slide

Chapter 1Chemistry and Measurement

?

Copyright © Houghton Mifflin Company. All rights reserved. 1 | 2

Example of The law of conservation of mass: Aluminum powder burns in oxygen to produce a substance called aluminum oxide. A sample of 2.00 grams of aluminum is burned in oxygen and produces 3.78 grams of aluminum oxide. How many grams of oxygen were used in this reaction?

aluminum + oxygen = aluminum oxide

2.00 g + oxygen = 3.78 g

oxygen = 1.78 g

?


Potassium is a soft, silvery-colored metal that melts at 64°C. It reacts vigorously with water, with oxygen, and with chlorine. Identify all of the physical properties and chemical properties given in this description.

Reacts with chlorineMelting point (64°C)

Reacts with oxygenSilvery-colored

Reacts with waterSoft

Chemical PropertyPhysical Property

?


Perform the following calculation and round your answer to the correct number of significant figures:

Calculator answer:

3.23225000

The answer should be rounded to two significant figures (92.35 X 0.035):

3.2

)421.0456.0(35.92

?


In winter, the average low temperature in interior Alaska is -30.°F (two significant figures). What is this temperature in degrees Celsius and in kelvins?

F9

C5F32

ο

οο

FC tt

F9

C5F62

ο

οο

C t

F9

C5F32F.30

ο

οοο

C t

C34.4444444οC t

C34οC t


K15.273C1

K1οCK

tt

K15.273C1

K1C34

οο

K

t

K15.273K34K t

K15.239K t

K239K t

?


Oil of wintergreen is a colorless liquid used as a flavoring. A 28.1 g sample of oil of wintergreen has a volume of 23.7 mL. What is the density of oil of wintergreen?

V

md

mL23.7

g28.1

V

m

mL23.7

g28.1d

mL

g1.18565491d

mL

g1.19d

?


The dimensions of Noah’s ark were reported as 3.0 × 102 cubits by 5.0 × 101 cubits. Express this size in units of feet and meters. (1 cubit = 1.5 ft)

(exact)m0.9144yd1

yd1ft3

ft1.5cubit1

cubit1

ft1.5cubits103.0 2

ft104.5000000 2byft104.5 2

cubit1

ft1.5cubits105.0 1

ft107.5000000 1ft75ft107.5 1


(exact)m0.9144yd1

yd1ft3

ft1.5cubit1

m101.37160000 2

yd1

m0.9144

ft3

yd1ft104.5 2

bym101.4 2 m23

yd1

m0.9144

ft3

yd1ft75

m22.8600000

byft104.5 2 ft75ft107.5 1


Chapter 2Atoms,

Molecules, and Ions


Write the nuclide symbol for the atom that has 19 protons and 20 neutrons.

K3919

Atomic number: Z = 19The element is potassium, K.

Mass number: A = 19 + 20 = 39

The nuclide symbol is


An element has four naturally occurring isotopes. The mass and percentage of each isotope are as follows:

Percentage Abundance Mass (amu)

1.48 203.973

23.6 205.9745

22.6 206.9759

52.3 207.9766

What is the atomic weight and name of the element?


To find the portion of the atomic weight due to each isotope, multiply the fraction by the mass of

the isotope. The atomic weight is the sum of these products.

Fractional Abundance

Mass (amu) Mass From Isotope

0.0148 203.973 3.01880040

0.236 205.9745 48.6099820

0.226 206.9759 46.7765534

0.523 207.9766 108.771762

207.177098The atomic weight is 207 amu; the element is lead.

Total =


What is formula of the ionic compound of Mg2+ and N3-?

The common multiple of the charges is 6, so we need three Mg2+ and two N3-. The resulting formula is

Mg3N2


Common Monatomic Ions of the Main-Group Elements

Bi3+Pb2+Tl3+, Tl+

Ba2+Cs+

I-Te2-Sn2+In3+Sr2+Rb+

Br-Se2-Ga3+Ca2+K+

Cl-S2-Al3+Mg2+Na+

F-O2-N3-Be2+Li+

H-

6

5

4

3

2

1

VIIAVIAVAIVAIIIAIIAIAPeriod


What are the names of the following ionic compounds?

– BaO

– Cr2(SO4)3

BaO is barium oxide.Cr2(SO4)3 is chromium(III) sulfate or chromic sulfate.


What are the chemical formulas for the following ionic compounds?

– potassium carbonate

– manganese(II) sulfate

The ions K+ and CO32- form K2CO3

The ions Mn2+ and SO42- form MnSO4


What are the names of the following compounds?

– OF2

– S4N4

– BCl3

OF2 is oxygen difluoride

S4N4 is tetrasulfur tetranitride

BCl3 is boron trichloride


What are the formulas for the following binary molecular compounds?

– carbon disulfide

– nitrogen tribromide

– dinitrogen tetrafluoride

The formula for carbon disulfide is CS2.

The formula for dinitrogen tetrafluoride is N2F4.

The formula for nitrogen tribromide is NBr3.


A compound whose common name is green vitriol has the chemical formula FeSO4·7H2O. What is the chemical name of this compound?

FeSO4·7H2O is iron(II) sulfate heptahydrate.


Calcium chloride hexahydrate is used to melt snow on roads. What is the chemical formula of the compound?

The chemical formula for calcium chloride hexahydrate is CaCl2·6H2O.


Balance the following equations:

NH3 + O2 NO + H2O

C2H5OH + O2 CO2 + H2O

4NH3 + 5O2 4NO + 6H2O

C2H5OH + 3O2 2CO2 + 3H2O

Chapter 3


Calculate the formula weight of the following compounds from their formulas. Report your answers to three significant figures.

– calcium hydroxide, Ca(OH)2

– methylamine, CH3NH2


3 significant figures74.1 amu

3 significant figures31.1 amu

Total 74.095

2 O 2(16.00) = 32.00 amu

Ca(OH)2

1 Ca 1(40.08) = 40.08 amu

2 H 2(1.008) = 2.016 amu

CH3NH2

1 C 1(12.01) = 12.01 amu

5 H 5(1.008) = 5.040 amu1 N 1(14.01) = 14.01 amu

Total 31.060

Calculate the formula weight of the following compounds from their formulas. calcium hydroxide, Ca(OH)2

methylamine, CH3NH2


A sample of nitric acid, HNO3, contains 0.253 mol HNO3. How many grams is this?

First, find the molar mass of HNO3:

1 H 1(1.008) = 1.0081 N 1(14.01) = 14.013 O 3(16.00) = 48.00

63.018 (2 decimal places)63.02 g/mol


mole1

g63.02xmole0.253

figures)tsignifican(3

g 15.9

Next, using the molar mass, find the mass of 0.253 mole:

= 15.94406 g


The average daily requirement of the essential amino acid leucine, C6H14O2N, is 2.2 g for an adult. What is the average daily requirement of leucine in moles?

First, find the molar mass of leucine:

6 C 6(12.01) = 72.062 O 2(16.00) = 32.001 N 1(14.01) = 14.0114 H 14(1.008) = 14.112

132.1822 decimal places

132.18 g/mol


g132.18

mole1xg2.2

mol10x1.6643 2


mol0.017ormol10x1.7 2

Next, find the number of moles in 2.2 g:


mol1

atoms10x6.02x

g51.996

mol1xg10x1.0

236

First, find the molar mass of Cr:1 Cr 1(51.996) = 51.996 g/mol

Now, convert 1.0 x 10-6 grams to moles:

=1.157781368 x 1016 atoms

1.2 x 1016 atoms(2 significant figures)

The daily requirement of chromium in the human diet is 1.0 × 10-6 g. How many atoms of chromium does this represent?


Lead(II) chromate, PbCrO4, is used as a paint pigment (chrome yellow). What is the percentage composition of lead(II) chromate?

First, find the molar mass of PbCrO4:

1 Pb 1(207.2) = 207.21 Cr 1(51.996) = 51.9964 O 4(16.00) = 64.00

323.196 (1 decimal place)323.2 g/mol


64.11%100%xg323.20

g207.2:Pb

16.09%100%xg323.20

g51.996:Cr

19.80%100%xg323.20

g64.00:O

Now, convert each to percent composition:

Check:64.11 + 16.09 + 19.80 = 100.00 %


Benzene has the empirical formula CH. Its molecular weight is 78.1 amu. What is its molecular formula?

613.02

78.1

amu3.02

amu 1.008) 12.01weightformulaEmpirical

1

(

Molecular formulaC6H6


Hexamethylene is one of the materials used to produce a type of nylon. It is composed of 62.1% C, 13.8% H, and 24.1% N. Its molecular weight is 116 amu. What is its molecular formula?

Nmol1.720Ng14.01

Nmol1X Ng24.1

Hmol13.69Hg1.008

Hmol1X Hg13.8

Cmol5.171Cg12.01

Cmol1X Cg62.1

11.720

1.720

81.720

13.69

31.720

5.171

Empirical formulaC3H8N


The empirical formula is C3H8N.

Molecular formula = (Empirical formula)n

Find the empirical formula weight:

3(12.01) + 8(1.008) + 1(14.01) = 58.104 amu

Molecular formula: C6H16N2

258.10

116 n


Propane, C3H8, is normally a gas, but it is sold as a fuel compressed as a liquid in steel cylinders. The gas burns according to the following equation:

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

How many grams of CO2 are produced when 20.0 g of propane is burned?


Molar massesC3H8: 3(12.01) + 8(1.008) = 44.094 gCO2: 1(12.01) + 2(16.00) = 44.01 g

59.9 g CO2

(3 significant figures)

2

2

83

2

83

8383 COmol1

COg44.01

HCmol1

COmol3

HCg44.094

HCmol1HCg20.0 XXX

2COg359.8856987

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)


Urea, CH4N2O, is used as a nitrogen fertilizer. It is manufactured from ammonia and carbon dioxide at high pressure and high temperature:

2NH3 + CO2(g) CH4N2O + H2O

In a laboratory experiment, 10.0 g NH3 and 10.0 g CO2 were added to a reaction vessel. What is the maximum quantity (in grams) of urea that can be obtained? How many grams of the excess reactant are left at the end of the reactions?



Molar masses: NH3 1(14.01) + 3(1.008) = 17.02 g CO2 1(12.01) + 2(16.00) =

44.01 g CH4N2O 1(12.01) + 4(1.008) + 2(14.01) + 1(16.00) = 60.06 g

CO2 is the limiting reactant. 13.6 g CH4N2O will be produced.

ONCHg13.6

ONCHmol1

ONCHg60.06

COmol1

ONCHmol1

COg44.01

COmol1COg10.0

24

24

24

2

24

2

22

XXX

ONCHg17.6

ONCHmol1

ONCHg60.06

NHmol2

ONCHmol1

NHg17.02

NHmol1NHg10.0

24

24

24

3

24

3

33

XXX


10.0 g at start-7.73 g reacted 2.27 g remains

2.3 g NH3 is left unreacted.(1 decimal place)

reactedNHg7.73 3

3

3

2

3

2

22 NHmol1

NHg17.02

COmol1

NHmol2

COg44.01

COmol1COg10.0 XXX

3NHg7.73460577

To find the excess NH3, we find how much NH3 reacted:

Now subtract the amount reacted from the starting amount:




When 10.0 g NH3 and 10.0 g CO2 are added to a reaction vessel, the limiting reactant is CO2. The theoretical yield is 13.6 g of urea. When this reaction was carried out, 9.3 g of urea was obtained. What is the percent yield?

Theoretical yield = 13.6 gActual yield = 9.3 g

= 68% yield(2 significant figures)

100%x g 13.6

g 9.3

Chapter 4

Chemical Reactions


Determine whether each of the following reactions occurs. If it does, write the molecular, ionic, and net ionic equations.

KBr + MgSO4 NaOH + MgCl2 K3PO4 + CaCl2


Classify the following as strong or weak acids or bases:

a. KOH

b. H2S

c. CH3NH2

d. HClO4


Write the molecular, ionic, and net ionic equations for the neutralization of sulfurous acid, H2SO3, by potassium hydroxide, KOH.


Write the molecular, ionic, and net ionic equations for the reaction of copper(II) carbonate with hydrochloric acid.


Potassium permanganate, KMnO4, is a purple-colored compound; potassium manganate, K2MnO4, is a green-colored compound. Obtain the oxidation numbers of the manganese in these compounds.


What is the oxidation number of Cr in dichromate, Cr2O7

2-?


Balance the following oxidation-reduction reaction:

FeI3(aq) + Mg(s) Fe(s) + MgI2(aq)


You place a 1.52 g of potassium dichromate, K2Cr2O7, into a 50.0 mL volumetric flask. You then add water to bring the solution up to the mark on the neck of the flask. What is the molarity of K2Cr2O7 in the solution?

Molar mass of K2Cr2O7 is 294 g.

L 10 x 50.0

g 294

mol 1 g 1.52

3- 0.103 M


A solution of sodium chloride used for intravenous transfusion (physiological saline solution) has a concentration of 0.154 M NaCl. How many moles of NaCl are contained in 500. mL of physiological saline? How many grams of NaCl are in the 500. mL of solution?

NaClmol0.0770

L0.5000.154

Vmol

M

M

NaClg4.50

mol1

g58.4mol0.0770

g58.4NaClmassMolar


A saturated stock solution of NaCl is 6.00 M. How much of this stock solution is needed to prepare 1.00 L of physiological saline solution (0.154 M)?

i

ffi

ffii

M

VMV

VMVM

mL25.7orL0.0257

6.00

L))(1.00(0.154

i

i

V

M

MV


A soluble silver compound was analyzed for the percentage of silver by adding sodium chloride solution to precipitate the silver ion as silver chloride. If 1.583 g of silver compound gave 1.788 g of silver chloride, what is the mass percent of silver in the compound?


Agmol1

Agg107.9 x

AgClmol1

Agmol1 x

AgClg143.32

AgClmol1 x AgClg1.788

= 1.346 g Ag in the compound

100% Xcompoundsilverg1.583

Agg1.346

= 85.03% Ag

Molar mass of silver chloride (AgCl) = 143.32 g


Zinc sulfide reacts with hydrochloric acid to produce hydrogen sulfide gas:

ZnS(s) + 2HCl(aq) ZnCl2(aq) + H2S(g)

How many milliliters of 0.0512 M HCl are required to react with 0.392 g ZnS?


Molar mass of ZnS = 97.47 g

= 0.157 L = 157 mL HCl solution

HClmol0.0512

solutionL1 x

ZnSmol1

HClmol2 x

ZnSg97.47

ZnSmol1 x ZnSg0.392

ZnS(s) + 2HCl(aq) ZnCl2(aq) + H2S(g)


A dilute solution of hydrogen peroxide is sold in drugstores as a mild antiseptic. A typical solution was analyzed for the percentage of hydrogen peroxide by titrating it with potassium permanganate:

5H2O2(aq) + 2KMnO4(aq) + 6H+(aq) 8H2O(l) + 5O2(g) + 2K+(aq) + 2Mn2+(aq)

What is the mass percent of H2O2 in a solution if 57.5 g of solution required 38.9 mL of 0.534 M KMnO4 for its titration?


22

22

4

2243

OHmol1

OHg34.01 x

KMnOmol2

OHmol5 x

L1

KMnOmol0.534 x L10x38.9

Molar mass of H2O2 = 34.01 g

= 1.77 g H2O2

100% Xsolutiong57.5

OHg1.77 22

= 3.07% H2O2

5H2O2(aq) + 2KMnO4(aq) + 6H+(aq) 8H2O(l) + 5O2(g) + 2K+(aq) + 2Mn2+(aq)

Chapter 5


mmHg)(359

mmHg)mL)(751(38.7f V

= 81.0 mL(3 significant figures)

f

iif P

VPV

A volume of oxygen gas occupies 38.7 mL at 751 mmHg and 21°C. What is the volume if the pressure changes to 359 mmHg while the temperature remains constant?


You prepared carbon dioxide by adding HCl(aq) to marble chips, CaCO3. According to your calculations, you should obtain 79.4 mL of CO2 at 0°C and 760 mmHg. How many milliliters of gas would you obtain at 27°C?


Vi = 79.4 mLPi = 760 mmHgTi = 0°C = 273 K

Vf = ?Pf = 760 mmHgTf = 27°C = 300. K

K)(273

mL)K)(79.4(300.f V

= 87.3 mL(3 significant figures)

i

iff T

VTV


Divers working from a North Sea drilling platform experience pressure of 5.0 × 101 atm at a depth of 5.0 × 102 m. If a balloon is inflated to a volume of 5.0 L (the volume of the lung) at that depth at a water temperature of 4°C, what would the volume of the balloon be on the surface (1.0 atm pressure) at a temperature of 11°C?


Vi = 5.0 LPi = 5.0 × 101 atmTi = 4°C = 277 K

Vf = ?Pf = 1.0 atmTf = 11°C = 284. K

fi

iiff PT

VPTV

atm)K)(1.0(277

L)atm)(5.010xK)(5.0(284 1

f V

= 2.6 x 102 L(2 significant figures)


Ideal Gas Law

The ideal gas law is given by the equation

PV=nRT

The molar gas constant, R, is the constant of proportionality that relates the molar volume of a gas to T/P.


A 50.0 L cylinder of nitrogen, N2, has a pressure of 17.1 atm at 23°C. What is the mass of nitrogen in the cylinder?

V = 50.0 LP = 17.1 atmT = 23°C = 296 K

RT

PVn

mol20.35

K)(296

Kmol

atmL0.08206

L)atm)(50.0(17.1n

mass = 986 g(3 significant figures)mol

g28.01X mol35.20mass


What is the density of methane gas (natural gas), CH4, at 125°C and 3.50 atm?

Mm = 16.04 g/molP = 3.50 atmT = 125°C = 398 K

RT

PMd m

K)(398Kmol

atmL0.08206

atm))(3.50mol

g(16.04

dfigures)tsignifican(3L

g1.72d


A 500.0-mL flask containing a sample of octane (a component of gasoline) is placed in a boiling water bath in Denver, where the atmospheric pressure is 634 mmHg and water boils at 95.0°C. The mass of the vapor required to fill the flask is 1.57 g. What is the molar mass of octane? (Note: The empirical formula of octane is C4H9.) What is the molecular formula of octane?


P = 634 mmHg = 0.8342 atm

P

dRTM m

atm)(0.8342

K368.2Kmol

atmL0.08206

L

g3.140

m

M

figures)tsignifican(3mol

g114mM

d = 1.57 g/0.5000 L = 3.140 g/L

T = 95.0°C = 368.2 K


2

mol

g57

mol

g114

n

Molecular formula: C8H18

Molar mass = 114 g/molEmpirical formula: C4H9

Empirical formula molar mass = 57 g/mol


When a 2.0-L bottle of concentrated HCl was spilled, 1.2 kg of CaCO3 was required to neutralize the spill. What volume of CO2 was released by the neutralization at 735 mmHg and 20.°C?


First, write the balanced chemical equation:

CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)

3

2

3

33

3

CaCOmol1

COmol1

CaCOg100.09

CaCOmol1CaCOg10x1.2 XX

Moles of CO2 produced = 12 mol

Second, calculate the moles of CO2 produced:

Molar mass of CaCO3 = 100.09 g/mol


n = 12 molP = 735 mmHg = 0.967 atmT = 20°C = 293 K

P

nRTV

atm)(0.967

K)(293Kmol

atmL0.08206mol12

V

= 3.0 × 102 L(2 significant figures)


A 100.0-mL sample of air exhaled from the lungs is analyzed and found to contain 0.0830 g N2, 0.0194 g O2, 0.00640 g CO2, and 0.00441 g water vapor at 35°C. What is the partial pressure of each component and the total pressure of the sample?


mL10

L1mL100.0

K308Kmol

atmL0.08206

Ng28.01

Nmol1Ng0.0830

2

3

2

22

NP

mL10

L1mL100.0

K308Kmol

atmL0.08206

Og32.00

Omol1Og0.0194

2

3

2

22

OP

mL10

L1mL100.0

K308Kmol

atmL0.08206

COg44.01

COmol1COg0.00640

2

3

2

22

COP

mL10

L1mL100.0

K308Kmol

atmL0.08206

OHg18.01

OHmol1OHg0.00441

2

3

2

22

OHP

atm0.749

atm0.153

atm0.0368

atm0.0619


atm0.7492N P

atm0.1532O P

atm0.03682CO P

atm0.0619OH2P

OHCOON 2222PPPPP

P = 1.00 atm


The partial pressure of air in the alveoli (the air sacs in the lungs) is as follows: nitrogen, 570.0 mmHg; oxygen, 103.0 mmHg; carbon dioxide, 40.0 mmHg; and water vapor, 47.0 mmHg. What is the mole fraction of each component of the alveolar air?

OHCOON 2222PPPPP

570.0 mmHg103.0 mmHg

40.0 mmHg47.0 mmHg

P = 760.0 mmHg


Mole fraction of N2

Mole fraction of H2OMole fraction of CO2

Mole fraction of O2

mmHg760.0

mmHg47.0

mmHg760.0

mmHg40.0

mmHg760.0

mmHg103.0

mmHg760.0

mmHg570.0

Mole fraction N2 = 0.7500

Mole fraction O2 = 0.1355

Mole fraction CO2 = 0.0526

Mole fraction H2O= 0.0618


You prepare nitrogen gas by heating ammonium nitrite:

NH4NO2(s) N2(g) + 2H2O(l)

If you collected the nitrogen over water at 23°C and 727 mmHg, how many liters of gas would you obtain from 5.68 g NH4NO2?


P = 727 mmHgPvapor = 21.1 mmHgPgas = 706 mmHgT = 23°C = 296 K P

nRTV

Molar mass NH4NO2

= 64.04 g/mol

24

2

24

24 24

NONHmol1

Nmol1

NONHg64.04

NONHmol1NONHg5.68 XX

= 0.08869 mol N2 gas


P = 727 mmHgPvapor = 21.1 mmHgPgas = 706 mmHgT = 23°C = 296 Kn = 0.08869 mol

P

nRTV

mmHg760

atm1 xmmHg706

K)(296Kmol

atmL0.08206mol0.0887

V

= 2.32 L of N2

(3 significant figures)


mol

kg0.04401

K296Kmol

s

mkg

8.314532

2

rmsu

2

25

rmss

m1.68x10u

s

mx104.10 2

rms u

2

2

s

mkgJ

Recall

mM

RTu

3rms

What is the rms speed of carbon dioxide molecules in a container at 23°C?


Both hydrogen and helium have been used as the buoyant gas in blimps. If a small leak were to occur, which gas would effuse more rapidly and by what factor?

Hydrogen will diffuse more quickly by a factor of 1.4.

2.016

4.002

4.002

12.016

1

HeRate

HRate 2

Chapter 6


A person weighing 75.0 kg (165 lbs) runs a course at 1.78 m/s (4.00 mph). What is the person’s kinetic energy?

2

K s

m1.78kg)(75.0

2

1

E

m = 75.0 kgV = 1.78 m/s

EK = ½ mv2


J119s

mkg119

2

2

K

E


In an endothermic reaction:

The reaction vessel cools.

Heat is absorbed.

Energy is added to the system.

q is positive.

In an exothermic reaction:

The reaction vessel warms.

Heat is evolved.

Energy is subtracted from the system.

q is negative.


S8(s) + 8O2(g) 8SO2(g)

S8(s) + 8O2(g) 8SO2(g); H = –2.39 × 103 kJ

kJ10 2.39Δ

Smol1

Sg256.52

Sg1

kJ9.31Δ

3

8

8

8

H

H

Sulfur, S8, burns in air to produce sulfur dioxide. The reaction evolves 9.31 kJ of heat per gram of sulfur at constant pressure. Write the thermochemical equation for this reaction.


8

3

8

88 Smol1

kJ10x2.39

Sg256.5

Smol1Sg15.0

q

q = –1.40 × 102 kJ

Molar mass of S8 = 256.52 g

You burn 15.0 g sulfur in air. How much heat evolves from this amount of sulfur? The thermochemical equation is

S8(s) + 8O2(g) 8SO2(g); H = -2.39 x 103 kJ


The daily energy requirement for a 20-year-old man weighing 67 kg is 1.3 x 104 kJ. For a 20-year-old woman weighing 58 kg, the daily requirement is

8.8 x 103 kJ. If all this energy were to be provided by the combustion of glucose, C6H12O6, how many grams of glucose would have to be consumed by the man and the woman per day?

C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l);Ho = -2.82 x 103 kJ


C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l);

H = -2.82 x 103 kJ

glucosemol1

glucoseg180.2

kJ2.82x10

glucosemol1kJ1.3x10

34

glucose m

For a 20-year-old man weighing 67 kg:

= 830 g glucose required(2 significant figures)

glucosemol1

glucoseg180.2

kJ2.82x10

glucosemol1kJ8.8x10

33

glucose m

For a 20-year-old woman weighing 58 kg:

= 560 g glucose required(2 significant figures)


A piece of zinc weighing 35.8 g was heated from 20.00°C to 28.00°C. How much heat was required? The specific heat of zinc is 0.388 J/(g°C).

m = 35.8 gs = 0.388 J/(g°C)t = 28.00°C – 20.00°C = 8.00°C

q = m · s · t

C8.00Cg

J0.388g35.8

q

q = 111 J(3 significant figures)


Nitromethane, CH3NO2, an organic solvent burns in oxygen according to the following reaction:

CH3NO2(g) + 3/4O2(g) CO2(g) + 3/2H2O(l) + 1/2N2(g)

You place 1.724 g of nitromethane in a calorimeter with oxygen and ignite it. The temperature of the calorimeter increases from 22.23°C to 28.81°C. The heat capacity of the calorimeter was determined to be 3.044 kJ/°C. Write the thermochemical equation for the reaction.


We first find the heat evolved for the 1.724 g of nitromethane, CH3NO2.

Now, covert that to the heat evolved per mole by using the molar mass of nitromethane, 61.04 g.

23

23

23rxn NOCHmol1

NOCHg61.04

NOCHg1.724

kJ 20.03- q

H = –709 kJ

kJ 20.03 C22.23C28.81C

kJ3.044

Δ

rxn

calrxn

q

tCq

CH3NO2(l) + ¾O2(g) CO2(g) + 3/2H2O(l) + ½N2(g); H = –709 kJ


What is the enthalpy of reaction, H, for the reaction of calcium metal with water?

Ca(s) + 2H2O(l) Ca2+(aq) + 2OH-(aq) + H2(g)

This reaction occurs very slowly, so it is impractical to measure H directly. However, the following facts are known:

H+(aq) + OH-(aq) H2O(l); H = –55.9 kJCa(s) + 2H+(aq)

Ca2+(aq) + H2(g); H = –543.0 kJ


Ca(s) + 2H2O(l) Ca2+(aq) + 2OH-(aq) + H2(g)

2H2O(l) 2H+(aq) + 2OH-(aq); H = +111.8 kJ

Ca(s) + 2H+(aq) Ca2+(aq) + H2(g); H = –543.0 kJ

Ca(s) + 2H2O(l) Ca2+(aq) + 2OH-(aq) + H2(g); H = –431.2 kJ


We want H° for the reaction:CH3OH(l) CH3OH(g)

mol

kJ200.7Δ:methanolgaseousFor

mol

kJ238.7Δ:methanolliquidFor

f

f

H

H

reactants

fproducts

freaction ΔΔΔ HnHnH

mol

kJ238.7 mol 1

mol

kJ200.7 mol 1Δ vapH = +38.0 kJ

What is the heat of vaporization of methanol, CH3OH, at 25°C and 1 atm?

Use standard enthalpies of formation (Appendix C).


Methyl alcohol, CH3OH, is toxic because liver enzymes oxidize it to formaldehyde, HCHO, which can coagulate protein. Calculate Ho for the following reaction:

2CH3OH(aq) + O2(g) 2HCHO(aq) + 2H2O(l)

Standard enthalpies of formation, :ofΔH

CH3OH(aq): -245.9 kJ/molHCHO(aq): -150.2 kJ/molH2O(l): -285.8 kJ/mol


We want H° for the reaction:2CH3OH(aq) + O2(aq) 2HCHO(aq) + 2H2O(l)

reactants

fproducts

freaction ΔΔΔ HnHnH

mol

kJ0mol 1

mol

kJ245.9 mol 2

mol

kJ285.8-mol 2

mol

kJ150.2 mol 2Δ o

reactonH

kJ 491.8kJ 571.6kJ 300.4 Δ oreaction H

kJ 491.8kJ 872.0Δ oreaction H

kJ 380.2Δ oreaction H

Chapter 7

Quantum Theory of An Atome


What is the wavelength of blue light with a frequency of 6.4 × 1014/s?

= 6.4 × 1014/sc = 3.00 × 108 m/s

c = so = c/

= 4.7 × 10-7 m

s

110x6.4

s

m10x3.00

λ14

8

c


What is the frequency of light having a wavelength of 681 nm?

= 681 nm = 6.81 × 10-7 mc = 3.00 × 108 m/s

c = so = c/

v = 4.41 × 1014 /s

m10x6.81s

m10x3.00

7

8

c


The range of frequencies and wavelengths of electromagnetic radiation is called the electromagnetic spectrum.


The blue–green line of the hydrogen atom spectrum has a wavelength of 486 nm. What is the energy of a photon of this light?

= 486 nm = 4.86 × 10-7 mc = 3.00 × 108 m/sh = 6.63 × 10-34 J · s

E = h andc = so E = hc/

E = 4.09 × 10-19 J

m10x4.86

s

m10x3.00Js6.63x10

λ 7

834

hc

E


What is the wavelength of the light emitted when the electron in a hydrogen atom undergoes a transition from n = 6 to n = 3?

ni = 6nf = 3RH = 2.179 × 10-18 J

J10x1.816-

s

m10x2.998sJ10x6.626

λ19

834

= -1.816 x 10-19 J

1.094 × 10-6 m

soλ

Δhc

E E

hc

Δλ

22E

6

1

3

1J10x2.179Δ 18

2i

2f

H

11Δ

nnRE


Compare the wavelengths of (a) an electron traveling at a speed that is one-hundredth the speed of light and (b) a baseball of mass 0.145 kg having a speed of 26.8 m/s (60 mph).


Electronme = 9.11 × 10-31 kg; v = 3.00 × 106 m/s

Baseballm = 0.145 kg; v = 26.8 m/s

s

m10x3.00kg10x9.11

sJ10x6.63λ

631

34

2.43 × 10-10 m

s

m26.8kg0.145

sJ10x6.63λ

34

1.71 × 10-34 m

Comment: This is such an exceedingly small wavelength that the wave properties of a baseball cannot be detected by any existing measuring device.


Which of the following are permissible sets of quantum numbers?

n = 4, l = 4, ml = 0, ms = ½

n = 3, l = 2, ml = 1, ms = -½

n = 2, l = 0, ml = 0, ms = ³/²

n = 5, l = 3, ml = -3, ms = ½(a) Not permitted. When n = 4, the maximum

value of l is 3.(b) Permitted.(c) Not permitted; ms can only be +½ or –½. (b) Permitted.

Chapter 8

Electron Configuration and Periodicity


The building-up principle (or aufbau principle) is a scheme used to reproduce the ground-state electron configurations by successively filling subshells with electrons in a specific order (the building-up order).

This order generally corresponds to filling the orbitals from lowest to highest energy. Note that these energies are the total energy of the atom rather than the energy of the subshells alone.


1s

2s 2p

3s 3p 3d

4s 4p 4d4f

5s 5p 5d5f

6s 6p 6d

7s 7pThis results in the following order:

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p


Another way to learn the building-up order is to correlate each subshell with a position on the periodic table.

The principal quantum number, n, correlates with the period number.

Groups IA and IIA correspond to the s subshell; Groups IIIA through VIIIA correspond to the p subshell; the “B” groups correspond to the d subshell; and the bottom two rows correspond to the f subshell. This is shown on the next slide.


Write the complete electron configuration of the arsenic atom, As, using the building-up principle.

1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p3

For arsenic, As, Z = 33.


What are the electron configurations for the valence electrons of arsenic and zinc?

Arsenic is in period 4, Group VA.Its valence configuration is 4s24p3.

Zinc, Z = 30, is a transition metal in the first transition series. Its noble-gas core is Ar, Z = 18.Its valence configuration is

4s23d10.


Write an orbital diagram for the ground state of the nickel atom.

3s 3p

1s 2s 2p

4s 3d

For nickel, Z = 28.


Which of the following electron configurations or orbital diagrams are allowed and which are not allowed by the Pauli exclusion principle? If they are not allowed, explain why?

a. 1s22s12p3

b. 1s22s12p8

c. 1s22s22p63s23p63d8 d. 1s22s22p63s23p63d11

e.

1s 2s

a. Allowed; excited.b. p8 is not allowed.c. Allowed.d. d11 is not allowed.e. Not allowed; electrons in

one orbital must have opposite spins.


A representation of atomic radii is shown below.


Refer to a periodic table and arrange the following elements in order of increasing atomic radius: Br, Se, Te.

Te is larger than Se.Se is larger than Br.

Br < Se < Te


The size of each sphere indicates the size of the ionization energy in the figure below.


Refer to a periodic table and arrange the following elements in order of increasing ionization energy: As, Br, Sb.

Sb is larger than As.As is larger than Br.

Ionization energies: Sb < As < Br


The element is in Group IIA.

The electron affinity is > 0, so the element must be in Group IIA or VIIIA.

The dramatic difference in ionization energies is at the third ionization.


R is arsenic, As.

For R2O5 oxides, R must be in Group VA.R is a metalloid, so R could be As or Sb.The oxide is acidic, so

Chapter 9

Ionic and Covalent Bonding


Table 9.1 illustrates the Lewis electron-dot symbols for second- and third-period atoms.


Give the electron configuration and the Lewis symbol for the chloride ion, Cl-.

Cl -][

For chlorine, Cl, Z = 17, so the Cl- ion has 18 electrons. The electron configuration for Cl- is

1s2 2s2 2p6 3s2 3p6

The Lewis symbol for Cl- is


Give the electron configurations of Mn and Mn2+.

Manganese, Z = 25, has 25 electrons;. Its electron configuration is

1s2 2s2 2p6 3s2 3p6 4s23d5

Mn2+ has 23 electrons. When ionized, Mn loses the 4s electrons first; the electron configuration for Mn2+ is

1s2 2s2 2p6 3s23p6 3d5


Using the periodic table only, arrange the following ions in order of increasing ionic radius: Br-, Se2-, Sr2+. These ions are isoelectronic,

so their size decreases with increasing atomic number:

Sr2+ < Br- < Se2-


Using electronegativities, arrange the following bonds in order by increasing polarity:

C—N

Na—F

O—H

For Na—F, the difference is 4.0 (F) – 0.9 (Na) = 3.1.

For C—N, the difference is 3.0 (N) – 2.5 (C) = 0.5.

For O—H, the difference is 3.5 (O) – 2.1 (H) = 1.4.

C—N <Bond polarities: O—H < Na—F


Writing Lewis Electron-Dot Formulas

Calculate the number of valence electrons.

Write the skeleton structure of the molecule or ion. The central atom is the one with less electronegativity.

Distribute electrons to the atoms surrounding the central atom or atoms to satisfy the octet rule.

Distribute the remaining electrons as pairs to the central atom or atoms.


Write the electron dot formulas for the following:

a. OF2

b. NF3

c. NH2OH, hydroxylamine


a. Count the valence electrons in OF2:

O 1(6)

F 2(7)

20 valence electrons

O is the central atom (it is less electronegative). Now, we distribute the remaining 16 electrons, beginning with the outer atoms. The last four electrons go on O.

OF F


b. Count the valence electrons in NF3:

N 1(5)

F 3(7)


N is the central atom (it is less electronegative). Now, we distribute the remaining 20 electrons, beginning with the outer atoms. The last two electrons go on N.

F FN

F


c. Count the electrons in NH2OH:

N 1(5)

H 3(1)

O 1(6)


N is the central atom (it is less electronegative than O). Now, we distribute the remaining six electrons, beginning with the outer atoms. The last two electrons go on N.

H

NH HO


Write electron-dot formulas for the following:

a. CO2

b. HCN


a. Count the electrons in CO2:

C 1(4)

O 2(6)


C is the central atom. Now, we distribute the remaining 12 electrons, beginning with the outer atoms.

Carbon does not have an octet, so two of the lone pairs shift to become a bonding pair, forming double bonds.

OO CO O


b. Count the electrons in HCN:H 1(1)C 1(4)N 1(5)

10 valence electrons.

C is the central atom. The remaining electrons go on N.

Carbon does not have an octet, so two of the lone pairs shift to become a bonding pair, forming a triple bond.

CH NN


Phosphorus pentachloride exists in solid state as the ionic compound [PCl4]+[PCl6]-; it exists in the gas phase as the PCl5 molecule. Write the Lewis formula of the PCl4+ ion.


Count the valence electrons in PCl4+:

P 1(5)

Cl 4(7)

-1

32

P is the central atom. The remaining 24 nonbonding electrons are placed on Cl atoms. Add square brackets with the charge around the ion.

Cl

Cl

ClCl P

+


Count the valence electrons in IF5:

I 1(7)

F 5(7)


I is the central atom. Thirty-two electrons remain; they first complete F octets. The remaining electrons go on I.

F

F

F

I

F

F

Give the Lewis formula of the IF5 molecule.


Compare the formal charges for the following electron-dot formulas of CO2.

CO O

For the left structure: For the right structure:

O: 6 – 2 – 4 = 0C: 4 – 4 – 0 = 0

The left structure is better.

CO O

Formal charge = group number – (number of bond pairs) – (number of nonbonding electrons)

C: 4 – 4 – 0 = 0O: 6 – 1 – 6 = –1O: 6 – 3 – 2 = +1

Chapter 10


The diagrams below illustrate molecular geometry and the impact of lone pairs on it for linear and trigonal planar electron-pair arrangements.


Molecular geometries with a tetrahedral electron-pair arrangement are illustrated below.


Molecular geometries for the trigonal bipyramidal electron-pair arrangement are shown on the next slide.


Molecular geometries for the octahedral electron-pair arrangement are shown below.


Use the VSEPR model to predict the geometries of the following molecules:

a. AsF3

b. PH4+

c. BCl3

a. Trigonal pyramidal.b. Tetrahedral.c. Trigonal planar.


Using the VSEPR model, predict the geometry of the following species:

a. ICl3b. ICl4-

a. T-shaped.b. Square planar.


Which of the following molecules would be expected to have a zero dipole moment?

a. GeF4

b. SF2

c. XeF2

d. AsF3

a. GeF4 tetrahedral molecular geometryzero dipole moment

b. SF2 bent molecular geometrynonzero dipole moment

c. XeF2 linear molecular geometryzero dipole moment

d. AsF3 trigonal pyramidal molecular geometrynonzero dipole moment


Hybrid orbitals have definite directional characteristics, as described in Table 10.2.


Use valence bond theory to describe the bonding about an N atom in N2H4.

N N

F

F

F

F


The sp3 hybridized N atom is

1s sp3

Consider one N in N2F4: the two N—F bonds are formed by the overlap of a half-filled sp3 orbital with a half-filled 2p orbital on F. The N—N bond forms from the overlap of a half-filled sp3 orbital on each. The lone pair occupies one sp3 orbital.


Use valence bond theory to describe the bonding in the ClF2

- ion.


3dsp3d

The sp3d hybridized orbital diagram for the Cl- ion is

Two Cl—F bonds are formed from the overlap of two half-filled sp3d orbitals with half-filled 2p orbitals on the F atom. These use the axial positions of the trigonal bipyramid.

Three lone pairs occupy three sp3d orbitals. These are in the equatorial position of the trigonal bipyramid.


Describe the bonding about the C atom in formaldehyde, CH2O, using valence bond theory.

C

H

O

H


1s sp2 2p

After hybridization, the orbital diagram for C is

The C—H bonds are formed from the overlap of two C sp2 hybrid orbitals with the 1s orbital on the H atoms.

The C—O bond is formed from the overlap of one sp2 hybrid orbital and one O half-filled p orbital.

The C—O bond is formed from the sideways overlap of the C 2p orbital and an O 2p orbital.


Give the orbital diagram and electron configuration of the F2 molecule.

Is the molecular substance diamagnetic or paramagnetic?

What is the order of the bond in F2?


F2 has 18 electrons. The KK shell holds 4 electrons so 14 remain.

The molecular electron configuration isKK(2s)2(2s)2(2p)4(2p)2 (*2p)4

The bond order is ½(8 - 6) = 1.The molecule is diamagnetic.

For F2 and O2, 2p is lower in energy than 2p. This order would not affect the determination of bond order and magnetic properties for these molecules.

s2 s2*

p2 p2 p2 p2


A number of compounds of the nitrosonium ion, NO+, are known, including nitrosonium hydrogen sulfate, (NO+)(HSO4

-). Use the molecular orbitals similar to those of homonuclear diatomic molecules and obtain the orbital diagram, electron configuration, bond order, and magnetic characteristics of the NO+ ion.

Note: The stability of the ion results from the loss of an antibonding electron from NO.


NO+ has 14 electrons. The KK shell holds 4 electrons, leaving 10 electrons for bonding.

The molecular electron configuration isKK(2s)2(2s)2(2p)4(2p)2

The bond order is ½(8 - 2) = 3.

The ion has a diamagnetic molecular orbital electron configuration.

s2 s2*

p2 p2 p2 p2


The valence-shell electron-pair repulsion (VSEPR) model predicts the shapes of molecules and ions by assuming that the valence-shell electron pairs are arranged about each atom so that electron pairs are kept as far away from one another as possible, thereby minimizing electron pair repulsions.

The diagram on the next slide illustrates this.


Two electron pairs are 180° apart ( a linear arrangement).

Three electron pairs are 120° apart in one plane (a trigonal planar arrangement).

Four electron pairs are 109.5° apart in three dimensions (a tetrahedral arrangement).


Five electron pairs are arranged with three pairs in a plane 120° apart and two pairs at 90°to the plane and 180° to each other (a trigonal bipyramidal arrangement).

Six electron pairs are 90° apart (an octahedral arrangement).

This is illustrated on the next slide.


These arrangements are illustrated below with balloons and models of molecules for each.

slide

Technology

f t3 f t

ft by7

ft3 ft

c tc tf32 f

aluminum oxygen

tk34 c

g oxygen

oil of wintergreen