Download Chapter 2: Atoms and Elements

Chapter 2: Atoms and Elements
25
Chapter 2: Atoms and Elements
Teaching for Conceptual Understanding
The terms atom, molecule, element, and compound can be confusing to students and this confusion can persist
beyond the introductory chemistry level. It is important to introduce each term clearly and to use many examples
and non-examples. Ask students to draw particulate level diagrams of 1) atoms of an element, 2) molecules of an
element, 3) atoms of a compound (NOT possible), and 4) molecules of a compound.
The idea of thinking about matter on three levels (macroscopic, particulate, and symbolic) introduced in chapter
1 can be reinforced now by bringing to class a variety of element samples. When presenting each element, write
the symbol, draw a particulate representation in its physical state at room temperature, and show a ball-and-stick
or space-filled model.
The mole concept is one of the most important and most difficult concepts in a first course of introductory
chemistry. Because Avogadro’s number is so large, it is impossible to show students a mole quantity of
anything with distinct units that they can see and touch. Give numerous examples of mole quantities or have
students think up some themselves, e.g., the Pacific Ocean holds one mole of teaspoons of water, or the entire
state of Pennsylvania would be a foot deep in peas if one mole of peas were spread out over the entire state.
Many students have the misconception that a mole is a unit of mass; so, be on guard for this.
Suggestions for Effective Learning
Students will remember the seven diatomic elements if you point out to them that the elements H, N, O, F, Cl, Br,
and I form the number 7 in the periodic table. Another learning strategy is the mnemonic: I Bring Clay For Our
New House.
Metric relationships can be confusing because they can be represented two ways, e.g., 1 m = 100 cm or 0.01 m =
1 cm. Research has shown that students are more successful when decimal quantities are not used. It is easier
for them to visualize a large number of small units equaling a larger one than to visualize what fraction of a large
unit equals a smaller unit.
When doing mole calculations, many students get answers that are off by a power of ten. Usually the problem
can be traced back to an error in a calculator entry. Avogadro’s number is erroneously entered as: 6.022,
24
multiplication key, 10, exponent key, 23, which results in the value 6.022 x 10 . Take the time to teach students
how to correctly enter exponential numbers into their calculators. Stokes Publishing Company caries an
inexpensive overhead projector calculator that works well for this type of instruction.
In addition to showing representative samples of the elements, demonstrate physical properties, e.g., sublimation
of iodine, to link back to material in Chapter 1 and chemical changes, e.g., Li, Na, and K reacting with water, to
link to future material.
If you plan on having students learn the names and symbols for a set number of elements you will use most
often in your course, consider giving them a blank periodic table and having them do the exercise below. The
psychomotor activity will help them better remember the characteristics of these elements and their location in
the periodic table.
1.
In the upper-right hand corner of each block, write the atomic number for all the elements.
2.
Write in the center of each block the symbol for the elements you are responsible for knowing. Below
each symbol write the atomic mass (to two decimal places) of the element.
Chapter 2: Atoms and Elements
26
3.
Write the group number above each column and the periodic number beside each row (don’t forget the
inner transition elements).
4.
The elements are sometimes classified as metals, nonmetal, and metalloids. Draw a line through the
periodic table separating the metals from the nonmetals. Circle the atomic number of those elements
referred to as metalloids.
5.
Using a colored pen (pencil or crayon), outline the blocks of the two elements which are liquids at room
temperature. Use a different colored pen to outline the blocks of the eleven elements which are gases at
room temperature.
6.
Label the groups known as the: halogens, noble gases, alkali metals, and alkaline earth metals.
7.
Label the sections of the periodic table referred to as the: main-group elements, transition elements, and
the inner-transition elements.
Cooperative Learning Activities
•
Have students list elements (not compounds) they interact with each day.
•
Questions for Review and Thought: 86, 90, 100.
•
Concept map terms: atom, atomic number, atomic structure, atomic weight, Avogadro’s number,
electron, isotopes, mass number, metal, metalloid, molar mass, mole, molecule, neutron, nonmetal,
proton.
Chapter 2: Atoms and Elements
27
Solutions for Chapter 2
Questions for Review and Thought
Review Questions
1.
The coulomb (C) is the fundamental unit of electrical charge.
2.
Millikan devised an experiment to determine the charge of an electron. It consisted of a chamber with
electrically charged plates on the top and bottom. (Figure 2.2). Tiny oil droplets were sprayed into the
chamber. As the droplets settled slowly through the chamber, they were exposed to x-rays. This caused
electrons from gas molecules in the air to be transferred to the oil droplets. Using a small telescope to
observe the tiny droplets, he then adjusted the electrical charge on the plates so that the upward pull of the
electrostatic charges just balanced the downward motion of the droplet due to gravity. From this, he was
able to calculate the charge on the droplets. Different droplets had different charges, but these charges
were all multiples of the same small number, which he decided must represent the smallest fundamental
negative charge: the charge of one electron. With the mass-to-charge ratio determined by Thomson, and
the charge of the electron, the mass of the electron could also be calculated.
3.
(a) The proton is about 1800 times heavier than the electron.
(b) The charge on the proton has the opposite sign of the charge of the electron.
4.
(a) The surprising results in Rutherford’s gold foil experiment were that some of the alpha particles were
deflected backwards. Rutherford compared this surprise to how you’d feel if you shot a cannon at a
piece of paper and had the paper deflect the cannon ball back at you!
(b) Rutherford calculated that the nucleus is about 10,000 times smaller than the atom.
5.
In a neutral atom, the number of protons is equal to the number of electrons.
Units and Unit Conversions
6.
Define the problem: If the nucleus were scaled to a diameter of 4 cm, determine the diameter of the atom.
Develop a plan: Find the accepted relationship between the size of the nucleus and the size of the atom.
Use size relationships to get the diameter of the “artificially large” atom.
Execute the plan: The atom is about 10,000 times bigger than the nucleus.
10,000 × 4 cm = 40,000 cm
Check your answer: A much larger nucleus means a much larger atom. This large atomic diameter result
looks right.
7.
Define the problem: A piece of paper is exactly 11 cm high. Use conversion factors to change the units to
centimeters, millimeters, and meters.
Develop a plan: Use the exact relationship between inches and centimeters to convert between in and cm.
Use metric relationship between cm and m, and m and mm to convert cm into mm. Use the metric
relationship between cm and m to convert cm into m.
Execute the plan:
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Chapter 2: Atoms and Elements
11 in ×
27.94 cm ×
2.54 cm
= 27.94 cm
1 in
1 m
1000 mm
×
= 279 .4 mm
100 cm
1 m
27.94 cm ×
1 m
= 0.2794 m
100 cm
Check your answers: The number of centimeters should be larger than the number of inches, since an inch
is larger than a centimeter. The number of millimeters should be larger than the number of centimeters, since
a millimeter is smaller than a centimeter. The number of meters should be smaller than the number of
centimeters, since a meter is larger than a centimeter. Looking at a pencil, it’s not hard to imagine that it’s 7.5
inches.
8.
Define the problem: The pole vault record is a height of 6.14 m. Use conversion factors to change the units
to centimeters, feet, and inches.
Develop a plan: Use the metric relationship between m and cm to convert m into cm. Then, use the metric
relationship between cm and inches to convert cm into inches. Then, use the relationship between inches
and feet to convert from inches to feet.
Execute the plan: If any one of these questions were asked separately, we would start with the given
information and apply the appropriate conversion factors.
6.14 m ×
6.14 m ×
6.14 m ×
100 cm
= 614 cm
1 m
100 cm
1 in
×
= 242 cm
1 m
2.54 cm
100 cm
1 in
1 ft
×
×
= 20.1 ft
1 m
2.54 cm 12 in
When answering all three questions, we can use the results of the previous calculation to make the next
calculation faster:
614 cm ×
242 in ×
1 in
= 242 in
2.54 cm
1 ft
= 20.1 ft
12 in
Check your answers: The number of centimeters should be larger than the number of meters, since
“centimeter” is a smaller unit of measure. The number of inches should be smaller than the number of
centimeters, since “inch” is a larger unit of measure. The number of feet should be smaller than the number
of inches, since “foot” is a larger unit of measure. These answers make sense.
9.
Define the problem: Given the speed in miles per hour, determine the speed in kilometers per hour.
Develop a plan: Use conversion factor between miles and kilometers to change the units from miles per
hour to kilometers per hour.
Execute the plan:
Chapter 2: Atoms and Elements
29
65 mi 1.6093 km
km
km
×
= 104 .6
≅ 1.0 × 102
1 hr
1 mi
hr
hr
Keep in mind that the original information has only two significant figures, so the result of these
multiplications and divisions must also have two significant figures.
Check your answer: Miles are longer than kilometers, so the number of kilometers should be larger than the
number of miles. This answer looks right.
10. Define the problem: Given the distance to the fence in feet, determine the distance in meters.
Develop a plan: Use conversion factors between feet and inches, then inches and centimeters, then
centimeters and meters to change the units from feet to meters.
404 ft ×
Execute the plan:
12 in 2.54 cm
1 m
×
×
= 123 m
1 ft
1 in
100 cm
Check your answer: Feet are shorter than meters, so the number of meters should be smaller than the
number of feet. This answer looks right.
11. Define the problem: Given the distance of a basketball hoop to the floor in feet, determine the distance in
centimeters and meters.
Develop a plan: Use conversion factors between feet and inches, then inches and centimeters, then
centimeters and meters to change the units from feet to centimeters and to meters.
10 ft ×
Execute the plan:
12 in 2.54 cm
1 m
×
×
= 3.048 m
1 ft
1 in
100 cm
10 ft ×
12 in 2.54 cm
×
= 304 .8 cm
1 ft
1 in
Check your answers: Feet are shorter than meters, so the number of meters should be smaller than the
number of feet. Feet are longer than centimeters, so the number of meters should be larger than the number
of feet. These answers looks right.
12. Define the problem: Given displacement volume of 120 in 3, determine the volume in cm3 and liters.
Develop a plan: Use conversion factor between inches and centimeters to change the units from cubic
inches to cubic centimeters. Then use conversion factors between cubic centimeters and milliliters, then
milliliters and liters to change the units from cubic centimeters to liters.
Execute the plan: NOTE: Each of the three “inch” units in “cubic inches” needs to be converted into
centimeters:
3
2.54 cm 2.54 cm 2.54 cm
3  2.54cm 
3
3
120 in ×
×
×
= 120 in × 
 = 2.0 × 10 cm
1 in
1 in
1 in
1in


3
3
3
2.0 × 10 cm ×
1 mL
1 cm 3
×
1 L
= 2.0 L
1000 mL
Notice: The conversion between liters and milliliters used here indicates a larger number (1000) of small
things (mL) equal to a smaller number (1) of large things (L). Conversely, you can use the conversion
–3
relating a smaller number (10 ) of larger things (L) to a larger number (1) of small things.
30
Chapter 2: Atoms and Elements
3
1 mL
3
2.0 × 10 cm ×
1 cm 3
×
−3
10 L
= 2.0 L
1 mL
Check your answer: Centimeters are smaller than inches, so a cubic centimeter is much smaller than a cubic
inch. It makes sense that the number of cubic centimeters is larger than the number of cubic inches. A liter
is larger than a cubic centimeter, so it makes sense that the number of liters is smaller than the number of
cubic inches. These answers look right.
13. Define the problem: Given displacement volume of 250. in 3, determine the volume in cm3 and liters.
Develop a plan: Use conversion factor between inches and centimeters to change the units from cubic
inches to cubic centimeters. Then use conversion factors between cubic centimeters and milliliters, then
milliliters and liters to change the units from cubic centimeters to liters.
Execute the plan:
3  2.54 cm 
250. in × 

 1 in 
 2.54 cm 
250. in × 

 1 in 
3
= 4.10 × 10
3
3
×
1 mL
1 cm 3
3
cm
3
1L
= 4.10 L
1000 mL
×
Notice: The conversion between liters and milliliters used here indicates that a larger number (1000) of
small things (mL) are equal to a smaller number (1) of large things (L). Conversely, you can use the
–3
conversion relating a smaller number (10 ) of larger things (L) to a larger number (1) of small things.
 2.54 cm 
250. in × 

 1 in 
3
3
×
1 mL
1 cm 3
×
−3
10
L
= 4.10 L
1 mL
Check your answers: Centimeters are smaller than inches, so a cubic centimeter is much smaller than a cubic
inch. It makes sense that the number of cubic centimeters is larger than the number of cubic inches. A liter
is larger than a cubic centimeter, so it makes sense that the number of liters is smaller than the number of
cubic inches. These answers look right.
14. Define the problem: Given one square meter, determine the number of square inches.
Develop a plan: Use the metric conversion between meters and centimeters, then use the relationship
between centimeters and meters.
2
Execute the plan:
2
 1 cm 
2  100 cm 
2
1 m ×
 ×
 = 1550 in
1
m
2.54
cm




Check your answer: A square meter is larger than a square inch, so it makes sense that the number of
square inches is larger.
15. Define the problem: Given one cubic inch, determine the number of cubic centimeters.
Develop a plan: Use the relationship between centimeters and inches.
Execute the plan:
3  2.54 cm 
1 in × 

 1 in 
3
= 16.387 cm
3
Chapter 2: Atoms and Elements
31
Check your answer: a cubic centimeter is smaller than a cubic inch, so it makes sense that the number of
square centimeters will be larger.
16. Define the problem: Given the length, width, and height of a room, determine the volume in metric units.
Develop a plan: Using the conversion factor between inches and feet, determine the height of the room in
feet. Use those lengths to calculate the volume in cubic feet. Use conversion factors between feet and
inches and between inches and centimeters to determine the volume in cubic centimeters. Use conversion
factors between cubic centimeters and milliliters and between milliliters and liters to determine the volume in
liters.

1 ft 
Execute the plan: height = 8 ft +  6 in ×
 = 8.5 ft
12 in 

3
V = (length) × (width) × (height)
(then convert to m )
3
3
 12 in 
 2.54 cm 
 1 m 
V = (18 ft) × (15 ft) × (8.5 ft)× 
 ×
 ×

 1 ft 
 1 in 
 100 cm 
3
= 65 m
3
3
1 mL
1L
3  100 cm 
4
65 m × 
×
= 6.5 × 10 L
 ×
1
m
3
1000
mL


1 cm
Check your answers: A cubic meter is larger than a liter, so the number of cubic meters should be smaller
than the number of liters. These answers make sense.
17. The symbols for the elements in this crystal are Ca (calcium) and F (fluorine).
Define the problem: Given the mass of the crystal as 2.83 grams, find the mass in kilograms and pounds.
Develop a plan: Using the conversion factor between grams and kilograms, determine the mass in kilograms.
Using the conversion factor between grams and pounds, determine the mass in pounds.
Execute the plan:
2.83 g ×
1 kg
= 0.00283 kg
1000 g
2.83 g ×
1 lb
= 0.00624 lb
453 .59 g
Check your answers: Pounds and kilograms are both larger units than grams, so it makes sense that the
number of kilograms and the number of pounds would be smaller than the number of grams.
Scanning Tunneling Microscope
18. It is true that the scanning tunneling microscope enables scientists to obtain images of individual atoms on
surface. This is described in some detail in “Tools of Chemistry” in Section 2.3 of the textbook on pages 46
and 47.
19. Flow is used to control the height adjustment of the probe tip, which is monitored to produce a
topographical map of the surface being investigated.
Significant Figures
32
Chapter 2: Atoms and Elements
20. Define the problem: Given several measured quantities, determine the number of significant figures.
Develop a plan: Use rules given in Section 2.4, summarized here: All non-zeros are significant. Zeros that
precede (sit to the left of) non-zeros are never significant (e.g., 0.003). Zeros trapped between non-zeros are
always significant (e.g., 3.003). Zeros that follow (sit to the right of non-zeros are (a) significant if a decimal
point is explicitly given (e.g., 3300) OR (b) not significant, if a decimal point is not specified (e.g., 3300.).
Execute the plan:
(a) 1374 kg has four significant figures (The 1, 3, 7, and 4 digits are each significant.)
(b) 0.00348 s has three significant figures (The 3, 4, and 8 digits are each significant. The zeros are all
before the first non-zero-digit 3 and therefore they are not significant.)
(c) 5.619 mm has four significant figures (The 5, 6, 1, and 9 digits are each significant.)
(d) 2.475×10–3 cm has four significant figures (The 2, 4, 7, and 5 digits are each significant.)
(e) 33.1 mL has three significant figures (The 3, 3, and 1 digits are each significant.)
Check your answers: Only once answer had zeros in it, and those (in (b)) were to the left of the first nonzero digit, so none of the zeros here were significant.
21. Define the problem: Given several measured quantities, determine the number of significant figures.
Develop a plan: Use rules given in Section 2.4, summarized here: All non-zeros are significant. Zeros that
precede (sit to the left of) non-zeros are never significant (e.g., 0.003). Zeros trapped between non-zeros are
always significant (e.g., 3.003). Zeros that follow (sit to the right of non-zeros are (a) significant if a decimal
point is explicitly given (e.g., 3300) OR (b) not significant, if a decimal point is not specified (e.g., 3300.).
Execute the plan:
(a) 1.022×103 km has four significant figures (The 1, 0, 2, and 2 digits are each significant. The zero is
trapped between the non-zeros 1 and 2 and therefore is significant.)
(b) 34 m2 has two significant figures (The 3 and 4 are each significant.)
(c) 0.042 L has two significant figures (The 4 and 2 digits are each significant. The zeros are all before the
first non-zero-digit 4 and therefore they are not significant.)
(d) 28.2 °C has three significant figures (The 2, 8, and 2 digits are each significant.)
(e) 323. mg has three significant figures (The 3, 2, and 3 digits are each significant.)
Check your answer: Two answers had zeros in them. In (a), it was trapped and significant; in (c), they were
to the left of the first non-zero digit, so not significant.
22. Define the problem: Given several measured quantities, round them to three significant figures and write
them in scientific notation.
Develop a plan: Use rules for rounding given in Section 2.4. If the last digit is below 5, then rounding does
not change the digit before it. If the last digit is above a five, the digit before it is made one larger. If the last
digit is exactly five, round the digit before it to an even number, up if odd and down if even. After rounding,
adjust the appearance of the number to scientific notation (i.e., to a number between 1 and 9.999… that is
multiplied by ten to a whole-number power.)
Execute the plan:
(a) 0.0004332 has four significant figures. To round it the three significant figures, we need to remove the
Chapter 2: Atoms and Elements
33
fourth significant figure, the 2. Since 2 is below 5, the result is 0.000433. Putting this value in scientific
notation, we get 4.33× 10 – 4.
(b) 44.7337 has six significant figures. To round it the three significant figures, we need to remove the
fourth, fifth and sixth significant figures. The removal of the last digit 7 rounds the 3 next to it up to a 4,
but the removal of 4 doesn’t change the 3 next to it, and the removal of 3 doesn’t change the 7 next to it,
the result is 44.7. Putting this value in scientific notation, we get 4.47× 10 1 .
(c) 22.4555 has six significant figures. To round it the three significant figures, we need to remove the
fourth, fifth and sixth significant figures. Since the removal of the last 5 rounds the 5 next to it up to a 6,
and the removal of 6 rounds the 5 next to it up to a 6, and the removal of 6 rounds the 4 next to it to a 5,
the result is 22.5. Putting this value in scientific notation, we get 2.25× 10 1 .
(d) 0.0088418 has five significant figures. To round it the three significant figures, we need to remove the
fourth and fifth significant figures. Since the removal of the last 8 rounds the 1 next to it up to a 2, and
the removal of 2 doesn’t change the 4 next to it, the result is 0.00884. Putting this value in scientific
notation, we get 8.84× 10 –3.
Check your answer: Small numbers are still small, large numbers are still large, what remains after the
rounding is the larger part of the value. Each answer has three significant figures and each number is
represented in proper scientific notation.
23. Define the problem: Given several measured quantities, round them to four significant figures and write
them in scientific notation.
Develop a plan: Use rules for rounding given in Section 2.4. If the last digit is below 5, then rounding does
not change the digit before it. If the last digit is above a five, the digit before it is made one larger. If the last
digit is exactly five, round the digit before it to an even number, up if odd and down if even. After rounding,
adjust the appearance of the number to scientific notation (i.e., to a number between 1 and 9.999… that is
multiplied by ten to a whole-number power.)
Execute the plan:
(a) 247.583 has six significant figures. To round it the four significant figures, we need to remove the fifth
and sixth significant figures. Since the removal of the last digit 3 doesn’t change the 8 next to it, but the
removal of 8 rounds the 5 next to it up to a 6, the result is 247.6. Putting this value in scientific notation,
we get 2.476× 10 2.
(b) 100578 has six significant figures. To round it the four significant figures, we need to remove the fifth
and sixth significant figures. The removal of the last digit 8 rounds the 7 next to it up to a 8, and the
removal of 8 rounds the 5 next to it up to a 6, so the result is 100600. Notice, the places left behind by
the rounding still exist, we just don’t know what digit goes there. The “place holder” is a zero, hence
the ones -place and the tens-place are now being held by non-significant zeros. Putting this value in
scientific notation, we get rid of the non=significant zeros: 1.006× 10 5.
(c) 0.000348719 has six significant figures. To round it the four significant figures, we need to remove the
fifth and sixth significant figures. The removal of the last digit 9 rounds the 1 next to it up to a 2, and the
removal of 2 doesn’t change the 7 next to it, so the result is 0.0003487. Putting this value in scientific
notation, we get 3.487× 10 –5.
(d) 0.004003881 has seven significant figures. To round it the four significant figures, we need to remove
the fifth, sixth and seventh significant figures. The removal of the last digit 1 doesn’t change the 8 next
to it, the removal of the 8 rounds the 8 next to it up to a 9, and the removal of 9 rounds the 3 next to it up
34
Chapter 2: Atoms and Elements
to a 4, so the result is 0.004004. Putting this value in scientific notation, we get 4.004× 10 –3.
Check your answer: Small numbers are still small, large numbers are still large, what remains after the
rounding is the larger part of the value. Each answer has three significant figures and each number is
represented in proper scientific notation.
24. Define the problem: Given some numbers combined using calculations, determine the result with proper
significant figures.
Develop a plan: Perform the mathematical steps according to order of operations, applying the proper
significant figures (addition and subtraction retains the least number of decimal places in the result;
multiplication and division retain the least number of significant figures in the result). Note: if operations are
combined that use different rules, it is important to stop and determine the intermediate result any time the
rule switches.
Chapter 2: Atoms and Elements
35
Execute the plan:
4.850 g − 2.34 g
1.3 mL
(a)
The numerator uses the subtraction rule. The first number has three decimal places (the 8, the 5, and
the 0 are all decimal places -- digit that follow the decimal point to the right) and the second number has
two decimal places (the 3 and the 4 are both decimal places), so the result of the subtraction has two
decimal places.
2.51 g
1.3 mL
The ratio uses the division rule. The numerator has three significant figures and the denominator has
two significant figures, so the answer will have two significant figures. Therefore, we get 1.9 g/mL.
V = πr3 = (3.1415926) × (4.112 cm) 3
(b)
This whole calculation uses the multiplication rule, with four significant figures, limited by the
measurement of r. The value of π should be carried to more than four significant figures, such as
3.14159… The answer comes out 218.4 cm3.
(4.66 × 10–3) × 343.2
(c)
This calculation uses the multiplication rule. The first number, 4.66 × 10–3, has three significant figures
and the second number, 343.2, has four significant figures, so the answer has three significant figures
0.0217.
0.003400
65.2
(d)
This calculation uses the division rule. The numerator has four significant figures and the denominator
has three significant figures, so the answer has three significant figures 0.0000521.
Check your answer: The proper significant figures rules were used. The size and units of the answers are
appropriate.
25. Define the problem: Given some numbers combined using calculations, determine the result with proper
significant figures.
Develop a plan: Perform the mathematical steps according to order of operations, applying the proper
significant figures (addition and subtraction retains the least number of decimal places in the result;
multiplication and division retain the least number of significant figures in the result). Note: if operations are
combined that use different rules, it is important to stop and determine the intermediate result any time the
rule switches.
Execute the plan:
(a)
2221.05 −
3256 .5
3.20
The ratio uses the division rule. The numerator has five significant figures and the denominator has
three significant figures, so the result of division will have three significant figures. Therefore, we get
3
3
2221.05 − 1.02 × 10 = 2.22105 × 10 − 1.02 × 10
3
Chapter 2: Atoms and Elements
36
The next calculation uses the subtraction rule. To count decimal places, both numbers must have the
same power of ten:
=
3
2.22105 × 10 − 1.02 × 10
3
In the ×103 power of ten, the first number has five decimal places (the 2, the 2, the 1, the 0, and the 5 are
all decimal places) and the second number has two decimal places (the 0 and the 2 are both decimal
places). The result of the subtraction has two decimal places, in the ×103 power of ten: 1.20× 10 3.
Another way to look at this is that the uncertainty in the second number is in the tens place, so the
answer must be rounded to the tens place.
(343.2) × (2.01×10–3)
(b)
This calculation uses the multiplication rule. The first number, 343.2, has four significant figures and the
second number, 2.01×10–3, has three significant figures, so the answer has three significant figures
0.690.
S = 4πr2 = 4 × (3.1415926) × (2.55 cm) 2
(c)
This whole calculation uses the multiplication rule, with three significant figures, limited by the
measurement of r. The numeral 4 is exact, in this context. The value of π should be carried to more than
three significant figures, such as 3.14159… The answer comes out 81.7 cm3.
(
2802
− 0.0024 × 10, 000.
15
(d)
)
The ratio uses the division rule. The numerator has four significant figures and the denominator has
two significant figures, so the result has two significant figures. The multiplication also uses the same
rule. The first number has two sigrnificant figures and the second number has 5 significant figures. So,
the result of the multiplication will have two significant figures.
1.9×102 – 25
The difference uses the subtraction rule. We need these numbers in the same power of ten to be able to
count decimal places.
1.9×102 – 0.25×102
In the ×102 power of ten, the first number has one decimal places (the 9) and the second number has
two decimal places (the 2 and the 5 are both decimal places). The result of the subtraction has one
decimal places, in the ×102 power of ten: 1.6× 10 2.
Another way to look at this is that the uncertainty in the first number is in the tens place, so the answer
must be rounded to the tens place, 160, with the last zero being non-significant.
Check your answer: The proper significant figures rules were used. The size of the answers are appropriate.
Uncertainty is carried through the calculations properly.
Percent
26. Define the problem: Given a 17.6-gram bracelet that contains 14.1 grams of silver and the rest copper,
Chapter 2: Atoms and Elements
37
determine the percentage silver and the percentage copper.
Develop a plan: Determine the percentage of silver by dividing the mass of silver by the total mass and
multiplying by 100 %. Since the metal is made up of only silver and copper, determine the percentage of
copper by subtracting the percentage of silver from 100 %.
14 .1 g silver
× 100 % = 80.1 % silver
17.6 g bracelet
Execute the plan:
100 % total – 80.1 % silver = 19.9 % copper
NOTICE: When you are using grams of different substances, be careful to carry enough information in the
units so you don’t confuse one mass with another.
Check your answers: A significant majority of the metal in the bracelet is silver, so it makes sense that the
percentage of silver is larger than the percentage of copper.
27. Define the problem: Given a 1.00-lb block of solder, and the percentages of lead and tin in the solder,
determine the mass of lead and the mass of tin in the block.
Develop a plan: Always start with the sample. Using the conversion factor between pounds and grams,
determine the mass of the block in grams, then use the mass percentage of lead in solder as a conversion
factor to determine the mass of lead in the sample. Repeat the same process for tin using tin’s mass
percentage.
Execute the plan:
Every 100 g of this solder contains 67 g lead and 33 g tin.
1.00 lb solder ×
1.00 lb solder ×
453 .59 g solder
67 g lead
×
= 3.0 × 102 g lead
1 lb solder
100 g solder
453 .59 g solder
33 g tin
×
= 1.5 × 10 2 g tin
1 lb solder
100 g solder
NOTICE: When you are using grams of different substances, be careful to carry enough information in the
units so you don’t confuse one mass with another.
Check your answers: One third of the block’s mass is tin and two thirds of the block’s mass is lead, as
described by the mass percentages. This looks right.
28. Define the problem: Given the volume of a battery acid sample, the density and the percentage of sulfuric
acid in the battery acid by mass, determine the mass of acid in the battery.
Develop a plan: Always start with the sample. Use the density of the solution to create a conversion factor
between milliliters and grams, so you can determine the mass of the battery acid in grams. Then use the
mass percentage of sulfuric acid in the battery acid as a conversion factor to determine the mass of sulfuric
acid in the sample.
Execute the plan:
Every 1.000 mL of battery acid solution weighs 1.285 grams.
Every 100.00 g of battery acid solution contains 38.08 grams of sulfuric acid.
500 . mL solution ×
1.285 g solution
38.08 g sulfuric acid
×
= 245 g sulfuric acid
1.000 mL solution
100 .00 g solution
NOTICE: When you are using grams of different substances, be careful to carry enough information in the
Chapter 2: Atoms and Elements
38
units so you don’t confuse one mass with another.
Check your answer: Only about a third of the sample is sulfuric acid, so the mass of sulfuric acid should be
smaller than the volume of the solution. This looks right.
29. Define the problem: Given the mass of the popcorn kernel before and after popping, determine the
percentage of mass lost.
Develop a plan: Determine the mass lost by taking the difference between the final and initial masses. Then
determine the percentage mass lost by dividing the mass loss by the initial mass and multiplying by 100 %.
Execute the plan: mass before – mass after = 0.125 g – 0.106 g = 0.019 g lost
NOTICE: The number of significant figures drops during this subtraction.
0.019 g lost
× 100 % = 15 % mass lost
0.125 g initial
Check your answer: A small amount of mass was lost, so it makes sense that the percentage lost is low.
30. Define the problem: Given the mass of a sample of cereal and the number of milligrams of sodium in that
sample, determine the percentage of sodium in the cereal.
Develop a plan: Determine the mass of sodium in the same units as the sample mass. Then determine the
percentage sodium by dividing the mass of sodium by the sample mass and multiplying by 100 %.
Execute the plan: 280 mg Na ×
1 g Na
= 0.28 g Na
1000 mg Na
0.28 g Na
× 100 % = 0.9 % Na in cereal
30 g cereal
Check your answer: A very small amount of sodium is present in the cereal, so it makes sense that the
percentage mass is so low.
Isotopes
31. (a) FALSE. Atoms of the same element might have different masses due to the different number of
neutrons.
(b) FALSE. Atoms of the same element MUST have the same atomic number, since the number of protons
is equal to the atomic number.
32. Define the problem: Given the identity of an element (cobalt) and the atom’s mass number (60), find the
number of electrons, protons, and neutrons in the atom.
Develop a plan: Look up the symbol for cobalt and find that symbol on the periodic table. The periodic
table gives the atomic number. The atomic number is the number of protons. The number of electrons is
equal to the number of protons since the atom has no charge. The number of neutrons is the difference
between the mass number and the atomic number.
Execute the plan: The element technetium has the symbol Co. On the periodic table, we find it listed with
the atomic number 27. So, the atom has 27 protons, 27 electrons and (60 – 27 =) 33 neutrons.
Check your answers: The number protons and electrons must be the same (27=27). The sum of the protons
and neutrons is the mass number (27 + 33 = 60). This is correct.
Chapter 2: Atoms and Elements
39
33. Define the problem: Given the identity of an element (technetium) and the atom’s mass number (99), find the
number of electrons, protons, and neutrons in the atom.
Develop a plan: Look up the symbol for technetium and find that symbol on the periodic table. The
periodic table gives the atomic number. The atomic number is the number of protons. The number of
electrons is equal to the number of protons since the atom has no charge. The number of neutrons is the
difference between the mass number and the atomic number.
Execute the plan: The element technetium has the symbol Tc. On the periodic table, we find it listed with
the atomic number 43. So, the atom has 43 protons, 43 electrons and (99 – 43 =) 56 neutrons.
Check your answers: The number protons and electrons must be the same (43=43). The sum of the protons
and neutrons is the mass number (56 + 43 = 99). This is correct.
34. Atoms of the same element have the same number of protons in the nucleus, and therefore all the atoms for
any given element would have the same atomic number.
35. The “atomic mass unit” is defined as
1
12
of the mass of a carbon atom having six protons and six neutrons in
the nucleus.
36. To estimate an atom’s mass, one only has to add up the number of protons and neutrons in the nucleus.
This estimate, which is a whole number, refers to the mass number for a given atom. The mass of an
electron is 1800 times smaller than the mass of a proton or neutron. That makes its contribution to the
atom’s mass negligible.
37. Mass number is the sum of the number of protons and neutrons. Atomic number is the number of protons.
So, the difference between the mass number and the atomic number of an atom is the number of neutrons.
38. Mass number is the sum of the number of protons and neutrons. Atomic number is the number of protons.
So, when you subtract the atomic number from the mass number, you obtain the number of neutrons.
39. Atomic number is the number of protons. Look up the atomic number on the periodic table to find the
element. Here, the atom has 20 protons. The periodic table gives calcium as the element with atomic
number 20.
40. Define the problem: Using the mass of two isotopes and the average atomic weight, determine the
abundance of the isotopes.
Develop a plan: Establish variables describing the isotope percentages. Set up two relationships between
these variables. The sum of the percents must be 100 %, and the weighted average of the isotope masses
must be equal to the reported atomic weight.
Execute the plan: X % 35Cl and Y %37Cu, This means: X + Y = 100 %
That gives us one equation with two unknowns.
Every 100 atoms of chlorine contains X atoms of the 35Cl isotope and Y atoms of the 37Cu isotope. The
atomic weight for Cl from the periodic table is 35.453 amu/atom.
X atoms 35 Cl  34.9689 amu 
Y atoms 37Cu  36.9659 amu 
amu
 +
 = 35.453
×
×

100 Cl atoms  1 atom 35Cl 
100 Cl atoms
37 
Cl atom


 1 atom Cl 
That gives us a second equation with the same two unknowns.
40
Chapter 2: Atoms and Elements
X
Y
× 34.9689 +
× 36.9659 = 35.453
100
100
Solve the first equation for Y in terms of X, then substitute back into the second equation:
Y = 100 – X
X
100 − X
× 34.9689 +
× 36.9659 = 35.453
100
100
Now, solve for X then Y:
0.349689X + 36.9659− 0.369659X = 35.453
36.9659− 35.453= 0.369659X − 0.349689X
(
)
1, 529 = 0.019970 X
X = 75.76 %
Y = 100 – X = 100 – 75.76 = 24.24 %
Therefore the abundances for these isotopes are: 75.76 % 35Cl and 24.24 % 37Cl
Check your answers: The periodic table gives the atomic weight as closer to 35 than to 37, so it makes sense
that the percentage of 35Cl is larger than 37Cl. This answer looks right.
41. Define the problem: Given the identity of an element and the number of neutrons in the atom, determine the
mass number of the atom.
Develop a plan: Look up the symbol for the element and find that symbol on the periodic table. The
periodic table gives the atomic number, which represents the number of protons. Add the number of
neutrons to the number of protons to get the mass number.
Execute the plan:
(a) The element beryllium has the symbol Be. On the periodic table, we find it listed with the atomic number
4. The given number of neutrons is 5. So, (4 + 5 =) 9 is the mass number for this beryllium atom.
(b) The element titanium has the symbol Ti. On the periodic table, we find it listed with the atomic number
22. The given number of neutrons is 26. So, (22 + 26 =) 48 is the mass number for this titanium atom.
(c) The element gallium has the symbol Ga. On the periodic table, we find it listed with the atomic number
31. The given number of neutrons is 39. So, (31 + 39 =) 70 is the mass number for this gallium atom.
Check your answers: Mass number should be close to (but not exactly the same as) the atomic weight also
given on the periodic table. Beryllium’s atomic weight (9.0122) is close to the 9 mass number. Titanium’s
atomic weight (47.88) is close to the 48 mass number. Gallium’s atomic weight (69.72) is close to the 70 mass
number. These numbers seem reasonable.
42. Define the problem: Given the identity of an element and the number of neutrons in the atom, determine the
mass number of the atom.
Develop a plan: Look up the symbol for the element and find that symbol on the periodic table. The
periodic table gives the atomic number, which represents the number of protons. Add the number of
neutrons to the number of protons to get the mass number.
Execute the plan:
Chapter 2: Atoms and Elements
41
(a) The element iron has the symbol Fe. On the periodic table, we find it listed with the atomic number 26.
The given number of neutrons is 30. So, (26 + 30 =) 56 is the mass number for this iron atom.
(b) The element americium has the symbol Am. On the periodic table, we find it listed with the atomic
number 95. The given number of neutrons is 148. So, (95 + 148 =) 243 is the mass number for this
americium atom.
(c) The element tungsten has the symbol W. On the periodic table, we find it listed with the atomic number
74. The given number of neutrons is 110. So, (74 + 110 =) 184 is the mass number for this tungsten
atom.
Check your answers: Mass number should be close to (but not always the same as) the atomic weight given
on the periodic table. Iron’s atomic weight (55.85) is close to its 56 mass number. Americium’s atomic
weight (243) is the same as its 243 mass number. Tungsten’s atomic weight (183.85) is close to its 184 mass
number. These look right.
43. Define the problem: Given the identity of an element and the number of neutrons in the atom, determine the
A
atomic symbol Z X .
Develop a plan: Look up the symbol for the element and find that symbol (X) on the periodic table. The
periodic table gives the atomic number (Z), which represents the number of protons. Add the number of
neutrons to the number of protons to get the mass number (A).
Execute the plan:
(a) The element sodium has the symbol Na. On the periodic table, we find it listed with the atomic number
11. The given number of neutrons is 12. So, (11 + 12 =) 23 is the mass number for this sodium atom. Its
23
atomic symbol looks like this: 11 Na .
(b) The element argon has the symbol Ar. On the periodic table, we find it listed with the atomic number 18.
The given number of neutrons is 21. So, (18 + 21 =) 39 is the mass number for this argon atom. Its
39
atomic symbol looks like this: 18 Ar .
(c) The element gallium has the symbol Ga. On the periodic table, we find it listed with the atomic number
31. The given number of neutrons is 38. So, (31 + 38 =) 69 is the mass number for this gallium atom. Its
69
atomic symbol looks like this: 31Ga .
Check your answers: Mass number should be close to (but not exactly the same as) the atomic weight also
given on the periodic table. Sodium’s atomic weight (22.99) is close to the 23 mass number. Argon’s atomic
weight (39.95) is close to the 39 mass number. Gallium’s atomic weight (69.72) is close to the 69 mass
number. These numbers seem reasonable.
44. Define the problem: Given the identity of an element and the number of neutrons in the atom, determine the
A
atomic symbol Z X .
Develop a plan: Look up the symbol for the element and find that symbol (X) on the periodic table. The
periodic table gives the atomic number (Z), which represents the number of protons. Add the number of
neutrons to the number of protons to get the mass number (A).
Execute the plan:
(a) The element nitrogen has the symbol N. On the periodic table, we find it listed with the atomic number 7.
The given number of neutrons is 8. So, (7 + 8 =) 15 is the mass number for this nitrogen atom. Its
42
Chapter 2: Atoms and Elements
15
atomic symbol looks like this: 7 N .
(b) The element zinc has the symbol Zn. On the periodic table, we find it listed with the atomic number 30.
The given number of neutrons is 34. So, (30 + 34 =) 64 is the mass number for this zinc atom. Its atomic
64
symbol looks like this: 30 Zn .
(c) The element xenon has the symbol Xe. On the periodic table, we find it listed with the atomic number 54.
The given number of neutrons is 75. So, (54 + 75 =) 129 is the mass number for this xenon atom. Its
129
atomic symbol looks like this: 54 Xe .
Check your answers: Mass number should be close to (but not exactly the same as) the atomic weight also
given on the periodic table. Nitrogen’s atomic weight (14.01) is close to the 15 mass number. Zinc’s atomic
weight (65.38) is close to the 64 mass number. Xenon’s atomic weight (131.3) is close to the 129 mass
number. These numbers seem reasonable.
A
45. Define the problem: Given the atomic symbol Z X of the isotope, determine the number of electrons,
protons, and neutrons.
Develop a plan: The atomic number (Z) represents the number of protons. In neutral atoms, the number of
electrons is equal to the number of protons. To get the number of neutrons, subtract the number of protons
from the mass number (A).
Execute the plan:
40
(a) The isotope given is 20Ca . That means A = 40 and Z = 20. So, the number of protons is 20, the
number of electrons is 20, and the number of neutrons is (40 – 20 =) 20.
119
(b) The isotope given is 50Sn . That means A = 119 and Z = 50. So, the number of protons is 50, the
number of electrons is 50, and the number of neutrons is (119 – 50 =) 69.
244
(c) The isotope given is 94 Pu . That means A = 244 and Z = 94. So, the number of protons is 94, the
number of electrons is 94, and the number of neutrons is (244 – 94 =) 150.
Check your answers: The number of protons and electrons must be equal in neutral atoms. The mass
number must be the sum of the protons and neutrons. These answers look okay.
A
46. Define the problem: Given the atomic symbol Z X of the isotope, determine the number of electrons,
protons, and neutrons.
Develop a plan: The atomic number (Z) represents the number of protons. In neutral atoms, the number of
electrons are equal to the number of protons. To get the number of neutrons, subtract the number of
protons from the mass number (A).
Execute the plan:
13
(a) The isotope given is 6C . That means A = 13 and Z = 6. So, the number of protons is 6, the number of
electrons is 6, and the number of neutrons is (13 – 6 =) 7.
50
(b) The isotope given is 24Cr . That means A = 50 and Z = 24. So, the number of protons is 24, the number
of electrons is 24, and the numb er of neutrons is
Chapter 2: Atoms and Elements
(50 – 24 =) 26.
205
(c) The isotope given is 83 Bi . That means A = 205 and Z = 83. So, the number of protons is 83, the
number of electrons is 83, and the number of neutrons is
(205 – 83 =) 122.
Check your answers: The number of protons and electrons must be equal in neutral atoms. The mass
number must be the sum of the protons and neutrons. These answers look okay.
47. Define the problem: Fill in an incomplete table with Z, A, number of neutrons and element identity.
Develop a plan: The atomic number (Z) represents the number of protons. The mass number (A) is the
number of neutrons and protons. The element’s identity can be determined using the periodic table by
looking up the atomic number and getting the symbol.
43
44
Chapter 2: Atoms and Elements
Execute the plan:
Number of Neutrons
Z
A
Element
35
81
(a)
(b)
(c)
(d)
62
Pd
77
(e)
115
(f)
(g)
151
(h)
Eu
(a) Number of neutrons = A – Z = 81 – 35 = 46
(b) Z = 35. Look up element #35 on periodic table: Element = Br
(c) Look up Pd on the periodic table: Z = 46
(d) A = Z + number of neutrons = 46 + 62 = 108
(e) A = Z + number of neutrons = 77 + 115 = 192
(f) Z = 77. Look up element #77 on periodic table: Element = Ir
(g) Look up Eu on the periodic table: Z = 63
(h) Number of neutrons = A – Z = 151 – 63 = 88
So, the complete table looks like this:
Number of Neutrons
Z
A
Element
35
81
46
Br
46
108
62
Pd
77
192
115
Ir
63
151
88
Eu
Check your answers: The atomic number and the symbol must match what is shown on the periodic table.
The mass number must be the sum of the atomic number and the number of neutrons. These answers look
okay.
48. Define the problem: Fill in an incomplete table with Z, A, number of neutrons and element identity.
Develop a plan: The atomic number (Z) represents the number of protons. The mass number (A) is the
number of neutrons and protons. The element’s symbol can be determined by looking up the atomic
number on the periodic table.
Execute the plan:
Number of Neutrons
Z
A
Element
60
144
(a)
(b)
(c)
(d)
12
Mg
64
(e)
94
(f)
Chapter 2: Atoms and Elements
(g)
37
(h)
45
Cl
(a) Number of neutrons = A – Z = 144 – 60 = 84
(b) Z = 60. Look up element #60 on periodic table: Element = Nd
(c) Look up Mg on the periodic table: Z = 12
(d) A = Z + number of neutrons = 12 + 12 = 24
(e) A = Z + number of neutrons = 64 + 94 = 158
(f) Z = 64. Look up element #64 on periodic table: Element = Gd
(g) Look up Cl on the periodic table: Z = 17
(h) Number of neutrons = A – Z = 37 – 17 = 20
So, the complete table looks like this:
Number of Neutrons
Z
A
Element
60
144
84
Nd
12
24
12
Mg
64
158
94
Gd
17
37
20
Cl
Check your answers: The atomic number and the symbol must match what is shown on the periodic table.
The mass number must be the sum of the atomic number and the number of neutrons. These answers look
okay.
A
49. The atomic symbol general form is Z X , where A is the atomic mass and Z is the atomic number. Isotopes
18
20
15
will have Z the same, but A will be different. Therefore, 9 X, 9 X, and 9 X are all isotopes of the same
element.
50. Define the problem: Knowing the identity of the element and the number of neutrons, determine the atomic
A
symbol Z X .
Develop a plan: The element’s identity can be used to get the atomic number (Z). The atomic number
represents the number of protons. The mass number (A) is the number of neutrons and protons. Determine
the atomic number and mass number of each isotope and construct the atomic symbol.
Execute the plan: Look up the symbol for cobalt (Co), and find Co on the periodic table: Z = 27.
A = Z + number of neutrons = 27 + 30 = 57
57
27Co
A = Z + number of neutrons = 27 + 31 = 58
58
27Co
A = Z + number of neutrons = 27 + 33 = 60
60
27 Co
Check your answers: Isotopes must have the same symbols and the same Z values, but different A values.
These answers look okay.
46
Chapter 2: Atoms and Elements
Mass Spectrometry
51. The species that is moving through a mass spectrometer during its operation are ions (usually +1 cations)
that have been formed from the sample molecules by a bombarding electron beam.
52. Define the problem: In a mass spectrum the x-axis is the mass of the ions and the y-axis is the abundance of
the ions. The mass spectrum is a representation of the masses and abundances of the ions formed in the
mass spectrometer, which are directly related to the molecular structure of the sample atoms or molecules.
53. The ions in the mass spectrometer are separated from each other using a magnetic field which causes the
ions to have different paths through the instrument.
Atomic Weight
54. Define the problem: Using the exact mass and the percent abundance of several isotopes of an element,
determine the atomic weight.
Develop a plan: Calculate the weighted average of the isotope masses.
Execute the plan:
6
Every 1000 atoms of lithium contains 75.00 atoms of the Li isotope.
7
Every 1000 atoms of lithium contains 925.0 atoms of the Li isotope.
6
 6.015121 amu
75.00 atoms Li
× 
1000 Li atoms
6
 1 atom Li
7

925 .0 atoms Li  7.016003 amu
 +
× 

1000 Li atoms
7

 1 atom Li

 = 6.941 amu/Li atom


Check your answer: The periodic table value for the atomic weight is the same as this calculated value.
This answer looks right.
55. Define the problem: Using the exact mass and the percent abundance of several isotopes of an element,
determine the atomic weight.
Develop a plan: Calculate the weighted average of the isotope masses.
Execute the plan:
Every 10000 atoms of magnesium contains 7899 atoms of the
isotope, and 1101 atoms of the
26
24
Mg isotope, 1000 atoms of the
25
Mg
Mg isotope.
24
7899 atoms Mg  23 .985042 amu 

×
10000 Mg atoms  1 atom 24Mg 


25
1000 atoms Mg  24.98537 amu 

× 
10000 Mg atoms
25Mg 
1
atom


+
+
 25.982593 amu
× 
10000 Mg atoms  1 atom 26Mg
1101 atoms
26
Mg

amu
 = 24 .31

Mg atom

Note: The given percentages limit each term to four significant figures, therefore the first term has only
two decimal places. So, this answer is rounded off significantly.
Check your answer: The periodic table value for the atomic weight is the same this calculated value. This
answer looks right.
Chapter 2: Atoms and Elements
47
56. Define the problem: Using the exact mass of several isotopes and the atomic weight, determine the
abundance of the isotopes.
Develop a plan: Establish variables describing the isotope percentages. Set up two relationships between
these variables. The sum of the percents must be 100 %, and the weighted average of the isotope masses
must be the reported atomic mass.
69
71
Execute the plan: X % Ga and Y % Ga, This means:
Every 100 atoms of gallium contains X atoms of the
69
Every 100 atoms of gallium contains Y atoms of the
71
Ga isotope.
Ga isotope.
X + Y = 100 %
69
 68.9257 amu
X atoms Ga
× 
100 Ga atoms
69
 1 atom Ga
71

Y atoms Ga  70.9249 amu 
amu
 +
 = 69 .723
× 


100 Ga atoms
71
Ga atom

 1 atom Ga 
We now have two equations and two unknowns, so we can solve for X and Y algebraically. Solve the first
equation for Y
Y = 100 – X
Plug that in for Y in the second equation. Then solve for X:
X
100 − X
× 68.9257 +
× 70 .9249 = 69.723
100
100
(
)
(
)
0.689257X + 70.9249− 0.709249X = 69.723
(
)
70.9249− 69.723 = 0.709249X − 0.689257X = 0.709249− 0.689257 X
(
)
1.202= 0.01999 X
X = 60.13, so there is 60.13 % 69Ga
Now, plug the value of X in the first equation to get Y.
Y = 100 – X = 100 – 60.12 = 39.88, so there is 39.87 % 69Ga
Therefore the abundances for these isotopes are: 60.12 % 69Ga and 39.87 % 71Ga
Check your answers: The periodic table value for the atomic weight is closer to 68.9257 than it is to 70.9249,
so it makes sense that the percentage of 69Ga is larger than 71Ga. This answer looks right.
57. Define the problem: Using the exact mass of several isotopes and the atomic weight (from the periodic table
of the elements), determine the abundance of the isotopes.
Develop a plan: Establish variables describing the isotope percentages. Set up two relationships between
these variables. The sum of the percents must be 100 %, and the weighted average of the isotope masses
must be the reported atomic mass.
Execute the plan: X % 107Ag and Y % 109Ag, This means:
Every 100 atoms of silver contains X atoms of the 107Ag isotope.
48
Chapter 2: Atoms and Elements
Every 100 atoms of silver contains Y atoms of the 109Ag isotope.
X + Y = 100 %
107
109
 106 .90509 amu 
X atoms
Ag
Y atoms
Ag  108 .90476 amu
 +
× 
×
100 Ag atoms
107 Ag 
100 Ag atoms  1 atom 109 Ag
 1 atom



amu
 = 107 .8682

Ag atom

We now have two equations and two unknowns, so we can solve for X and Y algebraically. Solve the first
equation for Y
Y = 100 – X
Plug that in for Y in the second equation. Then solve for X:
X
100 − X
× 106.90509 +
× 108.90476 = 107.8682
100
100
(
)
(
)
1.0690509X + 108.90476− 1.0890476X = 107.8682
108.90476− 107.8682 = 1.0890476 X − 1.0690509 X
(
)
1.0366= 0.0199967 X
X = 51.837
Now, plug the value of X in the first equation to get Y.
Y = 100 – X = 100 – 51.837 = 48.163
Therefore the abundances for these isotopes are: 51.837 % 107Ag and 348.163 % 109Ag
Check your answers: The periodic table value for the atomic weight about half way between the mass of the
107Ag isotope and the 109Ag isotope. This answer looks right.
58. The two isotopes of lithium are 6Li and 7Li. The mass of 6Li is close to 6 amu and the mass of 7Li is close to
7
7 amu. Because lithium’s atomic weight (6.941 amu) is much closer to 7 amu than to 6 amu, the isotopic Li is
6
more abundant than the isotope Li.
40
59. Knowing that almost all of the argon in nature is Ar, a good estimate for the atomic weight of argon is a
little less than 40 amu/atom.
Define the problem: Using the exact mass and the percent abundance of several isotopes of an element,
determine the atomic weight.
Develop a plan: Calculate the weighted average of the isotope masses.
Execute the plan: Every 100000 atoms of argon contains 337 atoms of the 36Ar isotope, 63 atoms of the
38Ar isotope, and 99600 atoms of the 40Ar isotope.
337 atoms 36Ar  35.968 amu 
63 atoms 38Ar  37.963 amu 
 +

×
×
100000 Ar atoms  1 atom 36Ar 
100000 Ar atoms  1 atom 38Ar 




99600 atoms 40Ar  39.962 amu 
amu
 = 39.95
+
×
100000 Ar atoms  1 atom 40Ar 
Ar atom


Chapter 2: Atoms and Elements
49
Check your answer: This calculated answer matches the estimate. Also, the periodic table value for the
atomic weight is the same as this calculated value.
The Mole
60. A few common counting units are: pair (2), dozen (12), gross (144), hundred (100), million (1,000,000), billion
(1,000,000,000), etc.
61. Define the problem: Determine how long will it take for all the people in the United States to count 1 mole of
pennies if they spend their entire working days counting.
Develop a plan: Calculate the number of pennies each person has to count. Then calculate how many
working days that person would spend counting their share.
Execute the plan:
23
6.022 × 10
pennies
= 2.11×1015 pennies/person
285, 000, 000 people
15
2.11 × 10 pennies
1s
1 min . 1 hour 1 work day
×
×
×
×
= 7.34 × 1010 work days
person
1 penny
60 s
60 min .
8 hours
If we give everyone weekends off and two weeks off per year for vacation, there are approximately 250 work
days per year. Assuming no one quits the job or dies without being replaced, it would take over 300 billion
years to count this one mole of pennies.
Check your answer: The quantity of pennies in one mole is huge. It should take people a LONG time to
count that many pennies.
62. Counting the individual molecules is inconvenient for two reasons. Individual molecules are too small, and
in samples large enough to see, their numbers are so great that not even normal “large number” words are
very convenient. For example, a common “large number” word is “trillion”. That’s 1,000,000,000,000 or
1×1012. One mole of molecules is almost a trillion times more than a trillion!
63. Define the problem: Determine mass in grams from given quantity in moles.
Develop a plan: Look up the elements on the periodic table to get the atomic weight (with at least four
significant figures). If necessary, calculate the molecular weight. Use that number for the molar mass (with
units of grams per mole) as a conversion factor between moles and grams.
Note: Whenever you use physical constants that you look up, it is important to use more significant
figures than the rest of the numbers, to prevent causing inappropriate round-off errors.
Execute the plan:
(a) Boron (B) has atomic number 5 on the periodic table. Its atomic weight is 10.81, so the molar mass is
10.81g/mol.
2.5 mol B ×
10 .81 g B
= 27 g B
1 mol B
(b) O2 (diatomic molecular oxygen) is made with two atoms of element with atomic number 8 on the periodic
table. Its atomic weight is 16.00; therefore, the molecular weight of O2 is 2×16.00=32.00, and the molar
mass is 32.00 g/mol.
50
Chapter 2: Atoms and Elements
0.015 mol O2 ×
32.00 g O2
= 0.48 g O2
1 mol O2
(c) Iron (Fe) has atomic number 26 on the periodic table. Its atomic weight is 55.85, so the molar mass is
55.85 g/mol.
1.25 × 10−3 mol Fe ×
55.85 g Fe
= 6.98 × 10 −2 g Fe
1 mol Fe
(d) Helium (He) has atomic number 2 on the periodic table. Its atomic weight is 4.003, so the molar mass is
4.003 g/mol.
653 mol He ×
4.003 g He
= 2.61 × 103 g He
1 mol He
Check your answers: The “moles” units cancel when the factor is multiplied, leaving the answer in grams.
64. Define the problem: Determine the mass in grams from given quantity in moles.
Develop a plan: Look up the elements on the periodic table to get the atomic weight (with at least four
significant figures). Use that number for the molar mass (with units of grams per mole) as a conversion
factor between moles and grams.
Note: Whenever you use physical constants that you look up, it is important to use more significant
figures than the rest of the numbers, to prevent causing inappropriate round-off errors.
Execute the plan:
(a) Gold (Au) has atomic number 79 on the periodic table. Its atomic weight is 197.0, so the molar mass is
197.0 g/mol.
6.03 mol Au ×
197 .0 g Au
= 1.19 × 103 g Au
1 mol Au
(b) Uranium (U) has atomic number 92 on the periodic table. Its atomic weight is 238.0, so the molar mass is
238.0 g/mol.
0.045 mol U ×
238 .0 g U
= 11 g U
1 mol U
(c) Neon (Ne) has atomic number 10 on the periodic table. Its atomic weight is 20.18, so the molar mass is
20.18 g/mol.
15 .6 mol Ne ×
20 .18 g Ne
= 315 g Ne
1 mol Ne
(d) Radioactive plutonium (Pu) has atomic number 94 on the periodic table. The atomic weight given on the
periodic table is the weight of its most stable isotope 244, so the molar mass is 244 g/mol.
3.63 × 10 −4 mol Pu ×
244 g Pu
= 0.0886 g Pu
1 mol Pu
Check your answers: Notice that the “moles” units cancel when the factor is multiplied, leaving the answer
in grams.
65. Define the problem: Determine the quantity in moles from given mass in grams.
Chapter 2: Atoms and Elements
51
Develop a plan: Look up the elements on the periodic table to get the atomic weight (with an appropriate
number of significant figures). Use that number for the molar mass (with units of grams per mole) as a
conversion factor between grams and moles.
Note: Whenever you use physical constants that you look up, it is important to use more significant
figures than the rest of the numbers, whenever possible, to prevent causing inappropriate round-off errors.
Execute the plan:
(a) Copper (Cu) has atomic number 29 on the periodic table. Its atomic weight is 63.546, so the molar mass
is 63.546 g/mol.
1 mol Cu
= 1.9998 mol Cu
63.546 g Cu
127 .08 g Cu ×
(b) Calcium (Ca) has atomic number 20 on the periodic table. Its atomic weight is 40.08, so the molar mass is
40.08 g/mol.
20.0 g Ca ×
1 mol Ca
= 0.499 mol Ca
40.08 g Ca
(c) Aluminum (Al) has atomic number 13 on the periodic table. Its atomic weight is 26.982, so the molar
mass is 26.982 g/mol.
16 .75 g Al ×
1 mol Al
= 0.6208 mol Al
26.982 g Al
(d) Potassium (K) has atomic number 19 on the periodic table. Its atomic weight is 39.1, so the molar mass
is 39.1 g/mol.
0.012 g K ×
1 mol K
= 3.1 × 10−4 mol K
39 .1 g K
(e) Radioactive americium (Am) has atomic number 95 on the periodic table. The atomic weight given on
the periodic table is the weight of its most stable isotope 243, so the molar mass is 243 g/mol.
Convert milligrams into grams, first.
1 g Am
1 mol Am
5.0 mg Am ×
×
= 2.1 × 10 −5 mol Am
1000 mg Am 243 g Am
Check your answers: Notice that the “grams” units cancel when the factor is multiplied, leaving the answer
in moles.
66. Define the problem: Determine the quantity in moles from given mass in grams.
Develop a plan: Look up the elements on the periodic table to get the atomic weight (with an appropriate
number of significant figures). Use that number for the molar mass (with units of grams per mole) as a
conversion factor between grams and moles.
Execute the plan:
(a) Sodium (Na) has atomic number 11 on the periodic table. Its atomic weight is 22.99, so the molar mass is
22.99 g/mol.
16 .0 g Na ×
1 mol Na
= 0.696 mol Na
22 .99 g Na
52
Chapter 2: Atoms and Elements
(b) Platinum (Pt) has atomic number 78 on the periodic table. Its atomic weight is 195.1, so the molar mass
is 195.1 g/mol.
0.0034 g Pt ×
1 mol Pt
= 1.7 × 10−5 mol Pt
195 .1 g Pt
(c) Phosphorus (P) has atomic number 15 on the periodic table. Its atomic weight is 30.97, so the molar
mass is 30.97 g/mol.
1.54 g P ×
1 mol P
= 0.0497 mol P
30.97 g P
(d) Arsenic (As) has atomic number 33 on the periodic table. Its atomic weight is 74.92, so the molar mass
is 74.92 g/mol.
0.876 g As ×
1 mol As
= 0.0117 mol As
74 .92 g As
Chapter 2: Atoms and Elements
53
(e) Xenon (Xe) has atomic number 54 on the periodic table. Its atomic weight is 131.3, so the molar mass is
131.3 g/mol.
0.983 g Xe ×
1 mol Xe
= 7.49 × 10−3 mol Xe
131 .3 g Xe
Check your answers: Notice that the “grams” units cancel when the factor is multiplied, leaving the answer
in moles.
67. Define the problem: Determine the quantity in moles from the given mass in grams.
Develop a plan: Look up the Na on the periodic table to get the atomic weight (with an appropriate number
of significant figures). Use that number for the molar mass (with units of grams per mole) as a conversion
factor between grams and moles.
Execute the plan: Sodium (Na) has atomic number 11 on the periodic table. Its atomic weight is 22.99, so
the molar mass is 22.99 g/mol.
50.4 g Na ×
1 mol Na
= 2.19 mol Na
22.99 g Na
Check your answer: Notice that the “grams” units cancel when the factor is multiplied, leaving the answer
in moles. The resulting moles should be smaller than the mass.
68. Define the problem: Determine the quantity in moles from given mass in grams.
Develop a plan: Look up krypton on the periodic table to get the atomic weight (with an appropriate
number of significant figures). Use that number for the molar mass (with units of grams per mole) as a
conversion factor between grams and moles.
Execute the plan:
Krypton (Kr) has atomic number 36 on the periodic table. Its atomic weight is 83.80, so the molar mass is
83.80 g/mol.
0.00789 g Kr ×
1 mol Kr
= 9.42 × 10−5 mol Kr
83.80 g Kr
Check your answer: Notice that the “grams” units cancel when the factor is multiplied, leaving the answer
in moles.
69. Define the problem: Given a mass of helium, determine the moles.
Develop a plan: Use molar mass of He as a conversion factor between moles and grams.
Execute the plan: Helium (He) has atomic number 2 on the periodic table. Its atomic weight is 4.0026, so the
molar mass is 4.0026 g/mol.
4.6 × 10−3 g ×
1 mol He
= 1.1 × 10−3 mol He
4.0026 g He
Check your answer: Notice that the “grams” units cancel when the factor is multiplied, leaving the answer
in moles. The resulting moles should be smaller than the mass.
70. Define the problem: A sample of sodium has a given density. Determine the volume of a cube of sodium
and the length of each side.
54
Chapter 2: Atoms and Elements
Develop a plan: Always start with the sample; here, start with the given moles. Use the molar mass of
sodium as a conversion factor between moles and grams. Then use the density to convert between grams
and cubic centimeters. Then take the cube-root of the volume to find the length of each side of the cube.
Execute the plan:
0.125 mol Na ×
length =
3
3
22.99 g Na
1 cm
×
= 2.97 cm 3
1 mol Na
0.968 g Na
3
volume = 2.97 cm
3
= 1.45 cm
Check your answer: The problem makes it sound like this cube will be manageable by a normal strength
chemist with a simple knife. This cube’s size doesn’t look too big or too small. This answer looks right.
71. Define the problem: A sample of chromium has a known mass. Determine the number of atoms in the
sample.
Develop a plan: Start with the mass. Use the molar mass of chromium as a conversion factor between
grams and moles. Then use Avogadro’s number as a conversion factor between moles of chromium atoms
and the actual number of chromium atoms.
Execute the plan:
35.67 g Cr ×
23
1 mol Cr atoms 6.0221 × 10 Cr atoms
×
= 4.131 × 1023 Cr atoms
51.996 g Cr
1 mol Cr atoms
Check your answer: A sample of chromium that a person can see and hold is macroscopic. It will contain a
very large number of atoms. This answer seems right.
72. Define the problem: A ring of gold has a known mass. Determine the number of atoms in the sample.
Develop a plan: Start with the mass. Use the molar mass of gold as a conversion factor between grams and
moles. Then use Avogadro’s number as a conversion factor between moles of gold atoms and the actual
number of gold atoms.
Execute the plan:
1.94 g Au ×
23
1 mol Au atoms 6.022 × 10 Au atoms
×
= 5.93 × 1021 Au atoms
197.0 g Au
1 mol Au atoms
Check your answer: A ring of gold is something that a person can see and hold. It will contain a very large
number of atoms. This answer seems right.
73. Define the problem: Given the number of copper atoms, determine the mass.
Develop a plan: Start with the quantity. Use Avogadro’s number as a conversion factor between the
number of copper atoms and moles of copper atoms. Then use the molar mass of copper as a conversion
factor between moles and grams.
Execute the plan:
1 Cu atom ×
1 mol Cu atoms
6.022 × 1023 Cu atoms
×
63.546 gCu
−22
= 1.055 × 10
gCu
1 mol Cu atoms
Check your answer: An atom of copper is NOT something that a person can see or hold. It will have a very
small mass. This answer seems right.
Chapter 2: Atoms and Elements
55
74. Define the problem: Given the number of titanium atoms, determine the mass.
Develop a plan: Start with the quantity. Use Avogadro’s number as a conversion factor between the
number of titanium atoms and moles of titanium atoms. Then use the molar mass of titanium as a conversion
factor between moles and grams.
Execute the plan:
1 Ti atom ×
1 mol Ti atoms
6.022 × 10 23 Ti atoms
×
47.88 g Ti
−23
= 7.951 × 10
g Ti
1 mol Ti atoms
Check your answer: An atom of titanium is NOT something that a person can see or hold. It will have a
very small mass. This answer seems right.
The Periodic Table
75. A group on the periodic table is the collection of elements that share the same vertical column, whereas a
period on the periodic table is the collection of elements that share the same horizontal row.
76. (a) Most of the elements on the lower left two-thirds of the periodic table are metals. Some common answers
from Group 1A: sodium (Na, Period 3) and potassium (K, Period 4), and from Group 2A: magnesium
(Mg, Period 3) and calcium (Ca, Period 4), etc. There are many metallic elements to choose from.
(b) Non-metals are found on the upper right side of the periodic table. Some common answers are in Period
2: carbon (C, Group 4A), nitrogen (N, Group 5A), oxygen (O, Group 6A) and neon (Ne, Group 8A).
There are many non-metallic elements to choose from.
(c) Metalloids are found on the periodic table next to the diagonal zigzag line that is drawn starting
between Al and B. Except for Al, all stable elements whose periodic table box shares a “side” with this
zigzag line are metalloids. Looking at the periodic table: boron (B), silicon (Si), germanium (Ge), arsenic
(As), antimony (Sb), and tellurium (Te).
77. New College Edition of the American Heritage Dictionary of the English Language (©1978) gives the
following information:
Titanium: A strong, low-density, highly corrosion resistant, lustrous white metallic element that occurs
widely in igneous rocks and is used to alloy aircraft metals for low weight, strength, and high-temperature
stability. Atomic number 22, atomic weight 47.90, melting point 1,675°C, boiling point 3,260°C, specific
gravity 4.54, valences 2, 3, 4. [New Latin, from Greek Titan, TITAN , So named by Klaproth, who had also
named uranium after the Planet Uranus. Uranus, in Greek Mythology, is the father of the Titans.](page 1348)
Chromium: A lustrous, hard, steel-gray metallic element, resistant to tarnish and corrosion, and found
primarily in chromite. It is used as a catalyst, to harden steel alloys, to produce stainless steel, in corrosionresistant decorative platings, and as a pigment in glass. Atomic number 24, atomic weight 51.996, melting
point 1,890°C, boiling point 2,482°C, specific gravity 7.18, valences 2, 3, 6. [New Latin, from French Chrome,
CHROM(E).] (page 240)
Iron: A silvery-white, lustrous, maleable, ductile, magnetic or magnetizable, metallic element occuring
abundantly in combined forms, notably in hematite, limonite, magnetite, and taconite, and used alloyed in a
wide range of important structural materials. Atomic number 26, atomic weight 55.847, melting point 1,535°C,
boiling point 3,000°C, specific gravity 7.847 (20°C), valences 2, 3, 4, 6. [Middle English yren, yron, iren, Old
English iren, earlier isern, isen] (page 691)
Copper: A ductile, maleable, reddish-brown metallic element that is an excellent conductor of heat and
56
Chapter 2: Atoms and Elements
electricity and is widely used for electrical wiring, water piping, and corrosion-resistant parts either pure or
in alloys such as brass and bronze. Atomic number 29, atomic weight 63.54, melting point 1,083°C, boiling
point 2,595°C, specific gravity 8.96, valences 1, 2. [Middle English coper, Old English coper,copor, from
common Germanic kupar (unattested), from Late Latin cuprum, from Latin Cyprium, “(copper) of Cyprus”
(Cyprus was known in ancient times as the source of the best copper”, from Cyprus, of Cyprus, from Greek
Kuprios, from Kupros, CYPRUS] (page 294)
78. New College Edition of the American Heritage Dictionary of the English Language (©1978) gives the
following information:
Chlorine: A highly irritating, greenish-yellow gaseous halogen, capable of combining with nearly all other
elements, produced principally by electrolysis of sodium chloride and used widely to purify water, as a
disinfectant, a bleaching agent, and in the manufacture of many important compounds including chloroform
and carbon tetrachloride. Atomic number 17, atomic weight 35.45, freezing point – 100.98°C, boiling point –
34.6°C, specific gravity 1.56 (– 33.6°C), valences 1, 3, 5, 7. [CHLORO- + -INE] (page 236)
Fluorine: A pale-yellow, highly corrosive, highly poisonous, gaseous halogen element, the most
electronegative and most reactive of all elements. It is used in a wide variety of industrially important
compounds. Atomic number 9, atomic weight 18.9984, freezing point – 219.62°C, boiling point – 188.94°C,
specific gravity of liquid 1.108, valences 1. [French, from New Latin fluor, generic name for a group of
minerals used as fluxes, FLUOR] (page 506)
79. Common transition elements are: iron, copper, chromium (Period 4). Halogens are the collection of elements
in Group 7A. Common halogens are: fluorine and chlorine.
Alkali metals are the collection of metallic elements in Group 1A (note: metallic means excluding hydrogen).
A common alkali metal is sodium.
80. Alkali metals are the collection of metallic elements in Group 1A (note: metallic means excluding hydrogen).
A common alkali metal is sodium. Alkaline earth metals are the collection of metallic elements in Group 2A.
A common alkaline earth metal is magnesium. Halogens are the collection of elements in Group 7A. A
common halogen is chlorine.
81. Madame Curie isolated radium and polonium. Radium (Ra) has an atomic number of 88. Polonium (Po) has
an atomic number of 84. New College Edition of the American Heritage Dictionary of the English Language
(©1978) gives the following word origin of “radium”: New Latin, from Latin radius ray (radium emits rays
that penetrate opaque matter). (page 1077) It gives the word origin of “polonium”: From Latin, Polonia,
Poland native country of its discoverers, the Curies.
82. There are five elements in Group 4A of the periodic table. They are non-metallic carbon (C), metalloids
silicon (Si) and germanium (Ge), and metals tin (Sn) and lead (Pb).
83. The fourth period of the periodic table has eighteen elements. They are the following metals: potassium (K),
calcium (Ca), scandium (Sc), titanium (Ti), vanadium (V), chromium (Cr), manganese (Mn), iron (Fe), cobalt
(Co), nickel (Ni), copper (Cu), zinc (Zn), gallium (Ga); the following metalloids: germanium (Ge) and arsenic
(As); and the following non-metals selenium (Se), bromine (Br), and krypton (Kr).
84. In the current periodic table, the period with the most known elements is the sixth period. It contains a full
transition metal series of ten elements, as well as the lanthanide series of 14 elements. The seventh period
will have as many elements as the sixth period, if and when the elements above atomic number 112 are ever
isolated.
85. Two periods of the periodic table have eight elements (periods 2 and 3). Two periods of the periodic table
have 18 elements (Periods 4 and 5). One period of the current periodic table has 32 elements (Period 6).
Period 7 is currently six elements short of the total of 32 elements – this is not because there aren’t 32, but
Chapter 2: Atoms and Elements
57
because no one has yet isolated and characterized them.
86. There are many answers to this Question. The following are examples.
(a) An element in Group 2A is magnesium (Mg).
(b) An element in the third period is sodium (Na).
(c) The element in the second period of Group 4A is carbon (C).
(d) The element in the third period in Group 6A is sulfur (S).
(e) The halogen in the fifth period is iodine (I).
(f) The alkaline earth element in the third period is magnesium (Mg).
(g) The noble gas element in the fourth period is krypton (Kr).
(h) The non-metal in Group 6A and the third period is sulfur (S).
(i) A metalloid in the fourth period is germanium (Ge).
87. There are many answers to this question. These are examples.
(a) An element in Group 2B is zinc (Zn).
(b) An element in the fifth period is xenon (Xe) (or any other element whose atomic number is within the
range 37–54).
(c) The element in the sixth period in Group 4A is lead (Pb).
(d) The element in the third period in Group 6A is sulfur (S).
(e) The alkali metal in the third period is sodium (Na).
(f) The noble gas in the fifth period is xenon (Xe).
(g) The element in Group 6A and the fourth period is selenium (Se). It is a non-metal.
(h) A metalloid in the fifth period is antimony (Sb) or tellurium (Te).
88. (a) According to the chart, the most abundant metal (of the first 36 elements) has atomic number 26. That
is iron.
(b) According to the chart, the most abundant nonmetal (of the first 36 elements) has atomic number 1.
That is hydrogen.
(c) According to the chart, the most abundant metalloid (of the first 36 elements) has atomic number 14.
That is silicon.
(d) According to the chart, the most abundant transition element (of the first 36 elements) has atomic
number 26. That is iron.
(e) Among the first 36 elements, three halogens are considered: fluorine (9), chlorine (17), and bromine (35).
Of these three, the most abundant is chlorine.
89. The chart showing the plot of relative abundance of the first 36 elements is given in Question 88. There is a
general trend showing that the lighter elements are more abundant than the heavier ones. Looking more
closely, even numbered elements are fairly consistently more abundant than the odd numbered elements on
either side of them. (Both of these general trends have exceptions.)
58
Chapter 2: Atoms and Elements
General Questions
90. Zinc (Zn, atomic number = 30, atomic weight = 65.39) Zinc is in the fourth period and in Group 2B of the
periodic table. When neutral, an atom of zinc has 30 protons and 30 electrons. Metallic elemental zinc is
gray. Salts of zinc are colorless. Zinc is non-toxic. It was discovered after the element copper was.
According to the New College Edition of the American Heritage Dictionary of the English Language, the
uses of zinc are “to form a wide variety of alloys including brass, bronze, German silver, various solders, and
nickel silver, in gavanizing iron and other metals, for electric fuses, anodes, and meter cases, and in roofing,
gutters, and various household objects.” (page 1489)
91. Define the problem: A known mass of sulfuric acid is in a solution with a given density and mass
percentage. Determine the volume in mL.
Develop a plan: Always start with the sample – here, the mass of sulfuric acid. Use the mass percentage as
a conversion factor between grams of sulfuric acid and grams of solution. Then use the density of the
solution as a conversion factor between grams and cubic centimeters. Then convert the cubic centimeters
into milliliters.
Execute the plan:
Notice: When you have masses of different things, make sure that you identify
clearly what the unit “grams” refers to.
100 grams of the solution contains 30.08 grams H2SO4
1 cubic centimeter of the solution weighs 1.285 grams.
125 gH2 SO 4 ×
100 gsolution
1cm3 solution 1mL solution
×
×
= 255 mL solution
38.08g H2SO 4 1.285 gsolution 1cm 3 solution
Check your answer: The units “g H2SO4” cancel properly, as do the units “g solution”. The method looks
right. Multiplying a number by approximately 2.5, then dividing by approximately half that, should give a
numerical answer that’s about twice as big. The numbers look to be the right size.
92. Define the problem: A distance is given in angstroms (Å), which are defined. Determine the distance in
nanometers and picometers.
Develop a plan: Use the given relationship between angstroms and meters as a conversion factor to get
from angstroms to meters. Then use the metric relationships between meters and the other two units to find
the distance in nanometers and picometers.
Execute the plan:
1.97 • ×
−10
1× 10
m
1 nm
×
= 0.197 nm
1•
1 × 10−9 m
1.97 • ×
−10
1× 10
m
1 pm
×
= 197 pm
1 •
1 × 10−12 m
Check your answers: The unit nanometer is larger than an angstrom, so the distance in nanometers should
be a smaller number. The unit picometer is smaller than an angstrom, so the distance in picometers should
be a larger number. These answers look right.
93. Define the problem: A distance is given nanometers. Determine the distance in meters and angstroms,
Chapter 2: Atoms and Elements
59
which is defined.
Develop a plan: Use the metric relationship to convert nanometers into meters, then use the given
relationships between meters and angstroms to determine the length in angstroms.
Execute the plan:
0.154 nm ×
0.154 nm ×
−9
1 × 10 m
= 1.54 × 10−10 m
1 nm
−9
1 × 10 m
1•
×
= 1.54 •
1 nm
1 × 10−10 m
Check your answers: The unit meter is much larger than a nanometer, so the distance in meters should be a
much smaller number. The unit angstrom is smaller than an nanometer, so the distance in angstroms should
be a larger number. These answers look right.
94. Define the problem: The edge length of a cube is given nanometers. Determine the volume of the cube in
cubic nanometers and in cubic centimeters.
Develop a plan: Cube the edge length in nanometers to get the volume of the cube in cubic nanometers.
Use metric relationships to convert nanometers into meters, then meters into centimeters. Cube the edge
length in centimeters to get the volume of the cube in cubic centimeters.
Execute the plan:
(
)
3
V = (edge length)3 = 0.563 nm = 0.178 nm 3
0.563 nm ×
−9
1 × 10
m 100 cm
×
= 5.63× 10−8 cm
1 nm
1m
3

−8

−22
3
V = (edge length)3 =  5.63 × 10 cm  = 1.78 × 10
cm


Check your answers: Cubing fractional quantities makes the number smaller. The first volume makes sense.
The unit centimeter is larger than a nanometer, so the volume in cubic centimeter should be a very small
number. These answers look right.
95. Define the problem: Given the mass of a sample of cisplatin along with its percentage platinum, determine
the grams of platinum.
Develop a plan: Always start with the sample. Convert the mass of compound to mass of platinum using
the percentage platinum as a conversion factor.
Execute the plan: 100 grams of cisplatin contains 65.0 grams of Pt.
1.53 g cisplatin ×
65.0 g Pt
= 0.995 g Pt
100 g cisplatin
Check your answer: The mass of Pt should be about
2
3
of the mass of the compound cisplatin.
96. Define the problem: Given the mass of a bag of fertilizer along with mass percentage of nitrogen-containing
compounds, phosphorus-containing compounds, and potassium-containing compounds in the fertilizer and
60
Chapter 2: Atoms and Elements
the mass percentage of phosphorus in the phosphorus-containing compounds, determine the grams of
phosphorus-containing compounds.
Develop a plan: Always start with the sample. Convert the bag’s mass in pounds to grams. Then use the
percentage of phosphorus-containing compounds in the fertilizer as a conversion factor to determine the
mass of phosphorus-containing compounds in the bag. Then use the mass percentage of phosphorus in
the phosphorus-containing compounds to determine how much phosphorus is in the bag.
Execute the plan: 100 grams fertilizer contains 4.0 grams of phosphorus-containing compounds (PCCs).
40.0 lb fertilizer ×
453 .59 g fertilizer
4.0 g PCCs
×
= 7.3 × 102 g PCCs
1 lb fertilizer
100 g fertilizer
7.3 × 102 g PCCs ×
43.64 g P
= 3.2 × 10 2 g P
100 g PCCs
Check your answers: The significant figures are limited to two by the 4.0 % figure. The mass units cancel
appropriately. The mass of phospohrus is less than half the mass of PCCs. These answers make sense.
Chapter 2: Atoms and Elements
61
97. Define the problem: Given the number of people in the city who use water, the volume of water each person
needs per day, the number of days in a year, the mass of one gallon of water, the fluoride concentration, and
the mass percentage of fluoride in sodium fluoride, determine the number of tons of sodium fluoride needed
per year.
Develop a plan: Always start with the sample. Given the number of people in the city, use the volume of
water each person needs per day, then calculate everyone’s water needs for the day. Using the number of
days in a year calculate the total volume of water used per year. Then using the mass of one gallon of water
as a conversion factor, determine the grams of water. Convert the grams to tons. Then, using the fluoride
concentration as a conversion factor, determine the number of tons of fluoride, and using the mass
percentage of fluoride in sodium fluoride to determine the number of tons.
Execute the plan:
175 gal water 365 days
×
1 day
1 year
8.34 lb water
1 ton water
150, 000 people ×
×
×
1 person
1 gal water
2000 lb water
×
1 ton fluoride
100 tons sodium fluoride
tons sodium fluoride
×
= 89
1, 000, 000 tons water
45 .0 tons flouride
year
Check your answer: The significant figures are limited to two by the 150,000 figure. The mass units are
appropriately labeled. The units cancel appropriately to give tons per year. This is a large number of people
using a large amount of water so the large quantity of sodium fluoride makes sense.
98. These are a few common elements seen in some people’s daily life:
Iron, in cast iron frying pan (Fe, Period 4, first column of Group 8B)
Carbon, in charcoal brickets (C, Period 2, Group 4A)
Copper, in a copper tea kettle (Cu, Period 4, Group 1B)
Gold in a ring (Au, Period 6, Group 1B)
Silver in a necklace (Ag, Period 5, Group 1B)
Platinum in a ring (Pt, Period 6, third column of Group 8B), etc.
99. Potassium’s atomic weight is 39.0983. The isotopes that contribute most to this mass are 39K and 41K,
since the problem tells us that 40K has a very low abundance. Since the atomic mass is closer to 39 than 41,
that confirms that the 39K isotope is more abundant.
100
The symbols of the isotopes are 20Ne, 21Ne, and 22Ne. The most abundant isotope is 20Ne, with 10
protons, 10 neutrons and 10 electrons. Because the most abundant isotope is 20Ne (90.92%), the
approximate atomic weight of neon is 20.
101. (a) Ti has atomic number = 22 and atomic weight = 47.88.
(b) Titanium is in Period 4 and Group 4B. The other elements in its group are zirconium (Zr), hafnium
(Hf), and rutherfordium (Rf).
(c) Titanium is light-weight and strong, making it a good choice for something that needs to be sturdy
and small.
(d) According to the New College Edition of the American Heritage Dictionary of the English Language
62
Chapter 2: Atoms and Elements
(©1978): Titanium: A strong, low-density, highly corrosion resistant, lustrous white metallic element
that occurs widely in igneous rocks and is used to alloy aircraft metals for low weight, strength, and
high-temperature stability. (page 1348)
102.
Helium-4 atom has 2 protons and 2 neutrons in the nucleus and 2 electrons outside the nucleus. (This
picture is not to scale – the nucleus is actually much smaller than indicated here.)
eG
+
2p
2 n0
eG
103. Define the problem: Given the carat mass of a diamond and the relationship between carat and milligrams,
determine how many moles of carbon are in the diamond.
Develop a plan: Always start with the sample. Diamond is an allotropic form of pure carbon. Given the
carats of the diamond, use the relationship between carats and milligrams as a conversion factor to
determine milligrams of carbon. Then using metric relationships to determine grams of carbon, and the
molar mass of carbon to determine the moles of carbon.
Execute the plan:
2.3 carats C ×
200 mg C
1gC
1 mol C
×
×
= 0.038 mol C
1 carat C 1000 mg C 12.01 g C
Check your answer: The significant figures are limited to two by the 2.3 figure. The units cancel
appropriately to give moles of carbon. This looks right.
104.
Define the problem: Given the relationship between troy ounces and grams, and the price of a troy ounce
of platinum, (a) determine the moles of platinum in 1 troy ounce and (b) determine the grams and moles of
platinum you bought after spending a certain amount of money.
(a) Develop a plan: The sample is 1 troy ounce of platinum. Use the given relationship between troy
ounces and grams as a conversion factor to get the grams of platinum and the molar mass of platinum
to get the moles of platinum.
Execute the plan:
1 troy ounce Pt ×
31 .1 g Pt
1 mol Pt
×
= 0.159 mol Pt
1 troy ouncePt 195 .08 gPt
(b) Develop a plan: The starting quantity is $5000, since that’s how much is spent. Use the given
relationship between dollars and troy ounces as a conversion factor to determine troy ounces of
platinum. Use the given relationship between troy ounces and grams as a conversion factor to get
the grams of platinum and then the molar mass of platinum to get the moles of platinum.
Execute the plan:
$5000 ×
1 troy ounce Pt
31.1 g Pt
×
= 478 g Pt
$325
1 troy ouncePt
Chapter 2: Atoms and Elements
478 g Pt ×
63
1 mol Pt
= 2.45 mol Pt
195 .08 g Pt
Check your answers: The units cancel properly and the calculated values appear to be the right size.
105. Define the problem: Given the moles of gold you want to buy, the relationship between troy ounces and
grams, and the price of a troy ounce of gold, determine the amount of money you must spend.
Develop a plan: The sample is the 1.00 mole of gold. Use the molar mass of gold to determine the grams
of gold and the given relationship between grams and troy ounces to determine the troy ounces of gold,
then use the price per troy ounce to determine how much that would cost.
Execute the plan:
1.00 mol Au ×
197 .0 g Au 1 troy ounce Au
$ 338 .70
×
×
= $2, 145 ≈ $2, 150
1 mol Au
31.1 g Au
1 troy ounceAu
Check your answer: The number of moles has three significant figures, so the answer must be reported
with three significant figures. The units cancel properly. A mole of gold is pretty heavy, so this seems like
a reasonable amount of money to spend.
106.
Define the problem: Given the mass of copper in the Statue of Liberty and the relationship between
pounds and grams, determine the total grams and moles of copper in the Statue of Liberty.
Develop a plan: The sample is the 2.00 × 105 lb copper. Use the given relationship between grams and
pounds to determine grams of copper, then use the molar mass of copper to determine the number of
moles of copper.
Execute the plan:
2.00 × 10 5 lb Cu ×
9.08 × 10 7 g Cu ×
454 g Cu
= 9.08 × 10 7 g Cu
1 lb Cu
1 mol Cu
= 1.43 × 10 6 mol Cu
63 .55 g Cu
Check your answers: Grams is a smaller unit of mass than pounds, so the mass in grams should be a
larger number. The number of moles should be smaller than the number of grams. The number of pounds
has three significant figures, so the answer must be reported with three significant figures. The units
cancel properly. These answers look right.
107. Define the problem: Given the dimensions of a piece of copper wire and the density of copper, determine
the moles of copper and the number of atoms of copper in the wire.
Develop a plan: Use the metric and English length relationships to convert the wire’s dimensions to
centimeters. Then use those dimensions to find its volume. Then use the density of copper to determine
the mass of the wire. Then use the molar mass of copper to determine the moles of copper and
Avogadro’s number to determine the actual number of copper atoms.
Execute the plan:
wire length in centimeters = L = 25 ft long ×
12 in 2.54 cm
×
= 7.6 × 102 cm long
1 ft
1 in
64
Chapter 2: Atoms and Elements
wire diameter in centimeters = d = 2.0 mm diameter ×
wire radius in centimeters = r =
1 m
100 cm
×
= 0.20 cm
1000 mm
1 m
d 0.20 cm
=
= 0.10 cm
2
2
cylindrical wire’s volume = V = A × L=(pr2) × L
V= (3.14159) × (0.10 cm) 2 × (7.6 × 102 cm) = 24 cm3
3
24 cm Cu ×
8.92 g Cu
1 cm 3 Cu
3.4 mol Cu atoms ×
×
1 mol Cu atoms
= 3.4 mol Cu atoms
63.55 g Cu
23
6.022 × 10 Cu atoms
= 2.0 × 1024 Cu atoms
1 mol Cu atoms
Check your answers: The length and diameter of the wire both have two significant figures, so the answer
must be reported with two significant figures. The units cancel properly. The number of atoms in a wire
you can see and hold is conveniently represented in the quantity unit “moles”. The actual number of
atoms is very large. These answers make sense.
Applying Concepts
108.
Check to see if the following are true: The atomic number must be the same as the number of protons and
the mass number must be the sum of the number of protons and neutrons.
(a) These values are not possible, since the atomic number (42) is not the same as the number of protons
(19), and the sum of protons and neutrons (19+23=42) is not the same as the mass number (19).
(b) These values are possible.
(c) These values are not possible. The sum of protons and neutrons (131+79=210) is not the same as the
mass number (53).
(d) These values are not possible. The sum of protons and neutrons (15+15=30) is not the same as the
mass number (32).
(e) These values are possible.
(f) These values are not possible. The sum of protons and neutrons (18+40=58) is not the same as the
mass number (40).
109. A particle is an atom or a molecule. The unit “mole” is a convenient way of describing a large quantity of
particles. It is also important to keep in mind that 1 mol of particles contains 6.022×1023 particles and the
molar mass describes the mass of 1 mol of particles.
(a) A sample containing 1 mol of Cl has the same number of particles as a sample containing 1 mol Cl2,
since each sample contains 1 mole of particles. (The masses of the samples are different, the numbers
of atoms contained in the samples are different, but those statements do not answer this question.)
(b) A sample containing 1 mol of O2 contains 6.022×1023 molecules of O2. This 1-mol sample has more
particles than a sample containing just 1 molecule of O2.
(c) Each sample contains only one particle. These samples have the same number of particles.
Chapter 2: Atoms and Elements
65
(d) A sample containing 6.022×1023 molecules of F2 contains 1 mol of F2 molecules. This sample has the
same number of particles as 6.022×1023 molecules of F2.
(e) The molar mass of Ne is 20.18 g/mol, so a sample of 20.2 grams of neon contains 1 mol of neon. Note
that, with the degree of certainty limited to one decimal place, the slightly smaller molar mass is
indistinguishable from the sample mass. This 20.2-gram sample has the same number of particles as
the sample with 1 mol of neon.
(f) The molar mass of bromine (Br2) is (79.9 g/mol Br)×(2 mol Br) = 159.8 g/mol, so a sample of 159.8
grams of bromine contains 1 mol bromine which equals 6.022×1023 molecules. This 159.8-gram
sample has more particles than a sample containing just 1 molecule of Br2.
(g) The molar mass of Ag is 107.9 g/mol, so a sample of 107.9 grams of Ag contains 1 mol of Ag. The
molar mass of Li is 6.9 g/mol, so a sample of 6.9 grams of Li contains 1 mol of Li. These samples have
the same number of particles, since each sample contains 1 mole of particles.
(h) The molar mass of Co is 58.9 g/mol, so a sample of 58.9 grams of Co contains 1 mol of Co. The molar
mass of Cu is 63.55 g/mol, so a sample of 58.9 grams of Cu contains less than 1 mol of Cu. The 58.9-g
sample of Co has more particles than the 58.9-g sample of Cu.
(i) A sample containing 6.022×1023 atoms of Ca contains 1 mol of Ca atoms. The molar mass of Ca is
23
40.1 g/mol, so a sample of 1 gram of Co contains less than 1 mol of Co. The 6.022×10 -atoms sample
has more particles than the 1-gram sample.
(j) A chlorine molecule contains two chlorine atoms. Since chlorine atoms all weigh the same, a chlorine
molecule weighs twice as much as a chlorine atom. If the two samples of chlorine both weigh the
same, the atomic chlorine (Cl) sample will have more particles in it than in the molecular sample (Cl2).
110. The unit “mole” is a convenient way of describing a large quantity of particles. It is also important to
keep in mind that 1 mol of particles contains 6.022×1023 particles and that the molar mass gives the grams
in one mol of particles.
(a) The molar mass of iron (Fe) is 55.85 g/mol. The molar mass of aluminum (Al) is 27.0 g/mol, so a 1 mol
sample of Fe has a greater mass than a 1 mol sample of Al.
(b) A sample of 6.022×1023 lead atoms contains 1 mol of lead. This sample will have the same mass as a
sample containing 1 mol of lead.
(c) 1 mol of copper contains 6.022×1023 copper atoms. This sample will have a greater mass than a
sample containing only 1 copper atom.
(d) A Cl2 molecule has twice the mass of one Cl atom. So, comparing samples with the same number of
particles, 1 mol, the Cl2 sample will have a greater mass than the Cl sample.
(e) A “gram” is a unit of mass. If both samples weigh 1 gram, then they have the same mass.
66
Chapter 2: Atoms and Elements
(f) The molar mass of magnesium (Mg) is 24.3 g/mol, so a sample weighing 24.3 grams contains 1 mol of
Mg. This sample will have the same mass as a sample containing 1 mol of Mg.
(g) The molar mass of Na is 23.0 g/mol, so a 1 mol sample of Na will weigh 23.0 grams. This sample has a
greater mass than a sample containing 1 gram of Na.
23
(h) The molar mass of He is 4.0 g/mol, so a 1 mol sample of He weighs 4.0 grams. A sample of 6.022×10
He atoms contains 1 mol of He. These two samples have the same mass.
23
(i) A sample with 1 mol of I2 contains 6.022×10 I2 molecules. This sample weighs more than a sample
that only contains 1 I2 molecule.
(j) An oxygen molecule (O2) has twice the mass of one O atom, so the O2 sample will have a greater
mass than the O sample.
More Challenging Problems
111. (a) Define the problem: Given the volume of a bottle containing water, the density of water and ice (at
different temperatures), determine the volume of ice.
Develop a plan: Always start with the sample. Convert the volume of water to grams using the
density of water. The entire mass of water freezes at the new temperature. Use the density of water as
a conversion factor to determine the volume of ice.
Execute the plan:
250 mL water ×
0.997 g water
1 g ice
1 mL ice
×
×
= 272 mL ice
1 mL water
1 g water 0.917 g ice
Check your answer: The density of ice is lower than the density of water, so the volume should be
greater once the water freezes.
(b) The ice could not be contained in the bottle. The volume of ice exceeds the capacity of the bottle.
112.
(a) To make a periodic table of these elements, arrange the atoms according to increasing atomic mass,
then line them up in columns according to similar properties.
R
A
E
Q
M
1.02
3.2
5.31
8.97
11.23
gas
solid
solid
liquid
gas
colorless
silvery
golden
colorless
colorless
very low
high
very high
very low
very low
very high
medium
medium
medium
very low
D
G
L
X
13.5
15.43
see
21.57
23.68
gas
solid
(c)
liquid
gas
colorless
silvery
below
colorless
colorless
very low
high
very low
very low
very high
medium
medium
very low
J
Ab
T
Z
see
27.89
29.85
33.85
36.2
Chapter 2: Atoms and Elements
(b)
below
solid
silvery
high
medium
solid
golden
very high
medium
solid
colorless
very low
medium
67
gas
colorless
very low
medium
(b) A new element, X, with atomic weight 25.84 would probably fit in the table under the D element; hence,
it would have properties similar to that group – namely, it would be a gaseous, colorless element with
very low electrical conductivity and very high reactivity.
(c) An element with an atomic weight about 17 is still missing. It will be a golden solid element, with very
high electrical conductivity and medium electrical reactivity.
113. Define the problem: Using the mass of several diatomic molecules of bromine that differ by iostopic
composition and the relative heights of their spectral peaks on a mass spectrum, determine the identity of
isotopes in the molecules, the mass of each isotope of bromine, the average atomic mass of bromine, and
the abundance of the isotopes.
Develop a plan: Identify which of the two isotopes contribute to each of the mass spectrum peaks. Then
use that information to find that atomic mass of each isotope. Use the relative peak heights and the
isotope masses to find the average molar mass of Br2 and the average atomic mass of Br, then establish
variables describing the isotope percentages. Set up two relationships between these variables. The sum
of the percents must be 100 %, and the weighted average of the molecular masses must be the reported
atomic mass.
Execute the plan:
(a) The peak representing the diatomic molecule with the lightest mass must be composed of two atoms
of the lightest isotopes. The peak representing the diatomic molecule with the largest mass must be
composed of two atoms of the heavier isotopes. The middle peak must represent molecules made
with one of each.
(b) The mass of the molecule composed of two lightweight atoms is 157.836 g/mol, so each atom must
weigh 0.5×(157.836 g/mol) = 78.918 g/mol. The mass of the molecule composed of two heavy weight
atoms is 161.832 g/mol, so each atom must weigh 0.5×(161.832 g/mol) = 80.916 g/mol.
(c) The average mass of the molecules would be the weighted average of these three variants:
79


79 81
 159.834 amu
25.54 atoms Br2
157.836 amu
49.99 atoms Br Br
 +

× 
×


100 Br2 molecules
79
100 Br2 molecules
79 81
 1 atom Br Br
 1 atom Br 2 

 +


81


24.46 atoms Br2
161.832 amu
amu
 = 159.803
× 

100 Br2 molecules
81
Br 2 molecules
 1 atom Br 2 
The diatomic molecule mass gets divided by two determine the atomic mass.0.5×(159.803 g/mol) =
79.902 g/mol.
(d) The X % 79Br and Y % 81Br, This means:
Every 100 atoms of silver contains X atoms of the 79Br isotope.
Every 100 atoms of silver contains Y atoms of the 81Br isotope.
X + Y = 100 %
Chapter 2: Atoms and Elements
68
79
 78.918 amu
X atoms Br
× 
100 Br atoms
79
 1 atom Br
91

Y atoms Br  80.916 amu 
amu
+
 = 79.902
×

100 Ag atoms  1 atom 91Br 
Br atom



We now have two equations and two unknowns, so we can solve for X and Y algebraically. Solve
the first equation for Y
Y = 100 – X
Plug that in for Y in the second equation. Then solve for X:
X
100 − X
× 78.918 +
× 80.916 = 79.902
100
100
(
)
(
)
0.78918 X + 80.916 − 0.80916 X = 79.902
80.916 − 79.902= 0.80916 X − 0.78918 X
(
)
1.014= 0.01998 X
X = 50.75
Now, plug the value of X in the first equation to get Y.
Y = 100 – X = 100 – 50.75 = 49.25
Therefore the abundance for these isotopes are: 50.75 % 89Br and 49.25 % 91Br
Check your answer: These isotopes are about equally abundant as indicated by the near 25:50:25 ratio of
the isotopes in the three mss spectrum peaks.
114. (a) K is an alkali metal. (Group 1A)
(b) Ar is a nobel ga.s (Group 8A)
(c) Cu is a transition metal. (Group 1B)
(d) Ge is a metalliod. (Group 4A)
(e) H is a group 1 nonmetal.
(f) Al is a metal that forms a 3+ ion. (Group 3A)
(g) O is a nonmetal that forms a 2– ion. (Group 6A)
(h) Ca is an alkaline earth metal. (Group 2A)
(i) Br is a halogen. (Group 7A)
(j) P is a nonmetal that is a solid. (Group 5A)
115. Total of ten molecules eight (80%) are N2 molecules (white dots) and two are O2 (20%) molecules (black
dots).
116. (a) O is a nonmetal in Group 6A. The atomic number is the same as the number of protons. The number
of protons is the same as the number of electrons in an uncharged atom. The element in Group 6A
that has 34 electrons is Se.
(b) The alkali metal group is Group 1A. To get the number of neutrons, subtract the atomic number (Z)
Chapter 2: Atoms and Elements
69
from the mass number (A). The element in Group 1A that has A - Z = 39 – 19 = 20 is 39K.
(c) A halogen is an element in Group 7A. The atomic number is the number of protons, and the sum of
the protons and the neutrons is the mass number. The element in Group 7A with Z=35 and A = (35 +
44 =) 79 is 79Br.
(d) A noble gas is an element in Group 8A. The atomic number is the number of protons, and the sum of
the protons and the neutrons is the mass number. The element in Group 8A with Z=10 and A = (10 +
10 =) 20 is 20Ne.
70
Chapter 2: Atoms and Elements
Conceptual Challenge Problems
CP-2.A. The ratio of the stacks should give whole number proportions.
ratio 2:1 =
15 .96 g
7
= 1.75 =
9.12 g
4
ratio 3:1 =
If Stack 1 has four dimes: mass of one dim e =
27 .36 g
3
= 3.00 =
9.12 g
1
27.36 g
12
= 1.714 =
15 .96 g
7
9.12 g
= 2.280 g
4
If Stack 2 has seven dimes: mass of one dim e =
If Stack 3 has 12 dimes: mass of one dim e =
ratio 3:2 =
15.96 g
= 2.280 g
7
27.36 g
= 2.280 g
12
The proposed mass of the dime is 2.280 g (or some integer multiple of this mass). All three masses are accounted
for in this description.
CP-2.B. 18 billion years ×
5.7 × 10
1,000 ,000 , 000 years 365 .25 days 24 h 3600 s
×
×
×
= 5.7 × 10 17 s
1 billion years
1 year
1 day
1h
17
C atoms ×
1 mole C
6.022 × 10
23
C atoms
×
12.0107 g C
= 0.000011 g C
1 mole C
This mass it to small to be detected on a balance that detects masses as small as 0.0001 g.