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Chemistry 11 Solutions

978‐0‐07‐105107‐1 Chapter 5 The Mole: A Chemist’s Counter • MHR | 1

Chapter 5

The Mole: A Chemist’s Counter Section 5.1 The Mole and the Avogadro Constant Solutions for Practice Problems Student Edition page 228 1. Practice Problem (page 228)

An average refrigerator has a volume of 0.6 m3. If a grain of salt has a volume of 9.39 × 10−11 m3, how any refrigerators would 1 mol of salt grains fill? What Is Required? You need to determine the number of refrigerators that would be filled by 1 mol of grains of sand. What Is Given? You know the volume of one refrigerator: 0.6 m3

You know the volume of one grain of sand: 9.39 × 10–11 m3

Plan Your Strategy The Avogadro constant, NA, is the number of particles in 1 mol: 6.02 × 1023 Multiply the volume of one grain of sand by the Avogadro constant to determine the total volume of 6.02 × 1023 grains of sand. Then divide the total volume of the grains of sand by the volume of one refrigerator. Act on Your Strategy total volume of 1 mol of grains of sand 236.02 × 10 grains= 11 3 9.39 10 m / grain −× ×

13 35.5986 10 m= ×

number of refrigerators13 35.5986 10 m×

=30.6 m

13

13

/refrigerator9.331 10 refrigerators9 10 refrigerators

= ×

×=



Therefore, 6.02 × 1023 grains of sand would fill 9 × 1013 refrigerators. Check Your Solution The units cancel properly. The answer seems reasonable and the number of significant digits in the answer agrees with the given data having the least number of significant digits.

2. Practice Problem (page 228) If you drove for 6.02 × 1023 days at a speed of 110 km/h, how far would you travel? What Is Required? You need to calculate the total distance travelled. What Is Given? You know the speed you are travelling at: 110 km/h You know the total time elapsed during the trip: 6.02 × 1023 days Plan Your Strategy Use the conversion factor of 24 h/day to determine the total time for the trip in hours. Calculate the distance travelled using the following relationship: distance = speed × time. Act on Your Strategy time elapsed (in hours) 236.02 10 days= × 24 h/ day×

251.4448 10 h= ×

distance travelled speed time

110 km/ h

= ×

= 25 1.4448 10 h× ×27

27

1.58928 10 km1.6 10 km

= ×

= ×

The distance travelled in 6.02 × 1023 days is 1.6 × 1027 km. Check Your Solution The units cancel properly. The answer seems reasonable and the number of significant digits in the answer agrees with the given data having the least number of significant digits.



3. Practice Problem (page 228) How long would it take to count 6.02 × 1023 raisins, if you counted at a rate of one raisin per second? What Is Required? You need to calculate the time needed to count 6.02 × 1023 of raisins. What Is Given? You know the total number of raisins to be counted: 6.02 × 1023 You know the rate at which the raisins are counted: 1 raisin per second Plan Your Strategy Divide the total number of raisins by the rate of counting. To express the answer in days or years, multiply the time in seconds by the appropriate

conversion factors: 1 h3600 s

, 1 day24 h

, and 1 year

365 days.

Act on Your Strategy

time (in seconds) to count raisins

2

23

3

number of raisinsrate of counting

6.02 10 raisins1 raisin

6.02 10/ss×

×

=

=

=

To convert the time from seconds to days:

23time (in days) 6.02 10 s= × h 1 ×

3600 s1 day 24 h

×

186.9676 10 days×=

To convert the time from days to years:

18time (in years) 6.9676 10 days= ×1 year

365 days×

16

16

1.90892 10 years1.9 10 years

= ×

= ×

It would take 6.97 × 1018 days, or 1.9 × 1016 years, to count 6.02 × 1023 raisins.



Check Your Solution The units cancel properly. The answer seems reasonable and the number of significant digits in the answer agrees with the given data having the least number of significant digits.

4. Practice Problem (page 228) A ream of paper (500 sheets) is 4.8 cm high. If you stacked 6.02 × 1023 sheets of paper on top of each other, how high would the stack be, in kilometres? What Is Required? You need to calculate the height (in kilometres) of 6.02 × 1023 sheets of paper. What Is Given? You know the total number of sheets of paper: 6.02 × 1023 sheets You know the height of 500 sheets of paper: 4.8 cm Plan Your Strategy Divide the total number of sheets by 500. Multiply this amount by the height (in centimetres) of 500 sheets. Convert this height to kilometres: 1 cm = 1 × 10–5 km Act on Your Strategy

total height (in centimetres) 236.02 10 sheets= ×4.8 cm

500 s

s

heet×

215.779 10 cm= ×

total height (in kilometres) 215.779 10 cm= × –5 1 10 km/ cm× ×

16

16

5.779 10 km5.8 10 km= ×

= ×

The height of 6.02 × 1023 sheets of paper is 5.8 × 1016 km. Check Your Solution The units cancel properly. The answer seems reasonable and the number of significant digits in the answer agrees with the given data having the least number of significant digits.



5. Practice Problem (page 228) The total volume of the Rogers Centre is 1.6 × 10 m3. If the volume of 100 peas is about 55 cm3, how many Rogers Centres would 6.02 × 1023 peas fill? What Is Required? You need to determine the number of Rogers Centres that would be filled by 6.02 × 1023 peas. What Is Given? You know the volume of one Rogers Centre: 1.6 × 106 m3 You know the volume of 100 peas: 55 cm3 You know the total number of peas: 6.02 × 1023 Plan Your Strategy Divide the total number of peas by 100. Multiply this answer by 55 to calculate the total volume of the peas in cm3. Convert the total volume of peas to m3: 1 m3 = 1 × 106 cm3 Divide the total volume of peas in m3 by the volume of one Rogers Centre. Act on Your Strategy

total volume (in cm3) of 6.02 × 1023 peas 236.02 × 10 peas=355 cm

100 peas×

23 33.311 10 cm= ×

total volume (in km3) of 6.02 × 1023 peas 23 33.311 10 cm= ×3

6 3

1 m1 × 10

cm

×

17 33.311 10 m= ×

number of Rogers Centres 17 33.311 × 10 m

=6 31.6 × 10 m

11

11

2.069 102.1 10

×

= ×

=

Therefore, 6.02 × 1023 peas would fill 2.1 × 1011 Rogers Centres. Check Your Solution The units cancel properly. Comparing the powers of ten in the calculation indicates that the answer is reasonable. The number of significant digits in the answer agrees with the given data having the least number of significant digits.



6. Practice Problem (page 228) Canada’s coastline is 243 042 km long. If you laid 6.02 × 1023 metre sticks end to end along the coast of Canada, how many rows of sticks would there be? What Is Required? You need to determine the number of rows of metre sticks that would line the coastline. What Is Given? You know the length of the coastline (which is one row of sticks): 243 042 km You know the total number of metre sticks: 6.02 × 1023 Plan Your Strategy Convert the length of the coastline to metres: 1 km = 1 × 103 m Calculate the number of rows of metre sticks by dividing the total number of metre sticks by the length of the coastline. Act on Your Strategy length of coastline (in metres) 243 042 km= 3 1 10 m/ km× ×

82.43042 10 m= ×

number of rows of metre sticks 236.02 × 10 metre sticks

= 82.43042 × 10 metre sticks15

15

2.4769 10 rows2.48 10

/ro

w

rows=

= ×

×

There would be 2.48 × 1015 rows of metre sticks. Check Your Solution The units cancel properly. The answer seems reasonable and the number of significant digits in the answer agrees with the given data having the least number of significant digits.

7. Practice Problem (page 228) Earth’s oceans contain about 1.31 × 109 km3 of water. One tablespoon is equal to 15 cm3. If you could remove 6.02 × 1023 tablespoons of water from Earth’s oceans, would you completely drain them? What Is Required? You need to determine if the removal of 6.02 × 1023 tablespoons of water would drain the oceans of water.



What Is Given? You know the total volume of water in the ocean: 1.31 × 109 m3 You know the volume of one tablespoon: 15 cm3 Plan Your Strategy Convert the volume of water in the oceans from km3 to cm3: 1 km3 = 1 × 1015 cm3 Divide the total volume of water in the oceans by the volume of one tablespoon. Compare the number of tablespoons of water in the ocean with 6.02 × 1023 tablespoons. Act on Your Strategy

volume (in cm3) of water in the ocean 9 31.31 10 km= ×15 3

3

1 × 10 1

cm km

×

24 31.31 10 cm= ×

total number of tablespoons of water in the oceans 24 31.31 × 10 cm

=315 cm

22

/tablespoon8.73 10 tablesp

oons×=

Since Earth’s oceans contain only 8.73 × 1022 tablespoons of water, removing any more than that would completely drain the oceans, which would be the case if removing 6.02 × 1023 tablespoons. Alternative Solution Plan Your Strategy The 6.02 × 1023 tablespoons equates to 1 mol of tablespoons. Calculate the volume (in cm3) of 1 mol of teaspoons. Convert the volume of 1 mol of tablespoons to km3: 1 cm3 = 1 × 10–15 km3

Compare the volume of 1 mol of tablespoons to the total volume of the oceans. Act on Your Strategy Total volume, V, of 1 mol of tablespoons:

236.02 10 tablespoonsV = × 3 15 cm / tablespoons×24 3

24 3

9.03 10 cm

9.03 10 cm

= ×

= × –15 3 3 1 10 km / cm× ×9 39.03 10 km= ×



This is greater than the total volume of water in the oceans. The oceans would be drained by the removal of 1 mol of tablespoons. Check Your Solution The units cancel properly. A check of the powers of ten indicates that the answer is reasonable.

8. Practice Problem (page 228) Suppose that you were given 6.02 × 1023 pennies when you were born and you lived for 100 years. How much money, in dollars, would you need to spend each second if you wanted to spend all this money in your lifetime? What Is Required? You need to determine how much money (in dollars) to spend per second. What Is Given? You know the total amount of money to be spent: 6.02 × 1023 pennies You know that the money must be spent in 100 years. Plan Your Strategy

Convert the time of 100 years to seconds using1 year

365 days,1 day

24 h, and 1 h

3600 s.

Divide the number of pennies by the time in seconds to determine the rate of spending. Divide the rate of spending in pennies by 100 to determine the rate of spending in dollars. Act on Your Strategy

time (in seconds) of 100 years 100 years=365

× days

1 year24 h ×

1 day3600 s ×

h93.1536 10 s×=

rate of spending (in pennies) 23

14

1

9

4

1.908

6.02

9 10 pennie

× 10 pennies3.1536 ×

s/s1.91

10

10 pen i

s

s

n es/

=

= ×

= ×



rate of spending (in dollars) 141.91 10 pennies= ×1 dollar/s

100 pennies×

121.91 10 dollars/s= ×

The 6.02 × 1023 pennies must be spent at a rate of 1.91 × 1012 dollars per second to be used up in 100 years. Check Your Solution The units cancel properly. The answer seems reasonable and the number of significant digits in the answer agrees with the given data having the least number of significant digits.

9. Practice Problem (page 228) How would the mass of pennies you were given in question 8 compare with the mass of Earth? The mass of 10 pennies is about 24 g. The mass of Earth is 5.98 × 1024 kg. What Is Required? You need to compare the mass of Earth with the mass of 6.02 × 1023 of pennies. What Is Given? You know the total number of pennies: 6.02 × 1023 You know the mass of 10 pennies: 24 g You know the mass of Earth: 5.98 × 1024 kg Plan Your Strategy Calculate the mass of 6.02×1023 pennies (in grams) by dividing the number of pennies by 10. Then multiply this number by 24. Convert the mass of pennies to kilograms: 1 g = 1 × 10–3 kg Compare the mass of Earth with the mass of 6.02 × 1023 pennies, expressed in kilograms. Act on Your Strategy Mass, m, (in grams) of 6.02 × 1023 pennies:

236.02 × 10 penniesm =24 g

10 pe

s

nnie×

241.4448 × 10 g=



Mass, m, (in kilograms) of 6.02 × 1023 pennies: 241.4448 10 gm = × –3 1 10 kg/ g× ×211.4448 10 kg= ×

Ratio of mass of Earth to mass of pennies:

24

23

5.98 × 10 kgmass of Earth mass of 6.02 × 10 pennies

= 211.4448 × 10 kg3

3

3

4.139 101

4.139 104.1 10

×=

= ×

= ×

The mass of Earth is about 4.1 × 102 times the mass of 6.02 × 1023 pennies. Check Your Solution The units cancel properly. The answer seems reasonable.

10. Practice Problem (page 228) If a row of approximately 5.0 × 107 atoms measured 1.0 cm, how long would a row of 6.02 × 1023 atoms be? What Is Required? You need to determine the length of a row containing 6.02 × 1023 atoms. What Is Given? You know the total number of atoms: 6.02 × 1023 You know the number of atoms in a length of 1 cm: 5.0 × 107 Plan Your Strategy Divide the total number of atoms by the number of atoms in a 1 cm length. Act on Your Strategy

length of 6.02 × 1023 atoms 23 6.02 × 10 atoms

= 75.0 × 10 atoms161.2 1

/0 cm

cm×=

The length of a row of 6.02 × 1023 of atoms would be 1.2 × 1016 cm.




Section 5.1 The Mole and the Avogadro Constant Solutions for Practice Problems Student Text page 230

11. Practice Problem (page 230)

A Canadian penny contains 0.106 mol of copper. How many atoms of copper are in a Canadian penny? What Is Required? You need to calculate the number of atoms, N, of copper in a Canadian penny. What Is Given? You know that the penny contains 0.106 mol of copper, Cu(s), atoms. Plan Your Strategy Use the Avogadro constant: 236.02 10AN = × Calculate the number of atoms of Cu(s) using the relationship AN n N= × . Act on Your Strategy Number of atoms, N, of Cu(s):

0.106 molAN n N= ×

= 23 6.02 10 atoms/ mol× ×22

22

6.3812 10 atoms6.38 10 atoms

= ×

= ×

There are 6.38 × 1022 atoms of Cu(s) in the Canadian penny. Check Your Solution The units cancel properly. The answer seems reasonable and the number of significant digits in the answer agrees with the given data having the least number of significant digits.



12. Practice Problem (page 230) The head of a small pin contains about 8 × 10–3 mol of iron. How many iron atoms are in the head of the pin? What Is Required? You need to determine the number of atoms, N, of iron in the head of a pin. What Is Given? You know that the head of the pin contains 8 × 10–3 mol of iron, Fe(s), atoms. Plan Your Strategy Use the Avogadro constant: 236.02 10AN = × Calculate the number of atoms of Fe(s) using the relationship AN n N= × . Act on Your Strategy Number of atoms, N, of Fe(s):

3

8 10 molAN n N−

= ×

= × 23 6.02 10 atoms/ mol× ×21

21

4.816 10 atoms5 10 atoms

= ×

= ×

There are 5 × 1021 atoms of Fe(s) in the head of a pin. Check Your Solution The units cancel properly. The answer seems reasonable and the number of significant digits in the answer agrees with the given data having the least number of significant digits.

13. Practice Problem (page 230) How many molecules of oxygen gas are in a room that contains 8.5 × 103 mol of oxygen gas? What Is Required? You need to determine the number of molecules, N, of oxygen gas, O2(g), in a room. What Is Given? You know that there is 8.5 × 103 mol of oxygen molecules in the room.



Plan Your Strategy Use the Avogadro constant: 236.02 10AN = × Calculate the number of molecules of O2(g) using the relationship AN n N= × . Act on Your Strategy Number of molecules, N, of O2(g):

3

8.5 10 molAN n N= ×

= × 23 6.02 10 molecules/ mol× ×27

27

5.117 10 molecules5.1 10 molecules

= ×

= × There are 5.1 × 1027 molecules of O2(g) in the room. Check Your Solution The units cancel properly. The answer seems reasonable and the number of significant digits in the answer agrees with the given data having the least number of significant digits.

14. Practice Problem (page 230) If a marble countertop contains 849 mol of calcium carbonate, CaCO3(s), how many formula units of calcium carbonate are in the countertop? What Is Required? You need to determine the number of formula units, N, of calcium carbonate in a countertop. What Is Given? You know that the countertop contains 849 mol of calcium carbonate. Plan Your Strategy Use the Avogadro constant: 236.02 10AN = × Calculate the number of formula units of CaCO3(s) using the relationship AN n N= × . Act on Your Strategy Number of formula units, N, of CaCO3(s):

849 molAN n N= ×

= 23 6.02 10 formula units/ mol× ×265.11 10 formula units = ×



There are 5.11 × 1026 formula units of CaCO3(s) in the countertop. Check Your Solution The units cancel properly. The answer seems reasonable and the number of significant digits in the answer agrees with the given data having the least number of significant digits.

15. Practice Problem (page 230) A recipe calls for half a teaspoon of salt, which contains 5.23 × 10–2 mol of sodium chloride. How many formula units of sodium chloride are needed? What Is Required? You need to determine the number of formula units, N, of sodium chloride, NaCl(s), in half a teaspoon of salt. What Is Given? You know that half a teaspoon contains 5.23 × 10–2 mol of sodium chloride. Plan Your Strategy Use the Avogadro constant: 236.02 10AN = × Calculate the number of formula units of NaCl(s) using the relationship AN n N= × . Act on Your Strategy Number of formula units, N, of NaCl(s):

–2

5.23 10 molAN n N= ×

= × 23 6.02 10 formula units/ mol× ×22

22

3.1485 10 formula units3.15 10 formula units

= ×

= ×

A total of 3.15 × 1022 formula units of NaCl(s) are needed. Check Your Solution The units cancel properly. The answer seems reasonable and the number of significant digits in the answer agrees with the given data having the least number of significant digits.



16. Practice Problem (page 230) A window-cleaning solution contains 3.86 mol of acetic acid, CH3COOH(ℓ). How many molecules of acetic acid are in the solution? What Is Required? You need to determine the number of molecules, N, of acetic acid in a window- cleaning solution. What Is Given? You know that there are 3.86 mol of acetic acid in the solution. Plan Your Strategy Use the Avogadro constant: 236.02 10AN = × Calculate the number of molecules of CH3COOH(ℓ) using the relationship AN n N= × . Act on Your Strategy Number of molecules, N, of CH3COOH(ℓ):

3.86 molAN n N= ×

= 23 6.02 10 molecules/ mol× ×242.32 10 molecules = ×

There are 2.32 × 1024 molecules of CH3COOH(ℓ) in the window-cleaning solution. Check Your Solution The units cancel properly. The answer seems reasonable and the number of significant digits in the answer agrees with the given data having the least number of significant digits.


A fuel tank used in a barbecue contains 2.0 × 102 mol of propane, C3H8(g). What is the total number of atoms in the tank? What Is Required? You need to determine the total number of atoms in a fuel tank containing propane. What Is Given? You know that there are 2.0 × 102 mol of propane molecules in the fuel tank.



Plan Your Strategy Use the Avogadro constant: 236.02 10AN = × Calculate the number of molecules, N, of C3H8(g) using the relationship AN n N= × . The chemical formula for propane shows that one molecule of C3H8(g) contains 3 carbon atoms and 8 hydrogen atoms for a total of 11 atoms. Multiply the number of molecules of C3H8(g) by the number of atoms per molecule. Act on Your Strategy Number of molecules, N, of C3H8(g):

2

2.0 10 molAN n N= ×

= × 23 6.02 10 molecules/ mol× ×261.204 10 molecules= ×

total number of atoms 261.204 10 molecules= × 11 atoms/ molecule×

271.3 10 atoms= ×

There are 1.3 × 1025 atoms in the fuel tank. Check Your Solution The units cancel properly. The answer seems reasonable. The number of atoms is about 10 times the number of molecules. The number of significant digits in the answer agrees with the given data having the least number of significant digits.

18. Practice Problem (page 230) Freon™, CCl2F2 (g), is a refrigerant that is no longer used in car air conditioners because it damages the ozone layer. A sample contains 4.82 mol of Freon™. a. How many molecules of Freon™ are in the sample? b. How many atoms, in total, are in the sample? What Is Required? a. You need to determine the total number of molecules, N, of Freon™ in a sample. b. You need to determine the total number of atoms in the sample of Freon™. What Is Given? You know that there are 4.82 mol of Freon™ molecules in the sample.



Plan Your Strategy a. number of molecules Use the Avogadro constant: 236.02 10AN = × Calculate the number of molecules of CCl2F2(g) using the relationship AN n N= × . b. number of atoms The chemical formula, CCl2F2(g), shows that one molecule of Freon™ contains 1 carbon atom, 2 chlorine atoms, and 2 fluorine atoms for a total of 5 atoms. Multiply the number of molecules of CCl2F2(g) by the number of atoms per molecule. Act on Your Strategy a. number of molecules Number of molecules, N, of CCl2F2(g):

A

4.82 mol

N n N= ×

= 23 6.02 10 molecules/ mol× ×242.90 10 molecules= ×

There are 2.90 × 1024 molecules of Freon™ in the sample. b. number of atoms

24number of atoms 2.9016 10 molecules= ×5 atoms

1 molecule×

25 1.45 10 atoms= ×

There are 1.45 × 1025 atoms in the Freon™ sample. Check Your Solution The units cancel properly. The answers seem reasonable. The total number of atoms is about 5 times the number of molecules. The number of significant digits in the answers agrees with the given data having the least number of significant digits.



19. Practice Problem (page 230) Glauber’s salt is a common name for sodium sulfate decahydrate, Na2SO4•10H2O(s). It is used in the manufacture of detergents. Suppose that a sample of 36.2 mol of sodium sulfate decahydrate is required. a. What number of sodium atoms would be in the sample? b. What number of water molecules would be in the sample? What Is Required? a. You need to determine the total number of sodium, Na(s), atoms in a sample of sodium sulfate decahydrate. b. You need to determine the total number of molecules of water in a sample of sodium sulfate decahydrate. What Is Given? You know that there are 36.2 mol of sodium sulfate decahydrate in the sample. Plan Your Strategy a. number of sodium atoms Use the Avogadro constant: 236.02 10AN = × Calculate the number of formula units of Na2SO4•10H2O using the relationship

.AN n N= × Determine the number of atoms of Na(s) per formula unit of Na2SO4•10H2O. Calculate the number of Na(s) atoms by multiplying the total number of formula units by the number of Na(s) atoms per formula unit. b. number of water molecules Calculate the number of water molecules by multiplying the total number of formula units by the number of water molecules per formula unit. Act on Your Strategy a. number of sodium atoms Number of formula units of Na2SO4•10H2O(s):

A

36.2 mol

N n N= ×

= 23 6.02 10 formula units/ mol× ×252.179 10 formula units= ×

From the chemical formula, there are 2 sodium atoms in one formula unit of Na2SO4•10H2O.



number of Na(s) atoms sample 252.179 10 formula units= ×

Na atomsformul

a 2

unit×

254.36 10 atoms

⎛ ⎞⎜ ⎟⎝ ⎠

= ×

There are 4.36 × 1025 sodium atoms in the sample of sodium sulfate decahydrate. b. number of water molecules From the chemical formula, there are 10 water molecules in one formula unit of Na2SO4•10H2O. number of water molecules in sample 252.179 10 formula units= ×

2H O moleculesformu

la t

10uni

×

22

62.18 10 molecuH lO es

⎛ ⎞⎜ ⎟⎝ ⎠

= ×

There are 2.18×1026 water molecules in the sample of sodium sulfate decahydrate. Check Your Solution The units cancel properly. The answers seem reasonable. The number of water molecules is about 5 times the number of sodium atoms. The number of significant digits in the answers agrees with the given data having the least number of significant digits.

20. Practice Problem (page 230) A sample of sucrose, C12H22O11 (s), contains 0.16 mol of oxygen atoms. a. What amount in moles of sucrose is in the sample? b. How many atoms of carbon are in the sample? What Is Required? a. You need to determine the total amount in moles (N) of molecules of sucrose in a sample. b. You need to determine the total number of atoms of carbon in the sucrose sample. What Is Given? You know there are 0.16 moles of oxygen atoms in the sample of sucrose.



Plan Your Strategy a. moles of sucrose Determine the amount in moles of atoms of oxygen per mole of C12H22O11(s). Calculate the number of moles of sucrose using the

ratio 12 22 11

12 22 11

1 mol C H O11 mol O per mol C H O

.

b. atoms of carbon Use the Avogadro constant: 236.02 10AN = × Calculate the number of molecules, N, of C12H22O11 using the relationship

.AN n N= × Determine the number of carbon atoms per molecule of C12H22O11(s). Calculate the number of carbon atoms in the sample of C12H22O11(s). Act on Your Strategy a. moles of sucrose From the chemical formula for sucrose, there are 11 moles of oxygen atoms per mole of C12H22O11(s). Amount in moles, n, of C12H22O11(s):

12 22 111 mol C H O11 mol oxygen atoms0.16 mol oxygen atoms

0.16 mol oxygen atoms

n

n

=

= 12 22 111 mol C H O11 mol oxygen a

s

tom

×

12 22 11

12 22 11

0.014545 mol C H O0.015 mol C H O

==

There is 0.015 mol of C12H22O11(s) in the sample. b. atoms of carbon Number of molecules, N, of C12H22O11:

0.014545 molAN n N= ×

= 23 6.02 10 molecules/ mol× ×21

12 22 118.756 10 molecules C H O (s)= ×

From the chemical formula for sucrose, there are 12 carbon atoms per molecule of C12H22O11(s).



Number of carbon atoms, N, in the sample:

21

21

12 C atoms8.756 10 molecules 1 molecule

8.756 10 molecules

N

N

=×

= ×12 C atoms 1 molecule

×

23

23

1.051 10 C atoms 1.05 10 C atoms

= ×

= ×

There are 1.04 × 1023 atoms of carbon in the sample. Check Your Solution The units cancel properly. The answers seem reasonable and the number of significant digits in the answers agrees with the given data having the least number of significant digits. Changing the number of atoms of carbon to moles gives approximately the same number as the original number of moles oxygen. This is reasonable since the number of C atoms and the number of O atoms in C12H22O11(s) are approximately the same.

Section 5.1 The Mole and the Avogadro Constant Solutions for Practice Problems Student Edition page 231


A gold coin contains 9.51 × 1022 atoms of gold. What amount in moles of gold is in the coin? What Is Required? You need to determine the amount in moles of gold, Au(s), is in a gold coin. What Is Given? You know the number of atoms, N, of gold in a gold coin: 9.51 × 1022 Plan Your Strategy Use the Avogadro constant: 236.02 10AN = ×

Calculate the amount in moles of gold using the relationship A

NnN

= .



Act on Your Strategy Amount in moles, n, of Au(s):

22

A

u

9.51 × 10 atomsA

NnN

=

= 236.02 × 10 atoms0.15797 mol0.15

/mo

ol

l

8 m==

There is 0.158 mol of gold in the coin. Check Your Solution The units cancel properly. The answer seems reasonable and the number of significant digits in the answer agrees with the given data having the least number of significant digits.

22. Practice Problem (page 231) A patient in a dentist’s office is given 1.67 × 1023 molecules of dinitrogen monoxide (laughing gas), N2O(g), during a procedure. What amount in moles of dinitrogen monoxide is the patient given? What Is Required? You need to determine the amount in moles of dinitrogen monoxide a patient receives at the dentist’s office. What Is Given? You know the number of molecules, N, of dinitrogen monoxide: 1.67 × 1023

Plan Your Strategy Use the Avogadro constant: 236.02 10AN = ×

Calculate the amount in moles of N2O(g) using the relationship A

NnN

= .

Act on Your Strategy Amount in moles, n, of N2O(g):

2

23

N

O

1.67 × 10 moleculesA

NnN

=

= 236.02 × 10 molecules0.277 l

/mol mo=



There is 0.277 mol of dinitrogen monoxide given to the patient at the dentist’s office. Check Your Solution The units cancel properly. The answer seems reasonable and the number of significant digits in the answer agrees with the given data having the least number of significant digits.


A sheet of drywall contains 1.2 × 1026 formula units of gypsum (calcium sulfate dihydrate), CaSO4•2H2O(s). What amount in moles of gypsum is in the sheet of drywall? What Is Required? You need to determine the amount in moles of calcium sulfate dihydrate in a sheet of drywall. What Is Given? You know the number of formula units, N, of calcium sulfate dihydrate: 1.2 × 1026

Plan Your Strategy Use the Avogadro constant: 23

A 6.02 10N = × Calculate the amount in moles of CaSO4•2H2O(s) using the relationship

.A

NnN

=

Act on Your Strategy Amount in moles, n, of CaSO4•2H2O(s):

4 2CaSO •2H O

26 1.2 × 10 forumula unitsA

NnN

=

= 236.02 × 10 formula units

2

199.34 mol2.0 1

/

0

mol

mol=

= ×

The sheet of drywall contains 2.0 × 102 mol of calcium sulfate dihydrate.




24. Practice Problem (page 231) Limewater, a weak solution of calcium hydroxide, Ca(OH)2(s), is used to detect the presence of carbon dioxide gas. Suppose that you are given a solution that contains 8.7 × 1019 formula units of calcium hydroxide. What amount in moles of calcium hydroxide is in the solution? What Is Required? You need to determine the amount in moles of calcium hydroxide in a solution of limewater. What Is Given? You know the number of formula units, N, of calcium hydroxide: 8.7 × 1019

Plan Your Strategy Use the Avogadro constant: 23

A 6.02 10N = ×

Calculate the amount in moles of Ca(OH)2(s) using the relationship A

NnN

= .

Act on Your Strategy Amount in moles, n, of Ca(OH)2(s):

( )2

19

Ca OH

8.7 × 10 formula unitsA

NnN

=

= 236.02 × 10 formula units–4

–4

1.445 10 mol1.4 10

/mo

m l

l

o= ×

= ×

The limewater solution contains 1.4 × 10–4 mol of calcium hydroxide. Check Your Solution The units cancel properly. The answer seems reasonable and the number of significant digits in the answer agrees with the given data having the least number of significant digits.



25. Practice Problem (page 231) If there are a total of 7.3 × 1029 atoms in a sample of glucose, C6H12O6(s), what amount in moles of glucose is in the sample? What Is Required? You need to determine the amount in moles of glucose in a given sample. What Is Given? You know the total number of atoms in the sample of glucose: 7.3 × 1029 Plan Your Strategy Determine the total number of atoms in one molecule of C6H12O6(s). Use the Avogadro constant: 236.02 10AN = ×

Calculate the amount in moles of C6H12O6(s) using the relationship A

NnN

= .

Act on Your Strategy From the chemical formula for glucose, there are 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms for a total of 24 atoms in one molecule of C6H12O6(s). Total number of molecules, NT, of C6H12O6(s):

T29

29T

1 molecule7.3 10 atoms atoms

7.3 10 atom

24

s

N

N

=×

= ×1 molecule 24 a

s

tom

×

283.0416 10 molecules= ×

Amount in moles, n, of C6H12O6(s):

6 12 6

T

28

C

H O


NnN

=

= 236.02 × 10 molecules4

4

5.05249 10 mol5.1 1

/

0 mol

mol×

= ×

=

There is 5.1 × 104 mol of glucose in the sample.



Check Your Solution The units cancel properly. Estimating using the powers of ten in the calculation, the answer seems reasonable:

423

29 1 4.8 106

17 10 24 10

= ×× × ××

The number of significant digits in the answer agrees with the given data having the least number of significant digits.

26. Practice Problem (page 231) A sample of aluminum oxide, Al2O3(s), contains 8.29 × 1025 total atoms. Calculate the amount in moles of aluminum oxide in the sample. Hint: This is a two-step problem. Calculate the number of formula units first. What Is Required? You must determine the amount in moles of aluminum oxide in a sample. What Is Given? You know the total number of atoms, N, in the sample of aluminum oxide: 8.29 × 1025 Plan Your Strategy Determine the number of atoms in one formula unit of Al2O3(s). Calculate the number of formula units of Al2O3(s). Use the Avogadro constant: 236.02 10AN = ×

Calculate the amount in moles of Al2O3(s) using the relationship A

NnN

= .

Act on Your Strategy From the chemical formula for aluminum oxide, there are 2 atoms of aluminum and 3 atoms of oxygen for a total of 5 atoms in one formula unit of Al2O3(s). Number of formula units, N, of Al2O3(s):

25

25

1 formula unit8.29 10 total atoms total 5 atoms

8.29 10 total atoms

N

N

=×

= ×1 formula unit5 total

s

atom

×

251.658 × 10 formula units=



Amount in moles, n, of Al2O3(s):

2 3

2

Al OA

51.658 × 10 formula units

NnN

=

= 236.02 × 10 formula units27.541 mol27.

/

5

mol

mol==

There is 27.5 mol of aluminum oxide in the sample. Check Your Solution The units cancel properly. Estimating using the powers of ten in the

calculation, the answer seems reasonable: 223

5

6 101 18 10 27 mol5

× × =×

×


27. Practice Problem (page 231) Trinitrotoluene, or TNT for short, has the chemical formula C7H5N3O6(s). If a stick of dynamite is pure TNT and it contains 2.5 × 1025 atoms in total, what amount in moles of TNT does it contain? What Is Required? You must determine the amount in moles of trinitrotoluene that the stick of dynamite contains. What Is Given? You know the total number of atoms in the sample of trinitrotoluene: 2.5×1025 Plan Your Strategy Determine the total number of atoms in one molecule of C7H5N3O6(s). Calculate the number of molecules, N, of C7H5N3O6(s). Use the Avogadro constant: 23

A 6.02 10N = ×

Calculate the amount in moles of C7H5N3O6(s) using the relationship A

NnN

= .



Act on Your Strategy From the chemical formula for trinitrotoluene, there are 7 carbon atoms, 5 hydrogen atoms, 3 nitrogen atoms, and 6 oxygen atoms for a total of 21 atoms in one molecule of C7H5N3O6(s). Number of molecules, N, of C7H5N3O6(s):

25

25

1 molecule212.5 10 total atoms total atoms

2.5 10 tota

l toms

a

N

N

=×

= ×1 molecule

21 total a

s

tom×

241.1905 × 10 molecules=

Amount in moles, n, of C7H5N3O6(s):

7 5 3 6C H N OA

241.1905 10 molecules

NnN

=

=×

236.02 10 molecules×1.977 mol2.0 mol

/mol==

There is 2.0 mol of trinitrotoluene in the stick of dynamite. Check Your Solution The units cancel properly. Estimating using the powers of ten in the calculation, the answer seems reasonable:

2523

1 12.5 10 2 mol2 11 6 0

× × × =×


28. Practice Problem (page 231) A sample of rubbing alcohol solution contains ethanol, C2H5OH(ℓ). If the sample contains 1.25 × 1023 atoms of hydrogen in the ethanol, what amount in moles of ethanol is in the sample? What Is Required? You must determine the amount in moles of ethanol in a sample of rubbing alcohol.



What Is Given? You know the total number of hydrogen atoms in the sample (N) of ethanol: 1.25×1023 Plan Your Strategy Determine the number of hydrogen atoms in one molecule of C2H5OH(ℓ). Calculate the number of molecules of C2H5OH(ℓ). Use the Avogadro constant: 236.02 10AN = ×

Calculate the amount in moles of C2H5OH(ℓ) using the relationship A

NnN

= .

Act on Your Strategy From the chemical formula for ethanol, there are 6 atoms of hydrogen in one molecule of C2H5OH(ℓ). Number of molecules, N, of C2H5OH(ℓ):

2 523

23

C H OH1 molecule 61.25 10 H atoms H atoms

1.25 10 s

H atom

N

N

=×

= × 2 5C H1 mo Ole Hcule 6 H

atoms

×

222 52.0833 10 molecules C H OH = ×

Amount in moles, n, of C2H5OH(ℓ):

2 5C H OH


NnN

=

= 236.02 × 10 molecules0.0346

/mol mol=

There is 0.0346 mol of ethanol in the sample. Check Your Solution The units cancel properly. Estimating using the powers of ten in the calculation, the answer seems reasonable:

223

3 1 16 10

1.3 10 0.36 mol6

× × =×

×




29. Practice Problem (page 231) A cleaning solution contains 7.9 × 1026 molecules of ammonia, NH3(aq). What amount in moles of ammonia is in the solution? What Is Required? You need to determine the amount in moles of ammonia in a cleaning solution. What Is Given? You know the total number of molecules, N, of ammonia: 7.9 × 1026

Plan Your Strategy Use the Avogadro constant: 236.02 10AN = ×

Calculate the amount in moles of ammonia using the relationship A

NnN

= .

Act on Your Strategy Amount in moles, n, of NH3(aq):

3

26

NH

A

7.9 10 molecules

NnN

=×

=

236.02 10 molecules×3

3

1.312 10 mol1.3 1

/

0 mol

mol=

=

×

×

There is 1.3 × 103 mol of ammonia in the cleaning solution. Check Your Solution The units cancel properly. Comparing the powers of ten in the solution, the answer seems reasonable. The number of significant digits in the answer agrees with the given data having the least number of significant digits.

30. Practice Problem (page 231) A muffin recipe calls for cream of tartar, or potassium hydrogen tartrate, KHC4H4O6(s). The amount of cream of tartar that is required contains 2.56 × 1023 atoms of carbon. What amount in moles of potassium hydrogen tartrate is required? What Is Required? You need to determine the amount in moles of potassium hydrogen tartrate required in a recipe.



What Is Given? You know the number of atoms, N, of carbon in the sample: 2.56 × 1023 Plan Your Strategy Determine the number of carbon atoms in one molecule of KHC4H4O6(s). Calculate the number of molecules of KHC4H4O6(s). Use the Avogadro constant: 236.02 10AN = ×

Calculate the amount in moles of KHC4H4O6(s) using the relationship A

NnN

= .

Act on Your Strategy From the chemical formula of potassium hydrogen tartrate, there are 4 carbon atoms in one molecule of KHC4H4O6(s). Number of molecules, N, of KHC4H4O6(s):

4 423

23

KHC H O1 molecule 42.56 10 C atoms C atoms

2.56 10 s

C atom

N

N

=×

= × 4 4KHC H O 1 molecule 4 C atoms

×

224 46.40 10 molecules KHC H O= ×

Amount in moles, n, of KHC4H4O6(s):

4 4 6KHC H O


NnN

=

= 236.02 × 10 molecules0.106 l

/mol mo=

There is 0.106 mol of potassium hydrogen tartrate in the sample. Check Your Solution

The units cancel properly. The 0.106 mol of KHC4H4O6 is about 14

the number

of moles of C, which is consistent. The number of significant digits in the answer agrees with the given data having the least number of significant digits.



Section 5.1 The Mole and the Avogadro Solutions for Selected Review Questions Student Edition page 232

4. Review Question (page 232)

If the volume of one mole of water at room temperature is 18.02 mL, what is the volume of one molecule of water, in litres? What Is Required? You need to determine the volume (in litres) of one molecule of water. What Is Given? You know the volume of 1 mol of water molecules: 18.02 mL Plan Your Strategy Use the Avogadro constant: 23

A 6.02 10N = × Divide the volume of 1 mol of molecules by the number of molecules in 1 mol to determine the volume of one molecule. Convert the volume from millilitres to litres: 1 mL = 1 × 10–3 L Act on Your Strategy volume, V, (in mL) of 1 molecule of H2O(ℓ):

23

23

18.02 mL6.02 × 10 molecules2.993 × 10 mL/molecule

V

−

=

=

volume, V, (in L) of 1 molecule of H2O(ℓ):

–232.993 10 mLV = × –3/molecule 1 10 L/ mL× ×–26 2.993 10 L/molecule= ×

The volume of one molecule of water is 2.993×10–26 L. Check Your Solution The units cancel properly. Estimating using rounded numbers gives an answer that is close to the calculated answer:

23

206 × 10

× 10–3 = 3 × 10−26 L

The answer is reasonable and has the correct number of significant digits.



6. Review Question (page 232) Calculate the number of formula units of sodium chloride in 0.0578 mol. What is Required? You need to determine the number of formula units of sodium chloride, NaCl(s). What is Given? You know the amount in moles of sodium chloride: 0578 mol Plan Your Strategy Use the Avogadro constant: 236.02 10AN = × Calculate the number of formula units of NaCl(s) using the relationship

AN n N= × . Act on Your Strategy Number of formula units, N, of NaCl(s):

0.0578 mol

AN n N= ×

=236.02 × 10 formula units 1 mol

×

22

22


= ×

= ×

There are 3.48 × 1022 formula units of sodium chloride in the sample. Check Your Solution The units cancel properly. The answer has the correct number of significant digits. Converting the answer from formula units to moles gives the original amount in moles.

7. Review Question (page 232) Calculate the number of particles in each sample. Indicate the correct type of particle (atom, molecule, ion, or formula unit). a. 0.156 mol Au(s) b. 7.8 mol MgCl2(s) c. 15.2 mol H2O2(ℓ) What Is Required? You need to determine the number of particles and indicate the correct type of particle in each sample.



What Is Given? You know the chemical formula for each sample. You know the amount of each substance: a. 0.156 mol of Au(s) b. 7.8 mol of MgCl2(s) c. 15.2 mol of H2O2(ℓ) Plan Your Strategy Identify the type of particle that makes up each sample. Use the Avogadro constant: 236.02 10AN = × Calculate the number of particle units of each sample using the relationship

AN n N= × . Act on Your Strategy a. Au(s) is gold, and is made up of atoms Number of atoms, N, of Au(s):

0.156 mol

AN n N= ×

=236.02 × 10 atoms

1 mol×

229.39 10 atoms= ×

There are 9.39× 1022 atoms of Au(s) in the sample. b. MgCl2(s) is the ionic compound magnesium chloride, and the solid form is made up of formula units Number of formula units, N, of MgCl2(s):

7.8 mol

AN n N= ×

=236.02 × 10 formula units 1 mol

×

244.7 10 formula units= ×

There are 4.7 × 1024 formula units of MgCl2(s) in the sample.



c. H2O2(ℓ) is hydrogen peroxide, and is made up of molecules. Number of molecules, N, of H2O2(ℓ):

15.2 mol

AN n N= ×

=236.02 × 10 molecules

1 mol×

249.15 10 molecules= ×

There are 9.15 × 1024 molecules of H2O2(ℓ) in the sample. Check Your Solution In each case, the units cancel properly. The answers seem reasonable and have the correct number of significant digits.

8. Review Question (page 232) Calculate the amount in moles of molecules for each substance listed. a. a sample of ammonia, NH3(g), containing 8.1 × 1020 atoms of hydrogen b. a sample of diphosphorus pentoxide, P2O5(s), containing a total of 4.91 × 1022 atoms of phosphorus and oxygen What Is Required? You need to determine the amount in moles of molecules of each sample. What Is Given? a. You know the sample of ammonia contains 8.1 × 1020 atoms of hydrogen. b. You know the sample of diphosphorus pentoxide contains a total of 4.91 × 1022 atoms of phosphorus and oxygen. Plan Your Strategy a. ammonia From the chemical formula, NH3, determine the number of hydrogen atoms per molecule of ammonia. Calculate the number of molecules of the NH3(g). Use the Avogadro constant: 236.02 10AN = ×

Calculate the amount in moles of NH3(g) using the relationship A

NnN

= .



b. diphosphorus pentoxide Calculate the number of molecules of P2O5(s) using the information that there is a total of 7 atoms per molecule. Use the Avogadro constant: 236.02 10AN = ×

Calculate the amount in moles of the P2O5(s) using the relationship .A

NnN

=

Act on Your Strategy a. ammonia From the chemical formula, NH3, there are 3 atoms of hydrogen per molecule of ammonia. Number of molecules, N, of NH3(g):

20

20

1 molecule8.1 10 H atoms 3 H atoms

8.1 10 H atoms

N

N

=×

= ×1 molecule 3 H atoms

×

202.7 10 molecules= ×

Amount in moles, n, of the NH3(g):

3NHA

202.7 × 10 molecules

NnN

=

= 236.02 × 10 molecules–4

/mol4.5 10 mol = ×

There is 4.5 × 10–4 mol of NH3(g) in the sample. b. diphosphorus pentoxide From the chemical formula, P2O5, there are 2 phosphorus atoms and 5 oxygen atoms for a total of 7 atoms per molecule. Number of molecules, N, of the P2O5(s):

22

22

1 molecule4.91 10 atoms 7 atoms

4.91 10 atoms

N

N

=×

= ×1 molecule 7 atoms

×

227.014 10 molecules = ×



Amount in moles, n, of P2O5(s):

2 5P OA


NnN

=

= 236.02 × 10 molecules–2

–2

/mol1.165 10 mol1.17 10 mol

= ×

= ×

There is 1.17 × 10–2 mol of P2O5(s) in the sample. Check Your Solution In each case, the units cancel properly. Using rounded numbers to estimate the answer gives a result that is close to the calculated answers.

a. 8 × 1020 × 13

× 23

16 × 10

= 4.4 × 10–4

b. 5 × 1022 × 23

206 × 10

× 23

16 × 10

= 1.2 × 10–2

The answers are reasonable.

9. Review Question (page 232) A sample of ethanol, C2H5OH(ℓ), contains 2.49 × 1023 molecules of ethanol. A sample of carbon contains 1.65 mol of carbon atoms. Which sample, the compound or the element, contains the greater amount in moles of carbon? What Is Required? You need to compare the amount in moles of carbon in samples of the compound ethanol and element carbon. What Is Given? You know the amount in moles of carbon, C, atoms in the carbon sample: 1.65 mol You know the number of molecules of C2H5OH(ℓ) in the ethanol sample: 2.49 × 1023 Plan Your Strategy Use the Avogadro constant: 236.02 10AN = × Calculate the amount in moles of C2H5OH(ℓ) in the sample of ethanol using

the relationship A

NnN

= .



From the chemical formula, C2H5OH, determine the amount in moles of carbon atoms per mole of C2H5OH(ℓ) molecules. Calculate the amount in moles of C atoms in the C2H5OH(ℓ). Compare this amount in moles of C atoms to the given amount in the sample of the element. Act on Your Strategy Amount in moles, n, of C2H5OH(ℓ):

2 5C H OHA


NnN

=

= 236.02 × 10 molecules /mol0.4136 mol=

From the chemical formula, C2H5OH, there are 2 mol of carbon atoms per mole of ethanol. Amount in moles, n, of C atoms in C2H5OH(ℓ):

C

2 5 2 5

C 2 5

2 mol C atoms0.4136 mol C H OH 1 mol C H OH

0.4136 mol C H OH

n

n

=

=2 5

2 mol C atoms 1 mol C H OH

×

0.8272 mol C atoms0.827 mol C atoms

==

The sample of ethanol contains 0.827 mol of carbon atoms. The sample of the element carbon containing 1.65 mol of carbon atoms has the greater amount of carbon atoms. Check Your Solution The units cancel properly and the answers have the correct number of significant digits. The answer is reasonable.

10. Review Question (page 232) If there are 4.28 × 1021 atoms of hydrogen and oxygen in a sample of water, what amount in moles of water is in the sample? What is Required? You need to determine the amount in moles of water in a sample.



What Is Given? You know the total number of atoms in the sample of water: 4.28 × 1021 Plan Your Strategy Use the total number of atoms per molecule of H2O(ℓ) to calculate the number of molecules. Use the Avogadro constant: 23

A 6.02 10N = ×

Calculate the amount in moles of water using the relationship A

NnN

= .

Act on Your Strategy From the chemical formula, H2O, there are 2 hydrogen atoms and 1 oxygen atom for a total of 3 atoms per molecule. Number of molecules, N, of H2O(ℓ)

21

21

1 molecule4.28 × 10 atoms 3 atoms

4.28 × 10 atoms

N

N

=

=1 molecule 3 atoms

×

211.4266 10 molecules = ×

Amount in moles, n, of H2O(ℓ):

2H O

211.4266 10 moleculesA

NnN

=

×= 236.02 × 10 molecules

–3

–3

/mol2.369 10 mol2.37 10 mol

= ×

= ×

There is 2.37 × 10–3 mol of water molecules. Check Your Solution The units cancel properly. Using rounded numbers to estimate the answer:

4 × 1021 × 13

× 23

16 × 10

= 2 × 10–3 mol

The estimated answer is close to the calculated answer. The answer is reasonable and has the correct number of significant digits.



11. Review Question (page 232) Aluminum oxide, Al2O3(s), forms a thin coating on an aluminum surface, such as the body of an airplane, when aluminum is exposed to oxygen in air. This coating helps to reduce further corrosion of the aluminum. Suppose that a sample contains 2.6 mol of aluminum oxide. a. How many formula units are in the sample? b. How many atoms, in total, are in the sample? c. How many aluminum atoms are in the sample? What Is Required? a. You need to determine the number of formula units of Al2O3(s). b. You need to determine the total number of atoms in the sample of Al2O3(s). c. You need to determine the number of aluminum atoms in the sample. What Is Given? You know the amount in moles of the Al2O3(s): 2.6 mol Plan Your Strategy a. number of formula units Use the Avogadro constant: 236.02 10AN = × Calculate the number of formula units of the Al2O3(s) using the relationship AN n N= × . b. total number of atoms From the chemical formula, Al2O3, determine the number of atoms in one formula unit. Use the total number of atoms in one formula unit to find the number of atoms in the sample. c. number of aluminum atoms Use the number of Al atoms in one formula unit of the Al2O3(s) to find the number of Al atoms in the sample. Act on Your Strategy a. number of formula units Number of formula units, N, of the Al2O3(s):

A

2.6 mol

N n N= ×

= 23 6.02 10 formula units/ mol× ×24

24


= ×

= ×



There are 1.6 × 1024 formula units of aluminum oxide in the sample. b. total number of atoms From the chemical formula, Al2O3, there are 2 aluminum atoms and 3 oxygen atoms for a total of 5 atoms in one formula unit of Al2O3(s). Total number of atoms, N, in the Al2O3(s):

24

24

5 atoms1.5652 10 formula units 1 formula unit

1.5652 10 formula units

N

N

=×

= ×5 atoms

1 formula unit×

24

24

7.826 10 atoms7.8 10 atoms

= ×

= ×

There are 7.8 × 1024 atoms in total in the sample of aluminum oxide. c. number of aluminum atoms From the chemical formula, Al2O3, there are 2 aluminum atoms in one formula unit of Al2O3(s). Number of Al atoms, N, in the Al2O3(s):

24

24

2 Al atoms1.5652 10 formula units 1 formula unit


N

N

=×

= ×2 Al atoms

1 formula unit×

24

24

3.1304 10 Al atoms 3.1 10 Al atoms

= ×

= ×

There are 3.1 × 1024 Al atoms in the sample of aluminum oxide. Check Your Solution The units cancel properly. The answer is reasonable and has the correct number of significant digits.



12. Review Question (page 232) Zinc chloride hexahydrate, ZnCl2•6 H2O(s), is used in the textile industry, often to prepare fireproofing agents. In a particular fireproofing solution, 0.46 mol of zinc chloride hexahydrate is used. a. How many formula units of zinc chloride hexahydrate are in the solution? b. How many atoms of chlorine are in the solution? What Is Required? a. You need to determine the number of formula units of ZnCl2•6H2O(s). b. You need to determine the number of chlorine, Cl, atoms in the sample of ZnCl2•6H2O(s). What Is Given? You know the amount in moles of the ZnCl2•6H2O(s): 0.46 mol Plan Your Strategy a. number of formula units Use the Avogadro constant: 236.02 10AN = × Calculate the number of formula units of the ZnCl2•6H2O(s) using the relationship AN n N= × . b. number of chlorine atoms From the chemical formula, ZnCl2•6H2O, determine the number of chlorine atoms per formula unit of zinc chloride hexahydrate. Calculate the number of chlorine atoms in the sample. Act on Your Strategy a. number of formula units Number of formula units, N, of the ZnCl2•6H2O(s):

0.46 molAN n N= ×

= 23 6.02 10 formula units/ mol× ×23

23


= ×

= ×

There are 2.8 × 1023 formula units of ZnCl2•6H2O(s) in the sample. b. number of chlorine atoms From the chemical formula, ZnCl2•6H2O, there are 2 chlorine atoms in one formula unit of zinc chloride hexahydrate.



Number of Cl atoms, N, in ZnCl2•6H2O(s):

23

23

2 Cl atoms2.769 10 formula units 1 formula unit


N

N

=×

= ×2 Cl atoms

1 formula unit×

235.538 10 Cl atoms= ×

There are 5.5 × 1023 chlorine atoms in the sample. Check Your Solution

a. Estimating, you have approximately 12

mol (0.46 mol) of ZnCl2•6H2O that

would be equivalent to about 12

of the Avogadro constant of formula units, or

2.8 × 1023. The units cancel properly. The answer is reasonable and has the correct number of significant digits. b. An estimate of the answer is approximately 3 × 2 atoms, which is close to the answer that was calculated. The units cancel properly. The answer is reasonable and has the correct number of significant digits.

14. Review Question (page 232) Arrange the following three samples from largest to smallest in terms of their numbers of representative particles: • 3.92 mol of octane, C8H18(ℓ) • 6.52 × 1023 atoms of copper, Cu(s) • 1.25 × 1024 formula units of sodium hydrogen carbonate, NaHCO3(s) What Is Required? You need to arrange the number of particles of three different samples in order from the largest number to the smallest. What Is Given? You know the amount in moles of octane: 3.92 mol You know the number of atoms of copper: 6.52 × 1023 You know the number of formula units of sodium hydrogen carbonate: 1.25 × 1024



Plan Your Strategy Compare the number of atoms in each sample. Use the Avogadro constant: 23

A 6.02 10N = × Calculate the number of molecules in 3.92 mol of C8H18(ℓ) using the relationship A N n N= × . Use the chemical formulas for the compounds octane and sodium hydrogen carbonate to determine the number of atoms per unit. List the compounds from the largest number of atoms to the smallest. Act on Your Strategy • octane Number of molecules, N, of the C8H18(ℓ):

A

3.92 mol

N n N= ×


From the chemical formula, C8H18, there are 8 carbon atoms and 18 hydrogen atoms for a total of 26 atoms per molecule. Total number of atoms, N, in the C8H18(ℓ):

24

24

26 atoms2.3598 10 molecules 1 molecule

2.3598 10 molecules

N

N

=×

= ×26 atoms

1 molecule×

256.14 10 atoms= ×

The sample of octane contains 6.14 × 1025 atoms. • sodium hydrogen carbonate From the chemical formula, NaHCO3, there are 1 sodium atom, 1 hydrogen atom, 1 carbon atom, and 3 oxygen atoms for a total of 6 atoms per molecule. Total number of atoms, N, in the NaHCO3(s):

24

24

6 atoms1.25 10 formula units 1 formula unit


N

N

=×

= ×6 atoms

1 formula unit×

247.50 10 atoms= ×



The sample of sodium hydrogen carbonate contains 7.50 × 1024 atoms. • copper Number of atoms in the Cu(s): 6.52 × 1023 atoms (given) Samples arranged from largest number of representative particles (atoms) to the smallest: C8H18(ℓ) (6.14 × 1025 atoms) > NaHCO3(s) (7.50 × 1024 atoms ) > Cu(s) (6.52 × 1023 atoms) Check Your Solution In each case, the units cancel properly. Using rounded numbers to estimate the calculated amounts: • octane: 4 × 6 × 1023 × 26 = 6 × 1025 atoms • sodium hydrogen carbonate 1 × 1024 × 6 = 6 × 1024 atoms The estimated answers are close to the calculated answers. The answers are reasonable and have the correct number of significant digits.

15. Calculate the number of atoms in 6.0 mol of fluorine, F2(g), molecules. What Is Required? You need to determine the number of atoms of fluorine, F, in a sample of F2(g). What Is Given? You know the amount in moles of the F2(g): 6.0 mol Plan Your Strategy Use the Avogadro constant: 236.02 10AN = × Calculate the number of molecules of F2(g) using the relationship N = n × NA. Use the number of F atoms (2) in one molecule of F2(g) to find the number of F atoms in the sample. Act on Your Strategy Number of molecules, N, of the F2(g):

6.0 molAN n N= ×

= 23 6.02 10 molecules/ mol × ×243.612 10 molecules= ×



Number of F atoms, N, in the F2(g):

24

24

2 F atoms3.612 10 molecules 1 molecule

3.612 10 molecules

N

N

=×

= ×2 F atoms

1 molecule×

24

24

7.224 10 atoms7.2 10 atoms

= ×

= ×

There are 7.2 × 1023 F atoms in the sample. Check Your Solution The units cancel properly. Using rounded numbers to estimate the answer: 6 × 6 × 1023 × 2 = 7.2 × 1024 atoms This estimate agrees with the calculated answer. The answer is reasonable and has the correct number of significant digits.

Section 5.2 Mass and the Mole Solutions for Practice Problems Student Edition page 235


State the molar mass of each element. a. sodium b. tungsten c. xenon d. nickel What Is Required? You need to determine the molar mass of a. sodium. b. tungsten. c. zenon. d. nickel. What Is Given? You are the given the names of the elements. Plan Your Strategy The molar mass of any element is numerically equal to the atomic mass of the element found on the periodic table.



Act on Your Strategy a. MNa = 22.99 g/mol b. MW = 183.84 g/mol c. MXe = 131.29 g/mol d. MNi = 58.69 g/mol Check Your Solution Check to see that the atomic masses have been recorded correctly.

32. Practice Problem (page 235) Calculate the molar mass of phosphorus, P4(s). What Is Required? You need to determine the mass of 1 mol of phosphorus. What Is Given? You know the chemical formula for phosphorus: P4 Plan Your Strategy Use the periodic table to determine the total mass contributed by 4 mol of phosphorus atoms. Act on Your Strategy Molar mass, M, of P4(s):

( )4P P 4

4 30.97 g/mol123.88 g/mol

M M=

=

=

The molar mass of phosphorus is 123.88 g/mol. Check Your Solution Check that the atomic mass has been recorded and multiplied correctly.

33. Practice Problem (page 235) Determine the molar mass of calcium phosphate, Ca3(PO4)2(s). What Is Required? You need to determine the mass of 1 mol of calcium phosphate. What Is Given? You know the chemical formula for calcium phosphate: Ca3(PO4)2



Plan Your Strategy Use the periodic table to determine the atomic molar mass of each element in Ca3(PO4)2. Multiply each element’s atomic molar mass by the number of atoms of the element in the compound. Add the molar masses of the atoms of each element to get the molar mass of the compound. Act on Your Strategy Molar mass, M, of Ca3(PO4)2(s):

( ) ( ) ( )3 4 2Ca (PO ) Ca P O3 2 + 8

3 40.08 g/mol + 2 30.97 g/mol + 8 16.00 g/mol120.24 g/mol 61.94 g/mol 128 g/mol

310.18 g/mol

M M M M= +

=

= + +=

The molar mass of calcium phosphate is 310.18 g/mol. Check Your Solution Check that the atomic masses have been recorded correctly. Rounded values for the atomic masses can be used to estimate the answer: (3 × 40) + (2 × 30) + (8 × 16) = 308 g/mol

34. Practice Problem (page 235) Calculate the molar mass of lead(II) nitrate, Pb(NO3)2(s). What Is Required? You need to determine the mass of 1 mol of lead nitrate. What Is Given? You know the chemical formula for lead nitrate: Pb(NO3)2 Plan Your Strategy Use the periodic table to determine the atomic molar mass of each element in Pb(NO3)2. Multiply each element’s atomic molar mass by the number of atoms of the element in the compound. Add the molar masses of the atoms of each element to get the molar mass of the compound.



Act on Your Strategy Molar mass, M, of Pb(NO3)2:

3 2Pb(NO ) Pb N O1 + 2 + 6

1(207.2 g/mol) + 2(14.01 g/mol) + 6(16.00 g/mol)207.2 g/mol + 28.02 g/mol 96.00 g/mol331.2 g/mol

M M M M=

== +=

The molar mass of lead nitrate is 331.2 g/mol. Check Your Solution Check that the atomic masses have been recorded correctly. Rounded values for the atomic masses can be used to estimate the answer: (1 × 210) + (2 × 14) + (6 × 16) = 334 g/mol

35. Practice Problem (page 235) Determine the molar mass of iron(III) thiocyanate ion, FeSCN2+(aq). What Is Required? You need to determine the mass of 1 mol of iron(III) thiocyanate ion. What Is Given? You know the chemical formula for iron(III) thiocyanate ion: FeSCN2+ Plan Your Strategy Use the periodic table to find the atomic molar mass of each element in FeSCN2+(aq). Multiply each element’s atomic molar mass by the number of atoms of the element in the compound. Add the molar masses of the atoms of each element to get the molar mass of the compound. Note: The fact that this is an ion rather than a neutral species has insignificant impact on the molar mass. Act on Your Strategy Molar mass, M, of FeSCN2+(aq):

2 Fe S C NFeSCN1 1 1 1

1(55.85 g/mol) 1(32.07 g/mol) 1(12.01 g/mol) 1(14.01 g/mol)55.85 g/mol 32.07 g/mol 12.01 g/mol 14.01 g/mol

=113.94 g/mol

M M M M M+ = + + +

= + + += + + +

The molar mass of FeSCN2+(aq) is 113.94 g/mol.



Check Your Solution Check that the atomic masses have been recorded correctly. Rounded values for the atomic masses can be used to estimate the answer: (1 × 55) + (1 × 32) + (1 × 12) + (1 × 14) = 113 g/mol

36. Practice Problem (page 235) Calculate the molar mass of sodium stearate, NaC17H35COO(s). What Is Required? You need to determine the mass of 1 mol of sodium stearate. What Is Given? You know the chemical formula for sodium stearate: NaC17H35COO Plan Your Strategy Use the periodic table to determine the atomic molar mass of each element in NaC17H35COO(s). Multiply each element’s atomic molar mass by the number of atoms of the element in the compound. Add the molar masses of the atoms of each element to get the molar mass of the compound. Act on Your Strategy Molar mass, M, of NaC17H35COO(s):

( ) ( ) ( ) ( )17 35NaC H COO Na C H O1 18 35 2

1 22.99 g/mol 18 12.01 g/mol 35 1.01 g/mol 2 16.00 g/mol22.99 g/mol 216.18 g/mol 35.35 g/mol 32.00 g/mol306.52 g/mol

M M M M M= + + +

= + + +

= + + +=

The molar mass of sodium stearate is 306.52 g/mol. Check Your Solution Check that the atomic masses have been recorded correctly. Rounded values for the atomic masses can be used to estimate the answer: (1 × 23) + (18 × 12) + (35 × 1) + (2 × 16) = 306 g/mol



37. Practice Problem (page 235) Calculate the molar mass of barium hydroxide octahydrate, Ba(OH)2•8H2O(s). Hint: The molar mass of a hydrate must include the water component. What Is Required? You need to determine the mass of 1 mol of barium hydroxide octahydrate. What Is Given? You know the chemical formula for barium hydroxide octahydrate: Ba(OH)2•8H2O Plan Your Strategy Use the periodic table to determine the atomic molar mass of each element in Ba(OH)2•8H2O(s). Multiply each element’s atomic molar mass by the number of atoms of the element in the compound. Add the molar masses of the atoms of each element to get the molar mass of the compound. Act on Your Strategy Molar mass, M, of Ba(OH)2•8H2O(s):

( ) ( ) 222 2H OBa OH •8H O Ba OH

Ba O H H O

1 8

1[1 + 2 + 2 ] + 8[2 + 1 ]1[1(137.33 g/mol) + 2(16.00 g/mol) + 2(1.01 g/mol)]

+ 8[2(1.01 g/mol) + 1(16.00 g/mol)]137.33 g/mol 32.00 g/mol 2.02 g/mol 8 18.02

M M M

M M M M M

= +

==

= + + + ( ) g/mol137.33 g/mol 32.00 g/mol 2.02 g/mol 144.16 g/mol315.51 g/mol

= + + +=

The molar mass of barium hydroxide octahydrate is 315.51 g/mol. Check Your Solution Check that the atomic masses have been recorded correctly. Rounded values for the atomic masses can be used to estimate the answer: (1 × 140) + (2 × 16) + (2 × 1) + (8 × 18) = 318 g

38. Practice Problem (page 235) Determine the molar mass of tetraphosphorus decoxide, P4O10(s). What Is Required? You need to determine the molar mass of 1 mol of tetraphosphorus decoxide.



What Is Given? You know the chemical formula for tetraphosphorus decoxide: P4O10 Plan Your Strategy Use the periodic table to determine the atomic molar mass of each element in P4O10(s). Multiply each element’s atomic molar mass by the number of atoms of the element in the compound. Add the molar masses of the atoms of each element to get the molar mass of the compound. Act on Your Strategy Molar mass, M, of P4O10(s):

4 10P O P O4 + 10

4(30.97 g/mol) + 10(16.00 g/mol)123.88 g/mol 160.00 g/mol283.88 g/mol

M M M=

== +=

The molar mass of tetraphosphorus decoxide is 283.88 g/mol. Check Your Solution Check that the atomic masses have been recorded correctly. Rounded values for the atomic masses can be used to estimate the answer: (4 × 30) + (10 × 16) = 280 g/mol

39. Practice Problem (page 235) Calculate the molar mass of iron(II) ammonium sulfate hexahydrate, (NH4)2Fe(SO4)2•6H2O(s). What Is Required? You need to determine the mass of 1 mol of iron(II) ammonium sulfate octahydrate. What Is Given? You know the chemical formula for iron(II) ammonium sulfate hexahydrate: (NH4)2Fe(SO4)2•6H2O(s)



Plan Your Strategy Use the periodic table to determine the atomic molar mass of each element in (NH4)2Fe(SO4)2•6H2O(s). Multiply each element’s atomic molar mass by the number of atoms of the element in the compound. Add the molar masses of the atoms of each element to find the molar mass of the compound. Act on Your Strategy Molar mass, M, of (NH4)2Fe(SO4)2•6H2O(s):

( ) ( ) ( ) ( )

( ) ( ) ( )( ) ( )( ) ( )

24 4 2 4 42 2 2 2H ONH Fe SO •6H O NH Fe SO

N H Fe S O H O

1 + 6

1[2 8 1 2 8 ] 6[2 1 ]1[2 14.01 g/mol 8 1.01 g/mol + 1 55.85 g/mol

2 32.07 g/mol + 8 16.00 g/mol ]

6 2 1.01 g/mol 1 16.00 g/mol

28.02 g/mol

M M M

M M M M M M M

=

= + + + + + +

= +

+

⎡ ⎤+ +⎣ ⎦=

( ) 8.08 g/mol 55.85 g/mol 64.14 g/mol

128.00 g/mol 6 18.02 g/mol28.02 g/mol 8.08 g/mol 55.85 g/mol 64.14 g/mol

128.00 g/mol 108.12 g/mol392.21 g/mol

+ + +

+ +

= + + ++ +

=

The molar mass of ammonium iron(II) sulfate hexahydrate is 392.21 g/mol. Check Your Solution Check that the atomic masses have been recorded correctly. Rounded values for the atomic masses can be used to estimate the answer: (2 × 14) + (8 × 1) + (1 × 55) + (2× 30) + (8 × 16) + (6 × 18) = 387 g/mol

40. Practice Problem (page 235) The formula for a compound that contains an unknown element, A, is A2SO4. If the molar mass of the compound is 361.89 g/mol, what is the atomic molar mass of A? What Is Required? You need to determine the atomic molar mass of element A in 1 mol of A2SO4(s).



What Is Given? You know the chemical formula of the compound: A2SO4 You know the molar mass of the compound: 361.89 g/mol Plan Your Strategy Let y be the atomic molar mass of element A. Use the periodic table to determine the atomic molar mass of the two named elements in A2SO4(s). Multiply each element’s atomic molar mass by the number of atoms of the element in the compound. Subtract the molar mass of each of the named elements from the total molar mass of the compound to determine the molar mass of element A. Solve for y by dividing the molar mass of element A by 2. Act on Your Strategy Molar mass, M, of A2SO4(s):

( ) ( ) ( )( ) ( ) ( )

( ) [ ) ( )( )( )( )

2 4A SO A S O2 1 4

2 1 32.07 g/mol + 4 16.00 g/mol

361.89 g/mol 2 1 32.07 g/mol 4 16.00 g/mol

2 361.89 g/mol 1(32.07 g/mol 4 16.00 g/mol ]

2 361.89 g/mol 96.01 g/mol

2 265.82 g/mol

2 265.82 g/m2

M M M M

y

y

y

y

y

y

= + +

= +

= + +

= − +

= −

=

=ol

2132.91 g/moly =

The atomic molar mass of element A is 132.91 g/mol. Check Your Solution Check that the atomic masses have been recorded correctly. The atomic molar mass of y corresponds to the element cesium in the periodic table.





Calculate the mass of 3.57 mol of vanadium. What Is Required? You need to determine the mass of a sample of vanadium, V(s). What Is Given? You know the amount in moles of vanadium: 3.57 mol Plan Your Strategy Use the periodic table to determine the atomic molar mass of V(s). Use the relationship .m n M= × Multiply the amount in moles of the V(s) by its molar mass. Act on Your Strategy Molar mass of V(s): M = 50.94 g/mol (from the periodic table) Mass, m, of V(s):

V

3.57 mol

m n M= ×

= 50.94 g/ mol×181.8558 g182 g

==

The mass of the sample of vanadium is 182 g. Check Your Solution The units cancel properly. Using rounded values to estimate the answer: 3.5 × 50 = 175 g This value is close to the calculated answer.

42. Practice Problem (page 237) Calculate the mass of 0.24 mol of carbon dioxide. What Is Required? You need to determine the mass of carbon dioxide.



What Is Given? You know the amount in moles of the carbon dioxide: 0.24 mol You know the chemical formula for carbon dioxide: CO2 Plan Your Strategy Use the periodic table to find the atomic molar masses of carbon and oxygen. Multiply each element’s atomic molar mass by the number of atoms of the element in the compound. Add the molar masses of each element to find the molar mass of CO2(g). Use the relationship .m n M= × Multiply the amount in moles of CO2(g) by its molar mass. Act on Your Strategy Molar mass, M, of CO2(g):

2CO C O1 2

1(12.01 g/mol) + 2(16.00 g/mol)12.01 g/mol 32.00 g/mol 44.01 g/mol

M M M= +

== +=

Mass, m, of the CO2(g):

2CO

0.24 mol

m n M= ×

= 44.01 g/ mol ×10.56 g11 g

==

The mass of the carbon dioxide is 11 g. Check Your Solution

The units cancel properly. Estimating, 0.24 is approximately 14

mol, and 11 is

approximately 14

the molar mass. The answer is reasonable.


Calculate the mass of 1.28 × 10–3 mol of glucose, C6H12O6(s). What Is Required? You need to determine the mass of a sample of glucose.



What Is Given? You know the amount in moles of glucose: 1.28 × 10–3 mol You know the chemical formula for glucose: C6H12O6 Plan Your Strategy Use the periodic table to find the atomic molar masses of the elements in C6H12O6(s). Multiply each element’s atomic molar mass by the number of atoms of the element in the compound. Add the molar masses of each element to find the molar mass of the compound. Use the relationship .m n M= × Multiply the amount in moles of C6H12O6(s) by its molar mass. Act on Your Strategy Molar mass, M, of C6H12O6(s):

6 12 6C H O C H O6 + 12 + 6

6(12.01 g/mol) + 12(1.01 g/mol) + 6(16.00 g/mol)72.06 g/mol 12.12 g/mol 96.00 g/mol 180.18 g/mol

M M M M=

== + +=

Mass, m, of the C6H12O6(s):

6 12 6C H O

–3

1.28 10 mol

m n M= ×

= × 180.18 g/ mol× 0.2306 g 0.231 g==

The mass of the glucose is 0.231 g. Check Your Solution The units cancel properly. Use rounded values for the atomic molar masses to determine an estimate of the answer: 0.001 × 200 = 0.200 g This estimate is close to the answer that was calculated.



44. Practice Problem (page 237) Calculate the mass of 0.0029 mol of magnesium bromide, MgBr2(s), in milligrams. What Is Required? You need to determine the mass (in milligrams) of magnesium bromide. What Is Given? You know the amount in moles of magnesium bromide: 0.0029 mol

You know the chemical formula for magnesium bromide: MgBr2 Plan Your Strategy Use the periodic table to find the atomic molar masses of the elements in MgBr2(s). Multiply each element’s atomic molar mass by the number of atoms of the element in the compound. Add the molar masses of each element to find the molar mass of the compound. Use the relationship .m n M= × Multiply the amount in moles of MgBr2(s) by its molar mass. Convert the mass from grams to milligrams: 1 g = 1 × 103 mg Act on Your Strategy Molar mass, M, of MgBr2(s):

2MgBr Mg Br1 + 2

1(24.31 g/mol) + 2(79.90 g/mol)24.31 g/mol 159.80 g/mol184.11 g/mol

M M M=

== +=

Mass, m, of the MgBr2(s):

2MgBr

0.0029 mol

m n M= ×

= 184.11 g/ mol ×0.533919 g

0.533919 g

=

= 3 1 10 mg/ g× ×

2

533.919 mg5.3 10 mg

=

= ×

The mass of the magnesium bromide is 5.3 × 102 mg.



Check Your Solution The units cancel properly. Use rounded values for the atomic molar masses to get an estimate of the answer: 0.003 × 180 × 103 = 540 mg This estimate is close to the calculated answer.

45. Practice Problem (page 237) Name each compound, and then calculate its mass. Express this value in scientific notation. a. 4.5 × 10–3 mol of Co(NO3)2(s) b. 29.6 mol of Pb(S2O3)2(s) What Is Required? a. You need to name and determine the mass of Co(NO3)2(s). b. You need to name and determine the mass of Pb(S2O3)2(s). What Is Given? a. You know the amount in moles of the Co(NO3)2(s): 4.5 × 10–3 mol b. You know the amount in moles of the Pb(S2O3)2(s): 29.6 mol Plan Your Strategy Refer to Chapter 2, Section 2.2 (beginning at page 64) to name the compounds. Use the periodic table to find the atomic molar masses of the elements in each compound. For each compound, multiply each element’s atomic molar mass by the number of atoms of the element in the compound. Add the molar masses of each element to find the molar mass of the compound. Use the relationship .m n M= × Multiply the amount in moles of each compound by the compound’s molar mass. Act on Your Strategy a. Co(NO3)2(s) The name of the compound is cobalt(II) nitrate. Molar mass, M, of Co(NO3)2(s):

( ) ( ) ( )3 2Co(NO ) Co N O1 2 6

1 58.93 g/mol 2 14.01 g/mol 6 16.00 g/mol 58.93 g/mol 28.02 g/mol 96.00 g/mol182.95 g/mol

M M M M= + +

= + +

= + +=



Mass, m, of the Co(NO3)2(s):

( )3 2Co NO

–3

4.5 10 mol

m n M= ×

= × 182.95 g/ mol×

–1

0.82328 g 8.2 10 g

=

= ×

The mass of the cobalt(II) nitrate is 8.2 × 10–1 g. b. Pb(S2O3)2(s) The name of the compound is lead(IV) thiosulfate. Molar mass, M, of Pb(S2O3)2(s):

( ) ( ) ( )2 3 2Pb(S O ) Pb S O1 4 6

1 207.2 g/mol 4 32.07 g/mol 6 16.00 g/mol207.2 g/mol 128.28 g/mol 96.00 g/mol431.48 g/mol

M M M M= + +

= + +

= + +=

Mass, m, of the Pb(S2O3)2(s):

( )2 3 2Pb S O

29.6 mol

m n M= ×

= 431.48 g/ mol×4

4

1 .2772 10 g1 .28 10 g

= ×

= ×

The mass of the lead(IV) thiosulfate is 1.28 × 104 g. Check Your Solution The units cancel properly. Check that the correct atomic molar masses have been used. Use rounded values for the atomic molar masses to get an estimate of the answers: a. 5 × 10–3 × 200 = 1 g b. 30 × 400 = 1.2 × 104 g These estimates are reasonably close to the answers that were calculated.



46. Practice Problem (page 237) Determine the chemical formula for each compound, and then calculate its mass. a. 4.9 mol of ammonium nitrate b. 16.2 mol of iron(III) oxide What Is Required? a. You need to determine the chemical formula for ammonium nitrate, and then calculate its mass. b. You need to determine the chemical formula for iron(III) oxide, and then calculate its mass. What Is Given? a. You know the amount in moles of ammonium nitrate: 4.9 mol

b. You know the amount in moles of iron(III) oxide: 16.2 mol Plan Your Strategy Refer to Chapter 2, Section 2.2 (beginning at page 64) to write the chemical formulas. Use the periodic table to find the atomic molar masses of the elements in each compound. For each compound, multiply each element’s atomic molar mass by the number of atoms of the element in the compound. Add the molar masses of each element to find the molar mass of the compound. Use the relationship .m n M= × Multiply the amount in moles of each compound by the compound’s molar mass. Act on Your Strategy a. ammonium nitrate The chemical formula for ammonium nitrate is NH4NO3. Molar mass, M, of NH4NO3(s):

( ) ( ) ( )4 3NH NO N H O2 4 3

2 14.01 g/mol 4 1.01 g/mol 3 16.00 g/mol 28.02 g/mol 4.04 g/mol 48.00 g/mol 80.06 g/mol

M M M M= + +

= + +

= + +=



Mass, m, of the NH4NO3(s): 4 3NH NO

4.9 mol

m n M= ×

= 80.06 g/ mol×

2

392.294 g 3.9 10 g

=

= ×

The mass of the ammonium nitrate is 3.9 × 102 g. b. iron(III) oxide The chemical formula for iron(III) oxide is Fe2O3. Molar mass, M, of Fe2O3(s):

( ) ( )2 3Fe O Fe O2 3

2 55.85 g/mol 3 16.00 g/mol111.7 g/mol 48.00 g/mol159.7 g/mol

M M M= +

= +

= +=

Mass, m, of the Fe2O3(s):

2 3Fe O

16.2 mol

m n M= ×

= 159.7 g/ mol×3

3

2.587 10 g2.59 10 g

= ×

= ×

The mass of the iron(III) oxide is 2.59 × 103 g. Check Your Solution The units cancel properly. Check that the correct atomic molar masses have been used. Use rounded values for the atomic molar masses to get an estimate of the answers: a. 80 × 5 = 400 g b. 16 × 160 = 2.6 × 103 g These estimates are reasonably close to the answers that were calculated.



47. Practice Problem (page 237) What is the mass of 1.6 × 10–3 mol of calcium chloride dihydrate, CaCl2•2H2O(s), in milligrams? What Is Required? You need to determine the mass (in milligrams) of calcium chloride dihydrate. What Is Given? You know the chemical formula for calcium chloride dihydrate: CaCl2•2H2O You know the amount in moles of CaCl2•2H2O(s): 1.6 × 10–3 mol

Plan Your Strategy Use the periodic table to find the atomic molar masses of the elements in CaCl2•2H2O(s). Multiply each element’s atomic molar mass by the number of atoms of the element in the compound. Add the molar masses of each element to find the molar mass of the compound. Use the relationship .m n M= × Multiply the amount in moles of CaCl2•2H2O(s) by its molar mass. Convert the mass from grams to milligrams: 1 g = 1 × 103 mg Act on Your Strategy Molar mass, M, of CaCl2•2H2O(s):

( ) ( )( )

2 2 2 2CaCl •2H O CaCl H O

Ca Cl H O

1 2

1[1 2 ] 2[2 1 ]

1[1(40.08 g/mol) 2(35.45 g/mol)] 2 2 1.01 g/mol 1 16.00 g/mol

40.08 g/mol 70.90 g/mol 2 18.02 g/mol40.08 g/mol 70.90 g/mol 36.04 g/m

M M M

M M M M

= +

= + + +

⎡ ⎤= + + +⎣ ⎦= + +

= + + ol147.02 g/mol=

Mass, m, of the CaCl2•2H2O(s):

2 2CaCl •2H O

–3

1.6 10 mol

m n M= ×

= × 147.02 g/ mol× 0.23523 g

0.23523 g

=

= 3 1 10 mg/ g× ×

2

235.23 mg2.4 10 mg

=

= ×



The mass of calcium chloride dihydrate is 2.4 × 102 mg. Check Your Solution The units cancel properly. Use rounded values for the atomic molar masses to get an estimate of the answer: 0.0016 × 150 × 103 = 240 mg This estimate is close to the answer that was calculated.

48. Practice Problem (page 237) A litre of water contains 55.56 mol of water molecules. What is the mass of a litre of water, in kilograms? What Is Required? You need to determine the mass (in kilograms) of 1 L of water. What Is Given? You know the amount in moles of 1 L of water: 55.56 mol You know the chemical formula for water: H2O Plan Your Strategy Use the periodic table to find the atomic molar masses of hydrogen and oxygen. Multiply each element’s atomic molar mass by the number of atoms of the element in H2O(ℓ). Add the molar masses of each element to find the molar mass of H2O(ℓ). Use the relationship .m n M= × Multiply the amount in moles of H2O(ℓ) by its molar mass. Convert the mass from grams to milligrams: 1 g = 1 × 103 mg Act on Your Strategy Molar mass, M, of H2O(ℓ):

2H O H O2 1

2(1.01 g/mol) 1(16.00 g/mol)2.02 g/mol 16.00 g/mol18.02 g/mol

M M M= +

= += +=



Mass, m, of H2O(ℓ): 2H O

55.56 mol

m n M= ×

= 18.02 g/ mol×3

3

1.001 10 g

1.001 10 g

= ×

= × –3 1 10 kg/ g× ×

1 .001 kg =

The mass of 1 L of water is 1.001 kg. Check Your Solution The units cancel properly. Check that the correct atomic molar masses have been used. Use rounded values for the atomic molar masses to get an estimate of the answer: 55 × 20 × 10–3 = 1.1 kg This is a reasonably good estimate of the calculated answer.

49. Practice Problem (page 237) For each group of three samples, determine the sample with the largest mass. a. 2.34 mol of bromine, Br2(ℓ); 9.80 mol of hydrogen sulfide, H2S(g); 0.568 mol of potassium permanganate, KMnO4(s) b. 13.7 mol of strontium iodate, Sr(IO3)2(s); 15.9 mol of gold(III) chloride, AuCl3(s); 8.61 mol of bismuth silicate, Bi2(SiO3)3(s) What Is Required? a. You need to determine the masses of samples of bromine, hydrogen sulfide, and potassium permanganate and identify which has the largest mass. b. You need to determine the masses of samples of strontium iodate, gold(III) chloride, and bismuth silicate and identify which has the largest mass. What Is Given? a. You know the chemical formulas for the samples and the amount in moles of each: bromine, Br2; n = 2.34 mol hydrogen sulfide, H2S; n = 9.80 mol potassium permanganate, KMnO4; n = 0.568 mol b. You know the chemical formulas for the samples and the amount in moles of each: strontium iodate, Sr(IO3)2; n = 13.7 mol gold(III) chloride, AuCl3; n = 15.9 bismuth silicate, Bi2(SiO3)3; n = 8.61 mol



Plan Your Strategy For each of a. and b.: Use the periodic table to find the atomic molar masses of the elements in each compound. Multiply each element’s atomic molar mass by the number of atoms of the element in each compound. For each compound, add the molar masses of each element to find the molar mass. Use the relationship .m n M= × Multiply the amount in moles of each compound by the compound’s molar mass. Identify the sample in each group that is the largest in mass. Act on Your Strategy a. samples of bromine, hydrogen sulfide, and potassium permanganate Molar mass, M, of Br2(ℓ):

2Br Br2

2(79.90 g/mol)=159.8 g/mol

M M=

=

Mass, m, of the Br2(ℓ):

2Br

2.34 mol

m n M= ×

= 159.8 g/ mol×23.74 10 g= ×

The mass of the sample of bromine is 3.74 × 102 g. Molar mass, M, of H2S(g):

( ) ( )2H S H S2 1

2 1.01 g/mol 1 32.07 g/mol 2.02 g/mol 32.07 g/mol34.09 g/mol

M M M= +

= +

= +=

Mass, m, of the H2S(g):

2H S

2

9.8 mol 34.09 g/mol3.3 10 g

m n M= ×

= ×

= × The mass of the sample of hydrogen sulfide is 3.74 × 102 g.



Molar mass, M, of KMnO4(s):

( ) ( ) ( )4KMnO K Mn O 1 1 4

1 39.1 g/mol 1 54.94 g/mol 4 16.00 g/mol 39.10 g/mol 54.94 g/mol 64.00 g/mol158.04 g/mol

M M M M= + +

= + +

= + +=

Mass, m, of the KMnO4(s):

4KMnO

0.568 mol 158.04 g/mol89.8 g

m n M= ×

= ×=

The mass of the sample of potassium permanganate is 89.8 g. The sample of bromine has the largest mass. b. samples of strontium iodate, gold(III) chloride, and bismuth silicate: Molar mass, M, of Sr(IO3)2(s):

3 2Sr(IO ) Sr I O1 2 + 6

1(87.62 g/mol) + 2(126.9 g/mol) + 6(16.00 g/mol)87.62 g/mol 253.8 g/mol 96.00 g/mol437.42 g/mol

M M M M= +

== + +=

Mass, m, of the Sr(IO3)2(s):

( )3 2Sr IO

13.7 mol

m n M= ×

= 437.42 g/ mol×35.99 10 g= ×

The mass of the sample of strontium iodate is 5.99 × 103 g. Molar mass, M, of AuCl3(s):

( ) ( )3AuCl Au Cl1 3

1 196.97 g/mol 3 35.45 g/mol196.97 g/mol 106.35 g/mol 303.32 g/mol

M M M= +

= +

= +=



Mass, m, of the AuCl3(s): 3AuCl ×

15.9 mol

m n M=

= 303.32 g/ mol×3 4.82 10 g= ×

The mass of the sample of gold(III) chloride is 4.82 × 103 g. Molar mass, M, of Bi2(SiO3)3(s):

( ) ( ) ( )2 3 3Bi (SiO ) Bi Si O2 + 3 + 9


M M M M=

= + +

= + +=

Mass, m, of the Bi2(SiO3)3(s):

( )2 3 3Bi SiO

8.61 mol

m n M= ×

= 646.23 g/ mol×3

5.56 10 g= ×

The mass of the sample of bismuth silicate is 5.56 × 103 g. The sample of strontium iodate has the largest mass. Check Your Solution The units cancel properly. Check that the correct atomic molar masses have been used. The answers seem reasonable. However, the comparisons are too close to estimate using rounded numbers. The calculations should be rechecked.

50. Practice Problem (page 237) Which has the smallest mass: 0.215 mol of potassium hydrogen sulfite, KHSO3(s); 1.62 mol of sodium hydrogen sulfite, NaHSO3(s); or 0.0182 mol of aluminum iodate, Al(IO3)3(s)? What Is Required? You need to determine the masses of samples of potassium hydrogen sulfite, sodium hydrogen sulfite, and aluminum iodate and identify which has the smallest mass.



What Is Given? You know the chemical formulas for the compounds and their amounts in moles: potassium hydrogen sulfite, KHSO3; n = 0.215 mol sodium hydrogen sulfite, NaHSO3; n = 1.62 mol aluminum iodate, Al(IO3)3; n = 0.0182 mol Plan Your Strategy Use the periodic table to find the atomic molar masses of the elements in each compound. Multiply each element’s atomic molar mass by the number of atoms of the element in each compound. For each compound, add the molar masses of each element to find the molar mass. Use the relationship .m n M= × Multiply the amount in moles of each compound by the compound’s molar mass. Identify the sample that has the smallest mass. Act on Your Strategy Molar mass, M, of KHSO3(s):

3KHSO K H S O1 1 1 3

1(39.10 g/mol) 1(1.01 g/mol) 1(32.07 g/mol) 3(16.00 g/mol)39.10 g/mol 1.01 g/mol 32.07 g/mol 48.00 g/mol120.18 g/mol

M M M M M= + + +

= + + += + + +=

Mass, m, of the KHSO3(s):

3KHSO

0.215 mol

m n M= ×

= 120.18 g/ mol×25.8 g=

The mass of the sample of potassium hydrogen sulfite is 25.8 g. Molar mass, M, of NaHSO3(s):

( ) ( ) ( ) ( )3NaHSO Na H S O 1 1 1 3

1 22.99 g/mol 1 1.01 g/mol 1 32.07 g/mol 3 16.00 g/mol22.99 g/mol 1.01 g/mol 32.07 g/mol 48.00 g/mol104.07 g/mol

M M M M M= + + +

= + + +

= + + +=



Mass, m, of the NaHSO3(s): 3NaHSO

1.62 mol

m n M= ×

= 104.07 g/ mol×168 g=

The mass of the sample of sodium hydrogen sulfite is 168 g. Molar mass, M, of Al(IO3)3(s):

( ) ( ) ( )3 3Al(IO ) Al I O1 3 9


M M M M= + +

= + +

= + +=

Mass, m, of the Al(IO3)3(s):

( )3 3Al IO

0.0182 mol

m n M= ×

= 551.68 g/ mol×10.0 g=

The mass of the sample of aluminum iodate is 10.0 g. The sample of aluminum iodate has the smallest mass. Check Your Solution The units cancel properly. Check that the correct atomic molar masses have been used. The answers seem reasonable. However, the comparisons are too close to estimate using rounded numbers. The calculations should be rechecked.



Convert 29.5 g of ammonia to the amount in moles. What Is Required? You need to convert 29.5 g of ammonia to the amount in moles. What Is Given? You know the mass of the ammonia: 29.5 g



Plan Your Strategy Determine the chemical formula for ammonia. Use the periodic table to find the atomic molar masses of the elements in ammonia. Multiply the atomic molar masses by the number of atoms of each element in the compound. Add these values to calculate the molar mass of ammonia. Divide the mass by the molar mass to determine the amount in moles of

ammonia using the relationship .mnM

=

Act on Your Strategy The chemical formula for ammonia is NH3. Molar mass, M, of NH3(g):

( ) ( )3NH N H1 3

1 14.01 g/mol + 3 1.01 g/mol17.04 g/mol

M M M= +

=

=

Amount in moles, n, of the NH3(g):

3NH

29.5 g

mnM

=

=17.04 g

1.73

/mol

mol=

There is 1.73 mol of ammonia in the sample. Check Your Solution The units are correct and substitutions have been made correctly. The answer seems reasonable and shows the correct number of significant digits.

52. Practice Problem (page 239) Determine the amount in moles of potassium thiocyanate, KSCN(s), in 13.5 kg. What Is Required? You need to determine the amount in moles of 13.5 kg of potassium thiocyanate.



What Is Given? You know the chemical formula for potassium thiocyanate: KSCN You know the mass of the KSCN(s): 13.5 kg Plan Your Strategy Use the periodic table to find the atomic molar masses of the elements in KSCN(s). Multiply the atomic molar masses by the number of atoms of each element in the compound. Add these values to calculate the molar mass of KSCN(s). Convert the mass of the KSCN(s) from kilograms to grams: 1 kg = 1 × 103 g Divide the mass by the molar mass to determine the amount in moles, n, of

KSCN(s) using the relationship .mnM

=

Act on Your Strategy Molar mass, M, of KSCN(s):

( ) ( ) ( ) ( )KSCN K S C N1 + 1 + 1 + 1

1 39.10 g/mol + 1 32.07 g/mol + 1 2.01 g/mol + 1 14.01 g/mol97.19 g/mol

M M M M M=

=

=

Mass (in grams), m, of the KSCN(s):

KSCN 13.5 kgm = 3 1 10 g/ kg× ×

13 500 g=

Amount in moles, n, of KSCN(s):

KSCN

13 500 g

mnM

=

=97.19 g

138.9 mol1

/m

3 ol

ol

9 m==

There is 139 mol of potassium thiocyanate in the sample. Check Your Solution The units are correct and substitutions have been made correctly. The answer seems reasonable and shows the correct number of significant digits.



53. Practice Problem (page 239) Determine the amount in moles of sodium dihydrogen phosphate, NaH2PO4(s), in 105 mg. What Is Required? You need to determine the amount in moles of 105 mg of sodium dihydrogen phosphate. What Is Given? You know the chemical formula for sodium dihydrogen phosphate: NaH2PO4 You know the mass of the NaH2PO4(s): 105 mg Plan Your Strategy Use the periodic table to find the atomic molar masses of the elements in NaH2PO4(s). Multiply the atomic molar masses by the number of atoms of each element in the compound. Add these values to calculate the molar mass of NaH2PO4(s). Convert the mass of the NaH2PO4(s) from kilograms to grams: 1 kg = 1 × 103 g Divide the mass by the molar mass to determine the amount in moles, n, of

NaH2PO4(s) using the relationship mnM

= .

Act on Your Strategy Molar mass, M, of NaH2PO4(s):

( ) ( ) ( ) ( )2 4NaH PO Na H P O1 + 2 + 1 + 4

1 22.99 g/mol + 2 1.01 g/mol + 1 30.97 g/mol + 4 16.00 g/mol 119.98 g/mol

M M M M M=

=

= Mass (in grams), m, of the NaH2PO4(s):

2 4NaH PO 105 mgm = –3 1 10 g/ mg× ×

0.105 g=

Amount in moles, n, of NaH2PO4(s):

2 4NaH PO

0.105 g

mnM

=

=

119.98 g–4

–4

8.7514 10 mol8.75 10

/m

ol

l

m

o

= ×

= ×



There is 8.75 × 10–4 mol of sodium dihydrogen phosphate in the sample. Check Your Solution The units are correct and substitutions have been made correctly. The answer seems reasonable and shows the correct number of significant digits.

54. Practice Problem (page 239) Determine the amount in moles of xenon tetrafluoride, XeF4(s), in 22 mg. What Is Required? You need to determine the amount in moles of 22 mg of xenon tetrafluoride. What Is Given? You know the chemical formula for xenon tetrafluoride: XeF4 You know the mass of the XeF4(s): 22 mg Plan Your Strategy Use the periodic table to find the atomic molar masses of the elements in XeF4(s). Multiply the atomic molar masses by the number of atoms of each element in the compound. Add these values to calculate the molar mass of XeF4(s). Convert the mass of the XeF4(s) from kilograms to grams: 1 kg = 1 × 103 g Divide the mass by the molar mass to determine the amount in moles, n, of

XeF4(s) using the relationship mnM

= .

Act on Your Strategy Molar mass, M, of XeF4(s):

( ) ( )4XeF Xe F1 4

1 131.29 g/mol 4 19.00 g/mol207.29 g/mol

M M M= +

= +

=

Mass (in grams), m, of the XeF4(s):

4XeF 22 mgm = –3 1 × 10 g/ mg×

0.022 g=



Amount in moles, n, of XeF4(s):

4XeF

0.022 g

mnM

=

=207.29 g

–4

–4

1.0613 10 mol1.1 10

/mo

m l

l

o= ×

= ×

There is 1.1 × 10–4 mol of xenon tetrafluoride in the sample. Check Your Solution The units are correct and substitutions have been made correctly. The answer seems reasonable and shows the correct number of significant digits.

55. Practice Problem (page 239) Write the chemical formula for each compound, and then calculate the amount in moles in each sample. a. 3.7 × 10–3 g of silicon dioxide b. 25.38 g of titanium(IV) nitrate c. 19.2 mg of indium carbonate d. 78.1 kg of copper(II) sulfate pentahydrate What Is Required? You need to write the chemical formula for, and determine the amount in moles of, samples of a. silicon dioxide. b. titanium(IV) nitrate. c. indium carbonate. d. copper(II) sulfate pentahydrate. What Is Given? a. You know the mass of the silicon dioxide: 3.7 × 10–3 g b. You know the mass of the titanium(IV) nitrate: 25.38 g c. You know the mass of the indium carbonate: 19.2 mg d. You know the mass of the copper(II) sulfate pentahydrate: 78.1 kg Plan Your Strategy For each compound: Write the chemical formula. Use the periodic table to find the atomic molar masses of the elements in the compound.



Multiply the atomic molar masses by the number of atoms of each element in the compound. Add these values to calculate the molar mass of the compound. Convert the mass in milligrams to grams: 1 mg = 1 × 10–3 g Convert the mass in kilograms to grams: 1 kg = 1 × 103 g Divide the mass by the molar mass to determine the amount in moles of each

compound using the relationship .mnM

=

Act on Your Strategy a. silicon dioxide The chemical formula for silicon dioxide is SiO2. Molar mass, M, of SiO2(s):

( ) ( )2SiO S O1 2

1 28.09 g/mol 2 16.00 g/mol60.09 g/mol

M M M= +

= +

=

Amount in moles, n, of SiO2(s):

2

–

SiO

33.7 10 g

mnM

=×

=

60.09 g–5

–5

6.157 10 mol6.2 10

/mo

m l

l

o= ×

= ×

There is 6.2 × 10–5

mol of silicone dioxide in the sample. b. titanium(IV) nitrate The chemical formula for titanium(IV) nitrate is Ti(NO3)4. Molar mass, M, of Ti(NO3)4(s):

( ) ( ) ( )3 4Ti(NO ) Ti N O1 4 + 12

1 47.87 g/mol 4 14.01 g/mol 12 16.00 g/mol 295.91 g/mol

M M M M= +

= + +

=



Amount in moles, n, of the Ti(NO3)4(s):

( )3 4Ti NO

25.38 g

mnM

=

=295.91 g

–2

–2

8.5769 10 mol8.577 10

/mol

mol= ×

= ×

There is 8.577 × 10–2 mol of titanium(IV) nitrate in the sample. c. indium carbonate The chemical formula for indium carbonate is In2(CO3)3. Molar mass, M, of In2(CO3)3(s):

( ) ( ) ( )2 3 3In (CO ) In C O2 + 3 + 9


M M M M=

= + +

=

Mass (in grams), m, of In2(CO3)3(s):

( )2 3 3In CO 19.2 mgm = –3 1 10 g/ mg× ×

0.0192 g=

Amount in moles, n, of In2(CO3)3(s):

( )2 3 3In CO

0.0192 g

mnM

=

=409.67 g

–5

–5

4.68669 10 mol4.69 10

/mol

mol= ×

= ×

There is 4.69 × 10–5 mol of indium carbonate in the sample.



d. copper(II) sulfate pentahydrate The chemical formula for copper(II) sulfate pentahydrate is CuSO4•5H2O. Molar mass, M, of CuSO4•5H2O(s):

( ) ( ) ( ) ( )

4 2 4 2CuSO •5H O CuSO H O

Cu S O H O

1 5

1[1 + 1 + 4 ] + 5[2 1 ]1[1 63.55 g/mol + 1 32.7 g/mol + 4 16.00 g/mol ] + 5 18.02 g/mol249.62 g/mol

M M M

M M M M M

= +

= +

=

= Mass (in grams), m, of CuSO4•5H2O(s):

4 2CuSO •5H O 78.1 kgm = 3 1 10 g/ kg× ×4 7.81 10 g= ×

Amount in moles, n, of CuSO4•5H2O(s):

4 2CuSO •5H O

47.81 × 10 g

mnM

=

=249.62 g

2

2

3.12875 10 mol3.13

/

10 mol

mol

×

×

=

=

There is 3.13 × 102 mol of copper(II) sulfate pentahydrate in the sample. Check Your Solution Check that the correct atomic molar masses have been used and substitutions have been made correctly. The units are correct. The answers seem reasonable and show the correct number of significant digits.

56. Practice Problem (page 239) The characteristic odour of garlic comes from allyl sulfide, (C3H5)2S(ℓ). Determine the amount in moles of allyl sulfide in 168 g. What Is Required? You need to determine the amount in moles of 168 g of allyl sulfide. What Is Given? You know the chemical formula for allyl sulfide: (C3H5)2S You know the mass of the (C3H5)2S(ℓ): 168 g



Plan Your Strategy Use the periodic table to find the atomic molar masses of the elements in (C3H5)2S(ℓ). Multiply the atomic molar masses by the number of atoms of each element in the compound. Add these values to calculate the molar mass of (C3H5)2S(ℓ). Divide the mass by the molar mass to determine the amount in moles of

(C3H5)2S(ℓ) using the relationship mnM

= .

Act on Your Strategy Molar mass, M, of (C3H5)2S(ℓ):

( ) ( ) ( )3 5 2(C H ) S C H S6 + 10 + 1

6 12.01 g/mol + 10 1.01 g/mol + 1 32.07 g/mol114.23 g/mol

M M M M=

=

=

Amount in moles, n, of (C3H5)2S(ℓ):

( )3 5 2C H S

168 g

mnM

=

=114.23 g

1.47

/mol

mol=

There is 1.47 mol of allyl sulfide in the sample. Check Your Solution The units are correct and substitutions have been made correctly. The answer seems reasonable and shows the correct number of significant digits.

57. Practice Problem (page 239) Road salt, CaCl2(s), is often used on roads in the winter to prevent the build-up of ice. What amount in moles of calcium chloride is in a 20.0 kg bag of road salt? What Is Required? You need to determine the amount in moles of 20.0 kg of calcium chloride. What Is Given? You know the chemical formula for calcium chloride: CaCl2 You know the mass of the CaCl2(s): 20.0 kg



Plan Your Strategy Use the periodic table to find the atomic molar masses of the elements in CaCl2(s). Multiply the atomic molar masses by the number of atoms of each element in the compound. Add these values to calculate the molar mass of CaCl2(s). Convert the mass from kilograms to grams: 1 kg = 1 × 103 g Divide the mass by the molar mass to determine the amount in moles of

CaCl2(s) using the relationship .mnM

=

Act on Your Strategy Molar mass, M, of CaCl2(s):

( ) ( )2CaCl Ca Cl1 2

1 40.08 g/mol + 2 35.45 g/mol110.98 g/mol

M M M= +

=

=

Mass (in grams), m, of CaCl2(s):

2CaCl 20.0 kgm = 3 1 10 g/ kg× ×4

2.00 10 kg= ×

Amount in moles, n, of CaCl2(s):

2

4

CaCl

2.00 × 10 g

mnM

=

=110.98 g

21.80 10

/m l

o

o

m l= ×

There is 1.80 × 102 mol of calcium chloride in the bag. Check Your Solution The units are correct and substitutions have been made correctly. The answer seems reasonable and shows the correct number of significant digits.



58. Practice Problem (page 239) Calculate the amount in moles of trinitrotoluene, C7H5(NO2)3(s), an explosive, in 3.45 × 10–3 g. What Is Required? You need to determine the amount in moles of 3.45 × 10–3 g of trinitrotoluene. What Is Given? You know the chemical formula for trinitrotoluene: C7H5(NO2)3 You know the mass of the C7H5(NO2)3(s): 3.45 × 10–3 g Plan Your Strategy Use the periodic table to find the atomic molar masses of the elements in C7H5(NO2)3(s). Multiply the atomic molar masses by the number of atoms of each element in the compound. Add these values to calculate the molar mass of C7H5(NO2)3(s). Divide the mass by the molar mass to determine the amount in moles of

C7H5(NO2)3(s) using the relationship .mnM

=

Act on Your Strategy Molar mass, M, of C7H5(NO2)3(s):

( ) ( ) ( ) ( )7 5 2 3C H (NO ) C H N O7 + 5 + 3 + 6

7 12.01 g/mol + 5 1.01 g/mol + 3 14.01 g/mol + 6 16.00 g/mol 227.15 g/mol

M M M M M=

=

= Amount in moles, n, of the C7H5(NO2)3(s):

( )7 5 2 3C H NO

–33.45 × 10 g

mnM

=

=227.15 g

–5

–5

1.51880 10 mol1.52 10

/mol

mol= ×

= ×

There is 1.52 × 10–5 mol of trinitrotoluene in the explosive. Check Your Solution The units are correct and substitutions have been made correctly. The answer seems reasonable and shows the correct number of significant digits.



59. Practice Problem (page 239) Arrange the following substances in order from largest to smallest amount in moles: • 865 mg of Ni(NO3)2(s) • 9.82 g of Al(OH)3(s) • 10.4 g of AgCl(s) What Is Required? You need to determine the amount in moles of samples of Ni(NO3)2(s), Al(OH)3(s), and AgCl(s) and arrange these amounts from largest to smallest. What Is Given? You know the mass of each substance: Ni(NO3)2(s) = 865 mg Al(OH)3(s) = 9.82 g AgCl(s) =10.4 g Plan Your Strategy For each compound: Use the periodic table to find the atomic molar masses of the elements in the compound. Multiply the atomic molar masses by the number of atoms of each element in the compound. Add these values to calculate the molar mass of the compound. Convert the mass from milligrams to grams: 1 mg = 1 × 10–3 g Divide the mass by the molar mass to determine the amount in moles of each

compound using the relationship .mnM

=

Arrange the amounts in moles from largest to smallest. Act on Your Strategy Molar mass, M, of Ni(NO3)2(s):

( ) ( ) ( )3 2Ni(NO ) Ni N O1 + 2 + 6


M M M M=

=

=

Mass (in grams), m, of the Ni(NO3)2(s): ( )3 2Ni NO 865 mgm = –3 1 10 g/ mg× ×

0.865 g =



Amount in moles, n, of the Ni(NO3)2:

( )3 2Ni NO

0.865 g

mnM

=

=182.71 g

–3

–3

4.734 10 mol4.73 10

/mo

m l

l

o= ×

= ×

The amount in moles of the Ni(NO3)2 is 4.73 × 10−3 mol. Molar mass, M, of Al(OH)3(s):

( )

( ) ( ) ( )3

Al O HAl OH 1 + 3 + 3


M M M M=

= + +

=

Amount in moles, n, of the Al(OH)3(s):

( )3Al OH

9.82 g

mnM

=

=78.10 g

0.12574 mol0.126

/mo

o

l

m l==

The amount in moles of the Al(OH)3(s) is 0.126 mol. Molar mass, M, of AgCl(s):

( ) ( )AgCl Ag Cl1 + 1

1 107.87 g/mol + 1 35.45 g/mol143.32 g/mol

M M M=

=

=



Amount in moles, n, of the AgCl(s):

AgCl

10.4 g

mnM

=

=143.32 g

–2

–2

/mol

7.2565 10 mol 7.26 10 mol

= ×

= ×

The amount in moles of the AgCl(s) is 7.26 × 10−2 mol. Amounts in moles from largest to smallest: Al(OH)3(s) (0.126 mol) > AgCl(s) (0.0726 mol) > Ni(NO3)2(s) (0.00473 mol) Check Your Solution Check that the correct atomic molar masses have been used and substitutions have been made correctly. The units are correct. The answers seem reasonable and show the correct number of significant digits. The amounts in moles are arranged from largest to smallest.

60. Practice Problem (page 239) Place the following substances in order from smallest to largest amount in moles, given 20.0 g of each: • glucose, C6H12O6(s) • barium perchlorate, Ba(ClO4)2(s) • tin(IV) oxide, SnO2(s) What Is Required? You need to determine the amount in moles of samples of glucose, barium perchlorate, and tin(IV) oxide and arrange these amounts from smallest to largest. What Is Given? You know the chemical formula for each compound: glucose, C6H12O6; barium perchlorate, Ba(ClO4)2; tin(IV) oxide, SnO2 You know the mass of each compound: 20.0 g Plan Your Strategy For each compound: Use the periodic table to find the atomic molar masses of the elements in the compound.



Multiply the atomic molar masses by the number of atoms of each element in the compound. Add these values to calculate the molar mass of the compound. Divide the mass by the molar mass to determine the amount in moles of each

compound using the relationship mnM

= .

Arrange the amounts in moles from smallest to largest. Act on Your Strategy Molar mass, M, of C6H12O6(s):

( ) ( ) ( )6 12 6C H O C H O6 + 12 + 6

6 12.01 g/mol 12 1.01 g/mol 6 16.00 g/mol180.18 g/mol

M M M M=

= + +

=

Amount in moles, n, of the C6H12O6(s):

6 12 6C H O

0 g2 .0

mnM

=

=180.18 g

0.111 l

/ mol

mo=

The amount in moles of glucose is 0.111 mol. Molar mass, M, of Ba(ClO4)2(s):

( ) ( ) ( )4 2Ba(ClO ) Ba Cl O1 + 2 +8


M M M M=

= + +

=

Amount in moles, n, of the Ba(ClO4)2(s):

( )4 2Ba ClO

20.0 g

mnM

=

=336.23 g

0.0595

/mol

mol=

The amount in moles of barium perchlorate is 0.0595 mol.



Molar mass, M, of SnO2(s):

( ) ( )2SnO Sn O1 + 2

1 118.71 g/mol 2 16.00 g/mol 150.71 g/mol

M M M=

= +

=

Amount in moles, n, of the Sn2O4(s):

2 4Sn O

20.0 g

mnM

=

=150.71 g

0.133 l

/mol

mo=

The amount in moles of tin(IV) oxide is 0.133 mol. Amounts in moles from smallest to largest: Ba(ClO4)2(s) (0.0595 mol) < C6H12O6(s) (0.111 mol) < SnO2(s) (0.133 mol) Check Your Solution Check that the correct atomic molar masses have been used and substitutions have been made correctly. The units are correct. The answers seem reasonable and show the correct number of significant digits. The amounts in moles are arranged from smallest to largest.



Calculate the mass of each sample. a. 1.05 × 1026 atoms of neon, Ne(g) b. 2.7 × 1024 molecules of phosphorus trichloride, PCl3(ℓ) c. 8.72 × 1021 molecules of karakin, C15H21N3O15(s) d. 6.7 × 1027 formula units of sodium thiosulfate, Na2S2O3(s) What Is Required? You need to determine the mass of a sample of a. neon. b. phosphorus trichloride. c. karakin. d. sodium thiosulfate.



What Is Given? a. You know the number of atoms of neon, Ne(g): 1.05 × 1026 b. You know the number of molecules of phosphorus trichloride, PCl3(ℓ): 2.7 × 1024 c. You know the number of molecules of karakin, C15H21N3O15(s): 8.72 × 1021 d. You know the number of formula units of sodium thiosulfate, Na2S2O3(s): 6.7 × 1027 Plan Your Strategy Use the Avogadro constant: 23

A 6.02 10N = × Calculate the amount in moles of each compound using the relationship

A

NnN

= .

Use the periodic table to find the atomic molar masses of the elements in the compound. Multiply the atomic molar masses by the number of atoms of each element in the compound. Add these values to calculate the molar mass of the compound. Calculate the mass of each compound using the relationship .m n M= × Alternative Solution The steps can be completed in one line, using the units to calculate the answer. Calculate the mass of each compound by multiplying the given amount of the

compound by 23

1 mol6.02 10 units×

and then by the molar mass, M.

Act on Your Strategy a. neon M Ne = 20.18 g/mol (from the periodic table) Amount in moles, n, of Ne(g):

2

Ne

6

A

1.05 10 o

at ms

NnN

=

=×

236.02 10 atoms×21.7442 10

/m lol

om= ×



Mass, m, of the Ne(g): Ne

2

1.7442 10 mol

m n M= ×

= × 20.18 g/ mol×33.52 10 g= ×

The mass of the neon is 3.52 × 103 g. Alternative solution M Ne = 20.18 g/mol (from the periodic table) Mass, m, of the Ne(g):

26Ne 1.05 10 atomsm = × l 1 mo

× 236.02 10 atoms×20.18 g × 1 mol

33.52

10 g×=

The mass of the neon is 3.52 × 103 g. b. phosphorus trichloride Molar mass, M, of PCl3(ℓ):

( ) ( )3PCl P Cl1 + 3

1 30.97 g/mol 3 35.45 g/mol 137.32 g/mol

M M M=

= +

=

Amount in moles, n, of PCl3(ℓ):

3PClA

242.7 10 molecules

NnN

×

=

= 236.02 10 moledules×4.485 l

/mol mo=

Mass, m, of the PCl3(ℓ):

3PCl

4.485 mol

m n M= ×

= 137.32 g/ mol×2

2

6.15887 10 g6.2 10 g

= ×

= ×

The mass of the phosphorus trichloride is 6.2 × 102 g.



Alternative Solution Molar mass, M, of PCl3(ℓ):

( ) ( )3PCl P Cl1 + 3

1 30.97 g/mol 3 35.45 g/mol 137.32 g/mol

M M M=

= +

=

Mass, m, of the PCl3(ℓ):

3

24PCl 2.7 10 moleculesm = × l 1 mo

× 236.02 10 molecules×137.32 g

1 mol×

2

2

6.15887 10 g6.2 10 g

×

= ×

=

The mass of the phosphorus trichloride is 6.2 × 102 g. c. karakin Molar mass, M, of C15H21N3O15(s):

( ) ( ) ( ) ( )15 21 3 15C H N O C H N O15 21 + 3 + 15


M M M M M= +

=

=


15 21 3 15C H N OA

218.72 10

molecules

NnN

=

×= 236.02 10 moledules×

–21.4485 10/m l

oo

m l= ×

Mass, m, of the C15H21N3O15(s):

15 21 3 15C H N O

–2

1.4485 10 mol

m n M= ×

= × 483.39 g/ mol×7.0019 g7.00 g

==

The mass of the karakin is 7.00 g.



Alternative Solution Molar mass, M, of C15H21N3O15(s):

( ) ( ) ( ) ( )15 21 3 15C H N O C H N O15 21 + 3 + 15


M M M M M= +

=

=

Mass, m, of the C15H21N3O15(s):

15 21 3 15

21C H N O 8.72 10 moleculesm = × 1 mol

× 236.02 × 10 molecules483.39 g

1 mol×

7.0019 g7. 00 g=

=

The mass of the karakin is 7.00 g. d. sodium thiosulfate Molar mass, M, of Na2S2O3(s):

( ) ( ) ( )2 2 3Na S O Na S O2 2 + 3


M M M M= +

=

=

Amount in moles, n, of the Na2S2O3(s):

2 2 3Na S OA

276.7 10 form

ula units

NnN

×

=

= 236.02 10 formula units×41.11295 10

/ m lol

om= ×

Mass m, of the Na2S2O3(s):

2 2 3Na S O

4

1.1129 10 mol

m n M= ×

= × 158.12 g/ mol×6

6

1.7598 10 g1.8 10 g

= ×

= ×

The mass of the sodium thiosulfate is 1.8 × 106 g.



Alternative Solution Molar mass, M, of Na2S2O3(s):

( ) ( ) ( )2 2 3Na S O Na S O2 2 + 3


M M M M= +

=

=

Mass, m, of the Na2S2O3(s):

2 2 3

27Na S O 6.7 10 formula unitsm = × l 1 mo

× 236.02 10 formula units×158.12 g

1 mol×

6

6

1.7598 10 g1.8 10 g

×

= ×

=

The mass of the sodium thiosulfate is 1.8 × 106 g. Check Your Solutions In each case, check that the correct atomic molar masses and formulas have been used. Using rounded numbers, estimate the answers:

a. 1 × 1026 × 23

1 6.02 10×

× 20 = 3 × 103 g

b. 3 × 1024 × 23

1 6.02 10×

× 140 = 7 × 102 g

c. 9 × 1021 × 23

1 6.02 10×

× 480 = 7.2 g

d. 7 × 1027 × 23

1 6.02 10×

× 160 = 1.9 × 106 g

All of the estimated answers are close to the calculated answers. The answers are reasonable.

62. Practice Problem (page 242) Determine the number of molecules or formula units in each sample. a. 32.4 g of lead(II) phosphate, Pb3(PO4)2(s) b. 8.62 × 10–3 g of dinitrogen pentoxide, N2O5(s) c. 48 kg of molybdenum(VI) oxide, MoO3(s) d. 567 g of tin(IV) fluoride, SnF4(s) What Is Required? You need to determine the number of a. formula units of lead(II) phosphate. b. molecules of dinitrogen pentoxide.



c. formula units of molybdenum(VI) oxide . d. formula units of tin(IV) fluoride. What Is Given? You know the mass of each of the samples: a. Pb3(PO4)2(s) = 32.4 g b. N2O5(s) = 8.62 × 10–3 g c. MoO3(s) = 48 kg d. SnF4(s) = 567 g Plan Your Strategy For each compound: Use the periodic table to find the atomic molar masses of the elements in the compound. Multiply the atomic molar masses by the number of atoms of each element in the compound. Add these values to calculate the molar mass of the compound. Solution 1: Convert the mass in kilograms to grams: 1 kg = 1 × 103 g Calculate the amount in moles of each compound using the relationship

mnM

= .

Use the Avogadro constant: 236.02 10AN = × Calculate the number of molecules or formula units of each compound using

the relationship A

NnN

= .

Solution 2: The following steps can be completed in one line using the units to calculate the answer. Calculate the number of formula units or molecules of the compounds by

multiplying the mass by 1molar mass

and then by the Avogadro constant, NA.

Act on Your Strategy a. lead(II) phosphate Molar mass, M, of Pb3(PO4)2(s):

( ) ( ) ( )3 4 2Pb (PO ) Pb P O3 + 2 + 8

3 207.2 g/mol + 2 30.97 g/mol + 8 16.00 g/mol 811.54 g/mol

M M M M=

=

=



Solution 1 Amount in moles, n, of the Pb3(PO4)2(s):

( )3 4 2Pb PO

32.4 g

mnM

=

=811.54 g

–23.99 10

/mo

ol

l

m= ×

Number of formula units, N, of Pb3(PO4)2(s):

–2

3.99 10 mol

AN n N= ×

= ×23 formul6.02 × 10 unitsa1

mol

×


There are 2.41 × 1022 formula units of lead(II) phosphate in the sample. Solution 2 Number of formula units, N, of Pb3(PO4)2(s):

32.4 gN =1 l mo

×811.54 g

236.02 × 10 formula units 1 l

mo

×


There are 2.41 × 1022 formula units of lead(II) phosphate in the sample. b. dinitrogen pentoxide Molar mass, M, of N2O5(s):

( ) ( )2 5N O N O2 + 5

2 14.01 g/mol + 5 16.00 g/mol108.02 g/mol

M M M=

=

=

Solution 1 Amount in moles, n, of the N2O5(s):

2 5

–

N O

38.64 1 0 g

mnM

×

=

=108.02 g

–57.995 10

/m l

o

o

m l= ×



Number of molecules, N, of N2O5(s): A

–5

7.995 10 mol

N n N= ×

= ×23 mo6 l.0 ec2 1 ules 0

1 ol

m×

×

194.82 10 molecules×=

There are 4.82 × 1019 molecules of dinitrogen pentoxide. Solution 2 Number of molecules, N, of N2O5(s):

–38.64 10 gN = ×1 l mo

×108.02 g

23 6.02 1 0 molecules 1 mol

××

194.82 10 molecules×=

There are 4.82 × 1019 molecules of dinitrogen pentoxide in the sample. c. molybdenum(VI) oxide Molar mass, M, of MoO3(s):

3MoO Mo O1 3

1(95.96 g/mol) + 1(16.00 g/mol)143.96 g/mol

M M M= +

==

Solution 1 Mass (in grams), m, of the MoO3(s):

3MoO 48 kgm = 3 1 10 g/ kg× ×

48 000 g=

Amount in moles, n, of the MoO3(s):

3MoO

48 000 g

mnM

=

=143.96 g

23.3343 10

/m l

ol

o

m= ×



Number of formula units, N, of the MoO3(s):

2

3.3343 10 mol

AN n N= ×

= ×23 formul6.02 a uni1 0 ts1

mol

××

262.0 10 formula units= × There are 2.0 × 1026 formula units of molybdenum(VI) oxide in the sample. Solution 2 Number of formula units, N, of the MoO3(s):

48 000 gN =1 l mo

×143.96 g

236.02 1 0 formula units 1 ol m

××

26 10 formula u2.0 nits= ×

There are 2.0 × 1026 formula units of molybdenum(VI) oxide in the sample. d. tin(IV) fluoride Molar mass, M, of the SnF4(s)

( ) ( )4SnF Sn F1 4

1 118.71 g/mol 4 19.00 g/mol194.71 g/mol

M M M= +

= +

=

Solution 1 Amount in moles, n, of the SnF4(s):

4SnF

5 g67

mnM

=

=194.71 g

2.912 l

/mol

mo=

Number of formula units, N, SnF4(s):

2.912 mol

AN n N= ×

=23 formul6.02 a uni1 0 ts1

mol

××


There are 1.75 × 1024 formula units of tin(IV) fluoride in the sample.



Solution 2 Number of formula units, N, of the SnF4(s):

567 gN =1 l mo

×194.71 g

236.02 × 10 formula units 1 l

mo

×

24 10 formula 1. u t75 ni s= × There are 1.75 × 1024 formula units of tin(IV) fluoride in the sample. Check Your Solutions In all cases, check that the correct atomic molar masses have been used. Using rounded numbers, estimate the answers:

a. 30 × 1 800

× 6 × 1023 = 2 × 1022 formula units of Pb3(PO4)2(s)

b. 9 × 10−3 × 1 100

× 6 × 1023 = 5 × 1019 molecules of N2O5(s)

c. 5 × 104 × 1 150

× 6 × 1023 = 2 × 1026 formula units of MoO3(s)

d. 560 × 1 200

× 6 × 1023 = 1.7 × 1024 formula units of SnF4(s)

All of the estimated answers are close to the calculated answers. The answers are reasonable.

63. Practice Problem (page 242) Sodium hydrogen carbonate, NaHCO3(s), is the principal ingredient in many stomach-relief medicines. a. A teaspoon of a particular brand of stomach-relief medicine contains 6.82 × 1022 formula units of sodium hydrogen carbonate. What mass of sodium hydrogen carbonate is in the teaspoon? b. The bottle of this stomach-relief medicine contains 350 g of sodium hydrogen carbonate. How many formula units of sodium hydrogen carbonate are in the bottle? a. mass of sodium hydrogen carbonate What Is Required? You have to determine the mass of sodium hydrogen carbonate in a teaspoon of stomach-relief medicine. What Is Given? You know there are 6.82 × 1022 formula units of sodium hydrogen carbonate. You know the chemical formula for sodium hydrogen carbonate: NaHCO3



Plan Your Strategy Use the periodic table to find the atomic molar masses of the elements in NaHCO3(s). Multiply the atomic molar masses by the number of atoms of each element in the compound. Add these values to calculate the molar mass of the compound. Solution 1 Use the Avogadro constant: 236.02 10AN = ×

Calculate the number of moles of NaHCO3(s) using the relationship A

NnN

= .

Calculate the mass of NaHCO3(s) using the relationship m n M= × . Solution 2 The following steps can be completed in one line using the units to calculate the answer. Calculate the mass by multiplying the given amount of formula units of

NaHCO3(s) by the ratio 23

1 mol6.02 10 formula units×

and then by the molar

mass, M. Act on Your Strategy Molar mass, M, of NaHCO3(s):

( ) ( ) ( ) ( )3NaHCO Na H C O1 1 1 3

1 22.99 g/mol 1 1.01 g/mol 1 12.01 g/mol 3 16.00 g/mol84.01 g/mol

M M M M M= + + +

= + + +

= Solution 1 Amount in moles, n, of NaHCO3(s):

3NaHCO

226.82 10 formula unitsA

NnN

=

×= 236.02 10 formula units× /mol

0.113289 mol=



Mass, m, of the NaHCO3(s):

3NaHCO

0.113289 mol

m n M= ×

= 84.01 g/ mol×9.5174 g 9.52 g

==

The mass of the sodium carbonate is 9.52 g. Solution 2 Mass, m, of the NaHCO3(s):

3

22NaHCO 6.82 10 formula unitsm = ×

1 mol × 236.02 10 formula units× 84.01 g/ mol×

9.52 g =

The mass of the sodium carbonate is 9.52 g. Check Your Solution Using rounded numbers to estimate the answer:

7 × 1022 × 23

1 6.02 10×

× 80 = 9 g

The answer is reasonable. b. formula units of sodium hydrogen carbonate What Is Required? You have to determine the number of formula units of sodium hydrogen carbonate in the bottle of stomach-relief medicine. What Is Given? You know the mass of the sodium hydrogen carbonate: 350 g Plan Your Strategy Use the periodic table to find the atomic molar masses of the elements in NaHCO3(s). Multiply the atomic molar masses by the number of atoms of each element in the compound. Add these values to calculate the molar mass of the compound. Solution 1

Calculate the amount in moles of NaHCO3(s) using the relationship .mnN

=



Calculate the mass of the NaHCO3(s) using the relationship m n M= × . Use the Avogadro constant: 236.02 10AN = × Calculate the number of formula units of NaHCO3(s) using the relationship

AN n N= × . Solution 2 The following steps can be completed in one line using the units to calculate the answer. Calculate the number of formula units by multiplying the mass by

1molar mass


Act on Your Strategy Molar mass, M, of NaHCO3(s):

( ) ( ) ( ) ( )3NaHCO Na H C O1 1 1 3


M M M M M= + + +

= + + +

= Solution 1 Amount in moles, n, of the (s):

3NaHCO

350 g

mnN

=

=84.01 g /mol

4.166 mol=

Number of formula units, N, of the NaHCO3(s):

A

4.166 mol

N n N= ×

=236.02 10 formula units

1 mol×

×


There are 2.51 × 1024 formula units of sodium hydrogen carbonate.



Solution 2 Number of formula units, N, of NaHCO3(s):

350 gN =1 mol ×

84.01 g

236.02 10 formula units 1 mol

××

242.51 10 formula units = × There are 2.51 × 1024 formula units of sodium hydrogen carbonate. Check Your Solution Check to see that the correct atomic molar masses have been used. Use rounded numbers to estimate the answer:

350 × 1 80

× 6 × 1023 = 2.6 × 1024 formula units of NaHCO3(s)

The answer is reasonable.

64. Practice Problem (page 242) Riboflavin, C17H20N4O6(s), is an important vitamin in the metabolism of fats, carbohydrates, and proteins in your body. a. The current recommended dietary allowance (RDA) of riboflavin for adult men is 1.3 mg/day. How many riboflavin molecules are in this RDA? b. The RDA of riboflavin for adult women contains 1.8 × 1018 molecules of riboflavin. What is the RDA for adult women, in milligrams? a. RDA for men What Is Required? You need to determine the number of molecules of riboflavin in a recommended dietary allowance. What Is Given? You know the RDA: 1.3 mg/day You know the chemical formula for riboflavin: C17H20N4O6 Plan Your Strategy Use the periodic table to find the atomic molar masses of the elements in C17H20N4O6(s). Multiply the atomic molar masses by the number of atoms of each element in the compound. Add these values to calculate the molar mass of C17H20N4O6(s).



Solution 1 Convert the mass in a daily dosage of C17H20N4O6(s) from milligrams to grams: 1 mg = 1× 10–3 g

Calculate the number of moles of C17H20N4O6(s) using the relationship mnN

= .

Use the Avogadro constant: 236.02 10AN = × Calculate the number of molecules of riboflavin using the relationship

AN n N= × . Solution 2 The following steps can be completed in one line using the units to calculate the answer. Calculate the number of molecules of C17H20N4O6(s) by multiplying the mass

by 1molar mass


Act on Your Strategy Molar mass, M, of C17H20N4O6(s):

( ) ( ) ( ) ( )17 20 4 6C H N O C H N O17 20 4 6


M M M M M= + + +

= + + +

=

Solution 1 Mass (in grams), m, of the C17H20N4O6(s):

17 20 4 6C H N O 1.3 mgm = –3 1 10 g/ mg× ×–31.3 10 g= ×

Amount in moles, n, of the C17H20N4O6(s):

17 20 4 6C H N O

–31.3 10 g

mnN

=

×=

376.41 g–6

/mol

3.4536 10 mol= ×



Number of molecules, N, of the C17H20N4O6(s):

–6

3.4536 10 mol

AN n N= ×

= ×236.02 10 molecules

1 mol×

×

18

18

2.079 10 molecules2.1 10 molecules

= ×

= ×

There are 2.1 × 1018 molecules of riboflavin in the RDA for men. Solution 2 Number of molecules, N, of C17H20N4O6(s):

–31.3 10 gN = ×

1 mol ×376.41 g

236.02 10 molecules 1 mol

××

182.1 10 molecules= × There are 2.1 × 1018 molecules of riboflavin in the RDA for men. Check Your Solution Check to see that the correct atomic molar masses have been used. Use rounded numbers to estimate the answer:

1 × 10–3 × 1380

× 6 × 1023 = 1.6 × 1018

The estimate is close to the calculated answer. The answer is reasonable. b. RDA for women What Is Required? You need to determine the recommended dietary allowance of riboflavin (in milligrams) for adult women. What Is Given? You know there are 1.8 × 1018 molecules of riboflavin in the RDA. You know the chemical formula for riboflavin: C17H20N4O6 Plan Your Strategy Solution 1 Use the periodic table to find the atomic molar masses of the elements inC17H20N4O6(s). Multiply the atomic molar masses by the number of atoms of each element in the compound. Add these values to calculate the molar mass of C17H20N4O6(s).



Use the Avogadro constant: 236.02 10AN = × Calculate the number of moles of C17H20N4O6(s) using the relationship

A

NnN

= .

Calculate the mass of the C17H20N4O6(s) using the relationship m n M= × . Convert the mass from grams to milligrams: 1 g = 1 × 103 mg Solution 2 The following steps can be completed in one line using the units to calculate the answer. Calculate the mass of C17H20N4O6(s) by multiplying the number of molecules

by 23

1 mol6.02 10 molecules×

and then by the molar mass, M.

Convert the mass from grams to milligrams: 1 g = 1 × 103 mg Act on Your Strategy Molar mass, M, of riboflavin, C17H20N4O6(s):

( ) ( ) ( ) ( )17 20 4 6C H N O C H N O17 20 4 6

1 12.01 g/mol 20 1.01 g/mol 4 14.01 g/mol 6 16.00 g/mol376.41 g / mol

M M M M M= + + +

= + + +

=


17 20 4 6

18

C H N O1.8 1 0 moleculesn ×

= 236.02 10 molecules×–6

/mol2.990 10 mol= ×

Mass, m, of C17H20N4O6(s):

17 20 4 6C H N O

–6

2.990 10 mol

m n M= ×

= × 376.41 g/ mol×–3

–3

–3

1.12547 10 g1.1 10 g

1.1 10 g

= ×

= ×

= × 3 1 10 mg/ g× ×

1.1 mg= The daily RDA of riboflavin for adult women is 1.1 mg.



Solution 2 Mass, m, of the C17H20N4O6(s):

17 20 4 6

18C H N O 1.8 10 moleculesm = ×

1 mol × 236.02 10 molecules× 376.41 g/ mol×

–3

–3

1.1 10 g

1.1 10 g

= ×

= × 3 1 10 mg/ g× ×

1.1 mg= The daily RDA of riboflavin for adult women is 1.1 mg. Check Your Solution Using rounded numbers to estimate the answer:

2 × 1018 × 236 110

× × 380 × 103 = 1.3 mg


65. Practice Problem (page 242) What is the mass, in grams, of a single atom of platinum? What Is Required? You must find the mass of one platinum atom. What Is Given? You know the number of atoms, N, of platinum, Pt(s): 1 Plan Your Strategy Use the atomic molar mass in the periodic table to calculate the molar mass of Pt(s). Solution 1 Use the Avogadro constant: 236.02 10AN = ×

Calculate the number of moles, n, of the Pt(s) using the relationship A

NnN

= .

Calculate the mass of the Pt(s) using the relationship m n M= × . Solution 2 The following steps can be completed in one line using the units to calculate the answer.



Calculate the number of atoms of Pt(s) by multiplying the mass

by 23

1 mol6.02 10 molcules×

, and then by the molar mass, M.

Act on Your Strategy Molar mass, M, of Pt(s): 195.08 g/mol (from the periodic table) Solution 1 Amount of moles, n, of the Pt(s):

PtA

1 atom

NnN

=

= 236.02 10 atoms×–24

/mol1.661 10 mol= ×

Mass, m, of the Pt(s):

Pt

–24

1.661 10 mol

m n M= ×

= × 195.08 g/ mol×–223.24 10 g= ×

The mass of a single platinum atom is 3.24 × 10–22 g. Solution 2 Mass, m, of the Pt(s):

Pt

1 atom

m n M= ×

=1 mol × 236.02 10 atoms×

195.08 g/ mol×

–223.24 10 g= × The mass of a single platinum atom is 3.24 × 10–22 g. Check Your Solution Use rounded numbers to estimate the answer:

1 × 236 110

× × 200 = 3.3 × 10–22 g




66. Practice Problem (page 242) Rubbing alcohol often contains propanol, C3H7OH(ℓ). Suppose that you have an 85.9 g sample of propanol. a. How many carbon atoms are in the sample? b. How many hydrogen atoms are in the sample? c. How many oxygen atoms are in the sample? What Is Required? a. You need to determine the number of carbon atoms in a sample of propanol. b. You need to determine the number of hydrogen atoms in a sample of propanol. c. You need to determine the number of oxygen atoms in a sample of propanol. What Is Given? You know the mass of the propanol: 85.9 g You know the chemical formula for propanol: C3H7OH Plan Your Strategy Use the periodic table to find the atomic molar masses of the elements in C3H7OH(ℓ). Multiply the atomic molar masses by the number of atoms of each element in the compound. Add these values to calculate the molar mass of C3H7OH(ℓ). Solution 1

Calculate the number of moles of C3H7OH(ℓ) using the relationship mnN

= .

From the chemical formula for C3H7OH(ℓ), determine the amount in moles of atoms of each element in 1 mol of molecules of C3H7OH(ℓ). Calculate the total number of moles of atoms of each element. Use the Avogadro constant: 236.02 10AN = × Calculate the number of atoms of each element in the C3H7OH(ℓ) using the relationship AN n N= × . Solution 2 The following steps can be completed in one line using the units to calculate the answer. Calculate the number of atoms of each element by multiplying the mass

by 1molar mass

, by the number of moles of carbon atoms in 1 mol of molecules

of C3H7OH(ℓ), and then by the Avogadro constant, NA.



Act on Your Strategy Molar mass, M, of C3H7OH(ℓ):

( ) ( ) ( )3 7C H OH C H O 3 8 1


M M M M= + +

= + +

=

a. number of carbon atoms Solution 1 Amount in moles, n, of C3H7OH(ℓ):

3 7C H OH

85.9 g

mnM

=

=60.11 g /mol

1.429 mol=

From the chemical formula, C3H7OH, there are 3 mol of carbon atoms per mole of propanol molecules. Amount in moles, n, of carbon atoms:

C

3 7 3 7

3 7 3 7

C 3 7

3 mol C atoms1.429 mol C H OH 1 mol C H OH

× 1 mol C H OH 1.429 mol C H OH ×3 mol C atoms

1.429 mol C H OH

n

n

n

=

=

=3 7

3 mol C atoms 1 mol C H OH

×

4.287 mol C atoms=

Number, N, of carbon atoms:

A

4.287 mol

N n N= ×

=236.02 10 C atoms

1 mol×

×

242.58 10 C atoms= ×

There are 2.58 × 1024 carbon atoms in the sample of propanol.



Solution 2 Number, N, of carbon atoms:

85.9 gN =1 mol ×

60.11 g3 mol × C atoms

1 mol molecules

236.02 10 molecules ××

1 mol242.58 10 C atoms= ×

There are 2.58 × 1024 carbon atoms in the sample of propanol. b. number of hydrogen atoms Solution 1 Amount in moles, n, of C3H7OH(ℓ):

3 7C H OH

85.9 g

mnM

=

=60.11 g /mol

1.429 mol=

From the chemical formula, C3H7OH, there are 8 mol of hydrogen atoms per mole of propanol molecules. Amount in moles, n, of H atoms in C3H7OH(ℓ):

H

3 7 3 7

3 7 3 7

H 3 7

8 mol H atoms1.429 mol C H OH 1 mol C H OH 1 mol C H OH 1.429 mol C H OH × 8 mol H atoms

1.429 mol C H OH

n

n

n

=

× =

=3 7

8 mol H atoms 1 mol C H OH

×

11.432 mol H atoms=

Number, N, of hydrogen atoms:

A

11.432 mol

N n N= ×

=236.02 10 H atoms

1 mol×

×

246.88 10 H atoms= ×

There are 6.88 × 1024 hydrogen atoms in the sample of propanol.



Solution 2 Number, N, of hydrogen atoms:

85.9 gN =1 mol ×

60.11 g8 mol × H atoms

1 mol molecules


1 mol246.88 10 H atoms= ×

There are 6.88 × 1024 hydrogen atoms in the sample of propanol. c. number of oxygen atoms Solution 1 Amount in moles, n, of C3H7OH(ℓ):

3 7C H OH

85.9 g

mnM

=

=60.11 g /mol

1.429 mol=

From the chemical formula, C3H7OH, there is 1 mol of oxygen atoms per mole of propane molecules. Amount in moles, n, of O atoms in C3H7OH(ℓ):

O

3 7 3 7

3 7 3 7

O 3 7

1 mol O atoms1.429 mol C H OH 1 mol C H OH 1 mol C H OH 1.429 mol C H OH × 1 mol O atoms

1.429 mol C H OH

n

n

n

=

× =

=3 7

1 mol O atoms 1 mol C H OH

×

1.429 mol O atoms=

Number, N, of oxygen atoms:

A

1.429 mol

N n N= ×

=236.02 10 O atoms

1 mol×

×

23 8.60 10 O atoms= ×

There are 8.60 × 1023 oxygen atoms in the sample of propanol.



Solution 2 Number, N, of oxygen atoms:

85.9 gN =1 mol ×

60.11 g1 mol × O atoms

1 mol molecules


1 mol238.60 10 O atoms= ×

There are 8.60 × 1023 oxygen atoms in the sample of propanol. Check Your Solution Use rounded numbers to estimate each answer: a. carbon atoms

90 × 1 60

× 3 × 6 × 1023 = 2.7 × 1024 carbon atoms

b. hydrogen atoms

90 × 1 60

× 8 × 6 × 1023 = 7.2 × 1024 hydrogen atoms

c. oxygen atoms

90 × 1 60

× 1 × 6 × 1023 = 9 × 1023 oxygen atoms

Each estimate is close to the calculated answer. The answers are reasonable.

67. Practice Problem (page 242) a. How many formula units are in a 3.14 g sample of aluminum sulfide, Al2S3(s)? b. How many ions (aluminum and sulfur), in total, are in this sample? a. number of formula units What Is Required? You must find the number of formula units of aluminum sulfide in a sample. What Is Given? You know the mass of the aluminum sulfide: 3.14 g You know the chemical formula for aluminum sulfide: Al2S3 Plan Your Strategy a. formula units Solution 1 Use the periodic table to find the atomic molar masses of the elements in Al2S3(s). Multiply the atomic molar masses by the number of atoms of each element in the compound.



Add these values to calculate the molar mass of Al2S3(s).

Calculate the number of moles of Al2S3(s) using the relationship mnM

= .

Use the Avogadro constant: 236.02 10AN = × Calculate the number of formula units in the sample of Al2S3(s) using the relationship AN n N= × . Solution 2 The following steps may be completed in one line using the units to calculate the answer. Calculate the number of formula units by multiplying the mass

by 1molar mass


Act on Your Strategy Solution 1 Molar mass, M, of Al2S3(s):

( ) ( )2 3Al S Al S2 3

2 26.98 g/mol 3 32.07 g/mol150.17 g/mol

M M M= +

= +

=

Amount in moles, n, of the Al2S3(s):

2 3Al S

3.14 g

mnM

=

=150.17 g

–2

/mol

2.0906 10 mol= ×

Number of formula units, N, of the Al2S3(s):

A

–2

2.0906 10 mol

N n N= ×

= ×236.02 × 10 formula units 1 mol

×

22

22


= ×

= ×

There are 1.26 × 1022 formula units of aluminum sulfide in the sample.



Solution 2 Number, N, of formula units of the Al2S3(s):

3.14 gN =1 mol ×

150.17 g

236.02 × 10 formula units 1 mol

×


There are 1.26 × 1022 formula units of aluminum sulfide in the sample. Check Your Solution Use rounded numbers to estimate the answer:

3 × 1 150

× 3 × 6 × 1023 = 1.2 × 1022 formula units

The estimate is close to the calculated answer. The answer is reasonable. b. total number of ions What Is Required? You must find the total number of ions in the sample of aluminum sulfide. What Is Given? You know from the answer to Part a that there are 1.26 × 1022 formula units of Al2S3(s). Plan Your Strategy Determine the total number of ions in one formula unit of Al2S3(s). Multiply the number of formula units of Al2S3(s) by the total number of ions per formula unit. Act on Your Strategy From the chemical formula, Al2S3, there are 2 aluminum ions and 3 sulfide ions for a total of 5 ions in one formula unit of aluminum sulfide. Total number of ions, N, in the Al2S3(s):

22

22

5 ions1.26 10 formula units 1 formula unit


N

N

=×

= ×5 ions

1 formula units×

22

6.29 10 ions= × There are a total of 6.29 × 1022 ions in the sample of aluminum sulfide.



Check Your Solution The answer matches the ratio shown in the chemical formula for aluminum sulfide. There are 5 times as many ions as there are formula units. The chemical formula for aluminum sulfide, Al2S3(s), also shows 5 ions in one formula unit. 68. Practice Problem (page 242) Which of the following two substances contains the greater mass? • 6.91 × 1022 molecules of nitrogen dioxide, NO2(g) • 6.91 × 1022 formula units of gallium arsenide, GaAs(s) What Is Required? You must find which of two given samples has a greater mass. What Is Given? You know that one sample has 6.91 × 1022 molecules of nitrogen dioxide. You know the other sample contains 6.91 × 1022 formula units of gallium arsenide. Plan Your Strategy Use the Avogadro constant: 236.02 10AN = ×

Calculate the amount in moles of each sample using the relationship A

NnN

= .

Use the periodic table to find the atomic molar masses of the elements in each of the NO2(g) and the GaAs(s). For each sample, multiply the atomic molar masses by the number of atoms of each element in the compound. Add these values to calculate the molar mass of each compound. Calculate the mass of each sample using the relationship .m n M= × Act on Your Strategy Amount in moles, n, of each sample:

A

226.91 10 particles

NnN

=

×= 236.02 10 particles× /mol

0.11478 mol=



Molar mass, M, of NO2(g):

( ) ( )2NO N O1 + 2

1 14.01 g/mol 2 16.00 g/mol 46.01 g/mol

M M M=

= +

=

Molar mass, M, of GaAs(s):

( ) ( )GaAs Ga As1 +

1 69.72 g/mol 1 74.92 g/mol 144.64 g/mol

M M M=

= +

=

Mass, m, of the NO2(g ):

2NO

0.11478 mol 46.01 g

m n M= ×

= × /mol

5.28 g =

The mass of the nitrogen dioxide is 5.28 g. Mass, m, of the GaAs(s):

GaAs

0.11478 mol

m n M= ×

= 144.64 g/ mol× 16.6 g =

The mass of the gallium arsenide is 16.6 g. The sample of gallium arsenide has the greater mass. Check Your Solution Since the number of moles of each sample is the same, the compound having the greater molar mass will have the greater mass. Gallium arsenide has the greater molar mass and the greater mass.



69. Problem (page 242) Many common dry-chemical fire extinguishers contain ammonium phosphate, (NH4)3PO4(s), as their principal ingredient. If a sample of ammonium phosphate contains 4.5 × 1021 atoms of nitrogen, what is the mass of the sample? What Is Required? You need to determine the mass of a sample of ammonium phosphate in a dry-chemical fire extinguisher. What Is Given? You know the number of atoms of nitrogen in the sample: 4.5 × 1021 You know the chemical formula for ammonium phosphate: (NH4)3PO4 Plan Your Strategy Solution 1 From the chemical formula, (NH4)3PO4, determine the number of nitrogen atoms in one formula unit of ammonium phosphate. Calculate the number of formula units of (NH4)3PO4(s). Use the Avogadro constant: 236.02 10AN = ×

Calculate the amount in moles of (NH4)3PO4(s) using the relationship A

NnN

= .

Use the periodic table to find the atomic molar masses of the elements in (NH4)3PO4(s). Multiply the atomic molar masses by the number of atoms of each element in the compound. Add these values to calculate the molar mass of (NH4)3PO4(s). Calculate the mass of (NH4)3PO4(s) using the relationship m n M= × . Solution 2 The following two steps can be completed in one line using the units to calculate the answer. Calculate the mass of (NH4)3PO4(s) by multiplying the number of atoms of

nitrogen by 1 formula unitnumber of N atoms per formula unit

, and then by

23

1 mol6.02 × 10 formula units

, and finally by the molar mass, M.



Act on Your Strategy From the chemical formula, (NH4)3PO4, there are 3 nitrogen atoms in one formula unit of ammonium phosphate. Number of formula units, N, of (NH4)3PO4(s)

4 3 421

21

1 formula unit (NH ) (PO)4.5 10 N atoms 3 N atoms

4.5 10 N atoms

N

N

=×

= × 4 3 41 formula unit (NH ) (PO) 3 N atoms

×

214 3 4

1 .5 10 formula units (NH ) (PO)= ×

Amount in moles, n, of the (NH4)3PO4(s):

( )4 43NH POA

211.5 × 10 formula units

NnN

=

= 236.02 × 10 formula units–32.4916 10

/m lo

om l= ×

Molar mass, M, of (NH4)3PO4(s):

( )

( ) ( ) ( ) ( )4 43

N H P ONH PO 3 12 1 4


M M M M M= + + +

= + + +

=

Mass, m, of the (NH4)3PO4(s):

( )4 43NH PO

–3

2.4916 10 mol

m n M= ×

= × 149.12 g/ mol×0.371547 g0.37 g

==

The mass of the sample of ammonium phosphate is 0.37 g.



Solution 2 Mass, m, of the (NH4)3PO4(s):

( )4 43

21NH PO 4.5 × 10 N atomsm =

1 formula unit × 3 N atoms

1 mol × 236.02 × 10 formula units149.12 g × 1 mol

0.371547 g0.37 g

==

The mass of the sample of ammonium phosphate is 0.37 g. Check Your Solution Use rounded numbers to estimate the answer:

5 × 1021 × 1 3

× 236 × 101 × 150 = 0.42 g

The estimate is close to the calculated answer. The answer is reasonable.

70. Problem (page 242) Place the following three substances in order, from greatest to smallest number of hydrogen atoms: • 268 mg of sucrose, C12H22O11(s) • 15.2 g of hydrogen cyanide, HCN(ℓ) • 0.0889 mol of acetic acid, CH3COOH(ℓ) What Is Required? You need to determine the number of hydrogen atoms in three different samples and identify which has the smallest number of hydrogen atoms. What is Given? You know the mass of the sucrose: 268 mg You know the mass of the hydrogen cyanide: 15.2 g You know the amount in moles of the acetic acid: 0.0889 mol Plan Your Strategy Use the periodic table to find the atomic molar masses of the elements in C12H22O11(s) and in HCN(ℓ). Multiply the atomic molar masses by the number of atoms of each element in each compound. Add these values to calculate the molar masses of C12H22O11(s) and in HCN(ℓ). Convert the mass of C12H22O11(s) from milligrams to grams: 1 mg = 1 × 10–3 g



Calculate the amount in moles of C12H22O11(s) and of HCN(ℓ) using the

relationship mnM

= .

From the chemical formulas for each of the three compounds, determine the amount in moles of hydrogen atoms in 1 mol of each compound. Using the amount in moles of each compound, calculate the total amount in moles of hydrogen atoms of each compound. Identify which compound has the smallest amount in moles of hydrogen atoms. The number of hydrogen atoms is the same proportion as the amount in moles of hydrogen atoms. Act on Your Strategy • sucrose Molar mass, M, of C12H22O11(s):

( ) ( ) ( )12 22 11C H O C H O12 22 11


M M M M= + +

= + +

=

Mass (in grams), m, of the C12H22O11(s):

12 22 11

–3C H O 268 mg 1 10 g/mg

0.268 g

m = × ×

=

Amount in moles, n, of the C12H22O11(s):

12 22 11C H O

0.268 g

mnM

=

=342.34 g

–4

/mol

7.828 10 mol= × From the chemical formula, C12H22O11, there is 22 mol of hydrogen atoms per mole of sucrose.



Total amount in moles, n, of H atoms in the C12H22O11(s): H

–412 22 11 12 22 11

–4H 12 22 11

22 mol H atoms7.828 10 mol C H O 1 mol C H O

7.828 10 mol C H O

n

n

=×

= ×12 22 11

22 mol H atoms 1 mol C H O

×

0.0172 mol H atoms=

There is a total of 0.0172 mol of hydrogen atoms in the sucrose sample. • hydrogen cyanide Molar mass, M, of HCN(ℓ):

( ) ( ) ( )HCN H C N 1 1 1


M M M M= + +

= + +

=

Amount in moles, n, of the HCN(ℓ):

HCN

15.2 g

mnM

=

=27.03 g /mol

0.562 mols=

From the chemical formula, HCN, there is 1 mol of hydrogen atoms per mole of hydrogen cyanide. Total amount in moles, n, of H atoms in the HCN(ℓ):

H

H

1 mol H atoms0.562 mol HCN 1 mol HCN

0.562 mol HCN

n

n

=

=1 mol H atoms 1 mol HCN

×

0.562 mol H atoms=

There is a total of 0.562 mol of hydrogen atoms in the hydrogen cyanide sample.



• acetic acid From the chemical formula, CH3COOH, there is 4 mol of hydrogen atoms per mole of acetic acid. Total amount in moles of H atoms in the CH3COOH(ℓ):

H

3 3

H 3

4 mol H atoms0.0889 mol CH COOH 1 mol CH COOH

0.0889 mol CH COOH

n

n

=

=3

4 mol H atoms 1 mol CH COOH

×

0.3556 mol H atoms=

There is a total of 0.3556 mol of hydrogen atoms in the acetic acid sample. The listing of the compounds from the one with the most hydrogen to the least hydrogen will be the same regardless of whether the units compared are amount in moles of H atoms or number of H atoms. Listed from the largest number to the smallest number of H atoms: HCN(ℓ) (0.562 mol H atoms) > CH3COOH(ℓ) (0.3556 mol) > C12H22O11(s) (0.0172 mol) Check Your Solution The units cancel properly and an estimate of the amount in moles of hydrogen is consistent with the calculated values. The answer is reasonable.

Section 5.2 Mass and the Mole Solutions for Selected Review Questions Student Edition page 243

4. Review Question (page 243)

Determine the amount in moles of gallium oxide, Ga2O3(s), in a 45.2 g sample. What Is Required? You need to determine the amount in moles of a sample of gallium oxide. What Is Given? You know the chemical formula for gallium oxide: Ga2O3 You know the mass of the Ga2O3(s): 45.2 g



Plan Your Strategy Use the periodic table to find the atomic molar masses of the elements in Ga2O3(s). Multiply the atomic molar masses by the number of atoms of each element in the compound. Add these values to calculate the molar mass of Ga2O3(s).

Calculate the amount in moles of the Ga2O3(s) using the relationship mnM

= .

Act on Your Strategy Molar mass, M, of Ga2O3(s):

( ) ( )2 3Ga O Ga O 2 3

2 69.72 g/mol 3 16.00 g/mol 187.44 g/mol

M M M= +

= +

=

Amount in moles, n, of Ga2O3(s):

2 3Ga O

45.2 g

mnM

=

=187.44 g /mol

0.241 mol=

There is 0.241 mol of gallium oxide in the sample. Check Your Solution The molar mass has been determined correctly. The answer seems reasonable and shows the correct number of significant digits.

5. Review Question (page 243) What is the mass of 3.2 × 102 mol of cerium nitrate, Ce(NO3)3(s)? What Is Required? You need to determine the mass of a sample of cerium nitrate. What Is Given? You know the amount in moles of cerium nitrate: 3.2 × 102 mol You know the chemical formula for cerium nitrate: Ce(NO3)3



Plan Your Strategy Use the periodic table to find the atomic molar masses of the elements in Ce(NO3)3(s). Multiply the atomic molar masses by the number of atoms of each element in the compound. Add these values to calculate the molar mass of Ce(NO3)3(s). Calculate the mass of Ce(NO3)3(s) using the relationship m n M= × . Act on Your Strategy Molar mass, M, of Ce(NO3)3(s):

( )

( ) ( )3 3

Ce N OCe NO 1 3 9

1(140.12 g/mol) 3 14.01 g/mol 9 16.00 g/mol326.15 g/mol

M M M M= + +

= + +

=

Mass, m, of the Ce(NO3)3(s):

( )3 3Ce NO

2

3.2 10 mol

m n M= ×

= × 326.15 g/ mol×5

5

1.04368 10 g1.0 10 g

= ×

= ×

The mass of the cerium nitrate is 1.0 × 105 g. Check Your Solution. The units cancel properly. Check to see that correct atomic masses have been used. Using rounded numbers: 3 × 102 × 330 = 9.9 × 104

The estimate is close to the calculated value. The answer is reasonable.

6. Review Question (page 243) Calculate the amount in moles of strontium chloride, SrCl2(s), in a 28.6 kg sample. What Is Required? You need to determine the amount in moles of a sample of strontium chloride. What Is Given? You know the chemical formula for strontium chloride: SrCl2 You know the mass of the strontium chloride: 28.6 kg



Plan Your Strategy Use the periodic table to find the atomic molar masses of the elements in SrCl2(s). Multiply the atomic molar masses by the number of atoms of each element in the compound. Add these values to calculate the molar mass of SrCl2(s). Convert the mass of the SrCl2(s) from kilograms to grams: 1 kg = 1 × 103 g

Calculate the amount in moles of the SrCl2(s) using the relationship mnM

= .

Act on Your Strategy Molar mass, M, of SrCl2(s):

( ) ( )2SrCl Sr Cl 1 2

1 87.62 g/mol 2 35.45 g/mol 158.52 g/mol

M M M= +

= +

=

Mass (in grams), m, of the SrCl2(s):

2SrCl 28.6 kgm = 3 1 10 g/ kg× ×4 2.86 10 g= ×

Amount in moles, n, of the SrCl2(s):

2SrCl

42.86 10 g

mnM

=

×=

158.52 g2

/mol

1.80 10 mol= ×

There is 1.80 × 102 mol of strontium chloride in the sample. Check Your Solution The molar mass has been determined correctly. The answer seems reasonable and shows the correct number of significant digits.

7. Review Question (page 243) What is the mass of 0.68 mol of iron(III) sulfate, Fe2(SO4)3(s)? What Is Required? You need to determine the mass of a sample of iron(III) sulfate.



What Is Given? You know the chemical formula for iron(III) sulfate: Fe2(SO4)3 You know the amount in moles of the iron(III) sulfate: 0.68 mol Plan Your Strategy Use the periodic table to find the atomic molar masses of the elements in Fe2(SO4)3(s). Multiply the atomic molar masses by the number of atoms of each element in the compound. Add these values to calculate the molar mass of the Fe2(SO4)3(s). Calculate the mass of the Fe2(SO4)3(s) using the relationship m n M= × . Act on Your Strategy Molar mass, M, of Fe2(SO4)3(s):

( ) ( ) ( )2 4 3Fe (SO ) Fe S O 2 3 12


M M M M= + +

= + +

=

Mass, m, of the Fe2(SO4)3(s):

( )2 4 3Fe SO

0.68 mol

m n M= ×

= 399.91 g/ mol×2

2

2.719 10 g2.7 10 g

= ×

= ×

The mass of the iron(III) sulfate is 2.7 × 102 g. Check Your Solution The units cancel properly. Check to see that correct atomic masses have been used. Using rounded numbers: 7 × 10–1 × 330 = 231 g The estimate is close to the calculated value. The answer is reasonable.

8. Review Question (page243) What is the mass of 2.9 × 1026 molecules of dinitrogen pentoxide, N2O5(g)? What Is Required? You need to determine the mass of a sample of dinitrogen pentoxide.



What Is Given? You know the chemical formula for dinitrogen pentoxide: N2O5 You know the number of molecules of the dinitrogen pentoxide: 2.9 × 1026 Plan Your Strategy Use the periodic table to find the atomic molar masses of the elements inN2O5(g). Multiply the atomic molar masses by the number of atoms of each element in the compound. Add these values to calculate the molar mass of the compound. Solution 1

Calculate the amount in moles of the N2O5(g) using the relationship A

NnN

=.

Calculate the mass of the N2O5(g) using the relationship m n M= × . Solution 2 The following steps can be completed in one line using the units to calculate the answer. Calculate the mass of N2O5(g) by multiplying the given number of molecules

of dinitrogen pentoxide by 23

1 mol6.02 10 molecules×

and then by the molar mass,

M. Act on Your Strategy Molar mass, M, of N2O5(g):

( ) ( )2 5N O N O 2 5

2 14.01 g/mol 5 16.00 g/mol108.02 g/mol

M M M= +

= +

=

Solution 1 Amount in moles, n, of the N2O5(g):

2 5N OA


NnN

=

= 236.02 10 molecules×2

/mol4.817 10 mol= ×



Mass, m, of the N2O5(g): 2 5N O

2

4.817 10 mol

m n M= ×

= × 108.02 g/ mol×4

4

5.2033 10 g5.2 10 g

= ×

= ×

The mass of the sample of dinitrogen pentoxide is 5.2 × 104 g. Solution 2 Mass, m, of the N2O5(g):

2 5

26N O 2.9 10 moleculesm = ×

1 mol × 236.02 10 molecules×108.02 g 1 mol

×

4

4

5.2033 10 g5.2 10 g

= ×

= ×

The mass of the dinitrogen pentoxide is 5.2 × 104 g. Check Your Solution Use rounded numbers to estimate the answer:

3 × 1026 × 23

1 6 10×

× 110 = 5.5 × 104 g


9. Review Question (page243) Calculate the number of oxygen atoms in 15.2 g of trinitrotoluene, C7H5(NO2)3(s). What Is Required? You need to determine the number of oxygen atoms in a sample of trinitrotoluene. What Is Given? You know the mass of the trinitrotoluene: 15.2 g You know the chemical formula for trinitrotoluene: C7H5(NO2)3



Plan Your Strategy Use the periodic table to find the atomic molar masses of the elements inC7H5(NO2)3(s). Multiply the atomic molar masses by the number of atoms of each element in the compound. Add these values to calculate the molar mass of the compound. Solution 1 Calculate the amount in moles of the C7H5(NO2)3(s) using the relationship

mnN

= .

From the chemical formula, C7H5(NO2)3, determine the amount in moles of oxygen in 1 mol of trinitrotoluene. Calculate the amount in moles of oxygen atoms. Use the Avogadro constant: 236.02 10AN = × Calculate the number of oxygen atoms using the relationship .AN n N= × Solution 2 The following steps can be completed in one line using the units to calculate the answer. Calculate the number of oxygen atoms by multiplying the mass

by 1molar mass

, then by number of moles of oxygen atoms1 mol of molecules

, and then by the

Avogadro constant 23A( 6.02 10 )N = × .

Act on Your Strategy Molar mass, M, of C7H5(NO2)3(s):

( ) ( ) ( ) ( )7 5 2 3C H (NO ) C H N O 7 5 3 6


M M M M M= + + +

= + + +

= Solution 1 Amount in moles, n, of the C7H5(NO2)3(s):

( )7 5 2 3C H NO

15.2 g

mnN

=

=227.15 g

–2

/mol

6.6916 10 mol= ×



From the chemical formula, C7H5(NO2)3, there are 6 mol of oxygen atoms per mole of trinitrotoluene. Amount in moles, n, of O atoms in C7H5(NO2)3(s):

O–2

7 5 2 3 7 5 2 3

–2O 7 5 2 3

6 mol O atoms6.6916 10 mol C H (NO ) 1 mol C H (NO )

6.6916 10 mol C H (NO )

n

n

=×

= ×7 5 2 3

6 mol O atoms 1 mol C H (NO )

×

0.40149 mol O atoms=

Number, N, of oxygen atoms in the C7H5(NO2)3(s):

0.40149 molAN n N= ×

= 23O atoms 6.02 10 atoms/ mol× ×232.42 10 O atoms= ×

There are 2.42 × 1023 atoms of oxygen in the sample of trinitrotoluene. Solution 2 Number of oxygen atoms, N, in the C7H5(NO2)3(s):

15.2 gN =

1 mol ×227.15 g

6 mol × O atoms1 mol

236.02 10 atoms 1 mol×

×

232.42 10 atoms= ×

There are 2.42 × 1023 atoms of oxygen in the sample of trinitrotoluene. Check Your Solution Use rounded numbers to estimate the answer:

15 × 1230

× 6 × 6 × 1023 = 2.3 × 1023 oxygen atoms


10. Review Question (page243) Which has more sulfur atoms: 13.4 g of potassium thiocyanate, KSCN(s), or 0.067 mol of aluminum sulfate, Al2(SO4)3(s)? What Is Required? You need to determine the number of sulfur atoms in two samples and identify which sample has the greater number.



What Is Given? You know the chemical formula for potassium thiocyanate: KSCN You know the chemical formula for aluminum sulfate: Al2(SO4)3 You know the mass of the potassium thiocyanate: 13.4 g You know the amount in moles of the aluminum sulfate : 0.067 mol Plan Your Strategy Use the periodic table to find the atomic molar masses of the elements in each of KSCN(s) and Al2(SO4)3(s). For each compound, multiply the atomic molar masses by the number of atoms of each element in the compound. Add these values to calculate the molar mass of each compound. Solution 1

Calculate the amount in moles of the KSCN(s) using the relationship mnM

= .

From the chemical formulas, determine the amount in moles of sulfur atoms in 1 mol of each compound. Calculate the amount in moles of sulfur atoms in KSCN(s) and Al2(SO4)3(s). The amount in moles of sulfur atoms is in the same proportion as the number of sulfur atoms. Compare the amounts of sulfur atoms in the two compounds and identify which has the greater number. Solution 2 The following steps can be completed in one line using the units to calculate the answer.

Calculate the number of sulfur atoms by multiplying the mass by 1molar mass

,

then by number of moles of sulfur atoms1 mol of molecules

, and then by the Avogadro constant 23

A( 6.02 10 )N = × . Act on Your Strategy Molar mass, M, of KSCN(s):

( ) ( ) ( ) ( )KSCN K S C N 1 1 1 1


M M M M M= + + +

= + + +

=



Molar mass, M, of Al2(SO4)3(s):

( ) ( ) ( )2 4 3Al (SO ) Al S O 2 3 12


M M M M= + +

= + +

=

Solution 1 Amount in moles, n, of the KSCN(s):

KSCN

13.4 g

mnM

=

=97.19 g /mol

0.1387 mol=

From the chemical formula, KSCN, there is 1 mol of sulfur atoms in 1 mol of the compound. Amount in moles, n, of S atoms in the KSCN(s):

S

S

1 mol S atoms0.138787 mol KSCN 1 mol KSCN

0.138787 mol KSCN

n

n

=

=1 mol S atoms

mol KSCN×

0.13787 mol S atoms0.138 mol S atoms

==

There is 0.138 mol of sulfur atoms in the sample of potassium cyanide. From the chemical formula, Al2(SO4)3, there is 3 mol of sulfur atoms in 1 mol of aluminum sulfate. Amount in moles, n, of S atoms in the Al2(SO4)3(s):

S

2 4 3 2 4 3

S 2 4 3

3 mol S atoms0.067 mol Al (SO ) 1 mol Al (SO )

0.067 mol Al (SO )

n

n

=

=2 4 3

3 mol S atoms 1 mol Al (SO )

×

0.201 mol S atoms=

There is 0.201 mol of sulfur atoms in the sample of aluminum sulfate.



Solution 2 Amount in moles, n, of S atoms in the KSCN(s):

S1 mol 1 mol S atoms 13.4 g

97.19 g 1 mol KSCN(s)0.138 mol S atoms

n = × ×

=

Amount in moles, n, of S atoms in 0.067 mol of the Al2(SO4)3

S 0.067 moln =3 mol S atoms

mol×

0.201 mol S atoms=

The amount in moles of sulfur atoms is greater in the sample of aluminum sulfate. Check Your Solution Use rounded numbers to estimate the amount in moles of S atoms in the sample of KSCN(s):

13 × 1100

× 1 = 0.13 mol of sulfur atoms

Use rounded numbers to estimate the amount in moles of S atoms in the sample of Al2(SO4)3: 0.07 × 3 = 0.21 mol of sulfur atoms Each estimate is close to the calculated answer. The answer is reasonable.

11. Review Question (page 243) Which has a greater amount in moles: a sample of sulfur trioxide containing 4.9 × 1022 atoms of oxygen or a 4.9 g sample of carbon dioxide? What Is Required? You need to determine the amount in moles of two samples and identify which sample has the greater amount. What Is Given? You know the chemical formula for sulfur trioxide: SO3 You know the chemical formula for carbon dioxide: CO2 You know the number of oxygen atoms in the sample of sulfur trioxide: 4.9 × 1022 atoms You know the mass of the carbon dioxide: 4.9 g



Plan Your Strategy Solution 1 For the sulfur trioxide: From the chemical formula, SO3, determine the number of oxygen atoms per molecule of sulfur trioxide. Calculate the number of SO3(g) molecules. Use the Avogadro constant: 236.02 10AN = × .

Calculate the amount in moles of the SO3(g) using the relationship A

NnN

= .

For the carbon dioxide: Use the periodic table to find the atomic molar masses of the elements in CO2(g). Multiply the atomic molar masses by the number of atoms of each element in the compound. Add these values to calculate the molar mass of the compound.

Calculate the amount in moles of the CO2(g) using the relationship mnM

= .

Compare the amounts in moles of the two compounds and identify which has the greater amount. Solution 2 The following steps can be completed in one line using the units to calculate the answer. For the sulfur trioxide: Calculate the amount in moles of SO3(g) by multiplying the number of atoms

of oxygen by 31 molecule SOnumber of O atoms per molecule

, and then by

23

1 mole6.02 × 10 molecules

.

For the carbon dioxide: Calculate the amount in moles by multiplying the mass of CO2(g)

by 1molar mass

.

Act on Your Strategy Solution 1 For the sulfur trioxide: From the chemical formula, SO3, there are 3 atoms of oxygen in one molecule of sulfur trioxide.



Number of molecules, N, of SO3(g):

2

22

32

1 molecule S4.9 × 10 O atoms 3 O atoms

4.9 × 10 O

O

atoms

N

N =

=

31 molecule 3 O a

SOtoms

×

2231.6333 10 molecules SO= ×

Amount in moles, n, of SO3(g):

3A

2

O

2

S


NnN

=

= 23 6.02 × 10 molecules0.027131 mol0.02

/m

7

ol

mol ==

The amount in moles of the sulfur trioxide is 0.027 mol. For the carbon dioxide: Molar mass, M, of CO2(g):

( ) ( )2CO C O1 2

1 12.01 g/mol 2 16.00 g/mol44.01 g/mol

M M M= +

= +

=

Amount in moles, n, of the CO2(g):

2CO

4.9 g

mnM

=

=44.01g /mol

0.11 mol=

The amount in moles of the carbon dioxide is 0.11 mol. The amount in moles of the sample of carbon dioxide is greater.



Solution 2 For the sulfur trioxide: Amount in moles, n, of the SO3(g)

3

22SO 4.9 10 O atomsn = × 31 molecule SO

×3 O atoms 23

3

1 mol 6.02 10 molecules SO

××

0.027131 mol 0.027 mol

==

The amount in moles of the sulfur trioxide is 0.027 mol. For the carbon dioxide: Amount in moles, n, of the CO2(g):

2CO

4.9 g

mnM

=

=44.01g /mol

0.1113 mol 0.11 mol==

The amount in moles of the carbon dioxide is 0.11 mol. The amount in moles of the sample of carbon dioxide is greater. Check Your Solution Use rounded numbers to estimate the amount in moles of the SO3(g):

5 × 1022 × 13

× 23

16 10×

= 2.8 × 10–2 mol

Use rounded numbers to estimate the amount in moles of the CO2(g): 540

= 0.125 mol

In each case, the estimate is close to the calculated answer. The answer is reasonable.



12. Review Question (page 243) Imagine that you are an environmental chemist who is testing drinking water for water hardness. A sample of water you test has 3.5 × 10–2 mg of dissolved calcium carbonate, CaCO3(aq). a. How many formula units of calcium carbonate are in the sample? b. How many oxygen atoms are in the calcium carbonate in the sample? What Is Required? a. You have to find the number of formula units of calcium carbonate in the sample. b. You have to find the number of oxygen atoms in the calcium carbonate in the sample. What Is Given? You know the chemical formula for calcium carbonate: CaCO3 You know the mass of the calcium carbonate: 3.5 × 10–2 mg a. formula units of calcium carbonate Plan Your Strategy Use the periodic table to find the atomic molar masses of the elements in CaCO3(aq). Multiply the atomic molar masses by the number of atoms of each element in the compound. Add these values to calculate the molar mass of the compound. Convert the mass of the CaCO3(aq) from milligrams to grams: 1 mg = 1 × 10−3 g Solution 1

Calculate the number of moles of CaCO3(aq) using the relationship mnM

= .

Use the Avogadro constant: 236.02 10AN = × Calculate the number of formula units in the sample of CaCO3(aq) using the relationship AN n N= × . Solution 2 The following steps may be completed in one line using the units to calculate the answer. Calculate the number of formula units by multiplying the mass

by 1molar mass




Act on Your Strategy Molar mass, M, of CaCO3(aq):

( ) ( ) ( )3CaCO Ca C O1 1 3


M M M M= + +

= + +

=

Mass (in grams), m, of the CaCO3(aq):

3

–2CaCO 3.5 10 mgm = × 3 1 10 g/ mg−× ×

53.5 10 g−= ×

Solution 1 Amount in moles, n, of the CaCO3(aq):

3CaCO

–53.5 10 g

mnM

=

×=

100.09 g–7

/mol

3.4968 10 mol= ×

Number of formula units, N, of the CaCO3(aq):

A

–7

3.4968 10 mol

N n N= ×

= ×236.02 × 10 formula units 1 mol

×


There are 2.1 × 1017 formula units of calcium carbonate in the sample. Solution 2 Number of formula units, N, of the CaCO3(aq):

–53.5 10 gN = ×1

100.09 g×

/ mol

236.02 × 10 formula units 1 mol

×

17 2.1 10 formula units= ×

There are 2.1 × 1017 formula units of calcium carbonate in the sample.



b. number of oxygen atoms Plan Your Strategy Multiply the number of formula units of the CaCO3(s) by the number of oxygen atoms in one formula unit. Act on Your Strategy From the chemical formula, CaCO3, there are 3 oxygen atoms for every formula unit of calcium carbonate. Number of oxygen atoms, N, in the CaCO3(s):

173 3

173

3 O atoms2.1 10 formula units CaCO 1 formula unit CaCO

2.1 10 formula units CaCO

N

N

=×

= ×3

3 O atoms 1 formula unit CaCO

×

176.3 10 O atoms= ×

There are 6.3 × 1017 atoms of oxygen in the calcium carbonate sample. Check Your Solution Check to see that the correct atomic molar masses have been used. Using rounded numbers, estimate the answers:

a. 3.5 × 10–2 × 10–3 × 1100

× 6 × 1023 = 2.1 × 1017 formula units of calcium

carbonate b. 2 × 1017 × 3 = 6 × 1017 oxygen atoms The estimates are close to the calculated answers. The answers are reasonable.

13. Review Question (page 243) Again, imagine that you are the chemist who is testing water samples in the previous question. This time, however, you are testing for lead content downstream from a battery manufacturing plant. Health Canada suggests that drinking water should have a maximum lead content of 0.010 mg/L of water. If your test reveals that a 1 L water sample contains 3.1 × 1017 atoms of lead, is the water you tested safe to drink? Explain. What Is Required? You need to determine the mass of lead in a sample of water to see if it exceeds the limit of 0.010 mg/L of water.



What Is Given? You know the number of atoms of lead, Pb(s), in the sample: 3.1 × 1017 atoms You know the maximum safe content for lead in drinking water: 0.010 mg/L Plan Your Strategy Solution 1 Use the periodic table to determine the atomic molar mass of Pb(s).

Calculate the amount in moles of Pb(s) using the relationship A

NnN

= .

Calculate the mass of the sample of Pb(s) using the relationship m n M= × . Convert the mass from grams to milligrams: 1 g = 1 × 103 mg. Solution 2 The following steps may be completed in one line using the units to calculate the answer.

Multiply the number of Pb atoms by 23

1 mol6.02 × 10 atoms

, then by molar mass1 mol

,

and then by 31 × 10 mg

1 g.

Act on Your Strategy Molar mass, M, of Pb(s): 207.2 g/mol (from the periodic table) Amount in moles, n, of the Pb(s):

PbA

173.1 × 10 atoms

NnN

=

= 236.02 × 10 atoms–7

/mol5.149 10 mol= ×

Mass, m, of the Pb(s):

Pb

–7

5.149 10 mol

m n M= ×

= × 207.2 g/ mol×–4

–4

1.06692 10 g1.066 10 g

= ×

= ×

The mass (in milligrams), m, of the lead in the sample:



–4Pb 1.066 10 gm = ×

31 × 10 mg 1 g

×

–1

1.1 10 mg= ×

Solution 2 Mass, m, of the Pb(s):

Pb1 molm = 236.02 × 10 atoms

207.2 g ×

1 mol

31 × 10 mg × 1 g

–11.06692 10 g= ×31 × 10 mg

1 g×

–11.1 10 mg= ×

The amount of lead exceeds the safe limit of 1.0 × 10–2 mg and the water is not safe to drink. Check Your Solution Use rounded numbers to estimate the answer:

3 × 1017 × 23

16 × 10

× 210 × 103 = 0.10 mg of lead atoms


15. Review Question (page 243) Methyl salicylate, C6H4(OH)COOCH3(s), is used in many consumer products, such as mouthwash, as a flavouring. A mouthwash sample contains 1.38 × 1018 molecules of methyl salicylate. What is the mass of the methyl salicylate in the sample? What Is Required? You need to determine the mass of methyl salicylate in a sample of mouthwash. What Is Given? You know the chemical formula for methyl salicylate: C6H4(OH)COOCH3 You know the number of molecules of the methyl salicylate: 1.38 × 1018 molecules



Plan Your Strategy Solution 1 Use the periodic table to find the atomic molar masses of the elements in C6H4(OH)COOCH3(s). Multiply the atomic molar masses by the number of atoms of each element in the compound. Add these values to calculate the molar mass of the compound. Use the Avogadro constant: 236.02 10AN = × Calculate the amount in moles of the C6H4(OH)COOCH3(s) using the

relationship A

NnN

= .

Calculate the mass of the C6H4(OH)COOCH3(s) using the relationship m n M= × .

Solution 2 The following steps can be completed in one line using the units to calculate the answer. Calculate the mass of the methyl salicylate by multiplying the given number of

molecules of the C6H4(OH)COOCH3(s) by 23


and then by

the molar mass, M. Act on Your Strategy Molar mass, M, of C6H4(OH)COOCH3(s):

( )

( ) ( ) ( )6 4 3 C H OC H OH COOCH 8 8 3


M M M M= + +

= + +

=

Amount in moles, n, of the C6H4(OH)COOCH3(s):

( )6 4 3C H OH COOCHA


NnN

=

= 236.02 × 10 molecules–6

/mol2.292 10 mol= ×



Mass, m, of the C6H4(OH)COOCH3(s):

( )6 4 3C H OH COOCH

–6

2.292 10 mol

m n M= ×

= × 152.16 g/ mol×–4

–4

3.488 10 g3.49 10 g

= ×

= ×

The mass of the methyl salicylate in a sample of mouthwash is 3.49 × 10–4 g. Solution 2 Mass, m, of the C6H4(OH)COOCH3(s):

( )6 4 3

18C H OH COOCH 1.38 10 moleculesm = ×

1 mol × 236.02 × 10 molecules152.16 g 1 mol

×

–43.49 10 g = × The mass of the methyl salicylate in a sample of mouthwash is 3.49 × 10–4 g. Check Your Solution Use rounded numbers to estimate the answer:

1.4 × 1018 × 23

16.02 × 10

× 150 = 3.5 × 10–4 g


16. Review Question (page 243) Determine the order of the following three substances, from smallest to greatest number of carbon atoms: 5.6 × 1023 molecules of benzoic acid, C6H5COOH(s); 1.3 mol of acetic acid, CH3COOH(ℓ); 0.17 kg of oxalic acid, HOOCCOOH(s). What Is Required? You need to determine the number of carbon atoms in three different samples and list the compounds from the smallest to greatest number of carbon atoms. What Is Given? You know the number of molecules of benzoic acid: 5.6 × 1023 molecules You know the amount in moles of acetic acid:1.3 mol You know the mass of the oxalic acid): 0.17 kg



Plan Your Strategy Solution 1 For the benzoic acid: From the chemical formula C6H5COOH, determine the number of carbon atoms per molecule of benzoic acid. Calculate the number of carbon atoms in 5.6 × 1023 molecules of C6H5COOH(s). For the acetic acid: From the chemical formula CH3COOH, determine the amount in moles of carbon atoms per mole of acetic acid molecules. Calculate the number of carbon atoms in the CH3COOH(ℓ) using the relationship AN n N= × . For the oxalic acid: Convert the mass of HOOCCOOH(s) from kilograms to grams: 1 kg = 1 × 103 g Use the periodic table to find the atomic molar masses of the elements in HOOCCOOH(s). Multiply the atomic molar masses by the number of atoms of each element in the compound. Add these values to calculate the molar mass of the compound. Calculate the amount in moles of the HOOCCOOH(s) using the relationship

mnM

= .

Calculate the number of molecules of the HOOCCOOH(s) using the

conversion factor 236.02 10 molecules

1 mol× .

From the chemical formula, determine the number of carbon atoms in one molecule of HOOCCOOH(s). Calculate the total number of carbon atoms by multiplying the number of molecules of HOOCCOOH(s) by the number of carbon atoms per molecule. Compare the number of carbon atoms in each compound and list the compounds from smallest to greatest number of carbon atoms. Solution 2 Calculate the number of carbon atoms using the conversion factors

236.02 10 units1 mol× and 1 mol

molar mass.



Act on Your Strategy Solution 1 For the benzoic acid: From the chemical formula, C6H5COOH, there are 7 carbon atoms per molecule of benzoic acid. Number of C atoms, N, in the C6H5COOH(s):

23

23


5.6 10 molecules

N=

×


×

243.9 10 C atoms = ×

There are 3.9 × 1024 carbon atoms in the sample of benzoic acid. For the acetic acid: From the chemical formula CH3COOH, there are 2 carbon atoms in one molecule of acetic acid. Number of molecules, N, of the CH3COOH(ℓ):

1.3 molAN n N= ×


Number of C atoms, N, in the CH3COOH(ℓ):

23

23


7.826 10 molecules

N

N

=×

= ×2 C atoms

1 molecule×

241.6 10 C atoms= ×

There are 1.6 × 1024 carbon atoms in the sample of acetic acid. For the oxalic acid: Mass (in grams), m, of the HOOCCOOH(s):

HOOCCOOH 0.17 kgm = 3 1 10 g/ kg× ×21.70 10 g= ×



Molar mass, M, of HOOCCOOH(s):

( ) ( ) ( )HOOCCOOH C H O 2 2 4


M M M M= + +

= + +

=

Amount in moles, n, of the HOOCCOOH(s):

HOOCCOOH

21.70 10 g

mnN

=

×=

90.04 g /mol

1.888 mol=

Number of molecules, N, of the HOOCCOOH(s):

A

1.888 mol

N n N= ×


From the chemical formula, HOOCCOOH(s), there are 2 carbon atoms in one molecule of oxalic acid. Number of C atoms, N, in HOOCCOOH(s):

24

24


1.1366 10 molecules

N

N

=×

= ×2 C atoms

1 molecule×

242.3 10 C atoms= ×

There are 2.3 × 1024 carbon atoms in the sample of oxalic acid. Solution 2 For the benzoic acid: Number of C atoms, N, in 5.6 × 1023 molecules of C6H5COOH(s):

23

23


5.6 10 molecules

N=

×


×

243.9 10 C atoms = ×



There are 3.9 × 1024 carbon atoms in the sample of benzoic acid. For the acetic acid: Number of C atoms, N, in 1.3 mol of CH3COOH(ℓ):

1.3 molN =2 mol × C atoms

1 mol

236.02 10 atoms 1 mol×

×

241.6 10 C atoms= ×

There are 1.6 × 1024 carbon atoms in the sample of acetic acid. For the oxalic acid: Mass (in grams), m, of the HOOCCOOH(s):

HOOCCOOH 0.17 kgm = 3 1 10 g/ kg× ×21.70 10 g= ×

Number of C atoms, N, in the HOOCCOOH(s):

21.70 10 gN = ×1 mol ×

90.04 g2 mol ×

C atoms1 mol

236.02 10 atoms 1 mol×

×

242.3 10 C atoms= ×

There are 2.3 × 1024 carbon atoms in the sample of oxalic acid. Listing the compounds from the smallest number of carbon atoms to the greatest number: CH3COOH(ℓ)(1.6 × 1024) < HOOCCOOH(s)(2.3 × 1024) < C6H5COOH(s)(3.9 × 1024) Check Your Solution Use rounded numbers to estimate the number of carbon atoms: In the sample of C6H5COOH(s): 6 × 6 × 1023 = 3.6 × 1024

In the sample of CH3COOH(ℓ): 1 × 2 × 6 ×1023 = 1.2 × 1024 In the sample of HOOCCOOH(s):

2 × 102 × 190

× 2 × 6 × 1023 = 2.7 × 1024

In each case, the units cancel properly and an estimate of the number of atoms of carbon is consistent with the calculated value. Each answer seems reasonable.

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