MAD SCIENCE HOMESCHOOL PRODUCTIONSLab Manual for Illustrated Guide to Home Chemistry Elizabeth Moore 7/5/2011 Honors level high school chemistry laboratory manual produced by and used in our home school, Renaissance Classical Academy. Table of Contents Required Materials ....................................................................................................................................... 1 Required Book........................................................................................................................................... 1 Safety Equipment ...................................................................................................................................... 1 Other Suggested Books and Materials...................................................................................................... 1 Lab Session Procedures................................................................................................................................. 2 Before the First Lab Session ...................................................................................................................... 2 Before Each Lab Session............................................................................................................................ 2 Lab Safety ...................................................................................................................................................... 3 General Chemistry Lab Safety ................................................................................................................... 3 Attire ..................................................................................................................................................... 4 Conduct ................................................................................................................................................. 4 Proper Handling of Chemicals and Equipment ..................................................................................... 4 Safety Exams: ............................................................................................................................................ 5 Materials Data Sheets: .............................................................................................................................. 5 Lab Experiments............................................................................................................................................ 6 Introduction to Lab Weeks 1-5: Techniques for Separating Mixtures ...................................................... 6 Week 1: 6.1 Differential Solubility: Separate Sugar and Sand .................................................................. 6 Mixtures ................................................................................................................................................ 6 Differing Solubilities .............................................................................................................................. 6 Week 2: 6.2 Distillation: Purify Ethanol .................................................................................................... 7 Distillation ............................................................................................................................................. 7 Week 3: 6.3 Recrystallization: Purify Copper Sulfate ............................................................................... 8 Crystallization........................................................................................................................................ 8 How to do a crystallization.................................................................................................................... 8 Week 4: 6.4 Solvent Extraction ................................................................................................................. 9 Extraction in the chemistry teaching labs ............................................................................................. 9 How to do an extraction ....................................................................................................................... 9 Week 5: 6.5 Chromatography: Two-Phase Separation of Mixtures ....................................................... 11 Chromatography ................................................................................................................................. 11 Weeks 1-5 Summary ........................................................................................................................... 13 Introduction to Lab Weeks 6-8: Solutions .............................................................................................. 13 Homogeneous Mixtures...................................................................................................................... 13 Varying Concentrations of Ingredients Produces Different Solutions ................................................ 13 Types of Solutions ............................................................................................................................... 14 Similar Structures Allow Solutions to Occur ....................................................................................... 14 Water: The Universal Solvent ............................................................................................................. 17 Week 6: Solutions of Solid Chemicals ..................................................................................................... 17 7.1 Make Up a Molar Solution of a Solid Chemical............................................................................. 17 Introduction ........................................................................................................................................ 17 Molarity ............................................................................................................................................... 18 7.2 Make Up a Molal Solution of a Solid Chemical ............................................................................. 20 Week 7: Solution of Liquid Chemicals and Mass-to-Volume Percentage ............................................... 21 7.3 Make Up a Molar Solution of a Liquid Chemical ........................................................................... 21 7.4 Make Up a Mass-to-Volume Percentage Solution ........................................................................ 22 Week 8: Colligative Properties of Solutions ............................................................................................ 23 8.1 Determine Molar Mass by Boiling Point Elevation ....................................................................... 27 8.2 Determine Molar Mass by Freezing Point Depression ................................................................. 27 8.3 Observe the Effects of Osmotic Pressure ..................................................................................... 27 Introduction to Lab Weeks 9-11: Intro to Chemical Reactions and Stoichiometry ................................ 28 Chemical Reactions ............................................................................................................................. 28 Week 9: 9.2 Observe a Decomposition Reaction.................................................................................... 41 Week 10: 9.3 Observe a Single-Displacement Reaction ......................................................................... 43 Week 11: 9.4 Stoichiometry of a Double Displacement Reaction .......................................................... 43 Introduction to Lab Weeks 12-13: Reduction-Oxidation (Redox) Reactions .......................................... 43 Week 12: 10.1 Reduction of Copper Ore to Copper Metal .................................................................... 43 Week 13: 10.2 Observe the Oxidation States of Manganese ................................................................. 43 Introduction to Lab Weeks 14-17: Acid-Base Chemistry ........................................................................ 43 Week 14: 11.1 Determine the Effects of Concentration on pH .............................................................. 43 Week 15: 11.2: Determine the pH of Aqueous Salt Solutions ................................................................ 43 Week 16: 11.3 Observe the Characteristics of a Buffer Solution ........................................................... 43 Week 17: 11.4 Standardize Hydrochloric Acid Solution by Titration ...................................................... 43 Introduction to Lab Weeks 18-19: Chemical Kinetics ............................................................................. 43 Week 18: Determine Effects of Temperature and Surface Area on Reaction Rate................................ 44 12.1 Determine the Effects of Temperature on Reaction Rate .......................................................... 44 12.2 Determine the Effects of Surface Area on Reaction Rate ........................................................... 44 Week 19: 12.3 Determine the Effects of Concentration on Reaction Rate ............................................ 44 Week 20: 13.1 Observe Le Chatelier's Principle in Action ...................................................................... 44 Week 21: 13.2 Quantify the Common Ion Effect .................................................................................... 44 Week 22: 14.1 Observe the Volume-Pressure Relationship of Gasses (Boyle's Law)............................. 44 Week 23: 14.2 Observe the Volume-Temperature Relationship of Gasses (Charles's Law) .................. 44 Week 24: 14.3 Observe the Pressure -Temperature Relationship of Gases (Gay-Lussac's Law) ........... 44 Week 25: 14.4 Use the Ideal Gas Law to Determine the Percentage of Acetic Acid in Vinegar............. 44 Week 26: 15.1 Determine Heat of Solution ............................................................................................ 44 Week 27: 15.2 Determine the Specific Heat of Ice ................................................................................. 44 Week 28: 15.3 Determine the Specific Heat of a Metal ......................................................................... 44 Week 29: 16.1 Produce Hydrogen and Oxygen by Electrolysis of Water ............................................... 44 Week 30: 18.1 Observe Some Properties of Colloids and Suspensions ................................................. 44 Week 31: 18.2 Produce Firefighting Foam.............................................................................................. 44 Week 32: 18.3 Prepare a Gelled Sol........................................................................................................ 44 Week 33: Discriminate Metal Ions .......................................................................................................... 44 19.1 Using Flame Tests to Discriminate Metal Ions ............................................................................ 44 19.2 Using Borax Bead Tests to Discriminate Metal Ions ................................................................... 44 Week 34: 20.1 Quantitative Analysis of Vitamin C by Acid-Base Titration ............................................. 44 Week 35: 20.2 Quantitative Analysis of Chlorine Bleach by Redox Titration ......................................... 44 Week 36: 21.1 Synthesize Methyl Salicylate from Aspirin...................................................................... 44 Solutions to Pre-Lab Problems .................................................................................................................... 45 Week 6: Solutions of Solid Chemicals ..................................................................................................... 45 Week 7: Solution of Liquid Chemicals and Mass-to-Volume Percentage ............................................... 45 Week 8: Colligative Properties of Solutions ............................................................................................ 46 Introduction to Lab Weeks 9-11: Intro to Chemical Reactions and Stoichiometry ................................ 47 Week 9: 9.2 Observe a Decomposition Reaction.................................................................................... 47 Lab Manual for Illustrated Guide to Home Chemistry Required Materials Required Book Illustrated Guide to Home Chemistry Experiments by Robert Bruce Thompson. (O’Reilly, 2008) Safety Equipment For EACH student, supervising or observing parent, and observing sibling: Splash Goggles Vinyl Lab Apron (or chemical resistant lab coat) Chemical resistant safety gloves (These materials are available from http://www.hometrainingtools.com inexpensively.) Other Suggested Books and Materials Bridge Math by Durrell C. Dobbins (Beginnings Publishing House, 2008) http://beginningspublishing.com/version2/bridge.htm Note: Completion of this program prior to starting the course should help ensure student success with the math associated with several of the labs. Calculations in Chemistry http://www.chemreview.net/ Use of these tutorials in conjunction with the course can provide further instruction and practice in the math associated with chemistry problem solving. Successful Lab Reports: A Manual for Science Students by Christopher S. Lobban, MarLa Schefter (Cambridge University Press, 1993) Note: available used inexpensively from Amazon.com Chemistry Problem Solver (Problem Solvers) by A. Lamont Tyler (REA, 2008) Note: While not required, this may help with math associated with several of the labs. Relevant problems are listed for labs involving math or math-like problem solving in the reading section. Each problem should be copied out on paper and attempted before checking the solution and moving to the next problem. The McGraw-Hill Dictionary of Chemistry online at http://survival-training.info/Library/Chemistry/Chemistry%20-%20MCGraw-Hill%20Osborne%20Dictionary%20of%20Chemistry%20-%20Unknown.pdf Hardbound Quad Ruled Lab Notebook for each student in the class Scientific calculator (We like the TI-30XS, but any scientific calculator will work.) Uncle Tungsten: Memories of a Chemical Boyhood by Oliver Sacks (Unassigned individual reading or family read-aloud. (Suggested Schedule: One chapter per week) Neuroscientist Oliver Sachs recalls his childhood, back in the days when drug stores sold chemicals to inquiring boys. He has great stories of his experiments (and explosions) in his backyard "lab".) 1 Suggestions for lab safety quizzes are at the end of that section. and suggested readings in other texts followed by a short explanation of the theory or techniques to be used in the lab. The explanations in this lab manual may be read prior to the text readings if desired. These should be done after the problems assigned. Any pre-lab problems should also be completed prior to the lab. One student may be assigned to prepare a flowchart. It is strongly suggested that students key word outline the discussion of the technique and brief overview of theory prior to the lab and use that to write a rough draft of the Introduction section of their lab report directly into their lab notebook as part of their pre-lab preparation. such as quantities of particular compounds needed to create solutions of specified concentrations. each student should be familiar with the lab procedures and warnings for all labs planned for the day prior to the start of the session. Safety equipment should be worn by all participants and observers in the labs at all times. and safety information for each planned lab in the session to present to the group prior to beginning the lab. This will also greatly speed up completion of lab reports if they are assigned by the student’s instructor. However. Read the Lab Safety section of this manual prior to the first lab session in addition to familiarizing yourself with the lab to be performed. Good pre-lab preparation and fast turnover of lab reports are good habits to establish for students intending to major in science or engineering in college. Each week and/or experiment begins with any required reading in the lab text. This presentation should not take more than 5-10 minutes. Answers for worksheets are in the back of this manual. as well as disposal instructions. Prepare a flow chart of the lab. Perform all calculations needed prior to the start of the lab. Suggested problems are provided to ensure student proficiency with quantitative methods used in the lab. The lab itself is not specifically assigned.Lab Session Procedures Before the First Lab Session Familiarize yourself with the style of this lab manual. Read the section of this manual for the planned lab session and all assigned readings in that section as well as the lab itself. Take special note of warnings and hazards associated with the lab and any chemicals used or created in the process. discussion. but must also be read prior to the lab session. and contact lenses should not be worn during lab sessions. Illustrated Guide to Home Chemistry. In some cases worksheets may be provided in the manual. Before Each Lab Session Review lab safety procedures. 2 . Additional safety precautions will be announced in class prior to experiments where a potential danger exists.Lab Safety Read: "Keeping a Laboratory Notebook” pp 5-8 Chapter 2: Laboratory Safety General Chemistry Lab Safety The chemistry laboratory can be a place of discovery and learning. and carcinogenic substances from the experiments which you will perform. there is a certain unavoidable hazard associated with the use of a variety of chemicals and glassware. However. 3 . While every effort has been made to eliminate the use of explosive. by the very nature of laboratory work. it can be a place of danger if proper common-sense precautions aren't taken. Students who fail to follow all safety rules may endanger themselves. their parent instructors. highly toxic. and the laboratory space in which we are working. You are expected to learn and adhere to the following general safety guidelines to ensure a safe laboratory environment for both yourself and the people you may be working near. their siblings. waft a small sample to your nose and ONLY when instructed to do so. Learn how to interpret these labels. Chemicals in the lab are marked with NFPA hazardous materials diamond labels. Contact lenses are not allowed. Put these safely in another room. Always wash your hands before leaving lab. Coats. fire blankets. Never smell anything unless instructed to do so. etc. Even when worn under safety goggles. Closed toe shoes and long pants must be worn in the lab. Appropriately label any stock solutions made during lab or pre-lab exercises. Notify the instructor immediately in case of an accident. Material Safety Data Sheets (MSDS) are available in lab or via internet for all chemicals in use. No unauthorized experiments are to be performed. Lab aprons must be worn at all times. backpacks. Keep food and beverages covered and put away during lab activities and until after the lab space has been thoroughly cleaned up and lab materials put away. Chemical resistant lab gloves must be worn when working with or observing experiments involving hazardous chemicals. If you take too much. Make sure no tripping hazards are in the lab space or exits from the lab space. Never taste anything. Do not inhale these vapors but take in only enough to detect an odor if one exists.. Conduct Eating or drinking are strictly prohibited in the laboratory space while it is in use or has open chemical bottles in it even if the kitchen is doubling as laboratory space. This includes fire extinguishers.Attire Safety goggles must be worn at all times while in the laboratory. If you are curious about trying a procedure not covered in the experimental procedure. dispose of the excess. Excess reagents are never to be returned to stock bottles. Carefully read the label twice before taking anything from a bottle. Beware that lab chemicals can destroy personal possessions. Old T-shirts are suggested as dedicated labwear underneath lab aprons. 4 . should not be left on the lab surfaces or nearby floors. Know what chemicals you are using. Sandals and shorts are not allowed. and eye-wash stations. or physical joking around of any kind is permitted in the lab. cartwheels. Never directly smell the source of any vapor or gas. No handstands. instead by means of your cupped hand. consult with your laboratory instructor. various fumes may accumulate under the lens and cause serious injuries or blindness. This rule must be followed whether you are actually working on an experiment or simply writing in your lab notebook while chemicals and equipment are unpacked in the space. Proper Handling of Chemicals and Equipment Consider all chemicals to be hazardous unless you are instructed otherwise. Long hair must be tied back at all times in the laboratory space. mock fighting. especially when using open flames. even just to go to the bathroom. Never pipette by mouth. Learn where the safety and first-aid equipment is located. These will inform you of any hazards and precautions of which you should be aware. flinnsci. Beware of hot glass--it looks exactly like cold glass.flinnsci. alcohols and acetone. sometimes violently.pdf http://www. and the acid will splatter. Do not use them anywhere near open flames. Always pour acids into water. Adapted from: http://chemlabs. Some spills. If you pour water into acid. Clean up all broken glassware immediately and dispose of the broken glass properly with adult assistance and supervision. Never point a test tube or any vessel that you are heating at yourself or your neighbor--it may erupt like a geyser. flush immediately with copious amounts of water and consult with your instructor. Dispose of chemicals properly.sciencebyjones. should be neutralized prior to wipe up and disposal.html Safety Exams: http://www. Never leave burners unattended. If chemicals come into contact with your skin or eyes.com/Documents/miscPDFs/Safety_exam_HS.htm Materials Data Sheets: http://www.com/search_MSDS.edu/Safety/GeneralInstructions. for example. Turn them off whenever you leave your workstation. the heat of reaction will cause the water to explode into steam.asp 5 . are highly flammable. Be sure that the gas is shut off at the bench rack when you leave the lab. such as acids. Consult with instructor for clean-up of any chemical spills. Many common reagents.uoregon.com/lab_safety_quiz. Follow instructions in our lab manual for disposal of waste and chemicals used in the lab under adult supervision. a sample is added to water and heated to boiling. (From CK12 Chemistry 2nd Edition p 408) Define the following words in your lab notebook. catalysts. nor are involved quantitative methods required. and biological sciences classes will be sufficient for students to understand the main points of why each method is successful for separating certain kinds of mixtures. the substance of interest I generally found in a very complex mixture with many other substances. Differing Solubilities Mixtures of solids may often be separated on the basis of differences in their solubilities. When a substance must be isolated from a natural biological source. the water-soluble component can be removed from the mixture by dissolving the mixture in water and filtering the mixture through filter paper. for example.5.1 Differential Solubility: Separate Sugar and Sand Read: "Separating Mixtures” p 93 “Using a Balance” p 73 “Filtration” pp 83-84 “Using Heat Sources” pp 85-88 Successful Lab Reports: Part I: Writing the First Draft: Format (pp 1-49) Mixtures Mixtures occur very commonly in chemistry. 6. dissolve soluble solution solubility 6 . all of which must be removed.1 – 6. Week 1: 6. familiarizes students with basic laboratory techniques to separate mixtures. When a new substance is synthesized. Often times. it may also be possible to separate the components by controlling the temperature at which the solution occurs or at which the filtration is performed. Younger siblings observing can be expected to follow along sufficiently to appreciate larger scientific principles and laboratory techniques involved. Chemists have developed a series of standard methods for the separation of mixtures. the new substance usually must be separated from various side-products. physical. If one component of the mixture is soluble in water while the other components are insoluble in water.Lab Experiments Introduction to Lab Weeks 1-5: Techniques for Separating Mixtures The first group of labs. the separation of mixtures into their constituent substances defines an entire sub-field of chemistry referred to as separation science. These crystals can then be separated by another filtration. Use the glossary in your Chemistry theory text or a resource such as the McGraw-Hill Dictionary of Chemistry. Students are not expected to have detailed knowledge of the theory behind these techniques. Since the solubility of substances is greatly influenced by temperature. In fact. which causes crystallization of those substances whose solubilities are very temperature dependent. Background knowledge from previous general. and any excess reagent still present. The component dissolved in water will pass through the filter while the undissolved solids will be caught in the filter. and the filtrate (the material that went through the filter) will then contain only those substances whose solubilities are not as temperature dependent. The hot sample is then filtered to remove completely insoluble substances. The sample is then cooled to room temperature or below. (From CK12 Chemistry 2nd Edition pp 408-409) Define the following words in your lab notebook. it may be possible to separate the mixture simply by monitoring the temperature of the vapor produced as the mixture is heated. distillation involves heating a liquid to its boiling point. the separation is not as easy. As the solution boils. it is a solution. This is a quick method of determining if a liquid is a pure substance or a solution: start boiling the solution. and if it continues to boil at the same temperature.2 Distillation: Purify Ethanol Read: “Measuring Liquids by Volume” pp 74-78 (Burette's will be covered later. Use the glossary in your Chemistry theory text or a resource such as the McGraw-Hill Dictionary of Chemistry. then collecting. increased temperature is necessary to keep the solution boiling because its boiling point has increased. and condensing the vapor produced into a separate container. a separation can be accomplished. In solutions of non-volatile (resistant to vaporization) solid solutes in liquid solvent. By changing the receiving flask at the correct moment. whereas if its boiling point increases. This process is known as fractional distillation.Week 2: 6. Since the concentration increases. solute solvent volatile distillation distillate 7 . A common distillation setup is illustrated to the right. with a sharp rise in the temperature of the vapor being distilled indicating when a new component of the mixture has begun to boil. Liquid components of a mixture will each boil in turn as the temperature is gradually increased. cooling. As the solvent vaporizes and all of the solute remains behind. In general. the boiling point of the solution is also increasing. the same amount of solute is now dissolved in less solvent. only the solvent boils off and all of the solid remains in the solution. when the solution is boiled. For a mixture of liquids in which several components of the mixture are likely to be volatile (easily vaporized). it is a pure substance. If the components of the mixture differ reasonably in their boiling points.) “Using Heat Sources” p85-88" Successful Lab Reports: Revising Your Paper pp 53-61 Distillation Homogeneous solutions are most commonly separated by distillation. Understanding the process of crystallization in itself will not make a student a master crystallizer. How to do a crystallization To crystallize an impure.html on March 22. Impurities are excluded from the growing crystals and the pure solid crystals can be separated from the dissolved impurities by filtration. it will be more likely to remain on the crystal than it is to go back into the solution. After the solution has come to room temperature.edu/hndbksupport/cryst/crystproc.colorado.colorado. in which solute molecules – both the desired compound and impurities – move freely among the hot solvent molecules. Crystallization is based on the principles of solubility: compounds (solutes) tend to be more soluble in hot liquids (solvents) than they are in cold liquids. Use the glossary in your Chemistry theory text or a resource such as the McGraw-Hill Dictionary of Chemistry. The flask then contains a hot solution. Source: http://orgchem. recrystallization 8 . successful crystallization relies on a blend of science and art. Detailed photos from start to finish: http://orgchem. observation. Therefore. the solute is no longer soluble in the solvent and forms crystals of pure compound.html Crystallization Movie: http://video. If the geometry of the molecule fits that of the crystal. it is carefully set in an ice bath to complete the crystallization process. This simplified scientific description of crystallization does not give a realistic picture of how the process is accomplished in the laboratory. its success depends more on experimentation. 2011 Define the following word in your lab notebook. and skill than on mathematical and physical predictions.google. solid compound. If a saturated hot solution is allowed to cool. The chilled solution is then filtered to isolate the pure crystals and the crystals are rinsed with chilled solvent. the solute. As the solution cools.Week 3: 6.com/videoplay?docid=5617753349611003526 On GoogleVideo choose "smoothing" and "original size" from the lower right pull-down menu for best video.edu/hndbksupport/cryst/cryst.3 Recrystallization: Purify Copper Sulfate Read: “Using a Balance” p 73 “Filtration” pp 83-84 “Using Heat Sources” p85-88 Successful Lab Reports: Citations and Reference List pp 62-72 Crystallization Crystallization is a technique which chemists use to purify solid compounds. and they begin to leave the solution and form solid crystals. each growing crystal consists of only one type of molecule. rather. each solute molecule in turn approaches a growing crystal and rests on the crystal surface. During this cooling. Rather. the solvent can no longer “hold” all of the solute molecules. imagination. this understanding must be combined with laboratory practice to gain proficiency in this technique. add just enough hot solvent is added to it to completely dissolve it. It is one of the fundamental procedures each chemist must master to become proficient in the laboratory. which is methylene chloride.google.) Some compounds are more soluble in the organic layer (the "oil") and some compounds are more soluble in the aqueous layer (the "vinegar"). you will be able to figure out which layer. Recall last year’s discussion in Biology of lipid bilayers in cell membranes and cell membrane permeability. This is determined by the density of the two solvents. and red food coloring and allow the layers to separate.com/videoplay?docid=945061828857075635&hl=en On GoogleVideo choose "smoothing" and "original size" from the lower right pull-down menu for best video. Water is immiscible with the other liquid. As we do have a separatory funnel. You will also need to know which layer will be on top in the separatory funnel. The "liquid-liquid" phrase means that two liquids are mixed in the extraction procedure.edu/hndbksupport/ext/extprocedure. water. therefore. like oil and vinegar do in dressing.Week 4: 6.colorado.colorado. and red food coloring? Go to the next page for the answer. What would you expect to see if you mixed diethyl ether. (A demonstration of this fact using red wine or balsamic vinegar and oil may be useful for students and/or younger siblings. Methylene chloride is heavier (denser) than water. aqueous or organic. Densities are listed in various sources of scientific data. The red layer is simply red food coloring in water. The photo at the right illustrates how two liquid layers separate. The photo at the right illustrates what you see if you mix methylene chloride. we will not be following adapted procedures in The Illustrated Guide to Home Chemistry that would not allow quite as good a separation. 9 .html How can you determine the density of a solvent you are working with if you did not have access to sources of scientific data? How to do an extraction Extraction Procedure (separatory funnel): http://orgchem. In a particular experiment in simple extraction or in chemically active extraction. as referenced on the Chem Info page on this orgchem site: Hazard and Physical Data for Compounds: http://orgchem. aqueous food coloring layer.4 Solvent Extraction Read: "Separations" pp 84-85 Successful Lab Reports: Sample Lab Report and Revision pp 73-91 Extraction in the chemistry teaching labs Liquid-liquid extractions using a separatory funnel are essentially the only kind of extraction performed in the most chemistry teaching labs.edu/cheminfo/cheminfo. water.html Extraction Movie: http://video. will contain the compound you want to isolate. The liquids must be immiscible: this means that they will form two layers when mixed together. the clear methylene chloride layer is under the red. Quiz: Red food coloring is soluble in water and not in methylene chloride or diethyl ether. immiscible extraction 10 . The food color remains dissolved in the water. measure volume into glassware. Use the glossary in your Chemistry theory text or a resource such as the McGraw-Hill Dictionary of Chemistry. and methylene chloride in the separatory funnel on the right. The following data is from the Tables in the Handbook: diethyl ether d = 0. the water will be on the bottom. you need to know the densities. Compare with the separated mixture of water. if you mix diethyl ether. food color. Determining densities: D=M/V so you could measure the mass of a measured volume (weigh volumetric glassware.00 Therefore. since water is more dense than diethyl ether. and red food coloring.colorado. shake it in a sep funnel and allow the layers to separate. html on March 22. 2011 Define the following words in your lab notebook.71 methylene chloride d = 1.33 water d = 1. it will look like the mixture in the sep funnel on the left in the photo below (big yellow check mark). weigh glassware with solvent) and divide to determine density.edu/hndbksupport/ext/ext. Adapted From: http://orgchem.Quiz Answer: In order to make a prediction as to which layer is on the top and which on the bottom. water. A single drop of the unknown mixture to be separated is applied about half an inch from the end of a strip of filter paper or TLC slide. various effects can occur. The separation of a mixture by chromatography is not only a function of the solubility in the solvent used. Since the filter paper and the TLC slide coating are permeable to liquids.Week 5: 6. The methods are very similar in operation and principle. Each component of the mixture is likely to interact with the medium to a different extent. Rf. The Rf value is characteristic of a substance when the same solvent and the same type of 11 . depending on the constituents of the spot. as seen in the illustration of a paper chromatography strip to the right. thus slowing the components of the mixture differentially depending on the level of interaction. They differ primarily in the medium used. In this way.5 Chromatography: Two-Phase Separation of Mixtures Successful Lab Reports: Anatomy of a Scientific Paper pp 93-99 Chromatography Chromatography is another method for separating mixtures. Paper and thin-layer chromatography are simple techniques that can be used to separate mixtures into the individual components. the original mixture spot is spread out into a series of spots or bands. Thin-layer chromatography (TLC) uses a thin coating of aluminum oxide or silica gel on a glass microscope slide or plastic sheet to which the mixture is applied. the solvent begins rising up the paper by capillary action. Paper chromatography uses ordinary filter paper as the medium upon which the mixture to be separated is applied. The word chromatography means colorwriting. with each spot representing one single component of the mixture. Those components that are not at all soluble will be left behind at the original location of the spot. As the solvent rises to the level of the mixture spot. In chromatography analysis. Chromatography in its various forms is perhaps the most important known method for the chemical analysis of mixtures. The filter paper or TLC slide is then placed in a shallow layer of solvent in a jar or beaker. Those components of the spot that are completely soluble in the solvent will be swept along with the solvent front as it continues to rise. The primary interaction between the mixture components and the medium is due to the polarity of the components and that of the medium. The filter paper or TLC coating consists of molecules that may interact with the molecules of mixture as they are carried up the medium. there is a mathematical function called the retention factor. Most components of the mixture will move up the paper or slide at an intermediate speed somewhat less than the solvent front speed. The name was chosen around 1900 when the method was first used to separate colored components from plant leaves. The retention factor depends on what solvent is used and on the specific composition of the filter paper or slide coating used. The retention factor. is defined as: Rf is the ratio of the distance a substance moves up the stationary phase to the distance the solvent have moved. In this form.stationary phase is used. a set of known substances can be analyzed at the same time under the same conditions. and column chromatography with UV active materials. As the liquid drips out the bottom of the column. Therefore.com/watch?v=gzp2S0e9o8s (From CK12 Chemistry 2nd Edition pp 409-412) 12 . Another form of chromatography is column chromatography. is used to carry the mixture of gases through the tube. the components are separated.youtube. Mixtures of gases are commonly separated by gas chromatography. Separation of the components occurs as the mixture moves through the tube. As with paper chromatography. This video presents thin layer chromatography with fluorescent materials. As the mixture flows down the column. the Rf for the green spot is Paper chromatography and TLC are only two examples of many different chromatographic methods. In this method. again. In the case shown below. the mixture is poured in at the top. usually helium. components of the solution will exit at different times and can be collected. a mixture of liquids are vaporized and passed through a long tube of solid absorbent material. There is no narration on the video so it would be advantageous to watch with a chemistry teacher (6f): http://www. the components of the mixture will have different solubilities and different attractions for the solid absorbent. a vertical column is filled with solid absorbent. A carrier gas. and a carrier solvent is added. by differing solubilities in the carrier solvent and different absorbencies to the solid packing. The individual components exit the tube one by one and can be collected. the particles of salt would be so 13 . Thinking of the prefix “homo” meaning “sameness”. Mixtures of liquids with reasonably different boiling points can also be separated by distillation. a solution would form. the definition is referring to properties at the particle level. this definition makes perfectly good sense. Vinegar is approximately 5% acetic acid in water. In the next group of labs we will learn about different ways to make up solutions and describe their composition. Well. Compounds may be separated based on their relative solubilities in two different immiscible liquids. Solid compounds may be separated from impurities – purified – by recrystallization. We will also look at some properties of solutions that are not obvious as the sum of their component parts. To the naked eye this brass coin seems like it is just one substance but at a particle level two substances are present (copper and zinc). An alloy is a homogeneous solution formed when one solid is dissolved in another. Gas chromatography and column chromatography are also used to separate the components of a solution (Adapted from CK12 Chemistry 2nd Edition pp 409-412) Introduction to Lab Weeks 6-8: Solutions Read: “Solubility and Solutions” pp 121-125 We talked about solution earlier in conjunction with our separation labs in weeks 1 and 2. Now. The solution would have the same properties throughout. Margarine is a combination of a number of substances at the molecular level but to the naked eye it is a homogeneous solution that looks like just one substance. Components of a solution composed of a non-volatile solid solute and a liquid solvent can be separated by distillation. Another example of a solution is margarine. so take extra care in measurement and recording while doing these labs. So the brass represents a homogeneous mixture. vinegar that is used in cooking. Homogeneous Mixtures A homogeneous mixture is a solution of the same appearance or composition throughout. Varying Concentrations of Ingredients Produces Different Solutions The point should be made that because solutions have the same composition throughout does not mean you cannot vary the composition.Weeks 1-5 Summary Mixtures of solids may be separated by differing solubilities of the solids in a single solvent. Thus the combination of zinc filings and copper pieces in a pile does not represent a homogeneous solution. what does this mean? Let's consider brass as an example. by solvent extraction. consider a handful of zinc filings and copper pieces. Is this now a homogeneous solution? The properties of any scoop of the “mixture” you are holding would not be consistent with any other scoop you removed from the mixture. for example. A point should be made here that when a solution is said to have uniform properties throughout. Homogeneous solutions carry the same properties throughout the solution. contains 5% acetic acid and 95% water. The brass is an alloy made from copper and zinc. Solutions with several components can be separated by paper or thin-layer chromatography. Many of the solutions we will make in the next two weeks will be used in later labs. Take. usually water and an organic solvent. This means that every teaspoon of vinegar that is removed from the container. If you were to take one cup of water and dissolve ¼ teaspoon of table salt in it. air is a solution made up of mostly oxygen gas and nitrogen gas. to seawater. Both methanol and water are polar molecules and form a solution because they both have permanent dipoles (positive and 14 . suppose you are dissolving methanol in water.small that they would not be seen and the composition of every milliliter of the solution would be the same. Look at the Table below. These are shown in boldface. take a deep breath. Recall that in chemistry. The others. we can predict when solutions will form and others won’t using a little saying … “like dissolves like”. during the study of Valence Shell Electron Pair Repulsion Theory (VSEPR). There are limits to the amount of substance that can be dissolved into another substance and still remain homogeneous. we have nine possibilities. a polar molecule is one that has a positive end and a negative end while nonpolar molecules have charges that are evenly distributed throughout the molecule. In solution chemistry. But you can vary the composition of this solution to a point. If we think about solutions and the possibilities of combining these states together to form solutions. In fact. You still have a solution where the salt particles are so small that they would not be seen and the solution has the same properties throughout. For example. the solution has passed its limit as to the amount of salt it can dissolve and it would no longer be a homogeneous solution. For a gas in a liquid solution. are less common in everyday lives. it would not. Finally. liquid. What would happen if you tried to dissolve ½ cup of salt in the water. Types of Solutions Solid Liquid Gas Solid Solid in a Solid Solid in a Liquid Solid in a Gas Liquid Liquid in a Solid Liquid in a Liquid Liquid in a Gas Gas Gas in a Solid Gas in a Liquid Gas in a Gas In the table (above). to understand the gas in a gas solution. So solutions have constant composition but you can vary the composition up to a point. you would make another solution. Would the solution stay homogeneous? No. but this time there would be a different composition than the last. These are examples of solid – solid solutions. thus it is homogeneous. Types of Solutions There are three states of matter: solid. although still solutions. That’s right. the most common example is soda pop: carbon dioxide dissolved in water (with lots of sugar!) Another example is the ammonia solution you may use (or have seen used) to clean in the home. A solid in a solid solution is less common but still we see a lot of steel and brass around in our everyday world. If you were to add another ½ teaspoon of salt to the cup of water. there are really only four that are common types of solutions. a solid in a liquid solution can be anything from salt or sugar solution. The other types of solutions are less common but do exist in the world of solution chemistry. Liquid in liquid solutions include the antifreeze/coolant we use for our cars and vinegar. The “like dissolves like’ saying helps us to predict solubility based on the two parts of a solution having similar intermolecular forces. For example. Similar Structures Allow Solutions to Occur Over the course of your study in chemistry you have learned the terms polar and non-polar. At this point. and gas. you learned that the chemical structures themselves have built in molecular polarity. Since these molecules are both polar. The same is true for the case of a non-polar substance such as carbon tetrachloride being dissolved in another nonpolar substance such as pentane. The reason for this is again explained with the structures of the substances. London-dispersion forces are the intermolecular bonds that hold the carbon tetrachloride together as a liquid. you can see that there are also permanent dipoles making it a polar molecule. which means these two liquids will make a solution. they will form a solution when mixed together. a solution is made. at any given time. Notice in the representation of the individual molecules of methanol and water how the methanol has a permanent dipole due to the bonds and is a polar molecule. Since carbon tetrachloride (or pentane) has no permanent dipoles in its molecules. Let’s look at the individual structure of the water and methanol molecules. However. London-dispersion forces are also the forces that allow pentane to be a liquid at room temperature. (See Figure below). no permanent dipole. The two sides of Velcro allow the pieces of fabric to be fastened together because the Velcro has similar structure that “attract” each other. to either carbon tetrachloride or pentane.negative parts of the molecules) that allow the molecules of each of the substances to be attracted to the other. there would be place for the charges particles in a crystal of NaCl to be attracted. Since both of these substances have the same intermolecular forces. the non-polar molecules have. The intermolecular forces for both of these molecules are dipole-dipole attractions. When this occurs. Unlike the polar molecules. A way to understand this is to think about why Velcro is used to hold two different pieces of fabric together. 15 . NaCl. The hydration of ions in a polar solvent. In the representation of the water molecule. a solution will be formed. when they are mixed together. If we were to add table salt. we would find that the salt would not dissolve. one side of Velcro would not stay together with a piece of silk since the silk doesn’t have any part of its structure with which the Velcro can connect. We say they are miscible. Therefore. As ionic solids dissolve into solution. are held together with covalent bonds. and oxygen ions. the electronegativity value is 2. when you dissolve a spoonful of sugar into a glass of water. they can pull the electrons closer to themselves and away from the element that has a smaller electronegativity. the intermolecular forces between the sugar molecules are disrupted but the intramolecular bonds are not. However. Recall that elements with a greater electronegativity have a stronger attraction for shared electrons. The sugar molecules remain intact.so a solution does not form. covalent compounds result from the sharing of electrons between atoms. This attraction allows the two different types of molecules to form a solution. and for oxygen it is 3. The resultant structure is represented below. but the molecule is non-polar overall because the shifting of the shared electrons toward the oxygen atoms are in equal but opposite directions. each oxygen atom shares two of its electrons with carbon and the carbon shares two of its electrons with each oxygen atom.5. the positive polar ends of the solute molecules attract the negative polar ends of the solvent molecules and vice versa. When a polar solute is added. there are no distinct charges associated with the atoms in covalent compounds.In a polar solvent. hydrogen ions. In CO2. For carbon. which result from the transfer of electrons. they can interact with the polar water molecules to form a solution. For example. there is no overall dipole moment on the molecule. Molecular compounds. the non-polar solute particles cannot attract the solvent molecules away from each other . the molecules of solvent are attracted to each other by the partial charges on the ends of the molecules. If a non-polar solute was added to a polar solvent. which are not readily broken. however. The bonds in this molecule are polar. the electrons are not shared equally. Covalent compounds have a different type of attraction occurring between the solute and solvent molecules. As a result. Unlike ionic compounds. You can write the following equation for the dissolution of sugar in water: C12H22O11(s) → C12H22O11(aq) (Bonus: Why doesn’t the formula for table sugar conform to the standard CnH2nOn for carbohydrates that you learned last year in Biology?) 16 . The result in this molecule is that the electrons are pulled closer to oxygen than carbon. but because of their polar properties. these solids separate into ions. Look at the figure below: This sharing of valence electrons represents covalent bonding. As a result. The sugar will not separate into carbon ions.5. The term solvent is used to represent the medium that is used to produce the solution. does dissolve in water and it is this oxygen that the fish take in through their gills. In this case. (From CK12 Chemistry 2nd Edition pp 374-387) (Bonus question answer: Sucrose is formed by dehydration synthesis of glucose and fructose.1 Make Up a Molar Solution of a Solid Chemical Read: Introductory materials above. and mole fraction (6d): http://www. For instance. Have you ever wondered why fish are able to breathe? Oxygen gas.5ppm and say to ourselves. water: the universal solvent. including chemistry. watered down. And from your own experience and the description above you know that table sugar.com/watch?v=9br3XBjFszs 7. Pepsi-cola and all the other sodas have carbon dioxide gas.Polar solvents will dissolve polar and ionic solutes because of the attraction of the opposite charges on the solvent and solute particles. Or. But that’s another part of the story! Although qualitative observations are necessary and have their place in every part of science. The term universal is used to describe the fact that water. What does this mean? So being able to deal with the quantitative side of solutions helps us to move toward a deeper understanding of solutions. A concentrated solution is one in which there is a large amount of solute in a given amount of solvent. the sodas are kept under pressure. A dilute solution is one in which there is a small amount of solute in a given amount of solvent. also dissolves in water. Even some nonpolar substances dissolve is water but only to a limited degree. one more example of a nonpolar compound that dissolves in water is the reason we can enjoy carbonated sodas. can dissolve many types and kinds of substances. along with many of its other unique aspects. Solutions can be said to be dilute or concentrated. A dilute solution is a concentrated solution that has been. Therefore. is an ionic compound but easily makes a solution with water. a polar covalent compound. CO2. NaCl. to keep as much gas in solution as possible. you are diluting the concentrated 17 . Non-polar solvents will only dissolve non-polar solutes because they cannot attract the dipoles or the ions. a nonpolar molecule. a nonpolar compound. It shows how to calculate molarity. what does that mean? Is it important? We read labels in the grocery store that are in weight percent. In order to make juice. dissolved in a sugar-water solution.) Week 6: Solutions of Solid Chemicals Here is a link to a video of a teacher writing on an electronic blackboard. We might read in the headlines that the amount of mercury found in the fish is up by 0. table salt.youtube. This is true for many ionic compounds. In the next two labs we will explore some of the different quantitative applications of solution chemistry. we have seen throughout our study of science that there is a definite need for quantitative measurements in science. one that involves not only a numerical analysis but a critical analysis as well. Introduction Concentration is the measure of how much a given substance is mixed with another substance. Complete tables in book. you mix the frozen juice from inside these containers with about 3 or 4 times the amount of water. This is particularly true in solution chemistry. One H2O molecule is lost in this reaction. And this is also true for other polar compounds such as vinegar and corn syrup. Think of the frozen juice containers you buy in the grocery store. Water: The Universal Solvent Think of the title of this section. in essence. molality. 500 liter) = 4. the expression [Ag+] refers to the molarity of the silver ion. Use the glossary in your Chemistry theory text or a resource such as the McGraw-Hill Dictionary of Chemistry. The symbol given for molarity is M.25 mol/L)(0. Molarity Of all the quantitative measures of concentration. The terms “concentrated” and “dilute. In this lesson.625 mol mol = . (From CK12 Chemistry 2nd Edition pp 387-388) Define the following words in your lab notebook.34 mol) / (0.500 L) = 0.68 M The concentration of the NaCl solution is 4.3 g/mol) = 109 g Therefore. so mol = M × L = (1. Example: What would be the mass of potassium sulfate in 500 mL of a 1.juice. or moles/liter. molarity is the one used most frequently by chemists.68 mol/L. only provide a qualitative way of describing concentration. Molarity is defined as the number of moles of solute per liter of solution. we will explore some quantitative methods of expressing solution concentration.” however. For example. Chemists also use square brackets to indicate a reference to the molarity of a substance. when 2. molarity = Example: What is the concentration. the mass of the K2SO4 that dissolves in 500 mL of H2O to make this solution is 109 g.34 mol of NaCl has been dissolved in 500 mL of H2O? Solution: [NaCl] = (2. so mass = (mol)·(molar mass) mass = (0.625 mol)(174. concentration concentrated dilute molarity Pre-Lab Problems next page: 18 .25 mol/L potassium sulfate solution? Solution: M= . in mol/L. What is the volume of NaCl solution he will make? (from edHelper. He weighs 20 g of NaCl crystals.154 M NaCl solution.1. What is the molarity of a 100 ml solution prepared by dissolving 30 g of NaCl? 2) A laboratory assistant prepares to make 0.com) 19 . 2 Make Up a Molal Solution of a Solid Chemical Read: Introductory materials above for this set of labs. If the density of this glucose solution is 1. molality Pre-Lab Problem: 3) 18 g glucose (C6H12O6) is dissolved in 100 g of water. Use the glossary in your Chemistry theory text or a resource such as the McGraw-Hill Dictionary of Chemistry.com) 20 . Molality has the symbol m. and mass is not temperature dependent. if you recall. there is a slight advantage to using molality over molarity when temperatures move away from standard conditions.07 g/ml. Molality is another way to measure concentration of a solution.7. Volume is temperature dependent.5 g of hydrochloric acid has been dissolved in 115 g of water. Molality does not involve volume. Complete tables in book. Solution: mol HCl = molality HCl = = 0.98 m (From CK12 Chemistry 2nd Edition pp 389) Define the following word in your lab notebook. 1) What is the molarity of this solution? 2) What is the molality of this solution? (from edhelper. the molarity of the solution will actually decrease slightly because the volume will increase slightly. As the temperature rises. It is calculated by dividing the number of moles of solute by the number of kilograms of solvent.343 mol = = 2. Thus. is the number of moles of solute per volume of solution. molality (m) = Molarity. Example: Calculate the molality of a solution of hydrochloric acid where 12. Complete tables in book.2 M HCl solution.Week 7: Solution of Liquid Chemicals and Mass-to-Volume Percentage Read: Review introductory materials in this manual for this set of labs. student C combines these two solutions.45 g) 2) Student A has 100 ml of 2 M HCl solution. atomic mass of Cl = 35. What mass of NaCl will the student have to weigh? ( atomic mass of Na = 23 g .3 Make Up a Molar Solution of a Liquid Chemical Pre-Lab Problems: 1) A student at a chemical laboratory wants to make 500 ml of a 2 M NaCl solution.com) 21 . What is the molarity of this new solution? (from edHelper. Accidentally. 7. and student B has 50 ml of 3. Use the glossary in your Chemistry theory text or a resource such as the McGraw-Hill Dictionary of Chemistry.com) Mass to volume percent uses a similar calculation with the formula: Mass-volume percent = × 100% 22 .18 g/ml.4 Make Up a Mass-to-Volume Percentage Solution Mass to volume percent is similar to another measure of concentration. It has the formula: percent by mass = or percent by mass = Example: An alloy is prepared by adding 15 g of zinc to 65 g of copper. What is the mass percent of zinc? Solution: percent by mass = × 100% = × 100% = 19% × 100% × 100% (From CK12 Chemistry 2nd Edition pp 388) Define the following terms in your lab notebook. Mass Percent Mass percent is the number of grams of the solute in the number of grams of solution. mass percent mass to volume percent Pre-Lab Problem: 3) A concentrated HCl solution is 36% by mass and has a density of 1.7. mass percent. 1) What is the molarity of this concentrated HCl? 2) What volume of concentrated HCl is needed to make 5 liters of 1 M HCl solution? 3) What volume of H2O is needed to make 100 ml of 4 M HCl? (from edHelper. Mass percent is a term frequently used when referring to solid solutions. 20 m solution of KNO3. When table salt is added to water. A number of solutions are used.0 atm is 100. requiring the temperature to decrease further before the solution will solidify. the boiling point of pure water at 1. Here the salt is put on the roads so that the water on the roads will not freeze at the normal 0◦C but at a lower temperature. The Mathematics of Boiling Point and Freezing Point Changes The amount to which the boiling point increases or the freezing point decreases depends on the amount solute that is added to the solvent. Therefore. the boiling point elevation would be 2◦C. In comparison. we can see similarities between the two. The de-icing of planes is another common example of freezing point depression in action. This is true for any solute added to a solvent. Essentially. Boiling point elevation is another example of a colligative property. but salt water does not. Remember the solution has a higher boiling point. the solute particles take up space at the solvent/air interface. Recall that the freezing point is the temperature at which the liquid changes to a solid. but commonly a solution such as ethylene glycol or a less toxic monopropylene glycol is used to de-ice an aircraft. meaning that the change in boiling point is related only to the number of solute particles in solution. if a substance is added to a solvent (such as water). then. the resulting solution has a higher boiling point than water alone.Week 8: Colligative Properties of Solutions Read: “Colligative Properties of Solutions” pp 147-148 Colligative properties are properties that are due only to the number of particles in solution and not to the chemical properties of the solute. while the boiling point of a 2% salt-water solution is about 102◦C. The freezing point depression. At a given temperature. A common example is found when salt is used on icy roadways. physically blocking some of the more energetic water molecules from escaping into the gas phase.◦C. the solute-solvent interactions prevent the solvent from going into the solid phase. the freezing point depression is found by subtracting the freezing point of the solution from the freezing point of the pure solvent. A mathematical equation can be used to calculate the boiling point elevation or the freezing point depression. Boiling Point Elevation The boiling point of a liquid occurs when the vapor pressure above the surface of the liquid equals the surrounding pressure. Freezing Point Depression The effect of adding a solute to a solvent has the opposite effect on the freezing point of a solution as it does on the boiling point. Both the boiling point elevation and the freezing point depression are related to the molality of the solutions. so to find the boiling point elevation you would subtract the boiling point of the solvent from the boiling point of the solution. 23 . The aircrafts are sprayed with the solution when the temperature is predicted to drop below the freezing point. pure water boils at 100◦C. For example. At 1 atm of pressure. In this week labs we will investigate two such properties of solutions. regardless of what those particles are.20 m solution of table salt would have the same change in boiling point as a 0. as low as -9◦C. Looking at the formulas for the boiling point elevation and freezing point depression. A 0. is the difference between the freezing points of the solution and the pure solvent. radiators. Freezing point depression: ΔT f = Kfm where ΔT f = T f (solution) − T f (pure solvent) Kf = freezing point depression constant m = molality of the solution.86◦C. You can find these constants for hundreds of solvents listed in data reference publications for chemistry and physics.86◦C/m. for water is -1. Since water expands when it freezes. The main component in antifreeze is ethylene glycol.1 g/mol) = 9. The boiling point and freezing point constants are different for every solvent and are determined experimentally in the lab. Example: Antifreeze is used in automobile radiators to keep the coolant from freezing. C2H4(OH)2. what mass of ethylene glycol was used? The freezing point constant. Kf . If the addition of an unknown amount of ethylene glycol to 150 g of water dropped the freezing point of the solution by −1.150 kg) = 0.00 mol/kg)(0. = = 1. 9. using pure water as the coolant could allow the water to freeze. freezing coolant could crack engine blocks. Solution: ΔT f = Kf m m= m= moles solute = (molality)(kg solvent) = (1.86◦C.150 mol)(62.150 mol mass C2H4(OH)2 = (mol)(molar mass) = (0. In geographical areas where winter temperatures go below the freezing point of water.Boiling point elevation: ΔTb = Kbm where ΔTb = Tsolution − Tpure solvent Kb = boiling point elevation constant m = molality of the solution.32 g Therefore.32 g of ethylene glycol would have been added to the 150 g of water to lower the freezing point by 1. and coolant lines.00 m 24 . For lithium nitrate (LiNO3). adding 10 molecules of solute to the solution will produce 20 ions (solute particles) in the solution.Remember that colligative properties are due to the number of solute particles in the solution.20 m solution. so the van’t Hoff factor for NaCl is i = 2. Adding 10 molecules of sugar to a solvent will produce 10 solute particles in the solution. Freezing point depression: ΔT f = iKfm where i = van’t Hoff factor ΔT f = T f (solution) − T f (puresolvent) Kf = freezing point depression constant m = molality of the solution. We can use this formula for both electrolyte and non-electrolyte solutions since the van’t Hoff factor for nonelectrolytes is always 1 because they do not dissociate. i = 3. adding enough NaCl solute to a solvent to produce a 0. However. For example. Boiling point elevation: ΔTb = iKbm where i = van’t Hoff factor ΔTb = Tsolution − Tpure solvent Kb = boiling point elevation constant m = molality of the solution. Therefore. and for calcium chloride (CaCl2). The van’t Hoff factor (i) is the number of particles that the solute will dissociate into upon mixing with the solvent. i = 2. when the solute is an electrolyte. such as NaCl. We can now rewrite our colligative properties formulas to include the van’t Hoff factor.20 m solution will have twice the effect of adding enough sugar to a solvent to produce a 0. sodium chloride (NaCl) will dissociate into two ions. 25 . 26 . What is the boiling point of the solution? Solution: ΔTb = iKbm mol NaCl = molality = = = = 0.78◦C Therefore.0 g of water in an attempt to elevate the boiling point. the boiling point of the solution of 10 g of NaCl in 100 g of water is 102◦C.71 m For NaCl.0 g of sodium chloride is added to 100.86 0C/m.52 0C/m and Kf(H2O) = 1. what will the melting and boiling points be of the resulting solution? Kb(H2O) = 0.71 m) = 1.52◦C/m)(1.78◦C = 101. i = 2 (NaCl → Na+ + Cl−) Kb(water) = 0.171 mol = 1.78◦C Tb(solution) = Tb(puresolvent) + ΔTb Tb(solution) = 100◦C + 1.Example: A solution of 10. (From CK12 Chemistry 2nd Edition pp 402-404) Optional Pre-Lab Questions: 1) If I add 45 grams of sodium chloride to 500 grams of water.52◦C/m ΔTb = iKbm ΔTb = (2)(0. including those that make up the membranes around living cells. Certain materials. Recall our discussions of osmosis and osmotic pressure during our Biology studies last year. 453-454) Define the following terms in your lab notebook. passage from the pure solvent side to the solution side is more favored and occurs faster. That is. Include optional activities in lab if possible. As a result.2 Determine Molar Mass by Freezing Point Depression Read: Introductory material above. solvent molecules pass through the membrane in a process called osmosis. Include optional activities in lab if possible. 8. the amount of liquid on the pure solvent side decreases.1 Determine Molar Mass by Boiling Point Elevation Read: Introductory material above. Use the glossary in your Chemistry theory text or a resource such as the McGraw-Hill Dictionary of Chemistry. When a solution and a pure solvent (or two solutions of different concentration) are separated by the right kind of semipermeable membrane. Van’t Hoff factor Colligative property Osmosis Osmotic pressure 27 .html ) 8.brinkster. 8. (from Chemistry (McMurray. Although the passage of solvent through the membrane takes place in both directions. and the concentration of the solution decreases. Include optional activities in lab if possible. are semipermeable. Fay.2) Which solution will have a higher boiling point: A solution containing 105 grams of sucrose (C12H22O11) in 500 grams of water or a solution containing 35 grams of sodium chloride in 500 grams of water? (from http://misterguch.3 Observe the Effects of Osmotic Pressure Read: Introductory material above and discussion below.net/pra_solutionworksheets. 4th Edition) pp. but they block the passage of large solute molecules or ions. the amount of liquid on the solution side increases. they allow water or other small molecules to pass through. we call it a chemical reaction. There are other symbols. sulfur trioxide is produced. we need to indicate what substances are present at the beginning and what substances are present at the end. the chemical bonds of the substances break. The general equation for a reaction is: Reactants → Products There are a few special symbols that we need to know in order to communicate in chemical shorthand. and the atoms that make up the substances separate and re-arrange themselves into new substances with new chemical bonds.Introduction to Lab Weeks 9-11: Intro to Chemical Reactions and Stoichiometry Read: “Introduction to Chemical Reactions and Stoichiometry” pp 161-162 We have been using chemical equations and formula for a while already. In order to describe a chemical reaction. but these are the main ones that we need to know. When this process occurs. and the substances present at the end are called products. but now we will study the in more depth. sulfur dioxide and oxygen (SO2 and O2) are reactants. 28 . In the following table is a summary of the major symbols used in chemical equations. Writing Chemical Equations When sulfur dioxide is added to oxygen. The substances that are present at the beginning are called reactants. In the chemical equation shown below. and sulfur trioxide (SO3) is the product. In order for this to occur. A chemical reaction is the process in which one or more substances are changed into one or more new substances. Chemical Reactions In a chemical change. new substances are formed. Cu(s) + AgNO3(aq)→ Cu(NO3)2(aq) + Ag(s) The description of this reaction might read something like “solid copper reacts with an aqueous solution of silver nitrate to produce a solution of copper(II) nitrate and solid silver. chemical formulas are used instead of chemical names for reactants and products. They could draw a picture of the chemical reaction. Solution: 29 .Chemists have a choice of methods for describing a chemical reaction. they could describe the reaction in words. however. Chemists could also write the equation in chemical shorthand. It should be apparent that the chemical shorthand method is the quickest and clearest method for writing chemical equations. Look at the following reaction in shorthand notation and describe the reaction in words. 2H2(g) + O2(g)→ 2H2O(g) In the symbolic equation. and symbols are used to indicate the phase of each substance.” Example: Transfer the following symbolic equations into verbal descriptions or vice versa. Alternatively. Gaseous propane. we could simply write: Ca(NO3)2(aq) + 2NaOH(aq) → Ca(OH)2(s) + 2NaNO3(aq) How much easier is that to read? Let’s try it in reverse. 1. HCl(aq) + NaOH(aq)→ NaCl(aq) + H2O(l) 2. burns in oxygen gas to produce gaseous carbon dioxide and liquid water. The image above can be described as two molecules of hydrogen gas reacting with one molecule of oxygen gas to produce two molecules of water vapor. like in the image shown below. liquid water. 3. For example. In shorthand. C3H8. and gaseous carbon dioxide. we could write out that an aqueous solution of calcium nitrate is added to an aqueous solution of sodium hydroxide to produce solid calcium hydroxide and an aqueous solution of sodium nitrate. Hydrogen fluoride gas reacts with an aqueous solution of potassium carbonate to produce an aqueous solution of potassium fluoride. it is understood that only one atom is present. HF(g) + K2CO3(aq)→ KF(aq) + H2O(l) + CO2(g) Even though chemical compounds are broken up to form new compounds during a chemical reaction. the two sides of the equation must be balanced. An aqueous solution of hydrochloric acid reacts with an aqueous solution of sodium hydroxide to produce an aqueous solution of sodium chloride and liquid water. and write a word equation for the reaction. There are subscripts. for the time being. the subscripts may not be changed. and once the formulas for the reactants and products are determined. Even if you saw them react. in a balanced chemical equation. which are part of the chemical formulas of the reactants and products.1. The same atoms that were present in the reactants are present in the products. C3H8(g) + O2(g)→ CO2(g) + H2O(l) 3. There are two types of numbers that appear in chemical equations. In a complete chemical equation. and there are coefficients that are placed in front of the formulas to indicate how many molecules of that substance are used or produced. you would not know what the products are without running any tests to identify them. Balancing Equations The process of writing a balanced chemical equation involves three steps. If only one atom or molecule is present. you will be told both the reactants and products in any equation you are asked to balance. That is. Step 2: Write the formulas for all the reactants and products. you will not know whether or not two given compounds will react or not. Recall that a subscript of 1 is not written . atoms in the reactants do not disappear. In chemical reactions. As a beginning chemistry student. Step 1: Know what the reactants and products are. The atoms are merely re-organized into different arrangements. The coefficients indicate the mole ratios of each substance involved in the 30 . nor do new atoms appear to form the products. The equation above indicates that one mole of solid copper is reacting with two moles of aqueous silver nitrate to produce one mole of aqueous copper(II) nitrate and two moles of solid silver. Step 3: Adjust the coefficients to balance the equation. Therefore. In the chemical formula shown below. the coefficients and subscripts are labeled. The subscripts are part of the formulas.when no subscript appears for an atom in a formula. 2. the coefficient of 1 is omitted. the same number of each atom must be present on the reactant and product sides of the equation. atoms are never created or destroyed. The same is true in writing balanced chemical equations. we can balance the bromine atoms. Al2(SO4)3 + CaBr2 → 2AlBr3 + CaSO4 31 . chlorine + sodium bromide yields bromine + sodium chloride Step 2: Substitute the correct formulas into the equation. never by changing subscripts. Example: Write a balanced equation for the reaction that occurs between chlorine gas and aqueous sodium bromide to produce liquid bromine and aqueous sodium chloride. By placing a coefficient of 2 in front of NaBr. Cl2 + 2NaBr → Br2 + 2NaCl A final check (always do this) shows that we have the same number of each atom on the two sides of the equation. Cl2 + NaBr → Br2 + NaCl Step 3: Insert coefficients where necessary to balance the equation. In order to balance the aluminum atoms. Coefficients are inserted into the chemical equation in order to make the total number of each atom on both sides of the equation equal. we can balance the chlorine atoms. Step 1: Write the word equation (keeping in mind that chlorine and bromine refer to the diatomic molecules). Al2(SO4)3 + CaBr2 → AlBr3 + CaSO4 Step 3: Insert coefficients to balance the equation. aluminum sulfate + calcium bromide yields aluminum bromide + calcium sulfate Step 2: Replace the names of the substances in the word equation with formulas.reaction and may be changed in order to balance the equation. Example: Write a balanced equation for the reaction between aluminum sulfate and calcium bromide to produce aluminum bromide and calcium sulfate. we must insert a coefficient of 2 in front of the aluminum compound in the products. Step 1: Write the word equation. so this equation is properly balanced. Note that equation balancing is accomplished by changing coefficients. We have also used the smallest whole numbers possible as the coefficients. Recall that polyatomic ions usually remain together as a unit throughout a chemical reaction. By placing a coefficient of 2 in front of the NaCl. ) Fe(NO3)3 + NaOH → Fe(OH)3 + NaNO3 (skeletal equation) Solution: We can balance the hydroxide ion by inserting a coefficient of 3 in front of the NaOH on the reactant side. you should not only check to make sure it is balanced. Since it is required that an equation be balanced with the lowest whole number coefficients. three calcium atoms. Al2(SO4)3 + 3CaBr2 → 2AlBr3 + 3CaSO4 The insertion of the 3 in front of the reactant CaBr2 also balances the calcium atoms in the product CaSO4. We could divide each of the coefficients in this equation by 2 to get another set of coefficients that still balance the equation and are whole numbers. and six bromine atoms on each side. When you have finished balancing an equation.In order to balance the sulfate ions. we must insert a coefficient of 3 in front of the product CaSO4. Chemical equations should be balanced with the simplest whole number coefficients.” While this set of coefficients does “balanced” the equation. This equation is balanced. (The term “skeletal equation” refers to an equation that has the correct chemical formulas but does not include the proper coefficients. the equation above is not properly balanced. Al2(SO4)3 + CaBr2 → 2AlBr3 + 3CaSO4 In order to balance the bromine atoms. twelve oxygen atoms. Fe(NO3)3 + 3 NaOH → Fe(OH)3 + NaNO3 We can then balance the nitrate ions by inserting a coefficient of 3 in front of the sodium nitrate on the product side. three sulfur atoms. we must insert a coefficient of 3 in front of the reactant CaBr2. Note that this equation would still have the same number of atoms of each type on each side with the following set of coefficients: 2Al2(SO4)3 + 6CaBr2 → 4AlBr3 + 6CaSO4 Count the number of each type of atom on either side of the equation to confirm that this equation is “balanced. Fe(NO3)3 + 3 NaOH → Fe(OH)3 + 3NaNO3 Counting the number of each type of atom on the two sides of the equation will now show that this equation is balanced. you should also check to make sure that it is balanced with the simplest set of whole number coefficients possible. A final check shows that there are two aluminum atoms. 32 . Example: Balance the following skeletal equation. they are not the lowest set of coefficients possible. 0 daltons) = 255 daltons Total mass of products = 106. and combustion. balance them. Ba(NO3)2(aq) + Na2CO3(aq) → BaCO3(aq) + NaNO3(aq) 4. Chemical reactions are classified into types to help us analyze them and to help us predict what the products of the reaction will be.9 daltons + 120. although we may not actually observe a composition reaction.9 daltons 3 molecules of NaOH × molecular weight = (3) · (40.9 daltons 3 molecules of NaNO3 × molecular weight = (3) · (85. Stoichiometry involved calculating the quantities of reactants or 33 . We will look at composition and decomposition reactions in the next lab. In the following weeks we will study single and double replacement reactions.Pre-Lab Problems: Given the following skeletal (un-balanced) equations. This is logically similar to saying that a group of 20 objects stacked in different ways will still have the same total mass no matter how you stack them.9 daltons) = 106. Fe(NO3)3 + 3 NaOH → Fe(OH)3 + 3 NaNO3 Verify to yourself that this equation is balanced by counting the number of each type of atom on each side of the equation.9 daltons) = 241.9 daltons As you can see.9 daltons Product Side Mass: 1 molecule of Fe(OH)3 × molecular weight = (1) · (106. Mass of the Reactant Side: 1 molecule of Fe(NO3)3 × molecular weight = (1) · (241.9 daltons + 255 daltons = 361. Next we will explore the quantitative relationships that exist between the reactants and products in a balanced equation. decomposition. both the number of atoms and mass are conserved during chemical reactions. The five major types of chemical reactions are synthesis. daltons = 361. H2SO4(aq) + Al(OH)3(aq) → Al2(SO4)3(aq) + H2O(l) 3. single replacement. C2H6(g) + O2(g) → CO2(g) + H2O(l) Conservation of Mass in Chemical Reactions We already know from the law of conservation of mass that mass is conserved in chemical reactions. CaCO3(s) → CaO(s) + CO2(g) 2. double replacement. daltons Total mass of reactants = 241.0 daltons) = 120. 1. This is known as stoichiometry. We can demonstrate that mass is conserved by determining the total mass on both sides of the equation. But what does this really mean? Consider the following reaction. we can see that the coefficient in front of the hydrogen is 2. and metron. while the coefficient in front of water is also 2. which means measure. the mole ratio can be written as: or Similarly. the ratio of oxygen molecules to water molecules would be: or In the following example.” Example: Four moles of solid aluminum are mixed with three moles of gaseous oxygen to produce two moles of solid aluminum oxide. mole ratio of aluminum to oxygen = or 2. mole ratio of oxygen to aluminum oxide = or 34 . which means element. one way we could read the following reaction is that 2 moles of H2(g) react with 1 mole of O2(g) to produce 2 moles of H2O(l). Mole Ratios A mole ratio is the relationship between two components of a chemical reaction. and (3) oxygen to aluminum oxide? Solution: Balanced chemical equation: 4 Al(s) + 3 O2(g) → 2 Al2O3(s) 1. mole ratio of aluminum to aluminum oxide = or 3. What is the mole ratio of (1) aluminum to oxygen. (2) aluminum to aluminum oxide. let’s try finding the mole ratios by first writing a balanced chemical equation from a “chemical sentence.products in a chemical reaction using the relationships found in the balanced chemical equation. The word stoichiometry actually comes from two Greek words: stoikheion. For instance. 2 H2(g) + O2(g) → 2 H2O(l) The mole ratio of H2(g) to O2(g) would be: or What is the ratio of hydrogen molecules to water molecules? By examining the balanced chemical equation. Therefore. the equation must be balanced. H3PO4. while what we know is in the denominator. Looking at the unbalanced equation for the reaction of phosphorus trihydride with oxygen. is: Keep in mind that before any mathematical calculations are made relating to a chemical equation. we need to do calculations using mole ratios.50 moles of magnesium hydroxide. however. Therefore. find the mole ratios for (1) calcium carbide to water and (2) calcium carbide to calcium hydroxide. When written. we rarely work with exactly one mole of a chemical. would be required for the reaction? 2 H3PO4 + 3 Mg(OH)2 → Mg3(PO4)2 + 6 H2O Step 1: To determine the conversion factor. how many moles of phosphoric acid. the chemical equation must always be balanced before the mole ratios are used for calculations. PH3(g) + O2(g) → P4O10(s) + H2O(g) Once the equation is balanced. mole ratio of calcium carbide to calcium hydroxide = or The correct mole ratios of the reactants and products in a chemical equation are determined by the balanced equation. the conversion factor is: mole ratio = Note that what we are trying to calculate is in the numerator. it is difficult to guess the correct mole ratio of phosphorus trihydride to oxygen gas.Example: Write the balanced chemical equation for the reaction of solid calcium carbide (CaC2) with water to form aqueous calcium hydroxide and acetylene (C2H2) gas. the mole ratio of phophorus trihydride to oxygen gas is apparent. Mg(OH)2. 35 . Therefore. Balanced chemical equation: 4 PH3(g) + 8 O2(g) → P4O10(s) + 6 H2O(g) The mole ratio of phophorus trihydride to oxygen gas. Look at the following equation. Mole-Mole Calculations In the chemistry lab. Solution: Balanced chemical equation: CaC2(s) + 2 H2O(l) → Ca(OH)2(aq) + C2H2(g) 1. are present. If only 0. then. In order to determine the amount of reagent (reacting chemical) necessary or the amount of product expected for a given reaction. mole ratio of calcium carbide to water = or 2. we want to convert from moles of Mg(OH)2 to moles of H3PO4. (0. we would need 0. Step 1: Write and balance the chemical equation. the amount of H3PO4 required would have been calculated incorrectly. and we would have concluded that 0.5 mol of H3PO4 were required. 36 .33 mol H3PO4 Therefore.42 mol of carbon disulfide. Step 4: Convert the moles of unknown to the requested units.33 mol of H3PO4 to react with all of the magnesium hydroxide. Using Proportion to Solve Stoichiometry Problems All methods for solving stoichiometry problems contain the same four steps. CS2.36 mol NaNO3) · ) = 7. The ratio would have been 1:1. Step 3: Convert the moles of known to moles of unknown.08 mol Na2O Example: How many moles of sulfur are required to produce 5.42 mol CS2) ·( ) = 10. if we have 0.84 mol S Mass-Mass Calculations A mass-mass calculation would allow you to solve one of the following types of problems: Determine the mass of reactant necessary to product a given amount of product Determine the mass of product that would be produced from a given amount of reactant Determine the mass of reactant necessary to react completely with a second reactant As was the case for mole ratios.Step 2: Use the conversion factor to answer the question. using the balanced chemical equation below? C + 2 S → CS2 Solution: (5. Notice if the equation was not balanced.50 mol Mg(OH)2) · = 0. Step 2: Convert the given quantity to moles.50 mol of Mg(OH)2. it is important to double check that you are using a balanced chemical equation before attempting any calculations. Example: How many moles of sodium oxide (Na2O) can be formed from 2.36 mol of sodium nitrate (NaNO3) using the balanced chemical equation below? 10 Na + 2 NaNO3 → 6 Na2O + N2O Solution: (2. we want to determine the x moles of oxygen needed to react with 0. We can then solve the proportion by multiplying the denominator from the left side to both sides of the equal sign. is already in moles. usually grams.50 mole of hydrogen. To find the number of moles in x grams of a substance. reacts with oxygen gas to produce carbon dioxide and water. Note that if some other unit of mass is used. The balanced equation below shows the reaction between hydrogen gas and oxygen gas to produce water. we will be solving “mass-mass problems. we multiply by the molar mass. but not for grams.50 mol C5H12 37 . C5H12 + 8 O2 → 5 CO2 + 6 H2O Step 2: Convert the given quantity to moles. we divide by the molar mass. In this case. We set up a proportion to help complete Step 3. If you were to write out a mathematical equation to describe this problem. In this section.Step 1 has been covered in previous sections. Similarly. Since the equation is already balanced.50 moles of hydrogen. This is because the molar ratios keep the same proportion. 2 H2(g) + O2(g) → 2 H2O(l) The molar ratio in this equation is two moles of hydrogen react with one mole of oxygen to produce two moles of water. Since you are using a quarter as much hydrogen. The set up proportion would be similar to the one above.” which means that both the given value and the requested answer will both be in units of mass. Example: Pentane. We also just saw how to complete Step 3 in the previous section by using mole ratios. you should convert to grams first. so let’s look at some sample problems. we simply need to know how to convert between moles and the given or requested units. Step 1 is already completed. From the balanced equation. and use that value for further calculations. This process is best illustrated through examples. and to go back from moles to grams. How many grams of carbon dioxide will be produced by the reaction of 108. Remember that the coefficients in the balanced equation are true for moles or molecules. you would also know the moles of oxygen required for the reaction and the moles of water that would be produced. The conversion factor between grams and moles is the molar mass (g/mol). we know that 1 mole of oxygen reacts with 2 moles of hydrogen. In order to complete the remaining two steps.00 moles of hydrogen in this reaction.0 grams of pentane? Step 1: Write and balance the equation. you would need a quarter as much oxygen and produce a quarter as much water.25 moles of O2.50 mole of hydrogen. so Step 2 is also completed. you will find that x = 0. C5H12. you would set up the following proportion: = The given quantity. If you were told that you were going to use 2. = 1. It is only slightly more difficult to determine the moles of oxygen required and the moles of water produced if you were told that you will be using 0. 0. = 38 . x mol Al(OH)3 = 2.50.0 g/mol) = 330 grams Example: Aluminum hydroxide reacts with sulfuric acid to produce aluminum sulfate and water.00 mol) · (78. grams Al(OH)3 = (2. 2 Al(OH)3 + 3 H2SO4 → Al2(SO4)3 + 6 H2O Step 2: Convert the given quantity to moles. x mol CO2 = 7.0 g/mol) = 156 grams Example: 15.Step 3: Set up and solve the proportion to find moles of unknown. = Therefore. = 6. grams CO2 = (7. Step 4: Convert the moles of unknown to grams. How many grams of aluminum hydroxide are necessary to produce 108 grams of water? Step 1: Write and balance the equation. Cl2(g) + 2 S(l) → S2Cl2(l) Step 2: Convert the given quantity to moles. How much product is produced in grams? Step 1: Write and balance the chemical equation. = 0.00 mol H2O Step 3: Set up and solve the proportion to find moles of unknown.0 grams of chlorine gas is bubbled through liquid sulfur to produce liquid disulfur dichloride. Step 4: Convert the unknown moles to requested units (grams).50 mol) · (44.00.212 mol Step 3: Set up and solve the proportion to find moles of unknown. = Therefore. how much carbon dioxide is produced? Step 1: Write and balance the chemical equation. g of iron is produced in the reaction. 39 . = Therefore. grams Fe2O3 = (4. how much iron(III) oxide was used as reactant? Step 1: Write and balance the chemical equation.48 mol) · (160. Step 4: Convert the moles of unknown to grams. The reaction releases enough heat to melt the iron that is produced. If 200 g of ibuprofen is combusted.Therefore. Fe2O3(s) + 2 Al(s) → 2 Fe(l) + Al2O3(s) Step 2: Convert the given quantity to moles. If 500. = Therefore.967 mol Step 3: Set up and solve the proportion to find moles of unknown. = 0. It has the formula C13H18O2.212 mol) · (135 g/mol) = 28.6. grams S2Cl2 = (0. x mol Fe2O3 = 4. 2 C13H18O2(s) + 33 O2(s) → 26 CO2(g) + 18 H2O(l) Step 2: Convert the given quantity to moles. g/mol) = 717 grams Example: Ibuprofen is a common painkiller used by many people around the globe. x mol CO2 = 12. x mol S2Cl2 = 0. = 8.95 mol Step 3: Set up and solve the proportion to find moles of unknown.48. Step 4: Convert the moles of unknown to grams.6 grams Example: A thermite reaction occurs between elemental aluminum and iron(III) oxide to produce aluminum oxide and elemental iron.212. youtube.463 mol Step 3: Set up and solve the proportion to find moles of unknown.1 g/mol) = 22.com/watch?v=EdZtSSJecJc (9:21). grams CO2 = (12.6 mol) · (44. How much sulfuric acid must be reacted to produce 12. 268-277) Extra Practice Problems: Balance the following equations: 1) 2) 3) 4) 5) ___ N2 + ___ F2 ___ NF3 ___ C6H10 + ___ O2 ___ CO2 + ___ H2O ___ HBr + ___ KHCO3 ___ H2O + ___ KBr + ___ CO2 ___ GaBr3 + ___ Na2SO3 ___ Ga2(SO3)3 + ___ NaBr ___ SnO + ___ NF3 ___ SnF2 + ___ N2O3 40 .0 g/mol) = 554 grams Example: If sulfuric acid is mixed with sodium cyanide. Step 4: Convert the moles of unknown to grams. 2 NaCN(s) + H2SO4(aq) → Na2SO4(aq) + 2 HCN(g) Step 2: Convert the given quantity to moles.232 mol) · (98.5 grams of hydrogen cyanide? Step 1: Write and balance the chemical equation. grams H2SO4 = (0.7 grams A blackboard type discussion of stoichiometry (3e) is available at http://www. = Therefore. (from CK12 Chemistry 2nd Edition pp 250-258. the deadly gas hydrogen cyanide is produced.232. x mol H2SO4 = 0.Step 4: Convert the moles of unknown to grams. = 0. answer the following questions: 6) If I do this reaction with 35 grams of C6H10 and 45 grams of oxygen. 2 H2(g) + O2(g) → 2 H2O(l) You should also be able to write the chemical equation for a synthesis reaction if you are given a product by picking out its elements and writing the equation. how many grams of carbon dioxide will be formed? (from http://misterguch. Predict the products for the following reaction: Li2O(s) + CO2(g). Write the chemical equation for the synthesis reaction of silver bromide. you should be able to predict the products.html ) Week 9: 9.Using the equation from problem 2 above. 3.brinkster. Two elements (hydrogen and oxygen) combine to form one product (water). 2.net/pra_equationworksheets. If you are given elemental reactants and told that the reaction is a synthesis reaction. 2 Na(s) + Cl2(g) → 2 NaCl(s) Pre-Lab Problems: 1. Predict the products for the following reaction: CO2(g) + H2O(l). we can write the synthesis reaction for sodium chloride just by knowing the elements that are present in the product. consider the equation below. For example.2 Observe a Decomposition Reaction 9. AgBr. The general equation for a synthesis reaction is: A + B → AB Synthesis reactions occur as a result of two or more simpler elements or molecules combining to form a more complex molecule. 41 . As a result. We can always identify a synthesis reaction because there is only one product.1 is skipped as impractical due to smell and toxicity Synthesis Reactions A synthesis reaction is one in which two or more reactants combine to make one product. MgO (from CK12 Chemistry 2nd Edition pp 258-260) Define the following terms in your lab notebook.Decomposition Reactions When one type of reactant breaks down to form two or more products. Al2O3 5. Use the glossary in your Chemistry theory text or a resource such as the McGraw-Hill Dictionary of Chemistry. we have a decomposition reaction. Mg3N2(s) → 3 Mg(s) + N2(g) Notice there is only one reactant. NH4NO3(s) → N2O(g) + 2H2O(g) Notice that there is only one reactant. we can predict the reactants in a similar manner as we did for synthesis reactions. Mg3N2. Pre-Lab Problems: Write the chemical equation for the decomposition of the following compounds into their individual elements: 4. there is only one reactant. The general equation for a decomposition reaction is: AB → A + B Look at the equation below for an example of a decomposition reaction. Now we can write a decomposition reaction for magnesium nitride. Ag2S 6. on the left of the arrow and that there is more than one on the right side of the arrow. When studying decomposition reactions. The best way to remember a decomposition reaction is that for all reactions of this type. ammonium nitrate breaks down to form dinitrogen oxide and water. Look at the formula for magnesium nitride. In this reaction. NH4NO3. This is the exact opposite of a synthesis reaction. What elements do you see in this formula? You see magnesium and nitrogen. chemical reaction products reactants stoichiometry composition reaction decomposition reaction 42 . 4 Stoichiometry of a Double Displacement Reaction Introduction to Lab Weeks 12-13: Reduction-Oxidation (Redox) Reactions Week 12: 10. this will not present a problem for the next group of labs as these experiments are more qualitative in nature and will not involve detailed mathematical determination of reaction rates as taught in most chemistry textbooks. which is that temperature. and concentration all affect the speed of reactions.2: Determine the pH of Aqueous Salt Solutions Week 16: 11. surface area.1 Reduction of Copper Ore to Copper Metal Week 13: 10.3 Observe a Single-Displacement Reaction Week 11: 9.1 Determine the Effects of Concentration on pH Week 15: 11.2 Observe the Oxidation States of Manganese Introduction to Lab Weeks 14-17: Acid-Base Chemistry Week 14: 11. students will not have difficulty understanding the main point.4 Standardize Hydrochloric Acid Solution by Titration Introduction to Lab Weeks 18-19: Chemical Kinetics Although you may not have covered chemical kinetics in the lecture portion of course yet. 43 .Week 10: 9.3 Observe the Characteristics of a Buffer Solution Week 17: 11. By this point. 1 Observe Some Properties of Colloids and Suspensions Week 31: 18.1 Synthesize Methyl Salicylate from Aspirin 44 .3 Determine the Specific Heat of a Metal Week 29: 16.2 Determine the Specific Heat of Ice Week 28: 15.1 Determine the Effects of Temperature on Reaction Rate 12.1 Observe the Volume-Pressure Relationship of Gasses (Boyle's Law) Week 23: 14.1 Quantitative Analysis of Vitamin C by Acid-Base Titration Week 35: 20.1 Determine Heat of Solution Week 27: 15.2 Quantify the Common Ion Effect Week 22: 14.1 Using Flame Tests to Discriminate Metal Ions 19.Week 18: Determine Effects of Temperature and Surface Area on Reaction Rate 12.3 Prepare a Gelled Sol Week 33: Discriminate Metal Ions 19.1 Observe Le Chatelier's Principle in Action Week 21: 13.2 Determine the Effects of Surface Area on Reaction Rate Week 19: 12.2 Produce Firefighting Foam Week 32: 18.4 Use the Ideal Gas Law to Determine the Percentage of Acetic Acid in Vinegar Week 26: 15.1 Produce Hydrogen and Oxygen by Electrolysis of Water Week 30: 18.3 Determine the Effects of Concentration on Reaction Rate Week 20: 13.2 Using Borax Bead Tests to Discriminate Metal Ions Week 34: 20.3 Observe the Pressure -Temperature Relationship of Gases (GayLussac's Law) Week 25: 14.2 Quantitative Analysis of Chlorine Bleach by Redox Titration Week 36: 21.2 Observe the Volume-Temperature Relationship of Gasses (Charles's Law) Week 24: 14. Solutions to Pre-Lab Problems Week 6: Solutions of Solid Chemicals 1) What is the molarity of a 100 ml solution prepared by dissolving 30 g of NaCl? 2) A laboratory assistant prepares to make 0.07 g/ml.45 g) Molar mass of NaCl = 23 g + 35. He weighs 20 g of NaCl crystals.9 g in 1000 ml solution To make 500 ml of 2 M NaCl solution.45 g NaCl 45 . If the density of this glucose solution is 1. atomic mass of Cl = 35. 1) What is the molarity of this solution? 2) What is the molality of this solution? Week 7: Solution of Liquid Chemicals and Mass-to-Volume Percentage 1) A student at a chemical laboratory wants to make 500 ml of a 2 M NaCl solution.45 g = 58. What mass of NaCl will the student have to weigh? ( atomic mass of Na = 23 g .45 g 1 M = 1 mole solute in 1000 ml solution = 58. What is the volume of NaCl solution he will make? 3) 18 g glucose (C6H12O6) is dissolved in 100 g of water.9 g / 2 =58.154 M NaCl solution. the student has to weigh 116.45 g in 1000 ml solution = 116.45 g NaCl in 1000 ml solution 2 M NaCl = 2 x 58. 600 C. When you do this calculation.600 C.52 0C/m and Kf(H2O) = 1.86 0C/m. What is the molarity of this new solution? Total volume of new solution = 100 ml + 50 ml = 150 ml 2 M x 100 ml + 3.2 M HCl solution.2) Student A has 100 ml of 2 M HCl solution. the effective molality for the purposes of colligative property calculations is twice this. you’ll first need to find the concentration of the solution.500 kg H2O).4 M 3) A concentrated HCl solution is 36% by mass and has a density of 1. The final answers: Melting point = -5.54 m. 1) What is the molarity of this concentrated HCl? 2) What volume of concentrated HCl is needed to make 5 liters of 1 M HCl solution? 3) What volume of H2O is needed to make 100 ml of 4 M HCl? Week 8: Colligative Properties of Solutions 1) If I add 45 grams of sodium chloride to 500 grams of water. and student B has 50 ml of 3. When you do this. what will the melting and boiling points be of the resulting solution? Kb(H2O) = 0. 46 . To find the melting and boiling points. To find the molality. it’s a matter of plugging the molality into the equations for freezing point depression and boiling point elevation. student C combines these two solutions. Accidentally. or 3.730 C and the change in boiling point is 1. Because sodium chloride breaks into two particles when it dissociates in water.18 g/ml.769 moles) and divide by the number of kilograms of solvent (0. you should get that the change in melting point is 5. convert grams of sodium chloride to moles (it turns out to be 0. the molality is 1. From here.2 M x 50 ml = y M x 150 ml y = 2.08 m.730 C. boiling point = 101. 47 . you find that 113 grams of CO2 will be formed. When you do the calculation for 45 grams of oxygen. 3H2SO4(aq) + 2Al(OH)3(aq) → Al2(SO4)3(aq) + 6H2O(l) 3.400 C.32 0 C and the boiling point of the sodium chloride solution is 101. 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l) Week 9: 9. Clearly. CO2(g) + H2O(l) → H2CO3(aq) 3. you find that the boiling point of the sucrose solution is 100. CaCO3(s) → CaO(s) + CO2(g) (sometimes the skeletal equation is already balanced) 2. the sodium chloride solution has a higher boiling point. you find that 43. Because 43.2) Which solution will have a higher boiling point: A solution containing 105 grams of sucrose (C12H22O11) in 500 grams of water or a solution containing 35 grams of sodium chloride in 500 grams of water? Using the same calculations as in question 1. Ba(NO3)2(aq) + Na2CO3(aq) → BaCO3(aq) + 2 NaNO3(aq) 4. 2 Ag(s) + Br2(l) → 2 AgBr(s) 2. Introduction to Lab Weeks 9-11: Intro to Chemical Reactions and Stoichiometry 1. 2 Al2O3 → 4 Al + 3 O2 5. answer the following questions: 6) If I do this reaction with 35 grams of C6H10 and 45 grams of oxygen. oxygen is the limiting reagent.7 grams is the smaller number. Ag2S → 2 Ag + S 6.7 grams of CO2 will be formed. forming 43.2 Observe a Decomposition Reaction 1. 2 MgO → 2 Mg + O2 Extra Practice Problems Balance the following equations: 1) 2) 3) 4) 5) 1 N2 + 3 F2 2 NF3 2 C6H10 + 17 O2 12 CO2 + 10 H2O 1 HBr + 1 KHCO3 1 H2O + 1 KBr + 1 CO2 2 GaBr3 + 3 Na2SO3 1 Ga2(SO3)3 + 6 NaBr 3 SnO + 2 NF3 3 SnF2 + 1 N2O3 Using the equation from problem 2 above. how many grams of carbon dioxide will be formed? When you do this calculation for 35 grams of C6H10.7 grams of product. Li2O(s) + CO2(g) → Li2CO3(s) 4. 48 .
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