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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 2 Evenson 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. 3 Evenson 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. 4 Evenson 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 5 Evenson 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 6 Evenson 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. 7 Evenson 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. 8 Evenson 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. 9 Evenson 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 10 Evenson 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. 11 Evenson Quick Tips for Boosting Grades Rewrite all lecture notes 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 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. 12 Evenson 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. 13 Evenson 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 14 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) 15 Evenson 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. 16 Evenson 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) 17 Evenson 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. 18 Evenson 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 19 Evenson 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. 21 Evenson 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. 22 Evenson 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? 23 Evenson 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. 24 Evenson 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. 25 Evenson Question 1. Develop a stepwise (1., 2., 3.,…) procedure for determining the density (g/ml) of 100.0 ml of water. 26 Evenson 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? 27 Evenson 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: 28 Evenson 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: 29 Evenson % 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 30 Evenson 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. 31 Evenson 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. 32 Evenson 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. 33 Evenson 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. 34 Evenson 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. 35 Evenson 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 41 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? 42 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? 43 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? 44 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? 45 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? 46 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. 47 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. 53 Evenson 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) 54 Evenson 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? 55 Evenson 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) 56 Evenson Sample problems Element H Sn Proton (+) 1 50 Electron (-) 1 50 Neutron 0 69 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, 57 Evenson 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 58 Evenson 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 59 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 60 Evenson 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 61 Evenson 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)? 62 Evenson 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 63 Evenson 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? 64 Evenson 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. 65 Evenson 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. 66 Evenson 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? 67 Evenson 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? 68 Evenson 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 69 Evenson 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 70 Evenson 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 71 Reason / Why Evenson 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. 72 Evenson 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+BC 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. 73 Evenson 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. 74 Evenson 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) 75 Evenson 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) 76 Evenson 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) 77 alka-seltzer Evenson 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? 78 Evenson 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? 79 Evenson 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. 80 Evenson 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 81 Evenson 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. 82 Evenson 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. 83 Evenson 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. 84 Evenson 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 85 SO4-2 Evenson 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. 86 Evenson 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 87 Evenson 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) 88 Evenson 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. 89 Evenson 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 90 Evenson 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 91 Evenson 92 Evenson 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 93 Evenson 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? 94 Evenson 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? 95 Evenson 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 % 96 Evenson 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. 97 Evenson 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? 98 Evenson 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. 99 Evenson 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 100 Evenson 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. 101 Evenson 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. 102 Evenson 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. 103 Evenson % 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 104 Evenson 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 105 Evenson 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 106 Evenson 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. 107 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 108 Evenson 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 109 Evenson 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? 110 Evenson 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? 111 Evenson 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? 112 Evenson 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? 113 Evenson 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 2 114 Evenson 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. 115 Evenson 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 116 Evenson 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 117 Evenson 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. 118 Evenson 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 119 Evenson 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? 120 Evenson Unit 6 Objectives for Kinetic Theory of Matter Kinetic Energy relationship to temperature Kinetic theory Solids Liquids Gasses Plasmas Grahm's Law Calculations Application 121 Evenson 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. 122 Evenson 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 123 Evenson 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. 124 Evenson 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. 125 Evenson 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? 126 Evenson 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 127 Evenson 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. 128 Evenson 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. 129 Evenson 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 130 Evenson 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. 131 Evenson 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? 132 Evenson 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? 133 Evenson 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? 134 Evenson 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? 135 Evenson 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. 136 Evenson 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? 137 Evenson 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 138 Evenson 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. 139 Evenson 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. 140 Evenson 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 141 Evenson 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? 142 Evenson 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? 143 Evenson 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? 144 Evenson 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: 145 Evenson 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 146 Evenson 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? 147 Evenson 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. 148 Evenson 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. 149 Evenson 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. 150 Evenson 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) 151 Evenson 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) 152 Evenson 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 153 Evenson 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) 154 Evenson 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? 155 Evenson 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. 156 Evenson 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 157 Evenson 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. 158 Evenson 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) 159 Evenson 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. 160 Evenson 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) 161 Evenson 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. 162 Evenson 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.” 163 Evenson 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. 164 Evenson 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. 165 Evenson 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. 166 Evenson 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 167 Evenson 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) 168 Evenson 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 10-6 169 Keq = 3.89x10-9 Evenson 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 170 Evenson 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. 171 Evenson 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 172 Evenson 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. 173 Evenson 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). 174 Evenson 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 = lc 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 = lc) 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. 175 Evenson 176 Evenson 177 Evenson 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) 178 1.3 1.5 Evenson 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) 179 9 Evenson 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? 180 Evenson 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. 181 Evenson 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) 182 440 460 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? 183 Evenson 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. 184 Evenson 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 185 Evenson 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? 186 Evenson 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 187 Evenson 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 188 Evenson 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: 189 Evenson 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 190 Evenson 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. 191 Evenson 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 192 Evenson 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 30 193 Evenson 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 194 + 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. 195 Evenson 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 196 Evenson 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 -8 197 Evenson 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 198 Evenson 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? 199 Evenson 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) 200 Evenson 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. 201 Evenson 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 202 Evenson 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? 203 Evenson 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? 204 Evenson 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? 205 Evenson % 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? 206 Evenson 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- 207 Evenson 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. 208 Evenson 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 209 Evenson 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 210 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. 211 Evenson 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 212 Evenson 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. 213 Evenson 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. 214 Evenson 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. 215 Evenson 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) 216 Evenson 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? 217 Evenson 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 218 Evenson 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 219 Evenson 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? 220 Evenson 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 221 Evenson 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 222 Evenson 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 223 Evenson 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 224 Evenson 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 225 Evenson 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 - TS 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 226 Evenson 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? 227 Evenson 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 228 Evenson 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) 229 Evenson 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) 230 Evenson 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- 231 Evenson 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 232 Evenson 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 233 Evenson 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) 234 Evenson 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. 235 Evenson 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. 236 Evenson 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) 237 Evenson 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) 238 Evenson 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)? 239 Evenson 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 240 Evenson o 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 241 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. 242 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) 243 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) 0.96 244 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) - - 245 Evenson 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) 246 Evenson 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? 247 Evenson 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. 532-536 248 Evenson 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. 249 Evenson 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. 250 Evenson 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 251 Evenson Unit 14 Objectives for Organic Chemistry Nomenclature Isomers Geometric Structural Enatiomeric Biological effects Extractions techniques Polarity Pharmaceutical Ethnobotanical 252 Evenson 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 253 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 254 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 255 Evenson 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. 256 Evenson 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 257 Evenson 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 258 Evenson 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? 259 Evenson 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 260 Evenson 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 261 Evenson 262 Evenson 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. 263 Evenson 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? 264 Evenson 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? 265 Evenson 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? 266 Evenson 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? 267 Evenson 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 268 Evenson 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) 269 Evenson 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 50 60 Evenson 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*** 271 Evenson 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. 272 Evenson 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. 273 Evenson 274 Evenson 275 Evenson 276 Evenson 277 Evenson 278 Evenson 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) 279 40 Evenson 280 Evenson 281 Evenson 282 Evenson 283 Evenson 284 Evenson 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 285 Evenson 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 286 Evenson 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 287 Evenson 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 288 Evenson 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 Evenson 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 291 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 292 Evenson 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 293 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 300 Evenson This page is for your records please write LATE or REDO on the top of assignment being handed in. Then fill in the information below so that you remember that the slip has been used. Quarter 1 Late / Redo Slip Quarter 2 Late / Redo Slip 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 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 Date used: Date used: Assigned date: Assigned date: Assignment (page # / date): Assignment (page # / date): Quarter 3 Late / Redo Slip Quarter 4 Late / Redo Slip 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 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 Date used: Date used: Assigned date: Assigned date: Assignment (page # / date): Assignment (page # / Date): 301 Evenson 302 Evenson 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 303