MAD SCIENCE HOMESCHOOL PRODUCTIONS

Lab 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

1

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".)

2

Lab Session Procedures

Before the First Lab Session Familiarize yourself with the style of this lab manual. Each week and/or experiment begins with any required

reading in the lab text, Illustrated Guide to Home Chemistry, 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. The lab itself is not specifically assigned, but must also be read prior to

the lab session. Take special note of warnings and hazards associated with the lab and any chemicals used or

created in the process, as well as disposal instructions. Safety equipment should be worn by all participants and

observers in the labs at all times, and contact lenses should not be worn during lab sessions.

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. Suggestions for lab safety quizzes are at the end of that section.

Before Each Lab Session Review lab safety procedures. Read the section of this manual for the planned lab session and all assigned

readings in that section as well as the lab itself. Prepare a flow chart of the lab. Perform all calculations needed

prior to the start of the lab, such as quantities of particular compounds needed to create solutions of specified

concentrations. One student may be assigned to prepare a flowchart, discussion, and safety information for

each planned lab in the session to present to the group prior to beginning the lab. This presentation should not

take more than 5-10 minutes. However, 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.

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. This will also greatly speed up completion of lab

reports if they are assigned by the student’s instructor.

Any pre-lab problems should also be completed prior to the lab. Suggested problems are provided to ensure

student proficiency with quantitative methods used in the lab. In some cases worksheets may be provided in

the manual. These should be done after the problems assigned. Answers for worksheets are in the back of this

manual. 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.

3

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. However, by the very nature of laboratory

work, it can be a place of danger if proper common-sense precautions aren't taken. While every effort has been

made to eliminate the use of explosive, highly toxic, 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. 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. Additional safety

precautions will be announced in class prior to experiments where a potential danger exists. Students who fail to

follow all safety rules may endanger themselves, their siblings, their parent instructors, and the laboratory space

in which we are working.

4

Attire

Safety goggles must be worn at all times while in the laboratory. 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.

Contact lenses are not allowed. Even when worn under safety goggles, various fumes may accumulate

under the lens and cause serious injuries or blindness.

Closed toe shoes and long pants must be worn in the lab. Sandals and shorts are not allowed. Lab

aprons must be worn at all times. Old T-shirts are suggested as dedicated labwear underneath lab

aprons.

Chemical resistant lab gloves must be worn when working with or observing experiments involving

hazardous chemicals.

Long hair must be tied back at all times in the laboratory space, especially when using open flames.

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. 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.

No handstands, cartwheels, mock fighting, or physical joking around of any kind is permitted in the lab.

No unauthorized experiments are to be performed. If you are curious about trying a procedure not

covered in the experimental procedure, consult with your laboratory instructor.

Never taste anything. Never smell anything unless instructed to do so. Never directly smell the source

of any vapor or gas; instead by means of your cupped hand, waft a small sample to your nose and ONLY

when instructed to do so. Do not inhale these vapors but take in only enough to detect an odor if one

exists.

Coats, backpacks, etc., should not be left on the lab surfaces or nearby floors. Put these safely in

another room. Make sure no tripping hazards are in the lab space or exits from the lab space. Beware

that lab chemicals can destroy personal possessions.

Always wash your hands before leaving lab, even just to go to the bathroom.

Learn where the safety and first-aid equipment is located. This includes fire extinguishers, fire blankets,

and eye-wash stations.

Notify the instructor immediately in case of an accident.

Proper Handling of Chemicals and Equipment

Consider all chemicals to be hazardous unless you are instructed otherwise. Material Safety Data Sheets

(MSDS) are available in lab or via internet for all chemicals in use. These will inform you of any hazards

and precautions of which you should be aware.

Never pipette by mouth.

Know what chemicals you are using. Carefully read the label twice before taking anything from a bottle.

Chemicals in the lab are marked with NFPA hazardous materials diamond labels. Learn how to interpret

these labels. Appropriately label any stock solutions made during lab or pre-lab exercises.

Excess reagents are never to be returned to stock bottles. If you take too much, dispose of the excess.

5

Many common reagents, for example, alcohols and acetone, are highly flammable. Do not use them

anywhere near open flames.

Always pour acids into water. If you pour water into acid, the heat of reaction will cause the water to

explode into steam, sometimes violently, and the acid will splatter.

If chemicals come into contact with your skin or eyes, flush immediately with copious amounts of water

and consult with your instructor.

Never point a test tube or any vessel that you are heating at yourself or your neighbor--it may erupt like

a geyser.

Dispose of chemicals properly. Follow instructions in our lab manual for disposal of waste and chemicals

used in the lab under adult supervision.

Clean up all broken glassware immediately and dispose of the broken glass properly with adult

assistance and supervision.

Consult with instructor for clean-up of any chemical spills. Some spills, such as acids, should be

neutralized prior to wipe up and disposal.

Never leave burners unattended. Turn them off whenever you leave your workstation. Be sure that the

gas is shut off at the bench rack when you leave the lab.

Beware of hot glass--it looks exactly like cold glass.

Adapted from: http://chemlabs.uoregon.edu/Safety/GeneralInstructions.html

Safety Exams: http://www.flinnsci.com/Documents/miscPDFs/Safety_exam_HS.pdf

http://www.sciencebyjones.com/lab_safety_quiz.htm

Materials Data Sheets: http://www.flinnsci.com/search_MSDS.asp

6

Lab Experiments

Introduction to Lab Weeks 1-5: Techniques for Separating Mixtures The first group of labs, 6.1 – 6.5, familiarizes students with basic laboratory techniques to separate mixtures.

Students are not expected to have detailed knowledge of the theory behind these techniques, nor are involved

quantitative methods required. Background knowledge from previous general, physical, 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. Younger siblings observing can be expected to follow along sufficiently to

appreciate larger scientific principles and laboratory techniques involved.

Week 1: 6.1 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. When a new substance is synthesized, for example, the new

substance usually must be separated from various side-products, catalysts, and any excess reagent still present.

When a substance must be isolated from a natural biological source, the substance of interest I generally found

in a very complex mixture with many other substances, all of which must be removed. Chemists have developed

a series of standard methods for the separation of mixtures. In fact, the separation of mixtures into their

constituent substances defines an entire sub-field of chemistry referred to as separation science.

Differing Solubilities

Mixtures of solids may often be separated on the basis of differences in their solubilities. If one component of

the mixture is soluble in water while the other components are insoluble in water, the water-soluble component

can be removed from the mixture by dissolving the mixture in water and filtering the mixture through filter

paper. The component dissolved in water will pass through the filter while the undissolved solids will be caught

in the filter. Since the solubility of substances is greatly influenced by temperature, 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. Often times, a sample is added to water and heated to boiling. The hot sample is then filtered to

remove completely insoluble substances. The sample is then cooled to room temperature or below, which

causes crystallization of those substances whose solubilities are very temperature dependent. These crystals can

then be separated by another filtration, and the filtrate (the material that went through the filter) will then

contain only those substances whose solubilities are not as temperature dependent.

(From CK12 Chemistry 2nd Edition p 408)

Define the following words in your lab notebook. Use the glossary in your Chemistry theory text or a resource

such as the McGraw-Hill Dictionary of Chemistry.

dissolve

solution

soluble

solubility

7

Week 2: 6.2 Distillation: Purify Ethanol Read: “Measuring Liquids by Volume” pp 74-78 (Burette's will be covered later.)

“Using Heat Sources” p85-88"

Successful Lab Reports: Revising Your Paper pp 53-61

Distillation

Homogeneous solutions are most commonly separated by distillation.

In general, distillation involves heating a liquid to its boiling point, then

collecting, cooling, and condensing the vapor produced into a separate

container. A common distillation setup is illustrated to the right.

In solutions of non-volatile (resistant to vaporization) solid solutes in

liquid solvent, when the solution is boiled, only the solvent boils off and

all of the solid remains in the solution. As the solvent vaporizes and all

of the solute remains behind, the same amount of solute is now

dissolved in less solvent. Since the concentration increases, the boiling

point of the solution is also increasing. As the solution boils, increased

temperature is necessary to keep the solution boiling because its boiling point has increased. 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, it is a pure substance, whereas if its boiling point increases, it is a solution.

For a mixture of liquids in which several components of the mixture are likely to be volatile (easily vaporized),

the separation is not as easy. If the components of the mixture differ reasonably in their boiling points, it may be

possible to separate the mixture simply by monitoring the temperature of the vapor produced as the mixture is

heated. Liquid components of a mixture will each boil in turn as the temperature is gradually increased, with a

sharp rise in the temperature of the vapor being distilled indicating when a new component of the mixture has

begun to boil. By changing the receiving flask at the correct moment, a separation can be accomplished. This

process is known as fractional distillation.

(From CK12 Chemistry 2nd Edition pp 408-409)

Define the following words in your lab notebook. Use the glossary in your Chemistry theory text or a resource

such as the McGraw-Hill Dictionary of Chemistry.

solute

solvent

volatile

distillation

distillate

8

Week 3: 6.3 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. It is one of the fundamental

procedures each chemist must master to become proficient in the laboratory. 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. If a saturated hot solution is allowed to cool, the solute is no longer soluble in the solvent and forms

crystals of pure compound. Impurities are excluded from the growing crystals and the pure solid crystals can be

separated from the dissolved impurities by filtration.

This simplified scientific description of crystallization does not give a realistic picture of how the process is

accomplished in the laboratory. Rather, successful crystallization relies on a blend of science and art; its success

depends more on experimentation, observation, imagination, and skill than on mathematical and physical

predictions. Understanding the process of crystallization in itself will not make a student a master crystallizer,

rather, this understanding must be combined with laboratory practice to gain proficiency in this technique.

How to do a crystallization

To crystallize an impure, solid compound, add just enough hot solvent is added to it to completely dissolve it.

The flask then contains a hot solution, in which solute molecules – both the desired compound and impurities –

move freely among the hot solvent molecules. As the solution cools, the solvent can no longer “hold” all of the

solute molecules, and they begin to leave the solution and form solid crystals. During this cooling, each solute

molecule in turn approaches a growing crystal and rests on the crystal surface. If the geometry of the molecule

fits that of the crystal, it will be more likely to remain on the crystal than it is to go back into the solution.

Therefore, each growing crystal consists of only one type of molecule, the solute. After the solution has come to

room temperature, it is carefully set in an ice bath to complete the crystallization process. The chilled solution is

then filtered to isolate the pure crystals and the crystals are rinsed with chilled solvent.

Detailed photos from start to finish: http://orgchem.colorado.edu/hndbksupport/cryst/crystproc.html

Crystallization Movie: http://video.google.com/videoplay?docid=5617753349611003526 On GoogleVideo -

choose "smoothing" and "original size" from the lower right pull-down menu for best video.

Source: http://orgchem.colorado.edu/hndbksupport/cryst/cryst.html on March 22, 2011

Define the following word in your lab notebook. Use the glossary in your Chemistry theory text or a resource

such as the McGraw-Hill Dictionary of Chemistry.

recrystallization

9

Week 4: 6.4 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. The "liquid-liquid" phrase means that two liquids are mixed in the extraction

procedure. The liquids must be immiscible: this means that they will form two layers when mixed together, like

oil and vinegar do in dressing. (A demonstration of this fact using red wine or balsamic vinegar and oil may be

useful for students and/or younger siblings.) Some compounds are more soluble in the organic layer (the "oil")

and some compounds are more soluble in the aqueous layer (the "vinegar"). Recall last year’s discussion in

Biology of lipid bilayers in cell membranes and cell membrane permeability.

The photo at the right illustrates how two liquid layers separate. The red layer is

simply red food coloring in water. Water is immiscible with the other liquid, which is

methylene chloride. Methylene chloride is heavier (denser) than water, therefore, the

clear methylene chloride layer is under the red, aqueous food coloring layer.

Quiz: Red food coloring is soluble in water and not in methylene chloride or diethyl

ether. The photo at the right illustrates what you see if you mix methylene chloride,

water, and red food coloring and allow the layers to separate. What would you expect

to see if you mixed diethyl ether, water, and red food coloring? Go to the next page

for the answer.

In a particular experiment in simple extraction or in chemically active extraction, you

will be able to figure out which layer, aqueous or organic, will contain the compound

you want to isolate. You will also need to know which layer will be on top in the

separatory funnel. This is determined by the density of the two solvents. Densities are

listed in various sources of scientific data, as referenced on the Chem Info page on this

orgchem site:

Hazard and Physical Data for Compounds: http://orgchem.colorado.edu/cheminfo/cheminfo.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.colorado.edu/hndbksupport/ext/extprocedure.html

Extraction Movie: http://video.google.com/videoplay?docid=945061828857075635&hl=en On GoogleVideo -

choose "smoothing" and "original size" from the lower right pull-down menu for best video.

As we do have a separatory funnel, we will not be following adapted procedures in The Illustrated Guide to

Home Chemistry that would not allow quite as good a separation.

10

Quiz Answer: In order to make a prediction as to which layer is on the top and which on the bottom, you need

to know the densities. The following data is from the Tables in the Handbook:

diethyl ether d = 0.71

methylene chloride d = 1.33

water d = 1.00

Therefore, if you mix diethyl ether, water, and red

food coloring, shake it in a sep funnel and allow the

layers to separate, it will look like the mixture in the

sep funnel on the left in the photo below (big yellow

check mark). The food color remains dissolved in the

water; since water is more dense than diethyl ether,

the water will be on the bottom.

Compare with the separated mixture of water, food

color, and methylene chloride in the separatory

funnel on the right.

Determining densities: D=M/V so you could measure

the mass of a measured volume (weigh volumetric

glassware, measure volume into glassware, weigh

glassware with solvent) and divide to determine

density.

Adapted From:

http://orgchem.colorado.edu/hndbksupport/ext/ext.

html on March 22, 2011

Define the following words in your lab notebook. Use the glossary in your Chemistry theory text or a resource

such as the McGraw-Hill Dictionary of Chemistry.

immiscible

extraction

11

Week 5: 6.5 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. The word chromatography means colorwriting. The

name was chosen around 1900 when the method was first used to separate colored components from plant

leaves. Chromatography in its various forms is perhaps the most important known method for the chemical

analysis of mixtures. Paper and thin-layer chromatography are simple techniques that can be used to separate

mixtures into the individual components. The methods are very similar in operation and principle. They differ

primarily in the medium used.

Paper chromatography uses ordinary filter paper as the medium upon which the mixture to be separated is

applied. 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. 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. The filter paper or TLC

slide is then placed in a shallow layer of solvent in a jar or beaker. Since the filter paper and the TLC slide coating

are permeable to liquids, the solvent begins rising up the paper by capillary

action. As the solvent rises to the level of the mixture spot, various effects

can occur, depending on the constituents of the spot. 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. Those components that are

not at all soluble will be left behind at the original location of the spot.

Most components of the mixture will move up the paper or slide at an

intermediate speed somewhat less than the solvent front speed. In this

way, the original mixture spot is spread out into a series of spots or bands,

with each spot representing one single component of the mixture, as seen

in the illustration of a paper chromatography strip to the right.

The separation of a mixture by chromatography is not only a function of the solubility in the solvent used.

The filter paper or TLC coating consists of molecules that may interact with the molecules of mixture as they are

carried up the medium. The primary interaction between the mixture components and the medium is due to the

polarity of the components and that of the medium. Each component of the mixture is likely to interact with the

medium to a different extent, thus slowing the components of the mixture differentially depending on the level

of interaction.

In chromatography analysis, there is a mathematical function called the retention factor. The retention factor, Rf,

is defined as:

Rf is the ratio of the distance a substance moves up the stationary phase to the distance the solvent have moved.

The retention factor depends on what solvent is used and on the specific composition of the filter paper or slide

coating used. The Rf value is characteristic of a substance when the same solvent and the same type of

12

stationary phase is used. Therefore, a set of known substances can be analyzed at the same time under the

same conditions.

In the case shown below, the Rf for the green spot is

Paper chromatography and TLC are only two examples of many different chromatographic methods. Mixtures of

gases are commonly separated by gas chromatography. In this method, a mixture of liquids are vaporized and

passed through a long tube of solid absorbent material. A carrier gas, usually helium, is used to carry the mixture

of gases through the tube. As with paper chromatography, the components of the mixture will have different

solubilities and different attractions for the solid absorbent. Separation of the components occurs as the mixture

moves through the tube. The individual components exit the tube one by one and can be collected.

Another form of chromatography is column chromatography. In this form, a vertical column is filled with solid

absorbent, the mixture is poured in at the top, and a carrier solvent is added. As the mixture flows down the

column, the components are separated, again, by differing solubilities in the carrier solvent and different

absorbencies to the solid packing. As the liquid drips out the bottom of the column, components of the solution

will exit at different times and can be collected.

This video presents thin layer chromatography with fluorescent materials, and column chromatography with UV

active materials. There is no narration on the video so it would be advantageous to watch with a chemistry

teacher (6f): http://www.youtube.com/watch?v=gzp2S0e9o8s

(From CK12 Chemistry 2nd Edition pp 409-412)

13

Weeks 1-5 Summary

Mixtures of solids may be separated by differing solubilities of the solids in a single solvent.

Components of a solution composed of a non-volatile solid solute and a liquid solvent can be separated

by distillation.

Mixtures of liquids with reasonably different boiling points can also be separated by distillation.

Solid compounds may be separated from impurities – purified – by recrystallization.

Compounds may be separated based on their relative solubilities in two different immiscible liquids,

usually water and an organic solvent, by solvent extraction.

Solutions with several components can be separated by paper or thin-layer chromatography.

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. In the next group of

labs we will learn about different ways to make up solutions and describe their composition. We will also look

at some properties of solutions that are not obvious as the sum of their component parts.

Many of the solutions we will make in the next two weeks will be used in later labs, so take extra care in

measurement and recording while doing these labs.

Homogeneous Mixtures

A homogeneous mixture is a solution of the same appearance or composition throughout. Thinking of the prefix “homo” meaning “sameness”, this definition makes perfectly good sense. Homogeneous solutions carry the same properties throughout the solution. Take, for example, vinegar that is used in cooking. Vinegar is approximately 5% acetic acid in water. This means that every teaspoon of vinegar that is removed from the container, contains 5% acetic acid and 95% water.

A point should be made here that when a solution is said to have uniform properties throughout, the definition is referring to properties at the particle level. Well, what does this mean? Let's consider brass as an example. The brass is an alloy made from copper and zinc. 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. So the brass represents a homogeneous mixture. Now, 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. Thus the combination of zinc filings and copper pieces in a pile does not represent a homogeneous solution. Another example of a solution is margarine. 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.

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. If you were to take one cup of water and dissolve ¼ teaspoon of table salt in it, a solution would form. The solution would have the same properties throughout, the particles of salt would be so

14

small that they would not be seen and the composition of every milliliter of the solution would be the same. But you can vary the composition of this solution to a point. If you were to add another ½ teaspoon of salt to the cup of water, you would make another solution, but this time there would be a different composition than the last. 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, thus it is homogeneous. What would happen if you tried to dissolve ½ cup of salt in the water. Would the solution stay homogeneous? No, it would not. At this point, the solution has passed its limit as to the amount of salt it can dissolve and it would no longer be a homogeneous solution.

So solutions have constant composition but you can vary the composition up to a point. There are limits to the amount of substance that can be dissolved into another substance and still remain homogeneous.

Types of Solutions

There are three states of matter: solid, liquid, and gas. If we think about solutions and the possibilities of combining these states together to form solutions, we have nine possibilities. Look at the Table below.

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), there are really only four that are common types of solutions. These are shown in boldface. The others, although still solutions, are less common in everyday lives. For example, a solid in a liquid solution can be anything from salt or sugar solution, to seawater. Liquid in liquid solutions include the antifreeze/coolant we use for our cars and vinegar. For a gas in a liquid solution, 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. Finally, to understand the gas in a gas solution, take a deep breath. That’s right, air is a solution made up of mostly oxygen gas and nitrogen gas.

A solid in a solid solution is less common but still we see a lot of steel and brass around in our everyday world. These are examples of solid – solid solutions. The other types of solutions are less common but do exist in the world of solution chemistry.

Similar Structures Allow Solutions to Occur

Over the course of your study in chemistry you have learned the terms polar and non-polar. Recall that in chemistry, 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 fact, during the study of Valence Shell Electron Pair Repulsion Theory (VSEPR), you learned that the chemical structures themselves have built in molecular polarity.

In solution chemistry, we can predict when solutions will form and others won’t using a little saying … “like dissolves like”. The “like dissolves like’ saying helps us to predict solubility based on the two parts of a solution having similar intermolecular forces. For example, suppose you are dissolving methanol in water. Both methanol and water are polar molecules and form a solution because they both have permanent dipoles (positive and

15

negative parts of the molecules) that allow the molecules of each of the substances to be attracted to the other. When this occurs, a solution is made.

A way to understand this is to think about why Velcro is used to hold two different pieces of fabric together. The two sides of Velcro allow the pieces of fabric to be fastened together because the Velcro has similar structure that “attract” each other. However, 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.

Let’s look at the individual structure of the water and methanol molecules. 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. In the representation of the water molecule, you can see that there are also permanent dipoles making it a polar molecule. The intermolecular forces for both of these molecules are dipole-dipole attractions. Since these molecules are both polar, they will form a solution when mixed together. We say they are miscible, which means these two liquids will make a solution. (See Figure below).

The hydration of ions in a polar solvent.

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. London-dispersion forces are the intermolecular bonds that hold the carbon tetrachloride together as a liquid. London-dispersion forces are also the forces that allow pentane to be a liquid at room temperature. Since both of these substances have the same intermolecular forces, when they are mixed together, a solution will be formed.

Unlike the polar molecules, the non-polar molecules have, at any given time, no permanent dipole. If we were to add table salt, NaCl, to either carbon tetrachloride or pentane, we would find that the salt would not dissolve. The reason for this is again explained with the structures of the substances. Since carbon tetrachloride (or pentane) has no permanent dipoles in its molecules, there would be place for the charges particles in a crystal of NaCl to be attracted.

16

In a polar solvent, the molecules of solvent are attracted to each other by the partial charges on the ends of the molecules. When a polar solute is added, the positive polar ends of the solute molecules attract the negative polar ends of the solvent molecules and vice versa. This attraction allows the two different types of molecules to form a solution. If a non-polar solute was added to a polar solvent, the non-polar solute particles cannot attract the solvent molecules away from each other - so a solution does not form.

Covalent compounds have a different type of attraction occurring between the solute and solvent molecules. Unlike ionic compounds, which result from the transfer of electrons, covalent compounds result from the sharing of electrons between atoms. As a result, there are no distinct charges associated with the atoms in covalent compounds. In CO2, each oxygen atom shares two of its electrons with carbon and the carbon shares two of its electrons with each oxygen atom. Look at the figure below:

This sharing of valence electrons represents covalent bonding. However, the electrons are not shared equally. Recall that elements with a greater electronegativity have a stronger attraction for shared electrons. Therefore, they can pull the electrons closer to themselves and away from the element that has a smaller electronegativity. For carbon, the electronegativity value is 2.5, and for oxygen it is 3.5. The result in this molecule is that the electrons are pulled closer to oxygen than carbon. The resultant structure is represented below.

The bonds in this molecule are polar, but the molecule is non-polar overall because the shifting of the shared electrons toward the oxygen atoms are in equal but opposite directions. As a result, there is no overall dipole moment on the molecule.

As ionic solids dissolve into solution, these solids separate into ions. Molecular compounds, however, are held together with covalent bonds, which are not readily broken. For example, when you dissolve a spoonful of sugar into a glass of water, the intermolecular forces between the sugar molecules are disrupted but the intramolecular bonds are not. The sugar will not separate into carbon ions, hydrogen ions, and oxygen ions. The sugar molecules remain intact, but because of their polar properties, they can interact with the polar water molecules to form a solution. 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?)

17

Polar solvents will dissolve polar and ionic solutes because of the attraction of the opposite charges on the solvent and solute particles. Non-polar solvents will only dissolve non-polar solutes because they cannot attract the dipoles or the ions.

Water: The Universal Solvent

Think of the title of this section, water: the universal solvent. The term solvent is used to represent the medium that is used to produce the solution. The term universal is used to describe the fact that water, along with many of its other unique aspects, can dissolve many types and kinds of substances. For instance, table salt, NaCl, is an ionic compound but easily makes a solution with water. This is true for many ionic compounds. And from your own experience and the description above you know that table sugar, a polar covalent compound, also dissolves in water. And this is also true for other polar compounds such as vinegar and corn syrup.

Even some nonpolar substances dissolve is water but only to a limited degree. Have you ever wondered why fish are able to breathe? Oxygen gas, a nonpolar molecule, does dissolve in water and it is this oxygen that the fish take in through their gills. Or, one more example of a nonpolar compound that dissolves in water is the reason we can enjoy carbonated sodas. Pepsi-cola and all the other sodas have carbon dioxide gas, CO2, a nonpolar compound, dissolved in a sugar-water solution. In this case, to keep as much gas in solution as possible, the sodas are kept under pressure. But that’s another part of the story!

Although qualitative observations are necessary and have their place in every part of science, including

chemistry, we have seen throughout our study of science that there is a definite need for quantitative

measurements in science. This is particularly true in solution chemistry. We might read in the headlines that the

amount of mercury found in the fish is up by 0.5ppm and say to ourselves, what does that mean? Is it important?

We read labels in the grocery store that are in weight percent. 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, one that

involves not only a numerical analysis but a critical analysis as well. In the next two labs we will explore some of

the different quantitative applications of solution chemistry.

(From CK12 Chemistry 2nd Edition pp 374-387)

(Bonus question answer: Sucrose is formed by dehydration synthesis of glucose and fructose. One H2O molecule

is lost in this reaction.)

Week 6: Solutions of Solid Chemicals Here is a link to a video of a teacher writing on an electronic blackboard. It shows how to calculate molarity,

molality, and mole fraction (6d): http://www.youtube.com/watch?v=9br3XBjFszs

7.1 Make Up a Molar Solution of a Solid Chemical

Read: Introductory materials above. Complete tables in book.

Introduction

Concentration is the measure of how much a given substance is mixed with another substance. Solutions can be

said to be dilute or concentrated. A concentrated solution is one in which there is a large amount of solute in a

given amount of solvent. A dilute solution is one in which there is a small amount of solute in a given amount of

solvent. A dilute solution is a concentrated solution that has been, in essence, watered down. Think of the

frozen juice containers you buy in the grocery store. In order to make juice, you mix the frozen juice from inside

these containers with about 3 or 4 times the amount of water. Therefore, you are diluting the concentrated

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juice. The terms “concentrated” and “dilute,” however, only provide a qualitative way of describing

concentration. In this lesson, we will explore some quantitative methods of expressing solution concentration.

Molarity

Of all the quantitative measures of concentration, molarity is the one used most frequently by chemists.

Molarity is defined as the number of moles of solute per liter of solution. The symbol given for molarity is M, or

moles/liter. Chemists also use square brackets to indicate a reference to the molarity of a substance.

For example, the expression [Ag+] refers to the molarity of the silver ion.

molarity =

=

Example:

What is the concentration, in mol/L, when 2.34 mol of NaCl has been dissolved in 500 mL of H2O?

Solution:

[NaCl] = (2.34 mol) / (0.500 liter) = 4.68 M

The concentration of the NaCl solution is 4.68 mol/L.

Example:

What would be the mass of potassium sulfate in 500 mL of a 1.25 mol/L potassium sulfate solution?

Solution:

M =

, so mol = M × L

(1.25 mol/L)(0.500 L) = 0.625 mol

mol =

, so mass = (mol)·(molar mass)

mass = (0.625 mol)(174.3 g/mol) = 109 g

Therefore, the mass of the K2SO4 that dissolves in 500 mL of H2O to make this solution is 109 g.

(From CK12 Chemistry 2nd Edition pp 387-388)

Define the following words in your lab notebook. Use the glossary in your Chemistry theory text or a resource

such as the McGraw-Hill Dictionary of Chemistry.

concentration

concentrated

dilute

molarity

Pre-Lab Problems next page:

19

1. What is the molarity of a 100 ml solution prepared by dissolving 30 g of NaCl?

2) A laboratory assistant prepares to make 0.154 M NaCl solution. He weighs 20 g of NaCl crystals. What is the

volume of NaCl solution he will make?

(from edHelper.com)

20

7.2 Make Up a Molal Solution of a Solid Chemical

Read: Introductory materials above for this set of labs. Complete tables in book.

Molality is another way to measure concentration of a solution. It is calculated by dividing the number of moles

of solute by the number of kilograms of solvent. Molality has the symbol m.

molality (m) =

Molarity, if you recall, is the number of moles of solute per volume of solution. Volume is temperature

dependent. As the temperature rises, the molarity of the solution will actually decrease slightly because the

volume will increase slightly. Molality does not involve volume, and mass is not temperature dependent. Thus,

there is a slight advantage to using molality over molarity when temperatures move away from standard

conditions.

Example:

Calculate the molality of a solution of hydrochloric acid where 12.5 g of hydrochloric acid has been dissolved in

115 g of water.

Solution:

mol HCl =

= 0.343 mol

molality HCl =

=

= 2.98 m

(From CK12 Chemistry 2nd Edition pp 389)

Define the following word in your lab notebook. Use the glossary in your Chemistry theory text or a resource

such as the McGraw-Hill Dictionary of Chemistry.

molality

Pre-Lab Problem:

3) 18 g glucose (C6H12O6) is dissolved in 100 g of water. If the density of this glucose solution is 1.07 g/ml,

1) What is the molarity of this solution?

2) What is the molality of this solution?

(from edhelper.com)

21

Week 7: Solution of Liquid Chemicals and Mass-to-Volume Percentage Read: Review introductory materials in this manual for this set of labs. Complete tables in book.

7.3 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. What mass of NaCl will the

student have to weigh? ( atomic mass of Na = 23 g ; atomic mass of Cl = 35.45 g)

2) Student A has 100 ml of 2 M HCl solution, and student B has 50 ml of 3.2 M HCl solution. Accidentally,

student C combines these two solutions. What is the molarity of this new solution?

(from edHelper.com)

22

7.4 Make Up a Mass-to-Volume Percentage Solution

Mass to volume percent is similar to another measure of concentration, mass percent.

Mass Percent

Mass percent is the number of grams of the solute in the number of grams of solution. Mass percent is a term

frequently used when referring to solid solutions. It has the formula:

percent by mass =

× 100%

or

percent by mass =

× 100%

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%

(From CK12 Chemistry 2nd Edition pp 388)

Define the following terms in your lab notebook. Use the glossary in your Chemistry theory text or a resource

such as the McGraw-Hill Dictionary of Chemistry.

mass percent

mass to volume percent

Pre-Lab Problem:

3) A concentrated HCl solution is 36% by mass and has a density of 1.18 g/ml.

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.com)

Mass to volume percent uses a similar calculation with the formula:

Mass-volume percent =

× 100%

23

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. In this week labs we will investigate two such properties of solutions.

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. Recall that the freezing point is the temperature at which the liquid changes to a solid. At a

given temperature, if a substance is added to a solvent (such as water), the solute-solvent interactions prevent

the solvent from going into the solid phase, requiring the temperature to decrease further before the solution

will solidify. A common example is found when salt is used on icy roadways. 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, as low as -9◦C. The

de-icing of planes is another common example of freezing point depression in action. A number of solutions are

used, but commonly a solution such as ethylene glycol or a less toxic monopropylene glycol is used to de-ice an

aircraft. The aircrafts are sprayed with the solution when the temperature is predicted to drop below the

freezing point. The freezing point depression, then, is the difference between the freezing points of the solution

and the pure solvent.

Boiling Point Elevation

The boiling point of a liquid occurs when the vapor pressure above the surface of the liquid equals the

surrounding pressure. At 1 atm of pressure, pure water boils at 100◦C, but salt water does not. When table salt

is added to water, the resulting solution has a higher boiling point than water alone. Essentially, the solute

particles take up space at the solvent/air interface, physically blocking some of the more energetic water

molecules from escaping into the gas phase. This is true for any solute added to a solvent. Boiling point

elevation is another example of a colligative property, meaning that the change in boiling point is related only to

the number of solute particles in solution, regardless of what those particles are. A 0.20 m solution of table salt

would have the same change in boiling point as a 0.20 m solution of KNO3.

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. A mathematical equation can be used to calculate the boiling point elevation or the

freezing point depression. Remember the solution has a higher boiling point, so to find the boiling point

elevation you would subtract the boiling point of the solvent from the boiling point of the solution. For example,

the boiling point of pure water at 1.0 atm is 100.◦C, while the boiling point of a 2% salt-water solution is about

102◦C. Therefore, the boiling point elevation would be 2◦C. In comparison, the freezing point depression is

found by subtracting the freezing point of the solution from the freezing point of the pure solvent.

Both the boiling point elevation and the freezing point depression are related to the molality of the solutions.

Looking at the formulas for the boiling point elevation and freezing point depression, we can see similarities

between the two.

24

Boiling point elevation:

ΔTb = Kbm

where

ΔTb = Tsolution − Tpure solvent

Kb = boiling point elevation constant

m = molality of the solution.

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.

The boiling point and freezing point constants are different for every solvent and are determined experimentally

in the lab. You can find these constants for hundreds of solvents listed in data reference publications for

chemistry and physics.

Example:

Antifreeze is used in automobile radiators to keep the coolant from freezing. In geographical areas where winter

temperatures go below the freezing point of water, using pure water as the coolant could allow the water to

freeze. Since water expands when it freezes, freezing coolant could crack engine blocks, radiators, and coolant

lines. The main component in antifreeze is ethylene glycol, C2H4(OH)2. If the addition of an unknown amount of

ethylene glycol to 150 g of water dropped the freezing point of the solution by −1.86◦C, what mass of ethylene

glycol was used? The freezing point constant, Kf , for water is -1.86◦C/m.

Solution:

ΔT f = Kf m

m =

=

= 1.00 m

m =

moles solute = (molality)(kg solvent) = (1.00 mol/kg)(0.150 kg) = 0.150 mol

mass C2H4(OH)2 = (mol)(molar mass) = (0.150 mol)(62.1 g/mol) = 9.32 g

Therefore, 9.32 g of ethylene glycol would have been added to the 150 g of water to lower the freezing point by

1.86◦C.

25

Remember that colligative properties are due to the number of solute particles in the solution. Adding 10

molecules of sugar to a solvent will produce 10 solute particles in the solution. However, when the solute is an

electrolyte, such as NaCl, adding 10 molecules of solute to the solution will produce 20 ions (solute particles) in

the solution. Therefore, adding enough NaCl solute to a solvent to produce a 0.20 m solution will have twice the

effect of adding enough sugar to a solvent to produce a 0.20 m solution.

The van’t Hoff factor (i) is the number of particles that the solute will dissociate into upon mixing with the

solvent. For example, sodium chloride (NaCl) will dissociate into two ions, so the van’t Hoff factor for NaCl is i = 2.

For lithium nitrate (LiNO3), i = 2, and for calcium chloride (CaCl2), i = 3. We can now rewrite our colligative

properties formulas to include the van’t Hoff factor.

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.

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 non-

electrolytes is always 1 because they do not dissociate.

26

Example:

A solution of 10.0 g of sodium chloride is added to 100.0 g of water in an attempt to elevate the boiling point.

What is the boiling point of the solution?

Solution:

ΔTb = iKbm

mol NaCl =

=

= 0.171 mol

molality =

=

= 1.71 m

For NaCl, i = 2 (NaCl → Na+ + Cl−)

Kb(water) = 0.52◦C/m

ΔTb = iKbm

ΔTb = (2)(0.52◦C/m)(1.71 m) = 1.78◦C

Tb(solution) = Tb(puresolvent) + ΔTb

Tb(solution) = 100◦C + 1.78◦C = 101.78◦C

Therefore, the boiling point of the solution of 10 g of NaCl in 100 g of water is 102◦C.

(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, what will the melting and boiling points be

of the resulting solution? Kb(H2O) = 0.52 0C/m and Kf(H2O) = 1.86 0C/m.

27

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.brinkster.net/pra_solutionworksheets.html )

8.1 Determine Molar Mass by Boiling Point Elevation

Read: Introductory material above. Include optional activities in lab if possible.

8.2 Determine Molar Mass by Freezing Point Depression

Read: Introductory material above. Include optional activities in lab if possible.

8.3 Observe the Effects of Osmotic Pressure

Read: Introductory material above and discussion below. Include optional activities in lab if possible.

Recall our discussions of osmosis and osmotic pressure during our Biology studies last year. Certain materials,

including those that make up the membranes around living cells, are semipermeable. That is, they allow water or

other small molecules to pass through, but they block the passage of large solute molecules or ions. When a

solution and a pure solvent (or two solutions of different concentration) are separated by the right kind of

semipermeable membrane, solvent molecules pass through the membrane in a process called osmosis.

Although the passage of solvent through the membrane takes place in both directions, passage from the pure

solvent side to the solution side is more favored and occurs faster. As a result, the amount of liquid on the pure

solvent side decreases, the amount of liquid on the solution side increases, and the concentration of the

solution decreases.

(from Chemistry (McMurray, Fay, 4th Edition) pp. 453-454)

Define the following terms in your lab notebook. Use the glossary in your Chemistry theory text or a resource

such as the McGraw-Hill Dictionary of Chemistry.

Van’t Hoff factor

Colligative property

Osmosis

Osmotic pressure

28

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, but now we will study the in more

depth.

Chemical Reactions

In a chemical change, new substances are formed. In order for this to occur, the chemical bonds of the

substances break, and the atoms that make up the substances separate and re-arrange themselves into new

substances with new chemical bonds. When this process occurs, we call it a chemical reaction. A chemical

reaction is the process in which one or more substances are changed into one or more new substances.

In order to describe a chemical reaction, we need to indicate what substances are present at the beginning and

what substances are present at the end. The substances that are present at the beginning are called reactants,

and the substances present at the end are called products.

Writing Chemical Equations

When sulfur dioxide is added to oxygen, sulfur trioxide is produced. In the chemical equation shown below,

sulfur dioxide and oxygen (SO2 and O2) are reactants, and sulfur trioxide (SO3) is the product.

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. In the

following table is a summary of the major symbols used in chemical equations. There are other symbols, but

these are the main ones that we need to know.

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Chemists have a choice of methods for describing a chemical reaction. They could draw a picture of the chemical

reaction, like in the image shown below.

Alternatively, they could describe the reaction in words. 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.

Chemists could also write the equation in chemical shorthand.

2H2(g) + O2(g)→ 2H2O(g)

In the symbolic equation, chemical formulas are used instead of chemical names for reactants and products, and

symbols are used to indicate the phase of each substance. It should be apparent that the chemical shorthand

method is the quickest and clearest method for writing chemical equations. For example, 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. In shorthand, however, 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. Look at the following reaction in shorthand notation and

describe the reaction in words.

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.”

Example:

Transfer the following symbolic equations into verbal descriptions or vice versa.

1. HCl(aq) + NaOH(aq)→ NaCl(aq) + H2O(l)

2. Gaseous propane, C3H8, burns in oxygen gas to produce gaseous carbon dioxide and liquid water.

3. Hydrogen fluoride gas reacts with an aqueous solution of potassium carbonate to produce an aqueous

solution of potassium fluoride, liquid water, and gaseous carbon dioxide.

Solution:

30

1. 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.

2. C3H8(g) + O2(g)→ CO2(g) + H2O(l)

3. 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, atoms in

the reactants do not disappear, nor do new atoms appear to form the products. In chemical reactions, atoms are

never created or destroyed. The same atoms that were present in the reactants are present in the products. The

atoms are merely re-organized into different arrangements. In a complete chemical equation, the two sides of

the equation must be balanced. That is, in a balanced chemical equation, the same number of each atom must

be present on the reactant and product sides of the equation.

Balancing Equations

The process of writing a balanced chemical equation involves three steps. As a beginning chemistry student, you

will not know whether or not two given compounds will react or not. Even if you saw them react, you would not

know what the products are without running any tests to identify them. Therefore, for the time being, you will

be told both the reactants and products in any equation you are asked to balance.

Step 1: Know what the reactants and products are, and write a word equation for the reaction.

Step 2: Write the formulas for all the reactants and products.

Step 3: Adjust the coefficients to balance the equation.

There are two types of numbers that appear in chemical equations. There are subscripts, which are part of the

chemical formulas of the reactants and products, and there are coefficients that are placed in front of the

formulas to indicate how many molecules of that substance are used or produced. In the chemical formula

shown below, the coefficients and subscripts are labeled.

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. Recall that a subscript of 1 is

not written - when no subscript appears for an atom in a formula, it is understood that only one atom is present.

The same is true in writing balanced chemical equations. If only one atom or molecule is present, the coefficient

of 1 is omitted.

The subscripts are part of the formulas, and once the formulas for the reactants and products are determined,

the subscripts may not be changed. The coefficients indicate the mole ratios of each substance involved in the

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reaction and may be changed in order to balance the equation. Coefficients are inserted into the chemical

equation in order to make the total number of each atom on both sides of the equation equal.

Note that equation balancing is accomplished by changing coefficients, 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.

Step 1: Write the word equation (keeping in mind that chlorine and bromine refer to the diatomic molecules).

chlorine + sodium bromide yields bromine + sodium chloride

Step 2: Substitute the correct formulas into the equation.

Cl2 + NaBr → Br2 + NaCl

Step 3: Insert coefficients where necessary to balance the equation.

By placing a coefficient of 2 in front of NaBr, we can balance the bromine atoms. By placing a coefficient of 2 in

front of the NaCl, we can balance the chlorine atoms.

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. We have also used the smallest whole numbers possible as the coefficients, so this equation is

properly balanced.

Example:

Write a balanced equation for the reaction between aluminum sulfate and calcium bromide to produce

aluminum bromide and calcium sulfate. Recall that polyatomic ions usually remain together as a unit throughout

a chemical reaction.

Step 1: Write the word equation.

aluminum sulfate + calcium bromide yields aluminum bromide + calcium sulfate

Step 2: Replace the names of the substances in the word equation with formulas.

Al2(SO4)3 + CaBr2 → AlBr3 + CaSO4

Step 3: Insert coefficients to balance the equation.

In order to balance the aluminum atoms, we must insert a coefficient of 2 in front of the aluminum compound in

the products.

Al2(SO4)3 + CaBr2 → 2AlBr3 + CaSO4

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In order to balance the sulfate ions, we must insert a coefficient of 3 in front of the product CaSO4.

Al2(SO4)3 + CaBr2 → 2AlBr3 + 3CaSO4

In order to balance the bromine atoms, we must insert a coefficient of 3 in front of the reactant CaBr2.

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. A

final check shows that there are two aluminum atoms, three sulfur atoms, twelve oxygen atoms, three calcium

atoms, and six bromine atoms on each side. This equation is balanced. 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.” While this set of coefficients does “balanced” the equation, they are not the lowest set of

coefficients possible. Chemical equations should be balanced with the simplest whole number coefficients. 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. Since it is required that an equation be balanced with the lowest whole

number coefficients, the equation above is not properly balanced. When you have finished balancing an

equation, you should not only check to make sure it is balanced, you should also check to make sure that it is

balanced with the simplest set of whole number coefficients possible.

Example:

Balance the following skeletal equation. (The term “skeletal equation” refers to an equation that has the

correct chemical formulas but does not include the proper coefficients.)

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.

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.

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.

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Pre-Lab Problems:

Given the following skeletal (un-balanced) equations, balance them.

1. CaCO3(s) → CaO(s) + CO2(g)

2. H2SO4(aq) + Al(OH)3(aq) → Al2(SO4)3(aq) + H2O(l)

3. Ba(NO3)2(aq) + Na2CO3(aq) → BaCO3(aq) + NaNO3(aq)

4. 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. But what

does this really mean? Consider the following reaction.

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. We can demonstrate that mass is conserved by determining the total mass on both sides of the

equation.

Mass of the Reactant Side:

1 molecule of Fe(NO3)3 × molecular weight = (1) · (241.9 daltons) = 241.9 daltons

3 molecules of NaOH × molecular weight = (3) · (40.0 daltons) = 120. daltons

Total mass of reactants = 241.9 daltons + 120. daltons = 361.9 daltons

Product Side Mass:

1 molecule of Fe(OH)3 × molecular weight = (1) · (106.9 daltons) = 106.9 daltons

3 molecules of NaNO3 × molecular weight = (3) · (85.0 daltons) = 255 daltons

Total mass of products = 106.9 daltons + 255 daltons = 361.9 daltons

As you can see, both the number of atoms and mass are conserved during chemical reactions. 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.

Chemical reactions are classified into types to help us analyze them and to help us predict what the products of

the reaction will be. The five major types of chemical reactions are synthesis, decomposition, single replacement,

double replacement, and combustion. We will look at composition and decomposition reactions in the next lab,

although we may not actually observe a composition reaction. In the following weeks we will study single and

double replacement reactions.

Next we will explore the quantitative relationships that exist between the reactants and products in a balanced

equation. This is known as stoichiometry. Stoichiometry involved calculating the quantities of reactants or

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products in a chemical reaction using the relationships found in the balanced chemical equation. The word

stoichiometry actually comes from two Greek words: stoikheion, which means element, and metron, which

means measure.

Mole Ratios

A mole ratio is the relationship between two components of a chemical reaction. For instance, 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).

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, we

can see that the coefficient in front of the hydrogen is 2, while the coefficient in front of water is also

2. Therefore, the mole ratio can be written as:

or

Similarly, the ratio of oxygen molecules to water molecules would be:

or

In the following example, let’s try finding the mole ratios by first writing a balanced chemical equation from a

“chemical sentence.”

Example:

Four moles of solid aluminum are mixed with three moles of gaseous oxygen to produce two moles of solid

aluminum oxide. What is the mole ratio of (1) aluminum to oxygen, (2) aluminum to aluminum oxide, 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 oxygen =

or

2. mole ratio of aluminum to aluminum oxide =

or

3. mole ratio of oxygen to aluminum oxide =

or

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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. When written, find the mole ratios for (1) calcium carbide

to water and (2) calcium carbide to calcium hydroxide.

Solution:

Balanced chemical equation: CaC2(s) + 2 H2O(l) → Ca(OH)2(aq) + C2H2(g)

1. mole ratio of calcium carbide to water =

or

2. 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. Therefore, the chemical equation must always be balanced before the mole ratios are used for

calculations. Looking at the unbalanced equation for the reaction of phosphorus trihydride with oxygen, it is

difficult to guess the correct mole ratio of phosphorus trihydride to oxygen gas.

PH3(g) + O2(g) → P4O10(s) + H2O(g)

Once the equation is balanced, however, the mole ratio of phophorus trihydride to oxygen gas is apparent.

Balanced chemical equation: 4 PH3(g) + 8 O2(g) → P4O10(s) + 6 H2O(g)

The mole ratio of phophorus trihydride to oxygen gas, then, is:

Keep in mind that before any mathematical calculations are made relating to a chemical equation, the equation

must be balanced.

Mole-Mole Calculations

In the chemistry lab, we rarely work with exactly one mole of a chemical. In order to determine the amount of

reagent (reacting chemical) necessary or the amount of product expected for a given reaction, we need to do

calculations using mole ratios.

Look at the following equation. If only 0.50 moles of magnesium hydroxide, Mg(OH)2, are present, how many

moles of phosphoric acid, H3PO4, would be required for the reaction?

2 H3PO4 + 3 Mg(OH)2 → Mg3(PO4)2 + 6 H2O

Step 1: To determine the conversion factor, we want to convert from moles of Mg(OH)2 to moles of H3PO4.

Therefore, the conversion factor is:

mole ratio =

Note that what we are trying to calculate is in the numerator, while what we know is in the denominator.

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Step 2: Use the conversion factor to answer the question.

(0.50 mol Mg(OH)2) ·

= 0.33 mol H3PO4

Therefore, if we have 0.50 mol of Mg(OH)2, we would need 0.33 mol of H3PO4 to react with all of the magnesium

hydroxide. Notice if the equation was not balanced, the amount of H3PO4 required would have been

calculated incorrectly. The ratio would have been 1:1, and we would have concluded that 0.5 mol of H3PO4 were

required.

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:

(2.36 mol NaNO3) ·

) = 7.08 mol Na2O

Example:

How many moles of sulfur are required to produce 5.42 mol of carbon disulfide, CS2, using the balanced

chemical equation below?

C + 2 S → CS2

Solution:

(5.42 mol CS2) ·(

) = 10.84 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, it is important to double check that you are using a balanced chemical equation

before attempting any calculations.

Using Proportion to Solve Stoichiometry Problems

All methods for solving stoichiometry problems contain the same four steps.

Step 1: Write and balance the chemical equation.

Step 2: Convert the given quantity to moles.

Step 3: Convert the moles of known to moles of unknown.

Step 4: Convert the moles of unknown to the requested units.

37

Step 1 has been covered in previous sections. We also just saw how to complete Step 3 in the previous section

by using mole ratios. In order to complete the remaining two steps, we simply need to know how to convert

between moles and the given or requested units. In this section, we will be solving “mass-mass problems,”

which means that both the given value and the requested answer will both be in units of mass, usually grams.

Note that if some other unit of mass is used, you should convert to grams first, and use that value for further

calculations. The conversion factor between grams and moles is the molar mass (g/mol). To find the number of

moles in x grams of a substance, we divide by the molar mass, and to go back from moles to grams, we multiply

by the molar mass. This process is best illustrated through examples, so let’s look at some sample problems.

The balanced equation below shows the reaction between hydrogen gas and oxygen gas to produce water.

Since the equation is already balanced, Step 1 is already completed. Remember that the coefficients in the

balanced equation are true for moles or molecules, but not for grams.

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. If you were told that you were going to use 2.00 moles of hydrogen in this reaction, you would also

know the moles of oxygen required for the reaction and the moles of water that would be produced. 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 0.50 mole of hydrogen. Since you are using a quarter as much hydrogen, you would

need a quarter as much oxygen and produce a quarter as much water. This is because the molar ratios keep the

same proportion. If you were to write out a mathematical equation to describe this problem, you would set up

the following proportion:

=

The given quantity, 0.50 mole of hydrogen, is already in moles, so Step 2 is also completed. We set up a

proportion to help complete Step 3. From the balanced equation, we know that 1 mole of oxygen reacts with 2

moles of hydrogen. Similarly, we want to determine the x moles of oxygen needed to react with 0.50 moles of

hydrogen. The set up proportion would be similar to the one above. We can then solve the proportion by

multiplying the denominator from the left side to both sides of the equal sign. In this case, you will find that x =

0.25 moles of O2.

Example:

Pentane, C5H12, reacts with oxygen gas to produce carbon dioxide and water. How many grams of carbon dioxide

will be produced by the reaction of 108.0 grams of pentane?

Step 1: Write and balance the equation.

C5H12 + 8 O2 → 5 CO2 + 6 H2O

Step 2: Convert the given quantity to moles.

= 1.50 mol C5H12

38

Step 3: Set up and solve the proportion to find moles of unknown.

=

Therefore, x mol CO2 = 7.50.

Step 4: Convert the unknown moles to requested units (grams).

grams CO2 = (7.50 mol) · (44.0 g/mol) = 330 grams

Example:

Aluminum hydroxide reacts with sulfuric acid to produce aluminum sulfate and water. How many grams of

aluminum hydroxide are necessary to produce 108 grams of water?

Step 1: Write and balance the equation.

2 Al(OH)3 + 3 H2SO4 → Al2(SO4)3 + 6 H2O

Step 2: Convert the given quantity to moles.

= 6.00 mol H2O

Step 3: Set up and solve the proportion to find moles of unknown.

=

Therefore, x mol Al(OH)3 = 2.00.

Step 4: Convert the moles of unknown to grams.

grams Al(OH)3 = (2.00 mol) · (78.0 g/mol) = 156 grams

Example:

15.0 grams of chlorine gas is bubbled through liquid sulfur to produce liquid disulfur dichloride. How much

product is produced in grams?

Step 1: Write and balance the chemical equation.

Cl2(g) + 2 S(l) → S2Cl2(l)

Step 2: Convert the given quantity to moles.

= 0.212 mol

Step 3: Set up and solve the proportion to find moles of unknown.

=

39

Therefore, x mol S2Cl2 = 0.212.

Step 4: Convert the moles of unknown to grams.

grams S2Cl2 = (0.212 mol) · (135 g/mol) = 28.6 grams

Example:

A thermite reaction occurs between elemental aluminum and iron(III) oxide to produce aluminum oxide and

elemental iron. The reaction releases enough heat to melt the iron that is produced. If 500. g of iron is produced

in the reaction, how much iron(III) oxide was used as reactant?

Step 1: Write and balance the chemical equation.

Fe2O3(s) + 2 Al(s) → 2 Fe(l) + Al2O3(s)

Step 2: Convert the given quantity to moles.

= 8.95 mol

Step 3: Set up and solve the proportion to find moles of unknown.

=

Therefore, x mol Fe2O3 = 4.48.

Step 4: Convert the moles of unknown to grams.

grams Fe2O3 = (4.48 mol) · (160. g/mol) = 717 grams

Example:

Ibuprofen is a common painkiller used by many people around the globe. It has the formula C13H18O2. If 200 g of

ibuprofen is combusted, how much carbon dioxide is produced?

Step 1: Write and balance the chemical equation.

2 C13H18O2(s) + 33 O2(s) → 26 CO2(g) + 18 H2O(l)

Step 2: Convert the given quantity to moles.

= 0.967 mol

Step 3: Set up and solve the proportion to find moles of unknown.

=

Therefore, x mol CO2 = 12.6.

40

Step 4: Convert the moles of unknown to grams.

grams CO2 = (12.6 mol) · (44.0 g/mol) = 554 grams

Example:

If sulfuric acid is mixed with sodium cyanide, the deadly gas hydrogen cyanide is produced. How much sulfuric

acid must be reacted to produce 12.5 grams of hydrogen cyanide?

Step 1: Write and balance the chemical equation.

2 NaCN(s) + H2SO4(aq) → Na2SO4(aq) + 2 HCN(g)

Step 2: Convert the given quantity to moles.

= 0.463 mol

Step 3: Set up and solve the proportion to find moles of unknown.

=

Therefore, x mol H2SO4 = 0.232.

Step 4: Convert the moles of unknown to grams.

grams H2SO4 = (0.232 mol) · (98.1 g/mol) = 22.7 grams

A blackboard type discussion of stoichiometry (3e) is available at

http://www.youtube.com/watch?v=EdZtSSJecJc (9:21).

(from CK12 Chemistry 2nd Edition pp 250-258, 268-277)

Extra Practice Problems:

Balance the following equations:

1) ___ N2 + ___ F2 ___ NF3

2) ___ C6H10 + ___ O2 ___ CO2 + ___ H2O

3) ___ HBr + ___ KHCO3 ___ H2O + ___ KBr + ___ CO2

4) ___ GaBr3 + ___ Na2SO3 ___ Ga2(SO3)3 + ___ NaBr

5) ___ SnO + ___ NF3 ___ SnF2 + ___ N2O3

41

Using the equation from problem 2 above, answer the following questions:

6) If I do this reaction with 35 grams of C6H10 and 45 grams of oxygen, how many grams of carbon dioxide

will be formed?

(from http://misterguch.brinkster.net/pra_equationworksheets.html )

Week 9: 9.2 Observe a Decomposition Reaction

9.1 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. 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. We can always identify a synthesis reaction because there is only one product.

If you are given elemental reactants and told that the reaction is a synthesis reaction, you should be able to

predict the products. For example, consider the equation below. Two elements (hydrogen and oxygen) combine

to form one product (water).

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. As a result, we can write the synthesis reaction for sodium

chloride just by knowing the elements that are present in the product.

2 Na(s) + Cl2(g) → 2 NaCl(s)

Pre-Lab Problems:

1. Write the chemical equation for the synthesis reaction of silver bromide, AgBr.

2. Predict the products for the following reaction: CO2(g) + H2O(l).

3. Predict the products for the following reaction: Li2O(s) + CO2(g).

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Decomposition Reactions

When one type of reactant breaks down to form two or more products, we have a decomposition reaction. The

best way to remember a decomposition reaction is that for all reactions of this type, 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. In this reaction, ammonium nitrate

breaks down to form dinitrogen oxide and water.

NH4NO3(s) → N2O(g) + 2H2O(g)

Notice that there is only one reactant, NH4NO3, on the left of the arrow and that there is more than one on the

right side of the arrow. This is the exact opposite of a synthesis reaction.

When studying decomposition reactions, we can predict the reactants in a similar manner as we did for

synthesis reactions. Look at the formula for magnesium nitride, Mg3N2. What elements do you see in this

formula? You see magnesium and nitrogen. Now we can write a decomposition reaction for magnesium nitride.

Mg3N2(s) → 3 Mg(s) + N2(g)

Notice there is only one reactant.

Pre-Lab Problems:

Write the chemical equation for the decomposition of the following compounds into their individual elements:

4. Al2O3

5. Ag2S

6. MgO

(from CK12 Chemistry 2nd Edition pp 258-260)

Define the following terms in your lab notebook. Use the glossary in your Chemistry theory text or a resource

such as the McGraw-Hill Dictionary of Chemistry.

chemical reaction

products

reactants

stoichiometry

composition reaction

decomposition reaction

43

Week 10: 9.3 Observe a Single-Displacement Reaction

Week 11: 9.4 Stoichiometry of a Double Displacement Reaction

Introduction to Lab Weeks 12-13: Reduction-Oxidation (Redox) Reactions

Week 12: 10.1 Reduction of Copper Ore to Copper Metal

Week 13: 10.2 Observe the Oxidation States of Manganese

Introduction to Lab Weeks 14-17: Acid-Base Chemistry

Week 14: 11.1 Determine the Effects of Concentration on pH

Week 15: 11.2: Determine the pH of Aqueous Salt Solutions

Week 16: 11.3 Observe the Characteristics of a Buffer Solution

Week 17: 11.4 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, 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. By this point,

students will not have difficulty understanding the main point, which is that temperature, surface area, and

concentration all affect the speed of reactions.

44

Week 18: Determine Effects of Temperature and Surface Area on Reaction Rate

12.1 Determine the Effects of Temperature on Reaction Rate

12.2 Determine the Effects of Surface Area on Reaction Rate

Week 19: 12.3 Determine the Effects of Concentration on Reaction Rate

Week 20: 13.1 Observe Le Chatelier's Principle in Action

Week 21: 13.2 Quantify the Common Ion Effect

Week 22: 14.1 Observe the Volume-Pressure Relationship of Gasses (Boyle's Law)

Week 23: 14.2 Observe the Volume-Temperature Relationship of Gasses (Charles's

Law)

Week 24: 14.3 Observe the Pressure -Temperature Relationship of Gases (Gay-

Lussac's Law)

Week 25: 14.4 Use the Ideal Gas Law to Determine the Percentage of Acetic Acid in

Vinegar

Week 26: 15.1 Determine Heat of Solution

Week 27: 15.2 Determine the Specific Heat of Ice

Week 28: 15.3 Determine the Specific Heat of a Metal

Week 29: 16.1 Produce Hydrogen and Oxygen by Electrolysis of Water

Week 30: 18.1 Observe Some Properties of Colloids and Suspensions

Week 31: 18.2 Produce Firefighting Foam

Week 32: 18.3 Prepare a Gelled Sol

Week 33: Discriminate Metal Ions

19.1 Using Flame Tests to Discriminate Metal Ions

19.2 Using Borax Bead Tests to Discriminate Metal Ions

Week 34: 20.1 Quantitative Analysis of Vitamin C by Acid-Base Titration

Week 35: 20.2 Quantitative Analysis of Chlorine Bleach by Redox Titration

Week 36: 21.1 Synthesize Methyl Salicylate from Aspirin

45

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 0.154 M NaCl solution. He weighs 20 g of NaCl crystals. What is the

volume of NaCl solution he will make?

3) 18 g glucose (C6H12O6) is dissolved in 100 g of water. If the density of this glucose solution is 1.07 g/ml,

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. What mass of NaCl will the

student have to weigh? ( atomic mass of Na = 23 g ; atomic mass of Cl = 35.45 g)

Molar mass of NaCl = 23 g + 35.45 g = 58.45 g

1 M = 1 mole solute in 1000 ml solution = 58.45 g NaCl in 1000 ml solution

2 M NaCl = 2 x 58.45 g in 1000 ml solution = 116.9 g in 1000 ml solution

To make 500 ml of 2 M NaCl solution, the student has to weigh 116.9 g / 2 =58.45 g NaCl

46

2) Student A has 100 ml of 2 M HCl solution, and student B has 50 ml of 3.2 M HCl solution. Accidentally,

student C combines these two solutions. 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 x 50 ml = y M x 150 ml

y = 2.4 M

3) A concentrated HCl solution is 36% by mass and has a density of 1.18 g/ml.

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, what will the melting and boiling points be of

the resulting solution? Kb(H2O) = 0.52 0C/m and Kf(H2O) = 1.86 0C/m.

To find the melting and boiling points, you’ll first need to find the concentration of the solution. To find the

molality, convert grams of sodium chloride to moles (it turns out to be 0.769 moles) and divide by the

number of kilograms of solvent (0.500 kg H2O). When you do this calculation, the molality is 1.54 m.

Because sodium chloride breaks into two particles when it dissociates in water, the effective molality for the

purposes of colligative property calculations is twice this, or 3.08 m.

From here, it’s a matter of plugging the molality into the equations for freezing point depression and boiling

point elevation. When you do this, you should get that the change in melting point is 5.730 C and the change

in boiling point is 1.600 C.

The final answers: Melting point = -5.730 C, boiling point = 101.600 C.

47

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, you find that the boiling point of the sucrose solution is 100.320

C and the boiling point of the sodium chloride solution is 101.400 C. Clearly, the sodium chloride solution

has a higher boiling point.

Introduction to Lab Weeks 9-11: Intro to Chemical Reactions and Stoichiometry

1. CaCO3(s) → CaO(s) + CO2(g) (sometimes the skeletal equation is already balanced)

2. 3H2SO4(aq) + 2Al(OH)3(aq) → Al2(SO4)3(aq) + 6H2O(l)

3. Ba(NO3)2(aq) + Na2CO3(aq) → BaCO3(aq) + 2 NaNO3(aq)

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

Week 9: 9.2 Observe a Decomposition Reaction 1. 2 Ag(s) + Br2(l) → 2 AgBr(s)

2. CO2(g) + H2O(l) → H2CO3(aq)

3. Li2O(s) + CO2(g) → Li2CO3(s)

4. 2 Al2O3 → 4 Al + 3 O2

5. Ag2S → 2 Ag + S

6. 2 MgO → 2 Mg + O2

Extra Practice Problems

Balance the following equations:

1) 1 N2 + 3 F2 2 NF3

2) 2 C6H10 + 17 O2 12 CO2 + 10 H2O

3) 1 HBr + 1 KHCO3 1 H2O + 1 KBr + 1 CO2

4) 2 GaBr3 + 3 Na2SO3 1 Ga2(SO3)3 + 6 NaBr

5) 3 SnO + 2 NF3 3 SnF2 + 1 N2O3

Using the equation from problem 2 above, answer the following questions:

6) If I do this reaction with 35 grams of C6H10 and 45 grams of oxygen, how many grams of carbon dioxide

will be formed?

When you do this calculation for 35 grams of C6H10, you find that 113 grams of CO2 will be formed.

When you do the calculation for 45 grams of oxygen, you find that 43.7 grams of CO2 will be formed.

Because 43.7 grams is the smaller number, oxygen is the limiting reagent, forming 43.7 grams of product.

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