CHEMISTRY 133 LECTURE / STUDY GUIDE FOR R.H. …faculty.sfasu.edu/langleyricha/Chem133/guide133.pdfwithout written permission ... with approximately equal numbers of students using

CHEMISTRY 133

LECTURE / STUDY GUIDE

FOR

R.H. LANGLEY

Note: Make sure you read the introduction.

© 2016 Sevagram Enterprises No part of this book may be reproduced, in any form or by any means,

without written permission

© 2016 Sevagram Enterprises

1

INTRODUCTION

Note: If you did not get this until after the beginning of the term, do not just start using it

where you currently are in class. You should, at the very minimum, look over all

comments from the beginning of these notes. You do not want to miss any of the

comments, because one of them may be the most important one for you.

This guide is to supplement the lecture. Its usefulness will depend on how much you add.

This package outlines the class notes and gives additional comments. Not everything said in

class will be in these notes; you will need to add extra material. Numerous studies show that

the more you add the more helpful class notes are. There is additional information in these

notes to help you understand the material, many of these items go beyond what is in the

lecture, but you have it here to help you. You should intersperse class notes and problems in

the appropriate places. Simply reading these notes or reading what someone else has written,

is a total waste of your time. What you write in here and what you do here is the only way to

make these notes of any benefit to you.

This guide will not replace the Syllabus. Everything you need to know to get through this

class is in the Syllabus. Many students have had their course grade significantly lowered by

not reading the Syllabus. This booklet will help you implement many of the suggestions

given in the Syllabus. Anytime that you seem to be having trouble in class you should re-

read the Syllabus and/or the notes contained in this supplement.

You may use a calculator on exams. You must use your own calculator. A

fancy/complicated calculator is not required. If you buy one and are going to use it for

Chemistry 134 get one that does square roots, logs, and lns. You should use the same

calculator on the homework that you will use on the test.

You may use a programmable calculator. However, you should be aware of the fact that on a

recent exam, with approximately equal numbers of students using programmable and

nonprogrammable calculators, there was a significant difference in their average grades. The

average for those students using programmable calculators was 49.50, while that of students

using nonprogrammable calculators was 66.56. Thus, students using a programmable

calculator averaged over 17 points below the other students.

THERE IS NO SUBSTITUTE FOR CLASS ATTENDANCE – DO NOT MISS CLASS.

During a regular semester, expect each absence to lower your grade on the next test by 10-12

points. During a summer session, expect this value to double. The basis for these

predictions is class performance on the exams during the past several terms.


2

Note: Enter notes on these pages or, less effectively, on notebook paper to insert into the

appropriate place, recopying may help.

Note: All chapter numbers, section numbers, problems etc. are keyed to N.J. Tro,

Chemistry, Pearson/Prentice Hall, 2008.

Note: If at any point you begin to feel overwhelmed, you should see the instructor.

CHAPTER 1

1.1 (This, and all other section numbers, refer to the appropriate sections in BLBMWS.)

Read

1.2-1.3 – Classification of Matter/Properties of Matter

Matter:

Mass:

Note: While mass and weight are technically different, they are interchangeable in this

course.

Matter normally occurs in one of three states:

1.

2.

3.

Law of Conservation of Mass

It is possible to describe Matter by changes/properties:

Physical Change:

Chemical Change (Chemical Reaction):

Energy


3

Intensive Property:

Extensive Property:

Matter may occur as a pure substance or as a mixture.

Pure Substance:

Mixture:

Element:

Compound:

Mixture:

Heterogeneous Mixture:

Homogeneous Mixture (Solution):

1.4 Units of Measurement

Units: SI

NOTE: you are responsible for the base units in the textbook.

BASE UNITS:

Mass

Amount of a substance

Length

Electric Current

Time

Temperature

Luminous Intensity


4

A combination of units may be necessary:

Speed =

Sometimes it is necessary to use the same unit more than once:

Area =

Temperature scales: °F, °C, K (no degree symbol on K)

K = °C + 273.15 (or K = °C + 273)

If the base unit is too large or too small for convenient use, it is possible to add a

multiplier.

A prefix indicates the presence of a multiplier.

i.e., 1 000 000 m =

NOTE: you are responsible for prefixes, except as discussed in class.

KNOW: M = mega =

k = kilo =

d = deci =

c = centi =

m = milli =

= micro =

n = nano =

p = pico =

In order to simplify expressing some numbers use scientific notation.

2 457 000 000 = 2.457 × 109 0.000 000 239 8 = 2.398 × 10–7


5

NOTE: 1.00 Liter = 1.00 mL =

NOTE: You should also know:

All SI-SI conversions (mostly from prefixes)

All common English-English length and mass relations

K = °C + 273.15 (or K = °C + 273)

One English-SI conversion for length, and one for mass

Note: A very common, and expensive (point wise), mistake is to try to learn more than these

do. Study for the test; do not waste your time on unnecessary conversions.

NOTE: See the PROBLEM SOLVING section at the very end of this guide.

Note: If you have not completed problems for the previous section, do so as soon as

possible.

1.5 Uncertainty in Measurements

All calculated values consist of two parts: a number and a unit.

BOTH must be present to receive full credit on an exam.

Exact Numbers: Measured Numbers:

Example: Multiple measurements:

6.75

6.74

6.77

6.76

6.73

6.76


6

Significant Figures:

Rules:

Rules for calculations:

Exact numbers have no effect on the number of significant figures.

Multiplication & Division –

1456

3.16x2.4

Addition & Subtraction –

55.1|5

23.2|

27.1|3

Do not worry about rounding (as long as you are reasonable). DO NOT make the mistake of

rounding too soon.


7

1.6 Dimensional Analysis


The key to the ease/difficulty of this course is the Factor-Label Method (Dimensional

Analysis). The sooner you confine your problem solving to this method the easier the

remainder of the course will be. Resist any attempt to solve problems by other methods.

You may find that you are more comfortable with other methods, but they will let you down

in the end. For example, you will not be able to do most of Chapter 6 without knowing the

Factor-Label Method, and if you wait until then to learn it you will be impossibly far behind.

Do not waste your time learning extra conversions. The time that these extras will save you

is time lost from studying important material for the exam. A common example is to learn

the prefixes (as required) followed by a number of superfluous conversions such as: 10 mm

= 1 cm, 100 cm = 1 m, 1000 mm = 1 m, 10 cm = 1 dm, and 10 dm = 1 m. While these

conversions are correct, once you have learned the prefixes none of these are necessary, and

a waste of your time. You should spend your time studying for the test not learning

superfluous conversions.

Example: How many cm are in 2 inches?

Example: How many meters are in 2.00 miles?


8

Density =

Note: For this course Density = Mass/Volume is a definition NOT a mathematical

relationship. Using it as a mathematical relation may make a few Chapter 1 problems

easier – but it will put you significantly behind in preparing for other chapters.

Example: What is the density of a liquid for which 2.00 ft3 weighs 125 lbs?

Example: Express this density as g/cm3.

Example: How many grams do 500. mL of a liquid weigh if the density of the liquid

is 1.53 g/cm3?

Example: What is the volume of a 125 g sample of a liquid if the density of the liquid

is 0.8372 g/cm3?



9

Study Guide Example – not covered in class:

Try to solve this problem without looking at the solution.

How many millimeters thick is a piece of aluminum foil if a piece measuring 18.0

inches by 20.0 inches weighs 9.875 grams. The density of aluminum is 2.70 g / cm3.

001.0

m

c

01.0

in0.200.18

1

cm54.2

in1

g70.2

cmg875.9

2

23

= 1.57 × 10–2 mm

Note that no equations appear in the solution, and the only information needed that

was not in the problem were simple conversions. This is how a factor label

problem should be set-up. As with any factor label problem, you could have

started with any step, and it is still correct. Any step(s) that you added, or any

equations you used, or any conversions you did elsewhere, or any different

conversions that you used were a waste of your time.

Do not forget the number of assigned problems has been trimmed to the absolute minimum

for the exams. Unfortunately, there is no way to assign fewer problems and assure good

performance on the exams.

Note: The final review for the test needs to be as close to test conditions as possible.

Thus, you should work the problems during this final review by yourself and

“closed book.”


10

Reminder: Enter notes on these pages or, less effectively, on notebook paper to insert into

the appropriate place, recopying may help.

Reminder: If at any point you begin to feel overwhelmed, you should see the instructor.

CHAPTER 2

2.1 The Atomic Theory of Matter (This, and all other section numbers, refer to the appropriate sections in BLBMWS.)

Dalton's Atomic Theory

1.

2.

3.

4.

5.

Law of definite proportions (Law of constant composition)

Law of multiple proportions

2.2 The Discovery of Atomic Structure

Particle charge mass

(Coulombs) (Grams)

Electron

Proton

Neutron


11

Rutherford (1911)

2.3 The Modern View of Atomic Structure

Atomic Number:

Atom:

A chemical symbol designates an element – an abbreviation of one or two letters.

1st (or only) must be capitalized.

2nd (if present) must not be capitalized.

NOTE: Be careful on exams.


12

Examples: Chlorine Cl Oxygen O

2

1

4

3Cl Area 1:

Area 2:

Area 3:

Area 4

Mass Number

Isotope:

This information must be given to you; it is NOT normally given on

the Periodic Table. (Many students get into trouble by assuming that

the mass number is present.)

For Cl, the common values are 35 and 37

2.4 Atomic Weights

The masses will be very important. A mass scale has been set up to deal with atoms.

The basic unit is the Atomic Mass Unit (amu). (1 amu = 1 Dalton)

Atomic Mass Unit

Atomic Mass (Weight):

Example: Natural lead contains 1.4% lead-204 (203.973 amu), 24.1% lead-206

(205.9745 amu), 22.1% lead-207 (206.9759 amu), and 52.4% lead-208 (207.9766

amu). Calculate its atomic weight.


13


Fill in the gaps in the following table:

Symbol __3122Sb __ __________ __________ __________

Protons __________ ____74____ __________ ____90____

Neutrons __________ ___110____ ___125____ ___142____

Electrons __________ ____71____ ____80____ __________

Net Charge __________ __________ ____2+____ ____4+____

Answers:

Symbol __3122Sb __ __

3184 W _ _2207 Pb __ __ 4232 Th _

Protons ____51____ ____74____ ____82____ ____90____

Neutrons ____71____ ___110____ ___125____ ___142____

Electrons ____54____ ____71____ ____80____ ____86____

Net Charge ____–3____ ____+3____ ____2+____ ____4+____

2.5 Periodic Table The symbols of the elements are gathered together on the Periodic Table (a summary

of the various elements grouped by their properties).

Groups (Families):

Period:

The elements on the left side (except H) are _______________

The elements on the right side are _______________

The elements between these are called _______________

Metal:

Nonmetal:

Metalloid (Semimetal):

NOTE: Hydrogen is often an exception.


14

Certain columns (families) have names.

Alkali Metal:

Alkaline Earth Metal:

Halogen:

Noble Gas:

2.6-2.7 Chemical Substances

Molecule:

For elements, a few of these are very important.

KNOW: Hydrogen Nitrogen Oxygen Fluorine

Chlorine Bromine Iodine

You should not only know that these elements are diatomic, but you should

also know their symbols and their names (spelled correctly).

Note: You should precede to Nomenclature I on the website and begin studying from it.

See the notes in the text also.

Allotrope:

Compounds may contain ions instead of molecules. If they do contain ions, there

must be equal numbers of positive and negative charges.

Isolated ions always have a charge written: Cl– OH– Na+

Compounds are neutral so there is no charge written even though ions are

present: NaCl

Metals usually react with nonmetals to produce ions not molecules.

Ion:

Cation:

Anion:


15

Monatomic Ion:

Polyatomic Ion:

Polyatomic ions are like molecules except for the fact that they

have charges.

2.8-2.9 Nomenclature

Note: The nomenclature covered in these two sections will be distributed over the first three

exams. For this reason, you only need to know a portion of this material by the next exam.

If you have not begun Nomenclature I on the website, you should do so immediately. This

guide will let you know when you need to move on to the other nomenclature sections on the

website.

Inorganic Nomenclature

Note: Plan to look at nomenclature daily. This takes time, it is not possible to learn all of the

nomenclature that you will need by "cramming," and it needs to be in long-term memory.

Note: You will only be responsible for the names put on the board in class, but they will be

comprehensive.

Note: It will be necessary for you to learn the names of certain elements, the names,

and charges of certain ions, along with the rules for predicting the charges of

some elements. You will also need to learn how to put everything together in

a compound. Simple memorization is not sufficient; you must practice naming

compounds.

Note: If you have not started Nomenclature I on the website, do so as soon as possible.

Compounds:

Molecular Formula (Chemical Formula):

Structural Formula:

Empirical Formula:


16

Inorganic Nomenclature

See Nomenclature I on Website


17

Reminder: Enter notes on these pages or, less effectively, on notebook paper to insert into

the appropriate place, recopying may help.


CHAPTER 3

3.1 Chemical Equations (This, and all other section numbers, refers to the appropriate sections in BLBMWS.)

Reactant:

Product:

Example: H2(g) + Cl2(g) 2 HCl(g)

Equations must be balanced – they are not much help otherwise.

Balancing equations – General method

1. Set up reactants and products (If all are not given then they must be

determined) – use correct formulas- do not change the formulas.

C4H10 + O2

___ C4H10 + ___O2 ___ CO2 + ___ H2O

2. Pick an element and balance it (postpone balancing elements appearing

more than once on a side).

i.e., C (or H,) but not O

___C4H10 + ___O2 4 CO2 + ___ H2O


18

3. Adjust another element ("forced")

In this case H

___C4H10 + ___O2 ___ CO2 + ___ H2O

4. Repeat step 3 until all elements are done.

___C4H10 + ___O2 ___ CO2 + ___ H2O

(Fractions are allowed temporarily, but not in the final solution.)

5. Clear fractions (if present)

___ C4H10 + ___ O2 ___ CO2 + ___ H2O

NOTE: Also, make sure the coefficients are reduced to their lowest whole number values.

6. Check to make sure that it really is balanced.

(An often-neglected step)

NOTE: If it is not balanced, check for obvious mistakes.

NOTE: If not found quickly it is probably best to start over (sometimes with another

element).

Study Guide Example – not covered in class: Write balanced chemical equations for each of the following reactions. a. Water

reacts with phosphorus trichloride to produce hydrogen chloride and H3PO3.

b. Water reacts with diboron trisulfide to produce H3BO3 (boric acid) and hydrogen

sulfide. c. Zinc metal (Zn) reacts with hydrochloric acid to produce hydrogen gas and

zinc chloride (ZnCl2). d. PH3 burns in oxygen gas to produce water and diphosphorus

pentoxide. e. silicon dioxide reacts with carbon at high temperatures to produce

elemental silicon and carbon monoxide gas.

Answers: Note: this problem also tests your nomenclature.

a. 3 H2O + PCl3 3 HCl + H3PO3

b. 6 H2O + B2S3 2 H3BO3 + 3 H2S

c. Zn + 2 HCl H2 + ZnCl2

d. 4 PH3 + 8 O2 6 H2O + 2 P2O5

e. SiO2 + 2 C Si + 2 CO


19

3.2 Types of Reactions

Be able to classify the various types.

Some reactions may fit into more than one category; it will not be important

which one you choose (as long it is one of the choices).

Combination Reactions –

Decomposition Reactions –

Displacement Reactions –

Metathesis Reactions –

Neutralization Reactions –

Combustion Reactions –

3.3 Formula Weights

Molecular (Formula) Weight:

Example: N2

Example: HCN amu H

amu C

amu N

amu HCN

Example: [Cu(NH3)4]SO4·.2H2O

1 Cu amu

4 N amu

16 H amu

1 S amu

6 O amu ________

amu


20

An alternative to formulas is percent composition. It may be derived from a formula,

or more importantly, a formula may be derived from it.

Example: C6H12O6 FW = 180.158 amu

% C =

% H =

% O =

Note: the same answers would result if grams and moles appeared instead of amu

and atoms.

Reminder: If you have had trouble with any of the preceding problems, you should get help

immediately.

3.4 Avogadro’s Number and the Mole

The SI unit for the quantity of a substance is a mole.

Mole:

Avogadro's Number:

NOTE: MOST PROBLEMS FROM THIS POINT ON WILL INVOLVE MOLES IN

AT LEAST ONE STEP.

NOTE: AVOGADRO'S NUMBER IS USED ONLY RARELY IN CALCULATIONS.

IT IS NECESSARY ONLY WHEN THE NUMBER OF ATOMS OR

MOLECULES IS ASKED FOR OR GIVEN.

Example: How many moles are present in 1.20 × 1025 silver atoms?


21

Example: If one N2 molecule weighs 28.01 amu, calculate the number of grams in

one mole of N2 molecules.

Molar Mass:

Example: How many grams of CO2 are in 4.00 moles of CO2?

Example: How many oxygen atoms are in 22.0 grams of CO2?

3.5 Empirical Formulas from Analysis

Determining the formula from the composition

Example: Analysis of a sample of a gas found 2.34g of nitrogen and 5.34g of

oxygen. What was the formula of this gas?

Use N and O atoms not N2 and

O2 molecules. Molecules do

not contain other molecules.

NOTE: DO NOT CONFUSE THIS RATIO WITH LATER.


22

Example: Analysis of a sample of Vitamin C found it to contain: 40.9% carbon,

4.58% hydrogen and the remainder was oxygen. Its formula weight is about

180g/mole. What are its empirical and molecular formulas?

BE CAREFUL TO MAINTAIN SUFFICIENT SIG. FIGS.

Note: this method will always give the empirical formula; only the additional information

(MW) allows the molecular formula to be determined.

Example: A 1.778 g sample of an unknown solid was burned in oxygen, and 2.842 g

of CO2, 0.2609 g of H2O, and 0.4056 g of N2 were produced. The FW of the

unknown was about 740 g/mole. The compound contained C, H, N, and O.

Determine its molecular formula.

Note: Most students believe the preceding type of question to be the most difficult one

possible on Exam 1. Do not expect it to be perfectly clear immediately in class.


23




3.6 Quantitative Information from Balanced Equations

Calculations involving chemical reactions

Stoichiometry:

NOTE: 1. You must have a balanced chemical equation.

2. The reactants/products are related by a mole ratio.

Example: If 40.0 g of Cl2 and excess H2 are combined, HCl will be produced. How

many grams of HCl will form?

3.7 Limiting Reactants

Limiting Reagent (Limiting Reactant):

Example: The following reaction will produce bismuth(III) fluoride:

2 Bi(s) + 3 F2(g) 2 BiF3(s)

During one experiment, the reaction mixture contained 183.9 grams of bismuth and

105.6 grams of fluorine. What was the maximum number of grams of BiF3 formed?

HINT: Anytime the quantities of more than one reactant are given, it is probably a L. R.

problem.

Do not confuse with earlier


24

Study Guide Example – not covered in class: You are a bicycle manufacturer. You have an inventory of 3165 handlebars, 6000

wheels, and 4175 frames. a. How many bicycles can you make before you have to

order more parts? b. Which part will you need to order first? c. How much of each

of the parts will you still have on hand when you order new parts?

a. You could make 3165 bicycles from the handlebars, or 3000 bicycles from the

wheels, or 4175 bicycles from the frames. The smallest value is 3000 bicycles – thus

the wheels behave like a limiting reagent, and manufacture must stop not matter how

many handlebars and frames are remaining.

b. You will need to order wheels first, because they are the limiting part.

c. You will have 3165 – 3000 = 165 handlebars, 6000 – 2(3000) = 0 wheels, and 4175

– 3000 = 1175 frames.

Various things may occur in a reaction to prevent the reactants from complete

conversion to products.

Actual Yield:

Theoretical Yield:

Percent Yield =

Example: The heating of a 25.0 g sample of calcium oxide with excess hydrogen

chloride produces water and 37.5 g of calcium chloride. What is the percent yield?


25

Reminder: You are not expected to understand the lecture material while you are in the

lecture room, you are expected to have some understanding (or specific questions)

when the next lecture begins.





Reminder: The final review for the test needs to be as close to test conditions as possible.

Thus, you should work the problems during this final review by yourself and

“closed book.” In addition, work though a practice exam under “exam conditions”

about 24 hours before the scheduled exam time. This will give you sufficient time

to correct your studying.

END OF MATERIAL FOR EXAM 1

Note: As soon as you finish with this test, begin Nomenclature II on the website.

Note: If you feel that, you did not do as well as you would have liked on the exam, you

should begin immediately preparing for the retest. Do not wait until you get the test

back before you start looking at the material and do not use the test you just

completed as a study guide. Also, make sure you find out if there is a curve, before

you make any decisions about this class. Persons with B’s or better in the class have

dropped the class because they did not bother to find out.


26

The following is the key to an Exam 1 that was used in the past. The format has been slightly

altered to allow the addition of comments. The answers are given in italics and the

comments will be given in italics and in parentheses. Comments are based on the class that

took this exam.

EXAM 1 CHEMISTRY 133 SIL 1, 2, 3 R.H. Langley

(There are not a lot of directions – make sure you follow them.)

Put your answers in the appropriate blanks. Name (Print) ___________________

Make sure your exam has 10 different questions. Section _____

1.(10) Define each of the following, as they apply to this course, in a sentence or two.

(a) Chemical Property Any change based upon a change in composition.

(b) Group A vertical column on the periodic table.

(c) Metathesis Reaction A double displacement reaction.

(d) Empirical FormulaThe simplest formula giving the lowest ratio of the atoms

present.

(e) Density Mass/Volume

(Approximately 50% of the students did not pass this question. Less than 5% got all 10

points. A question is worth a letter grade, time must be spent on the definitions. In addition,

you cannot understand the material if you do not understand the terms.)

2.(10) Name or give the formula (symbol) for each of the following.

(a) Arsenic ____As____ (f) Bi ______Bismuth_________________

(b) Oxygen ____O_____ (g) IF7 ___Iodine heptafluoride__________

(c) Antimony trichloride____SbCl3_ (h) SiBr4 __Silicon tetrabromide_____

(d) Disulfur dibromide ___S2Br2___ (i) Cl2O5 _Dichlorine pentaoxide_____

(e) Hydrochloric acid ___HCl____ (j) NH3 ________Ammonia______________

(Only 5% of the students got all 10 of these correct. This was the best performance on any of

the nomenclature questions all semester. A low performance on this question (below an 8)

indicates that significantly more time is required. One student managed to lose a total of 131

points to nomenclature over the semester (22% of the possible points for the term), the best

student only lost 18 points (3%) during the term. The average student lost 77 points (13%).

Something this important requires significant amounts of study time, and the students in this

particular class did not study enough nomenclature. Do not forget – spelling counts.)


27

Exam 1

3.(10) The Morgan silver dollar, which contains 90. % silver, has a mass of 26.73 g. When

the coin was first minted in 1878, silver was worth $1.18 per troy ounce (31.103 g).

What was the value of the silver in the silver dollar? (From Problem ....)

(On a regular exam, this

(This answer is only worth 2 points.) would be a homework

(Make sure the answer is in this blank.) problem)

___$0.91___

(This is the only

TroyOz

18.1$

g103.31

.TroyOz

%100

%90g73.26 = $0.91

totally correct

answer – it must.)

look exactly like (As with any factor label problem the sequence is not

this or $.91. important. For full credit ALL steps, including the,

Additional “trivial” ones must be included. Thus, for example,

conversions converting 90 % to 0.90 in your head will not be given

could be credit. You will receive credit for the work shown, not what

added, you did in your head. While this may seem extreme, this

but are will get you in the habit for later on in the term.)

unnecessary.)

4.(10) A cesium (Cs) atom has a diameter of about 4.7 Å. How many cesium atoms would

be needed to equal 1.0 cm? (From Problem ....)

(Again, on a regular exam this would

(Make sure the answer in the blank.) be a homework problem.)

(Wrong number of significant figures = –1)

_2.1 × 107 atoms__

Å7.4

atom1

m10

Å1

c

01.0cm0.1

10= 2.1 × 107 atoms

(This is the proper

number of significant

figures.) (The Å to m conversion was given to the students

with the exam, they did not need to memorize it.)

(Even though this was directly out of the homework 15% of the students got 0 points for it,

and less than one fourth of the class got full credit. It is bad enough to get 0 points on a

problem (a letter grade loss on the exam), but a 0 is worse when it is on a homework

problem. A score of 0 on a homework problem indicates no studying effort on the part of the

student (whether this is true or not). Also, if a student understood the problem so poorly,

then they should have been in the instructor’s office (if they cared at all about their grade).)


28

Exam 1

5.(10) How many vitamin C (C6H8O6)molecules are in a 500-mg tablet of vitamin C.

(From Problem ...)

(Three chapters on this test =

(No units = –1) three homework problems.)

_1.71 × 1021 molecules_

686

23

686

686

OHmoleC1

molecules10x022.6

OHgC12.176

OHmoleC1

m

001.0mg500 = 1.709 × 1021 molecules

(An answer of 1.70 × 1021 (Answer to be rounded

molecules would also be acceptable.) to the proper number of

significant figures.)

C 6(12.01)

H 8(1.008)

O 6(16.00)

176.12 g/mole

(It is not necessary to show this, but it is safer.)

(On the average, people did worse on this homework problem than on the preceding one.

Over 20 % of the students received 0 points on this problem. The comments on question 4

apply here.)

(At this point, the students taking the exam have had two “memorization” problems

(definitions and nomenclature) and three homework problems. This amounts to a total of 50

points based only on things that the student should have seen before, and discussed with the

instructor if there was any problem in understanding the material.)

6.(10) The density of germanium, Ge, is 5.35 g/cm3. How many pounds does 2.00 ft3 of

germanium weigh?

g59.453

lb1

cm

g35.5

in1

cm54.2

ft1

in12ft00.2

3

33

3 = 667.9827 lbs

(To be rounded)

__668 lbs__ (A common mistake in the calculation is in the two cubed

conversions. Many people enter the first conversion into their

calculator as 12 /1. This is incorrect; it should be entered as 123/1

(technically as 123/13).

(This is a factor label problem, thus the order is irrelevant. In addition, if you learned

different conversions in place of the ones used above they will be acceptable. A common

problem occurs when a student attempts to learn more than the required number of

conversions. A student learning too many conversions often gets them mixed. Nearly 30% of

the students taking this test could not do a single conversion for this problem (i.e., they did

not even write down that 12 inches = 1 foot).)


29

Exam 1

7.(10) Fill in the gaps in the following table:

Symbol F18

9

_6Li1+_ _224Ra 232

16S

3 88

Protons _9___ __3__ _88__ _16__

Neutrons _9___ __3__ _136_ _16__

Electrons _9___ __2__ _88__ __18_

Mass Number _18__ __6__ _224_ _32__

(One person managed to get a 0 on this problem.)

8.(10) Balance the following equations. Enter ALL coefficients, EVEN ONES.

(a) _1_ H2 + _1_ Cl2 _2_ HCl

(b) _1_ XeF6 + _3_ H2O _1_ XeO3 + _6_ HF

(c) _3_ Ra(OH)2 + _2_ H3VO4 _1_ Ra3(VO4)2 + _6_ H2O

(d) _2_ C2H6 + _7_ O2 _4_ CO2 + _6_ H2O

(e) Methane reacts with fluorine to form carbon tetrafluoride and hydrogen fluoride.

1 CH4 + 4 F2 1 CF4 + 4 HF

(Only 15% of the students balanced all five equations correctly. The most common mistakes

were not following directions (leaving out the ones), not checking, and not knowing the

nomenclature for the last equation.)

(For the class as a whole this was their best page.)


30

Exam 1

9.(10) How many grams of ClO2 could be formed if 9.010 g of H2C2O4, 9.810 g of H2SO4,

and 12.200 g of KClO3 are mixed and allowed to react as follows:

H2C2O4 + H2SO4 + 2 KClO3 K2SO4 + 2 H2O + 2 CO2 + 2 ClO2

422

422

422

OCgH038.90

OCmolH1OCgH010.9 =

1

10006889.0= 0.100068859

6.7150 g ClO2_

(Correct sig. figs.)

42

42

42

SOgH08.98

SOmolH1SOgH810.9 =

1

100020391.0 = 0.100020391

3

3

3

gKClO55.122

molKClO1gKClO200.12 =

2

099550391.0 = 0.049775195

(Limiting Reagent)

2

2

3

2

3

molClO1

gClO453.67

molKClO2

molClO2molKClO099550391.0 = 6.71497 g ClO2

(Extra sig. figs.)

(Most people did not do well on this problem because they did not recognize it as a limiting

reagent problem. Even without this recognition, everyone should have been able to get some

credit by simply converting everything to moles. Many people simply left the problem blank.

Leaving a problem blank is the only way to guarantee no partial credit.)

(All molecular weights should have at least as many significant figures as the grams given.)

10.(10) Under certain conditions the compound S4N4 may be explosive. One way to destroy

this compound safely is by the following reaction:

S4N4 + 6 NaOH + 3 H2O Na2S2O3 + 2 Na2SO3 + 4 NH3

In one test reaction, 1.000 gram of S4N4 generated 0.298 grams of NH3. What was

the percent yield in the test reaction?

3

3

44

3

44

44

44

molNH1

gNH031.17

NmolS1

molNH4

NgS28.184

NmolS1NgS000.1 = 0.36967 g

___80.6 %___ (Extra sig. fig.)

%100xg36967.0

g298.0

= 80.600% (Leaving out the × 100% [doing it in

your head] will result in a point

deduction)

(This was question with the lowest average grade on this test. Over a third of the class

received 0 points. If they had simply written out the definition of percent yield they would

have gotten 2 points – never leave a question blank).

(As with all tests, the average grade on this test was a C. Point wise a C on this test was a

56.)


31

Note: It is exceedingly important that, while it is fresh in your mind, you make sure that you

have the correct solution to every problem on the first exam (and re-test). Not doing

so will have a strong negative influence on your grade on the final exam.

Note: If your grade on the test is not what you want, you should immediately refer to the

syllabus and/or see the instructor.

Reminder: If you feel that, you did not do as well as you would have liked on the exam,

you should begin immediately preparing for the retest. Do not wait until you get the

test back before you start looking at the material and do not use the test you just

completed as a study guide. Also, make sure you find out if there is a curve, before

you make any decisions about this class. Persons with B’s or better in the class have

dropped the class because they did not bother to find out.

Note: Many people get the same grade on the retest because they continue to make the

same studying mistakes. If you do not fix whatever it was that caused you to get a

low grade, you will continue getting the same grades. See the syllabus for

suggestions for improving your study habits.

Note: A few people do not do as well on the retest as the original test. These people

have consciously, or subconsciously, used the first test as a study guide. You must

prepare for the retest like a fresh test.

Reminder: Enter notes on these pages and/or on notebook paper inserted later in the

appropriate place, recopying may help.

Reminder: Begin on Nomenclature II on the website.


CHAPTER 10

Note: The textbook gives a large number of additional gas law equations. You only need

to know four. Learning additional gas laws will not help you on the exam, and they

take away from your study time. In addition, if you attempt to learn unnecessary

material you are more likely to make mistakes.

10.1 Gases (This, and all other section numbers, refers to the appropriate sections in BLBMWS.)

States of Matter

Condensed Phases:

Intermolecular Forces:


32

10.2 Pressure

Pressure = Units of Pressure =

Pascal (Pa) =

Standard Atmosphere:

Note: You need to know the atm to torr (mmHg) conversion (Note: 1 mmHg = 1 torr)

10.3-10.4 Gas Laws

Boyle's Law – P and V

Charles's Law – V and T

Avogadro's Law – V and n

Other combinations are possible – however they lead to the same overall

relationships.

Boyle's Law

Charles's Law

Avogadro's Law


33

Temperature Problem:

Absolute zero = 0 K = –273.15 °C

NOTE: For all gas laws the temperatures must be expressed on an absolute

temperature scale. (Normally Kelvins are used (Not °K).)

TK = T°C + 273(.15)

The Molar Volume is the volume of 1 mole of gas at STP.

= mole

L41.22 at 0°C and 1 atm.

Boyle's, Charles's and Avogadro's Laws may be combined.

KNOW: R = K•mol

atm•L08206.0

PV = nRT (Ideal Gas Equation = Ideal Gas Law)

For others rearrange and cancel what does not change.

R = nT

PV

11

11

Tn

VP = R =

22

22

Tn

VP

Combined Gas Law

NOTE: A common mistake is to use 22.4 L/mol under conditions that are not STP.


34

Example: A sample of a gas occupies 25 liters under a pressure of 1.0 atm. What

will be the volume of the gas after increasing the pressure to 2.0 atm?

Example: A 500.0 mL sample of a gas is confined at a pressure of 1 atm at 25°C.

What temperature is necessary for the sample to occupy 600.0 mL?

Example: A sample of krypton gas occupies 50.0 L at 27°C and 1000. torr.

Determine its volume at 0°C and 1.10 atm.


35

Example: How many moles of helium gas are present in a 20.0 L container if the

pressure is 14.0 atm at a temperature of 27°C?


In 1937, the German Zeppelin Hindenburg exploded while landing in New Jersey.

The Hindenburg contained 7.1 × 106 ft3 of hydrogen gas. If, at the time of the

explosion, the hydrogen was at a temperature of 20.°C at a pressure of 745 torr, how

many grams of hydrogen were present?

Answer: n = RT

PV =

2

2

3

333336

molH1

gH0.2

dm1

L1

1.0

d

c

01.0

in1

cm54.2

ft1

in12

torr760

atm1

K293K•mol

atm•L08216.0

ft10x1.7torr745

= 1.6 × 107 grams H2 (Both your factor label and nomenclature are

being tested.)

10.5-10.6 More Gas Laws


When dealing with gas mixtures:

ntotal =

Ptotal =

Dalton's Law of Partial Pressures:

Partial Pressures:

NOTE: This is the only significant gas law in this section.


36

Also of use is the Mole Fraction:

Xa =

X may also be related to pressure.

Xa =

Example: The mole fraction of nitrogen in air is 0.79, and the mole fraction of

oxygen is 0.20. What is the partial pressure of each of these gases, if the total

pressure is 2.00 atm.?

Example: (a) The vaporization of a sample of an unknown liquid produced 0.8001

grams of vapor. At a temperature of 99°C and a pressure of 750. mmHg, the vapor

occupied 0.250 L. What was the molecular weight of the unknown liquid?

(b) Analysis of the unknown liquid found the following: 24.2% C, 4.04% H,

and 71.7%Cl. Using these data and the above, determine the molecular

formula of this compound.


37

Gas Stoichiometry Need to be able to use mole relationships

Example: A 10.0 L container is filled with 600. g of ethane (C2H6) and 2080 g of

oxygen. When ignited the mixture reacts as follows:

2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(g)

After the reaction, the mixture has a temperature of 150.°C. Determine the

partial pressure of each gas after the reaction, and the total pressure.

CO2

H2O

NOTE: Since O2 is the limiting reagent then, by definition, it is entirely gone, so it can

yield no pressure.

O2

NOTE: The C2H6 is a reactant, not a product, so its treatment must be different.

C2H6


38

Example: Determine the mole fraction of each of the components after the

reaction described in the last example.

Two methods:

Pressure Method Mole Method

atm C2H6 moles

atm O2 moles

atm CO2 moles

atm H2O moles

atm total moles

Total

10.7 Kinetic Molecular Theory

Kinetic Molecular Theory Present explanation of the behavior of gases.

Basic Postulates:

1.

2.

3.

4.

Kinetic Energy (molecular speed)

The pressure is due to the number of molecules hitting the walls, and how

hard the molecules hit the walls.

Decrease V –

Increase n –

Increase T –

Two gases –


39

Ideal Gas:

10.8 Applications of Kinetic Molecular Theory

Diffusion:

Effusion:

Graham's Law of Diffusion

NOTE: Rates will ALWAYS have a time unit in the denominator, i.e., mL/s or g/hr.

10.9 Real Gases

van der Waal's Equation:




Note: The textbook gives a large number of additional gas law equations. You only need to

know four. Learning additional gas laws will not help you on the exams, and they take away

from your study time for the exam. In addition, if you attempt to learn unnecessary material

you are more likely to make mistakes.



40

Reminder: Enter notes on these pages and/or on notebook paper inserted later in the

appropriate place, (recopying may help).

Note: If you ask, most students who have already completed this class “which is the hardest

chapter?” most will say chapter 4. Plan to adjust your studying accordingly. Also,

consider, if this is the hardest chapter, the remaining chapters must be easier.


CHAPTER 4

4.1 General Properties of Aqueous Solutions (This, and all other section numbers, refers to the appropriate sections in BLBMWS.)

Aqueous Solution:

NOTE: Water is assumed the solvent unless otherwise specified.

Solute:

Solvent:

Saturated:

Unsaturated:

Supersaturated:

Nonelectrolyte:

Electrolyte:

Dissociation:


41

Types of electrolytes: acids, bases, and salts

Acid:

Base:

Strong Electrolyte:

Weak Electrolyte:

Two important types: weak acids and weak bases

There is only a partial separation into ions: HNO2(aq) H+(aq) + NO2–(aq)

NOTE: You should see the Net Ionic Equations section on the website.

Reminder: You are not expected to understand the lecture material while you are in the

lecture room, you are expected to have some understanding (or specific questions)

when the next lecture begins.

4.2 Precipitation Reactions

Solubility rules: Ionic compounds in water.

The rules depend on the strength of the ionic bonds, with a few exceptions.

1.

2.

3.


42

Note: See the Solubility Rules on the website.

The insoluble materials will not dissolve (to a great degree); also, any reaction that

forms them will give a precipitate.

Various driving forces:

Form a gas CO2, NH3, SO2, H2S

Form a weak/nonelectrolyte H2O, a weak acid, or a weak base

Form a precipitate See solubility rules

4.3 Acids, Bases, and Neutralization Reactions

Strong Acid:

KNOW:

Strong Base:

KNOW:

What are ionic compounds?


43

Reactions may be written in more than one way.

Molecular Equation:

Complete Ionic (Total) Equation:

Net Ionic Equation:

Spectator Ion:

NOTE: THIS IS WHERE MOST PEOPLE HAVE TROUBLE ON THE EXAM.

PLAN ON EXTRA STUDY TIME.


Write balanced net ionic equations for the reactions that occur when each of the

following are mixed. Assume all reactions are in aqueous solution. It is possible that

no reaction will occur.

Answers

a. H2SO4 + BaCl2 a. SO42– + Ba2+ BaSO4

b. AgNO3 + K2CO3 b. 2 Ag+ + CO32– Ag2CO3

c. ZnS + HCl c. ZnS + 2 H+ Zn2+ + H2S

d. Li3PO4 + HBr d. PO43– + 3 H+ H3PO4

e. RbCl + NH4NO3 e. RbCl + NH4NO3 No Reaction

f. Sr(OH)2 + HI f. OH– + H+ H2O

g. NaOH + HC2H3O2 g. OH– + HC2H3O2 H2O + C2H3O2–

h. Cr(OH)3 + HClO3 h. Cr(OH)3 + 3 H+ Cr3+ + 3 H2O


44

4.4 Oxidation-Reduction (Redox) Reactions

Oxidation-Reduction (Redox):

Oxidation:

Reduction:

Example: 2 Mg + O2 2 MgO

Reducing Agent:

Oxidizing Agent:

This is very useful with the activity series.

Study Guide Example – not covered in class: Based on the Activity Series, write balanced molecular and net ionic equations for

each of the following.

Answers:

a. Hydrobromic acid with iron a. 2 HBr + Fe FeBr2 + H2

2 H+ + Fe Fe2+ + H2

b. Sulfuric acid with nickel b. H2SO4 + Ni NiSO4 + H2

2 H+ + Ni Ni2+ + H2

c. Hydrochloric acid with chromium c. 6 HCl + 2 Cr 2 CrCl3 + 3 H2

6 H+ + 2 Cr 2 Cr3+ + 3 H2

d. Cu + CoCl2 d. Cu + CoCl2 No Reaction

e. Pb + AgNO3 e. Pb + 2 AgNO3 Pb(NO3)2 + 2 Ag

Pb + 2 Ag+ Pb2+ + 2 Ag

f. H2 + Hg(NO3)2 f. H2 + Hg(NO3)2 2 HNO3 + Hg

H2 + Hg2+ 2 H+ + Hg

g. H2 + MnCl2 g. H2 + MnCl2 No Reaction

h. Li + H2O h. Li + 2 H2O 2 LiOH + H2

Li + 2 H2O 2 Li+ + 2 OH– + H2

i. Zn + HC2H3O2 i. Zn+2HC2H3O2 Zn(C2H3O2)2 + H2

Zn+2HC2H3O2 Zn2++2C2H3O2–

+ H2

j. Sn + CuSO4 j. Sn + CuSO4 SnSO4 + Cu

Sn + Cu2+ Sn2+ + Cu


45

4.5 Concentrations of Solutions

Concentration may be expressed in many ways – one of which is molarity.

Molarity = M = (Recall 1.00 L = 1.00 dm3)

Note: Many students get into trouble because they incorrectly use m and the abbreviation

for moles. The correct abbreviation is mol. The abbreviation m means molality,

which appears in chapter 13. The use of an incorrect abbreviation has resulted in

students making very significant factor label mistakes, and losing several points on an

exam.

Example: A solution of NaCl contains 39.12 g of this compound in 100.0 mL of

solution. Calculate the molarity of NaCl.

Example: A maximum of 85.4 g of Na2SO4 will dissolve in 200. mL of water. It

completely separates into sodium ions and sulfate ions. What is the

concentration of sodium ions in the solution?

Example: How many moles of ammonium ions are in 0.100 L of a 0.20 M

ammonium sulfate solution?

Example: How many mL of 0.20 M lithium carbonate solution are necessary to

supply 1.0 mole of lithium ions?


46

Example: How would you prepare 650.0 mL of a 0.1000 M ammonium sulfate

solution?

Sometimes it is necessary to change the concentration of a solution. The

concentration may be changed by changing the amount of solute or solvent.

M1V1 = M2V2

BE CAREFUL OF WHEN TO APPLY THIS.

Example: How much 16.0 M NaOH solution is necessary to prepare 1.000 L of

1.000 M NaOH?

Example: To 350. mL of a 0.250 M sodium chloride solution is added 450. mL of

water. What is the final concentration?

4.6 Solution Stoichiometry and Chemical Analysis

Quantitative Analysis:

Two types discussed here.

1.

2.


47

Example: How many grams of KCl must be added to precipitate all the mercury(I)

ions in 0.10L of a 0.020 M mercury(I) nitrate solution? The mercury will precipitate

as mercury(I) chloride.

Example: The sulfur in a 0.3054 g sample containing the mineral chalcopyrite

(CuFeS2) was converted to sulfate. The precipitation of this sulfate with barium gave

0.6525 g of barium sulfate. What was the percentage of CuFeS2 in the sample?

Titration:

Titrant:

Equivalence Point

End Point:

Indicator:

Any type of reaction may be used.

A neutralization reaction

Acid + Base Salt + (Water)


48

The net ionic equation:

H+(aq) + OH–(aq) H2O(l)

Example: Using 100.0 mL of 0.100 M acid plus x mL of 0.100 M RbOH, what will

be the concentration of acid after each of the following?

(a) The acid is HNO3 and 100.0 mL of RbOH is used.

(b) The acid is HNO3 and 75.0 mL of RbOH is used.

(c) The acid is H2SO4 and 100.0 mL of RbOH is used.

(a)

(b)


49

(c)

Example: A Na2CO3 solution was reacted with an HCl solution. The sodium

carbonate solution contained 0.5015 g of Na2CO3 in 100.0 mL of water. A total of

48.47mL of acid was necessary for the titration. What was the original concentration

of the acid? The reaction was: Na2CO3 + 2 HCl 2 NaCl + H2O + CO2

Example: If 35.00 mL of a 0.1500 M KOH solution is required to titrate 40.00 mL

of a phosphoric acid solution, what is the concentration of the acid? The reaction

is: 2 KOH + H3PO4 K2HPO4 + 2 H2O


50





END EXAM 2 MATERIAL

Note: After the exam, you should move immediately to Oxidation Numbers and

Nomenclature III on the website.

Reminder: It is exceedingly important that, while it is fresh in your mind, to make sure that

you have the correct solution to every problem on the exam (and re-test). Not doing so will

have a strong negative influence on your grade on the final exam.


51

Reminder: Start the Oxidation Number and Nomenclature III sections on the website.

Reminder: If your grade on the test is not what you want, you should immediately refer to

the syllabus and/or see the instructor.

CHAPTER 5

Note: Many students consider these end-of-chapter problems to be very difficult to interpret.

However, most of them are easy to solve. Factor label will be exceedingly

important. This chapter REQUIRES the extensive utilization of factor label, if you

have postponed learning factor label you need to catch up.

Note: Other than, the equations given in class do not waste your time on any other

equations given in the textbook. It is more important to learn the necessary material

than to waste time on unnecessary equations. Of course, if you have plenty of extra

time to memorize extraneous material, you can use these equations (though probably

not on an exam).


Thermochemistry:

Thermodynamics:

5.1-5.2 The Nature of Energy/The First Law of Thermodynamics (This, and all other section numbers, refers to the appropriate sections in BLBMWS.)

Energy:

Potential Energy:

Kinetic Energy:

Energy – Joule = J = – calories = cal

1 cal = 4.184 J (exactly)

Heat:

Law of Conservation of Energy (First Law of Thermodynamics):


52

System:

Surroundings:

State:

State Functions – Example: T, P

Energy in the system: Internal Energy = E

Changes in E can be measured = E (final – initial)

E =

Work done on the system + Work done by the system –

There are many types of work

5.3 Enthalpy

Enthalpy:

Exothermic:

Endothermic:


53

5.4 Enthalpies of Reaction

Chemical equations may be written with energy changes included. These equations

specifically deal with moles (never molecules), and are called Thermochemical

Equations.

Thermochemical Equation:

NOTE: They are the only equations that allow fractions.

5.5 Calorimetry

Calorimetry:

Need: Energy

Temperature change (T)

Conversion factor:

Specific Heat:

The units tell exactly

how to solve the problem. (J/g°C or cal/g°C) (Mass is needed)

Heat Capacity:

(J/°C or cal/°C)

Molar Heat Capacity:

(J/mole°C or cal/mole°C)

Heat of Reaction:

For problems factor label to get the proper units


54

Example: The ignition of a 1.5886 g sample of glucose (C6H12O6) in a bomb

calorimeter resulted in a temperature increase of 3.682°C. The heat capacity of the

calorimeter was 3.562kJ/°C, and the calorimeter contained 1.000 kg of water. Find

the molar heat of reaction for:

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

NOTE: Watch the signs; they depend on definitions and not on mathematics.

5.6 Hess’s Law

Thermochemical equations may be combined to produce new thermochemical

equations.

Hess's Law –

Example: Given the following information:

C(s) + O2(g) CO2(g) –393.5 kJ

H2(g) + (1/2) O2(g) H2O(l) –285.8 kJ

C2H2(g) + (5/2) O2(g) 2 CO2(g)+ H2O(l) –1299.8 kJ

Find the enthalpy change for: 2 C(s) + H2(g) C2H2(g)


55

5.7 Enthalpies of Formation

Reactions may be carried out in many different ways.

Standard Heat of Reaction:

H° (Note ° sign)

The heat of reaction is measured under Standard Conditions.

NOT STP Standard Conditions:

Standard State:

Standard Heat (Enthalpy) of Formation:

H°f ° = standard conditions

f = formation (only one-way to write the equation.)

NOTE: For an element, it is 0.000

Hrxn = H°f(products) – H°f(reactants)

Example: Calculate Hrxn for:

6 H2O(g) + 4 NO(g) 5 O2(g) + 4 NH3(g)

NOTE: A common mistake is not to subtract all the reactants from all the products.

5.8 Read





56

CHAPTER 6

6.1 The Wave Nature of Light (This, and all other section numbers, refers to the appropriate sections in BLBMWS.)

Electromagnetic Radiation (Electromagnetic Energy):

It is possible to treat light as waves.

= c

= Wavelength:

= Frequency:

c = the Speed of Light

6.2-6.3 Quantized Energy and Photons/Line Spectra and the Bohr Atom

Photon:

The energy of a photon may be calculated:

E = h h =

Bohr Theory of the atom:

Energy terms – not a physical three-dimensional arrangement.


57

Ground State:

Excited State:

6.4 The Wave Behavior of Matter

De Broglie first predicted the wavelike properties of matter:

de Broglie:

NOTE: Do not mistake v for .

Electron behavior wave behavior Schrödinger Equation

Heisenberg Uncertainty Principle:

Only the probability of locating an electron is possible.

Only a region in space: Probability Density

6.5 Quantum Mechanics and Atomic Orbitals

Quantum Mechanics Results from the Schrödinger Equation – Quantum Numbers

Quantum Numbers:

1. Principle Quantum Number = n =

Shell:


58

2. Angular Momentum Quantum Number = l =

Subshell:

Orbital:

3. Magnetic Quantum Number = ml =

4. Electron Spin Quantum Number = ms =

6.6 Representations of Orbitals

What does an atomic orbital look like? Only a probability

n size l shape ml orientation

s Orbitals: (n, l = 0, ml = 0 )

p Orbitals: (n, l = 1, ml = –1, 0, 1)

d Orbitals: (n, l = 2, ml = –2, –1, 0, 1, 2)

f Orbitals: (n, l = 3, ml = –3, –2, –1, 0, 1,2,3)





59

6.7 Many Electron Atoms

Pauli Exclusion Principle:

Effective Nuclear Charge:

6.8 Electron Configurations

This may be expressed as the Electron Configuration or as an Orbital Diagram.

Electron Configuration:

Orbital Diagram:

NOTE: Many students get into trouble by giving the wrong one of these on an exam (both is

not a substitute).

Aufbau Principle:

Each orbital in a set of orbitals may hold 2 electrons (ms = +1/2 and –1/2)

First n = 1 so: l = 0 ml = 0 ms = +1/2 or –1/2

H

He

n l ml ms

H, He 1 0 0 +1/2

He 1 0 0 –1/2

The Pauli Principle leaves no other choices, so the next electron must

go into the next available orbital.

3 electrons (Li)

May be drawn vertically:

4 electrons (Be)


60

5 electrons (B)

6 electrons (C)

Hund's Rule:

7 electrons (N)

8 electrons (O)

9 electrons (F)

10 electrons (Ne)

11 electrons (Na)

Note: H, Li, B, C, N, O, F, and Na are Paramagnetic

Paramagnetic:

He, Be, and Ne are Diamagnetic.

Diamagnetic:

NOTE: Do not use abbreviated electron configurations when the complete one is

needed.


61

Ordering: ___ ___ ___ ___ ___

4d

___ ___ ___

4p

___ ___ ___ ___ ___ ___

4s 3d

___ ___ ___

3p

___

3s

___ ___ ___

2p

___

2s

___

1s

Predictions are complicated, but:

1s

2s 2p

3s 3p 3d

4s 4p 4d 4f People get in trouble with this because

5s 5p 5d 5f they either draw it sloppy or they add

6s 6p 6d 6f impossible orbitals (i.e., 2d).

7s 7p

Also: s 2 electrons d 10 electrons

p 6 electrons f 14 electrons

Example: Fe 26e–

Example: Ra 88e–

KNOW: There are some exceptions:

Cr

Cu


62

Example: List the quantum numbers for each of the electrons in the outer shell of

fluorine.

6.9 Electron Configurations and the Periodic Table

NOTE: H

Li

Na

Valence Shell:

Valence Electrons:

Core Electrons:

Main Group Elements (Representative Elements):

Transition Metals:

Inner Transition Metals:

Lanthanide Series:

Actinide Series:


63

Relating to the periodic table:

The Periodic Table gives a lot of important information. Such as:

1. Number of electrons in the outer shell (valence electrons) = Group Number

2. Electron configuration (Position)

3. Group Names:

4. Metals give cations Nonmetals give anions

Electron configurations of ions

Anions: simply continue to add electrons (aufbau)

Cations: remove electrons from the outer shell (maximum n, not aufbau)

Isoelectronic:


64

CHAPTER 7

7.1-7.2 (This, and all other section numbers, refers to the appropriate sections in BLBMWS.)

Read

7.3 Sizes of Atoms and Ions

Atomic Radius

The downward increase is because there are more shells.

The leftward increase is because there are fewer electrons in the outer shell, so

there is less shielding.

There are some minor variations, especially in the transition elements.

The conversion of an atom to an ion not only changes the number of electrons but

also the size.

Atom atomic radius Ion ionic radius

In an atom, there is a balance between the nuclear attraction (shielded) and the

electron repulsion.

Cations are ______________ than their parent atoms.

Anions are _______________ than their parent atoms.

The greater the number of electrons lost or gained the greater the change in radii.


65

7.4 Ionization Energy

Ionization Energy:

A(g) + energy A+(g) + e– (First) ionization energy

A+(g) + energy A2+(g) + e– Second ionization energy

There are periodic trends

In general, the value is higher for elements to the right and to the top of the

periodic table.

Larger radii (further from the nucleus)

Low ionization energies metals (cations)

Specifics:

Filled shells (noble gases) high

Filled subshell (IIA) high

Half-filled subshell (VA) high

Removal of more than one electron:

7.5 Electron Affinities

Electron Affinity:

A(g) + e– A–(g) + energy The energy may be endothermic or exothermic

A–(g) + e– A2–(g) + energy Second electron affinity (always endothermic)

It is difficult to add more electrons than are required to fill the valence shell.

Best for small atoms – electrons enter nearer the nucleus, also for nearly filled shells.

The Noble gases have no affinity.

There are some variations in the trend (c.f. ionization energy)


66

7.6-7.8 Chemical Properties

Read


Write balanced chemical equations for each of the following.

a. Sodium is added to water

b. Strontium is added to water

c. Magnesium is heated in nitrogen to produce magnesium nitride

d. Zinc burns in oxygen

e. Lithium reacts with chlorine

f. Hydrogen reacts with rubidium

g. Barium oxide is added to water

h. Iron(II) oxide reacts with hydrochloric acid

i. Selenium trioxide reacts with water

j. Carbon dioxide reacts with potassium hydroxide

k. Tetraphosphorus decaoxide reacts with water

l. Calcium hydride reacts with water

Answers:

a. 2 Na + 2 H2O 2 NaOH + H2

b. Sr + 2 H2O Sr(OH)2 + H2

c. 3 Mg + N2 Mg3N2

d. 2 Zn + O2 2 ZnO

e. 2 Li + Cl2 2 LiCl

f. H2 + 2 Rb 2 RbH

g. BaO + H2O Ba(OH)2

h. FeO + 2 HCl FeCl2 + H2O

i. SeO3 + H2O H2SeO4

j. CO2 + 2 KOH K2CO3 + H2O

k. P4O10 + 6 H2O 4 H3PO4

l. CaH2 + 2 H2O Ca(OH)2 + 2 H2





END EXAM 3 MATERIAL


67



have a strong negative influence on your grade on the final exam.



CHAPTER 8 Bonding – increased stability exhibited by the Bond Energy

CHANGE ORDER (Cover 8.4 first, and then cover in order)

(This, and all other section numbers, refers to the appropriate sections in BLBMWS.)

8.4 Bond Polarity and Electronegativities

Ionization energy and electron affinity both deal with electron attraction – sometimes

one or the other is important. More often, a combination is necessary:

Electronegativity

Electronegativity:

The highest electronegativity is _____.

In general, the closer an element is to _____ the higher is its electronegativity.

For two elements the same distance away, the one __________ on the periodic table

has the higher value.

Example: H — H Na — Cl H — Cl

The degree of polarity depends on just how different the electronegativities are:

Identical atoms

Different nonmetals

Metal + nonmetal


68

8.1 Lewis Symbols

Lewis Symbol:

Use only the valence electrons

They only work well for representative elements

The number of valence electrons equals the group number (except helium)

The maximum for an element is 8

Example: Se

Most ions have the same number of electrons in their outer shells as a noble gas. For

all noble gases except He this is 8.

The formation of ionic bonds involves a transfer of electrons.

8.2 Ionic Bonding

Example: 2 Na + F2 2 NaF

Example: Li + S what compound will form?


69

Lattice Energy:

It is always endothermic.

The Born-Haber Cycle is a method for determining lattice energies. This is an

application of Hess's Law

8.3 Covalent Bonding

The atomic orbitals may overlap and allow electrons to be shared:

The electrons can more through the overlap region and are shared by both atoms.

The atoms can only get so close bond length

Example: Cl2

(Single) Covalent Bond:

Do not use lines for bonds yet

Example: O2

Double (Covalent) Bond:

Bonding Pair:

Nonbonding Pairs (Lone Pairs):

Example: CN–

Triple Bond:

The maximum number of pairs between two atoms is 3.


70

Octet Rule:

8.5 Drawing Lewis Structures

Based on the Octet Rule

Example: HBr

Note: If the Lewis structure that you draw does not match one given in class, do not

simply correct your result. You should write the correct answer next to your

result, then when you review you can analyze your mistake and make sure that is

does not happen again. Finding out why you made a mistake is more helpful than

simply correcting you mistakes.

Example: H2S

Example: C + F

Example: NF3

Example: HCN

Example: NH4+


71

Example: NO

Example: HNO2

Note: Watching Lewis structures being drawn often leads to overconfidence on the part of

the student. It takes a lot of practice before they become as easy as they appear in class.

Start practicing these as soon, and as often as possible.

Formal charge may be used to check to see if a structure is reasonable.

Formal Charge =

The formal charges always sum to the charge on the molecule or ion.

Elements with a high electronegativity usually do not have a positive formal

charge (and vice versa for low electronegativity).

Formal charges are usually +1, 0 or –1

Adjacent atoms are not normally both + or both –.

8.6 Resonance Structures

Example: SO3

Resonance:

Resonance Hybrid:

Resonance Structures:

Example: NO2–


72

8.7 Exceptions to the Octet Rule

There are exceptions to the octet rule.

Less than an octet:

Near He (He structure = 2 electrons)

Many elements with less than 4 valence electrons

Sometimes it is possible to complete the octet through reaction with

other molecules.

Coordinate Covalent Bond:

Odd electron molecules

If there is not an even number of electrons, there must be an odd

electron.

Example: NO2


73

More than an octet

This may occur for elements in the third period or below.

It is the most obvious for elements with 5 or more bonds.

Example: XeF4 (Xe already has an octet, but the fluorines are attached

so it will get more electrons)

The most electronegative element will get its octet.

NOTE: Go to Lewis Structures (see the website).

8.8 Strengths of Covalent Bonds

Bond Energy =

For a reaction: H = bonds broken – bonds formed

The types of bonds are important (single, double or triple)

It takes more energy to break a bond if resonance is present.





74

CHAPTER 9

9.3 Molecular Shape and Molecular Polarity (This, and all other section numbers, refers to the appropriate sections in BLBMWS.)

Polar Molecule:

Dipole:

Dipole-Dipole Force:

Dipole Moment:

Ion-Dipole Force:

Instantaneous Dipole:

The instantaneous dipole creates an induced dipole.

This is related to the number of electrons and how tightly they

are held (polarizability).

London Dispersion Force:

van der Waals Force:

If only one covalent bond is present, a molecule is polar if the bond is polar.

If more than one covalent bond is present, a molecule may or may not be

polar. Both the polar covalency and the shape are important considerations.

Example: HCl and H2O


75

Kinetic Molecular Theory applied to solids and liquids

Same as for gases, except there are different numbers

There is a competition between the kinetic energy and the intermolecular forces.

K.E. >> Intermolecular forces

K.E. Intermolecular forces

K.E. << Intermolecular forces

Changes of state

Freezing (Solidification):

Fusion:

Vaporization (Evaporation):

Condensation:

Sublimation:

Deposition:

9.1-9.2 and 9.4-9.5 VSEPR – VB Structures of molecules

Two basic methods (for representative elements)

Both give the geometry. The geometry is fundamental to many physical and

chemical properties.

A. VSEPR: All electrons in the valence shell of the central atom are

significant as are all electrons donated to the central atom. These are

determined from a correct Lewis structure.

Electrons normally occur in pairs. Bonding pairs (bp) and Lone pairs (lp)

Repulsion ranking: lp-lp >> lp-bp > bp-bp

The Lewis structure gives the Orbital Geometry (Electron-Pair Geometry).


76

Orbital Geometry (Electron-Pair Geometry):

The orbital geometry plus the atoms gives the Molecular Geometry.

Molecular Geometry:

B. VB Theory: The electron orbitals may rearrange to form a more

stable structure. Orbitals overlap and electrons are shared.

In order to make more orbitals available (for more bonds), hybridization

(mixing) may be required.

Set up a table (on a separate page)

e–pairs | lone pairs | orbital | molecular | hybridization | polar |

| | geometry | geometry | | |

Polar

A diatomic molecule is polar if the two atoms differ in electronegativity, and

is nonpolar if they have the same electronegativity.

For polyatomic molecules, the polarity depends on the electronegativity

differences and the geometry.

Example:

BeI2

Compounds with this type of Lewis structure are linear. This is commonly

observed for a small group of covalent compounds containing the metals Be,

Zn, Cd, and Hg. It is also observed for some compounds containing multiple

bonds.

VSEPR

VB

Hybridization:


77

Remember: If the Lewis structure that you draw does not match one given in class, do not

simply correct your result. You should write the correct answer next to your result, then

when you review you can analyze your mistake and make sure that is does not happen again.

Finding out why you made a mistake is more helpful than simply correcting you mistakes.

Example: BCl3 and NO2–

VSEPR

VB

The double bond is special – an unhybridized p orbital is required for it. (2 are

required for a triple bond.)

p2 (the other p is not hybridized)

This would give an angle of 90°, which is too small. The observed

angle is near 120°, so sp2.

Example: CCl4 NCl3 OF2

VSEPR

Polarity:

However, if one atom is changed:

CHCl3


78

VB

Note: for a larger central atom (i.e., S) and a smaller substituent (i.e., H) the

opening is not needed.

H2O 105° H2S 90°

Example: SbF5 ClF4+ ClF3 KrF2

VSEPR – Orbital Geometry – trigonal bipyramid

VB

Example: ClF6+ IF5 BrF4

–

VSEPR

VB

NOTE: Go to VSEPR on the website.


79

9.6 Multiple Bonds

Direct overlap:

H2

HF

F2

All are bonds

bond:

Formation of a double bond

CO2

There is no more room between the C and the O's for additional electron pairs.

Bond:

Delocalized Orbitals

NO2–

Lewis


80

9.7 Molecular Orbitals

Molecular Orbital Theory:

Molecular Orbital Energy Level Diagrams –

Example: H2

Bonding Molecular Orbital:

Antibonding Molecular Orbital:

For p orbitals both and * bonds are possible.

The electrons are indicated as in an orbital diagram ( || ) the orbitals are filled

beginning with the lowest. Hund's rule and the Pauli Exclusion Principle still hold.

H2 __

*

__ __

1s 1s

__

Bond Order =

In general, all s combinations give the same pattern (incomplete as shown):

__

*

__ __

ns ns

__


81

p orbitals are in a set, so they must be dealt with together (incomplete as shown).

__

*

__ __

* *

__ __ __ __ __ __

np np

__

__ __

The patterns for O, F, and Ne are slightly different (incomplete as shown):

__

*

__ __

* *

__ __ __ __ __ __

np np

__ __

__

He2

ions:

H2+

H2–


82

9.8 Period 2 Diatomic Molecules

Li2

Only use those parts of the valence shell that are occupied, but make sure all are used.

Be2 bond order =

B2 =

C2 =

N2 =

O2 =

F2 =

Ne2 =


83

For ions fill the molecular orbitals as expected, then add or subtract the appropriate

number of electrons.

Heteronuclear diatomics

To a certain extent, they are like homonuclear cases.

Example: CN–




END OF EXAM 4 MATERIAL


84



have a strong negative influence on your grade on the final exam. The Final is getting

close.



Note: Many people get into trouble on this last material because the final not only contains

this new material, but it also contains a comprehensive portion. The cause of the difficulty is

that the students budget their normal amount of study time here, however this is only

sufficient time to prepare the new material OR review for the comprehensive portion. In the

past a large percentage of the persons taking the test have done well on one portion of the test

and exceedingly poorly on the other. From now until the end of the term your study sessions

should contain at least some review of the old exams.

CHAPTER 11

11.1 (This, and all other section numbers, refers to the appropriate sections in BLBMWS.)

Read

11.2 Intermolecular Forces

London Dispersion Forces attract nonpolar molecules to each other.

Types of intermolecular forces:

1. "Normal" bonding – Ionic, Covalent(C, BN, metalloids and SiO2),

and Metallic

2. Ion-Dipole Forces – an intermediate case normally involving two

substances. Also stronger than a simple dipole-dipole

3a. Dipole-Dipole Forces – between polar molecules, not as strong as a

regular bond (only partial charges are involved)

3b. Hydrogen Bond: A special case of dipole-dipole involving H directly

bonded to N, O, or F.

4. Ion-Induced Dipole


85

5. London Dispersion Forces – always present

Instantaneous Dipole (Temporary dipole)

The instantaneous dipole creates an induced dipole.

This depends on the number of electrons and how tightly they

are held (polarizability).

These non-bonding intermolecular forces are labeled van der Waal's Forces.

The distance between nonbonded atoms (minimum) is equal to the sum of the

van der Waal's radii.

11.3 Select Properties of Liquids

Surface Tension:

Capillary Action:

Viscosity:

Amorphous Solid (Glass):


86

11.4 Phase Changes

The main difference between gases vs. liquids and solids is due to Intermolecular Forces.

Evaporation (Vaporization, Condensation) Evaporation, Vaporization

Liquid <==================> Gas

Condensation

Melting Point (Freezing, Melting, Solidification) Freezing

Liquid <=======> Solid

Melting

Solid-Gas (Sublimation, Deposition) Sublimation

Solid <===========> Gas

Deposition

At the boiling point, a lot more energy is necessary to overcome the intermolecular

forces and to convert the liquid to gas Molar Heat of Vaporization (energy

required to vaporize a mole of a substance).

Hvap CH4 9.2 kJ/mole

Et2O 26.0 it is related to the

H2O 40.8 intermolecular forces

The reverse is the Molar Heat of Condensation (energy released when a mole

of a substance condenses) – the same numerical value, but the opposite sign.

For other transitions

Molar Heat of Fusion (Crystallization, Solidification)

Molar Heat of Sublimation (Deposition)

Heat of ....: Energy change when a substance undergoes the indicated change.


87

Heating curve:

Cooling follows the same curve in the opposite direction..

Superheating:

Supercooling:

11.5 Vapor Pressure

Vapor Pressure

Evaporation goes for a while, and then it seems to stop.

Vapor Pressure:

Dynamic Equilibrium:

Le Châtelier's Principle:

Volatile:

The vapor pressure is due to molecules escaping from the surface – below the

surface, they are trapped.

When Pvap = Pext bubbles can form below the surface and not be

crushed.

Boiling Point:

Normal Boiling Point:

Variations in the vapor pressure with temperature may be calculated:


88

ln (P1

P2) =

∆Hvap

R[

1

T2−

1

T1]

The T's are in Kelvins

NOTE: R = 8.314 J/mole K = 1.987 cal/mole K

Example: The normal boiling point of ethyl alcohol is 78.4°C. The heat of vaporization of this

compound is 40.5 kJ / mole. Calculate the vapor pressure of ethyl alcohol at a

temperature of 55.0°C.

11.6 Phase Diagrams

Phase Diagram: A summary of the phase relationships.

P

T

All phase diagrams look basically the same (for one-component)

Triple Point:

Critical Point:

11.7 Liquid Crystals

Read


89



have a strong negative influence on your grade on the final exam. The Final is getting

close.



Note: Many people get into trouble on this last material because the final not only contains

this new material, but it also contains a comprehensive portion. The cause of the difficulty is

that the students budget their normal amount of study time here, however this is only

sufficient time to prepare the new material OR review for the comprehensive portion. In the

past a large percentage of the persons taking the test have done well on one portion of the test

and exceedingly poorly on the other. From now until the end of the term, your study sessions

should contain at least some review of the old exams.

CHAPTER 12

12.1 Classification of Solids (This, and all other section numbers, refers to the appropriate sections in BLBMWS.)

These attractions are not as strong as “normal” bonding: ionic, covalent, and metallic.

“Normal” bonding occurs in Ionic Solids, Network (Covalent) Solids, and

Metallic Solids. (“Normal” bond is a simplification – it is never an

acceptable answer to any problem on an exam.)

Ionic Solid:

Network (Covalent) Solid:

Metallic Solid:

All solids not held together by “normal” bonds are Molecular Solids,

Molecular Solid:

Any attraction, other than a “normal” bond, is a van der Waal’s Force.

All attractions of any kind are Intermolecular Forces.


90

12.2 Structures of Solids

Crystalline Solid (true solid):

Amorphous Solids supercooled liquids (undercooled liquids)

Crystalline solids ordered array = Lattice

Lattice:

• • • • • • •

• • • • • • •

• • • • • • •

• • • • • • •

• • • • • • •

Unit Cell:

There are only 7 different types of lattices with only 14 types of unit cells

(Bravais Lattices) needed to describe every solid.

A unit cell is like a "brick." How do you describe a brick?

Six things are necessary:

Length of sides = a, b, and c

Angles between = , , and

Cubic

Tetragonal

Orthorhombic

Monoclinic

Triclinic

Hexagonal

Rhombohedral


91

Three of the Bravais lattices are cubic:

Simple (Primitive) Body-Centered Face-centered

Closest Packed Structure:

12.3/12.5/12.6/12.7 Types of Solids

Types of crystals

Ionic Solids

Contain ions ionic bonds

Covalent Solids (Network Solids)

Carbon Allotropes – graphite diamond


92

Silicon dioxide

Metallic Solids Metallic bonds

Bonding – Band Theory a modification of molecular orbital theory

Molecular Solids/Atomic Solids

Contain individual molecules and in some cases only atoms

Amorphous "Solids" Glass is a common example, so are some plastics.

Summary:

Property Ionic Molecular Covalent Metallic

m. p.

Solubility

in H2O(polar)

in nonpolar

Conducts electricity

Hard and Brittle

Bonding


93

12.4 (12.7) Metallic Bonding

Metal:

Band Theory:

Bonding – Band Theory a modification of molecular orbital theory

Li [He]2s1

Some network solids may be semiconductors. Band theory still applies. With

covalent bonds, there is a definite separation of bonding and antibonding

molecular orbitals.

Conduction Band

Band Gap:

Valence Band:

Most semiconductors are doped to improve their conductivity.

n-type –

p-type –


94




12.8-12.9

Read


95

CHAPTER 13

Reminder: Many people get in trouble on this last material because the final not only

contains new material, but it also contains a comprehensive portion. The cause of the

difficulty is that the students budget their normal amount of study time here, however this is

only sufficient time to prepare the new material OR review for the comprehensive portion.

In the past a large percentage of the persons taking the test have done well on one part of the

test and exceedingly poorly on the other. From now until the end of the term, your study

sessions should contain at least some review of the old exams. (This is why you went over

the exams and re-tests as soon as they were returned, and while they were fresh in your mind.

If you did not do this, you have to catch up now.)

Solutions are homogeneous mixtures.

The solvent may be a solid, liquid or a gas. The solute may be a solid, liquid or a gas.

13.1 The Solution Process (This, and all other section numbers, refers to the appropriate sections in BLBMWS.)

Examples of various solutions

Example: Water plus a polar solute

Solvent-solvent Solute-solute Solvent-solute

Example: Water plus nonpolar solute



96

Example: Water plus a gas


Example: Water plus an ionic compound


Example: Nonpolar plus nonpolar


Polar substances will dissolve in each other.

Nonpolar substances will dissolve in each other.


97

Two considerations

Heat of Solution:

Disorder (entropy) –

13.2-13.3 Saturated Solutions and Solubility

Solutions will only hold so much:

Saturated – Unsaturated – Supersaturated

Solute-Solvent Interactions

"Like dissolves like"

Temperature effects

Pressure effects

Henry's Law:

13.4 Concentrations

Molarity =

Mole fraction = X =

Weight Percent (Mass Percent) =

Volume Percent =

Molality = m =

Reminder: Many students get into trouble because they incorrectly use “m” and the

abbreviation for moles. The correct abbreviation is “mol.” The abbreviation “m”

means molality. The use of an incorrect abbreviation has resulted in students

making very significant factor label mistakes, and losing several points on an exam.


98

Example: At room temperature, 13.20 g of ammonia will dissolve in 100.0 mL of

ethyl alcohol (C2H5OH). Calculate the mass percent, m, and X of ammonia. (The

density of ethyl alcohol is 0.7893 g/cm3.)

Example: A solution contains 85.4 g of sodium sulfate dissolved in 200.0 g of water

at 100°C. What is the total molality of the ions in this solution?

13.5 Colligative Properties

Colligative Properties:

Vapor pressure of solutions Ptotal < P°pure

Two cases: 1. Volatile solvent with a nonvolatile solute

P = XP°solvent Raoult's Law:

2. Both solvent (A) and solute (B) are volatile.

Ptotal = Psolvent + Psolute

Ptotal = XAP°A + XBP°B


99

Still Raoult's Law

Ideal Solution:

PB°

PA°

0 XB 1

Boiling point elevation and Freezing point depression

Boiling Point Elevation:

Freezing Point Depression:

T = Kbm or T = Kfm

The T's refer to how much the boiling point in increased or to how much the

freezing point is lowered.

van't Hoff Factor (i):

Given the following setup:


100

Osmosis:

Osmotic Pressure = =

For electrolytes, the van't Hoff factor is needed:

=

Reverse Osmosis –

Example: A 225.0 gram sample of benzene, C6H6, has a 10.0-gram sample of lauryl

alcohol, extracted from coconut oil, dissolved in it. A 1.22°C depression of the

freezing point of the benzene occurred. Determine the molar mass of lauryl alcohol.

The value of Kf for benzene is 5.08°C/m.

13.6 Colloids

There are also systems intermediate between homogeneous mixtures and

heterogeneous mixtures

Colloid (Colloidal Dispersion):

Tyndall Effect:





END OF 133 MATERIALS


101

PROBLEM SOLVING This is a problem-solving course. You must be able to solve many types of problems.

Nearly every student, who has trouble with problem solving, has trouble because that student

makes the class more difficult than it is. Even students who do not have trouble with

problem solving make the class more difficult. A common example of making the class

more difficult concerns conversions. You are required to learn one English-SI length

conversion. Many students go ahead and learn two or three conversions instead. The

additional conversions may make one or two problems, out of all the problems done during

the term, easier. Therefore, for the sake of maybe one or two problems these students have

doubled or tripled the work required. In addition, the more conversions you memorize the

greater the chance of becoming confused. A very common mistake on exams is to find

mixed conversions. A mixed conversion is occurs when a student combines two or more

conversions. Now, simply learning an extra length conversion is not going to take that much

time, however, students do not stop there; they learn extra items every time. Not all these

extra items make you a better student; they just clog up your mind. Another example occurs

in the Gas Law chapter. Every problem assigned, presented in class, or given on an exam

requires only four equations. However, students insist on learning the 15-20 equations given

in most textbooks; again, there are serious problems with wasting time on the 11-16 extra

equations. The following discussion emphasizes many points that will help simplify this

class.

One of two methods (or a combination of the two) will solve the problems in this course.

The first method is the “plug-in” method, and the other is Factor Label. The plug-in method

concerns the rearrangement of one of a small number of equations given you, and then

numbers given to you are “plugged-in.” The Factor Label method involves tracking the units

so that all unwanted units cancel to leave only the desired unit(s). Finally, some problems

may involve a plug-in followed by one or more Factor Label steps. Many students make this

last case difficult by treating this type of problem as two or more different problems.

The plug-in method requires you to learn one or more of the following equations:

Exam 1 (Chapters 1-3) NONE

Exam 2 (Chapters 4-10): PV = nRT RateA

RateB = √

MWB

MWA

M1V1 = M2V2 P1V1

n1T1 =

P2V2

n2T2 Ptotal = Pa + Pb + Pc + ....

Exam 3 (Chapters 5-7): E = q + w = h

mv = c E = h =

hc

E = 1/2 mv2

Exam 4 (Chapters 8-9): NONE

Exam 5 (Chapters 11-13): ln (P1

P2) =

∆Hvap

R[

1

T2−

1

T1]

P = P°solvent V = nRT = iMRT T = i Kbm or T = i Kfm


102

As may be seen for the above list, this entire course consists of 16 equations. Some of these

equations are simple variations of others on the list. There is one additional equation, which,

if you need it, will be given to you in the problem. You only need these equations. The most

equations that you need to learn for any exam is five. Do not be foolish and try to learn

more. You would be surprised to see how many equations, besides those listed above;

people try to learn for Exam 1.

Every problem not requiring one of the above equations requires the Factor Label method.

This method may appear in addition to one or more of the above equations. Factor Label

involves SI prefixes, specifically designated conversions and constants, and definitions. If

you want to avoid making this course more difficult, stay very strictly within each of these

categories.

The SI prefixes and their numerical values are in a table in Chapter 1. Learn these exactly as

they appear in this table. Thus if you learn that c = 0.01 then by a simple substitution: 1 cm =

0.01 m. Many people learn this relationship and then waste their time learning that 100 cm =

1 m. You must learn the table for the Exams; why learn extra superfluous conversions? In

addition, during a normal term 60-70% of the students in the class will make at least one

mistake on the exam using the second type of conversion.

You will need to learn a minimum number of other conversions and constants. In some

cases, it will be your option, such as with the English-SI length and mass conversions, while

in other cases there are very specific conversions. The constants have different numerical

values depending on the units specified; learn only the ones given in class do not waste your

time on others.

Finally, definitions are very important to the Factor Label method. There are two ways of

setting up a definition for Factor Label. The first type uses very specific units, for example, J

= kg m2 / s2 or Pa = kg / m s2. The second type uses classes of measurements, for example,

speed = distance / time or density = mass / volume. The second type means that speed, for

example, may be expressed several ways; some of these would be miles / hour, kilometers /

hour, inches / second, and meters / minute. All that is necessary in this type of definition is

any measurement unit that fits the classification. Many people first see this type of definition

as if it were a mathematical relation; however, the sooner these people leave this limitation

behind the better they will be. The elementary level treatment of, for example, the density

relation as a mathematical relation retards many students. The sooner you go beyond this the

easier other problems will be. The advantage of using density as an equation makes a few

introductory problems easier, and impedes your progress towards excellence in the Factor

Label method. Do not let anything slow your work on understanding the Factor Label

method – you do not want to be 4-5 weeks behind when you get to Chapter 6, because you

will not get pass this chapter without Factor Label. Force yourself to use the Factor Label

method even when you want to fall back on elementary methods.

Documents

CHEMISTRY 133 LECTURE / STUDY GUIDE FOR R.H. …faculty.sfasu.edu/langleyricha/Chem133/guide133.pdfwithout written permission ... with approximately equal numbers of students using