CHEM108 Lab Manual - Saddleback College If a brick has a length of 13.77 cm, a width of 8.50 cm, and a height of 5.12 cm: a) What is the volume of the brick? b) If the brick has a

CHEM108 Lab Manual Includes Lab Reports and

Pre-Lab Assignments

Required for all Saddleback CHEM108 Classes

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NAME: ____________________

Pre-lab #1: Introduction to Lab Techniques Introduction to Measurements There are numerous aspects to chemistry, but a common thread between them all is the process of collecting data and observations through field studies or in a laboratory. It is critical to understand how to make and read measurements, and we will focus on the basic types in this experiment: mass, length, volume and temperature. All quantitative measurements (involving a numerical value) are meaningless without an associated unit. It is necessary to always include a unit with a number, and that unit is determined by the measuring device. Since there are many different units for mass, volume, length, etc., we will look at the relation between some common units from the English system and the metric system during lab. The other major goal is to learn how to use, read and interpolate the basic measuring devices found in the laboratory, such as the balances, graduated cylinders, rulers and thermometers. The concept of significant figures will be used to aid in understanding the limitations inherent in any measurement and later calculations involving that value. Points to keep in Mind: There are many basic concepts that will be carried throughout the entire semester in lab which are briefly covered below. More details will be given as the semester progresses. 1) Safety: All safety rules must be followed at all times, with specific ones covered in the prelab lecture for each class. Make sure you are on time! Safety goggles are required to be worn at all times while doing experimental work in the lab. No flip-flops or open-toe shoes are to be worn in the lab. Shoes must cover the tops of your feet. 2) Pre-lab Assignments: These are due at the beginning of lab, no exceptions. You may not perform the lab if the prelab is not complete. You can download these from the course website and have an entire week to ask questions if you are unsure of anything. Be familiar with the proper instructions and calculations for the lab before you try anything! 4) Problem Solving and Calculations: All work must be shown for mathematical calculations, and all unit conversions should be set up as unit line equations for clarity. Be neat! Always include units for any number. For example, to calculate the density (D) of a liquid with a measured volume (V) of 9.83 mL and a mass (M) of 12.90 g, show the formula, insert the primary data with units, show the unrounded answer, and then box the rounded answer to the appropriate number of significant figures with the correct units. D = M = 12.90 g = 1.31¦23 g ------> V 9.83 mL mL If this value was to be converted into units of grams per Liter (L), use the unrounded answer in the Unit Line Equation with the conversion factor to change the units of mL to L: 1.31¦23 g 1000 mL = 131¦2.3 g ------> or 1.31 x 103 g/L mL L L in scientific notation

1.31 g/mL

1310 g/L

Making Measurements: Each measuring device has its own limitations and techniques. For digital readings, 50 there is no estimation involved and all digits shown are recorded except for any zero preceding a non – zero number (i.e.: 3.20 g instead of 03.20 g). For non – digital instruments, the first step is to determine the value represented by the increment, or the space between the smallest divisions (lines) on the measuring device (represented by dashed lines in the figure to the right). This is most easily done by dividing the difference between two labeled lines by the number of lines between them. For the example shown to the right (assume it is a graduated cylinder), the difference between the labeled lines is: 50 mL – 40 mL = 10 mL. There are 10 divisions between 40 these labeled lines, and so the value of the increment is 10 mL ÷ 10 = 1 mL. WHEN USING NON-DIGITAL MEASURING DEVICES, REPORT MEASURED VALUES ONE DEMICAL PLACE MORE PRECISE THAN THE INCREMENT. Metric Rulers: A portion of a metric ruler is shown on the right. The smallest division (the increment) is a tenth (0.1) of a centimeter (cm). You must estimate the reading between these lines to hundredth (0.01) of a cm in order to report the measured value one decimal place more precise than the increment. The position marked by the dashed line is read as 5.38 cm (or 5.39 cm). 5 6 Thermometers: All temperature measurements are made in degrees Celsius (oC) 40 for the thermometers used in this lab. You do not need to shake the thermometer before use, and do not hold the bulb in your hand when making any measurements. If the red liquid in the thermometer is segmented or the thermometer broken, turn it in to your instructor to be fixed and get a new one. A portion of a thermometer is shown to the right in units of oC. Note the (increment) is one degree, and the measurement is estimated to a tenth of a degree (10 mental divisions), which is one decimal place more precise than the increment.

• The position marked by the dashed line is read as 33.5 oC. 30 Graduated Cylinders: There are two main factors to watch out for when reading a graduated cylinder: reading from the meniscus and avoiding parallax error. Since the liquid surface is not level (concave or convex), then a standard 3 must by agreed upon for consistency. This point is at the bottom of the curve, called the meniscus. Line up a burette reading card behind the cylinder so that the upper edge of the black line almost touches the bottom of the meniscus. This is shown on the figure to the right. Parallax error involves orienting your line of sight even with the meniscus. For very large cylinders, leave them on the counter and adjust your height to even your line of sight. For smaller cylinders, you can pick them up and hold them like a plumb bob between your fingers in line with your eyes. 2 There are some tricks involved with determining whether you are oriented properly which will be covered in lab as a demonstration. For the example on the right, the increment is 0.1 mL (3 mL – 2 mL = 1 mL divided by 10 divisions). Measurements in this cylinder is are estimated to the nearest hundredth of a mL (± 0.01 mL), (one decimal place more precise than the increment). Since the meniscus falls exactly on the line, the value would be 2.60 mL. (You must include the last zero since this cylinder is estimated to 2 decimal places).

Balances: We use top loader digital balances in this lab, and there are some important rules to follow so as to not damage them. Unless instructed otherwise, report all of the numbers displayed in the digital readout. a) Never move a balance or turn them off. b) Never put chemicals directly on a balance (use weighing paper or a container). c) Clean up any spilled chemicals immediately!

Practice Problems: For the following scales (no units), determine the size of the increment, and then estimate the readings at the lines given to the appropriate number of digits.

Answers to the Practice problems: 1) Increment: _1 __ 2) Increment: _0.1

Reading at X: 67.4 Reading at X: 7.74 Reading at Y: 61.0 Reading at Y: 7.10

Density Introduction: Matter is anything that has a mass associated with it, as well as encompassing a particular volume. The more matter there is in a certain volume (i.e.: the more mass within that volume), the more dense that matter is. So the density of matter, D, is the mass of that matter, M, divided by its volume, V. D = M V Solids normally have densities recorded in units of g/cm3 whereas liquids are g/mL. Gases are much less dense than liquids or solids and so typically have densities given in units of g/L. Density is constant for a particular species, unless the temperature changes. Under thermal expansion, the volume of a species increases when it is heated or contracts when cooled. If the mass remains constant while the volume changes upon heating, then the density changes based on the equation above. In this experiment, the densities of various solids and liquids will be measured with the particular techniques used explained in each individual section (i.e.: regularly shaped vs. irregularly shaped solids vs. liquids). You will also be given an unknown liquid and are to determine its density as accurately as possible. The Density of a Liquid: In determining the density of any liquid, the volume is easily measured directly from a graduated cylinder or burette. The mass, though, must be measured indirectly through a method called “measuring by difference”. Using this method, the mass of a dry graduated cylinder is subtracted from the mass of that cylinder filled with a measured volume of liquid. This gives the liquid’s mass, and when divided by its volume, its density.

The Density of a Regularly Shaped Solid: If a solid is square or rectangular, the volume can be directly measured using a ruler or other similar measuring device. Knowing that the volume of a square or rectangle is length (l) x width (w) x height (h), and with the mass being directly measured on a balance, then the density is easily obtained. Other solid volumes can be determined if their equation for volume is known (i.e.: Cones, cylinders, spheres). The Density of an Irregularly Shaped Solid: It can be very difficult or impossible to directly measure the volume of an irregularly shaped object (like an egg or tooth). A common method used is to measure the volume by water displacement. In this method, the solid object of known mass is placed in a graduated cylinder or other volume measuring device with a pre-recorded volume of water in it. The increase in volume due to addition of the solid is the volume of the solid.

Prelab Questions: These are due at the beginning of lab and must be completed to start the lab. 1) For the following scales (no units), determine the size of the increment, and then estimate the readings at the lines given to the appropriate number of digits.

100

2) If a brick has a length of 13.77 cm, a width of 8.50 cm, and a height of 5.12 cm:

a) What is the volume of the brick?

b) If the brick has a mass of 895.3 g, what is its density?

3) If 8.15 mL of water is placed in a graduated cylinder (dry mass = 56.98 g), and the combined mass of both the cylinder + water is 65.11 g:

a) What is the mass of the water:

b) What is the density of the water:

4) If a non-polar organic liquid with a density of 0.88 g/mL were mixed with the water from question #3, would the water or the organic liquid float on top?

Chemistry 108 Lab #1 !

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Name_____________________________

Lab #1: Introduction to Lab Techniques INTRODUCTION Our goals in this experiment are (1) to make some measurements using a metric ruler, (2) to learn how to determine volumes with a graduated cylinder, and (3) to determine the density of an unknown liquid and an unknown solid. PROCEDURE A. The Metric Ruler i) On the image of the metric ruler below, draw a small arrow pointing to 2.00 cm and write “A”, do the same and write “B" at 12.00 cm, "C" at 3.50 cm, "D" at 8.55 cm, and “E" at 10.60 cm. NOTE: The numbers that are displayed are cm, the increment is 0.1 cm.

ii) Use a metric ruler to determine the length and width of this piece of paper. Metric rulers are on the counter in the front of the lab. Use the correct number of decimal places base on the ruler increments (as you did in the pre-lab). Increment on metric ruler = ______________ cm Length of the page = ___________cm Do calculations to convert the length of the page to mm and meter (show your calculations for full credit) = ______mm = _______m. Width of the page = ___________cm Do calculations to convert the width of the page to mm and meter (show your calculations for full credit) = ______mm = _______m.


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B. The Graduated Cylinder There are 3 graduated cylinders set up in the lab. Each line on the 1000mL graduated cylinder represents ten milliliters (note that 1000 mL = 1 L). Each line on the 100mL graduated cylinder represents one milliliter. Each line on the 10 mL graduated cylinder represents 0.1 milliliters. Observe the top of the liquid in the 100 mL cylinder. Note that the liquid surface is curved, not level. The curved surface is called the meniscus. The volume is always read at the lowest point of the meniscus. Hold the graduated cylinder so the meniscus is exactly at eye level. Now raise and lower the graduated cylinder and observe that the volume reading changes as the cylinder is raised and lowered. Only when your eye is at exactly the same level as the bottom of the meniscus can you obtain an accurate volume reading. (The error introduced if your eye is high or low is called parallax.) Using the correct number of decimal places (as you did in the pre-lab), determine the volume of liquid in each of the 3 cylinders and record the data below. Graduate Cylinder Increment of

Graduated Cylinder Volume of Liquid in Cylinder

1000 mL (= 1L)

100 mL

10 mL

Remember to use units whenever you write a number.


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C. Measuring the Density of a Solid and of a Liquid i) Density of a metal cylinder 1) Each student will be given a metal cylinder, record the unknown number in the table below. 2) Weigh the metal cylinder, record the mass (in table below) to at least three places past the decimal. 3) Determine the increment of the 100 mL graduated cylinder and record it in the table below. 4) Place about 30 ml of water in your 100 mL graduated cylinder, record the volume to one decimal place more precise than the increment (as done in pre-lab) in the table below. 5) While holding the graduated cylinder at an angle, carefully slide your metal slug into the graduated cylinder. The metal cylinder must be completely submerged. Record the new volume to one decimal place more precise than the increment in the table below. 6) Calculate the volume of your metal cylinder. 7) Go to page 5 and calculate the density of your metal cylinder. Show your calculations, using the correct number of significant figurers in the appropriate box on page 5.

Metal Cylinder: Unknown number: ……………...…._______________________

Mass of metal……………………………..…….…_____________ g Increment on a 100 ml graduated cylinder………….. ___________mL (Figure it out just as you did on the prelab or ruler) Volume before submersion of metal ….…..….….______________mL (report volume to one decimal place more precise than the increment) Volume after submersion of metal ….…..……….______________mL (report volume to one decimal place more precise than the increment) Calculate the volume of the metal (Use the data above and the “volume by displacement method” as discussed in the powerpoint prelab introduction to calculate the volume of the metal) = _____________mL


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ii) Density of a liquid 1) Obtain an unknown liquid from the instructor and record the number of the unknown on the data table below. 2) Determine the increment of the 10 mL graduated cylinder and record it in the table below. Note that the increment is different for the 10ml graduated cylinder than it was for the 100 ml graduated cylinder that was used for the metal slug. 3) Measure and record, in the table below, the mass of a clean, dry 10 mL graduated cylinder using a balance. Record the mass to at least three places past the decimal. 4) Pour between 5 and 10 mL of the unknown liquid in the 10 mL graduated cylinder. Re-weigh and record the mass, to at least three places past the decimal, in the table below. 5) Next, record the volume of the unknown liquid in the 10 mL cylinder to one decimal place more precise than the increment. If you need to redo the density of your unknown liquid, the cylinder must be completely dry before you weigh it. The graduated cylinder can be dried using a rolled-up paper towel. 6) Calculate the mass of your unknown liquid. 7) Go to page 5 and calculate the density of your unknown liquid. Show your calculations, using the correct number of significant figurers in the appropriate box on page 5.

Unknown liquid: Unknown number: …………………….______________________ Increment on a 10 ml graduated cylinder………….. ___________mL (Figure it out just as you did on the prelab or ruler) Mass of 10 mL graduated cylinder…..………………____________g Mass of 10 mL graduated cylinder + liquid…………____________g

Volume of liquid ……………………………….…_____________mL (Report volume to one decimal place more precise than the increment.) Calculate the mass of the liquid (Calculated using “weight by difference” method) =____________ g


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DENSITY CALCULATIONS In the area below, calculate the densities of the metal cylinder and the unknown liquid. Remember: Every number in a measurement must have a number and a unit) i) Calculation of the unknown solid’s density

Unknown solid number_______ Density of your unknown solid ___________________ Did you use the correct number of significant figures? ii) Calculation of the unknown liquid’s density Unknown liquid number__________ Density of your unknown liquid_______________ Did you use the correct number of significant figures?

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Chemistry 108 prelab Name_________________________

Prelab #2 Coffee Cup Calorimetry

Heat is a form of energy, sometimes called thermal energy that will pass spontaneously from an object at a high temperature to an object at a lower temperature. If the two objects are in contact they will, given sufficient time, both reach the same temperature. Heat flow is ordinarily measured in a device called a calorimeter. A calorimeter is simply a container with insulating walls, made so that essentially no heat is exchanged between the contents of the calorimeter and the surroundings. A coffee cup or thermos is an example of containers designed to prevent heat from escaping its contents. Within the calorimeter chemical reactions may occur or heat may pass from one part of the contents to another, but no heat flows into or out of the calorimeter from or to the surroundings. A. Specific Heat. When heat energy flows into a substance, the temperature of that substance will increase. The quantity of heat energy (Q) required to cause a temperature change in any substance is equal to the specific heat (S) of that particular substance times the mass (m) of the substance times temperature change (∆T), as shown in Equation 1. Q = S x m x (∆T) Equation (1) The specific heat can be considered to be the amount of heat required to raise the temperature of one gram of the substance by one degree Celsius. Amounts of heat energy are measured in either joules or calories. To raise the temperature of 1 gram of water by 1 degree Celsius, 4.184 joules of heat must be added to the water. The specific heat of water is therefore 4.184 joules/goC. Since 4.184 joules equals one calorie, we can also say that the specific heat of water is 1 calorie/goC. That is not a coincidence, this is how the calorie was originally defined- the amount of energy needed to raise 1g of water by 1oC! B. Calorimetry The specific heat of a metal can readily be measured in a calorimeter. A calorimeter is a container with a known heat capacity. When a heating or cooling process in a calorimeter occurs, the amount of energy (heat) gained or lost in the process is absorbed by the calorimeter. This energy can be calculated from equation (1) if the calorimeter’s change in temperature is measured, since its heat capacity (S) and mass (m) are known. Very often, a calorimeter is composed of pure water in an insulating container. If one assumes that a negligible amount of heat escapes the insulated container and there is a negligible change in temperature of the container during the process, then all the energy released during the process of interest goes in to the water. Since we know the heat capacity and the mass of the water used in calorimeter and can measure the temperature change of the water, then we can use equation (1) to calculate the energy gained or lost by the water. The key point is that the energy gained (or lost) by the water is equal to the energy lost (or gained) in the process (such as cooling of the metal) that occurs in the calorimeter.

Chemistry 108 Heat Capacity Prelab

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A pre-weighed amount of metal is heated to some known temperature and is then quickly transferred into a calorimeter that contains a measured amount of water at a known temperature. Energy (heat) flows from the metal to the water, and the two eventually equilibrate (come to the same temperature) at some temperature between the initial temperatures of the water and metal. Assuming that no heat is lost from the calorimeter to the surroundings, and that a negligible amount of energy is absorbed by the calorimeter walls, the amount of energy that flows from the metal as it cools is equal to the amount of energy absorbed by the water. In other words, the energy that the metal loses is equal to the energy that the water gains. Since the metal loses energy (Tfinal is less than Tinitial; therefore ∆T is negative) Qmetal is negative. The water in the calorimeter gains energy (Tfinal is greater than Tinitial; therefore ∆T is positive) and Qwater is positive. Since the total energy is always conserved (cannot disappear), we can write equation (2): Qmetal + Qwater = 0 equation (2) Rearranging equation (2) gives (note the negative sign): Qmetal = - Qwater equation (3) In this experiment we measure the mass of water in the calorimeter and mass of the unknown metal and their initial and final temperatures. Using equation (1), the heat energy gained by the water and lost by the metal can be written: Qwater = Swaterx mwater x ∆Twater equation (4) Qmetal = Smetalx mmetal x ∆Tmetal equation (5) (Note that ∆Tmetal< 0 and ∆Twater > 0, since ∆T = Tfinal - Tinitial) We can calculate Qwater from the experimental data; we measured mwater, ∆Twater, and we know Swater= 4.184 J/goC. We know that the heat energy lost by the metal is equal (but opposite sign) as the heat energy gained by the water. To determine the specific heat of your unknown metal (Smetal), substitute Qmetal (in equation (5)) with – Qwater: -Qwater = Smetalx mmetal x ∆Tmetal equation (6) In this equation, we know 3 of the 4 variables: we calculated Qwater from experimental data; we measured mmetal and ∆Tmetal. We can now solve equation (6) for Smetal. Equation (6) can be rearranged to give Smetal as a function of the known variables (Qwater, mmetal, and ∆Tmetal): !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!"#$% = !!!"#$%

!!"#$%!×∆!!"#$% equation (7)

You will use this procedure to obtain the specific heat of an unknown metal. You will then compare the specific heat of your unknown metal to a table containing values of specific heats for several metals in order to determine the identity of your metal.


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Prelab Questions:

1) Why can we substitute Qmetal (in equation (5)) with – Qwater?

2) Why is ∆Tmetal< 0? 3) Why is ∆Twater > 0 ? 4) Should Qmetal be positive or negative? Why? 5) Should Qwater be positive or negative? Why? 6) A metal sample weighing 45.2 g and at a temperature of 100.0°C was placed in 38.6 g of water in a calorimeter at 25.2°C. At equilibrium the temperature of the water and metal was 32.4oC. a. What was ∆T for the water? (∆T = Tfinal - Tinitial) ____ °C b. What was ∆T for the metal? _____oC c. Using the specific heat of water (4.184 J/goC), calculate how much heat flowed into the water? _____joules d. Calculate the specific heat of the metal, using equation (7). _____________joules/goC


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Chemistry 108 lab Name_________________________

Lab #2: Coffee Cup Calorimetry

INTRODUCTION In this experiment, you will determine the specific heat for an unknown metal. The metal sample will be heated to a high temperature (100oC) then placed into a coffee cup calorimeter containing a known amount of water. If you can find out how much heat was gained by the water in the calorimeter then you will know how much heat was lost by the metal. You can then calculate and compare the specific heat of your unknown metal to known values of metal specific heats and identify your metal. EXPERIMENTAL

1) Set up a calorimeter. The instructor will give each student a sample of an unknown metal. Each student must do their own unknown. Record your unknown number in the DATA section below. (The thermometer is very expensive, so be careful when handling it.) The calorimeter consists of two polystyrene coffee cups fitted with a styrofoam cover (placed in a 400 ml beaker for balance). There are two holes in the cover for a thermometer and a glass stirring rod. Do not fill the calorimeter with water yet. Assemble the experimental setup as shown in the figure below.

2) One of the two lab partners do the following: Fill a 600 ml beaker two-thirds full of water, add a couple of boiling chips, and begin heating it to boiling on a hotplate. Hotplates are located in the common drawers. This boiling water will be shared by both lab partners. 3) Each person will weigh an empty, extra large test tube and stopper. Record the mass of the empty, stoppered tube your data table (use all of the digits displayed on the balance).

400 ml beaker

Chemistry 108 Heat Capacity Lab

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4) Transfer your unknown metal to the extra large test tube that you just weighed and replace the stopper. (the boiling water bath would remove the labels from the tubes your metals came in). Weigh your sample of unknown metal in the extra large, stoppered test tube. Record the mass of the stoppered tube and metal in your data table. Record all of the units displayed on the balance read-out.

• You will return your metal back to the original tube at the end of the experiment, so keep track of your own tube and do not mix it up with your partner’s.

5) Place the loosely stoppered tube with the metal into the boiling water in the beaker. Do not put water in the tube with the metal! Do not put the stopper tightly in to the tube or the tube may explode as the air in the tube is heated. The water level in the beaker should be high enough so that the top of the metal is below the water surface. Continue heating the metal in the water for at least 15 minutes after the water begins to boil to ensure that the metal attains the temperature of the boiling water (100.0oC). Add water to the beaker as necessary to maintain the water level. 6) While the water is boiling, weigh the empty calorimeter (both styrofoam cups and cover only; no water, no stirring rod, no thermometer, no 400ml beaker). Record the mass of the empty calorimeter in your data table, record this mass using three numbers after the decimal point. 7) Place about 30 ml of tap water in the calorimeter and weigh it again. Record the mass of the calorimeter and water in your data table. (styrofoam cups, water, and cover only; no stirring rod, no thermometer, no 400ml beaker) Do not be concerned that the last digit on the balance is not stable, you are seeing the water evaporate! Because of evaporation occurring, record this mass using three numbers after the decimal point. 8) Measure the initial temperature of the water contained in calorimeter. Note that you will need to hold the calorimeter at an angle so that the thermometer bulb is completely under the water. Record, to 0.1°C (one place to the right of the decimal), the temperature of the water (Tinitial water) in the calorimeter. 9) Insert the stirrer and thermometer into the calorimeter through the 2 holes in the cover. 10) Take the test tube out of the beaker of boiling water, remove the stopper, and pour the metal into the water in the calorimeter. Replace the calorimeter cover and stir the water/metal mixture as best you can with the glass stirrer. Record, to 0.loC, the maximum temperature reached by the water in the calorimeter (this is the Tfinal of the water and the metal). 11) OPTIONAL If you have 40 more minutes before the end of lab time, if you wish, you can repeat the experiment a second time (trial 2). Be sure to dry your metal before reusing it; this can be done using several paper towels. Be sure to dry the metal completely or you will introduce error to your measurements.


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WHEN FINISHED: The metal used in this experiment is to be dried with paper towels and returned to the front counter in the test tube in which you obtained it.

DATA Your unknown number ________________________ Trial 1 Trial 2 Mass of empty test tube and stopper

g

g

Mass of stoppered test tube plus metal

g

g

Mass of empty calorimeter

g

g

Mass of calorimeter and water

g

g

Initial temperature of water in calorimeter (Tinitial water)

°C

°C

Initial temperature of metal (assume 100.0°C) (Tinitial metal)

100.0 °C

100.0 °C

Equilibrium temperature of metal and water in the calorimeter (Tfinal water = Tfinal metal)

°C

°C

CALCULATIONS Be sure to use the correct number of significant figures and to use units with every number you write!!!! 1) Calculate ∆Tmetal

Trial 1 Trial 2

2) Calculate ∆Twater

Trial 1 Trial 2


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3) Calculate the mass of the water in the calorimeter (mwater). (Think about how you get this from the data table values….you have the mass of the calorimeter with the water in it and the mass of the empty calorimeter……)

Trial 1 Trial 2 (OPTINAL)

4) Calculate the mass of the metal (mmetal). (Think about how you get this from the data table values….you have the mass of the metal in the tube and the mass of the empty tube……)

Trial 1 Trial 2

5) Calculate the heat energy gained by the water (Qwater).

Trial 1 Trial 2 Did you use the correct number of sig. figs.?

6) Knowing that the heat gained by the water is equal and opposite the heat lost by the metal, use equation (6) or (7) in the prelab to calculate the specific heat of your metal.

Trial 1 Trial 2 Did you use the correct number of sig. figs.?

7) If you did two trials, take the average of the specific heats calculated in the two trials (above).


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Average specific heat of metal (if you did 2 trials only)……..________________________ (Did you use the correct number of significant figures and the correct units?)

CONCLUSION Re-write your unknown number here: ____________________ Your unknown metal is one of the metals in the table below. Use this table of specific heats to determine the identity of your metal. Choose the metal with the specific heat that is closest to your experimental value.

Metal Specific Heat (J/goC)

Aluminum 0.900 Bismuth 0.126 Tin 0.226 Nickel 0.443 My metal is (circle one): Aluminum Bismuth Tin Nickel

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Chemistry 108 Lab #3

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Name_____________________________

Prelab #3 Gases: Percent Yield of Hydrogen Gas from Magnesium and Hydrochloric Acid

Introduction For chemical reactions involving gases, gas volume measurements provide a convenient means of determining stoichiometric relationships. A gaseous product is collected in a long, thin graduated glass tube, called a eudiometer, by displacement of a liquid, usually water. Magnesium reacts with hydrochloric acid, producing hydrogen gas:

Mg (s) + 2HCl (aq) o MgCl2 (aq) + H2 (g)

Note: for every mole of Mg (s) that is reacted, one mole of H2(g) is produced. If we know the mass of Mg(s) we can convert to moles of Mg(s). Then, since we get 1 mole of H2(g) for every mole of Mg(s), we can predict how many moles of H2(g) could be made (theoretical yield). We use an excess of HCl so that we would react all the Mg(s) before we used all of the HCl.

Eudiometer Tube Set-up Reaction Set-up (after inverting eudiometer tube)


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When the magnesium reacts with the acid, the evolved hydrogen gas is collected by water displacement and its volume is measured. The temperature of the gas is taken to be the same as the temperature of the water it is in contact with because, given a sufficient amount of time, the two will reach thermal equilibrium. The level of water in the eudiometer is adjusted so that it is equal to the level of water outside the eudiometer. This insures that the pressure in the eudiometer is equal to the prevailing atmospheric pressure. Because the hydrogen gas was collected above water, and water has a significant vapor pressure, to get the pressure of pure hydrogen (dry hydrogen), we must subtract the vapor pressure of water. The pressure of the dry hydrogen gas (PH2) is calculated from Dalton's Law of Partial Pressures:

Ptotal = PH2 + PH2O so PH2 = Ptotal - PH2O where Ptotal (the pressure in the eudiometer) is atmospheric pressure, and PH2O (the water vapor pressure) is the pressure exerted by water vapor that has evaporated into the eudiometer. We will get the vapor pressure of water from the table below of vapor pressure vs. temperature.

The number of moles of hydrogen gas collected can then be calculated from the ideal gas law:

(n = # moles H2) n = PV (Use PH2 here, not Ptotal)

RT This will give you the experimental # moles of hydrogen gas collected. The theoretical # of moles of H2(g) can be calculated by converting the mass of Mg to moles Mg, and understanding that we get 1 mole of H2 from every mole of Mg(s). From the theoretical yield and the experimental yield, one can calculate the percent yield: Percent Yield = experimental # moles H2 u 100% theoretical #mole H2


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PRELAB QUESTIONS: 1. Suppose you did this experiment and obtained the following data. The initial mass of magnesium metal was 0.0426 g. The volume of the gas produced in the eudiometer was 41.75 mL, the atmospheric pressure is 763.2 torr, and the water temperature is 24.1 qC. As if you carried out this experiment and obtained the following data, fill in the data table and do the calculations below.

DATA TABLE:

Mass of Magnesium Metal . g

Volume of Gas . mL

Temperature of Gas (assumed to be the same temp. as the water) . qC

Atmospheric Pressure . torr

Water Vapor Pressure (at the above temperature, see table on last page) . torr

CALCULATIONS: 1. Theoretical (calculated) yield of H2 gas (# moles H2).

a) convert mass Mg to #moles Mg b) Convert # moles Mg reacted to moles of H2 that could be produced. (1 mole H2 is produced for every 1 mole Mg reacted- this comes from the balanced chemical equation):

Mg (s) + 2HCl (aq) o MgCl2 (aq) + H2 (g) # moles H2 = _____________________________________ (theoretical yield of H2)

2. Experimental yield of H2 gas (# moles H2).

a) Determine pressure of dry H2 (PH2) by subtracting the vapor pressure of water from the total pressure,

Ptotal = PH2 + PH2O so PH2 = Ptotal - PH2O

where Ptotal (the pressure in the eudiometer) is atmospheric pressure, and PH2O (the water vapor pressure) is the pressure exerted by water vapor that has evaporated into the eudiometer. Use the table provided on the previous page to find the vapor pressure of water as a function of temperature. PH2 = _____________________torr b) Convert this pressure from torr to atm (760.0 torr = 1atm) PH2 = _____________________atm


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c) Use the Ideal Gas Equation to calculate the number of moles (n) of H2 that you produced in your experiment (experimental yield). Make sure to use the correct units so that they match the units in the gas constant (R).

3. Calculate the Percent Yield.

% Yield = experimental # moles H2 u 100% theoretical #mole H2

% Yield =______________________________

OTHER PRE-LAB QUESTIONS 1. When the volume of gas is measured in a eudiometer, the water levels are the same on the

inside and outside of the eudiometer. Why do we need to do this? __________________________________________________________________________ __________________________________________________________________________ __________________________________________________________________________ __________________________________________________________________________ 2. If a gas is collected by water displacement in a eudiometer, the atmospheric pressure is known, and the water vapor pressure is known, what equation will allow you to calculate the pressure of the dry gas? __________________________________________________________________________ __________________________________________________________________________ __________________________________________________________________________ __________________________________________________________________________


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Name_____________________________

Lab # 3: Gases Percent Yield of Hydrogen Gas from Magnesium and Hydrochloric Acid

Introduction For chemical reactions involving gases, gas volume measurements provide a convenient means of determining stoichiometric relationships. A gaseous product is collected in a long, thin graduated glass tube, called a eudiometer, by displacement of a liquid, usually water. Magnesium reacts with hydrochloric acid, producing hydrogen gas:

Mg (s) + 2HCl (aq) o MgCl2 (aq) + H2 (g)

Note: for every mole of Mg (s) that is reacted, one mole of H2(g) is produced. If we know the mass of Mg(s) we can convert to moles of Mg(s). Then, since we get 1 mole of H2(g) for every mole of Mg(s), we can predict how many moles of H2(g) could be made (theoretical yield). We use an excess of HCl so that we would react all the Mg(s) before we used all of the HCl. When the magnesium reacts with the acid, the evolved hydrogen gas is collected by water displacement and its volume is measured. The temperature of the gas is taken to be the same as the temperature of the water it is in contact with because, given a sufficient amount of time, the two will reach thermal equilibrium. The level of water in the eudiometer is adjusted so that it is equal to the level of water outside the eudiometer. This insures that the pressure in the eudiometer is equal to the prevailing atmospheric pressure. Because the hydrogen gas was collected above water, and water has a significant vapor pressure, to get the pressure of pure hydrogen (dry hydrogen), we must subtract the vapor pressure of water. The pressure of the dry hydrogen gas (PH2) is calculated from Dalton's Law of Partial Pressures:

Ptotal = PH2 + PH2O so PH2 = Ptotal - PH2O where Ptotal (the pressure in the eudiometer) is atmospheric pressure, and PH2O (the water vapor pressure) is the pressure exerted by water vapor that has evaporated into the eudiometer. We will get the vapor pressure of water from the table below of vapor pressure vs. temperature.


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The number of moles of hydrogen gas collected can then be calculated from the ideal gas law:

(n= # moles H2) n = PV (Use PH2 here, not Ptotal)

RT This will give you the experimental # moles of hydrogen gas collected. The theoretical # of moles of H2(g) can be calculated by converting the mass of Mg to moles Mg, and understanding that we get 1 mole of H2 from every mole of Mg(s). From the theoretical yield and the experimental yield, one can calculate the percent yield: Percent Yield = experimental # moles H2 u 100% theoretical #mole H2

PROCEDURE:

1.Fill the largest beaker in your drawer (400 ml or a 600mL beaker) about 2/3 full of water and allow it to sit on the base of a ring stand so that the temperature of the water may adjust to room temperature. Place a double buret clamp on the ring stand well above the beaker.

2.Obtain a 4-5 cm length of magnesium ribbon from the back counter of the

lab room. With sand paper, sand the magnesium ribbon until it is shiny. Determine its mass on the side-loading balance, and record its mass in your data table. Your magnesium should have a mass no larger than 0.0450 g. Roll the magnesium ribbon into a loose coil. Obtain a piece of thread 25 cm in length, and tie it to one end of the magnesium ribbon in such a way that all the loops of coil are tied together (see Figure 1).

3.Obtain a eudiometer from the stockroom. Always carry a eudiometer in a vertical position. The eudiometer will contain water to keep dust out. Empty out the water

into your sink and temporarily attach it to the buret clamp, open end up.

4.Measure out 10 mL of hydrochloric acid in a graduated cylinder and pour it into

your eudiometer. Remove the eudiometer from the buret clamp, hold it on a slight slant, and add enough water to the eudiometer to fill it completely. Try to mix the water and the acid as little as possible. Reattach the eudiometer to the buret clamp, open end up (see Figure 2).

5.Obtain a one-hole rubber stopper from the back counter. Take your magnesium coil and lower it into the water of the eudiometer to a depth of about 5 cm. Have the thread attached to the coil hang over the lip and out of the eudiometer. Insert the one-hole rubber stopper into the eudiometer so the thread is held firmly against the edge, and when water squirts out of the hole in the stopper, cover the hole firmly with your thumb (see Figure 1).

Figure 1

Figure 2


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6. PUT ON RUBBER GLOVES. Taking care that no air enters, remove the eudiometer from the buret

clamp, invert it, and place its open end underwater in the beaker. Re-clamp the eudiometer to the buret clamp so that the bottom of the eudiometer is about 1 cm below the surface of the water in the beaker. The acid will flow down the eudiometer and react with the magnesium.

7.When the magnesium has disappeared entirely and the reaction has stopped, tap the tube with your finger to dislodge any bubbles you see attached to the side of the eudiometer. Measure the temperature of the water in your beaker; this will be the temperature of the hydrogen gas in the eudiometer. Record this value, to the nearest 0.1oC, in your data table. Because your thermometer reads to a tenth of a degree Celsius, add 273.2 when converting to Kelvin.

8.Place your finger over the hole in the

stopper and remove the eudiometer from the beaker. Lower the eudiometer into the leveling tank and remove your finger. Raise or lower the eudiometer until the water level inside the eudiometer is the same as the water level in the leveling tank. Read the volume of gas in the eudiometer and record it in your data table. Record the volume to the nearest 0.01 mL.

9.Your instructor will write the

barometer for today’s atmospheric pressure on the chalk board. Record this value in your data table. The water vapor pressure can be found in the table on page 2. Record this value for the vapor pressure of water in your data table.

10.When finished with the experiment, clean the eudiometer with water, dry the outside, fill it with

deionized water, and return it to the stockroom.

Figure 3. Inverted eudiometer illustrating gas collection and water displacement.


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DATA TABLE:

Mass of Magnesium Metal = g (must be less than 0.0450 g) Volume of Gas = mL Temperature of Gas = Temperature of the water = = qC Atmospheric Pressure = torr Water Vapor Pressure (at the above temperature, see table on last page) = torr

CALCULATIONS: 1. Theoretical (calculated) yield of H2 gas (# moles H2).

a) Convert mass Mg to #moles Mg b) Convert # moles Mg reacted to moles of H2 that could be produced (1 mole H2 is produced for every 1 mole Mg reacted- this comes from the balanced chemical equation):

Mg (s) + 2HCl (aq) o MgCl2 (aq) + H2 (g) # moles H2 = _____________________________________ (theoretical yield of H2)

2. Experimental yield of H2 gas (# moles H2).

a) Determine pressure of dry H2 (PH2) by subtracting the vapor pressure of water from the total pressure,

Ptotal = PH2 + PH2O so PH2 = Ptotal - PH2O

where Ptotal (the pressure in the eudiometer) is atmospheric pressure, and PH2O (the water vapor pressure) is the pressure exerted by water vapor that has evaporated into the eudiometer. Use the table provided on the last page to find the vapor pressure of water as a function of temperature. PH2 = _____________________torr b) Convert this pressure from torr to atm (760.0 torr = 1atm) PH2 = _____________________atm


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c) Use the Ideal Gas Equation to calculate the number of moles (n) of H2 that you produced in your experiment (experimental yield). Make sure to use the correct units so that they match the units in the gas constant (R).

3. Calculate the Percent Yield.

% Yield = experimental # moles H2 u 100% theoretical #mole H2

% Yield =______________________________

4. What are the possible sources of error in this experiment?

!

Chemistry 108 Chemical Reactions Prelab 4

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Pre-Lab #4: Chemical Reactions Many chemical reactions can be placed into one of two categories: oxidation-reduction reactions and double-replacement reactions. Oxidation-reduction reactions are ones in which electrons are transferred from one species to another. There are four types of oxidation-reduction reactions that we will investigate: synthesis, decomposition, single-replacement, and combustion. A) SYNTHESIS REACTIONS A synthesis reaction is one in which a single compound is formed from two or more substances:

A + B → AB An example of a synthesis reaction (one type of oxidation-reduction reaction) is the reaction that occurs between sodium metal and oxygen gas (O2):

Na(s) + O2(g) → Na2O(s)

The equation above is not balanced. Balance this equation:

_______Na(s) + ________O2(g) → _________Na2O(s) As stated earlier, oxidation-reduction reactions are ones in which electrons are transferred from one species to another. Let’s see how this transfer of electrons occurred in our reaction of sodium with oxygen. First, we must remember that the total charge of any pure element or compound is always ZERO! This fact will help you to determine the charge of each atom in a compound. Na(s) and O2(g) are pure elements (not combined with other elements) therefore the charge of each of the atoms in a piece of pure sodium metal and a sample of pure oxygen gas is equal to ZERO. Next, let’s consider the charges on the Na and O in the Na2O(s) product. We once again must understand that the total charge of any pure element or compound is always ZERO! Since sodium oxide (Na2O(s)) is a compound, it has a total charge = ZERO. We also know that sodium oxide is an ionic compound, and that although the total charge is ZERO, the sodium cations and oxygen anions are charged particles. What is the charge of an oxide ion? _________________ What is the charge of a sodium ion? ________________ Note that ions combine in a ratio such that the total charge of a compound = ZERO, that is why sodium oxide has the formula Na2O. This discussion of charge can be found in detail in chapter 3, it was repeated here as a review. Now, back to our discussion as to why this synthesis reaction is classified as an oxidation-reduction reaction. Let us consider how electrons are removed from one reactant and transferred to the other. We will do so by considering what happens to the charge of sodium and oxygen as they are converted from reactants to products.


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Let’s consider Na first. In the reactant, Na exists as Na(s) and has a charge of ZERO (charges are shown below as superscripts). In the product, Na has a charge of 1+.

Na0 (sodium has 0 charge in Na2O) → Na+ (sodium has a +1 charge in Na2O) In this chemical reaction, sodium went from ZERO charge to 1+ charge…. THIS IS ONLY DONE BY LOSING AN ELECTRON! When a species loses electron(s), we call that oxidation.

Na0 → Na+ + e-

Where does the electron go? It must get transferred to the oxygen! Let’s consider the charge on oxygen as it is converted from reactant to product. In the reactant, oxygen exists as O2 and has a charge of ZERO (charges are shown below as superscripts). In the product, O has a charge of 2-. .

O0 (oxygen has 0 charge in O2) → O2- (oxygen has a 2- charge in Na2O)

We see in this chemical reaction, oxygen went from ZERO charge to 2-…. THIS IS ONLY DONE BY GAINING ELECTRONS! When a species gains electron(s), we call that reduction.

O0 + 2e- → O2- The fact that oxygen gains 2 electrons and sodium only loses 1 electron is accounted for in the balanced chemical equation, note that 4 sodium atoms react with 2 oxygen atoms (O2)….making the electron donation ration equal 2:1, two sodium atoms are required to reduce each oxygen atom. Practice Problems: 1) Predict the product and balance the following synthesis reaction:

______ Li(s) + _______ O2(g) → _________________________

What is the charge of the lithium in the reactant Li(s)? ________________ What is the charge of the lithium ion in the product? ________________ Did lithium gain or lose electron(s) in this reaction: _______________ if so, how many? ______ Was lithium oxidized or reduced? ________________ What is the charge of the oxygen in the reactant O2(g) ? _________________ What is the charge of the oxide ions in the product? _________________ Did oxygen gain or lose electron(s) in this reaction: ______________ if so, how many? __________ Was oxygen oxidized or reduced? ________________


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2) Predict the product and balance the following synthesis reaction:

Mg(s) + Cl2(g) → _________________________

What is the charge of the magnesium in the reactant Mg(s)? ________________ What is the charge of the magnesium ion in the product? ________________ Did magnesium gain or lose electron(s) in this reaction: _______________ if so, how many? _____ Was magnesium oxidized or reduced? ________________ What is the charge of the chlorine in the reactant Cl2(g) ? _________________ What is the charge of the chloride ion in the product? _________________ Did chlorine gain or lose electron(s) in this reaction: _______________ if so, how many? ________ Was chlorine oxidized or reduced? ________________ B) DECOMPOSITON REACTIONS A decomposition reaction is an oxidation-reduction reaction in which a single reactant breaks down into two or more substances:

AB → A + B An example of a decomposition reaction is the thermal (heat induced) decomposition of mercury(II) oxide:

HgO (s) → Hg (l) + O2 (g) (not balanced, you will balance this reaction in the problem below)

Note that the key to identifying a decomposition reaction is that one reactant species is converted to two or more products species. In our example, we start the reaction with just one compound present, HgO (s), after the reaction there are two different chemical species present, Hg (l) and O2 (g).


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Practice Problems: 3) Balance our example decomposition reaction:

________ HgO (s) → _______ Hg (l) + __________ O2 (g)

What is the charge of the mercury ion in the reactant (HgO (s)) ? ________________ What is the charge of the mercury in the product (Hg (l))? ________________ Did mercury gain or lose electron(s) in this reaction: _______________ if so, how many? ______ Was mercury oxidized or reduced? ________________ What is the charge of the oxygen ion in the reactant (HgO (s))? _________________ What is the charge of the oxygen in the product (O2 (g))? _________________ Did oxygen gain or lose electron(s) in this reaction: ______________ if so, how many? __________ Was oxygen oxidized or reduced? ________________ C) SINGLE-REPLACEMENT REACTIONS A single-replacement reaction is an oxidation-reduction reaction in which an element replaces another element from a compound:

A + BX → AX + B

For example, copper metal (Cu(s)) will replace the silver ion (Ag) in silver nitrate: Cu(s) + AgNO3 (aq) → Cu(NO3)2 (aq) + Ag (s)

(not balanced, you will balance this reaction in the problem below) Practice Problems: 4) Balance our example single-replacement reaction: ______ Cu(s) + _____AgNO3 (aq) → _____Cu(NO3)2 (aq) + _____Ag(s)

What is the charge of the copper in the reactant Cu(s) ? ________________ What is the charge of the copper in the product (Cu(NO3)2 )? ________________ Did copper gain or lose electron(s) in this reaction: _______________ if so, how many? ______


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Was copper oxidized or reduced? ________________ What is the charge of the silver in the reactant (AgNO3)? _________________ What is the charge of the silver in the product (Ag(s)) ? _________________ Did silver gain or lose electron(s) in this reaction: ______________ if so, how many? __________ Was silver oxidized or reduced? ________________ D) COMBUSTION REACTIONS A combustion reaction is an oxidation reduction reaction in which a compound reacts with oxygen. In the case of organic compounds, complete combustion will always result in the formation of CO2 (g) and H2O (g). Incomplete combustion of organic molecules often results in the formation of C(s), often in the form of small particles of pure carbon called “soot” and other small organic molecules like carbon monoxide. Soot can be thought of as tiny pieces of charcoal. In chemistry 108, we only focus on complete combustion of organic molecules to form the products CO2 (g) and H2O (g). As an example, let us consider the combustion of pentane: Write in the products and balance the reaction for the complete combustion of pentane:

C5H12 (l) + O2(g) → __________ + ____________ In part A of the prelab, we learned how to identify an oxidation reduction reaction by understanding how electrons are transferred from one species to another. We were able to see which species was oxidized and which species was reduced by keeping track of the charges of atoms in reactants and products. Seeing how the charges of atoms changed in an inorganic chemical reaction allows us to determine how electrons were transferred, which species gained electrons, and which species lost electrons. In the case of the oxidation and reduction of organic compounds, that process is not as easily applied, so we learned a couple of rules (in chapter 6) to determine if a species is oxidized or reduced:

Oxidation and Reduction of Organic Molecules

Method for determining oxidation or reduction in organic compounds:

An atom in a molecule is oxidized if it: • gains oxygen (is attached to more oxygen atoms in the product than in the reactant) OR • loses hydrogen (is attached to fewer hydrogen atoms in the product than in the reactant)

An atom in a molecule is reduced if it:

• loses oxygen (is attached to fewer oxygen atoms) OR • gains hydrogen (is attached to more hydrogen atoms)

Let’s apply these rules to our example of the combustion of pentane by asking: is carbon oxidized or is it reduced in this reaction?


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• When the carbon in C5H12 is converted to CO2, carbon is oxidized– not only does it lose hydrogen but it

also gains oxygen. Carbons are bound to hydrogens in pentane (reactant) and are bound to oxygen in CO2 (product). What is reduced? Where did the electrons that carbon lost go to?

• O2 is reduced – each oxygen in O2“lost” an oxygen in the formation of water. The two oxygen atoms in O2 that were bound to each other are no longer bound to an oxygen in the product. In addition, each oxygen gained 2 hydrogens in the formation of water. Practice Problems: 5) In aerobic organisms, in the absence of sufficient oxygen, the pyruvate made in glycolysis is not able to enter the citric acid cycle and is converted to lactate (using hydrogen supplied by a molecule called nicotinamide adenine dinucleotide, NADH):

Is this an oxidation reduction reaction? _____________, if so, is pyruvate being oxidized or reduced in the reaction? ________________ Is this a combustion reaction?__________________ E) DOUBLE REPLACEMENT REACTIONS Double replacement reactions are ones in which two substances in solution switch partners. They are not oxidation-reduction reactions as were the first 4 reactions discussed. There are two types of double replacement reactions that we will investigate: precipitation reactions and gas forming reactions. A precipitation reaction is one in which, when the partners of the two reactant compounds are switched, one of the products is insoluble in water:

AX + BY → AY + BX (one of the products, AY or BX is insoluble in water = solid = precipitate)


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A gas forming reaction is one in which, when the partners of the two reactant compounds are switched, the products include water, an ionic compound, and a gas. The only gas forming reaction that we are concerned with in chemistry 108 is the following reaction:

HCl (aq) + NaHCO3 → H2O(l) + CO2(g) + NaCl(aq)

• See your chapter 7 notes for more information on precipitation reactions and gas forming reactions. Practice Problems: 6) For each double replacement reaction below, determine if and what type of reaction will occur (precipitation, gas forming, or no reaction), determine the appropriate products for the reaction and then write a complete balanced chemical equation. For those which do not react, indicate this by writing "No Reaction." Hint: See chapter 7 lecture notes on double replacement reactions. REACTION: REACTION TYPE Sodium bromide + silver nitrate…………………………………………____________________ Iron (III) chloride + sodium carbonate……………………………….______________________ Copper (II) sulfate + ammonium hydroxide…………………………______________________ Sodium bicarbonate + hydrochloric acid (HCl)……………………._______________________ In this experiment each of these six types of reactions will be performed. The evidence that reactions are occurring may be (1) a gas is produced (2) a precipitate (a solid produced when two solutions are mixed) is formed, (3) a color change is observed, (4) a temperature change is noted, or (5) a flame is produced.


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Chemistry 108 Chemical Reactions lab

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Name__________________________________

Lab #4: Chemical Reactions Many chemical reactions can be placed into one of two categories: oxidation-reduction reactions and double replacement reactions. Oxidation-reduction reactions are ones in which electrons are transferred from one species to another. There are four types of oxidation-reduction reactions that we will investigate: synthesis, decomposition, single replacement, and combustion. There are two types of double replacement reactions that we will investigate: precipitation reactions and gas forming reactions. In this experiment, these six types of reactions will be performed. The evidence that reactions are occurring may be (1) a gas is produced (2) a precipitate (a solid produced when two solutions are mixed) is formed, (3) a color change is observed, (4) a temperature change is noted, or (5) a flame is produced.

The purpose of this lab is to give the learner hands-on experience in classification, observation, and analysis of various types of chemical reactions.

PROCEDURE SAFETY NOTE: WEAR GLOVES AND GOGGLES FOR THE ENTIRE EXPERIMENT. PART A - SYNTHESIS REACTIONS

Obtain a piece of magnesium ribbon. If the ribbon is dull, sand it with a piece of sand paper. In order not to damage the bench top, perform the next step of the experiment over your evaporating dish (an evaporating dish looks like a small, white ceramic bowl and is located in your locker drawer). Holding one end of the magnesium ribbon with your tongs, ignite the other end in the burner flame. Look for just a moment at the burning magnesium; do not stare at the magnesium while it is burning because of the intensity of the light. Record any evidence of chemical change in the Part A Observation Data Table. NOTE: The evidence that reactions are occurring may be (1) a gas is produced (2) a precipitate (a solid produced when two solutions are mixed) is formed, (3) a color change is observed, (4) a temperature change is noted, or (5) a flame is produced. Dispose of the ash from this part of the experiment in the trashcan. PART B - DECOMPOSITION REACTIONS

Obtain a plastic cup with two platinum wires fused into it and fill the container with the catalyst-containing water (the water used for this reaction contains an acid catalyst) until the water is about 1cm from the top of the container. Next, fill an extra-small test tube completely with the catalyst-containing water. Wearing a disposable glove, cover the top of the test tube with your finger, so that there are no bubbles in it, invert it into the water in the container, and position it over one of the platinum wires. The extra-small test tubes are the smallest test tubes in your drawer. Repeat the same procedure with a second extra-small test tube, positioning it over the second platinum wire. Obtain a 9-volt battery, and connect it to the battery clip on the wires sticking out of the bottom of the container. Keep the battery connected for two minutes. Record any evidence of chemical change in the Part B Observation Data Table. The reactant is water; think about what the identity of the two products could be. If you need more information, review the decomposition reaction discussion in your prelab. Dispose of the solution from this part of the experiment down the sink.


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PART C – SINGLE REPLACEMENT REACTIONS

Part C-1. Place 3 mL of silver nitrate solution in a medium test tube. Add a piece of copper metal and observe the test tubes for several minutes. Record any evidence of chemical change in the Part C-1 Observation Data Table. IMPORTANT: Pour the solution into the largest beaker from your lab drawer. This beaker will be used for other chemical waste throughout the lab. Keep the waste beaker on your lab bench. At the end of the experiment, the entire contents of your waste beaker will be disposed of in the waste bottle in the fume hood near the instructor’s desk. Clean the test tube.

Part C-2. Place 3 mL of 6M hydrochloric acid (HCl) into a medium test tube. Add a small piece of magnesium metal (labeled "magnesium turnings") to the test tube. Record any evidence of chemical change in the Part C-2 Observation Data Table. If a gas is produced, hold your thumb over the mouth of the test tube for about one minute, light a wooden splint with a Bunsen burner, then remove your thumb and hold the burning splint to the mouth of the test tube. H2 a combustible gas, is produced in the reaction. In this reaction, magnesium replaces the hydrogen in HCl. Place the solution in the waste beaker on your lab bench. Clean the test tube.

Part C-3. Place 3 mL of deionized water into a medium test tube, which is resting in your test tube rack. Using the forceps provided, add a small piece of calcium metal to the test tube. CAUTION: Do not touch calcium metal, and handle with care.

Record any evidence of chemical change in the Part C-3 Observation Data Table. Obtain a piece of red litmus paper. Dip a stirring rod in the solution and touching it to the piece of red litmus paper. Red litmus paper turning blue indicates that a product of the chemical reaction is an ionic compound where the anion is hydroxide (OH-). In this reaction, the calcium replaces one of the hydrogens in H2O. Place the solution in a waste beaker on your lab bench. Clean the test tube. PART D – COMBUSTION REACTIONS

Light your laboratory burner to burn the methane gas coming from the gas outlet at your lab station. Remove any flammable material from your lab bench. Holding the burner by its base, turn the burner over and have the flame come in contact with the lab bench for 2 or 3 seconds. Place the burner back on the lab bench and turn it off. Record your observations of the product seen on the lab bench in the Part D Observation Data Table.


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PARTS E and F : DOUBLE REPLACEMENT REACTIONS PART E – PRECIPITATION REACTIONS

Place about 2 mL of mercury(II) nitrate solution into a clean, medium test tube, and place about 2 mL of aluminum nitrate solution into a second clean, medium test tube. Add about 2 mL of potassium iodide solution to both of the test tubes. Record any evidence of chemical change in the Part E-1 and E-2 Observation Data Tables. If a precipitate forms, write the balanced equations in the Part E–1 or Part E–2 sections of the Reaction Table. If no reactions occurred, write NR (NR means No Reaction). Place both solutions in the waste beaker on your lab bench.

Place about 2 mL of mercury(II) nitrate solution into a clean, medium test tube, and place about 2 mL of aluminum nitrate solution into a second clean, medium test tube. Add about 2 mL of sodium phosphate solution to both. Record any evidence of chemical change in the Part E-3 and E-4 Observation Data Tables. If a precipitate forms, write the balanced equations in the Part E–3 or Part E–4 sections of the Reaction Table. If no reactions occurred, write NR. Place both solutions in a waste beaker on your lab bench. PART F – GAS FORMING REACTIONS

Place about 0.50 grams of solid sodium bicarbonate (NaHCO3) in the bottom of a large test tube. Add several drops of 6M HCl. Record any evidence of chemical change in the Part F Observation Data Table. The contents of the tube can be washed down the sink with water. Empty the entire contents of your waste beaker in the waste bottle in the fume hood near the instructor’s desk. COMPLETE THE REACTION TABLE ON THE LAST PAGE OF THE HANDOUT. Write the balanced equations for each of these reactions in the Reaction Table. Remember to write the phase [(g),(l),(s), or (aq)] after each reactant or product for all of the chemical equations in your Reaction Table. THE REACTION TABLE CAN BE DONE AT HOME. TURN THE LAB AND PRELAB IN AT THE BEGINNING OF THE NEXT LAB PERIOD.


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DATA TABLE PART A OBSERVATIONS EVIDENCE OF REACTION Magnesium + Oxygen PART B OBSERVATIONS EVIDENCE OF REACTION Water PART C-1 OBSERVATIONS EVIDENCE OF REACTION Silver Nitrate + Copper PART C-2 OBSERVATIONS EVIDENCE OF REACTION Hydrochloric Acid + Magnesium PART C-3 OBSERVATIONS EVIDENCE OF REACTION Water + Calcium

PART D OBSERVATIONS EVIDENCE OF REACTION


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Methane + Oxygen PART E-1 OBSERVATIONS EVIDENCE OF REACTION Mercury (II) Nitrate + Potassium Iodide PART E-2 OBSERVATIONS EVIDENCE OF REACTION Aluminum Nitrate + Potassium Iodide PART E-3 OBSERVATIONS EVIDENCE OF REACTION Mercury (II) Nitrate + Sodium Phosphate

PART E-4 OBSERVATIONS EVIDENCE OF REACTION Aluminum Nitrate + Sodium Phosphate PART F OBSERVATIONS EVIDENCE OF REACTION Hydrochloric Acid (HCl) + Sodium Bicarbonate


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REACTION TABLE: Write a balanced chemical equation for each of these reactions. Remember to write the phase

[(g),(l),(s), or (aq)] after each reactant and product.

PART A Hint: One of the reactants is O2, and the product is an ionic compound. PART B Note: The only reactant is water; the acid is just a catalyst and not a reactant or product. Since the only reactant is water, think about what the identity of the two products are for a decomposition reaction. PART C-1 Hints: (1) See the single replacement discussion and examples in your pre-lab, and (2) the copper is being oxidized to Cu2+. PART C-2 PART C-3 In this reaction, the calcium replaces one of the hydrogens in H2O. PART D PART E-1 Mercury (II) Nitrate + Potassium Iodide PART E-2 Aluminum Nitrate + Potassium Iodide PART E-3 Mercury (II) Nitrate + Sodium Phosphate PART E-4 Aluminum Nitrate + Sodium Phosphate PART F HCl + Sodium Bicarbonate

Chemistry 108 Name_______________________________

Lab #5 Prelab: EXTRACTION AND SEPARATION OF PLANT PIGMENTS

Purpose of the lab: The purpose of this lab activity is for the student to learn about extraction and chemical separation technology. Specifically, the student will learn how to do a liquid phase-extraction and Thin Layer Chromatography in order to separate a mixture of molecules. Method: In the first step of this lab, the student will remove some of the water from spinach using a methanol rinse. In the second step, liquid-phase extraction will be used to extract pigment molecules from dried, pulverized spinach. In the third and final step, the pigment molecules will be separated from each other using Thin Layer Chromatography. Introduction Chromatography is a useful analytical technique that allows various components of a mixture to be separated based on their polarity and/or size. There many different types of chromatographic separation apparatuses and methods. We will use a technique called thin-layer chromatography (TLC). The thin-layer chromatographic surfaces that we will use in this experiment are plastic slides (TLC slides) with a coating of very small porous silica (SiO2) particles. The particles are so small that when they are not adhered to a surface, they behave like dust. A small drop of the mixture to be separated is applied (spotted) near one end of the TLC slide. The spotted end of the TLC slide is then dipped into a developing solvent (see figure 1a), called the mobile phase¸ which flows up the TLC surface by capillary action. As the developing solvent flows up the TLC surface, it can carry along the components of the mixture. Each component of the spotted mixture will move upward at a different rate.

The fact that different molecules move along the chromatographic surface/solvent interface at different rates is the key feature of chromatographic separation. Think of the component molecules as constantly moving back-and-forth from being adsorbed to the TLC surface to being carried upward with the mobile phase solvent. In the case of TLC, the more soluble the component is in the solvent, the faster

Chemistry 108 Plant Pigments

it travels up the surface. If it is not very soluble, it will remain adsorbed on the TLC surface (stationary phase) longer and therefore travel at a slower rate. In addition to the component’s affinity for the solvent, the component’s affinity for the stationary phase also plays a part in the rate at which the component moves. If there is a strong affinity between a component molecule and the stationary phase surface, the molecule will move upward more slowly. The affinity to the surface is determined by (1) non covalent interactions such as dipole-dipole forces and (2), in cases of porous stationary phases, the size of the molecule. If a molecule has a size such that it is easily “stuck” in a pore (hole), it will move upward more slowly since it takes more time to become “un-stuck”. The rates of flow are measure in terms of Rf (retardation factor) values. An Rf is the relative distance that a sample component has moved relative to the distance moved by the mobile phase solvent. The following illustration will demonstrate how the Rf is calculated. Rf is measured by dividing the distance the component traveled by the distance the solvent traveled. Therefore, an Rf value can never be greater than 1. The Rf value for a particular component is characteristic for that component in that particular solvent. Therefore, it will always be the same (considering that the mobile and stationary phases are the same) and can be identified in other mixtures

Rf = Distance component traveled Distance solvent traveled

Example: Calculation of TLC Rf values for a sample containing 4 components:

Solute Distance Traveled

(cm)

Rf value

A

2.0 cm

2.0 cm = 0.33 6.0 cm

B

3.0 cm

3.0 cm = 0.50 6.0 cm

C 3.5cm

3.5 cm = 0.58 6.0 cm

D 5.5cm

5.5 cm = 0.92 6.0 cm

Chemistry 108 Plant Pigments

Prelab Questions

1) Consider the following TLC chromatograph. Calculate the Rf factors for each of the corresponding spots.

Solute Distance Traveled

(cm)

Rf value

A

3.10 cm

B

2.58 cm

C

4.20 cm

D

5.08 cm

2) In the first step of this lab you will crush the wash the spinach with methanol. Crushing breaks the plant cells open and the methanol wash removes much of the water from the spinach. Make a drawing showing how methanol molecules (CH3OH, methyl alcohol) are attracted to water molecules (and vice versa). Name the attraction and explain why it is a relatively strong intermolecular attraction.

Chemistry 108 Plant Pigment Lab

4

In the second step of the lab, we will extract the pigment molecules in a technique called liquid-phase extraction. In this step you will separate the hydrophobic plant pigment molecules from other hydrophilic component molecules and solids. This is done by placing the pulverized spinach in a flask that contains an extremely hydrophobic liquid called “petroleum ether” and a hydrophilic liquid. These two liquids will not mix and form two layers because one is hydrophobic and the other hydrophilic. When the flask is shaken vigorously, then left to settle so that the two liquids separate into layers again, the spinach pigment molecules will be extracted into the hydrophobic petroleum ether layer. These various pigment molecules that are extracted into the petroleum ether layer will be separated from each other in a final TLC Chromatography step of the lab described earlier. 3) Petroleum ether contains molecules like CH3CH2CH2CH2CH2CH3 and CH3C(CH3)2CH2CH(CH3)CH2CH3. (Note: Although the name implies that petroleum ether is in the ether family, it is not; it is a mixture of non-aromatic hydrocarbons. The name is historical from when “ether” was the same as “spirits” and had to do with the fact that the hydrocarbon mixture had a high vapor pressure.) Look at the structure of β-carotene (one of our plant pigments) below and thoroughly explain why β-carotene is much more soluble in petroleum ether than in water. Do not simply say that “like dissolves like” (although it is true), name and explain the nature of the intermolecular force that is involved here.

Chemistry*108*lab** Name_________________________*

Lab #5: EXTRACTION AND SEPARATION OF PLANT PIGMENTS

Purpose of the lab: The purpose of this lab activity is for the student to learn about extraction and chemical separation technology. Specifically, the student will learn how to do a liquid phase-extraction and Thin Layer Chromatography in order to separate a mixture of molecules. Method: In the first step of this lab, the student will remove some of the water from spinach using a methanol rinse. In the second step, liquid-phase extraction will be used to extract pigment molecules from the partially dehydrated pulverized spinach. In the third and final step, the pigment molecules will be separated from each other using thin layer chromatography. PROCEDURE 1. Vigorously crush about 10 spinach leafs with a mortar and pestle for 30 seconds.

2. Add 30 mL of methanol and continue to crush the spinach-methanol mixture for 30 seconds. Pour off

as much of the methanol extract as possible and discard it in the waste container in the hood. It is ok if a small amount of crushed spinach is lost in the waste.

3. Locate the nearest fire extinguisher. There is a fire extinguisher near three of the four corners of the lab, so one should be nearby. Pick it up, study it, and decide exactly what you would do if you needed to use it. If you have any questions, ask the instructor. Replace the fire extinguisher on its mount on the wall.

4. Working under your bench-top hood to contain the vapors, add 25 mL of methanol and 35 mL of petroleum ether to the spinach in the mortar (Caution: petroleum ether is extremely flammable).

5. Crush for two minutes, and filter the liquid with vacuum as follows: • Use a 250 ml side-arm flask,

Buchner funnel, and a piece of #3- filter paper in the funnel. Clamp the vacuum flask to a ring-stand. Clamps are in the common drawers.

• Turn on the vacuum then pour the spinach-liquid mixture from the mortar into the funnel.

6. Place the liquid that you collected in to a clean 125 mL Erlenmeyer flask.

Buchner Funnel

Line to vacuum

250 ml side-arm flask

Figure 1. Filtration apparatus.

Chemistry*108* Plant*Pigment*Lab* *

2**

7. Add 25 mL of water to the liquid in the 125 ml Erlenmeyer flask from step 6. Stopper, then shake slowly (try not to get foaming) for 30 seconds, and then allow the two liquid layers to separate. At this point you should have a dark, intensely green, clear layer on top and a lighter green, cloudy layer on the bottom. (If you don't have a dark green layer on top, add 10 mL of petroleum ether and stopper, then shake again. If that doesn't work, consult with the instructor.)

8. Use a Pasteur pipette (located on the counter) to transfer about 1 ml of the dark green, top layer to your smallest, Erlenmeyer flask. Be sure the flask is dry; do not pre-rinse it.

9. Obtain two chromatography slides. You will be working in pairs. Each student should spot one slide. Spot using the capillary tubes provided and develop the slides in the capped jars as instructed below. 9.1. The spots must be small and dark. Spot the slide about 3 times (in the same place): ¾ of an inch

from the bottom of the slide, waiting about 30 seconds between each spotting so the previous spot dries. DRAW A LINE NEXT TO YOUR SPOT (NOT TOUCHING THE SPOT) SO YOU WILL BE ABLE TO CALCULATE THE Rf VALUES LATER.

9.2. Use the developing troughs at the end of the lab benches. Place both partners’ slides in the trough, the developing solvent will work best if you keep the lid on trough. Note your slot number so you do not exchange your TLC slide with someone else’s.

9.3. Stop the development when the solvent comes to within 1.5 inches of the top of the TLC slide or when you can clearly see good separation of 5 spots. Immediately draw a line near the top of the slide where the solvent reached for calculation of Rf values.

DISCUSSION/DATA

The visible spots are, in order of decreasing Rf values, due to β-carotene (orange to orange-yellow with the highest Rf value), pheophytin a & b (grey, the pheopthytin b spot is often too weak to be seen), chlorophyll a & b (green to blue-green), and the xanthophylls (yellow with the lowest Rf value). Xanthophyll is a general term that encompasses many different compounds. They are molecules of !-carotene that have been oxidized, so they contain many OH groups. Pheophytin a & b are identical to chlorophyll a & b except the Mg ions are replaced by hydrogen ions.

Chlorophyll(a( Chlorophyll(b((


3**

Measure the distance traveled by each spot and by the solvent. START ALL MEASUREMENT FROM WHERE THE SLIDE WAS ORIGNALLEY SPOTTED. Make a drawing in the box (below and to right) of your TLC slide, identify the spots, and record the distance traveled for the solvent and for each pigment molecule in the data table below. The pigment molecules are ordered from the top of your slide to the bottom, with spot #1 being the spot that travels the furthest (β-carotene). Note that you may not be able to see spot # 3, #7, #8 because of the small amounts present in most spinach samples.

DATA: If you do not see all these spots, just write “not visible “for those not seen. Start all distance measurements, including the distance the solvent traveled from your original spot.

Start your Rf measurements where you originally spotted the slide. Then measure to the center of each spot.

Spot Molecule Distance traveled(cm)

solvent

distance the solvent traveled from your original spot

1 top spot, highest on slide

β-carotene color = yellow/orange

2

Pheophytin a color = grey

3

Pheophytin b color = grey if visible

4

Chlorophyll a color = blue/green

5

Chlorophyll b color = green

less intense than chlor. a

6

Xanthophyll l color = yellow

7

Xanthophyll 2 color = yellow if visible

8


Drawing*of*developed*slide*goes*here:*


4**

Calculate the Rf values:

QUESTION 1) Silica gel (the coating on your TLC slide) is very polar and the developing solvent is quite non-polar. Based on your TLC slide data (Rf values), which of the plant pigments that you see on your slide is the most polar? Explain.

Spot Molecule Calculation Rf value

1 top spot, highest on slide

β-carotene color = yellow/orange

note: pay attention to significant figures and units

2

Pheophytin a color = grey

3

Pheophytin b color = grey if visible

4

Chlorophyll a color = blue/green

5

Chlorophyll b color = green

less intense than chlor. a

6

Xanthophyll l color = yellow

7


8


Chemistry 108 Carboxylic Acids Pre-lab

1

Lab#6 Prelab CARBOXYLIC ACIDS

Background: O || Carboxylic acids contain the carboxyl functional group: – C – O – H * carboxyl + hydroxyl groups. Also written as – COOH These are found in amino acids, proteins, fatty acids and oils, and as intermediates of carbohydrate metabolism. Four to twenty C-atom-chain carboxylic acids are obtained from animal and plant fats or oils and so are often called fatty acids. Nomenclature: IUPAC Names:

• Take the base name from longest C atom chain containing the carboxyl group, with that acting as C #1. Drop the "e" ending from the parent alkane name and add the suffix "–oic acid".

O || ex: ethanoic acid : CH3 — C — OH Common Names: • Use the suffix " - oic acid" added to the Latin/Greek name origin. Examples: formic acid (methanoic acid) and acetic acid (ethanoic acid). Acidity of Carboxylic Acids: • The hydroxyl H atom is weakly acidic (ionizes ~ 0.5 %). RCOOH + H2O (l) RCOO

– (aq) + H3O+ (aq)

• RCOO– is the carboxylate ion, and is much more soluble than the parent acid because it has a charge (it is an ion). • Nomenclature of carboxylate ions is done by changing the "–ic acid" suffix of the parent

acid to "–ate ion". • Note that the reaction is reversible, the carboxylate ion can be converted to the carboxylic

acid form under acidic conditions (neutralization of a base with an acid):

RCOO– (aq) + H3O+ (aq) RCOOH + H2O (l)

Or RCOO– (aq) + HCl (aq) RCOOH + Cl- (aq)

• Likewise, The acid form can be converted to the base form (carboxylate ion) under basic conditions (neutralization of an acid with a base):

RCOOH + OH- (aq) RCOO– (aq) + H2O (l)


2

Preparation (synthesis) of Carboxylic Acids: 1. Oxidation of an Aldehyde: Addition of an oxygen atom to an aldehyde creates a carboxylic acid. O O || || R – C – H + O R – C – O – H

• The source of the oxygen atom can be from oxygen gas (O2), or another oxygen containing compound such as the permanganate ion (MnO4-).

2. Oxidation of a primary (1o) alcohol: Oxidation of a primary alcohol creates a carboxylic acid. O || R – C– O – H + O R – C – O – H

• The source of the oxygen atom is from a strong oxidizing agent such as the dichromate ion (Cr2O7

2-). Carboxylic Acid Reactions:

1) Neutralization: A carboxylate salt is created upon addition of a strong base to a carboxylic. • Note: neutralization of a long chain carboxylic acid (fatty acid) is called Saponification.

RCOOH + NaOH RCOO– Na+ + H2O 2) Esterification: Addition of an alcohol with a strong acid catalyst creates an ester. O O || H+ || R – C – O – H + H – O – R' R – C – O – R' + H2O 3) Amide Formation: Amides are formed by reaction of an amine with a carboxylic acid. This is an endothermic reaction (requires energy). In biological systems, where it is not possible to use heat, a specific enzyme is needed

a) Reaction of carboxylic acids and ammonia (NH3) O O || || R’ – C – OH + H – N– H R’ – C – NH2 + H2O | H


3

b) Reaction of carboxylic acids and primary (1o) amines (NH2R) O O || || R’ – C – OH + H – N – R R’ – C – N–R + H2O | | H H

c) Reaction of carboxylic acids and secondary (2o) amines (NHR2)

O O || || R’ – C – OH + H – N– R’ R’ – C – N –R’ + H2O | | R” R”

d) Reaction of carboxylic acids and tertiary (3o) amines (HR3) NO REACTION occurs with tertiary (3o) amines

Solubility of Carboxylic Acids and their Conjugates Bases:

Acid Form O || R – C – OH

Carboxylic Acid

Base Form O || R – C – O-

Carboxylate ion If R has less than 5 carbons:

x Soluble (aq)

If R has less than 12 carbons:

x Soluble (aq)

If R has 5 or more carbons:

x Insoluble (s)

If R has 12 or more carbons:

x Amphipathic x Forms monolayers or micelles

For the acid (HA) and its conjugate base (A-), HA + H2O A- + H3O+

• There is more HA when the pH is lower than the pKa. • There is more A- when the pH is higher than the pKa. • There are equal amounts of HA and A- when the pH = pKa.


4

Questions: 1) Name the following carboxylic acids: a) CH3CH2CH2COOH:

IUPAC: ___________________________

b) CH3CHCH2CH2COOH: | CH3

IUPAC: ___________________________

2) Draw line bond structures for the following carboxylic acids:

a) 2, 3 – dibromohexanoic acid b) 3 – methylpentanoic acid:

3) Write the reaction when acetic acid reacts with 1-butanol (CH3CH2 CH2 CH2―OH) 4) Write the reaction when propanoic acid reacts with ethanamine: 5) Write the reaction when the butanoate ion is neutralized by hydrochloric acid:


Lab #6: CARBOXYLIC ACIDS LAB PART I: Preparation of Carboxylic Acids (a) Oxidation of an Aldehyde by Oxygen from the Air: Benzaldehyde is an aromatic aldehyde with a familiar odor. On a clean, dry watch glass, place 3 or 4 drops of the pure liquid benzaldehyde (this is 100% benzaldehyde, it is not a solution). Spread the benzaldehyde out on the glass so as to expose more of it to the oxygen in the air and allow it to remain for most of the lab period. Near the end of the period, examine the material on the watch glass. Record your observations describing fully the starting material and any changes that occur. Write a balanced equation for the reaction of benzaldehyde and oxygen gas to produce the carboxylic acid product: Move on to part (b) and the rest of the experiment, then come back to this question: Based your observation of the product, how do you know a chemical reaction occurred? ______________________________________________________________________________ ______________________________________________________________________________ (b) Oxidation of an Aldehyde by Potassium Permanganate Solution: Place about 2 drops of benzaldehyde in a small test tube and add about 4 drops of 6 M sodium hydroxide solution and about 12 drops of 0.1 M potassium permanganate (KMnO4) solution. MIX contents of the test tube. Allow the mixture to react for 10 minutes. (You can do a model from Part 3 while you are waiting.) Describe your observations: Note: The oxidizing agent is the permanganate ion (MnO4

-). MnO4- is reduced to MnO2, a brown

precipitate, in the reaction.

Chemistry 108 Carboxylic Acids Lab

2

Does the acid-form (benzoic acid) or the base-form (benzoate ion) dominate the equilibrium in this tube? Since the solution is very basic, the pH is much greater than 7 and the pKa of benzoic acid is about 5, you should know if the acid- or base-form dominates based on what we learned in lecture (chapter 9 and 10, comparing pH and pKa). Draw the line bond structure of that predominant form in the box to the right.

Obtain a piece of filter paper from the supplied for the experiment. Folding it properly, first, place it in your funnel and then moisten the filter paper with DI water. Put your funnel in a clean large test tube and pour the brown suspension from the small test tube into the filter/funnel. Wait until at least a few drops of liquid pass through the funnel and into the large test tube. Then, REMOVE THE FUNNEL, and add 4 drops of 12 M HCl to the filtrate (liquid that went through the filter paper and is now in the large test tube). The HCl will make the solution acidic. Describe what you see. If you do not see a change occur when you add the HCl, see the instructor.

Observations: Think about what you observed based on the solubility of carboxylic acids vs. carboxylate ions! Write a balanced chemical equation for the reaction that occurred when you added the HCl.

• By adding the acid, you are converting your product formed (you drew it in the box in the upper right hand corner of this page) in the oxidation reaction to its acid form (reacting with the HCl, HINT: see the prelab neutralization equations).

Equation:


3

Part II: Solubility of a Carboxylic Acid and Its Salt in Water:

Neutralization of a Carboxylic Acid: The solubility of carboxylic acids in water is very low when they contain more than 5 or 6 carbon atoms. Knowing this, predict the water solubility of benzoic acid and explain your answer. Circle one: predicted to be soluble or predicted to be insoluble Explain_______________________________________________________________________ Now test your prediction by adding benzoic acid crystals (0.01g to 0.02 g) to 2 ml of water in a large test tube. Shake well and record your observations (what observation tells you if it was soluble or not): Test the pH of the mixture in your test tube by touching the tip of your stirring rod to the suspension in the test tube and then touching a spot of liquid from the tip of the stirring rod onto a piece of red litmus paper, and then place another spot of the liquid onto blue litmus paper. The spot on the red litmus paper should look red (indicating that the solution is acidic, pH<7). The spot on the blue litmus paper should look also look red (indicating that the solution is acidic, pH<7). Record the spot color on the red litmus paper: __________________________________ Record the spot color on the blue litmus paper: _________________________________ Note that although carboxylic acids are “weak acids”, they do produce enough H3O+ to make the solution acid. Write a balanced chemical equation for the acid reacting with H2O (see prelab):

• Is the acid form or the base form predominant at pH < 5? __________________. Now add, one drop at a time, 1 M NaOH solution, stirring the contents of the test tube after each addition of the NaOH and testing the pH of the solution using the red litmus paper. Continue adding NaOH solution in this manner until the solution pH causes red litmus paper to turn blue, indicating that the pH >7.


4

Check the contents of the test tube and record your observations (is the solid still present?): Write the chemical equation describing this reaction (see prelab the reaction of a carboxylic acid and a base): Explain why you no longer see a solid present. What kind of reaction is this (see prelab)? _____________________________ .


5

Part III: Models Examine the numbered models set out by the instructor in the balance room. Draw condensed structural formulas for the compounds represented. Write the organic family name of the molecule (for example: alkane, alkene, amine, carboxylic acid etc.….) Also write the systematic name for each molecule. Atoms Colors: Black = carbon, White = hydrogen, Red = oxygen, Blue = nitrogen

MODEL # CONDENSED STRUCTURAL FORMULA FAMILY SYSTEMATIC NAME

1

2

3

4 For this model, do not draw the condensed form here; instead, draw a side view. See you Chapter 4 Lecture Notes if you do not recall side-view representations.

5

6

7


6

Post-Lab Questions: Draw the products for the following reactions:

H3C

C

O

OH CH3

CH

CH2

H2C

CH3

HO

+

Acid Catalyst

Hint: Check your prelab CH3CH2CH2COOH + NaOH CH3CH2CH2COO- + HCl

NAME: ____________________

Lab #7: Carbohydrates Prelab Introduction: Of the three main sources of energy and raw materials for cellular function, carbohydrates are normally ingested in the largest portions, typically 60 to 70% of our diet. They are also the primary fuel, or source of energy, for the cells in our body. The main categories of carbohydrates are sugars, starches and cellulose. Carbohydrates were originally thought to be hydrates (containing water) of carbon, and this was the derivation of the name. It turned out that they were not hydrates, but did contain the elements C, H and O expressed in the ratio: Cx(H2O)y. They are now usually classified as polyhydroxy (many OH groups) aldehydes or ketones. The simple sugars are the monosaccharides, with the three main dietary ones being glucose (blood sugar), galactose and fructose (fruit sugar). They all have the general formula C6H12O6, and are isomers that exist in both open and closed forms. The open chain forms are regularly shown as Fischer projections and the closed ring forms as Haworth diagrams. These are shown below for the monosaccharides.

C

HC

CH

HC

H2C

OH

HO

OH

OH

CH2OH

OH

CH2OH

C

CH

HC

H2C

O

HO

OH

CH2OH

OH

CH2OH

CH2OH

OH

OH

OHO

O

OH

H

OH

OH

OH

CH2OH

Glucose(an aldohexose)

α-D-Glucose

Fructose(a ketohexose)

1

2

34

beta-D-Fructose

OOH OH

H

OH

OH

CH2OHC

HC

CH

CH

H2C

OH

HO

OH

CH2OH

OH

HO

β-D-GalactoseGalactose

D-Glucose (an aldohexose)

D-Galactose (an aldohexose)

D-Fructose (an ketohexose)

Β-D-Fructose

The position of the OH group on the anomeric carbon on the far right of the ring determines whether the sugar is an alpha (α) or a beta (β) sugar. The OH group points down below the ring in α sugars and up above the ring in β sugars. These have important consequences later for digestion purposes. The other sugars are the disaccharides, formed when two monosaccharides join together, releasing a molecule of water in the process. If the O-link (glycosidic bond) between the monosaccharide units is in the down position, it is an α O-bridge, or a β O-bridge if pointing up. The three main dietary disaccharides are maltose, sucrose and lactose. Maltose (malt sugar) forms when two glucose units join with an α O-bridge. Sucrose (table sugar) is a glucose with fructose, and lactose (milk sugar) is glucose with a galactose. To digest these disaccharides, the necessary digestive enzymes, maltase, sucrose, and lactase, must be present in the small intestine. If many glucose monosaccharides are joined together, then a polysaccharide forms. Sometimes called complex carbohydrates, the polysaccharides include plant and animal starch, as well as cellulose. Starch is the main energy storage form for plants, usually in the seeds. The glucose units are joined by α O-bridges and are therefore digestible by humans with the aid of the amylase enzyme in saliva and pancreatic juices. Starch consists of about 20% amylose and 80 % amylopectin. Amylose is linear and hundreds to several thousand glucose units long, whereas amylopectin is branched and up to 100,000 glucose units. Glycogen is the main energy storage form for animals, mainly in their liver and muscles. It is often called animal starch. It consists of up to 1 million glucose units joined by α O-bridges and is even more highly branched than amylopectin. Cellulose is the main structural component of plant cell walls. It consists of several thousand glucose units in a linear fashion just like amylose, but has β O-bridges and therefore is not digestible by humans. Some ruminant animals contain special bacteria in their stomachs that allow digestion of cellulose β bridges. For humans, it acts as dietary fiber, or “roughage”, and cleans out the digestive tract. In this experiment we will become acquainted with some of the common tests/reactions for carbohydrates. We will look at the reactions of fructose, sucrose, cellulose, lactose, and starch. We will use these reactions to determine which of the five sugars is present in an unknown. Part I: Benedict's Test Benedict’s reagent is a copper compound that will oxidize only aldehyde groups (aldoses) and not alcohols. If you consider cyclic forms of carbohydrates, hemiacetals give positive tests while acetals give negative tests. The reason for this is that the cyclic form interconverts (is in equilibrium) with the linear form that contains an aldehyde!

A sugar that reacts with Benedict’s solution is called a reducing sugar since it reduces the ion Cu2+ ! Cu+

Part II: Seliwanoff’s Test The Seliwanoff reagent contains hydrochloric acids which converts fructose to S-hydroxymethylfurfural. The reagent also contains resorcinol which reacts with the S-hydroxymethylfurfural to give a red color. Complex carbohydrates which contain fructose units can also give a positive test. Aldohexoses react similarly, but more slowly.

Part III: Iodine test One of the two molecular components of starch are shaped like very long spiral staircases, inside of which is just enough space to accommodate iodine molecules. The blue color arises when the electrons of the entrapped iodine molecules interact with the electrons of the starch molecule and the resulting complex absorbs visible light (appears dark). Questions: 1) Why do some disaccharides give positive Benedict’s tests while other disaccharides give negative tests? 2) Fructose is not an aldose. Why does it give a positive Benedict's test? 3) Is the aldehyde group on a sugar oxidized or reduced in the Benedict’s test? Explain.

4) A carbohydrate that gives a positive Benedict's test is said to be a reducing sugar. Why? What's being reduced?

5) Cu2+ in Benedicts tests is a weak oxidizing agent. Why don't we use a stronger oxidizing agent?

6) Starch is made up of two components. a) What are the two components, and how do the two components differ from each other? b) Which of the two components of starch is similar to glycogen? How does the component differ from glycogen? c) Draw a section of a amylose molecule containing two glucose units. Label the linkages (for example β (1→4)) and place an asterisk (*) next a carbon where branching would occur if it was a amylopectin molecule.


Lab #7: CARBOHYDRATES LAB INTRODUCTION In this experiment we will become acquainted with some of the common carbohydrate reactions of fructose, sucrose, cellulose, lactose, and starch. We will use these reactions to determine which of the five is present in an unknown. PROCEDURE Part I: Benedict's Test Benedict’s reagent is a copper compound that will oxidize only aldehyde groups (aldoses) and not alcohols. If you consider cyclic forms of carbohydrates, hemiacetals give positive tests while acetals give negative tests. The reason for this is that the cyclic form interconverts (is in equilibrium) with the linear form that contains an aldehyde, for example:

1) Add about 1-2 inches of water and 3 boiling chips to a 600 mL beaker (it is the largest beaker in your drawer). Place the beaker and water on a hot plate, turn the hot plate on, and begin to boil the water.

2) Label 7 medium size test tubes (from your drawer) for each of the 5 sugars, your unknown, and your partner’s unknown. DO NOT ADD THE SUGAR SOLUTIONS UNTIL LATER

3) Using the metering dispenser, place 2 mL of Benedict's solution in EACH OF THE 7 labeled, medium size test tubes and heat in gently-boiling water bath for 2 minutes. NOTE: groups of 3 students will have 8 tubes.

4) Remove the tube that is labeled for the fructose solution. Add 10 drops of the 5% fructose solution to the Benedict's solution, mix thoroughly, and set the tube in the boiling water bath for 60 more seconds. Remove the tube from the water bath and check for a color change. A color change from clear blue to cloudy green, yellow or to brick red indicates the carbohydrate is a reducing sugar. The brick green, yellow, or red color is copper(I) oxide.

5) Repeat step 4 for your other 6 sample test tubes, using sucrose, cellulose, starch, lactose and each partner’s unknown instead of fructose. When using dropper to add your unknown solutions to the tubes, be sure to rinse the droppers after each partner’s unknown to prevent cross-contamination of unknowns. Record your results in the DATA TABLE.

Chemistry 108 Carbohydrates Lab

2

Part II: Seliwanoff’s Test The Seliwanoff reagent contains hydrochloric acids which converts fructose to S-hydroxymethylfurfural. The reagent also contains resorcinol which reacts with the S-hydroxymethylfurfural to give a red color. Complex carbohydrates which contain fructose units can also give a positive test. Aldohexoses react similarly, but more slowly.

1) Maintain the boiling water bath used in Part II, you may need to add more water because of evaporation. 2) Put 5 mL of freshly prepared Seliwanoff’s reagent in a medium test tube, add 10 drops of 5% fructose solution and place in a boiling water bath. (Caution: This is a strongly acidic solution.) A red color within two minutes is indicative of a ketohexose.

• Record your results in the DATA TABLE. 3) Repeat step 2 for each of your other six carbohydrate solutions (instead of fructose). Part III: Iodine test Some of the starch molecules are shaped like very long spiral staircases, inside of which is just enough space to accommodate iodine molecules. The blue color arises when the electrons of the entrapped iodine molecules interact with the electrons of the starch molecule and the resulting complex absorbs visible light (appears dark). Be sure to rinse your graduated cylinder between sugar additions so you do not cross contaminate your samples! 1) Add 2 mL of the 5% fructose solution to a small test tube (from your drawer), add two drops of iodine solution.

• A deep blue color is a positive test for starch. • Record your results in the DATA TABLE.

2) Repeat for the other 6 carbohydrate solutions (instead of fructose).


3

DATA TABLE (write (+) for a positive test and (–) for a negative test)

Benedict’s

Test Seliwanoff’s Test

Iodine Test

Fructose

Sucrose

Cellulose

Starch

Lactose

Unknown #

Conclusion: Unknown Number _________________ is __________________________.


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Part IV: Representing 3D structures with two-dimensional drawings. (a) Fischer projection formulas of open-chain monosaccharides: • Select model #1 of an open-chain monosaccharide. If the model comes apart when you are

manipulating it and you are not sure how to put it back together, bring it to the instructor!

Does the model represent an aldose or ketose? ______________ Does the model represent a triose, tetrose, etc.?______________ Does the model represent a L- or a D- sugar?______________ Classify this monosaccharide as a combination of the previous 3 answers ___________________. Determine and draw the Fischer projection for this model below, being sure to view each C atom so that the –H and – OH groups are facing up towards you (above the plane). Replace model #1 and repeat the same procedure for models #2-4. MODEL# 2 MODEL #3 MODEL#4 Classification: ________________ Classification: ________________ Classification: ________________ Fischer Projection Fischer Projection Fischer Projection (b) Haworth projection formulas of cyclic monosaccharides. View model #5 of the cyclic monosaccharide and answer the following: • Is the ring structure a pyranose (6 sides) or a furanose (5 sides): ___________________

• Is the structure α or β?__________________

Draw the Haworth projection for the compound represented by this model.


Lab #8: Protein, Triglycerides, and Esters Lab Purposes of the Lab: In Part 1, you make cheese by separating the caseins from the other components of milk. In Part 2, you will make soap using the saponification of triglycerides reaction. In Part 3, you will synthesize two esters that smell very good! Safety Precautions: 1. Use goggles at all times. 2. Wear gloves at all times. 3. Never put room temperature glass on a hot hotplate surface. 4. Do not eat the cheese that you make; you did not use sanitary glassware! 5. Do not use the soap that you make; it was made in lab glassware that may have contained other chemical residues. Part 1 –Cheese Making: Separating the Caseins from Milk Procedure: 1. Pour 30 mL of nonfat milk into your 100mL beaker. 2. Place the 100 mL beaker on your hotplate. Turn the heat to the 50% (half way) position. 3. Monitor the temperature until the milk reaches 40oC. Swirl the milk every few minutes. See the instructor’s apparatus for proper thermometer installation. 4. When the temperature of the milk has reached 40oC, TURN OFF THE HOTPLATE. Remove the beaker from the hotplate. Add glacial acetic acid drop-by-drop to the warm milk, while slowly stirring with your glass stir-rod, until the caseins flocculate (become solid), then add two more drops of glacial acetic acid. The caseins may solidify on your stir-rod, that is ok. 5. Obtain a piece of cheesecloth. Fold the cheesecloth in half two times so that it is 4 layers thick. You will filter the flocculate using cheesecloth. While one lab partner holds the cheesecloth above the sink, the other partner pours the cheese and liquid onto the cheesecloth. Next, rinse the cheese by running tap water over cheese while it is still on the cheesecloth. 6. After rinsing, wrap the cheesecloth around the cheese then squeeze any excess liquid into the sink. 7. Remove the cheesecloth to examine the cheese you made. 8. Dispose of the cheese and cheesecloth using any trashcans in the room

• You can make cheese at home in the same manner using some vinegar or lemon juice as a substitute for the glacial acetic acid.

Chemistry 108 Protein, Triglycerides, and Ester Lab

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Part 2: Synthesis of Triglycerides: Soap Making • SAFETY WARNING: YOU WILL BE USING A VERY STRONG BASE (NaOH)

SOLUTION. STRONG BASES CAN BE JUST AS DANGEROUS AS STRONG ACIDS.

PROCEDURE:

1. Using the graduated cylinder that is located next to the olive oil, measure 20.0 mL of olive oil into your 100mL beaker.

2. OPTIONAL: Add 6 drops of any fragrant oil extract or one drop of perfume/cologne to your 100 mL beaker that contains the olive oil.

3. Using the graduated cylinder that is provided and labeled for the NaOH solution, measure 5.0 mL of the NaOH solution. SLOWLY add it to the 100ml beaker containing your olive oil while stirring with your glass stir-rod.

4. Obtain a magnetic stir-bar and add it to the beaker containing the olive oil and NaOH. 5. Place the beaker on the stir-plate (it is the hotplate that you used for heating in previous labs). If

your hotplate surface is still hot from Part 1, use another hotplate (there are two hotplates at each lab station). Set the stir speed to ‘medium’ and stir for 5 minutes. SAFETY ALERT: BE SURE TO USE THE STIR CONTROL KNOB AND NOT THE HEAT CONTROL KNOB!!!

6. While the mixture is being stirred, get a paper cup (for use as a soap mold) and write your initials on the cup.

7. After 5 minute of stirring, turn off the stir-plate. While wearing gloves, remove the stir-bar. 8. Pour the mixture of NaOH and oil into your paper cup mold. 9. Place your mold with the soap in the area designated by the instructor for storage. 10. OPTIONAL: If both partners want to make a bar of soap, using gloves and paper towels, wipe

the excess oil from your 100 mL beaker and stir-bar then repeat the procedure for lab partner #2. You can put the used paper towels into any of the trashcans in the lab.

11. OUR LEGAL DEPARTMENT FORBIDS THE USE OF THE SOAP MADE IN LAB. 12. Use soap and water to thoroughly clean your 100 mL beaker and stir bar. 13. Store you soap in your lab locker until we meet for the lab locker checkout, it will be solidified

by then.

Part 3: That Smells Good: Esterification Reactions Esterification: Addition of an alcohol with a strong acid catalyst creates an ester. O O || H3O

+ || R – C – O – H + H – O – R' R – C – O – R' + H2O

(a) Esterification of Acetic Acid with 1-pentanol: 1) Start a boiling hot water bath by adding about 2 inches of water and a few boiling chips to your 600 mL beaker. Place the beaker on a hotplate and set the temperature to the highest setting until the water begins to boil. (Note, after the water bath begins to boil, turn the heat control down to about half-power to maintain a gentle boil). Go on to step #2 while waiting for the water to boil.

2) While the water is coming to a boil, wearing protective gloves, add about 15 drops of glacial acetic acid and about 6 drops of 1-pentanol to a large dry test tube. (Glacial acetic acid is the name given to pure acetic acid, which is a liquid above 16.6 °C). Mix the contents of the tube, and then add 5 drops of concentrated sulfuric acid, one drop at a time, with mixing after each drop is added.


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3) After the H2SO4 has been added, heat the contents by placing the test tube in the beaker of boiling water for 3 minutes. 4) Now pour the contents of the test tube into a small beaker that is half-filled with tap water (not the beaker that you used for boiling). Does there appear to be any "oil" on top of the water? ________________. (The liquid ester appears like oil does on water because it is insoluble.) Cautiously determine/describe the odor of the product. (Can you identify the aroma?): Write the chemical equation that describes the formation of this ester: Name this ester: _______________________________________________ (see your chapter 10 lecture notes for naming esters) (b) Esterification of Salicylic Acid with Methanol: 1) Place about 0.5 grams of salicylic acid (2-hydroxybenzoic acid) in a clean dry test tube. 2) Add 3 ml of methanol to the salicylic acid. 3) When the entire solid has dissolved, wearing protective gloves, slowly add 10 drops of concentrated sulfuric acid, mixing the contents after each addition. 4) After all of the sulfuric acid has been added, heat the test tube in the beaker of boiling water for 3 minutes. Next, pour the contents of the test tube into a small beaker of tap water (not the beaker that you used for boiling). You should see a solid form in the cool water. The solid is the ester that you synthesized. Cautiously determine/describe the odor of the product: Write the chemical equation that describes the formation of this ester. Name this ester: _______________________________________________


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Part 4: Models Structures and names of some organic compounds Examine the numbered models found in throughout the lab. Draw a condensed structural formula for each molecule represented. In the case of alcohols or amines, classify each as 1o, 2o or 3o. Name each compound. If you wish to draw the condensed formulas on a separate page, that is fine. Black spheres are carbon, white are hydrogen, blue are nitrogen, and red are oxygen. Model Condensed Formula Classification Name 1

2

3

4

5

6

7

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Chemistry*108* Protein,*Triglycerides,*and*Esters*Lab*

Lab #8 Prelab: Protein, Triglycerides, and Esters Lab

Purpose of the lab: In Part 1, you will make cheese by separating the caseins from milk. In part 2, you will make soap using the saponification of triglycerides reaction. In part 3, you will synthesize two esters that smell very good!

Part 1: Proteins in milk: Separating the Caseins from Milk The name protein is taken from the Greek protelos, which means first. This name is well chosen. Of all chemical compounds, proteins could certainly be ranked first, for they are the substance of life. Proteins make up a large part of the animal body, they hold it together, and they run it. They are found in all living cells. They are the principal material of skin, muscle, tendons, nerves, and blood; of enzymes, antibodies, and many hormones. (Only the nucleic acids, which control heredity, can challenge the position of proteins; and the nucleic acids are important because they direct the synthesis of proteins.) "Chemically, proteins are high polymers. They are polyamides, and the monomers from which they are derived are the α-amino carboxylic acids. A single protein molecule contains hundreds or even thousands of amino acids units; these units can be of twenty-odd different kinds. The number of different protein molecules that are possible, is almost infinite. It is likely that tens of thousands of different proteins are required to make up and run an animal body; and this set of proteins is not identical with the set required by an animal of a different kind." - Morrison & Boyd's Organic Chemistry In this lab activity we will study proteins - described so eloquently in the above quote from Morrison & Boyd's Organic Chemistry- from several different viewpoints. In Part 1, we will separate some amino acids (including an unknown) by using paper chromatography. In Part 2 we will use a pH meter to

measure the acidity of solutions of three amino acids. In Part 3 we will separate the caseins from milk. Caseins*(from Latin caseus "cheese") make up about 80% of the proteins in milk. Caseins are phosphoproteins, that is, there are phosphate groups attached to some of the amino acid side chains. It is interesting that proline (see figure), an unsymmetrical amino acids makes up 20% of the amino acids residues in some caseins. In proline, the side chain makes a ring structure with the amino group. The result is a lack of ordered internal protein secondary structure probably and facilitates attack by digestive enzymes in infants. The rest of the milk protein is a mixture of soluble proteins that includes the albumins and globulins; these are commonly called the whey proteins.

Casein consists of four different proteins, usually referred to as α, β. γ, and κ caseins. The α, β, and κ forms are most abundant, making up roughly 50%, 33%, and 15% of the casein, respectively. Casein exists in milk as the calcium salt, calcium caseinate. The protein has a negative overall charge and its charge is balanced by the positive calcium ions. Surprisingly, though, it is found that calcium ions, at the concentration of Ca2+ normally found in milk, cause α and β casein, singly or in combination to precipitate. However, in milk, the κ casein is soluble and is thought to surround the α and β casein forming a soluble micelle. Disruptions, such as denaturation or enzymatic bond cleavage, of the micelles cause the casein proteins to coagulate; this is what the cheese-making process does. This can be done with acids, bases, temperature, mechanical agitation, or with certain enzymes (as in the case of yogurt); we will use acetic acid in the lab activity.


Part 2: Saponification of Triglycerides: Soap Making

Soaps are amphipathic molecules used for cleaning and bathing. Soaps molecules are the base form, the carboxylate ions, of fatty acids. Soaps are made using the saponification reaction that converts a triglyceride molecule into 3 carboxylate ions (soap molecules) and a glycerol molecule. Since carboxylate ions have a polar head and a nonpolar tail, they are able to emulsify oil and nonpolar substances in the cleaning and bathing processes. An example of the saponification reaction is given below.

Note that the fatty acid residues contained in the reactant triglyceride are arbitrary; any fatty acid residues could have been used.

This reaction is simply three "hydrolysis of an ester" reactions. We first saw the hydrolysis of an ester reaction in chapter 6 then once again in chapter 10. In the saponification of triglycerides, the reaction is catalyzed by a strong base. Since the reaction occurs in a strong base (pH > pKa of fatty acids), it is the base form of the fatty acids, the carboxylate ions, which are produced. The base used for catalysis is usually sodium hydroxide. Sodium hydroxide is also called "lye". When one makes soap using sodium hydroxide, the soap is called "lye soap".


History of Soap Making (From Wikipedia):

(Note: I have removed the references from the literature cited, however, they can be found on the Wikipedia site if you want more information or you wish to use this information in a report for another class)

Early history. The earliest recorded evidence of the production of soap-like materials dates back to around 2800 BC in ancient Babylon. A formula for soap consisting of water, alkali, and cassia oil was written on a Babylonian clay tablet around 2200 BC.

The Ebers papyrus (Egypt, 1550 BC) indicates the ancient Egyptians bathed regularly and combined animal and vegetable oils with alkaline salts to create a soap-like substance. Egyptian documents mention a soap-like substance was used in the preparation of wool for weaving.

In the reign of Nabonidus (556–539 BC), a recipe for soap consisted of uhulu [ashes], cypress [oil] and sesame [seed oil] "for washing the stones for the servant girls".

Ancient Rome. The word sapo, Latin for soap, first appears in Pliny the Elder's Historia Naturalis, which discusses the manufacture of soap from tallow and ashes, but the only use he mentions for it is as a pomade for hair; he mentions rather disapprovingly that the men of the Gauls and Germans were more likely to use it than their female counterparts. Aretaeus of Cappadocia, writing in the first century AD, observes among "Celts, which are men called Gauls, those alkaline substances that are made into balls, called soap".

A popular belief claims soap takes its name from a supposed Mount Sapo, where animal sacrifices were supposed to have taken place; tallow from these sacrifices would then have mixed with ashes from fires associated with these sacrifices and with water to produce soap, but there is no evidence of a Mount Sapo in the Roman world and no evidence for the apocryphal story. The Latin word sapo simply means "soap"; it was likely borrowed from an early Germanic language and is cognate with Latin sebum, "tallow", which appears in Pliny the Elder's account. Roman animal sacrifices usually burned only the bones and inedible entrails of the sacrificed animals; edible meat and fat from the sacrifices were taken by the humans rather than the gods.

Zosimos of Panopolis, circa 300 AD, describes soap and soapmaking. Galen describes soap-making using lye and prescribes washing to carry away impurities from the body and clothes. According to Galen, the best soaps were Germanic, and soaps from Gaul were second best. This is a reference to true soap in antiquity.

Ancient China. Soap, or more accurately a detergent similar to soap, was manufactured in ancient China from vegetation and herbs. True soap, made of animal fat, did not appear in China until the modern era. Soap-like detergents were not as popular as ointments and creams.

Middle East. A 12th-century Islamic document describes the process of soap production. It mentions the key ingredient, alkali, which later becomes crucial to modern chemistry, derived from al-qaly or "ashes".

By the 13th century, the manufacture of soap in the Islamic world had become virtually industrialized, with sources in Nablus, Fes, Damascus, and Aleppo.

Medieval Europe. Soapmakers in Naples were members of a guild in the late sixth century, and in the eighth century, soap-making was well known in Italy and Spain. The Carolingian capitulary De


Villis, dating to around 800, representing the royal will of Charlemagne, mentions soap as being one of the products the stewards of royal estates are to tally. Soapmaking is mentioned both as "women's work" and as the produce of "good workmen" alongside other necessities such as the produce of carpenters, blacksmiths, and bakers.

15th–19th centuries. In France, by the second half of the 15th century, the semi-industrialized professional manufacture of soap was concentrated in a few centers of Provence— Toulon, Hyères, and Marseille — which supplied the rest of France. In Marseilles, by 1525, production was concentrated in at least two factories, and soap production at Marseille tended to eclipse the other Provençal centers. English manufacture tended to concentrate in London.

Finer soaps were later produced in Europe from the 16th century, using vegetable oils (such as olive oil) as opposed to animal fats. Many of these soaps are still produced, both industrially and by small-scale artisans. Castile soap is a popular example of the vegetable-only soaps derived by the oldest "white soap" of Italy.

Modern Times. In modern times, the use of soap has become universal in industrialized nations due to a better understanding of the role of hygiene in reducing the population size of pathogenic microorganisms. Industrially manufactured bar soaps first became available in the late 18th century, as advertising campaigns in Europe and the United States promoted popular awareness of the relationship between cleanliness and health.

Until the Industrial Revolution, soapmaking was conducted on a small scale and the product was rough. Andrew Pears started making a high-quality, transparent soap in 1789 in London. His son-in-law, Thomas J. Barratt, opened a factory in Isleworth in 1862. William Gossage produced low-priced, good-quality soap from the 1850s. Robert Spear Hudson began manufacturing a soap powder in 1837, initially by grinding the soap with a mortar and pestle. American manufacturer Benjamin T. Babbitt introduced marketing innovations that included sale of bar soap and distribution of product samples. William Hesketh Lever and his brother, James, bought a small soap works in Warrington in 1886 and founded what is still one of the largest soap businesses, formerly called Lever Brothers and now called Unilever. These soap businesses were among the first to employ large-scale advertising campaigns.

Liquid Soap (Detergent). Liquid soap was not invented until the 1800s. In 1865, William Shepphard patented liquid soap. In 1898, B.J. Johnson developed a soap formula, and his company (the B.J. Johnson Soap Company) introduced Palmolive soap the same year. This new soap was made of palm and olive oils and became popular in a short amount of time; Palmolive became so popular that B.J. Johnson Soap Company changed its name to Palmolive. At the turn of the century, Palmolive was the world's best-selling soap.

In the early 1900s, other companies began to develop their own liquid soap. Products such as Pine-Sol and Tide appeared on the market, making the process of cleaning clothing, counters and bathrooms easier.

As a detergent, liquid soap tends to be more effective than flake soap, and there is a smaller chance of residue being left on clothing with liquid soap. Liquid soap also works better for more traditional washing methods, such as using a washboard.


Part 3: Esterification When an carboxylic acid (R-COOH) and an alcohol (R-OH) are mixed together and heated in the presence of an acid catalyst (such as H2SO4), the two will react to form an ester (plus H2O). This process is called esterification. Each ester has its own unique odor, and with a discriminating nose, one can use this fact to help identify them. In this lab you will be reacting various organic acids (acetic acid & salicylic acid) with various alcohols (1-pentanol & ethanol). You will make two different esters with odors that should be familiar to you. You may have noticed that esterification is the reverse reaction of the hydrolysis of esters seen in the saponification reaction. Esterification: Addition of an alcohol with a strong acid catalyst creates an ester. O O || H+ || R – C – O – H + H – O – R' R – C – O – R' + H2O

PRELAB QUESTIONS: 1. Draw the structure of di-peptide Tyr-Phe as it would exist at pH = 1, 7, and 14. pH = 1

pH = 7

pH = 14

Chemistry*108* Protein*and*Esters*Lab*

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2) Draw a triglyceride (do not draw the triglyceride used in the example on page 2 of this prelab, you can draw any other triglyceride) : 3) Draw the products of the saponification of the triglyceride that you drew in question #2 above:

4. Draw and name the ester that would be formed in the reaction of propanoic acid with ethanol: (see chapter 10 lecture notes if you need a reminder about naming esters)

Documents

CHEM108 Lab Manual - Saddleback College If a brick has a length of 13.77 cm, a width of 8.50 cm, and a height of 5.12 cm: a) What is the volume of the brick? b) If the brick has a