Issuu

Solutions Manual for Chemistry 14th Edition By Jason Overby, Raymond Chang (All Chapters 1-25) CHAPTER 1 CHEMISTRY: THE STUDY OF CHANGE Problem Categories Biological: 1.12, 1.36, 1.75, 1.76, 1.84, 1.90, 1.100, 1.101, 1.109, 1.114. Conceptual: 1.3, 1.4, 1.51, 1.52, 1.57, 1.58, 1.60, 1.68, 1.95, 1.98, 1.105, 1.107, 1.116. Environmental: 1.43, 1.76, 1.93, 1.95, 1.102, 1.112. Industrial: 1.39, 1.61, 1.87.

1.3

(a)

Quantitative. This statement clearly involves a measurable distance.

(b)

Qualitative. This is a value judgment. There is no numerical scale of measurement for artistic excellence.

(c)

Qualitative. If the numerical values for the densities of ice and water were given, it would be a quantitative statement.

(d)

Qualitative. Another value judgment.

(e)

Qualitative. Even though numbers are involved, they are not the result of measurement.

1.4

(a)

hypothesis

1.9

density =

1.10

Strategy: We are given the density and volume of a liquid and asked to calculate the mass of the liquid.

(b)

law

(c)

theory

mass 586 g = = 3.12 g /mL volume 188 mL

Rearrange the density equation, Equation (1.1) of the text, to solve for mass. density =

mass volume

Solution: mass = density ´ volume mass of methanol =

0.7918 g ´ 89.9 mL = 71.2 g 1 mL

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

1.11

?°C = (°F - 32°F)´

(a) (b) (c) (d) (e)

5°C 9°F

5°C = 35°C 9°F 5°C ?°C = (12 - 32)°F ´ = -11°C 9° F 5°C ?°C = (102 - 32)°F ´ = 39°C 9°F 5°C ?°C = (1852 - 32)°F ´ = 1011°C 9°F æ 9°F ö÷ ?°F = çç°C´ ÷ + 32°F çè 5°C ÷ø ?°C = (95 - 32)°F ´

æ 9°F ö÷ ?°F = çç-273.15°C ´ ÷ + 32°F = -459.67°F çè 5°C ÷ø

1.12

Strategy: Find the appropriate equations for converting between Fahrenheit and Celsius and between

Celsius and Fahrenheit given in Section 1.7 of the text. Substitute the temperature values given in the problem into the appropriate equation. (a)

Conversion from Fahrenheit to Celsius. ?°C = (°F - 32°F)´

5°C 9°F

?°C = (105 - 32)°F ´

(b)

5°C = 41°C 9°F

Conversion from Celsius to Fahrenheit.

æ 9°F ö÷ ?°F = çç°C´ ÷ + 32°F çè 5°C ÷ø æ 9°F ö÷ ?°F = çç-11.5 °C ´ ÷ + 32°F = 11.3°F 5°C ÷ø èç

(c)

Conversion from Celsius to Fahrenheit.

æ 9°F ö÷ ?°F = çç°C´ ÷ + 32°F çè 5°C ÷ø æ 9°F ö÷ ?°F = çç6.3´103°C ´ ÷ + 32°F = 1.1´10 4 °F 5°C ÷ø èç

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

(d)

Conversion from Fahrenheit to Celsius. ? °C = (°F - 32°F)´

5°C 9°F

? °C = (451 - 32)°F ´

1.13

1.14

K = (°C + 273°C)

5°C = 233°C 9°F

1K 1°C

(a)

K = 113°C + 273°C = 386 K

(b)

K = 37°C + 273°C = 3.10 ´ 10 K

(c)

K = 357°C + 273°C = 6.30 ´ 10 K

(a)

K = (°C + 273°C)

1°C °C = K - 273 = 77 K - 273 = -196°C

(b)

°C = 4.2 K - 273 = -269°C

(c)

°C = 601 K - 273 = 328°C

1.17

(a)

2.7 ´ 10-

1.18

(a)

10- indicates that the decimal point must be moved two places to the left.

(b)

3.56 ´ 10

(c)

4.7764 ´ 10

9.6 ´ 10-

(d)

1.52 ´ 10- = 0.0152 2

(b)

10- indicates that the decimal point must be moved 8 places to the left. 8

7.78 ´ 10- = 0.0000000778 8

1.19

(a)

145.75 + (2.3 ´ 10- ) = 145.75 + 0.23 = 1.4598 ´ 10

(b)

79500 7.95´10 4 = = 3.2´10 2 2.5´102 2.5´102

(c)

(7.0 ´ 10- ) - (8.0 ´ 10- ) = (7.0 ´ 10- ) - (0.80 ´ 10- ) = 6.2 ´ 10-

(d)

(1.0 ´ 10 ) ´ (9.9 ´ 10 ) = 9.9 ´ 10

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

1.20

(a)

Addition using scientific notation. n

Strategy: Let’s express scientific notation as N ´ 10 . When adding numbers using scientific notation, we

must write each quantity with the same exponent, n. We can then add the N parts of the numbers, keeping the exponent, n, the same. Solution: Write each quantity with the same exponent, n. n

Let’s write 0.0095 in such a way that n = -3. We have decreased 10 by 10 , so we must increase N by 10 . Move the decimal point 3 places to the right. 0.0095 = 9.5 ´ 10-

Add the N parts of the numbers, keeping the exponent, n, the same. 9.5 ´ 10-

+ 8.5 ´ 10-

18.0 ´ 10-

The usual practice is to express N as a number between 1 and 10. Since we must decrease N by a factor of 10 n

to express N between 1 and 10 (1.8), we must increase 10 by a factor of 10. The exponent, n, is increased by 1 from -3 to -2. 18.0 ´ 10- = 1.8 ´ 103

(b)

Division using scientific notation. n

Strategy: Let’s express scientific notation as N ´ 10 . When dividing numbers using scientific notation,

divide the N parts of the numbers in the usual way. To come up with the correct exponent, n, we subtract the exponents. Solution: Make sure that all numbers are expressed in scientific notation.

653 = 6.53 ´ 10

Divide the N parts of the numbers in the usual way. 6.53 ¸ 5.75 = 1.14 Subtract the exponents, n.

1.14 ´ 10+ - - = 1.14 ´ 10+ + = 1.14 ´ 10 2

(

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

(c)

Subtraction using scientific notation. n

Strategy: Let’s express scientific notation as N ´ 10 . When subtracting numbers using scientific notation,

we must write each quantity with the same exponent, n. We can then subtract the N parts of the numbers, keeping the exponent, n, the same. Solution: Write each quantity with the same exponent, n.

Let’s write 850,000 in such a way that n = 5. This means to move the decimal point five places to the left. 850,000 = 8.5 ´ 10

Subtract the N parts of the numbers, keeping the exponent, n, the same. 8.5 ´ 10

-9.0 ´ 10

-0.5 ´ 10

The usual practice is to express N as a number between 1 and 10. Since we must increase N by a factor of 10 n

to express N between 1 and 10 (5), we must decrease 10 by a factor of 10. The exponent, n, is decreased by 1 from 5 to 4. 5

-0.5 ´ 10 = -5 ´ 10

(d)

Multiplication using scientific notation. n

Strategy: Let’s express scientific notation as N ´ 10 . When multiplying numbers using scientific notation,

multiply the N parts of the numbers in the usual way. To come up with the correct exponent, n, we add the exponents. Solution: Multiply the N parts of the numbers in the usual way.

3.6 ´ 3.6 = 13 Add the exponents, n.

13 ´ 10- + + = 13 ´ 10 4

(

The usual practice is to express N as a number between 1 and 10. Since we must decrease N by a factor of 10 n

to express N between 1 and 10 (1.3), we must increase 10 by a factor of 10. The exponent, n, is increased by 1 from 2 to 3. 2

13 ´ 10 = 1.3 ´ 10

1.21

(a)

four

(b)

two

(c)

five

(d)

two, three, or four

(e)

three

(f)

one

(g)

one

(h)

two

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

(a)

one

(b)

three

(c)

three

(e)

two or three

(f)

one

(g)

one or two

1.23

(a)

10.6 m

1.24

(a) Division

1.22

(b)

0.79 g

16.5 cm

(c)

(d)

four

1 × 10 g/cm

Strategy: The number of significant figures in the answer is determined by the original number having the

smallest number of significant figures. Solution: 7.310 km = 1.283 5.70 km

The 3 (bolded) is a nonsignificant digit because the original number 5.70 only has three significant digits. Therefore, the answer has only three significant digits. The correct answer rounded off to the correct number of significant figures is: 1.28 (b)

(Why are there no units?)

Subtraction

Strategy: The number of significant figures to the right of the decimal point in the answer is determined by

the lowest number of digits to the right of the decimal point in any of the original numbers. Solution: Writing both numbers in decimal notation, we have

0.00326 mg - 0.0000788 mg

0.0031812 mg The bolded numbers are nonsignificant digits because the number 0.00326 has five digits to the right of the decimal point. Therefore, we carry five digits to the right of the decimal point in our answer. The correct answer rounded off to the correct number of significant figures is: 0.00318 mg = 3.18 ´ 10- mg 3

(c)

Addition

Strategy: The number of significant figures to the right of the decimal point in the answer is determined by

the lowest number of digits to the right of the decimal point in any of the original numbers.

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

Solution: Writing both numbers with exponents = +7, we have 7

(0.402 ´ 10 dm) + (7.74 ´ 10 dm) = 8.14 ´ 10 dm 7

Since 7.74 ´ 10 has only two digits to the right of the decimal point, two digits are carried to the right of the decimal point in the final answer. (d)

Subtraction, addition, and division

Strategy: For subtraction and addition, the number of significant figures to the right of the decimal point in

that part of the calculation is determined by the lowest number of digits to the right of the decimal point in any of the original numbers. For the division part of the calculation, the number of significant figures in the answer is determined by the number having the smallest number of significant figures. First, perform the subtraction and addition parts to the correct number of significant figures, and then perform the division. Solution:

(7.8 m - 0.34 m) 7.5 m = = 3.8 m /s (1.15 s + 0.82 s) 1.97 s

1.25

Calculating the mean for each set of data, we find: Student A: 87.6 mL Student B: 87.1 mL Student C: 87.8 mL From these calculations, we can conclude that the volume measurements made by Student B were the most accurate of the three students. The precision in the measurements made by both students B and C are fairly high, while the measurements made by student A are less precise. In summary: Student A: neither accurate nor precise Student B: both accurate and precise Student C: precise, but not accurate

1.26

Calculating the mean for each set of data, we find: Tailor X: 31.5 in Tailor Y: 32.6 in Tailor Z: 32.1 in From these calculations, we can conclude that the seam measurements made by Tailor Z were the most accurate of the three tailors. The precision in the measurements made by both tailors X and Z are fairly high, while the measurements made by tailor Y are less precise. In summary: Tailor X: most precise Tailor Y: least accurate and least precise Tailor Z: most accurate

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

1.27

1 dm = 226 dm 0.1 m

(a)

? dm = 22.6 m ´

(b)

? kg = 25.4 mg´

(c)

? L = 556 mL ´

(d)

g 10.6 kg 1000 g æçç1´10-2 m ö÷÷ 3 = ´ ´ç ? ÷ = 0.0106 g /cm 1 kg ççè 1 cm ÷÷ø 1 m3 cm 3

0.001g 1mg

1 ´ 10-3 L 1 mL

1kg 1000 g

= 2.54´10-5 kg

= 0.556 L 3

1.28

(a) Strategy: The problem may be stated as

? mg = 242 lb A relationship between pounds and grams is given on the end sheet of your text (1 lb = 453.6 g). This relationship will allow conversion from pounds to grams. A metric conversion is then needed to convert grams to milligrams (1 mg = 1 ´ 10- g). Arrange the appropriate conversion factors so that pounds and 3

grams cancel, and the unit milligrams is obtained in your answer. Solution: The sequence of conversions is

lb  grams  mg Using the following conversion factors,

453.6 g 1 lb

1 mg 1´10-3 g

we obtain the answer in one step: ? mg = 242 lb ´

453.6 g 1 mg ´ = 1.10´10 8 mg 1 lb 1´10-3 g

Check: Does your answer seem reasonable? Should 242 lb be equivalent to 110 million mg? How many

mg are in 1 lb? There are 453,600 mg in 1 lb. (b) Strategy: The problem may be stated as 3

? m = 68.3 cm

Recall that 1 cm = 1 ´ 10- m. We need to set up a conversion factor to convert from cm to m . 2

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

Solution: We need the following conversion factor so that centimeters cancel and we end up with meters.

1´10-2 m 1 cm Since this conversion factor deals with length and we want volume, it must therefore be cubed to give 1´10-2 m 1´10-2 m 1´10-2 m æç1´10-2 m ÷ö ÷ ´ ´ = çç çè 1 cm ÷ø÷ 1 cm 1 cm 1 cm

We can write æ1´10-2 m ö÷ ÷ = 6.83 ´10-5 m 3 ? m 3 = 68.3 cm 3 ´ççç çè 1 cm ÷÷ø 3

Check: We know that 1 cm = 1 ´ 10- m . We started with 6.83 ´ 10 cm . Multiplying this quantity by 3

1 ´ 10- gives 6.83 ´ 10- . 6

? L = 7.2 m

In Chapter 1 of the text, a conversion is given between liters and cm (1 L = 1000 cm ). If we can convert m

to cm , we can then convert to liters. Recall that 1 cm = 1 ´ 10- m. We need to set up two conversion 3

factors to convert from m to L. Arrange the appropriate conversion factors so that m and cm cancel, and the unit liters is obtained in your answer. Solution: The sequence of conversions is 3

m  cm  L Using the following conversion factors, æ 1 cm ö÷ çç ÷ çè1´10-2 m ÷ø

1L 1000 cm 3

the answer is obtained in one step: æ 1 cm ö÷ 1L = 7.2 ´10 3 L ? L = 7.2 m 3 ´çç ÷ ´ çè1´10-2 m ÷ø 1000 cm 3 3

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

Check: From the above conversion factors you can show that 1 m = 1 ´ 10 L. Therefore, 7 m would 3

equal 7 ´ 10 L, which is close to the answer. (d) Strategy: The problem may be stated as

? lb = 28.3 mg A relationship between pounds and grams is given on the end sheet of your text (1 lb = 453.6 g). This relationship will allow conversion from grams to pounds. If we can convert from mg to grams, we can then convert from grams to pounds. Recall that 1 mg = 1 ´ 10- g. Arrange the appropriate conversion factors so 6

that mg and grams cancel, and the unit pounds is obtained in your answer. Solution: The sequence of conversions is mg  g  lb

Using the following conversion factors,

1´10-6 g 1 mg

1 lb 453.6 g

we can write

? lb = 28.3 mg´

1´10-6 g 1 lb ´ = 6.24´10-8 lb 1 mg 453.6 g

Check: Does the answer seem reasonable? What number does the prefix m represent? Should 28.3 mg be a

very small mass?

1.29

3600 s 1255 m 1 mi ´ ´ = 2808 mi /h 1s 1609 m 1h

1.30

Strategy: The problem may be stated as

? s = 365.24 days You should know conversion factors that will allow you to convert between days and hours, between hours and minutes, and between minutes and seconds. Make sure to arrange the conversion factors so that days, hours, and minutes cancel, leaving units of seconds for the answer. Solution: The sequence of conversions is

days  hours  minutes  seconds

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

Using the following conversion factors,

24 h 1 day

60 min 1h

60 s 1 min

we can write

24 h 60 min 60 s ? s = 365.24 day´ ´ ´ = 3.1557 ´107 s 1 day 1h 1 min Check: Does your answer seem reasonable? Should there be a very large number of seconds in 1 year? 1.609 km 1000 m 1s 1 min ´ ´ ´ = 8.3 min 1 mi 1 km 3.00 ´108 m 60 s 1 mi 5280 ft 12 in 1 min = 118 in/s ? in/s = ´ ´ ´ 8.92 min 1 mi 1 ft 60 s

1.31

(93´106 mi)´

1.32

(a)

1.33

(b)

? m /min =

(c)

? km/h =

6.0 ft ´

1 mi 1609 m ´ = 1.80 ´10 2 m /min 8.92 min 1 mi

1 mi 1609 m 1 km 60 min = 10.8 km/h ´ ´ ´ 8.92 min 1 mi 1000 m 1h

1m = 1.8 m 3.28 ft

168 lb´

453.6 g 1 kg ´ = 76.2 kg 1 lb 1000 g

286 km 1 mi = 178 mph ´ 1h 1.609 km

1.34

? mph =

1.35

62 m 1 mi 3600 s ´ ´ = 1.4´10 2 mph 1 s 1609 m 1h

1.36

0.62 ppm Pb =

0.62 g Pb 1´106 g blood

6.0 ´103 g of blood ´

1.37

0.62 g Pb = 3.7 ´10-3 g Pb 1´106 g blood

365 day 24 h 3600 s 3.00´108 m 1 mi ´ ´ ´ ´ = 8.35´1012 mi 1 yr 1 day 1h 1s 1609 m

(a)

1.42 yr ´

(b)

32.4 yd ´

36 in 2.54 cm ´ = 2.96´103 cm 1 yd 1 in

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

1.38

(c)

3.0´1010 cm 1 in 1 ft ´ ´ = 9.8´108 ft /s 1s 2.54 cm 12 in

(a)

? lbs = 70 kg´

(b)

? s = 14´109 yr ´

(c)

? m = 90 in ´

(d)

æ 1 cm ö÷ 1L ? L = 88.6 m3 ´çç = 8.86´104 L ÷´ èç1´10-2 m ÷ø 1000 cm3

1 lb = 1.5´102 lbs 0.4536 kg 365 d 24 h 3600 s ´ ´ = 4.4´1017 s 1 yr 1d 1h

2.54 cm 1´10-2 m ´ = 2.3 m 1 in 1 cm 3

1.39

density =

2.70 g 1 kg æç 1 cm ö÷ ´ ´ç ÷ = 2.70´10 3 kg /m 3 1 cm 3 1000 g èç 0.01 m ø÷

1.40

density =

0.625 g 1L 1 mL ´ ´ = 6.25´10-4 g /cm 3 1L 1000 mL 1 cm3

1.41

Basic approach:

· Estimate the mass of one ant. In this problem a reasonable estimate is provided, but in future problems of this type you will need to provide such estimates and look up certain constants. See page 24 of your textbook for some general advice on looking up chemical information. 23

· Multiply by 6 ´ 10 to obtain the mass of one mole of ants, and convert mg to kg. The mass of one mole of ants in kilograms is 6 ´1023 ´3 mg ´

1´10 –3 g 1 kg ´ » 2 ´1018 kg 1 mg 1000 g

It is interesting to compare this mass with the total mass of Earth’s human population. The world population is very close to 6.9 billion people. Assuming an average body mass of 150 lb, we write

6.9´109 ´150 lb´

453.6 g 1 kg ´ » 5´1011 kg 1 lb 1000 g

Thus, one mole of ants outweigh all humans by 4 million times!

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

1.42

Basic approach: · Assume an average amount of time that a person sleeps in one night. · Multiple that value by the number of nights in 80 years. We assume that an average person sleeps eight hours a night. Although an infant or a young child sleeps more than eight hours a day, as one grows older, one tends to sleep less and less. Thus, the time spent sleeping in 80 years is

8 h sleep 365 day 1 day 1 yr ´ ´80 yr ´ ´ » 27 yr 1 day 1 yr 24 h 365 day So regardless of an adult’s age, on average, a third of his/her life is spent sleeping!

1.43

Basic approach: · List the major ways water is used indoors by a typical family. · Estimate the volume of water used according to the activity and the number of times that activity occurs in one day. · Calculate total volume of water used in one day. We can estimate the usage of water in the following categories: Showers Each person takes one shower per day; each shower lasts for 8 minutes; 2 gallons of water used per minute. 4 people ´ 1 shower/person ´ 8 minutes/shower ´ 2 gallons/minute = 64 gallons Washing Hands Each person washes hands six times a day; each washing uses 0.5 gallons of water. 4 people ´ 6 washings/person ´ 0.5 gallon/washing = 12 gallons Brushing Teeth Each person brushes teeth twice a day; each brushing uses 0.5 gallon of water. 4 people ´ 2 brushings/person ´ 0.5 gallon/brushing = 4 gallons Flushing Toilets Each person uses the toilet four times a day; each flushing uses 1.5 gallons of water. 4 people ´ 4 flushes/person ´ 1.5 gallons/flush = 24 gallons Dishwasher It is used twice a day; for newer dishwashers, each wash uses 6 gallons of water. 2 dishwashing ´ 6 gallons/dishwashing = 12 gallons

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

Laundry Seven loads of laundry per week; each load uses 40 gallons of water. 1 week ´ 7 loads/week ´ 40 gallons/load = 40 gallons 7 Other Miscellaneous Uses (washing dishes by hand, watering plants, drinking, etc.)

20 gallons Summing all of the activities gives 176 gallons. Obviously this is a very rough estimate, and the amount of water used by a family of four in one day will vary considerably depending on the habits of that family; however, this estimate is almost certainly within a factor of two, and it is noteworthy that the daily water consumption of a typical US family is considerable and rather extravagant by international standards.

1.44

Basic approach: 3

· Calculate the volume of a bowling ball in cm . · Convert the mass of the ball to grams. 3

· Calculate the density of the bowling ball and compare to the density of water (1 g/cm ). The volume of the bowling ball is given by

4 3 4 æç 8.6 in ö÷ æç 2.54 cm ö÷ pr = p ç ÷ ´ç ÷ = 5.5´103 cm3 3 3 èç 2 ÷ø èç 1 in ÷ø 3

Starting with an 8 lb bowling ball and assuming two significant figures in the mass, converting pounds to grams gives 8 lb ´

453.6 g = 3.6 ´103 g 1 lb

So the density of an 8 lb bowling ball would be

density =

mass 3.6 ´103 g = = 0.65 g/cm3 volume 5.5´103 cm3

Carrying out analogous calculations for the higher-weight bowling balls gives 3

ball weight (lb)

mass (g)

4.1 ´ 10

0.75

4.5 ´ 10

0.82

5.0 ´ 10

0.91

5.4 ´ 10

0.98

5.9 ´ 10

1.1

density (g/cm )

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

Therefore we would expect bowling balls that are 11 lb or lighter to float since they are less dense than water. Bowling balls that are 13 lb or heavier would be expected to sink since they are denser than water. The 12 lb bowling ball is borderline, but it would probably float. Note that the above calculations were carried out by rounding off the intermediate answers, as discussed in Section 1.8. If we carry an additional digit past the number of significant figures to minimize errors from rounding, the following densities are obtained: 3

ball weight (lb)

density (g/cm )

0.83

0.91

1.01

The differences in the densities obtained are slight, but the value obtained for the 12 lb ball now suggests that it might sink. This problem illustrates the difference rounding off intermediate answers can make in the final answers for some calculations.

1.49

Li, lithium; F, fluorine; P, phosphorus; Cu, copper; As, arsenic; Zn, zinc; Cl, chlorine; Pt, platinum; Mg, magnesium; U, uranium; Al, aluminum; Si, silicon; Ne, neon.

(a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

1.51

(a)

element

(b)

compound

(c)

1.52

(a)

homogeneous mixture

(b)

element

(c)

compound

(d)

homogeneous mixture

(e)

heterogeneous mixture

(f)

heterogeneous mixture

(g)

element

(a)

Chemical property. Oxygen gas is consumed in a combustion reaction; its composition and identity are

1.50

1.57

element

(d)

compound

changed. (b)

Chemical property. The fertilizer is consumed by the growing plants; it is turned into vegetable matter (different composition).

(c)

Physical property. The measurement of the boiling point of water does not change its identity or composition.

(d)

Physical property. The measurement of the densities of lead and aluminum does not change their composition.

(e)

Chemical property. When uranium undergoes nuclear decay, the products are chemically different substances.

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

1.58

1.59

1.60

(a)

Physical change. The helium isn’t changed in any way by leaking out of the balloon.

(b)

Chemical change in the battery.

(c)

Physical change. The orange juice concentrate can be regenerated by evaporation of the water.

(d)

Chemical change. Photosynthesis changes water, carbon dioxide, etc., into complex organic matter.

(e)

Physical change. The salt can be recovered unchanged by evaporation.

Substance

Qualitative Statement

Quantitative Statement

(a)

water

colorless liquid

freezes at 0°C

(b)

carbon

black solid (graphite)

density = 2.26 g/cm

(c)

iron

rusts easily

density = 7.86 g/cm

(d)

hydrogen gas

colorless gas

melts at -255.3°C

(e)

sucrose

tastes sweet

at 0°C, 179 g of sucrose dissolves in 100 g of H2O

(f)

table salt

tastes salty

melts at 801°C

(g)

mercury

liquid at room temperature

boils at 357°C

(h)

gold

a precious metal

density = 19.3 g/cm

(i)

air

a mixture of gases

contains 20% oxygen by volume

3 3

See Section 1.6 of your text for a discussion of these terms. (a)

Chemical property. Iron has changed its composition and identity by chemically combining with oxygen and water.

(b)

Chemical property. The water reacts with chemicals in the air (such as sulfur dioxide) to produce acids, thus changing the composition and identity of the water.

(c)

Physical property. The color of the hemoglobin can be observed and measured without changing its composition or identity.

(d)

Physical property. The evaporation of water does not change its chemical properties. Evaporation is a change in matter from the liquid state to the gaseous state.

(e)

Chemical property. The carbon dioxide is chemically converted into other molecules.

1 ton = 4.75´10 7 tons of sulfuric acid 2.0 ´10 3 lb

1.61

(95.0 ´10 9 lb of sulfuric acid)´

1.62

Volume of rectangular bar = length ´ width ´ height

density =

m 52.7064 g = = 2.6 g /cm 3 V (8.53 cm)(2.4 cm)(1.0 cm)

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

1.63

mass = density ´ volume é4 ù mass = (19.3 g/cm 3 )´ ê p(10.0 cm)3 ú = 8.08´10 4 g êë 3 úû 3 æ 1 cm ö÷ 3 ç (b) mass = (21.4 g/cm )´ç0.040 mm ´ ÷ = 1.4´10-6 g ÷ çè 10 mm ø

(a)

(c)

1.64

mass = (0.798 g/mL) (50.0 mL) = 39.9 g

You are asked to solve for the inner diameter of the bottle. If we can calculate the volume that the cooking oil 2

occupies, we can calculate the radius of the cylinder. The volume of the cylinder is, Vcylinder = pr h (r is the inner radius of the cylinder, and h is the height of the cylinder). The cylinder diameter is 2r.

volume of oil filling bottle =

mass of oil density of oil

volume of oil filling bottle =

1360 g = 1.43´103 mL = 1.43´103 cm3 0.953 g/mL

Next, solve for the radius of the cylinder. 2

Volume of cylinder = pr h

volume p´ h

1.43´103 cm 3 = 4.60 cm p´ 21.5 cm

The inner diameter of the bottle equals 2r. Bottle diameter = 2r = 2(4.60 cm) = 9.20 cm

1.65

From the mass of the water and its density, we can calculate the volume that the water occupies. The volume that the water occupies is equal to the volume of the flask. volume =

mass density

Mass of water = 87.39 g - 56.12 g = 31.27 g

Volume of the flask =

mass 31.27 g = = 31.35 cm 3 density 0.9976 g/cm 3

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

1.66

343 m 1 mi 3600 s ´ ´ = 767 mph 1s 1609 m 1h

1.67

The volume of silver is equal to the volume of water it displaces. 3

Volume of silver = 260.5 mL - 242.0 mL = 18.5 mL = 18.5 cm density =

1.68

194.3 g = 10.5 g /cm 3 18.5 cm 3

In order to work this problem, you need to understand the physical principles involved in the experiment in Problem 1.61. The volume of the water displaced must equal the volume of the piece of silver. If the silver did not sink, would you have been able to determine the volume of the piece of silver? The liquid must be less dense than the ice in order for the ice to sink. The temperature of the experiment must be maintained at or below 0°C to prevent the ice from melting.

1.69

The volume of a sphere is: 4 3 4 æç 48.6 cm ö÷ pr = p ç ÷÷ = 6.01´10 4 cm 3 3 3 èç 2 ø 3

density =

1.70

Volume =

mass density

Volume occupied by Li =

1.71

mass 6.852´105 g 3 = = 11.4 g/cm volume 6.01´10 4 cm 3

1.20 ´103 g = 2.3´10 3 cm 3 3 0.53 g/cm

For the Fahrenheit thermometer, we must convert the possible error of 0.1°F to °C. ?°C = 0.1°F ´

5°C = 0.056°C 9°F

The percent error is the amount of uncertainty in a measurement divided by the value of the measurement, converted to percent by multiplication by 100.

Percent error =

known error in a measurement ´100% value of the measurement

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

For the Fahrenheit thermometer,

percent error =

0.056°C ´100% = 0.1% 38.9°C

For the Celsius thermometer,

percent error =

0.1°C ´100% = 0.3% 38.9°C

Which thermometer is more accurate?

1.72

To work this problem, we need to convert from cubic feet to L. Some tables will have a conversion factor of 3

28.3 L = 1 ft , but we can also calculate it using the dimensional analysis method described in Section 1.9 of the text. First, converting from cubic feet to liters:

æ12 in ö æ 2.54 cm ö 1 mL 1´10-3 L (5.0´107 ft 3 )´ççèç 1 ft ÷÷÷ø ´èççç 1 in ø÷÷÷ ´1 cm3 ´ 1 mL = 1.42´109 L 3

The mass of vanillin (in g) is: 2.0 ´10-11 g vanillin ´(1.42 ´10 9 L) = 2.84 ´10-2 g vanillin 1L

The cost is:

(2.84 ´10

-2

1.73

g vanillin)´

$112 = $0.064 = 6.4 c 50 g vanillin

æ 9°F ö÷ ? °F = çç°C ´ ÷ + 32°F çè 5°C ÷ø Let temperature = t 9 t + 3 2° F 5 9 t - t = 32°F 5 4 - t = 32°F 5 t=

t = - 40°F = - 40°C

1.74

There are 78.3 + 117.3 = 195.6 Celsius degrees between 0°S and 100°S. We can write this as a unit factor.

æ195.6°C ö÷ çç ÷ çè 100°S ÷ø

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

Set up the equation like a Celsius to Fahrenheit conversion. We need to subtract 117.3°C, because the zero point on the new scale is 117.3°C lower than the zero point on the Celsius scale.

æ195.6°C ö÷ ? °C = çç ÷ (?°S ) -117.3°C çè 100°S ÷ø

1.75

Solving for ?° S gives:

æ 100°S ö÷ ? °S = (?°C +117.3°C) çç ÷ èç195.6°C ÷ø

For 25°C we have:

æ 100°S ö÷ ? °S = (25 +117.3)°C çç ÷ = 73°S çè195.6°C ÷ø

The key to solving this problem is to realize that all the oxygen needed must come from the 4% difference (20% - 16%) between inhaled and exhaled air. The 240 mL of pure oxygen/min requirement comes from the 4% of inhaled air that is oxygen. 240 mL of pure oxygen/min = (0.04)(volume of inhaled air/min) Volume of inhaled air/min =

240 mL of oxygen/min = 6000 mL of inhaled air/min 0.04

Since there are 12 breaths per min,

volume of air /breath =

1.76

1.77

6000 mL of inhaled air 1 min ´ = 5´102 mL /breath 1 min 12 breaths

(a)

6000 mL of inhaled air 0.001 L 60 min 24 h ´ ´ ´ = 8.6´103 L of air /day 1 min 1 mL 1h 1 day

(b)

8.6´103 L of air 2.1´10-6 L CO ´ = 0.018 L CO /day 1 day 1 L of air

Twenty-five grams of the least dense metal (solid A) will occupy the greatest volume of the three metals, and 25.0 g of the most dense metal (solid B) will occupy the least volume. We can calculate the volume occupied by each metal and then add the volume of water (20.0 mL) to find the total volume occupied by the metal and water.

Solid A:

1 mL = 8.6 mL 2.9 g A Total volume = 8.6 mL + 20.0 mL = 28.6 mL

25.0 g A ´

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

Solid B:

Solid C:

1 mL = 3.0 mL 8.3 g B Total volume = 3.0 mL + 20.0 mL = 23.0 mL

25.0 g B´

1 mL = 7.6 mL 3.3 g C Total volume = 7.6 mL + 20.0 mL = 27.6 mL

25.0 g C ´

Therefore, we have: (a) solid C, (b) solid B, and (c) solid A.

1.78

The diameter of the basketball can be calculated from its circumference. We can then use the diameter of a ball as a conversion factor to determine the number of basketballs needed to circle the equator. Circumference = 2pr

circumference 29.6 in = = 9.42 in p p The circumference of the Earth is 2pr = (2 × p × 6400 km) = 40212 km d = 2r =

1000 m 1cm 1in 1ball 40212 km´ ´ ´ ´ = 168,000,000 balls –2 1km 1´10 m 2.54 cm 9.42 in

We round up to an integer number of basketballs with 3 significant figures.

1.79

Assume that the crucible is platinum. Let’s calculate the volume of the crucible and then compare that to the volume of water that the crucible displaces. volume =

mass density

Volume of crucible =

860.2 g = 40.10 cm 3 21.45 g/cm 3

Volume of water displaced =

(860.2 - 820.2) g = 40.1 cm 3 0.9986 g/cm 3

The volumes are the same (within experimental error), so the crucible is made of platinum.

1.80

Volume = surface area ´ depth 3

Recall that 1 L = 1 dm . Let’s convert the surface area to units of dm and the depth to units of dm. æ1000 m ö÷ æ 1 dm ÷ö surface area = (1.8´108 km 2 )´çç ÷ ´çç ÷ = 1.8´1016 dm 2 èç 1 km ÷ø èç 0.1 m ÷ø 2

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

depth = (3.9´103 m)´

1 dm = 3.9´104 dm 0.1 m 16

Volume = surface area ´ depth = (1.8 ´ 10 dm ) (3.9 ´ 10 dm) = 7.0 ´ 10 dm = 7.0 ´ 10 L

1.81

31.103 g Au = 75.0 g Au 1 troy oz Au

(a)

2.41 troy oz Au´

(b)

1 troy oz = 31.103 g

? g in 1 oz = 1 oz ´

1 lb 453.6 g ´ = 28.35 g 16 oz 1 lb

A troy ounce is heavier than an ounce.

1.82

Volume of sphere =

4 3 pr 3

4 æ15 cm ö÷ Volume = pçç ÷÷ = 1.77´103 cm3 ç 3 è 2 ø 3

mass = volume ´ density = (1.77´103 cm 3 )´

4.0´101 kg Os´

1.83

1.84

22.57 g Os 1 kg ´ = 4.0´101 kg Os 1000 g 1 cm 3

2.205 lb = 88 lb Os 1 kg

(a)

0.798 g/mL - 0.802 g/mL ´100% = 0.5% 0.798 g/mL

(b)

0.864 g - 0.837 g ´100% = 3.1% 0.864 g

62 kg = 6.2 ´ 10 g 4

(6.2 ´ 10 g)(0.65) = 4.0 ´ 10 g O

(6.2 ´ 10 g)(0.18) = 1.1 ´ 10 g C

(6.2 ´ 10 g)(0.10) = 6.2 ´ 10 g H

(6.2 ´ 10 g)(0.03) = 2 ´ 10 g N 4

Ca: (6.2 ´ 10 g)(0.016) = 9.9 ´ 10 g Ca P:

(6.2 ´ 10 g)(0.012) = 7.4 ´ 10 g P

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

1.85

3 minutes 43.13 seconds = 223.13 seconds Time to run 1500 meters is:

1500 m ´

1.86

1 mi 223.13 s ´ = 208.01 s = 3 min 28.01 s 1609 m 1 mi

?°C = (7.3 ´ 10 - 273) K = 4.6 ´ 10 °C æ 9o F ö ?o F = ççç(4.6 ´10 2o C)´ o ÷÷÷ + 32 o F = 8.6´10 2 o F çè 5 C ÷ø

34.63% Cu 1000 g ´ = 1.77´106 g Cu 100% ore 1 kg

1.87

? g Cu = (5.11´103 kg ore)´

1.88

(8.0´104 tons Au)´ 2000 lb Au ´16 oz Au ´ $948 = $2.4´1012 or 2.4 trillion dollars

1.89

? g Au =

1 ton Au

1 lb Au

1 oz Au

4.0´10-12 g Au 1 mL ´ ´(1.5´1021 L seawater) = 6.0´1012 g Au 1 mL seawater 0.001 L

value of gold = (6.0´1012 g Au)´

1 lb 16 oz $948 ´ ´ = $2.0´1014 453.6 g 1 lb 1 oz

No one has become rich mining gold from the ocean, because the cost of recovering the gold would outweigh the price of the gold.

1.1´1022 Fe atoms = 5.4´1022 Fe atoms 1.0 g Fe

1.90

? Fe atoms = 4.9 g Fe

1.91

mass of Earth’s crust = (5.9´1021 tons)´

0.50% crust = 2.95´1019 tons 100% Earth 27.2% Si 2000 lb 1 kg mass of silicon in crust = (2.95´1019 tons crust)´ ´ ´ = 7.3´10 21 kg Si 100% crust 1 ton 2.205 lb

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

1.92

10 cm = 0.1 m. We need to find the number of times the 0.1 m wire must be cut in half until the piece left is equal to the diameter of a Cu atom, which is (2)(1.3 ´ 10- m). Let n be the number of times we can cut the 10

Cu wire in half. We can write: æ1ö 10 ççç ÷÷÷ ´ 0.1 m = 2.6 ´10 m è2ø n

æ 1 ö÷ çç ÷ = 2.6 ´10-9 m çè 2 ÷ø n

Taking the log of both sides of the equation:

æ1ö nlog çç ÷÷÷ = log(2.6 ´10-9 ) çè 2 ø n = 29 times

5000 mi 1 gal gas 9.5 kg CO2 ´ ´ = 5.9´1011 kg CO2 1 car 20 mi 1 gal gas

1.93

(250´106 cars)´

1.94

Volume = area ´ thickness. From the density, we can calculate the volume of the Al foil. Volume =

mass 3.636 g = = 1.3472 cm3 density 2.699 g/cm 3 2

Convert the unit of area from ft to cm . æ12 in ÷ö æ 2.54 cm ö÷ 1.000 ft ´çç ÷ ´ç ÷ = 929.03 cm 2 çè 1 ft ÷ø ççè 1 in ÷ø 2

thickness =

1.95

1.96

volume 1.3472 cm3 = = 1.450 ´10-3 cm = 1.450´10-2 mm area 929.03 cm 2

(a)

homogeneous

(b)

heterogeneous. The air will contain particulate matter, clouds, etc. This mixture is not homogeneous.

First, let’s calculate the mass (in g) of water in the pool. We perform this conversion because we know there is 1 g of chlorine needed per million grams of water.

(2.0´104 gollons H2 O)´ 3.79 L ´ 1 mL ´ 1 g = 7.58´107 g H2 O 1 gallon 0.001 L 1 mL

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

Next, let’s calculate the mass of chlorine that needs to be added to the pool. = 75.8 g chlorine (7.58´10 7 g H 2 O)´ 1 g chlorine 6 1´10 g H 2 O

The chlorine solution is only 6 percent chlorine by mass. We can now calculate the volume of chlorine solution that must be added to the pool.

75.8 g chlorine ´

1.97

100% soln 1 mL soln ´ = 1.3´103 mL of chlorine solution 6% chlorine 1 g soln

The volume of the cylinder is: 2

V = pr h = p(0.25 cm) (10 cm) = 2.0 cm

The number of Al atoms in the cylinder is: 2.0 cm 3 ´

1.98

(a)

2.70 g 1 Al atom ´ = 1.2 ´10 23 Al atoms 3 1 cm 4.48´10-23 g

The volume of the pycnometer can be calculated by determining the mass of water that the pycnometer holds and then using the density to convert to volume.

(43.1195 - 32.0764) g´

(b)

1 mL = 11.063 mL 0.99820 g

Using the volume of the pycnometer from part (a), we can calculate the density of ethanol.

(40.8051- 32.0764) g = 0.78900 g /mL 11.063 mL (c)

From the volume of water added and the volume of the pycnometer, we can calculate the volume of the zinc granules by difference. Then, we can calculate the density of zinc.

volume of water = (62.7728 - 32.0764 - 22.8476) g´

1 mL = 7.8630 mL 0.99820 g

volume of zinc granules = 11.063 mL - 7.8630 mL = 3.200 mL

density of zinc =

22.8476 g = 7.140 g/mL 3.200 mL

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

1.99

Let the fraction of gold = x, and the fraction of sand = (1 – x). We set up an equation to solve for x. 3

(x)(19.3 g/cm ) + (1 – x) (2.95 g/cm ) = 4.17 g/cm 19.3x – 2.95x + 2.95 = 4.17 x = 0.0746

Converting to a percentage, the mixture contains 7.46% gold.

1.100

First, convert 10 mm to units of cm.

10 mm ´

1´10-4 cm = 1.0´10-3 cm 1 mm

Now, substitute into the given equation to solve for time.

x2 (1.0 ´10-3 cm)2 = = 0.88 s 2 D 2(5.7´10-7 cm 2 /s)

It takes 0.88 seconds for a glucose molecule to diffuse 10 mm.

1.101

The mass of a human brain is about 1 kg (1000 g) and contains about 10 cells. The mass of a brain cell is: 1000 g = 1´10-8 g/cell 11 1´10 cells

Assuming that each cell is completely filled with water (density = 1 g/mL), we can calculate the volume of each cell. Then, assuming the cell to be cubic, we can solve for the length of one side of such a cell.

1´10-8 g 1 mL 1 cm3 ´ ´ = 1´10-8 cm3 /cell 1 cell 1 g 1 mL Vcube = a a = (V)

1/3

= (1 × 10- cm ) 8

3 1/3

= 0.002 cm 11

Next, the height of a single cell is a, 0.002 cm. If 10 cells are spread out in a thin layer a single cell thick, 11

the surface area can be calculated from the volume of 10 cells and the height of a single cell. V = surface area × height

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

The volume of 10 brain cells is:

1000 g´

1 mL 1 cm3 ´ = 1000 cm3 1 g 1 mL

The surface area is: æ1´10-2 m ö÷ V 1000 cm 3 ÷ = 5´101 m 2 = = 5´10 5 cm 2 ´ççç Surface area = çè 1 cm ø÷÷ height 0.002 cm 2

1.102

(a)

A concentration of CO of 800 ppm in air would mean that there are 800 parts by volume of CO per 1 million parts by volume of air. Using a volume unit of liters, 800 ppm CO means that there are 800 L of CO per 1 million liters of air. The volume in liters occupied by CO in the room is: æ 1 cm ö÷ 1L 17.6 m ´ 8.80 m ´ 2.64 m = 409 m 3 ´çç = 4.09 ´10 5 L air ÷ ´ çè1´10-2 m ÷ø 1000 cm 3 3

4.09´105 L air ´

(b)

8.00´102 L CO = 327 L CO 1´106 L air

1 mg = 1 × 10- g and 1 L = 1000 cm . We convert mg/m to g/L: 3

-3 0.050 mg 1 ´ 10 g æç1´10-2 m ö÷ 1000 cm 3 ÷ ´ ´ ´çç = 5.0 ´10-8 g /L çè 1 cm ÷÷ø 1 mg 1L 1 m3 3

(c)

1 mg = 1 × 10- mg and 1 mL = 1 × 10- dL. We convert mg/dL to mg/mL: 3

-2

1 ´ 10 dL 120 mg 1 mg ´ ´ = 1.20 × 10 3 mg /mL -3 1 dL 1 mL 1´10 mg

1.103

This problem is similar in concept to a limiting reagent problem. We need sets of coins with 3 quarters, 1 nickel, and 2 dimes. First, we need to find the total number of each type of coin.

Number of quarters = (33.871´103 g)´ Number of nickels = (10.432´103 g)´ Number of dimes = (7.990´103 g)´

1 quarter = 6000 quarters 5.645 g

1 nickel = 2100 nickels 4.967 g

1 dime = 3450 dimes 2.316 g

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

Next, we need to find which coin limits the number of sets that can be assembled. For each set of coins, we need 2 dimes for every 1 nickel.

2100 nickels´

2 dimes = 4200 dimes 1 nickel

We do not have enough dimes. For each set of coins, we need 2 dimes for every 3 quarters.

6000 quarters´

2 dimes = 4000 dimes 3 quarters

Again, we do not have enough dimes, and therefore the number of dimes is our “limiting reagent”. If we need 2 dimes per set, the number of sets that can be assembled is:

3450 dimes´

1 set = 1725 sets 2 dimes

The mass of each set is:

æ 5.645 g ö÷ çæ 4.967 g ö÷ æç 2.316 g ö÷ ÷ + ç1 nickel´ ÷ + ç2 dimes´ ÷ = 26.534 g/set ççç3 quarters´ ÷ ÷ 1 quarter ø÷ çè 1 nickel ø çè 1 dime ÷ø è Finally, the total mass of 1725 sets of coins is:

1725 sets´

1.104

26.534 g = 4.577´104 g 1 set

We wish to calculate the density and radius of the ball bearing. For both calculations, we need the volume of the ball bearing. The data from the first experiment can be used to calculate the density of the mineral oil. In the second experiment, the density of the mineral oil can then be used to determine what part of the 40.00 mL volume is due to the mineral oil and what part is due to the ball bearing. Once the volume of the ball bearing is determined, we can calculate its density and radius. From experiment one: Mass of oil = 159.446 g - 124.966 g = 34.480 g

Density of oil =

34.480 g = 0.8620 g/mL 40.00 mL

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

From the second experiment: Mass of oil = 50.952 g - 18.713 g = 32.239 g

Volume of oil = 32.239 g´

1 mL = 37.40 mL 0.8620 g

The volume of the ball bearing is obtained by difference. 3

Volume of ball bearing = 40.00 mL - 37.40 mL = 2.60 mL = 2.60 cm Now that we have the volume of the ball bearing, we can calculate its density and radius.

18.713 g = 7.20 g /cm 3 3 2.60 cm Using the formula for the volume of a sphere, we can solve for the radius of the ball bearing. Density of ball bearing =

4 3 pr 3

2.60 cm 3 =

4 3 pr 3

r = 0.621 cm r = 0.853 cm

1.105

It would be more difficult to prove that the unknown substance is an element. Most compounds would decompose on heating, making them easy to identify. For example, see Figure 4.13(a) of the text. On heating, the compound HgO decomposes to elemental mercury (Hg) and oxygen gas (O2).

1.106

We want to calculate the mass of the cylinder, which can be calculated from its volume and density. The 2

volume of a cylinder is pr l. The density of the alloy can be calculated using the mass percentages of each element and the given densities of each element. The volume of the cylinder is: 2

V = pr l 2

V = p(6.44 cm) (44.37 cm) 3

V = 5781 cm

The density of the cylinder is: 3

density = (0.7942)(8.94 g/cm ) + (0.2058)(7.31 g/cm ) = 8.605 g/cm

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

Now, we can calculate the mass of the cylinder. mass = density × volume 3

mass = (8.605 g/cm )(5781 cm ) = 4.97 × 10 g The assumption made in the calculation is that the alloy must be homogeneous in composition.

1.107

Gently heat the liquid to see if any solid remains after the liquid evaporates. Also, collect the vapor and then compare the densities of the condensed liquid with the original liquid. The composition of a mixed liquid would change with evaporation along with its density.

1.108

The density of the mixed solution should be based on the percentage of each liquid and its density. Because the solid object is suspended in the mixed solution, it should have the same density as this solution. The density of the mixed solution is: (0.4137)(2.0514 g/mL) + (0.5863)(2.6678 g/mL) = 2.413 g/mL As discussed, the density of the object should have the same density as the mixed solution (2.413 g/mL).

Yes, this procedure can be used in general to determine the densities of solids. This procedure is called the flotation method. It is based on the assumptions that the liquids are totally miscible and that the volumes of the liquids are additive.

1.109

When the carbon dioxide gas is released, the mass of the solution will decrease. If we know the starting mass of the solution and the mass of solution after the reaction is complete (given in the problem), we can calculate the mass of carbon dioxide produced. Then, using the density of carbon dioxide, we can calculate the volume of carbon dioxide released.

1.140 g = 45.60 g 1 mL Mass of solution before reaction = 45.60 g + 1.328 g = 46.928 g Mass of hydrochloric acid = 40.00 mL ´

We can now calculate the mass of carbon dioxide by difference. Mass of CO2 released = 46.928 g - 46.699 g = 0.229 g Finally, we use the density of carbon dioxide to convert to liters of CO2 released.

Volume of CO2 released = 0.229 g´

1L = 0.127 L 1.81 g

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

1.110

As water freezes, it expands. First, calculate the mass of the water at 20°C. Then, determine the volume that this mass of water would occupy at -5°C.

Mass of water = 242 mL ´

0.998 g = 241.5 g 1 mL

Volume of ice at - 5°C = 241.5 g ´

1 mL = 264 mL 0.916 g

The volume occupied by the ice is larger than the volume of the glass bottle. The glass bottle would crack!

1.111

Basic approach: · Approximate the shape of the juncture. · Estimate the dimensions in nm and calculate the volume, converting to L. The volume could be approximated in several ways. One approach would be to model the junction as a cylinder with an internal diameter of 200 nm and a height of 200 nm. The volume of the cylinder would be 2 æ10 2 cm ö÷ 1 mL æ 200 nm ö÷ 1L ç ÷´ pr h = p ç ´ = 6 ´10-18 L ÷÷ ´ 200 nm ´ççç 9 ÷ 3 çè 2 ø ÷ èç10 nm ø 1 cm 1000 mL 3

To get a sense of that volume relative to a molecule, consider that the volume of a water molecule in liquid water is roughly 3 ´ 10- L. Therefore a volume of 6 ´ 10- L could still potentially contain (6 ´ 1026

L)/( 3 ´ 10- L) » 2 ´ 10 water molecules! Water molecules are very small, and they pack tightly in the 26

liquid state due to strong intermolecular forces (Chapter 11). For larger molecules such as those involved in biological processes, it might be possible to trap a much smaller number in a nanofiber juncture.

1.112

Basic approach:

· Look up or estimate the number of cars currently operating in the United States. · Estimate the average fuel efficiency (miles per gallon) and average miles driven by a car in one year. The information from the web shows there are about 250 million passenger cars in the US. Assume on average each car covers 10,000 miles at the gas consumption rate of 20 mpg (miles per gallon). The total number of gallons of gasoline consumed in a year is 250 ´10 6 cars´

1.113

10000 mi 1 gal ´ » 1´1011 gal 1 car 20 mi

Basic approach:

· Look up the percentage of Earth’s surface covered by oceans and the average depth of the ocean. · Calculate volume based on the surface area of the oceans and the average depth.

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

About 70 percent of Earth is covered with water. Useful information from the web: Average depth of ocean is 8

4000 m; radius of Earth = 6400 km or 6.4 ´ 10 cm. Using the formula for the surface area of a sphere of 2

radius r as (4pr ), we calculate the area of ocean water as follows: 2

4pr ´ 0.7 = 4p(6.4 ´ 10 cm) ´ 0.7 = 3.6 ´ 10 cm Volume is area ´ depth so the volume of ocean water is 18

(3.6 ´ 10 cm ) ´ (4 ´ 10 cm) » 1 ´ 10 cm = 1 ´ 10 L

1.114

Basic approach:

· Model the shape and approximate the dimensions of an average human body, and calculate volume. · Look up the percent water in the body, and estimate the fraction of that water contained in the blood stream. One way to proceed is to assume that the human body can be treated as a cylinder of height h and diameter d. 2

The volume of the body is then given by pr ´ h where r is the radius. For a 6 ft tall person with an average width of 15 in, the volume in cubic inches is 2

p(15 in/2) ´ 6 ft ´ 12 in/ft = 1.3 ´ 10 in Expressed in liters,

æ 2.54 cm ö÷ 1L = 210 L 1.3´104 in3 ´çç ÷´ çè 1 in ÷ø 1000 cm3 3

About 60 percent of a human body is water, so the volume of water is 0.6 ´ 210 L = 130 L Assuming that one-tenth of the water is blood, the volume of blood would be 13 L. This calculation turns out to be a rather rough estimate, where most of our error probably comes from our assumption that 10% of the water in our bodies is contained in the blood stream. In reality, the volume of blood in an adult is around 5 L.

1.115

Basic approach:

· Break up period of time over which a professional basketball game is played according to the level of activity of the players. · Estimate the time spent engaged in that activity and the average running speed. · Calculate the total distance traveled. An NBA game lasts 48 minutes. We can divide this time period into three parts: (1) dashing, (2) running, and (3) movement during offense and defense under the basket. Reasonable estimates for time and speed during these activities are: dashing (3 min at 10 m/s); running (5 min at 7 m/s); movement under basket (40 min at

CHAPTER 1: CHEMISTRY–THE STUDY OF CHANGE

3 m/s). (Note that both the 10 m/s and 7 m/s estimates are near world track records.) Summing up, we calculate the average distance covered by the player as (3 min ´ 60 s/min ´10 m/s) + (5 min ´ 60 s/min ´ 7 m/s) + (40 min ´ 60 s/min ´ 3 m/s) » 1´10 4 m

Expressed in miles, 1´10 4 m ´

1 km 1 mi ´ » 7 mi 1000 m 1.61 km

Alternatively, one can make a quicker (and cruder) estimate by assuming that the players traverse the length of the court (94 ft) every 24 seconds based on the shot clock. Over 48 minutes, the distance traveled would be 48 min ´

60 s 94 ft 1 mi ´ ´ » 2 mi 1 min 24 s 5280 ft

Taking the higher estimate, this distance would correspond to quite a bit of running, with an average speed of roughly 1 mile every 7 minutes. Keep in mind, however, that hardly any player is on the court for a full 48 minutes of a game, and all players get a break during the timeouts.

1.116

Basic approach:

· Calculate the thickness of the oil layer from the volume and surface area, and assume that to be the length of one molecule. We assume that the thickness of the oil layer is equivalent to the length of one oil molecule. We can calculate the thickness of the oil layer from the volume and surface area. æ 1 cm ö÷ 40 m 2 ´çç ÷ = 4.0 ´10 5 cm 2 çè 0.01 m ÷ø 2

Given that 0.10 mL = 0.10 cm , we write volume = surface area ´ thickness and rearranging that equation gives thickness =

volume 0.10 cm 3 = = 2.5´10-7 cm surface area 4.0 ´10 5 cm 2

Converting to nm: 2.5´10 –7 cm ´

0.01 m 1 nm ´ = 2.5 nm 1 cm 1 ´ 10 –9 m

CHAPTER 2 ATOMS, MOLECULES, AND IONS Problem Categories Biological: 2.79, 2.80. Conceptual: 2.31, 2.32, 2.33, 2.34, 2.61, 2.65, 2.71, 2.72, 2.81, 2.82, 2.93, 2.103, 2.113. Descriptive: 2.24, 2.25, 2.26, 2.51, 2.52, 2.66, 2.70, 2.76, 2.84, 2.85, 2.86, 2.87, 2.88, 2.90, 2.91, 2.92, 2.94, 2.98, 2.101, 2.114. Environmental: 2.122. Organic: 2.47, 2.48, 2.69, 2.105, 2.107, 2.108, 2.115.

2.7

First, convert 1 cm to picometers. 1 cm 

0.01 m 1 pm   11010 pm 1 cm 11012 m

? He atoms  (11010 pm)

2.8

1 He atom  1108 He atoms 1102 pm

Note that you are given information to set up the unit factor relating meters and miles. ratom  10 4 rnucleus  10 4  2.0 cm 

2.13

1m 1 mi   0.12 mi 100 cm 1609 m

For iron, the atomic number Z is 26. Therefore the mass number A is: A  26  28  54

2.14

Strategy: The 239 in Pu-239 is the mass number. The mass number (A) is the total number of neutrons and protons present in the nucleus of an atom of an element. You can look up the atomic number (number of protons) on the periodic table. Solution: mass number  number of protons  number of neutrons number of neutrons  mass number  number of protons  239  94  145

2.15

Isotope

3 2

4 2

24 12

25 12

48 22

79 35

195 78

No. Protons

No. Neutrons

117

CHAPTER 2: ATOMS, MOLECULES, AND IONS

2.16

15 7

33 16

No. Protons

No. Neutrons No. Electrons

Isotope

23 11

(b)

84 38

112

122

64 28

63 29

130 56

186 74

202 80

2.17

(a)

2.18

The accepted way to denote the atomic number and mass number of an element X is as follows:

AX Z where, A  mass number Z  atomic number (a)

2.23

186 74

(b)

201 80

Helium and selenium are nonmetals whose name ends with ium. (Tellerium is a metalloid whose name ends in ium.)

2.24

(a)

Metallic character increases as you progress down a group of the periodic table. For example, moving down Group 14, the nonmetal carbon is at the top and the metal lead is at the bottom of the group.

(b)

Metallic character decreases from the left side of the table (where the metals are located) to the right side of the table (where the nonmetals are located).

2.25

2.26

The following data were measured at 20°C. 3

H2O (0.98 g/cm )

Hg (13.6 g/cm )

(a)

Li (0.53 g/cm )

K (0.86 g/cm )

(b)

Au (19.3 g/cm )

Pt (21.4 g/cm )

(c)

Os (22.6 g/cm )

(d)

Te (6.24 g/cm )

F and Cl are Group 17 elements; they should have similar chemical properties. Na and K are both Group 1 elements; they should have similar chemical properties. P and N are both Group 15 elements; they should have similar chemical properties.

2.31

(a)

This is a polyatomic molecule that is an elemental form of the substance. It is not a compound.

(b)

This is a polyatomic molecule that is a compound.

(c)

This is a diatomic molecule that is a compound.

CHAPTER 2: ATOMS, MOLECULES, AND IONS

2.32

(a)

This is a diatomic molecule that is a compound.

(b)

This is a polyatomic molecule that is a compound.

(c)

This is a polyatomic molecule that is the elemental form of the substance. It is not a compound.

2.33

2.34

2.35

2.36

Elements:

N2, S8, H2

Compounds:

NH3, NO, CO, CO2, SO2

There are more than two correct answers for each part of the problem. (a)

H2 and F2

(b)

(d)

H2O and C12H22O11 (sucrose) 

Ion

No. protons No. electrons

11 10

2

HCl and CO

20 18

3

13 10

(c)

2

26 24



53 54

S8 and P4

2

3

9 10

16 18

8 10

7 10

The atomic number (Z) is the number of protons in the nucleus of each atom of an element. You can find this on a periodic table. The number of electrons in an ion is equal to the number of protons minus the charge on the ion. number of electrons (ion)  number of protons  charge on the ion 

Ion

No. protons No. electrons

19 18

12 10



3

Fe 26 23

3



35 36

3

25 23

2

4

6 10

2

29 27

4

F and N

2.38

(a)

52 25

2.45

(a)

Sodium ion has a 1 charge and oxide has a 2 charge. The correct formula is Na2O.

(b)

The iron ion has a 2 charge and sulfide has a 2 charge. The correct formula is FeS.

(c)

The correct formula is Co2(SO4)3

(d)

Barium ion has a 2 charge and fluoride has a 1 charge. The correct formula is BaF2.

(b)

22 10

(c)

(18 electrons), Fe

2.37

(10 electrons), Ar and P

2

107 47

(d)



and V (23 electrons), Sn and Ag (46 electrons).

127 53

(e)

239 94

CHAPTER 2: ATOMS, MOLECULES, AND IONS

(a)

The copper ion has a 1 charge and bromide has a 1 charge. The correct formula is CuBr.

(b)

The manganese ion has a 3 charge and oxide has a 2 charge. The correct formula is Mn2O3.

(c)

We have the Hg2

(d)

Magnesium ion has a 2 charge and phosphate has a 3 charge. The correct formula is Mg3(PO4)2.

2.47

(a)

2.48

Strategy: An empirical formula tells us which elements are present and the simplest whole-number ratio of

2.46

2

(b)



ion and iodide (I ). The correct formula is Hg2I2.

(c)

C9H20

(d)

P2O5

(e)

BH3

their atoms. Can you divide the subscripts in the formula by some factor to end up with smaller wholenumber subscripts? Solution: (a)

Dividing both subscripts by 2, the simplest whole number ratio of the atoms in Al2Br6 is AlBr3.

(b)

Dividing all subscripts by 2, the simplest whole number ratio of the atoms in Na2S2O4 is NaSO2.

(c)

The molecular formula as written, N2O5, contains the simplest whole number ratio of the atoms present. In this case, the molecular formula and the empirical formula are the same.

(d)

The molecular formula as written, K2Cr2O7, contains the simplest whole number ratio of the atoms present. In this case, the molecular formula and the empirical formula are the same.

2.49

The molecular formula of glycine is C2H5NO2.

2.50

The molecular formula of ethanol is C2H6O.

2.51

Compounds of metals with nonmetals are usually ionic. Nonmetal-nonmetal compounds are usually molecular.

2.52

Ionic:

LiF, BaCl2, KCl

Molecular:

SiCl4, B2H6, C2H4

Compounds of metals with nonmetals are usually ionic. Nonmetal-nonmetal compounds are usually molecular.

2.59

Ionic:

NaBr, BaF2, CsCl.

Molecular:

CH4, CCl4, ICl, NF3

(a)

sodium chromate

(h)

phosphorus trifluoride

(b)

potassium hydrogen phosphate

(i)

phosphorus pentafluoride

(c)

hydrogen bromide (molecular compound)

(j)

tetraphosphorus hexoxide

CHAPTER 2: ATOMS, MOLECULES, AND IONS

2.60

(d)

hydrobromic acid

(k)

cadmium iodide

(e)

lithium carbonate

(l)

strontium sulfate

(f)

potassium dichromate

(m) aluminum hydroxide

(g)

ammonium nitrite

(n)

sodium carbonate decahydrate

Strategy: When naming ionic compounds, our reference for the names of cations and anions is Table 2.3 of the text. Keep in mind that if a metal can form cations of different charges, we need to use the Stock system. In the Stock system, Roman numerals are used to specify the charge of the cation. The metals that have only 

2

one charge in ionic compounds are the alkali metals (1), the alkaline earth metals (2), Ag , Zn , Cd , 3

and Al . When naming acids, binary acids are named differently than oxoacids. For binary acids, the name is based on the nonmetal. For oxoacids, the name is based on the polyatomic anion. For more detail, see Section 2.7 of the text. Solution: (a)



This is an ionic compound in which the metal cation (K ) has only one charge. The correct name is potassium hypochlorite. Hypochlorite is a polyatomic ion with one less O atom than the chlorite ion, 

ClO2 . (b)

silver carbonate

(c)

This is an ionic compound in which the metal can form more than one cation. Use a Roman numeral to specify the charge of the Fe ion. Since the chloride ion has a 1 charge, the Fe ion has a 2 charge. The correct name is iron(II) chloride.

(d)

potassium permanganate

(e)

cesium chlorate

(f)

hypoiodous acid

(g)

This is an ionic compound in which the metal can form more than one cation. Use a Roman numeral to specify the charge of the Fe ion. Since the oxide ion has a 2 charge, the Fe ion has a 2 charge. The correct name is iron(II) oxide.

(h)

iron(III) oxide

(i)

This is an ionic compound in which the metal can form more than one cation. Use a Roman numeral to specify the charge of the Ti ion. Since each of the four chloride ions has a 1 charge (total of 4), the Ti ion has a 4 charge. The correct name is titanium(IV) chloride.

(j)

sodium hydride

(k)

lithium nitride

(l)

sodium oxide

CHAPTER 2: ATOMS, MOLECULES, AND IONS



2

(m) This is an ionic compound in which the metal cation (Na ) has only one charge. The O2

ion is called

the peroxide ion. Each oxygen has a 1 charge. You can determine that each oxygen only has a 1 charge, because each of the two Na ions has a 1 charge. Compare this to sodium oxide in part (l). The correct name is sodium peroxide.

2.61

2.62

(n)

iron(III) chloride hexahydrate

(a)

RbNO2

(b)

K2S

(c)

NaHS

(d)

Mg3(PO4)2

(e)

CaHPO4

(f)

KH2PO4

(g)

IF7

(h)

(NH4)2SO4

(i)

AgClO4

(j)

BCl3

Strategy: When writing formulas of molecular compounds, the prefixes specify the number of each type of atom in the compound. When writing formulas of ionic compounds, the subscript of the cation is numerically equal to the charge of the anion, and the subscript of the anion is numerically equal to the charge on the cation. If the charges of the cation and anion are numerically equal, then no subscripts are necessary. Charges of common cations and anions are listed in Table 2.3 of the text. Keep in mind that Roman numerals specify the charge of the cation, not the number of metal atoms. Remember that a Roman numeral is not needed for some metal cations, 

because the charge is known. These metals are the alkali metals (1), the alkaline earth metals (2), Ag , 2

2

3

Zn , Cd , and Al . When writing formulas of oxoacids, you must know the names and formulas of polyatomic anions (see Table 2.3 of the text). Solution: (a)

The Roman numeral I tells you that the Cu cation has a 1 charge. Cyanide has a 1 charge. Since the charges are numerically equal, no subscripts are necessary in the formula. The correct formula is CuCN.

(b)



Strontium is an alkaline earth metal. It only forms a 2 cation. The polyatomic ion chlorite, ClO2 , has a 1 charge. Since the charges on the cation and anion are numerically different, the subscript of the cation is numerically equal to the charge on the anion, and the subscript of the anion is numerically equal to the charge on the cation. The correct formula is Sr(ClO2)2.

(c)



Perbromic tells you that the anion of this oxoacid is perbromate, BrO4 . The correct formula is HBrO4(aq). Remember that (aq) means that the substance is dissolved in water. 

(d)

Hydroiodic tells you that the anion of this binary acid is iodide, I . The correct formula is HI(aq).

(e)

Na is an alkali metal. It only forms a 1 cation. The polyatomic ion ammonium, NH4 , has a 1 charge



3

and the polyatomic ion phosphate, PO4 , has a 3 charge. To balance the charge, you need 2 Na



cations. The correct formula is Na2(NH4)PO4.

CHAPTER 2: ATOMS, MOLECULES, AND IONS

(f)

The Roman numeral II tells you that the Pb cation has a 2 charge. The polyatomic ion carbonate, 2

CO3 , has a 2 charge. Since, the charges are numerically equal, no subscripts are necessary in the formula. The correct formula is PbCO3. (g)

The Roman numeral II tells you that the Sn cation has a 2 charge. Fluoride has a 1 charge. Since the charges on the cation and anion are numerically different, the subscript of the cation is numerically equal to the charge on the anion, and the subscript of the anion is numerically equal to the charge on the cation. The correct formula is SnF2.

(h)

This is a molecular compound. The Greek prefixes tell you the number of each type of atom in the molecule. The correct formula is P4S10.

(i)

The Roman numeral II tells you that the Hg cation has a 2 charge. Oxide has a 2 charge. Since, the charges are numerically equal, no subscripts are necessary in the formula. The correct formula is HgO.

(j)

The Roman numeral I tells you that the Hg cation has a 1 charge. However, this cation exists as 2

2

Hg2 . Iodide has a 1 charge. You need two iodide ions to balance the 2 charge of Hg2 . The correct formula is Hg2I2. (k)

This is a molecular compound. The Greek prefixes tell you the number of each type of atom in the molecule. The correct formula is SeF6.

2.63

Let’s compare the ratio of the fluorine masses in the two compounds.

3.55 g F  1.50 2.37 g F This calculation indicates that there is 1.5 times more fluorine by mass in SF6 compared to the other compound. The value of n is:

n

6 4 1.5

This is consistent with the Law of Multiple Proportions which states that if two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in ratios of small whole numbers. In this case, the ratio of the masses of fluorine in the two compounds is 6:4 or 3:2.

2.64

(a)

dinitrogen pentoxide (N2O5)

(b)

boron trifluoride (BF3)

(c)

dialuminum hexabromide (Al2Br6)

2.65

Uranium is radioactive. It loses mass because it constantly emits alpha () particles.

2.66

Changing the electrical charge of an atom usually has a major effect on its chemical properties. The two electrically neutral carbon isotopes should have nearly identical chemical properties.

CHAPTER 2: ATOMS, MOLECULES, AND IONS

2.67

The number of protons  65  35  30. The element that contains 30 protons is zinc, Zn. There are two 2

fewer electrons than protons, so the charge of the cation is 2. The symbol for this cation is Zn .

2.68

Atomic number  127  74  53. This anion has 53 protons, so it is an iodide ion. Since there is one more 

electron than protons, the ion has a 1 charge. The correct symbol is I .

2.69

(a)

molecular, C3H8

(b)

empirical, C3H8

molecular, C2H2

(c)

empirical, CH

molecular, C2H6

(d)

empirical, CH3

molecular, C6H6 empirical, CH

2.70

NaCl is an ionic compound; it doesn’t form molecules.

2.71

Yes. The law of multiple proportions requires that the masses of sulfur combining with phosphorus must be in the ratios of small whole numbers. For the three compounds shown, four phosphorus atoms combine with three, seven, and ten sulfur atoms, respectively. If the atom ratios are in small whole number ratios, then the mass ratios must also be in small whole number ratios.

2.72

2.73

2.74

2.75

The species and their identification are as follows: (a)

SO2

molecule and compound

(g)

element and molecule

(b)

element and molecule

(h)

CH4

molecule and compound

(c)

element

(i)

KBr

compound

(d)

N2O5

molecule and compound

(j)

element

(e)

element

(k)

element and molecule

(f)

element and molecule

(l)

LiF

compound

(a)

Species with the same number of protons and electrons will be neutral. A, F, G.

(b)

Species with more electrons than protons will have a negative charge. B, E.

(c)

Species with more protons than electrons will have a positive charge. C, D.

(d)

A: 105 B

(a)

Ne, 10 p, 10 n

(b)

Cu, 29 p, 34 n

(c)

Ag, 47 p, 60 n

(d)

W, 74 p, 108 n

(e)

Po, 84 p, 119 n

(f)

Pu, 94 p, 140 n

(a)

BaO, barium oxide

(b)

Ca3P2, calcium phosphide

(c)

Al2S3, aluminum sulfide

(d)

Li3N, lithium nitride

B: 147 N 3

C: 39 K 19

D: 66 Zn 2  30

81 E: 35 Br 

F: 115 B

G: 199 F

CHAPTER 2: ATOMS, MOLECULES, AND IONS

2.76

(a)

2.77

When an anion is formed from an atom, you have the same number of protons attracting more electrons. The

(b)

(c)

(d)

(e)

(f)

(g) Cl

(h)

electrostatic attraction is weaker, which allows the electrons on average to move farther from the nucleus. An anion is larger than the atom from which it is derived. When a cation is formed from an atom, you have the same number of protons attracting fewer electrons. The electrostatic attraction is stronger, meaning that on average, the electrons are pulled closer to the nucleus. A cation is smaller than the atom from which it is derived.

2.78

(a)

Rutherford’s experiment is described in detail in Section 2.2 of the text. From the average magnitude of scattering, Rutherford estimated the number of protons (based on electrostatic interactions) in the nucleus.

(b)

Assuming that the nucleus is spherical, the volume of the nucleus is:

4 4 V   r 3   (3.041013 cm)3  1.1771037 cm 3 3 3 The density of the nucleus can now be calculated. d

m 3.82 1023 g   3.25 1014 g /cm 3 V 1.177 1037 cm 3

To calculate the density of the space occupied by the electrons, we need both the mass of 11 electrons, and the volume occupied by these electrons. The mass of 11 electrons is:

11 electrons

9.1095  1028 g 1 electron

 1.002051026 g

The volume occupied by the electrons will be the difference between the volume of the atom and the volume of the nucleus. The volume of the nucleus was calculated above. The volume of the atom is calculated as follows: 186 pm 

11012 m 1 cm   1.86 108 cm 1 pm 1102 m

4 4 Vatom  r 3   (1.86 108 cm)3  2.6951023 cm3 3 3 Velectrons  Vatom  Vnucleus  (2.695  10

23

cm )  (1.177  10

37

cm )  2.695  10

23

CHAPTER 2: ATOMS, MOLECULES, AND IONS

As you can see, the volume occupied by the nucleus is insignificant compared to the space occupied by the electrons. The density of the space occupied by the electrons can now be calculated. 26 m 1.0020510 g d   3.72 104 g /cm 3 V 2.6951023 cm 3

The above results do support Rutherford’s model. Comparing the space occupied by the electrons to the volume of the nucleus, it is clear that most of the atom is empty space. Rutherford also proposed that the nucleus was a dense central core with most of the mass of the atom concentrated in it. Comparing the density of the nucleus with the density of the space occupied by the electrons also supports Rutherford’s model.

2.79

The molecular formula of caffeine is C8H10N4O2. The empirical formula is C4H5N2O.

2.80

The empirical and molecular formulas of acetaminophen are C8H9NO2.

2.81

(a)

Iodate ion is IO32 . The correct formula is Mg(IO3)2.

(b)

The formula shown is phosphorous acid. The correct formula for phosphoric acid is H3PO4.

(c)

Sulfite ion is SO 23  . The correct formula is BaSO3.

(d)

NH 4 is the ammonium ion. The correct formula is NH4HCO3.

(a)

The charge on the tin cation needs to be specified. The correct name is tin(IV) chloride.

(b)

The charge on the copper ion is 1. The correct name is copper(I) oxide.

(c)

The charge on the cobalt cation needs to be specified. The correct name is cobalt(II) nitrate.

(d)

Cr2 O 27 is the dichromate ion. The correct name is sodium dichromate.

2.82

2.83

2.84

31 15

P 3

11 5

Protons

Neutrons

117

136

Electrons

Net Charge

2

3

54 26

Fe 2 

Symbol

196 79

222 86

(a)

Ionic compounds are typically formed between metallic and nonmetallic elements.

(b)

In general the transition metals, the actinides, and the lanthanides have variable charges.

CHAPTER 2: ATOMS, MOLECULES, AND IONS

2.85

(a)

Li , alkali metals always have a 1 charge in ionic compounds

(b)

(c)

I , halogens have a 1 charge in ionic compounds

(d)

(e)

Al , aluminum always has a 3 charge in ionic compounds

(f)

Cs, alkali metals always have a 1 charge in ionic compounds

(g)

Mg , alkaline earth metals always have a 2 charge in ionic compounds.

2.86



2



3 3

2

The symbol Na provides more information than 11Na. The mass number plus the chemical symbol identifies a specific isotope of Na (sodium) while combining the atomic number with the chemical symbol tells you nothing new. Can other isotopes of sodium have different atomic numbers?

2.87

The binary Group 17 element acids are: HF, hydrofluoric acid; HCl, hydrochloric acid; HBr, hydrobromic acid; HI, hydroiodic acid. Oxoacids containing Group 17 elements (using the specific examples for chlorine) are: HClO4, perchloric acid; HClO3, chloric acid; HClO2, chlorous acid: HClO, hypochlorous acid. Examples of oxoacids containing other Group 13-17 elements are: H3BO3, boric acid (Group 13); H2CO3, carbonic acid (Group 14); HNO3, nitric acid and H3PO4, phosphoric acid (Group 15); and H2SO4, sulfuric acid (Group 16). Hydrosulfuric acid, H2S, is an example of a binary Group 16 acid while HCN, hydrocyanic acid, contains both a Group 14 and 15 element.

2.88

Mercury (Hg) and bromine (Br2)

2.89

(a)

(b)

Isotope

4 2

20 10

40 18

84 36

132 54

No. Protons

No. Neutrons

neutron/proton ratio

1.00

1.22

1.33

1.44

The neutron/proton ratio increases with increasing atomic number.

2.90

H2, N2, O2, F2, Cl2, He, Ne, Ar, Kr, Xe, Rn

2.91

Cu, Ag, and Au are fairly chemically unreactive. This makes them specially suitable for making coins and jewelry, that you want to last a very long time.

CHAPTER 2: ATOMS, MOLECULES, AND IONS

2.92

They do not have a strong tendency to form compounds. Helium, neon, and argon are chemically inert.

2.93

Magnesium and strontium are also alkaline earth metals. You should expect the charge of the metal to be the same (2). MgO and SrO.

2.94

All isotopes of radium are radioactive. It is a radioactive decay product of uranium-238. Radium itself does not occur naturally on Earth.

2.95

(a)

Berkelium (Berkeley, CA); Europium (Europe); Francium (France); Scandium (Scandinavia); Ytterbium (Ytterby, Sweden); Yttrium (Ytterby, Sweden).

(b)

Einsteinium (Albert Einstein); Fermium (Enrico Fermi); Curium (Marie and Pierre Curie); Mendelevium (Dmitri Mendeleev); Lawrencium (Ernest Lawrence).

(c)

Arsenic, Cesium, Chlorine, Chromium, Iodine. 77

2

2.96

The atomic number is 77  43  34. The symbol for the anion is Se .

2.97

The mass of fluorine reacting with hydrogen and deuterium would be the same. The ratio of F atom to hydrogen (or deuterium) is 1:1 in both compounds. This does not violate the law of definite proportions. When the law of definite proportions was formulated, scientists did not know of the existence of isotopes.

2.98

(a)

NaH, sodium hydride

(b)

B2O3, diboron trioxide

(c)

Na2S, sodium sulfide

(d)

AlF3, aluminum fluoride

(e)

OF2, oxygen difluoride

(f)

SrCl2, strontium chloride

2.99

(a)

2.100

All of these are molecular compounds. We use prefixes to express the number of each atom in the molecule.

(b)

(c)

(d)

(e)

The names are nitrogen trifluoride (NF3), phosphorus pentabromide (PBr5), and sulfur dichloride (SCl2). 2.101

The metalloids are shown in gray.

CHAPTER 2: ATOMS, MOLECULES, AND IONS

2.102 Cation Mg Sr

2

Fe  3

2

4

2

Hg2

2



Cu Li



2.103

2.104

Anion

Formula

Name

Mg(HCO3)2

Magnesium bicarbonate

SrCl2

Strontium chloride

Fe(NO2)3

Iron(III) nitrite

Mn(ClO3)2

Manganese(II) chlorate

SnBr4

Tin(IV) bromide

Co3(PO4)2

Cobalt(II) phosphate

Hg2I2

Mercury(I) iodide

Cu2CO3

Copper(I) carbonate

3

Li3N

Lithium nitride

2

Al2S3

Aluminum sulfide

HCO3 Cl



NO2 ClO3 Br

PO4 I



 3

 2

CO3 N

3



(a)

CO2(s), solid carbon dioxide

(f)

Ca(OH)2, calcium hydroxide

(b)

NaCl, sodium chloride

(g)

NaHCO3, sodium bicarbonate

(c)

N2O, nitrous oxide

(h)

Na2CO3·10H2O, sodium carbonate decahydrate

(d)

CaCO3, calcium carbonate

(i)

CaSO4·2H2O, calcium sulfate dihydrate

(e)

CaO, calcium oxide

(j)

Mg(OH)2, magnesium hydroxide

The change in energy is equal to the energy released. We call this E. Similarly, m is the change in mass. Because m 

E , we have c2

E m  2  c

1000 J 1 kJ  1.911011 kg  1.91108 g (3.00 108 m/s)2

(1.715103 kJ)

 1 kg  m 2  Note that we need to convert kJ to J so that we end up with units of kg for the mass. 1 J    s2 

We can add together the masses of hydrogen and oxygen to calculate the mass of water that should be formed. 12.096 g  96.000  108.096 g The predicted change (loss) in mass is only 1.91  10

8

g which is too small a quantity to measure. Therefore,

for all practical purposes, the law of conservation of mass is assumed to hold for ordinary chemical processes.

CHAPTER 2: ATOMS, MOLECULES, AND IONS

2.105

CH4, C2H6, and C3H8 each only have one structural formula.

H H

C4H10 has two structural formulas.

HH H C H

H H

C5H12 has three structural formulas.

H H

2.106

(a)

HH H C

HH H

HH H C H H

H H H

The volume of a sphere is V

4 3 r 3

Volume is proportional to the number of nucleons. Therefore, V  A (mass number) 3

r A 1/3

rA (b)

Using the equation given in the problem, we can first solve for the radius of the lithium nucleus and then solve for its volume. 1/3

r  r0A

15

r  (1.2  10

1/3

m)(7)

r  2.3  10 m 15

CHAPTER 2: ATOMS, MOLECULES, AND IONS

V

4 3 r 3

4 Vnucleus  (2.31015 m)3  5.11044 m 3 3 (c)

In part (b), the volume of the nucleus was calculated. Using the radius of a Li atom, the volume of a Li atom can be calculated. Vatom 

4 3 4 r  (152 1012 m)3  1.47 1029 m 3 3 3

The fraction of the atom’s volume occupied by the nucleus is: Vnucleus 5.11044 m 3   3.51015 Vatom 1.47 1029 m 3

Yes, this calculation shows that the volume of the nucleus is much, much smaller than the volume of the atom, which supports Rutherford’s model of an atom.

2.107

Two different structural formulas for the molecular formula C2H6O are:

H O

C H

H O

In the second hypothesis of Dalton’s Atomic Theory, he states that in any compound, the ratio of the number of atoms of any two of the elements present is either an integer or simple fraction. In the above two compounds, the ratio of atoms is the same. This does not necessarily contradict Dalton’s hypothesis, but Dalton was not aware of chemical bond formation and structural formulas.

2.108

(a)

Ethane

Acetylene

2.65 g C 0.665 g H

4.56 g C 0.383 g H

Let’s compare the ratio of the hydrogen masses in the two compounds. To do this, we need to start with the same mass of carbon. If we were to start with 4.56 g of C in ethane, how much hydrogen would combine with 4.56 g of carbon?

0.665 g H 

4.56 g C  1.14 g H 2.65 g C

CHAPTER 2: ATOMS, MOLECULES, AND IONS

We can calculate the ratio of H in the two compounds.

1.14 g 3 0.383 g This is consistent with the Law of Multiple Proportions which states that if two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in ratios of small whole numbers. In this case, the ratio of the masses of hydrogen in the two compounds is 3:1. (b)

For a given amount of carbon, there is 3 times the amount of hydrogen in ethane compared to acetylene. Reasonable formulas would be:

2.109

(a)

Ethane

Acetylene

CH3

C2H6

C2H2

The following strategy can be used to convert from the volume of the Pt cube to the number of Pt atoms. 3

cm  grams  atoms 1.0 cm 3 

(b)

21.45 g Pt 1 atom Pt   6.610 22 Pt atoms 3 1 cm 3.240 1022 g Pt 22

Since 74 percent of the available space is taken up by Pt atoms, 6.6  10 atoms occupy the following volume: 3

0.74  1.0 cm  0.74 cm

We are trying to calculate the radius of a single Pt atom, so we need the volume occupied by a single Pt atom. volume Pt atom 

0.74 cm 3  1.12 1023 cm 3 /Pt atom 6.6 10 22 Pt atoms

4 3 r . Solving for the radius: 3 4 V  1.12 1023 cm3  r 3 3

The volume of a sphere is

24

r  2.67  10 r  1.4  10

8

CHAPTER 2: ATOMS, MOLECULES, AND IONS

Converting to picometers: radius Pt atom  (1.4 108 cm)

2.110

0.01 m 1 pm   1.4 10 2 pm 1 cm 11012 m

The mass number is the sum of the number of protons and neutrons in the nucleus. Mass number  number of protons  number of neutrons Let the atomic number (number of protons) equal A. The number of neutrons will be 1.2A. Plug into the above equation and solve for A. 55  A  1.2A A  25 The element with atomic number 25 is manganese, Mn.

2.111

2.112

The acids, from left to right, are chloric acid, nitrous acid, hydrocyanic acid, and sulfuric acid.

2.113

From the equation derived in Problem 2.106, we can first solve for the radius of the iron nucleus and then solve for its volume. 1/3

r  r0A

15

r  (1.2  10

15

r  4.6  10

V

1/3

m)(56)

4 3 r 3

4 Vnucleus  (4.61015 m)3  4.11043 m 3  4.11037 cm3 3 The density of an iron-56 nucleus is: density 

mass 9.229 1023 g   2.3  10 14 g cm 3 volume 4.11037 cm 3

CHAPTER 2: ATOMS, MOLECULES, AND IONS

The density of an iron nucleus is 230 trillion grams per 1 cm . Because of this very high density, neutrons are needed to stabilize the nucleus, keeping the protons from repelling each other. It has been postulated that the exterior of the neutron is negatively charged and the interior is positively charged. The negative exteriors of the neutrons attract the protons holding the particles together in this extremely dense arrangement.

2.114

The formula of the ionic compound is XY2. Element X is most likely in Group 4 and element Y is most likely in Group 16. A possible compound is TiO2, titanium(IV) oxide. Other choices are elements in Group 14: SnO2 [tin(IV) oxide] and PbO2 [lead(IV) oxide].

2.115

Let’s compare the ratio of the hydrogen masses in the three compounds. To do this, we need to start with the same mass of carbon. If we were to start with 0.749 g of C in ethane, how much hydrogen would combine with 0.749 g of carbon? If we were to start with 0.749 g of C in propane, how much hydrogen would combine with 0.749 g of carbon? 0.749 g C  0.188 g H 0.799 g C 0.749 g C propane : 0.183 g H   0.168 g H 0.817 g C

ethane : 0.201 g H 

We can calculate the ratio of H between methane and propane, and ethane and propane. 0.251 g  1.49 0.168 g 0.188 g ethane : propane :  1.12 0.168 g methane : propane :

This is consistent with the Law of Multiple Proportions which states that if two elements combine to form more than one compound, the masses of one element (in this case hydrogen) that combine with a fixed mass of the other element (in this case carbon) are in ratios of small whole numbers. The ratio of the masses of hydrogen in the three compounds, propane, ethane, and methane is 1:1.12:1.49 or 8:9:12. 2.116

Basic approach:  Determine the charge of an alpha particle (helium nucleus) and gold nucleus in coulombs.  Use Coulomb’s law to solve for the distance at which the electrical potential energy is equal to the initial kinetic energy. The charge of a proton is 1.6022  10

19

Q1, of 3.2044  10

19

CHAPTER 2: ATOMS, MOLECULES, AND IONS

17

A gold nucleus which contains 79 protons has a charge, Q2, of 1.2657  10

C. We solve for r, the distance of

separation in meters between the alpha particle and the gold nucleus.

r

2.117

kQ1Q2 E



3  9.010 9 kg  m  (3.2044 1019 C)(1.26571017 C)   s2  C 2  2  6.0 1014 kg  m    s2 

 6.11013 m

Basic approach:  Consider that almost all of the volume in an atom or monatomic ion is occupied by the electrons.  Determine the number of electrons in each species, and rank their size according the number of electrons. The volume of each species depends on the outer boundary of the electrons. The number of protons remains the same in all three cases. Li has the most electrons and therefore the greatest repulsion among them, leading to a larger outer boundary compared to the Li atom, which has one fewer electron. Li has the fewest number of electrons and the smallest outer boundary. Thus, the size decreases as follows: Li  Li  Li In general for a given element, the anion is the largest, the atom is second, and the cation is the smallest. In Chapter 8 we will consider a more sophisticated way of predicting the relative sizes of atoms and monatomic ions based on electron configurations and nuclear charge.

2.118

Isotopes of the same element have the same number of electrons, hence following the reasoning in Problem 2.117, they would have the same size. The number of neutrons affects the size of the nucleus, but not the size of the atom.

2.119

Basic approach:  Look up the size of a silver atom.  Calculate the number of silver atoms required to reach the minimum length visible to the human eye, and then calculate the number of silver atoms required to give a surface area that would be visible. A Web search shows that the atomic radius of an Ag atom is 144 pm or 1.44  10

8

cm. The number of Ag

atoms that must be lined up along one dimension is given by the minimum length humans can see divided by the diameter of a silver atom 2 10 –5 cm  700 Ag atoms 2 1.44 10 –8 cm/Ag atom

CHAPTER 2: ATOMS, MOLECULES, AND IONS

But to actually be seen by the human eye, a grouping of silver atoms would need to have both length and width. Assuming a square array of 700 Ag atoms by 700 Ag atoms 700 Ag atoms

700 Ag atoms

the number of silver atoms required to be visible would be 700  700  500,000 Ag atoms.

2.120

Basic approach:  Estimate the radius of a pea.  Look up the radius of a typical atom and its nucleus. The radius of the atom provides a range of possible distances of the electron from the nucleus.  Use the ratio of the radii for the pea and the atom to calculate the distance of the electron from the nucleus if the nucleus were the size of a pea. The diameter of a pea is about 0.5 cm. Radius of nucleus is about 5  10

13

cm. Therefore, the expansion

factor for the radius of the nucleus is 0.5 cm/2  51011 times 510 –13 cm

A typical atom has a radius of about 100 pm (10

8

cm). Using that value as the distance of the electron from

the nucleus in an atom, then if the atom was scaled to the size of a pea, the electron would be (1  10 11

8

cm)

 (5  10 )  5  10 cm or 50 m (roughly half a football field) from the nucleus!

2.121

Basic approach:  Consider the relative composition of sodium and potassium ions in biological systems (for starters see the Chemistry in Action on p. 49, and then look up the relative compositions in other biological systems). Over long periods of time (on a geological scale), minerals containing potassium and sodium are slowly 



decomposed by wind and rain, and their K and Na ions are converted to more soluble compounds. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

CHAPTER 2: ATOMS, MOLECULES, AND IONS

Eventually, rain leaches these compounds out of the soil and carries them to the sea. Plants take up many of 



the K ions along the way, while the Na ions are free to move on to the sea because they are not needed for biological functions by the plants.

2.122

Basic approach:  Estimate the diameter of a typical carbon microsphere and calculate the volume.  Estimate for the density of carbon in the microsphere.  Look up the most common isotope of carbon and calculate the mass of that isotope.  Calculate the number of atoms in the microsphere. A typical microsphere appears to be about 3 m in diameter, giving a volume of 3 3 4  d  4  3 m   10 2 cm  11 3         6   110 cm 3  2  3  2  10 m  3

It is reasonable to expect that the density of carbon in the microspheres would be somewhere between that of 3

graphite (2.2 g/cm ) and diamond (3.5 g/cm ), so we assume a density of 3 g/cm for the carbon in the microsphere. The mass of a typical carbon microsphere would be 11011 cm 3 

3gC  31011 g C 3 cm

While this may seem like an exceedingly small mass, it still represents a very large number of carbon atoms which are much smaller and far less massive than the microspheres. In order to determine the number of atoms in a typical microsphere, we need to approximate the mass of an individual carbon atom. An online search tells us that the most common isotope of carbon is 126 C . Taking this isotope as a typical carbon atom and using the data in Table 2.1, we can calculate the mass of one atom by summing the masses of six electrons, six protons, and six neutrons. 6 electron (9.109381028 g/electron)  6 proton(1.67262 1024 g/proton)  6 neutron(1.674931024 g/neutron)  2 1023 g

Therefore, the number of carbon atoms in a typical microsphere would be 31011 g C 

1 C atom  2 1012 C atoms 2 1023 g C

This calculation gives us a feel for the incredibly small size of an atom. It should be noted that the mass of the carbon atom used in the calculation is only an approximation. Besides the assumption that all carbon atoms are carbon-12 (only about 99% of carbon atoms are carbon-12), we will see in Chapter 19 that the mass of an atom is actually somewhat less than the sum of the masses of the individual particles. In Chapter 3 we will introduce “atomic mass” which is the average mass of the naturally occurring isotopes of a particular element.

CHAPTER 3 MASS RELATIONSHIPS IN CHEMICAL REACTIONS Problem Categories Biological: 3.28, 3.29, 3.38, 3.72, 3.111, 3.117, 3.118, 3.123, 3.124, 3.129, 3.131. Conceptual: 3.33, 3.34, 3.61, 3.62, 3.81, 3.82, 3.132, 3.135, 3.137, 3.162. Descriptive: 3.70, 3.76, 3.78, 3.99, 3.100, 3.102, 3.115, 3.133. Environmental: 3.50, 3.69, 3.117, 3.142, 3.146, 3.152, 3.153, 3.155, 3.159. Industrial: 3.39, 3.40, 3.45, 3.65, 3.89, 3.91, 3.92, 3.94, 3.103, 3.116, 3.152, 3.153, 3.158, 3.160, 3.164. Organic: 3.101, 3.141, 3.161.

3.5

(34.968 amu)(0.7553)  (36.956 amu)(0.2447)  35.45 amu

3.6

Strategy: Each isotope contributes to the average atomic mass based on its relative abundance. Multiplying the mass of an isotope by its fractional abundance (not percent) will give the contribution to the average atomic mass of that particular isotope. 6

It would seem that there are two unknowns in this problem, the fractional abundance of Li and the fractional 7

abundance of Li. However, these two quantities are not independent of each other; they are related by the 6

fact that they must sum to 1. Start by letting x be the fractional abundance of Li. Since the sum of the two abundances must be 1, we can write 7

Abundance Li  (1  x) Solution: Average atomic mass of Li  6.941 amu  x(6.0151 amu)  (1  x)(7.0160 amu) 6.941  1.0009x  7.0160 1.0009x  0.075 x  0.075 6

x  0.075 corresponds to a natural abundance of Li of 7.5 percent. The natural abundance of Li is (1  x)  0.925 or 92.5 percent.

3.7

 6.0221023 amu   The unit factor required is   1g  ? g  13.2 amu 

1g  2.191023 g 23 6.02210 amu

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

3.8

 6.0221023 amu   The unit factor required is   1g  ? amu  8.4 g 

3.11

In one year: (7.2 10 9 people)

Total time 

3.12

6.022 10 23 amu  5.110 24 amu 1g

2 particles 3600 s 24 h 365 days     4.51017 particles/yr 1 person each second 1h 1 day 1 yr

6.0221023 particles  1.3106 yr 4.51017 particles/yr

The thickness of the book in miles would be: 0.0036 in 1 ft 1 mi   (6.022 10 23 pages)  3.42 1016 mi 1 page 12 in 5280 ft

The distance, in miles, traveled by light in one year is: 1.00 yr 

365 day 24 h 3600 s 3.00 108 m 1 mi      5.881012 mi 1 yr 1 day 1h 1s 1609 m

The thickness of the book in light-years is: (3.42 1016 mi)

1 light-yr  5.8103 light - yr 5.881012 mi 3

It will take light 5.8  10 years to travel from the first page to the last one!

6.0221023 S atoms  3.07 1024 S atoms 1 mol S

3.13

5.10 mol S

3.14

(6.00 10 9 Co atoms)

3.15

77.4 g of Ca 

3.16

Strategy: We are given moles of gold and asked to solve for grams of gold. What conversion factor do we

1 mol Co  9.961015 mol Co 23 6.022 10 Co atoms

1 mol Ca  1.93 mol Ca 40.08 g Ca

need to convert between moles and grams? Arrange the appropriate conversion factor so moles cancel, and the unit grams is obtained for the answer.

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

Solution: The conversion factor needed to covert between moles and grams is the molar mass. In the periodic table (see inside front cover of the text), we see that the molar mass of Au is 197.0 g. This can be expressed as 1 mol Au  197.0 g Au From this equality, we can write two conversion factors. 1 mol Au 197.0 g Au

and

197.0 g Au 1 mol Au

The conversion factor on the right is the correct one. Moles will cancel, leaving the unit grams for the answer. We write ? g Au  15.3 mol Au

197.0 g Au  3.01103 g Au 1 mol Au

Check: Does a mass of 3010 g for 15.3 moles of Au seem reasonable? What is the mass of 1 mole of Au?

3.17

3.18

(a)

200.6 g Hg 1 mol Hg   3.3311022 g /Hg atom 23 1 mol Hg 6.022 10 Hg atoms

(b)

20.18 g Ne 1 mol Ne   3.3511023 g /Ne atom 1 mol Ne 6.02210 23 Ne atoms

(a) Strategy: We can look up the molar mass of arsenic (As) on the periodic table (74.92 g/mol). We want to find the mass of a single atom of arsenic (unit of g/atom). Therefore, we need to convert from the unit mole in the denominator to the unit atom in the denominator. What conversion factor is needed to convert between moles and atoms? Arrange the appropriate conversion factor so mole in the denominator cancels, and the unit atom is obtained in the denominator. Solution: The conversion factor needed is Avogadro’s number. We have 23

1 mol  6.022  10 particles (atoms) From this equality, we can write two conversion factors. 1 mol As 6.0221023 As atoms

and

6.0221023 As atoms 1 mol As

The conversion factor on the left is the correct one. Moles will cancel, leaving the unit atoms in the denominator of the answer.

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

We write ? g /As atom 

(b)

74.92 g As 1 mol As   1.2441022 g /As atom 23 1 mol As 6.022 10 As atoms

Follow same method as part (a). ? g /Ni atom 

58.69 g Ni 1 mol Ni   9.7461023 g /Ni atom 1 mol Ni 6.0221023 Ni atoms

Check: Should the mass of a single atom of As or Ni be a very small mass? 1 mol Pb 207.2 g Pb   3.441010 g Pb 23 1 mol Pb 6.02210 Pb atoms

3.19

1.00  1012 Pb atoms

3.20

Strategy: The question asks for atoms of Cu. We cannot convert directly from grams to atoms of copper. What unit do we need to convert grams of Cu to in order to convert to atoms? What does Avogadro’s number represent? Solution: To calculate the number of Cu atoms, we first must convert grams of Cu to moles of Cu. We use the molar mass of copper as a conversion factor. Once moles of Cu are obtained, we can use Avogadro’s number to convert from moles of copper to atoms of copper. 1 mol Cu  63.55 g Cu The conversion factor needed is 1 mol Cu 63.55 g Cu

Avogadro’s number is the key to the second conversion. We have 23

1 mol  6.022  10 particles (atoms) From this equality, we can write two conversion factors. 1 mol Cu 6.0221023 Cu atoms

and

6.0221023 Cu atoms 1 mol Cu

The conversion factor on the right is the one we need because it has the number of Cu atoms in the numerator, which is the unit we want for the answer.

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

Let’s complete the two conversions in one step. grams of Cu  moles of Cu  number of Cu atoms ? atoms of Cu  0.063 g Cu 

23 1 mol Cu 6.022 10 Cu atoms   6.010 20 Cu atoms 63.55 g Cu 1 mol Cu

Check: Should 0.063 g of Cu contain fewer than Avogadro’s number of atoms? What mass of Cu would contain Avogadro’s number of atoms?

3.21

1 mol H 6.022 1023 H atoms   6.571023 H atoms 1.008 g H 1 mol H

For hydrogen:

1.10 g H

For chromium:

14.7 g Cr 

1 mol Cr 6.0221023 Cr atoms   1.7010 23 Cr atoms 52.00 g Cr 1 mol Cr

There are more hydrogen atoms than chromium atoms.

3.22

2 Pb atoms 

1 mol Pb 207.2 g Pb   6.8811022 g Pb 23 6.022 10 Pb atoms 1 mol Pb

(5.11023 mol He) 

4.003 g He  2.0 1022 g He 1 mol He 23

2 atoms of lead have a greater mass than 5.1  10

3.23

3.24

mol of helium.

Using the appropriate atomic masses, (a)

CH4

12.01 amu  4(1.008 amu)  16.04 amu

(b)

NO2

14.01 amu  2(16.00 amu)  46.01 amu

(c)

SO3

32.07 amu  3(16.00 amu)  80.07 amu

(d)

C6H6

6(12.01 amu)  6(1.008 amu)  78.11 amu

(e)

NaI

22.99 amu  126.9 amu  149.9 amu

(f)

K2SO4

2(39.10 amu)  32.07 amu  4(16.00 amu)  174.27 amu

(g)

Ca3(PO4)2

3(40.08 amu)  2(30.97 amu)  8(16.00 amu)  310.18 amu

Strategy: How do molar masses of different elements combine to give the molar mass of a compound? Solution: To calculate the molar mass of a compound, we need to sum all the molar masses of the elements in the molecule. For each element, we multiply its molar mass by the number of moles of that element in one

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

mole of the compound. We find molar masses for the elements in the periodic table (inside front cover of the text).

3.25

(a)

molar mass Li2CO3  2(6.941 g)  12.01 g  3(16.00 g)  73.89 g

(b)

molar mass CS2  12.01 g  2(32.07 g)  76.15 g

(c)

molar mass CHCl3  12.01 g  1.008 g  3(35.45 g)  119.37 g

(d)

molar mass C6H8O6  6(12.01 g)  8(1.008 g)  6(16.00 g)  176.12 g

(e)

molar mass KNO3  39.10 g  14.01 g  3(16.00 g)  101.11 g

(f)

molar mass Mg3N2  3(24.31 g)  2(14.01 g)  100.95 g

To find the molar mass (g/mol), we simply divide the mass (in g) by the number of moles. 152 g  409 g /mol 0.372 mol

3.26

Strategy: We are given grams of ethane and asked to solve for molecules of ethane. We cannot convert directly from grams of ethane to molecules of ethane. What unit do we need to obtain first before we can convert to molecules? How should Avogadro’s number be used here? Solution: To calculate number of ethane molecules, we first must convert grams of ethane to moles of ethane. We use the molar mass of ethane as a conversion factor. Once moles of ethane are obtained, we can use Avogadro’s number to convert from moles of ethane to molecules of ethane. molar mass of C2H6  2(12.01 g)  6(1.008 g)  30.068 g The conversion factor needed is

1 mol C2 H6 30.068 g C2 H6 Avogadro’s number is the key to the second conversion. We have 23

1 mol  6.022  10 particles (molecules) From this equality, we can write the conversion factor: 6.0221023 ethane molecules 1 mol ethane

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

Let’s complete the two conversions in one step. grams of ethane  moles of ethane  number of ethane molecules

? molecules of C2 H6  0.334 g C2 H6 

1 mol C2 H6 6.0221023 C2 H6 molecules  30.068 g C2 H6 1 mol C2 H6

 6.69  10 C2H6 molecules Check: Should 0.334 g of ethane contain fewer than Avogadro’s number of molecules? What mass of ethane would contain Avogadro’s number of molecules?

3.27

1 mol glucose 6.0221023 molecules glucose 6 C atoms 1.50 g glucose    180.2 g glucose 1 mol glucose 1 molecule glucose 22

 3.01  10 C atoms The ratio of O atoms to C atoms in glucose is 1:1. Therefore, there are the same number of O atoms in 22

glucose as C atoms, so the number of O atoms  3.01  10 O atoms. The ratio of H atoms to C atoms in glucose is 2:1. Therefore, there are twice as many H atoms in glucose as C 22

atoms, so the number of H atoms  2(3.01  10 atoms)  6.02  10 H atoms.

3.28

Strategy: We are asked to solve for the number of C, S, H, and O atoms in 7.14  10 g of dimethyl sulfoxide (DMSO). We cannot convert directly from grams DMSO to atoms. What unit do we need to obtain first before we can convert to atoms? How should Avogadro’s number be used here? How many atoms of C, S, H, or O are in 1 molecule of DMSO? 3

Solution: Let’s first calculate the number of C atoms in 7.14  10 g of dimethyl sulfoxide. First, we must convert grams of DMSO to number of molecules of DMSO. This calculation is similar to Problem 3.26. The molecular formula of DMSO shows there are two C atoms in one DMSO molecule, which will allow us to convert to atoms of C. We need to perform three conversions: grams of DMSO  moles of DMSO  molecules of DMSO  atoms of C The conversion factors needed for each step are: 1) the molar mass of DMSO, 2) Avogadro’s number, and 3) the number of C atoms in 1 molecule of DMSO. We complete the three conversions in one calculation.

7.14 103 g DMSO

1 mol DMSO 6.0221023 DMSO molecules 2 C atoms   78.14 g DMSO 1 mol DMSO 1 molecule DMSO

 1.10  10 C atoms

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

The above method utilizes the ratio of molecules (DMSO) to atoms (carbon). We can also solve the problem by reading the formula as the ratio of moles of DMSO to moles of carbon by using the following conversions: grams of DMSO  moles of DMSO  moles of C  atoms of C Try it. 3

Check: Does the answer seem reasonable? We have 7.14  10 g DMSO. How many atoms of C would 78.14 g of DMSO contain? We could calculate the number of atoms of the remaining elements in the same manner, or we can use the atom ratios from the molecular formula. The sulfur atom to carbon atom ratio in a DMSO molecule is 1:2, the hydrogen atom to carbon atom ratio is 6:2 or 3:1, and the oxygen atom to carbon atom ratio is 1:2. ? atoms of S  (1.1010 26 C atoms) 

3.29

1 S atom  5.50 10 25 S atoms 2 C atoms

? atoms of H  (1.10 10 26 C atoms)

3 H atoms  3.30  10 26 H atoms 1 C atom

? atoms of O  (1.1010 26 C atoms)

1 O atom  5.50 10 25 O atoms 2 C atoms

The molar mass of C19H38O is 282.5 g.

1.0 1012 g 

1 mol 6.0221023 molecules   2.110 9 molecules 282.49 g 1 mol 12

Notice that even though 1.0  10

g is an extremely small mass, it still is comprised of over a billion

pheromone molecules!

3.30

Mass of water  2.56 mL 

1.00 g  2.56 g 1.00 mL

Molar mass of H2O  (16.00 g)  2(1.008 g)  18.016 g/mol

? H 2 O molecules  2.56 g H2 O

1 mol H2 O 6.02210 23 molecules H2 O  18.016 g H2 O 1 mol H2 O

 8.56  10 molecules

3.33



Since there are only two isotopes of carbon, there are only two possibilities for CF4 . 12 6

C 199 F4 (molecular mass 88 amu) and 136 C 199 F4 (molecular mass 89 amu)

There would be two peaks in the mass spectrum.

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

3.34

Since there are two hydrogen isotopes, they can be paired in three ways: H H, H H, and H H. There will then be three choices for each sulfur isotope. We can make a table showing all the possibilities (masses in amu): 32

There will be seven peaks of the following mass numbers: 34, 35, 36, 37, 38, 39, and 40. 1

Very accurate (and expensive!) mass spectrometers can detect the mass difference between two H and one 2

3.37

H. How many peaks would be detected in such a “high resolution” mass spectrum?

Molar mass of SnO2  (118.7 g)  2(16.00 g)  150.7 g %Sn 

(2)(16.00 g/mol) 100%  21.23% 150.7 g/mol

%O 

3.38

118.7 g/mol 100%  78.77% 150.7 g/mol

Strategy: Recall the procedure for calculating a percentage. Assume that we have 1 mole of CHCl3. The percent by mass of each element (C, H, and Cl) is given by the mass of that element in 1 mole of CHCl3 divided by the molar mass of CHCl3, then multiplied by 100 to convert from a fractional number to a percentage. Solution: The molar mass of CHCl3  12.01 g/mol  1.008 g/mol  3(35.45 g/mol)  119.4 g/mol. The percent by mass of each of the elements in CHCl3 is calculated as follows: %C 

12.01 g/mol 100%  10.06% 119.4 g/mol

%H 

1.008 g/mol 100%  0.8442% 119.4 g/mol

%Cl 

3(35.45) g/mol 100%  89.07% 119.4 g/mol

Check: Do the percentages add to 100%? The sum of the percentages is (10.06%  0.8442%  89.07%)  99.97%. The small discrepancy from 100% is due to the way we rounded off.

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

3.39

The molar mass of cinnamic alcohol is 134.17 g/mol. (a)

(b)

%C 

(9)(12.01 g/mol) 100%  80.56% 134.17 g/mol

%H 

(10)(1.008 g/mol) 100%  7.51% 134.17 g/mol

%O 

16.00 g/mol 100%  11.93% 134.17 g/mol

1 mol C9 H10 O 6.022 1023 molecules C9 H10 O 0.469 g C9 H10 O  134.17 g C9 H10 O 1 mol C9 H10 O 21

 2.11  10 molecules C9H10O

3.40

Compound

Molar mass (g)

N% by mass

(a)

(NH2)2CO

60.06

2(14.01 g) 100%  46.65% 60.06 g

(b)

NH4NO3

80.05

2(14.01 g) 100%  35.00% 80.05 g

(c)

HNC(NH2)2

59.08

3(14.01 g) 100%  71.14% 59.08 g

(d)

NH3

17.03

14.01 g 100%  82.27% 17.03 g

Ammonia, NH3, is the richest source of nitrogen on a mass percentage basis.

1 mol Fe2 O3 2 mol Fe   0.308 mol Fe 159.7 g Fe 2 O3 1 mol Fe 2 O3

3.41

24.6 g Fe2 O3 

3.42

Strategy: Tin(II) fluoride is composed of Sn and F. The mass due to F is based on its percentage by mass in the compound. How do we calculate mass percent of an element? Solution: First, we must find the mass % of fluorine in SnF2. Then, we convert this percentage to a fraction and multiply by the mass of the compound (24.6 g), to find the mass of fluorine in 24.6 g of SnF2. The percent by mass of fluorine in tin(II) fluoride is calculated as follows:

mass % F 

mass of F in 1 mol SnF2 100% molar mass of SnF2 

2(19.00 g) 100%  24.25% F 156.7 g

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

Converting this percentage to a fraction, we obtain 24.25/100  0.2425. Next, multiply the fraction by the total mass of the compound. ? g F in 24.6 g SnF2  (0.2425)(24.6 g)  5.97 g F Check: As a ball-park estimate, note that the mass percent of F is roughly 25 percent, so that a quarter of the mass should be F. One quarter of approximately 24 g is 6 g, which is close to the answer. Note: This problem could have been worked in a manner similar to Problem 3.46. You could complete the following conversions: g of SnF2  mol of SnF2  mol of F  g of F

3.43

In each case, assume 100 g of compound. (a)

2.1 g H 

1 mol H  2.08 mol H 1.008 g H

65.3 g O

32.6 g S

1 mol O  4.081 mol O 16.00 g O

1 mol S  1.017 mol S 32.07 g S

This gives the formula H2.08S1.017O4.081. Dividing by 1.017 gives the empirical formula, H2SO4.

(b)

20.2 g Al 

1 mol Al  0.7487 mol Al 26.98 g Al

79.8 g Cl 

1 mol Cl  2.251 mol Cl 35.45 g Cl

This gives the formula, Al0.7487Cl2.251. Dividing by 0.7487 gives the empirical formula, AlCl3.

3.44

(a) Strategy: In a chemical formula, the subscripts represent the ratio of the number of moles of each element that combine to form the compound. Therefore, we need to convert from mass percent to moles in order to determine the empirical formula. If we assume an exactly 100 g sample of the compound, do we know the mass of each element in the compound? How do we then convert from grams to moles? Solution: If we have 100 g of the compound, then each percentage can be converted directly to grams. In this sample, there will be 40.1 g of C, 6.6 g of H, and 53.3 g of O. Because the subscripts in the formula

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

represent a mole ratio, we need to convert the grams of each element to moles. The conversion factor needed is the molar mass of each element. Let n represent the number of moles of each element so that nC  40.1 g C  nH  6.6 g H 

1 mol C  3.339 mol C 12.01 g C

1 mol H  6.55 mol H 1.008 g H

nO  53.3 g O

1 mol O  3.331 mol O 16.00 g O

Thus, we arrive at the formula C3.339H6.55O3.331, which gives the identity and the mole ratios of atoms present. However, chemical formulas are written with whole numbers. Try to convert to whole numbers by dividing all the subscripts by the smallest subscript (3.331).

3.339 1 3.331

6.55 2 3.331

3.331 1 3.331

This gives the empirical formula, CH2O. Check: Are the subscripts in CH2O reduced to the smallest whole numbers? (b)

Following the same procedure as part (a), we find: nC  18.4 g C 

1 mol C  1.532 mol C 12.01 g C

nN  21.5 g N 

1 mol N  1.535 mol N 14.01 g N

nK  60.1 g K 

1 mol K  1.537 mol K 39.10 g K

Dividing by the smallest number of moles (1.532 mol) gives the empirical formula, KCN.

3.45

The molar mass of CaSiO3 is 116.17 g/mol. %Ca 

%Si 

%O 

40.08 g  34.50% 116.17 g

28.09 g 116.17 g

 24.18%

(3)(16.00 g)  41.32% 116.17 g

Check to see that the percentages sum to 100%. (34.50%  24.18%  41.32%)  100.00%

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

3.48

The empirical molar mass of CH is approximately 13.018 g. Let’s compare this to the molar mass to determine the molecular formula. Recall that the molar mass divided by the empirical mass will be an integer greater than or equal to one. molar mass  1 (integer values) empirical molar mass

In this case, molar mass 78 g  6 empirical molar mass 13.018 g

Thus, there are six CH units in each molecule of the compound, so the molecular formula is (CH)6, or C6H6.

3.49

Find the molar mass corresponding to each formula. For C4H5N2O:

4(12.01 g)  5(1.008 g)  2(14.01 g)  (16.00 g)  97.10 g

For C8H10N4O2:

8(12.01 g)  10(1.008 g)  4(14.01 g)  2(16.00 g)  194.20 g

The molecular formula is C8H10N4O2. 3.50

METHOD 1: Step 1: Assume you have exactly 100 g of substance. 100 g is a convenient amount, because all the percentages sum to 100%. The percentage of oxygen is found by difference: 100%  (19.8%  2.50%  11.6%)  66.1% In 100 g of PAN there will be 19.8 g C, 2.50 g H, 11.6 g N, and 66.1 g O. Step 2: Calculate the number of moles of each element in the compound. Remember, an empirical formula tells us which elements are present and the simplest whole-number ratio of their atoms. This ratio is also a mole ratio. Use the molar masses of these elements as conversion factors to convert to moles. nC  19.8 g C 

1 mol C  1.649 mol C 12.01 g C

nH  2.50 g H  nN  11.6 g N 

1 mol H  2.480 mol H 1.008 g H

1 mol N  0.8280 mol N 14.01 g N

nO  66.1 g O 

1 mol O  4.131 mol O 16.00 g O

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

Step 3: Try to convert to whole numbers by dividing all the subscripts by the smallest subscript. The formula is C1.649H2.480N0.8280O4.131. Dividing the subscripts by 0.8280 gives the empirical formula, C2H3NO5. To determine the molecular formula, remember that the molar mass/empirical mass will be an integer greater than or equal to one. molar mass  1 (integer values) empirical molar mass

In this case, molar mass 120 g  1 empirical molar mass 121.05 g

Hence, the molecular formula and the empirical formula are the same, C2H3NO5. METHOD 2: Step 1: Multiply the mass % (converted to a decimal) of each element by the molar mass to convert to grams of each element. Then, use the molar mass to convert to moles of each element. nC  (0.198)  (120 g) 

1 mol C  1.98 mol C  2 mol C 12.01 g C

nH  (0.0250) (120 g)  nN  (0.116) (120 g)  nO  (0.661) (120 g) 

1 mol H  2.98 mol H  3 mol H 1.008 g H

1 mol N  0.994 mol N  1 mol N 14.01 g N

1 mol O  4.96 mol O  5 mol O 16.00 g O

Step 2: Since we used the molar mass to calculate the moles of each element present in the compound, this method directly gives the molecular formula. The formula is C2H3NO5. Step 3: Try to reduce the molecular formula to a simpler whole number ratio to determine the empirical formula. The formula is already in its simplest whole number ratio. The molecular and empirical formulas are the same. The empirical formula is C2H3NO5.

3.51

Assume you have exactly 100 g of substance. nC  44.4 g C 

1 mol C  3.697 mol C 12.01 g C

nH  6.21 g H 

1 mol H  6.161 mol H 1.008 g H

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

nS  39.5 g S

1 mol S  1.232 mol S 32.07 g S

nO  9.86 g O

1 mol O  0.6163 mol O 16.00 g O

Thus, we arrive at the formula C3.697H6.161S1.232O0.6163. Dividing by the smallest number of moles (0.6163 mole) gives the empirical formula, C6H10S2O. To determine the molecular formula, divide the molar mass by the empirical mass. molar mass 162 g  1 empirical molar mass 162.28 g

Hence, the molecular formula and the empirical formula are the same, C6H10S2O.

3.52

METHOD 1: Step 1: Assume you have exactly 100 g of substance. 100 g is a convenient amount, because all the percentages sum to 100%. In 100 g of MSG there will be 35.51 g C, 4.77 g H, 37.85 g O, 8.29 g N, and 13.60 g Na. Step 2: Calculate the number of moles of each element in the compound. Remember, an empirical formula tells us which elements are present and the simplest whole-number ratio of their atoms. This ratio is also a mole ratio. Let nC, nH, nO, nN, and nNa be the number of moles of elements present. Use the molar masses of these elements as conversion factors to convert to moles. nC  35.51 g C  nH  4.77 g H 

1 mol C  2.9567 mol C 12.01 g C

1 mol H  4.732 mol H 1.008 g H

nO  37.85 g O  nN  8.29 g N 

1 mol O  2.3656 mol O 16.00 g O

1 mol N  0.5917 mol N 14.01 g N

nNa  13.60 g Na 

1 mol Na  0.59156 mol Na 22.99 g Na

Thus, we arrive at the formula C2.9567H4.732O2.3656N0.5917Na0.59156, which gives the identity and the ratios of atoms present. However, chemical formulas are written with whole numbers.

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

Step 3: Try to convert to whole numbers by dividing all the subscripts by the smallest subscript.

2.9567  4.9981  5 0.59156 0.5917 N:  1.000 0.59156

4.732  7.999  8 0.59156 0.59156 Na : 1 0.59156

2.3656  3.9989  4 0.59156

This gives us the empirical formula for MSG, C5H8O4NNa. To determine the molecular formula, remember that the molar mass/empirical mass will be an integer greater than or equal to one. molar mass  1 (integer values) empirical molar mass

In this case, molar mass 169 g  1 empirical molar mass 169.11 g

Hence, the molecular formula and the empirical formula are the same, C5H8O4NNa. It should come as no surprise that the empirical and molecular formulas are the same since MSG stands for monosodiumglutamate. METHOD 2: Step 1: Multiply the mass % (converted to a decimal) of each element by the molar mass to convert to grams of each element. Then, use the molar mass to convert to moles of each element. nC  (0.3551) (169 g)

1 mol C  5.00 mol C 12.01 g C

nH  (0.0477) (169 g)

1 mol H  8.00 mol H 1.008 g H

nO  (0.3785) (169 g)

1 mol O  4.00 mol O 16.00 g O

nN  (0.0829) (169 g) 

1 mol N  1.00 mol N 14.01 g N

nNa  (0.1360) (169 g)

1 mol Na  1.00 mol Na 22.99 g Na

Step 2: Since we used the molar mass to calculate the moles of each element present in the compound, this method directly gives the molecular formula. The formula is C5H8O4NNa.

3.57

The balanced equations are as follows: (a)

2C  O2  2CO

(c)

H2  Br2  2HBr

(b)

2CO  O2  2CO2

(d)

2K  2H2O  2KOH  H2

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

3.58

3.61

(e)

2Mg  O2  2MgO

(j)

S8  8O2  8SO2

(f)

2O3  3O2

(k)

2NaOH  H2SO4  Na2SO4  2H2O

(g)

2H2O2  2H2O  O2

(l)

Cl2  2NaI  2NaCl  I2

(h)

N2  3H2  2NH3

(m) 3KOH  H3PO4  K3PO4  3H2O

(i)

Zn  2AgCl  ZnCl2  2Ag

(n)

CH4  4Br2  CBr4  4HBr

The balanced equations are as follows: (a)

2N2O5  2N2O4  O2

(h)

2Al  3H2SO4  Al2(SO4)3  3H2

(b)

2KNO3  2KNO2  O2

(i)

CO2  2KOH  K2CO3  H2O

(c)

NH4NO3  N2O  2H2O

(j)

CH4  2O2  CO2  2H2O

(d)

NH4NO2  N2  2H2O

(k)

Be2C  4H2O  2Be(OH)2  CH4

(e)

2NaHCO3  Na2CO3  H2O  CO2

(l)

3Cu  8HNO3  3Cu(NO3)2  2NO  4H2O

(f)

P4O10  6H2O  4H3PO4

(m) S  6HNO3  H2SO4  6NO2  2H2O

(g)

2HCl  CaCO3  CaCl2  H2O  CO2

(n)

2NH3  3CuO  3Cu  N2  3H2O

On the reactants side there are 8 A atoms and 4 B atoms. On the products side, there are 4 C atoms and 4 D atoms. Writing an equation, 8A  4B  4C  4D Chemical equations are typically written with the smallest set of whole number coefficients. Dividing the equation by four gives, 2A  B  C  D The correct answer is choice (c).

3.62

On the reactants side there are 6 A atoms and 4 B atoms. On the products side, there are 4 C atoms and 2 D atoms. Writing an equation, 6A  4B  4C  2D

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

Chemical equations are typically written with the smallest set of whole number coefficients. Dividing the equation by two gives, 3A  2B  2C  D The correct answer is choice (d).

3.63

The mole ratio from the balanced equation is 2 moles CO2: 2 moles CO. 3.60 mol CO 

3.64

2 mol CO2  3.60 mol CO 2 2 mol CO

Si(s)  2Cl2(g)   SiCl4(l) Strategy: Looking at the balanced equation, how do we compare the amounts of Cl2 and SiCl4? We can compare them based on the mole ratio from the balanced equation. Solution: Because the balanced equation is given in the problem, the mole ratio between Cl2 and SiCl4 is known: 2 moles Cl2

1 mole SiCl4. From this relationship, we have two conversion factors.

2 mol Cl2 1 mol SiCl 4

and

1 mol SiCl 4 2 mol Cl2

Which conversion factor is needed to convert from moles of SiCl4 to moles of Cl2? The conversion factor on the left is the correct one. Moles of SiCl4 will cancel, leaving units of "mol Cl2" for the answer. We calculate moles of Cl2 reacted as follows:

2 mol Cl 2 ? mol Cl 2 reacted  0.507 mol SiCl4   1.01 mol Cl2 1 mol SiCl4 Check: Does the answer seem reasonable? Should the moles of Cl2 reacted be double the moles of SiCl4 produced?

3.65

Starting with the amount of ammonia produced (6.0 moles), we can use the mole ratio from the balanced equation to calculate the moles of H2 and N2 that reacted to produce 6.0 moles of NH3. 3H2(g)  N2(g)  2NH3(g)

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

3.66

? mol H2  6.0 mol NH3 

3 mol H2  9.0 mol H2 2 mol NH3

? mol N2  6.0 mol NH3 

1 mol N2  3.0 mol N2 2 mol NH3

Starting with the 9.8 moles of CH3OH, we can use the mole ratio from the balanced equation to calculate the moles of H2O formed. 2CH3OH(l)  3O2(g)  2CO2(g)  4H2O(l)

? mol H 2 O  9.8 mol CH3 OH

3.67

4 mol H2 O  20 mol H2 O  2.0101 mol H 2 O 2 mol CH3 OH

The balanced equation is: 2Al(s)  3I2(s)   2AlI3(s) Using unit factors, we convert: g of Al  mol of Al  mol of I2  g of I2

20.4 g Al

3.68

1 mol Al 3 mol I2 253.8 g I 2    288 g I 2 26.98 g Al 2 mol Al 1 mol I 2

Using unit factors we convert: g of Hg  mol Hg  mol S  g S ? g S  246 g Hg 

3.69

1 mol Hg 1 mol S 32.07 g S    39.3 g S 200.6 g Hg 1 mol Hg 1 mol S

It is convenient to use the unit ton-mol in this problem. We normally use a g-mol. 1 g-mol SO2 has a mass of 64.07 g. In a similar manner, 1 ton-mol of SO2 has a mass of 64.07 tons. We need to complete the following conversions: tons SO2  ton-mol SO2  ton-mol S  ton S.

(2.6107 tons SO2 )

3.70

1 ton-mol SO2 1 ton-mol S 32.07 ton S    1.310 7 tons S 64.07 ton SO2 1 ton-mol SO2 1 ton-mol S

(a)

2NaHCO3   Na2CO3  H2O  CO2

(b)

Molar mass NaHCO3  22.99 g  1.008 g  12.01 g  3(16.00 g)  84.008 g Molar mass CO2  12.01 g  2(16.00 g)  44.01 g

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

The balanced equation shows one mole of CO2 formed from two moles of NaHCO3.

mass NaHCO 3  20.5 g CO2 

1 mol CO2 2 mol NaHCO3 84.008 g NaHCO3   44.01 g CO2 1 mol CO2 1 mol NaHCO3

 78.3 g NaHCO3

3.71

The balanced equation shows a mole ratio of 1 mole NH3

1 mole NCl3.

1 mol NH3 1 mol NCl3 120.4 g NCl3 2.94 g NH3     20.8 g NCl3 17.03 g NH3 1 mol NH3 1 mol NCl3

3.72

C6H12O6   2C2H5OH  2CO2 glucose

ethanol

Strategy: We compare glucose and ethanol based on the mole ratio in the balanced equation. Before we can determine moles of ethanol produced, we need to convert to moles of glucose. What conversion factor is needed to convert from grams of glucose to moles of glucose? Once moles of ethanol are obtained, another conversion factor is needed to convert from moles of ethanol to grams of ethanol. Solution: The molar mass of glucose will allow us to convert from grams of glucose to moles of glucose. The molar mass of glucose  6(12.01 g)  12(1.008 g)  6(16.00 g)  180.16 g. The balanced equation is given, so the mole ratio between glucose and ethanol is known; that is 1 mole glucose

2 moles ethanol.

Finally, the molar mass of ethanol will convert moles of ethanol to grams of ethanol. This sequence of three conversions is summarized as follows: grams of glucose  moles of glucose  moles of ethanol  grams of ethanol

? g C2 H 5 OH  500.4 g C6 H12 O6 

1 mol C6 H12 O6 2 mol C 2 H5 OH 46.068 g C2 H5OH   180.16 g C6 H12 O6 1 mol C6 H12 O6 1 mol C2 H5 OH

 255.9 g C2H5OH Check: Does the answer seem reasonable? Should the mass of ethanol produced be approximately half the mass of glucose reacted? Twice as many moles of ethanol are produced compared to the moles of glucose reacted, but the molar mass of ethanol is about one-fourth that of glucose. The liters of ethanol can be calculated from the density and the mass of ethanol.

volume 

mass density

Volume of ethanol obtained 

255.9 g  324 mL  0.324 L 0.789 g/mL

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

3.73

The mass of water lost is just the difference between the initial and final masses. Mass H2O lost  15.01 g  9.60 g  5.41 g

moles of H2 O  5.41 g H2 O

3.74

The balanced equation shows that eight moles of KCN are needed to combine with four moles of Au. ? mol KCN  29.0 g Au

3.75

The balanced equation is:

1.0 kg CaCO3 

3.76

1 mol H2 O  0.300 mol H 2 O 18.016 g H2 O

1 mol Au 8 mol KCN   0.294 mol KCN 197.0 g Au 4 mol Au

CaCO3(s)   CaO(s)  CO2(g)

1000 g 1 mol CaCO3 1 mol CaO 56.08 g CaO     5.610 2 g CaO 1 kg 100.09 g CaCO3 1 mol CaCO3 1 mol CaO

(a)

 N2O(g)  2H2O(g) NH4NO3(s) 

(b)

Starting with moles of NH4NO3, we can use the mole ratio from the balanced equation to find moles of N2O. Once we have moles of N2O, we can use the molar mass of N2O to convert to grams of N2O. Combining the two conversions into one calculation, we have: mol NH4NO3  mol N2O  g N2O

1 mol N 2 O 44.02 g N2 O ? g N 2 O  0.46 mol NH 4 NO3    2.0101 g N 2 O 1 mol NH 4 NO3 1 mol N2 O 3.77

The quantity of ammonia needed is:

1.00 108 g (NH4 )2 SO4 

1 mol (NH 4 )2 SO4 2 mol NH3 17.034 g NH3 1 kg    132.15 g (NH 4 )2 SO 4 1 mol (NH 4 )2 SO4 1 mol NH 3 1000 g

 2.58  10 kg NH3 3.78

The balanced equation for the decomposition is:

 2KCl(s)  3O2(g) 2KClO3(s)  ? g O2  46.0 g KClO3 

1 mol KClO3 3 mol O2 32.00 g O2    18.0 g O2 122.55 g KClO3 2 mol KClO3 1 mol O2

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

3.81

2A  B  C (a)

The number of B atoms shown in the diagram is 5. The balanced equation shows 2 moles A

1 mole B.

Therefore, we need 10 atoms of A to react completely with 5 atoms of B. There are only 8 atoms of A present in the diagram. There are not enough atoms of A to react completely with B. A is the limiting reagent. (b)

There are 8 atoms of A. Since the mole ratio between A and B is 2:1, 4 atoms of B will react with 8 atoms of A, leaving 1 atom of B in excess. The mole ratio between A and C is also 2:1. When 8 atoms of A react, 4 molecules of C will be produced.

B C

3.82

N2  3H2  2NH3 9 moles of H2 will react with 3 moles of N2, leaving 1 mole of H2 in excess. The mole ratio between N2 and NH3 is 1:2. When 3 moles of N2 react, 6 moles of NH3 will be produced.

H2 NH3

3.83

This is a limiting reagent problem. Let’s calculate the moles of NO2 produced assuming complete reaction for each reactant. 2NO(g)  O2(g)  2NO2(g)

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

0.886 mol NO

0.503 mol O2 

2 mol NO 2  0.886 mol NO2 2 mol NO

2 mol NO2  1.01 mol NO2 1 mol O2

NO is the limiting reagent; it limits the amount of product produced. The amount of product produced is 0.886 mole NO2.

3.84

(a)

2NH3(g)  H2SO4(aq)

(b)

Sulfuric acid is in excess. First, let’s calculate the moles of ammonia reacted to produce 20.3 g of

(NH4)2SO4(aq)

ammonium sulfate.

20.3 g (NH 4 )2 SO4 

1 mol (NH 4 )2 SO4 2 mol NH3   0.307 mol NH3 reacted 132.15 g (NH4 )2 SO4 1 mol (NH4 )2 SO4

The number of moles of sulfuric acid reacted will be half the moles of ammonia reacted (see mole ratio from the balance equation). The starting mass of each reactant is:

0.307 mol NH3 

17.03 g NH3  5.23 g NH 3 1 mol NH3

0.154 mol H2 SO4 

98.09 g H2 SO4  15.1 g H2 SO4 reacted 1 mol H 2 SO4

15.1 g H2SO4  5.89 g H2SO4 unreacted  21.0 g H2SO4

3.85

 3CO2(g)  4H2O(l) C3H8(g)  5O2(g) 

(a)

The balanced equation is:

(b)

The balanced equation shows a mole ratio of 3 moles CO2 : 1 mole C3H8. The mass of CO2 produced is:

3.65 mol C3 H8 

3.86

3 mol CO2 44.01 g CO2   482 g CO2 1 mol C3 H8 1 mol CO2

This is a limiting reagent problem. Let’s calculate the moles of Cl2 produced assuming complete reaction for each reactant.

0.86 mol MnO2  48.2 g HCl 

1 mol Cl2  0.86 mol Cl2 1 mol MnO2

1 mol Cl2 1 mol HCl   0.3305 mol Cl2 36.458 g HCl 4 mol HCl

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

HCl is the limiting reagent; it limits the amount of product produced. It will be used up first. The amount of product produced is 0.3305 mole Cl2. Let’s convert this to grams.

? g Cl 2  0.3305 mol Cl2 

3.89

The balanced equation is given:

70.90 g Cl2  23.4 g Cl 2 1 mol Cl2

 CaSO4  2HF CaF2  H2SO4 

The balanced equation shows a mole ratio of 2 moles HF : 1 mole CaF2. The theoretical yield of HF is:

(6.00103 g CaF2 )

1 mol CaF2 2 mol HF 20.008 g HF 1 kg     3.075 kg HF 78.08 g CaF2 1 mol CaF2 1 mol HF 1000 g

The actual yield is given in the problem (2.86 kg HF).

3.90

(a)

% yield 

actual yield 100% theoretical yield

% yield 

2.86 kg 100%  93.0% 3.075 kg

Start with a balanced chemical equation. It’s given in the problem. We use NG as an abbreviation for nitroglycerin. The molar mass of NG  227.1 g/mol.

 6N2  12CO2  10H2O  O2 4C3H5N3O9  Map out the following strategy to solve this problem. g NG  mol NG  mol O2  g O2 Calculate the grams of O2 using the strategy above.

? g O 2  2.00 102 g NG

(b)

1 mol NG 1 mol O2 32.00 g O2    7.05 g O2 227.1 g NG 4 mol NG 1 mol O2

The theoretical yield was calculated in part (a), and the actual yield is given in the problem (6.55 g). The percent yield is: % yield 

actual yield 100% theoretical yield

% yield 

6.55 g O2 100%  92.9% 7.05 g O2

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

3.91

The balanced equation shows a mole ratio of 1 mole TiO2 : 1 mole FeTiO3. The molar mass of FeTiO3 is 151.73 g/mol, and the molar mass of TiO2 is 79.88 g/mol. The theoretical yield of TiO2 is:

8.00 106 g FeTiO3 

1 mol FeTiO3 1 mol TiO2 79.88 g TiO 2 1 kg    151.73 g FeTiO3 1 mol FeTiO3 1 mol TiO2 1000 g

 4.21  10 kg TiO2

The actual yield is given in the problem (3.67  10 kg TiO2).

% yield 

3.92

actual yield 3.67103 kg 100%  100%  87.2% theoretical yield 4.21103 kg

The actual yield of ethylene is 481 g. Let’s calculate the yield of ethylene if the reaction is 100 percent efficient. We can calculate this from the definition of percent yield. We can then calculate the mass of hexane that must be reacted. % yield 

actual yield 100% theoretical yield

42.5% yield 

481 g C 2 H 4 100% theoretical yield 3

theoretical yield C2H4  1.132  10 g C2H4 The mass of hexane that must be reacted is:

(1.132103 g C2 H 4 )

3.93

1 mol C2 H4 1 mol C6 H14 86.172 g C6 H14    3.4810 3 g C6 H14 28.052 g C 2 H4 1 mol C 2 H4 1 mol C6 H14

This is a limiting reagent problem. Let’s calculate the moles of Li3N produced assuming complete reaction for each reactant. 6Li(s)  N2(g)  2Li3N(s)

12.3 g Li

1 mol Li 2 mol Li3 N   0.5907 mol Li3 N 6.941 g Li 6 mol Li

33.6 g N 2 

1 mol N 2 2 mol Li3 N   2.398 mol Li3 N 28.02 g N 2 1 mol N2

Li is the limiting reagent; it limits the amount of product produced. The amount of product produced is 0.5907 mole Li3N. Let’s convert this to grams.

? g Li3 N  0.5907 mol Li3 N

34.833 g Li3 N  20.6 g Li 3 N 1 mol Li3 N

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

This is the theoretical yield of Li3N. The actual yield is given in the problem (5.89 g). The percent yield is:

% yield 

3.94

actual yield 5.89 g 100%  100%  28.6% theoretical yield 20.6 g

This is a limiting reagent problem. Let’s calculate the moles of S2Cl2 produced assuming complete reaction for each reactant. S8(l)  4Cl2(g)  4S2Cl2(l) 4.06 g S8 

1 mol S8 256.56 g S8

6.24 g Cl2 



4 mol S2 Cl 2  0.0633 mol S2 Cl 2 1 mol S8

1 mol Cl2 4 mol S2 Cl2   0.0880 mol S2 Cl2 70.90 g Cl2 4 mol Cl2

S8 is the limiting reagent; it limits the amount of product produced. The amount of product produced is 0.0633 mole S2Cl2. Let’s convert this to grams.

135.04 g S2 Cl2 ? g S2 Cl2  0.0633 mol S2 Cl2   8.55 g S2 Cl2 1 mol S2 Cl2 This is the theoretical yield of S2Cl2. The actual yield is given in the problem (6.55 g). The percent yield is: % yield  3.95

6.55 g actual yield 100%  100%  76.6% theoretical yield 8.55 g

This is an atom economy problem. Remember that atom economy is the ratio of the total mass of the desired product(s) over the total mass of all of the reactants. Let’s start by determining the total mass of desired product (C4H2O3) and the total mass of all reactants in the balanced equation. 2C6H6  9O2 156.22 287.99

H2O3  4CO2  4H2O

192.12

The atom economy is atom economy 

3.96

192.12 100%  43.250% 156.22  287.99

This is an atom economy problem. Remember that atom economy is the ratio of the total mass of the desired product(s) over the total mass of all of the reactants. Let’s start by determining the total mass of desired product (CH3CO2H) and the total mass of all reactants in the balanced equation. CH3OH  CO 32.04

28.01

CO2H

60.05

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

The atom economy is atom economy 

3.97

60.05 100%  100% 32.04  28.01 69

Start by letting x be the fractional abundance of Ga. Since the sum of the two abundances must be 1, we can write: 71

Abundance Ga  (1  x) Average atomic mass of Ga  69.72 amu  x(68.9256 amu)  (1  x)(70.9247 amu) 69.72  1.9991x  70.9247 x  0.603 69

x  0.603 corresponds to a natural abundance of Ga of 60.3 percent. The natural abundance of Ga is (1  x)  0.397 or 39.7 percent.

3.98

Start by letting x be the fractional abundance of Rb. Since the sum of the two abundances must be 1, we can write: 87

Abundance Rb  (1  x) Average atomic mass of Rb  85.47 amu  x(84.912 amu)  (1  x)(86.909 amu) 85.47  1.997x  86.909 1.997x  1.44 x  0.721 85

x  0.721 corresponds to a natural abundance of Rb of 72.1 percent. The natural abundance of Rb is (1  x)  0.279 or 27.9 percent.

3.99

All the carbon from the hydrocarbon reactant ends up in CO2, and all the hydrogen from the hydrocarbon reactant ends up in water. In the diagram, we find 4 CO2 molecules and 6 H2O molecules. This gives a ratio between carbon and hydrogen of 4:12. We write the formula C4H12, which reduces to the empirical formula CH3. The empirical molar mass equals approximately 15 g, which is half the molar mass of the hydrocarbon. Thus, the molecular formula is double the empirical formula or C2H6. Since this is a combustion reaction, the other reactant is O2. We write: C2H6  O2  CO2  H2O Balancing the equation, 2C2H6  7O2  4CO2  6H2O

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

3.100

2H2(g)  O2(g)  2H2O(g) We start with 8 molecules of H2 and 3 molecules of O2. The balanced equation shows 2 moles H2

1 mole

O2. If 3 molecules of O2 react, 6 molecules of H2 will react, leaving 2 molecules of H2 in excess. The balanced equation also shows 1 mole O2

2 moles H2O. If 3 molecules of O2 react, 6 molecules of H2O will be

produced. After complete reaction, there will be 2 molecules of H2 and 6 molecules of H2O. The correct diagram is choice (b).

3.101

First, let’s calculate the theoretical yield.

4.66 g C2 H4 

1 mol C2 H 4 1 mol C2 H 5Cl 64.51 g C 2 H5 Cl    10.7 g C2 H5 Cl 28.05 g C2 H4 1 mol C2 H 4 1 mol C2 H5 Cl

The mass of ethyl chloride produced at 89.4% yield is: (0.894)(10.7 g)  9.57 g C2H5Cl.

3.102

3.103

5CO2(g)  6H2O(l)

(a)

C5H12(l) 8O2(g)

(b)

NaHCO3(s) HCl(aq)

(c)

6Li(s) N2(g)

(d)

PCl3(l) 3H2O(l)

(e)

3CuO(s) 2NH3(g)

CO2(g)  NaCl(aq)  H2O(l)

2Li3N(s) H3PO3(aq)  3HCl(g) 3Cu(s)  N2(g)  3H2O(l)

First, let’s convert to moles of HNO3 produced.

1.00 ton HNO3 

1 mol HNO3 2000 lb 453.6 g    1.44 10 4 mol HNO3 1 ton 1 1b 63.018 g HNO3 3

Now, we will work in the reverse direction to calculate the amount of reactant needed to produce 1.44  10 mol of HNO3. Realize that since the problem says to assume an 80% yield for each step, the amount of reactant needed in each step will be larger by a factor of

100% , compared to a standard stoichiometry 80%

calculation where a 100% yield is assumed. Referring to the balanced equation in the last step, we calculate the moles of NO2.

2 mol NO2 100% (1.44 104 mol HNO3 )   3.6010 4 mol NO2 1 mol HNO3 80%

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

Now, let’s calculate the amount of NO needed to produce 3.60  10 mol NO2. Following the same procedure as above, and referring to the balanced equation in the middle step, we calculate the moles of NO.

(3.6010 4 mol NO2 )

1 mol NO 100%   4.50104 mol NO 1 mol NO2 80% 4

Now, let’s calculate the amount of NH3 needed to produce 4.5  10 mol NO. Referring to the balanced equation in the first step, the moles of NH3 is:

(4.50 104 mol NO)

4 mol NH 3 100%   5.625  104 mol NH3 4 mol NO 80%

Finally, converting to grams of NH3:

17.034 g NH3 5.625104 mol NH3   9.58105 g NH 3 1 mol NH3

3.104

We assume that all the Cl in the compound ends up as HCl and all the O ends up as H2O. Therefore, we need to find the number of moles of Cl in HCl and the number of moles of O in H2O.

mol Cl  0.233 g HCl

mol O  0.403 g H2 O

1 mol HCl 1 mol Cl   0.006391 mol Cl 36.458 g HCl 1 mol HCl

1 mol H2 O 1 mol O   0.02237 mol O 18.016 g H2 O 1 mol H2 O

Dividing by the smallest number of moles (0.006391 mole) gives the formula, ClO3.5. Multiplying both subscripts by two gives the empirical formula, Cl2O7.

3.105

The balanced equation is: 2C4H10(g)  13O2(g)

26.7 g C 4 H10 

3.106

8CO2(g)  10H2O(l).

1 mol C 4 H10 10 mol H2 O 18.02 g H2 O    41.4 g H 2 O 58.12 g C 4 H10 2 mol C 4 H10 1 mol H2 O

This problem can be solved by two different methods.

90.04 g H 2 C 2 O 4 26.2 g H2 C2 O4  2H2 O  18.7 g H 2 C 2 O4 126.1 g H2 C2 O4  2H2 O

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

Or,

1 mol H2 C2 O4  2H 2 O 1 mol H 2 C 2 O 4 90.04 g H2 C2 O4 26.2 g H 2 C2 O4  2H 2 O    126.1 g H2 C2 O4  2H2 O 1 mol H2 C 2 O4  2H2 O 1 mol H2 C 2 O4  18.7 g H2C2O4

3.107

The number of moles of Y in 84.10 g of Y is:

27.22 g X 

1 mol X 1 mol Y   0.81448 mol Y 33.42 g X 1 mol X

The molar mass of Y is: molar mass Y 

84.10 g Y  103.3 g/mol 0.81448 mol Y

The atomic mass of Y is 103.3 amu.

3.108

The symbol “O” refers to moles of oxygen atoms, not oxygen molecule (O2). Look at the molecular formulas given in parts (a) and (b). What do they tell you about the relative amounts of carbon and oxygen? (a) (b)

3.109

1 mol O  0.212 mol O 1 mol C 2 mol O 0.212 mol C   0.424 mol O 1 mol C 0.212 mol C 

The observations mean either that the amount of the more abundant isotope was increasing or the amount of the less abundant isotope was decreasing. One possible explanation is that the less abundant isotope was undergoing radioactive decay, and thus its mass would decrease with time.

3.110

This is a calculation involving percent composition. Remember, percent by mass of each element 

mass of element in 1 mol of compound 100% molar mass of compound

The molar masses are: Al, 26.98 g/mol; Al2(SO4)3, 342.17 g/mol; H2O, 18.016 g/mol. Thus, using x as the number of H2O molecules,   2(molar mass of Al) 100% mass % Al    molar mass of Al 2 (SO 4 )3  x (molar mass of H2 O)    2(26.98 g) 100% 8.10%    342.17 g  x(18.016 g) 

(0.081)(342.17)  (0.081)(18.016)(x)  53.96 x  17.98 Rounding off to a whole number of water molecules, x  18. Therefore, the formula is Al2(SO4)318 H2O.

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

3.111

There are 3 N atoms per molecule of nitroglycerin. Therefore, 3 NO molecules can be released per molecule. The mass percent of NO in nitroglycerin is:

mass % NO 

3.112

molar mass of 3 NO 3(30.01 g) 100%  100%  39.64% molar mass of nitroglycerin 227.1 g

The number of carbon atoms in a 24-carat diamond is: 23 200 mg C 0.001 g C 1 mol C 6.022  10 atoms C 24 carat      2.4 1023 atoms C 1 carat 1 mg C 12.01 g C 1 mol C

3.113

The amount of Fe that reacted is:

1 664 g  83.0 g reacted 8

The amount of Fe remaining is:

664 g  83.0 g  581 g remaining

Thus, 83.0 g of Fe reacts to form the compound Fe2O3, which has two moles of Fe atoms per 1 mole of compound. The mass of Fe2O3 produced is: 83.0 g Fe 

1 mol Fe 1 mol Fe 2 O3 159.7 g Fe 2 O3    119 g Fe 2 O 3 55.85 g Fe 2 mol Fe 1 mol Fe 2 O3

The final mass of the iron bar and rust is:

3.114

581 g Fe  119 g Fe2O3  700 g

The mass of oxygen in MO is 39.46 g  31.70 g  7.76 g O. Therefore, for every 31.70 g of M, there is 7.76 g of O in the compound MO. The molecular formula shows a mole ratio of 1 mole M : 1 mole O. First, calculate moles of M that react with 7.76 g O.

mol M  7.76 g O 

molar mass M 

1 mol O 1 mol M   0.485 mol M 16.00 g O 1 mol O

31.70 g M  65.4 g/mol 0.485 mol M

Thus, the atomic mass of M is 65.4 amu. The metal is most likely Zn. 3.115

(a)

 ZnSO4(aq)  H2(g) Zn(s)  H2SO4(aq) 

(b)

We assume that a pure sample would produce the theoretical yield of H2. The balanced equation shows a mole ratio of 1 mole H2 : 1 mole Zn. The theoretical yield of H2 is:

3.86 g Zn 

1 mol Zn 1 mol H2 2.016 g H2    0.119 g H2 65.39 g Zn 1 mol Zn 1 mol H2

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

percent purity  (c)

3.116

0.0764 g H2 100%  64.2% 0.119 g H2

We assume that the impurities are inert and do not react with the sulfuric acid to produce hydrogen.

The wording of the problem suggests that the actual yield is less than the theoretical yield. The percent yield will be equal to the percent purity of the iron(III) oxide. We find the theoretical yield:

(2.62103 kg Fe2 O3 )

1000 g Fe2 O3 1 mol Fe2 O3 2 mol Fe 55.85 g Fe 1 kg Fe     1 kg Fe 2 O3 159.7 g Fe2 O3 1 mol Fe2 O3 1 mol Fe 1000 g Fe

 1.833  10 kg Fe

percent yield 

actual yield 100% theoretical yield

percent yield 

3.117

1.64 103 kg Fe 100%  89.5%  purity of Fe2 O3 1.833103 kg Fe

 6CO2  6H2O The balanced equation is: C6H12O6  6O2  6 mol CO2 44.01 g CO2 365 days 5.0 10 2 g glucose 1 mol glucose      (7.210 9 people) 1 person each day 180.16 g glucose 1 mol glucose 1 mol CO2 1 yr 15

 1.9  10 g CO2/yr 3.118

The carbohydrate contains 40 percent carbon; therefore, the remaining 60 percent is hydrogen and oxygen. The problem states that the hydrogen to oxygen ratio is 2:1. We can write this 2:1 ratio as H2O. Assume 100 g of compound. 40.0 g C 

1 mol C  3.331 mol C 12.01 g C

60.0 g H 2 O

1 mol H 2 O  3.330 mol H 2 O 18.016 g H 2 O

Dividing by 3.330 gives CH2O for the empirical formula. To find the molecular formula, divide the molar mass by the empirical mass. molar mass 178 g  6 empirical mass 30.026 g

Thus, there are six CH2O units in each molecule of the compound, so the molecular formula is (CH2O)6, or C6H12O6.

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

3.119

The molar mass of chlorophyll is 893.48 g/mol. Finding the mass of a 0.0011-mol sample: 0.0011 mol chlorophyll 

893.48 g chlorophyll  0.98 g chlorophyll 1 mol chlorophyll

The chlorophyll sample has the greater mass.

3.120

If we assume 100 g of compound, the masses of Cl and X are 67.2 g and 32.8 g, respectively. We can calculate the moles of Cl. 67.2 g Cl 

1 mol Cl  1.896 mol Cl 35.45 g Cl

Then, using the mole ratio from the chemical formula (XCl3), we can calculate the moles of X contained in 32.8 g. 1.896 mol Cl 

1 mol X  0.6320 mol X 3 mol Cl

0.6320 mole of X has a mass of 32.8 g. Calculating the molar mass of X: 32.8 g X  51.9 g /mol 0.6320 mol X

The element is most likely chromium (molar mass  52.00 g/mol). 3.121

This is an atom economy problem. Remember that atom economy is the ratio of the total mass of the desired product(s) over the total mass of all of the reactants. Let’s start by determining the total mass of desired product (C9H8O4) and the total mass of all reactants in the balanced equation. C7H6O3  C4H6O3 138.12

102.09

H8O4  C2H4O2

192.12

The atom economy is atom economy 

3.122

192.12 100%  79.980% 138.12  102.09

This is an atom economy problem. Remember that atom economy is the ratio of the total mass of the desired product(s) over the total mass of all of the reactants. Let’s start by determining the total mass of desired product (N2H4) and the total mass of all reactants in the balanced equation. 2NH3 NaOCl

N2H4  NaCl  H2O

32.06

32.05

74.44

The atom economy is atom economy 

32.05 100%  30.09% 32.06  74.44

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

3.123

(a)

The molar mass of hemoglobin is: 2952(12.01 g)  4664(1.008 g)  812(14.01 g)  832(16.00 g)  8(32.07 g)  4(55.85 g) 4

 6.532  10 g (b)

To solve this problem, the following conversions need to be completed: L  mL  red blood cells  hemoglobin molecules  mol hemoglobin  mass hemoglobin We will use the following abbreviations: RBC  red blood cells, HG  hemoglobin

6.532104 g HG 1 mL 5.0 109 RBC 2.8108 HG molecules 1 mol HG 5.00 L      0.001 L 1 mL 1 RBC 1 mol HG 6.02210 23 molecules HG 2

 7.6  10 g HG

3.124

A 100 g sample of myoglobin contains 0.34 g of iron (0.34% Fe). The number of moles of Fe is: 0.34 g Fe 

1 mol Fe  6.09 103 mol Fe 55.85 g Fe

Since there is one Fe atom in a molecule of myoglobin, the moles of myoglobin also equal 6.09  10

3

mole.

The molar mass of myoglobin can be calculated. molar mass myoglobin 

3.125

(a)

0.764 g CsI 

100 g myoglobin  1.6 10 4 g /mol 6.09 103 mol myoglobin

1 mol CsI 6.022 10 23 CsI 1 Cs ion    1.77 10 21 Cs ions 259.8 g CsI 1 mol CsI 1 CsI 





Since there is one I for every one Cs , the number of I ions  1.77  10 I ions

(b)

72.8 g K 2 Cr2 O 7 

1 mol K 2 Cr2 O 7 294.2 g K 2 Cr2 O7



6.02210 23 K 2 Cr2 O 7 1 mol K 2 Cr2 O7



2 K  ions  2.9810 23 K  ions 1 K 2 Cr2 O7



Since there are two K for every one Cr2 O 72  , the number of Cr2 O27 ions  1.49  10

(c)

6.54 g Hg 2 (NO3 )2 

1 mol Hg2 (NO3 )2 525.2 g Hg2 (NO3 )2



6.0221023 Hg2 (NO3 )2 1 mol Hg2 (NO3 )2



Cr2 O72 ions

1 Hg 22 ion 1 Hg2 (NO3 )2

 7.50  10 Hg 22  ions 22

Since there is one Hg 22  for every two NO3 ions, the number of NO3 ions is  1.50  10 NO3 ions Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

3.126

If we assume 100 g of the mixture, then there are 29.96 g of Na in the mixture (29.96% Na by mass). This amount of Na is equal to the mass of Na in NaBr plus the mass of Na in Na2SO4. 29.96 g Na  mass of Na in NaBr  mass of Na in Na2SO4 To calculate the mass of Na in each compound, grams of compound need to be converted to grams of Na using the mass percentage of Na in the compound. If x equals the mass of NaBr, then the mass of Na2SO4 is 100 x. Recall that we assumed 100 g of the mixture. We set up the following expression and solve for x. 29.96 g Na  mass of Na in NaBr  mass of Na in Na2SO4

 22.99 g Na   (2)(22.99 g Na)  29.96 g Na   x g NaBr   (100  x ) g Na 2 SO4   102.89 g NaBr   142.05 g Na 2 SO4   29.96  0.22344x  32.369  0.32369x 0.10025x  2.409 x  24.03 g, which equals the mass of NaBr. The mass of Na2SO4 is 100  x which equals 75.97 g. Because we assumed 100 g of compound, the mass % of NaBr in the mixture is 24.03% and the mass % of Na2SO4 is 75.97%.

3.127

Based on the stoichiometry of the problem, reactant A is the limiting reagent. First, we calculate the theoretical yield of the reaction. 4.0 mol A 

3 mol C  4.0 mol C 3 mol A

The percent yield of the reaction is: % yield 

3.128

C3H8(g)  5O2(g)

3.129

(a)

0.400 g aspirin 

(b)

0.307 g salicylic acid

2.8 mol C 100%  70%  7.0 10 1% 4.0 mol C

3CO2(g)  4H2O(l) 1 mol aspirin 1 mol salicylic acid 138.12 g salicylic acid    0.307 g salicylic acid 180.15 g aspirin 1 mol aspirin 1 mol salicylic acid

1  0.410 g salicylic acid 0.749

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

If you have trouble deciding whether to multiply or divide by 0.749 in the calculation, remember that if only 74.9% of salicylic acid is converted to aspirin, a larger amount of salicylic acid will need to be reacted to yield the same amount of aspirin.

(c)

9.26 g salicylic acid 

1 mol salicylic acid 1 mol aspirin   0.06704 mol aspirin 138.12 g salicylic acid 1 mol salicylic acid

8.54 g acetic anhydride 

1 mol acetic anhydride 1 mol aspirin   0.08365 mol aspirin 102.09 g acetic anhydride 1 mol acetic anhydride

The limiting reagent is salicylic acid. The theoretical yield of aspirin is: 0.06704 mol aspirin 

180.15 g aspirin  12.1 g aspirin 1 mol aspirin

The percent yield is: % yield 

3.130

10.9 g 100%  90.1% 12.1 g

The mass percent of an element in a compound can be calculated as follows: percent by mass of each element 

mass of element in 1 mol of compound 100% molar mass of compound

The molar mass of Ca3(PO4)2  310.18 g/mol % Ca 

3.131

(a)

(3)(40.08 g) 100%  38.76% Ca 310.18 g

%P

(2)(30.97 g) 100%  19.97% P 310.18 g

%O

(8)(16.00 g) 100%  41.27% O 310.18 g

First, calculate the mass of C in CO2, the mass of H in H2O, and the mass of N in NH3. For now, we will carry more than 3 significant figures and then round to the correct number at the end of the problem. ? g C  3.94 g CO 2 

1 mol CO 2 1 mol C 12.01 g C    1.075 g C 44.01 g CO 2 1 mol CO 2 1 mol C

? g H  1.89 g H 2 O 

1 mol H 2 O 2 mol H 1.008 g H    0.2114 g H 18.02 g H 2 O 1 mol H 2 O 1 mol H

? g N  0.436 g NH 3 

1 mol NH 3 1 mol N 14.01 g N    0.3587 g N 17.03 g NH 3 1 mol NH 3 1 mol N

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

Next, we can calculate the %C, %H, and the %N in each sample, then we can calculate the %O by difference. %C 

1.075 g C 100%  49.43% C 2.175 g sample

%H 

0.2114 g H 100%  9.720% H 2.175 g sample

%N 

0.3587 g N 100%  19.15% N 1.873 g sample

The % O  100%  (49.43%  9.720%  19.15%)  21.70% O Assuming 100 g of compound, calculate the moles of each element. ? mol C  49.43 g C 

1 mol C  4.116 mol C 12.01 g C

? mol H  9.720 g H 

1 mol H  9.643 mol H 1.008 g H

? mol N  19.15 g N 

1 mol N  1.367 mol N 14.01 g N

? mol O  21.70 g O

1 mol O  1.356 mol O 16.00 g O

Thus, we arrive at the formula C4.116H9.643N1.367O1.356. Dividing by 1.356 gives the empirical formula, C3H7NO. (b)

The empirical molar mass is 73.10 g. Since the approximate molar mass of lysine is 150 g, we have: 150 g 2 73.10 g

Therefore, the molecular formula is (C3H7NO)2 or C6H14N2O2.

3.132

Yes. The number of hydrogen atoms in one gram of hydrogen molecules is the same as the number in one gram of hydrogen atoms. There is no difference in mass, only in the way that the particles are arranged. Would the mass of 100 dimes be the same if they were stuck together in pairs instead of separated?

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

3.133

The mass of one fluorine atom is 19.00 amu. The mass of one mole of fluorine atoms is 19.00 g. Multiplying the mass of one atom by Avogadro’s number gives the mass of one mole of atoms. We can write: 19.00 amu  (6.022 10 23 F atoms)  19.00 g F 1 F atom

or, 23

6.022  10 amu  1 g This is why Avogadro’s numbers has sometimes been described as a conversion factor between amu and grams.

3.134

Since we assume that water exists as either H2O or D2O, the natural abundances are 99.985 percent and 0.015 percent, respectively. If we convert to molecules of water (both H2O or D2O), we can calculate the molecules that are D2O from the natural abundance (0.015%). The necessary conversions are: mL water  g water  mol water  molecules water  molecules D2O

400 mL water 

23 1 g water 1 mol water 6.022  10 molecules 0.015% molecules D2 O    1 mL water 18.02 g water 1 mol water 100% molecules water

 2.01  10 molecules D2O

3.135

There can only be one chlorine per molecule, since two chlorines have a combined mass in excess of 70 amu. 35

Since the Cl isotope is more abundant, let’s subtract 35 amu from the mass corresponding to the more intense peak. 50 amu  35 amu  15 amu 12

15 amu equals the mass of one C and three H. To explain the two peaks, we have: 12

molecular mass C H3 Cl  12 amu  3(1 amu)  35 amu  50 amu molecular mass C H3 Cl  12 amu  3(1 amu)  37 amu  52 amu 35

Cl is three times more abundant than Cl; therefore, the 50 amu peak will be three times more intense than

the 52 amu peak.

3.136

First, we can calculate the moles of oxygen. 2.445 g C 

1 mol C 1 mol O   0.2036 mol O 12.01 g C 1 mol C

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

Next, we can calculate the molar mass of oxygen. molar mass O 

3.257 g O  16.00 g/mol 0.2036 mol O

If 1 mole of oxygen atoms has a mass of 16.00 g, then 1 atom of oxygen has an atomic mass of 16.00 amu.

3.137

The molecular formula for Cl2O7 means that there are 2 Cl atoms for every 7 O atoms or 2 moles of Cl atoms for every 7 moles of O atoms. We can write: mole ratio 

3.138

1 mol Cl 2 2 mol Cl  7 mol O 3.5 mol O 2

(a)

The mass of chlorine is 5.0 g.

(b)

From the percent by mass of Cl, we can calculate the mass of chlorine in 60.0 g of NaClO3. mass % Cl 

35.45 g Cl 100%  33.31% Cl 106.44 g compound

mass Cl  60.0 g  0.3331  20.0 g Cl (c)

0.10 mol of KCl contains 0.10 mol of Cl. 0.10 mol Cl 

(d)

35.45 g Cl  3.5 g Cl 1 mol Cl

From the percent by mass of Cl, we can calculate the mass of chlorine in 30.0 g of MgCl2. mass % Cl 

(2)(35.45 g Cl) 100%  74.47% Cl 95.21 g compound

mass Cl  30.0 g  0.7447  22.3 g Cl (e)

The mass of Cl can be calculated from the molar mass of Cl2. 0.50 mol Cl 2 

70.90 g Cl  35.45 g Cl 1 mol Cl 2

Thus, (e) 0.50 mol Cl2 contains the greatest mass of chlorine. 3.139

The mass percent of Cl is given. From the mass of the compound and the number of hydrogen atoms given, we can calculate the mass percent of H. The mass percent of carbon is then obtained by difference. Once the mass percentages of each element are known, the empirical formula can be determined. 4.19 10 23 H atoms

1 mol H 1.008 g H   0.7013 g H 6.022 10 23 H atoms 1 mol H

CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS

mass % H 

0.7013 g H 100%  7.792% H 9.00 g compound

mass % C  100%  (55.0%  7.792%)  37.21% C To determine the empirical formula, assume 100 g of compound and convert to moles of each element present. mol C  37.21 g C 

1 mol C  3.098 mol C 12.01 g C

mol H  7.792 g H 

1 mol H  7.730 mol H 1.008 g H

mol Cl  55.0 g Cl 

1 mol Cl  1.551 mol Cl 35.45 g Cl

Thus, we arrive at the formula C3.098H7.730Cl1.551, which gives the identity and the mole ratios of atoms present. However, chemical formulas are written with whole numbers. Try to convert to whole numbers by dividing each of the subscripts by the smallest subscript (1.551). This gives the empirical formula C2H5Cl.

3.140

Both compounds contain only Pt and Cl. The percent by mass of Pt can be calculated by subtracting the percent Cl from 100 percent. Compound A: Assume 100 g of compound. 26.7 g Cl 

73.3 g Pt 

1 mol Cl  0.753 mol Cl 35.45 g Cl

1 mol Pt  0.376 mol Pt 195.1 g Pt

Dividing by the smallest number of moles (0.376 mole) gives the empirical formula, PtCl2. Compound B: Assume 100 g of compound. 42.1 g Cl 

1 mol Cl  1.19 mol Cl 35.45 g Cl

57.9 g Pt 

1 mol Pt  0.297 mol Pt 195.1 g Pt

Dividing by the smallest number of moles (0.297 mole) gives the empirical formula, PtCl4.