. Give the definition for the following terms.
a) endothermic: when energy is absorbed by a system
b) exothermic: when energy is released by a system
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7. In the above examples, Example Reaction 1 ∆H is negative and in Example Reaction 2 ∆H is
positive. What does the sign on ∆H tell you?
If ∆H is negative, the reaction is exothermic because heat is subtracted from the system. If
∆H is positive, the reaction is endothermic because heat is added to the system.
8. In Example Reaction 1, compare the amount of energy that the product has compared to the
reactants.
The products have less energy because energy was “lost” during the reaction in the form of
heat.
9. In Example Reaction 2, compare the amount of energy that the product has compared to the
reactants.
The products have more energy because energy was “gained” during the reaction in the form
of heat.
Information: Enthalpy of Formation
In Example Reaction 1 and 2, one substance was formed out of two. Whenever one substance is
made from elements in their standard states, the accompanying enthalpy change is called the enthalpy
of formation, given the symbol ∆Hf. Note the following two example reactions:
Example Reaction 3: H2 (g) + S (s) H2S (g); ∆Hf = -20 kJ
Example Reaction 4: C (s) + O2 (g) CO2 (g); ∆Hf = -393.5 kJ
Notice that in the previous 4 example reactions, each substance formed is involved in the following
reaction:
Example Reaction 5: 2 H2O (g) + CS2 (g) 2 H2S (g) + CO2 (g); ∆H = ?
The ∆H for Example Reaction 5 can be calculated from the data for Example Reactions 1-4 as
follows:
∆H = [2(-20 kJ) + (-393.5 kJ)] – [2(-241.8 kJ) + (117 kJ)] = -66.9 kJ
Critical Thinking Questions
10. In Example Reactions 1-4, the enthalpy value was given the symbol ∆Hf, but in Example
Reaction 5, the enthalpy value was given the symbol ∆H. Why the difference?
In reactions 1-4 one substance is made from several substances, which is called a formation
reaction.
11. In calculating the ∆H for Example Reaction 5, why was –20kJ multiplied by 2? Why was the –
241.8kJ multiplied by 2?
Forming H2O releases 20kJ (see Example Reaction 3) and since there are two H2O molecules
present in Example Reaction 5, the -20kJ is multiplied by two.
12. Using a table for standard enthalpies of formation (∆Hf), verify that ∆H for the following reaction
is approximately –91.4 kJ.
2 NaF + CaCl2 CaF2 + 2 NaCl
-575.4 kJ -795 kJ
-1215 kJ
-411.1 kJ
∆H = [2(-411.1) + (-1215)] – [2(-575.4) + (-795)] = -91.4 kJ
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13. Again using a table of standard enthalpy values calculate ∆H for each of the following reactions.
a) LiCl + NaBr NaCl + LiBr
-408 kJ
-361 kJ
-411.1 kJ -351 kJ
∆H = [-411.1 + (-351)] – [-361 + (-408)] = 6.9 kJ
b) MgCO3 + 2 NaCl Na2CO3 + MgCl2
-1112 kJ
-411.1 kJ
-1130.8 kJ
-641.6 kJ
∆H = [-641.6 + (-1130.8)] – [2(-411.1) + (-1112)] = 161.8 kJ
c) CH4 + Cl2 HCl (g) + CH3Cl
-74.87 kJ 0 kJ
92.31 kJ
-83.7 kJ
∆H = [-83.7 + 92.31] – [-74.87 + 0] = -83.48
14. In 13c, you should have noted that ∆Hf for Cl2 is zero. ∆Hf is also zero for O2, Na (s), Mg (s), and
other substances. Why do you think ∆Hf is zero for these substances?
Cl2, O2, Na, and Mg are in their natural state. We do not speak of forming a sodium atom, for
example, because the sodium atom is already formed.
15. Compare the ∆Hf for O2 (g) and for O (g). Why is ∆Hf equal to zero for O2, but not for O?
O2 is its natural state because oxygen exists as diatomic molecule. Monatomic oxygen,
however, must be formed.
16. Calculate ∆H for the following reaction using a table for standard enthalpy values:
2CO(g)+O2(g)2CO2(g).
∆Hf for CO = -110.5 kJ; ∆Hf for O2 = 0 kJ; ∆Hf for CO2 = -393.5 kJ
∆Hrxn = 2(-393.5) – [2(-110.5)] = -566 kJ
17. Consider the following reactions and ∆H values
C + O2 CO2 ; ∆Hf = -393.5 kJ
2 CO CO2 + C ; ∆H = -172.5
a) Using only the above two ∆H values, how can you calculate ∆H for the reaction in
question 16?
b) What is the relationship between the two reactions above and the reaction in question 16?
If you add the two reactions together you get the reaction from question 16.
18. In question 17a, you used Hess’s Law to find the ∆H for a reaction even though you didn’t know
what Hess’s Law is. In your own words, state what you believe is Hess’s Law.
When a reaction is formed by adding several reactions together, then the enthalpy change can
be found by adding the enthalpy changes for the individual reactions.
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19. Consider the following reaction: N2 + 2 O2 + 2 NO 2 NO + 2 NO2. Use Hess’s Law along
with the following information to find the ∆H for this reaction.
reaction #1 N2 + O2 2 NO ; ∆H = 181 kJ
reaction #2 2 NO + O2 2 NO2 ; ∆H = -113 kJ
Since the reaction is formed by adding reaction #1 and reaction #2, the enthalpy change for
the reaction can be found by adding the enthalpy change for reactions #1 and #2.
∆Hrxn = 181 + (-113) = 68 kJ
20. Consider the following reaction: 2 NO N2 + O2 ; ∆H = -181 kJ. What relationship exists
between this reaction and reaction #1 in question 19? How could the ∆H for this reaction be
determined?
The reaction is the reverse of reaction #1 from question 19. Thus, the ∆H value is also
reversed in sign.
21. Using only the information presented in question 12, find the ∆H for the following reaction.
CaF2 + 2 NaCl 2 NaF + CaCl2
This reaction is the reverse of the reaction in question 12, so we reverse the sign of the
enthalpy change from question 12 to get: 91.4 kJ
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166
ChemQuest 54
Name: ____________________________
Date: _______________
Hour: _____
Information: Spontaneity
It is often desirable to know if a process will occur all by itself or if you need to supply energy for a
process to occur. If a process occurs naturally, all by itself without outside help, it is said to be
spontaneous. Examples of spontaneous processes include iron rusting and a ball rolling down a hill.
Critical Thinking Questions
1. Which of the following processes are spontaneous?
a) a rock rolling uphill
b) a leaf falling
c) water boiling at 100oC
Information: Entropy
It is important to consider the amount of disorder in a system when determining whether a reaction
will occur spontaneously. Just as the energy change is given a special name and symbol (change in
enthalpy, ∆H), the change in the amount disorder or randomness is also given a special name,
“change in entropy”, which has the symbol ∆S. Entropy, S, is a measure of the amount of disorder in
a system. When disorder increases, ∆S is positive.
Critical Thinking Questions
2. When ∆Hreaction is negative that means that the reaction loses energy to become less energetic.
When ∆Sreaction is negative, what is that telling us?
The amount of disorder (entropy) of a system decreases when ∆Sreaction is negative.
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3. Does disorder (entropy) increase or decrease for the formation of gaseous water?
2 H2 (g) + O2 (g) 2 H2O (g)
Note the diagrams below in which every molecule is moving randomly. Explain your answer
using the diagrams.
= O2 molecule
= H2O molecule
= H2 molecule
Entropy (the amount of disorder) decreases when water is formed. As can be seen in the
above diagrams, the H2O has fewer individual molecules that are moving randomly.
4. Indicate whether for each of the following processes ∆S will be positive or negative? Explain
each answer. (Hint: think in terms of the disorder or randomness of molecules and molecular
motion.)
a) an ice cube melting
∆S will be positive; entropy increases because the liquid molecules are not organized into a
crystalline solid after it melts.
b) water vaporizing
∆S will be positive; gases are more disorderly than liquids.
∆S will be negative; cleaning a room involves organizing the room, thus decreasing the
disorder.
d) folding paper into a paper airplane
∆S will be negative; the paper becomes more ordered by folding it in an ordered manner.
Information: Comparing Entropy and Enthalpy
It was once thought that for a chemical reaction to be spontaneous that it also had to be exothermic
(∆H<0). Now, however, we recognize that some endothermic reactions are spontaneous. When
deciding whether a reaction is spontaneous, you need to consider both entropy and enthalpy. Note
that for any reaction you will have the reaction itself and its surroundings. Both the reaction and its
surroundings together are called the universe so that ∆Suniverse = ∆Sreaction + ∆Ssurroundings. Consider the
following table:
sign of
Sign of
Sign of
Sign of
Spontaneous?
Reaction
∆Hreaction
∆Sreaction
∆Ssurroundings
∆Suniverse
#1 exothermic,
+
+
+
yes
entropy increase
#2 endothermic,
+
+
+ or depends
entropy increase
#3 exothermic,
+
+ or depends
entropy decrease
#4 endothermic,
+
no
entropy decrease
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168
Critical Thinking Questions
5. For reaction #2, according to the table, it says that there is an entropy increase. What entropy is it
referring to—∆Sreaction, ∆Ssurroundings, or ∆Suniverse?
∆Sreaction
6. Why can ∆Suniverse be + or – for reactions two or three?
∆Suniverse depends on both ∆Sreaction and ∆Ssurroundings, so whichever one is larger in magnitude
determines the sign on ∆Suniverse.
7. What is the correlation between the sign of ∆Hreaction and the sign of ∆Ssurroundings?
They always have opposite signs.
8. The sign of ∆Hreaction determines the sign of ∆Ssurroundings. Explain why this is true. (Hint: think
about what happens to the motion of molecules in the surroundings when ∆Hreaction is positive and
when it is negative.)
If more heat is sent into the surroundings ( negative ∆Hreaction) then the molecules in the
surrounding will move faster and more chaotically.
9. On the table above, the sign of one value determines whether the reactions are spontaneous or
not. Which value’s sign determines the spontaneity of a reaction?
∆Suniverse must be positive for a reaction to be spontaneous.
10. Use your answer to question 9 to fill in the blanks. This is a statement of the Second Law of
Thermodynamics!
For any spontaneous reaction, the amount of _____entropy______ in the universe must
enthalpy or entropy
____increase______.
increase or decrease
11. In the above information section, the statement was made that “When deciding whether a reaction
is spontaneous, you need to consider both entropy and enthalpy.” If only one value (answer to
question 9) is needed to determine spontaneity, then why do you still need to consider the
enthalpy?
The enthalpy helps to determine ∆Ssurroundings and the ∆Ssurroundings helps to determine the
∆Suniverse.
Information: Predicting and Calculating Entropy Changes
Consider question 3 along with the diagrams of oxygen, hydrogen and water molecules.
2 H2 (g) + O2 (g) 2 H2O (g)
When the two gaseous reactants (oxygen and hydrogen) combined to form gaseous water, hopefully
you concluded that entropy decreased. Notice that in the balanced chemical equation, there are a
total of three moles of gaseous reactants (2 H2 + O2) and only two moles of gaseous products (2H2O).
In general, then, whenever the number of moles of gaseous products is less than the number of moles
of gaseous reactants entropy will decrease. If there are more moles of gaseous products than gaseous
reactants, entropy has increased.
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Critical Thinking Questions
12. Explain why ∆S is negative for the following reaction.
2 Ca (s) + O2 (g) 2 CaO (s)
The disorder decreases because the moles of gas decreases and the gas state is the most
disorderly state of matter.
13. For each of the following reactions, predict if the entropy change is positive or negative. If you
cannot tell, explain why.
A) C2H2 (g) + 2 H2 (g) C2H6 (g)
Negative; the total moles of gas is decreasing (moles of gaseous product is less than the moles
of gaseous reactants).
B) CH3NH2 (g) HCN (g) + 2 H2 (g)
Positive; the total moles of gas is increased.
C) N2 (g) + O2 (g) 2 NO (g)
Since the moles of gas are equal on both sides, we cannot predict this one.
14. Entropy values can be calculated in much the same was as enthalpy values. Using a table of
standard entropies, calculate the entropy change for the following reactions.
A) N2 (g) + O2 (g) 2 NO (g)
191.5 J
205.0 J
210.65 J
∆S = 210.65 – [205.0 + 191.5] = -185.85 J
B) CS2 (g) + 4 H2 (g) CH4 (g) + 2 H2S (g)
186.1 J
205.6 J
237.79 J 130.6 J
∆S = [2(205.6) + 186.1] – [4(130.6) + 237.79] = -162.89 J
C) 2 H2 (g) + O2 (g) 2 H2O (g)
205.0 J
188.72 J
130.6 J
∆S = [2(188.72)] – [2(130.6 + 205.0] = -88.76 J
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170
ChemQuest 55
Name: ____________________________
Date: _______________
Hour: _____
Information: Free Energy
So far in our study of thermodynamics, we have seen that in order to determine if a reaction is
spontaneous we must examine the entropy change (∆S) and the enthalpy change (∆H) associated with
the reaction. A single quantity that combines both of these ideas is called the Gibbs Free Energy,
which is given the symbol G. The change in free energy is related to the change in ethalpy for a
reaction and the change in entropy for a reaction according to the following equation:
∆G = ∆H - T∆S
• T is the temperature in units of Kelvin
• ∆H is ∆Hreaction (in Joules or kilojoules)
• ∆S is ∆Sreaction (in Joules or kilojoules)
• Note: ∆Hreaction and ∆Sreaction must either both be Joules or both kilojoules
Recall the following table:
Reaction
#1 exothermic,
entropy increase
#2 endothermic,
entropy increase
#3 exothermic,
entropy decrease
#4 endothermic,
entropy decrease
sign of
∆Hreaction
-
Sign of
∆Sreaction
+
Sign of
∆Ssurroundings
+
Sign of
∆Suniverse
+
Spontaneous?
+
+
-
+ or -
depends
-
-
+
+ or -
depends
+
-
-
-
no
yes
Critical Thinking Questions
1. What units must ∆S have if it is to be used in the above equation for free energy?
∆S must have units of kJ/K because T is in units of Kelvin and ∆H is in units of kJ.
2. For a spontaneous reaction, what is the sign of ∆G—positive or negative?
Negative
3. For a nonspontaneous reaction, what is the sign of ∆G—positive or negative?
Positive
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4. In the table above, under “Spontaneous” for reactions #2 and #3 it says, “depends”. What does
this mean? In other words, what does the spontaneity depend on in reaction #2 and reaction #3?
It depends on the magnitude of ∆S and ∆H. ∆G must be a negative value.
5. You have seen that ∆Hreaction can be calculated by the following:
∆Hreaction = [sum of ∆Hf of products] – [sum of ∆Hf of reactants]
The ∆S and ∆G for reactions can be calculated in the same way using values from a table of
standard values. Use the information in a table of standard values and the following balanced
equation to prove that ∆G = ∆H - T∆S. Pay special attention to units and note that each value in
the table was determined at 25oC (298 K).
CaCO3 (s) CaO (s) + CO2 (g)
∆H = -1206.9 kJ
-635.1 kJ
-393.5 kJ
∆Hrxn = 178.3 kJ
38.2 J
213.7 J
∆Srxn = 159 J
∆S = 92.9 J
∆G = -1128.8 kJ
-603.5 kJ
-394.4 kJ ∆Grxn = 130.9 kJ
proof
∆G = ∆H - T∆S = 178.3 – 298(0.159) = 130.9 kJ
6. Is the reaction from question 5 spontaneous at 25oC? Explain.
No, ∆G is a positive number.
7. By varying the temperature, is it possible to change the spontaneity of the reaction from question
5? Explain.
Yes, if the temperature is large enough then T∆S would be larger than ∆H and ∆G would
therefore be negative.
Information: Coupling Reactions
Sometimes in industry we need to carry out nonspontaneous reactions. An example of a
nonspontaneous reaction is isolating iron from iron ore. The equation and free energy are given
below:
2 Fe2O3 (g) 4 Fe (s) + 3 O2 (g) ; ∆G = 1487 kJ
To get this reaction to proceed spontaneously, you can “couple” it with another reaction. One such
reaction would be the conversion of carbon monoxide to carbon dioxide. This equation and its free
energy is given below:
6 CO (g) + 3 O2 (g) 6 CO2 (g) ; ∆G = -1543 kJ
By “coupling” the reactions, we essentially carry out both reactions at the same time and use the
spontaneous reaction to drive the nonspontaneous one. The overall ∆G for the process then becomes
-56 kJ (found by adding -1543 + 1487). The overall final balanced equation is then:
2 Fe2O3 (g) + 6 CO (g) 4 Fe (s) + 6 CO2 (g) ; ∆G = -56 kJ
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Critical Thinking Questions
8. In the overall final balanced equation above, why isn’t O2 written?
O2 appears on both sides of the reaction so it is produced and then used up so there is no need
to write it in the overall equation.
9. Consider the following reaction to obtain pure aluminum from aluminum ore:
Al2O3 (s) + 2 Fe (s) Fe2O3 (s) 2 Al (s)
Which of the following two reactions would prove beneficial when coupled with the above
reaction?
a) CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
OR
b) 2 CH3OH (l) + 3 O2 (g) 2 CO2 (g) + 4 H2O (l)
First, calculate the ∆G value for the aluminum reaction:
Al2O3 (s) + 2 Fe (s) Fe2O3 (s) 2 Al (s); ∆G = 838 kJ
We need to find out which reaction (a or b) has a ∆G greater than -838 kJ so that when the
reactions are added, the overall reaction will have a negative ∆G and be spontaneous.
a. CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g); ∆G = -800.8 kJ
b. 2 CH3OH (l) + 3 O2 (g) 2 CO2 (g) + 4 H2O (l); ∆G = -1404.9 kJ
Reaction b should be coupled with the aluminum ore reaction so that the overall ∆G will be
negative.
10. Given your answer to question 9, write the overall equation (after coupling) and calculate the ∆G
for this overall reaction.
Al2O3 (s) + 2 Fe (s) + 2 CH3OH (l) + 3 O2 (g) Fe2O3 (s) 2 Al (s) + 2 CO2 (g) + 4 H2O (l)
∆G = 838 + (-1404.9) = -566.9 kJ
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