04 Unit 1 SB - Quia · PDF fileThere are 3.24 moles of methane molecules ... There are 1.2 1024 atoms in one mole of a ... There are 4.81 10–1 mol of hydrogen bromide molecules

NEL Section 2.1 Student Book Solutions 39

GETTING STARTED

TRY THIS ACTIVITY: THE MINERAL CONTENT OF MILK

(Page 79)

Observations (answers will vary):

Whole milk Skim milk

Mass of evaporating dish (g) 32.791 32.794

Mass of milk residue (minerals) + evaporating dish (g) 32.810 32.809

Mass of milk residue (minerals) (g) 0.019 0.018

(a) There does not seem to be a significant difference between the mineral content of whole milk and that of skim milk. (b) Whole milk and skim milk have approximately the same mineral content by mass.

2.1 AMOUNTS IN CHEMISTRY: MASS, MOLES, AND MOLAR MASS

PRACTICE

(Page 81)

Understanding Concepts 1.

The relative atomic mass of hydrogen is 1.01 u. 2. The atomic mass unit is the standard mass used to determine the relative masses of all elements. The atomic mass

unit, u, is equal to one-twelfth the mass of a carbon-12 atom. 3. (a)

The molecular mass of hydrogen bromide is 80.91 u.

(b)

The molecular mass of glucose is 180.18 u. (c)

The formula unit mass of sodium hydrogen carbonate is 84.01 u.

H

H

(percent abundance hydrogen-1 atomic mass hydrogen-1)

(percent abundance hydrogen-2 atomic mass hydrogen-2)

(0.99985 1.01 u) (0.00015 2.01 u)

1.01 u 0.000302 u

1.01 u

�

�

= ×+ ×

= × + ×= +=

HBr H Br

HBr

1.01 u 79.90 u

80.91 u

� � �

�

= += +=

6 12 6

6 12 6

C H O C H O

C H O

6( ) 12( ) 6( )

6(12.01 u) 12(1.01 u) 6(16.00 u)

72.06 u 12.12 u 96.00 u

180.18 u

� � � �

�

= + +

= + += + +=

3

3

NaHCO Na H C O

NaHCO

3( )

22.99 u 1.01 u 12.01 u 3(16.00 u)

84.01 u

� � � � �

�

= + + +

= + + +=

40 Unit 2 Student Book Solutions NEL

(d)

The formula unit mass of ammonium phosphate is 149.12 u.

PRACTICE

(Page 85)

Understanding Concepts 4. (a)

The molar mass of benzene is 78.12 g.

(b)

The molar mass of barium nitrate is 261.35 g.

(c)

The molar mass of potassium phosphate is 212.27 g.

(d)

The molar mass of ammonia is 17.04 g.

PRACTICE

(Page 86)


The molar mass of tin is 118.7 g/mol.

4 3 4

4 3 4

(NH ) PO N H P O

(NH ) PO

3( ) 12( ) ( ) 4( )

3(14.01 u) 12(1.01 u) (30.97 u) 4(16.00 u)

42.03 u 12.12 u 30.97 u 64.00 u

149.12 u

� � � � �

�

= + + +

= + + += + + +=

6 6

6 6

C H C H

C H

6( ) 6( )

6(12.01 g) 6(1.01 g)

72.06 g 6.06 g

78.12 g

� � �

�

= +

= += +=

3 2

3 2

Ba(NO ) Ba N O

Ba(NO )

( ) 2( ) 6( )

(137.33 g) 2(14.01 g) 6(16.00 g)

137.33 g 28.02 g 96.00 g

261.35 g

� � � �

�

= + +

= + += + +=

3 4

3 4

K PO K P O

K PO

3( ) ( ) 4( )

3(39.10 g) (30.97 g) 4(16.00 g)

117.3 g 30.97 g 64.00 g

212.27 g

� � � �

�

= + +

= + += + +=

3

3

NH N H

NH

( ) 3( )

(14.01 g) 3(1.01 g)

14.01 g 3.03 g

17.04 g

� � �

�

= +

= += +=

Sn

Sn

massmolar mass

amount119.88 g

1.010 mol118.7 g/mol

�

�

=

=

=


6.

The molar mass of platinum is 195.1 g/mol.

7.

The molar mass of the element is 12.01 g/mol. From the periodic table, this element is carbon.

PRACTICE

(Page 88)

Understanding Concepts 8. mass of 1 package of cotton swabs = 240 g (key value)

mass of 1 cotton swab = 0.480 g (conversion factor equation)

There are 500 cotton swabs in a package that contains 240 g of cotton swabs. 9. mass of 1 package of rubber bands = 729 g (key value)

mass of 1 rubber band = 0.81 (conversion factor equation)

There are 900 rubber bands in a package that contains 729 g of rubber bands. 10. Method 1:

mass of 1 box of playing cards = 558 g (key value) mass of 1 playing card = 1.50 g mass of 1 dozen playing cards = 1.50 g � 12 mass of 1 dozen playing cards = 18 g (conversion factor equation)

There are 31 dozen playing cards in a box that contains 558 g of playing cards.

Method 2: mass of 1 box of playing cards = 558 g (key value) mass of 1 playing card = 1.50 g (conversion factor value)

number of playing cards in box = 558 g � 1 playing card

1.50 g = 372 playing cards (key value)

1 dozen playing cards = 12 playing cards (conversion factor equation)

There are 31 dozen playing cards in a box that contains 558 g of playing cards.

Pt

Pt

massmolar mass

amount49.74 g

0.2550 mol195.1 g/mol

�

�

=

=

=

massmolar mass

amount26.66 g

2.220 mol12.01 g/mol

�

�

=

=

=

1 cotton swabnumber of cotton swabs in a package = 240 g

0.480 g

number of cotton swabs in a package = 500 cotton swabs

×

1 rubber bandnumber of rubber bands in package 729 g

0.81 g

number of rubber bands in package 900 rubber bands

= ×

=

1 dozen playing cardsdozens of playing cards in box 558 g

18 g

dozens of playing cards in box 31 dozen playing cards

= ×

=

dozens of playing cards in a box 372 playing cards= 1 dozen playing cards

12 playing cards×

dozens of playing cards in a box 31 dozen playing cards=


PRACTICE

(Page 91)

Understanding Concepts 11. mfeathers = 200.4 g (key value)

6 feathers = 2.40 g (conversion factor equation)

There are 501 feathers in a pillow that contains 200.4 g of feathers.

12. mpennies = 1.605 kg (key value)

3 pennies = 7.50 g (conversion factor equation)

Convert the key value, 1.605 kg, to a value in grams by multiplying it by the metric conversion factor 1000 g

1.000 kg:

There are 642 pennies in a piggybank that contains 1.605 kg of pennies.

13. mS = 48.0 g S (key value)

6.02 � 1023 atoms S = 32.06 g S (conversion factor equation) There are 9.01 � 1023 atoms of sulfur in a 48.0-g chunk of sulfur.

14. MCH4 = 52.0 g CH4

1 mol CH4 = 52.0 g CH4

There are 3.24 moles of methane molecules in 52.0 g of methane.

TRY THIS ACTIVITY: COUNTING BY MASS

(Page 91)

Observations and Calculations

Procedural Step

1 Mass of one dozen paper clips 4.9 g

4 Mass of paper clips in cup (g) 33.0 g

feathers

feathers

6 feathers200.4 g

2.40 g

501 feathers

�

�

= ×

=

pennies 1.605 kg� = 1000 g

1.000 kg×

3pennies

3pennies

pennies

1.605 10 g (key value)

3 pennies1.605 10 g

7.50 g

642 pennies

�

�

�

= ×

= × ×

=

S 48.0 g S� =236.02 10 atomsS

32.06 g S

××

23S 9.01 10 atomsS� = ×

4CH 52.0 g CH� = 4

4

1 mol CH

16.05 g CH×

4

4

CH 43.24 mol CH� =


Step 5: Npaper clips in cup = �12 paper clips

33.0 g4.9 g

Npaper clips in cup = 81 paper clips

According to the calculation, there should be 81 paper clips in the cup. Step 6: There are 81 paper clips in the cup. (a) The number of paper clips calculated in step 5 equals the number of paper clips counted in step 6. There are no

differences. (b) Steps 1 and 6 would not be possible. (c) Step 6 empirically (visually) confirms the value obtained in the calculation performed in step 5. (d) A chemist uses the molar masses of the elements (published in a reliable source such as a periodic table) to

determine the mass of a particular number of chemical entities (step 1). A chemist cannot count the number of chemical entities in the sample (as in step 6) to verify that the number of entities calculated equals the actual number in the sample. The chemist assumes that the molar masses given in reliable sources are accurate.

(e) Observations and Calculations

Mass of one paper clip 0.41 g

Mass of paper clips in cup (g) 33.0 g

Npaper clips in cup = �1 paper clip

33.0 g0.41 g

Npaper clips in cup = 80 paper clips

According to the calculation, there should be 80 paper clips in the cup. The procedure outlined in steps 1 to 5 seems to be better than the procedure outlined in step (e) because the

number of paper clips calculated in step 5 (81) equals the counted number of paper clips (81). The number of paper clips calculated in step (e) (80) does not equal the number counted (81). The procedure outlined in steps 1 to 5 is more like the work of a chemist because a chemist uses the mass (in grams) of one mole of entities (6.02 � 1023 entities) (the molar mass) when calculating the number of entities in a given sample.

SECTION 2.1 QUESTIONS

(Page 92)

Understanding Concepts 1. (a) The coefficients of nitrogen, hydrogen, and ammonia are 1, 3, and 2, respectively.

(b) One molecule of nitrogen is used to produce two molecules of ammonia. (c) One dozen molecules of nitrogen is used to produce two dozen molecules of ammonia. (d) One mole of nitrogen is used to produce two moles of ammonia

2. (a) Nucleons are protons and neutrons. They are found in the nucleus of an atom.

(b) The mass of an electron is approximately 1

2000of the mass of a proton or a neutron and is

therefore usually ignored when determining the mass of an atom. (c) An atom of carbon is approximately 12 times heavier than an atom of hydrogen because hydrogen contains one

nucleon (a proton), and carbon contains 12 nucleons (6 protons and 6 neutrons). 3. Student answers will vary. A sample of magnesium taken from nature contains 79% magnesium-24 by mass. The

rest of the sample is composed of other isotopes of magnesium. 4. The symbol for Avogadro’s constant is NA. Its value is 6.02 � 1023.

5. Student answers will vary. A mole is 6.02 � 1023 entities. 6. Atomic mass is the mass of one atom of an element, measured in atomic mass units. Molar mass is the mass, in

grams, of one mole (6.02 � 1023) of an element's atoms. The atomic mass and molar mass of an element have the same numerical value but different units (u for atomic mass; g/mol for molar mass).

7. (a) The term “diatomic” means “composed of two atoms.” (b) hydrogen, nitrogen, oxygen, fluorine, chlorine, bromine, iodine, astatine (c) There are 6.02 � 1023 molecules in one mole of a diatomic molecule. (d) There are 1.2 � 1024 atoms in one mole of a diatomic molecule.

8.

7 dimes = 12.30 g dimes (conversion factor equation)

dimes 7.800 kg� = 1000 g

1 kg× 37.8 10 g (key value)= ×


There are 4439 dimes in a bag that contains 7.800 kg of dimes. 9. mHBr = 38.40 g HBr (key value)

2 mol HBr = 159.8 g HBr (conversion factor equation)

There are 4.81 � 10–1 mol of hydrogen bromide molecules in the cylinder. 10. (a) Molar mass is the mass in grams per mole of one mole (6.02 � 1023) of chemical entities. The symbol for molar

mass is M and its SI unit is g/mol. (b) MC = 12.01 g/mol (c) The mass of 6.02 � 1023 atoms of zinc is 65.38 g. This value is equal to the molar mass of zinc (65.38 g/mol). (d) The element could be gold.

Applied Inquiry Skills 11. (a) 1. Measure the total mass of the coin mixture (in grams).

2. Separate one penny, one nickel, one dime, and one quarter from the mixture. 3. Measure the mass of one penny, one nickel, one dime, and one quarter together (in grams). 4. To estimate the number of coins in the mixture, divide the mass of the mixture measured in step 1 by the

mass of the four coins measured in step 3, and multiply the answer by 4. (b) To estimate the number of dimes, divide the answer in part 4 by 4.

Making Connections 12. (a) Chemists measure and mix chemicals to produce products with properties that differ from the properties of the

reactants. Similarly, chefs measure and mix food ingredients to produce food products with properties that differ from the properties of the ingredients. Chemists balance the reactants and products in a chemical equation. Similarly, accountants prepare balance sheets to balance the income and expenditures of a person or a business.

(b) Chemists work with entities and cannot see the particles. Chefs work with ingredients they can see and even count individually. Chemists balance reactants and products in a chemical equation to determine the quantities of reactants needed and the products formed in a chemical reaction. The balance between reactants and products is always equal. Accountants balance the income and expenditures for clients to determine if a profit or loss has occurred. The relationship between the income and expenditures do not always balance (i.e., one can be greater or less than the other).

13. (a) Student answers will vary. The scanning tunnelling microscope (STM) is able to display the protrusions and depressions caused by atoms on the surface of substances like metals. The shapes of atoms are detected by changes in electric current that occur when the fine tip of a piezoelectric tube moves over the material being observed.

(b) The STM will most likely not replace our need to calculate quantities in chemistry because atoms and molecules are so small that we will always have to deal with huge numbers of a chemical entity than to count individual entities.

2.2 CALCULATIONS INVOLVING THE MOLE

PRACTICE

(Page 94)

Understanding Concepts 1. nAl = 1.60 mol Al

MAl = 26.98 mol/Al 1 mol Al = 26.98 g Al

3dimes

dimes

7 dimes7.800 10 g

12.30 g

4439 dimes

�

�

= × ×

=

HBr 38.40 g HBr� = 2 mol HBr

159.8 g HBr×

1HBr 4.81 10 mol HBr� −= ×


The mass of 1.60 mol aluminum atoms is 43.2 g. 2. nS = 0.25 mol S

MS = 32.06 g/mol S 1 mol S = 32.06 g S

The mass of 0.25 mol sulfur is 8.0 g.

PRACTICE

(Page 95)

Understanding Concepts 3. mFe = 3.30 g Fe

MFe = 55.85 g/mol Fe 1 mol Fe = 55.85 g Fe

There is 0.059 mol iron in a 3.30-g iron nail. 4. mAg = 23.6 g Ag

MAg = 107.87 g/mol Ag 1 mol Ag = 107.87 g/mol Ag

There is 0.219 mol silver in the silver coin. 5. mCu = 7.65 g Cu

MCu = 63.55 g/mol Cu 1 mol Cu = 63.55 g Cu

There is 0.120 mol copper in the bracelet.

PRACTICE

(Page 98)

Understanding Concepts 6. mC = 3.30 g C

MC = 12.01 g/mol C 1 mol C = 12.01 g C 1 mol C = 6.02 � 1023 atoms C

Al 1.60 mol Al� = 26.98 g Al

1 mol Al×

Al 43.2 g Al� =

S 0.25 mol S� = 32.06 g S

1 mol S×

S 8.0 g S� =

Fe 3.30 g Fe� = 1 mol Fe

55.85 g Fe×

Fe 0.059 mol Fe� =

Ag 23.6 gAg� = 1 molAg

107.87 gAg×

Ag 0.219 molAg� =

Cu 7.65 g Cu� = 1 mol Cu

63.55 g Cu×

Cu 0.120 mol Cu� =


C 3.30 gC� �

1 mol C�

12.01 g C

236.02 10 atoms C

1 mol C

��

23C 1.65 10 atoms C� � �

There are 1.65 � 1023 atoms of carbon in a 3.30-g diamond. 7. mNe = 6.80 g Ne

MNe = 20.18 g/mol Ne 1 mol Ne = 20.18 g Ne 1 mol Ne = 6.02 � 1023 atoms Ne

Ne 6.80 g Ne� �

1 mol Ne�

20.18 g Ne

236.02 10 atoms Ne

1 mol Ne

��

23Ne 2.03 10 atoms Ne� � �

There are 2.03 � 1023 atoms of neon in a sign containing 6.80 g of neon. 8. mHg = 78.2 g Hg

MHg = 200.59 g/mol Hg 1 mol Hg = 200.59 g Hg 1 mol Hg = 6.02 � 1023 atoms Hg

Hg 78.2 g Hg� �

1 mol Hg�

200.59 g Hg

236.02 10 atoms Hg

1 mol Hg

��

23Hg 2.35 10 atoms Hg� � �

There are 2.35 � 1023 atoms of mercury in the thermometer.

PRACTICE

(Page 99)


The molar mass of octane is 114.26 g/mol.

10.

The molar mass of acetylsalicylic acid is 180.17 g/mol.

11.

The molar mass of calcium sulfate is 136.14 g/mol.

( ) ( )( ) ( )

8 18

8 18

C H C H

C H

8 18

8 12.01 g/mol 18 1.01 g/mol

96.08 g/mol 18.18 g/mol

114.26 g/mol

� � �

�

= +

= +

= +=

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

9 8 4

9 8 4

C H O C H O

C H O

9 8 4

9 12.01 g/mol 8 1.01 g/mol 4 16.00 g/mol

108.09 g/mol 8.08 g/mol 64.00 g/mol

180.17 g/mol

� � � �

�

= + +

= + +

= + +=

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

4

4

CaSO Ca S O

CaSO

1 1 4

1 40.08 g/mol 1 32.06 g/mol 4 16.00 g/mol

40.08 g/mol 32.06 g/mol 64.00 g/mol

136.14 g/mol

� � � �

�

= + +

= + +

= + +=


PRACTICE

(Page 100)


1 mol NH3 = 17.04 g NH3

The mass of the ammonia is 15.3 g.

13.

1 mol CCl2F2 = 120.91 g CCl2F2

The mass of 3.60 mol freon-12 is 435 g.

PRACTICE

(Page 102)


1 mol Mg(OH)2 = 58.33 g Mg(OH)2

There is 3.497 mol magnesium hydroxide in 204.0 g of magnesium hydroxide.

15.

� � � � � �12 22 11

12 22 11

C H O

C H O 12 22 11

12 22 11 12 22 11

12 12.01 g/mol 22 1.01 g/mol 11 16.00 g/mol

342.34 g/mol C H O

1 mol C H O 342.34 g C H O

�

�

� � �

�

�

There is 2.92 mol sucrose in a bag that contains 1.00 kg of sucrose.

3

3

3

NH 3

NH

NH 3

= 0.900 mol NH

= 1(14.01 g/mol) + 3(1.01 g/mol)

= 17.04 g/mol NH

�

�

�

3NH 30.900 mol NH� = 3

3

17.04 g NH

1 molNH×

3NH 315.3 g NH� =

2 2

2 2

2 2

CCl F 2 2

CCl F

CCl F 2 2

= 3.60 mol CCl F

= 1(12.01 g/mol) + 2(35.45 g/mol) + 2(19.00 g/mol)

= 120.91 g/mol CCl F

�

�

�

2 2CCl F 2 23.60 mol CCl F� = 2 2

2 2

120.91 g CCl F

1 mol CCl F×

2 2CCl F 2 2435 g CCl F� =

2

2

2

Mg(OH) 2

Mg(OH)

Mg(OH) 2

= 204.0 g Mg(OH)

= 1(24.31 g/mol) + 2(16.00 g/mol) + 2(1.01 g/mol)

= 58.33 g/mol Mg(OH)

�

�

�

2Mg(OH) 22.040 g Mg(OH)� = 2

2

1 mol Mg(OH)

58.33 g Mg(OH)×

2Mg(OH) 23.497 mol Mg(OH)� =

12 22 11

12 22 11

12 22 11C H O 12 22 11

12 22 11

3C H O 12 22 11

1000 g C H O1.00 kg C H O

1 kg C H O

1.00 10 g C H O

�

�

= ×

= ×

12 22 11

3C H O 12 22 111.00 10 gC H O� = × 12 22 11

12 22 11

1 mol C H O

342.34 g C H O×

12 22 11C H O 12 22 112.92 mol C H O� =


PRACTICE

(Page 103)


1 mol H2O = 18.02 g/mol H2O

2H O 2250.0 g H O� = 21 mol H O×

218.02 g H O

232

2

6.02 10 molecules H O

1 mol H O

××

2

24H O 28.35 10 moleculesH O� = ×

There are 8.35 � 1024 molecules of water in a bottle that contains 250.0 g of water. 17.

There are 2.59 � 1024 formula units of cobalt(III) dichromate in a 3.30-kg sample.

PRACTICE

(Page 105)


1 mol F2 = 38.00 g F2

2F 24.4 g F� = 21 mol F×

238.00 g F

2326.02 10 molecules F×

×21 mol F 2

2 atoms F

1 molecule F×

2

23F 1.4 10 atoms F� = ×

There are 1.4 � 1023 atoms of fluorine in 4.4 g of fluorine gas. 19.

2

2

2

H O 2

H O

H O 2

= 250.0 g H O

= 2(1.01 g/mol) + 1(16.00 g/mol)

= 18.02 g/mol H O

�

�

�

2N 21.26 kg N� = 2

2

1000 g N

1 kg N×

2

3N 21.26 10 g N� = ×

2 2 7 3Co (Cr O ) 2 2 7 33.30 kg Co (Cr O )� =3

2 2 7 3

2 2 7 3

10 g Co (Cr O )

1 kg Co (Cr O )×

2 2 7 3

3Co (Cr O ) 2 2 7 33.30 10 g Co (Cr O )� = ×

( ) ( ) ( )2 2 7 3

2 2 7 3

Co (Cr O )

Co (Cr O ) 2 2 7 3

2 2 7 3 2 2 7 3

2 58.93 g/mol 6 52.00 g/mol 21 16.00 g/mol

765.86 g/mol Co (Cr O )

1 mol Co (Cr O ) 765.86 g Co (Cr O )

�

�

= + +

=

=

2 2 7 3

3

Co (Cr O ) 2 2 7 33.30 10 g Co (Cr O ) � = × 2 2 7 31 mol Co (Cr O )

×2 2 7 3765.86 g Co (Cr O )

23

2 2 7 3

2 2 7 3

6.02 10 formulaunitsCo (Cr O )

1 mol Co (Cr O )

××

2 2 7 3

24

Co (Cr O ) 2 2 7 32.59 10 formula units Co (Cr O ) � = ×

2

2

2

F 2

F

F 2

= 4.4 g F

= 2(19.00 g/mol)

= 38.00 g/mol F

�

�

�


� �

2

2

N

N 2

2 2

2 14.01 g/mol

28.02 g/mol N

1mol N 28.02 g N

�

�

�

�

�

3N 21.26 10 g N� = × 21 mol N

×228.02 g N

2326.02 10 molecules N×

×21 mol N 2

2 atoms N

1 molecule N×

25N 5.41 10 atoms N� = ×

There are 5.41 � 1025 atoms of nitrogen in 1.26 kg of nitrogen gas.

20.

H 2 429.5 g C H� = 2 41 mol C H×

2 428.06 g C H

232 46.02 10 molecules C H×

×2 41 mol C H 2 4

4 atoms H

1 molecule C H×

24H 2.53 10 atoms H� = ×

There are 2.53 � 1024 atoms of hydrogen in 29.5 g of ethane. 21.

( ) ( ) ( )2

2

Sr(OH)

Sr(OH) 2

2 2

1 87.62 g/mol 2 16.00 g/mol 2 1.01 g/mol

121.64 g/mol Sr(OH)

1 mol Sr(OH) 121.64 g Sr(OH)

�

�

= + +

=

=

4O 21.70 10 g Sr(OH)� � −= 21 mol Sr(OH)

×2121.64 g Sr(OH)

2326.02 10 molecules Sr(OH)×

×21 mol Sr(OH) 2

2 atoms O

1 molecule Sr(OH)×

18O 1.68 10 atoms O� = ×

There are 1.68 � 1018 atoms of oxygen in 0.170 mg of strontium hydroxide.

TRY THIS ACTIVITY: COUNTING ATOMS, MOLECULES, AND OTHER ENTITIES

(Page 105)

1. mass of 50 drops of water = 1.95 g

mass of 1 drop of water = ��

��

mass of 1 drop of water = 0.039 g

The mass of 1 drop of water is 0.039 g.

2. time of 1 drop of water to evaporate = 65 min, or 3.9 � 103 s

2H O 20.039 g H O� = 21 mol H O×

218.02 g H O

232

2

6.02 10 molecules H O

1 mol H O

××

2

21H O 21.30 10 molecules H O� = ×

There are 1.30 � 1021 molecules of water in every drop.

( ) ( )2 4

2 4

2 4

C H 2 4

C H

C H 2 4

2 4 2 4

= 29.5 g C H

2 12.01 g/mol 4 1.01 g/mol

28.06 g/mol C H

1 mol C H 28.06 g C H

�

�

�

= +

=

=

2Sr(OH) 20.170 mg Sr(OH)� =3

2

2

10 g Sr(OH)

1 mg Sr(OH)

−

×

2

4Sr(OH) 21.70 10 g Sr(OH)� −= ×


Water molecules evaporate at the average rate of 3.34 � 1017 molecules/s. 3. mass of 1 penny = 2.44 g

Cu 2.44 g Cu� = 1 mol Cu×63.55 g Cu

236.02 10 atoms Cu

1 mol Cu

××

22Cu 2.31 10 atoms Cu� = ×

There are 2.31 � 1022 copper atoms in a copper penny.

22

23

1 centvalue of a copper atom

2.31 10 atoms

value of a copper atom 4.33 10 cents−

=×

= ×

Each copper atom in a penny has a value of 4.33 � 10–23 cents.

4. 12 22 11C H O 12 22 11 = 337.34 g/mol C H O�

When poured into a graduated cylinder, the volume of half a mole of sucrose is 185 mL.

5. One mole of sucrose, C12H22O11(s), contains 12 mol carbon atoms. Therefore, you would need one-sixth of a mole of sucrose to get 2 mol carbon atoms.

337.34 grequired mass of sucrose

1 mol= 1 mol×

6

required mass of sucrose 56.22 g=

When poured into a graduated cylinder, the volume of sucrose is 60 mL.

6. initial mass of chalk = 14.23 g final mass of chalk = 14.15 g change in mass of chalk = 0.08 g

3

3

CaCO

CaCO 3

= 100.09 g/mol

0.08 g CaCO

�

� = 31 mol CaCO×

3100.09 g CaCO

233

3

6.02 10 formula units CaCO

1 mol CaCO

××

3

20CaCO 34.81 10 formula units CaCO� = ×

Since there are 5 atoms in each formula unit of calcium carbonate, we must multiply the answer by 5 atoms.

20number of atoms 4.81 10 formula unit= × 5 atoms

1 formula unit×

21number of atoms 2.40 10 atoms= ×

The number of atoms needed to write my name in chalk is 2.40 � 1021 atoms. 7. mNaCl = 3.00 g NaCl

MNaCl = 58.44 g/mol NaCl

337.34 gmass of half a mole of sucrose

1 mol= 1 mol×

2

mass of half a mole of sucrose 168.67 g=

212

3

17

1.30 10 molecules H Oevaporation rate

3.9 10 s

evaporation rate 3.34 10 molecules/s

×=

×= ×


NaCl 3.00 g NaCl� = 1 mol NaCl×57.44 g NaCl

236.02 10 formula units NaCl

1 mol NaCl

××

22NaCl 3.09 10 formula units NaCl� = ×

Since there is one sodium ion in every formula unit of sodium chloride, the number of sodium ions is 3.09 � 1022. 8. mnail = 4.16 g Fe

MFe = 55.85 g/mol Fe

Fe 4.16 g Fe� = 1 mol Fe×55.85 g Fe

236.02 10 atoms Fe

1 mol Fe

××

22Fe 4.48 10 atoms� = ×

There are 4.48 � 1022 atoms in the iron nail. 9.

It takes 1.91 � 1016 years to span one mole of seconds.


(Page 106)


The molar mass of magnesium hydroxide is 58.33 g/mol.

2.

The molar mass of ozone is 48.00 g/mol.

3. Student answers will vary. Let the substance be represented by Z. MZ = 67.2 g/mol

Z

1.0 molZ 67.2 g Z

8.0 molZ 67.2 g Z 8

538 g Z�

== ×=

The mass of 8.0 mol of the substance is 538 g. 4. The mass of one mole of sucrose is equal to its molar mass.

� � � � � �

� � �

� � �

�

12 22 11

12 22 11

C H O C H O

C H O 12 22 11

12( ) 22( ) 11( )

12 12.01 g/mol 22 1.01 g/mol 11 16.00 g/mol

342.34 g/mol C H O

� � � �

�

Since the molar mass of sucrose is equal to 342.34 g/mol, the mass of one mole of sucrose is 342.34 g.

60 snumber of seconds in one year

1 min= 60 min×

7

24 h 365 d

1 h 1 d 1 year

number of seconds in one year 3.15 10 s/year

× ×

= ×

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

2

2

Mg(OH) Mg O H

Mg(OH)

1 2 2

1 24.31g/mol 2 16.00 g/mol 2 1.01 g/mol

58.33 g/mol

� � � �

�

= + +

= + +

=

3

3

O O

O 3

3( )

3(16.00 g/mol)

48.00 g/mol O

� �

�

=

==

23

7

16

6.02 10 s 1 yearyears in one mole of seconds

1 mol 3.15 10 s

years in one mole of seconds 1.91 10 year/mol

×= ××

= ×


5. (a) nAl = 0.1 mol Al 1 mol Al = 6.02 � 1023 atoms Al

Al 0.1 molAl� =236.02 10 atomsAl

1 molAl

××

22Al 6.02 10 atomsAl� = ×

There are 6.02 � 1022 atoms of aluminum in 0.1 mol aluminum.

(b) 2MgCl 2 = 3.5 mol MgCl�

1 mol MgCl2 = 6.02 � 1023 formula units MgCl2

2MgCl 23.5 mol MgCl� =

232

2

6.02 10 formula units MgCl

1 mol MgCl

××

2

24MgCl 22.1 10 formula units MgCl� = ×

There are 2.1 � 1024 formula units of magnesium chloride in 3.5 mol magnesium chloride. 6. (a)

( ) ( ) ( )12 22 11

12 22 11

C H O

C H O 12 22 11

12 22 11 12 22 11

12 12.01 g/mol 22 1.01 g/mol 11 16.00 g/mol

342.34 g/mol C H O

1 mol C H O 342.34 g C H O

�

�

= + +

=

=

There is 14.6 mol sucrose in 5.00 kg of table sugar.

(b)

10 8C H 10 8250 g C H� = 10 8

10 8

1 mol C H

128.18 g C H×

10 8C H 10 81.95 mol C H� =

There is 1.95 mol naphthalene in 250 g of naphthalene moth balls.

(c)

3 8C H 3 835.0 g C H� = 3 8

3 8

1 mol C H

44.11 g C H×

3 8C H 3 80.793 mol C H� =

There is 0.793 mol propane in the stove cylinder.

12 22 11C H O 12 22 115.00 kg C H O� = 12 22 11

12 22 11

1000 g C H O

1 kg C H O×

12 22 11

3C H O 12 22 115.00 10 g C H O� = ×

( ) ( )10 8

10 8

10 8

C H 10 8

C H

C H 10 8

10 8 10 8

250 g C H

10 12.01 g/mol 8 1.01 g/mol

128.18 g/mol C H

1 mol C H 128.18 g C H

�

�

�

=

= +

=

=

( ) ( )3 8

3 8

3 8

C H 3 8

C H

C H 3 8

3 8 3 8

35.0 g C H

3 12.01 g/mol 8 1.01 g/mol

44.11 g/mol C H

1 mol C H 44.11 g C H

�

�

�

=

= +

=

=

12 22 11

3C H O 12 22 115.00 10 g C H O� = × 12 22 11

12 22 11

1 mol C H O

342.34 g C H O×

12 22 11C H O 12 22 1114.6 mol C H O� =


(d)

9 8 4C H O 9 8 40.275 g C H O� = 9 8 4

9 8 4

1 mol C H O

180.17 g C H O×

9 8 4

3C H O 9 8 41.53 10 mol C H O� −= ×

There is 1.53 � 10–3 mol acetylsalicylic acid in a 275-mg headache relief tablet. (e)

3 8C H O 3 8240 g C H O� = 3 8

3 8

1 mol C H O

60.11 g C H O×

3 8C H O 3 84.0 mol C H O� =

There is 4.0 mol 2-propanol in 240 g of rubbing alcohol. 7. (a)

3NH 32.67 mol NH� = 3

3

17.04 g NH

1 mol NH×

3NH 345.5 g NH� =

The mass of 2.67 mol ammonia is 45.5 g.

(b)

NaOH 0.965 mol NaOH� = 40.00 g NaOH

1 mol NaOH×

NaOH 38.6 g NaOH� =

The mass of 0.965 mol sodium hydroxide is 38.6 g. (c)

2

2

2

H O 2

H O

H O 2

2 2

19.7 mol H O

2(1.01 g/mol) 1(16.00 g/mol)

18.02 g/mol H O

1 mol H O 18.02 g H O

�

�

�

=

= +

=

=

( ) ( ) ( )

9 8 4

9 8 4

9 8 4

9 8 4

39 8 4

C H O 9 8 49 8 4

C H O 9 8 4

C H O

C H O 9 8 4

9 8 4 9 8 4

10 g C H O275 mg C H O

1 mg C H O

0.275 gC H O

9 12.01 g/mol 8 1.01 g/mol 4 16.00 g/mol

180.17 g/mol C H O

1 molC H O 180.17 g/mol C H O

�

�

�

�

−

= ×

=

= + +

=

=

( ) ( ) ( )3 8

3 8

3 8

C H O 3 8

C H O

C H O 3 9

3 8 3 8

240 gC H O

3 12.01 g/mol 8 1.01 g/mol 1 16.00 g/mol

60.11 g/mol C H O

1 mol C H O 60.11 g C H O

�

�

�

=

= + +

=

=

3

3

3

NH 3

NH

NH 3

3 3

2.67 mol NH

1(14.01 g/mol) 3(1.01 g/mol)

17.04 g/mol NH

1 mol NH 17.4 g NH

�

�

�

=

= +

=

=

NaOH

NaOH

NaOH

0.965 mol NaOH

1(22.99 g/mol) 1(16.00 g/mol) 1(1.01 g/mol)

40.00 g/mol NaOH

1 mol NaOH 40.00 g NaOH

�

�

�

== + +=

=


2H O 219.7mol H O� = 2

2

18.02 g H O

1 mol H O×

2H O 2355 g H O� =

The mass of 19.7 mol water vapour is 355 g. (d)

4KMnO 43.85 mol KMnO� = 4

4

158.04 g KMnO

1 mol KMnO×

4KMnO 4608 g KMnO� =

The mass of 3.85 mol potassium permanganate is 608 g. (e)

4 2 4(NH ) SO 4 2 40.47 mol (NH ) SO� = 4 2 4

4 2 4

132.16 g NH ) SO

1 mol (NH ) SO×

4 2 4(NH ) SO 4 2 462 g (NH ) SO� =

The mass of 0.47 mol ammonium sulfate is 62 g. 8. (a)

2CO 22.5 mol CO� =23

2

2

6.02 10 molecules CO

1 mol CO

××

2

24CO 21.5 10 molecules CO� = ×

There are 1.5 � 1024 molecules OF solid carbon dioxide in 2.5 mol carbon dioxide. (b)

3NH 32.5 g NH� = 31 mol NH×

317.04 g NH

233

3

6.02 10 molecules NH

1 mol NH

××

3

22NH 38.8 10 molecules NH� = ×

There are 8.8 � 1022 molecules of ammonia in 2.5 g of ammonia gas.

4

4

4

KMnO 4

KMnO

KMnO 4

4 4

3.85 mol KMnO

1(39.10 g/mol) 1(54.94 g/mol) 4(16.00 g/mol)

158.04 g/mol KMnO

1 mol KMnO 158.04 g KMnO

�

�

�

=

= + +

=

=

4 2 4

4 2 4

4 2 4

(NH ) SO 4 2 4

(NH ) SO

(NH ) SO 4 2 4

4 2 4 4 2 4

0.47 mol (NH ) SO

2(14.01 g/mol) 8(1.01 g/mol) 1(32.06 g/mol) 4(16.00 g/mol)

132.16 g/mol (NH ) SO

1 mol (NH ) SO 132.16 g (NH ) SO

�

�

�

=

= + + +

=

=

2CO 2

232 2

2.5 mol CO

1 mol CO 6.02 10 molecules CO

� =

= ×

3

3

3

NH 3

NH

NH 3

3 3

233 3

2.5 g NH

1(14.01 g/mol) 3(1.01 g/mol)

17.04 g/mol NH

1 mol NH 17.04 g NH

1 mol NH 6.02 10 molecules NH

�

�

�

=

= +

=

=

= ×


(c)

HCl 2.5 g HCl� �1 mol HCl

�

36.46 g HCl

236.02 10 molecules HCl

1 mol HCl

��

22HCl 4.1 10 molecules HCl� � �

There are 4.1 � 1022 molecules in 2.5 g of hydrogen chloride gas.

9. (a)

2CO 20.10 mol CO� �2

2

44.01 g CO

1 mol CO�

2CO 24.4 g CO� �

The mass of 0.10 mol carbon dioxide is 4.4 g. (b)

6 12 6C H O 6 12 60.10 mol C H O� �6 12 6

6 12 6

180.18 g C H O

1 mol C H O�

6 12 6C H O 6 12 618 g C H O� �

The mass of 0.10 mol glucose is 18 g. (c)

2O 20.10 mol O� �2

2

32.00 g O

1 mol O�

2O 23.2 g O� �

The mass of 0.10 mol oxygen gas is 3.2 g. 10.

2O 22.7 mol O� �

232

2

6.02 10 molecules O

1 mol O

�

�

2

24O 21.6 10 molecules O� � �

There are 1.6 � 1024 oxygen molecules in 2.7 mol oxygen gas.

HCl

HCl

HCl

23

2.5 g HCl

1(1.01 g/mol) 1(35.45 g/mol)

36.46 g/mol HCl

1 mol HCl 36.46 g HCl

1 mol HCl 6.02 10 molecules HCl

�

�

�

== +=

== ×

=

= +

=

=

2

2

2

CO 2

CO

CO 2

2 2

0.10 mol CO

1(12.01 g/mol) 2(16.00 g/mol)

44.01 g/mol CO

1 mol CO 44.01 g CO

�

�

�

6 12 6

6 12 6

6 12 6

C H O 6 12 6

C H O

C H O 6 12 6

6 12 6 6 12 6

0.10 mol C H O

6(12.01g/mol) 12(1.01 g/mol) 6(16.00 g/mol)

180.18 g/mol C H O

1 mol C H O 180.18 g C H O

�

�

�

=

= + +

=

=

=

=

=

=

2

2

2

O 2

O

O 2

2 2

0.10 mol O

2(16.00 g/mol)

32.00 g/mol O

1 molO 32.00 g O

�

�

�

2O 2

232 2

2.7 mol O

1 molO 6.02 10 molecules O

� =

= ×


11.

= + +

=

=

= ×

6 8 6

6 8 6

C H O

C H O 6 8 6

6 8 6

236 8 6 6 8 6

6(12.01 g/mol) 8(1.01 g/mol) 6(16.00 g/mol)

176.14 g/mol C H O

1 mol C H O 176.14 g/mol

1 mol C H O 6.02 10 molecules C H O

�

�

6 8 6

2C H O 6 8 69 10 g C H O� �

� �6 8 61 mol C H O

�

6 8 6176.14 g C H O

236 8 6

6 8 6

6.02 10 molecules C H O

1 mol C H O

�

�

6 8 6

20C H O 6 8 63.1 10 molecules C H O� � �

The number of molecules of vitamin C taken each day would be 3.1 � 1020 molecules.

12. water

2

2

2

H O 2

H O

H O 2

2 2

450 gH O

2(1.01 g/mol) 1(16.00 g/mol)

18.02 g/mol H O

1 mol H O 18.02 g H O

�

�

�

�

� �

�

�

2H O 2450 gH O� �2

2

1 mol H O

18.02 g H O�

2H O 225 mol H O� �

sugar

12 22 11

12 22 11

12 22 11

C H O 12 22 11

C H O

C H O 12 22 11

12 22 11 12 22 11

100 g C H O

12(12.01 g/mol) 22(1.01 g/mol) 11(16.00 g/mol)

342.34 g/mol C H O

1 mol C H O 342.34 g C H O

�

�

�

�

� � �

�

�

12 22 11C H O 12 22 11100 g C H O� �12 22 11

12 22 11

1 mol C H O

342.34 g C H O�

12 22 11C H O 12 22 110.3 mol C H O� �

vinegar

2 3 2

2 3 2

2 3 2

HC H O 2 3 2

HC H O

HC H O 2 3 2

2 3 2 2 3 2

2.4 g HC H O

4(1.01 g/mol) 2(12.01 g/mol) 2(16.00 g/mol)

60.06 g/mol HC H O

1 mol HC H O 60.06 g HC H O

�

�

�

�

� � �

�

�

2 3 2HC H O 2 3 22.4 g HC H O� �2 3 2

2 3 2

1mol HC H O

60.06 g HC H O�

2 3 2HC H O 2 3 20.040 mol HC H O� �

6 8 6

6 8 6

36 8 6

C H O 6 8 66 8 6

2C H O 6 8 6

10 g C H O90 mg C H O

1 mg C H O

9 10 g C H O

�

�

−

−

= ×

= ×


salt

NaCl

NaCl

NaCl

2 g NaCl

1(22.99 g/mol) 1(35.45 g/mol)

58.44 g/mol NaCl

1 molNaCl 58.44 g NaCl

�

�

�

�

� �

�

�

NaCl 2 g NaCl� �1 mol NaCl

58.44 g NaCl�

NaCl 0.03 mol NaCl� �

The recipe calls for 25 mol water, 0.3 mol sugar, 0.040 mol acetic acid (vinegar), and 0.03 mol salt.

Applied Inquiry Skills 13. Materials

eye protection stirring rod lab apron coil of copper wire 250-mL beaker filter paper silver nitrate scoopula distilled water

Warning: Silver chloride may stain skin and cause skin irritation. Avoid contact with skin. Do not taste.

Procedure 1. Prepare a solution of silver ions by dissolving 0.1 g of silver nitrate in 100 mL water in a 250-mL beaker. 2. Place a coil of copper wire into the silver nitrate solution and let sit for 24 h. 3. Measure the mass of a piece of filter paper, mfilter paper. 4. Carefully remove the copper coil from the silver nitrate solution and scrape all the silver crystals from the coil

onto the filter paper. 5. Allow the silver crystals to dry. 6. Determine the mass of the filter paper and dry silver crystals, mfilter paper + Ag. 7. Calculate the mass of the silver crystals by subtracting the mass of the filter paper (Step 4) from the mass of the

filter paper and silver crystals (step 7).

mAg = mfilter paper + Ag – mfilter paper 8. Calculate the amount of silver recovered, nAg, as follows.

� �( )

1 mol

107.87 g��

Making Connections 14. Student answers will vary. The mole deserves to win this award because without the mole, chemists could not

calculate the number of chemical entities they use in chemical reactions. Knowing the number of entities used in reactions enables chemists to maximize the amount of product formed, and minimize the amount of reactants that remain unreacted. When chemicals are mixed haphazardly, excess reactants mix with products and must be separated in order to purify the products. This effect increases the cost associated with the industrial preparation of chemicals, and may lead to increased levels of chemical waste and chemical pollution.


2.3 DETERMINING CHEMICAL FORMULAS

TRY THIS ACTIVITY: WHAT MAKES POPCORN POP?

(Page 110)

Type of popcorn kernel

Initial mass (g)

Final mass (g)

Change in mass (g)

Percentage water (%)

Percentage of popped corn (%)

whole 18.16 16.30 1.86 10.2 ~90

split crosswise 15.34 13.90 1.45 9.45 ~20

split lengthwise 15.04 13.54 1.50 9.97 <1

(a) In general, the results seem to confirm that popcorn pops because the moisture inside the large part (endosperm) of the seed vaporizes, builds up the internal pressure, and suddenly breaks the hard outer coating (pericarp). The evidence of the popping percentage of the whole seeds versus the lengthwise-split seeds clearly illustrates this. The results for the crosswise-split seeds are inconclusive, perhaps because this splitting will, in some cases, include some of the endosperm and, in other cases, be restricted only to the bottom, germinating part of the seed.

PRACTICE

(Page 114)

Understanding Concepts 1. H = 2.20%

C = 26.7% O = 71.1%

� �

�

H

H

2.20100 g

1002.20 g

�

�

� �

�

C

C

26.7100 g

10026.7 g

�

�

� �

�

O

O

71.1100 g

10071.1 g

�

�

H 2.20 g H� �1 mol H

1.01 g H�

H

C

2.18 mol H

26.7 g C

�

�

�

�1 mol C

12.01 g C�

C 2.22 mol C� �


O 71.1 g O� �1 mol O

16.00 g O�

O 4.44 mol O� �

�

�

�

�

H C O

H C O

: : 2.18 : 2.22 : 4.44

2.18 2.22 4.44: :

2.18 2.18 2.181.00 :1.02 : 2.04

: : 1:1: 2

� � �

� � �

The empirical formula of this compound is HCO2. 2. Al = 35.9% S = 64.1%

Al

Al

S

S

35.9100 g

10035.9 g Al

64.1100 g

10064.1 g S

�

�

�

�

� �

�

� �

�

Al 35.9 g Al� �1 molAl

26.98 gAl�

Al

S

1.33 mol Al

64.1 g S

�

�

�

�1 mol S

32.06 g S�

S 2.00 mol S� �

�

�

�

�

�

�

: 1.33 : 2.00

1.33 2.00:

1.33 1.331.00 :1.50

2(1.00) : 2(1.50)

: 2.00 : 3.00

: 2 : 3

� �

� �

� �

� �

� �

� �

The empirical formula of the compound is Al2S3.

PRACTICE

(Page 118)

Understanding Concepts 3. C = 68.54%

H = 8.63% O = 22.83% Mcompound = 140.20 g/mol


�

�

�

�

�

�

�

�

�

C

C

H

H

O

O

68.54100 g

100

68.54 g

8.63100 g

100

8.63 g

22.83100 g

100

22.83 g

�

�

�

�

�

�

C 68.54 g C � �

1 mol C

12.01 g C�

C 5.707 mol C� �

H 8.63 g C � �

1 mol C

1.01 g C�

H 8.54 mol C� �

O 22.83 g O � �

1 mol O

16.00 g O�

O 1.427 mol O� �

C H O

C H O

: :

: :

5.707 : 8.54 :1.427

5.707 8.54 1.427: :

1.427 1.427 1.4273.999 : 5.98 :1.00

4 : 6 :1

� � �

� � �

�

�

�

�

The empirical formula of the compound is C4H6O.

C H O4 6

C H O C H O4 6 4 6

4(12.01 g/mol) 6(1.01 g/mol) 1(16.00 g/mol)

70.10 g/mol

�

�

� � �

�

compound

C H O4 6

compound

C H O4 6

140.20 g/mol

70.10 g/mol

2

�

�

�

�

�

�

� 4 6

8 12 2

molecular formula = 2(empirical formula)

2(C H O)

molecular formula = C H O

The molecular formula of the compound is C8H12O2.


4. C = 76.5% H = 12.2% O = 11.3% Mcompound = 706.3 g/mol

� �

�

� �

�

� �

�

C

C

H

H

O

O

76.5100 g

10076.5 g

12.2100 g

10012.2 g

11.3100 g

10011.3 g

�

�

�

�

�

�

C 76.5 g C� �1 mol C

12.01 g C�

C

H

6.37 mol C

12.2 g H

�

�

�

�1 mol H

1.01 g H�

H

O

12.1 mol H

11.3 g O

�

�

�

�1 mol O

16.00 g O�

O 0.706 mol O� �

�

�

�

�

C H O

C H O

: : 6.37 :12.1: 0.706

6.37 12.1 0.706: :

0.706 0.706 0.7069.02 :17.1:1.00

: : 9 :17 :1

� � �

� � �

The empirical formula of the compound is C9H17O.

9 17

9 17

C H O

C H O 9 17

9(12.01 g/mol) 17(1.01 g/mol) 1(16.00 g/mol)

141.26 g/mol C H O

�

�

� � �

�

9 17

compound

C H O

706.3 g/mol�

��

141.26 g/mol

9 17

compound

C H O

5�

��


9 17

45 85 5

molecular formula 5(empiricalformula)

5(C H O)

molecular formula C H O

�

�

�

The molecular formula of the fat is C45H85O5. 5. K = 26.65%

Cr = 35.33% O = 38.02% Mcompound = 294.20 g/mol

K

K

Cr

Cr

O

O

K

26.65100 g

10026.65 g

35.33100 g

10035.33 g

38.02100 g

10038.02 g

26.65 g K

�

�

�

�

�

�

�

� �

�

� �

�

� �

�

�1 mol K

39.10 g K�

K

Cr

0.6816 mol K

35.33 g Cr

�

�

�

�1 mol Cr

52.00 g Cr�

Cr

O

0.6794 mol Cr

38.02 g O

�

�

�

�1 mol O

16.00 g O�

O 2.376 mol O� �

�

�

�

�

K Cr O

C H O

: : 0.6816 : 0.6794 : 2.376

0.6816 0.6794 2.376: :

0.6794 0.6794 0.67941.003 :1.000 : 3.497

: : 1:1: 3.5

� � �

� � �

Multiply the ratio by 2 to obtain whole numbers.

�

�

C H O

C H O

: : 2(1) : 2(1) : 2(3.5)

: : 2 : 2 : 7

� � �

� � �

The empirical formula of the compound is K2Cr2O7.

2 2 7

2 2 7

K Cr O

K Cr O 2 2 7

2(39.10 g/mol) 2(52.00 g/mol) 7(16.00 g/mol)

294.20 g/mol K Cr O

�

�

� � �

�


�

2 2 7

substance

K Cr O

294.20 g/mol�

� 294.20 g/mol

�

2 2 7

substance

K Cr O

1�

�

�

� 2 2 7

2 2 7

molecular formula 1(empirical formula)

1(K Cr O )

molecular formula = K Cr O

The molecular formula of the substance is K2Cr2O7.

6. C = 74.0% H = 8.7% N = 17.3% Mnicotine = 162.26 g/mol

C

C

H

H

N

N

74.0100 g

10074.0 g

8.7100 g

1008.7 g

17.3100 g

10017.3 g

�

�

�

�

�

�

� �

�

� �

�

� �

�

C 74.0 g C� �1 mol C

12.01 g C�

C

H

6.16 mol C

8.7 g H

�

�

�

�1 mol H

1.01 g H�

H

N

8.6 mol H

17.3 g N

�

�

�

�1 mol N

14.01 g N�

N 1.23 mol N� �

C H N

C H N

: : 6.16 : 8.6 :1.23

6.16 8.6 1.23: :

1.23 1.23 1.235.01: 7.0 :1.00

: : 5 : 7 :1

� � �

� � �

�

�

�

�

The empirical formula of the compound is C5H7N.


5 7

5 7

C H N

C H N 5 7

5(12.01 g/mol) 7(1.01 g/mol) 1(14.01 g/mol)

81.13 g/molC H N

�

�

� � �

�

�

5 7

nicotine

C H N

162.26 g/mol�

� 81.13 g/mol

�

5 7

nicotine

C H N

2�

�

�

� 5 7

10 14 2


2(C H N)

molecular formula = C H N

The molecular formula of the compound is C10H14N2.

PRACTICE

(Page 119)

Understanding Concepts 7. (a) mC = 12.01 u

mH = 1.01 u mO = 16.00 u

� � �

� � �

� � �

�

6 8 6

6 8 6

C H O C H O

C H O

6( ) 8( ) 6( )

6(12.01 u) 8(1.01 u) 6(16.00 u)

72.06 u 8.08 u 96.00 u

176.14 u

� � � �

�

�

�

�

72.06 u% C = 100%

176.14 u

% C = 40.91%

8.08 u% H = 100%

176.14 u

% H = 4.59%

96.00 u% O = 100%

176.14 u

% O = 54.50%

The percentage composition by mass of C6H8O6 is 40.91% carbon, 4.59% hydrogen, and 54.50% oxygen.

(b) mAl = 26.98 u mO = 16.00 u

� �

� �

� �

�

2 3

2 3

Al O Al O

Al O

2( ) 3( )

2(26.98 u) 3(16.00 u)

53.96 u 48.00 u

101.96 u

� � �

�


�

�

53.96 u% Al = 100%

101.96 u

% Al = 52.92%

48.00 u% O = 100%

101.96 u

% O = 47.08%

The percentage composition by mass of aluminum oxide is 52.92% aluminum and 47.08% oxygen.

(c) mZn = 65.38 u mN = 14.01 u mO = 16.00 u

� � � �

� � � �

� � �

� � �

� � �

�

3 2

3 2

Zn(NO ) Zn N O

Zn(NO )

2 6

65.38 u 2 14.01 u 6 16.00 u

65.38 u 28.02 u 96.00 u

189.4 u

� � � �

�

�

�

�

65.38 u% Zn = 100%

189.4 u

% Zn = 34.52%

28.02 u% N = 100%

189.4 u

% N = 14.79%

96.00 u% O = 100%

189.4 u

% O = 50.69%

The percentage composition by mass of zinc nitrate is 34.52% zinc, 14.79% nitrogen, and 50.69% oxygen.


(Page 120)

Understanding Concepts 1. The law of constant composition states that compounds contain the same ratio of elements by mass, regardless of

their source. 2. (a) A small sample of a compound is placed in the mass spectrometer where it is vapourized, ionized, and possibly

broken into a number of fragments. The resulting charged fragments are accelerated by an electric field and deflected by a magnetic field. The amount of deflection of the fragments is used to calculate the molar mass of the original compound.

(b) A combustion analyzer is used to determine the percentages of carbon, hydrogen, oxygen, and possibly nitrogen in a compound containing these elements. An accurately known mass of the compound is burned in a stream of pure oxygen gas to form water and carbon dioxide gas. The resulting water and carbon dioxide are trapped in water and carbon dioxide absorbers. Using the mass of water and carbon dioxide absorbed by the traps, and the mass of the original sample, the percentage composition of the original compound may be calculated.

3. (a) You need to know the percentage composition of a compound to determine the empirical formula. (b) You need to know the molar mass of a compound (in addition to its empirical formula) to determine the

molecular formula of a compound.


4. (a) CH2O (b) NH3 (c) CH

5. K = 26.6% Cr = 35.4% O = 38.1%

� �

�

� �

�

� �

�

K

K

Cr

Cr

O

O

26.6100 g

10026.6 g

35.4100 g

10035.4 g

38.1100 g

10038.1 g

�

�

�

�

�

�

K 26.6 g K� �1 mol K

39.10 g K�

K

Cr

0.680 mol K

35.4 g Cr

�

�

�

�1 mol Cr

52.00 g Cr�

Cr

O

0.681 mol Cr

38.1 g O

�

�

�

�1 mol O

16.00 g O�

O 2.38 mol O� �

�

�

�

K Cr O

K Cr O

: : 0.680 : 0.681: 2.30

0.680 0.681 2.38: :

0.680 0.680 0.680: : 1:1: 3.5

� � �

� � �


�

�

K Cr O

K Cr O

: : 2(1:1: 3.5)

: : 2 : 2 : 7

� � �

� � �

The empirical formula of the compound is K2Cr2O7.

6. (a) C = 40.87% H = 3.72% N = 8.67% O = 24.77% Cl = 21.98%


� �

�

� �

�

� �

�

� �

�

� �

�

C

C

H

H

N

N

O

O

Cl

Cl

40.87100 g

10040.87 g

3.72100 g

1003.72 g

8.67100 g

1008.67 g

24.77100 g

10024.77 g

21.98100 g

10021.98 g

�

�

�

�

�

�

�

�

�

�

O 24.77 g O� �1 mol O

16.00 g O�

O

Cl

1.548 mol O

21.98 g Cl

�

�

�

�1 mol Cl

35.45 g Cl�

Cl 0.6200 mol Cl� �

�

�

�

�

C H N O Cl

C H N O Cl

: : : : 3.403 : 3.68 : 0.619 :1.548 : 0.6200

3.403 3.68 0.619 1.548 0.6200: : : :

0.619 0.619 0.619 0.619 0.6195.498 : 5.95 :1.00 : 2.501:1.002

: : : : 5.5 : 6 :1: 2.5 :1

� � � � �

� � � � �

C 40.87 g C� �1 mol C

12.01 g C�

C

H

3.403 mol C

3.72 g H

�

�

�

�1 mol H

1.01 g H�

H

N

3.68 mol H

8.67 g N

�

�

�

�1 mol N

14.01 g N�

N 0.619 mol N� �



�

�

C H N O Cl

C H N O Cl

: : : : 2(5.5) : 2(6) : 2(1) : 2(2.5) : 2(1)

: : : : 11:12 : 2 : 5 : 2

� � � � �

� � � � �

The empirical formula of chloromycetin is C11H12N2O5Cl2.

(b) C = 41.86% H = 4.65% N = 16.28% O = 18.60% S = 18.60%

� �

�

� �

�

� �

�

C

C

H

H

N

N

41.86100 g

10041.86 g

4.65100 g

1004.65 g

16.28100 g

10016.28 g

�

�

�

�

�

�

� �

�

� �

�

O

O

S

S

18.60100 g

10018.60 g

18.60100 g

10018.60 g

�

�

�

�

C 41.86 g C� �1 mol C

12.01 g C�

C

H

3.485 mol C

4.65 g H

�

�

�

�1 mol H

1.01 g H�

H

N

4.60 mol H

16.28 g N

�

�

�

�1 mol N

14.01 g N�

N

O

1.162 mol N

18.60 g O

�

�

�

�1 mol O

16.00 g O�

O 1.163 mol O� �


S 18.60 g S� �1 mol S

32.06 g S�

S 0.5802 mol S� �

�

�

�

�

C H N O S

C H N O S

: : : : 3.485 : 4.60 :1.162 :1.163 : 0.5802

3.485 4.60 1.162 1.163 0.5802: : : :

0.5802 0.5802 0.5802 0.5802 0.58026.007 : 7.93 : 2.00 : 2.00 :1.00

: : : : 6 : 8 : 2 : 2 :1

� � � � �

� � � � �

The empirical formula of sulfanilamide is C6H8N2O2S. 7. (a) mH = 1.01 u

mO = 16.00 u

� �

� �

� �

� �

� �

�

2

2

H O H O

H O

2

2 1.01 u 16.00 u

2.02 u 16.00 u

18.02 u

� � �

�

�

�

2.02 u% H = 100%

18.02 u

% H = 11.21%

16.00 u% O = 100%

18.02 u

% O = 88.79%

The percentage composition by mass of water is 11.21% hydrogen and 88.79% oxygen.

(b) mCa = 40.08 u mO = 16.00 u mH = 1.01 u

� � � �

� � � �

� � �

� � �

� � �

�

2

2

Ca(OH) Ca O H

Ca(OH)

2 2

40.08 u 2 16.00 u 2 1.01 u

40.08 u 32.00 u 2.02 u

74.10 u

� � � �

�

40.08 u

% Ca = 100%74.10 u

% Ca = 54.09%

32.00 u% O = 100%

74.10 u

% O = 43.18%

�

�


2.02 u

% H = 100%74.10 u

% H = 2.73%

�

The percentage composition by mass of calcium hydroxide is 54.09% calcium, 43.18% oxygen, and 2.73% hydrogen.

Applying Inquiry Skills 8. Na = 21.9%

C = 45.7% H = 1.9% O = 30.5% Mcompound = 210 g/mol

� �

�

� �

�

� �

�

� �

�

Na

Na

C

C

H

H

O

O

21.9100g

10021.9 g

45.7100g

10045.7 g

1.9100 g

1001.9 g

30.5100 g

10030.5g

�

�

�

�

�

�

�

�

Na 21.9 g Na� �1 mol Na

22.99 g Na�

Na

C

0.953 mol Na

45.7 g C

�

�

�

�1 mol C

12.01 g C�

C

H

3.81 mol C

1.9 g H

�

�

�

�1mol H

1.01 g H�

H

O

1.88 mol H

30.5 g O

�

�

�

�1 mol O

16.00 g O�

O 1.91 mol O� �


�

�

�

�

Na C H O

Na C H O

: : : 0.953 : 3.81:1.88 :1.91

0.953 3.81 1.88 1.91: : :

0.953 0.953 0.953 0.9531.00 : 4.00 :1.97 : 2.00

: : : 1: 4 : 2 : 2

� � � �

� � � �

The empirical formula of the compound is NaC4H2O2.

4 2 2

4 2 2

NaC H O

NaC H O 4 2 2

1(22.99 g/mol) 4(12.01 g/mol) 2(1.01 g/mol) 2(16.00 g/mol)

105.05 g/mol NaC H O

�

�

� � � �

�

�

4 2 2

compound

NaC H O

210 g/mol�

� 105.05 g/mol

�

4 2 2

compound

NaC H O

2�

�

�

�

�

4 2 2

2 8 4 4


2(NaC H O )

molecular formula Na C H O

The molecular formula of the compound is Na2C8H4O4.

9. C = 49.38% H = 3.55% O = 9.40% S = 37.67% M

compound = 170.2 g/mol

C

C

H

H

O

O

S

S

C

49.38100 g

10049.38 g

3.55100 g

1003.55 g

9.40100 g

1009.40 g

37.67100 g

10037.67 g

49.38 g C

�

�

�

�

�

�

�

�

�

� �

�

� �

�

� �

�

� �

�

�1 mol C

12.01 g C�

C 4.112 mol C� �


H 3.55 g H� �1 mol H

1.01 g H�

H

O

3.51 mol H

9.40 g O

�

�

�

�1 mol O

16.00 g O�

O

S

0.588 mol O

37.67 g S

�

�

�

�1 mol S

32.06 g S�

S 1.175 mol S� �

�

�

�

�

C H O S

C H O S

: : : 4.112 : 3.51: 0.588 :1.175

4.112 3.51 0.588 1.175: : :

0.588 0.588 0.588 0.5886.99 : 5.97 :1.00 : 2.00

: : : 7 : 6 :1: 2

� � � �

� � � �

The empirical formula is C7H6OS2.

7 6 2

7 6 2

C H OS

C H OS 7 6 2

7(12.01 g/mol) 6(1.01 g/mol) 1(16.00 g/mol ) 2(32.06 g/mol)

170.25 g/mol C H OS

�

�

� � � �

�

�

7 6 2

compound

C H OS

170.2 g/mol�

� 170.25 g/mol

� �

7 6 2

compound

C H OS

0.9997 1�

�

�

�

�

7 6 2

7 6 2


1(C H OS )

molecular formula C H OS

The molecular formula of the newly discovered compound is C7H6OS2.


Evaluation (f) If some of the magnesium oxide had escaped from the crucible, the mass of the product left in the crucible would be

less than the Prediction. The mass of magnesium would not be affected, however. Therefore, the percentage composition of magnesium would be too high.

(g) If the other component of air is nitrogen, the percentage composition of magnesium would be higher than the Prediction. Nitrogen has a lower molar mass than oxygen. Also, the formula of magnesium nitride, Mg3N2, means that the ratio of magnesium to nitrogen is 1 : 0.67 versus 1:1 for magnesium oxide. Therefore, the mass of product would be less for the same mass of magnesium and the percentage composition would be higher than the predicted value.

(h) Magnesium metal, like many metals, oxidizes in air to form an oxide layer. The oxide layer needs to be removed to more accurately determine the mass of pure magnesium.

(i) The best modification would be to cause the magnesium to react inside a container of pure oxygen. This method would require a substantial modification to the materials and procedure, however.

(j) The Prediction is inconclusive because the certainty of the experimental answer is only one significant digit. The accuracy of the balance is probably one or two hundredths of a gram. Therefore, the answer can vary widely. The law of definite proportions remains valid until it is tested with an improved experiment.

Synthesis (k) Some of the educational requirements to become a gemologist include courses such as Diamonds, Diamond

Grading, Coloured Stones, Coloured Stone Grading, and Gem Identification. A Math and Science background, liberal arts education, and the knowledge of foreign languages are also useful assets. Gemologists work with many different instruments, including a gem microscope, which is more powerful than an average microscope, a refractrometer to measure the refractive index of gems, and a balance beam to measure specific gravity.

2.5 QUANTITATIVE ANALYSIS: CONCENTRATION OF SOLUTIONS

PRACTICE

(Page 126)


� �

� �

�

ethanolsolution

solution

solution

100%

4.1 L100%

55 L7.5% V/V

�

�

The ethanol concentration of a typical gasohol mixture is 7.5% V/V.

2.

� �

� �

�

2ZnClsolution

solution

solution

100%

16 g100%

500 mL3.2%W/V

�

�

The concentration of the zinc chloride in solder flux solution is 3.2% W/V.

ethanol

solution

solution

4.1 L

55 L

?

�

�

�

�

�

�

�

�

2ZnCl

solution

solution

16 g

500 mL

?

�

�

Jim

Rectangle


3.

� �

� �

�

6 12 6C H Osolution

solution

solution

100%

27.5 g100%

550 mL5.0%W/V

�

�

The concentration of glucose in the IV solution is 5.0% W/V.

PRACTICE

(Page 128)

NaOCl

NaOCl

NaOCl

4. 5.25 g NaOCl

74.44 g/mol NaOCl

100.0 mL

�

�

�

�

�

�1 L

1000 mL�

NaOCl

NaOCl

0.1000 L

?

�

�

�

NaOCl 5.25 g NaOCl� �1 mol NaOCl

74.44 g NaOCl�

NaOCl 0.0705 mol NaOCl� �

NaOClNaOCl

NaOCl

NaOCl

0.0705 mol NaOCl

0.100 L0.705 mol/LNaOCl

�

�

�

�

�

The molar concentration of the sodium hypochlorite solution is 0.705 mol/L. 5.

NaCl 235 g NaCl� �1 mol NaCl

58.44 g NaCl�

NaCl 4.02 mol NaCl� �

�

�

�

NaClNaCl

NaCl

NaCl

4.02 mol NaCl

3.00 L1.34 mol/LNaCl

�

�

The molar concentration of the salt solution is 1.34 mol/L.

�

�

�

6 12 6C H O

solution

solution

27.5 g

550 mL

?

�

�

NaCl

NaCl

NaCl

NaCl

235 g NaCl

58.44 g/mol NaCl

3.00 L

?

�

�

�

�

�

�

�


6.


36.46 g HCl�

HCl 0.210 mol HCl� �

HClHCl

HCl

HCl

0.210 mol HCl

1.50 L0.140 mol/L HCl

�

�

�

�

�

The molar concentration of the hydrochloric acid stock solution is 0.140 mol/L.

7.

�2 2 7K Cr O 2 2 7102.9 g K Cr O� �

2 2 7

2 2 7

1 mol K Cr O

294.20 g K Cr O

�2 2 7K Cr O 2 2 70.3498 mol K Cr O�

�

�

�

2 2 7

2 2 7

2 2 7

2 2 7

K Cr OK Cr O

K Cr O

2 2 7

K Cr O 2 2 7

0.3498 mol K Cr O

1.75 L0.200 mol K Cr O

�

�

The molar concentration of the potassium dichromate solution is 0.200 mol/L.

PRACTICE

(Page 130)

Understanding Concepts

8.

�

�

�

� � �

�

3

3

3

3 3 3

3

AgNOAgNO

AgNO

AgNO AgNO AgNO

23

AgNO 3

0.570 molAgNO5.00 10 L

L

0.0285 molAgNO

�

�

� �

�

There is 0.0285 mol silver nitrate in the solution.

HCl

HCl

HCl

HCl

7.66 g HCl

36.46 g/mol HCl

1.50 L

?

�

�

�

�

�

�

�

�

�

�

�

2 2 7

2 2 7

2 2 7

2 2 7

K Cr O

K Cr O 2 2 7

K Cr O

K Cr O

102.9 g

294.20 g/mol K Cr O

1.75 L

?

�

�

�

3AgNO 50.0 mL� �1 L

1000 mL�

3

3

3

2AgNO

AgNO

AgNO

5.00 10 L

0.570 mol/L

?

�

�

�

� �

�

�


9.

HClHCl

HCl

HCl HCl HCl

HCl

12.4 mol HCl1.50 L

L

18.6 mol HCl

�

�

� �

�

�

�

� �

�

There is 18.6 mol hydrochloric acid in 1.50 L of the solution.

10.

hemoglobinhemoglobin

hemoglobin

hemoglobin hemoglobin hemoglobin

32

5hemoglobin

1.90 10 mol hemoglobin2.50 10 L

L

4.75 10 mol hemoglobin

�

�

� �

�

�

�

�

�

�

��

� �

There is 4.75 � 10–5 mol hemoglobin in the solution.

11.

KOHKOH

KOH

KOH KOH KOH

22

4KOH

1.76 10 mol KOH3.58 10 L

L

6.30 10 mol KOH

�

�

� �

�

�

�

�

�

�

��

� �

The technician added 6.30 � 10–4 mol potassium hydroxide to the mixture.

PRACTICE

(Page 131)


�

�

�

HCl

HCl

HCl

12.4 mol/L

1.50 L

?

�

�

�hemoglobin 25.0 mL� �1 L

1000 mL�

�

� �

� �

�

2hemoglobin

3hemoglobin

hemoglobin

2.50 10 L

1.90 10 mol/L

?

�

�

KOH

2KOH

2KOH

KOH

1 L35.8 mL

1000 mL

3.58 10 L

1.76 10 mol/L

?

�

�

�

�

�

� �

� �

� �

�

2

2

2

MgCl

MgCl 2

MgCl

0.055 mol/L

4.1 mol MgCl

?

�

�

�

�

�


2

2

2

2

2

2

2

MgClMgCl

MgCl

MgClMgCl

MgCl

2

MgCl

4.1 mol MgCl

0.055 mol/L75 L

�

�

��

�

�

�

�

�

Therefore, 75 L of magnesium chloride solution contains 4.1 mol magnesium chloride. 13.

HClHCl

HCl

HClHCl

HCl

HCl

0.050 mol HCl

7.6 mol/L0.0066 L

�

�

��

�

�

�

�

�

Therefore, 6.6 mL (0.0066 L) of hydrochloric acid solution must be poured into the flask.

14.

3

3

3

3

3

3

3

FeClFeCl

FeCl

FeClFeCl

FeCl

3

FeCl

1.25 mol FeCl

6.00 mol/L0.208 L

�

�

��

�

�

�

�

�

Therefore, 208 mL (0.208 L) of stock solution must be poured into the flask.

15.

2 2 7

2 2 7

2 2 7

2 2 7

2 2 7

2 2 7

2 2 7

Na Cr ONa Cr O

Na Cr O

Na Cr ONa Cr O

Na Cr O

2 2 7

Na Cr O

5.0 mol Na Cr O

0.0020 mol/L2500 L

�

�

��

�

�

�

�

�

A volume of 2500 L of the sodium dichromate solution contains 5.0 mol sodium dichromate.

HCl

HCl

HCl

7.6 mol/L

0.050 mol HCl

?

�

�

�

�

�

3

3

3

FeCl 3

FeCl

FeCl

1.25 mol FeCl

6.00 mol/L

?

�

�

�

�

�

2 2 7

2 2 7

2 2 7

Na Cr O

Na Cr O 2 2 7

Na Cr O

0.0020 mol/L

5.0 mol Na Cr O

?

�

�

�

�

�


PRACTICE

(Page 133)


�

�

�

� �

2

2

2

2

CH OCH O

CH O

3CH O

3.2 mg

500 L

6.4 10 mg/L

�

�

The concentration of formaldehyde in the 200-L sample is 6.4 � 10–3 ppm.

17.

5Cu 3.0 10 g Cu� �

� �1000 mg Cu

1 g Cu�

2Cu 3.0 10 mg Cu� �

� �

�

�

�

��

� �

�

CuCu

Cu

2

2

Cu

3.0 10 mg

1.0 L

3.0 10 mg/L

0.030 ppm

�

�

The concentration of copper in the drinking water is 0.030 ppm.

18.

�

�

�

�

2

2

2

2

COCO

CO

CO

1.8 mg

0.35 L5.1 mg/L

5.1 ppm

�

�

The concentration of dissolved carbon dioxide in the pond water is 5.1 ppm.

2

2

2

CH O 2

CH O

CH O

3.2 mg CH O

500 L

?

1 ppm 1 mg/L

�

�

�

�

�

�

�

� �

�

�

�

5Cu

Cu

Cu

3.0 10 gCu

1.0 L

?

1 ppm 1 mg/L

�

�

�

�

�

�

�

2

2

2

2

CO

CO

CO

CO

1.8 mg

350 mL

?

1 ppm 1 mg/L

350 mL

�

�

� �1 L

1000 mL

�2CO 0.035 L�


PRACTICE

(Page 136)


� �

�

�

�

i i f f

i if

f

15% 240 mL

� �

�

�

� �300.0 mL

�f 12% W/V

The final concentration of the glucose solution is 12% W/V.

20.

�

�

�

i i f f

f fi

i

0.100 mol/L

� �

��

� �� 500.0 mL

16.0 mol/L

�i 3.13 mL�

Therefore, 3.13 mL of sulfuric acid is needed to make the sulfuric acid solution.

21.

�

�

�

i i f f

f fi

i

1.00 mol/L

� �

��

� �� 100.0 mL

14.8 mol/L

�i 6.76 mL�

Therefore, 6.76 mL of 14.8-mol/L ammonia solution is needed to form 100 mL of 1.00-mol/L ammonia solution.

22.

�

�

�

�

i

i

f

f

15% W/V

240 mL

300.0 mL

?

�

�

�

�

�

�

i

f

f

i

16.0 mol/L

0.100 mol/L

500.0 mL

?

�

�

�

�

�

�

i

f

f

i

14.8 mol/L

1.00 mol/L

100.0 mL

?

�

�

�

�

�

�

i

i

f

f

0.400 mol/L

125 mL

500.0 mL

?

�

�


� �

�

�

�

i i f f

i if

f

0.400 mol/L 125 mL

� �

�

�

� �500.0 mL

�f 0.100 mol/L

The final concentration of the barium chloride solution is 0.1 mol/L.


(Page 137)


1.

� �

�

�

2 2

2 2

H Osolution

solution

solution solutionH O

100%

100%

(6.0 g/ mL

�

�

��

�5 %)(1.2 10 mL

�2 2H O 2 2

)

100%7200 gH O�

Therefore, 7200 g of hydrogen peroxide is required.

2.

�

�

�

� � �

�

�

�

�

�

FF

F

F F F

F

(1.5 mg/L)(0.25 L)

0.38 mg

�

�

� �

�

The maximum mass of fluorine in the glass of water is 0.38 mg.

3.

�solution 500 bottles� �250 mL

1 bottle

� �

�

�2 2

5solution

solution

H O

1.2 10 mL

6.0% W/V

?

�

�

F250 mL� � �

1 L

1000 mL�

F

F

F

0.25 L

1.5 ppm or 1.5 mg/L

?

1 ppm 1 mg/L

�

�

�

�

�

�

�

�

�

�12 22 11C H O 50.0 mL� �

1 L

1000 mL

�

�

�

12 22 11

12 22 11

12 22 11

C H O

C H O

C H O

0.050 L

0.50 mol/L

?

�

�


12 22 11

12 22 11

C H O

C H O

12 22 11 12 22 11

12(12.01 g/mol) 22(1.01 g/mol) 11(16.00 g/mol)

342.34 g/mol

1 mol C H O 342.34 g C H O

�

�

� � �

�

�

� ��

12 22 11

12 22 11

12 22 11

12 22 11 12 22 11 12 22 11

12 22 11

C H OC H O

C H O

C H O C H O C H O

12 22 11

C H O 12 22 11

0.50 mol C H O ( )(0.050 L)

L

0.025 mol C H O

�

�

� �

�

�

�

�

�

12 22 11C H O 12 22 110.025 mol C H O� �12 22 11

12 22 11

342.34 g C H O

1 mol C H O�

12 22 11C H O 12 22 118.6 g C H O� �

Therefore, 8.6 g of sucrose is in the sucrose solution.

4. 3 4H PO = 40.2 g�

�3 4H PO 250 mL� �

1 L

1000 mL

�3 4

3 4

H PO

H PO

0.25 L

= ?

�

� � �

�

3 4

3 4

H PO

H PO

3(1.01 g/mol) (30.97 g/mol) 4(16.00 g/mol)

98.00 g/mol

�

�

�3 4H PO 3 440.2 g H PO� �

3 4

3 4

1 mol H PO

98.00 g H PO

�3 4H PO 3 40.4100 mol H PO�

�

�

�

3 4

3 4

3 4

3 4

H POH PO

H PO

3 4

H PO

0.4100 mol H PO

0.25 L1.64 mol/L

�

�

The molar concentration of the phosphoric acid solution is 1.64 mol/L.

5. (a)

� �

�

�

�

i i f f

i if

f

1.50 mol/L 45.5 mL

� �

�

�

� �200.0 mL

�f 0.341 mol/L

The concentration of the dilute sodium sulfate solution is 0.341 mol/L.

�

�

�

�

i

i

f

f

1.50 mol/L

45.5 mL

200.0 mL

?

�

�


(b)

�

�

�

i i f f

i if

f

3.50 mol/L

� �

��

� �� 50.0 mL

2.50 mol/L

�f 70.0 mL�

The final volume of the diluted nitric acid solution is 70.0 mL.

Applying Inquiry Skills 6. (a)

�

�

�

i i f f

f fi

i

0.010 mol/L

� �

��

�

� �(250 mL)

0.25 mol/L

�i 10 mL�

Therefore, 10 mL of the sodium carbonate stock solution is needed to prepare the diluted solution.

(b) Step 1. Place 10.0 mL of sodium carbonate stock solution, Na2CO3(aq), into a clean, dry 250-mL volumetric flask. Step 2. Add enough distilled water to bring the volume to the 250-mL mark on the volumetric flask. Step 3. Place the stopper on the volumetric flask and invert several times to mix.

Making Connections 7. LD50 is a test that measures the concentration of a substance that kills 50% of the animals tested. The LD50 value is

obtained by administering increasing concentrations of a test product until half of the test animals die within 14 days of a single administration. Test animals are usually mice, rats, rabbits, or hamsters. A substance with an LD50 between 1 to 50 ppm is considered highly toxic, and a substance with an LD50 between 500 to 5000 ppm is considered slightly toxic.

8. Concentrations must be clearly communicated to health care workers to avoid errors when dispensing or administering medications to patients. Concentration units commonly used in the health care field include percentage concentrations in W/V and V/V, mg/100 mL, mg/dL, mol/L, �mol/L, and ppm.

2.6 TECH CONNECT: THE SPECTROPHOTOMETER

CAREER CONNECTION: CHEMICAL LABORATORY TECHNICIAN (i) Student answers will vary. Educational requirements for admission to a chemical laboratory technician program at

Seneca College of Applied Arts and Technology are: Ontario Secondary School Diploma with a majority of senior credits at the College Preparation (C), University

Preparation (U) or University/College Preparation (M) level Grade 12 English: ENG4 (C) or ENG4 (U) Grade 11 Biology (C)

�

�

�

�

i

f

i

f

3.50 mol/L

2.50 mol/L

50.0 mL

?

�

�

�

�

�

�

i

f

f

i

0.25 mol/L

0.010 mol/L

250 mL

?

�

�

Jim

Rectangle


Grade 12 Chemistry (C) Grade 12 Mathematics: MCT4 (C) or any Grade 12 (U) Mathematics (ii) Ontario Job Futures 2000 rates the outlook for chemical technicians as good over the next five years. Employment

for this occupation is expected to grow through the year 2005. Demand for chemical laboratory technicians should increase as chemical companies continue to research and develop new chemicals and as older workers who retire are replaced.

(iii) Student answers will vary. Chemical laboratory technicians have to know about math, chemistry, environmental science, and laboratory safety rules and methods. Chemical laboratory technicians should be patient, adaptable, self-disciplined, and observant. They should also be able to work with a team, work well under pressure, and have creative ability to find solutions to laboratory problems.

TECH CONNECT 2.6 QUESTIONS

(Page 139)

Understanding Concepts 1. (a) Gravimetric analysis is a method used to determine the concentration of a solute in a solution. Gravimetric

analysis involves precipitating solutes out of solution, determining the mass of each solute, converting the mass to amount, and calculating the concentration of the solution.

(b) Gravimetric analysis is not very useful when the concentration of an unknown solute is very low. 2. (a) A spectrophotometer measures the percentage of light of a particular colour (wavelength) that a solution absorbs

or transmits. In a spectrophotometer, a beam of white light is dispersed into its various wavelengths (colours). A movable slit allows individual wavelengths to pass through a sample of the solution of unknown concentration. Light that is transmitted by the solution strikes a photocell. The photocell converts the light energy into an electric current, which is then measured and converted to a percent transmittance value. The percent transmittance value is proportional to the concentration of the solution.

(b) The percent absorbance or transmittance values are used to produce a standard curve of concentration versus percent transmittance or percent absorbance. The percent transmittance or percent absorbance of a sample of the solution of unknown concentration is measured using the spectrophotometer, and this value and the standard curve are used to determine the concentration of the solution.

3. (a) A spectrophotometer measures the percentage of light of a particular colour absorbed or transmitted by a solution. Only coloured solutions absorb or transmit some of the light. Colourless solutions absorb 0% of light and transmit 100% of light. Therefore, they cannot be used in a spectrophotometer.

(b) Percent transmittance or absorbance values are used to determine the concentration of the solution. Cuvettes are colourless because the container holding the solution in a spectrophotometer should transmit all of the light, so that only the solution being tested absorbs light.

Making Connections 4. (a) Forensic laboratories investigate crimes and determine the identity of drugs and other chemicals recovered from

the scene of a crime. (b) Forensic researchers use infrared spectrophotometers to identify compounds such as drugs and poisons. (c) Spectrophotometers are used to determine the concentrations of substances such as drugs or poisons in blood

and other body fluids to determine if people or animals contain abnormal concentrations. They are also used to identify compounds that may have been used in crimes.

2.7 ACTIVITY: DETERMINING THE CONCENTRATION OF A SOLUTION

(Pages 140–141)

Observations The spectrophotometer wavelength was set at 640 nm.

Jim

Rectangle

Jim

Rectangle


Table 1 Observation Table

Test tube [Cu2+ ] (mol/L) Absorbance

1 0.50 1.61

2 0.25 0.84

3 0.14 0.53

4 0.07 0.36

5 0.04 0.16

6 0.02 0.12

7 0.002 0.10

8 0.0002 0.08

9 0 0

10 ? 1.02

Analysis (a) (b) This graph tells us that absorbance is proportional to concentration. (c) By interpolation in the Absorbance versus [Cu ] graph, the concentration of the copper sulfate solution is

approximately 0.3 mol/L.

Evaluation (d) It is necessary to zero the spectrophotometer when a blank is used because the blank is a colourless solution that

does not absorb light at the set wavelength (640 nm). Since the difference in colour between the copper(II) sulfate solutions is compared to the blank, the absorbance of the blank is set to zero on the absorbance scale of the spectrophotometer.

(e) Measurement error during the dilution process may have produced copper(II) sulfate solutions with actual concentrations that were different than those calculated.

(f) More precise instruments, such as volumetric pipettes, could be used to measure the volume of the liquids used in this activity to help reduce measurement error.

(aq)

Jim

Rectangle


2.9 THE MOLE AND CHEMICAL EQUATIONS: STOICHIOMETRY

PRACTICE

(Page 148)


Balanced equation 2 Al2O3(s) → 4 Al(s) + 3 O2(g)

Given mass (g) 125 g

Molar mass (g/mol) 101.96 g/mol 26.98 g/mol 32.00 g/mol

2 3Al O 2 3125 g Al O� �2 3

2 3

1 molAl O

101.96 gAl O�

2 3Al O 2 3

Al 2 3

1.226 molAl O

1.226 molAl O

�

�

�

�

2 3

4 molAl

2 molAl O�

Al

Al

2.452 molAl

2.452 molAl

�

�

�

�26.98 gAl

1 molAl�

Al 66.2 gAl� �

Therefore, 66.2 g of aluminum is produced from 125 g of aluminum oxide.

The combined calculation is as follows:

�Al 2 3125 gAl O� �2 31 molAl O

2 3101.96 gAl O�

4 molAl

2 32 molAl O�

26.98 gAl

1 molAl

�Al 66.2 gAl�

Therefore, 66.2 g of aluminum is produced from 125 g of aluminum oxide. 2.

Balanced equation 2 K(s) + 2 HCl(aq) → 2 KCl(aq) + H2(g)

Given mass (g) 5.00 g

Molar mass (g/mol) 39.16 g/mol 1.01 g/mol

(2.02 g/mol)


2H 25.00 g H� �2

2

1 mol H

2.02 g H�

2H 2

K 2

2.48 mol H

2.48 mol H

�

�

�

�

2

2 mol K

1 mol H�

K

K

4.96 mol K

4.96 mol K

�

�

�

�39.10 g K

1 mol K�

K 194 g K� �

Therefore, 194 g of potassium is required to produce 5.00 g of hydrogen gas.


�K 25.00 gH� �21 molH

22.02 gH�

2 molK

21 molH�

39.10 gK

1 molK

�K 194 gK�

Therefore, 194 g of potassium is required to produce 5.00 g of hydrogen gas. 3.

Balanced equation 2 KClO3(s) → 2 KCl(s) + 2 O2(g)


1 mol KClO3 = 6.02 � 1023 formula units KClO3

2O 20.96 g O� �2

2

1 mol O

32.00 g O�

2

3

O 2

KClO 2

0.030 mol O

0.030 mol O

�

�

�

�3

2

2 mol KClO

3 mol O�

3

3

KClO 3

KClO 3

0.020 mol KClO

0.020 mol KClO

�

�

�

�

233

3

6.02×10 formula units KClO

1 mol KClO�

3

22KClO 31.2 10 formula units KClO� � �

Therefore, 1.2 � 1022 formula units of potassium chlorate must decompose.


3KClO 20.96 g O� �21 mol O

�

232.00 g O

32 mol KClO�

23 mol O

233

3

6.02 10 formula units KClO

1 mol KClO

�

�

3

22KClO 31.2 10 formulaunits KClO� � �

Therefore, 1.2 � 1022 formula units of potassium chlorate must decompose.



(Page 148)

Understanding Concepts 1. A mole ratio is the ratio of moles of two or more entities in a balanced chemical equation. 2. Yes, the total amount (in moles) of atoms in the reactants equals the total amount (in moles) of atoms in the products

because one mole of any type of atom is 6.02 � 1023 atoms. 3. 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g)

The mole ratio of the reactants is 3:1. The mole ratio of the products is 2:1. 4. Stoichiometry is the procedure for calculating quantities of reactants or products in a chemical reaction. In

stoichiometry, mole ratios in balanced equations are used to calculate quantities of reactants of products produced. 5.

Balanced equation 2 H2O(l) → 2 H2(g) + O2(g)


1 mol H2 = 6.02 � 1023 molecules H2

2H O 212.0 g H O� �2

2

1 mol H O

18.02 g H O�

2

2

H O 2

H 2

0.666 mol H O

0.666 mol H O

�

�

�

�2

2

2 mol H

2 mol H O�

2

2

H 2

H 2

0.666 mol H

0.666 mol H

�

�

�

�

232

2

6.02 10 molecules H

1 mol H

�

�

2

23H 24.01 10 molecules H� � �

Therefore, 4.01 � 1023 molecules of hydrogen gas are produced from the decomposition of 12.0 g of water.


2H 212.0 g H O� �21 mol H O

�

218.02 g H O

22 mol H�

22 mol H O

232

2

6.02 10 molecules H

1 mol H

�

�

2

23H 24.01 10 molecules H� � �

Therefore, 4.01 � 1023 molecules of hydrogen gas are produced from the decomposition of 12.0 g of water.


2.10 LIMITING AND EXCESS REAGENTS

PRACTICE

(Page 153)


1. (a)

Balanced equation 2 H2(g) + O2(g) → 2 H2O(l)

Before reaction 10.0 g 32.00 g 0 g

Reaction according to balanced equation

2 mol

4.04 g

1 mol

32.00 g

2 mol

After reaction 5.96 g 0 g ? g

According to the chart, oxygen is the limiting reagent, and hydrogen is the excess reagent.

(b)

2O 232.00 g O� �2

2

1 mol O

32.00 g O�

2

2

O 2

H O 2

1.000 mol O

1.000 mol O

�

�

�

�2

2

2 mol H O

1 mol O�

2

2

H O 2

H O 2

2.000 mol H O

2.000 mol H O

�

�

�

�2

2

18.02 g H O

1 mol H O�

2H O 236.04 g H O� �

Therefore, 36.04 g of water is produced.


2H O 232.00 g O� �21 mol O

�

232.00 g O

22 mol H O�

21 mol O2

2

18.02 g H O

1 mol H O�

2H O 236.04 g H O� �

Therefore, 36.04 g of water is produced.

2.

2

2

H 2

O 2

10.0 g H

32.00 g O

�

�

�

�

4

2

CH 4

O 2

32.1 g CH

160.0 g O

�

�

�

�


Balanced equation 2 CH4(g) + 3 O2(g) → 2 CO(g) + 4 H2O(g)

Before reaction 32.1 g 160.0 g 0 g 0 g


2 mol

(32.10 g)

3 mol

(96.00 g)

After reaction 0 g 64.0 g ?

According to the chart, methane is the limiting reagent and oxygen is the excess reagent.

4CH 432.1 g CH� �4

4

1 mol CH

16.05 g CH�

4CH 4

CO 4

2.00 mol CH

2.00 mol CH

�

�

�

�

4

2 mol CO

2 mol CH�

CO

CO

2.00 mol CO

2.00 mol CO

�

�

�

�28.01 g CO

1 mol CO�

CO 56.0 g CO� �

Therefore, 56.0 g of carbon monoxide is produced during incomplete combustion. The combined calculation is as follows:

CO 432.1 g CH� �41 mol CH

�

416.05 g CH

2 mol CO�

42 mol CH

28.01 g CO

1 mol CO�

CO 56.0 g CO� �

Therefore, 56.0 g of carbon monoxide is produced during incomplete combustion. 3. (a)

Balanced equation 2 SO2(g) + O2(g) → 2 SO3(g)



2 mol

(128.00 g)

1 mol

(32.00 g)

After reaction 64.18 g 0 g ?

According to the chart, oxygen is the limiting reagent and sulfur dioxide is the excess reagent. (b)

2O 232.00 g O� �2

2

1 mol O

32.00 g O�

2

3

O 2

SO 2

1.000 mol O

1.000 mol O

�

�

�

�3

2

2 mol SO

1 mol O�

3SO 32.00 molSO� �

Therefore, 2.00 mol sulfur trioxide is produced.

2

2

SO 2

O 2

192.18 g SO

32.00 g O

�

�

�

�


(c)

Therefore, 160.1 g of sulfur trioxide is produced.

4.

Balanced equation P4(s) + 10 Cl2(g) → 4 PCl5(s)



1mol

(123.88 g)

10 mol

(709.0 g)


4P 4123.88 g P� �4

4

1 mol P

123.88 g P�

4

5

P 4

PCl 4

1.000 mol P

1.000 mol P

�

�

�

�5

4

4 mol PCl

1 mol P�

5

5

PCl 5

PCl 5

4.000 mol PCl

4.000 mol PCl

�

�

�

�5

5

208.22 g PCl

1 mol PCl�

5PCl 5832.9 g PCl� �

The mass of phosphorus pentachloride produced from the reaction is 832.9 g.


5PCl 4123.88 g P� �41 mol P

�

4123.88 g P

54 mol PCl�

41 mol P5

5

208.22 g PCl

1 mol PCl�

5PCl 5832.9 g PCl� �

The mass of phosphorus pentachloride produced from the reaction is 832.9 g.


(Page 153)

Understanding Concepts 1. The excess reagent is the reactant that is not completely consumed in a chemical reaction. The limiting reagent is the

reactant that is completely consumed in a chemical reaction. The limiting reagent limits the amount of product that can be formed.

2. No, there does not need to be a limiting reagent in a chemical reaction. If reactants are mixed in amounts equal to, or proportional to, the coefficients in the balanced reaction equation, then all reagents will be completely consumed.

4

2

2

P 4

2Cl 2

2

3Cl 2

123.88 g P

1000 g Cl1.00 kg Cl

1 kg Cl

1.00 10 g Cl

�

�

�

�

� �

� �

3SO 32.00 mol SO� �3

3

80.06 g SO

1 mol SO�

3SO 3160.1 g SO� �


3. Substance A must be the limiting reagent, and substance B must be in excess. All of substance A must react to determine the amount of product formed and, therefore, the initial amount of substance A. The only way to be sure that all of substance A has reacted is to use an excess of substance B.

4. (a)

Balanced equation 2 CH4→(g) + 3 O2(g) → 2 CO(g) + 4 H2O(g)

Given mass (g) 6.4 g ?


4CH 46.4 g CH� �4

4

1 mol CH

16.05 g CH�

4

2

CH 4

O 4

0.40 mol CH

0.40 mol CH

�

�

�

�2

4

3 mol O

2 mol CH�

2

2

O 2

O 2

0.60 mol O

0.60 mol O

�

�

�

�2

2

32.00 g O

1 mol O�

2O 219 g O� �

The mass of oxygen required to react with 6.4 g of methane is 19 g.


2O 46.4 g CH� �41 mol CH

�

416.05 g CH

23 mol O�

42 mol CH2

2

32.00 g O

1 mol O�

2O 219 g O� �

The mass of oxygen required to react with 6.4 g of methane is 19 g.

(b)

Balanced chemical equation

2 CH4(g) + 3 O2(g) → 2 CO(g) + 4 H2O(g)



4

2

CH 4

O

6.4 g CH

?

�

�

�

�

4CH 4

CO

6.4 g CH

?

�

�

�

�


From (a):

4CH 4

CO 4

0.40 mol CH

0.40 mol CH

�

�

�

�

4

2 mol CO

2 mol CH�

CO

CO

0.40 mol CO

0.40 mol CO

�

�

�

�28.01 g CO

1 mol CO�

CO 11 g CO� �

The mass of carbon monoxide formed is 11 g.


CO 46.4 g CH� �41 mol CH

�

416.05 g CH

2 mol CO�

42 mol CH

28.01 g CO

1 mol CO�

CO 11 g CO� �

The mass of carbon monoxide formed is 11 g.

5.

Balanced chemical equation 2 Al(s) + 3 H2SO4(aq) → Al2(SO4)3(aq) + 3 H2(g)


Molar mass (g/mol) 26.98 g/mol 342.14 g/mol

Al 6.71 g Al� �1 mol Al

26.98 g Al�

2 4 3

Al

Al (SO )

0.2487 mol Al

0.2487 mol Al

�

�

�

�2 4 31 mol Al (SO )

2 mol Al�

2 4 3

2 4 3

Al (SO ) 2 4 3

Al (SO ) 2 4 3

0.1244 mol Al (SO )

0.1244 mol Al (SO )

�

�

�

�2 4 3

2 4 3

342.14 g Al (SO )

1 mol Al (SO )�

2 4 3Al (SO ) 2 4 342.5 g Al (SO )� �

The mass of aluminum sulfate formed is 42.5 g.


�2 4 3Al (SO ) 6.71 gAl� �

1 molAl

26.98 gAl�

2 4 31 molAl (SO )

2 molAl�

2 4 3

2 4 3

342.14 gAl (SO )

1 molAl (SO )

�2 4 3Al (SO ) 2 4 342.5 gAl (SO )�

The mass of aluminum sulfate formed is 42.5 g.

2 4 3

Al

Al (SO )

6.71 g Al

?

�

�

�

�


6.

Balanced chemical equation 2 Cu2O(s) + Cu2S(s) → 6 Cu(s) + SO2(g)

Before reaction 286.2 g 286.2 g 0 g 0 g

Reaction according to balanced chemical equation

2 mol

(286.20 g)

1 mol

(159.16 g)

6 mol

(381.30 g)

1 mol

(64.06 g)


From the table, the limiting reagent is Cu2O(s) and the excess reagent is Cu2S(s).

2Cu O 2286.2 g Cu O� �2

2

1 mol Cu O

143.1 g Cu O�

2Cu O 2

Cu 2

2.000 mol Cu O

2.000 mol Cu O

�

�

�

�

2

6 mol Cu

2 mol Cu O�

Cu

Cu

6.000 mol Cu

63.55 g Cu6.000 mol Cu

1 mol Cu

�

�

�

� �

Cu 381.3 g Cu� �

The mass of copper obtained from the reaction is 381.3 g.


Cu 2286.2 g Cu O� �21 mol Cu O

�

2143.1 g Cu O

6 mol Cu�

22 mol Cu O

63.55 g Cu

1 mol Cu�

Cu 381.3 g Cu� �

The mass of copper obtained from the reaction is 381.3 g.

2.11 INVESTIGATION: THE LIMITING REAGENT IN A CHEMICAL REACTION

(Page 154)

Prediction (a) According to the stoichiometric method, 2.32 g of strontium sulfate precipitate will be produced from the reaction of

2.00 g of strontium chloride with excess copper(II) sulfate. The reaction equation is shown below.

�

�

�

2

2

Cu

Cu O 2

Cu S 2

?

286.2 g Cu O

286.2 g Cu S

�

�

�

Jim

Rectangle


Observations (e) mass of 250-mL beaker = 102.41 g

mass of 250-mL beaker + SrCl2(s) = 104.41 g mass of 150-mL beaker = 68.83 g mass of 150-mL beaker + CuSO4·5H2O(s) = 71.39 g mass of filter paper = 0.90 g mass of filter paper and precipitate = 3.24 g The blue copper(II) sulfate solution mixed with the strontium chloride solution to produce a white precipitate and a blue solution.

Analysis (f) mass of precipitate = 3.24 g – 0.90 g = 2.34 g (g) According to the observations collected in this experiment, the mass of the strontium sulfate precipitate from the

reaction of strontium chloride and copper(II) sulfate solutions was determined to be 2.34 g.

Evaluation (h) The major source of error is likely caused by the number of mass measurements made. Making more measurements

would help to reduce the error. (i) The filtration design is adequate, and there are no obvious improvements to be made. Adequate care was taken in

filtering and washing the precipitate. �

�2.34 g 2.32 g

% difference = 1002.32 g

% difference = 0.86%

Based on the very low percent difference, the Prediction is valid. (j) The stoichiometric method is valid because the Prediction was verified.

2.12 PERCENTAGE YIELD

PRACTICE

(Pages 158–159)


7 6 3C H O 7 6 3

8 8 3

used 2.00 g C H O

actual yield 1.65 g C H O

� �

�

Balanced equation C7H6O3(s) + CH3OH(l) → C8H8O3(l) + H2O(l)

Given mass (g) 2.00 1.65

Molar mass (g/mol)

138.13 32.05 152.16 18.02

Jim

Rectangle

Jim

Rectangle


7 6 3C H O 7 6 32.00 g C H O� �7 6 3

7 6 3

1 mol C H O

138.13 g C H O�

7 6 3

8 8 3

C H O 7 6 3

C H O 7 6 3

0.01448 mol C H O

0.01448 mol C H O

�

�

�

�8 8 3

7 6 3

1 mol C H O

1 mol C H O�

8 8 3

8 8 3

C H O 8 8 3

C H O 8 8 3

0.01448 mol C H O

0.01448 mol C H O

�

�

�

�8 8 3

8 8 3

152.16 g C H O

1 mol C H O�

8 8 3C H O 8 8 32.203 g C H O� �


8 8 3C H O 7 6 32.00 g C H O� �7 6 31 mol C H O

�

7 6 3138.13 g C H O

8 8 31 mol C H O�

7 6 31 mol C H O8 8 3

8 8 3

152.16 g C H O

1 mol C H O�

8 8 3C H O 8 8 32.203 g C H O� �

� �

� �

�

actualyieldpercentageyield 100%

theoreticalyield

1.65 g100%

2.203 g

percentageyield 74.9%

The percentage yield of wintergreen is 74.9%. 2. (a) The balanced chemical equation is

Al(s) + Br2(l) → AlBr2(s) (b)

Balanced chemical equation 2 Al(s) + 3 Br2(l) → 2 AlBr3(s)


Molar mass (g/mol) 26.98 159.80 266.68

2Br

3

53.7 g Br

actual yield 48.4 g AlBr

� �

�


2Br 253.7 g Br� �2

2

1 mol Br

159.80 g Br�

2

3

Br 2

AlBr 2

0.336 mol Br

0.336 mol Br

�

�

�

�3

2

2 mol AlBr

3 mol Br�

3

3

AlBr 3

AlBr 3

0.224 mol AlBr

0.224 mol AlBr

�

�

�

�3

3

266.68 g AlBr

1 mol AlBr�

3AlBr 359.74 g AlBr� �


3AlBr 253.7 g Br� �21 mol Br

�

2159.80 g Br

32 mol AlBr�

23 mol Br3

3

266.68 g AlBr

1 mol AlBr�

3AlBr 359.74 g AlBr� �

� �

� �

�

actual yieldpercentage yield 100%

theoretical yield

48.4 g100%

59.74 g

percentage yield 81.0%

The percentage yield of aluminum bromide is 81.0%.

3.

Balanced equation Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)

Given mass (g) 0.999 1.541

Molar mass (g/mol) 36.46 136.28


36.46 g HCl�

2

HCl

ZnCl

0.0274 mol HCl

0.0274 mol HCl

�

�

�

�21 mol ZnCl

2 mol HCl�

2

2

ZnCl 2

ZnCl 2

0.0137 mol ZnCl

0.0137 mol ZnCl

�

�

�

�2

2

136.28 g ZnCl

1 mol ZnCl�

2ZnCl 21.867 g ZnCl� �

HCl

2

0.999 g HCl

actualyield 1.541 g ZnCl

� �

�



2ZnCl 0.999 g HCl� �1 mol HCl

�

36.46 g HCl

21 mol ZnCl�

2 mol HCl2

2

136.28 g ZnCl

1 mol ZnCl�

2ZnCl 21.867 g ZnCl� �

2


theoretical yield

1.541 g ZnCl

� �

�

21.867 g ZnCl100%


�

�

The percentage yield of zinc chloride is 82.5%. 4. (a)

Balanced equation

C7H6O3(s) + C4H6O3(aq) → C9H8O4 (s) + HC2H3O2(aq)

Given mass (g) 213.0 189.3

Molar mass (g/mol)

138.13 180.17

7 6 3C H O 7 6 3213.0 g C H O� �7 6 3

7 6 3

1 mol C H O

138.13 g C H O�

7 6 3

9 8 4

C H O 7 6 3

C H O 7 6 3

1.542 mol C H O

1.542 mol C H O

�

�

�

�9 8 4

7 6 3

1 mol C H O

1 mol C H O�

9 8 4

9 8 4

C H O 9 8 4

C H O 9 8 4

1.542 mol C H O

1.542 mol C H O

�

�

�

�9 8 4

9 8 4

180.17 g C H O

1 mol C H O�

9 8 4C H O 9 8 4277.8 g C H O� �

The theoretical yield of aspirin is 277.8 g.


9 8 4C H O 7 8 3213.0 g C H O� �7 6 31 molC H O

�

7 6 3138.13 gC H O

9 8 41 mol C H O�

7 6 31 mol C H O9 8 4

9 8 4

180.17 g C H O

1 mol C H O�

9 8 4C H O 9 8 4277.8 g C H O� �

The theoretical yield of aspirin is 277.8 g.

7 6 3C H O 7 6 3

9 8 4

213.0 g C H O

actualyield 189.3 g C H O

� �

�


(b) The percentage yield of aspirin is 68.14%.


(Page 159)

Understanding Concepts 1. Actual yield is the mass of a product measured when a chemical reaction is carried out. Theoretical yield is the mass

of a product calculated using the information provided by a balanced chemical equation. It is the mass that is expected to be obtained if there is no experimental error.

2. No, the actual yield cannot truly be greater than the theoretical yield. The actual yield may be greater than the theoretical due to experimental error, such as errors in determining the mass or including by-products as part of the final mass.

3.

(a) The balanced chemical equation for the reaction is:

AgNO3(s) + NaBr(s) → AgBr(s) + NaNO3(s) (b)

Balanced equation AgNO3(s) + NaBr(s) → AgBr(s) + NaNO3(s)



3AgNO 35.00 g AgNO� �3

3

1 mol AgNO

169.88 g AgNO�

3AgNO 3

AgBr 3

0.02943 mol AgNO

0.02943 mol AgNO

�

�

�

�

3

1 mol AgBr

1 mol AgNO�

AgBr

AgBr

0.02943 mol AgBr

0.02943 mol AgBr

�

�

�

�187.77 g AgBr

1 mol AgBr�

AgBr 5.53 g AgBr� �

The theoretical yield of silver bromide is 5.53 g.


AgBr 35.00 g AgNO� �31 mol AgNO

�

3169.88 g AgNO

1 mol AgBr�

31 mol AgNO

187.77 g AgBr

1 mol AgBr�

AgBr 5.53 g AgBr� �

The theoretical yield of silver bromide is 5.53 g.


theoretical yield

189.3 g100%

277.8 g


� �

� �

�

3AgNO 35.00 gAgNO

actualyield 5.03 gAgBr

� �

�


(c) The actual yield of silver bromide is given as 5.03 g.

(d) The percentage yield of silver bromide is 91.0%.

4.

Balanced equation 4 FeS(s) + 7 O2(g) → 2 Fe2O3(s) + 4 SO2(g)



FeS 16.1 g FeS� �1 mol FeS

87.91 g FeS�

2 3

FeS

Fe O

0.1831 mol FeS

0.1831 mol FeS

�

�

�

�2 32 mol Fe O

4 mol FeS�

2 3

2 3

Fe O 2 3

Fe O 2 3

0.09155 mol Fe O

0.09155 mol Fe O

�

�

�

�2 3

2 3

159.70 g Fe O

1 mol Fe O�

2 3Fe O 2 314.62 g Fe O� �


2 3Fe O 16.1 g FeS� �1 mol FeS

�

87.91 g FeS

2 32 mol Fe O�

4 mol FeS2 3

2 3

159.70 g Fe O

1 mol Fe O�

2 3Fe O 2 314.62 g Fe O� �

� �

� �

�


theoretical yield

14.1 g100%

14.62 g


The percentage yield of iron(III) oxide is 96.4%.

Applying Inquiry Skills 5. The percentage yield may be increased by measuring the mass of the precipitate in the evaporating dish instead of

using weighing paper. (The mass of the evaporating dish would have to be known and subtracted from the total mass of the evaporating dish with the precipitate.) Aqueous reactants are lost because they stick to the surfaces of their containers. Limiting the number of container transfers may also increase percentage yield.


theoretical yield

5.03 g AgBr100%

5.53 g AgBr


� �

� �

�

FeS

2 3

16.1 g FeS

actualyield 14.1 g Fe O

� �

�


Making Connections 6. (a) The typical percentage yield of the carmine extraction process is 23% of 62% pure carmine. Carmine is

extracted by boiling the insects in water, followed by filtration, precipitation, and washing and drying the final product.

(b) Cochineal processing plants have boosted local employment in Peru. While they earn only about 10% of the revenue generated from cochineal processing, an estimated 50 000 people harvest the insects by hand, dry them in the sun, and sell them to carmine processors in Lima, Peru's capital city. It is estimated that as many as 400 000 rural families depend on this industry for their livelihood.

2.13 INVESTIGATION: PERCENTAGE YIELD OF A CHEMICAL REACTION

(Page 160)

Prediction (a) mass of copper(II) chloride dihydrate used = 2.00 g

�Cu 2 2 2.00 g CuCl 2H O� �

�2 21 mol CuCl 2H O�

�2 2170.49 g CuCl 2H O

3 mol Cu�

�2 23 mol CuCl 2H O

63.55 g Cu

1 mol Cu�

Cu 0.745 g Cu� �

The theoretical yield of copper is 0.745 g. It is predicted that the actual yield will be approximately 10% lower than the theoretical yield, or 0.671 g.

Qualitative Observations A reddish residue of copper formed immediately when the foil was added to the solution. Vigorous bubbling was observed on the aluminum surface as the reaction started. Some of the copper changed to a green colour after it was dried overnight. The aluminum foil broke up into many small pieces as the reaction occurred.

Quantitative Observations mass of copper(II) chloride dihydrate = 2.00 g mass of 150-mL beaker and copper residue = 86.38 g mass of empty 150-mL beaker = 85.97 g mass of copper residue collected = 86.38 g – 85.97 g mass of copper residue collected = 0.410 g

Analysis (b) There were many small pieces of aluminum left in the beaker when the reaction had come to a halt. Therefore,

aluminum is the excess reagent and copper(II) chloride dihydrate is the limiting reagent. (c) 0.410 g of copper was formed when 2.00 g of copper(II) chloride dihydrate reacted with excess aluminum. (d)

The percentage yield of copper is 55.0%.

actual yieldpercentage yield = 100%

theoretical yield

0.410 g = 100%

0.745 g

percentage yield = 55.0%

�

�

Jim

Rectangle

NEL Unit 2 Review Student Book Solutions 109

UNIT 2 REVIEW

(Pages 169–173)

Understanding Concepts 1. (a) The coefficients of aluminium, oxygen and aluminium oxide are 4, 3, and 2 respectively.

(b) The equation states that 4 mol solid aluminium reacts with 3 mol of gaseous oxygen to yield 2 mol solid aluminium oxide.

2. (a) There are 12 (twelve) doughnuts in one dozen doughnuts. There are 6.02 � 1023 doughnuts in one mole of doughnuts.

(b) 1 doughnut = 70 g 1 mol doughnuts = 6.02 � 1023 doughnuts

doughnuts 1 mol doughnuts� �

236.02 10 doughnuts

1 mol doughnuts

��

23doughnuts

23doughnuts

6.02 10 doughnuts

6.02 10 doughnuts

�

�

� �

� �70 g

1 doughnut�

25doughnuts 4 10 g� � �

The mass of one mole of doughnuts is 4 � 1025 g.


doughnuts 1 mol doughnuts� �

236.02 10 doughnuts�

�

1 mol doughnuts

70 g

1 doughnut�

25doughnuts 4 10 g� � �

The mass of one mole of doughnuts is 4 � 1025 g. (c) Student answers will vary. One mole of doughnuts is not a reasonable number of doughnuts because this

number is far too large. This many doughnuts would be unmanageable to do anything with, such as count, distribute, or eat.

3. (a)

Hg 1 mol Hg� �

236.02 10 atoms Hg

1 mol Hg

��

23Hg 6.02 10 atoms Hg� � �

There are 6.02 � 1023 atoms of mercury in 1 mol mercury.

(b)

Hg 1 mol Hg� �200.59 g Hg

1 mol Hg�

Hg 200.59 g Hg� �

One mole of mercury atoms has a mass of 200.59 g. (c) One mole of mercury atoms is a reasonable number of mercury atoms because it is a quantity that chemists can

measure, observe, and work with. 4. We use Avogadro's constant because it allows a chemist to know how many chemical entities (atoms, molecules, or

formula units) there are in a particular mass of the chemical.

Hg

23

1 mol Hg

1 mol Hg 6.02 10 atoms Hg

� �

� �

Hg

Hg

1 mol Hg

200.59 g Hg

�

�

�

�


5. (a)

C 0.50 carat C� �0.2 g C

1 carat C�

C

C

0.10 g C

0.10 g C

�

�

�

�1 mol C

12.01 g C�

3C 8.3 10 mol C� �

� �

The amount of carbon in the diamond is 8.3 � 10–3 mol.

3C 8.3 10 mol C� �

� �

236.02 10 atoms C

1 mol C

��

21C 5.0 10 atoms C� � �

There are 5.0 � 1021 atoms of carbon in the diamond.

The combined calculation for amount is as follows:

C 0.50 carat C� �

0.2 g C�

1 carat C

1 mol C

12.01 g C�

3C 8.3 10 mol C� �

� �

The amount of carbon in the diamond is 8.3 � 10–3 mol.

The combined calculation for number of atoms is as follows:

C 0.50 carat C� �

0.2 g C�

1 carat C

1 mol C�

12.01 g C

236.02 10 atoms C

1 mol C

��

21C 5.0 10 atoms C� � �

There are 5.0 � 1021 atoms of carbon in the diamond.

(b)

Au 6.50 g Au� �1 mol Au

196.97 g Au�

Au 0.0330 mol Au� �

There is 0.0330 mol gold in the ring.

Au 0.0330 mol Au� �

236.02 10 atoms Au

1 mol Au

��

22Au 1.99 10 atoms Au� � �

There are 1.99 � 1022 atoms of gold in the ring.

C

23

0.50 carat C

1 carat C 0.2 g C

1 mol C 12.01 g C

1 mol C 6.02 10 atoms C

� �

�

�

� �

Au

23

6.50 g Au

1 mol Au 196.97 g Au

1 mol Au 6.02 10 atoms Au

� �

�

� �


6. (a)

The molar mass of 1,4-benzenedicarboxylic acid is 166.14 g/mol.

(b)

8 9 2

3C H NO 8 9 21.5 10 mol C H NO� �

� �8 9 2

8 9 2

151.18 g C H NO

1 mol C H NO�

8 6 4C H O 8 9 20.23 g C H NO� �

The patient should take 0.23 g of Tylenol.

(c)

4 10C H 4 100.95 g C H� �4 10

4 10

1 mol C H

58.14 g C H�

4 10C H 4 100.016 mol C H� �

There is 0.016 mol butane in the lighter. (d)

6 8 6C H O 6 8 60.5 g C H O� �6 8 6

6 8 6

1 mol C H O

176.14 g C H O�

6 8 6

C H O6 8 6

3C H O 6 8 6

36 8 6

2.8 10 mol C H O

2.8 10 mol C H O

�

�

�

�

� �

� �

236 8 6

6 8 6


1 mol C H O

�

�

6 8 6

21C H O 6 8 6

21C 6 8 6



�

�

� �

� �

6 8 6

6 atoms C

1 molecule C H O�

22C 1.0 10 atomsC� � �

The vitamin C table contains 1.0 � 1022 atoms of carbon. 7. A molecular element contains two or more atoms of the same element attached by covalent bonds. Examples include

oxygen, O2(g), and bromine, Br2(l). A compound contains two or more different elements per molecule. Examples include carbon dioxide, CO2(g), and ammonia, NH3(g).

8. (a) A mass spectrometer provides the molar mass of the compound. (b) The mass of the carbon dioxide and water traps are measured before and after they trap carbon dioxide and

water, respectively, produced as a result of a combustion reaction. Subtracting the mass of the traps before the reaction from the mass of the traps after the reaction determines the masses of carbon dioxide and water produced in the combustion reaction.

8 6 4

8 6 4

C H O C H O

C H O

6( ) 6( ) 4( )

8(12.01 g/mol) 6(1.01 g/mol) 4(16.00 g/mol)

96.08 g/mol 6.06 g/mol 64.00 g/mol

166.14 g/mol

� � � �

�

� � �

� � �

� � �

�

8 9 2

8 9 2

3C H NO 8 9 2

C H NO 8 9 2

1.5 10 mol C H NO

151.18 g/mol C H NO

�

�

�

� �

�

4 10

4 10

C H 4 10

C H 4 10

0.95 g C H

58.14 g/mol C H

�

�

�

�

6 8 6

6 8 6

C H O 6 8 6

C H O 6 8 6

236 8 6 6 8 6

6 8 6

0.5 g C H O

176.14 g/mol C H O

1 mol C H O 6.02 10 molecules C H O

1 molecule C H O 6 atoms C

�

�

�

�

� �

�


9. (a) C = 38.72% C H = 9.72% H O = 51.56% O

C

C

H

H

O

O

38.72100%

10038.72 g C

9.72100%

1009.72 g H

51.56100%

10051.56 g O

�

�

�

�

�

�

� �

�

� �

�

� �

�

C 38.72 g C� �1 mol C

12.01 g C�

C

H

3.223 mol C

9.72 g H

�

�

�

�1 mol H

1.01 g H�

H

O

9.62 mol H

51.56 g O

�

�

�

�1 mol O

16.00 g O�

O

C H O

C H O

3.222 mol O

: : 3.223 : 9.62 : 3.222

3.223 9.62 3.222 : :

3.222 3.222 3.2221.000 : 2.99 :1.000

: : 1: 3 :1

�

� � �

� � �

�

�

�

�

�

The empirical formula of the compound is CH3O. (b) Two possible molecular formulas for the compound are CH3O or C2H6O2. (c) You need to know the molar mass of the compound to determine its molecular formula.

10. Compound A �

�

�

�compound

C 64.6%C

H 10.8%H

O 24.6%O

260.0 g/mol�


C

C

H

H

O

O

64.6100%

10064.6 g C

10.8100%

10010.8 g H

24.6100%

10024.6 g O

�

�

�

�

�

�

� �

�

� �

�

� �

�

C 64.6 g C� �1 mol C

12.01 g C�

C

H

5.38 mol C

10.8 g H

�

�

�

�1 mol H

1.01 g H�

H

O

10.7 mol H

24.6 g O

�

�

�

�1 mol O

16.00 g O�

O 1.54 mol O� �

C H O

C H O

: : 5.38 :10.7 :1.54

5.38 10.7 1.54 : :

1.54 1.54 1.543.49 : 6.95 :1.00

2(3.49) : 2(6.95) : 2(1.00)

6.98 :13.9 : 2.00

: : 7 :14 : 2

� � �

� � �

�

�

�

�

�

�

The empirical formula of the compound is C7H14O2.

� � �

�

7 14 2

7 14 2

C H O

C H O

7(12.01 g/mol) 14(1.01 g/mol) 2(16.00 g/mol)

130.21 g/mol

�

�

7 14 2

compound

C H O

260.0 g/mol�

��

130.21 g/mol

7 14 2

compound

C H O

1.997 2�

��

�

�

�

7 14 2

14 28 4


2(C H O )

molecular formula C H O

The molecular formula of compound A is C14H28O4.


Compound B C = 38.67% C H = 16.22% H N = 45.11% N Mcompound = 31.06 g/mol

C

C

H

H

N

N

38.67100%

10038.67 g C

16.22100%

10016.22 g H

45.11100%

10045.11 g N

�

�

�

�

�

�

� �

�

� �

�

� �

�

C 38.67 g C� �1 mol C

12.01 g C�

C

H

3.22 mol C

16.22 g H

�

�

�

�1 mol H

1.01 g H�

H

N

16.06 mol H

45.11 g N

�

�

�

�1 mol N

14.01 g N�

N 3.220 mol N� �

C H N

C H N

: : 3.220 :16.06 : 3.220

3.220 16.06 3.220: :

3.220 3.220 3.2201.00 : 4.988 :1.00

: : 1: 5 :1

� � �

� � �

�

�

�

�

The empirical formula of the compound is CH5N.

� � �

�

5

5

CH N

CH N

1(12.01 g/mol) 5(1.01 g/mol) 1(14.01 g/mol)

31.07 g/mol

�

�

5

compound

CH N

31.06 g/mol�

��

31.07 g/mol

5

compound

CH N

1�

��


�

�

�

5

5


1(CH N)

molecular formula CH N

The molecular formula of compound B is CH5N. 11. The molar concentration of a solution is measured in moles per litre of solution instead of moles per litres of water

because solutes occupy some of the volume of a solution. If 1.0 mol NaCl(s) is dissolved in 100 mL of water, the final volume of solution will be slightly greater than 100 mL. Since different solutes occupy different volumes, a solution containing 1.0 mol NaCl(s) in 100 mL of water will have a different concentration than a solution containing 1.0 mol NH3NO3(s) in 100 mL of water.

12. A dilute solution has a relatively smaller amount of solute per unit volume of solution than a concentrated solution. 13.

3

3

3

3 3 3

3

NO

NONO

NO NO NO

3

3NO

2.3 mg NO0.25 L

L

0.58 mg NO

�

�

� �

�

�

�

�

� � �

�

�

�

�

�

� �

�

The mass of nitrate in the drinking water is 0.58 mg. 14. (a)

NaOH 12.0 g NaOH� �1 mol NaOH

40.00 g NaOH�

NaOH 0.300 mol NaOH� �

�

�

�

NaOHNaOH

NaOH

NaOH

0.300 mol

2.5 L0.12 mol/L

�

�

The molar concentration of the sodium hydroxide is 0.12 mol/L. (b)

4 5 6KC H O 4 5 62.28 g KC H O� �4 5 6

4 5 6

1 mol KC H O

188.19 g KC H O�

4 5 6KC H O 4 5 60.0121 mol KC H O� �

3

3

NO

NO

2.3 ppm or 2.3 mg/L

250 mL

�

�

�

�

�1 L

1000 mL�

3

3

NO

NO

0.25 L

?

�

�

�

�

�

�

NaOH

NaOH

NaOH

NaOH

12.0 g NaOH

40.00 g/mol NaOH

2.5 L

?

�

�

�

�

�

�

�

4 5 6

4 5 6

4 5 6

KC H O

KC H O 4 5 6

KC H O

2.28 g

188.19 g/mol KC H O

100.0 mL

�

�

�

�

�

�1 L

1000 mL�

4 5 6KC H O 0.1000 L� �


�

�

�

4 5 6

4 5 6

4 5 6

4 5 6

KC H OKC H O

KC H O

4 5 6

KC H O

0.0121 molKC H O

0.1000 L0.121 mol/L

�

�

The molar concentration of potassium hydrogen tartrate is 0.121 mol/L. (c)

2 6C H O 2 60.08 g C H O� �2 6

2 6

1 mol C H O

46.08 g C H O�

2 6

3C H O 2 61.7 10 mol C H O� �

� �

2 6

2 6

2 6

2 6

C H OC H O

C H O

32 6

2C H O

1.7 10 mol C H O

0.1 L

1.7 10 mol/L

�

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�

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�

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� �

The legal limit of blood alcohol concentration in Canada is 1.7 � 10–2 mol/L. 15. mNaOH = 0.40 g

solution 100 mL� �1 L

1000 mL�

solution 0.1 L� �

(a) csolution = ? W/V = ?

molar concentration � � �

� � �

�

NaOH Na O H

NaOH

22.99 g/mol 16.00 g/mol 1.01 g/mol

40.00 g/mol

� � � �

�

NaOH 0.40 g NaOH� �1 mol NaOH

40.00 g NaOH�

NaOH 0.010 mol NaOH� �

�

�

�

NaOHsolution

solution

solution

0.010 mol NaOH

0.1 L0.1 mol/L

�

�

The molar concentration is 0.1 mol/L NaOH.

2 6

2 6

2 6

C H O 2 6

C H O 2 6

C H O

0.08 g C H O

46.08 g/mol C H O

100 mL

�

�

�

�

�

�1 L

1000 mL�

2 6C H O 0.1 L� �


weight by volume

� �

� �

�

solutesolution

solution

solution

100%

0.40 g100%

100 mL0.4% W/V

�

�

The concentration of the sodium hydroxide solution is 0.4% W/V. (b) molar concentration ci = 0.1 mol/L vi = 10 mL vf = 50 mL cf = ?

� ��

�

�

�

�

i i f f

i if

f

f

0.1 mol/L 0.010 mL

0.050 mL

0.02 mol/L NaOH

� �

�

�

The final molar concentration is 0.02 mol/L of sodium hydroxide solution.

weight by volume ci = 0.4% W/V vi = 10 mL vf = 50 mL cf = ?

� ��

�

�

�

�

i i f f

i if

f

f

0.4% 10 mL

50 mL

0.08% W/V

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�

The final concentration of the sodium hydroxide solution is 0.08% W/V. 16. (a) 3 Fe + 4 H2O → Fe3O4 + 4 H2

(b) H2SO4 + 2 NaOH → 2 H2O + Na2SO4 (c) 4 Cu + O2 → 2 Cu2O (d) Fe2(SO4)3 + 12 KSCN → 2 K3Fe(SCN)6 + 3 K2SO4

17. (a) C2H6O(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g) (b) C3H8O3 + 3 HNO3 → C3H5N3O9 + 3 H2O(l)

18. (a) 2 3

2Fe O 2 33.00 10 g Fe O� � �

Balanced equation Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)

Given mass (g) 3.00 � 102 ?



2 3

2Fe O 2 33.00 10 g Fe O� � �

2 3

2 3

1 mol Fe O

159.70 g Fe O�

2 3Fe O 2 3

Fe 2 3

1.88 mol Fe O

1.88 mol Fe O

�

�

�

�

2 3

2 mol Fe

1 mol Fe O�

Fe

Fe

3.76 mol Fe

3.76 mol Fe

�

�

�

�55.85 g Fe

1 mol Fe�

2Fe 2.10 10 g Fe� � �

The reaction produces 3.76 mol iron, which is 2.10 � 102 g of iron.


2Fe 2 33.00 10 g Fe O� � �

2 31 mol Fe O�

2 3159.70 g Fe O 2 3

2 mol Fe

1 mol Fe O�

Fe

Fe 2 3

3.76 mol Fe

300 g Fe O

�

�

�

�2 31 mol Fe O

�

2 3159.70 g Fe O

2 mol Fe�

2 31 mol Fe O

55.85 g Fe

1 mol Fe�

2Fe 2.10 10 g Fe� � �

The reaction produces 3.76 mol iron, which is 2.10 � 102 g of iron. (b) actual yield = 178 g Fe theoretical yield = 2.1 � 102 g Fe

� �

� �

�

�

2


theoretical yield

178 g100%

2.10 10 g


The percentage yield of iron is 84.8%. 19. MAl = 5.00 g Al

Balanced equation 4 Al(s) + 3 O2(g) → 2 Al2O3(s)

Given mass (g) 5.00 ?




26.98 g Al�

2

Al

O

0.185 mol Al

0.185 mol Al

�

�

�

�23 mol O

4 mol Al�

2

2

O 2

O 2

0.139 mol O

0.139 mol O

�

�

�

�2

2

32.00 g O

1 mol O�

2O 24.45 g O� �

The reaction requires 0.139 mol oxygen, which is 4.45 g of oxygen.

The combined calculation can be done as follows:

2O 5.00 g Al� �1 mol Al

�

26.98 g Al23 mol O

4 mol Al�

2

2

O 2

O

0.139 mol O

5.00 g Al

�

�

�

�1 mol Al

�

26.98 g Al

23 mol O�

4 mol Al2

2

32.00 g O

1 mol O�

2O 24.45 g O� �

reaction requires 0.139 mol oxygen, which is 4.45 g of oxygen. 20. (a)

Balanced equation Fe2O3(s) + 2 Al(s) → 2 Fe(l) + Al2O3(s)

Given mass (g) 135.0 ?



26.98 g Al�

2 3

Al

Al O

5.004 mol Al

5.004 mol Al

�

�

�

�2 31 mol Al O

2 mol Al�

2 3

2 3

Al O 2 3

Al O 2 3

2.502 mol Al O

2.502 mol Al O

�

�

�

�2 3

2 3

101.96 g Al O

1 mol Al O�

2 3Al O 2 3255.1 g Al O� �

The maximum mass of aluminum oxide that can be produced is 255.1 g.

Al 135.0 g Al� �


The combined calculation can be done as follows:

2 3Al O 135.0 g Al� �1 mol Al

26.98 g Al�

2 31 mol Al O�

2 mol Al2 3

2 3

101.96 g Al O

1 mol Al O�

2 3Al O 2 3255.1 g Al O� �

The maximum mass of aluminum oxide that can be produced is 255.1 g.

(b) percentage yield = 87% theoretical yield = 255.1g Al2O3

actual yield = ?

� ��

� ��

� �

�

�

� �

2 3

22 3

actualyieldpercentageyield 100%

theoreticalyield

percentageyield theoreticalyieldactualyield

100%

87% 255.1 gAl O

100%

actualyield 2.22 10 gAl O

The mass of aluminum oxide actually produced is 2.22 � 102 g.

Applying Inquiry Skills 21. Analysis

(a) rainwater H2(g): O2(g) = 23.72 : 11.80 H2(g): O2(g) = 2 : 1 tap water H2(g): O2(g) = 8.39 : 4.18 H2(g): O2(g) = 2 : 1 The hydrogen-to-oxygen ratio is 2:1 for both rainwater and tap water. (b) Yes, the law of constant composition holds for water molecules.

Evaluation (c) We assume that the gases produced in the Hoffman apparatus are hydrogen and oxygen only, and that the

hydrogen and oxygen gas in the collection tubes are only produced by the decomposition of water molecules. (d) The Observations support the Prediction so the Prediction is valid.

22. (a) A small sample of calcium of known mass is placed in a crucible and heated strongly over a Bunsen burner flame until it has completely reacted with oxygen in air to produce calcium oxide. The mass of the calcium oxide is measured. The difference in mass between the original sample of calcium and the calcium oxide gives the mass of oxygen that has combined with calcium in the reaction. The percentage composition by mass of calcium oxide is determined by dividing the mass of calcium by the mass of calcium oxide, and dividing the mass of oxygen by the mass of calcium oxide.

(b) No, the same experimental design cannot be used because the carbon and hydrogen atoms of the octane molecules separate and form different product molecules (carbon dioxide and water).

23. (a) Step 1: Place 2.8 g (0.015 mol) of Cu(NO3)2(s) into a clean, dry 100-mL volumetric flask. Step 2: Add a small volume (approximately 25 mL) of distilled water to the flask. Step 3: Stopper the flask and shake until all of the solid copper(II) nitrate crystals have dissolved. Step 4: Remove the stopper and add enough distilled water to reach the 100-mL mark on the flask. Step 5: Stopper the flask and invert several times to mix.

(b) Step 1: Place 200 mL of the 0.15-mol/L Cu(NO3)2(aq) solution into a clean, dry 1.0-L volumetric flask. Step 2: Add enough distilled water to reach the 1.0 L mark on the flask. Step 3: Stopper the flask and invert several times to mix.


24. (a)

(b) The concentration of atropine solution is approximately 4 g/L. (c) No, every chemical has a different standard curve associated with it, so the curve in (a) may not be used.

25. We cannot be sure that all of the lead(II) nitrate has reacted to form the lead(II) sulfate precipitate. The student must continue adding sodium sulfate solution to the lead(II) nitrate solution until no more precipitate forms, which indicates that all of the lead(II) nitrate in solution has reacted. Only then may the mass of the lead(II) sulfate precipitate formed be used to calculate the concentration of the original lead(II) nitrate solution.

Making Connections 26. (a) To measure alcohol concentration in breath, a driver breathes into the Breathalyzer. The breath bubbles into a

vial containing a mixture of sulfuric acid, potassium dichromate, silver nitrate, and water. The reaction between these reactants and the ethyl alcohol in the person's breath changes the colour of the solution in the vial from red-orange to green. The degree of the colour change is directly related to the concentration of alcohol in the person's blood. The machine compares the colour of the reacted mixture to the colour of an unreacted solution in a second vial using a photocell system. The needle in the meter moves according to the intensity of the colour difference.

(b) The legal BAC limit for fully qualified drivers in Ontario is 0.08 g of alcohol per 100 mL of blood. However, for drivers with G1 or G2 licences, it is 0 g of alcohol per 100 mL of blood.

(c) A portable Breathalyzer is a hand-held device that is small and compact. It operates on the chemical reactions described in (a). A stationary Breathalyzer is a larger desktop device, usually located in a laboratory. It is a spectrophotometer that identifies and measures the concentration of molecules based on the way they absorb infrared light.

(d) Common defences for drunk driving convictions include having been given drinks laced with alcohol, inhaling alcohol vapours at work, use of skin antiseptics containing alcohol, alleged mix-up of blood specimens, consumption of elixirs or health tonics containing alcohol, and faulty Breathalyzer readings caused by interference between the Breathalyzer machine and the police officer's radio or cell phone.

(e) Student papers will vary. Students should give clear, concise arguments and should argue for or against the issue only. Students may wish to use questions from Workbook 2.6 Alternative Exercise: Tech Connect: The Breathalyzer to help them get started.

27. (a) Student answers will vary. Possible sources of hydrogen include water and methane. One possible source of nitrogen is atmospheric nitrogen.

(b) Hydrogen for the Haber process is produced by reacting methane, CH4(g), with steam, H2O(g), which produces carbon dioxide gas, CO2(g), and hydrogen gas, H2(g). The nitrogen gas, N2(g), is produced by the fractional distillation of air (air is approximately 78% nitrogen).

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