Download chem textbook 2015 - Manitowoc Public School District

General Chemistry
Reference Manual
My goal for the year is to prepare you for a successful collegiate level
chemistry program, while validating chemical theory through the application
of practical chemistry techniques.
Second Edition—2015-2016
A. Evenson
Evenson
Table of Contents
…………………………………………….3
Course policies and expectations
Exam 1
Measurement and Graphical (unit 1)
Analysis
Periodic Table (unit 2)
……………………………………………19
Reactions (unit 3)
……………………………………………73
Bonding (unit 4)
……………………………………………96
Limiting Reactants (unit 5)
……………………………………………108
Kinetic Theory (unit 6)
……………………………………………124
Phase Changes (unit 7)
…………………………………………….133
Gas laws (unit 8)
……………………………………………141
Equilibrium (unit 9)
……………………………………………160
Solutions (unit 10)
……………………………………………171
Acid and Base (unit 11)
……………………………………………188
Thermodynamics (unit 12)
……………………………………………210
RedOx reactions (unit 13)
……………………………………………229
Organic (unit 14)
……………………………………………253
Nuclear (unit 15)
……………………………………………261
……………………………………………52
Exam 2
Exam 3
Exam 4
Exam 5
Exam 6
Exam 7
Exam 8
Appendix
Resource reactions
Lab work Samples
……………………………………………269
……………………………………………270
……………………………………………274
Glossary
……………………………………………286
2015-2016 edition
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CHEMISTRY
Mr. Evenson
[email protected]
Introduction and Objectives
Chemistry may be defined as the study of the elements and the compounds they form. This course will
give you the necessary foundation upon which to build higher level theoretical concepts and practice in
higher level thinking allowing you to make informed decisions on your surroundings. Chemistry builds
upon itself; therefore it is important that you fully understand a concept before you move to the next idea.
This class does run with the expectation that the students enrolled have strong math skills and have had
a successful experience in Algebra. Bearing this in mind do not hesitate to come in outside of regular
class hours to get clarification. Also feel free to send relative e-mail to the address given above.
Course Philosophy
I consider this class to be a superior chemistry course as it is taught with the following objectives in mind.
1. That students develop critical thinking and problem solving skills, not only to use in
chemistry but, by extension, to use in everyday experiences.
2. That the students learn the facts, formulas, and principles that compose a solid foundation for
a college chemistry course.
3. That the students understand the basic concepts underlying the facts, formulas, and
principles in the course.
I cannot teach you chemistry, I can only teach you how to learn chemistry. This is an important distinction
and understanding the difference means that you understand that you have reached a point in your
academic career in which you are in charge of your own education, and must dictate where it will take
you.
How to succeed
Before each class it is a good idea to skim through the section in the text that will be covered that day.
Read the section headings, look at the tables and figures, read the figure legends and the problems at the
end of the chapter. This will get you generally orientated for what we will discuss in class. After the class
you will know exactly what material was covered in class. Read the assigned sections, elaborating on
and filling in any missing information from your class notes. Most importantly ask questions and see me
as soon as something is unclear.
Notes are required and may be periodically graded, take notes as if you are taking notes for someone
else. You will find you get a better quality of notes this way.
Follow directions on all assignments and projects. You will find that my expectations are very clear and
precise, every assignment has a format to be followed. I give you an example of that format and expect it
to by followed exactly.
Learn from your mistakes. Read comments on returned assignments so that you do not make the same
mistake in the future.
Late Work
Grades will be posted in class on Monday mornings. You are given one late work pass to use each
quarter. These passes are not handed in en lieu of an assignment but are handed in with a completed
late assignment. Additional missing assignments may be handed in within a week of original due day for
50% credit. It will be nearly impossible to pass the course with excessive late or missing work. Work/labs
missed due to illness are allowed one week to make up. There is no late work accepted one week before
the end of a quarter or two weeks before the end of a semester.
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Evaluation
You will be evaluated on homework, quizzes, lab work and exams. The point distribution is
predominately laboratory and homework. Extra credit will be all but nonexistent. There are no make up
or “redos,” come prepared for class. Grades and missing work are updated and posted in class at the
beginning of each week and will be updated during midterms on Infinite Campus. Infinite Campus is not
the official class grade, but rather a mechanism to check for missing work. Each quarter has
approximately 430-480 points and the grade policy will run as follows with +/- fitting in appropriately:
100-90
A
89-80
B
79-70
C
69-60
D
< 60
F
Cheating Policy
Cheating is not learning. This class is about learning therefore if you are caught cheating you will receive
an automatic zero (0) on that assignment. If you are caught copying from another student you will both
receive zeros. If you are caught cheating for a second time your QUARTER grade will be dropped one
grade level (i.e. A- will become a B-). While I encourage collaboration I despise copying. Remember that
work with someone can help to crystallize your ideas but you are required to turn in original work.
Identical papers will be considered copying.
Tardy Policy
If you are tardy enter the room quietly and in a non-disruptive manner take your seat, begin to work on
the problem of the day. If you accumulate three tardies you will serve 15 minutes of detention, doubling
for every additional three (15, 30, 60, 120…) If you have a pass, place it on the front desk as you enter
and then take your seat. If tardiness becomes a persistent problem (even with a pass) I will mark you
unexcused for the hour. Any unexcused absences are not allowed to make up missed work for those
days (including tests and quizzes).
Books and Materials
You will be provided with a Chemistry Textbook, authored by Chang. This is to be used as a resource to
supplement classroom lectures.
It is highly encouraged that you have a separate folder or three-ringed binder for only chemistry. It will
also be to your benefit to have two notebooks (one for class notes and the other for assignments).
Taking notes will be required for this class. Color-coding your notes is often useful, I typically use blue,
black, red and green in my lectures and attempt to color-code whenever possible. A scientific calculator
that you know how to use is required for this course, programmable calculators are not allowed on
exams. A computer pass and cloud storage will be required to store your lab data and run graphical
analysis programs. Students should also have an e-mail account that can be used for this class.
Audio or Video taping of any lecture content without written permission is a violation of copyright
restrictions. If permission is given it is granted with an understanding of being used solely by the student
for the purpose of content mastery.
Teacher Contact
You are expected to contact me outside of class if you need additional help. My e-mail address is posted
on the front page. My phone number is unlisted, "Evensons" in the phone book are not me.
Audio and/or video taping of lectures is a copyright violation and requires written permission. If
permission is granted it is understood by both parties to be used by the individual student only and for the
purposes of content learning only.
General Expectations
I expect that you are a mature young adult and will treat you as such until given a reason to treat you
otherwise. I expect you to think and make a good use of your resources. There is a list of general
expectations in this text (page 5), familiarize yourself with these expectations.
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General Policies and Expectations for Evenson
There is to be no talking while the teacher or any other person has the floor
The student is required to find out what was missed during an absence
Labs can be made up for one week after the original lab date, but should be attempted
the morning immediately after the absence.
You are allowed three tardies and then detention is assigned in 15 minute intervals,
doubling with each occurrence
Cheating, copying, or the appearance of such activities will not be tolerated and all
parties will receive a zero grade
You are to bring writing utensils, notebook, and calculator to each class period
Respect each other, there is to be no derogatory, sexual or mean spirited comments
uttered to or at other students or the teacher
A signed safety contract needs to be signed by both the student and the parental unit
before any labs can be conducted
There is no food allowed in the Laboratory area all food will be confiscated and kept by
the teacher.
Drinking is a privilege that can be taken away on a group basis—do not leave your trash
around
The sinks are sinks not trash cans
There is no milling around the door at the end of class like cattle, remain seated until
teacher dismisses you.
Leaving class for a drink or bathroom is not permitted—take care of this between
classes
An absence in this class to participate in another class cannot be required by another
teacher—do so at your own risk
If you need help on the course you need to see me ASAP
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Skills Needed by Prospective Chemistry Students
Students enrolled in Chemistry are expected to have the following skills set.
Biology / IPS
Scientific method
Laboratory independence
Independent vs. Dependent variables
Understanding of graph options as dictated by data to be displayed
Linear regression models for trendline interpretation
Math
Multiplication of Fractions (Factor-Label method)
Manipulation of algebraic equations involving:
Negative terms
Squared terms
Distributed terms
Graphing skills:
Independent vs. Dependent variables
Proper scaling
Determine the equation of a line
Interpret the equation of a line
Understanding of graph options as dictated by data to be displayed
Linear regression models for trendline interpretation
Computation and use of a linear equation
English / Composition
Writing skills of proper grammar and semantics
Possession of a vocabulary to adequately communicate ideas in written and/or oral media
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ACT Science Standards
Liberal
Score Range 13 –15
Select a single piece of textual (nonnumerical) information from a table
Select the highest/lowest value from a specified column or row in a table
Select a single data point from a simple table, graph or diagram
Score Range 16 – 19
Select data from a simple table, graph, or diagram
e.g. table or graph with two or three variables, a food web
Identify basic features from a table or graph
e.g. headings, units of measurement, axis labels
Understand basic scientific terminology
Find basic information in a brief body of text
Identify a direct relationship between variables in a simple table, graph or diagram
Traditional
Score Range 20 – 23
Compare data from a simple table graph or diagram
Determine whether a relationship exists between two variables
Identify an inverse relationship between variables in a simple table, graph or diagram
Translate information (data or text) into graphic form
Select data from a complex table, graph, or diagram
e.g. a table or graph with more than three variables; a topographic map
Understand simple lab procedures
Identify the control in an experiment
Selective
Score Range (24 –27)
Compare data from a complex table, graph, or diagram
Interpolate between data points in a table or graph
Identify or use a simple mathematical relationship that exists between data
Identify a direct or inverse relationship between variables in a complex table, graph or diagram
Compare or combine data from two simple data sets
Compare new simple information (data or text) with given information (data or text)
Identify strengths and weaknesses in one or more viewpoints
Identify similarities and differences in two or more viewpoints
Identify key issues or assumptions in an argument or viewpoint
Determine whether new information supports of weakens a viewpoint or hypothesis
Understand moderately complex lab procedures
Understand simple experimental designs
Select a simple hypothesis, prediction, or conclusion that is supported by one or more data sets or viewpoints
Highly Selective
Score Range (28 –32)
Identify or use a complex mathematical relationship that exists between data
Extrapolate from data points in a table or graph
Compare or combine given text with data from tables, graphs, or diagrams
Understand complex lab procedures
Determine the hypothesis for an experiment
Understand moderately complex experimental designs
Identify an alternate method for testing a hypothesis
Select a complex hypothesis, prediction, or conclusion that is supported by a data set or viewpoint
Select a set of data or a viewpoint that supports or contradicts a hypothesis, prediction, or conclusion
Predict the most likely or least likely result based on a given viewpoint
Score Range (33 –36)
Compare or combine data from two complex data sets
Combine new, complex information (data or text) with given information (data or text)
Understand precision and accuracy issues
Predict how modifying an experiment or study (adding a new trial or changing a variable) will affect results
Identify new information that could be collected from a new experiment or by modifying an existing experiment
Select a complex hypothesis, prediction, or conclusion that is supported by two or more data sets or viewpoints
Determine why given information (data or text) supports or contradicts a hypothesis or conclusion.
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Note taking
I feel learning how to take notes is an important and valuable study skill, not only for
chemistry but in all disciplines. I will check notebooks a maximum of twice a quarter.
After the initial check, improvement in note taking will be considered in the grading. The
following is a summary of what I will look for.
Date: The first thing you should write in your notes is the date.
Terms, Rules, and Laws: Should be underlined and the definition should follow. Include
when it is appropriate to use the term, rule or law.
History: Important people should have life span and their contributions to science.
Verbalized note taking
You will need to be able to take notes based on the in-class discussions. Writing down
only what is on the board will be of little value without notations on the discussion
leading up to the concept.
Most Important:
I want to see evidence of not only doing class problems but also listing the steps that
were taken to convert a problem to an answer.
If for some reason you are uncomfortable asking a question in class write the question
in the margin and ask outside of class.
I would also like to see evidence of your notes being rewritten—this will help reinforce
the ideas discussed in class. Rewriting your notes is not required but is highly
recommended for you to achieve the highest grade possible.
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Template for Rewritten notes
Date
Question of the day
Solution to question of the day
Steps followed to obtain the solution to
the question of the day
Any additional non-mathematical
thought processes that were required for
the solution
Class discussion and problems / concepts
Again add any and all thought
processes and mathematical steps
required, to solve these problems
Include any Demo’s that took place and the explanations about the demos
Always include any historical aspects about the people who did the original work or for
who a law or theory is named after.
Elaborate on the class discussion including and supplemental information from the text
or handouts.
Include practice problems of your own that are similar to what was done in class.
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Corrections for the next lab or homework assignment
The following is a two-column table. The left side is for the items in which you lost
points on an assignment. The right side is for how you are going to prevent this from
happening on the next assignment.
Incorrect
|
|
|
|
|
|
|
|
|
|
Correction
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Student Trouble Shooting Guide (a.k.a. FAQ)
The intent of the following information is to give answers and suggestions to questions that students often
ask, it is meant to work in conjunction with Suggestions for Boosting Grades.
“It makes sense in class but not when I get home.”
This generally means that your notes are incomplete, meaning that you wrote down much of what
was on the board but did not record any of the verbal discussion or rationale used to explain what
was taking place. It is important that your notes include your thoughts rather than just what I right
on the board. It was your thoughts that made sense in class so give yourself a reminder (notes)
as to what you were thinking at the time. That way when you look back at your notes you are
reminded of what your thoughts were and the concept should also make sense again.
“I panic on test.” or “I am just not a good test taker.”
Tests represent a very small portion of the overall point distribution (generally less than 15%).
Test anxiety is created due to poor preparation and/or ineffective study habits. When you
prepare for an exam you need to study small amounts over a long period of time, do not cram the
night before. As you study you need to ask yourself questions that you think I may ask, along this
line your notes need to include the verbal discussions that are asked in class so that you have
multiple examples of the types of questions that can be asked regarding the content. In this
manner the questions on an exam will not seem foreign to you and therefore are less likely to
cause an stress during the exam.
“I read it but I don’t get it.”
If you read text passage and find yourself half way down the page and no idea how you got there
then you are not engaged with the text. Reading should create an internal dialogue in which you
are constantly asking questions about the meaning of the text and making predictions about what
the text is going to tell you as your reading continues. I would recommend not using a highlighter,
it will be your tendency to highlight to stay awake or highlight a passage because it sounds
important, even though you have not taken the time to determine what is important or why it is
important. When you read, read with a pencil. Either take notes in the text or a separate
notebook in which you record the questions you have about the text and how the text connects to
other ideas. Most of what I ask you in class is about making connections between multiple ideas
so when you read you need to continually ask how this information connects to other chemistry
topics and to the world at large.
“Can I get a chemistry tutor?”
At times it may be beneficial to work with other members of your class or prior chemistry students
as they may present material in a different manner than I that could resonate with you and
enhance your learning. However I do caution that this relationship can quickly degrade to a
scenario in which the tutor is giving you answers without enhancing your true understanding of
the concept or required thought process to solve the problem. I would encourage you to make us
of me before that of a tutor. I am available early in the mornings and working one on one we can
accomplish great things as I will be able to determine specifically where you are thinking or
computation is erroneous. If that approach does not work for you then I would also suggest
contacting the student advisor of National Honor Society at Lincoln to recommend an NHS
student that may be willing to tutor you.
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Quick Tips for Boosting Grades
Rewrite all lecture notes
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
This will reduce studying time by allotting small increments of time to the subject rather than a
marathon cram session.
The material is reviewed and elaborated on while the topic is still fresh in the student’s mind.
Allows the student to ask questions right away about misunderstandings in the course.
Preventing a snowball effect of misconceptions and confusion.
Use the left hand side of your open notebook to record each QUESTION I ask during class,
this will give you a series of questions that require the same logic as exam and lab questions.
Use the teacher as a resource outside of normal class time.



This allows for excellent one on one time and a large amount of material can be quickly
covered with such a ratio.
If this doesn’t work due to scheduling use e-mail to ask questions and receive timely answers
([email protected])
I can review homework and labs before they are due to prevent missing easy points.
Keep Parents informed about what is happening in class




Parents can and should ask to see the student’s notebook every night.
Have the student reteach the material to the parent, misunderstandings will quickly be
uncovered and again can be cleared up the following day in class.
Parents should be made aware by the student of assignments and upcoming exams. This
may help provide structure at home for completing the material.
Parents are welcome to contact me as to the student’s progress and/or upcoming
assignments
Homework

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All assignments need to be done.
The purpose of an assignment is to provide practice and instill competence in the topics of
the course.
The objective of exams mirrors the objectives of the assignments.
Read comments on homework and learn from previous mistakes.
Missing work will always have a detrimental effect on a student’s grade.
Grades and missing assignments are posted every Monday so the student always knows
where they stand.
Following Directions


Both verbal and written directions must be followed
Many assignments (i.e. lab reports, graphs, projects) require a set format to follow. These
formats must be followed in order to obtain credit for the work.
Read Recreationally

Helps develop the spatial reasoning, it is important to visualize many aspects of chemistry
that are otherwise abstract.
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Science Safety Rules
The following list pertains to safety in the science classroom. Your Signature at the end
of this list indicates that you have read, understand and agree to abide by each of the
rules stated below.
1. I will practice safe conduct in the classroom. I have been informed that horseplay will not be tolerated
and can lead to my removal from laboratory settings.
2. I will follow written and verbal instructions concerning procedures. I realize the procedures are for my
protection and will not change a procedure without permission.
3. I may not work in the laboratory without the permission and presence of the instructor. Experiments
done in class are for instructional purposes. They are planned in order to teach and idea. I will
perform only authorized experiments.
4. I will handle only those chemicals or equipment for which I have received instructions or training.
5. I will not taste, smell or mix unknown chemicals unless instructed to do so by the instructor.
6. I will always read all stock bottle labels carefully.
7. To avoid serious burns I will never reach across a flame or use flammable substances near an open
flame. I will confine long hair and loose clothing.
8. I have been shown and trained on how to use the fire equipment in the classroom and to dial 9911 in
case of an emergency.
9. I know the locations and how to operate the safety equipment in the classroom. Such as safety
showers, eyewash stations, fire blankets, fire extinguishers and first aid kits.
10. I will use caution when working with hot materials (glass, crucibles, evaporating dishes…) to avoid
burns to myself and others.
11. I will wear my goggles anytime I am in the laboratory area of the classroom regardless of the lab
procedure. Contact lenses are not advisable even with goggles on. If I don’t wear my goggles I
realize I will need to bake 1 dozen chocolate chip cookies or be deducted 20% of the lab grade for
that lab.
12. I will wear all other protective clothing (gloves, aprons, mask..) and have been instructed as to there
whereabouts.
13. I understand that open toed shoes can be dangerous and will avoid wearing them on lab days. If I
wear open toed shoes I understand that I may not be able to participate in some laboratory exercises.
14. I will pick up broken glass with a broom and dustpan and put the pieces in a specified glass only
container. I will then fill out a breakage report to be signed by the instructor. I understand I can by
charged the replacement cost of the damaged equipment.
15. If an accident or injury occurs I will report it immediately to the instructor.
16. I will not dispose of any chemicals down the sink without prior approval from the instructor. I will also
carefully read the labels on all waste containers to ensure proper disposal of hazardous materials.
17. I will clean the lab area, sink and my hands thoroughly before leaving the laboratory area of the
classroom.
18. While compounds used during labs are unlikely to teratogenic, they are not routinely checked for
teratogenic potential and if I become pregnant during the course I will advise my councilor or
instructor.
19. When provided I will read the lab procedures BEFORE entering the laboratory and will make sure I
have all the necessary materials before I proceed with the lab exercise.
20. I know I am responsible for any missing or damaged equipment after the first five minutes of
laboratory time. After that time I may be charged for missing or broken equipment.
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Chemistry Breakage and Replacement
Fees
Item
ac/dc power supply
analytical balance (200g)
beaker (100 ml)
beaker (1000ml)
beaker (250 ml)
beaker (400 ml)
beaker (50 ml)
beaker (600 ml)
beaker tongs
buchner funnel
bunsen burner
buret
buret clamp (double)
ceramic triangle
condensing tube
conductivity meter
crucible
crucible cover
crucible tongs
deflagration spoon
distilling flask (250 ml)
erlenmeyer flask (125 ml)
erlenmeyer flask (250 ml)
eudiometer (50 ml)
evaporating dish
filtering flask (250 ml)
florence flask (250 ml)
forceps
funnel support
grad cylinder (100 ml)
grad. Cylinder (10 ml)
grad. Cylinder (25ml)
grad. Cylinder (50 ml)
iron ring
medicine dropper
Mortar
pestle
pH Checker
pinch clamp
plactic wash bottle
porous cup
ring stand
separatory funnel (250 ml)
short stem funnel
spatula
Cost
$201.50
$98.00
$2.49
$7.09
$3.49
$4.76
$2.31
$5.76
$8.12
$34.35
$44.00
$64.00
$27.95
$1.90
$54.60
$16.95
$5.30
$1.10
$6.99
$3.10
$17.47
$3.19
$3.30
$66.47
$8.45
$11.75
$5.95
$1.05
$13.75
$8.45
$5.25
$5.40
$6.30
$10.00
$0.22
$6.35
$4.29
$39.95
$2.20
$4.95
$16.04
$13.69
$72.88
$3.70
$3.85
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Item
test tube (large)
test tube (small)
test tube brush
test tube clamp (for ring stand)
test tube holder
test tube rack
thermometer (Hg)
triangular file
volumetric flask (10 ml)
volumetric flask (100ml)
volumetric flask (50 ml)
watch glass (large)
watch glass (small)
wire gauze
Cost
$0.75
$0.35
$0.99
$1.65
$1.65
$4.95
$10.90
$3.95
$8.75
$37.50
$15.74
$3.75
$2.50
$6.95
Stirrer / hotplate
Stirring rod
$316.00
$0.35
Evenson
Grading Rubric for Chemistry lab Reports (20pts)
Correct lab format
Title
Objective
Data
Calculations
Questions/ conclusions
Graphs stapled to the back
All sections must be properly labeled
(i.e. Objective: It is the purpose of this lab…)
All necessary data included with the correct number
of digits and units shown in labeled data table
All necessary calculations clearly documented and
with the correct number of significant figures
Lab questions answered completely, legibly and correctly
Explained for understanding by a non-science person
5 pts
5 pts
5 pts
5 pts
The lab write up should include the above headings (title, objective, data, calculations, and questions/
conclusions) the following is a description of what each of the headings will include. On the reverse side
is a short example of what your lab report should look like. Notice each section is clearly labeled and the
organized nature of the report includes a large amount of white space.
Title:
Is the experiment name, the date it was begun and finished. It may also be appropriate to record dates in
the data section for multiple lab days were data was recorded over a period of time.
Objective:
The objective of the experiment-what it is one hopes to prove or determine-should be clearly but briefly
stated. It is unlikely that the objective will exceed two sentences.
Data:
If you measure it write it here. All data, raw and calculated, should be recorded here. This includes
unknown numbers, weight and volume measurements taken or derived from chart recordings and
computer printouts. It is also necessary to include any qualitative (not directly measured) observations.
You may asterisk data to include a footnote about an error in a procedure, spill, mis-measurement or any
information that makes that lab data vary from the rest. Then run a second trial.
Calculations:
This section includes the calculations required to compute the results of the experiment, as well as the
results. If a number of similar results are to be reported (i.e. the temperature change for each of 4 trials) it
will be appropriate to show those results in a neatly labeled table. If a number of similar calculations are
required it is usually proper to only show one example calculation. This will be determined by your
teacher and depends on the lab. While graphs should be recorded in this section we will affix (staple)
them to the back.
Questions / Conclusions:
All of the questions that accompany a lab need to be answered and usually follow in a logical order,
answering question 1 will help you answer question 2 and so forth. This is the section that the
calculations from the data are interpreted and analyzed.
The use of slang terms and phrases (i.e. ‘cuz, kinda, ya’know, , …) have no place in formal writing of
this type and will lower your lab grade.
Common Short hand: IA (inaccurate)
SF (wrong number of significant figures)
Uts (missing units)
M.O. (Missed Objective)
P (wrong precision)
HE (human error instead of procedural error)
circle and arrow (answers says nothing)
M.C. (Missed Calculations)
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Requirements for Chemistry Lab Notebook
Written communication is the most important method by which chemists transmit their findings.
The laboratory notebook is a crucial element in this communication. The laboratory notebook is
regarded as a legal document in court and has historically been used to determine who made
an initial discovery when two or more chemists independently discover a new entity. The
following are the requirements for your lab notebook.
1. Must be hard bound with a stitched binding (not spiral or perforated pages) with EVERY page
numbered.
2. The first several pages 4 to 5 numbered with lower-case roman numerals (i, ii, iii, iv, v) are to
be reserved for a table of contents.
3. Each experiment will be titled, as it is in the table of contents.
4. An objective will be given for each experiment (Objective: The following experiment will be
used to find…)
5. When it applies a reaction mechanism will follow the objective to show the net equations and
overall chemistry taking place.
6. The Procedure will be written out in full, explaining any novel techniques or corrections to a
proposed procedure. This is the only procedure allowed in lab.
7. Characteristics (physical and chemical) of both the reactants and products should be noted
with special attention given to special handling requirements. This is also where disposal of
excess material should be recorded.
8. Data tables should record precise measurements and units in properly labeled data tables.
9. All calculations should be proceeded with a brief statement about what the calculation is
showing or why it was done.
10. Any graphs must be taped into the lab notebook and signed across the edge and dated.
11. Any calculations done in a spreadsheet format should also be referenced in the Calculations
section of the lab notebook with the disk number and filename.
12. Your signature is required at the end of each day an often the signature of a witness, as
again this is a legal document.
Other requirements that are not specific sections in the lab notebook
1. All entries are done in black ink and handwritten.
2. Nothing is ever erased, and only a single line is drawn through an error.
3. A diagonal line is drawn through a page that is not filled and the line is signed and dated
4. Each new day a new page is started and DATED
5. Everything is recorded in your lab notebook, nothing is ever done on scrap paper--If I see you
using scrap paper I will take it from you and your data will be lost!
Your lab notebook is a complete record of what you did and when you did it. Its purpose is to
clearly and concisely convey what you did to another chemist in so much as that they may
repeat your exact experiment, write legibly and be aware of grammar and spelling.
Answers must be explicit and elaborate on your complete thought. Responses of “mess up the
data” and “Change the results” do not tell the reader HOW the data will be effected.
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Lab Notebook Grading (25 pts)
General layout (10 pts)
_______Table of contents (2pts)
_______ Pages Numbered (2 pts)
_______Black Pen (2 pts)
_______Date (2 pts)
_______Signature (2 pts)
Format and Mechanics (15 pts)
______ Title (1pt)
______ Reaction Mechanism/Reaction (1 pt)
______ Objective (1 pt)
______Procedure (2 pts)
Cautions
Disposal
______Data Table (5 pts)
Units (1 pt)
Precision (1 pt)
Organization (2 pts)
Labeling of all columns (1 pt)
_____Calculations (5 pts)
Short explanation of the calculations
Graphs signed
Quick Check List for Lab Reports (this is a good list to tape inside of your lab manual to avoid missing
easy points)
Data Section
Vertical table
All units
Proper precision
Data values (not calculated value)
All data is provided to reach objective
Calculations section
Equations from graph displayed with proper units as a MODEL
Model is used to reach objective
Calculations show the objective been met
All derived values are supported by clear calculations
Algebra shows unit/value determination
% error calculation includes a theoretical / physical constant
Questions/analysis section
No unidentified pronouns
Answer do not begin with “Yes, no, because”
Answers do not include options (i.e. …..increases or decreases…)
Full Sentences (generally multiple sentences to fully explain phenomenon)
Evidence from Data is cited in responses
All answers explain phenomenon from a MOLECULAR level of reason
Source of error question do not include human errors (i.e. miscalculations, spilling material, etc)
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Types of Questions
Communication is required to increase the collective knowledge of the class. Unshared knowledge is not
worth having. In order to learn you must ask questions and in turn your understanding will be deepened
by answering the questions of others. In asking questions you clarify your misunderstandings and
continue to grow as a student. By answering questions you put your ideas and understanding before the
judgment of the class and receive personalized feedback that is both critically corrective and helps
crystallize your understanding. Below are several classifications, albeit not an exhaustive list, of the types
of questions that you should ask and prepare to answer during class.
Comments of Concurrence: If you concur, that means that you agree with what a person is saying.
During class discussions it is appropriate for you to indicate that you agree with what a person is saying
or depicting. If you agree it is also important that you state why you agree with their interpretation,
generally by citing your own data or other corroborating evidence.
Clarifying questions:
Clarification questions are asked when you do not understand a drawing or term that is provided. For
example if you don’t understand a person’s penmanship or are confused by what a drawing is attempting
to show, then you would want to ask a question to clarify and correct your misunderstanding. Such as
“Can you clarify what your drawing is showing? I don’t understand what your picture is depicting.”
Explanation questions:
Explanation questions are similar to clarification questions but more in depth. An explanation question is
asking for someone to show you the relationship between two or more entities and describe how they are
connected or why they are related.
Application questions:
An application question is asking the presenter to apply their conclusion to a new situation, “How does
your conclusion relate to…?” Application questions can center around an experience you have had in the
past in which you now wonder if the concept being discussed applies to your anecdotal story.
Extension questions:
An extension question is similar to an application question in that the question is asking for a new use of
the idea presented. An extension question is asking the presenter to go beyond the immediate work that
they conducted and make a speculative prediction about a new situation based on the work they are
presenting.
Defensive questions:
A defense question is asked when you want the reporter to defend what they have said/written. If another
student has chosen to quantify their data with units that differ from yours, then you are in your rights to
ask them to justify why those units were chosen.
Lab Design Questions:
Lab Design Questions are generally too big of a question to expect a single presenter to answer, but that
does not mean they should not be asked. The contrary the questions should be asked to the class as a
whole. Situations will arise in which we do not have the information to answer the questions and we need
to then design an experiment that will allow us to determine the value or answer needed.
Debate (“Stump the Chump Challenge”):
A debate question is generally a very high-level question in which you cite your conclusion and evidence
for your conclusion and point out that it differs from the conclusion of the presenter. This must be done in
a civil tone and is used to start a debate amongst the class as a whole about the merits of each
conclusion.
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Unit 1 Objectives for Measurements and Graphical Analysis
 Metric Terms
Mass
(grams)
Distance
(meters)
Pressure
(Pascal, atmospheres, mm Hg)
Volume
(liter or m3)
Weight
(Newton)
Time
(second)
Temperature (Kelvin, Celsius)
Energy
(joule)








Metric Prefixes
Symbols and magnitude from Mega to nano
Foundations in dimensional analysis
Significant Figures
Water displacement
Density
Precision reading of lab equipment
Identification and use of lab apparatus
Graphical Analysis
 Identification of variables
 Development and application of mathematical models
 Interpolation vs. extrapolation
 Linearize non-linear data
 Design laboratory procedures based on units of objective
 Quantify and apply the area under the curve
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Measurements and Dimensional Analysis (Factor-Label method)
As you know the sciences use the SI (International System of units) or commonly called
metrics. Metrics make calculations easy, they are all on a base ten system and you only
need to memorize a few prefixes to know the magnitude of the number.
Base units
unit
symbol
Length
Volume
Mass
Temperature
Time
Pressure
Energy
meter
cubic meter
gram
Kelvin
second
pascal
joule
m
m3 (the liter is often used, 1 ml = 1cm3)
g
K
s
Pa (atm or mm Hg often used)
J
Prefix
Abbreviation
Meaning (factor)
YottaY
ZettaZ
ExaE
PetaP
TeraT
GigaG
MegaM
Kilok
Hectoh
Dekada
BASE UNITS
Decid
Centic
Millim
Micro
Nanon
Picop
Femtof
Attoa
Zeptoz
Yoctoy
1024
1021
1018
1015
1012
109
106
103
102
101
10-1
10-2
10-3
10-6
10-9
10-12
10-15
10-18
10-21
10-24
20
*adding a prefix to a base
unit tells a scientist the
size of the base unit, a
kilogram therefore is 1000
grams.
Evenson
Dimensional analysis
It is necessary that we can convert between units of the same base unit and from one
unit to another. The method we use for this is called dimensional analysis or factor
label method. This method is based on basic algebra and the fact that multiplying or
dividing any number by one does not change the number.
Remember these two foundational steps:
1. We can cancel any unit that is both in the denominator and the numerator.
2. We can multiply anything by one with out changing the value of the number
(one is simply a number divided by itself).
We can change 460.0 cm into meters
We know that 1 meter = 100 cm
We also know that if we multiply 460.0 by 1 it is still 460.0
We can change the appearance of 1 by dividing 1 meter by an equal value (100 cm)
1m 1
100 cm =
So this is also true
460.0 cm x 1 m = 4.60 m
100 cm
Notice that because of rule number 1 above we can cancel out the centimeters because
the unit appears in both the numerator (top) and the denominator (top).
These are linear equations where the whole problem is written out and solved in one
final calculation. As you will see in the next example you can not go from a prefixed unit
to another prefixed unit, you must first convert to the base unit and then back to the
prefixed unit.
Convert 14000 ml to kl.
Start with the term you are given, 14000 ml, and convert it to the base unit, liters.
14000 ml x 1 liter x
1 kl = 0.014 kl
1000 ml
1000 liter
Again notice this is not two separate calculations but one continuous line of
mathematics. This prevents rounding errors and will become more important as you
become familiar with significant figures.
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Counting Significant Figures (Sig. Figs.)
These are the rules for Sig. Figs. Do not lose this paper or destroy it until you have
memorized these rules.
1. All nonzero digits are significant. There are three significant figures in 2.68 cm and
four significant figures in 9.648 cm.
2. Zeros between nonzero digits are significant. There are three significant figures in
106 ml and in 1.07 cm.
3. Zeros beyond the decimal point at the end of a number are significant. When we
say that the volume of a liquid is 8.00 ml we imply that the two zeros are
experimentally meaningful. The quantity 8.00 ml carries the same degree of
precision as 8.13 or 1.07 cm.
4. Zeros preceding the first nonzero digit in a number are not significant. In a mass
measurement of 0.002 g, there is only one significant figure--the "2" at the end. The
zeros serve only to fix the position of the decimal point. This becomes obvious if we
express the mass in exponential (scientific) notation in that case, we would write
0.002g as 2 x 10-3.
When dividing or multiplying:
The number of significant figures in the result is the same as the in the quantity with the
smallest number of significant figures.
(437 x 0.0003)/(0.435 x 2.1)=0.1434139573  0.1
When adding or subtracting :
The number of digits beyond the decimal point in the result is the same as that in the
quantity with the smallest number of digits beyond the decimal point.
(437-0.0003)+(2.1-0.435)=438.6647 439
Exact numbers and definitional numbers:
These are not figured in when determining the number of significant figures.
2.87 L x 1000ml / L x 1 g / 1 ml 2870 g
** even though 1000 and both 1s have only one sig fig these are considered exact
numbers--there are exactly 1000 ml in one liter and there is exactly 1 g of mass in 1 ml
of water. Exact numbers can have as many sig figs as are needed, for example we
could write the conversion between 1 liter and 1000 ml as follows:
1.0000000000000000000000000000 Liters in 1000.000000000000000000000000000
ml.
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PROBLEM SOLVING FOR SCIENCE STUDENTS.
A general approach to solving numeric (and non-numeric) problems. The method for
using this general approach is strictly linear: Step 1 before Step 2 before Step 3 before
Step 4 before Step 5.
STEP 1. ANALYSE.
The aims of this step are:
1.to understand the problem and to create a mental image of the problem.
2.to extract the given data and to understand the nature of the unknown.
3.to estimate an answer to the problem.
STEP 2. BRAINSTORM FOR A PLAN.
At this step you should aim at finding relationships (equations) by which the unknown
may be related to the known.
STEP 3. CALCULATE.
At this step the route found in step 2 is used to calculate the solution.
STEP 4. DEFEND, BY CHECKING AND PRESENTING A SOLUTION.
At this step the aim is to:
1.make sure that the solution obtained in Step 3 is acceptable.
2.present the solution in a reasonable format.
STEP 5. EVALUATE
The problem is solved, a satisfactory solution has been presented: what has been
learned?
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Metric Measurement Lab
Background:
Rulers and Meter sticks
Meter sticks are used to measure a distance and is similar in length to a yard (3 feet).
The meter is the SI unit of distance.
1. Examine the ruler and determine the scale of the ruler. This is usually inches (in) or
centimeters (cm). This class will always deal in the vernacular of science, metrics.
2. Determine the length that each of the smallest marks represents. This is typically 1
millimeter (mm).
3. Align one end of the meter stick with the edge of the object being measured and
read the length measurement, estimating the last reported digit. The space between
the smallest two markings on any lab apparatus can be estimated. For example, if
the reading is exactly half way between 1.0 and 1.1 cm then the proper
measurement to be recorded for that length is 1.05 cm. Do not just round to the
nearest value displayed on the meter stick or ruler.
4. Always include the units that you are using. There is a great deal of difference
between 100 dollars and 100 cents; the same is true for all measurements.
Graduated cylinders and Burettes
Graduated cylinders and burettes are used to accurately measure volumes of liquids.
The SI measurement of volume is the liter and is the measure of space any matter
occupies.
1. Again determine the scale on the graduated cylinder or burette. Remember that you
can estimate one decimal place beyond the smallest increment given.
2. Notice how the liquid’s surface is curved upward at the edges of the graduated
cylinder. This curved surface is called a meniscus and is caused by the surface
tension of the liquid.
3. Read the volume of the liquid, using the bottom of the meniscus.
4. When reading the meniscus it is very important that your eye is at the same level as
the meniscus. Do not pick up the graduated cylinder, but rather place it on a flat
surface and lower your head to the level of the graduated cylinder. It is sometimes
beneficial to hold a portion of colored paper behind a clear liquid to add contrast.
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Triple beam balances
Triple beam balances are used to measure the SI unit of mass, grams. Mass unlike
weight is not effected by gravity and therefore is constant regardless of location.
1. There are two main parts to a triple beam balance, the beams and the pan.
2. The pan is the large circular area where an unknown mass is placed and the beams
are where known masses are hung.
3. Behind the pan (left hand side) is a small screw that is used to zero or calibrate the
balance. With all of the hanging masses pushed as far to the left as possible adjust
the screw until the pointer line is equal with the stabile line on the far right hand side
of the balance.
4. Once a mass is placed on the pan begin to adjust the hanging masses until the two
lines are equal. First adjust the small masses and then move your way up to the
larger masses.
5. To record the mass of the unknown object be sure to add up all of the masses used.
Again estimating one decimal point beyond what is readable on the scale.
Electronic Balances
Electronic balances are sensitive, delicate instruments and must be treated
appropriately if they are to continue to be available for your use. Always mass objects
(including chemicals) in a container or weigh paper. Never add chemicals to a
container that is on the balance. Remove the container then add and then remass. Hot
objects are never placed on the balance; this will melt the plastic portions of the scale.
You have been warned and will be charged.
1. Turn the balance on by using the ON button.
2. Wait until the balance beeps and then press tare. This will zero out the balance.
3. Check the right-hand side of the balance’s read out for the units that the scale is
measuring in. It should always be in grams.
4. If it is not in grams gently press the mode button until the small pointer is on “g” for
grams.
Objective: Gain an understanding of metric values and the size relationships between
the prefixes, as well as obtain some experience with proper lab techniques for
measuring and recording accurate and precise values.
Directions on how to properly record and use the various laboratory materials.
Procedure:
1. Follow the placards at each lab station.
2. Although you do not have to do the stations in numerical order, your answers must be
reported in numerical order.
Data
Make a data table with four columns. The first column will be the station number and
the second column will be the measurement. The last two columns will be for
measurements recorded in different metric units (i.e. grams, milligrams, and
kilograms…). Be sure to include the proper number of digits past the decimal point and
the units.
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Question
1. Develop a stepwise (1., 2., 3.,…) procedure for determining the density (g/ml) of
100.0 ml of water.
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Discovering and Using Conversion Factors
Objective:
Determine some of the commonly used metric to standard conversions using the given
materials.
Materials:
Weight plate
meter stick
gallon jug
Procedure:
1. Use a meter stick for each group
2. The whole stick is one meter, the longest lines (at the whole numbers) are
centimeters and the smaller lines are millimeters.
3. Count the number of centimeters and the number of millimeters in a meter and record
them in your data table.
4. Also record your weight in pounds, the relationship between pounds and kilograms
and liters to gallons.
5. Having done this you should also be able to find the relationships between fluid
ounces (fl. oz.) and millliliters (ml) as well as the between inches and centimeters
Data:
Record all relevant information in a clear and concise data TABLE.
Calculations:
Determine the conversion factor (ratio) of centimeters to meters as well as the following:
millimeters to meters, gallons to liters, pounds to kilograms.
Show the calculations to determine your:
Mass in kilograms
Height in centimeters
Questions:
1. The metric prefix symbolizes the ratio (the same way pre in the word prefix means
"before") what do "centi" and mill" mean?
2. The prefixes always mean the same amount, so what is the size of 1.00 milliliter if it
were in liters?
3. "Kilo" means 1000, what is the size of a kilometer?
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Laboratory Apparatus Identification Exercise
You are to draw pictures of the apparatus indicated and tell where the item is found and what it is used
for.
1.
Test tube
Use:
Location:
2. Crucible and Crucible cover
Use:
Location:
3. Evaporating dish
Use:
Location:
4. Watch Glass
Use:
Location:
5. Beaker
Use:
Location:
6. Crucible tongs
Use:
Location:
7. Test tube holder (2 types)
Use:
Location:
8. Florence flask
Use:
Location:
9.Graduated cylinder
Use:
Location:
10.Scoopula
Use:
Location:
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11. Short stem funnel
Use:
Location:
12. Test tube brush
Use:
Location:
13.Ceramic triangle
Use:
Location:
14. Bunsen burner
Use:
Location:
15. Ring stand
Use:
Location:
16. Iron rings (for ring stand)
Use:
Location:
17. Beaker tongs (rubber-coated ends)
Use:
Location:
18. Test tube rack
Use:
Location:
19. Wire Gauze
Use:
Location:
20. Forceps
Use:
Location:
21. Erlenmeyer flask
Use:
Location:
22. Striker
Use:
Location:
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% of Sugar removed from Gum over Time 1
Objective: Determine the extraction rate of sugar from chewing gum.
Procedure:
1. Obtain one piece of bubble gum of the same brand and flavor for you and your
partner. You will provide your own gum.
2. While the wrapper is still on the gum record the precise mass of gum and wrapper.
3. Without touching the gum, remove the wrapper and chew the gum.
4. SAVE your gum wrapper
5. While you are chewing your gum record the mass of your wrapper.
6. Re-mass your gum (and the original wrapper) every 2 minutes for a total of 15
minutes.
Data:
Data must be in a table with columns and rows. The columns must be labeled Time,
mass of wrapper, mass of gum and wrapper, mass of gum.
Calculations:
1. Calculate the % of sugar in the gum
2. Graph the loss of mass as a function of time (staple to the back of proper lab writeup.
3. Determine the extraction rate (this is the objective of the lab)
Questions:
1. How does your % of sugar differ compared to other lab groups?
2. What mass of sugar is required to create a 100.0 pound piece of gum (show work).
3. Besides sugar, what else could be responsible for the mass change in the gum?
4. What is your explanation for the shape of the graph? (This question does not ask
you to explain WHAT the graph looks like, It is asking to explain WHY the graph
appears as it does.)
1
The lab adapted from work done by Mrs. Carol Christiansen
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Graphing Instructions
What follows are general rules that must be adhered to when making a formal graph.
1. Scale
The scale of the graph needs to be easily used numbers. Each square equals some commonly used
numbers (whole numbers, factors of 5, factors of 10, 0.5, 0.25. 0.1). The scale must be sequential,
starting with a low value and ending with a high value. The scale of the graph must be arranged so
that the graph takes up ¾ to the whole page. Your scale does not need to start at zero for an origin,
nor can you ever have a scale with a break point in it. Graph paper is a must, choose graph paper in
which the deviations of scale fit your need.
2. Axis
Graphs are set up on an X, Y coordinate plan. Your X axis is your independent variable and your Y
axis is your dependent variable. The independent variable is the variable that you control. How you
adjust the independent variable will determine the value of the dependent variable. For example if
you were graphing the mass of pennies according to their year, then years is your independent
variable. This is because the year of the penny will determine the mass of the penny. The mass of
the penny is dependent on the year. Both axes must be drawn using a straight edge.
3. Labels
All axes must be labeled with what is being measured and the units the measurement was taken in
parenthesis (i.e. Density (g/ml)).
4. Adding lines
Most graphs we use are XY scatter graphs (connect the dots). Whether the dots are to be
connected or a trend line (best fit) is used depends on the function of the graph. If the graph is
looking for an average or general occurrence then a trend line should be used. A trend line gets
as close to as all of the points as possible. It may or may not touch any of the data points.
If you are going to need to find data that occurs between data points then you need to connect
the dots. If your data will make a stochastic graph (many peaks and valleys) then you also want
to connect the dots. All lines should be drawn using a straight edge.
Many of our graphs will be graphs of time versus some unit of measurement. The slope of these lines will
be a rate.
5. Title
The title of graph is what the line(s) on the graph represent. The title is not a renaming of the XY axis
on the top of the graph.
6. Legend:
A legend is an indication as to what each line on the graph represents. A legend is required if you
have more than one line per graph. However a legend is not required when only one line is present.
The title will then indicate what the line represents.
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Graphing in MS Excel
Computer generated graphs are not more accurate than hand graphs, but can be made much faster.
Excel is on of the more common and powerful graphing/spreadsheet programs available. I encourage
you not to use MS Works for any computer graphing – in short it does not. There are open source
versions of Excel available for both Mac and PC from Sun Microsystems under the program name of
NeoOffice and Open Office respectively.
The following is a brief tutorial for graphing in Excel.
The screen shown to the left is what
appears when Excel is first opened up.
There are vertical columns labeled with
alphabetic notations and numerical rows.
Each box is called a cell and the intersection
of a column and a row produces a cell with
the name such as “A1.” To the left cell A1 is
highlighted.
A computer graphing program is only as
good as the user and you must understand
what graphs mean and how they are
constructed in order to make use of a
computer program for graphing. When
graphing the independent variable is on the
X axis. The independent variable is the
variable in the experiment that you have
chosen. For example if you want to
determine the relationship between hours of
studying and your grade then your
independent variable should be the hours you study. This allows you to ask questions of you’re your
graph such as “If I study for 3.5 hours my grade should be what? The dependent variable in this example
is the grade and will be represented on the Y axis of the graph. In Excel the Independent variable data is
entered in the A column. Dependent variables are entered
into successive alphabetic columns.
Row one is used to label the data columns. Notice that the
label includes what is being measured (Study Time) and
the units in which that entity is being measured (Hour). It is
also important that the units are not placed next to the
data, this is only true for spreadsheet use as it
alphanumeric components are not graphed properly. Our
data has been represented with the appropriate
independent and dependent variables and now we can use
the “Chart Wizard” to help us create the actual graph.
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Before using the chart wizard
we must tell the program which
data to graph. Left click on the
letter A and then slide over to
the letter B. This will highlight
the entire columns. Notice that
the highlighted portion extends
beyond the data, and that even
portions of the columns without
data are highlighted—this is
required. Once that is done
The chart wizard icon is the on
the top tool bar and looks like a
small graph. To the left the
Chart wizard is shown as the
button above Column F. second
button from the right. With the
data highlighted, click on the
Chart wizard icon, doing so will
bring up a Dialog box.
With only a few exceptions our graphs will
always be XY (Scatter) graphs, so click the
option for XY Scatter and the sub type without
the lines connected (we will add a trend line
later). Then Click next. Screen 2 of 4 will
appear and again just click next.
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The next screen in the chart
wizard allows you to enter
your labels for each axis and
the title of your graph.
Remember that your axis
require labels with and units in
parenthesis as shown. If you
click on the “Gridlines” tab you
can also unclick “major
gridlines” which gives your
graph a cleaner look. Then
click next. On the last dialog,
box just click Finish.
Now we can add a trendline,
left click on a datum point,
and all of the data points will
be highlighted. Then Right
click and select “add
trendline.” A new dialogue
box will appear.
Most of our graphs will be linear
regressions although the current data for
our example graph is showing an
exponential curve. Highlight the type of
graph and then select the options tab from
the top of the dialogue box.
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Under the options tab select
“Display Equation on Chart” and
“Display R-squared value on Chart”
This will tell the program to give the
equation for your line. The equation
for the line is the reason for making
the graph – the equation is the
mathematical model that will allow
use to make predictions – science is
all about mathematical models and
predictions. Click on OK and your
graph is done.
There are some
additional “clean-up”
issues that can be
done, such as deleting
the legend. There is no
need for a legend as the
title of the graph tells us
what the line is
showing. Only include
a legend if you have
more than one line.
You can also double
click on the background
of your graph to change
it from an ink wasting
light gray to an ink
saving white
background.
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Investigation of Factors that Affect the Density of Water
Objective: To collect and graph the data required to determine the density (g/ml) of water at various
temperatures.
Procedure:
Room Temperature Sample:
1. Record the precise temperature of your Distilled water sample.
2. Record the mass of a clean and dried 50-ml beaker.
3. Using the micropipette add 1.00 ml of water to the beaker and record the correlating mass of
the water.
4. Using the micropipette Add an additional 3.00 ml of water to the beaker and record the
correlating mass of the water (This should be a cumulative total of 4.00 ml).
5. Continue adding aliquots of water and record the mass of 8.00 ml, 12.00 ml, 16.00 ml and
20.00ml.
Hot or Cold water sample:
even stations: Hot water bath are to use boiling water. Repeat steps 1-5
odd stations:
Cold water will make an ice bath
Repeat steps 1-5
Data:
In a well organized data TABLE record the masses, volumes and temperatures of all water samples
analyzed. You will need data for all three temperatures; this means if you did the cold water samples
then you must get the hot water data from the group across from you and vice-versa.
Calculations:
1. Graph the data for each temperature of water (room temperature, hot and cold) on the same graph
AND determine the equation of each line.
2. Calculate the average density of water at each temperature.
3. Use the information generated from your initial graph to determine an expansion rate of water
(density/temp).
4. Calculate your percent error based on the room temperature water data. Theoretical densities for a
range of temperatures are on the next page.
(theoretica l  actual )
x 100 = % error
theoretica l
Questions:
1. Write the equation of your lines in terms of your axis; be sure to include all units and indicate the
temperature that each line corresponds with.
2. Will the cooling of your water as you record the mass of the water have an affect on the calculated
density of the sample?
3. What is happening to water molecules as temperature is changed that causes a change in the density
of the sample?
4. Human errors are the errors that can be corrected (e.g. miscalculating, misreading a scale, etc.) while
procedural errors occur do to limitations of equipment or other requirements of the lab. What is one
procedural error that would help to explain your percent error?
36
Evenson
Water Density Table
Values are in grams per cubic centimeter
o
o
o
o
o
o
o
o
o
o
0.0 C
0.1 C
0.2 C
0.3 C
0.4 C
0.5 C
0.6 C
0.7 C
0.8 C
0.9 C
o
0.99862
0.99860
0.99858
0.99856
0.99855
0.99853
0.99851
0.99849
0.99847
0.99845
o
0.99843
0.99841
0.99839
0.99837
0.99835
0.99833
0.99831
0.99829
0.99827
0.99825
o
0.99823
0.99821
0.99819
0.98817
0.99815
0.99813
0.99811
0.99808
0.99806
0.99804
o
0.99802
0.99800
0.99798
0.99795
0.99793
0.99790
0.99788
0.99786
0.99784
0.99782
o
0.99780
0.99777
0.99774
0.99772
0.99769
0.99766
0.99764
0.99762
0.99776
0.99758
o
0.99756
0.99754
0.99752
0.99750
0.99747
0.99744
0.99741
0.99738
0.99736
0.99734
o
0.99732
0.99729
0.99726
0.99724
0.99721
0.99719
0.99716
0.99713
0.99711
0.99709
o
0.99707
0.99704
0.99701
0.99699
0.99696
0.99694
0.99691
0.99688
0.99686
0.99683
o
0.99681
0.99678
0.99676
0.99673
0.99670
0.99668
0.99665
0.99663
0.99660
0.99657
o
0.99654
0.99651
0.99649
0.99646
0.99643
0.99641
0.99638
0.99635
0.99632
0.99629
o
0.99626
0.99624
0.99621
0.99618
0.99615
0.99612
0.99609
0.99606
0.99603
0.99600
o
0.99597
0.99594
0.99591
0.99588
0.99585
0.99582
0.99579
0.99576
0.99573
0.99570
o
0.99567
0.99564
0.99561
0.99558
0.99555
0.99552
0.99549
0.99546
0.99543
0.99540
18 C
19 C
20 C
21 C
22 C
23 C
24 C
25 C
26 C
27 C
28 C
29 C
30 C
37
Evenson
Burn Baby, Burn!
Objective: Determine the rate of combustion for the candle
Procedure:
1. Record the height of your candle.
2. Adhere the candle to a watch glass with a small amount of melted wax.
3. Place the candle on the balance and record the initial mass of the candle.
4. Leave the candle on the balance and light the candle, record the mass of the
candle every 30 seconds.
Calculations:
Graph your data appropriately making sure to include the equation of your line with
all your units.
Use the equation of your line to determine the rate of combustion.
Questions:
1. What factors will likely lead to a variety of different graphs among various lab
groups?
2. There is an old Chinese clock designed to divide the day into three parts, 1/3 for
play, 1/3 for work and 1/3 for sleeping. The clock worked by having a candle
inside of a box and as the candle burned down the flame would be visible in one
of the three regions indicating the ‘time’ of day (i.e. work, play, sleep). How tall
would your candle need to be in order to be used in such a clock?
3. What would you need to do if you wanted to use a shorter clock, candle?
4. Each graph will have a y-intercept, what does the y-intercept mean in regards to
your candle?
38
Evenson
Determination of Sugar Concentration
Background:
Concentration is a measure of the amount of solute dissolved into some amount of solvent, in short
concentration is the amount of stuff in a solution. Concentration can be measured in a wide variety of
ways such as Normality, Molality, Molarity, m/m%, m/v%, v/v% …. Each unit of concentration
measurement has it own application and each is therefore useful in a given situation. Density can be
thought of as a concentration. Density is the amount of grams that are in a milliliter of material (g/ml).
Objective: Determine the concentration (g/ml) of sucrose (C12H22O11) for each of the given samples of
aqueous sucrose.
Procedure:
Develop a procedure to determine the concentration of five sucrose solution samples if only a single
concentration is known. (This procedure must be included in your lab notebook and lab report)
Data:
Record all relevant data and units to the proper precision
Calculations:
Show all mathematics required and use a short logical statement to explain how your calculations have
been used to achieve the objective.
Questions:
1. If 500.0 ml of each solution was made for this lab, what is the total mass (kg) of sucrose needed?
2. When the density of each solution is graphed as the dependent variable and the mass of each solution
is graphed as the independent variable, what does the slope represent?
3. If the axis in question two were reversed (mass on Y and Density on X) what would the slope
represent, would it change?
4. The density of a golf ball is 1.15 g/ml. Assuming that the water hazard on hole #4 has a 165 kL of
water, what mass of sucrose must be added to the water hazard so that a golfer’s ball will float when hit
into the water hazard. Use the equation of you line to determine this (watch your units).
39
Evenson
Energy is Energy (Temperature Conversions)
Background: The most common temperature scales (i.e. Fahrenheit, and Celsius) are arbitrary scales
based on the physical properties of water. The purpose of each scale is to communicate a relative
amount of energy within a system or surroundings.
Objective: Develop a procedure to graphically determine the mathematical model required to convert
between Celsius and Fahrenheit scales.
Procedure:
Develop a procedure to determine the relationship between the two temperature scales of interest. (This
procedure must be included in your lab notebook and lab report)
Data:
Record all relevant data and units to the proper precision
Calculations:
Show all mathematics required and use a short logical statement to explain how your calculations have
been used to achieve the objective. A proper graph and mathematical model is required.
Questions:
1. The graph is a required tool to determine the equation of the line. What is the equation of your line (be
sure to include all units).
2. Kelvin is another temperature scale and is based on absolute zero. The magnitude of a Kelvin is the
o
o
same magnitude as a C, Although the zero Kelvin occurs at -273.15 C. Sketch a graph of this
relationship and the equation of the line for that graph.
3. The change in temperature for boiling water to freezing water is the same for both Celsius and
Fahrenheit however the change in the units are different. There are more units of Fahrenheit than there
are units of Celsius changed, therefore a degree Celsius is larger than a degree Fahrenheit. How many
times larger is the magnitude of a Degree Celsius than a degree Fahrenheit, According to your data?
4. Theoretically a Celsius unit is 9/5 the size of a Fahrenheit unit. What is your percent error based on the
magnitude of units?
(theoretica l  actual )
100=% error
theoretica l
40
Evenson
Single Serving Kool-Aid®
Objective:
Make a single serving (8.00 oz.) of Kool-Aid.
Directions to make Kool-Aid:
Add 0.14 oz. of Kool-Aid powder to 1 cup of C12H22O11 and 2.00 quarts of H2O to make 8 servings
of Kool-Aid.
Procedure:
1. Convert the amount of sugar needed to make a single serving of Kool-Aid (1.00 cup/8 servings
needed for full batch).
2. Convert the amount of Kool-Aid needed to make a single serving of Kool-Aid (
needed for a full batch).
0.14 oz
/8 servings.
3. Convert the amount of water needed to make a single serving of Kool-Aid (2.00 qt/ 8 servings
Needed for a full batch).
Data:
Data table must show the actual amounts of sugar, Kool-aid powder and water that were added to the
correct precision of digits. Remember that calculated values are not your data, your data is what you
collect for values during the laboratory exercise.
Questions:
1. You could easily combine sugar and kool-aid powder in the proper ratios to make a single
serving of kool-aid that could be added directly to a half-liter water bottle (e.g. Aquafina,
Dasani, Ice Mountain), calculate the mass of sugar and kool-aid powder needed for a 500 ml
water bottle.
2. A half-liter bottle of kool-aid is more than a single serving; if a single serving provides you
with 10% of your daily vitamin C requirments then what percent of Vitamin C would be
provided if you drank the whole half-liter bottle?
3. Why is it necessary to only change one variable at a time to determine how to improve the
taste?
Some possibly helpful conversions:
1.00 g / 0.036 oz.
Mol / 6.02x1023
1.00cup / 8.00 fl oz.
1.00 ml / 0.030 fl oz.
1.00 g / 1.00ml
1.00 L / 1.06 qt
1.00 pt / 16.00 fl oz
60 sec. / min
10 paychecks / yr.
1.00 dry cup / 0.204 Kg
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Evenson
Dimensional Analysis Practice
Directions: Show all work in legible dimensional analysis.
1. If Drew Carey makes 1.500 million dollars for every episode and each episode is
28.00 minutes long how much does he make per second?
a.) What information are you looking for?
b.) What information do you know?
2. What is the density (g/ml) of a car if it has a volume of 98.50 kL and a mass of 20.90
Mg?
3. If a child is born every six seconds in the US how many new children are born in a
year?
4. A wire has a mass of 75.90 g and a density of 45.90 kg/kL what is the volume (ml) of
this portion of wire?
5. A man can make 12 toy monkeys in an hour and he works an eight hour day, a
woman who is faster makes the same number of monkeys a day but only works 6.50
hours a day. How many monkeys can the lady make in an hour?
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Evenson
Practice with dimensional analysis and conversions of metric prefixes
1. How many grams are in 2 kilograms?
2. How many ml in 6 liters?
3. Which is a larger volume 10 ml or 10 cm3?
4. How many millimeters in one centimeter?
5. How many centimeters in one kilometer?
6. How many nanoseconds in a minute?
7. My dog weighs 2.5 pounds (very small dog) how many grams is that?
8. If a wire has a mass of 685 kg/km, what is the mass of 0.50 m in grams?
9. How many times larger is an 8 gigabyte hard drive than 8 megabyte hard drive?
10. What would the mass of a penny be measured in?
11. What would the length of a Barack Obama’s hair be measured in?
12. How many miles does a car have on it, if it is advertised as 12 da?
13. A can of pop has 355 ml, how many microliters is that?
14. In my pocket I have a ratio of 4 nickels for every 3 quarters, 16 pennies for every 3
dollar bills, 12 dollar bills for each 10 nickels, and 1 quarter for each 5 pennies. If I
have 15 nickels how many dollar bills do I have? (Hint: answers will vary depending
on the route you take with the mathematics because these are fictitious numbers--I
don't have this much money.)
Using the same ratio as in number 15, If I only have 5 pennies in my pocket how much
money do I have?
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Evenson
More Dimensional Analysis Practice
Directions: Show all work in dimensional analysis format. This will show your cancelled
units and your line of logic through the mathematics. All illegible answers will be
counted incorrect.
1. How many grams are in 2.00 Kilograms?
2. How many millimeters are in 1.00 centimeters?
3. How many Centimeters are in 1.00 Kilometers?
4. In a computer that has 64 megabytes of RAM, how many bytes of RAM is that?
5. If a wire has a mass of 685.00 Kg/Km, what is the mass in grams of 0.50 m?
6. The density of a sample of mercury is 13.7 g/ml. If the volume of that sample of
mercury is 25.00 ml, what is the mass in grams of the sample?
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Evenson
Dimensional Analysis Practice
Directions: Show all work, including the cancellation of units, on a separate sheet of paper. Any and all
illegible answers will be considered wrong.
1. How many centigrams are in 63.958 kilograms?
7
8
2. If the distance from the Earth to the Sun is 9.2 x 10 miles and the speed of light is 3.0 x 10 meter/sec.
Calculate the time that it takes for light to travel from the Sun to the Earth (1 mile = 1.609 km).
3. A farmer needs to give a 155.0 lb calf a shot of an antibiotic. The bottle is labeled as having 250.0 mg
per mL and says the dosage is 12.50 mg per pound of body weight. The syringe is calibrated in cc's —
cubic centimeters. How many cc's should the calf be given?
4. LD50 for a drug is the dose that will be lethal for 50% of the population given that amount of drug. LD50
for aspirin in rats is 1.75 grams per kilogram of body weight. How many aspirin tablets weighing 325 mg
each would the average 70-kg human have to consume to achieve this dose?
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Evenson
Dimensional Analysis Practice
Directions: Show all work, including the cancellation of units, on a separate sheet of paper. Any and all
illegible answers will be considered wrong.
1. A student was given three pieces of metal that looked like brass. Sample I had a volume of 17.3 mL
and weighed 154.3 g, sample II had a volume of 28.4 mL and weighed 547.2 g, and sample III had a
volume of 22.2 mL and weighed 186.1 g. Which sample(s) was (were) brass? (Brass: d ≈ 8 g/cm3)
2. The density of sugar is 1.59 g/mL. Calculate the volume of a teaspoon of sugar if this quantity of sugar
weighs 6.98 grams.
3. The water-company charges 23.4 ¢ for every Dekaliter of water used. Your shower has a 1.25 L/min
shower head. What does it cost in dollars to shower for 1638 seconds?
4. What will be the volume of a 25.0 gram sample of concentrated sulfuric acid if the density is 1.84
grams per cubic centimeter?
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Evenson
Density Comparison Lab
Objective: Determine a rule as to how varying densities interact with each other (what
floats on what).
Materials:
Corn syrup
Vegetable oil
Glycerin
water
plastic
steel
rubber stopper
cork
Procedure:
Using the balances determine the masses of the given objects and the volumes of the
given masses. From this information calculate what the density of each object is.
Data:
Set up a data table as follows
Material
Mass
Volume
Density
Also record your observations about the beaker set up at the front of the room that
contains all of the materials, be sure to record the positions of each material.
Question:
Come up with a rule (feel free to name it after yourself) that explains how an object will
react when it comes into contact with an object of different density. It may be helpful if
you rank the objects according to density, either increasing or decreasing.
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Evenson
Put in your two cents
Background: Over the years the United States Government has replaced some of the
copper in pennies with zinc. This was done cut costs as the price of copper began to
increase. We can use density to determine when these changes were done.
Objective: Determine the mint year that the government started to plate zinc based on
the density of the pennies.
Materials:
100-ml graduated cylinder
Pennies of various years
Balance
Procedure:
1. Record the mint date of each of your pennies into your data table. Be sure you have
pennies from 1973-1988.
2. Weigh out the mass of each individual penny and record the mass in your data table.
3. Place about 50.0 ml of water into the 100-ml graduated cylinder and add 10 pennies
to determine the average volume of a penny and record in your data table. The
increase in volume will be the volume of ten pennies (divide this increase by ten to
get the average volume).
4. Graph results from your data table.
Data:
Mint date
Mass (g)
Average volume (ml)
Density (g/ml)
Calculations:
When multiple calculations of similar calculations are done, it is normal to only show
one sample calculation. Show one sample calculation for your determination of density.
Questions:
1. What factors do you think might lead to error in your density measurements?
2. Which of these factors could not be corrected by improved technique?
3. Explain how you might determine the density of an irregularly shaped solid that is
soluble in water.
4. How would the density of an unknown sample help to identify what the sample is
made of?
5. How does the density of zinc compare to the density of copper?
6. Why were all pennies minted in 1943 made of all zinc-plated steel?
48
Evenson
Numismatist Nirvana
Background: Over the years the United States Government has replaced some of the copper in pennies
with zinc. This was done cut costs as the price of copper began to increase. We can use density to
determine when these changes were done. The objective of this exercise is not to find when the change
occurred but rather what the relative percentages of zinc and copper are in the penny. The cost of
producing a penny is about $0.0097 which includes the metal costs, fabrication, labor/overhead and
transportation. As of April 28, 2006 the zinc was selling for $3.28/Kg and Copper was selling for
$7.12/Kg. As metal prices continue to rise is possible that the value of the metal in a penny will exceed
the face value of coin. This is not uncommon and is actually the rule for gold coins such as American
Eagles which have a face value of $50, but as a 1 oz. gold piece are worth more than $750. Currently a
penny is comprised of 97.5% zinc and 2.5% copper. Copper has a density of 8.94 g/ml while Zinc has a
density of 7.140 g/ml.
Objective: Determine the percent composition of a bimetallic penny.
Procedure: Develop a laboratory procedure to determine the % copper and % zinc that composes a
penny, be sure to include in your procedure a method of determining that the penny is bimetallic. Include
the procedure in your lab manual and your lab write report.
Data:
Record all necessary data as prescribed by your procedure.
Calculations:
Show all your mathematics used to determine the objective, be sure to include a logical statement to
accompany your calculations as an explanation.
Questions:
1. Why were all pennies minted in the year 1943 constructed of zinc plated steel discs?
2. Does your procedure exploit a physical or chemical property of the penny?
3. What does the price of copper have to reach before the copper alone in a bimetallic penny has a value
greater than the face value of the coin?
4. Given your procedure what is your percent error in metal composition, be sure to state what you are
comparing.
49
Evenson
Unit 2 Objectives for Periodic Table

History of Periodic table
Dalton
Crookes
Thomson
Rutherford
Becquerel
Miliken
Chadwick
Dobereiner
Mendelev
Mosely

Family Trends
Metals and Nonmetals

Periodic Trends (periodic law)
Electronegativity
Atomic radii
Ionic radii
Electronaffinity
Atomic mass

Subatomic Calculations
Valence numbers
Oxidation numbers (ions)
Isotopes

The Mole
Molar Mass

Electron configurations
Hybrid orbitals
Electron emissions

Polarity
50
Evenson
Graphing Instructions
What follows are general rules that must be adhered to when making a formal graph.
4. Scale
The scale of the graph needs to be easily used numbers. Each square equals some commonly used
numbers (whole numbers, factors of 5, factors of 10, 0.5, 0.25. 0.1). The scale must be sequential,
starting with a low value and ending with a high value. The scale of the graph must be arranged so
that the graph takes up ¾ to the whole page. Your scale does not need to start at zero for an origin,
nor can you ever have a scale with a break point in it. Graph paper is a must, choose graph paper in
which the deviations of scale fit your need.
5. Axis
Graphs are set up on an X, Y coordinate plan. Your X axis is your independent variable and your Y
axis is your dependent variable. The independent variable is the variable that you control. How you
adjust the independent variable will determine the value of the dependent variable. For example if
you were graphing the mass of pennies according to their year, then years is your independent
variable. This is because the year of the penny will determine the mass of the penny. The mass of
the penny is dependent on the year. Both axes must be drawn using a straight edge.
6. Labels
All axes must be labeled with what is being measured and the units the measurement was taken in
parenthesis (i.e. Density (g/ml)).
4. Adding lines
Most graphs we use are XY scatter graphs (connect the dots). Whether the dots are to be
connected or a trend line (best fit) is used depends on the function of the graph. If the graph is
looking for an average or general occurrence then a trend line should be used. A trend line gets
as close to as all of the points as possible. It may or may not touch any of the data points. If you
are going to need to find data that occurs between data points then you need to connect the dots.
If your data will make a stochastic graph (many peaks and valleys) then you also want to connect
the dots. All lines should be drawn using a straight edge.
Many of our graphs will be graphs of time versus some unit of measurement. The slope of these
lines will be a rate.
7. Title
The title of graph is what the line(s) on the graph represent. The title is not a renaming of the XY axis
on the top of the graph.
8. Legend:
A legend is an indication as to what each line on the graph represents. A legend is required if you
have more than one line per graph. However a legend is not required when only one line is present.
The title will then indicate what the line represents.
51
Evenson
Elemental Data
Symbol
H
He
Li
Be
B
C
N
O
F
Ne
Na
Mg
Al
Si
P
S
Cl
Ar
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Atomic
number
1.00
2.00
3.00
4.00
5.00
6.00
7.00
8.00
9.00
10.00
11.00
12.00
13.00
14.00
15.00
16.00
17.00
18.00
19.00
20.00
21.00
22.00
23.00
24.00
25.00
26.00
27.00
28.00
29.00
30.00
31.00
32.00
33.00
34.00
35.00
36.00
37.00
38.00
39.00
40.00
41.00
42.00
43.00
44.00
45.00
46.00
Atomic
mass
1.01
4.00
6.94
9.01
10.81
12.01
14.01
16.00
19.00
20.17
22.94
24.31
26.98
28.08
30.97
32.06
35.45
39.95
39.10
40.08
44.96
47.90
50.94
52.00
54.94
55.85
58.93
58.71
63.55
65.38
69.74
72.59
74.92
78.96
79.90
83.80
85.47
87.62
88.91
91.22
92.91
95.94
98.91
101.07
102.91
106.40
Electron
affinity
72.80
59.60
0.00
26.70
153.90
7.00
141.00
328.00
36.10
52.80
0.00
42.50
133.60
72.00
200.00
349.00
0.00
48.40
0.00
18.10
7.60
50.60
64.30
0.00
15.70
63.70
112.00
118.40
0.00
28.90
119.00
78.00
195.00
390.00
0.00
46.90
0.00
29.60
41.10
86.10
71.90
53.00
101.30
109.70
53.70
Atomic
radius
74.10
31.00
145.00
112.00
158.90
70.00
75.00
73.00
71.00
38.00
190.00
145.00
118.00
110.00
98.00
100.00
90.00
71.00
220.00
180.00
160.00
176.00
171.00
166.00
140.00
140.00
152.00
149.00
145.00
142.00
136.00
125.00
114.00
103.00
94.00
88.00
265.00
219.00
212.00
155.00
198.00
190.00
183.00
178.00
173.00
169.00
52
Ionic
Radius
52.90
30.10
164.10
108.50
80.70
29.00
132.00
124.00
119.00
Electronegativity
2.20
3.49
0.98
1.81
2.04
2.55
3.19
3.65
4.19
113.00
71.00
53.00
40.00
31.00
26.00
49.00
0.93
1.31
1.61
1.90
2.19
2.58
3.16
151.00
114.00
160.00
56.00
49.50
48.50
53.00
63.00
72.00
69.00
74.00
88.00
61.00
53.00
47.50
42.00
39.00
0.82
1.00
1.36
1.54
1.63
1.66
1.55
1.83
1.88
1.91
1.90
1.65
1.81
2.01
2.18
2.55
2.96
3.00
0.82
0.95
1.22
1.33
1.60
2.16
1.90
2.20
2.28
2.20
166.00
132.00
104.00
73.00
86.00
60.00
78.50
52.00
80.50
78.00
Evenson
Symbol
Ag
Cd
In
Sn
Sb
Te
I
Xe
Atomic
number
47.00
48.00
49.00
50.00
51.00
52.00
53.00
54.00
Atomic
mass
107.87
112.41
114.82
118.69
121.75
127.60
126.90
131.30
Electron
affinity
125.60
0.00
28.90
107.30
103.20
190.20
440.00
0.00
Atomic
radius
160.00
161.00
156.00
145.00
145.00
140.00
115.00
108.00
Ionic
Radius
114.00
92.00
94.00
83.00
90.00
80.00
56.00
54.00
Electronegativity
1.93
1.69
1.78
1.96
2.05
2.10
2.66
2.60
Assignment Objectives:
You are to hand graph at least one of the five graphs and computerize at least one of the five graphs.
You do need a total of five graphs. This will allow you to gain skills in both hand scaling, computerized
graphing and graphical interpretations. For each graph you will explain the periodic trend in terms of an
increase and the reason you feel that is the correct trend based on graphical information. Do not simply
describe the graph (i.e. "the line goes up and then down"). Yes, each data point must be graphed.
You will have five (5) graphs:
Atm # vs. atm mass
Atm # vs. electron affinity
Atm # vs. atomic radius
Atm # vs. ionic radius
Atm # vs. Electronegativity
Example of a periodic trend: Elemental density INCREASES from top to bottom and left to right.
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Chemical Family Trends
Families of groups are the vertical columns on the periodic table. Each of the elements
in a group are placed according to their behaviors. All elements in the same group have
similar chemical behaviors. This is beneficial because if you know how one chemical
behaves that can be an indicator of the behaviors of other elements within the same
family. For your assigned family or group you need to find the following information.
You will be responsible for the trends that exist within all of the families. You will
present your information and other students will present theirs, in this way you are able
to get information on the behaviors of all the families. You must find information on all
of the following categories for your assigned family.
Location on periodic table
Family name
Physical Properties
Density
Color
Hardness
Chemical properties
Reactivity
Uses
Atomic structure
Valence number
Oxidation number (outer electrons / electron configuration)
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Trends in Metallic Reactivity
Objective:
Measure the reactivities of a variety of metals and classify the metals into an activity
series.
Materials:
0.05 M AgNO3
0.05 M Pb(NO3)2
steel wool
0.05 M NaCl
0.05 M KCl
magnesium strips
0.05 M ZnCl2 0.05 M MgCl2
copper strips zinc strips
Procedure:
1. Polish the metal strips to a shiny luster.
2. To a polished strip of copper metal add a drop of silver nitrate and in another area of
the strip add a drop of lead nitrate. Record positive or negative reaction.
3. To a polished strip of zinc metal add a drop of lead nitrate and in another area add a
drop of magnesium chloride and copper sulfate.
4. To a polished area of magnesium, likewise add drops of sodium chloride, zinc
chloride and potassium chloride.
5. Wipe all metal strips clean and return to the proper place.
Data:
In a data table record the metal ions and solid metals in a neat and organized tables
with a record of whether or not a reaction took place.
Questions:
1. Why is it necessary to polish the metal strips before the experiment?
2. Write balanced chemical equations for those reactions that occurred, include the
states of matter.
3. Using your experimental data, list the metals in order of increasing activity. Explain
how you arrived at this list.
4. Using your results, do you think there would be a reaction if strips of copper of zinc
were placed in solutions of potassium chloride or sodium chloride?
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Sub-Atomic Particle Calculations.
The phrase sub-atomic particle means a particle that is smaller than an atom. The
prefix meaning beneath or under, as it does in submarine. As you recall there are three
subatomic particles that are most common, proton, neutron, and electron. Nucleus,
center of the atom, is comprised of the protons and neutrons. The electrons orbit
around the nucleus. To visualize this relationship it maybe beneficial to think of the
solar system in which the sun (nucleus) has many planets (electrons) orbiting around it.
This visual picture is called the Bohr model of the atom. It is only a model, which means
it is useful in calculations but may not exist exactly as is represented by the model.
On the periodic table is all of the information needed to calculate the numbers of
protons, neutrons and electrons. The atomic number is the large whole number above
the atomic symbol. The atomic symbol is the one or two letters that serve as an
abbreviation for the element. The same atomic symbols are used world wide, so not all
of the atomic symbols start with the same letters as the element name, some are from
Latin names, Russian, Greek or so forth. The average atomic mass is the decimal
number under the atomic symbol. The average atomic mass is the mass of the
nucleus, and therefore the number of neutrons and protons. Knowing this, and a little
logical thinking it is possible to calculate all of the sub-atomic particles.
Ground State atoms
Ground state atoms are atoms that have no electrical charge. Remember that protons
have a positive charge and electrons have a negative charge. If an atom has no charge
or is at ground state, that means that the number of electrons is equal to the number of
protons. As you know the atomic number is equal to the number of protons in an atom,
therefore it is also equal to the number of electrons in a ground state atom. For
example if hydrogen is a ground state atom it would have 1 proton and 1 electron
because the atomic number is one.
Sample problems
Element
Ca
Mg
Protons (+)
20
12
electrons (-)
20
12
Now to find the number of neutrons in an atom we must remember that the average
atomic mass is the number of neutrons and the number protons. The number of
protons is designated by the atomic number. So the number of neutrons must be equal
to the average atomic mass minus the atomic number.
Neutrons = Atomic mass - atomic number
There can be no partial neutrons, so round the decimal points appropriately.
The atomic mass of calcium is 40.08 and the atomic number is 20, therefore the
number of neutrons is 20 (40.08 - 20 = 20.08 rounded to 20)
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Sample problems
Element
H
Sn
Proton (+)
1
50
Electron (-)
1
50
Neutron
0
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Ions
Ions are charged atoms, atoms in which the number of electrons are not equal to the
number of protons. Remember that electrons are negatively charged and protons are
positively charged. It is not possible to change the number of protons and keep the
same element. If you change the number of protons you change the atomic number
and therefore the element has changed. The only way to change the charge on an
atom is to manipulate the number of electrons. This can become counterintuitive
because you will be adding and subtracting negative charges. If an ion is positively
charged that means it has more protons than electrons. The converse is true as well
that if an ion has a negative charge it has more electrons than protons. The magnitude
of the charge (size, number) is the difference between the electrons and protons. For
example calcium is often Ca+2 (pronounced calcium plus two). This means that the
calcium ion has two more protons than electrons because there are two more positive
charges, and positive charges are protons. If the ion charge (oxidation state) is positive
the ion has that many more protons than electrons if the ion charge (oxidation state) is
negative it has that many more electrons than protons.
Sample problem
Ion
Zn+2
S-2
Proton (+)
30
16
electron (-)
28
18
Neutron
35
16
Zinc has 30 protons that cannot be changed, however the ion has two more protons
than electrons so it is necessary to remove two electrons from the ground state number
of 30 this gives the number of 28 electrons. Sulfur is a negative two charge this means
the ion has two more electrons than protons. Again we cannot change the number of
protons (16) so it is necessary to add two electrons (adding two negative charges) to
the 16 electrons that are already present from the ground state atom to end up with 18
total electrons on the S-2 ion.
Isotopes
As you may have noticed the atomic mass has been referred to as the average atomic
mass, this is because some atoms of the same element have different masses. Protons
and neutrons have about the same mass, neutrons being slightly heavier. Electrons
weigh 1/1846 the mass of a proton changing the number of electrons will not effect the
mass of an atom. Changing the number of protons will change the atomic number and
therefore change the element. This leaves only the neutrons to be manipulated. It is
possible to change the number of neutrons without changing the element or the charge,
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neutrons are neutrally charged. Neutrons are actually comprised of a proton and an
electron forced together so that their charges cancel, this also explains why a neutron is
slightly heavier than a proton, due to the additional mass of the electron.
When specific atoms are discussed their atomic mass is represented in the upper left
hand corner of the atomic symbol. You have undoubtedly heard of carbon-14, this is
written chemically as 14C. The fourteen in this atom is now a specific atomic mass and
is used to calculate the number of neutrons rather than the average atomic mass in the
periodic table. The atomic number of carbon is 6, the atomic mass of the isotope is 14,
so the number of neutrons is 8 (14 - 6 = 8).
Sample problems
Isotope
3
H
20
F
Protons (+)
1
9
electrons (-)
1
9
Neutrons
2
11
Charged isotopes
All of these situations can be combined into ionic isotopes, these are isotopes with
charges. All of the previous rules would still hold true. Manipulate the electrons
according to the charge and manipulate the neutrons according to the specific atomic
mass.
Sample problems
Ion / isotope
3 +
H
31 -2
S
Proton (+)
1
16
electrons (-)
0
18
Neutrons
2
15
Try these for practice:
Find the number of protons, electrons and neutrons for the following
11
B
Se
P-3
96
Mo+4
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Subatomic Calculations Practice
Directions: Fill in all of the charts according to the type of atom you are dealing with
(ground state, ion or isotope).
Ground state atoms
Element / Symbol Protons
Oxygen (O)
Neutrons
Electrons
86
Copper (Cu)
16
Ions
Ion / Symbol
Cl-
Protons
Neutrons
46
electrons
44
Ca+2
O-2
Isotopes
Isotope / Symbol
235
U
Proton
13
Neutrons
14
3
H
120
Sb
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electrons
Evenson
From Loschmidt to Avogadro and a Connection between the Macroscopic and
Microscopic.
Count Lorenzo Romano Amedeo Carlo Avogadro di Quaregna e Cerrato, Amedeo
Avogadro to his friends, was both on August 9, 1776 and died on July 9, 1856. He
was born in Turin, Italy into a family of antique nobility. While little is know of his
personal life, there is a general idea about his professional and political activities. At he
age of twenty he earned a degree in Ecclesiastical Law. Ecclesiastical Law, Canon Law,
is the legal realm that governs the Roman Catholic Church. His practice of
Eccesiastical law was short lived and he soon dedicated himself to studying both
physics and mathematics, positive philiosophy. At the age of 33 he began teaching
high school in Vercelli, Italy. Twelve years later he was granted a professorship to
teach at the University of Turin. He was only a professor there for three years as his
political leanings caused his dismissal from the department chair. Ten years later he
was asked to come back and teach at the University of Turin, and continued to teach
there for another twenty years. During this time he married Felicita Mazze and fathered
six children.
His professional and publicly visible assignments included work in statistics,
meterology, weights and measures and was a member of the Royal Superior Council on
Public Instruction. As a high school teacher at Vercelli that he determined that equal
volumes of gases at the same temperature and pressure, contain the same number of
molecules. This hypothesis was developed after Joseph Louis Gay-Lusac (whom we
will meet in a couple of months) had published his law on gas volumes in 1808.
Avogadro’s reasoning went something like this: Gasses move based on their mass, so if
a gas is light then it will move fast and if a gas is heavy it will move slower (I am
borrowing from the thoughts of Grahm). So if two different gasses are placed into
separate balloons and the balloons are the same size then the same number of gas
particles are in each balloon. The size of the balloon is determined by the particles
bouncing into the wall of the balloon. So the light gas must travel faster to posses
enough inertia to inflate the balloon, and the heavier gas although traveling slower has
the same inertia due to the higher particle mass. So if two gas volumes are the same
then the number of particles of a gas must be equal in the volume. His work on the
topic of gas particles and volumes was published on July 14, 1811.
Before the time of Avogadro, there was really no distinction given to atoms and
molecules. As we now know molecule is used to determine a compound that is
comprised of atoms while an atom is used to discuss the smallest particle of a pure
element. One of the more important contributions of Avogadro’s work was to clearly
differentiate between atoms and molecules. It was this distinction that helped his
hypothesis on gas volumes and particle relationships.
The ideas of Avogadro were not well received by Scientific Community, in part
because there were many exceptions to Avogadro’s Law. There were many examples
of equal volumes of gasses that do not create an equal number of particles in the solid
state. These discrepancies were later accounted for by Stanislao Cannizzaro as some
compounds when heated, are not the same compound, but rather thermally degrade
into multiple compounds. The creation of multiple compounds when heated creates an
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additional variable in which a conclusion cannot be determined. As the ideas of
Avogadro were becoming accepted a value for the concept had yet to be determined.
In 1856 another high school teacher, Johann Josef Loschimdt (1821-1895)
determined a constant value between the macroscopic gas volumes and the
microscopic particles of gas within that volume. Using the results of Avogadro that any
gas under the same conditions has the same number of molecules per volume,
Loschmidt calculated the number of gas molecules in a cubic centimeter of any gas.
This number has most recently (2002) been calculated at 6.0221415 x 1023. While the
value is still referred, mostly in German speaking countries, as Loschmidt’s number
6.0221415 x 1023 is generally called Avogadro’s number. By any name the value has
great importance in chemistry and physics in relating the macroscopic volumes to the
actual number of particles, as it is the particles that actually react.
So by now, if you are still reading, you saying enough with the history—what is
this number good for? Avogadro’s number tells us how many particles are in a mole.
Which means we first must discuss a mole. A mole is simple a word that means a
number, such as a dozen or a gross. If you buy a dozen eggs you know that you have
twelve eggs. The word dozen means twelve. If you have gross of paper clips then you
have a 144 paperclips, because a gross means that a dozen-dozen or 144. A mole has
the same use. If you have a mole (mol) of something then you have 6.02x10 23 of them.
So if you have a mole of eggs then you have 602,000,000,000,000,000,000,000 eggs
(yes it is a big number). For some examples on the true size of a mole and by contrast
how truly small an atom is: a mole of marbles would spread evenly across the earth to a
depth of 50 miles, a mole of rain drops would contain more than 40 times the water in all
the world’s oceans, a mole of seconds would be an amount of time equivalent to
4,000,000 times longer than the earth has existed. A mole of atoms is also equal to the
atomic mass of an element. So if you had a mole of carbon atoms you would have
twelve grams of carbon. A mole of oxygen atoms would have a mass of 15.9994 grams
and a mole of iron atoms would have a mass of 55.847 grams. Notice of course that
this means that not all atoms have the same mass, however a mole of atoms is still a
mole of atoms and therefore always represents 6.02 x 1023 atoms (or particles or
whatever is being measured).
The other importance of Avogadro’s number, in case it was too subtle above, is
grams
that the units on atomic mass are g/mol (or
), that is right the atomic mass is a
mole
conversion factor that we can use for dimensional analysis. We can now calculate the
number of atoms in a specified mass of an element. For example if you have 20.658
grams of iron then we must have 1.24 x 10 23 atoms:
6.02 x10^ 23
1molFe
20.658 g Fe x
x
= 1.24 x 1023 atoms of iron
55.847 gFe
1mol
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Other mole calculations for Practice
Directions: Using dimensional analysis, show all work to solve the following questions.
1. How many atoms are in 6.798 grams of gold?
2. What is the mass of 5.63 x 1045 atoms of silicon?
3. 7.89 grams of Calcium contains how many electrons?
4. 56.34 Kg of Sulfur will contain how many protons?
5. What is the mass of Copper that has the equal number of atoms as 25.64 grams of Silver?
6. What is the mass of 8.97 x 1012 MOLECULES of carbon dioxide (CO2)?
7. How many moles of water are present in 78.90 ml of water at 4.00oC?
8. What is the expected mass of Cesium (Cs) if there are the same number of ATOMS that are present in
523.89 grams of sodium chloride (NaCl)?
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Introduction to Avogadro and moles
The mole (mol) is to chemists what the dozen is to an egg farmer. The mole is used to
easily convey the number of particles (atoms, molecules, ions…) in a mass or volume of
a substance. One mol is equal to 6.02 x 1023 particles.
1. How many moles of carbon are in 44.05 grams of carbon?
2. How many grams are in 25.36 moles of copper?
3. How many protons are in 21.9502 moles of xenon?
4. One mole of an element has an atomic mass of 238.0 grams, what element are we
dealing with.
5. How many moles of lithium are required to have a mass of 24.00 grams
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HEY! Avogadro I've Got Your Number
Objective:
Determine the masses of various samples of elements and calculate the number of
protons, neutrons and electrons present in the sample.
Procedure:
1. Record the mass of the empty vial
2. Record the masses of all ten elements given
3. Determine the mass of just the elements
Data:
Data table will be neat and organized. The table will have three columns with the
headings of Element, Sample Mass, and Element/material Mass.
Calculations: (setting up a spreadsheet will save you time when doing repetitive
calculations)
1. For each sample calculate the number of moles that were present.
2. Then calculate the number of atoms that were present in each sample.
3. From your calculations in step two calculate the number of protons, neutrons and
electrons present in each sample.
Questions:
1. Pick the smallest number of electrons that were present in any of your samples.
Assume you can count 5 electrons per second (try to count to five in 1 second to get
an idea how fast that would be). How long would it take you to count all of the
electrons in your smallest sample? Use an appropriate unit (years, decades, and
centuries…)
2. Using the number of electrons in your smallest sample, determine how many lifetimes
it would require to count all of those electrons assuming an average lifespan of 72 years
and the ability to count 3 electrons per second.
3. Why is a mole used?
4. What would you expect the mass of 1.00 mole of water (H 2O) to be?
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Copper for your thoughts?2
Objective: Determine the radius of Copper atom, and develop an appreciation for the
size of moles and atoms.
Materials:
Calipers
Copper wire
Analytical Balances
volume of Sphere = 4/3 πr3
2
Surface area of a Sphere = 4 πr
Surface area of Cylinder = 2B +Ch
Area of Circle = π r2
volume of Cylinder = (π r2)h
Procedure:
1. Measure the length of your copper wire in millimeters.
2. Determine the diameter of your copper wire.
3. Determine the mass of your copper wire on the analytical balances
Data:
Record all data in a neat an organized data TABLE with the proper headings, labels,
units and precision for all data collected.
Calculations:
1. Based on the mass of your wire determine the number of copper atoms that must be
present in the wire.
2. Once you know the number of atoms (and assume they all take up an equal amount
of space) use the dimensions of your wire to determine the volume of an individual
atom.
3. Once you know the volume that each copper atom must occupy then determine what
the radius of an average copper atom.
4. Calculate the percent error for your data: (theoretical radius = 145 picometers)
% Error = (Theoretical radius – Actual radius) x 100
Theoretical radius
Questions:
1. Excluding human error, what is a possible source of error in your procedure?
2. How would be the best way to eliminate the procedural error?
2
This laboratory exercise was developed in collaboration with Mr. Dennis Ewert.
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Mole Day Candy
Background:
There are 6.02 x 1023 M&Ms in a mole of M&Ms. Using any, all, or none of the equipment available in the
room determine the necessary information to answer the questions below. You should be able to
calculate the depth of M&Ms that would cover the Earth if one mole of M&Ms were evenly spread over the
Earth's surface.
Objective: Gain an appreciation for the size of a mole as well as potential limitations of graphical
extrapolation.
Procedure:
Review the questions and calculations that you need to answer and devise a way to obtain all of the
needed information, be sure to record the information in your data section. Be sure to carefully read the
question section so you can complete the whole lab.
Data:
In a neat and organized data table report all of the needed information to do the necessary calculations.
Calculations:
Showing all the necessary work answer the questions being asked.
1. What is the mass of one mole of M&Ms?
2. If one mole of M&Ms were evenly distributed on the Earth's surface, how deep would it be (surface
area of Earth = 9.53184 x 108 km2)?
3. Having graphed the time it took to eat various numbers (6 trials minimum) of your candy determine
your average rate of consumption.
4. Use your graph to determine the time it will take to consume 1 mole of your candy
Questions:
1. Fully Explain how you determined your average rate of consumption from your graph.
2. How does extrapolation differ from interpolation?
3. Explain the logical errors in this type of graphical extrapolation.
4. Give a single example from current events that uses this type of illogical extrapolation as an
attempt to change popular opinion.
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Murder She Wrote – A Crime Lab Investigation Police Forensics
A woman was found dead in her apartment. There was a suicide note near by but the family
even in their grief did not believe it to be written by the women. The police had to suspect foul play.
There was some evidence that pointed to a shifty boyfriend, enough evidence to support a search
warrant. The search turned up several pens that were of the same color used on the infamous letter.
The pens from the boyfriend’s house, the pens taken from the women’s apartment and the letter were
taken to the police forensics department----you are the police forensics department! Solve the mystery.
Background:
Paper chromatography is a method of separating mixtures by using a piece of absorbent paper. In the
process, the solution to be separated is placed on a piece of dry filter paper (the stationary phase). A
solvent (the mobile phase) is allowed to travel across the paper by capillary action. As the paper soaks
up the solvent, some of the components of the mixture are carried with it. The components of the mixture
that are most soluble in the solvent and least attracted to the paper travel the farthest. The resulting
pattern of molecules is called a chromatogram. In cases where the molecules are easily visible such as
inks, this method distinguishes the components of a mixture.
Objective:
Determine who wrote the letter found beside the woman.
Procedure:
Because compounds are soluble in various solvents, you will follow the procedure once for acetone and
once for water.
1. At no time must any of the Inks come in direct contact with the solvents nor may you use any marking
tool other than a pencil.
2. Obtain a small piece of the ink used to write the letter.
3. Place a small amount of solvent (less than 1 cm) in the bottom of a 50 or 100 ml beaker.
4. Without letting the ink and solvent come in direct contact gently place the piece of filter paper into the
solvent. Over time the ink mixture will separate out.
5. Do not let the solvent work on the ink for more than 15 minutes.
6. On a piece of filter paper draw a line in pencil about 2 cm above the bottom of the filter paper. Place
dot of ink from the one of the pens collected at the crime scene on the line. Be sure to label which
pen you are using and the solvent used. Be sure to use all of the pens.
7. Place the filter paper into the beaker with the solvent again making sure that the ink and solvent do
not come in direct contact. Do not allow separation to take place for more than 15 minutes.
Data:
You should have a chromatograph of each of the six pens and the letter as separated in water and
acetone, a total of 14 chromatograms in your jab journal.
The pigment and solvent distances also need to be in your data section.
Calculations:
Show the solvent front calculations (Rf) for the matching chromatographs. Then calculate the percent
error for both the polar TLC and the non-polar TLC.
Questions:
1. Some components of ink are minimally attracted to the stationary phase and very soluble in the
solvent. Where on the filter paper should these components be located in the final chromatogram?
2. Where on the filter paper do you expect the larger molecules to be located on the final
chromatogram?
3. Explain why all of the labels indicating the solvent used and the pen number had to be written in
pencil.
4. What would have happened to the chromatogram if the process were allowed to proceed longer than
15 minutes (over night)?
5. Why was it necessary that the ink and the solvent never come in direct contact?
6. Who did It and with which pen?
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Pyro-Forensics (Flame test of metals)
Background:
As energy is added to an element the electrons are promoted to a higher energy level. When the
electrons fall back to the ground state the energy that was absorbed must be given off. The energy that
is released is called a photon. The photon is a small packet of energy and is often in the visible spectrum
for the human eye. Each material has a characteristic color and emission spectrum.
Materials:
splints
Spectra scopes
NaCl
BaCl2
Bunsen burner
50 ml beakers
KCl
SrCl2
crucible tongs
LiCl
CaCl2
CuSO4
Objective:
Determine and prepare a known chart of metals and their respective colors when excited. This chart will
then be applied to a series of unknowns.
Procedure:
1. Soak a 3-cm piece of wooden splint in a known metal chloride. If atomizers are available spray the
mist into the flame and record data. Do not waste materials.
2. Holding the splint in a crucible tongs place the splint in the flame of the Bunsen burner. Be sure to
rinse off the crucible tong with water between each test to avoid contamination.
3. In your data chart record the name of the compound (formula) the visible color and emission lines
present.
4. Repeat for all of the known compounds.
5. Determine the unknown compounds based on visible color. While this should be quantified using the
emission spectral lines, I have deemed it too dangerous. Note some of the unknowns may be
mixtures of 2 or more metal chlorides.
Data:
Make a neat, usable table of data, known and unknowns should have separate data tables. Including the
compound, visible color and spectral emissions (if SE are visible). Be very specific in naming the visible
colors, this will make it easier to identify the unknowns (brick red versus sunrise magenta).
Questions:
1. Record the unknown numbers and the name(s) of the compound(s) it contained.
2. Why do different metal chlorides give off different emission spectrum and visible light color?
3. Which metal chloride gave off light with the highest frequency of visible light?
4. How can you be sure it was the metal ion and not the chloride that was responsible for the color
variations?
5. Why do the transition metals have a wide variety of colors in the solid crystalline state?
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Electron Configurations
Directions:
Legibly write the electron configurations for all elements given. Give the long version
unless otherwise specified.
1. Give the electron configuration for Calcium
2. Give the electron configuration for iodine.
3. Give the electron configuration for Cobalt
4. Give the short version of the electron configuration of germanium.
5. Give the electron configuration of Na + and the configuration of Neon
6. What is the following element?
1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p3
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Unit 3 Objectives for Reactions (Rxn)

Balancing reactions (Stoichiometry)

Types of chemical reactions
Synthesis
Decomposition
Combustion
Single replacement
Double replacement

Indicators of Chemical Reactions
Odor change
Gas evolution
Energy change
Exothermic
Endothermic
Color change
Precipitation

Ionic and Covalent Characteristics

The Mole

Polyatomics

Physical vs. Chemical reactions

Lewis Structures

Binary inorganic nomenclature (Word equations)

VSEPR theory
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Physical or Chemical Change Exercise
Directions: First and foremost you must have a signed and returned safety sheet to
complete this exercise and goggles must be worn at all times during the lab. Indicate
whether the activity at each station is a chemical or physical change and how you know
(i.e. state change, gas release, color change...). Be sure to list all of the reasons that
may apply to the activity. You may use the data sheet below for you data table to be
handed in.
Activity
Type of Change
1. Lighting a match
2. Boiling water
3. potassium Iodide and
Lead nitrate
4. KOH in Water
5. Aluminum in
Sodium hydroxide
6. Torn paper
7. Acetic acid and sodium bicarbonate
8. Electrolysis of water
9.Combustion of methane
10. Hydrogen peroxide on liver
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General Rules for Balancing Non-nuclear Reactions
___CH4 + __O2  ___H2O + ___CO2
Above is given a “skeleton equation.” A skeleton equation is an unblanced chemical
equation. In a skeleton equation all of the formulas are written correctly and no
molecules need to be added or subtracted.
When balncing an equation, the element on the left side of the arrow (reactants) MUST
equal the number of the same element on the right side of the arrow (products).
**Do not be confused by the numbers of molecules, it is not uncommon to have 2
reactants forming 1 product.**
When balancing you can only add coefficients. Coefficients are whole numbers to the
left of any molecule. Coefficients are where the blank lines are on the skeleton
equation.
The coefficient means the face value of the coefficient times all of the atoms in the
molecule. If we put a 2 outside of CH4 we would have 2 carbons and 8 hydrogens.
The two means we have 2 molecules of CH4. The subscript (small 4 in CH4) means that
we only have 4 hydrogens per molecule.
When parenthesis are encountered, as in the case of Pb(OH) 2 it means we have:
1 Pb
2O
2H
If a molecule such as the following occurs Fe 2(SO4)3 we then have the following:
2 Fe
3 S
12 O
The subscript 3 means we have 3 polyatomic ions of SO 4-2; therfore we have 3 Sulfurs
and 12 oxygens.
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Classifications of Chemical reactions
Generally speaking in this course the chemical reactions that we are involved in will fit into one of five
categories: synthesis, decomposition, combustion, single replacement or double replacement. These
types of chemical reactions and their names are important to know as they become part of the working
dialogue of the course. Saying that a particular reactant is combusted gives information about the
reactants involved as well as what products are likely to be made. Each of these types of chemical
reactions are classified based on what they have in common, so we can discuss the reaction types in
general terms and I will also give you specific examples.
Synthesis (or composition):
Synthesis is to make, and in general if a chemical reaction is a synthesis reaction that means that two or
more small reactants are combining to make a larger product. Much the same way you would add oil,
eggs, flour to make a cake.
A+BC
simple + simple  Complex
Synthesis reactions are easy to see as they usually have only a single product being formed.
4 Fe(s) + 3 O2(g)  2Fe2O3(s)
2 H2(g) + O2(g)  2 H2O(g)
Decomposition
Decomposition is to break down. This is when a complex substance breaks down into two or more
smaller products. A decomposition reaction is the reverse of a synthesis reaction. They are generally
easy to spot as they have only a single reactant and multiple products.
A B + C
Complex  Simple + simple
Decompositions are generally easy to spot as they have only a single reactant and multiple products.
H2CO3(aq)  H2O(l) + CO2(g)
.
CuSO4 5H2O(s)  CuSO4(s) + 5H2O(l)
The second example shows the decomposition of copper (II) sulfate pentahydrate. It is all one
compound. The small (s) at the end of the molecule means that the molecule is a solid. The ‘dot”
indicates that the copper (II) sulfate has holes that are filled with water molecules. The water molecules
are trapped inside of the crystal and are not considered a liquid, nor is the compound an dissolved in
water. The waters are called waters of hydration and rest in interstices. Interstices are the areas between
the copper (II) sulfate molecules. It may help you to think of racked pool balls the areas between the balls
represent interstices. It is in these interstices where the waters are located. They can be removed,
decomposed, by the addition of heat. Removing the water molecules also changes the crystal structures
interaction with light, and would show a color change.
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Combustion:
Combustion does not mean to burn, although they two terms are often used interchangeably. When an
object is burned it is combusted, oxygen is added. There are combustion reactions in which no burning
occurs. Combustion reactions only mean that oxygen (O2) is added. Sometimes combustion reactions
are also synthesis reactions, as the examples for synthesis show, this is not always the case however.
Reactant + OXYGEN  product(s)
CH4(g) + O2(g)  CO2(g) + H2O(g)
A hydrocarbon is a molecule that is comprised of carbon, hydrogen and sometimes oxygen. If a
hydrocarbon is combusted it always produces carbon dioxide (CO2) and water (H2O). It is this way that
the reaction classification becomes helpful in communication as the example above can be fully explained
by saying “the combustion of methane.” If we know it is combustion then oxygen is being added to the
methane and because methane is a hydrocarbon then the two products must be carbon dioxide and
water.
Single Replacement (single displacement)
In a single replacement reaction an element replaces an element that is part of a compound to form new
products.
Element A + compound B  element C + Compound D
These are relatively easy to spot as there will always be an element by itself in the reactants and a
different element by itself in the products.
K(s) + H2O(l)  KOH(aq) + H2(g)
Zn(s) + HCl(aq)  ZnCl2(aq) + H2(g)
Notice that zinc is initially by itself as a reactant but it replaces the hydrogen in the HCl to become ZnCl 2.
As this happens the hydrogen is now by itself as a product, H2. Remember that hydrogen is diatomic so
when it is alone it always has the form of H2.
Double Replacement (double displacement)
These are the longest reactions as they do not contain any single elements. Double replacements
require two changes and therefore require that two compounds are reacting to form two new compounds.
Compound A + Compound B  Compound C + Compound D
MgCO3(aq) + HCl(aq)  MgCl(aq) + H2CO3(aq)
Notice above that the Magnesium replaces the hydrogen and the chlorine is replaced by the polyatomic
ion of carbonate (CO3-2). Both replacements mean that it is a double replacement.
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Balance and Classify
Directions: balance the following chemical reactions and tell what type (classification)
the reaction is.
1. H2O(l)  H2(g) + O2(g)
2. CaCO3(s)  CaO(s) + CO2(g)
3. NaCl(aq) + Pb(NO3)2(aq)  NaNO3(aq) + PbCl2(s)
4. CaCO3(s) + HCl(aq)  CaCl2(aq) + H2O(l) + CO2(g)
5. FeCl3(aq) + KOH(aq)  Fe(OH)3(aq) + KCl(aq)
6. P4O10(s) + H2O(l)  H3O+(aq) + H2PO4-(aq)
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Balancing and Classifying Practice
Directions: Indicate the type of chemical reaction represented, synthesis,
decomposition, combustion, single replacement or double replacement.
1. Zn(s) + 2 HCl(aq)  ZnCl2(aq) + H2(g)
_________________________
2. Cl2(g) + 2 KBr(aq)  2 KCl(aq) + Br2(g)
_________________________
3. Pb(OH)2(aq)  PbO(s) + H2O(g)
_________________________
4. PbCl2(aq) + Li2SO4(aq)  2 LiCl(aq) + PbSO4(s) _________________________
5. NH3(g) + HCl(aq)  NH4Cl(s)
_________________________
6. CdCO3(s)  CdO(s) + CO2(g)
_________________________
Directions: Balance each of the following chemical equations.
7.
Zn(s)
+
HCl(aq) 
ZnCl2(aq)
8.
Fe(s)
+
O2(g)

Fe2O3(s)
9.
Zn(s)
+
CrCl3(aq)
10.
C12H22O11(s) 
11.
Al(s)
+
O2(g)
12.
Al(s)
+
HCl(aq)
+

CrCl2(s)
C(s)
+
H2O(g)

Al2O3(s)

AlCl3(aq)
H2(g)
+
ZnCl2(aq)
+
H2(g)
Directions: Fill in the missing chemical symbols or molecular formulas.
13. 2 ____ + Cl 2(g)  2NaCl(s)
14. ______ + F2(g)  2HF(g)
15. 4 _____ + O2(g)  2Ag2O(s)
16. Zn(s) + H2O(g)  ZnO(s) + _____
17. SnO2(s) + 2 C(s)  2 CO(g) + _____
18. 2 NO(g) + _____ 2 NO2(g)
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Equations for Balancing and Classification
Directions: On a separate sheet of paper, use this practice equations to hone your skills on balancing
and classification of types of Chemical reactions
KBr(s) + Al(NO3)3(aq)  AlBr3(aq) + KNO3(aq)
Pb(NO2)2(aq) + KI(aq)  PbI2(aq) + KNO2(aq)
C3H6(s)
+ O2(g)  CO2(g) + H2O(g)
Cu(CN)2(aq) + Fe(SO4)2(aq)  Fe(CN)4(aq) + CuSO4(aq)
.
NiCl2 6 H2O (s)  NiCl2(s) + H2O(g)
Fe2O3(s) + CO(g)  Fe(s) + CO2(g)
KOH(aq) + CO2(g)  K2CO3(s) + H2O(l)
C7H6O3(l) + C4H6O3(g)  C9H8O4(l) + H2O(g)
NaHCO3(aq) + C6H8O7(aq)  CO2(g) + H2O(l) + Na3C6H5O7(aq)
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It Takes a Rocket Scientist
Objective: Physically control and manipulate mole ratios for quantitative results
(distance) of the chemical reaction.
Materials:
Tesla coil
MnO2
Zinc
6.0 M HCl
3% H2O2
Gas Generators
grease pens
disposable pipettes
Procedure:
1. Setup two gas generators, one for Oxygen and the other for Hydrogen.
2. Place some type of graduations on your “rocket)
3. For hydrogen add a small amount of zinc (@ 1.0 grams) and about 3-4 ml of HCl,
and cap.
4. For Oxygen add a small amount of MnO2 (@ 1.0 grams) and about 4-5 ml of H2O2,
and cap. Manganese dioxide acts as a catalyst in the decomposition of hydrogen
peroxide.
5. Fill your rocket with water and use the displacement of the two gasses to determine
ratios. Be sure to record your amounts of gas in your rocket.
6. Leave a small amount of water in the bottom on the rocket to prevent gas escape.
7. Bring your rocket to the launch pad and record your distance.
Data:
Your data table should include the amount of oxygen, the amount of hydrogen and, the
distance of flight for each trial.
Questions:
1. Write and balance the three chemical reactions that took place (oxygen production,
hydrogen production and rocket propellant)
2. Indicate what classification of chemical reactions each of the reactions were.
3. What was the activation energy for the rocket propellant reaction?
4. What ratio do you find was the best for your distance.
5. How do you know what the actual ratio should be?
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Up in Smoke? (A Critical Thinking Lab with steel wool)
Objective: Determine the chemical reaction type that is taking place when steel wool
(iron) is burnt.
Procedure:
1. Record the mass of a small, loosely balled, 1.5-gram steel wool.
2. With a Bunsen burner allow the steel wool to burn (avoid any steel wool from
dropping on the lab table.
3. Record the mass of the COOLED steel wool.
4. Indicate the difference in mass in your data table.
5. You may do up to 3 trials.
Data:
Include a neat table for the data with each heading clearly labeled as well as trial
number.
Calculations:
Calculate the percent of iron in the new compound
Questions:
1. What are the indications that a chemical reaction took place?
2. Write a balanced chemical reaction for the reaction.
3. What type of chemical reaction took place?
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Ionic or Covalent
Objective:
Determine whether common substances are ionic or covalently bonded.
Materials:
Glass slides
50-ml beakers
grease pencils
stirring rod
hot plate
spatula
conductivity tester
Procedure:
1. Use a grease pencil and divide a glass slide into quadrants, labeled A, B, C, and D.
2. Place a small amount of the substance (size of a grain of rice) on one of the divisions.
3. With all of the quadrants containing material gently heat the slide over the hot plate
until two of the materials melt.
4. Again place a small amount of the material into a 50-ml beaker and stir to test
solubility.
5. Test the conductivity of the material with a clean conductivity tester.
Data: (Table should look similar)
Substance melt dissolve
conductivity
A
B
C
D
Questions:
1. List each substance and whether it was ionic or covalent.
2. Did all of the compounds melt at the same temperature?
3. Indicate the differences in properties of ionic and covalent bonds.
4. Explain how an electrolyte allows conduction of electricity.
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Writing chemical formulas and Naming Chemical Compounds
The first step to writing chemical formulas is understanding oxidation numbers.
Oxidation numbers are the charges on ions. This number tells how many electrons
were added (negative charge) or how many electrons were lost (positive charge).
Atoms will gain or lose electrons in order to become more stable. The stability of an
atom depends on the octet rule. The octet rule states that atoms will be most stable if
their outer valence shell is filled or has eight electrons. The valance shells (energy
levels) are referring to the Bohr model of the atom, in which the electrons orbit around
the nucleus in concentric rings, much like that of a bull's-eye target. Each ring or
valence shell is able to hold a specific number of electrons. Counting out from the
nucleus the maximum number of electrons is as follows: 2, 8,8,16,16,32, and 32.
Atoms will form ions according to how many electrons are present in the outer shell as a
ground state atom. For example calcium has twenty electrons that means there is 2 in
the first shell, 8 in the next shell, 8 in the next shell and 2 in the outer shell. Having
only 2 electrons in the outer shell, according to the octet rule, is not stable. Now we
must ask ourselves would it be easier to gain 6 electrons for a total of eight electrons in
the outer shell or would it be easier to remove 2 electrons. Removing 2 electrons will
eliminate the outer shell making the next ring the outer energy shell. This shell will then
have 8 electrons. Because it is easier to move two electrons than six, calcium will lose
two electrons giving it an oxidation number of +2. A plus two oxidation number means
that there are two more protons than electrons, or two electrons have been removed.
The oxidation number is easy to determine if the valence number is known. The
valence number is the number of electrons in the outer most shell. The periodic table
is arranged to quickly determine valence numbers. Each column (moving vertically) has
elements with the same valence number. Starting with hydrogen the valence number is
one. This means that all of the elements in the first column have the valence number of
1, or an oxidation number of +1. The oxidation number will be +1 because it is easier to
lose 1 electron than it is to gain seven. As you move from left to right on the periodic
table the valence numbers increase by one for each column. Each column is identified
by a group number, that is determined by the valence number. So group 2 (starting
with beryllium) has two outer electrons. Again it is easier to lose two electrons rather
than gain six and the oxidation number of all group twos is +2. We then skip the middle
section of the periodic table, these are transition metals and will follow some special
rules. Continue counting groups with Boron (group 3, oxidation number +3), Carbon
(group 4, oxidation number +/- 4), Nitrogen (group 5, oxidation number -3), Oxygen
(group 6, oxidation number -2) fluorine (group 7, oxidation number -1) helium (group 8,
oxidation number 0). With the group four elements (carbon family) it is just as easy to
gain four electrons as it is to lose four. Therefore the oxidation numbers becomes plus
or minus (+/-) four. The oxidation numbers then begin to count back down because it
becomes easier to gain electrons (negative charges) then to lose them.
In order to write a proper chemical formula the oxidation numbers must be known
for both species involved. The positive oxidation number is the cation (cat-eye-on) and
the negative ion is the anion (an-eye-on). A chemical formula has no overall charge,
therefore the cation and anion must balance each other out. For example regular table
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salt is sodium chloride, NaCl. The sodium cation is a positive one (Na +) and the
chlorine anion is a negative one. The positive one charge cancels out the negative one.
This allows for a one-to-one ration between the ions and the chemical formula is NaCl.
Sometimes the ions don't bond in a one-to-one ratio. However the overall charges must
still balance even if more of one ion than the other is needed. For example in calcium
chloride, calcium is a group two and therefore a positive two cation (Ca +2) and the
chlorine anion is a group 7 and therefore a negative one charge (Cl -). These two ions
can not come together in a one-to-one ratio because the positive charges would not
balance out. It will take two chlorine anions to cancel out the two positive charges of the
calcium.
Ca+2
Cl-
this becomes CaCl2
The subscript two on the chlorine means that there are two chlorine ions, each with a
negative one charge for a total of negative two charge. This will now balance out the
positive two charge on the calcium.
Here are some more examples
Cation
Anion
+
K
ClMg+2
F+3
Fe
O-2
Al+3
Br-
Formula
KCl
MgF2
Fe2O3
AlBr3
Criss-Cross shortcut
There is a short cut that is possible when writing chemical formulas from the oxidation
numbers. Look at the previous example for Iron and oxygen (Fe 2O3) notice that the
oxidation number of iron was three and the oxidation number for oxygen was a two.
The magnitude (just size not charge) of the oxidation numbers have cris-crossed and
the three is now on the oxygen as a subscript and the two is on the iron as a subscript.
This will always work, even with polyatomic ions providing you use parenthesis.
Chemical formulas with polyatomic ions
Polyatomic ions are chemical species that behave as a single ion but are comprised of
multiple atoms of different elements. A list of common polyatomic ions is on your
periodic table. It is important that if a chemical formula requires more than one
polyatomic to be written correctly that you use parenthesis to indicate there is multiple
polyatomics and not just multiples of one atom. For example OH- is a polyatomic anion,
hydroxide. If it forms a compound with an iron three, Fe +3, it must look like this:
Fe(OH)3. This indicates that we have three hydroxides. If the formula were written like:
FeOH3 it would indicate that there are three hydrogens and would be incorrect.
Sample Problems:
Cation
Anion
Na+
SO4-2
formula
Na2SO4
Here parenthesis are not needed
on the SO4 because there is only one.
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NH4+
Cr+4
As-3
NO3-
(NH4)3As
Cr(NO3)4
Nomenclature (naming chemical formulas)
Once chemical formulas are correctly written then it is possible to begin naming the
compounds. As a general rule cations retain their original element name and anions
drop the last few letters and add -ide. Polyatomic ions retain their name even if they are
anionic.
Examples
Cation
Na+
Cu+2
Sn+4
NH4+
Name
sodium
Copper
Tin
Ammonium
Anion
ClO-2
S-2
NO3-
Name
chlorine becomes chloride
Oxygen become Oxide
Sulfur becomes Sulfide
Nitrate retains the polyatomic
Name
Which letters are in the anionic name are kept and which are removed before adding ide goes by phonetics (what ever sounds better) Oxide sounds better than oxygide.
For complete Nomenclature rules see the additional page titled Inorganic nomenclature.
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Inorganic Binary Nomenclature
The naming of compounds fall into one of two categories, trivial or systematic. Trivial names are for
compounds that are common place and do not require a systematic name such as water, ammonia,
hydrazine, hydrogen peroxide... The rest of the compounds are either organic (carbon containing) or
inorganic (non-carbon centered). The following is a brief discussion on the systematic naming of
inorganic compounds.
The name of an ionic compound consists of two words: the first word gives the name of the positive ion
(cation) and the second gives the name of the negative ion (anion)
1. To assign names to individual ions take the name of the metal from which they are derived (elemental
name).
+
+
+2
Na sodium
K potassium Zn
zinc
There is one complication certain transition metals form more than one ion (oxidation state) and we must
indicate which ion is present. To distinguish between oxidation states (ion numbers) the charge of the
cation (positive ion) is written inside of parentheses as a Roman numeral.
Fe+2 Iron (II) Fe+3 Iron (III) Cu+ Copper (I) Cu+2 Copper (II)
*An older system used a series of suffixes in which the lower oxidation state was –ous and the higher
oxidation state was –ic
Fe+2 Ferrous
Fe+3 Ferric
Cu+ Cuprous Cu+2 Cupric
*There are some transition metals with fixed oxidation states (Zn+2, Ag+) and there are a handful of
common metals that have multiple oxidation states that are not transition metals (Pb, Sn)
2. The second part of the name tells us what the anion (negative charge) is. Monatomic (one atom)
negative ions are named by adding the suffix –ide. To the stem of the name from which it was
derived.
O-2 oxide S-2 sulfide
Cl- chloride
Te-2 telluride
3. If either a cation or an anion is a polyatomic ion (more than 1 atom with a charge) the name of the
polyatomic ion is used with out any changes.
NO3- nitrate
NH4+ ammonium
CrO4-2 chromate Cr2O7-2 dichromate
4. If there are multiple cations, anions, or polyatomics then the Latin system of numbers is used
Mono one
Hexa six
Di
two
Hepta seven
Tri
three
octa
eight
Tetra four
nano nine
Penta five
deca ten
Fe2O3 is named either Ferric oxide or Iron (III) oxide. Both Iron (III) and ferric communicate the idea that
iron is in a plus three state. Fe2O3 , however is generally not called diron trioxide because it is ionic
(metal and a nonmetal).
Covalent nomeclature is used when two nonmetals combine or metaloids and nonmetals. With the
nonmetal bonding the use of the Latin system is preferred as in the following examples of nonmetals
combined as compounds:
CCl4 Carbon tetrachloride
CO Carbon Monoxide
NO2 Nitrogen dioxide
Organic nomenclature follows similiar albeit different naming rules that we will address near the end of
the course.
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Formulas and Names
Complete the following table, then write the names of each of the
compounds at the bottom of the page.
I-1
S-2
N-3
Na+1
NO3-1
Ca+2
Al+3
Now write the chemical formula and the correct chemical name
Chemical Formula
Chemical Name
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Ion Bags
Name
Directions: fill in the table below for your compounds that you chose from your anion and cation bags.
Cation
Cation name
Anion
Anion name Formula
Formula Name
Directions: Cut the cations and anions on the following page into two respective groups. Randomly
draw a cation (+) and record it in the data table under cation. Record the name of the cation. Randomly
draw an anion (-) and record it in the data table under anion, record the name of the anion. Write the
proper chemical formula for a compound formed from the chosen cation and anion. Write the proper
name of the compound formed. Be cautious of polyatomics’ names as well as the proper use of
parenthesis around polyatomics when used in a formula. The reading on previous pages will help to
serve as a guide to naming and formula writing.
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H+
Be++
Mn+4
Hg++
Li+
Mg++
Fe++
Hg+
Na+
Ca++
Co3+
B+3
K+
Sr++
Ni3+
Al+3
Rb+
Ba++
Cu+
C+4
Cs+
Ra++
Cu++
Si+4
Fr+
Cr++
Zn++
NH4+
N-3
O-2
S-2
H-
F-
Cl-
Br -
I-
At -
CO3-2
ClO3-
ClO-
CrO4-2
Cr2O7-2 CN-
OH-
NO3-
NO2-
MnO4-
PO4-3
SiO3-2
SO4-2
SO3-2
SCN-
S2O3-2
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Word Equations
Directions: For the following problems convert the chemical equations to a word
equation or a word equation to a chemical equation depending on what is given. Make
all answers legible.
1. Iron (III) chloride plus Aluminum yields aluminum chloride and iron.
2. Calcium carbonate decomposes to yield calcium oxide and carbon dioxide.
3. NH4Cl(s) + Ba(OH)2(s)  BaCl2(aq) + H2O(l) + NH3(g)
4. Silver (I) nitrate and sodium chloride yields sodium nitrate and silver (I) chloride.
5. KI(aq) + Pb(NO3)2(aq)  PbI2(s) + KNO3(aq)
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Lewis structures
In1916 G. N. Lewis an American physical chemist was the first to suggest the idea of a covalent bond.
Knowing the stability of the noble gases, he proposed that other atoms could acquire similar
configurations by sharing electrons. Lewis structures are based on the idea of the Octet rule. The octet
rule roughly indicates that atoms, with few exceptions, are most stable with eight electrons in the outer
most level. Lewis structures uses an atom's valence electrons, the electrons in the outer most shell
(determined by the family number), and combines these with other atoms in such a way as to leave each
atom with eight electrons in orbit.
For very simple molecules, Lewis structures can often be written by inspection. Usually, though, you will
save time and avoid confusion by following these steps:
1. Count the number of valence electrons. For a molecule, simply sum up the valence electrons of
the atoms present. For a polyatomic anion, one electron is added for each unit of negative charge.
For a polyatomic cation, a number of electrons equal to the positive charge must be subtracted.
2. Draw a skeleton structure for the species, joining atoms by single bonds (--). In some cases,
only one arrangement of atoms is possible: in others, experimental evidence must be used to decide
between two or more alternative structures.
Most of the molecules and polyatomic ions that we will deal with consist of a central atom bonded
to two or more terminal atoms, located at the outer edges of the molecule or ion. For such
species (e.g. NH4+, SO2, CCl4), it is relatively easy to derive the skeleton structure. The central
atom is usually the one written first in the formula (N in NH4+, S in SO2, C in CCl4); put this in the
center of the molecule or ion. Terminal atoms are most often Hydrogen, oxygen, or a halogen
(group 7); bond these atoms to the central atom.
Deduct two valence electrons for each single bond written in step 2 Distribute the remaining
electrons as unshared pairs so as to give each atom eight electrons.
Example and thought processes
Draw a Lewis structure for H2CO
Total valence e- = 2 + 4 + 6 = 12. Carbon is the central atom, bonded to terminal oxygen and hydrogen
atoms. The skeleton is:
H
C---O
H
Valence electrons left = 12 – 6 = 6. We might put these as unshared pairs around the oxygen atom
H
. .
C
O
H
This however leaves carbon with only six valence electrons. To supply the two electrons to complete the
octet, an unshared pair form the oxygen is used to form a double bond with carbon. The correct Lewis
structure is:
H
C==O :
H
Note that the carbon here as in almost all cases forms four bonds, in this case 1 double bond and two
single bonds.
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Lewis Structure Practice
Directions: legibly draw the Lewis dot diagrams for the molecules listed in each
question.
1.
AsCl3
2.
N2
3.
CH4O
4.
SnBr2O
5.
C2H4
6.
CO2
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VSEPR Theory (Valence Shell Electron Pair Repulsion)
VSEPR theory is used with Lewis dot diagrams to predict the molecular shape (geometry) of a molecule.
The major features of molecular geometry can be predicted on the basis of VSEPR theory. N.V. Sigwick
and H.M. Powell first suggested VESPR theory in 1940. It was developed and expanded later by R.J.
Gillespie and R. S. Nyholm. According to the VSEPR theory, the density clouds associated with electron
pairs surrounding an atom repel one another and are oriented to be as far apart as possible. This is
logical when you recall that electrons have negative charges and like charges repel, therefore any
concentrations of negative charges will be as far from each other as possible, within the bonded
molecule.
Possible geometries:
o
Bent a molecule with three atoms (AX2) that is bent in an “L” shape with 109.5 angles. The electron
density is on the A (as in H2O)
Linear (AX2) the electron density is on the X, the molecule stays straight because the X atoms go the far
ends of the molecule.
The following table shows other geometries and was borrowed from Masterton & Hurley, Chemistry:
principles and reactions, 1989
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Unit 4 Objectives for Bonding

Polarity
Separation techniques

Interparticle forces
Hydrogen
Dipole interactions

Mass Percent composition
Waters of hydration
Percent Error

Empirical and Molecular Formulas
Stoichiometry
Mass to mass calculations
Balancing reactions

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Practice problems for Mass %
If you have 560.4 grams of CO2 what is the mass of carbon?
What is the percentage of water in Ba (OH)2 * 8 H2O?
What is the mass of iron that formed rust on a car if the car weighs 4500.0
Kg and is 24.5% covered in rust (Fe2O3)?
You have 45.9 grams of Al, however all of that Al is in the form of Al 2O3.
What is the mass of Al2O3 that you must have?
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Mass to Mass Practice
Directions: Show all work in legible, linear dimensional analysis with all units clearly
canceled out and equations balanced.
1. Mg(s) + ZnSO4(aq)  MgSO4(aq) + Zn(s)
How many grams of zinc will be produced if 253.0 g of ZnSO4 is allowed to react with an
excess of magnesium?
2.
Hg(NO3)2(aq) + AgCl(aq)  AgNO3(aq) + HgCl2(s)
How many grams of silver chloride are going to be needed to produce a theoretical yield
of 987.00 grams of mercury (II) chloride?
3.
FeBr3(aq) + H2S(g)  Fe2S3(aq) + HBr(aq)
How many grams of hydrogen sulfide are needed to create 1.00 Kg of Iron (III) sulfide?
4. Na2CrO4(aq) + AlCl3(aq)  NaCl(aq) + Al2(CrO4)3(aq)
If a lab reaction occurs with 56.897 grams of sodium chromate, what is the theoretical
yield of aluminum chromate?
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What is a Life Worth? (a percent composition exercise)
Directions: First remember that my tongue is firmly in my cheek. Below are the mass
percents of the most commonly found elements in your body. Using current prices of
each element, determine what the total value of a human life is based on elemental
composition.
Oxygen
65.0 %
Carbon
18.5 %
Hydrogen
9.4 %
Nitrogen
3.4 %
Calcium
1.5 %
Phosphorus 1.0 %
Potassium
0.4 %
Sulfur
0.3 %
Sodium
0.2 %
Chlorine
0.2 %
Magnesium 0.1 %
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Rules for Empirical and Molecular Formulas
To Find Empirical Formulas
1. Convert all % to grams by assuming you have a 100.00 gram sample, or
simply replace the % with grams
2. Convert all gram masses to moles. Using atomic masses from the
periodic table. Be careful of atoms that are normally diatomic, these are
now part of a compound and are treated as singular elements (e.g.
hydrogen = 1.01 g/mol).
3. Set the smallest mole amount to one by dividing all mole amounts by the
smallest number of moles.
4. These numbers are then rounded to the nearest whole number (or 0.5
number) and are the ratio of atoms in your empirical formula.
5. Write the elements in correct order and write the ratio numbers as
subscripts. This is your empirical formula.
To Find Molecular Formulas
1. First find the empirical formula (see above)
2. Calculate the molar mass of the empirical formula.
3. The molar mass of the compound must be known to find a molecular
formula. Divide the known molar mass by the molar mass of the
empirical formula.
4. Round this number to the nearest whole number and multiply by all
of the subscripts in the empirical formula. This is your molecular
formula.
5. The molar mass should be calculated to verify that the molecular formula
is correct.
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Mass to Mass and Empirical Formula Practice
Show all work for full credit making sure that reactions are balanced. Any and all
illegible work will be counted as incorrect.
1. How many grams of Li2CO3 are formed in the reaction with 94.37 grams of carbon
dioxide gas as it combines with solid lithium hydroxide to form solid lithium
carbonate and water?
(CO2(g) + LiOH(aq)  Li2CO3(s) + H2O(l) unbalanced)
2. How many grams of silver are formed if silver(I) nitrate is combined with 46.0 grams
of copper metal in a single replacement reaction( the new copper compound will be
a copper (II)?
3. If 69.37 grams of Iron are combined with Oxygen to form Iron (II) oxide, how many
grams of Iron (II) oxide will be formed?
4. Low Serotonin levels have been correlated with depression. Serotonin is a
compound that conducts nerve impulses in the brain. Serotonin is 68.25% carbon,
6.86 % hydrogen, 15.9 %nitrogen, and 9.08 % Oxygen, with a molar mass of 176.0
g / mol, what is the molecular formula?
5. Oleic acid is a component in olive oil. It is 76.5 % carbon, 12.1 % hydrogen, 11.3%
oxygen and a molar mass of the compound is approximately 282 g / mol. What is
the empirical formula?
6. Cocaine has a molecular mass of approximately 303 g / mol. Cocaine is 67.3%
carbon, 6.9 % hydrogen, 21.1% oxygen and 4.6 % nitrogen. What is the empirical
formula of cocaine?
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Empirical Formula of a Hydrate
Materials:
Epsom salt
Bunsen burner
Balance
Crucible and cover
Objective: Determine the Empirical formula of a hydrated salt by decomposition.
Procedure:
1. Determine the mass of a clean and dry crucible and cover. Intense heating may be
needed to clean the crucible. Allow the crucible to cool before a mass is recorded.
2. Add about 3 grams of the hydrated salt to your crucible and record the mass of the
crucible and the salt. Then record the mass of the salt alone.
3. Slowly heat the sample and gradually intensify the heat. Do not allow the crucible to
become red-hot. Heat for at least 10 minutes.
4. Allow the crucible to cool and then record the mass, also calculate the mass of the
anhydrous salt.
5. If time allows and to verify that all of the water has been removed reheat the crucible
for about five minutes, allow to cool and then remass.
Data:
Data table should be a neat and organized table with proper labels, columns and units.
Include only data collected there will be room for conclusions in the conclusion section.
Calculations:
Your original salt was a hydrated magnesium sulfate crystal with a molecular mass of
246.51g/mol. Calculate the empirical formula of your sample with the waters of
hydration. Show your calculations in the calculations section.
Questions:
1. The reaction was a decomposition reaction. Write the balanced chemical reaction,
based on your data, for the experiment.
2. What is one possible error that is likely to occur with this lab? (Do not included any
type of human error, such as incorrect calculations or erroneous reading a mass)
3. Calculate your percent error and indicate what theoretical and experimental values
were compared to determine this error.
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Mass Percent of H2O in CuSO4. 5H2O
Objective: Experimentally verify the empirical formula of a hydrated ionic crystal.
Procedure:
1. Calculate (showing all work in calculations section) the mass percent of H2O in CuSO4* 5H2O.
2. In a crucible weigh out 4.00 grams, and record mass in data table.
3. Calculate the theoretical mass of water in your sample according to your mass of CuSO 4* 5H2O.
4. Gently warm the crucible over a Bunsen burner until all of the crystals are white.
5. Allow crucible to cool with a crucible cover over the top and remass and record anhydrous mass.
Data:
Make a neat and legible data table with rows and columns that includes all relevant data to reach your
objective.
Calculations:
Show all calculations for theoretical and actual mass percent of water in your hydrated crystal. Show all
calculations for your empirical formula of the hydrated crystal
Show your percent error and identify what values you compared to determine your percent error.
Questions:
1. What are some possible sources of error?
2. What type of reaction did CuSO4* 5H2O undergo?
3. Write the chemical reaction based on your actual data/conclusion?
4. What was your % error? There are many comparisons that can be made to determine a percent error.
You can use any logical comparison so long as you indicate what you are comparing. Below is the
formula to use if you want to compare the masses.
% Error = (theoretical mass - experimental mass)
theoretical mass
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Empirical Formula Determination by Combination
Material:
Bunsen Burner
Crucible and Cover
Magnesium ribbon
Balance
Ring stand
Clay triangle
Iron Ring
Crucible tongs
Objective:
Experimentally determine the empirical formula of a synthesis reaction between oxygen and magnesium.
Procedure:
1. Clean a crucible and cover and record the mass. It may be necessary to flame clean the crucible. If
flame cleaning the crucible is necessary allow the crucible to cool before massing.
2. Polish 8-10 cm of magnesium ribbon with steel wool.
3. Curl the ribbon and mass the crucible, cover and ribbon as a whole.
4. Record the combined mass.
5. Heat the crucible and magnesium ribbon in the crucible. When the oxidation of magnesium
occurs rapidly DO NOT look directly at the flame.
6. When the reaction has ceased allow the crucible to cool with the cover on and then record mass.
Data:
Data table should be well organized, labeled, and include all units.
Calculations:
Calculate the empirical formula of the product according to the data you collected.
Questions / conclusions
1. Why is it necessary to polish the magnesium before beginning the reaction?
2. What is the chemical formula according to your calculations?
3. Write a balanced chemical reaction for the synthesis reaction.
4. Based on your experimental data (mass of oxygen) you can figure out what the percent oxygen was in
your product. How does that percent compare to the percent oxygen for magnesium oxide as determined
by valance/oxidation numbers? Use the % error method for this calculation.
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Determining the Empirical Formula Based on Decomposition
Materials:
Crucible
Balance
Bunsen Burner
Ring Stand
Clay Triangle
Crucible Cover
Objective:
Determine the Empirical formula for a pure compound based on the mass of the products in a
decomposition reaction.
Procedure:
1. Clean and mass a dry crucible and cover. Record the mass in the data table. It may be necessary
for the crucible to be flame cleaned if this is necessary allow the crucible to cool before massing.
2. Place about 1 gram of the unknown pure compound into the crucible and record the mass as both a
combined mass and the mass of the compound.
3. Heat the crucible slowly, gradually increasing the heat. Maintain intense heat for 5 minutes. Allow
the sample to cool and determine the mass of the sample. Again record the combined mass and
calculate the mass of the product.
4. Error will be lowest if the steps are repeated with a separate sample.
Data:
Data should be well organized and easily read in labeled columns with proper labels and units.
Calculations:
Calculations should be a separate section in this lab where all of the calculations are neatly written for the
conclusion section.
Questions:
1. The reaction produced carbon dioxide gas and solid calcium oxide. How many moles of each were
produced?
2. What is the empirical formula of the original pure compound?
3. Write the balanced decomposition reaction for this lab; include the states of matter.
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Empirical Formula determination from laboratory Analysis
Empirical formulas can be determined by synthesis or by decomposition. We have already determined
the empirical formula of MgO by a synthesis reaction (reacting magnesium with oxygen) and we have
decomposed a hydrated crystal to determine the waters of hydration. In both cases it was important to
develop a procedure in which the percent composition of the compound could be determined. We can
also determine the empirical formula by decomposition even if the compound is not a hydrated crystal—
so long as we can determine the percent composition of the compound.
Sample data
trial 1
mass crucible
11.0787g
mass crucible and sample 12.1277 g
mass of sample
1.0490 g
mass of CaO and Crucible 11.6664 g
mass of CaO
0.5877 g
Mass of CO2
0.4613 g
We can use the sample data provided by a sample that decomposes to form Carbon dioxide gas and
calcium oxide solid. We must find the percent composition of calcium, carbon and oxygen in the original
sample before we can find the empirical formula. Remember that for many low weight inorganic
compounds the empirical and molecular formulas are the same.
Find %calcium in Original Sample
We know that the mass of Calcium oxide is 0.5877 grams. Calcium does not exist in any other product
therefore all of the calcium that was in the original mass is know in the form of calcium oxide. If we find
the mass of the calcium in the calcium oxide we will know the mass of calcium in the original sample. We
can find the mass of calcium by multiplying the mass of calcium oxide by the percent of calcium in
calcium oxide.
% Ca in CaO
Atomic mass of Ca
= 40.078 g = 0.71469 or 71.469 % Ca
Molecular mass of CaO
56.0774 g
Mass of Ca in CaO
(0.71469)(0.5877g) = 0.42002 g Ca
If the original sample had 0.42002 grams of Calcium then the original % calcium must have been
mass of Calcium = 0.42002 g = 0.4004 or 40.04 % Ca
Mass of sample
1.0490 g
Mass of Oxygen in CaO
If the mass of Ca in CaO is 0.42002 g then the rest of the mass must be Oxygen
0.5877g – 0.42002 g = 0.16768 g (this would also be 28.531 % O in CaO)
Find % Carbon in original Sample
Carbon only exist as a product in a single compound, carbon dioxide. We can use the same thought
process to find the percent carbon in the original compound as we did for calcium. First we must find out
what mass of carbon is in the carbon dioxide. Because carbon dioxide is a gas it will represent the mass
that was “lost” by the sample.
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% Carbon in CO2
mass of Carbon = 12.011 g = 0.2729 (or 27.29% Carbon)
mass of CO2
44.0098g
Mass of Carbon in CO2
(0.2729)(0.4613 g) = 0.1259 g Carbon in CO2
The original sample must have also had 0.1259 grams of Carbon so the % of Carbon in the original
sample would then be:
Mass of Carbon = 0.1259 g = 0.1200 (or 12.00 % Carbon)
Mass of Sample 1.0490 g
Mass of Oxygen in CO2
If the mass of Carbon in CO2 was 0.1259 grams and all of the CO2 had a mass of 0.4613 grams the mass
of Oxygen in CO2 must have been:
0.4613 g – 0.1259 g = 0.3354 grams of Oxygen
% of Oxygen in original sample
We saved oxygen composition for last because it exists in both products. The mass of oxygen from both
products must be added together and then it can be determined what % oxygen was in the original
sample.
Mass of oxygen from CaO = 0.1677 g
+
Mass of Oxygen from CO2 = 0.3354 g
Mass of Oxygen in the original sample = 0.5031 g
There was 0.5031 grams of oxygen in the original sample, therefore oxygen was:
mass of oxygen = 0.5031 g = 0.4796 (or 47.96 % Oxygen )
mass of sample 1.0490 g
Empirical formula determination
We now know that the original sample had the following composition:
40.04 % Calcium
12.00 % Carbon
47.96 % Oxygen
Using these values it is now possible to determine the empirical formula of the original sample.
40.04 % Ca --> 40.04 g Ca x 1 mole Ca --> 0.9991 mole Ca = 1 mole Ca
40.078 g Ca 0.9991
12.00 % C --> 12.00 g C x 1 mole C --> 1.00 mole C = 1 mole C
12.00 g C
0.9991
47.96% O --> 47.96 g O x 1 mole O --> 2.998 mole O = 3 mole O
15.9994 g O 0.9991
Empirical formula must be: CaCO3
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Unit 5 Objectives for Limiting Reactants (reagents)

Limiting Reactant calculations
Mass to mole
Mole to mole
Mole to mass
Mass to mass
Volume to mass
Volume to mole
Volume to volume

Concentrations
Molarity
Molality
Normality

Standardization

Molar volume of gas

Percent Yield
Percent error
 Data Manipulation
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Limiting Reactants (or Reagents)
Limiting reactants are also referred to as limiting reagents. Reagents and reactants are
synonyms and therefore mean the same thing. They refer to the part (usually the left side) of a chemical
reaction that are used to make the reaction occur. The same way the right side of a chemical reaction
shows what is produced. Ammonia and oxygen represent the reactants while nitrogen monoxide and
water represent the products
4 NH3(g)
+
(Reactants)
5 O2(g) 
4 NO(g) +
6 H2O(g)
(Products)
(eq. 1)
All of the stoichiometric problems that we have done so far have had at least one reactant in
excess, meaning that there was more of that reactant than what was need to complete the reaction. The
stoichiometric problems (mass to mass) that we have done so far have only been concerned with the
mass of one reactant. In most laboratory situations this is not the case as definite amounts of two or
more reactants are known and added to the reaction vessel.
In chemical reactions, reactants collide and form new products. If there is no reactant to collide
then the chemical reaction does occur. This bit of logic must seem obvious to you. When ever the first
reactant to run out, does run out then the reaction stops.
For example, if I were to give the students in the class a candy sucker I would presumable make
those students happy:
Student + candy  happy student
(a synthesis reaction, balanced at 1:1:1)
Let us assume that there are 25 students in the class and that I have 14 candy suckers. Then in
chemistry 25 + 14 =14, because once I run out of suckers my reaction stops the amount of product made
is not determined by both reactants but only by the reactant the run out first.
Student + candy  happy student
(25)
(14)
(14)
It becomes irrelevant how many students I have, if I only have 14 suckers I can only produce 14 happy
students. The concept is the same in chemistry. Once the limiting reactant is used up the reaction stops
and the amount of product is then determined by the limiting reactant. If the amount of product is
determined by the limiting reactant then we must be able to determine which reactant is the limiting
reactant. The calculations are the same type of mass to mass calculations that you have been doing.
Example 1:
How many grams of aluminum oxide are theoretically formed from in a single replacement reaction in
which 45.00 grams of iron (III) oxide reacts with 26.00 grams of aluminum metal?
Follow all of the steps that you have learned so far, balance, analyze, brainstorm…
Fe2O3(s)
(45.00g)
+
2 Al(s) 
(26.00g)
2 Fe(l)
+
Al2O3(s)
(?)
(eq.2)
First step:
It is imperative that you find the limiting reactant before you do any other calculations. All of the
calculations revolve around the limiting reactant (LR) because once the LR is used up you can not make
any more product, regardless of how much you have. You have a choice when finding the limiting
reactant. This calculation is a straight mass to mass problem, simply pick one of the reactants (it does
not matter which one) and then solve for the units of the other reactant. If you pick the 45.00 grams of
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Fe2O3 then you need to determine how many grams of aluminum you will need to react with that amount
of iron (III) oxide.
45.00 grams Fe2O3 x
1 mol Fe2O3 x 2 mol Al
x
55.847 g Fe2O3 1 mol Fe2O3
26.982 g Al = 43.48 g Al
1 mol Al
It is very important that you are able to interpret what the answer to the problem (43.48 g Al)
means: It means that if all of your iron (III) oxide is used up you have to have at least 43.48 grams of
aluminum. According to the problem you only have 26.00 grams of aluminum, there is not enough
aluminum in to use all of the iron (III) oxide and therefore the aluminum will run out before the iron (III)
oxide. You can use this template sentence to help in your interpretation: I need______ and I have_____
is that enough? (I need 43.48 grams of aluminum and I have 26.00 grams of aluminum, is that enough?)
The aluminum is the limiting reactant for this scenario. We can confirm this by calculation how much iron
(III) oxide we would need to for the amount of aluminum given.
26.00 g Al
x
1 mol Al
26.982 g Al
x 1 mol Fe2O3 x 55.847 g Fe2O3 = 26.91 g Fe2O3
2 mol Al
1 mole Fe2O3
Be cautious in your interpretation. This calculation means that if all of the aluminum is used up
the reaction needs 26.91 grams of iron (III) oxide. The reaction scenario was that we had 45.00 grams of
iron (III) oxide and therefore we have more iron (III) oxide than we need. We have 45.00 grams and only
need 26.91 grams. The iron (III) oxide is in excess and the aluminum is the limiting reactant.
Consequently we could also determine how much extra iron (III) oxide we have be subtracting the amount
we have (45.00 grams) from the amount we need (26.91 grams) to determine how much iron (III) oxide
will remain, be in excess, when the reaction is completed.
Alas, the fun continues. As you have noticed this completes only step one and has not answered our
question. The question we are working on is how much aluminum oxide will be formed (eq. 2).
Step 2:
Once the limiting reactant has been determined then all calculations must be done with the original
amount of that reactant that is given in the scenario (laboratory of theoretical). Once the limiting reactant
is gone the reaction will end and no more product will be formed. So the amount of the reactant is excess
is virtually irrelevant to product formation. We had 26.00 grams of aluminum in the initial scenario.
Caution do not get carried away and use the amount of aluminum that you calculated (43.48 grams) you
do not have that many grams of aluminum you only have 26.00 grams of aluminum according to the
example scenario (ex.1). This again is just a straight mass to mass problem. You have 26.00 grams of
aluminum and are looking for the theoretical mass of aluminum oxide that will be formed as product.
Fe2O3(s)
+
2 Al(s) 
2 Fe(l)
+
26.00 g Al
x
1 mol Al
26.982 g Al
x 1 mol Al2O3
2 mol Al
x 101.961g Al2O3 = 49.13 g Al2O3
1 mole Al2O3
We will theoretically produce 49.13 grams of aluminum oxide.
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Al2O3(s)
(eq.2)
Evenson
Limiting Reaction Steps
1. Balance the reaction
2. Find the Limiting reactant
--Pick a reactant and solve for the units of the other reactant
--Be cautious in your interpretations of you calculations
3. Use the limiting reactant amount that you actually have to determine your product amount.
The concept of limiting reactions (and the steps to calculate) will not change: one reactant will be used up
before another. Except in very carefully controlled situations in which very precise measurement has
been done so that both reactants are very close to mutually limiting (run out at almost the same time).
Although the concepts will not change the complexity of the calculations can increase depending on the
units of the reactants and the states of matter of the reactants (l, g, aq). We will postpone treatment of
aqueous solutions for a while but we will review the treatment of reactants (and products) in the gas
stage. It will also be a good time to reinforce the idea that you cannot determine the limiting reactants
from their initial amounts.
Ex. 2:
What volume of steam (gaseous water) will be created if 45.00 ml of ammonia are reacted with 45.00 ml
of oxygen?
4 NH3(g)
(45.00 ml)
+
5 O2(g) 
(45.00 ml)
4 NO(g) +
6 H2O(g)
(?)
(eq. 1)
Step 1: Find the Limiting Reactant (pick a reactant and solve for the units of the other reactant)
As we are dealing a gas we must recall that any 1 mole of any gas at STP (standard temperature and
pressure) will occupy 22.4 liters of space.
45.00 ml NH3
x 1 L NH3 x
1mol NH3 x 5 mol O2 x 22.4 L O2 x 1000 ml O2 = 56.25 ml O2
1000 ml NH3 22.4 L NH3 4 mol NH3 1 mol O2 1 L O2
I need 56.25 ml of Oxygen and I have 45.00 ml, therefore I do not have enough oxygen to use all of my
ammonia and oxygen is my limiting reactant and ammonia is in excess.
Step 2: Use the Limiting reactant and determine the amount of product(s) made.
45.00 ml O2 x 1 L O2 x 1 mol O2 x 6 mol H2O x 22.4 L H2O x 1000 ml H2O = 54.0 ml H2O
1000 ml O2 22.4 L O2
5 mol O2
1 mol H2O
1 L H2O
Student Practice: Try this problem and compare your answer to the ones given.
29.75 grams of sodium reacts with 138.00 milliliters of liquid water* in a single replacement reaction that
produces what volume (ml) of hydrogen?
2 Na(s) + 2 H2O(l)  2 NaOH(aq) + H2(g)
Answers:
Limiting Reactant is sodium
Water is in excess by 114.68 ml
14.49 ml of hydrogen gas is produced
*remember that if you have liquid water (H2O(l)) then 1 gram = 1 ml
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Using Limiting Reactants Theory for Standardizations
The concept of limiting reactants is often exploited in such a way as that both reactants are consumed at
the same time. If both reactants are consumed at the same time then it is possible to determine their
absolute amounts based on their stoichiometric ratios. This process is called a standardization reaction.
Standardization reactions are used for all aqueous solutions in which a precise concentration is needed.
The best standardization reactions involve a solid compound, as using a solid eliminates potential errors
because it only requires a precise mass. We can use the standardization of hydrochloric acid as an
example.
Reacting hydrochloric acid and calcium carbonate yields, calcium chloride, water and carbon dioxide.
CaCO3(s) + 2HCl(aq) → H2O9l0 + CaCl2(aq) + CO2(g)
A known mass of anhydrous calcium carbonate is titrated with an unknown concentration of hydrochloric
acid. The reaction generates carbon dioxide gas as an effervescent. The titration continues until the
effervescence stops. When gas is no longer being generated the calcium carbonate and the hydrochloric
acid have reached the stoichiometric point (1 CaCO3 : 2 HCl). Dimensional analysis can now be used to
determine the molarity of the unknown hydrochloric acid because the following information is known:
mass of Calcium carbonate, volume of HCl.
Lets assume the following data was collected:
Mass of CaCO3
Volume of HCl
4.000g
78.00 ml
4.000g CaCO3 x 1 mole CaCO3
x
100.087 g CaCO3
2 mole HCl
=
1 mole CaCO3
0.07993 moles HCl
0.7993 moles where required to consume all (4.000g) of the calcium carbonate and it took 28.00 ml to
possess 0.7993 moles of HCl therefore:
0.7993 moles HCl x 1000 ml = 10.24 Mole HCl = 10.24 M HCl
78.00 ml
1L
L
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Limiting Reactant Practice
Determine which is the limiting reagent show all calculations to support your answer. If
all species will be totally consumed indicate this as well. Be sure all reactions are
balanced.
1. 2Pb(s) + O2(g) 2PbO(s)
If 2 moles of O2 are reacted with 2 moles of lead, which is the limiting reactant?
2. C3H8(g) + 5 O2(g)  3CO2(g) + 4 H2O(l)
If you begin with 67.00 grams of oxygen and 200.79 grams of propane which is the
limiting reactant?
3. 14.00 ml of pure water is reacted with 28.75 grams of solid sodium to form hydrogen
gas and aqueous sodium hydroxide. What mass of gas is formed?
4. We have 256.0 grams of lead (II) chloride combined with 432.7 grams of Zinc (II)
hydroxide in a double replacement reaction. Which is the limiting reagent?
5. A double replacement reaction takes place between 365.7 grams of silver (I) nitrate
and 74.3 grams of potassium iodide, What mass of AgI is produced?
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Limiting Reactant Problems
Directions: Answer the following questions on a separate piece of paper showing ALL
work including reactions, any and all illegible work will be marked incorrect.
Solid sodium reacts with water to form sodium hydroxide and hydrogen gas
according to the equation:
2Na(s) + 2 H2O(l) 2NaOH(aq) + H2(g)
1. If 90.0 g of sodium were dropped into 80.0 g of water, how many liters of hydrogen
at STP would be produced?
2. Which reactant is in excess and how much is left over?
Solid phosphorus burns in oxygen gas to produce phosphorus (V) Oxide (P 4O10).
3. If 2.50 g of phosphorus is ignited in a flask containing 750 ml of oxygen at STP, how
many grams of P4O10 is formed?
4. Which reactant is in excess and how much is left over?
Solid magnesium burns in oxygen gas to produce magnesium oxide.
5. If 1.00 g of magnesium is ignited in a flask containing 0.500 L of oxygen at STP, how
many grams of magnesium oxide are produced?
6. What is the name and amount of the reactant in excess?
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Heterogeneous Treatment of Limiting Reactant Scenarios
1.
When 5.67 grams of Hydrogen bromide reacts with 28.790 grams of anhydrous barium hydroxide
in a double replacement reaction; how many grams of barium bromide are expected?
2.
In a reaction 0.98 liters of Carbon dioxide reacts with an 1.43 grams of sodium to create carbon
and what mass of sodium carbonate?
3.
Zn(s) + H3PO4(aq)  H2(g) + Zn3(PO4)2(aq)
45.802 ml of 5.67 M phosphoric acid reacts with 4.85 grams of zinc to produce what volume of
hydrogen gas?
4.
165.320 grams of pentane when reacted with an excess of oxygen gas will theoretically produce
what mass of carbon dioxide?
5.
25.43 ml of 5.3 M Lead (II) nitrate reacts with 30.50 ml of 2.65 M sodium chromate to produce
sodium nitrate and what mass of lead (II) chromate?
6.
What is the expected concentration of sodium hydroxide formed if 0.453 grams of sodium is
combined with 55.6 ml of pure water?
Na(s) + H2O(l)  H2(g) + NaOH(aq)
7.
Pentane has a density of 0.6262 g/ml. If 67.89 ml of liquid pentane is combusted with only 5.00
Liters of oxygen available what is the expected volume of steam created?
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Additional Limiting reactant Practice Problems
28.57 ml of gaseous Hydrogen bromide reacts with 15.69 ml of 5.64 M barium hydroxide to produce water and barium bromide.
What is the expected volume of liquid water produced?
65.98 ml Carbon dioxide reacts with 5.89 grams of potassium at STP, producing potassium carbonate and carbon. What mass of
carbon is expected?
9.687 g of Calcium hydride reacts with 125.36 ml of steam to yield calcium hydroxide and hydrogen gas. If the resultant aqoueous
solution of calcium hydroxide is heated to dryness, what is the theoretical mass of Ca(OH) 2?
98.65 ml or 5.638 M Zinc (II) sulfide will react with 165.89 ml oxygen gas to for zinc (II) oxide and sulfur dioxide. How much excess
reactant must be treated for disposal?
65.63 grams of solid Carbon dioxide reacts with 9.63 g sodium producing sodium carbonate and carbon. What is the volume of
gaseous carbon dioxide that will sublime into the laboratory?
569.45 ml or 2.35 M Lead (II) nitrate reacts with 456.32 g solid sodium chromate to produce how many moles of lead (II) chrom ate?
The reaction also produces sodium nitrate. If the product is produced in 95.63 ml of pure water, then what is the concentration of
lead (II) nitrate created?
AgNO3(aq) + CaCl2(aq)  AgCl(s) + Ca(NO3)2(aq)
If the products are formed in 580.00 ml of water, what is the expected concentration of calcium nitrate produced when 36.25 m l of
9.60 M calcium chloride reacts with 75.36 ml of 0.359 M of silver (I) nitrate?
H2C2O4(g) + KMnO4(s)  H2O(g) + CO2(g) + MnO2(s) + KOH(s)
Can a reaction between 69.35 ml of H 2C2O4 and 5.36 grams of potassium permanganate be safely reacted in a sealed beaker that
has a total volume of 890.00ml?
C2H4(g) + O2(g)  H2O(g) + CO2(g)
Can a reaction between 869.45 ml of C 2H4 and 55.36 ml of oxygen be safely reacted in a sealed beaker that has a total volume of
590.00ml?
C8H18(l) + O2 (g)  H2O(g) + CO2(g)
What is the expected volume of condensed water formed from the reaction between 962.30 ml of oxygen and 562.045 ml of octane?
Zn(s) + H3PO4(aq)  H2(g) + Zn3(PO4)2(aq)
Assuming the final reaction vessel contains 650.00 ml of solution what is the expected concentration of zinc (II) phosphate w hen
12.30 grams of zinc is reacted with 86.63 ml of 11.5 M phosphoric acid?
Mg3N2(S) + H2O(l)  NH3(g) + Mg(OH)2(aq)
If exactly twelve grams of water reacts with fifteen and two tenths grams of magnesium nitride what is the expected volume of
ammonia produced?
C5H12(l) + O2(g)  H2O(g) + CO2(g)
What mass of oxygen gas was required to produce 56.34 ml of carbon dioxide, if the oxygen was reacted with 89.65 ml of pentane?
Fe2S3(s) + O2(g)  Fe2O3(s) + SO2(g)
65.38 grams of iron (III) sulfide was reacted to form 98.36 ml of sulfur dioxide. What volume of oxygen was required for this
reaction?
N2O(g) + NH3(g)  N2(g) + H2O(g)
89.00 ml of ammonia are reacted with 156.00 ml of nitrous oxide. How much of the excess reactant will remain after the reaction?
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Example Limiting Reactant problem for heterogeneous reactions.
What volume of hydrogen gas is created in a single replacement reaction between 3.78
grams of zinc and 17.95 ml of 2.3 M HCl?
Write a proper/balanced reaction:
2 HCl(aq) + Zn(s)  ZnCl2(aq) + H2(g)
First find Limiting Reagent:
1lHCl
2.3molHCl 1molZn 65.39 gZn
x
x
x
 1.4 gZn
17.95mlHCl x
1000mlHCl
1LHCl
2molHCl 1molZn
Or
3.78gZn
1molZn 2molHCl
1LHCl
1000mlHCl
x
x
x
 50.3mlHCl
65.39 gZn 1molZn 2.3MolHCl
1LHCl
HCl is the limiting reactant
Find Hydrogen Gas Produced:
17.95mlHCl x
1lHCl
2.3molHCl 1molH 2 22.4LH 2 1000mlH 2
x
x
x
x
 462 ml H2
1000mlHCl
1LHCl
2molHCl 1molH 2
1LH 2
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Standardization Lab
Objective:
Determine the concentration of an unknown acid to 3 significant figures.
Materials:
HCl
NaHCO3
Burette
Procedure:
1. Record the precise mass of 2.00 grams of sodium bicarbonate and place in a 250-ml
beaker.
2. Properly fill the Burrett as instructed using the cohesive and adhesive properties of
water.
3. Titrate the HCl into the evaporating dish with constant swirling
4. When the reaction ceases record the final volume of HCl with proper precision
Data:
Organize all data in a neat and concise data table with the proper headings and labels.
Include all volumes and masses required to meet the objective. Record your Data on
the computer as instructed.
Calculations:
With a balanced chemical reaction determine the molarity of the hydrochloric acid for
your trial
Questions:
1.
2.
3.
4.
Write a balanced chemical reaction for the reaction that took place.
What is the Molarity of your acid based on your calculations?
When is standardization necessary?
Explain in your own words how or why this method (calculations) works.
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Limiting Reagent Lab
NaHCO3(s) + HCl(aq)  NaCl(aq) + CO2(g) + H2O(l)
Materials:
Bunsen burner
Watch glass
NaHCO3
Wire gauze
Ring stand
6M HCl
Evaporating dish
Iron ring
Objective:
Experimentally prove what is mathematically shown for a limiting reagent, in a
neutralization reaction between sodium carbonate and hydrochloric acid.
Procedure:
1. Calculate which reactant is the limiting reactant if 1.50 grams of NaHCO 3 is
combined with 10.0 ml of 6.0 M HCl (equivalent to 2.2 grams of HCl). Show these
calculations in a separate calculation section.
2. Clean and mass an evaporating dish and watch glass. Record the data in the data
table.
3. Add 1.50 grams of sodium bicarbonate to the evaporating dish and record both the
combined mass and the mass of the sodium bicarbonate.
4. Add 10.0 ml of HCl acid to the evaporating dish, 1-ml at a time. Record the volume
and calculate the mass of HCl added.
Your volume of HCl x (6.0 M HCl / 1 L) x (1 L/ 1000ml) x (36.46 g HCl / 1 mol
HCl) = grams of HCl
5. After all 10.0 ml of HCl have been added, add one more drop and record
observations under your data table
6. Gently heat the evaporating dish and cover with watch glass.
Avoid spattering of the solid.
7. Allow evaporating dish and watch glass to cool, record combined mass. And
calculate the mass of the solid.
Data:
Data table should be a neat and organized table with proper columns, labels and units
displayed. All other non-numerical data (i.e. observations) should be placed under the
data table.
Calculations:
This lab will have a separate calculation section, which will include all of the calculations
asked for in the conclusion section.
Questions and Conclusions
1. What was proved by the addition of one more drop of HCl?
2. What indicated a chemical change?
3. Record the mass Calculated and found experimentally for both NaCl and Water
(label which is which)
4. Determine the % error for both water created and sodium chloride produced.
% Error = (Theoretical mass - Experimental mass ) x 100
Theoretical mass
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Limiting Reagent Lab3
Objective:
Determine which of two unknown reagents is the limiting reagent for a given ratio. From
this information it is also possible to determine the balanced equation in general terms.
Materials:
1 M NaOH (reagent A)
1 M Ca(NO3)2 (reagent B)
8 centrifuge tubes
Centrifuge
Eye dropper
Procedure:
1. Add 5 ml of reagent A to each of the eight test tubes/
2. Add reagent B as follows:
To test tubes 1, 5 add 1 ml
To test tubes 2, 6 add 2 ml
To test tubes 3, 7 add 3 ml
To test tubes 4 add 4 ml
3. Centrifuge the test tubes
4. Add one drop of reagent A to test tubes 1-4. Record results.
5. Add one drop of reagent B to test tubes 5-8. Record results.
Data:
In a neat and orderly table record all volumes and observations
Questions:
1. What was the limiting reactant in test tubes 1-4 and in 5-8?
2. If reagent A is Ax and reagent B is By, write a balanced equation for this double
replacement reaction (ml ratios will guide you).
3
Developed in collaboration with Rodney Johnson
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Limiting Reactant and Salt Production
HCl(aq) + NaOH(s)  H2O(l) + NaCl(aq)
Background
Hydrochloric acid and sodium hydroxide are a strong acid and a strong base, respectively. HCl is an
aqueous solution of hydrogen and chlorine gas atoms dissolved in water. NaOH is found in a pure solid.
Do not touch the sodium hydroxide with your skin it is corrosive and is deliquescent meaning it will absorb
water from both the air and from you skin to make concentrated NaOH. When a strong acid and a strong
base combine they form a salt and water. The pH of pure water is 7 and the added effect of the salt
should not affect the pH at a noticeable level. If all calculations are done correctly you should be able to
mix the exact amounts of both HCl and NaOH to form a neutral salt solution, where neither the HCl nor
the NaOH was the limiting reagent but both were completely reacted.
Objective
To calculate and then prove your calculations that neither the sodium hydroxide nor the hydrochloric acid
were the limiting reagent.
Procedure
1. Record the mass of a clean and dry evaporating dish.
2. In the evaporating dish record the mass of four pellets of NaOH.
3. Calculate the number of moles of HCl required to completely react all of the NaOH. Use the balance
equation given above.
4. The concentration of the HCl tells use how many moles are in one liter. In step three you calculated
how many moles you need now determine how many milliliters are necessary to obtain that many
moles of HCl.
5. Slowly add your calculated amount of HCl to the NaOH pellets with a burett. There will be gas
evolution and heat given off in this reaction. Do not add the HCl too fast or the reaction will overflow.
6. Once all of the HCl has been added stir the solution carefully and check the pH with litmus paper (the
litmus paper will be taped into your lab report).
7. Gently heat over a Bunsen burner (on a ring stand) the solution to drive off the water. Cover the
evaporating dish with a watch glass. Record the mass of the dry product. If the solution is not heated
gently the salt crystals will shatter and spatter out of the evaporation dish.
Data
Data table should include all of the following as well as any other information that you think may be
important.
Mass of NaOH
Volume of HCl
Mass of empty evaporation dish
Mass of Watch glass
Mass of evaporation dish and dried salt
Mass of salt
pH of solution before heating
concentration of HCl
Calculations
Show all calculations for determining the volume of HCl required to completely neutralize your amount of
NaOH.
Questions
1. Based on your mass of sodium hydroxide calculate the mass of NaCl that theoretically should have
been produced.
2. Based on your volume of HCl calculate the mass of NaCl that theoretically should have been
produced.
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3. If the pH range of a base is 7-14 and the pH range of an acid is 0-7, how is it possible to mix the two
and end up with a pH of near seven (answer this question using the balanced chemical equation on
the front page and the information given in the background section)
4. Determine the % error for the mass of NaCl produced based on both the volume of HCl (#2) and the
mass of NaOH (#1) used in your reaction.
% Error = (theoretical mass – experimental mass)
Theoretical mass
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Ah, Blow It Out Your Balloon
Objective: Determine the molar volume of a gas
Materials:
Balloons
Dry ice
Water displacement apparatus
Balance
Procedure:
1. Record the mass of an empty balloon.
2. Obtain a small piece (0.500 grams) of dry ice and record the mass. Dry ice will
sublime and any lost gas will affect your percent error.
3. The mass of dry ice needs to be quickly and carefully put into the balloon. Use a
forceps to transfer the dry ice. Any contact with flesh will cause severe frostbite.
4. Allow the balloon to inflate but not burst. A popped balloon will require you to start
over.
5. Determine the volume of the balloon using water displacement
Data:
Record mass of CO2(s) and volume of balloon, mass of balloon and mass of balloon
filled with gas. Also include the barometric pressure and the temperature. Before you
leave class be sure to enter your mass of CO2 and the volume of CO2 collected.
Calculations:
1. Use the course data posted online and create a graph of all the data, remember
that the objective is to find the molar volume of a gas.
2. From the equation of your line calculate the molar volume of Carbon dioxide in
(liters per mole).
Questions:
1. Should the mass of the gas and the solid be the same?
2. From you graph determine the equation of the line be sure to include the proper
units for the line.
3. Why is it more logical to calculate the molar volume of a gas rather than a mass to
volume ratio?
4. The accepted value is 22.4 L per mole, what is your percent error? You can
compare the mass of the gas, volume of the gas, or the mass of the solid.
(Theoretical value - experimental value) X 100 = % error
Theoretical value
5. What are two possible sources of errors for why your molar volume is not exactly
22.4 L/mol?
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Unit 6 Objectives for Kinetic Theory of Matter

Kinetic Energy relationship to temperature

Kinetic theory
Solids
Liquids
Gasses
Plasmas

Grahm's Law
Calculations
Application
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Graham’s Law Gas Diffusion Lab
Background
Diffusion is the process in which particles in a system move from an area of high concentration to
an area of low concentration until a uniform concentration of particles is reached throughout the system. It
has been shown that the rate at which gas molecules diffuse, at constant temperature, decreases as the
molecular mass of the gas increases. In fact, the rate of gas diffusion at constant temperature is inversely
proportional to the square root of the molecular mass of the gas. This proportionality is called Graham's
law of diffusion.
Rate of diffusion  1/  molecular mass
The relation of molecular mass to rate of diffusion can be understood by considering the kinetic
molecular theory as it applies to gases. According to this theory, two different gases at the same
temperature have the same average kinetic energy. But in order to have the same average kinetic
energy while having different molecular masses, the two gases must also have different average
velocities. The formula for average kinetic energy is given below:
2
Average kinetic energy = ½ mass x (average velocity)
For two gases, A and B at the same temperature the kinetic theory states that:
The average kinetic energy of gas A = average kinetic energy of gas B
By combining the two equations above, the following equation is obtained
½ mass of gas A x (average velocity of gas A)2 = ½ mass of gas B x (average velocity gas B)2
This equation can then be simplified to the following equality:
Mass gas A = (average velocity of gas B)2
mass gas B
( average velocity of gas A)2
According to the kinetic theory molecules of hydrogen gas for instance would have a higher average
velocity than molecules of oxygen gas at the same temperature because the hydrogen molecules are
lighter.
Graham’s law implies that lighter gas molecules diffuse faster than heavier gas molecules at the
same temperature. This is understandable in terms of the kinetic theory, which says that lighter gas
molecules have a higher average velocity than heavier gas molecules at the same temperature. The
relative rates of diffusion for any two gases can be expressed as the inverse ratio between the square
roots of their molecular masses. Consider the example of hydrogen and oxygen.
Diffusion rate of H2 = (molecular mass of O2) = 32 = 16 = 4
Diffusion rate of O2 (molecular mass of H2)
2
These equations indicate that hydrogen molecules will diffuse approximately 4 times faster than oxygen
molecules, which are 16 times heavier.
In this experiment, the relative rates of diffusion of two gases, having significantly different molecular
masses, will be determined. The gases that will be studied are ammonia, NH3(g), and hydrogen chloride,
HCI(g).
Objectives:
To measure the relative rates of diffusion of ammonia gas and hydrogen chloride gas.
To verify Graham's law of diffusion.
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Procedure:
1. In a clean and DRY diffusion tube put a cotton swab dipped into Fresh concentrated HCl in one end,
and another cotton swab dipped into concentrated NH3 into the other end.
2. Once the swabs are inserted begin timing and stop the timer when a white “smoke” ring of NH4Cl
forms. (hint: the ring will be easiest to see if the tube is on a black surface and if you determine which
gas will move fastest).
3. Clean the diffusion tube with a cotton swab, by running the total internal distance of the tube with the
Q-tip.
Data:
Record the total length of the diffusion tube and the distance traveled by each of the two gases and the
total reaction time.
Calculations:
1. Calculate the rate of diffusion for each gas by dividing the distance traveled (cm) by the time required
(seconds) for the appearance of the NH4Cl ring.
Rate of diffusion = distance traveled by gas
Time required for NH4Cl
2. Calculate the ratio between the rate of diffusion of NH3 and the rate of diffusion of HCl, using the rates
calculated above.
3. Using the molecular masses of NH3 and HCl, calculate the theoretical ratio between the rates of
diffusion of these gases. (The first equation in the background may be helpful)
Questions:
1. How would an increase in temperature affect the diffusion rates of each gas, how would the ratio of
the diffusion rates be affected?
2. The white substance is ammonium chloride, NH4Cl. It is the only product in the reaction. Write the
balanced equation for this reaction.
3. Why might it be necessary to use Fresh Solutions rather than those that have been set out in the
open for a long time?
4. Calculate the percent error for the theoretical ratio and the actual ratio of gas diffusion
% error = (theoretical value – experimental value)
theoretical value
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Goodie, Goodie It’s Graham's Law Practice
Complete all of the following problems in a legible manner. Show all necessary algebra.
1. An unknown gas is found to travel twice as fast as carbon dioxide, which traveled
14.00 cm in 12.00 sec. What is the molecular mass of the unknown gas?
2. In the lab an objective is set to determine the molecular mass of an unknown gas.
The following data was collected:
Time
3.87 min
Distance H2 traveled
2.75 meters
Distance of unknown
25.00 cm
3. Hydrogen is tested against effluent gas from a Uranium enriching plant. It is found
that after 25.00 minutes Hydrogen has traveled 14.00 meters while the unknown gas
has traveled 1.06 meters. What is the molecular mass of the unknown gas and is it
of any major concern for radioactivity?
4. Which of the two gases Carbon tetrafluoride or Sulfur trioxide is faster and by how
much?
5. Sarin, a nerve gas, has a molecular mass of 140.09 g/mol. If it is combined with
Pulegon, a peppermint smelling ketone, with a molecular mass of 152.23 g/mol, would
this prevent people dying of a nerve gas? Explain your answer.
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Reaction Rate Simulation Demo
Directions: Fill in the data table as we run the demo, show a sample calculation for
finding reaction rate and answer the questions that follow.
Data table
Condition
# Product made
Time
Rate (product/unit of time)
Low temp
High temp
Low concentration
High concentration
Low pressure
High pressure
Low surface area
High surface area
No catalyst
1 catalyst
2 catalyst
__
____
______
Calculations: Show a sample calculation for the rate of one of the conditions
Questions:
1. In your own words give a rule for each of the individual reaction control factors, put
all rules in terms of increasing reaction rates. You will have five rules.
2. Explain why the reaction rates change when the conditions change. Use the
collision theory to help explain yourself.
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How Fast? Calculating Reaction Rates
Directions: calculate the following reaction rates according to the situations given. Make
all calculations necessary and legible.
1. Determine the moles of product formed per second in the following reaction. 200.0
ml of Hydrogen combines with 650.7 ml of Oxygen to form gaseous water. The
reaction is very exothermic and is completed in only 0.000543 centiseconds.
2. Determine the moles of product formed per year when the following car rusts. A
reaction between 85.000 liters of Oxygen and 567.0 kg of iron slowly combines over
a 3.000 year time span to form iron (III) oxide.
3. Due to the low temperature of 45.00 K a reaction is slowed down. The reaction for
the combustion of 986.00 grams of liquid methane (CH 4) is combined with 89.347
grams of liquid Oxygen, which is a pretty blue color. What is the reaction rate if the
reaction is completed in 0.500 hour? Use a unit of product per minute.
4. If 14.00 grams of sodium combines with 98.00 grams of water at STP the reaction
requires 4.325 minutes to complete, and produces sodium hydroxide and hydrogen
gas. What is the reaction rate in moles of product formed per second?
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Reaction Rates
IO3-(aq) + 3 HSO3-(aq)  I-(aq) + 3 SO4-2(aq) + 3 H+(aq)
5 I-(aq) + 6 H+(aq) + IO3-(aq)  3 I2(aq) + 3 H2O(l)
Background:
The rate of a chemical reaction is the time required for a given quantity of reactant to be changed to
products. This is usually measured as moles per minute for a fast reaction. Pressure, temperature,
concentration of reactants, surface area, and presence of catalyst can affect the reaction rate. A
chemical reaction is the result of an effective collision between reactants, this is called the collision
theory. Temperature is a measure of the average kinetic energy of a system. Raising the temperature of
a system raises the average kinetic energy, this causes the reactant particles to move faster. An
increase in speed will increase the probability of collision and therefore product formation.
Increasing the concentrations of the reactants should have the same effect.
In this experiment, two solutions will be mixed and the completion of the reaction will be marked by a
color change. One solution contains the iodate ion (IO3-). The other contains the hydrogen sulfite ion
(HSO3-) and soluble starch. The entire reaction is a two step reaction shown above. Notice the products
of the first reaction are involved in the reactants of the second reaction, even though the mole ratios
required are different. These differences in mole ratio will determine which reaction is the rate
determining reaction.
Solution Preparation (1 liter per hour)
Soulution A (iodate solution)
0.00953 M KIO3 (2.0 g KIO3/liter)
Solution B (sulfate / starch solution)
3.0 g starch / liter—boiled and cooled (or use spray starch)
Under the hood add 0.80 g of NaHSO3 into 10 ml of 6.0 M H2SO4, add this after starch
solution has cooled then add 0.30 grams of NaHSO3 to cooled solution
Objective:
Determine the effect of temperature and concentration of the rate of reaction.
Materials:
250 ml beaker Thermometer
stopwatch
(2) 50 ml beaker
Large test tubes
Pipettes
Ice
Solution A (IO3-)
Solution B (HSO3- and Starch)
Procedure:
Temperature:
1. Place 10.00 ml of iodate solution in one test tube and 10.00 ml of hydrogen sulfite in another test
tube. Place both test tubes in an ice bath for 4 minutes to allow them to reach the ice bath
temperature. Record the temperature of the ice bath. Do not put your thermometer in the test tubes.
2. Add one test tube to the other and record the time for the reaction.
3. Repeat steps 1 and 2 but place the test tubes in a warm water bath (Do not allow the water to exceed
o
60 C) for 4 minutes. Then combine and record the time.
4. Record room temperature and use your data from step 1 under concentration for that temperature.
Concentration:
1. Add 10.00 ml of iodate solution to a 50 ml beaker
2. Add 10.00 ml of hydrogen sulfite solution to the same 50 ml beaker and begin the timing as soon as
the two solutions come in contact. Record the volumes and the time for the reaction to complete in
your data table
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3. Repeat the steps 1 and 2 but this time use 10.00 ml of hydrogen sulfite solution and 9.00 ml of iodate
solution and 1.00 ml of distilled water. Mix the distilled water and iodate solution before adding it to
the hydrogen solution. Record volumes and time.
4. Continue to record times and volumes for the following ratios of iodate solution to distilled water: 8.00
ml iodate to 2.00 ml distilled water, 7.00 ml iodate and 3.00 ml distilled water, 6.00 ml iodate and 4.00
ml distilled water, and 5.00 ml of iodate and 5.00 ml of distilled water. Be sure to rinse and dry the
graduated cylinder between each trial.
Data:
Data table should consist of rows and columns with proper headings. It would be appropriate to have
separate data tables for the concentration data and the temperature data.
Calculations:
Include a graph for each condition (temperature and concentration). Graphs are to be stapled to the
back of the lab report. Make sure dependent and independent variables are on the correct axes with
labels and scale that fits the paper. The graph is a tool and it used to generate the equation of the line.
Use the equation of the line for your calculations to determine the rates.
Questions:
1. According to your data and graphs make a generalized statement about the effect of temperature on
reaction rates, as well as the effect of concentration on reaction rates.
2. In your own words explain in terms of collision theory why the two variables above affected the
reaction rates.
3. Using the equation of your line (y=mx +b) determine at what volume of iodate in a total volume of 10
ml would give a reaction time of 4.35 sec.
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Reaction Rate Challenge Lab
IO3-(aq) + 3 HSO3-(aq)  I-(aq) + 3 SO4-2(aq) + 3 H+(aq)
5 I-(aq) + 6 H+(aq) + IO3-(aq)  3 I2(aq) + 3 H2O(l)
Objective:
Using the data your team collected for the Reaction Rate lab you will manipulate the
temperature or concentrations (volume of Iodate) to create a reaction rate of a specific
time (determined in class).
Procedure:
1. Develop a procedure and all supporting mathematics for a reaction rate of the
following two times:___________ and ___________ .
2. Your team will perform each reaction in front of me, as I will be the official timer.
Data:
Collect all relevant data needed to meet the objective. I will enter your time data on
your lab report and initial it as official.
Calculations:
Show all supporting calculations including your Percent errors (one for each time) based
on reaction rate time.
Questions:
1. Explain how it would be possible to use rate of the iodate reaction to quantify the
amount of energy (fully explain this procedure)?
2. Explain why the accuracy of your model differed (according to percent error) for
each expected time.
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Unit 7 Objectives for Phase changes

Phase change diagrams

Phase change calculations
Calories
Joules
Food calories

Absolute zero
Conversions from Celsius to Kelvin and vice-a-versa

Data Manipulation
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Phase Change Calculations
(T x mass x S. H. ice) + (Hformation x mass) + (T x mass x S. H. water) + (Hvaporization x
mass) + (T x mass x S. H. steam) = total energy required
All specific heats, heat of formation and heat of vaporization given are only for water in
various states. Different substances will have varying values.
The following steps are to determine the total amount of energy required to convert ice
at some temperature below 0C to some temperature above 100 0C. I you do not have
ice as a starting point then begin at some other step number depending on your starting
phase of matter.
T means the change in temperature (, delta, always means “change in”)
Specific heats
Ice
= 0.5 cal/g oC
Water = 1.0 cal/g0C
Steam = 0.5 cal/g oC
Heat of formation (used to go from solid to liquid)
80 cal / g
Heat of vaporization (used to go from liquid to gas)
540 cal / g
Ice below zero
1. Determine the original temperature of ice (i.e. –30 0C) add calculate the change in
temperature to convert that ice to 0 0C. For this example our T is 30 0 C.
2. Put your change in temperature in the delta T position multiply by the mass of ice
and the specific heat of ice.
Water from 0oC to 100 oC
3. That number is added to the heat of formation multiplied by the mass.
4. Again that number is added to the delta T to take water from 0 oC to water at 100oC.
The change in temperature is multiplied by the mass of water and multiplied by the
specific heat of water.
5. That number is then added to the product of the mass multiplied by the heat of
vaporization.
Steam at 100 oC
6. That number is added to the product of the mass of steam multiplied by the change
in temperature multiplied by the specific heat of steam.
7. This total sum is the required amount of energy to convert all of the to ice at some
temperature below zero, to all steam at another temperature above one hundred.
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Phase Change Practice
1. How many calories are required to convert 45.00 grams of liquid water at 30.0o C to
steam at 110.0o C?
2. How many calories must be released to convert 100.0 grams of summer lake water
at 65.0o C to winter ice at -10.0o C? How does this effect the local climate?
3. Melting ice in your mouth requires a lot of heat energy, this is why it cools you down.
How many calories will your body burn if you melt 6.0 grams of -25.0o C ice in your
mouth to body temperature of 37.0 o C?
4. A serving of Doritos contains 160.0 kilocalories (big C’s). Your mouth is at body
temperature (38.0o C). What mass of –25.00oC ice do you have to eat to burn all the
calories in one serving of Doritos?
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Phase Change Practice with Other Variables
Directions: Answer all questions clearly, remember that 1.0 calorie is the same as 4.184
Joules
1. What is the final temperature of a 50.00 g sample of ice at –10.00 oC if it is exposed
to 59.75 J of energy and remains ice?
2. What mass of ice was converted from –20.00 oC to liquid water at 78.00 0C by 7,560
calories?
3. A 150.00 g piece of –50.00 oC ice is exposed to enough energy (85.0 kJ) to convert
it to liquid water at what temperature?
4. 26.46 KJ of energy is released from 10.00 grams of steam as it condenses from
125.00oC . What is the final temperature of the condensed water?
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Phase Change Practice with Multiple variables
1. How much energy (cal) is released if a 27.75 ml of 85.00 oC water is solidified into
a -40.00oC ice?
2. What is the mass of ice that remains if a 30.00 gram block of -25.00oC ice is
exposed to 1575 calories of energy?
3. What is the final temperature of steam if 150.00 ml of 80.00 oC water is exposed
to 87.375 Kilocalories?
4. What volume (ml) of steam is created if 100.00 grams of 25.00 0C water is
heated by 21 Kilocalories?
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Just for Energy Nuts
Objective: Determine the efficiency of a pop can calorimeter
Materials:
Student made calorimeter
Matches
Water
Corks
Thermometer
Stick pins
Procedure:
1. Push the stickpin into the cork so that a peanut may be stuck to the sharp end of the
pin. Careful not to stick your self.
2. Record the mass of the peanut
3. Record the temperature and the volume of around 30 ml of water in the calorimeter.
4. With a match light the peanut (the oils in the peanut should ignite)
5. After the peanut has extinguished record the temperature of the water.
6. Record the mass of the peanut after it was burnt.
7. Repeat these steps several times (at least 5 trials). Be sure to record a NEW
starting temperature for the water for each trial.
8. If your initial water temperature is 75 0C or above then use new water.
Data:
Record all required measurements from the procedure section as well as the amount of
calories in a serving of peanuts, how many calories are from fat and how many grams of
peanuts are in one serving of peanuts.
Calculations:
1. Calculate the number of calories that were produced in each trial (specific heat of
water is 1 cal/ g oC.
2. Determine how much peanut mass was lost for each of the trials
3. Using the information in steps 1 and 2 determine the average number of calories per
gram of peanuts.
4. Also determine what the average number of calories per gram of peanuts is
according to the food label (determine both total calories and calories from fat)
Questions:
1. When comparing your results to the food label on the peanut container what must
you remember about the food industry and calories?
2. Of the two values that you calculated in number four, which is closest to your value?
Explain.
3. Based on question number 2 determine your percent error and explain why the error
may exist.
4. Why is it necessary to convert all units to calories per gram?
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Now You’re in Hot Water - a phase change lab
Background:
Water is one of the few compounds that are capable of existing as three states of matter under normal
conditions of temperature and pressure. Water will change from one state of matter to another with the
addition or subtraction of energy. As with any phase change however the temperature of a substance will
not change until the compound is all in the same state (not Wisconsin or Michigan but gas, liquid or solid).
This is because all of the energy that enters a compound is used to change the state of matter and not to
increase energy. This phenomenon explains maritime climates and the correlation between dew point
and low temperatures.
Objective:
To plot a phase change diagram for solid water to gaseous water and determine the amount of energy
required to make the conversion from solid to steam.
Procedure:
1. In a 500-ml beaker mass out about 100 grams of ice and add about 50.0 ml (record total initial
volume) of tap water. We will assume the water from the tap and the water used to make the ice was
pure water. Having made this assumption we can then use the density of pure water 1.0 grams /
milliliter. Record the mass of the ice and the volume of tap water in data section. Also determine the
total volume of water.
2. Record the temperature of the ice bath after the solution has stabilized.
3. Being sure to keep the thermometer off of the bottom of the beaker (a burette clamp may be helpful)
and constant stirring.
4. Over a Bunsen burner heat the ice bath recording the temperature every 30 seconds. You must stir
constantly. Continue recording temperature until the water is boiling heavily and the temperature has
not changed for more than 4 minutes.
5. Then record the volume of water remaining in the beaker (use a graduated cylinder for accuracy)
Data
Data table will include the mass of ice and the volume of water as well as the total volume of water and
the volume of water that was left in the beaker.
Make a smaller two-column table to record the temperature every 30 seconds.
Calculations
1. Plot a graph of time vs. temperature for your phase change data
2. Calculate the amount of energy required to convert the volume of water you convert from ice to
steam. The volume of water that you convert from ice to steam will be the volume of water lost as
steam.
3. Calculate the energy output of your Bunsen burner.
Questions:
1. Explain why the plateaus exist in the graph (why does the temperature not change).
2. What is the energy output of your Bunsen burner and how is that information of value?
2. Label the heats of vaporization and formation on the graph.
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Phase Changes “Suck”
Objective: Determine the phase diagram of a two-substance solution.
Materials:
White corn syrup
Sucker sticks
Stopwatch
Sugar
food coloring
FOOD USE material
600 ml beaker
25 ml Grad cylinder
10 ml grad cylinder
o
Thermometer with 200 C max
Stirring rods
Aluminum foil
flavoring
Wax paper
Disposable pipettes
Density of corn syrup = 1.30g/ml
Procedure:
1. You NOT will use any glassware from your lab drawer during this lab.
2. Make an aluminum ring mold for your sucker, and tape it securely on wax paper. Insert a sucker stick
through the aluminum mold.
3. Record the mass of 30.00 g of sugar and add to a 600-ml beaker.
4. Record the volume of 20.00 ml of water and add to a 600-ml beaker.
5. Record the volume of 20.00 ml of corn syrup and add to a 600-ml beaker. Record total mass of the
solution. Density of corn syrup is 1.30 g/ml
6. Heat contents of the 600-ml beaker over a Bunsen burner with constant stirring.
7. Record the temperature of the solution every 30 seconds.
8. Remove the beaker from heat when the temperature reaches 148 –154 oC.
9. Continue to stir as solution begins to cool and thicken. Do not allow the solution to harden in the
beaker.
10. Just before you pour your solution into your molds add 2 drops of food coloring and 0.5 ml of flavoring
(use disposable pipette) and stir.
11. Pour your solution into the molds and allow it to harden.
12. Immediately start to clean (with food use brushes) your equipment with hot water. You will not leave
until it is clean.
Data: Record all data in a neat and legible table with proper units and precision.
Your data will also include a graph of your phase change diagram for your solution. Follow the
graphing rules I have given you.
Calculations:
Show all work necessary to determine the specific heat of the sugar solution (sucrose and corn syrup
mixture). This is the question asked for question number 4.
Questions:
1. Your graph should have a plateau around 100 oC. Explain why the temperature quickly rises after
that plateau when there is still solution remaining.
2. Based on the temperature increase rates what can you determine about the specific heat of sugars
(sucrose and corn syrup) compared to the specific heat of water?
3. In your solution all of the water evaporated and is not part of your final product, why is it added to the
initial solution?
4. You know the rate (can be calculated) of energy input into your system and the mass of the corn syrup
and sugar mixture. What is the specific heat of this sugar solution?
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Unit 8 Objectives for Gas Laws

Application of Charles, Dalton, Gay-Lusac, and Boyles law

Theoretical calculations
Ideal Gas law
Real Gas law
Combined gas law

Molar mass of gas

Data manipulation
Convert story problems to data tables
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Gas Law Investigations
Three variables, temperature, pressure and volume effect gasses. Investigate the
following scenarios with these variables in mind
1. Erlenmeyer flask and Balloon
Record your observations:
Write your Gas law that explains this phenomenon.
2. Balloon in the vacuum Pump.
Record your observations:
Write your Gas law that explains this phenomenon.
3. Syringe and Stopper
Record your observations:
Write your Gas law that explains this phenomenon.
4. Hair Spray
Record your observations:
Write your Gas law that explains this phenomenon.
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Gas law Demos
1. Collapsing pop can:
a. What are the constants?
b. What are the variables?
c. Which law(s) are in play?
d. Explain what is happening using the correct scientific terminology.
2. Cartesian Diver:
a. What are the constants?
b. What are the variables?
c. Which law(s) are in play?
d. Explain what is happening using the correct scientific terminology.
3. Balloons and cup:
a. What are the constants?
b. What are the variables?
c. Which law(s) are in play?
d. Explain what is happening using the correct scientific terminology.
4. Pursed lips and black paper.
a. What are the constants?
b. What are the variables?
c. Which law(s) are in play?
d. Explain what is happening using the correct scientific terminology.
5. Egg in a bottle:
a. What are the constants?
b. What are the variables?
c. Which law(s) are in play?
e. Explain what is happening using the correct scientific terminology.
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Gas Laws
Gasses are governed by three variables: pressure, volume and temperature. A change in magnitude of
any of these three can effect the other two. It is easiest to measure pressures in atmospheres (atm),
volume in liters and temperature in Kelvin.
1 atm = 760 mmHg
o
C = K – 273.15
o
K = C + 273.15
In Chemistry and Physics there are 4 main gas laws, from which most others are derived.
Dalton’s law of Partial Pressure: States that the sum of the pressures of all gasses in a system
will be the total pressure of the system.
Mathematically: P1 + P2 + P3…= Ptotal
Example: If a gas canister is filled with 3.00 atm of CO2 and 7.00 atm of H2 then the total pressure to the
canister is 10.00 atm.
Boyle’s Law: Pressure and volume are indirectly proportional.
This means as pressure increases then volume decreases
Mathematically: P1V1=P2V2
Example: If the pressure around a balloon increases twofold then the volume will decrease to half its
original volume.
Charles’ Law: Temperature and volume are directly related.
This means that if the temperature increases then the volume will increase.
Mathematically: (V1/T1) = (V2/T2)
Example: The Temperature of a balloon dropped from 400 K to 200 K the volume will also be only half as
large.
Gay-Lusac’s Law: Temperature and Pressure are directly related.
This means that if the temperature decreases then pressure will also decrease.
Mathematically: (P1/T1) = (P2/T2)
Example: The Pressure of an aerosol deodorant can dropped from 2.0 atm to 1.0 atm the pressure of the
can will also drop from room temperature (25oC or 298 K) to 189 K. Remember that all temperatures
must be in Kelvin. It is always possible to convert back to Celsius after the calculation.
We can and will combine this laws to determine the effects of changing all variables of a system.
Combined gas law: determines the effects on a variable if the entire system is changing.
Mathematically: (P1V1)/T1 = (P2V2)/T2
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Conceptual Gas Law Practice
Answer all questions completely and in legible writing.
1. Using Dalton’s Laws of partial pressure calculate the following total pressures for
each situation:
a. Canister with 400.0 mm Hg of oxygen and 350.0 mm Hg of Carbon Dioxide
b. A balloon with 3.0 atm of breath from one person and 7.0 atm of breath from
another
2. Gauge pressure is the pressure read from a gauge (i.e. tire gauge) However that is
not the absolute or actual pressure exerted on an object. Gauge pressure is the
pressure above and beyond the atmospheric pressure. Using this information fill in the
following chart below. Assume atmospheric pressure is 1 atm or 760 mm Hg which
ever is applicable.
Gauge Pressure
6.0 atm
14 atm
900 mm Hg
Actual Pressure
3. The speed a gas travels increases the lighter the gas is. Which of the following is the
faster gas, Oxygen (O2) or Carbon dioxide (CO2)?
4. If someone stands on your chest and increases the pressure on your chest what
happens to the volume of your lungs?
5. Inhalants, as narcotics, pose another hazard other than brain damage, that of
frostbite. Why?
6. If a balloon is cooled to one half of its original temperature, if the original volume was
40.0 ml what is the final volume?
7. If two people take turns blowing up an air mattress and the total pressure at the end
is 35 atm and the first person blew a pressure of 25 atm what pressure did the other
person blow?
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More Gas Law Practice
Directions: On a separate sheet of paper legibly show all the mathematical work to
support your answer.
1. A balloon is filled with 2.0 atm of nitrogen, 4.0 atm of oxygen and 900.00 mm Hg of
carbon dioxide. What is the total pressure in the balloon?
2. How many grams of oxygen are in a 15.0 L cylinder if the pressure is 325.0 mm Hg
at 27.0 o C?
3. What is the pressure in a 15.0 L gas cylinder at 298.0 K if 3.00 moles of gas are
present?
4. A propane tank used for outdoor cooking is stored at 25.0 oC and pressurized at
3.00 atm. What is the final temperature when the cylinder is open (depressurized to
1.0 atm)?
5. What is the initial volume of a balloon if the temperature rises from 25.0 oC to 100.0
o
C and the final volume is 150.0 ml?
6. If the pressure is 1.80 atm where a fish lives and that fish burps and an air bubble
results with a volume of 2.60 ml, what is the volume of the burp when it reaches the
surface of the water?
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Even More Gas Law Practice--Gas Law Scenarios for Practice
1.The little moron’s balloon at STP is 4.00 L. What is the volume of the
balloon as it rises in the air (the reason he is a little moron) where the
pressure is 550.0 mm Hg and the temperature is –25.00 oC?
2. A fish burps at a depth where the pressure is 2.00 atm, the bubble has a
volume of 6.00 ml and a temperature of 5.00 oC. What is the temperature
of the burp at the surface if the volume has expanded to 15.00 ml?
3. A can of Ready-whip is pressurized at 12,000.0 mm Hg it is opened at
room temperature (22.00 oC) and allowed to depressurize (instantly). If the
volume remains constant what is the final temperature of the gas?
4. How many moles of nitrogen must be present in a 5.00 L balloon at
27.0oC and 758.00 mm Hg?
5. Would the pressure be dangerous 16.78 grams of solid CO2 (sublimes at
–77.00 oC) was stored at –35.00 oC, in a 55.00 ml steel canister?
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Short Discussion on Proper treatment of Gasses in Limiting Reactant Situations
In real world applications of chemistry all chemical reactions are a limiting reactant situation. Regardless
of the reactants involved or the stoichiometry of the reaction of interest, one of the reactants will be
consumed before all other reactants and in the process stop the production of additional products. We
have already learned how to treat solid reactants by converting them to moles using their molecular or
atomic masses and we can convert aqueous solutions to moles if we know their concentrations and the
volume of the solution. Liquids can be converted to moles if their densities are known, thereby converting
the volumes into a mass. Remember that it is number of moles that are important as the moles are a
direct function of the number of atoms or molecules present and it is the atoms or molecules that react,
regardless of the phase of matter their energy represents. Gasses are the only phase (commonly
encountered) that we have not yet adequately treated for proper theoretical calculations.
Calculating the actual number of molecules/atoms in a gas sample requires some additional work. For
example if you recall the kinetics of gasses allows a gas to fill its container, so a single particle of gas or
several million particles of a gas can occupy the same volume. Therefore volume alone is of little help in
determining the number of reactive particles present in the reagent. We can however use the ideal gas
law (PV=nRT) to determine the number of moles of a gas for any given set of conditions. As you continue
to advance in your chemical knowledge you will find modifications will be required to the ideal gas law,
but this is far outside the current scope of this course. When dealing with gasses as reactants you must
know the pressure, volume and temperature of the gas. The following examples may help.
Example 1 (gas as product):
What is the expected volume of hydrogen produced at 1548 mm Hg and 30.0 oC when 15.26 grams
of potassium reactats with 125.69 ml of water?
First write a balanced reaction:
2K(s) + 2H2O(l)  H2(g) + 2 KOH(aq)
Determine which reactant is the limiting reactant:
Notice neither is a gas so this should be review:
15.26 g K x
2molH 2O 18.02 gH 2O 1mlH 2O
1molK
X
X
X
= 7.03 ml H2O
2molK
1molH 2O
1gH 2O
39.0983gK
So potassium is the limiting Reactant
Use the limiting reactant (potassium) to determine the MOLES of gas created:
The number of moles is in a sense a constant and will not change regardless of the pressure, volume or
temperature. If we calculate the number of moles of hydrogen we can then determine the volume of
hydrogen at the pressure and temperature conditions specified.
15.26 g K x
1molH 2
1molK
X
= 0.1952 moles H2
2molK
39.0983gK
0.1952 moles of Hydrogen will be created at ANY condition. Use the ideal gas law to determine what the
volume of 0.1952 moles of gas will occupy at the conditions given.
Using the ideal gas law for final volume:
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PV=nRT
Convert pressure to atmospheres and temperature to Kelvin before calculating with the ideal gas law.
(2.037 atm)(V) = (0.1952 mole)(
0.0821Latm
)(303.15 K)
molK
solve for V: V = 2.39 L of hydrogen
Example 2 (Gas as a reactant):
253.67 ml of CO2 at 569.87 mm Hg and 25.00oC is used to oxidize 32.56 grams of magnesium. What is
the expected mass of carbon formed?
Write a balanced chemical reaction:
CO2(g) + 2 Mg(s)  C(s) + 2 MgO(s)
Convert gas to moles:
Using the ideal gas law, convert the carbon dioxide into moles – again it is the moles that react.
PV=nRT
(0.750 atm)(0.25367 L) = n(
0.0821Latm
)(298.15 K)
molK
n = 0.00777 moles CO2
**given the conditions it is logical that there is a small number of moles
Determine the limiting reagent:
0.00777 mol CO2 x
2molMg 24.3050 gMg
x
= 0.378 grams Mg
1molCO 2
1molMg
Carbon dioxide is the limiting reactant
Use the limiting reactant (CO2) to determine the product:
0.00777 mol CO2 x
1molC
12.011gC
x
= 0.0933 grams carbon
1molCO 2
1molC
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Gas Law Practice with Laboratory Application
Directions: On a separate sheet of paper answer the following questions showing all work. As always any
and all illegible answers will be marked incorrect.
1.
When a reaction occurs between 56.98 liters of Hydrogen stored at 25.00oC and pressurized to 567.98
o
torr and 25.98 liters of Oxygen stored at 25.00 C and pressurized to 980.45 torr (1 torr = 1 mm Hg) what
o
is the expected volume of product from this synthesis reaction if lab conditions are 23.5 C and the
barometric pressure is 758.30 mm Hg?
2.
The combustion of a hydrocarbon always yields Carbon dioxide and water. How many milliliters of liquid
water can be expected from the condensed water product if the following data was collected?
Data
Trial 1
Vol. C2H4(g)
65.29 L
Vol. O2(g)
189.25 L
Pressure C2H4(g)
Pressure O2(g)
Storage Temp.
Barometric Pressure
1198.75 torr
1500.00 torr
25.38 oC
775.25 mm Hg
3.
As a product, in a reaction between an excess of zinc and 75.98 milliliters of HCl, 98.50 ml of hydrogen is
produced when pressurized to 963.00 mm Hg and stored at laboratory temperatures of 22.87 o C. What
was the Molarity of the Hydrochloric acid?
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An Extreme Absence of Kinetic Energy-Proving the existence of absolute zero
Objective: Through graphical extrapolation determine the Celsius value of absolute
zero and why a temperature lower cannot physically exist.
Materials:
Bunsen burner
ring stand
250 ml Flask
One-hole stopper glass tubing
rubber hose
600 ml beaker
thermometer
ice bath
50 ml graduated cylinder
Procedure:
1. Fill a 1000-ml beaker with enough water to cover the bulb of a Florence flask and
start to boil.
2. Determine the exact volume of the flask when the stopper assembly is in place. This
will be your volume of the flask at the temperature of boiling water.
3. Place the one hole stopper and hose assembly in the round bottom or Florence flask
and place the flask into the boiling water for a least four minutes. This will ensure
that the air temperature inside of the flask is the same as the water temperature.
Should look like the figure shown.
4. Record temperature
of boiling water.
5. Remove the flask from the
hot water bath and submerge
in ice bath.
6. The unattached end of the rubber
hose needs to be in a beaker of
water before and during the cooling
of the flask.
7. Record the volume of the water collected
to the nearest 0.1 ml.
8. Record the temperature of the ice bath
9. You now have two temperatures and two volumes and
therefore a straight line.
10. This data will be graphed to determine the value of absolute zero.
Data:
Record the temperatures and the corresponding volumes as accurately as possible.
Remember it is possible to estimate one decimal point beyond the last increment value
on any instrument.
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Calculations:
Calculate the absolute zero temperature based on the mathematical model created from
your graph.
Questions:
1. Why did water flow into the flask when placed into the cold water bath?
2. From the graph, explain why temperature could not drop below absolute zero?
3. What methods could be used to establish another point on the line that would
improve the accuracy of this lab?
4. Indicate both your experimental value of absolute zero and the accepted theoretical
value and then determine your percent error.
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Graphing Instructions for Excel
We will be using a spreadsheet program, Microsoft Excel, to graph and extrapolate
data.
Your first need to open the Excel Program and enter your raw data from the lab.
Column A (vertical) is your independent variable column (temperature), and column B is
your dependent variable column (volume).
In Row 1 (horizontal) column A (A1) type the heading “Temperature” and type “Volume”
in B1. Then enter your data under the correct headings.
Once all of your data has been entered click on just the column letters and highlight all
of your data, this will highlight the entire column even where you do not have data
entered. When all of you data is highlighted click once on the small icon in the upper
right-hand side that looks like a bar graph.
When the dialog box opens choose XY (scatter) for chart type and under Chart sub
chart type choose the box, which has a graph without any points connected. Then click
next, and click next again.
You should be in a dialog box where you can enter chart titles and axis names. Enter
the names for these three things. Remember to give both units and the property that is
being measured (i.e. Temperature ( oC)).
Then click on the Gridlines tab and “unclick” the box that says major gridlines, and click
next and click finish.
You should now have a gray graph with two points and all of your axis labels showing.
Double click on the gray portion of the graph, this will give you another dialog box with
color choices. Under area choose the white color (right column, fourth from the top).
Then click OK.
Double click on your scale for the X-axis, then click on the Scale tab. Change the
minimum value to –300 and the “value (Y) crosses at” to –300. Then click OK.
Click once on the data points. They should become highlighted (yellow) Under the chart
heading on the tool bar choose “Add Trendline…” A dialog box will come up and under
“Options” click “Display R2 value on chart” and “Display Equation.” Then under forecast
Change the Backward value to 300, click OK.
Click and drag the equations to the upper right hand side of the paper.
Click about 1 inch from the left of the chart title, black boxes should appear at all of the
corners, then print.
To calculate your value for absolute zero use the equation to find the Celsius degrees
that corresponds to zero volume.
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Proving Molar Volume of a Gas
Objective: Determine the molar volume of a gas at STP by extrapolation of data collected at laboratory
conditions.
Procedure:
1. Record the Barometric reading for the day.
2. Record the temperature in the room.
3. Record the mass of a piece of magnesium ribbon 3.5 cm. This length is meant to give a mass value
near 4 hundredths of a gram.
4. Roll the magnesium ribbon and tie a small piece of string around it.
5. Place about 5.00 ml of 11.6 M HCl (concentrated) into a eudiometer.
6. Fill the remainder of the eudiometer with water from a wash bottle.
7. Place the Mg ribbon into the eudiometer about 2 cm from the opening and secure in place with a one
hole rubber stopper.
8. Set up a large Aquarium and record temperature of the water.
9. Invert the eudiometer in the aquarium and allow the more dense acid flow downward towards the
Mg.
10. When the reaction between the Mg and HCl is complete make sure that the water level inside of the
eudiometer is equal in height to the water level in the aquarium, and record the volume of gas
collected.
11. Record the vapor pressure of water, for the temperature of your water from the chart on the front
desk.
** If the volume in the eudiometer is above the water line in the aquarium you also need the distance
in mm that the water in the eudiometer is above the aquarium water line.
Data:
Record all specified information from the procedure in a neat, legible and organized data table.
Calculations:
Stay focused on your objective that will guide you on how to do the calculations. What follows is one
possible way to determine your objective of molar volume, but there are others.
1. Determine the number of moles of magnesium you used.
2. Find the number of moles of hydrogen with the use of a balanced equation corrected for temperature
and pressure.
3. Use the combined gas law to find the correct volume of gas at current conditions
4. Manipulate your information to determine the molar volume of your gas (what it would be if it where 1
mole at STP).
Questions
1. Why is it necessary to have the level of water in the aquarium equal to the level of water in the
eudiometer?
2. Using the values you calculated for molar volume of a gas and the theoretical value (22.4 L/ mol)
calculate you percent error.
3. Other than the established value for the molar volume of gas indicate three (3) other values that you
could use to determine a percent error.
Note: Density of mercury is 13.46 g/ml
PH = PBar – (PH2O + Dmm/13.46)
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Molar Volume Calculations
Data
Trial 1
Air Temp
25.0 C
Water Temp
21.2 C
Vapor Pressure of water 18.880 mm Hg
Vol. Of Hydrogen
42.620 ml
Mass of Mg
0.0400 g
Vol. Of HCl
10.00 ml
Conc. Of HCl
11.6 M
Barometric Pressure
29.00 in Hg
There are numerous ways to do the calculations for this laboratory. Remember the
objective and let the objective guide you. Below is one possible way of doing the
calculations and the logic that allows the calculations to be made.
We are looking for molar volume: ? moles of H2
? Liters of H2
Find Pressure of Hydrogen
You need to know what pressure hydrogen gas was exerting inside of the eudiometer.
Use Dalton’s Law of Partial Pressure
PBarometric= PHydrogen + Pwater
Barometric pressure (29.00 in Hg) is 0.969 atm
Vapor Pressure of water (18.880 mm Hg) is 0.0248 atm
PHydrogen = 0.969 atm – 0.0248 atm = 0.944 atm
Find Number of Moles of Hydrogen
Use the ideal gas law and the data collected, including the calculated pressure of
hydrogen to determine the number of moles of Hydrogen present in the eudiometer.
Remember to use the Temperature of Hydrogen (air temp), Volume of Hydrogen and
the Pressure of Hydrogen.
PV= nRT or n = PV/RT
n = (0.944 atm)(0.042620 L) mol K = 0.00164 mols H2
(0.0821 L atm) (298.15 K)
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Find the molar volume at lab Conditions (T = 298.15 K and P =0.969 atm)
0.04262 L H2
= 25.99 L/mol
0.00164 mol H2
Convert this molar volume for STP (T=273.15 K and P= 1.0 atm)
Use the combined gas law to determine what your laboratory molar volume would be if
it were subjected to the conditions of STP.
P1 V1 = P2 V2
T1
T2
(0.969 atm)(25.99 L/mol) = (1.0 atm)(V2)
(298.15 K)
(273.15 K)
V2 = (0.969 atm)(25.99 L/mol)(273.15K) = 22.5 L/mol
(1.0 atm)(298.15 K)
Percent error
There are many comparisons that can be made in this lab to determine your percent
error. Choose the one that will give you the best results.
(Theoretical – Actual) X 100 = % error
Theoretical
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Flick Your Bic It’s a Gas, Gas, Gas
Objective: Determine if the gas in a lighter is butane*
Procedure:
1. Soak your lighter and then dry it before recording the mass on an analytical balance.
2. Fill a eudiometer and invert it in a large beaker. Make sure there are no air bubbles
at the top of the eudiometer.
3. Under the water and under the open end of the eudiometer press the release button
and allow the eudiometer to fill with gas until the water level inside of the eudiometer
is equal to the water level in the beaker.
4. Read the volume of gas inside the eudiometer.
5. Dry and mass the lighter
6. Record the temperature of the water in the beaker and the air temperature (The gas
in the eudiometer will be at the same temperature as the air temp).
7. Record the atmospheric pressure for the day from the barometer.
Data:
Record all data in clearly labeled data table. Be mindful of units and precision.
Calculations:
Determine the number of moles of gas in the eudiometer. Remember the total pressure
inside of the eudiometer was the vapor pressure of water and the pressure of the gas—
you only want the pressure caused by the gas. Then determine the molar mass of the
gas to compare to a known value of butane (C4H10).
Questions:
1. Why was it important that the two water levels be equal in the procedure?
2. Why was it necessary to record the barometric pressure?
3. What was your percent error?
4. What was a possible cause for the error?
5. If your lighter did not contain butane what volatile organic compound did it have?
(Assume that it is a straight chain saturated hydrocarbon with the general formula of
CnH2n + 2)
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Additional Practice with Real-World Gas Scenarios
o
1. When 5.67 ml of gaseous Hydrogen bromide at 0.956 atm and 15.98 C reacts with 28.790 grams of
anhydrous barium hydroxide in a double replacement reaction; how many grams of barium bromide are
expected?
o
2. In a reaction 0.98 liters of Carbon dioxide under pressure of 965.5 torr and 29.65 C reacts with 1.43
grams of sodium to create solid carbon and what mass of sodium carbonate?
3. 45.802 ml of 5.67 M phosphoric acid reacts with 4.85 grams of zinc to produce what volume of
o
hydrogen gas at 19.8 C and 0.968 atm?
Zn(s) + H3PO4(aq)  H2(g) + Zn3(PO4)2(aq)
4. 165.320 ml of butane (C4H10) when reacted with an excess of oxygen gas at a barometric pressure of
30.18 in Hg and 21.6oC will theoretically produce what mass of carbon dioxide if the converted to dry ice?
5. 25.43 ml of 5.3 M Lead (II) nitrate reacts with 30.50 ml of 2.65 M sodium chromate to produce sodium
nitrate and what mass of lead (II) chromate?
6. Potassium hydroxide is formed during the production of hydrogen in a single replacement reaction
between water and potassium. What is the concentration of potassium hydroxide if 95.36 ml of the
solution is required to absorb and solidify 97.43 ml of carbon dioxide under 1.035 atm and stored at a
temperature of 23.5oC?
K (s) + H2O (l)  H2(g) + KOH(aq)
KOH(aq) + CO2(g)  K2CO3(s) +H2O(l)
7. Pentane (C5H12) has a density of 0.6262 g/ml. If 67.89 ml of liquid pentane is combusted with only 5.00
o
Liters of oxygen available from a pressurized 5.68 atm canister in the lab (22.3 C) what is the expected
volume of steam created?
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Review for Quiz on Gas laws and Applications
Know all three gas laws, Dalton’s law of partial pressures, combined gas law, and ideal
gas law.
Know the constants involved with gas laws
Know when STP conditions are applied and what STP conditions are
Now the conceptual information for Boyles’, Gay-Lusac’s, and Charles’ laws and how
they relate to our everyday experiences.
Remember that pressure is always measure in atmospheres (atm), Volume in liters (L)
and temperature in Kelvin (K) be able to convert from mm Hg, ml, and OC respectively.
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Unit 9 Objectives for Equilibrium

LeChatelier's Principle
Dynamic Equilibrium

Equilibrium Constants

Collision Theory

Reaction rates
Pressure
Temperature
Concentration
Catalyst

Activation energy
Catalyst
Endothermic vs. exothermic relationships

Data manipulation

Application of reaction controls
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Equilibrium, It’s a Give and Take2
Objective: To determine equilibrium in the transfer of materials under various simulated conditions.
Procedure:
Simulation 1: Low temperature no products formed
1. In your lab group determine who is the Reactants (R) and who is the Products (P).
2. The reactants will begin with 40 matchsticks and the products 0 record this in the data as initial.
3. Reactants will now hand over ½ of their total and the products will hand back ¼ of their total (ex. R
gives P 20 matches and P gives R back 5)
4. Record the number each person (P or R) transferred in the data table and the outcome (amount that
each person now has)
5. If a half number is encountered (3.5) round up (4).
6. When equilibrium has been reached shout “LeChatlier” and I will come over to verify then determine
your equilibrium constant (products/reactants).
Simulation 2: Low temperature some products formed
1. Repeat simulation 1 procedure, for a reaction in which some products already exist.
2. Initial values will be R=40 and P= 20, transfer rates are the same as for simulation 1.
3. Record all transfers and outcomes and determine equilibrium constant.
Simulation 3: High temperature no products formed
1. Repeat the initial values for simulation 1
2. Because temperature has increase so has the transfer rates at each transfer R will give up ¾ of their
matches and P will give up 1/8 of their matches.
3. Record all transfers and outcomes and calculate equilibrium constant for these conditions.
Data:
All data tables should have the following format, 1 data table for each simulation:
Reactants
# Of items
Products
# of items
# transferred
# transferred
Initial
st
1 transfer
nd
2 transfer
rd
3 transfer
th
4 transfer
(You will have more than 4 transfers!)
Calculations
Calculate the equilibrium product constant for each simulation
Keq = [Product]
[Reactant]
Questions
1. What is meant by a dynamic equilibrium?
2. Why did the transfer rate increase in simulation 3?
3. Is the equilibrium constant effected by whether or not there are already products in existence, give
proof of your answer?
2
Wilson, Audrey H., Journal of Chemical Education. "Equilibrium: A Teaching/Learning Activity." v. 75 n.9 Sept.
1998.
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Chemical Equilibrium
Chemical reactions are reversible, contrary to what you may have been told previously when discussing
physical and chemical changes. Another new twist is that not all chemical reactions go to completion,
meaning that reactants are not all converted into products. Until now we have viewed all chemical
reactions from a stoichiometric viewpoint, meaning that a reaction goes to completion until the limiting
reactant is consumed. While this is true for many reactions it is not universally true for all chemical
reactions. Many chemical reactions exist in a state of equilibrium in which there is an oscillation between
reactants and products. However while in a state of equilibrium there is no net change in the
concentrations of product or reactant.
In chemistry the definition of equilibrium is the exact balancing for two processes, one of which is the
opposite of the other. This means that the forward reaction (product formation) is balanced perfectly by
the reverse reaction (reactant formation). It is crucial that we dispel any misconceptions about what is
equal in an equilibrium reaction. In a chemical equilibrium (dynamic equilibrium) it is not the reactants
and the products that are equal, but rather the rate at which products are made and the rate at which
products are converted back into reactants.
Equilibrium reactions are denoted as a chemical reaction in which the yield arrow is pointing in both
directions, this is a symbolic representation that both product and reactant are being formed. Reactant
 Product. Again this does not mean that the concentration of reactants is equal to the concentration
of products. Equilibrium reactions are often referred to as being in a state of dynamic equilibrium. A
dynamic equilibrium means that the reaction is continuously taking place, reactants are forming products
and products are forming reactants. However because that rate of the forward reaction is the same as
the rate of the reverse reaction there is no net change in either the product concentration ([Product]) or
the reactant concentration ([Reactant]). Concentrations are shown with the use of brackets, [ ].
Equilibrium Constants and the Law of Chemical Equilibrium
The law of chemical equilibrium (often called the law of mass action) was derived by two Norwegian
chemists, Cato Maximillian Guldberg (1836-1902) and Peter Waage (1833-1900). The law of mass
action was proposed by the two Scandinavians in 1864 to explain mathematically the equilibrium
condition. The mathematical representation, like most things in chemistry, requires a balanced chemical
reaction.
Using the generic equation:
aA + bB  cC + dD
Where the capital letters represent chemical species and the lower case letters are the stoichiometric
mole ratios.
This allows an equilibrium expression as follows:
K= [C]c[D]d
a
b
[A] [B]
This is commonly refereed to as products over reactants.
Each species concentration is raised to the power of the stoichiometric coefficient. K is called the
equilibrium constant.
Sample problem:
What is the equilibrium constant for the following reaction?
4NO2(g) + 7O2(g)  4NO2(g) + 6H2O(g)
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Steps for solving equilibrium constant problems:
1. Balance reaction
2. Products over reactants
3. Substitute the concentrations in for the species concentrations
4. Calculate
4
6
K= [NO2] [H2O]
4
7
[NH3] [O2]
You could also calculate the equilibrium constant for the reverse reaction. In the reverse reaction the
products are now reactants and the reactants are now products, because the reaction is going the
`
direction. This is typically symbolized by a K prime (K )
`
4
7
K = [NH3] [O2]
4
6
[NO2] [H2O]
An equilibrium constant (K) is without units and are for specific temperatures, meaning if the temperature
changes the K will also change.
Calculations of K vs. Q
Q is the reaction quotient. The reaction quotient is obtained by using the law of chemical equilibrium with
the initial concentrations rather than the equilibrium concentrations. In a situation where it is unclear if the
reaction is at equilibrium it is possible to use the concentration of the reaction species and compare the
reaction quotient to the known equilibrium constant.
If K < Q then the reaction will form more reactants, this is a shift to the left.
If K > Q then the reaction will form more products, this is a shift to the right.
If K = Q then the reaction is at equilibrium.
As a pneumonic device, if the comparisons are written as K vs. Q then the direction of the less than or
greater than sign will show the direction of the shift.
Calculating equilibrium constants of heterogeneous equilibria
Heterogeneous equilibria are equilibrium conditions that have more than one state of matter present.
While homogeneous equilibria all have the same phase of matter. Different states of matter are treated
differently in an equilibrium condition. For example solids are usually omitted from the calculation. This is
because all of the calculations that are done use concentrations. The concentration of a solid will not
change. If there is a solid piece of copper it has a concentration of 100%. If you cut this piece of copper
in half it is still all copper, and therefore has a concentration of 100%. 100% is one. The is an idea
termed unity. Unity allows all solids and liquids to have the concentration of one in an equilibrium
constant calculation. Multiplying and dividing by one has no affect on K and therefore solids and liquids
are removed from the calculation.
Sample:
CaCO3(s)  CO2(g) + CaO(s)
K = [CO2] [CaO]
[CaCO3]
Both calcium carbonate and calcium oxide can be omitted by unity and the resultant equilibrium
expression is:
K = [CO2]
In truth gasses are often represented by their pressure rather than concentration, this idea will be fully
explored later.
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Relevance of Equilibrium Constants
At this point it is important that we understand why equilibrium constants are calculated. With an
equilibrium constant calculated it is possible to some extent to determine the tendency for a reaction to
occur. Remember just because a reaction can be written on a chalkboard and balanced does not mean it
will occur. An equilibrium constant can tell if the tendency is present, however it is not an indication of the
speed at which the reaction will occur. Calculating equilibrium constants is also an identifier of a reaction
that is at equilibrium or if not what direction it will shift, based on initial concentrations.
Tendencies based on K depend on the distance K is from one. If a K is much, much greater than one
(K>>1) then the reaction will have a tendency to proceed forward (form products). If K is much, much
less than (K<<) then the reaction will have a tendency not to proceed (form reactants). Again the
magnitude of K is not indicative of the speed of the reaction. The reaction rate and thermodynamics are
required to determine the speed of the reaction.
The importance of energy and states of matter for reactions in equilibrium
LeChatelier's Principle (Lay-shot-lee-air) states that when a change is imposed on a system at
equilibrium, the position of the equilibrium shifts in a direction that tends to reduce the effect of that
change. In English this means that in a chemical reaction if you add more reactants (stress the system)
the reaction will make more product to maintain equilibrium. If the system makes more product it is said
to shift to the right, and if the system makes more reactant it is said to shift to the left.
At this time it is important that we recognize that energy can be a product or a reactant. If the reaction is
endothermic the energy is a reactant, and if the reaction is exothermic then energy is a product.
Exothermic
Reactant  Product + energy
Endothermic
Energy + reactant  product
The effects of LeChatelier's principle can be seen by adding heat (energy) to a endothermic reaction. If
the system is stressed by adding heat it will form more products in order to maintain equilibrium, this is a
shift to the right. However if heat (energy) is added to an exothermic reaction, the heat will apply stress to
the product side and the system will maintain equilibrium by creating more reactants, a shift to the left.
States of matter are also effected and therefore important in equilibrium reactions. Solids and liquids are
considered to be at unity and ignored. Aqueous solutions are used with their concentrations. Gasses
however are typically defined in terms of their pressure, and not in concentration. As you recall the ideal
gas law states that, PV = nRT. Algebraically the following is also true:
(n/V) = (P/(RT)). n/V is replaced with C, C stands for the molar concentration of a gas. The ideal gas
law can be rewritten as P=CRT.
For the gaseous reactions equilibrium is still product over reactant, but now it is in terms of pressure, Kp.
Kp = (Pproduct)
(Preactant)
The pressure can affect equilibrium conditions when a gas is involved. When stressed a reaction at
equilibrium will shift in a direction that has the fewest number of moles of gas.
For Example:
2H2(g) + O2(g)  2H2O(g)
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The above reaction has 3 moles of gas on the reactant side and only 2 moles of gas on the product side.
Under a high pressure condition the reaction will shift to the right, product side. The right shift occurs
because all gasses take up the same volume (1 mole = 22.4 L at STP) under the same conditions. Two
moles of gas requires less space than three moles.
With this understanding of equilibrium and the effects of pressure on gaseous species we can now
explain why a can of pop will become "flat" after it is opened.
H2CO3(aq)  H2O(l) + CO2(g)
Carbonic acid is in a dynamic equilibrium with water and carbon dioxide. If you shake up a can of pop
you release gas (CO2) but if you allow that can to sit it doesn't explode when you open it. The increase in
pressure affects the equilibrium of the system. Under pressure the system will favor the side of the
reaction with the fewest number of moles of gas. In this case the pressure stresses the product side and
forces the carbon dioxide to react with water to form more carbonic acid. This is a shift to the left and
returns the pop to equilibrium. If the can is open pressure is removed. Under a low-pressure system an
equilibrium system will favor the reaction side with the largest number of gas moles. As your can of pop
sits open the carbonic acid is continually forming more product, because the carbon dioxide is being
removed and a right shift is required in an attempt to maintain equilibrium. This same idea on a larger
scale is what allows many industrial chemical processes to be economically feasible.
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IN WORLD
Clara Immerwahr was the courageous chemist who dared to take a lone stand against her husband and
an entire empire in opposition to chemical warfare. How will she be remembered in the International Year
of Chemistry? Last month marked the birth date of a very unique and special woman.
The date of June 21, marked the birthday of Clara Immerwahr.
Born on that date in 1870, Ms. Immerwahr was a Jewish-German chemist who is best remembered for
her stance against chemical weapons. Her position against the use of chemical weapons during World
War One placed her at odds with her husband, Fritz Haber, also a fellow chemist who not only supported
the use of such weapons but played the lead role in production of toxic gas in what was then known as
the German Empire. Her position also placed her at odds with the ruling establishment of that empire.
Her continued outspoken disdain for chemical warfare would earn her pariah status in German society as
her husband would deem her a traitor to the Fatherland. At one point Ms. Immerwahr even condemned
chemical warfare as a "perversion of science." Germany’s first major test with chemical weapons would
be the use of chlorine gas against Allied troops at Ypres on April 22, 1915. Haber directed the chemicalbased massacre that left 5,000 Allied troops dead and another 10,000 maimed. Germany's use of
chlorine on that day would spark a brutal chemical weapons arms race between it and the Allied forces
that would lead to the deaths of nearly 100,000 troops on both sides before the war was finally over.
Upon returning to the German home front from Ypres, Haber was celebrated nationally as a war hero.
The German establishment and the newspapers were full of praise while the Supreme War Council
promoted him to the rank of Captain. However, Haber did not receive such a warm homecoming from his
wife when he returned to their Berlin home. Ms. Immerwahr was absolutely livid and tore into Haber for
planning and then carrying out the chemical attack at Ypres while Haber in turn blasted her as a traitor to
the Fatherland. No longer able to convey in mere words to her husband or the empire the disdain she had
for chemical weapons, Ms. Immerwahr would now resort to a different but more darker and violent
medium to express her disapproval. In the darkness of the early morning hours of May 2, 1915, Ms.
Immerwahr borrowed her husband’s revolver as he slept and went into the garden of their home.
It was there that she leveled the firearm at her left breast and pulled the trigger.
Ms. Immerwahr shot herself through the heart and would bleed to death in the garden.
Her death is mired in controversy, not so much because of how she died and why, but rather because of
the manner in which Haber and the authorities handled her death.
Despite being the wife of Germany’s biggest war hero at that time there was only scant coverage of her
death in the local media. Even more suspicious and downright sinister was the destruction of a suicide
letter that she had left behind.
Today, Ms. Immerwahr is remembered as a model of civic courage and humanity. An example of the
social and moral responsibility of scientists. She is remembered as someone who took a stand against
injustice and never backed down or compromised her beliefs. Even in the face of severe social pressure
she never compromised her beliefs. She believed so passionately in the cause of science being used for
humanity that she would take her very own life in the name of it. To understand the phenomenal woman
that Ms. Immerwahr was, it is important to take a brief look back at a what appears to be a particularly
defining event in her life: her university graduation. It was on December 12, 1900, that Ms. Immerwahr
was awarded a doctorate in chemistry and graduated magna cum laude from the University of Breslau.
She was the first woman to ever receive a Ph.D in Germany. A local newspaper covering the historic
national achievement was there when Ms. Immerwahr recited a professional oath to, “never in speech or
writing to teach anything that is contrary to my beliefs. To pursue truth and to advance the dignity of
science to the heights which it deserves.” Ms. Immerwahr personally felt that chemistry should be a
science that is used to advanced and better the human condition. Ms. Immerwahr lived by those beliefs
and eventually died by them as well. Interestingly, those very same sentiments that she lived and died by
are also echoed in the International Year of Chemistry 2011 (IYC 2011) and are actually the basis for it.
A few months ago in April, United Nations Secretary General Ban ki-Moon delivered a speech at the
Remembrance Day for all Chemical Weapons Victims, an annual special event held by the Organization
for the Prohibition of Chemical Weapons at their Hague-based headquarters.
In his speech, Ban was quoted as “declaring 2011 the International Year of Chemistry in order to
celebrate chemistry as a science of peace and progress.”
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IYC 2011 is an international year-long event celebration of the science of chemistry and is being
coordinated by the International Union of Pure and Applied Chemistry and the United Nations
Educational, Scientific, and Cultural Organization. The event commemorates the achievements of
chemistry, and its contributions to humankind. Ms. Immerwahr’s personal beliefs and ideals on chemistry
mirror those of IYC 2011, especially the part about chemistry being a science of peace and celebrating
it’s contributions to humanity. I see much of Ms. Immerwahr’s presence in IYC 2011.
Writing this article has been a profoundly poignant and intimate experience for me.
Though Ms. Immerwahr and myself are several eras apart, her views and beliefs resonate with me.
admire her more than any other person. She is a personal role model of mine and I have always admire
her deeply for her convictions. I am currently studying in the hopes of one day being a counterproliferation specialist. Such a person works to keep Weapons of Mass Destruction (WMD) from those
who should never have it, namely, terrorist organizations and rouge nations as well as working to
significantly reduce such destructive armaments in nations that already have them.
As a counter-proliferation specialist I particularly hope to specialize in the section of WMD that involves
chemical weapons. I’m also interested in becoming a hazardous materials specialist. Though these are
normally two separate careers, they share many parallels and I strongly believe that I may be able to
marry them together. I have dedicated my future career in public safety/national security to Ms.
Immerwahr. I sometimes daydream about what it would have been like to follow her about as she went
about her day while heeding professional and scientific advice offered by her. Taking notes on a clipboard
or in a notebook with due diligence as any good intern or protégé would.
Under her tutelage and guidance I would have been an even more exceptional asset to the public safety
profession and society. I think you can tell a lot about a person by their personal heroes and role models
or those they seek to emulate. Obviously, a man or woman who seeks to emulate the lifestyle of a
notable missionary, human or civil rights advocate, social worker or an investigative news journalist who
highlights stories on injustice will have a different outlook and values than someone who seeks to emulate
a notable celebrity, socialite or business tycoon/mogul. Since the 70’s, Germany has sought to make
amends with Ms. Immerwahr for it’s poor treatment of her when she was alive.
While never truly admitting that the use of chemical warfare was wrong in the defense of national
interests during World War One, Germany has now acknowledged Ms. Immerwahr’s stance against the
use of such weapons as courageous.
In honor of her and in her name, the German state has significantly increased funding in it’s educational
system for science programs aimed at female students.
There is a street in Berlin that is named in her honor while the house that she shared with her infamous
husband has been declared a national landmark. Germany does not produce or stockpile any chemical
weapons today and is in fact a significant counter-proliferation force as it works with other European
Union nations and the United States to tamp down the spread of WMD – chemical, biological,
radiological/nuclear weapons. This is clearly a different Germany and a different era. Through it all Ms.
Immerwahr remained true to that oath she took so many years ago when she officially became a chemist,
the same oath that she recited in a local newspaper that would carry her words to the public on a national
level: “never in speech or writing to teach anything that is contrary to my beliefs. To pursue truth and to
advance the dignity of science to the heights which it deserves.” By remaining true to that oath, she
remained true to herself.
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Don't Rock the Boat (An Equilibrium Lab)
CuCl4(aq)  Cu+2(aq) + 4 Cl- + Heat
Green
Blue
Clear
Objective:
Determine the effects of temperature and concentration on an equilibrium reaction.
Materials:
dropper bottles of 4 M NaCl
6-13x 100 mm test tubes
2-250 ml beakers
8 well strips
dropper bottles of 1.5 M CuCl2
Ice
hot plate
3- pipettes
well plates
dropper bottles of 0.05 M AgNO3
Procedure:
Part A (temperature)
1. Start a hot water bath and an ice bath both in a 250-ml beaker.
2. Put 5.0 ml of 1.5 M Copper (II) chloride in a clean-labeled test tube place in a test tube rack this will
be the reference test tube.
3. Again put 5.0 ml of 1.5 M copper (II) chloride in a clean test tube warm the test tube in the hot water.
Record observations and equilibrium shift if present.
4. Place the hot test tube in the ice bath record observations then remove from the ice bath.
Part B (concentration)
1. Clean all well strips.
2. Place a few drops of 0.05 M silver nitrate in a clean-labeled well. Caution silver nitrate will stain
skin and clothing.
3. Place 3 drops of room temperature Copper (II) chloride solution in two wells. One well will be used as
a reference.
4. Add 2 drops of sodium chloride solution, a drop at a time, to one of the wells. Record observations.
5. Add 1 drop of silver nitrate to the same well. Record observations.
Data:
Arrange a neat, well-organized summation of you observations for each step of the procedure. The table
should be two columns one with chemical reactions and the other with observations. the headings should
include temperature and color.
Questions/conclusions:
1. Is the reaction endothermic or exothermic?
2. What effect does cooling the solution have on the color?
3. Why would sodium chloride affect the equilibrium?
4. How is it possible to have two ions of the same compound (Cu+2 and Cu+4) that have different colors?
5. Write a rule for predicting the effect of concentration change of one ion, has on the other chemical
species.
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Lab discussions for Equilibrium labs: Beaker lab and Don't rock the boat.
Beaker lab:
The overall objective of this lab is to again allow you the student with a visualization of what is equal in an
equilibrium reaction. As you saw demonstrated the reaction is said to be in equilibrium when the forward
reaction and the reverse reactions are equal. It is the amount being transferred between reactant and
product that is equal. The ability of a reaction to reach equilibrium is irrelevant to the initial concentrations
of product and reactant. Nor does the ability of a reaction to reach equilibrium depend on the speed of
the forward or reverse reactions (represented by the amount of water being transferred).
LeChatlier's principle that if a system at equilibrium is stressed it will form more products or more
reactants in such a way as to remove the stress and return to equilibrium. Although both of the
concentrations of product and reactant will change to reestablish equilibrium the overall equilibrium
constant will remain constant because both concentrations were changed. LeChatlier's principle was
demonstrated by stressing a system at equilibrium. The addition of water to either the products or the
reactants causes a stress on the system. The system will correct the stress in order to maintain
equilibrium by producing more products or more reactants. This stress on the equilibrium system will
require the reaction to shift to the left (form more reactants) if water was added to the product side, or
shift to the right (form more products) if water was added to the reactant side. While it is possible that the
water level in the product beaker and reactant beaker (representing the concentration of products and
reactants) was equal in some situations those concentrations were not equal in all situations. If the
volumes of water were not equal (which they were not) in all situations then it is not the concentrations of
product and reactant that are equal in an equilibrium reaction by the forward and reverse reactions that
are equal.
Don't Rock the Boat lab
CuCl4(aq) +  Cu+2(aq) + 4 Cl-(aq) + heat
(green)
(blue)
(clear)
This laboratory exercise is a demonstration in the effects of stressing a system at equilibrium. The
laboratory exercise shows the truism in LeChatlier's principle. The stress applied to the reaction is as
follows: Part A the presence of heat and the removal of heat. When the test tube is placed in hot water
the amount of heat is raised. The increase of heat applies stress on this exothermic reaction. By adding
heat the solution turns green because the reaction is required to form more reactants (shift to the left) in
order to offset the increase of heat which is a product. When the same test tube is placed in cold-water
heat is removed. If heat is removed the reaction will shift to the right and form more products, turning
blue. The conversion of reactants (copper IV chloride) to products diminishes the concentration of
reactants. Heat as a product is removed and in an effort to retain equilibrium the reaction produces more
product to achieve equilibrium. Part B is a manipulation of concentration. With the addition of aqueous
sodium chloride the product side of the reaction is stressed because there is an increase in the
concentration of chloride ions. The increase in the concentration of chloride ions creates a shift to the left
and more reactants are formed, a green color should be present. The addition of silver nitrate removes
chloride ions. Silver ions react with the chlorine ions and produce and insoluble precipitate. The removal
of chloride ions requires a shift to the right to produce more product. This shift creates a blue color. This
color can be a hard to see due to the white precipitate caused by the silver chloride. In conclusion, in any
chemical reaction at equilibrium changes in the concentration of one ionic species will effect the
concentrations of all ionic species in that system.
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Reaction Quotient and Equilibrium Constants
For each of the following conditions of the Haber-Bosch Process determine if the
scenario is at equilibrium OR which direction the reaction will proceed to each
equilibrium. If the concentrations are taken at 500 oC where the reaction has an
equilibrium constant of 6.02 x 10-2.
N2(g) + 3H2(g)  2NH3(g)
1. [H2]0 = 2.0 x 10-3 M
[N2]0 = 1.0 x 10-5 M
[NH3]0 = 1.0 x 10-3 M
2. [H2]0 = 3.54 x 10-1 M
[N2]0 = 1.50 x 10-5 M
[NH3]0 = 2.00 x 10-4 M
3. [H2]0 = 1.0 x 10-2 M
[N2]0 = 5.0 M
[NH3]0 = 1.0 x 10-4 M
4. [H2]0 = 1.197 M
[N2]0 = 3.99 x 10-1 M
[NH3]0 = 2.03 x 10-1 M
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Equilibrium Constants and Calculations
1. Write a correct equilibrium expression for the following reaction:
NOCl(g)
NO(g) + Cl2(g)
2. Calculate showing all steps the equilibrium constant for the combustion of sulfur
dioxide, SO2(g) + O2(g)
SO3(g), if the equilibrium concentrations are as
follows [SO2] = 0.590 M, [O2] = 0.0450 M, and [SO3] = 0.260 M
3. Calculate the equilibrium constant for the thermal decomposition of mercuric oxide,
HgO(s)
Hg(l) + O2(g), if the equilibrium concentrations are as follows: [Hg] =
38.3 M, [HgO]= 13.2 M and [O2] = 0.0446 M.
4. Determine the equilibrium constant for the following equilibrium concentrations (
[NH3] = 0.0446 M, [O2] = 0.0450 M, [NO2] = 0.0449,
[H2O] = 56.8 M, if the following reaction is at equilibrium:
NH3(g) + O2(g)
NO2(g) + H2O(l)
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Chemical Equilibrium Practice
Directions: Showing all work on a separate piece of paper determine the unknown concentrations based
on the equilibrium constants for the reactions provided.
1.
Cl2(g) + NaBr(aq)
NaCl(aq) + Br2(g)
Keq= 6.34x 103
[Cl2] = 5.36 x 10-3
[NaBr] = 1.25 x 102
[NaCl] = ?
-2
[Br2] = 2.70 x 10
2.
Ca3(PO4)2(aq) + H2SO4(aq)
CaSO4(aq) + H3PO4(aq)
Keq = 3.62x10-2
-5
[Ca3(PO4)2]= 5.36 x 10
[H2SO4] = 6.30
[CaSO4] = ?
[H3PO4] = 9.86 x 10-2
3.
Al(OH)3(aq) + H2SO4(aq)
H2O(g) + Al2(SO4)3(s)
Keq = 6.23x10-4
[Al(OH)3]= 2.36 x 10-3
[H2SO4] = 3.78
[H2O] = ?
[Al2(SO4)3] = 5.62 x 10-4
4.
H3PO4(aq) + H2O(l)
H3O+(aq) + PO4-3(aq)
[H3PO4] = 1.23 x 102
[H2O] = 5.65 x 101
[H3O+] = ?
[PO4-3] = 8.52 x 10-6
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Keq = 3.89x10-9
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Unit 10 Objectives for Solution Chemistry

Calculating concentrations
Molarity
Molality
Normality

Application of polarity
Hydrogen bonding
Surface tension
Capillary action

Saturation levels
Unsaturated
Saturated
Supersaturated

Solubilities
Calculating solubility constants

Heat of Solution
Enthalpy

Raoult's Law
Freezing point depression
Boiling point elevation

Beer's Law
Application of Spec 20

Data Manipulation

Story problems to data tables
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Solution Chemistry
Solutions are not just liquids. A solution can be a mixture of any two states of matter.
Examples:
Gas in a liquid
CO2(g) in water for carbonated water (pop)
Gas in a gas
Air is a mixture of N2(g), CO2(g), O2(g), SOx(g)
Solid in a solid
Alloy such as Brass, Bronze or steel
Solid in a liquid
Amalgams like mercury fillings in your teeth
Some basic definitions:
Solute: Medium being dissolved (less than 50%) such as the small amount of kool-aid powder that is
dissolved into a large amount of water. A solute could also be the small amount of carbon dioxide gas
that is dissolved in water of your pop or the oxygen dissolved in your blood.
Solvent: Medium doing the dissolving (greater than 50%) A solvent represents the larger amount of the
two components of a solution. The solvent is the blood into which the oxygen is dissolved or the water
into which the carbon dioxide is dissolved.
Classifications of Mixtures
Mixtures fall into one of three classifications depending on a set of rigid characteristics, based on the
properties of the particles in the mixture. There are solutions, colloids, and suspensions.
True solutions
Are homogenous mixtures, (Homo-meaning same) this means that the mixture is completely
uniform. The solute in a solution can not be seen, the solute is usually ions or molecules in solvent. They
will not settle out over time. A true solution does not exhibit the Tyndall effect. Tyndall effect is when a
particle in a mixture is able to scatter light. Common examples of the Tyndall effect are seen at laser
shows at concerts, or when the head lights of a car are seen on a foggy room. Some movies will show
the Tyndall effect as a thief tries to avoid an infrared laser by blowing smoke on the laser and then
stepping over the alarm.
Colloids
Colloids are also homogenous mixtures, they are not however a true solution. Colloids are not
considered solutions because their particle size is too large. While the particles of a colloid will not settle
out they will scatter light and therefore exhibit the Tyndall effect. The sky is blue because the air is a
colloidal mixture. The average particle size in the Earth's atmosphere is just the right size to scatter blue
light. When the sun shines on the earth all of the blue light is scattered, which is why the sky appears
blue everywhere you look. This is the same reason sun appears to be orange/red. With all of the blue
light removed by the colloidal particles looking directly at the sun only allows the red end of the spectrum
to come through, creating an orangish/red orb.
Suspensions
Suspensions are heterogeneous mixtures, the solute particles are larger than atoms, ions, and
molecules and can often be seen with the naked eye. Suspensions will settle out over time, much like hot
chocolate that is allowed to set or Tang. The settling out leaves a residue on the bottom of the container.
These large particles will show a Tyndall effect, however the mixtures are often so clouded the beam of
light can not penetrate the mixture, making the Tyndall effect hard to see.
Measuring the amount of solute in a solution (Concentration)
It is never enough that we know we have a solution, it is imperative that we know exactly how much
solute we have in a solution. The concentration of a solution is a measurement of, or ratio of solute to
solvent. In chemistry there are three types of concentrations measurements used most commonly,
molarity (M), molality (m) and normality (N). Of these three molarity is used most often.
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Molarity (M)
Molarity is a measure of the number of moles of solute per liter of solution.
M =
moles of solute
Liters of solution
The procedure to make all concentrations is important, for molar solutions the number of moles
(measured in mass) is added to a volumetric flask. The volumetric flask is then filled half way and the
solute is dissolved. Once all of the solute is dissolved then the remainder of the distilled water can be
added to the etched line on the volumetric flask. This must be done to prevent swelling of some
solutions.
Molality (m)
Molality is the moles of solute over the kilograms of solvent. The calculations are similar but the
preparation is different.
m =
Moles of solute
Kg of solvent
One liter of pure water at STP will have a density of 1 g/ml and therefore a mass of 1 kilogram. How the
solution is prepared is slightly but importantly different from molarity. The moles of solute is added
directly to the total mass of the solvent. For example is 1 kilogram of water was to be used then the mass
(number of moles) of solute would be added to 1 liter of water, the final volume of solution will more than
likely be greater than 1 liter. In dilute concentrations molarity and molality are almost identical.
Normality (N)
Normality is usually used only in acid base or titration chemistry. Normality is the number of equivalents
over liters
N=
number of equivalents
Liter
An equivalent is the mass of acid or base that can donate or accept 1 mole of hydrogen atoms. The
number of equivalents in one mole of H2SO4 (sulfuric acid) is two, because one mole of this compound
could donate 2 moles of hydrogen atoms if it were to undergo 100% dissociation.
Some other concentration measurements
Two other concentration measurements that you may come across are % m/m and %m/v. These are
percent mass to mass and percent mass to volume. The are calculate similar to any percentage, be
careful to divide by the total mass.
% m/m = mass solute
=
Xa
a
B
mass of solution
(X + X )
a
a
Where X is the mass of the solute and X + XB is the mass of the solute and the mass of the solvent.
Solubility
Solubility is the chemistry that explains why and how a compound is dissolved into another compound.
The cardinal rule of solubility is that like dissolves like. This rule is in reference to the polarity of the
substance. A polar compound will dissolve a polar compound and a nonpolar compound will only
dissolve a nonpolar compound. This is why Styrofoam (nonpolar) doesn’t dissolve in water (polar) but
does dissolve in gasoline (nonpolar).
The solubility of most solutes increases with temperature, as temperature increases you are able to
dissolve more solute in the solvent. Some exceptions to this rule are NaSO 4 (decreases) Cesium sulfate
(decreases) and sodium chloride (unaffected). Warm kool-aid will be able to hold more sugar than cold
kool-aid. Gasses however decrease in solubility as temperature increases. There is more oxygen gas
dissolved in cold water than in warm water. This is why fish that require high amounts of oxygen (brook
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trout, walleye) are found in cold water and fish that require very little oxygen or can surface breath (carp,
goldfish, beta) are found in warm water where there is less oxygen.
Saturations
Solubility is often discussed in terms of unsaturated, saturated, and supersaturated. All three
classifications are in terms of temperature because a solution that is saturated at one temperature may
be unsaturated at a higher temperature. Unsaturated solutions are able to dissolve more solute at that
temperature. Saturated solutions are unable to dissolve more solute at that temperature, and are already
"carrying" as much solute as they can. Supersaturated solutions are highly unstable; they have more
solute dissolved in the solvent than the solvent can hold at that temperature. A nonchemical
representations of this phenomenon is that of bench pressing. If a spotter helps you lift an amount of
weight that you cannot normally lift, and you are able to fully extend and lock your arms you can hold the
weight up. However, this is highly unstable and one wrong move and the weight will quickly come
crashing down. In order to make a supersaturated solution, the solution must be saturated at a higher
temperature and then slowly cooled. Supersaturated solutions will crystallize if stressed. They can be
stressed in one of two ways, scratched or seed crystal. Scratching the side of the reaction vessel creates
vibrations in the solution and the additional solute crystallizes out of solution. A seed crystal is a small
crystal of the solute that is added to the supersaturated solution. The solution will begin to grow crystals
around the seed crystal until the entire solution is either crystallized or at a point of saturation.
Factors effecting solubility
As previously discussed temperature can and does affect the solubility of solute in a solvent. Structure
and pressure will also affect solubility. The structure of a molecule, number of polar bonds or the number
of non-polar bonds will determine how well the material dissolves in a solvent. Some molecules that are
large and well branched also offer steric hindrance, which means that their atoms get in the way of other
atoms, limiting the amount of dissolving that can occur. Pressure will also affect solutes that are in the
gas phase. Henry's law states that the amount of a gas dissolved in a solution is directly proportional to
the pressure of the gas above the solution. In short, if the gas is under a high-pressure more of the gas
molecules will be forced into the solution. If the gas is under a low pressure more of the gas will be
allowed out of the solution and therefore solubility will decrease at low pressures for gasses.
Everyday Applications of solubility (other than fishing)
Hot water pipes are often plagued with a calcium plaque choking of the pipes. This phenomenon requires
two factors, hot water and hard water. Hard water is water that has a high mineral (rock) content. The
minerals are dissolved in the water. Typically this mineral is calcium and occurs in areas where there are
large limestone aquifers. In the hot water heater the water is heated up which increases the amount of
calcium ions that can dissolve in the water. As the water travels away from the hot water heater it is
cooled down and the solubility of calcium at the lower temperatures also decreases. As the calcium is
pulled out of solution it creates a rock hard mineral lining on the inside of the pipes. This is the same
white scale that can be seen on showerheads.
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Calculating and Making Solutions
Starting with a Solid and making a Molar concentration
Total volume wanted X Concentration wanted X molecular mass of solid
______ mL x
1 L x ____Mols solid x ______ g Solid = Mass of solid
1000 ml
1 Liter
1 Mol solid
The mass of solid is added to a volumetric flask determined by the total volume wanted
(i.e. if you want 250 ml then use a 250 volumetric flask). Distilled water is then added
up to the etched line on the volumetric flask. Only fill the volumetric flask half full,
dissolve the entire solid and then fill the remainder of the flask.
On some occasions as a substance dissolves it will increase the total volume. If you
start out by filling the flask up to the etched line your final volume will be more than what
you want and the concentration will be more dilute than what is required.
Starting with an aqueous and making a molar concentration
Desired molarity X desired total volume = concentrated molarity X volume of Concentrated
(M1V1) = (M2V2)
Or
V2 = (M1V1)
M2
You will be solving for the volume of concentrate (or high concentration) to be added to
water. This will make a more dilute concentration. Again you will use a volumetric flask
in the same manner as above. First add the concentrated solution (stock) and then
dilute to the etched line.
When working with strong acids it is important that the acid is added to the water. The
acid is typically denser, if water is added to the acid it can "bounce off" the acid surface
carrying some of the acid with it and cause facial damage. Therefore our technique will
not work with the volumetric flask. You need to measure out the acid in a graduated
cylinder and calculate the amount of water required to reach the desired total volume
(total volume = V2 +DIwater).
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Molar Absorptivity and Beer-Lambert Law (Spectral Analysis)
Spectroscopy is the technique most often used for modern chemical analysis. Spectroscopy is
the study of electromagnetic radiation emitted or absorbed by a given chemical species. Since
the quantity of radiation absorbed or emitted can be related to the quantity of the absorbing or
emitting species present, this technique can be used for quantitative analysis. There are many
spectroscopic techniques due to the width of the electromagnetic spectrum. The differences in
technique are really more of a distinction in the wavelength of light being used for the analysis.
In short spectroscopy is a measure of how much color a solution has.
If a solution is colored, it is because some portion of the solution (solute or solvent) is absorbing
some portion of the visible spectrum. The more the light absorbed the darker the color appears.
This does work in an antagonistic fashion where a color that is absorbed will show its
complimentary color. If a solution appears green it is absorbing red light.
The quantity of light being absorbed can be measured using a spectrophotometer. We will be
using the workhorse of all spectrophotometers, the Spec 20. This is a Milton - Roy product that
you will be most likely to see in your college quantitative analysis courses.
In general a Spec 20 will emit light in all frequencies of visible light and some infrared or
ultraviolet. A prism separates the light into specific frequencies and then only the specified
frequency is allowed to pass through the sample. The initial light beam is called the incident
light (Io) and the light after it has passed through the sample (I) are set up into a ratio called the
percent transmittance (I/Io or %T). The amount of light absorbed, absorbance (A) is the inverse
log of the transmittance.
A = -log (I/Io) or A = -log(%T)
Enter Beer - Lambert Law
The Beer - Lambert law (Beer's Law) is a mathematical expression to determine the
concentration of a solution based on the absorbance of a solution at a specific frequency of
light.
A = lc
Where:
A is the absorbance (ratio of incident light to transmittance light)
 is the molar absorptivity of a compound (L/(mol cm)
l is the path length (distance in cm light travels through sample)
c is the concentration of the sample ( mol/ L)
The Beer - Lambert expression (A = lc) will yield a linear expression except in extreme
concentrations. Therefore it is often combined with graphical analysis to determine unknown
concentrations or molar absorptivity, where the slope will be the molar absorptivity times the
path length of the sample.
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Applying Standardization curves for Spectroscopy
Standardization curves (or plots) created from known concentrations and empirically determined
(experimental) absorbances. Standardization curves are created for specific compounds at specific
wavelengths. Once the curve is developed it can be used to calculate any concentration (except
extremely low or extremely high) based on the absorbance of an unknown concentration. Use the
contrived standardization plots below to answer the questions posed.
1. Calculate the concentration for a solution of
Compound M if the absorbance is 7.34.
14
12
10
8
6
4
2
0
y = 1.9943x + 0.02
R² = 0.9997
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
5.5
6
6.5
Concentration (M)
Standard Absorption Curve for Compound P at 438 nm
2. What is the theoretical (expected) absorbance of a
4.358 M solution of Compound P?
3.0
2.5
2.0
1.5
1.0
0.5
0.0
y = 0.2429x + 0.9
R² = 0.9783
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
Concentration (M)
Standard Absorption Curve for Compound Q at 438 nm
3. What is the concentration of a
solution of Compound Q if the spec-20
shows a 0.41% transmittance?
Absorbance
Absorbance
Absorbance
Standard Absorption Curve for Compound M at 438 nm
3.0
2.5
2.0
1.5
1.0
0.5
0.0
y = 0.9593x + 0.901
R² = 0.9779
0.3
0.5
0.8
1.0
Concentration (M)
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1.3
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Beer-Lambert Practice
Directions: Answer all questions with the required math work shown. Any and all Illegible answers will be
marked incorrect.
1. What is the molar absorptivity of a 3.8 M Iron (III) sulfide solution if the absorbance is 0.879 in a spec
20 with a path length of 1.00cm?
-1
-1
2. What is the theoretical absorbance of CuCl 2 if the molar absorptivity is 0.687 L Mol cm and a
concentration of 4.50 M in a spec 20?
3. A sample of CuCl2 has a 22.0% transmittance if the molar absorptivity is 0.687 L/mol cm. What is the
concentration?
4. What is the molar absorptivity of a 4.175 M CoCl2 solution in a pathway of 1.00 cm if the absorption is
0.6576?
5. If the absorption of CoCl2 of 0.937 in the spec 20 (cuvette = 1.00 cm) what is the concentration of the
solution?
6. What is the concentration of compound X if an unknown absorbance of the compound is 0.453?
Standard Absorption graph of compound
X at 578 nm
Absorption at 578 nm
1
0.8
0.6
0.4
y = 0.1x + 0.1
0.2
0
0
1
2
3
4
5
6
7
8
Concentration of X (Molarity)
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Spectral Chemistry (Spec 20) A Beer's Law Lab
Objective:
Determine the molar absorbtivity of a copper ion in solution and determine your lab savvy in making
.
0.0400M CuSO4. Copper (II) Sulfate pentahydrate is a hydrated crystal: CuSO4 5H2O
Procedure:
1. Turn on the Spec 20 and allow at least 15 minutes for the machine to warm up. The power dial is on
the front left, it can be turned to the right until the needle on the read out is on the zero (use the
mirror behind the needle so that only one "needle" is visible.
2. Using proper technique you need to make a 0.0400 M CuSO4 solution. REMEMBER that you need
less than ten milliliters for analysis and that copper nitrate is a trihydrate. Copper sulfate pentahydrate
may be substituted for this lab.
3. Calibrate the spec twenty for 815 nm. First set the wavelength dial (top right) to 815. Then with
nothing in the sample slot use the left front dial to adjust the needle to zero. Place your blank in the
sample slot, wipe all of the fingerprints, water… from the outside of the cuvette with a kem-wipe. With
the sample door closed adjust the transmittance dial (right front) to 100%.
4. As you insert your samples of known concentration wipe them with kem-wipes and record their %
Transmittance. It is important that all of your samples are aligned properly in the Spec 20.
5. Record the % transmittance of your sample.
6. RETURN ALL SAMPLES to correct bottles
Data:
Your data will include a two-column table with the known concentrations and their percent transmittance,
as well as the absorbance of your 0.0400 M concentration made from the hydrated crystal and the
anhydrous crystal. Your data will need to include the mass of your CuSO4. Your graph will show
concentration versus absorbencies. Computerized graphing will allow you to find the equation of the line
more easily than hand graphing. The equation of the line needs to be determined regardless of the
method of graphing chosen.
Calculations:
Graph the known concentrations versus the absorbance.
Show all calculations including: Solution Prepration, absorbance calculations and determination of
concentration of prepared solution based on graphical analysis.
Using your graph determine the molar absorptivity.
Determine your percent error and why you chose the theoretical value that you did
Questions:
1. Why is the cuvette of distilled water placed in the sample holder when the transmittance is set at
100%?
2. Explain why a standardization curve for copper(II) nitrate could be used to determine concentration of
copper (II) sulfate.
3. Explain why it is your graph should be linear and positive.
4. What is a possible source of procedural error?
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Kool Spectroscopy Lab—A Beer’s Law Lab
Objective: Prepare a 0.045 M solution and determine your precision based on the standardization curve
you create for the compound. Use the standardization curve to calculate the molar absorptivity of the
unknown compound.
Materials:
Standard solutions:
0.01 M KA 65
0.025 M KA 65
0.05 M KA 65
0.10 M KA 65
0.25 M KA 65
Procedure:
1. Turn on the Spec 20 and allow at least 15 minutes for the machine to warm up. The power dial is on
the front left, it can be turned to the right until the needle on the read out is on the zero (use the mirror
behind the needle so that only one "needle" is visible.
2. Using proper technique you need to prepare a 0.0067 M solution. REMEMBER that you need less
than ten milliliters for analysis and that your unknown compound has a molar mass of 65.00 grams/mol.
3. Calibrate the spec twenty for 600 nm. First set the wavelength dial (top right) to 600. Then with
nothing in the sample slot use the left front dial to adjust the needle to zero. Place your blank in the
sample slot, wipe all of the fingerprints, water… from the outside of the cuvette with a Kem-wipe. With the
sample door closed adjust the transmittance dial (right front) to 100%.
4. As you insert your samples of known concentration wipe them with kem-wipes and record their %
Transmittance. It is important that all of your samples are aligned properly in the Spec 20.
5. Record the % transmittance of your sample.
6. RETURN ALL SAMPLES to correct bottles
Data:
Your data will include a two-column table with the known concentrations and their percent transmittance,
as well as the absorbance of your 0.045 M concentration. Your data will need to include the mass of your
unknown compound used in solution preparation. Your graph will show concentration versus
absorbencies. Computerized graphing will allow you to find the equation of the line more easily than
hand graphing. The equation of the line needs to be determined regardless of the method of graphing
chosen.
Calculations:
Graph the known concentrations versus the absorbance. Show all calculations including: Solution
Preparation, absorbance calculations and determination of concentration of prepared solution based on
graphical analysis.
Questions:
1. Why is it important that all of your cuvettes are properly aligned in the sample chamber?
2. Why is the cuvette of distilled water placed in the sample holder when the transmittance is set at
100%?
3. Explain why it is logical that your graph should be linear and positive.
4. What is a possible source of procedural error?
5. Determine your percent error and why you chose the theoretical value that you did.
6. Explain and show how you can determine the molar absorptivity from the graph.
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Optimization curve for Kool Spectroscopy
2.45
2.40
Absorbance
2.35
2.30
2.25
2.20
2.15
2.10
340
360
380
400
420
440
460
480
500
480
500
Wavelength (nm)
Optimization curve for Kool Spectroscopy
75.00%
70.00%
65.00%
Transmitance (%)
60.00%
55.00%
50.00%
45.00%
40.00%
35.00%
340
360
380
400
420
Wavelength (nm)
182
440
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Evenson
Sweet Tooth (Sugar Crystal lab)
Objective: To grow crystals of sugar by manipulation of solubility at various
temperatures.
Procedure:
Sucrose is soluble in a 2:1 ratio by mass (i.e. 2 grams of sugar will dissolve in every one
gram of water at 20oC)
1. Add 300.0 ml of water to a 500 ml beaker and calculate the mass of the water when
the water begins to boil, based on the density of your water when the water boils
(see chart).
2. Sucrose will dissolve at a 2:1 ratio by mass (2 grams of sugar for every 1 gram of
water). Based on your mass of water calculate the mass of sugar that will dissolve
in your water at your boiling temperature.
3. Add your calculated mass of sugar and stir to dissolve.
4. Attach a small piece of paper clip to a string and place in the sugar solution while the
solution cools.
5. Record the mass of the sugar crystals that formed.
Data:
Temp oC
Density of water (g/ml)
Data table should include:
19.0
20.0
21.0
22.0
23.0
24.0
25.0
26.0
97.0
98.0
99.0
100.0
101.0
102.0
103.0
0.9984082
0.9982071
0.9979955
0.9977735
0.9975415
0.9972995
0.9970479
0.9967867
0.9605025
0.9597951
0.9590831
0.9583665
0.9576450
0.9569200
0.9561890
temp of water:
mass of sugar added:
mass of string and clip:
mass of sugar crystal (dry):
difference in sugar added and
recaptured:
Questions:
1. Why did the sugar crystals form on the string?
2. What happened to the sugar that was added to the water but was not present in the
crystal?
3. What would happen to your crystal if you left it in your beaker and heated the water
back to the temperature recorded in your data table?
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Now You're in Hot Water! A boiling Point elevation lab
Background:
A French scientist Francois-Marie Raoult4 summarized his theories on vapor pressure in
Raoult's Law. Raoult's Law states that as the amount of solute increases in a solvent,
the vapor pressure decreases. This can be visualized with the understanding that if
solute is added to a solution it, the solute, will take up some of the space on the surface
of the solvent. The solvent can not occupy the space that is occupied by the solute and
therefore the amount of upward pressure (vapor pressure) that can be applied by the
solute is decreased. The amount of solute must be determined in moles of solute.
Simply using grams will not work because we are working at an atomic level. In order to
make comparisons between two compounds and the effects on boiling point elevation
or freezing point depression the same number of atoms must be used. Remember that
we are keeping the number of atoms (moles) constant not the mass. Molecules with
different ionization constants will produce different numbers of ions per mole when
dissolved in water.
Objective: Determine a graph of boiling point elevation for two different salts, NaCl and
CaCl2.
Materials:
NaCl
Thermometer
Weigh paper
100 ml Grad Cylinder
CaCl2
Scoopula
Graph paper
Bunsen Burner
Balance
400 ml Beakers
Procedure:
1. Without the addition of any salt bring 300 ml of water to a boil and record the
temperature.
2. Then add 1.0 gram of NaCl (0.01711 mols NaCl) to the water and record the
temperature after the salt has dissolved.
3. Add another 1.0 gram of NaCl (This is a total of 0.03422 mols) to the boiling water
and record the temperature. Repeat this process until you have added 5.0 grams of
NaCl (0.08555 mols) total.
4. Repeat the entire procedure using the same number of moles of CaCl 2. You will
need to add 1.90 grams of CaCl 2, and then consecutive amounts of: 3.80g, 5.70 g,
7.60g, and 9.50 g.
5. Under both Conditions (NaCl and CaCl 2) it is important to keep the volume of water
constant, 300 ml. You may add distilled water to the solution.
Data:
4
The background information on Raoult's Law can be referenced in : Atkins. P.W. and Beran, J.A. General
Chemistry. 2nd Ed. pp.427-432. 1992.
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Make a neat and legible data table with the following headings, including the mass of
sodium chloride, mass of calcium chloride, moles of Calcium chloride, moles of Sodium
chloride (0.01711 mols, 0.03422 mols, 0.05133 mols, 0.06844 mols, and 0.08555 mols)
and boiling temperature. You may include the number of ions for both salts, however
the calculations will be shown in the calculations section of the write-up.
SpeciesMoles Mass Temperature
NaCl 0.00000
NaCl 0.01711
NaCl 0.03422
NaCl 0.05133
NaCl 0.06844
NaCl 0.08555
Species
CaCl2
CaCl2
CaCl2
CaCl2
CaCl2
CaCl2
Moles Mass Temperature
0.00000
0.01711
0.03422
0.05133
0.06844
0.08555
Calculations:
1. Calculate the total change in temperature of each solution
2. Determine the temperature to mole ratio for each salt.
3. Graph the number of moles versus the temperature of each salt with a separate line
for each of the two salts (when using a spreadsheet us an exponential trend line).
Questions:
1. Why did the temperature increase with the addition of both NaCl and CaCl 2?
2. Did both of the two salts have temperature increases that were the same?
3. What would cause the difference in the temperatures of the boiling water if the same
amount (number of moles) of salt were being added
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Icy Hot Chem (Heat of Solution)
Objective:
Determine the heat of solution of in calories per mole for various ionic compounds.
Materials / Reagents:
Ammonium chloride, NH4Cl
Ammonium oxalate, (NH4)2C2O4
Calcium Chloride (dry), CaCl2
Ammonium nitrate, NH4NO3
Ammonium sulfate, (NH4)2SO4
Sodium Hydroxide, NaOH
Procedure:
1. Record the temperature of water to the nearest 0.01 degree of 25.0 ml of water (also recorded to the
nearest 0.1 ml) in a 100 ml beaker.
2. Add 1.00 grams (record exact mass) of NH4Cl to the water and stir with a stirring rod (never stir with a
thermometer) record the greatest change in temperature as the ionic compound dissolved.
3. Repeat with the other compounds available.
4. All solutions may be washed down the sink with excess water.
Data:
In a neat and organized data table record all of the required information as described in the procedure.
Calculations:
This lab requires a lot of simple calculations to be done a multiple of times due to the number of
compounds, use of a spread sheet may be helpful, if you are interested see Mr. Evenson.
1. Indicate which heats of solution were endothermic or exothermic. This will tell you if you have a
negative heat of solution or a positive heat of solution (any energy lost by the water is gained by the
crystal and vice-a-versa.
2. Determine the temperature change for all of your compounds.
3. Using the temperature change and the volume of water (converted to mass 1 g/ ml) determine the
number of calories absorbed or given off for each compound.
4. Divide your number of calories by the number of grams to get a cal / g ratio.
5. Determine what the calories per mole are for ammonium chloride, sodium hydroxide, and ammonium
nitrate.
6. Calculate the percent error for ammonium chloride, sodium hydroxide and ammonium chloride based
on the following accepted values:
cal
NH4Cl(s)
3533 /mol
NH4NO3(s)
6140 cal/mol
NaOH(s)
-10637 cal/mol
Questions:
1. What if any pattern resulted in comparing the compounds and the energy given off?
2. Why are all of the heat of solution magnitudes reported as cal/mol?
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Unit 11 Objectives for Acid and Base Chemistry

Acid and Base Characteristics

Neutralization

Strong Acid / Strong Base names

Titration curve
Determining Acid dissociation constants
Determining base dissociation constants
Equivalency point

Calculating pH
Dissociation of water
Calculate hydrogen (hydronium) ion concentration

Buffer systems
Conjugate acid and base pairs
Calculation pH of buffer systems

Saponification
History of soap
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Acid / Base Chemistry—an application to both equilibria and solutions
Acid and Base Characteristics
Acids and bases often offer people with their first glimpse into chemistry. Undoubtedly
you already have some prior knowledge (and misconceptions) about both acids and
bases. Lets start out with some to the general physical characteristics of both acids and
bases
Acids and bases are equally dangerous, but their properties are on opposite
ends of a continuum. Although you will not use taste any chemicals as a means of
identification in the laboratory acids do taste sour. Many of the foods you eat contain
acids and it is the acid that gives the tart or sour taste. All citrus fruits for example
contain citric acid. Oranges also have ascorbic acid (vitamin C). The pop that you drink
often contains phosphoric and carbonic acid, both of these help the tartness on your
tongue. The acid in your stomach (HCl) causes the sourness of vomit as it passes over
your tongue (and nose).
You may have some familiarity with the pH scale. First it is important to
remember that the pH scale is a scale from 0-14. This scale is an arbitrary scale based
on the pH of pure water. All pH readings are simply a comparison to water meaning
something is more acidic (less basic) than water or less acidic (more basic) than water.
Acids fall into a range of 0-7 on the pH scale. Bases are in the range of 7-14.
Obviously there are some materials that overlap in pH range of 7. Some materials such
as water are said to be amphoteric. The root of the word, ampho, means two lives and
is used to describe materials that can behave as an acid or a base depending on the
chemical conditions (waters, alcohols, some carboxylic acids).
Any material that has a pH between 0-7 (acids) are called corrosive. Corrosive
means that the material can degrade metals. All metals will react with acids, however
not all acids will react with all metals. Copper for example will not react with
hydrochloric acid (HCl) but is vigorously corroded if placed in nitric acid (HNO 3). The
noble metals are the metals that are often used in dentistry, gold, silver, palladium, and
platinum. These metals are relatively inactive. Gold only dissolves in a special acid
called “aqua regia.” You may recall my discussion on Niels Bohr dissolving his Noble
medal in a solution of aqua regia (loosely translates to water of the kings) before he
escaped from Nazi contrilled Copenhagen. Aqua regia is actually a mixture of nitric
and hydrochloric acid. The saliva in your mouth is acidic and the metals of dentistry
need to be resistant to corrosion caused by your acidic saliva. If your dental work were
copper your mouth would soon be a green drooling mess of Cu +2 ions.
As you recall ions in solution are electrolytes. Acids are a plethora of ions in
solution and therefore acids are electrolytes. Acids will conduct electricity. The electric
pickle demonstration from earlier in the year worked in part because of the sodium
chloride ions but was predominantly caused by the acetic acid (vinegar).
Undoubtedly you were able to do some experimentation with acids and bases in
the past. Acids and bases will react to indicators. Indicators (not a confusing name) will
indicate whether an acid or a base is present. There are many different types of
indicators each with its own pH range. The indicator that is probably most familiar to
you is litmus. Litmus is a plant protein that changes colors depending on the pH of a
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solution. Flowers of some plants will change color due to the acidity (pH) of the soil.
Hydrangeas will be blue if the soil is acidic and pinkish if the soil is basic (alkaline).
Litmus paper comes in two colors: red and blue. If red litmus paper is placed in acid
nothing happens however, if blue litmus paper is placed into an acid then the paper will
turn red. This is easy to remember Blue turns Red in Acid (BRA). This process is just
the opposite for a base. Blue litmus paper is unchanged in a base but red litmus paper
will turn blue in a base (no mnemonics). Seldom will we be using litmus paper in the
lab but the concept does seem to be asked often on standardized tests.
Bases are generally the opposite of acids. The difference in the litmus having
already been noted. Bases are alkaline and as we refer to the acidity of acids so to
shall we refer to the alkalinity of bases. Bases have a pH range of 7-14. Materials that
are seven are said to be neutral but often have the ability to be amphoteric. The prefix
ampho as in amphibian (frogs and salamanders) means two lives. Amphoteric
materials are those materials that can be either an acid or a base depending on the
reaction. Welcome to a world without black and white but several subtle shades of
gray. Bases are just as dangerous as acids, they are said to be caustic. Caustic
materials will degrade metals. The only significant difference between corrosive and
caustic is what material (acid or base) is doing the damage. Caustic materials, such as
bleach (sodium hypochlorite) will feel slippery. The slippery feeling is the result of a
chemical reaction between the base (bleach) and your fingers called Saponification.
Saponification is the process of making soap. Bases feel slippery because they are
turning the fats in your fingers into human soap. It would be a good idea not to touch
bases. As you recall I once encouraged you to drink bleach instead of Gatorade,
because bleach had more electrolytes (do not drink bleach). So you know that bases,
like acids, have ions in solution.
Bases also have a characteristic taste, like acids. Alkaline substances often
have alkaloid compounds. Alkaloid compounds are secondary metabolic products in
plants (i.e. nicotine, caffeine, cocaine). These compounds are all basic and have a very
bitter taste. When you watch a cop show where the bad guy gets caught with the little
bag of white powder, the hero detective always taste the evidence. This is a qualitative
test for narcotics, if it is bitter it is probably a drug—as most drugs are derived from plant
products. If you become a cop do not taste drugs—laxatives also taste bitter.
Acid and base chemistry will require you to memorize some information. You will
need to memorize the chemical formulas and names of all the strong acids and strong
bases (we will discuss the meaning of strong later). There are six strong acids in water.
It has been my intention to use many of these acids through out the year so you are
already familiar with many of them.
HCl hydrochloric acid
HNO3
nitric acid
HBr hydrobromic acid
H2SO4
sulfuric acid
HI
hydroiodic acid
HClO4
perchloric acid
The strong bases are all hydroxides (OH-) of the group I or II metals (e.g. NaOH,
Ca(OH)2, KOH).
Strength Versus Concentration in Acids and Bases:
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The two terms of strength and concentration are not synonymous when discussing
acids and bases. It is possible to have a dilute concentration of a strong acid (e.g.
0.09M HCl). The converse is also possible, a concentrated solution of a weak acid (e.g.
6.5 M H3PO4). Concentration is a measure of how much of the acid or base is dissolved
in water (review the section on Normality). When dealing with acids we are concerned
with how many hydronium ions (H3O+) are in the solution and with bases we are
concerned with how many hydronium ions can be neutralized by hydroxide ions (OH -).
Concentration deals with “How Much.” Strength deals with “How Often.” The strength
of an acid or base is determined by how often it completely dissociates. The six strong
acids have 100% dissociation. The hydroxides of the group I an II metals have 100%
dissociation. 100% dissociation means that the hydrogens or hydroxides are
completely removed from the acid/base compound in water.
H2SO4(aq)  2H+(aq) + SO4-2(aq)
100% broken apart = strong acid
Ca(OH)2(aq)  2OH-(aq) + Ca+2(aq) 100% broken apart = strong acid
Weak acids and bases do not completely break down. Some of the hydrogens are
removed from the acid but not all of them. The more hydrogens that are removed from
the acid the stronger the acid is, because its dissociation is greater. Carbonic acid is a
weak acid some of the hydrogens are removed but not all.
H2CO3(aq)  H+(aq) + CO3-2(aq) + H2CO3(aq) Some of the H2CO3 remains as
H2CO3 and was not broken down
The acid does not show 100%
Dissociation and is therefore a weak acid.
The same rules apply to bases and we could diagram out the same reactions with a
strong acid such as BeOH to show 100% dissociation or Fe(OH) 3 to show only partial
dissociation of a weak base. Remember acids and bases are equally dangerous.
Acid and base definitions:
As you will come to realize defining compounds as acids and bases becomes
increasingly more specific as your education increases. You were most likely taught
previously what is called the Arrenhius definition of acids and bases. The Arrenhius
definition is the most elementary definition of whether a compound is an acid or a base.
It is however also the most limited definition and we quickly find compounds that cannot
be defined according the Arrenhius. An Arrenhius acid is any compound that produces
hydronium ions (H3O+) in water. While an Arrenhius base is any compound that
produces hydroxide ions (OH-) in water. In short if a chemical formula begins with an
“H” than it is likely to be an Acid and if it ends in an “OH” than it is likely to be a base.
Examples:
NH3 –water NH4+ + OH-
Notice NH3 is a base, but does not have a
Hydroxide in the formula
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HI –water H3O+ + IThe hydronium ion forms when a water molecule removes a hydrogen ion (proton) from
an acid. A large number of compounds that you encounter can be determined
according to the Arrenhius definition, but there are many that you will not be able to.
In 1923 J.H. Bronstead, a Danish Chemist, and T.M. Lowry, an English chemist,
independently derived a new definition for acids and bases. The Bronstead-Lowry
definitions of acids and bases where written in terms of proton behavior, what happens
to the hydorgen ion. A Bronstead-lowry acid is any compound that donates a proton
(hydrogen atom) while a Bronstead-Lowry Base is any compound the accepts a proton.
All Arrenhius acids and bases must also be Bronstead-Lowry Acids and Bases.
Bronstead-Lowry definitions are to classify those compounds that do not fit the
Arrenhius definitions. Lets see if ammonia (NH3) is still a base (proton acceptor)
according to Bronstead-Lowry.
H
H
H
N
+
O
H
H
H
H
+
N
H
+
OH-
H
You can see that the oxygen hydrogen bond on the water molecule donates both
electrons in the bond back to the oxygen, this is what causes the negative charge on
the hydroxide ion as a product. The hydrogen ion (no longer associated with any
electrons) is only a proton, but can be accepted by the nitrogen in the ammonia
compound. The nitrogen has two unbound electrons that are used to make the N—H
bond. The presence of an additional hydrogen without any electrons causes a positive
charge on the ammonium product. The ammonia has accepted a proton and therefore
is base according to Bronstead-Lowry. Interestingly the water molecule is a BronsteadLowry acid because it has donated a hydrogen. This is yin-yang chemistry you can not
have an acid without a base any more than you are likely to have good without evil,
health without sickness. The Bronstead – Lowry definitions of acids and bases will
cover most of the compounds that we encounter. However there will be a few
compounds that show acidic or basic behavior without hydronium production or proton
donation. These are Lewis acid and bases.
G.N. Lewis, of Lewis dot structure fame, defined acids and bases in terms of
electrons. Lewis acids are electron acceptors and a Lewis base is an electron donor.
A substance that is an Arrhenius acid is also a Bronstead-Lowry acid and is also a
Lewis Acid. However a compound that is a Lewis acid may not be able to be defined as
an acid according to either Arrhenius or Bronstead-Lowry definitions.
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H
H
H
N
+
H
O
H
H
H
+
N
H
+
OH-
H
Using the ammonia-ammonium example again we can focus on the electron
movements that are required for Lewis definitions. A Lewis acid is any compound that
accepts an electron pair. The water is an acid because it has accepted an electron pair
from the ammonia. A Lewis base is any compound that donates an electron pair. The
ammonia donates an electron pair to form the new N—H bond and produce ammonium,
therefore ammonia is a base. Lewis acids and bases become crucial in the
explanations of biochemical models of metabolism and physiology.
Neutralization:
The term neutralization means that the end product of an acid base reaction has a pH of
seven. However just because an acid base reaction occurs does not mean a neutral pH
has been reached. A neutral endpoint is very much an exception rather than a rule.
There is one guarantee when an acid reacts with a base, the products will be a salt and
water. Salts are a class of chemical compounds and does not mean sodium chloride
only. A salt is any metal combined with a non-metal. The salt will be formed by reacting
the anions (negative ion) of the acid with the cation (positive ion) of the base to form an
ionic compound. The hydrogen from the acid (cation) then reacts with the hydroxide
(anion) from the base to form water.
Example
KOH(aq) + HNO3(aq)  KNO3(aq) + H2O(l)
Zn(OH)2(aq) + 2 HCl(aq)  ZnCl2(aq) + 2 H2O(l)
Base + Acid  Salt + Water
You will find several acid base reactions in the back of the text in which you can practice
the salt formation.
pH Scale
Sören Sörenson, a Danish Chemist, devised the pH scale in 1909 as a way of brewing consistent
beer. The scale is an arbitrary scale based on the dissociation of water. Water has a neutral pH
because the pH scale used water as a baseline against which all other materials could be measured. We
can now take a little closer look at the pH scale. The pH scale is a logarithmic scale. This means that the
quantities being measured (hydrogen ions) change on the scale in as an exponential function. The pH
scale has been devised to treat exponential growth (and decay) as a linear function rather than an
exponential one. There are many other common scales that are logarithms, such as the Richter scale
uses to measure earthquakes. The pH scale uses a base ten logarithm. This means that each step is
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ten times greater than the next. A pH of four is then 1000 times more acidic than a pH of seven. Seven
minus four is three and ten to the third is one thousand.
The “P” in pH is always lower case. This is because “p” is a mathematical symbol that means “log ( )” pH means the negative log of the hydronium ion concentration. We will see many of these “p”
functions very soon but here are two to get you started.
+
pH = -log (H3O )
pOH = -log (OH )
The larger discussion on calculating pH values
will follow.
Titration Curves
A titration is an analytical technique used to determine unknown concentrations. This technique
is not limited to acid base chemistry but can be used to find any unknown concentration. As base is
added to acid the pH will increase. A titration curve is a graphical representation of those pH changes.
All of the characteristics discussed for strong acid/strong base titration curves will also apply to weak
acid/strong base, strong acid/weak base and weak acid/weak base titration curves. The titrant is usually
the known concentration and is in the burrett while the analyte is the material of unknown concentration
being analyzed.
There are two reasons to compile data for a titration curve. The first is to determine the
stoichiometric point. The stoichiometric point is the point in which the mole ratios have been met for the
reaction taking place.
Ca(OH)2(aq) + 2 HCl(aq)  CaCl2(aq) + 2 H2O(l)
The stoichiometric point of a reaction between calcium hydroxide and hydrochloric acid will occur when
there is twice as many moles of HCl as there are of Ca(OH)2. Of course we are dealing with mole ratios
and not actual moles so it is possible to have a stoichiometric point occur when there are 0.050 moles of
HCl and 0.025 moles of Ca(OH)2 present.
The stoichiometric point can be determined by calculating the (calculus) where the slope is the
most vertical. In most of our work “eyeballing” the vertical portion is within the limitations of our other
instrumentation. For a strong acid strong base titration it is expected that the stoichiometric point be
close to a pH of 7 depending on what salt is also being formed. A stoichiometric point is only near neutral
for strong acid/strong base titrations for all other titrations the stoichiometric point will be skewed high or
low depending on the acid or base strength. Do not confuse the stoichiometric point (a chemical function)
with the endpoint. The endpoint is not a chemical function (of specific interest) but rather a function of the
indicator chosen. The endpoint is the point in which the indicator makes a drastic and conspicuous color
change. Always use an indicator that has an endpoint as near to your stoichiometric point as possible.
Once the stoichiometric point is determined on the titration curve (most vertical portion) then the
Ka value can be calculated. Ka is the acid dissociation constant, which gives the amount of the acid that
dissociates or breaks apart. Half the distance to the stoichiometric point is the pK a as read from the Y
axis of your titration curve. The pKa = -log (Ka) therefore Ka = 10-pKa.
All titration curves have similar characteristics. The
pH will rise very slowly until half of the acid is
consumed by the titrant. Once half the acid is
consumed you are also half the distance to your
stoichiometric point. Here your graph will show an
increased slope until the rapid incline as the graph
passes the stoichiometric point. The plateaus exist
on the acidic and basic areas of the graph because of
the large difference in acid and base concentrations.
When the titrant is first added (base in the example
shown) there is a very small amount of base and a
pH
Strong acid strong base titraton curve
13
12
11
10
9
8
7
6
5
4
3
2
1
0
0
5
10
15
20
Volume of Titrant (ml)
25
30
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large amount of acid therefore the pH remains very low. The converse is true in the upper region of the
titration curve. Titration curves are graphs and therefore are analytical tools the ability to determine the
Ka of an acid will help in identifying what an unknown acid is.
Buffers
What versus How: What is a definition of an object, how is a description of it ability to perform. A
buffer is anything that resists a change in pH. How a buffer works is much more involved but based on
simple principles that we have already determined. Buffers are able to prevent changes in pH because
there are weak acids or weak bases with their ionic salts. Using a weak acid or base means that there is
only partial dissociation and an equilibrium reaction is then established. The equilibrium reaction then
follows the rules of LeChatlier’s principles. Use the bicarbonate /carbonate buffer system as an example
HCO3- + H2O
CO3-2 + H3O+
If acid (H3O+) is added to this buffer the amount of product is increased and the reaction shifts to the left
making a weak acid (HCO3-) and water without any change in the pH. If a base is added to the buffer the
hydroxide reacts with the H3O+ which makes more water and prevents any change in the pH of the overall
system. The bicarbonate / carbonate buffer system is the one of the buffers that keeps a constant pH in
the worlds oceans and is also the buffer system that keeps your blood buffered.
Buffers cannot be made from strong acids or strong bases as they have 100% dissociation.
Complete dissociation will not leave any reactants and prevent a reversible reaction to balance an
equilibrium reaction.
Conjugate acids / base pairs
Conjugate means to change, as in conjugate verbs (e.g. see, saw, seen). In acid base chemistry a
chemical reaction occurs and the reactants are changed into products. The original reactant is changed
into a product (a conjugate pair).
O
H3C
+
C
H
O
OH
acetic acid
OH
H3C
H
Water
+
C
Acetate O-
+
O
H
H
Hydronium ion
H3C2O2-(aq) + H3O+(aq)
H4C2O2(aq) + H2O(aq)
We can use acetic acid as an example. The acetic acid is initially an “H4” compound but donates a
hydrogen to become an “H3” compound. This donation of a proton means that acetic acid is a BronsteadLowry acid. I have also shown the Lewis structures and electron pushing to show that acetic acid also
qualifies as a Lewis acid. Acetic acid is an acid and water which accepts a proton is a base. Both of
these compounds are changed as products. Products are conjugates because they have been changed.
Now notice another important feature of acid base reactions, the acidic reactant looses a hydrogen,
therefore the product of this lose can now accept a hydrogen (negative charge) and is therefore defined
as a base. The converse will be true for the reactant base. In the example water is a base but is
changed into a hydronium ion. The hydronium ion can donate a hydrogen and is therefore an acid. The
conjugate acids and conjugate bases are weaker acids and bases than the original acids and bases.
O
H3C
+
C
OH
Acid
H
O
OH
H3C
+
C
H
OBase
O
H
Conj ugate Base
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+
H
Conj ugate aci d
Evenson
It is important to be able to identify an acid and its conjugate base pair, as well as identifying a
base with its conjugate acid pair. The pH of a compound depends to some extent on the ratio of the
concentrations (product/reactant) of the conjugate pairs.
Example:
-
H2C2O4(aq) + H2O(aq)  HC2O4 (aq) + H3O
Acid:
Base:
Conjugate acid:
Conjugate Base:
+
(aq)
H2C2O4(aq) is converted to HC2O4-(aq) and has therefore lost a proton
+
H2O(aq) is converted to H3O (aq) and has therefore gained a proton
+
H3O (aq) could donate a proton
HC2O4-(aq) could accept a proton
pH calculations and rationale:
The pH scale is an arbitrary scale based on water. If something is “acidic” that means that it contains
more hydronium ions than water.
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Acid and Base practice
Directions: On a separate sheet of paper complete and balance the following acid base
reactions, as discussed the reaction between a strong acid and a strong base will yield
a salt and water.
1. Sodium hydroxide and hydrochloric acid
2. Calcium hydroxide and sulfuric acid
3. Hydroiodic acid and potassium hydroxide
4. Nitric acid and ammonium hydroxide
5. Magnesium hydroxide and Perchloric acid
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Practice Calculating pH of Various Scenarios
Directions: On a separate sheet of paper calculate the pH of the solution with the
information given; be sure to show all work in a clear manner. Any and all illegible work
will not be graded.
1. [H+] = 4.65 x 10-9
2. The concentration of Hydronium ion is 3.4 x 10 -4
3. The pOH is 5.6
4. The [OH-] = 2.0 x 10 –9
6. [H+] = 1.0 x 10 -8
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pH Calculations involving Acid Dissociation Constants
Directions: On a separate sheet of paper calculate the pH of the solution with the
information given; be sure to show all work in a clear manner. Any and all illegible work
will not be graded.
1. [H+] = 5.68 x 10-9
2. The concentration of Hydronium ion is 4.0 x 10 -3
3. The pOH is 4.5
4. What is the pH of 5.45 M Carbonic Acid (H2CO3) if the Ka is 4.467 x 10-7
O
C
HO
OH
carbonic acid
5. Oxalic acid is an organic acid that makes rhubarb leaves toxic (it forms crystals that
puncture the capillaries of the liver due to a pH change). What is the pH of 2.35 M
oxalic acid if the pKa is 2.67?
O
O
C
C
HO
OH
oxalic acid
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Buffer Systems and Conjugate Acid-Base Reactions
Directions: use arrows to connect the acid and the conjugate base in each reaction as
well as the base and the conjugate acid. Buffers are systems of weak acids or weak
bases and their salts. The ion products below could combine with a cation or anion to
create a salt. Also include the Equlibrium expression for each equation.
1. HIO3(aq) + H2O(l)  H3O+(aq) + IO3-(aq)
2. NH2NH2(aq) + H2O(l)  NH2NH3+(aq) + OH-(aq)
3. NH2CONH2(aq) + H2O(l)  NH2CONH3+(aq) + OH-(aq)
4. NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
5. Why are buffer systems always combinations of weak acids and weak bases, why
will strong acid – strong base combinations not be effective buffers?
6. Water is Amphoteric, what does this mean?
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pH Calculations of Buffer Solutions
Directions: Show all work on a separate sheet of paper. Any and all illegible work will
be marked wrong.
1. What is the pH of the buffer under the following conditions: 0.056 M (CH3)3N and 0.025 M (CH)3NH+ if
the Ka = 1.55 x10-10?
+
(CH3)3NH
(aq)
+ H2O(l)  H3O
+
(aq)
+ (CH3)3N(aq)
-2
2. What is the expected pH of a buffer made from 0.0023 M HClO 2 and 0.035 M KClO2 (Ka = 1.0x10 )
HClO2(aq) + H2O(l)  H3O+(aq) + ClO2-(aq)
(potassium is a spectator ion)
3. Find the pH of an ammonia/ammonium buffer if the pKa is 9.26 and there is a 0.065 M NH3 and 0.235
M NH4. Remember NH4OH is comprised of two polyatomic ions and that one of them is a spectator ion.
NH4OH(aq) + H2O(l)  NH3(aq) + H3O+(aq) + OH-(aq)
-
-2
4. What is the Hydronium ion concentration if 0.25 M H2PO4 is combined with 0.25 M HPO4 ?
H2PO4-(aq) + H2O(l)  HPO4-2(aq) + H3O+(aq) (pKa2 = 7.21)
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Morning Breath?
Objective: Determine the average calories burned per minute by blowing bubbles
Materials:
1-100 ml beaker
0.1 M, 0.5 M, 1 M NaOH
0.5 M HCl
Drinking straw
25-ml Graduated cylinder
Stop watch
125 ml Erlenmeyer flask
Procedure:
1. WEAR YOUR GOGGLES, Failure to do so will result in a ZERO for this lab!!!
2. Add 20 ml of water and 1.0 ml of 0.1 M KOH to a 125 ml Erlenmeyer flask and add two drops of
phenolphthalein
3. Blow through the straw and start the stopwatch, record the time required to change the solution from
pink to clear.
4. Repeat these steps for 0.5 M KOH, 1 M KOH and 0.5 M HCl solutions.
Data:
Record all necessary volumes, concentration, masses, pressures and other data of use in determining
your objective of caloric output.
Calculations:
Determine your CO2(g) output per minute and determine what mass of glucose you combust (by cellular
respiration) during a 24 hour period. Using the net equation for cellular respiration (C6H12O6(s) + 6 O2(g) 
6 CO2(g) + 6 H2O(g) + 164 Kcal) determine the caloric output the bubble blowing provides. Graphing the
moles of sodium hydroxide as a function of time will increase the accuracy of your calculations.
Questions
1. Write a balanced reaction for the carbon dioxide combining with the potassium hydroxide (hint: This
reaction is actually the summation of two other chemical reactions in which CO 2 first combines with
water.)
2. Why did the phenolphthalein not change color when added to the HCl?
3. Is the solution neutralized when the phenolthalein reaches the endpoint? Explain.
4. If moles/time can be determined without graphing then what is the purpose of graphing?
5. How does a stoichiometric point differ from an endpoint?
6. Design a lab procedure to determine the relationship between heart rate and carbon dioxide
production.
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Disappearing Ink
Objective: synthesize disappearing ink using the properties of indicators and neutralization reactions
Material:
125 ml Erlenmeyer flask
Ethyl alcohol
1 M NaOH
Dropper bottles
100 ml Grad cylinder
rubber stopper
Thymolphthalein
Stirring rod
White cotton
Procedure:
1. Add 50 ml of ethyl alcohol to the Erlenmeyer flask
2. Add 2 drops of thymolphthalein indicator.
3. Add just enough NaOH to produce a blue color (less than 4 drops)
4. Stir the solution and transfer to the dropper bottle.
5. Try your solution on the suspended cotton
Questions and conclusions:
1. Record observations and a qualitative estimate of the time lapse.
2. Write a balanced chemical reaction to explain your observations
 Na2CO3 +
3. Why is thymolphthalein used for the indicator?
H2O
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Soap on a Rope
Background
Soaps are the mettalic salts of fatty acids (oils). Soap is made by combining fats with a
solution of a strong base. This reaction is called saponification. Soaps are made by
boiling oils or liquid fats and react them with a strong base, such as NaOH the products
of the reaction are the formation of a soap and glycerol. Fats are esters formed from
glycerol (an alcohol) and a long-chain organic acid (fatty acid).
(RCOO)3C3H5 + 3 NaOH  3RCOONa + C3H5(OH)3
Objective:
Use general knowledge of organic chemistry and apply the principles of saponification
in the production of soap making.
Materials:
Oil
You can bring perfume/cologne if you want to add scent.
Litmus paper
11.4 M NaOH
Procedure:
1. Add 16.0 ml of oil to a clean and dry 100 ml beaker and 4.0 ml of 11.4 M NaOH to
the oil.
2. Over a wire gauze on a ring stand gently heat the mixture over a Bunsen burner.
3. Heat for 10-15 minutes stirring constantly to prevent spattering. Stop heating when
the odor of fat has disappeared and oil has dissolved. The solution should have
thickened at this point. Do not heat to dryness.
4. Form an aluminum mold for your soap about 2 square inches in size
5. Add any perfumes or dyes at this time.
6. Check the pH of the soap with litmus paper and discard the solution. If soap is too
basic slowly add 6 M HCl (drop-wise) and stir until pH is near neutral.
7. Pack the solid soap into the mold and allow to cool. (it is unlikely that you will have
enough supernatant to be concerned about filtering or discarding excess)
Data:
In a data table record all data collected, volumes, masses…
Questions
1. What was the pH of your soap?
2. Why would the pH of a commercially produced bar of soap intentionally not be
neutral?
3. How would you make soft soap, for pump dispensers?
4. Industrial production of soap includes the addition of sodium chloride to “salt-out” the
soap, why does this increase the yield of soap?
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Acid Rain Lab Exercise
Objective: Graphically display the different buffering capacities of a
carbonate/bicarbonate buffer system vs. the buffering capabilities of natural limestone
vs. distilled water.
Materials:
Limestone (chalk/sand)
0.1 M NaHCO3
0.1 M Na2CO3
Distilled water (boiled)
pH meters
50 ml beaker
Eye dropper
graph paper
Procedure:
Distilled water (control)
1. Add 10 ml of distilled water to a 50 ml beaker and record initial pH.
2. Add 2 drops of 1.0 M HCl and record pH
3. Repeat step two until pH reaches 2
Be sure to keep track of the number of drops added and the pH.
Na2CO3/ NaHCO3 buffer system
1. Place 10 ml of Na2CO3/ NaHCO3 buffer into a 50 ml beaker and record the initial pH.
2. Add 2 (two) drops of 1 M HCl and record pH
3. Repeat step two until pH reaches 2
4. Be sure to keep track of the number of drops added and the pH.
Lime stone (chalk/sand)
1. Add 1 (one) level spoon full of limestone to a 50 ml beaker.
2. Add 20 ml of distilled water and record the pH.
3. Add two drops of 1.0 M HCl and record the pH.
4. Repeat step 3 over and over again.
Data: (replicate data table for each reaction)
Record volumes, pH and Mass of limestone
Questions:
1. Graph the 3 different reactions of the same graph (each reaction should be in a
different color). A spreadsheet program may be helpful. Explain what the graph
represents.
2. Why did the pH drop so quickly in the distilled water?
3. What, if any correlations exist between the lime water reaction and the buffer
reaction?
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Graphing a Titration Curve of a Strong Acid and a Strong Base
Background:
Not all acid base titrations end with equal amounts of acid and base at a pH of 7. A
titration curve must be developed to determine when (at what pH) equal amounts of
acid and base are present. The point were equal amounts of acid and base are present
in a solution is called the equivalence point, or stoichiometric point. The equivalence
point must be determined for each acid-base titration so the correct indicator can be
used. The pH of the equivalence point must correspond with a color change of the
indicator at that pH. On a titration curve the equivalence point is the most vertical part
of the graph.
Objective:
To determine the equivalence point for a titration between, NaOH and HCl.
Procedure:
1. To a 250 ml Erlenmeyer flask add about 35 ml of 1 M HCl (record exact volume) and
add 2-3 drops of phenolphthalein.
2. Record the pH with a pH meter.
3. From a burette begin to add 1.00 ml of 1 M NaOH swirling after each addition and
recording the pH.
4. Continue after the pink appears but record the pH of the system when the pink
appears.
5. Continue until the pH is near 10, continue to add 6 more aliquots of 1.00 ml
Data:
In a very organized table record the exact volumes of NaOH used (it should be near 0.5
ml) and the exact volume of HCl used. Include the pH reading after the addition of each
0.5 ml of 1 M NaOH.
Questions:
1. Graph the pH versus the volume of NaOH added, a graphing program may be
helpful.
2. Indicate on the graph were the pink appeared by highlighting the data point.
3. What is the pKa of Hydrochloric acid?
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% Acetic acid in Vinegar
KOH(aq) + CH3COOH(aq)  H2O(l) + K+ -OOCCH3(aq)
Background:
Vinegar is formed by improper fermentation of fruit products, often apples. The must, fermenting fruit, is
exposed to oxygen and this allows for the formation of acetic acid. Yeast can only tolerate a certain level
of acidity. The acid levels eventually reach a level that kills the yeast and the fermentation ceases.
Therefore, natural vinegar has a set pH and the amount of acetic acid, HOOCH3C, can be determined.
Acetic acid, is a weak monoprotic acid and undergoes first ionization only.
Materials:
Phenolphthalein
Balances
250 ml volumetric flask
KOH pellets
Vinegar
Burette
Objective: Determine the molarity of acetic acid in vinegar.
Procedure:
1. Make 250 ml of titrant, 1.00 M KOH, using a 250 ml volumetric flask (record mass of KOH actually
used)
2. Fill a burette to the 0.00 ml mark with the 1.00 M KOH.
3. Pipette 35.0 ml of vinegar to a 250 ml Erlenmeyer flask and two drops of phenolphthalein.
4. Titrate the vinegar with KOH until the pink endpoint is reached. Record ml of titrant required.
5. Repeat the vinegar analysis a minimum of three times
Data:
Record all data in a neat legible TABLE (volume of vinegar, Mass of KOH, Molarity of KOH, volume of
titrant used). Record the percent acidity from the vinegar bottle.
Calculations:
1. Show all calculations for your solution preparation as well as the determination of the actual
concentration of your titrant.
2. Show all calculations required to meet the objective to determine the Molarity of acetic acid
3. Show the calculations for your percent error as well as a statement of what values you are comparing
Questions
1. What is the greatest source of error in the procedure?
2. What is the best way to correct this error?
3. When the endpoint is reached using phenolphthalein, is the solution neutral?
4. Why is it the procedurally best to use a strong base, such as KOH, when titrating a weak acid?
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Practice with acid base concepts
Complete and balance the following reactions
Iron (III) hydroxide and sulfuric acid
Potassium hydroxide and carbonic acid
Nitric acid and lead (II) hydroxide
Write dissociation reactions for the following
H2O
Ca(OH)2
Write the equilibrium product expression (Ksp) for the following reactions
K(s) + H2O(l)  KOH(aq) + H2 (g)
NaCl(s)  Na+(aq) + Cl-(aq)
Calculate the pH given the following information
[H3O+] = 6.90 x 10-4
Kb = 9.0 x 10-3
What is the Hydrogen ion concentration in a solution that has a pH of 8.36?
What is the Hydrogen ion concentration in a solution that has a pOH of 5.69?
Compare and contrast the three different definitions of acids and bases, Bronsted-Lowry, Lewis, and
Arrhenius.
Explain how a buffer system works using the Na 2CO3 / NaHCO3 system.
Using arrows label all of the acids, bases and conjugate pairs in the following reactions?
H2CO3 + H2O  H3O+ + HCO3C17H22O4N + H2O  HC17H22O4N+ + OH-
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Review for Acid / Base Unit
Be able to do the following or have a working knowledge of the conceptual ideas.
Write dissociation equations given a reaction, written or word equation
Write equilibrium product expression
Understand unity (pure liquids and solids)
Behavior of systems at equilibrium at a molecular level
Be able to solve for concentrations of aqueous ions given the Ksp (solubility product constant)
Strong acids and bases names why they are strong
Difference between strength and concentration
Various definitions of acids and bases (Arrhenius, Bronsted-Lowry, and Lewis)
Solve neutralization reactions
Buffer reactions and their behavior
Characteristics and uses of titration curves
Conjugate acids and bases
Solve for pH if Ka, Kb, pOH, [OH-], [H+] … are known
Be able to do all calculations associated with the labs; explain behavior of reactions in lab.
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Unit 12 Objectives for Thermodynamics

Application of concepts and vocabulary
Heat flow
System
Surrounding
State property
Specific heat
Heat capacity
Calorimeter
Enthalpy
Entropy
Free Energy

Thermochemical reactions
Energy as a product and a reactant

Hess's law
Additive property
Calculations

Spontaneous changes vs. instantaneous changes

Gibbs Free energy application and calculations
Coupled reactions
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Brief Introduction to Thermodynamics
Thermodynamics (e.g. thermochemistry) is the chemistry that deals with energy transfers, speeds
of reactions and probability of a reaction occurring. In this unit we will discuss the direction of energy
flow, the magnitude (i.e. size, amount) of energy flow, the concept of Enthalpy (∆H) and calculate
enthalpy by writing thermochemical equations.
Thermochemical reactions are chemical reactions in which the energy (also known as heat) is
included in the chemical reaction, we have seen many of these in earlier lectures:
Combustion of methane:
CH4(g) + O2(g)  CO2(g) + 2H2O(g) + Energy
Rocket fuel:
2N2H4(s) + N2O4(l)  3N2(g) + 4NO2(g) + Energy
Electrolysis of Water:
Energy +2 H2O(l)  O2(g) +2 H2(g)
In thermochemical reactions energy can be either a product or a reactant. If energy is given off
then the reaction is exothermic while a reaction that uses more energy than it produces (energy as
reactant) is a endothermic. The combustion of methane and the rocket fuel propulsion reaction are both
exothermic reactions, as they produce energy. Exothermic reactions have an enthalpy value that is
negative (-∆H) think of the energy as being released from the reaction. The electrolysis of water requires
energy to be added to the reactants (+∆H) this reaction is an endothermic reaction.
As an interesting side note, also notice that the rocket fuel example produces 7 moles of gas as
compared to reactants that contain no moles of gas—it is this rapid increase in gas volume, more than
the energy, that allows the reaction to be used for a propulsive force and the reason for its explosive
nature.
Thermodynamics is jargon rich and contains multiple equations each with its own symbolic
letters. It will be a good idea for you to keep a page or two in your notes that is dedicated to the symbols
used in thermodynamics so that all of the information is in one place. I often refer to this as “alphabet
soup.” These symbols are important as we develop a working vocabulary for the unit. The first concept I
would like to review is Heat flow (q). Heat flow is the movement of energy (heat) from one entity to
another. Passive heat flow is always from high concentration (high temperature) to low concentration
(low temperature). A body that has a high concentration of energy will have a high temperature and the
high speed of those molecules makes it
more probable that the high energy
molecules will move away and towards an
area with slower (cooler) molecules. The
molecule to the left shows two beakers on
hotplates. Scenario A shows a beaker at
o
o
85 C and a hotplate at 12 C. The beaker
has more energy and the heat will flow to the
hotplate (high to low) and warm up the
hotplate in the same fashion that a hot cup
of coffee will heat up your cold hands or the
way a warm kitten can give warmth to the
cold and icy heart of an old man (ah, but I
digress). In scenario B the heat flow is
reversed, and the energy will travel (flow)
from the high concentration in the hot plate
to the lower concentration of the beaker. Much like enthalpy (H) can be either positive or negative so to
can heat flow (q). But the heat flow is user defined. We must first indicate what our systems and our
surroundings are.
Generally the chemical reaction or the reaction vessel is the system. The system is your focal
point, the item of interest. The surroundings in theory are everything else in the universe, but
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Evenson
pragmatically the surroundings are the
close proximity to your system. For
example if I use the coffee in my cup as
my system then the cup, my hands and
even to some extent the classroom is
the surroundings. The coffee is at a
higher temperature than the
surroundings and so the heat will flow
out of the coffee and into the
surroundings. As the coffee loses
energy the surroundings must gain
energy, so again I only drink coffee to allow it to heat up the room. The heat flow in this example is from
the coffee to the rest of the room, from the system to the surroundings this ‘lose of heat” by the system
represents a negative heat flow (-q). Now let us revisit our original picture: If we identify our beaker as
our system and the hotplate as the surroundings then scenario A has a negative heat flow (-q) because
energy is leaving the system. Scenario B has a positive heat flow (+q) because energy will travel from
the surroundings into our system (the beaker), if the system gains energy then the heat flow is positive.
The heat flow tells the direction of energy movement but does not tell us the magnitude (amount) of
energy that is moving. Heat flow is usually combined with another concept called Heat Capacity (C). The
heat capacity can be thought of as the amount of energy an object can hold before it undergoes a
temperature change. It is a measure of how much energy must be added (or removed) before an object
will change temperature. Objects have different heat capacities. Metals generally have very low heat
capacities, which mean that they change temperature very easily. This is why you normally do not cook
with metal utensils and why you should not wear earrings (or any other visible piercing) when the weather
is very cold. The metal changes temperature very easily and can then cause frost bite and tissue
damage to the pierced tissue. Although different materials have different heat capacities, the heat
capacity alone is not responsible for total change in temperature. Much the same way that I can boil 50
ml of water in about 45 secs, I can not boil the water in Lake Michigan in same amount of time. There is
more water in the lake, this water has more mass and we have to heat up all the mass. The mass of an
object will play a role it the magnitude of the heat flow.
A brief review of Heat versus Temperature: The aforementioned boiling 50 ml of water has a
much higher temperature than the lake Michigan, but the beaker of water has much less heat than the
lake. Temperature is the average kinetic energy of an object, molecular speed (KE=1/2 mv2), while heat is
the TOTAL energy of an object. So while each molecule of water in the lake is moving very slowly the
collective movement of all the molecules is vastly greater than the total molecular movement in the 50 ml
beaker. This is another example of large bodies of water acting as a regulating force in local climates
(e.g. “cooler by the lake shore”) as the water acts as a heat (energy) reservoir.
Quantifying Heat flow:
Heat capacity is the heat flow over change in temperature (C = q/∆T or q = (C)( ∆T)). For a pure
substance the heat capacity is the specific heat and we know that the change in temperature is a factor of
the mass of the object so the heat capacity can be measured as C = Mass x specific heat. Combining
these two ideas/equations we return to a familiar equation q = mass x specific heat x ∆T. The heat flow
is what we measured when doing our phase change calculations. The calories that you continually
calculated out were the amount of energy and the direction it was flowing (heat flow). All of our chemical
reactions have a purpose, and that purpose is generally to do work. Thermodynamics is a study of the
energetic outputs of chemical reactions so that the energy can be harvested to do work. Work is a state
property (like enthalpy (H) and entropy (S) and free energy (G), temperature (T)). A state property is
any property in which only the initial and final states are important. State and phase are not synonymous,
state is a physical description of matter that includes the composition (phase), temperature, pressure and
mass. For example the state of a water sample could be described as being a 50.00 gram sample of
H2O(l) at 80.00 oC and 1.00 atm. With this description of the water sample we can determine the precise
amount of energy that the sample contains. Phase only tells us what the composition of the matter is
(solid, liquid…) while the state is a complete description of the energy of the matter. State properties are
concerned with the change in state. The change in state is the final minus the initial (∆ = final – initial)
with no regard given to how the state went from the initial to the final position.
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State property example:
Let us use a change in temperature as an example of a state property. If we start with a water sample at
25oC it has some energy associated with it which could flow (q). If the water heats up to 75 oC then is
o
o
o
cooled to 8 C and heated again to 60 C before it comes to a final temperature of 20 C then the total
o
o
o
change in energy is still only the final minus initial (20 C – 25 C = -5 C) which represents a loss of five
degrees. Energy was added to the sample to raise the temperature but energy was also released when
the water sample was cooled, so these energy values cancel out making only the initial and final values
important—this is true of all state properties.
Work (w)
All energy can be used to do work with the exception of heat energy. Heat energy is the lowest form of
energy and is too chaotic to allow for an organized flow through a system. Work can be done on the
surroundings by a system (-w). An explosion is an example of a system (chemical reaction) doing work
on the surroundings; while work can also be done on the system by the surroundings (+w). Notice that
work is always done ON something. If heat is the lowest form of energy then all energy is not equal in
quality. The quality of energy can change and in every conversion of energy some energy is changed to
heat energy. Heat energy mustn’t (homage to my great-grandmother) be confused with thermal energy.
Thermal energy is the energy of moving particles and is measure as temperature. Heat energy is the
motion of the atoms within a molecule as the bonds bend, twist and shimmy.
Laws of thermodynamics:
There are three laws of thermodynamics, these are the laws that govern all thermochemical reactions in
which energy and matter are involved (all of them). These laws are best summarized as: “you can’t win,
you can’t break even and you can’t get out of the game.” Please allow me to explain: The first law of
thermodynamics (you can’t win) involves the conservation of matter and energy, Energy is neither created
nor destroyed it can only be converted from one form to another– so you can not make energy ergo you
can not win. The second law of thermodynamics (you can’t break even) says that every time energy is
converted to another form some of the energy is given off in an unusable form (heat energy). When you
convert a gallon of gasoline (chemical energy) into mechanical energy to run your engine some of the
energy is converted into sound energy, light energy, electrochemical energy, thermal energy etc. The
chemical potential energy of your gasoline that is not converted in mechanical energy to run your engine
represents a loss of efficiency. Most gasoline engines only convert about 18-20% of the gasoline into
work to move your car from place to place. The third law of thermodynamics (you can’t get out of the
game) means that laws of thermodynamics govern the nature of chemical reactions and you are
comprised of chemical reactions in which the only reactions at equilibrium have no net affect from laws
one and two – however any biochemical reaction that is at equilibrium means that the organism is dead
(you can’t get out of the game).
Conservation of Energy (first law of thermodynamics)
Can also be used when quantify heat flow (q). When my coffee cup cools down it loses energy to the
surroundings. The surroundings must gain any energy given off by my coffee, and the surroundings must
gain the same amount of energy lost. Energy lost by system = energy gained by surroundings and the
vice versa. Mathematically E = -E.
Example:
When 1.00 gram of NH3NO3(s) is added to 50.0 grams of H2O(l) the temperature drops from 25.00o C to
23.32 oC (Specific heat of liquid water is 4.184 j/gC) What is the q for the water in the container? What is
the q for reaction?
Heat flow of water:
The water in this situation is the surroundings as our focus is reaction:
q = (50.0 g )(
4.184 j
)(23.32C  25.00)  -351 Joules (-∆H, water is releasing energy)
gC
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This means that 351 joules of energy are released by the water when ammonium nitrate dissolves. This
also means that the system (reaction) must be gaining the same amount of energy
∆Hreaction = +351 Joules (351 joules gained by reaction in order to dissolve)
Also remember that if E = -E then q = -q as well
-q means exothermic, the surroundings (water) is the exothermic so: qreaction = -(qwater) which can then be
expanded out to:
(massreaction)(Specific heatreaction)( ∆treaction) = -{(masswater)(Specific heatwater)( ∆twater)}
This relationship between the reaction and water (i.e. system and surroundings) will allow us to determine
myriad of variables about a given reaction such as the specific heat (heat of solution) for Ammonium
nitrate given the above information.
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Specifically Hot
Background:
The amount of heat energy that is required to raise the temperature of one gram of a substance by one
degree Celsius is called the specific heat, of that substance. Water for instance, has a specific heat of
1.00 calorie per gram degree Celsius (1.00 cal/g C). This value is high in comparison with the specific heats
of many other materials.
The amount of heat energy involved in changing the temperature of a sample of a particular substance
depends on three parameters: the specific heat of the substance, the mass of the sample, and the
magnitude of the temperature change. The Greek letter delta (Δ) is used to indicate a change.
ΔT = Finaltemp. - Initialtemp.
The amount of heat energy that is transferred in the process of producing a temperature change can be
calculated from this information, according to the following equation:
Change in =
Heat energy
specific heat
of sample
x
Mass of x
sample
ΔT of
sample
You will be using a crude calorimeter for this lab consisting of a styrofoam cup and a glass beaker. The
water and the metals will quickly obtain the same temperature. Heat cannot be lost outside of a system
because the Styrofoam is a good insulator. Therefore the heat lost by the metal is equal to the heat
gained by the water. All terms in the following equation will be known except the specific heat of the
metal, this can be calculated with algebra.
Objective:
Determine the specific heat of two metals.
Procedure:
1. In your data table record anything and everything that you measure either qualitatively or
quantitatively, as precisely as the apparatus will allow.
2. In a 250-ml beaker place a styrofoam cup with 50 ml of tap water, record the precise volume.
This is your calorimeter and your system.
3. Start a boiling water bath and place about “2-3 fingers worth” of metal into a large test tube
and then into the water bath. Record the precise temperature of the water and the precise
mass of the metal.
4. Allow the test tube to rest suspended in the water bath for at least three minutes, this will
allow your metal to equilibrate with the temperature of the water.
5. Quickly and carefully dump the metal from the test tube to the calorimeter and record the
highest temperature the solution reached as precisely as the thermometer allows. Stir the
solution with a glass-stirring rod NOT the thermometer.
6. Remove the metal, pat dry with paper towel and put in a designated place for drying.
7. Collect the mass of 10 pennies (post 1984) and run the procedure to determine the specific
heat of the alloy.
Data:
Record all observations in a TABLE (columns, rows, headings).
Calculations:
Pure samples
1. Calculate the change in temperature of both metals.
2. Calculate the heat gained by the water in each trial.
3. Calculate the specific heat of both metals (see background).
J
J
4. The accepted value (theoretical value) of copper is 0.3866 /g C and zinc is 0.388 /g C
J
.(aluminum is 0.91 /gC) Determine your percent error for your copper trials.
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Calculations
Pennies
1. Calculate the specific heat of the alloy.
2. Calculate the relative % composition of copper and zinc in the penny.
st
3. Using your data from 1 semester on penny composition, calculate a % difference between
the specific heat method and the density method to determine the composition of an alloy.
Questions:
1. Explain how you would determine the specific heat of a liquid (alcohol, oil…).
2. Explain how you would determine the specific heat of a material that is soluble in water.
3. Showing all work calculate the expected specific heat of rose gold (Russian gold) if the
alloyed jewelry is composed of 75 % Gold (SH = 0.129 J·g−1·K−1) and 25 % Copper.
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Practice with Thermochemical Equations
Directions: in each given situation place the energy (and quantity if applicable) on the
correct side of the equation.
1. H = 49.0 kJ
6 C (graphite) + 3 H2(g)  C6H6(l)
2. H = -393.5 kJ
C (graphite) + O2(g)  CO2(g)
3. +q
H2O(l)  H2O(g)
4. –q
CaCl2(s)  CaCl2(aq)
5. exothermic
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
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Heat Flow Practice
1. The specific heat of Bromine is 0.474 J/g oC. What is the heat flow if
10.00 grams of bromine is cooled from 50.00 oC to 30.00 oC?
2. If the temperature of the bromine cooled because a 10.00 oC, 5.25 gram
piece of aluminum was put into the bromine what was the specific heat of
the aluminum?
3. Liquid Benzene has a specific heat of 1.72 J/g oC, if 40.0 grams of
benzene (C6H6) is cooled from 40.0 oC to 10.0 oC by 232.7 grams of iron
that was at –10.0oC what was the specific heat of the iron?
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Specific Heat and Heat Capacity
Directions: On a separate sheet of paper answer all mathematics, showing all
necessary work in a legible manner. Remember that the magnitude of a change in
degrees Celsius is the same magnitude as the change in Kelvin.
1. How many kilojoules of heat energy are required to heat all the aluminum in a roll of aluminum foil
(500.0 g) from room temperature (25.00oC) to the temperature of a hot oven (250.00 oC)? specific
heat of aluminum 0.902 J/g K
2. One way to cool down your cup of coffee is to plunge an ice-cold piece of aluminum into it. Suppose
you store a 20.0 g piece of aluminum in the refrigerator at 4.4 oC and then drop it into your coffee.
o
o
The coffee temperature drops from 90.0 C to 55.0 C. How many kilojoules of heat energy did the
aluminum block absorb? (Ignore the cooling of the cup.)
3. Suppose you pick up a 16.0 kg ball of iron. The iron ball has the same temperature as the
o
atmosphere on a cool day (16.0 C). How many joules of heat energy must the iron ball absorb to
reach the temperature of your body (37.0oC)? Specific heat of iron 0.451 J/g K
4. Heat capacity is the amount of energy (Heat) required to raise the temperature of a system 1 degree
Celsius. What is the heat capacity if it takes 7540.0 J to raise a substance from 20.0 oC to 90.0 oC
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Heat Flow practice with the concept that E= -(E)
1. A 15.0 gram sample of metal is presumed to be nickel (SHNi = 0.44 cal/gC). IN the following lab
conditions the metal is taken from a steamer at 100.00 oC and placed into 150.00 grams of 23.78 oC
o
water. The final temperature of the water is 26. 99 C is the metal pure nickel?
2. A 50.00 ml sample of water is 26.79 oC if a 25.00 gram sample of 47.98oC gallium is added what is the
final temp of the gallium (SHGa = 0.371 cal/gC)?
3. A “boy” buys an engagement ring for his “sweetie” The ring has a 1.00 ct (200.00 mg) stone. If the
stone is at room temperature (22.00 oC) and is placed in 10.00 oC Bromine (SHBr=0.113 cal/gC) if 2.00
o
grams of bromine are raised to 10.673 C is the stone diamond or cubic zirconium?
0.124 cal/gC
0. 0671 cal/gC
Specific heat of diamond
Specific hear of CZ
4. 30.00 grams of –14.00 oC ice is added to 200.00 ml 20.00 oC water. What will the final temperature of
the water be?
5. What mass of –8.00 oC ice must be added to 1.00 L of 20.00 oC water to reach a drinking temperature
of 12.00 oC?
6. (Evenson’s new favorite problem)
A 20.00 gram sample of an unknown metal is placed in a steamer until it reaches 100.00 oC. The metal is
then placed in 46.38 grams of 20.00 oC water, which then rises to 25.00 oC. Identify the unknown metal
by its specific heat, then explain why this scenario is not possible.
Lead:
Gallium:
Calcium:
Mercury
Nickel
Tin
cal
0.0308 /gC
0.3710 cal/gC
0.1546 cal/gC
0.0335 cal/gC
0.1052 cal/gC
0.0545 cal/gC
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Hess Law Inquiry lab
Objective:
Determine what is meant by the additive properties of chemical reaction through an investigation of
exothermic heats of reaction.
Materials:
NaOH(s)
0.50 M HCl
Styrofoam cup
1.0 M NaOH
thermometer
1.0 M HCl
Procedure:
Reaction #1
1. With the Styrofoam cup inside of a 400-ml beaker add 100 ml of DI water to the Styrofoam cup (this
will serve as a very crude calorimeter).
2. Record the initial temperature of the DI water and then add 2.00 g of NaOH(s), stir the contents and
record the highest temperature reached. CAUTION: do not poke a hole through the cup and do not
prop the thermometer against the cup (it might tip over).
3. Dispose of solution properly.
Reaction #2
1. Place 100 ml of 0.50 M HCl into the same (but cleaned) styrofoam cup that has been set into the 400
ml beaker.
2. Take an initial temperature of the solution, then add 2.00 g of NaOH and with constant stirring record
the highest solution reached.
3. Determine the products made and what the proper disposal should be.
Reaction #3
1. Into a clean Styrofoam cup (again supported by the 400 ml beaker) add 50 ml of 1.0 M HCl.
2. Record the initial temperature and then add (with constant stirring) 50 ml of 1.0 M NaOH and record
the highest temperature achieved.
3. Determine the products made and what the proper disposal should be.
Data:
Record all relevant data in a neat and organized data table.
Calculations:
Assumptions can be made that the low concentrations of HCl and NaOH are close enough to pure water
to use the density of water and the specific heat of water.
1. Calculate the calories/mol NaOH of energy produced in each reaction.
2. Write the thermomchemical reactions for all of the reactions.
3. Arrange two of the thermochemical reactions (from calculation #2) so that the net reaction produces
the other reaction.
Questions:
1. Write out the chemical reactions (some are disassociation reactions) from reactions 1-3 include the
calories from each reaction in the correct spot.
2. What appears to be true about the energy released by the two thermochemical reactions that
produced the other thermochemical reaction?
3. What is your percent error based on the enthalpy (∆H) for the second reaction?
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Hess’s Law directions for calculations and an example to follow
We can use Hess’s law of additive enthalpies to determine the reaction enthalpy of
propane.
3C(s) + 4H2(g)  C3H8(g)
The thermochemical equations you will use are:
C3H8(g) + 5O2(g)  3 CO2(g) + 4H2O(l)
Ho = -2220 kJ (a)
C(s) + O2(g)  CO2(g)
Ho = -394 kJ
(b)
H2(g) + ½ O2(g)  H2O(l)
Ho = -286 kJ
(c)
If you reverse a thermochemical equation you must change the sign of the reaction
enthalpy (Ho), and if the stoichiometry of a reaction is changed the reactions enthalpy
must also be changed by the same factor. The overall reaction enthalpy is the sum of
the individual reaction enthalpies combined in the same way you combined the
individual chemical equations.
1. Start with a thermochemical equation that has at least one of the reactants on the
correct side of the arrow in the overall chemical equation. A good choice would be
either (b) or (c). I chose (b) and multiplied it by three, this means that the H= is
also multiplied by three.
3C(s) + 3O2(g)  3CO2(g)
Ho = (-394 kJ) x 3 = -1182 kJ
2. Add to it an equation that results in a desired product on the right of the arrow. To
obtain 1 mol of propane (C3H8) on the right, you reverse (a) and change the sign of
its reaction enthalpy. Add (a) to (b) as follows:
Ho = -1182 kJ
Ho = + 2220kJ
Ho = + 1038 kJ
3C(s) + 3O2(g)  3CO2(g)
3 CO2(g) + 4 H2O(l)  C3H8(g) + 5 O2(g)
3C(s) + 4H2O(l)  C3H8(g) + 2O2(g)
Notice that the final equation has been simplified, canceling out “items” that
appeared on both reactants and products. Also notice that 3 oxygens on the
reactants side and 5 oxygens on the products side becomes 2 oxygens on the
products side.
3. To cancel an unwanted reactant, you add an equation that has that substance as a
product (vice-a-versa for a product). In this example we want to get rid of the 4
waters, so we can add (c) after multiplying it by 4:
3C(s) + 4H2O(l)  C3H8(g) + 2O2(g)
4 H2(g) + 2O2(g)  4 H2O(l)
Ho = + 1038 kJ
Ho = 4 x (-286 kJ) = -1144 kJ
3C(s) + 4H2(g)  C3H8(g)
Ho = -106 kJ
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Hess’s Law Problems
Directions: Show all necessary work in a legible manner.
1. Calculate the enthalpy change (H) for the formation of 1 mole of strontium
carbonate (the material that gives the red color in fireworks) form its elements.
Sr(s) + C(graphite) + 3/2O2(g)  SrCO3(s)
You have the following information:
2Sr(s) + O2(g)  2SrO(s)
H = -1184 kJ
SrO(s) + CO2(g)  SrCO3(s)
H = -234 kJ
C(graphite) + O2(g)  CO2(g)
H = -394 kJ
2. We want to know the enthalpy change for the reaction of lead (II) chloride with
chlorine to give lead (IV) chloride.
PbCl2(s) + Cl2 (g)  PbCl4 (l)
H = ?
We already know that PbCl 2(s) can be formed from the metal and Cl 2(g)
Pb(s) + Cl2(g)  PbCl2(s)
H = -359.4 kJ
and that PbCl4(l) can be formed directly from the elements.
Pb(s) + 2Cl2(g)  PbCl4(l)
H = -329.3 kJ
From this information calculate the unknown H for this reaction.
3. Calculate the quantity of heat transfer at constant pressure when benzene, C 6H6,
burns in oxygen to give water and carbon dioxide. (*This type of reaction is called a
heat of combustion or enthalpy of combustion.)
C6H6(l) + 15/2 O2(g)  6 CO2(g) + 3 H2O(l)
H=?
6C(graphite) + 3H2(g)  C6H6(l)
C(graphite) + O2(g)  CO2(g)
H2(g) + ½ O2(g)  H2O(l)
H = +49.0 kJ
H = -393.5 kJ
H = -285.8 kJ
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Hess law Practice
1. Given:
C2H6(g) + 7/2 O2(g)  2 CO2(g) + 3 H2O(l)
ΔH= -1559.7 Kj
What is ΔH for:
2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6 H2O(l)
ΔH=?
2 CO2(g) + 3 H2O(l)  C2H6(g) + 7/2 O2(g)
ΔH=?
2. Carbon burning in a limited supply of oxygen produces a mixture of CO 2(g) and CO(g).
It is impossible to determine the enthalpy of the carbon monoxide produced directly, but
it can be calculated indirectly.
We do know the following:
C(s) + O2(g)  CO2
ΔH1 = -393.5 Kj
CO(g) + ½ O2(g)  CO2(g) ΔH2 = -283.0 Kj
Find the enthalpy for the production of Carbon monoxide.
3. Find the ΔH for: C2H4(g) + H2O(l)  C2H5OH(l)
Given:
C2H5OH(l) + 3 O2(g)  2 CO2(g) + 3 H2O(l)
C2H4(g) + 3 O2(g)  2 CO2(g) + 2 H2O(l)
ΔH1 = -1367 Kj
ΔH2 = -1411 Kj
4. Find the enthalpy for the dimerization of ethane.
3 C2H4(g)  C6H12(l)
ΔH = ?
Given:
C6H12(l) + 9 O2(g)  6 CO2(g) + 6 H2O(l)
C2H4(g) + 3 O2(g)  2 CO2(g) + 2 H2O(l)
ΔH1 = -3920 Kj
ΔH2 = -1411 Kj
5. What is the Enthalpy for the total decomposition of CaCO 3(s)?
CaCO3(s) C(gr) + Ca(s) + 3/2 O2(g)
C(gr) + O2(g)  CO2(g)
CaCO3(s)  CaO(s) + CO2(g)
CaO(s)  Ca(s) +1/2 O2(g)
ΔH1= -393.5 Kj
ΔH2= +178.3 Kj
ΔH3 = +278.9 Kj
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Hess’s Law Practice
Directions: On a separate sheet of paper calculate the enthalpy of each desired chemical reaction. Use
the thermochemical equations shown below to determine the enthalpy for the reaction:
3
1. COCl2(g) + 2H2O(l) → CH2Cl2(l) + H2(g) + /2O2(g)
CH2Cl2(l) + O2(g) →COCl2(g) + H2O(l)
1
/2H2(g) + 1/2Cl2(g) →HCl(g)
2HCl(g) + 1/2O2(g) →H2O(g) + Cl2(g)
2H2O(l)  2H2O(g)
ΔH = -33.2KJ
ΔH = -161KJ
ΔH = 73.5KJ
ΔH = +88 kJ
2. 2H2(g) + O2(g) →2H2O(l)
CH3COOH(l) + 2O2(g) →2CO2(g) + 2H2O(l)
C(graphite) + O2(g) →CO2(g)
2C(graphite) + 2H2(g) + O2(g) →CH3COOH(l)
ΔH = -435.5KJ
ΔH = -197 KJ
ΔH = -244.5KJ
3. N2H4(l) + H2(g) →2NH3(g)
N2H4(l) + CH4O(l) →CH2O(g) + N2(g) + 3H2(g)
N2(g) + 3H2(g) →2NH3(g)
CH4O(l) →CH2O(g) + H2(g)
ΔH = -18.5KJ
ΔH = -23KJ
ΔH = -32.5KJ
4. 2NH3(g) + 4H2O(l) →2NO2(g) + 7H2(g)
2NH3(g) →N2(g) + 3H2(g)
N2(g) + 2O2(g) →2NO2(g)
H2(g) + 1/2O2(g) →H2O(l)
ΔH = 138KJ
ΔH = -99KJ
ΔH = 52.5KJ
2NO2(g) →N2(g) + 2O2(g)
N2O4(g) →N2(g) + 2O2(g)
ΔH = -118.7KJ
ΔH = -16.8KJ
5. 2NO2(g) →N2O4(g)
6. C2H6(g) →C2H2(g) + 2H2(g)
C2H2(g) + 5/2O2(g) →2CO2(g) + H2O(g)
1
H2(g) + /2O2(g) →H2O(g)
7
C2H6(g) + /2 O2(g) →2CO2(g) + 3H2O(g)
ΔH = -470KJ
ΔH = -142.5KJ
ΔH = -566KJ
7. C2H6O(l) + 3O2(g) →2CO2(g) + 3H2O(l)
2C2H6O(l) + O2(g) →2C2H4O(l) + 2H2O(l)
C2H4O(l) + 5/2 O2(g) →2CO2(g) + 2H2O(g)
ΔH = -712.2KJ
ΔH = -2042.2KJ
8. H2O(l) + CO2(g)→H2CO(aq) + O2(g)
H2CO3(aq)→H2O(l) + CO2(g)
H2CO(aq) + O2(g)→H2CO3(aq)
ΔH = 139.5KJ
ΔH = -252KJ
9. H2O(l) +SO2(g) →H2SO3(l)
H2SO3(l) →H2S(g) + 3/2O2(g)
S(s) + O2(g) →SO2(g)
S(s) + H2O(l) →H2S(g) + 1/2O2(g)
ΔH = 51KJ
ΔH = -74.3KJ
ΔH = 38.7KJ
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Entropy Practice
o
o
Standard Molar Entropy (ΔS )—1.0 atm and 25 C
1. Is the following reaction spontaneous according to the reaction entropy?
NaOH(s) + HNO3(aq))  NaNO3(s) + H2O(l)
NaOH(s)
NaNO3(s)
H2O(l)
HNO3(aq)
j
64.5 /mol K
200.7 j/mol K
69.9 j/mol K
155.6 j/mol K
2. Is the oxidation of iron spontaneous according to the reaction entropy?
O2(g)
Fe(s)
Fe2O3(s)
j
205.0 /mol K
27.3 j/mol K
87.4 j/mol K
3. Is the thermite reaction entropically spontaneous?
Fe2O3(s) + 2 Al(s)  Al2O3(s) + 2 Fe(l)
Fe2O3(s)
Al2O3(s)
Al(s)
Fe(l)
j
87.4 /mol K
50.9 j/mol K
28.3 j/mol K
27.3 j/mol K
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Gibb’s Free Energy (G)

Spontaneous reactions take place as long as entropy is increased. To avoid doing two
separate calculations, one for the entropy of the system and one for the entropy of the
surroundings, it is possible to combine the two calculations to find a single indicator of
spontaneous reactions. The combination of enthalpy (H), temperature (T), and entropy
(S) in one equation is called Gibb’s free energy.
Go = Ho - TS
The superscript (o) means “standard” and denotes that the variable is at its standard
state.
Directions: Solve the following problems:(watch units, joules and kilojoules) and
indicate Spontaneity of reaction.
1. Predict the sign of G for the following reaction situations:
(a) an exothermic reaction with an increase in entropy
(b) an endothermic reaction with an increase in entropy
2. Calculate the standard free energy change for the formation of methane at 298 K.
C(graphite) + 2 H2(g)  CH4(g)
Horxn = -74 .81 kJ/mol
Sorxn = -80.81 J/mol K
3. Will the following reaction take place at 298 K, given the following conditions:
2H2(g) + O2(g)  2H2O(g)
Horxn = -483.6 kJ/mol
Sorxn = 1530 J/mol K
4. Calculate the free energy for the synthesis of ammonia gas, at 25 oC, from its
elements.
N2(g) + 3 H2(g)  2 NH3(g)
Horxn = 294.1 kJ/mol
Sorxn = 238.97 J/mol K
5. Calculate the free energy for the synthesis of KOH from potassium and water at
25.00oC.
2K(s) + 2H2O(l)  2KOH(aq) + H2(g)
Horxn= -482.37 kJ/mol
Sorxn = 91.6 J/mol K
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Gibbs Free Energy Practice
1. Is the dissociation of NaCl (s) Spontaneous at 25oC?
NaCl(s) –H2O NaCl(aq)
ΔHo = +3.97 Kj/mol ΔSo = +43.4 J/mol K
2. Is the production of PbO(s) Spontaneous when formed from its elements at 25.0 oC?
Pb(s) + O2(g)  PbO(s)
ΔHo = -215.0 Kj/mol
ΔSo = -92.0 J/mol K
3. At STP will elemental oxygen and nitrogen have a tendency to form NO (g)?
½ N2(g) + ½ O2(g)  NO(g)
ΔHo = 90.0 Kj/mol
ΔSo = -12.0 J/mol K
4. We calculated the dissociation of NaCl (s) at 25.0oC had a free energy of -9.0 kj/mol. At
what temperature will this reaction be non-spontaneous?
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Unit 13 Objectives for Oxidation Reduction reactions (RedOx)

Understanding Oxidation and Reduction in terms of electrons
LeO GeR

Balancing half reactions

Electrolytic cells
Electroplating

Galvanic / Voltaic cells
Batteries vs. cells
Salt bridges
Drawing cell diagrams

Antioxidants
Sacrifice metals
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Oxidation Reduction Chemistry (RedOx Chem)
Oxidation reduction chemistry or commonly called RedOx chem (pronounced reed-ox)
is a chemistry in which the electron transfers between atoms when bonding is crucial.
The oxidation numbers of atoms before and after a chemical reaction are of special
importance. Depending on whether the oxidation number increases or decrease
determines if the atom was reduced or oxidized. What is an oxidation number you say?
Oxidation numbers have been used up until this time to determine an ionic species.
The superscripted plus and minus numerals in the upper right hand side of an atomic
symbol are oxidation numbers.
Calculating oxidation numbers
Oxidation numbers are loosely based on electronegativites and can be easily calculated
from the valance number of the atom. If an atom is in the group one on the periodic
table it has one valence electron, outer electron. The valence numbers are lost and
gained in response to the atoms bonding behavior. For group one atoms it is easier to
lose one electron than to gain seven and with the lose of one negative charge (electron)
the atom acquires a positive charge. All group one atoms have a plus one oxidation
charge (Na+, Cs+, K+) and the following group trend ensues:
Group I
Plus one
(Li+)
Group II
Plus two
(Be+2)
Group III
Plus three
(B+3)
Group IV
Plus or minus four (C+/-4)
Group V
minus three
(N-3)
Group VI
Minus two
(O-2)
Group VII
Minus one
(F-)
Group VIII no oxidation charge (He)
That takes care of all of the elements except for the transition metals which have
multiple oxidation states due to the close proximity of energy levels p and d. The
oxidation number of transition metals is determined by what the metal is bound with.
For example in Fe2O3 oxygen is always a minus two (-2, group 6) if there are three
oxygens in the compound and each is minus two the total anionic charge is negative
six. The total cationic charge of the iron must then also equal a magnitude of six. If
there are two irons to have the six positive charges divided over the charge on each
individual iron must be a positive three (Fe +3). Using the same lines of logic, and
calculating with a know oxidation number (O -2) the iron in FeO is iron plus two (Fe+2).
As a general rule there will be at least one know element (Group I - Group VIII) in a
compound and as long as you know that one you can back calculate to find the others.
When assigning oxidation numbers the most electronegative element is given its most
common oxidation state. The most electronegative element is therefore determined
first, as it is most likely that it will possess the number of electrons normally associated
with it due to the difference in electronegativity. Using these known elements and
polyatomic ions it will be possible to find the oxidation numbers for the transition metals.
However beware that there are some exceptions, as with oxygen. Oxygen is typically
O-2, however under some situations it will exist as a minus one (O -1) peroxide. In
situations where oxygen is involved with a compound and has a stoichiometry of (--O2)
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it will be a negative two oxidation charge as it is in H 2O2 hydrogen peroxide because
each of the oxygens involved have a negative one oxidation charge. Hydrogen can also
exist as a negative one (hydride).
Calculating oxidation number examples
TiSi
Ti = +4
Si = -4
NaOH
Na = +1
O = -2
H = +1
Polyatomic has an overall charge of -1
BaClBa = +2
Cl = -1
Charge on molecule unbalanced oxidation numbers
CaO2
Ca = +2
O = -1
Peroxide ion, two of them each at a negative one oxidation
NO2 N= +4
O=-2
oxygen is more electronegative than nitrogen and therefore given its most
common oxidation number of –2, nitrogen then must be a +4
RedOx Chem
Oxidation is a term that is probably more familiar to you than reduction. Iron is often
oxidized to form Iron oxide (rust) and aluminum oxidizes to form a protective layer on
aluminum surfaces to prevent further oxidation and maintain the integrity of the metal.
Oxidation is the loss of electrons. If the oxidation charge of a compound changes in
such a way as to become more positive it has lost electrons and therefore been
oxidized. Fe+2 being changed to Fe+3 shows the oxidation of iron. Reduction is the
gaining of electrons and makes the oxidation number more negative (reduces the
oxidation number). O-1 being changed to O-2 shows the reduction of oxygen. If an atom
is oxidized another atom must be reduced. If one atom loses an electron another atom
will gain that electron. Reducing agents are the atoms that cause reduction and they
themselves are then oxidized. Reducing agents lose electrons (oxidation) so that other
atoms can gain electrons (be reduced). Oxidizing agents are atoms that cause the
oxidation of another atom and they will themselves be reduced in the process.
Oxidizing agents readily gain electrons (reduction) from other atoms that lose the
electrons (oxidation). At first the terms oxidizing agent and reducing agent seem
counterintuitive until you think about how they function. There is help in the form of a
pneumonic device LeO GeR . Put a little growl in it like a lion.
Loss of electrons Oxidation
Gain of electrons Reduction
Oxidation and Reduction reactions always occur together. Tracking the changes in
oxidation numbers allows determination of which material is being reduced and which is
being oxidized. Not all reactions are RedOx reactions, only if the oxidation states
change.
Sample RedOx reactions:
4 Fe(s) + 3 O2(g)  2 Fe2O3(s)
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Iron and oxygen as reactants both have an oxidation number of zero (0), this is called
the ground state of the atom. Any element not combined with another element has a
zero oxidation number. As products iron has a +3 oxidation number and oxygen has a 2 oxidation number. Iron has lost three electrons to become +3, and has therefore been
oxidized, while oxygen has gained two electrons to become -2. The difference in the
numbers of electrons gained and lost is balanced in the stoichiometry of the reaction.
The reaction: 4 Fe(s) + 3 O2(g)  2 Fe2O3(s) is written as half reactions to show the
electrons that are transferred (see separate page for balancing 1/2 reactions).
Oxidation reaction:
Reduction Reaction:
Fe0  Fe+3 + 3eO0 + 2e-  O-2
(reducing agent)
(Oxidizing agent)
When calculating the oxidation number of an element in a compound, it is the element
with the highest electronegativity that is assigned the charge first. You, of course, recall
that the periodic trend for electronegativity increases as you go from left to right and
bottom to top. Nonmetal species are therefore most often given there oxidation
numbers before metals. With covalent compounds of two nonmetals, rely on
electronegativies. For example NO2 oxygen is more electronegative than nitrogen and
is assigned the oxidation number of –2. As there are 2 oxygens, both at –2, for a total
of –4 the charge on nitrogen must be +4.
Sample Problem:
Identify the oxidation and reduction reactions for the following reaction:
2 Fe2O3(s) + 3C(gr)  4Fe(l) + CO2(g)
First determine the oxidation states for all of the species:
Reactant
Product
+3
Fe
Fe0
-2
O
O-2
0
C
C+4
Then determine which species lost and gained electrons
Reactant
Product
+3
Fe
Fe0 (gained three electrons)
O-2
O-2
(no change, not involved in RedOx Rxn)
0
+4
C
C
(Lost four electrons)
Write the half reactions
Reduction Rxn:
Oxidation Rxn:
Fe+3 + 3 e-  Fe0
C0  C+4 + 4e-
Balancing out the number of electrons gained or lost allows the entire reaction to be
balanced.
Reduction Rxn:
4Fe+3 + 12 e-  4Fe0
Oxidation Rxn:
3C0 3 C+4 + 12e-
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When we begin balancing half reactions you will quickly see that it is also important not
to separate liquids, gasses or solids but rather keep the whole species intact. This will
aid in the materials balance of the half reaction.
Balancing Reactions using the Half Reaction Method (RedOx Chemistry)
The method makes use of only those species, dissolved or otherwise, which actually
take part in the reaction. So called “spectator ions,” or ions which are present but play
no role in the chemistry, are not included in the balancing procedure. Also, the scheme
is slightly different for acid and base conditions. The step-wise procedure is as follows.
1. Look at the equation to be balanced and determine what is oxidized and what is
reduced. This involves checking the oxidation numbers and determining which
species have changed oxidation numbers. Remember an oxidation is a loss of
electrons (increase in positive charges) and a reduction is a gain in electrons
(decrease in positive charges).
Equation to be balanced:
MnO4-(aq) + Fe+2(aq)  Fe+3(aq) + Mn+2(aq)
Iron goes from a +2 to a + 3, loss of electrons, Iron is oxidized
Manganese is reduced from +7 to +2, gain of electrons
(Manganese is a plus 7 because the permanganate ion is a minus one, oxygen is a
minus two and there are four oxygens for a total negative charge of minus eight, In
order for the ion to be minus one manganese must be plus seven)
2. Write a “half-reaction” for both the oxidation process and the reduction process and
label as “oxidation” and “reduction.” These half-reactions show only the species
being oxidized or reduced and the appropriate electrons.
Oxidized: Fe+2  Fe+3 + eReduction: 5e- + MnO4-  Mn+2
3. If oxygen appears in any formula on either side in either equation, it is balanced by
writing H2O on the opposite side. This is possible since the reaction mixture is a
water mixture. The hydrogen in the water is then balanced on the other side by the
addition of H+; this is for an acid solution. If a basic solution was being used the
water oxygen is balanced using an OH- and the hydrogens are balanced using
water. Balance both half reactions by inspection.
Oxidized: Fe+2  Fe+3 + eReduction: 8H+ 5e-+ MnO4-  Mn+2 + 4H2O
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4. Balance the charges on both sides of the equation by adding the appropriate
number of electrons (e-) to whichever side is lacking negative charges. To increase
the number of electrons you can multiple as done with any other stoichiometric
process.
Oxidized: Fe+2  Fe+3 + eReduction: 5e- + 8H+ + MnO4-  Mn+2 + 4H2O
5. Multiplying through both equations by appropriate coefficients, so that the number of
electrons involved in both half-reactions is the same. This has the effect of making
the total charge loss equal to the total chare gain and thus eliminates electrons from
the balanced equation.
Oxidized: 5Fe+2  5Fe+3 + 5eReduction: 5e- + 8H+ + MnO4-  Mn+2 + 4H2O
6. Add the two equations together. The number of electrons, being the same on both
sides, cancels out and thus does not appear in the final result. One can also cancel
out any other species that exist on both sides of the equation (often water and
hydrogen ions) Note that Fe+2 and Fe+3 are not the same species and cannot be
canceled.
Oxidized:
Reduction:
5Fe+2  5Fe+3 + 5e5e + 8H+ + MnO4-  Mn+2 + 4H2O
5Fe+2 + 8H+ + MnO4- Mn+2 + 5Fe+3 + 4H2O
-
7. Make a final check to see that the equation is balanced.
5 Fe on each side
8 H on each side
1 Mn on each side
4 O on each side
+17 charge on each side
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Balancing RedOx Reactions in Acidic and Basic Environments
In chemical reactions the solvent plays a vital role in the reaction mechanism but is usually not shown in
the net chemical reaction. Such is the case with redox reactions that react in either acidic or basic
conditions. The acid (H+) or the base (OH-) is important but often overlooked.
Balancing acid/base redox reactions is very similar to the redox reactions we have done this far, except
+
we can add H or OH to the half reactions.
Redox Reactions in an Acidic Environment
The following example of the oxidation of sulfur dioxide by dichromate in an acidic environment should
help us see the steps and their purpose.
-2
_____H+___
>
(aq)
SO2(g) + Cr2O7
-2
(aq)
SO4
+ Cr
+3
(aq)
First assign oxidation numbers to all atoms in the system.
S+4O-22(g) + Cr+62O-27-2(aq)
_____H+___
> S+6O-24-2(aq) + Cr+3(aq)
Then write the appropriate half reactions. Remember that polyatomics, solids, liquids, and gases are all
represented in a half reaction. Only aqueous compounds can be separated into their components
(elemental or polyatomic)
Oxidation: S+4O-22(g)  S+6O-24-2
Reduction: Cr+62O-27-2(aq)  2Cr+3 ( two chromium are needed for a materials balance)
Show the electrons that must have been exchanged to make each half-reaction possible
Oxidation: S+4O-22(g)  S+6O-24-2 + 2eReduction: 6e- + Cr+62O-27-2(aq)  2Cr+3
The oxygens in both reactions need to be balanced by adding water (H2O(l)). This is a materials balance.
Oxidation: 2H2O(l) + SO-22(g)  SO4-2 + 2e-
(four oxygens on each side)
Reduction: 6e- + Cr2O7-2(aq)  2Cr+3 + 7H2O(l) (seven oxygens on each side)
+
Finally the hydrogens are balanced by the addition of hydrogen ions (H )
Oxidation: 2H2O(l) + SO-22(g)  SO4-2 + 2e- + 4H+
Reduction: 14H+ + 6e- + Cr2O7-2(aq)  2Cr+3 + 7H2O(l)
When each half reaction is balanced then make sure that the electrons are equal between both half
reactions by determining the lowest common multiple.
Oxidation: (2H2O(l) + SO-22(g)  SO4-2 + 2e- + 4H+) x 3
+
-2
+3
Reduction: 14H + 6e- + Cr2O7 (aq)  2Cr + 7H2O(l)
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Oxidation: (6H2O(l) + 3SO-22(g)  3SO4-2 + 6e- + 12H+)
Reduction: 14H+ + 6e- + Cr2O7-2(aq)  2Cr+3 + 7H2O(l)
Bold numbers indicate
multiplication products
Once the electrons are equal then the reactions can be added together with a removal of species that
exist as both product and reactant.
Oxidation: (6H2O(l) + 3SO-22(g)  3SO4-2 + 6e- + 12H+)
2 +
-2
+3
1
Reduction: 14 H + 6e- + Cr2O7 (aq)  2Cr + 7 H2O(l)
2H+ + 3SO-22(aq) + Cr2O7-2(aq)  2Cr+3 + H2O(l) + 3SO4-2(aq)
Balanced reaction for acidic
environments
Redox reactions in a Basic Environment
To balance redox reactions in a basic environment we can employ a very similar strategy to what was
used in acidic environments. Let’s use the oxidation of hydrogen peroxide by the permanganate ion in a
basic enviornment as an example.
MnO4-(aq) + H2O2(l)
______OH-___
> MnO2(s) + O2(g)
As before first assign oxidation numbers to all atoms. Remember that peroxides have the oxidation
number of –1.
Mn+7O-24-(aq) + H+12O-12(l)
______OH-___
> Mn+4O-22(s) + O02(g)
Then write half reactions for the oxidation and reduction reactions according to the oxidation numbers.
Remember polyatomics, solids, gasses and liquids remain as an entire species.
Oxidation: H+12O-12(l)
______OH-___
> O02(g) + 2e-
Reduction: 3e- + Mn+7O-24-(aq) ______OH-___> Mn+4O-22(s)
You can add the electrons into the half reactions as shown above, however as the complexity of the
reactions increase it may be difficult (or at least time consuming) to determine the oxidation state of every
species. Remember that the only purpose for showing the oxidation states within a complex is to
determine if the oxidation state changes. Many times you can tell which species will be changed although
you may not know whether it will be oxidized or reduced. In the reaction above it should be obvious that
the MnO4- will change into the MnO2 while the hydrogen peroxide will change into the oxygen. So we can
start with that information for our half-reactions:
H2O2(l)
______OH-___
> O2(g)
MnO4-(aq) ______OH-___> MnO2(s)
Balance the materials: balance oxygens by adding water. We can add water because the reactions often
involve at least one species as an aqueous solution. We balance hydrogens the same way as we did
with an acid, by the addition of hydrogen ions
H2O2(l)
______OH-___
> O2(g) + 2H+(aq)
4H+(aq) + MnO4-(aq) ______OH-___> MnO2(s) + 2H2O(l)
After the materials are balanced then add electrons appropriately to balance the charges.
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H2O2(l)
______OH-___
+
+ MnO4 (aq)
4H
+
> O2(g) + 2H
-
(aq)
H2O2(l)
______OH-___
+
3e- + 4H
(aq) +
(+2 charge on the Product side)
(aq)
______OH-___
> MnO2(s) + 2H2O(l) (+3 charge on the Reactant side)
> O2(g) + 2H+(aq) + 2 e- (we can now tell that this is an oxidation)
-
MnO4 (aq)
______OH-___
> MnO2(s) + 2H2O(l)
(we can now tell that this is a reduction)
Now that the materials and the charges are balance (i.e. the ½ reactions are balanced) we need to deal
with the hydrogen ions. In a basic environment all the Hydrogen ions will react with hydroxides to form
water. So add enough hydroxides to each equation to react with the hydrogen ions. Hydroxides must be
added to both sides in order to maintain a balanced reaction.
2 OH- + H2O2(l)
______OH-___
> O2(g) + 2H+(aq) + 2 e- + 2OH-
4OH- + 3e- + 4H+(aq) + MnO4-(aq) ______OH-___> MnO2(s) + 2H2O(l) + 4OHReacting the hydrogens with the hydroxides will form water. Also watch for any waters (or other
materials) that exists as a product and a reactant as these can be cancelled.
H2O2(l) ______OH-___> O2(g) + 2 e- + 2 H2O(l)
(2 H2O(l)) 4H2O(l) + 4e- + MnO4-(aq) ______OH-___> MnO2(s) + 2H2O(l) +2 OH-(aq)
2 H2O(l)+ 3e- + MnO4-(aq) ______OH-___> MnO2(s) + 4OH-(aq) (this is the ½ reaction after the waters are cancelled)
Once all species and charges are balanced in both ½ reactions then the lowest common multiple of the
electrons can be determined and each reaction multiplied by their respective numbers so that the
electrons are equal in both ½ reaction.
-
2 OH (aq) + H2O2(l)
electrons)
______OH-___
> O2(g) + 2 e- + 2 H2O(l) (Multiple this reaction by 3 so that each rxn has 6
2 H2O(l)+ 3e- + MnO4-(aq) ______OH-___> MnO2(s) + 4OH-(aq) (multiple this rxn by 2 so that each rxn has 6
electrons)
6 OH-(aq) + 3H2O2(l)
______OH-___
> 3O2(g) + 6 e- + 6 H2O(l)
4 H2O(l)+ 6e- + 2MnO4-(aq) ______OH-___> 2 MnO2(s) + 8OH-(aq)
Cancel any species, including electrons, that exist as products and reactants and write the final balanced
reaction.
6 OH-(aq) + 3H2O2(l) ______OH-___> 3O2(g) + 6 e- + 6 H2O(l)
______OH-___
4 H2O(l)+ 6e- + 2MnO4 (aq)
> 2 MnO2(s) + 8OH (aq)
Final rxn: 2MnO4-(aq)+ 3H2O2(l)
______OH-___
> 3O2(g) + 2 H2O(l) +2 MnO2(s) + 2OH-(aq)
The bolded coefficients in the final rxn are caused by partial cancellation due to unequal amounts in the
reactant and product.
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½ Reactions and Balancing ½ Reactions
Directions: Given the following chemical reactions show the ½ reactions for each one
and indicate which reaction is the oxidation reaction and which is the reduction reaction.
Use the ½ reactions to balance the original reaction, and label the oxidizing agent and
the reducing agent.
1. Al(NO3)3(aq) + Mg(s)  Al(s) + Mg(NO3)2(aq)
2. Cd(s) + HCl(aq)  CdCl2(aq) + H2(g)
3. Cu+2(aq) + Fe(s)  Cu(s) + Fe+2(aq)
4. Al(s) + Mn+2(aq)  Al+3(aq) + Mn(s)
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Acid Base RedOx Reactions
Directions: On a separate sheet of paper write balanced reactions for each unbalanced
reaction shown below. Pay special attention to the reaction conditions (e.g acidic or
basic) and show all of your work. Any and all illegible work will be incorrect.
1. H2S(l) + Cr2O7-2(aq) --H+ S(s) + Cr+3(aq)
Balance in an Acidic Solution
2. Bi(OH)3(s) + SnO2-2(aq) –OH- Bi(s) + SnO3-2(aq)
3. Zn(s) + VO3-(aq) --H+ V+2(aq) + Zn+2(aq)
4. MnO4-(aq) + Br-(aq) –OH- MnO2(s) + BrO3-(aq)
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It’s a Red Ox Charge
Materials:
Aluminum pop can (top cut off)
Cu wire
Carbon rod
Voltmeter or small appliance
NaCl solution
Porous Cup
Bleach
Balance
Objective: To make a voltaic cell from an aluminum pop can.
Procedure:
1. Cut the top off of an aluminum pop can. BE CAREFUL this will be a sharp edge, and you may want
to put masking tape around the edge to avoid any cuts.
2. Fill the aluminum can about half full with NaCl solution, You need to determine the concentration of
this solution and each group will have a different molarity of NaCl (0.25 M, 0.5 M or 1.0 M).
3. Carefully add Bleach to the porous cup (record volume), place a carbon rod in the cup as well. DO
NOT allow the bleach to overflow into the cup.
4. Place the cup into the can with the graphite rod inside of the porous cup. Do not allow the carbon rod
to come into direct contact with the side of the pop can.
5. Connect the wire leads to the edge of the pop can (remove some of the tape) and to the carbon rod.
This is a Galvanic / Voltaic cell so the anode is the negative and the cathode is the positive pole.
6.
Record your volts from the voltmeter (set at DC volts--20).
7. Continue to connect your cell with the group across from you, and then with the entire class for the
maximum voltage. Record the average voltage per cell.
Data:
Data should include, mass of NaCl, volume of H2O, Concentration of NaCl, volume of Bleach, number of
cell, and voltage.
Calculations
The following half reactions (notice both are reductions) were taking place:
+3
(anode)
3e- + Al → Al
Ered = - 1.68 V
(Cathode)
OCl + H2O + 2e- → Cl- + 2 OHEred = 0.890 V
Calculate the theoretical voltage of the cell you created based on standard reduction potentials. Find the
percent error in your cell.
Questions:
1. Diagram your cell showing the species in the correct places and the flow of electrons.
2. How could you increase the voltage of this cell?
3. What is the inherent flaw with a cell of this design (hint: be aware of where the ionic species are
coming from)?
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o
Electrolytic Cells and Standard Reduction Potentials (E )
Electrolytic Cells and Antioxidants
All cells are comprised of two or more materials in which an oxidation and a reduction is possible,
however not all cells are spontaneous (-ΔG). Galvanic cells (i.e. Voltaic cells) are spontaneous cells.
These cells are designed in such a way as to create a flow of electrons. The flow of electrons is with an
electrochemical gradient, meaning that the electrons flow from a high concentration of electrons (or
charge) to a low concentration (or charge).
Electrolytic cells (+ΔG) are non-spontaneous cells, these are cells that require an energy input. Yes,
there is a need for cells that require an energy input. Electrolytic cells use an input of energy to force the
+2
movement of electrons against the concentration gradient. If a galvanic cell such as Zn(s)|Zn (aq) ||
+2
+2
Cu (aq)|Cu(aq) produces 1.001 Volts (Erxn = +1.001v) then the electrolytic cell of Cu (aq)|Cu(aq)||Zn(s)|Zn+2(aq)
will require an input of 1.001 Volts (Erxn = -1.001 V). The input of energy causes the cathode to reduce on
the anode (plate out). This is how electroplating is done. In an electrolytic cell the cathode is still the site
of reduction and is where the aqueous material solidifies while the anode (site of oxidation) is where the
solid material dissolves into its aqueous form. The anode is slowly dissolved by this reaction and is often
referred to as the sacrificial anode. For this reason the anode is usually a cheaper metal than the
material being plated out.
Ever RedOx reaction has an anode that slowly looses electrons due to oxidation. The oxidation of a
metal usually causes a weakness in its structural integrity. Iron hulls of boats can oxidize (rust) to form
holes and wet sailors. Underground pipes can oxidize to the point of forming leaks and potentially
environmental hazards. These metals of structural importance are often paired with a sacrificial metal.
Sacrificial metals are metals that are intended to be a ready source of electrons and are more easily
oxidized than the metals needed for structural strength. Magnesium is often used as a sacrificial metal.
As an underground pipe is oxidized a wire connected to the pipe and a block of magnesium creates a
flow of electrons that replace (prevents) the oxidation of the iron in the pipe. As long as the pipe is not
oxidized then no holes will form. The same process can be accomplished in the hulls of ships, the
magnesium blocks are sacrificial and are ‘used up’ over time and need to be replaced.
Sacrificial metals are analogous to biological antioxidants. Vitamins E and C are antioxidants, these
materials prevent the oxidation of proteins and cellular elements because they are more easily oxidized.
Lemon juice works is sprinkled on fruit salads to prevent the browning of apples and bananas. Apples
and bananas turn brown as the quinine is oxidized, but if sprinkled with lemon juice (vitamin C) then the
vitamin C (ascorbic acid) is oxidized and the fruit remains white. Antioxidants are used in cosmetics for
much the same reason, to prevent the oxidation of proteins in the skin (elastin and collagen) that give skin
its elastic composition. When the elasticity of the skin is lost wrinkles are formed.
o
Standard Reduction Potentials(E ) & Spontaneity
Potential refers to voltage and so Standard reduction potential is really just the amount of voltage that a
material is reduced under standard conditions. Standards are need to compare one reduction to another,
so the standards are set at a 1.0 molar solution of the ion at 273.15 K and 1 atmosphere of pressure.
Voltage is a comparison, the potential for a charge to move. This is an important concept to remember
because a half reaction alone really has no voltage because it cannot move anywhere; there must be a
reduction and an oxidation. All voltages are arbitrary in the sense that they are compared to the Standard
Hydrogen Electrode (SHE). SHE is Pt(S)|H2(g)|H+(aq). All reductions are compared against the oxidation of
hydrogen gas on a platinum wire to hydronium. The oxidation of hydrogen gas to hydronium has been
defined as a 0.000 volt, so any voltage in a cell that involves the oxidation of hydrogen can be attributed
to the reduction of the other metal.
Sample of some Standard Reduction Potentials:
Ag+(aq) + e-  Ag(s)
Eored = 0.799 v
+
Li (aq) + e  Li(s)
Eored = -3.040 v
+2
Sn (aq) + 2e  Sn(s)
Eored = -0.141 v
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The greater the E red the stronger the OXIDIZING AGENT, this means that the when comparing two
substances the one with the greater reduction potential with oxidize (steal electrons) from the one with the
smaller Eored.
For Example:
When comparing silver and tin, the silver ion will oxidize the tin, because silver has a reduction potential
of 0.799 v while the tin is -0.141. This also means that a reaction between solid tin and aqueous silver is
likely to be spontaneous and create a positive voltage.
Voltage is the potential to move and is based on the materials being used, not on the amount of materials
being used in the reaction. If you multiple both sides of a reaction by 100 you have 100 times as much
material, however the difference between the charges has not changed because both sides have been
multiplied by 100. Voltage is not affected by the amount of material that is involved in the reaction.
Lets determine the voltage of your cell described above:
Ag(NO3)(aq)+ Sn(s)  Sn(NO3)2(aq) + Ag(s)
Ag+(aq) + e-  Ag(s)
Sn(s) Sn+2(aq) + 2e-
Eored = 0.799 v
Eored = +0.141 v (change direction change the Sign)
Balance the reaction by making sure that the electrons cancel out
2Ag+(aq) + 2e-  2Ag(s)
Sn(s) Sn+2(aq) + 2e+
+2
2 Ag (aq) + Sn(s  Sn (aq) + Ag(s)
Eored = 0.799 v (voltage is NOT multiplied by 2)
Eoox = +0.141 v
o
E rxn = + 0.94v
The reaction as discussed earlier is likely to be spontaneous because the summation of the voltages is
positive, and a cell that creates a positive flow of electrons is likely to increase the total entropy of the
universe and therefore be spontaneous.
Nernst Equation
Electrochemical spontaneity is actually determined by the Nernst Equation which calculates the free
energy of a cell. The Nernst equation is ΔG = -nFEorxn.
ΔG = Free energy
n= number of moles of electrons
F= Faraday’s Constant (96,485 J/molv)
Eorxn= Voltage from cell
Using the Nernst equation for our silver / tin cell gives the following:
ΔG = -(2 mol)(96485J/molv)(0.94 v) = -181392 J
A negative delta G, means that the cell is spontaneous.
The number of moles is always exact and positive, Faraday’s constant is also always positive so the
voltage dictates the spontaneity of the reaction; a positive voltage will produce a spontaneous cell, while a
negative voltage returns a positive delta G and is non-spontaneous.
Lab Discussion
The Pop can cell was comprised of an Aluminum anode and the reduction of hypochlorite ion in bleach.
Using the standard reduction potentials of aluminum and hypochlorite ion show that the theoretical
voltage of the cell was 2.57 volts and was therefore spontaneous (-ΔG). When the battery was
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Evenson
constructed the total voltage of the Cells did not change, however by constructing the battery we did
increase the voltage of the battery. I suspect that cell averages where around 1.5 volts. So the question
becomes where the error is. Why are our cells not producing the theoretical voltage expected? We can
rule out the concentration of the salt bridge, as everyone had a different molarity of NaCl and there was
little to no affect on the voltage produced. Aluminum and hypochlorite concentrations will not affect the
voltage because it will not affect the overall potential to move. One large source of error does come from
the can. The aluminum can is lined, it must be to prevent the aluminum from oxidizing in the presence of
acids in the pop. The thin plastic lining is acting as an insulator for the aluminum and could be a source
of error. It is also important to remember that your theoretical value is based on STANDARD reduction
potentials in which you have 1.0 molar concentration of all aqueous solutions. You started with solid
aluminum and did not have a 1.0 Molar solution of aluminum. As the reaction continued and your
concentration of aluminum became closer to 1 molar your voltage would become closer to your
theoretical voltage value. You may have noticed that when the battery was set up the voltage continued
to climb up slowly. Another aspect of the cell is that the can serves a dual purpose it is used both for the
structure of the cell as well as the anode of the cell. As the cell runs, the can will oxidize and the
continued oxidation of the aluminum will eventually produce holes in the aluminum causing the salt water
of the bridge to leak out and prevent a balancing of charges. At this point the cell will reach equilibrium
and cease to function with any net flow.
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Evenson
Standard Reduction Potentials
Directions: Using the Standard reduction potentials given, determine if the chemical
reaction is Spontaneous or non-spontaneous via the Nernst equation and what the
voltage output (positive or negative) will be.
Selected Standard Reduction Potentials
Eored(V)
0.890 V
1.360 V
0.339 V
0.000 V
0.908 V
-0.236 V
-1.68 V
-0.409 V
½ Reaction:
OCl-(aq) +H2O(l) + 2e-  Cl-(aq) + 2OH-(aq)
Cl2 + 2e-  2Cl-(g)
Cu+2(aq) + 2e-  Cu(s)
2H+(aq) +2 e-  H2(g)
Hg+2(aq) + 2 e-  Hg(l)
Ni+2(aq) + 2e-  Ni(s)
Al+3(aq) + 3e-  Al(s)
Fe+2(aq) + 2 e-  Fe(s)
1. Ni(s) + Cu+2(aq)  Ni+2(aq) + Cu(s)
2. Fe(s) +2HCl(aq)  FeCl2(aq) + H2(g)
3. Cu(s) +2 HCl(aq)  CuCl2(aq) + H2(g)
4. Hg(l) +2 HCl(aq)  HgCl2(aq) + H2(g)
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Evenson
Standard Electrode Potentials in Aqueous Solution at 25°C
Standard
Potential
E° (volts)
Reduction Half-Reaction
+
-
Li (aq) + e -> Li(s)
-3.04
K+(aq) + e- -> K(s)
-2.92
2+
-
Ca (aq) + 2e -> Ca(s)
+
-2.76
-
Na (aq) + e -> Na(s)
2+
-2.71
-
Mg (aq) + 2e -> Mg(s)
3+
-2.38
-
Al (aq) + 3e -> Al(s)
-1.66
-
-
2H2O(l) + 2e -> H2(g) + 2OH (aq)
-0.83
2+
-
-0.76
3+
-
-0.74
2+
-
-0.41
2+
-
-0.40
Zn (aq) + 2e -> Zn(s)
Cr (aq) + 3e -> Cr(s)
Fe (aq) + 2e -> Fe(s)
Cd (aq) + 2e -> Cd(s)
2+
-
Ni (aq) + 2e -> Ni(s)
-0.23
2+
-
-0.14
2+
-
Pb (aq) + 2e -> Pb(s)
-0.13
Fe3+(aq) + 3e- -> Fe(s)
-0.04
Sn (aq) + 2e -> Sn(s)
+
-
2H (aq) + 2e -> H2(g)
4+
-
0.00
2+
Sn (aq) + 2e -> Sn (aq)
2+
-
0.15
+
Cu (aq) + e -> Cu (aq)
ClO4-(aq)
0.16
-
-
-
+ H2O(l) + 2e -> ClO3 (aq) + 2OH (aq)
-
0.17
-
AgCl(s) + e -> Ag(s) + Cl (aq)
2+
0.22
-
Cu (aq) + 2e -> Cu(s)
ClO3-(aq)
0.34
-
-
-
+ H2O(l) + 2e -> ClO2 (aq) + 2OH (aq)
-
-
-
0.35
-
IO (aq) + H2O(l) + 2e -> I (aq) + 2OH (aq)
+
0.49
-
Cu (aq) + e -> Cu(s)
-
0.52
-
I2(s) + 2e -> 2I (aq)
ClO2 (aq)
0.54
-
-
-
+ H2O(l) + 2e -> ClO (aq) + 2OH (aq)
3+
-
0.59
2+
Fe (aq) + e -> Fe (aq)
0.77
Hg22+(aq)
0.80
+
-
+ 2e -> 2Hg(l)
-
Ag (aq) + e -> Ag(s)
2+
0.80
-
Hg (aq) + 2e -> Hg(l)
-
0.85
-
-
-
ClO (aq) + H2O(l) + 2e -> Cl (aq) + 2OH (aq)
2+
-
2Hg (aq) + 2e ->
-
+
0.90
Hg22+(aq)
0.90
-
NO3 (aq) + 4H (aq) + 3e -> NO(g) + 2H2O(l)
0.96
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Evenson
Br2(l) + 2e- -> 2Br-(aq)
+
1.07
-
O2(g) + 4H (aq) + 4e -> 2H2O(l)
Cr2O72-(aq)
+
1.23
-
3+
+ 14H (aq) + 6e -> 2Cr (aq) + 7H2O(l)
1.33
Cl2(g) + 2e- -> 2Cl-(aq)
4+
-
1.36
3+
Ce (aq) + e -> Ce (aq)
MnO4 (aq)
+
1.44
-
2+
+ 8H (aq) + 5e -> Mn (aq) + 4H2O(l)
+
1.49
-
1.78
2+
Co (aq) + e -> Co (aq)
1.82
2S2O8 (aq)
22SO4 (aq)
2.01
H2O2(aq) + 2H (aq) + 2e -> 2H2O(l)
3+
-
-
+ 2e ->
+
-
2.07
F2(g) + 2e -> 2F (aq)
2.87
O3(g) + 2H (aq) + 2e -> O2(g) + H2O(l)
-
-
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Standard Reduction Potentials and the Nernst Equation Additional Practice
Directions: balance the following given reaction using the ½ reaction method. Showing all work use the
Nernst equation to determine the spontaneity of each reaction given.
1.
CuS(aq) + HNO3(aq)
2.
2Fe
3.
KClO3(s)
4.
Fe3O4(s)+CO(g)
5.
K2(Cr2O7)(s)+HCl(aq)
+3
(aq)
CuSO4(aq) + NO(g) + H2O(l)
+2
(aq)
+2
(aq)
+ Sn
2Fe
+4
+ Sn
(aq)
KCl(s)+O2(g)
Fe(l)+CO2(g)
Cl2(g)+CrCl3(s)+H2O(l) + KCl(s)
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Red Ox Reactions (Redox)
Objective: Determine viable reactions by experimentation and then write the
corresponding half reactions.
Materials:
test tube rack
Steel Wool
Mg strips
0.05 M Pb(NO3)2
Zinc strips
Cu strips
0.05 M AgNO3
0.05 M ZnCl2
0.05 M KCl
0.05 M NaCl
0.05 M MgCl2
Procedure:
1. Polish metal strips of copper, zinc, and magnesium with steel wool until they are
clean / shiny.
2. All aqueous metal drops will be added to a new area of the metal strip.
3. On a strip of copper metal add 1 drop of each aqueous cation.
4. On a strip of zinc metal add 1 drop of each aqueous cation.
5. On a strip of magnesium metal add 1 drop of each aqueous cation.
6. Record observations.
7. Wipe all of the metal clean with a paper towel and return the metal to octagon.
Data:
Data Table should contain the metallic solids across the top and the aqueous cations
along the left side. The spectator ions are unimportant.
Calculations:
All of the reactions that occurred are single replacement reactions. Write balanced ½
reactions for those reactions that occurred, this needs to be done indicating the
reduction reaction and the oxidation reaction for each reaction.
Questions
1. Why is it necessary to polish the metal strips before the experiment?
2. What happened to the oxidation states of the materials (metals / aqueous metals) in
the reactions that did not occur?
3. How do you know that a chemical reaction occurred?
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Who’s been Boozin’
Background:5
Breathalyzer tests that are used by law enforcement agencies are based on oxidationreduction chemistry. A persons breath will contain an amount of alcohol that is related
to the total alcohol in the body. Determine the amount in the breath and you (a
computer in the Breathalyzer) can calculate the percent alcohol in the blood. The
reaction most often used involves the chromate ion (Cr 2O7-2). When this ion is in
solution it has an orange color, when it is oxidized by an organic alcohol, such as
ethanol, it changes to a blue green color. The intensity of the color change can be used
in another chemistry law (Beer’s Law) to determine the concentration of the alcohol.
Objective:
Determine which household chemicals contain alcohols.
Materials:
0.6 M Potassium dichromate (add H2SO4 to help dissolve)
Well plates
Household products
Procedure:
1. Label well plates with the house hold products (unknowns) to be tested.
2. Place 2 drops of unknown in each well plate.
3. Add 1 drop of potassium dichromate solution to each well to be tested.
4. Record any color changes in data table.
5. Choose one of the products that had a positive test for alcohol, and add 2 ml of that
substance to the three drops of potassium dichromate. What happens if more
alcohol is present? Record your observations.
Questions:
1. Was the presence of alcohol noted on the label of your product?
2. If the orange Cr2O7-2 ion reacts with the alcohol to form a blue-green color which
substance is the reducing agent and which is the oxidizing agent?
3. In step 5 you added more alcohol to the potassium dichromate how did that affect
the color?
4. Would it be possible for a law enforcement officer to get a false reading from a
Breathalyzer?
5
Thompson, Robert G. The Journal of Chemical Education. "The Thermodynamics of Drunk Driving." v. 74 n. 5
May 1997. p. 532-536
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Galvanized Currency
Cu(s) + ZnSO4(aq) → CuSO4(aq) + Zn(s)
Background:
Voltage is based on the difference between two charges and is therefore irrelevant to the amount of
material creating that difference. In other words change the stoichiometry of a reaction does not change
o
the voltage produced by that reaction (E Rxn). In reality the concentration the reactants does have a slight
effect on the voltage but not to an extent that it will be applied to our calculations. Free energy (ΔG) is
the amount of energy that can be used to do work and is directly related to the amount of material
involved in the reaction. If you double the stoichiometry of a reaction then the free energy is also
o
doubled. The Nernst equation {ΔG = - (n)(f)( E Rxn)} can be used to determine the amount of free energy
per mole for a given reaction. Remember that the free energy will have the units of Joules per mole. The
energy is per mole because unlike voltage the free energy is directly related to the quantities of the
reaction. A Joule is a derived unit that is the measure of energy required to move a 1 Newton force 1
meter (J = Nm). From this definition it can also be determined that (with substitution and basic algebra)
that a J = (amps)(volts)(seconds).
Objective: By constructing an electrolytic cell, galvanize zinc onto a copper penny and use empirical data
to calculate the input energy per mole for the cell.
Procedure:
1. Use steel wool to buff a penny and a zinc strip.
2. Record the initial mass of both the zinc strip and the penny.
3. Add 50.0 ml of zinc (II) acetate solution to a 100-ml beaker.
4. Attach the zinc strip and the penny to the electrodes of the power supply and an Ammeter so that zinc
will be reduced on the penny. Zinc coating is called Galvanization.
5. Record the required time for the penny to become coated with zinc. Do not allow the zinc to be
reduced on the alligator clip.
6. Record the final mass of both the zinc strip and the penny.
Data:
Record all data relevant to the objective including but not limited to initial and final masses, voltages,
amperage, concentrations and reaction time.
Calculations:
1. Determine the theoretical energy (J/mol) input that is required for the amount of zinc that was reduced
on your penny.
2. Calculate the actual energy (J/mol) input that was used for the mass of reduced zinc.
3. Calculate your percent error.
Questions:
1. Explain why the Zinc (II) acetate solution should have no NET change in concentration during the
electrolytic reaction.
2. Is the galvanized penny more or less prone to oxidation than an uncoated penny (use standard
reduction potentials to support your answer)?
3. Why is it necessary to use a DC current rather than an AC current for electrolytic reactions?
4. Use the energy exchange for the reaction to determine you percent error.
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Electroplating
Objective: To electroplate a paper clip with a thin layer of copper atoms and determine
the grams of reduced atoms plated.
Materials:
4.0 M CuSO4
Copper Strips
Paper clips
power supplies
Gator clips & leads
Steel wool
U-tubes
Cotton Plugs
Salt solution
Procedure:
1. Record the mass of the paper clip and zinc strips (polished).
2. The paper clip and the zinc strip needed to be setup so that the zinc strip is on the
anode and the paper clip is on the cathode. *Remember that the reduction of
copper will take place at the cathode.
3. Place the zinc strip in 50 ml of aqueous zinc solution and the paper clip in 50 ml of
aqueous copper solution.
4. Fill a U-tube with a salt solution and plug with cotton, use this as a salt bridge
between the zinc and copper solutions.
5. Use a moderate to low voltage setting to plate out the copper on to the paper clip.
6. Pat the clip dry being careful not to dislodge any reduced copper.
7. Remass the clip and the zinc strip to determine the change in mass attributed to both
elements.
8. Retain the paper clip and attach it to your lab report.
Data:
Construct a nice, neat, legible table including all masses, times and other observations.
Calculations:
Calculate the moles of copper (II) ions that were reduced and the moles of Zn +2 ions
that were oxidized. Calculate the moles of electrons that were transferred.
Determine your percent error based on the number of electrons used in the reduction
reaction as compared to the number of copper ions that were reduced.
Questions:
1. Write both the reduction and oxidation reactions for this reaction.
2. Write the balanced redox equation for this reaction.
3. Write the electrolytic cell name.
4. Clip your paperclip to your lab report.
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RedOx Chemistry Review
1. What is the oxidation number of the metal ion:
Cu(NO3)2
WF4
Cr2O5
2. Identify the oxidation reaction (reducing agent) and the reduction reaction (oxidizing agent) by use of
half reactions:
HCl(aq) + Zn(s)  ZnCl2(aq) + H2(g)
K(s) + H2S(g)  K2S(s) + H2(g)
3. Balance the following reaction using the half reaction method.
C12H22O21(s) + O2(g)  CO2(g) + H2O(g)
CH4(g) + S(s)  H2S(g) + CS2
Hg(l) + Cu(NO3)2(aq)  Hg(NO3)2 + Cu(s)
4. What is a sacrifice metal, and how does it work?
5. Why are antioxidants in cosmetics?
6. Why is vitamin E important during strenuous exercise?
7. Compare and contrast electrolytic and galvanic cells.
8. What is the purpose of a salt bridge?
9. Who is Luigi Galvani and where did he do his research?
10. Compare and contrast a cell and a battery?
+2
+4
11. Diagram what this cell would look like: C(gr)|Ag ||Zn|Cd
12. Why are standard reduction potentials important?
13. What is the theoretical voltage output for the following reaction (these are similar to Hess problems)
Cl2(g) + 2 Br-(aq)  2 Cl –(ag) + Br2(l)
-
o
Cl2 + 2 e-  2Cl
E = + 1.36 volts
Br2 + 2 e -  2 Br - Eo = + 1.09 volts
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Unit 14 Objectives for Organic Chemistry

Nomenclature

Isomers
Geometric
Structural
Enatiomeric
Biological effects

Extractions techniques
Polarity

Pharmaceutical
Ethnobotanical
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Organic Nomenclature Rules (severely abridged IUPAC)
The name of an organic compound is very descriptive. The organic name tells a chemist the exact
number of carbons, presence of any functional groups, how the bonds occur and on what carbon atom in
the carbon skeleton a functional group or bond other than a single bond occurs.
1.
Determine the number of carbons in the molecule. To do this you need to count the longest chain of carbon atoms.
# Carbons
1
2
3
4
5
6
7
8
9
10
Alkyl Group
methyl
ethyl
propyl
butyl
pentyl
hexyl
heptyl
octyl
nonanyl
decayl
# Carbons
11
12
13
14
15
16
17
18
19
20
Example: H3C--CH2==CH2--CH2
2.
Alkyl Group
undecyl
dodecyl
tridecyl
tetradecyl
pentadecyl
hexadecyl
heptadecyl
octadecyl
nonadecayl
eicosyl
4 carbons this is a Butane molecule
The type of bonding is also evident from the name. As well as where the bond occurs.
Single bonds end with
-ane
(these are alkanes)
Double bonds end with
-ene
(these are alkenes)
Triple bonds end with
-yne
(these are alkynes)
Rings begin with
cyclo
(such as cyclopentane)
In the name numbers will tell where the bond occurs
nd
Example: H3C--CH2==CH2--CH2
3.
rd
double bond (-ene ending) between 2 and 3
carbon the first bonded carbon is numbered
therefore: 2-butene
It is also necessary to indicate the presence of any functional groups, both by name and number (the R symbolizes irrelevant
carbons).
A functional group can be a branch carbon atoms not part of the main carbon skeleton (see example on bottom of next
page) these are named for the number of carbon atoms they posses.
Functional group
Name
Ending
R--OH
hydroxyl
R--C==O
l
H
Carbonyl
-ol (alcohols)
*polar
-al
(aldhyde)
1
Carbonyl
R--C--R
||
O
-one (pronounced own)
(ketone)
R--C==O
l
OH
Carboxyl these are acidic and usually end with –oic acid
R--N--H
l
H
Amines
R—C==O
|
NH2
Amides
-amine
*Basic
Remove the –oic acid from name and add -amide
*Basic
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Evenson
Functional group
Name
Ending
R--SH
Sulfhydryl
-thiol
-
O
l
R--O--P--O
l
O-
Phosphates
Phosphate
*Anionic
Example : H3C--CH--CH3
l
OH
our example has a hydroxyl group on the middle
carbon (carbon # 2) so the name will 2-propanol
Be aware that a little knowledge is a dangerous thing. I have given you a very rudimentary survey of
organic nomenclature. The complete IUPAC rules for nomenclature would fill a large phone book. For
example the numbering system depends on which functional groups have more control of the reactivity of
the molecule. We will attempt to keep our molecules simple and give you small look into the realm of
organic chemistry.
Some examples and their names
Cl
2,3 dichloropentane
CH
H3C
CH 2
CH
CH3
Cl
CH
H2C
CH2
HC
CH2
H3C
3-methyl-1-pentene (here a carbon group is the functional group)
CH3
CH2CH3 CH2
CH
CH2
CH3
OH
3 heptanol (functional groups are always counted from the side that will give them the lowest number)
CH 2
CH
NH2
H2C
CH 2
cyclobutanamine
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Evenson
H3C
H2C
O
CH 2
H3C
C
Ethanal
ethenol
O
methanoic acid
OH
OH
C
H
HO
O
O
C
H3C
Methanol
CH 2
H2C
CH3
CH 2 H C
3
C
CH 2
2 butanone
H3C
H
CH
propenone 2
H2C
CH 2
CH 2
CH 2
SH
CH 2 HC
1, 3, 5 cyclohexene
CH 2 CH3
3-hexanethiol
CH 2
C
H2C
O
H3C
CH 2
H
CH4
CH 2 C
CH 2 CH 2
cyclopentanone
NH2
HC
H3C
2-butylamine
CH3
CH 2
NH2
C
CH 2
methylamine
methane
H3C
NH2
O
butanal
H3C
H3C
CH3
CH HC
ethlyamide
H3C
O
CH
CH3
HO
3, 4 dimethyl-2-hexanol
H3C
CH 2
H3C
CH
1 butene
CH
H3C
CH2
CH
CH3
2 butene
CH 2
Cl
Cl
CH
CH
H2C
H3C
CH 2
CH 2
C
CH
CH
CH
HC
OH
H2N
CH
CH3
C
H3C
CH3
CH
OH
Br
H2C
3, 3 dihydroxy-2-butanamine
SH
CH3
CH3
3 bromo - 5 chloro-4 ethyl-4-methyl heptane
2 chloro-5-ethyl 4-cyclohexenethiol
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Organic Nomenclature Lab
Objective: Using a molecular model kit construct the following organic molecules. Making use of the
IUPAC naming system.
Procedure:
Construct the following molecules the draw the molecule.
Methane:
Ethane:
Propane:
Butane:
1,2-Butene
2- methyl propane
Construct 3 molecules of your own, draw them and give the correct chemical formula and name.
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Organic Nomenclature Practice
Following are several organic structures or names. If given a name draw the correct
structure, if given a structure write the correct name. You may use the molecular model
kits if you find them beneficial.
1. CH2==CH--CH2--CH3
2. CH3--OH
3. CH3--CH2--CH--CH2--CH2--CH3
l
CH2--CH3
CH2--CH3
l
4. CH3--CH2--C--CH2--CH2--CH3
l
CH2--CH3
5. Ethylene
6. Cyclohexanol
7. 4-ethyleneheptane
8. 2,4 dimethyl-2,3-hexene
9. propanone
10. Cyclooctamine
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Isomer Recognition Practice
Directions: For the following (1-5) identify which structural formulas are identical compounds or
structural (constitutional) isomers or not the same formula. Write all answers on a separate piece of
paper in blue, black ink or pencil.
1
H3C
H2C
CH 2
CH
CH3
CH 2
Cl
CH 2
CH 2
Cl
CH3
2
F
C
FC(CH 3)3
CH3
CH3
O
3
C
CH 3COCH 3
H3C
4
CH 3(CH 2)4CH 3
CH 2
H3C
CH 2
CH 2
5
CH 2
HO
CH 2
CH3
H2C
H3C
CH3
CH 2
CH3
CH3
CH
OH
6. Draw two structural isomers for aldehydes of the molecular formula: C4H8O
7. Draw three structural isomers for the ketone of molecular formula: C5H10O
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Caffeine Extraction from Tea leaves
Background:
Caffeine is a socially acceptable drug in the United States and most of the world. It is described as a
cardiac, respiratory, psycho-stimulant and as a diuretic. In the pure form caffeine is a white crystalline
solid with an LD50 of 14.8 mg / kg. Caffeine is thought to act on the brain by blocking adenosine receptors.
Adenosine, when bound to receptors of nerve cells, slows down nerve cell activity; this happens, among
other times, during sleep. The caffeine molecule, being similar to adenosine, binds to the same receptors
but doesn't cause the cells to slow down; instead, the caffeine blocks the receptors and thereby the
adenosine action. Caffeine has a melting point of 238oC and sublimes at 178oC.
Objective:
Isolate crystallized caffeine from tealeaves and determine a mass percent of caffeine for tealeaves.
Procedure:
Day one:
1. Place 100 ml of distilled water in a 500 ml beaker
2. Record the mass of tea in one tea bag and add to the distilled water.
3. If the tea is returned to the tea bag, then later filtering will be unnecessary.
4. Boil for 20 minutes with a constant stirring from a magnetic stir bar. Do not allow the mixture
to go to dryness. Distilled water may be used (conservatively) to wash mixture from sides of
beaker. Do not use any open flames if Chloroform is being used.
5. Manual stirring with a stirring rod may prevent the tea bag from rupturing.
6. Filter the hot mixture using gravity filtration.
7. Save the filtrate, covered with parafilm and store in a safe place with your name, hour, date
and caffeine extract labeled on beaker.
Day two:
1. Transfer the filtrate to a 250 ml Erlenmeyer flask and add 50 ml of chloroform.
2. Gently swirl the two layers together for about ten minutes. Note: any vigorous swirling or
mixing for longer than 10 minutes will make the separation difficult)
3. Pour chloroform layers into separatory funnel and separate the water from the chloroform.
The aqueous layer can be discarded and the chloroform layer retained in a premassed 100
ml beaker.
4. Evaporate the chloroform, under the hood, using a warm water bath. Do not allow
the level of the water in the warm water bath to pour into your chloroform layer. Caffeine will
sublime so do not allow the dry crystals to stay in the warm water bath too long.
5. The caffeine should be significantly pure to allow a purity determination based on melting point.
Load your sample via the bounce method and check the melting point in the Mel-Temp.
Data:
Make a legible table of all relevant masses and volumes, be sure to include the caffeine/serving and
serving size. The molecular structure of caffeine should also be part of your data section.
Calculations:
Show all of the calculations required in determining the mass percent of caffeine and the purity of your
product based on the percent error from the melting point. Determine the percent error based on the m/m
% of caffeine.
Questions:
1. Citing the molecular structure of Caffeine explain why both water and chloroform were necessary for
the extraction of caffeine.
2. Citing your mass percent of caffeine and its associated percent error, what could be responsible for
the error?
3. What is the cause of melting point deviations in your sample?
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Unit 15 Objectives for Nuclear Chemistry

Balancing Nuclear reactions
Alpha particles
Beta particles
Gamma particles

Radioactive Decay
Half lives
Dating

Fusion vs. Fission
Tokamak Reactors
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Balancing Nuclear Equations (Reactions)
In a nuclear reaction the nuclei of a specific isotope degrades, this will form new
elemental isotopes. For this reason both atomic mass (alpha decay, α) and atomic
number (beta decay, β) can change. The change of an element to another is called
transmutation. Mass is balanced out using beta particles (electrons, 0-1e), alpha
particles (helium nuclei, 42He), protons (11H), and neutrons (10n). In nuclear reactions
we are only concerned with conservation of mass. Do not worry about the elements
changing (transmutation) as long as the total mass and charge of each side is equal.
Most reactions we deal with are fission reactions, reactions in which the nucleus breaks
apart or decays. These reactions will follow a precise path from one isotope to the next
depending on the type of decay (alpha or beta) and the original isotope. A graph of
these predetermined decays is called decay series. I have included a decay series that
shows both alpha and beta decays as well as half-lives on the back page.
In short make sure that the total atomic masses (top number) on both sides is equal,
and make sure the total number of protons (bottom number) on both sides is equal.
Examples:
Alpha decay:
Beta decay:
238
92U
234
90Th
 23490Th + 42 He
 23492U + 2 0-1e-
The atomic mass (top number) is equal to 238 on both sides of equation
The atomic number, protons (bottom) number is equal to 92 on both sides of
equation.
Fusion reactions are reactions that create larger nuclei from smaller nuclei. These are
the reactions that power the sun (hydrogen fusion). Fusion reactions produce minimal
amounts of radioactive waste with 20 times the energy released as current fission
reactions. The high temperatures (10-15 million K) however make fusion reactions
difficult to produce. A reactor has been made called a Tokamak reactor. A Tokamak
reactor uses high electromagnetic fields to control the plasmic reaction. No net energy
has been created by these reactions however.
Fusion reaction:
2
1H
+ 31H  42He + 10n
Notice that the atomic mass and the atomic number are both increasing from reactant to
product
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Balancing Nuclear reaction Practice
1. Degradation of 238U to 234Th using alpha decay.
2.
234
Th undergoing beta decay to transmutate into
234
U.
3. Alpha decay of 226Ra to the most probable isotope.
4. Finish the nuclear reaction for beta decay:
218
84Po
 21886Rn
5. Balance the alpha decay reaction for
230
Th to 214Pb.
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Half-Life Lab
In this lab you will simulate radioactive decay with pennies. The pennies can be used to
discover the relationship between the passage of time and the number of radioactive
nuclei that decay.
A "heads-up" penny represents a radioactive isotope of the element Coinium; this
headsium isotope can decay to tailsium (a tails-up penny). You will be given
Objective:
Determine the half-life graph of coinium and explain the ½ life graph.
Procedure:
1. Begin by counting to verify that you have 100 Coinium atoms (pennies)
2. Place all of the pennies in the container so they are heads up
3. This represents the zero half-life and you have 100 pennies remaining, record this in
your data table.
4. Vigorously shake the container, keeping one hand on the lid so the pennies do not
fly everywhere.
5. Carefully remove all of the pennies that have decayed to become tailsium (tails-up
pennies). That was half-life one, record the number of headsium atoms remaining
and repeat.
6. Continue the half-lives until you have less than two headsium atoms remaining, that
is trial one. Repeat for a total of three trials.
Data:
In a neat and organized table record the half-life and the number of atoms remaining
after each half-life. Do this for each of the three trials.
Calculations:
Graph the average of your data for the three trials. The atoms remaining are dependent
variable and the half-lives are your independent variable.
Questions:
1. Make a general statement about the number of headsium nuclei that decayed after
each half-life.
2. In this simulation is there any way to predict when a particular penny will “decay”?
3. If you could follow the fate of an individual atom in a sample of radioactive material,
is there any way to predict when it would decay, explain?
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Isotopic Pennies
Objective:
Simulate the procedure of determining isotopic abundance of an element.
Procedure:
1. Record the average mass of 10 pre-1982 pennies and record (do not include
any 1982 pennies).
2. Record the average mass of 10 post 1982 pennies and record (do not include
any 1982 pennies).
3. Weight the mass of an empty canister and record.
4. Record the mass of a labeled canister and the ID # on the canister.
5. Record the ID # and mass of a second canister.
6. DO NOT OPEN THE CANISTERS.
Data:
In a neat, organized and labeled table record all of the information you collected.
Calculations:
Your calculations section for this lab will be lengthy. Calculate the isotopic
abundance of each type (pre-1982 and post-1982) pennies.
Questions:
1. Record your canister ID number and the numbers of each type of penny.
2. Hydrogen exists in three isotopic forms ( 1H, 2H, 3H) using the atomic mass
which of the three isotopes is the most common?
If an element exists in two isotopic forms ( 56X, 54X) and the atomic mass is 55.0000 g /
mol. What is the relative abundance of both of these two elements?
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Isotopic Abundance Practice
Directions: legibly show all work required to, answer the following questions.
1. What is the average atomic mass of an element if two isotopes exist in the following
compositions: 33.0% have an atomic mass of 94.0 and 67.0% have an atomic mass
of 87.0?
2. What is the atomic mass of an element whose isotopes exist in the following
concentrations 54.96 % of the isotopes have an atomic mass of 32.0 and the rest
have a mass of 45.0?
3. What are the isotopic frequencies of 87Sr and 88Sr?
4. What are the isotopic abundances of
59
Ni and 58Ni?
5. How many 85Rb atoms are in 253.68 grams of rubidium if the other rubidium isotope
has an atomic mass of 86.0?
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Radioactive Dating and Half Life Practice
Directions: Answer all questions completely, legible and showing all necessary work.
Half life values of various isotopes
Isotope
14
C
3
H
32
P
40
K
71
Se
226
U
238
U
Half life
5715 years
12.32 years
14.28 days
9
1.26 x 10 years
4.7 minutes
0.5 minutes
2.34 x107 years
1. How many half lives has a 100.0 gram mass of
mass of 800.0 grams?
226
U underwent if was originally a
2. How old is an artifact if it is found to contain 1/16 the amount of 40K as expected?
3. A researcher begins work with a 10.0 Kg sample of 71Se at 8:00 am, how much of
the sample will remain at noon when she goes to lunch?
4. A petrified tree is found to be about 22,860 years old how much of the trees original
14
C is still present?
5. How many years are required for a 200.0 mg sample of tritium ( 3H) to completely
decay?
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Appendices of additional projects and Resource Materials
Web Resources:
Student resources for Chemistry
http://chemed.chem.purdue.edu/genchem/
http://www.chemhelper.com/
http://www.chemtutor.com/
http://www.chem4kids.com/
http://www.chemistrycoach.com/tutorial.htm
http://chemistry.about.com/od/homeworkhelp/
http://dbhs.wvusd.k12.ca.us/webdocs/ChemTeamIndex.html
http://www.highergrades.com/
http://www.chemguide.co.uk/
http://xenon.che.ilstu.edu/genchemhelphomepage/
http://www.oaklandcc.edu/iic/iicah/ah_www_che.htm
http://www.utm.edu/departments/artsci/chemistry/HelpLinks.htm
http://www.chemcenter.org/
http://www.ouc.bc.ca/chem/probsol/ps_A-E.html
http://www.scs.uiuc.edu/~mainzv/HIST/index.htm#homepage
http://www.anachem.umu.se/jumpstation.htm
http://oswego.org/ExamPrep/chem.htm
http://www-sci.lib.uci.edu/HSG/GradChemistry.html
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Additional Equations for a Variety of Practices
These reactions can be used for a multitude of practice exercises as you begin to take
charge of your own education you will find more and more uses for this tool.
Hydrogen bromide reacts with barium hydroxide to produce water and barium bromide.
Carbon dioxide reacts with potassium, producing potassium carbonate and carbon.
Calcium hydride reacts with water to yield calcium hydroxide and hydrogen gas.
Zinc (II) sulfide will react with oxygen gas to for zinc (II) oxide and sulfur dioxide.
Carbon dioxide reacts with sodium producing sodium carbonate and carbon.
Lead (II) nitrate reacts with sodium chromate to produce lead (II) chromate and sodium
nitrate.
AgNO3(aq) + CaCl2(aq)  AgCl(s) + Ca(NO3)2(aq)
H2C2O4(g) + KMnO4(s)  H2O(g) + CO2(g) + MnO2(s) + KOH(s)
C2H4(g) + O2(g)  H2O(g) + CO2(g)
C8H18(l) + O2 (g)  H2O(g) + CO2(g)
Na(s) + H2O(l)  H2(g) + NaOH(aq)
Zn(s) + H3PO4(aq)  H2(g) + Zn3(PO4)2(aq)
Mg3N2(S) + H2O(l)  NH3(g) + Mg(OH)2(aq)
C5H12(l) + O2(g)  H2O(g) + CO2(g)
Fe2S3(s) + O2(g)  Fe2O3(s) + SO2(g)
N2O(g) + NH3(g)  N2(g) + H2O(g)
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Periodic trend for Atomic mass
Periodic trend for Electronegativity
Electronegativity (Pauling
values)
Atomic mass (g/mol)
140
120
100
80
60
40
20
0
0
5
10
15
20
25
30
35
40
45
50
55
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
60
0
10
Atomic number (protons)
500
450
400
350
300
250
200
150
100
50
0
10
20
30
40
50
60
Periodic trend for Atomic radius
Atomic radius (pm)
300
250
200
150
100
50
0
10
20
30
40
40
50
60
180
160
140
120
100
80
60
40
20
0
0
10
20
30
40
Atomic number (protons)
Atomic number (protons)
0
30
Periodic trend for Ionic radius
Ionic Radius (pm)
Electronaffinity
Periodic Trend for Electronaffinity
0
20
Atomic Number (protons)
50
60
Atomic number (protons)
270
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Oil Viscosity
Materials:
Various weights of oil
8 mm glass tubing
Corks to fit tubing
Stop watch
Thermometer
600 ml beaker
Hot plates
100 ml beaker
Ice bath
BBs
Ring stand
Test tube clamp
Objective:
Determine the viscosity of an unknown oil through the use of a student prepared graph
of time vs. temperature.
Procedure:
1. Determine the temperature of the oil in the ice bath and record. Then fill the glass
tube to about 3 cm from the top. Record the time it takes for the BB to fall to the
bottom.
2. With the same weight of oil determine the temperature of the oil in the hot water bath
and record. Then record the time it takes to descend the tube.
3. Plot this data on your graph.
4. Repeat this procedure with all three of the known oils.
5. For your unknown oil, which will be assigned, find its time of descent at room
temperature and determine what the weight of the oil is from your graph.
Questions:
1. Which has the higher viscosity rating, an oil of high SAE rating or one of low rating?
2. What is the SAE rating of your unknown oil?
3. How would you change your procedure to get more precise data?
4. How do your results compare with other lab teams who have the same unknown?
5. Hand in your graph.
*** Hand in: graph, data table, and questions-- title lab and record your name***
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Chlorine determination with sodium metal
Procedure:
1. In an evaporating dish place known volume of chlorinated water
2. Into the evaporating dish with the water sample place a small piece of sodium metal
of known mass. Remove the mineral oil from the sodium metal.
3. Determine the pH of the solution after the reaction
Alternatively it may be possible to evaporate the solution off after the sodium reacted
this should leave dry NaCl and the mass could be calculated that way and therefor the
concentration of Cl in the water
Calcs
pH+pOH = 14
pOH –log [OH-]
OH- present is a one to one ratio with sodium.
This shows the amount of sodium associated with the hydroxide ions.
The remaining sodium is associated with Chlorine in a one to one ration
Then determine the concentration of Chlorine.
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The cryogenic treatment of lipo-proteins in an ionic environment
Materials:
Stainless steel soup kettle (4 liters)
Big spoon
Liquid nitrogen or dry ice (crushed)
3 qts half & half
3 cups sugar
dash of salt (no more than the amount in a sugar packet)
1-2 oz of vanilla extract (more gives a better flavor)
Cool all ingredients in the refrigerator prior to use, then mix all ingredients and add small
amounts of liquid nitrogen with stirring.
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mass of sugar lost (g)
Rate of Sugar loss from Double Bubble Gum
0.06
0.05
0.04
0.03
0.02
0.01
0
0
10
20
30
Time (min)
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Chemistry Glossary
Abbreviated electron configuration
Brief notation in which only those electrons beyond the preceding
noble gas are shown. The abbreviated electron configuration of the Fe
atom is [18Ar] 4s23d6
Absorption spectrum
A graph showing the absorption of radiation by a sample over a range
of wavelengths
A nonmetal oxide that reacts with water to form an acidic solution
procedure used to determine the concentration of an acid or base.
The volume of a solution of an acid(or base)of known molarity
required to react with a known volume of base(or acid) measured
Acid anhydride
Acid-base titration
Acid-dissociation constant (Ka)
the equilibrium constant for the dissociation of a weak acid HB: HB
(aq)↔ H+(aq) + B-)((aq); Ka= [H+] x [B-] / [HB]
Acid rain
Rainfall with a pH of less than about 5.6, the value observed when
pure water is saturated with atmospheric CO2
Acidic ion
Ion that forms H+ ions in water. The NH4+ ion is acidic because of the
reaction:NH4+(aq) ↔ H+ (aq)+ NH3 (aq)
Acidic solution
Activated complex
An aqueous solution with pH less than 7 (at 25oC)
A species, formed by collision of energetic particles, that can react to
form products
The minimum energy that must be possessed by a pair of molecules if
collision is to result in reaction
Graph showing the relative energies of reactants, products, and
activated complex
Thermodynamic quantity, appearing in exact expressions for ΔG, ΔK,
ΔE. Which is approximately equal to molar concentration(M)
The amount of product obtained from reaction
Contamination of air by such species as, SO2, SO3, CO, NO, and NO2
Activation energy (Ea)
Activation energy diagram
Activity
Actual yield
Air pollution
Alcohol
A compound containing an OH group attached to a hydrocarbon
chain. Examples include methanol CH3OH and ethanol, C2H5OH
Alkali metal
Alkaline earth metal
Allotrope
A member in Group 1 of the Periodic Table. Examples: Li, Na, K
A member of Group 2. Examples: Be, Mg, Ca
One of two or more forms of an element in the same physical state.
Graphite and diamond are allotropes of carbon, and O2 and O3, are
allotropes of oxygen
Alloy
Metallic material made by melting together two or more elements, at
least one of which is a metal. Brass and steel are alloys
A solution of a metal in mercury
Rate of flow of electric current such that one coulomb passes a given
point in one second
Mixture of 3 volumes of 12 M HCl with 1 volume of 16M HNO3
Amalgam
Ampere (A)
Aqua regia
Atmosphere (atm)
Atomic spectrum
atomic theory
Standard unit of pressure, equal to 101.325 kPa; equivelent to the
pressure exerted by a column of mercury 760mm high
Diagram showing the wavelengths at which light is emitted by excited
electrons in an atom
Dalton's theory of the atomic nature of matter
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Aufbau (building up) principle
average speed(gas molecules)
balanced equation
Balmer series
base dissociation constant (Kb)
rule stating that sublevels are filled in order of increasing energy, from
one atom to another of higher atomic number
speed calculated using this equation: u= (3* RT / MM) * 1/2 more
elegantly (root mean square velocity)
an equation for a chemical reaction that has the same number of
atoms on each side
series of "lines" in the hydrogen spectrum resulting from the transition
of an electron from a higher energy level to the level: n=2
equilibrium constant for the dissociation of a weak base, B-: B- (aq) +
H2O ↔ OH- (aq) + HB (aq)
Kb= [OH-]*[HB] / [B-]
basic ion
an anion that forms OH- ions in water; the CO32 ion is basic because
of the reaction: CO32- (aq) + H2O ↔ OH – (aq) + HCO3 –(aq)
basic solution
belt of stability
an aqueous solution with pH greater than 7 (at 25 degrees C)
region of graph of number of neutrons versus atomic number within
which stable isotopes fall
molecule containing 3 atoms in which the bond angle is less than 180
degrees: ( ex: H2O)
a ligand, such as ethylenediamine, that forms 2 covalent bonds with a
metal atom
model of the hydrogen atom derived by Niels Bohr, which predicts that
the electronic energy (kJ/mol) is -1312/n2, where n is the principle
quantum number
Temp. at which tge vapor pressure of a liquid equals the applied
pressure, leading to the formation of vapor bubbles; when the applied
pressure is 1 atm, we refer to the normal boiling point
bent molecule
bidentate ligand
Bohr model
boiling point (bp)
boiling point elevation (ΔTb)
increase in the boiling point caused by addition of a nonvolatile solute
bomb calorimeter
device used to measure heat flow in which a reaction is carried out
within a sealed metal container.
the angle between 2 covalent bonds; the H-O-H bond angle in the
H2O molecule is 105 degrees
the distance between nuclei of 2 atoms joined by chemical bond
enthalpy change ΔH associated with a reaction in which 1 mole of a
particular type of bond is broken
number of bonds in a species the bond order is 1 in F-F, 2 in O- -O
an orbital associated with 2 electrons whose energy is less than in the
separated atoms
common atomic mass scale in which the carbon-12 isotope is
assigned a mass of exactly 12 amu
radioactive method of determining the age of an organic abject by
measuring the amount of carbon-14 present
species that affects reaction rate without being consumed
device inserted into the exhaust of an automobile, containing finely
divided Pt; this metal catalyst converts CO to CO2 unburned
hydrocarbons to CO2
bond angle
bond distance
bond energy
bond order
bonding orbital
carbon-12 scale
carbon-14 dating
catalyst
catalytic converter
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cation exchange
cell voltage
chiral center
cellulose
Celsius degrees (C)
centi- (abbreviate "c")
Charles law
chelating agent
chromatography
cis isomer
cloud seeding
coefficient rule
coffee cup calorimeter
coke
colligative property
common ion effect
condensation reaction
conductivity
conjugate acid
conjugate base
conservation of energy
Process by which a cation in water solution is "exchanged" for a
different cation, originally present in a solid resin. Used to soften by
exchanging Ca for Na ions
voltage associated with an electrochemical cell; can be calculated
from standard potentials and the Nernst equation
atom in a molecule that s bonded to 4 different groups; a source of
optical isomerism
complex polymeric carbohydrate, formed of glucose units
unit of temp. based on there being 100 degrees between the freezing
and boiling point of water
prefix on a metric unit indicating a multiple of 10
Relation stating that at constant P and n, the vol. Of a gas is directly
proportional to its absolute temp.
complexing ligand that forms more than one bond with a central metal
atom:
separation method in which the components of a solution are
adsorbed at different locations on a solid surface
Geometric isomer in which 2 identical bonded groups are relatively
close to one another.
technique used to bring about precipitation from a cloud by adding a
substance like CO2
rule that states that when the coefficients of a chemical equation are
multiplied by a number n, the nth power.
calorimeter in which essentially all of the heat given off by a reaction
absorbed by a known amount of the water.
solid material, mostly carbon, formed by the destructive distillation of
coal
a physical property of a solution that primarily depends on the
concentration of solute particles rather than the kind. Vapor pressure
lowering is a colligative property, color is not.
decrease in solubility of and ionic solution containing a high
concentration of one of the ions found in the solute itself
reaction in which 2 species combine to form a product by splitting out
a small molecule such as H2O
the relative ease with which a sample transmits electricity or heat
(should specify which). Because a much larger electrical current will
flow through an aluminum rod at a given voltage than through a glass
rod of the same shape, aluminum is a better electrical conductor than
glass
the acid formed by adding an H+ ion to a base NH4 + is the conjugate
acid of NH3+
the base formed by removing an H+ ion from an acid. F- is the
conjugate base of HF
the law that states that energy can neither be created nor destroyed
conservation factor
a ratio, numerically equal to 1, by which a quantity can be converted
to another equivalent quantity. To convert 0.202g H2O to moles,
multiply by the conversion factor 1 mol/18.0 g
coordinate covalent bond
covalent bond in which both electrons are furnished by 1 atom;
characteristics of the bonding in coordination compounds
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corrosion
coordination compound
coulomb's law
coupled reactions
covalent bond
cubic centimeter
curie
Dalton's law
daughter nucleus
deBroglie relation
a destructive chemical process; most often applied to the oxidation of
a metal ( the rusting of iron)
Compound in which either the cation or the anion is a complex ion.
(ex.: [Cu(NH3)4]Cl2
relation that states that the attractive energy between particles of
opposite charge is directly proportional to the charge product and
inversely proportional to the distance between them
2 reactions that add to give a third. By coupling a spontaneous
reaction to 1 that is nonspontaneous, it may be possible to obtain an
overall reaction that is spontaneous
a chemical link between 2 atoms, produced by sharing electrons in
the region between the atoms.
a volume unit equal to the volume of a cube 1 cm on an edge; a
milliliter
radiation corresponding to the decay of 3.700 * 10 to the tenth power
atoms per second of a radioactive species
a relation stating that the total pressure of a gas mixture is the sum of
the partial pressure of its components
nucleus produced by radioactive decomposition.
an equation used to describe the wave properties of matter Λ = h/mv
deliquescence
a process in which a soluble solid picks up water vapor from the air to
form a solution. For deliquescence to occur, the vapor pressure of
water in the air must be greater than that of saturated solution
delocalized orbital
molecule orbital in which the electron density is spread over the
entire molecule instead of being concentrated between 2 atoms
process in which small solute particles as well as solvent molecules
pass through a semipermeable membrane
a term indicating that a substance does not contain unpaired electrons
and so is not attracted into a magnetic field
a process by which 1 substance, by virtue of the kinetic properties of
its particles, gradually mixes with another
a species formed when 2 monomer units combine
species in which there is a separation of charge, I.e., a positive
charge at one point and a negative charge at a different point
an attractive force between polar molecules
an acid such as H2SO4 or H2CO3 that contains 2 ionizable H atoms
dialysis
diamagnetic
diffusion
dimer
dipole
dipole force
diprotic acid
dispersion force
disproportionation
distillation
double bod
dry cell
ductlity
dynamite
an attractive force between molecules that arises because of the
presence of temporary dipoles
reaction in which a species undergoes oxidation and reduction
simultaneously
procedure in which a liquid is vaporized and the vapors condensed
and collected
2 shared electron pairs between 2 bonded atoms
commercial voltaic cell that uses the reaction of Zn with MnO2
the ability of a solid to retain strength on being forced through an
opening; characteristics of metals
explosive made by adding nitroglycerine to a solid absorbent
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effective nuclear charge
effusion
Einstein equation
elastic collision
electrical neutrality
electrode
electrolysis
electron cloud
electron pair acceptor
electron pair donor
electron pair repulsion
electron sea model
elecrton spin
electronegativity
element
elimination reaction
empirical formula
enantiomer
end point
endothermic
energy(E)
enthalpy(H)
enthalpy change(Δ H)
entropy(S)
entropy change( Δ S)
equilibrium
positive charge felt by the outermost electrons in an atom; approx.
equal to the number of nuclear protons minus the number of electrons
in inner, complete levels
movement of gas molecules thruougha pinhole or capillary
the relation deltaE = delta mc squared; relating mass and energy
2
changes. ΔE=mC
collision in which the total kinetic energy of the colliding particles
remains constant
principle that, in any compound, the total positive charge must equal
the total negative charge
general name for an anode or cathode
passage of a direct electric current through a liquid containing ions,
producing chemical changes at the electrodes
region of negative charge around a nucleus, associated with an
atomic orbital
Lewis acid
Lewis base
principle used to predict the geometry of a molecule or polyatomic ion.
Electron pairs around a central atom tend to orient themselves to be
as far apart as possible
model of metallic bonding in which cations are as fixed points in a
mobile "sea" of electrons
property of an electron, loosely related to its spin about an axis;
described by the quantum number Ms, can be pos. or neg. 1/2
Property of an atom that increases with its tendency to attract the
electrons in a bond. Because Cl is more electronegative than H, the
bonding electrons in HCl is closer to CL than to H
substance whose atoms are all chemically the same, containing a
definite number of protons
reaction in which a small molecule like H2O or HCl is eliminated from
a reactant, forming a multiple bond
formula that gives the simplest whole- number atom ratio; often called
the simplest formula. H2O is the simplest formula of water
1 of a pair of optical isomers
point during a titration at which an indicator changes color
process in which heat is absorbed from the surroundings; delta H is
positive for an endothermic reaction
Property of a system related to its capacity to cause change in the
surroundings; can be altered only be exchanging heat or work with the
surroundings.
Property of a system that reflects its capacity to exchange heat, q,
with the surroundings. Defined so that Δ H = q for a constant
pressure process.
difference in enthalpy between products and reactants
property of a system related to its degree of organization; highly
ordered systems have low entropys
difference in entropy between products and reactants
a state of dynamic balance in which rates of forward and reverse
reactions are equal; system does not change with time
289
Evenson
equilibrium concentration
equilibrium constant (Kc)
concentration, in mols per liter, of a species at equilibrium;
represented by a symbol [ ]
a number characteristic of an equilibrium system at a particular temp.
equilibrium constant (Kp)
a quantity similar to Kc, except that partial pressure replaces molarity
equilibrium expression
equivalence point
mathematical formula for the equilibrium constant: Kc = [ ] / [ ]
The point during a titration between A and B when an amount of B
has been added that reacts with the A present. The equivalence point
in the titration of a strong acid with a strong base occurs when equal
numbers of the mols of OH- and H+ have been added
excited state
exclusion principle
an electronic state that is of a higher energy of the ground state
the rule stating that in an atom no 2 electrons can have the identical
set of 4 quantum numbers
process in which heat is evolved to the surrounding; delta H is neg. for
an exothermic reaction
more than 4 electron pairs about a central atom
a degree based on the temp. scale on which water freezes at 32
degrees and boils at 1 atm at 212 degrees
a group in the periodic table, such as the halogens (group 7)
solution that passes through a filter
process for separating a solid-liquid mixture by passing it through a
barrier with fine pours, such as filter paper
the energy that must be absorbed to remove the outermost electron
from a gaseous species, forming a +1 ion
empirical rule that the approximation a-x ~ a is valid if x is less than or
equal to 0.05a
a test carried out by observing a color imparted to a Bunsen flame by
a sample
separation process used to free finely divided ore from rocky
impurities
chart or sheet used to summarize the steps in a procedure, as in
qualitative analysis
the equilibrium constant for the formation of a complex ion
expression used to describe the relative numbers of atoms of different
elements present in a substance
sum of the atomic masses of the atoms in a formula
coal, petroleum, and natural gas
Process used to separate a pure solid from a solid mixture. The
mixture is dissolved in the minimum amount of hot solvent. Upon
cooling, the major component crystallizes while impurities should stay
in solution.
exothermic
expanded octet
Fahrenheit degrees (F)
family
filtrate
filtration
first ionization energy
five percent rule
flame test
flotation
flow chart (flow sheet)
formation constant (Kf)
formula
formula mass
fossil fuels
fractional crystallization
free energy
free energy change
free energy of formation
free radical
freezing point
a quantity, defined as H-TS, which decreases in a reaction that is
spontaneous at constant T and P.
the difference in free energy between products and reactants
delta G for the formation of a species from the elements
species having an unpaired electron, such as the H+ atom
the temperature at which a solid and liquid phase can coexist at
equilibrium
290
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freezing point depression
gas constant
genetic code
geometric isomerism
Gibbs-Helmholtz equation
Graham's Law
greenhouse effect
ground state
Haber process
half cell
half-equation
half-life
Hall process
Halogen
Heat
heat flow
heat of formation
heat of fusion
heat of sublimation
heat of vaporization
Henderson-Hasselbach equation
Henry's Law
Hertz(Hz)
Hess's law
heterogeneous
heterogeneous catalysis
heterogeneous equilibrium
Hybrid orbital
hybridization
hydrate
the decrease in the freezing point of a liquid caused by addition of a
solute
L atm
constant that appears in the Ideal Gas Law; R=0.0821
/mol K
sequence of three letter series that code the synthesis of proteins in
animals
type of isomerism that arises when two species have the same
molecular formulas but different geometric structures
the relationship delta G = delta H - T delta S
relation stating that the rate of effusion of a gas is inversely
proportional to the square root of it’s mass
the effect of water, carbon dioxide, and certain other gases in
absorbing IR radiation, thereby raising the earth's temperature
the lowest allowed energy state of an atom, ion, or molecule
an industrial process used to make ammonia from hydrogen and
nitrogen
half of a voltaic or electrolytic cell at which oxidation or reduction
occurs
an equation written to describe a half-reaction of oxidation or
reduction
the time required for a reaction to convert half of the initial reactant to
products
industrial process for electrolytic preparation of Al
and element in Group 7: F, Cl, Br, I
form of energy that flows between two samples because of their
difference in temperature
the amount of heat, q. passing into or out of a system; q is a positive if
flow is into system, negative if out of system
see enthalpy of formation
delta H for the conversion of unit amount(one gram or one mole) of a
solid to a liquid at constant P and T
delta H for the conversion of unit amount(one gram or one mole) of a
solid to a vapor at constant P and T
delta H for the conversion of unit amount(one gram or one mole) of a
liquid to a vapor at constant P and T
equation for calculating the pH of a buffer: pH = pKa + log10 [B-]/[HB]
Relation stating that the solubility of a gas in a liquid is directly
proportional to it's partial pressure
a unit of frequency: I cycle per second
relation stating that the heat flow in a reaction that is the sum of two
other reactions is equal to the sum of the heat flows in those two
reactions
having non-uniform composition
catalysis that occurs upon a solid surface
equilibrium system involving a solid or liquid as well as gases
an orbital made from a mixture of s, p, d, or f orbitals. An sp2 hybrid
orbital is made by mixing an s orbital with two p orbitals
mixing of two or more orbitals or structures
a compound containing bound water such as BaCl2
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Evenson
hydride
hydrogen bond
compound of hydrogen, specifically one containing H- ions
attractive force between molecules arising from interaction between a
hydrogen atom in one molecule and a strongly electronegative atom
hydronium ion
indicator, acid-base
the H3O ion characteristic of acidic water solutions
a chemical substance, usually a weak acid, that changes color with
pH; the weak acid and it's conjugate base have different colors
light having a wavelength greater than about 700nm
rate at the beginning of a reaction, before reactant concentration have
decreased appreciably
rat of reaction at a particular point in time
force between adjacent molecules(dipole,dispersion,hydrogen bond)
force between atoms within a molecule(covalent bond)
charged species
the product of the actual concentrations of cation and anion in a
solution, each raised to the appropriate power. If Q, upon mixing is
greater than Ksp, a precipitate forms
infrared
initial rate
instantaneous rate
intermolecular force
intramolecular force
ion
ion product
ionic bond
ionic compound
ionic radius
ionization energy
isoelectric point
isomer
isotope
Joule
Kelvin temperature scale
Kilo
lattice energy
law of combing volumes
law of constant composition
lead storage battery
Le Chatleir's Principle
limiting reactant
the electrostatic force of attraction between oppositely charged ions in
an ionic compound
compound made up of cations and anions
the radius of an ion, based on splitting up the interionic distance in a
somewhat arbitrary way
the energy that must be absorbed to remove an electron from a
species
pH at which the zwitterion of an amino acid has its highest
concentration
species with the same formulas as another species, but having
different properties. Structural, geometric, and optical isomers are
possible
an atom having the same number of nuclear protons as another, but
with a different number of neutrons
base SI unit of energy; equal to kinetic energy of a two-kilogram mass
moving at a speed of one meter per second
scale obtained by taking the lowest attainable temperature to be 0K;
size of degree is same as with Celsius scale
metric prefix indicating multiple of 1000
delta H for the process in which oppositely charged ions in the gas
phase combine to form an ionic, solid lattice
relation stating that relative volumes of different gases(at same T and
P) involved in reaction are in the same ratio as their coefficients I the
balanced equation
relation stating that the relative masses of the elements in a given
compound are fixed
commercial voltaic cell that uses the reaction between Pb and PbO 2,
in sulfuric acid solution to produce electrical energy
relation stating that, when a system at equilibrium is disturbed, it
responds in such a way as to counteract the change
the least abundant reactant, based on the equation, in a chemical
reaction; dictates the maximum of product that can be formed
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o
linear molecule
triatomic molecule in which the bond angle is 180 ; examples include
BeF2 and CO2
luster
Lyman series
characteristic shiny appearance of a metal surface
series of "lines" in the atomic spectrum of hydrogen resulting from
electron transitions from excited states to the ground state
capable of being shaped, as by pounding with a hammer
extensive property reflecting the amount of matter in a sample
difference between the mass of a nucleus and the sum of masses of
the neutrons and protons of which it is composed
integer equal to the sum of the number of protons and neutrons in an
atomic nucleus. The isotope of C1 that contains 17 protons and 18
neutrons has a mass number of 35
100 times the ration of the mass of a component to the total mass of a
sample
instrument used to determine the charge-to-mass ratio of a cation
relation describing the way in which molecular speeds or energies are
shared among the molecules in a gas
average distance traveled by an atom or molecule between collisions
malleable
mass
mass defect
mass number
mass percent
mass spectrometer
Maxwell distribution
mean free path
mechanism
metal
metallic character
metalloid
meter
millimixture
Molality
molar mass
molarity
mole
mole fraction
molecular formula
molecular geometry
molecular orbital
molecular substance
sequence of steps that occurs during the course of a reaction
substance having characteristic luster, malleability, and high electrical
conductivity; readily loses electrons to form cations
the extent to which a substance has the characteristics of a metal
an element such as Si that has properties intermediate between those
associated with metals from their ores
unit of length in the metric system
prefix on a metric unit indicating a multiple of 10-3
two or more substances combined so that each substance retains its
chemical identity
a concentration unit defined to be the number of moles of solute per
kilogram of solvent
mass of one mole of a substance
concentration unit defined to be the number of moles of solute per liter
of solution
23
Collection of 6.022 x 10 items. The mass in grams of one mole of a
substance is numerically equal to its formula mass. For example, one
mole of H2O weighs 18.02 g
concentration unit defined as the number of moles of a component
divided by the total number of moles of a sample
Formula in which the number of atoms of each type in a molecule is
indicated as a subscript after the symbol of the atom. The molecular
formula of hydrogen peroxide is H2O2
the shape of a molecule, describing the relative positions of atomic
nuclei
orbital involved in the chemical bond between two atoms, and taken to
be a linear combination of the orbitals on the two bonded atoms
Substance built up of discrete molecules. Hydrogen gas(H2) and
benzene(C6H6) are molecular substances
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Evenson
molecule
monatomic ion
monodentate ligand
mulitple bond
mulitple equilibria, rule of
aggregate of a few atoms that is the fundamental building block in all
gases and many solids and liquids; held together by covalent bonds
between the atoms
Ion formed from a single atom. Examples: Na+, Clligand that forms only one coordinate covalent bond with the central
metal in a complex ion
double or triple bond
rule stating that if Equation 1 + Equation 2 = Equation 3, then K1 x K2
= K3
n + l rule
rule that orbitals fill in order of increasing n + l.(For example, 4s fills
before 3d). For two orbitals with the same value n + l, the one with the
lower n value fills first(3p before 4s)
nano
natural logarithm
prefix on a metric unit indicating a multiple of 10-9
logarithm taken to the base e, where e = 2.7182818…; if logeX = Y,
Y
then e = X; log10X=logeX/2.303
Nernst equation
an equation relating cell voltage E to the standard voltage E0tot and
the concentrations of reactants and products: E=E0tot - 0.0257/n
net ionic equation
network covalent
a chemical equation for a reaction in which only those species that
actually react are included.
having a structure in which all the atoms in a crystal are linked by a
network of covalent bonds. Carbon and silicon dioxide have structures
of this type
neutral ion
ion that has no effect on the pH of water; it is neither acidic nor basic
neutral solution
neutralization
a water solution with pH = 7
reaction between a strong acid and a strong base to produce a neutral
solution
one of the particles in an atomic nucleus; mass = 1, charge = 0
an element in Group 8 at the far right of the Periodic Table
the ns2np6 outer electron structure in an atom or ion; a particularly
stable structure in atoms, ions, and molecules
one of the elements in the upper right corner of the Periodic Table that
does not show metallic properties; N, O, and Cl are nonmetals
a chemical bond in which the electrons are equally shared by two
atoms, so there are no positive and negative ends
molecule in which there is no separation of charge and hence no
negative and positive poles
compound in which the atom ratio is not exactly integral
the boiling point at 1 atm pressure
Symbol giving the atomic number and mass number of a nucleus.
Example 146C
neutron
Noble gas
Noble gas structure
nonmetal
nonpolar bond
nonpolar molecule
Non-stoichiometric compound
normal boiling point
nuclear symbol
nuclear winter
nucleon
nucleus
hypothesized cooling effect caused by dust particles in the
atmosphere generated by nuclear winter
proton or neutron
the small, dense, positively charged region at the center of the atom
294
Evenson
octahedral
octet rule
opposed spins
having the symmetry of a regular octahedron. In an octahedral
species, a central atom is surrounded by six other atoms one above,
one below, and four at the corners of the square
the principle that bonded atoms(except H) tend to have a share in
eight outermost electrons
a term that refers to electrons with different values of m2; two
electrons in a single orbital must have opposed spins
optical isomerism
Phenomenon in which two species, each with the same molecular
formula, rotate a beam of plane polarized light in opposite directions.
Such molecules have at least on chiral center
orbital
an electron cloud with an energy state characterized by given values
of n, l, and m1; has a capacity for two electrons of opposed spins
orbital diagram
a sketch showing electron populations of atomic orbitals, including
electron spins
an exponent to which the concentration of a reactant must be raised
to give the observed dependence of rate upon concentration; most
commonly 0, 1, or 2, but can be fractional
natural mineral deposit from which a metal can be extracted profitably
order of reaction
ore
outer electron configuration
Statement of the population of sublevels in outer principal levels. The
outer electron configuration of the I atom is 5s25p5
oxide
compound of oxygen; more specifically, a species containing the 02ion
oxidizing agent
ozone layer
a species that accepts electrons in a RedOx reaction
a region at about 30km where ozone, 03, has its maximum
concentration of about 10ppm
paired electrons
parallel spins
two electrons in the same orbital with opposed spins
a term that refers to electrons with the same m s value. Single
electrons in different orbitals of the same energy have parallel spins
paramagnetic
Having magnetic properties caused by unpaired electrons. The NO
and O2 molecules are paramagnetic
parent nucleus
partial ionic character
nucleus undergoing radioactive decay to form a "daughter" nucleus
Extent of polarity in a covalent bond. A bond with 50% ionic character
is midway between a pure ionic bond and a nonpolar covalent bond
partial pressure
the part of the total pressure in a gas mixture that can be attributed to
a particular component. The partial pressure of A is the pressure A
would exert if it were there by itself
parts per billion(ppb)
for gases, the number of moles of solute per billion moles of gas. For
liquids and solids, the number of grams of solute per billion grams of
sample
parts per million(ppm)
for gases, the number of moles of solute per million moles of gas. For
liquids and solids, the number of grams of solute per million grams of
sample
unit of pressure: 1.013 x 105 Pa = 1 atm
percentages by mass of the elements in a compound
Pascal
percent composition
295
Evenson
+
percent dissociation
for a weak acid HB, percent dissociation = 100 x [H ]/orginal
concentration HB
percent yield
period
Periodic Table
a quantity equal to 100 x actual yield/theoretical yield
a horizontal row of elements in the Periodic Table
an arrangement of the elements in rows and columns such that
elements with similar chemical properties fall in the same column
a binary compound containing an O-O bond; more specifically, a
2compound containing the peroxide ion, O2
peroxide
pH
defined as -log10[H+}]
pH meter
device for measuring pH; depends on fact that cell voltage is a
function of pH
for one-component systems, a graph of pressure vs. temperature,
showing the conditions for equilibrium between solid, liquid, and gas
phases
smoky fog produced by light-induced reactions in polluted air
an individual quantum of radiant energy
property of a substance related to its physical characteristics
a bond in which electrons are concentrated in orbitals located off the
internuclear axis; one bond in a double bond is a pi bond, and there
are two pi bonds in a triple bond
impure iron produced in a blast furnace
defined as -log10Ka, where Ka is the dissociation constant of a weak
acid
phase diagram
photochemical smog
photon
physical property
pi bond
pig iron
pKa
Planck's constant(h)
pOH
the constant in the equation E = hv =hc/λ.h = 6.626 x 10-34 J.s
defined as pOH = -log10[OH-]
polar bond
a chemical bond that has positive and negative ends; characteristic of
all bonds between unlike atoms
a molecule in which there is a separation of charge and hence a
positive and negative pole
a distortion of the electron distribution in a molecule, tending to
produce positive and negative poles
a charged species containing more than one atom
huge molecule made up of many small units linked together
chemically
a particular set of concentrations that satisfy the equilibrium constant
expression
lower members of Periodic Table Groups 3,4, and 5. Lead and
bismuth are post-transition metals
a solid that forms when two solutions are mixed
formation of an insoluble solid when two solutions are mixed
force per unit area
Non-rechargeable voltaic cell
a major species, present in relatively high concentration, formed by a
compound in water. In a water solution of sodium fluoride, the
principal species are Na+ and F-; in a water solution of hydrogen
fluoride, the principal species is HF
polar molecule
polarization
polyatomic ion
polymer
position of equilibrium
post-transition metal
precipitate
precipitation reaction
pressure
primary cell
principal species
product
a substance formed as a result of a reaction; appears on the right side
of the equation
296
Evenson
+
proton
the nucleus of a hydrogen atom, the H ion; a component of atomic
nuclei with mass = 1 charge = +1
p-type semiconductor
a semiconductor in which current is carried through a solid by electron
flow into "positive holes," caused by electron deficiencies
equation containing a second order term; an equation in x containing
2
an x term
quadratic equation
quadratic formula
the formula used to obtain the two roots of the general quadratic
equation: ax2 + bx + c = 0. The formula is x = -b (+ -) square root of
2
b - 4ac / 2a
qualitative analysis
the determination of the nature of the species present in a sample;
most often applied to cations or anions
the determination of how much of a given component is present in a
sample
approach used to calculate the energies and spatial distributions of
small particles confined to very small regions of space
a number used to describe the energy levels available to electrons in
atoms; there are four such numbers
a general theory that describes the allowed energies if atoms and
molecules
unit of absorbed radiation equal to 10-2 J absorbed per kilogram of
tissue
quantitative analysis
quantum mechanics
quantum number
quantum theory
rad
radioactivity
Raoult's Law
rate constant
rate-determining step
reactant
reaction rate
real gas
reciprocal rule
RedOx reaction
reducing agent
reduction
relative humidity
the ability possessed by some natural and synthetic isotopes to
undergo nuclear transformation to other isotopes
relation between the vapor pressure(P) of a component of a solution
and that of the pure component(Po) at the same temperature: P =
XPo, where X is the mole fraction
the proportionality constant in the rate equation for a reaction
the slowest step in a muli-tstep mechanism
the starting material in a reaction; appears on the left side of the
equation
the magnitude of the change in concentration of a reactant or a
product divided by the time required for the change to occur
gas that deviates measurably from the Ideal Gas Law; in practice all
gases are at least slightly non-ideal
the relation between equilibrium constants for forward (Kf) and reverse
(Kr) reactions: Kr = 1/Kf
a reaction involving oxidation and reduction
a species that furnishes electrons to another in a redox reaction
a half-reaction in which a species gains electrons or, more generally,
decreases in oxidation number
100 x P/Po, where P = pressure of water vapor in air, Po = equilibrium
vapor pressure of water at same temperature
resonance
model used to rationalize properties of octet rule species for which a
single Lewis structure is inadequate; resonance forms differ from one
another only in the distribution of electrons
reverse osmosis
process by which pure water is obtained from salt solution by applying
a pressure greater than the osmotic pressure to the solution
297
Evenson
Schrodinger equation
second order reaction
secondary cell
selective precipitation
semiconductor
wave equation that relates mass, potential energy, kinetic energy, and
coordinates of a particle
reaction whose rate depends upon the second power of reactant
concentration
rechargeable voltaic cell
precipitation of one ion in the presence of another, which also forms
an insoluble compound with the reagent
substance used in transistors, thermisters, etc., whose electrical
conductivity depends on the presence of tiny amounts of impurities in
a crystal
semipermeable membrane
a film that allows passage of solvent but not solute molecules or ions
shielding
term used to describe effect of inner electrons in decreasing the
attraction of an atomic nucleus on outermost electrons
a unit associated with the International System of Units
chemical bond in which electron density on the internuclear axis is
high, which is the case with all single bonds. Double bonds consist of
one sigma and one pi bond; triple bonds consist of one sigma and two
pi bonds
SI unit
sigma bond
significant figure
meaningful digit in a measured quantity; number of digits in a quantity
expressed in exponential notation is the number of significant figures
single bond
skeleton structure
solar energy
solubility
a pair of electrons shared between two bonded atoms
structure of a species in which only the sigma bonds are shown
direct generation of energy(heat, electrical energy) from sunlight
the amount of a solute that dissolves in a given amount of solvent at a
specified temperature
the equilibrium constant for the solution reaction of a slightly soluble
ionic compound. For the reaction: Ca(OH)2(s) → Ca2+(aq) + 2 OH(aq), Ksp = [Ca2+] x [OH-]2
solubility product constant(Ksp)
solubility rules
the rules used to classify ionic compounds as to their water solubility
solute
solution
the solution component present in smaller amount than the solvent
a phase(liquid, gas, or solid) containing two or more components
dispersed uniformly through the phase
voltage associated with an oxidation half-reaction, when all gases are
at 1 atm, all aqueous solutes at 1 M
identical with the standard reduction voltage
the voltage associated with a reduction half-reaction when all gases
are at 1 atm and all aqueous solutes are at 1 M
condition of a system when its properties are fixed; must specify
temperature, pressure, and chemical composition
a property of a system that is fixed when its temperature, pressure,
and composition are specified
the relationships between the masses(grams, moles) of reactants and
products in a chemical reaction
rechargeable voltaic cell, such as a lead storage battery
standard temperature and pressure(0oC, 1 atm for a gas)
standard oxidation voltage(E0ox)
standard potential
0
standard reduction voltage(E red)
state
state property
Stoichiometry
storage cell
STP
298
Evenson
+
strong acid
species that is completely dissociated to H ions in dilute water
solution
strong base
species that is completely dissociated to OH- ions in dilute water
solution
structural formula
structural isomers
formula showing the arrangement of atoms in a molecule
two or mores species having the same molecular formula but different
structural formulas. Example: C2H5OH and CH3 - O -CH3
solvent
a substance, usually a liquid, in which another substance, called the
solute, is dissolved
the amount of heat required to raise the temperature of one gram of a
substance one degree Celsius
an ion that, though present, takes no part in a reaction
in qualitative analysis, a test that can be used to identify an ion
without separating it from other ions
voltage of a cell in which all species are in their standard states(solids
and liquids are pure, solutes are at unit activity, taken to be 1 M, and
gases are at 1 atm)
a subdivision of an energy level designated by the quantum number 1
specific heat(S.H.)
spectator ion
spot test
0
tot)
standard cell voltage(E
sublevel
sublimation
substitution reaction
change in state from solid to gas
reaction in which one atom or group(e.g., -Cl, -NO2) takes the place of
another(e.g., -H_
successive approximation
technique used to solve quadratic or higher order equations. On the
basis of a reasonable assumption or an educated guess, a first,
approximate answer is obtained; that answer is used with the original
equation, to obtain a more nearly exact solution
superoxide
compound containing the O2- ion
supersaturated
containing more solute than allowed by equilibrium considerations;
unstable to addition of solute
everything outside the system being studied
a one-or two-letter abbreviation for the name of an element
the sample of matter under consideration
atom at one end of a molecule, in contrast to the central atom. The
most common terminal atoms are H, O, and the halogens
the amount of product obtained from the complete conversion of the
limiting reactant
ability to conduct heat
the exothermic reaction of Al with a metal oxide; the products are
Al2O3 and the free metal
surroundings
symbol
system
terminal atom
theoretical yield
thermal conductivity
thermite reaction
thermochemical equation
thermodynamics
third law of thermodynamics
titrant
titration
Titration curve
chemical equation in which the value of ΔH is specified
the study of heat, work, and the related properties of mechanical and
chemical systems
natural law that states that the entropy of a perfectly ordered, pure,
crystalline solid is 0 at 0 K
reagent added from a burette or similar measuring device in a titration
a process in which one reagent is added to another with which it
reacts; an indicator is used to determine the point at which equivalent
quantities of the two reagents have been added
Plot of some measurable variable, often pH, vs. volume of titrant
299
Evenson
Torr
Trans isomer
Translational energy
Triple bond
Triple point
Triprotic acid
Ultraviolet radiation
Unit cell
Unpaired electron
Useful work
Valence bond model
Valence electron
o
A unit of pressure equal to 1 mm Hg at 0 C
A geometric isomer in which two identical groups are as far apart as
possible.
The energy of motion through space. A falling raindrop has
translational energy
Three electron pairs shared between two bonded atoms
The temperature and pressure at which the solid, liquid, and vapor of
a pure substance can coexist in equilibrium
Acid, such as H3PO4, which has three ionizable H atoms per molecule
Light having a wavelength less than about 400nm but greater than
about 10nm
The smallest unit of a crystal that, if repeated indefinitely, could
generate the whole crystal
A single electron occupying an orbital by itself
Any work other than expansion work associated with a process
The theory that atoms tend to become bonded by pairing and sharing
their outer (valence) electrons
Electron in the outermost shell. Carbon with the electron configuration
1s22s22p2 has four valence electrons
van der Waals equation
An equation used to express the physical behavior of a real gas: (P +
a/V2 (V - b) = RT, where V = molar volume, a and b are constants
characteristic of a particular gas
Vapor pressure
The pressure exerted by a vapor when it is in equilibrium with a liquid
Vapor pressure lowering
The decrease in the vapor pressure of a liquid caused by addition of a
nonvolatile solute
Easily vaporized
A unit of electric potential: 1 V = 1 J/C
Device in which a spontaneous RedOx reaction produces electrical
energy
Quantitative chemical analysis in which the volume of a reagent is
measured
Valence Shell Electron Pair Repulsion model, used to predict
molecular geometry; states that electron pairs around a central atom
tend to be as far apart as possible
Volatile
Volt(V)
Voltaic cell
Volumetric analysis
VESPR model
Water softening
Wavelength
Weak acid
Weak base
Weak electrolyte
Work(w)
Yield
Zeolite
Zero-order reaction
Zwitterion
The removal of ions, particularly Ca2+ and Mg2+, from water
A characteristic property of light related to its color and equal to the
length of a full wave
+
Acid that is only partially dissociated to H ions in water solution
Base that is only partially dissociated to OH- ions in water solution
A species that, in water solution, forms an equilibrium mixture of
molecules and ions; includes weak acids and weak bases
Any form of energy except heat exchanged between system and
surroundings. Includes mechanical work, expansion work, electrical
work, and so on
The amount of product obtained from a reaction
A type of silicate mineral used to soften water by cation exchange
A reaction whose rate is independent of reactant concentration
Dipolar, electrically neutral form of an amino acid molecule
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Evenson
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rather to give you extra time or a second
chance (redo). Any other late assignments
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You are allowed one late slip per quarter;
this is not en lieu of an assignment but
rather to give you extra time or a second
chance (redo). Any other late assignments
will not be graded for credit
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301
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302
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Common polyatomic ions
Ammonium
Bicarbonate
Carbonate
Chromate
Chlorate
Chlorite
Cyanide
Dichromate
Hydroxide
Nitrate
Nitrite
Permanganate
Phosphate
Silicate
Sulphate
Sulphite
Thiocyanate
Thiosulphate
NH4+
HCO3CO3-2
CrO4-2
ClO3ClOCNCr2O7-2
OHNO3NO2MnO4PO4-3
SiO3-2
SO4-2
SO3-2
SCNS2O3-2
Common Aqueous Strong Acids
HCl
Hydrochloric acid
HI
Hydroiodic acid
HBr
Hydrobromic acid
H2SO4
Sulfuric acid
HNO3
Nitric Acid
HClO4
Perchloric Acid
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