Chapter 2: Atoms and Elements - University of Dayton...Chapter 2: Atoms and Elements28 11 in × 2.54 cm 1 in = 27.94 cm 27.94 cm × 1 m 100 cm × 1000 mm 1 m = 279.4 mm 27.94 cm ×

Chapter 2: Atoms and Elements

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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, multiplication key, 10, exponent key, 23, which results in the value 6.022 x 1024. 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.


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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.


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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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11 in ×

2.54 cm1 in

= 27.94 cm

27.94 cm ×

1 m100 cm

×1000 mm

1 m= 279.4 mm

27.94 cm ×

1 m100 cm

= 0.2794 m

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 ×

100 cm1 m

= 614 cm

6.14 m ×

100 cm1 m

×1 in

2.54 cm= 242 cm

6.14 m ×

100 cm1 m

×1 in

2.54 cm×

1ft12 in

= 20.1 ft

When answering all three questions, we can use the results of the previous calculation to make the next calculation faster:

614 cm ×

1 in2.54 cm

= 242 in

242 in ×

1 ft12 in

= 20.1 ft

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:


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65 mi1 hr

×1.6093 km

1 mi= 104.6

kmhr

≅ 1.0 × 102 kmhr

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.

Execute the plan: 404 ft ×

12 in1 ft

×2.54 cm

1 in×

1 m100 cm

= 123 m

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.

Execute the plan: 10 ft ×

12 in1 ft

×2.54 cm

1 in×

1 m100 cm

= 3.048 m

10 ft ×

12 in1 ft

×2.54 cm

1 in= 304.8 cm

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 in3, 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:

120 in3 ×

2.54 cm1 in

×2.54 cm

1 in×

2.54 cm1 in

= 120 in 3 ×2.54cm

1in

3

= 2.0 × 103 cm3

2.0 × 103 cm 3 ×

1 mL

1 cm 3×

1 L1000 mL

= 2.0 L

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 relating a smaller number (10–3) of larger things (L) to a larger number (1) of small things.


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2.0 × 103 cm 3 ×

1 mL

1 cm3×

10−3 L1 mL

= 2.0 L

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. in3, 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: 250. in3 ×

2.54 cm1 in

3

= 4.10 × 103 cm3

250. in3 ×

2.54 cm1 in

3

×1 mL

1 cm 3×

1 L1000 mL

= 4.10 L

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 conversion relating a smaller number (10–3) of larger things (L) to a larger number (1) of small things.

250. in3 ×

2.54 cm1 in

3

×1 mL

1 cm 3×

10−3 L1 mL

= 4.10 L

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.

Execute the plan: 1 m 2 ×

100 cm1 m

2

×1 cm

2.54 cm

2

= 1550 in 2

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: 1 in3 ×

2.54 cm1 in

3

= 16.387 cm3


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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.

Execute the plan: height = 8 ft + 6 in ×

1 ft12 in

= 8.5 ft

V = (length) × (width) × (height) (then convert to m3)

V = (18 ft) × (15 ft) × (8.5 ft) ×

12 in1 ft

3

×2.54 cm

1 in

3

×1 m

100 cm

3

= 65 m 3

65 m3 ×

100 cm1 m

3

×1 mL

1 cm3×

1 L1000 mL

= 6.5 ×104 L

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 kg1000 g

= 0.00283 kg

2.83 g ×

1 lb453.59 g

= 0.00624 lb

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


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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 non-zero 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


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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× 101.

(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× 101.

(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


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× 102.

(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× 105.

(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


(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


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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.


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Execute the plan:

(a)

4.850 g − 2.34 g1.3 mL

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 g1.3 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.

(b) V = πr3 = (3.1415926) × (4.112 cm)3

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.

(c) (4.66 × 10–3) × 343.2

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.

(d)

0.00340065.2

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 .53.20

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

2221.05 − 1.02 × 103 = 2.22105 × 103 − 1.02 × 103


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The next calculation uses the subtraction rule. To count decimal places, both numbers must have the same power of ten:

= 2.22105 × 103 − 1.02 × 103

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× 103.

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.

(b) (343.2) × (2.01×10–3)

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.

(c) S = 4πr2 = 4 × (3.1415926) × (2.55 cm)2

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.

(d)

280215

− 0.0024 × 10, 000.( )

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× 102.

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,


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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 %.

Execute the plan:

14 .1 g silver17.6 g bracelet

× 100 % = 80.1 % silver

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 ×

453.59 g solder1 lb solder

×67 g lead

100 g solder= 3.0× 102 g lead

1.00 lb solder ×

453.59 g solder1 lb solder

×33 g tin

100 g solder= 1.5× 102 g tin

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 solution1.000 mL solution

×38.08 g sulfuric acid

100.00 g solution= 245 g sulfuric acid

NOTICE: When you are using grams of different substances, be careful to carry enough information in the


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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 lost0.125 g initial

× 100 % = 15 % mass lost

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 Na1000 mg Na

= 0.28 g Na

0.28 g Na30 g cereal

× 100 % = 0.9 % Na in 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.


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 112

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 35Cl100 Cl atoms

×34.9689 amu

1 atom 35Cl

+

Y atoms 37Cu100 Cl atoms

×36.9659 amu

1 atom 37Cl

= 35.453

amuCl atom

That gives us a second equation with the same two unknowns.


40

X100

× 34.9689 +Y

100× 36.9659 = 35.453

Solve the first equation for Y in terms of X, then substitute back into the second equation:

Y = 100 – X

X100

× 34.9689 +100 − X

100× 36.9659 = 35.453

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:


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

atomic symbol ZA 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

atomic symbol looks like this: 1123Na .

(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

atomic symbol looks like this: 1839Ar .

(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

atomic symbol looks like this: 3169Ga .

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

atomic symbol ZA 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

atomic symbol looks like this: 715 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

symbol looks like this: 3064Zn .

(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

atomic symbol looks like this: 54129Xe .

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.

45. Define the problem: Given the atomic symbol ZA 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:

(a) The isotope given is 2040Ca . 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.

(b) The isotope given is 50119Sn . That means A = 119 and Z = 50. So, the number of protons is 50, the


(c) The isotope given is 94244Pu . That means A = 244 and Z = 94. So, the number of protons is 94, the


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.

46. Define the problem: Given the atomic symbol ZA 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:

(a) The isotope given is 613C . 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.

(b) The isotope given is 2450Cr . 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


43

(50 – 24 =) 26.

(c) The isotope given is 83205Bi . 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.


44

Execute the plan:

Z

A

Number of Neutrons 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:

Z

A Number of Neutrons

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:

Z


Element

60 144 (a) (b)

(c) (d) 12 Mg

64 (e) 94 (f)


45

(g) 37 (h) 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:

Z


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.

49. The atomic symbol general form is ZA X , where A is the atomic mass and Z is the atomic number. Isotopes

will have Z the same, but A will be different. Therefore, 918 X, 9

20X, and 915 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

symbol ZA 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 2757Co



Check your answers: Isotopes must have the same symbols and the same Z values, but different A values. These answers look okay.


46

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:

Every 1000 atoms of lithium contains 75.00 atoms of the 6Li isotope.

Every 1000 atoms of lithium contains 925.0 atoms of the 7Li isotope.

75.00 atoms 6Li1000 Li atoms

×6.015121 amu

1 atom 6Li

+

925.0 atoms 7Li1000 Li atoms

×7.016003 amu

1 atom 7Li

= 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.


Execute the plan:

Every 10000 atoms of magnesium contains 7899 atoms of the 24Mg isotope, 1000 atoms of the 25

Mg

isotope, and 1101 atoms of the 26Mg isotope.

7899 atoms 24 Mg10000 Mg atoms

×23.985042 amu

1 atom 24Mg

+

1000 atoms 25Mg10000 Mg atoms

×24.98537 amu

1atom 25Mg

+1101 atoms 26Mg

10000 Mg atoms×

25.982593 amu

1atom 26Mg

= 24.31

amuMg 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.


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.

Execute the plan: X % 69Ga and Y % 71Ga, This means:

Every 100 atoms of gallium contains X atoms of the 69Ga isotope.

Every 100 atoms of gallium contains Y atoms of the 71Ga isotope.

X + Y = 100 %

X atoms 69Ga100 Ga atoms

×68.9257 amu

1 atom 69Ga

+

Y atoms 71Ga100 Ga atoms

×70.9249 amu

1 atom 71 Ga

= 69.723

amuGa 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:

X100

× 68.9257( )+100 − X

100× 70.9249( )= 69.723

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

Every 100 atoms of silver contains Y atoms of the 109Ag isotope.

X + Y = 100 %

X atoms 107Ag100 Ag atoms

×106.90509 amu

1 atom 107Ag

+

Y atoms 109Ag100 Ag atoms

×108.90476 amu

1 atom 109Ag

= 107.8682

amuAg atom


Y = 100 – X


X100

× 106.90509( )+100 − X

100× 108.90476( )= 107.8682

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


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 amu. Because lithium’s atomic weight (6.941 amu) is much closer to 7 amu than to 6 amu, the isotopic 7Li is more abundant than the isotope 6Li.

59. Knowing that almost all of the argon in nature is 40Ar, 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.


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 36Ar100000 Ar atoms

×35.968 amu

1 atom 36Ar

+

63 atoms 38Ar100000 Ar atoms

×37.963 amu

1 atom 38Ar

+99600 atoms 40Ar100000 Ar atoms

×39.962 amu

1 atom 40Ar

= 39.95

amuAr atom


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:

6.022 ×1023 pennies285, 000, 000 people

= 2.11×1015 pennies/person

2.11 × 1015 penniesperson

×1s

1 penny×

1 min .60 s

×1 hour60 min .

×1 work day

8 hours= 7.34 × 1010 workdays

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 B1 mol B

= 27 g 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

0.015 mol O2 ×

32.00 g O21 mol O2

= 0.48 g 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 Fe1 mol Fe

= 6.98 × 10−2 g 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 He1 mol He

= 2.61 × 103 g 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 Au1 mol Au

= 1.19× 103 g 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 U1 mol U

= 11 g 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 Ne1 mol Ne

= 315 g 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 Pu1 mol Pu

= 0.0886 g 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.


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.

127.08 g Cu ×

1 mol Cu63.546 g Cu

= 1.9998 mol 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 Ca40.08 g Ca

= 0.499 mol 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 Al26.982 g Al

= 0.6208 mol 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 K39.1 g K

= 3.1 × 10−4 mol 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.

5.0 mg Am ×

1 g Am1000 mg Am

×1 mol Am243 g Am

= 2.1 × 10−5 mol Am

Check your answers: Notice that the “grams” units cancel when the factor is multiplied, leaving the answer in moles.


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 Na22.99 g Na

= 0.696 mol Na


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(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 Pt195.1 g Pt

= 1.7 × 10−5 mol 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 P30.97 g P

= 0.0497 mol 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 As74.92 g As

= 0.0117 mol As


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(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 Xe131.3 g Xe

= 7.49 × 10−3 mol 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.

50.4g Na ×

1 mol Na22.99 g Na

= 2.19 mol 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.


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 Kr83.80 g Kr

= 9.42 × 10−5 mol 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 He4.0026 g He

= 1.1× 10−3 mol 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

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 ×

22.99 g Na1 mol Na

×1 cm 3

0.968 g Na= 2.97 cm 3

length = volume3 = 2.97 cm 33 = 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 ×

1 mol Cr atoms51.996 g Cr

×6.0221 ×1023 Cr atoms

1 mol Cr atoms= 4.131 × 1023 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 ×

1 mol Au atoms197.0 g Au

×6.022 ×1023 Au atoms

1 mol Au atoms= 5.93 × 1021 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 gCu1 mol Cu atoms

=1.055 × 10−22 gCu

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.


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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 × 1023 Ti atoms×

47.88 g Ti1 mol Ti atoms

= 7.951 ×10−23 g Ti

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 corrosion-resistant 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

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


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.)


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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 gH2SO4 ×

100 gsolution38.08g H2SO4

×1cm3 solution

1.285 gsolution×

1mL solution

1cm 3 solution= 255 mL 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 • ×

1× 10−10 m1 •

×1 nm

1 × 10−9 m= 0.197 nm

1.97 • ×

1× 10−10 m1 •

×1 pm

1 × 10−12 m= 197 pm

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,


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 ×

1 ×10−9 m1 nm

= 1.54 × 10−10 m

0.154 nm ×

1 × 10−9 m1 nm

×1•

1 × 10−10 m= 1.54 •

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:

V = (edge length)3 = 0.563 nm( )3 = 0.178 nm 3

0.563 nm ×

1 ×10−9 m1 nm

×100 cm

1 m= 5.63× 10−8 cm

V = (edge length)3 =

5.63 × 10−8 cm

3= 1.78 × 10−22 cm 3

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 Pt100 g cisplatin

= 0.995 g Pt

Check your answer: The mass of Pt should be about 23

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

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 fertilizer1 lb fertilizer

×4.0 g PCCs

100 g fertilizer= 7.3 × 102 g PCCs

7.3 × 102 g PCCs ×

43.64 g P100 g PCCs

= 3.2 × 102 g P

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.


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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:

150, 000 people ×

175 gal water1 day

×365 days1 year

1 person×

8.34 lb water1 gal water

×1 ton water

2000 lb water

×

1 ton fluoride1, 000, 000 tons water

×100 tons sodium fluoride

45.0 tons flouride= 89

tonssodium fluorideyear

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


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(©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

2 p+

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 C1 carat C

×1 g C

1000 mg C×

1 mol C12.01 g C

= 0.038 mol 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 ×

31.1g Pt1 troy ouncePt

×1 mol Pt

195.08 gPt= 0.159 mol Pt

(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$325

×31.1 g Pt

1 troy ouncePt= 478 g Pt


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478 g Pt ×

1 mol Pt195.08 g Pt

= 2.45 mol 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 Au1 mol Au

×1 troy ounce Au

31.1 g Au×

$ 338.701 troy ounceAu

= $2, 145 ≈ $2, 150

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 × 105 lb Cu ×

454 g Cu1 lb Cu

= 9.08 × 107 g Cu

9.08 × 107 g Cu ×

1 mol Cu63.55 gCu

= 1.43 × 106 mol 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 in1 ft

×2.54 cm

1 in= 7.6 × 102 cm long


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wire diameter in centimeters = d = 2.0 mm diameter ×

1 m1000 mm

×100 cm

1 m= 0.20 cm

wire radius in centimeters = r =

d2

=0.20 cm

2 = 0.10 cm

cylindrical wire’s volume = V = A × L=(pr2) × L

V= (3.14159) × (0.10 cm)2 × (7.6 × 102 cm) = 24 cm3

24 cm3 Cu ×

8.92 g Cu

1 cm 3 Cu×

1 mol Cu atoms63.55 g Cu

= 3.4 mol Cu atoms

3.4 mol Cu atoms ×

6.022 × 1023 Cu atoms1 mol Cu atoms

= 2.0 ×1024 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.


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(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 40.1 g/mol, so a sample of 1 gram of Co contains less than 1 mol of Co. The 6.022×1023-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.


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(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.

(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 6.022×1023 He atoms contains 1 mol of He. These two samples have the same mass.

(i) A sample with 1 mol of I2 contains 6.022×1023 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 water1 mL water

×1 g ice

1 g water×

1 mL ice0.917 g ice

= 272 mL 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 1.02 gas

colorless very low very high

A 3.2

solid silvery high

medium

E 5.31 solid

golden very high medium

Q 8.97

liquid colorless very low medium

M 11.23 gas

colorless very low very low

D 13.5 gas

colorless very low very high

G 15.43 solid

silvery high

medium

see (c)

below

L 21.57 liquid

colorless very low medium

X 23.68 gas

colorless very low very low

see

J 27.89

Ab 29.85

T 33.85

Z 36.2


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(b) below

solid silvery high

medium

solid golden

very high medium

solid colorless very low medium

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:

25.54 atoms 79Br2100 Br2 molecules

×157.836 amu

1 atom 79Br2

+

49.99 atoms 79Br81Br100 Br2 molecules

×159.834 amu

1 atom 79Br81Br

+

24.46 atoms 81Br2100 Br2 molecules

×161.832 amu

1 atom 81Br2

=159.803

amuBr2 molecules

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 %


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X atoms 79Br100 Br atoms

×78.918 amu

1 atom 79Br

+

Y atoms 91Br100 Ag atoms

×80.916 amu

1 atom 91Br

= 79.902

amuBr atom


Y = 100 – X


X100

× 78.918( ) +100 − X

100× 80.916( )= 79.902

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


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)


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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.


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Conceptual Challenge Problems

CP-2.A. The ratio of the stacks should give whole number proportions.

ratio2:1 =

15.96 g9.12 g

= 1.75 =74

ratio3:1 =

27.36 g9.12 g

= 3.00 =31

ratio3:2 =

27.36 g15.96 g

= 1.714 =127

If Stack 1 has four dimes: mass of one dim e =

9.12 g4

= 2.280 g

If Stack 2 has seven dimes: mass of one dim e =

15.96 g7

= 2.280 g

If Stack 3 has 12 dimes: mass of one dim e =

27.36 g12

= 2.280 g

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 ×

1,000,000, 000 years1 billion years

×365.25 days

1 year×

24 h1 day

×3600 s

1 h= 5.7 × 1017 s

5.7 × 1017 C atoms ×

1 mole C

6.022 ×1023 C atoms×

12.0107 g C1 mole C

= 0.000011 g C

This mass it to small to be detected on a balance that detects masses as small as 0.0001 g.

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Chapter 2: Atoms and Elements - University of Dayton...Chapter 2: Atoms and Elements28 11 in × 2.54 cm 1 in = 27.94 cm 27.94 cm × 1 m 100 cm × 1000 mm 1 m = 279.4 mm 27.94 cm ×