chem 111 a09 general chemistry i laboratory reports fall...

CHEM 111 A09

GENERAL CHEMISTRY I

LABORATORY REPORTS

FALL 2016

General Chemistry I Lab Reports Class Portfolio

December 2016 • Volume 1, Issue 1

2

Letter from the Editor Welcome to the first issue of General Chemistry Student Lab Report Portfolio at IUP. We have made this journal to better introduce our exceptional student work to visitors. Professionals in science and engineering spend at least fifty percent of their time writing reports and memoranda according to recent nationwide surveys. By taking the CHEM 111 - general chemistry lab classes at IUP, our students are not only trained to learn technical skills in laboratory, but also trained to interpret experimental data and write good lab reports. Although report writing can be time intensive, the time is well spent because it provides our students with the opportunity to develop the skill that will be extremely valuable in their future careers. I give high grades to students who are able to describe their laboratory work in a clear, organized report than those who cannot. I grade lab reports based on the following four criteria: format (30%), results (30%), discussion (30%), and problems (10%). Students need to (1) adopt a given format and prepare research style reports; (2) convert their measurements to final results through calculations; (3) discuss the implication of their results and draw main conclusions; and (4) complete the pre-lab and after-lab assignments given in the lab manual. I’m glad to see the majority of our students did a very good job on their lab reports. Although it is very difficult to select 11 best reports from 264 submissions, I finally made it by giving a maximum number of students the chance of including their work in this class portfolio. I gratefully acknowledge the contributions of my students in CHEM 111 A09 section of Fall 2016. There has never been a more exciting time reading their reports. Please read this class portfolio and I believe you will have the same feeling with me. Sincerely,

Hao Tang, Ph.D., P.E. Assistant Professor of Environmental Engineering Indiana University of Pennsylvania

LAB

No.

Title Author Page

No.

1 Analysis of Aluminum Foil’s Thickness

and Effectiveness of Volumetric

Equipment

Mohammed Ali 3-7

2 Identification of Unknown Substance Gage J. Smith 8-15

3 Mass Relations Gage J. Smith 16-20

4 Determination of Phosphorus in

Fertilizer

Karlie Ebert 21-24

5 How Much Reactant Do I Need? Alexander

Ignatenko

25-28

6 Analysis of Antacids Alexander

Ignatenko

29-33

7 The Analysis of Eggshells for Calcium

Carbonate

Baily Blackburn 34-37

8 Gas Laws: Determination of the Molar

Mass of a Metal

Elizabeth Matusik 38-41

9 Formation of an Air Bag from the

Production of CO2

Tara M. Kenna 42-45

10 Causes of the Temperature Change Mohammed Ali 46-50

11 Creation of Hot Pack Using Calcium

Chloride

Mohammed Ali 51-55

3

TABLE OF CONTENT

CHEM 111 A09 FALL 2016, LAB 1, 3-7

CHEM 111 A09 GENERAL CHEMISTRY

LABORATORY FALL 2016

Indiana University of Pennsylvania

Analysis of Aluminum Foil’s Thickness and Effectiveness of Volumetric Equipment

Mohammed Ali

Department of Biology, Indiana University of Pennsylvania, Indiana, PA 15705

Academic Editor: H. Tang

Received: September 13, 2016 / Accepted: November 19, 2016 / Published: December 6, 2016

Abstract: This first procedure dealt with calculating the thickness of Reynold’s Heavy Duty aluminum foil. This was calculated by solving the derived density formula, d= m/lwh, for height. The average thickness of the foil was found to be .00228 cm ± .00007 cm with a percent error of 13.0%. The second part of this lab focused on analyzing the precision and accuracy of different sized beakers in comparison to a volumetric flask. 25 mL of distilled water was estimated in each container and then a graduated cylinder was used to determine actual volume. The glassware’s accuracy was calculated through percent error and precision through standard deviation. The volumetric cylinder was determined to be the most reliable as it had the lowest standard deviation and percent error. The beakers had less reliable accuracy and precision and these worsened as capacity increased. In both experiments, variations between the observed and actual values were likely due to human error, such as the measuring of the foil or the estimating of the water.

1. Introduction and Scope

This lab has two purposes: the first purpose is calculating the thickness of Reynold’s Heavy Duty aluminum foil. The foil’s thickness must be calculated indirectly as it is too small to be measured accurately with a ruler. This is important as many calculations in the laboratory, especially ones that are too small or difficult to quantify, must rely upon derivation to be calculated. In this lab, aluminum’s thickness was quantified from its known density and its measurable dimensions of length, width, and mass.

The second purpose is to analyze the accuracy and precision of different containers of volume found within the lab. This means that the results are close to the actual value for volume and can be

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CHEM 111 A09 FALL 2016, LAB 1, 3-7 4

reproduced with relatively little variation, respectively. This is important as accurate and reproducible results are crucial for valid conclusions in science. Measurements using the wrong glassware can seriously flaw the entire design of an experiment. This also explains the wide array of instruments used to measure volume in labs, including beakers, test tubes, graduated cylinders, or Erlenmeyer and volumetric flasks. This lab in particular deals with comparing two of these pieces: beakers and volumetric flasks. Various forms of beakers and a volumetric flask are used to estimate a target volume and find which one produces the most accurate results.

The scope of this paper is to attempt to calculate the thickness of aluminum foil and determine out of several beakers and a volumetric flask which is most accurate in measuring volume. This paper will show the methodology performed to come to both conclusions by focusing on the materials and procedures, data and analysis, and the error analysis and overall conclusions that can be drawn from the data.

2. Calculation of Aluminum Foil’s Thickness

Materials and Procedure This experiment required four pieces of Reynold’s Heavy Duty Aluminum Foil, one metric ruler,

and one electronic pan balance. It was performed by measuring the dimension of four pieces of aluminum foil that were pre-cut to certain dimensions. A ruler was used to measure both the length and width of the foil and these results were recorded. Following this, the foil was massed on an electronic pan balance and the results were recorded as well. This entire process was repeated four times with different pieces of foil to find the mass and dimension of each of them.

Results and Calculations

Foil Piece Length (cm) Width (cm) Mass (g) Thickness (cm)

Foil #1 14.20 6.20 0.538 0.00226

Foil #2 19.53 14.10 1.750 0.00235

Foil #3 17.17 17.70 1.895 0.00231

Foil #4 24.08 14.20 2.015 0.00218

Average ± Standard Deviation

18.75 ± 4.17 13.05± 4.86 1.550±0.683 0.00228± 0.00007

The thickness was calculated using two formulas, one for volume and the other for density. Density

is equal to mass divided by volume (d=m/v) and volume is the product of length, width, and height (V=lwh). It is important to note that volume appears in both formulas so the second formula for volume can be substituted in the first to get a derived formula. This equates the known value of density (given as of 2.70g/mL) to the quantity of mass divided by the product of length, width, and thickness (d= m/lwh. This formula can be used to input the data for each trial and algebraically solve for the height of the aluminum foil which is also the thickness of the foil. Mathematically, this is done by multiplying density by the length, width, and height and then dividing the mass by the entire quantity of length, width, and density so that the unknown height is isolated on one side and can be solved. This

CHEM 111 A09 FALL 2016, LAB 1, 3-7 5

process was repeated for each of the pieces of foil to find their thickness which on average was about .00228 ± 0.00007cm per piece. Error Analysis Percent error= [(|experimental value - accepted value|)/ (accepted value)]x100 The percent error allows the experimental value to be compared to an established, accepted value and to determine how much the two differ. The experimental value being used in this calculation is the average thickness of .00228 cm and the accepted value is .00262 cm for Reynold’s Heavy Duty aluminum foil, as determined by the VWR, which is a leading producer of laboratory equipment. The percent error is 13.0%, which means that there was a definite amount error present in the calculations of the aluminum’s thickness. The small standard deviation of .00007 suggests that the results were precise or low in variability, but the large percent error means that there was a problem with the accuracy of our measurements. This is likely due to the fact that when measuring the foil, the last digit of measurement had to be estimated. This leaves the possibility open to human error. Other areas of error are possible such as a density that is not exactly that of the given density (aluminum= 2.7 g/mL) because the box of foil said that it also included a coating of aluminum oxide which has a higher density of 3.95 g/mL. Errors in mass could be caused by not waiting for the balance to completely stabilize and the contamination of the foil after being handled by multiple people.

3. Comparison of Volumetric Equipment

Materials and Procedures This experiment required 300 mL of distilled water, one lab wash bottle, one 50 mL graduated

cylinder, one 25 mL volumetric flask, one 50 mL beaker, one 150 mL beaker, and one 400 mL beaker. It was performed by taking the volumetric flask and each of the beakers and squirting 25 mL of distilled water in each of them. For the volumetric flask, the water was added until the meniscus reached the bottom of the calibration mark, while in the beakers the water was estimated using the marked volumes to form a rough estimate. The volume of each container was then transferred into a graduated cylinder and recorded to ± 0.1 mL by estimating the final digit using the bottom of the meniscus. This process was repeated three times with each piece of glassware.

Results and Calculations

Equipment used: Trial #1 Trial #2 Trial #3 Average ± Standard Deviation Percent Error

50 mL beaker 22.9 24.0 22.5 23.1 ± 0.8 7.60%

150 mL beaker 25.5 22.0 19.4 22.3 ± 3.1 11.0%

400 mL beaker 36.1 37.8 33.2 35.7 ± 2.3 42.8%

25 mL volumetric flask 25.1 25.2 25.1 25.1 ± 0.1 .400%

Error Analysis The actual value that each of these volumetric instruments is being compared to is 25.0 mL. The

most reliable and effective instrument in the lab would be one that has high precision, indicated by a

CHEM 111 A09 FALL 2016, LAB 1, 3-7 6

low variability, and high accuracy, indicated by a low percent error. This is exhibited by the 25mL volumetric flask which in comparison to the other instruments has a very miniscule variation and percent error. This means that when it is used correctly for appropriate measurements, it will yield highly dependable results with less human error. The presence of human error can account for both precision and accuracy errors. For example, rash and unmethodical measuring, rough estimating of a solution, flawed human perception, or spills during transfer will both make the data varied and inaccurate in measuring a desired amount. Also, instrument standards may play a role in the effectivity of the container. Beakers are a lot cheaper and are mass produced so they are not required to meet the accuracy and quality standard that volumetric flasks are held to by the scientific community. Volumetric flasks are meant to be accurate to about 1%, but beakers are typically made to be accurate to only 5% so they are more likely to cause systematic error. The data also shows that in general human error increases as the capacity of the beaker increases. This means that the beakers became more inaccurate and varied in their measurements as volume increases. For example, the 50 mL beaker has a percent error of 7.60%, which increases to 11.0% in the 150 mL beaker, and dramatically rises to 42.8% in the 400mL beaker. This means that the 400 mL beaker is about five times more inaccurate then the 50 mL beaker when it comes to measuring a quantity. The variation follows a similar trend of increasing. For example, the variability increases about four times when using the 150 mL beaker as opposed to the 50 mL beaker. It is important to note that even though the 400 mL beaker has a lower variability than the 150 mL it is not more dependable. This just means that all of its measurements (36.1, 37.8, and 33.2 mL) are more uniform in overestimating the desired volume.

4. Summary and Discussion

Based on the results the average thickness of aluminum foil was calculated to be .00228 cm with a standard deviation of ± .00007 cm. This low variability means that the values for the thickness were relatively very similar or precise. Although, when compared to the true value of the thickness of Reynold’s Heavy Duty aluminum foil, the percent error was 13.0%. This means that there was error present in the calculations of the foil most likely due to different levels of human error.

The results of the experiment comparing volumetric equipment demonstrates that out of the glassware surveyed the volumetric flask offers the most accurate and precise readings when it comes to measuring an exact amount. This is supported by the low variability and percent error of its measurements. Beakers in comparison to volumetric flask lack precision and accuracy. This is shown by their significantly higher variabilities and percent errors. Moreover, the larger the beakers the more likely it will give a less accurate and precise measurement. This seems to be because beakers that hold a larger capacity also maximize the chances of human error by only offering a rough estimate for specific volumes. For example, a 400 mL beaker might only offer markers every 50 mL while a 50 mL beaker may offer markers every 10 mL. Although it would be a matter of perception in both cases to measure 25 mL, it would be more accurate in the smaller beaker as the volume can be to fine-tuned with confidence, leaving a smaller interval of uncertainty. Also from a technical standpoint, volumetric flasks are held to a higher standard of quality control and effectivity then beakers are normally subjected to which reduces the chance of systematic error. The overall conclusion that can be drawn is that a beaker should not be used for accurate or precise measurements but rather as vessels to carry

CHEM 111 A09 FALL 2016, LAB 1, 3-7 7

solutions. For more reliable measurements, scientists should opt for tools such as the volumetric flask that perform exceptionally well when it comes to precision and accuracy.

Acknowledgments

The author gratefully acknowledges his lab partner Brad O’Hara for his assistance in performing the experiment and all other aspects of the process, including data collection, statistical calculations, and lab clean-up to enumerate a few.

References and Notes

"Flasks, Beakers, & Graduated Cylinders." Dartmouth.edu. Dartmouth University, n.d. Web. 10 Sept. 2016. <https://www.dartmouth.edu/~chemlab/techniques/flasks.html>.

"Reynolds Wrap® Aluminum Foil." Us.vwr.com. VWR International, n.d. Web. 10 Sept. 2016. <https://us.vwr.com/store/product/4526537/reynolds-wrap-aluminum-foil>

© 2016 by the author. This article is an open access article.

CHEM 111 A09 FALL 2016, LAB 2, 8-15




Identification of Unknown Substance

Gage J. Smith

College of Natural Sciences and Mathematics, Indiana University of Pennsylvania, Indiana, PA 15705



Abstract: The purpose of this lab was to find the density of an unknown metal by utilizing water displacement and Archimedes Principle. Not only was the objective to identify an unknown metal but also to identify an unknown liquid by comparing collected data to a density of ethanol-water solutions chart to find the percentage of ethanol in the solution. Making sure you’re precise as possible when it comes to significant figures is very important because if you’re off, your results could be mistaken for a whole different substance. During the experiment, Tahmair played a crucial role of collecting the equipment and unknown substances in order to begin using the methods to determine density. As stated, the methods we used to complete this lab were the methods of water displacement and Archimedes Principle. Equipment used throughout this lab consisted of goggles, a scale, graduated cylinder, volumetric flask, 400 mL beaker, weight stand, calculator, the density equation, the percent error equation, densities of commercial metals at 25 degrees Celsius chart, density of ethanol-water solutions versus mass percent ethanol at 25 degrees Celsius chart, and of course the unknown metal and liquid. This report explains the guidelines you must know and understand when it comes to the concept of density. It isn’t just a simple calculation. There are several methods used to find the density of a particular substance. Understanding each method of water displacement is crucial when determining density. Methods of water displacement include the displacement method itself and Archimedes Principle. Just like any other calculations and results, significant figures reflect the precision of the values and are very important. Understanding significant figures and how they vary between equipment is key to a precise and accurate lab results.


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CHEM 111 A09 FALL 2016, LAB 2, 8-15 9

This paper is about understanding the concept of density by examining some solids using the methods of water displacement and Archimedes Principle. First, in order to determine ones’ density, you must know the density formula which is density = mass / volume. We can measure the mass directly by either using distance measurements or measure volume displacement, which are the two methods stated above in further detail. You need to know the density of water since it will be in your equation to determine the volume during Archimedes Principle.

Also, determining the density of an unknown liquid should be understood. You’re dealing with density, therefore once again you will be measuring mass and volume. Understanding how to use a scale and reading it along with other equipment such as, the graduated cylinder, volumetric flask, etc. I have stated this equipment specifically because these are the ones used in this lab experiment. Remember, when recording data, it is important to make sure you’re using the correct number of significant figures in your results. The more significant figures in your result, the more precise but depending on what apparatus you’re using will determine the number of significant figures.

2. Procedures

Materials and Procedure During our second lab, we went over the concept of density by examining a solid using the methods

of water displacement and Archimedes Principle. The unknown metal was chosen at random and the two methods were used to determine the density of the metal. When using the water displacement method, the metal was weighed and recorded. A graduated cylinder was then filled to a mark of your choice and recorded. The metal was then placed in the graduated cylinder. The water level was then recorded to get your final volume. The difference of the final volume and the volume without the metal gives you the volume of the metal. This method was completed another two times using the same unknown metal but different in mass and size. Then, Archimedes Principle was used. Using the same unknown metals from the first method, we put the metals securely on a stand and weighed them. Then, we used a beaker to submerge the mass and recorded the difference. With the collected data, the buoyancy calculation could be completed by taking the mass of the metal submerged on the stand minus the mass on the stand equal the density of water multiplied by Volume. Volume is the unknown in this case. Therefore, once calculated you use the density equation (rho=mass/volume). By using the mass taken from the displacement method and the volume from Archimedes Principle, density of the unknown metal can be determined. Average density of the three trails is then completed and the result is compared to a commercial metals chart to identify the metal according to our results. Percent error is then calculated and recorded.

After determining the unknown metal, we chose a liquid solution at random. We acquired 50 mL of the solution in a graduated cylinder. Another graduated cylinder and a volumetric flask were weighed dry on the scale and was recorded. After, we dispersed 25 mL of the solution in graduated cylinder we just weighed and the remaining 25 mL in the volumetric flask. This apparatus was weighed again with the solution. Mass of the liquid was then calculated from the difference. This process was completed three times in order to achieve an average mass. Density of the solution was then calculated by taking the difference of mass with the liquid minus the mass of the equipment dry divided by the volume content of 25.00 mL. An average of these densities was completed as well and compared to a density

CHEM 111 A09 FALL 2016, LAB 2, 8-15 10

of ethanol-water solutions versus mass percent ethanol at 25 degrees Celsius chart. The percent of ethanol for the solution of the graduated cylinder and volumetric flask were then averaged.

When it comes to significant figures, just like any other lab, you have to make sure you use the correct number of significant figures depending on the apparatus you’re using. The rules of significant figures should be known. There are four basic rules consisting of any non-zero digit is significant, a zero is significant if it’s between non-zero digits or if it’s after the decimal place for numbers greater than one. When it comes to adding and subtracting, the answer should have the same number of decimal places as the measurement with the fewest decimal places and when multiplying and dividing, the answer should have the same number of significant figures as the measurement with the fewest significant figures. When it came to mass measurements, we recorded six significant figures. Using the graduated cylinder, we recorded to + or – 0.1 and for calculations, they all had division in them so we rounded to the same number of significant figures as the one with the fewest sig figs that was within the equation. The more significant figures after the decimal, the more precise your answer is especially when it came to the density of ethanol-water solutions. Our density for our ethanol-water solution was five significant figures and it was incredible comparing our results to the chart supplied. The densities were so close and just with a little flex of your numbers could change your ethanol reading drastically. When determining the density of the unknown metal B, we first used the method of water displacement. We weighed the unknown metal and then used a 100 mL graduated cylinder to find the volume. We used a different size of unknown metal B for each of the three trials. Volume Displacement

Unknown Metal B

Mass of Object

Initial Volume

Final Volume Volume of Object

Density

Trail 1 18.140 g 89.0 mL 92.0 mL 3.0 mL 6.0g/mL Trail 2 54.596 g 76.0 mL 82.0 mL 6.0 mL 9.1 g/mL Trail 3 17.540 g 69.0 mL 72.0 mL 3.0 mL 5.8g/mL

When determining average and standard deviation, we ruled out trail 3 due to errors in measurement. We feel since we were using a 100 mL graduated cylinder, which has a scale of + or – 0.1 mL that is was less precise when we completed the buoyancy method.

Average 7.6 g/mL Standard Deviation

2.2

Average = (6.0 g/mL + 9.1 g/mL) / 2 = 7.6 g/mL To find density for the volume displacement method, we used equation: Mass of Object / Final Volume – Initial Volume = Density 18.140 g / 92.0 mL – 89.0 mL = Density 18.140 g / 3.0 mL = 6.0 g/mL After consulting the reference table to identify our unknown metal B, we used the buoyancy method density to compare considering it was more precise. Our density for unknown metal B was 7.6 g/mL, which was closet to the metal aluminum bronze that is 7.80 g/mL. We were off 0.2 g/mL in our results. Our percent error was 2.6%.

CHEM 111 A09 FALL 2016, LAB 2, 8-15 11

Percent error = ((experimental density – reference density) / reference density) x 100% Percent error = ((|7.6 g/mL – 7.80 g/mL|) / 7.80 g/mL) x 100% Percent error = 2.6% As stated earlier, we already knew the volume displacement was less precise than the buoyancy method because the number of significant figures in the calculations and results, therefore ruling it out. After completing the volume displacement method, also known as the water displacement method, we used Archimedes Principle, which is known as the buoyancy method. By using the given stand to hold our mass, we weighed the total mass with the weight on the stand. Then, using a beaker of water, we submerged the mass and recorded the mass again. Next, the buoyancy calculation was completed by taking the difference of the mass weights recorded and setting it equal to the density of water, which is 1.00 g/cm^3 and multiply by volume. Volume is what we are solving for in this case. We used the same three unknown metal B solids we used in the volume displacement method above. Buoyancy

Unknown Metal B

Mass of Object

Initial Mass of in Apparatus

Final (Submerged) Mass in Apparatus

Volume of Object

Density

Trail 1 18.140 g 64.747 g 62.206 g 2.54 mL 7.14 g/mL Trail 2 54.596 g 101.230 g 93.982 g 7.25 mL 7.53 g/mL Trail 3 17.540 g 64.180 g 61.254 g 2.93 mL 5.99 g/mL

When determining average and standard deviation, we ruled out trail 3 due to errors in measurement.


0.276

Average = (7.14 g/mL + 7.53 g/mL) / 2 = 7.34 g/mL To find density for the buoyancy method, we used the equation below to solve for Volume (V): Initial Mass of in Apparatus – Final (Submerged) Mass in Apparatus = Density of Water x (V) 64. 747 g – 62.206 g = 1.00 g/cm^3 x (V) 2.541 g / 1.00 g/cm^3 = V 2.54 cm^3 = V 1 cm^3 = 1 mL V = 2.54 mL

Buoyancy method, by far was the more precise out of the two methods to find density because the amount of significant figures in each recording. After consulting the reference table to identify our unknown metal B, we used the buoyancy method density to compare considering it was more precise. Our density for unknown metal B was 7.34 g/mL,

CHEM 111 A09 FALL 2016, LAB 2, 8-15 12

which was closet to the metal zinc that is 7.14 g/mL. We were off 0.20 g/mL in our results. Our percent error was 2.80%. Percent error = ((experimental density – reference density) / reference density) x 100% Percent error = ((7.34 g/mL – 7.14 g/mL) / 7.14 g/mL) x 100% Percent error = 2.80% After identifying our unknown metal B to be zinc, we chose unknown “Quarmby B” to determine the density and to identify the ethanol percentage within the solution. We used two pieces of equipment to determine this liquid. This equipment consisted of a graduated cylinder and volumetric flask. Graduated Cylinder

Unknown “Quarmby B”

Mass of Container

Mass of Container

Plus Liquid

Mass of Liquid

Volume of Liquid

Density

Trail 1 82.117 g 106.474 g 24.357 g 25.00 mL 0.9743 g/mL Trail 2 82.115 g 106.479 g 24.364 g 25.00 mL 0.9746 g/mL Trail 3 82.117 g 106. 472 g 24.355 g 25.00 mL 0.9742 g/mL


0.0002

Average = (0.9743 g/mL + 0.9746 g/mL + 0.9742 g/mL) / 3 = 0.9745 g/mL To find density of the alcohol sample for the unknown “Quarmby B”, we used equation: Rho = Mass of Liquid / Volume of Liquid Rho = 24.357 g / 25.00 mL Rho = 0.9743 g/mL Rho = Density Once we compare our result 0.9745 g/mL to the reference table, density of ethanol-water solutions, we found out in this sample using the graduated cylinder is closest to 0.97474 g/mL, which contains 14% ethanol in the water solution. Volume Percentage = Mass Percent x (Total Density / Density of pure ethanol) Volume Percentage = 14% x (0.9745 g/mL / 0.78507 g/mL) Volume Percentage = 17% Alcoholic proof = (2) x (% Volume of ethanol) Alcoholic proof = (2) x (17%)

CHEM 111 A09 FALL 2016, LAB 2, 8-15 13

Alcoholic proof = 34% Volumetric Flask

Unknown “Quarmby B”

Mass of Container

Mass of Container

Plus Liquid

Mass of Liquid

Volume of Liquid

Density

Trail 1 25.966 g 50.138 g 24.172 g 25.00 mL 0.9669 g/mL Trail 2 25.964 g 50.133 g 24.169 g 25.00 mL 0.9668 g/mL Trail 3 25.965 g 50.134 g 24.169 g 25.00 mL 0.9668 g/mL


7.071 x 10^-5

Average = (0.9669 g/mL + 0.9668 g/mL + 0.9668 g/mL) / 3 = 0.9668 g/mL To find density of the alcohol sample for the unknown “Quarmby B”, we used equation: Rho = Mass of Liquid / Volume of Liquid Rho = 24.172 g / 25.00 mL Rho = 0.9669 g/mL Rho = Density Once we compare our result 0.9668 g/mL to the reference table, density of ethanol-water solutions, we found out in this sample using the graduated cylinder is closest to 0.96640 g/mL, which contains 20% ethanol in the water solution. Since a volumetric flask is more precise than a graduated cylinder, we used our results for the density using the volumetric flask. Once again, our density was 0.9668 g/mL and when comparing, the closest density was 0.96640 g/mL. This concludes that there is 20% ethanol in the unknown “Quarmby B” solution. Volume Percentage = Mass Percent x (Total Density / Density of pure ethanol) Volume Percentage = 20% x (0.9668 g/mL / 0.78507 g/mL) Volume Percentage = 25% Alcoholic proof = (2) x (% Volume of ethanol) Alcoholic proof = (2) x (25%) Alcoholic proof = 50%

Students need to know the concept of density by examining a solid state using methods of water

displacement and Archimedes Principle. Determining liquids by using graduated cylinders and volumetric flasks by simply finding the mass of the containers, mass of the container with the solution, and taking the difference and dividing by volume should be known. Most important, understanding

CHEM 111 A09 FALL 2016, LAB 2, 8-15 14

and recognizing significant figures and determining the correct number depending on the circumstances should be learned and understood.

Conclusion questions:

1. We tried to find the densities of an unknown solid and ethanol-water solution to determine the

solids metal identity and the ethanol-water solutions ethanol percentage. We completed this by using two methods consisting of the volume displacement method and buoyancy method using mass and volume to calculate density and comparing our results to a reference table. When finding the density of the liquid, we used two methods which were the same except we used a graduated cylinder for one and a volumetric flask to measure masses in order to use the density equation to come up with a result. Our final results after long consideration of precise measurements and errors were that the unknown metal B was zinc and the unknown “Quarmby B” had 20 % ethanol in it.

2. In part 1, the buoyancy method was more precise because it gave more significant figures in our final answer and pretty much in every measurement to be exact. The standard deviation for the buoyancy method is also smaller than the standard deviation for the volume displacement method.

3. Using the more precise method, which was the buoyancy method, we compared our density

of 7.34 g/mL to the reference table and found it was closest to the density of zinc, which is 7.14 g/mL. If we used the less precise method, we would have not got the same answer. Instead, the method would have been aluminum.

4. In part 2, the volumetric flask produced the smaller standard deviation. The volumetric flask

has and advantage of being very precise when it comes to measuring exactly 25.00 mL. This produces an extremely low standard deviation, which everyone wants when recording results. When it comes to other measurements is basically a guessing game. The volumetric flask also only has a max capacity of 25.00 mL. The graduated cylinder can measure high water capacities for bulk volumes. The marks on a graduated cylinder a very easy to read as well. It does have its downfall though. Say you need to weigh the cylinder dry and the only one available is wet, you can’t get the whole graduated cylinder dry because the long narrow tube. Also, though the marks are easy to read. It is sometimes difficult to get a read when the solution is between marks. Always remember, you have to look at it level. If you’re too high, your volume will look greater than it is and vise versa.

5. Our lab instructor provided that unknown “Quarmby B” was 50% proof of alcohol solution.

With our volumetric flask method, we too determined the solution contained 20% ethanol. Ms. Quarmby seems to be watering down her beverages based on these results. Ms. Quarmby should recognize what she is doing wrong and fix it in order to increase the proof on her beverages so they are not under proof. The Herald Observer in this case can do as they please considering that the reports in which they were going to get sued over were correct. Knowing a newspaper nowadays, my guess is another headline would be reported now that the results have been processed and released.


In conclusion, we learned the concept of density in terms of a solid state and liquid state and the two methods to determine each. We used the volume displacement method and the Buoyancy method to

CHEM 111 A09 FALL 2016, LAB 2, 8-15 15

determine the density of the unknown metal B, which ended up being zinc. We ran three trial during each method. During the volume displacement method, we found the mass of the solid, as well as, establishing an initial volume in the graduated cylinder. Then, we dropped the solid into the graduated cylinder and recorded the final volume. All that was left was to calculate the difference and use the equation Rho = mass / volume to get the density. In the buoyancy method, you just weighed the stand with the solid and later submerged the solid in water and recorded the new mass. Mass difference was then computed and was set equal to the density of water multiplied by the volume we were determining. Then, once again the rho equation was used to find density. In my opinion, the Buoyancy method or Archimedes Principle was more precise because of the amount of significant figures in the final answer. When it came to determining the liquid, we used a graduated cylinder and a volumetric flask. To find the density, all we simply had to do for both methods was to weigh the containers dry, weigh the containers with the solution and take the difference, followed by taking the difference and dividing it by the volume to calculate the density. I felt that the volumetric flask was more precise because of the single mark at a long narrow neck making it easy to know where you need to stop. Finally, recognizing for each apparatus we were using, we had to make sure we were using the correct number of significant figures to support our results to get the most precise answer to compare to the reference tables.

Acknowledgments

The author gratefully acknowledges his lab partner Tahmir Smith for his assistance in performing the experiments.


CHEM 111 A09 FALL 2016, LAB 3, 16-20




Mass Relations

Gage J. Smith




Abstract: This report explains the guidelines you must know and understand when it comes to qualitative observations of the reaction of calcium carbonate and hydrochloric acid, as well as, quantitative exploration of the reaction for different masses of the reactants. This experiment shows you what different measurements and masses do during a chemical reaction when it comes to determining a limiting reagent.


This paper is about observing the chemical reaction between calcium carbonate and hydrochloric acid to determine the reactant that is completely used up in a chemical reaction and the point at which it occurs. This is just not seeing what it does but the different changes according to the mass difference of the calcium carbonate. By using three test tubes to put a different amount of calcium carbonate in each one, we can observe the different appearances of the to reactants, as well as, distinguish between the limiting and excess reagents in each reaction.

The purpose was to figure out relationships between reactant quantities and resulting products by examining the chemical reaction produced between calcium carbonate and hydrochloric acid. Also, understanding the bond between the two agents and knowing by observation, which chemical is the limiting reagents determining by amounts of mass and volume of fluids.

2. Procedures and Results

During the mass relationships part 1, dealing with how much reactant we need, we began a qualitative observation of the reaction between the calcium carbonate and hydrochloric acid by obtaining 50 mL of hydrochloric acid and recording the hydrochloric acid concentration. The

OPEN ACCESS

CHEM 111 A09 FALL 2016, LAB 3, 16-20 17

concentration is not the density of the acid and its measurement is 109 g/L. Step number two consisted of obtaining three test tubes and filling each with a different amount of calcium carbonate, which were 0.25 g, 0.60 g, and 1.0 g. Measuring was done using a scale. We placed the test tube for each in a beaker and recorded the difference when adding the calcium carbonate to determine the correct amount of calcium carbonate. After getting the correct mass of calcium carbonate, we added 5 mL of distilled water into each test tube. Then, we added about three or four drops of indicator in each test tube, which helps us detect when acid is present or not depending on the color after the reaction. In order, to begin with the chemical reaction, a calculation to determine how much hydrochloric acid had to be done. The calculation will be in the calculations section of the paper but the volume came out to 4.6 mL. We dumped 4.6 mL of hydrochloric acid in each test tube one at a time, observing the chemical reactions in each as we stirred it thoroughly. The evidence of a chemical change consisted of the calcium carbonate fizzing and releasing carbon dioxide, as well as, the change of color, which in the end helped us determine the limiting reagent.

Next, we did a quantitative exploration of the reaction for different masses of the reactants. First, we measured 5.0 mL of hydrochloric acid in a 10 mL graduated cylinder and then recorded the mass. Our group was assigned to record about 0.6 g of calcium carbonate. Our mass was recorded at 0.602 g of calcium carbonate. We used a 50 mL beaker to do the measurement. We then added about 10 mL of water and recorded the mass again. After, we added the hydrochloric acid to the beaker and mixed it with a stirring rod. Once the reaction was over, we weighed the content again. Everything was recorded into a data table to express the results.

Test tube with 0.25 g of calcium carbonate

• After dumping 4.6 mL of hydrochloric acid into the test tube with 0.25 g of calcium carbonate, the chemical reaction occurred instantly turning the solution pink

• The calcium carbonate started fizzing immediately releasing the gas carbon dioxide • After letting the reaction settle down and while mixing, the color of the solution remained

pink • The pink color change and pink end result determined the limiting reagent was the calcium

carbonate and all that was left was the hydrochloric acid

Test tube with 0.60 g of calcium carbonate • The test tube with 0.60 g of calcium carbonate had the same beginning chemical reaction as

the test tube with 0.25 g of calcium carbonate once the hydrochloric acid was added • The solution turned pink immediately • The calcium carbonate, once again, fizzed up instantly releasing carbon dioxide • When the chemical reaction started to settle, the color turned to yellow • After stirring and mixing the content of calcium carbonate thoroughly, the color started to

turn into a peach textured color • Once stirring was done and about 30 seconds went by, the solution returned to yellow • The limiting reagent in this test tube was the hydrochloric acid and all that was left was the

calcium carbonate

Test tube with 1.0 g of calcium carbonate • The test tube with 1.0 g of calcium carbonate turned pink like the rest at the beginning once

the hydrochloric acid was added

CHEM 111 A09 FALL 2016, LAB 3, 16-20 18

• Once again, the solution fizzed meaning the calcium carbonate gave of carbon dioxide • Before the solution even settled or mixed, it turned yellow • After mixing, there was no change • The limiting reagent in this test tube was hydrochloric acid and all that remained was

calcium carbonate Mass CaCO3 (g) Moles CaCO3 Mass CO2 formed Moles CO2

formed Moles CO2 per mole CaCO3

0.25 g 0.003 mol 7.5 x 10^-4 g 1.7 x 10^-5 mol 0.01 0.60 g 0.006 mol 3.6 x 10^-3 g 8.2 x 10^-5 mol 0.01 1.0 g 0.01 mol 1.0 x 10^-2 g 2.3 x 10^-4 mol 0.02

*Periodic table is needed to determine molecular mass for each compound Moles (n) CaCO3 = Mass/Molecular Mass = m/M = 0.25 g / ((1 x 40.08 g/mol) + (1 x 12.01 g/mol) + (3 x 16.00 g/mol) = 0.25 g / 100.09 g = 0.0025 mol Mass CO2 formed (M) = (Mass CaCO3 x Moles CaCO3) / 1 mol = (0.25g x 0.003 mol) / 1 mol = 7.5 x 10^-4 g CO2 = (1 x 12.01) + (2 x 16.00) = 44.01 g/mol Moles CO2 formed = Mass CO2 formed x (1 mol / 44.01 g) = 7.5 x 10^-4 g x (1 mol / 44.01 g) = 1.7 x 10^-5 mol Moles CO2 per mole CaCO3 = Moles CO2 formed/Moles CaCO3 = 1.7 x 10^-5 mol / 0.003 mol = 0.01 Calculation of moles HCl in 5.0 mL of solution: n = 5.0 mL x (1 L / 1000 mL) x (1.0 mol / 1 L) n = 5.0 x 10^-3 mols of HCl

0.0E+00

5.0E-05

1.0E-04

1.5E-04

2.0E-04

2.5E-04

0 0.002 0.004 0.006 0.008 0.01 0.012

Mol

e CO

2 (y

-axi

s)

Moles CaCO3 (x-axis)

Moles CaCO3 versus Moles CO2

CHEM 111 A09 FALL 2016, LAB 3, 16-20 19

Quantitative exploration of the reactant for different masses of reactants.

Group # CaCO3 (g) Cylinder (g)

Cylinder + 5 mL of HCl (g)

CaCO3 (g) Beaker + ~10 mL of Water and CaCO3 (g)

Beaker + ~10 mL of

Water, CaCO3,

and 5 mL of HCl (g)

B2 ~0.2 g 38.804 g 44.119 g 0.200 g 35.198 g 42.219 g B4 ~0.3 g 48.900 g 53.883 g 0.300 g 39.829 g 44.670 g B6 ~0.4 g 38.775 g 44.094 g 0.403 g 35.990 g 41.087 g B8 ~0.5 g 20.436 g 25.484 g 0.500 g 37.526 g 42.204 g B10 ~0.6 g 48.324 g 53.338 g 0.602 g 39.696 g 44.411 g B12 ~0.7 g 35.162 g 44.257 g 0.714 g 43.123 g 48.036 g B14 ~0.8 g 39.530 g 44.320 g 0.812 g 49.930 g 42.920 g B16 ~0.9 g 41.127 g 46.258 g 0.900 g 37.897 g 42.765 g B18 ~1.0 g 38.310 g 43.578 g 1.002 g 44.637 g 49.539 g B20 ~1.1 g 32.953 g 38.354 g 1.102 g 46.426 g 51.658 g B22 ~1.2 g 48.520 g 53.736 g 1.200 g 41.523 g 46.205 g B24 ~1.3 g 39.114 g 44.376 g 1.311 g 39.506 g 44.296 g

Other Calculations: HCl concentration = 109 g /L 1 mL = 109 mg 109 mg/mL 5 mL = 545 mg 4.6 mL = 500 mg

Students need to know the concept of determining which reactant limits a chemical reaction,

identification by appearances, and limiting and excess reagents in a chemical reaction.

Conclusion questions: 1. In this lab, we tried to determine which reactant limits a chemical reaction, distinguish between

different substances based on changes in appearance, and distinguish limiting and excess reagents in a reaction. We are trying to determine which reactant is completely used up in the chemical reaction and the point which it occurs. The final results were that CaCO3 that contained 0.25 g was the limiting reactant within seconds, CaCO3 that contained 0.60g was the excess reagent and the limiting reagent was the hydrochloric acid due to the pink color. This observation lasted around 30 seconds and changed colors a couple times but finally stayed at yellow. CaCO3 containing 1.0 g had a limiting agent of hydrochloric acid as well. Within 5 second of the initial color change to pink, it changed to yellow.

2. No, to determine the limiting reactant, you have to convert the reactants to one product using mole to mole ratio. The reactant that yields the smallest amount of product is the limiting reagent.

CHEM 111 A09 FALL 2016, LAB 3, 16-20 20

3. No, to determine the limiting reactant, you have to convert the reactants to one product using mole to mole ratio. The reactant that yields the smallest amount of product is the limiting reagent.

4. My largest value of moles CO2 per mole CaCO3 is 0.02. This value should conform to the ratio of moles for these two compounds, considering NaHCO3 (s) + CH3COOH (aq) H2O (l) + CO2 (g) + CH3COONa (aq).

5. At 0.25 g CaCO3, the limiting reagent was CaCO3 and was pink at the end. At 0.60g CaCO3, the limiting reagent was HCl and was yellow at the end. At 1.0 g CaCO3, the limiting reagent was HCl and was yellow at the end.

6. My plot of moles CaCO3 versus moles CO2 shows a single, reasonable straight line. Moles versus moles are a ratio based statistic and since it holds true, it produces a linear line.


In conclusion, when showing mass relationships, such as observing chemical reactions to determine reactant limits, differences between substances based on changes in appearance, distinguishing limiting and excess reagents, and just being able to see what reactant was completely used up. During the lab we used three different test tubes with three different masses in each. The lowest product of mass 0.25 g of CaCO3 was the limiting reagent and the remaining two, which consisted of 0.60 g and 1.0 g of CaCO3 were excess reagents. If the limiting reagent was CaCO3, the final color was pink since we used the indicator. If HCl was the limiting reagent, the final color was yellow. Signs of chemical change in this lab were the fact that when HCl was added, CaCO3 would instantly fizz and release CO2. Also, the color change of the solution indicates a chemical change. Overall, this quantitative relationships lab between reactant quantities and resulting product quantities were very interesting since it was about studying the chemical reaction between calcium carbonate and hydrochloric acid in terms of this mass relations lab experiment.

Acknowledgments

The author gratefully acknowledges his lab partner Tahmir Smith for his assistance in performing the experiments.


CHEM 111 A09 FALL 2016, LAB 4, 21-24

CHEM 111 A09

GENERAL CHEMISTRY



Determination of Phosphorus in Fertilizer

Karlie Ebert

Indiana University of Pennsylvania, Indiana, PA 15705


Received: October 11, 2016 / Accepted: December 13, 2016 / Published: December 13, 2016

Abstract: The goal of this lab is to analyze the amount of phosphorus in a given amount

and type of fertilizer. The results of the experiment will be compared to the labeling of the

package. Because fertilizers contain water soluble and insoluble products, the phosphorus

can be taken out by knowing that information. Then by using the idea of balanced

reactions, the mass of the element compared to the mass that was present in the original

compound can tell the percent mass of the element in the original compound.


A. Purpose

The objective of the experiment was to determine the amount phosphorus content in a given

fertilizer. The goal is to see if the found amount of phosphorus is the same amount found on the label.

B. Procedure

Major Roles:

Partners: Karlie Ebert and Logan Hutchinson

The roles included one student measuring and weighing while the other student was part of the

conducting the stirring and adding portion of the lab.

Separating soluble from insoluble:

The first step of the lab is to measure out 3.000 grams of the assigned fertilizer. In this case, it was

28 - 8 -16. The next step was to separate out the water soluble from the water insoluble. This was

approached by putting the 30.000 grams of fertilizer into a 150 mL beaker and adding 78 mL of water

in the fertilizer to dissolve it. This was then filtered into a 500 mL volumetric flask and the liquid that

was filtered out what added back to the 150 mL beaker. The calculated MgSO4 is then added to the

solution and the observations were then recorded. Then, 2 mL of ammonia was added to the mixture to

OPEN ACCESS

CHEM 111 A09 FALL 2016, LAB 4, 21-24 22

reach a pH of 9. The beaker was then put in an ice bath for 20 minutes. While this was happening,

damp filter paper was added to the funnel and the 500 mL volumetric flask. The cold mixture was then

poured in with the vacuum attached to the volumetric flask. After it was filtered, the filter paper is then

put onto an already weighed watch glass and is set aside to dry. After dried, the mass is calculated to

find the mass of phosphorus present. The percent mass of phosphorus then can be calculated and

compared to the label of the original fertilizer box.

Materials:

Fertilizer Water Watch glass Glass stirring rod 150 mL beaker

500 mL

volumetric flask

Vacuum rubber

tubing

Funnel Filter paper Ammonia

pH paper Magnesium

sulfate

Scale Dropper Ice bath

TYPE OF FERTILIZER: 24 - 8 -16

3.00 g of fertilizer x 0.08 = 0.24 grams P2O5

0.24 x

= 0.105 g P

= 0.0034 moles of P

0.0034 x 121 g/mol of MgSO4 = 0.408 g MgSO4

50% MORE

0.408 g MgSO4 x 1.5 = .612 g MgSO4

2. Main Body of the Paper

A. Data and Results

Mass of Fertilizer: 3.000 grams

Percent Composition of Fertilizer: 24 - 8 - 16

Description of Mixture After Addition of MgSO4: The original color of the blue was a little darker

than before.

Description of Mixture After Addition of NH3: When the ammonia was added, the mixture was

foggy until it was completely mixed and after it was completely mixed, it foggy throughout the entire

mixture.

Description of Product: The product was a light blue and it looked like a layer of slush.

Mass of Dried Product: 1.007 g.

B. Calculations

Moles of Product (Mg(NH4)(PO4)(H2O6)6):

(1.007 g)/(245.42 g/mol)= 0.0041 moles

Percent of P in Product (Mg(NH4)(PO4)(H2O6)6):

(30.97 g/mol P)/(245.42 g/mol(Mg(NH4)(PO4)(H2O6)6)x 100 = 12.62%

CHEM 111 A09 FALL 2016, LAB 4, 21-24 23

Mass of P in Product:

1.007 g x .1262 = 0.1271 g P

Mass of P2O5:

(0.1271 g P)/(30.97 g/mol P)x 142 g/mol P2O5 = 0.5828 g P2O5

Moles of P2O5:

(0.5828 g P2O5)/(142 g/mol P2O5)= 0.0041 moles P2O5

Experiment % P2O5 in Original Fertilizer Sample:

(0.5828 g P2O5)/(3.00 g)x 100 = 19.43%

Percent Error:

(19.43% - 8.00%)/(8.00%)= 142.9%.

Miscellaneous

Pre-Lab Assignment:

1.

a. 30.97 g/mol of P

b. (2 x 30.974 g/mol of P) + (5 x 16.00 g/mol of O) = 141.9 g/mol of P2O5

c. (1 x 24.31 of Mg) + (1 x 32.07 g/mol of S) + (4 x 16.00 g/mol of O)

= 120.4 g/mol of MgSO4

d. (1 x 24.31 g/mol of Mg) + (1 x 14.01 g/mol of N) + (16 x 1.008 g/mol of H)

+ (1 x 30.97 g/mol of P) + (10 x 16.00 g/mol of O)

= 245.4 g/mol of Mg(NH4)(PO4)(H2O)6

2.

a. 5% N, 30% P2O5, 15% K2O

b.

x 100 = 43.64 grams of P

c. 10.00 grams x 30% = 3 g P2O5

3 g x

= 1.31 g P

= .042 moles of P

d. 5.04 g of MgSO4

3.

a.

x 100 = 12.6% P

b. 11.503 g x .126 = 1.45 g of P

c.

x 142 g/mol = 3.32 g

d.

x 100 = 29%

CHEM 111 A09 FALL 2016, LAB 4, 21-24 24

3. Conclusion

The final percent mass found of P2O5 in the sample of fertilizer is 19.43% This did not match the

packaging because the packaging percent of P2O5 was written to be 8.00%. The sources of

experimental error could conclude to be inaccurate measurements, along with problems dealing with

the methods of making the precipitate. Another possible problem could have been the pH when

ammonia was added to the solution. If some of the solid MgKPO4(H2O)6 was in the solution, then the

calculated percentage would be higher because it would have added weight to the final precipitate.

There would be a difference in the results if the fertilizer was reformulated because if there is more of

the phosphorus, the weight of the final product would be higher than it would be with the actual

amount. Based on the results, the phosphorus could have been reformulated. The found amount of

P2O5 was over double the amount found in the final product.


The overall idea of the experiment was to determine the amount of P2O5 in a sample of fertilizer, in

order to compare it to the original fertilizer label. To do this, a method had to followed in order to

accomplish separating the products found in the fertilizer to be able to make a precipitate. Once this is

made and data is collecting, the percentage of the P2O5 can be found and compared to the original

label. Knowing that fertilizer companies can sometimes reformulate the amount of phosphorus is a

product, the exact percentage of P2O5 is likely not to be exactly the same as it is seen on the label. In

conclusion, the goal of this experiment was to understand the concept of mass relationships.

Acknowledgments

The author gratefully acknowledges her lab partner Logan Hutchinson for her assistance in

performing the experiments.


CHEM 111 A09 FALL 2016, LAB 5, 25-28

CHEM 111 A09

GENERAL CHEMISTRY



How Much Reactant Do I Need?

Alex Ignatenko



Received: September 27, 2016 / Accepted: December 13, 2016 / Published: December 13, 2016

Abstract: The purpose of this lab is to identify the limiting reactant through

experimentation of adding HCl to CaCO3 mixed with water. This lab is meant to educate

students about moles, and how much substance is required in order to achieve the ultimate

reaction. Frankie and I played key roles in gathering the material necessary for the lab, and

making sure the equipment was set up properly. Frankie did a fantastic job in gathering

HCl and CaCO3 for the lab, while I was in charge of recording the data, and observing the

reactions that took place. This report is going to explore the amount of reactant needed in

order to produce a reaction, how to find out the limiting reagent, or the amount of moles

that can be produced in the reaction. This lab involved multiple procedures in which

Calcium and Hydrochloric Acid were used. During this experiment, we had to make

calculations of moles, and determine which compounds were released due to the reactions

that took place.


The title of this lab speaks for itself, because it is a lab focused on how much reactant is needed in

order to produce the ultimate reaction, without having any leftover material. In this lab, multiple

reactions will be shown, and what their results produced. HCl is the main reactant used in this lab.

Different test tubes with various amounts of Calcium were used in order to find out which test tube

provides the best reaction. The second part of this lab will show how much CO2 was released in the air

by weighing the reactant before and after.

2. Main Body of the Paper

OPEN ACCESS

CHEM 111 A09 FALL 2016, LAB 5, 25-28 26

Part 1

The experimental procedure in this lab involved the use of Calcium and Hydrochloric Acid. My

partner, Frankie, acquired 50 mL of HCl from the fume hood, and he began gathering the 0.25 g, 0.60

g, and 1.0 g of CaCO3 into 3 different test tubes, as I began to set up the area for experimentation.

After the test tubes were set up, I added 5 mL of distilled water to all 3 beakers. Then, an indicator was

used in order to tell if the substance was basic or acidic. The indicator showed that the substance in the

beakers was in fact basic, because it appeared yellow when the indicator was added to the test tubes.

Me and my partner than added the calculated volume of HCl solution to the beakers. The first test

tube (0.25 g), turned pink as HCl was added to it. It slowly started to release bubbles, and it took less

than a minute for all of the CaCO3 to be reacted. The second test tube (0.6g) reacted much more

violently. Once the HCl was added, the solution reacted very quickly, and some of it managed to spill

out of the test tube. Once the HCl was added, the solution turned pink as well, and once the reaction

was completed, the solution became yellow again. Once the HCl was added to the third test tube (1.0

g), it turned pink as well, however it was bubbling very slowly, and after the reaction was completed,

CaCO3 was leftover at the bottom of the test tube.

The limiting reactant in this case is the HCl, because the CaCO3 was still left over in the third test

tube. If more HCl was added to the third test tube, all of CaCO3 would have reacted completely.

Part 2

The second part of this lab involved the use of HCl and CaCO3 again. I acquired a 10 mL graduated

cylinder, and weighed it. Our measurement of the cylinder was 41.127 g. After 5 mL of HCl was

added, the cylinder weighed 46.258 g. Therefore, the HCl itself weighed 5.131 g. Me and Frankie were

asked to approximately used 0.9 g of CaCO3 for the experiment. Frankie was able to gather 0.900 g of

CaCO3 exactly, and we put it in a 50 mL beaker with 10 mL of water mixed in. After the mixture was

created, we weighed the mass of the beaker and the solution, which came out to be 37.897 g. After the

HCl was added to the beaker, and the reaction was completed, the beaker weighed 42.765 g. The

difference between those 2 values is 4.868 g. So, if we compare the values 5.131 to 4.868, this

indicated that CO2 was released in the beaker with CaCO3 when the HCl was added. If we subtract

4.868 from 5.131, it equals 0.263, which is the amount of CO2 that was released during the reaction in

the beaker with CaCO3.

Part 3

Our instructor made sure that all the values were recorded in class so they could be compared. This

is the table of all the values that were recorded in the lab.

Group # approx.

CaCO3[g]

empty

cylinder

[g]

cylinder +

5ml HCl [g]

exact

CaCO3

[g]

Beaker+10ml

water+CaCO3

[g]

Beaker+10ml

water+CaCO3+

HCl [g]

B2 0.2 38.804 44.119 0.200 35.198 42.219

B4 0.3 48.900 53.883 0.300 39.829 44.670

B6 0.4 38.775 44.094 0.403 35.990 41.087

B8 0.5 20.436 25.484 0.500 37.526 42.204

B10 0.6 48.324 53.338 0.602 39.696 44.411

B12 0.7 35.162 44.251 0.714 43.123 48.036

B14 0.8 39.530 44.320 0.812 49.930 42.920

CHEM 111 A09 FALL 2016, LAB 5, 25-28 27

B16 0.9 41.127 46.258 0.900 37.897 42.765

B18 1.0 38.310 43.578 1.002 44.637 49.539

B20 1.1 32.953 38.354 1.102 46.426 51.658

B22 1.2 48.520 53.736 1.200 41.523 46.205

B24 1.3 39.114 44.376 1.311 39.506 44.296

Pre-lab Questions:

1. In my opinion, fuel would be the limiting reagent, because the oxygen is constantly in the air, while

fuel is going to be used up in order to start the fire.

2. The formula given in the pro-packet is a balanced chemical equation because all of the elements in

baking soda and acetic acid are conserved after the reaction is completed, and new compounds are

formed.

a. One of the observable changed that is expected in this reaction is the release of CO2.

b. A mole ratio that will react is one mole of acetic acid and one mole of baking soda.

c. Baking soda will be completely consumed if 2 moles of acetic acid and 1 mole of baking

soda is present in the reaction.

d. 3 moles of carbon dioxide will be produced because the moles of Carbon in the

equation is 5, and there is only 1 mole of Na in the equation. Therefore, there has to be 3

moles of carbon dioxide in order to have a balanced equation.

e. The moles of baking soda in (c.) were used up before carbon dioxided, therefore the baking

soda was the limiting reagent. Only 1 mole of NaCH3COO was produced. In (d.) 3 moles of

carbon dioxide was produced since the amount of moles of Carbon was 5, therefore in order for

the equation to be balanced, more moles of CO2 had to be produced.

3.

a. 0.119 moles

b. 0.119 moles

c. 0.119 moles

d. 9.76 g of NaCH3COO

e. Only 0.100 g of NaHCO3 would react, and only 0.100 moles of NaCH3COO would be

formed.

Conclusion Questions:

1) We tried to find out what the limiting reactant was for the procedure by using test tubes with

different amount of CaCO3 in them. We were able to conclude that the test tube with 0.6 g of CaCO3

was the one that produced an ultimate reaction, causing all of CaCO3 to react with HCl.

2) Not necessarily. The limiting reactant depends on the amount of molecules of HCl that are able to

bond to CaCO3 in order to produce a reaction.

3) Yes, the limiting reactant is going to be the one with the fewest moles. We concluded that HCl was

the reactant with fewest moles in test tube 3, because much of CaCO3 was still left over.

4) According to the formula provided at the beginning of the lab, 2 moles of HCl are required for 1

mole of CaCO3 in order to achieve the ultimate reaction, however CaCO3 weighs about 2.7 times

more than HCl. The reaction doesn't depend on the weight of the molecule, but on the chemicals that

are able to combine with each other.

5) In part 2 of this lab, the limiting reagent was CaCO3, because all of it was used up during the

experiment. CaCO3 would have been left over if HCl was the limiting reagent.

6) I was unable to include the plot in this document, however when I made the plot on paper, the line

was reasonably straight because difference between the values was roughly 5, except group B 14,

which seemed to make a measuring error.


CHEM 111 A09 FALL 2016, LAB 5, 25-28 28

To summarize this lab report, I would like to say that this was a very informative lab, and it taught

me and my lab partner Frankie about the difference between the mass and moles. By concluding the

experiment, I realized that the mass of the mole has nothing to do with it being the limiting reagent; it

all depends on the molecules that are able to bond to the other molecules. For example, at the

beginning of this lab, we were given an example that it takes 1 handlebar, 1 seat, 2 wheels, and 1 frame

in order to make one bike. So, if we are given 20 wheels, 18 frames, 18 seats, and 18 handlebars, we

are only able to make 10 bicycles, because 2 wheels are required for each bike. That leaves us with 8

frames, 8 seats, and 8 handlebars left over. This analogy can be used in the experiment that we

conducted. We used HCl and CaCO3, and the formula states that 2 HCl moles are required per 1 mole

of CaCO3 in order to have a perfect reaction.

In the first test tube, the limiting reagent was CaCO3, because the test tube was still pink at the end

of the reaction, which meant that it was still acidic. The second test tube was the ultimate reaction

produced, because both the HCl and CaCO3 reacted perfectly, since the test tube was yellow at the end

of the reaction. And in the third test tube, the limiting reagent was the HCl, because there was still

much of CaCO3 left over. I was able to conclude, according to the chart, that the line was fairly

straight, and that one group had made a calculation mistake, which was a simple error. I learned a lot

about moles and the calculation of limiting reagents because of this lab, and me and my partner will be

able to use this information in the future.

Acknowledgments

The author gratefully acknowledges his lab partner Francis Zwiercan for his assistance in



CHEM 111 A09 FALL 2016, LAB 6, 29-33

CHEM 111 A09

GENERAL CHEMISTRY



Analysis of Antacids

Alex Ignatenko




Abstract: The purpose of this lab was to analyze a brand of over-the-counter antacid by

the process of back titration. The process of back titration in this lab was dissolving the

antacid with excess HCl that was measured in order to ensure that all of the antacid is

dissolved completely. Then, by adding a chemical indicator, we were able to analyze the

amount of NaOH needed in order to achieve the equivalence point. As the equivalence

point was achieved, we were able to calculate the amount of moles of HCl and NaOH used,

and the amount of CaCO3 present in order to tell how cost effective our sample was

compared to other samples. During the experiment, Frankie did a great job of gathering the

antacid tablets, HCl liquid, and NaOH liquid. I was in charge of collecting the data, while

Frankie performed the experiment. We weighed 3 different samples of TUMS, and we

recorded that the active ingredient was calcium carbonate. We dissolved each of the

samples in exactly 25.00 ml of HCl, and we then added the chemical indicator and the

NaOH needed in order to achieve the equivalence point. After that, the procedure was

completed, and we had the proper data in order to do the calculations to figure out the

amount of active ingredient present. This report will carefully show and explain the

procedure and calculations that were performed during this experiment. The procedure

involved the use of 3 antacid samples (TUMS), HCl liquid, NaOH liquid, a scale, and

erlenmeyer flask, a volumetric pipette, and a mortar and pestle. It was crucial that the

measurements were as precise as possible in order to gather the most accurate data. We had

to use the process of back titration in order to figure out the amount of HCl that was used

up by the equivalence point of the reaction. After that was figured out, we were able to

calculate the amount of excess HCl that was left over after the reaction was completed.


OPEN ACCESS

CHEM 111 A09 FALL 2016, LAB 6, 29-33 30

This paper is about understanding the concept of back titration, and the use of HCl and NaOH in

order to find out the amount of excess HCl used in the reaction. The back titration is the concept where

excess amount of acid is added in order for all of the antacid to react. Then, the NaOH was added in

order to find out how much excess HCl was left over in order to calculate the amount of HCl that was

needed for the antacid to completely react. That allowed us to calculate the moles and grams of the

active ingredient. This paper will show the results of the procedure, and it will thoroughly explain the

calculations that we acquired.

2. Main Body

At first, we were told by our instructor to acquire 3 different samples of the same antacid. We

decided to pick 3 different samples of TUMS, and Frankie weighed each tablet as I recorded the

results. The first tablet was 1.215 g, the second tablet was 1.205 g, and the third tablet was 1.201 g.

The tablets were round and white, and their main ingredient was calcium carbonate. After we weighed

the tablets, I acquired 25.00 mL of 0.799 M HCl provided for us by the Dr. Tang. While I was doing

that, Frankie crushed the first tablet by using a mortar and pestle. The crushed tablet was relocated to

an Erlenmeyer flask. As the 25.00 mL of HCl was added to the flask, the solution began to bubble until

the reaction was completed.

Our next step was to heat the solution in order to reassure that all of the antacid was reacted. After

that, we let it cool for about 5 minutes, while Frankie gathered the NaOH and placed it in the buret. We

added 2 drops of phenophalyne indicator to the solution, and we began to add the NaOH solution to the

Erlenmeyer flask. We started with 7.60 mL of NaOH liquid, and the equivalence point was reached

when the NaOH liquid reached 22.00 mL in the buret. Therefore, by subtracting 7.60 from 22.00, we

were able to conclude that 14.40 mL of NaOH was used for tablet 1.

We have done the same procedure for tablet 2 and 3, and our results were slightly different. For the

second tablet we used 17.70 ml of NaOH, and for the third, we used 16.70 mL of NaOH.

Data and Results

1. Brand of antacid we analyzed: TUMS.

2. Active ingredients: calcium carbonate.

3. Neutralization reaction of active ingredients with HCl for antacid: CaCO3 + 2 HCl = CaCl2 + H2O +

CO2.

4. Cost and number of tablets per bottle: $3.34 for 150 tablets.

5. Cost of each antacid, in cents per gram: on average, each pill weighs (1.215 + 1.205 + 1.201) / 3 =

1.207 g/pill, so then 3.34 / 150 = 0.0223 = 2.23 cents/pill, therefore 2.23 x (1.000 / 1.207) = 1.85

cents/g

6. Concentration of HCl = 0.799 M

7. Concentration of NaOH = 0.601 M

8. Indicator used, and the color change: Phenophalyne, the color changed to purple.

9. Volume of HCl added, and moles of HCl added: 25.00 mL, (25.00 mL / 1000 ml/L) = 0.025 L x

0.799 mol/L = 0.0200 mol

10.

Antacid Amount (g) HCl (0.799 M) NaOH (0.601 M)

CHEM 111 A09 FALL 2016, LAB 6, 29-33 31

Tablet 1 1.215 25.00 mL 14.40 mL

Tablet 2 1.205 25.00 mL 17.70 mL

Tablet 3 1.201 25.00 mL 16.70 mL

11.

Antacid Cost

(cents)

Cost

(cents/g)

Moles

HCl

added

Moles

NaOH

added

Moles HCl

neutralized

Moles HCl

neutralized

per gram

Moles HCl

neutralized

per cent

Tablet 1 2.23 1.85 0.0200 0.00877 0.0112 0.00922 0.00498

Tablet 2 2.23 1.85 0.0200 0.0106 0.00940 0.00780 0.00422

Tablet 3 2.23 1.85 0.0200 0.0100 0.0100 0.00833 0.00450

12.

The calculation for "Moles NaOH added" (for tablet 1): 14.40 mL of NaOH x 0.000601 mol/mL =

0.00877 mol NaOh

The calculation for "Moles HCl neutralized" (for tablet 1): 0.0200 mol of HCl - 0.00877 mol of NaOH

= 0.0112 mol of HCl

The calculation for "Moles HCl neutralized per gram" (for tablet 1): 0.0112 x (1 / 1.215) = 0.00922

mol

The calculation for "Moles HCl neutralized per cent" (for tablet 1): 0.00922 x (1 / 1.85) = 0.00498

mol/cent

13.

Average value of moles HCl neutralized per gram tablet: (0.00922 + 0.00780 + 0.00833) / 3 = 0.00845

Average value of moles HCl neutralized per cent: (0.00498 + 0.00422 + 0.00450 ) / 3 = 0.00457

14.

Tablet 1: 0.0112 mol HCl x (1/2) = 0.00560 mol CaCO3



15.

Average of active ingredient (CaCO3): (0.00560 + 0.00470 + 0.00500) / 3 = 0.00510 mol

The calculation for grams of active ingredient: 0.00510 mol x 100 g/mol = 0.51 g = 510 mg.

The amount stated by the manufacturer was 500 mg. Therefore, percent error is calculated by: (510-

500)/ 500 x 100 = 2 % error

Pre-lab Questions:

1. Write the balanced reaction equations for the reaction of HCl with each of the active

ingredients for in Milk of Magnesia and TUMS.

CaCO3 + 2 HCl = CaCl2 + H2O + CO2

2. Why do you think this lab's procedure calls for you to titrate the excess HCl with NaOH

after reaction of the HCl and the antacid, as opposed to direct titration of the antacid with HCl?

Using excess HCl will make sure that the antacid is completely reacted.

3. In the titration used in this experiment, which substance goes in the buret and which one

goes in the Erlenmeyer flask? Would the experiment work if the substances placed the other way

around?

If they were switched, the CaCO3 wouldn't react completely. NaOH goes in the buret, and

the HCl goes in the Erlenmeyer flask for CaCO3 to react.

4. What is the precision of a buret and a volumetric pipette?

CHEM 111 A09 FALL 2016, LAB 6, 29-33 32

The precision is 2 digits past the decimal point.

5. Calculate the initial number of moles of HCl in 15.00 ml of the 1.00 M HCl solution.

15.00 L / 1000 ml = 0.015 L x 1.00 mol/L = 0.015 mol

6. Calculate the moles of NaOH added to the solution.

0.950 mol/L = 0.000950 mol/mL 5.8 mL x 0.000950 mol/mL = 0.00551 mol of NaOH

7. Calculate the moles of HCl neutralized by the antacid.

0.015 - 0.00551 = 0.00949 mol of HCl

8. Calculate the number of moles of HCl neutralized per gram of antacid.

0.00949 x 1/1.42 = 0.00668 mol/g

9. A box of the antacid used in the example above costs $2.38, contains 75 tablets and the

tablets have an average mass of 1.42 g/tablet. What is the cost of the antacid in cents per

tablet and in cents per gram?

2.38 / 75 = 0.00317 = 3.17 cents per tablet.

3.17 x (1.00 / 1.42) = 2.23 cents per gram.

Conclusion Questions:

1. Summarize what you tried to find out in the experiment.

Our ultimate goal was to find out the amount of CaCO3 per tablet by using the process of back

titration. We were able to conclude that the amount of CaCO3 that we calculated was 510 mg, and

therefore the percent error was 2%, which means we conducted the experiment quite well.

2. Compare the cost per gram of tablet with the cost of mole of H+

neutralized. Which of these values

is the better way to access the cost effectiveness of the tablet?

The cost per gram is 1.85cents, while the average cost of mole of H neutralized per cent is 0.00457.

Therefore, about 0.00845 moles of H+ will be neutralized per gram. The cost per gram doesn't provide

the buyer with the information on how effective the active ingredient is at neutralizing acid, however

the cost of mole of H +provides the buyer with the amount of moles of acid that will be neutralized by

the active ingredient. Therefore, the cost of mole of H+

neutralized is the better way to access the cost

effectiveness of the tablet.

4. Are there other considerations besides cost and neutralizing ability to consider? If so, what are they

and what impact would they have on your decision?

I don't think there are any other considerations that would impact my decision. I believe that the

experiments and calculations done in this lab have provided me with all the answers needed.

3. Discussion and Conclusion

In conclusion, this lab report provided me and my lab partner, Frankie, with important information

on how to calculate the amount of CaCO3, which was the active ingredient in the antacid sample that

we used. We were introduced to the process of back titration, which is the process where excess acid is

added to the solution with active ingredient to ensure complete reaction. Then, the solution is titrated

with NaOH and the help of chemical indicator in order to find out the equivalence point. Once the

equivalence point was calculated, we had the information needed in order to calculate the active

ingredient present.

To conclude our findings, we were able to calculate the amount of active ingredient present, which

in our case was CaCO3, and it came out to be 510 mg per tablet. On the TUMS box, it stated that the

active ingredient per tablet was 500 mg, so the percent error came out to be 2%, which means that our

lab experiment was conducted quite well. Also, by using the calculation methods we learned in

previous labs, we were able to figure out the amount of moles of acid that will be neutralized by the

active ingredient. Most importantly, this lab showed Frankie and I how to use the process of back

CHEM 111 A09 FALL 2016, LAB 6, 29-33 33

titration in order to find out the amount HCl that was neutralized, and Dr. Tang did a great job of

showing us the calculations that needed to be used in order to figure out the effectiveness of the

antacid sample that we were provided.

Acknowledgments

The author gratefully acknowledges his lab partner Francis Zwiercan for his assistance in



CHEM 111 A09 FALL 2016, LAB 7, 34-37

CHEM 111 A09

GENERAL CHEMISTRY



The Analysis of Eggshells for Calcium Carbonate

Baily Blackburn




Abstract: This paper will provide the findings of my laboratory experiments that were

based on the method of direct titration to find the amount of calcium carbonate in an

eggshell. This lab could be performed three ways: Molar mass and mathematical

calculation, direct titration, or back titration. I chose the direct titration to get a better

understanding of its use but I quickly found that since the solute is water, it has a huge

disadvantage when dissolving the eggshell to titrate. This paper will guide you through the

process of finding the necessary data to come to the proper conclusions, even though

they’re not the conclusions you’re looking for.


For this lab experiment, it’s almost and exact replica of the last two labs we have done. The only

difference here is that we are dealing with a new solute (CaCO3) and we get to choose what method

we want to use to find the amount of CaCO3 in the eggshell. I chose to use the direct titration method

simply because I felt more comfortable with it and I wanted to touch base with it again.

As we have done before, we need to dissolve the eggshell in water and titrate it with HCl acid that

has a known concentration. Keep in mind that you cannot overshoot the equivalence point when doing

this on your own; it will give you false results. To ensure our results are accurate, we will perform

three trials.

With this in mind, we can properly perform this experiment. It is important to know that when you

add a certain amount of base, you need to slowly drip the acid in to the solution. Once we find the

moles of CaCO3 we can then calculate the percentage of this substance in the eggshell.

OPEN ACCESS

CHEM 111 A09 FALL 2016, LAB 7, 34-37 35

2. Titration of an Eggshell with Known HCl using a Chemical Indicator

Problem

Use direct titration to acquire data about the CaCO3 content in an eggshell.

Data

Concentration of HCL: 0.799M Starting vol of HCl in buret: 47.5ml

Procedure

To start this lab, we had to crush half an eggshell (without the membrane) into a fine powder to

ensure it would dissolve in water. Once we did this, we had to find a suitable amount that would react

with 47.5ml or less of HCl. In our pre-lab, we found that 0.600g will give us proper results.

1.422g/ 100.09g/mol= 0.014mol

0.014mol x 2 = 0.028mol HCl

(0.799mol/1L)=(0.028mol/x) x= 0.035L or 35.00ml

This is the amount of HCl we expect the CaCO3 to react with to reach the equivalence point. After we

set up the solution to be titrated we found that the equivalence point was reached with only 1.00ml of

HCl added. We conducted two more trials to ensure our results.

Trial 2

Mass of eggshell: 1.598g

Starting amt. of HCl: 47.5ml

Expected equivalence point: 40.05ml

Amt added: 1ml

Trial 3

Mass of eggshell: 1.587g

Starting amt of HCl: 47.5ml

Expected equivalence point: 40.05ml

Amt added: 1ml

Discussion

We came nowhere close to our expected equivalence point in any of the trials. Going into this

experiment, we expected that not all the eggshell would dissolve into the water and that showed in our

observations. But we hypothesized that there wasn’t enough CaCO3 dissolved in the water to give a

proper equivalence point and we found that this was true.

PRE-LAB QUESTIONS

1. Balance equation for CaCO3 and HCl

CaCO3 + 2HCl = CaCl2 + H2CO3

2. If the eggshell is 100% calcium carbonate, how many moles of CaCO3 would be present?

0.053mol

b. Based on your balanced equation, how many moles of HCl would be needed to react with the moles

of calcium carbonate in 2a?

CHEM 111 A09 FALL 2016, LAB 7, 34-37 36

0.106mol

c. If HCl is 1.000M, what minimum vol. of HCl is needed to react with the CaCO3 in 2a? Is this a

volume of HCl that can be dispensed without refilling a 50.0ml buret?

0.106L or 106ml / no

d. What mass of eggshell should you aim to analyze assuming its 100% CaCO3?

1.25g

e. How will you ensure you won’t overshoot the endpoint in the titration?

Use an indicator

f. What are the advantages and disadvantages of direct titration?

ADV: Its simpler to use and requires less steps

DISADV: Not all the eggshell with dissolve in the water since it’s not a strong solvent

3. If you use a 25.00ml pipette to dispense 1.000M HCl, how many moles of HCl are present?

0.025mol

b. Find the maximum mass of CaCO3 that could be reacted with this amount of HCl.

1.25g

c. What would be a reasonable mass of eggshell to use?

0.600g

d. How many moles of HCl would be consumed by the eggshell? How many moles of HCl would be

leftover?

0.0120mol / 0.013mol

e. Write the balanced reaction between HCl and NaOH. What vol. of 1.000M NaOH would be needed

to neutralize the leftover HCl as determined in 3d?

HCl + NaOH = NaCl + H20 / 13.00ml

f. Does your answer in 3e require more than one filing of a 50ml buret? Does it take advantage of the

four sig figs of a buret?

No / Yes its 13.00 which is four sig figs

g. How will you find the endpoint without overshooting it?

Use an indicator

h. What are the advantages and disadvantages of back titration in this case?

ADV: It’s very precise

DISADV: There are a lot more calculations that could lead to error

POST-LAB QUESTIONS

1. Summarize what you tried to find and how you did it. What were the final results?

We tried to find the percentage of CaCO3 in an eggshell by using the direct titration method.

We found that it cannot be done with this method due to the water solvent.

2. Write a balanced equation for the reaction.

CaCO3 + 2HCl = CaCl2 + H2CO3

3. Include any other equations used.

None

4. Calculate the average % calcium carbonate in the eggshell.

CHEM 111 A09 FALL 2016, LAB 7, 34-37 37

We could not continue with the experiment due to our method. The water solvent greatly

affected the results.

5. List other businesses that might be interested in the calcium carbonate content of eggs.

Farmers and produce sellers could benefit from this information because the CaCO3 content in

the shell could affect the quality of the yolk.

6. How many grams of CaCO3 could be recouped if 37million pounds of eggshells were thrown away

and 70% of that can be recouped? How much is it worth (112.50/500g)?

37mill x .7= 2.59x10^7lbs x (453.592g/1lb)= 1.17x10^10g

1.17x10^10/500= 23496065.6 x $112.50= $2,643,307,380.00


The downside to direct titration is that you use water as a solvent. Although water can dissolve

many elements, it was unable to completely dissolve the eggshell and all the CaCO3. This posed a

problem because there was not enough CaCO3 to react with the HCl where we could calculate the

percentage in the eggshell. A better approach to this experiment would be to use the back-titration

method because it uses a strong acid as a solvent. Since we could not completely dissolve the

eggshell, we cannot complete the experiment with this method.

Acknowledgments

I gratefully acknowledge the constructive comments and suggestions of Mr. Ali, Dr. Tang, and Mr.

Brad, they significantly improved the quality of this paper and deserve recognition. Technical support

was provided by the Stapleton Library at Indiana University of Pennsylvania.


CHEM 111 A09 FALL 2016, LAB 8, 38-41

CHEM 111 A09

GENERAL CHEMISTRY



Gas Laws: Determination of the Molar Mass of a Metal

Elizabeth Matusik



Received: November 1, 2016 / Accepted: December 13, 2016 / Published: December 13, 2016

Abstract: This lab focuses on the concept of using the ideal gas law in order to determine

the unknown molar mass of a metal. This is done by solving for the other variables in the

ideal gas law equation, such as pressure, volume, and temperature. This data was collected

by reacted hydrochloric acid and the unknown metal to create a gas, and the electronic

equipment hooked up to the computer gave the data throughout the reaction. This data was

then used in combination with the chemical equation to calculate the moles of the metal,

and then converted into grams per mole.

1. Introduction

In order to be able to calculate the molar mass of the unknown metal that was used, we first must

know the balanced chemical reaction between hydrochloric acid and the metal, which have a one to

one mole ratio. Because of this ratio, the moles of hydrogen used are identical to the moles of the metal

used. The metal simply just needs to be weighed first, and then the reaction can be carried out. The

equipment that is hooked up to the computer then registers the pressure and temperature changes. With

all of this, we can then calculate the molar mass of the metal and compare it to the periodic table in

order to identify it.

2. Pre-lab Questions

1. What is the pressure change in this experiment, expressed in atm?

703 torr x (1 atm / 760 torr) = 0.925 atm

2. What is the temperature of the gas in degrees Kelvin?

C + 273.15 = 295.15 K

3. What is the volume of the gas in liters?

OPEN ACCESS

CHEM 111 A09 FALL 2016, LAB 8, 38-41 39

76.0 mL = 0.0760 L

4. Re-arrange the ideal gas law (PV = nrRT) so that it will solve for moles (n).

n = PV / RT 5. Use your answers for #1-4 above to find the moles of H2 gas produced by this reaction.

n = (0.925 atm)(.0760 L) / (0.0821)(295.15 K) = 0.00290 mol H2

6. How many moles of metal (M) reacted?

0.00290 mol H2 = 0.00290 mol M

7. If the sample of the metal weighed 0.345 g, what is the molar mass and the identity of

the metal?

0.345 g / 0.00290 mol = 118.97 g/mol = Tin

8. From the brief description in the procedure, explain how you will accurately and

precisely measure the volume of the gas in the flask.

In order to find the volume of gas in the flask, you must first find the total volume that

the flask can hold, and this is done by measuring out water in a graduated cylinder, and

then recording how many mL the flask can hold. Next, subtract 5 mL from this total

number, since this is the amount of HCl being added. This is then the total amount of

gas that can be held in the flask.

3. Measuring the Change in Pressure at Constant Volume

The first step in this procedure is to take a 125 mL Erlenmeyer flask and determine the volume of

hydrogen gas that it will be able to hold during the reaction. The process should be followed as stated

in pre-lab question #8 above. It was found that the volume that the flask would be able to hold would

be around 135 mL. The next step is to weigh a piece of the unknown metal, which is shown in the data

tables below. The average mass of the metal for all three trials was 0.011 g.

The equipment then was set up, as instructed by the lab manual. This involved hooking up the gas

pressure sensor and the T probe to the computer, and placing the gas pressure sensor into the rubber

stopper of the flask. Next, a large beaker was filled up with room temperature water, where the flask

would be placed in during the reaction. 5 mL of 3.0 M HCl solution was then added to the syringe, and

the rubber stopper was placed in the flask with the unknown metal in it. The syringe was placed on top

of the flask. Once the computer program was up and running and the probes and sensors were

connected, the start button was pressed. After about 10-15 seconds, the HCl was released into the

flask, and the reaction took place, being recorded on the graph on the computer screen.

It is also important to note that the pressure should be changed to be in atm. Once the reaction graph

has leveled off, you can then click for it to stop, and then analyze the data. The change in pressure can

be found by subtracting the point at the very bottom of the curve from the very top of the curve. The

data from all three trials is found in the table below. The temperature for all three of the trials was 295

K. The flask should be rinsed out, and this procedure should be repeated two more times to ensure

accurate results.

CHEM 111 A09 FALL 2016, LAB 8, 38-41 40

An example of how to solve for the molar mass of the metal is shown below.

n = PV / RT

n = (0.1153 atm)(0.135 L) / (0.0821)(295.25 K)

n = 6.42 x 10-4

0.013 g / 6.42 x 10-4 = 20.24 g/mol

Tables and Figures

Mass of Metal for Trial 1 of Experiment

Trial 1 Trial 2 Trial 3

Mass 0.013 g 0.012 g 0.013 g

Average 0.013 g

Standard Deviation 0.000577



Mass 0.010 g 0.011 g 0.010 g

Average 0.010 g




Mass 0.010 g 0.011 g 0.0012 g

Average 0.013 g


Change in Pressure


Pressure change 0.1153 atm 0.1284 atm 0.04300 atm

Average 0.09557 atm



Overall, we were able to accomplish the main goals of the lab. We weighed a piece of the unknown

metal, and using equipment and sensors hooked up to the computer and the beaker, were able to find

the variables in the ideal gas law equation. By recording this data and these variables, we were able to

solve for the number of moles, or n, in the equation. Once this variable was found, the original mass of

the metal was divided by the number of moles. This gives us the molar mass of the metal. We came to

the conclusion that the unknown metal was Magnesium. The actual molar mass on the periodic table

CHEM 111 A09 FALL 2016, LAB 8, 38-41 41

for this metal is 24.305 g/mol. Our average molar mass was 17.24 g/mol, however, our closest value

was trial one, which was 20.24 g/mol.

The numerical difference between the actual and experimental value is 4.065. The standard

deviation between these three trials was 3.90. This is less than two standard deviations between these

two values, which means that our data could be accepted and that the lab was successful.

Acknowledgments

Lydia Connelly was my lab partner and worked on this lab with me. Amanda and Symira also

worked on this lab procedure with us.


CHEM 111 A09 FALL 2016, LAB 9, 42-45

CHEM 111 A09

GENERAL CHEMISTRY



Formation of an Air Bag from the Production of CO2

Tara M. Kenna




Abstract: This paper presents the results of a model airbag designed to hold the maximum

amount of CO2 after reacting baking soda and hydrochloric acid in the following reaction:

NaHCO3(s) + HCl (aq)CO2 (g) + H2O (l) + NaCl (aq). The researchers used a medicine

cup, taped inside the Ziploc bag, to hold 4.40 g NaHCO3 and poured 17.30 mL of HCl into

the bottom of the bag. In order to react the two, they flipped the bag over and the reaction

began and the 0.052 moles of CO2 filled the bag 9.2 cm in width in 23.29 seconds.

1. Introduction

Chemists, in order to save millions of dollars in chemicals and disposal costs, carefully calculate

and balance the amount of reactants to make good economic and environmental sense. The invention

of the automobile airbag required not only chemistry it also consideration of economic and

environmental factors. When a crash occurs, an air bag must inflate rapidly, usually around 40

milliseconds, in an attempt to keep the occupants involved in the crash safe. The gas that is produced

from the inflation of the air bag is nontoxic, odorless, and cool enough to avoid burning the car

passengers. Chemists also look for compounds that don’t expand unexpectedly, and are easy and safe

to dispose of (General Chem. Lab Notebook, pg 62).

In this lab, in order to keep the reaction to be nontoxic, researchers will use the reation of sodium

bicarbonate, also known as baking soda, and react it with an aqueous solution of hydrochloric acid,

also known as stomach acid, to produce carbon dioxide gas to test the engineering design (General

Chem. Lab Notebook, pg 71). This reaction of hydrochloric acid and sodium bicarbonate is as follows:

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

OPEN ACCESS

CHEM 111 A09 FALL 2016, LAB 9, 42-45 43

In this reation, the CO2 produced obeys the Ideal Gas Law. All gases, at everyday temperatures and

pressures obey this law. This gas law combines together all major measurements when it comes to

gases: pressure, volume, temperature, and amount (moles) of gas (General Chem. Lab Notebook, pg

71). The equation is as follows:

PV = nRT

Where P is the pressure of the gas in atmospheres (atm), V is the volume in liters (L), n is the

amount in moles of the gas, R is the Ideal Gas Constant that has a value of 0.0821 atm-L/mole-K, and

T is the temperature of the gas in degrees kelvin (K). Using this equation, researchers can find any of

the variables using three of the other variables.

Within this paper, researchers will conduct experiments to design a model automobile airbag that

will expand to the largest possible volume without breaking the seal on the Ziploc bag. They will use

the ideal gas law to determine the proper amount of reactants so that there is no burst of the bag or

wastefulness of the chemicals.

2. Materials

Ziploc bag

Ruler

Calculator

Thermometer

10-15 g of NaHCO3

50 ml of 6 M HCl

Stopwatch

1 L graduated cylinder

Medicine cup

3. Design of the Air Bag

In order to make sure that the bag would not explode from the reaction, researchers had to determine

how much available volume there was within the Ziploc bag. In order to do this, they filled the bag

with water and then dumped the water into the 1 L graduated cylinder to determine the max volume of

the bag. The volume that the researchers found was 1260 mL.

With the researchers knowing the maximum volume, they next found the temperature change that

would occur within the reaction. They determined that the temperature change was not significant

enough so they kept the temperature at 303 K, the measured temperature after the reaction. Next,

researchers calculated the amount in moles of CO2 that would fill the bag after the reaction. They

determined this by using the Ideal Gas Law equation, solved for n (amount in moles).

n = PV/RT

= 0.052 moles CO2

CHEM 111 A09 FALL 2016, LAB 9, 42-45 44

Knowing the moles of carbon dioxide produced gives them a starting point on determining how much

sodium bicarbonate and hydrochloric acid they will need to produce this maximum amount of gas.

They do so in the following calculations:

0.052 mol CO2

= 4.40 g NaHCO3

0.052 mol CO2

= 17.30 mL HCl

Now that they have the proper amounts of reactants to form the air bag, they need the actual design. In

order to have minimal external parts or assistance and a lesser chance of having some carbon dioxide

escape the bag, the researchers decided that they would have to have both reactants inside the bag prior

to the reaction. Their final design was to tape a medicine cup to the inside of the sandwich bag so that

it sits upright towards the seal. They would put the 4.40 g of sodium bicarbonate inside this medicine

cup. Next, they would dump the 17.30 ml of hydrochloric acid into the bottom, away from the

medicine cup. When the timer said go, the researchers would flip the bag upside down and the two

reactants would mix and react. The reaction took 23.29 seconds to fill the bag with carbon dioxide to a

thickness of 9.2 cm. The researchers repeated this process three times.

4. Optimization of the Air Bag Design

Now that the researchers had their design in place, they needed to determine if changing the amount

of one or both of the reactants changed the speed of inflation of the bag or if it would create a better or

poorer cushion. To do this, they first, cut the measurements in half and using the same design, only

reacted 2.20 g of sodium bicarbonate and 8.65 ml of hydrochloric acid. This made less carbon dioxide

and created a poorer cushion but the reaction time did speed up. When the researchers doubled the

reactants to 8.80 g sodium bicarbonate and 34.60 ml of hydrochloric acid, they found that the rate at

which the bag filled up was the same if not slightly less than the ideal measurements they measured at

first. This amount of reactants, did in fact, create a thicker bag and a harder cushion.

5. Discussion

In conclusion, the researchers found that their design of air bag, complete with the medicine cup,

may have been a better way than other designs to separate the reactants until it was time for the

reactant but it was also worse than others. When the researchers flipped the bag over, the sodium

bicarbonate all fell in the same spot because it came from the small cup. The hydrochloric acid on the

other had, when flipped over, it went all along the seal of the bag because it was a liquid. If there was a

design that had all of the hydrochloric acid and the sodium bicarbonate to mix in the same spot, maybe

the reaction time would’ve been cut down even more.

By trying the different quantities of the reactants, the researchers found that if they added more

reactants, the bag would fill up with more air than maybe the bag could hold but the downside was,

that since there was more reactants, the time it took for the reactants to reacts and form the carbon

dioxide greatly increased. This was a plus for when less amounts of the reactants were tested because

CHEM 111 A09 FALL 2016, LAB 9, 42-45 45

there were less reactants to be reacted, the time it took to fill the bag was less but then again, the

cushion of the carbon dioxide formed from the less amount of reactants was weaker than the other

trials. The researchers settled on their original calculations of 17.30 ml of HCl and 4.40 g of NaHCO3.

Though it did not go off in 40 milliseconds like a normal car airbag does, it does create the optimal

cushion for the amount of reactants used.

This kind of lab is useful in everyday life and is very important when it comes to the safety of car

passengers. We all know too many people who have died or have been seriously injured in car crashes.

This Ideal Gas Law, and this experiment in particular, allows chemists to see the optimal amount of

reactants to produce the safest air bag for the safety of car passengers.

Acknowledgments

The author would like to acknowledge Jeffery Schmidt for assistance in performing the experiment.

References

Hoshin V. Gupta, Yuqiong Liu, Guidelines for the Preperation of Journal Style Written Reports,

January 19, 2005

Unknown Author, General Chemistry Laboratory Manual, Fall 2016


CHEM 111 A09 FALL 2016, LAB 10, 46-50

CHEM 111 A09

GENERAL CHEMISTRY



Causes of the Temperature Change

Mohammed Ali




Abstract: The purpose of this lab was to use a temperature probe to measure temperature

differences in reactions and what ultimately causes these changes in difference. This was

done specifically by determining the heat capacity of a calorimeter by mixing hot and cold

water, finding the specific heat capacity of sand by combining hot sand with cold water

inside a calorimeter ammonium nitrates dissociation is an endothermic or an exothermic

reaction by measuring it temperature change in a calorimeter. The class average for the

calorimeter’s heat capacity was found to be -13.123 J/ ºC with a deviation of 18.713 J/ ºC.

Moreover, it was found that the specific heat capacity of sand is 5.31 J/ ºC mean that it

takes about 5.31 joules of energy to raise one gram of the sand by one degree Celsius. The

results also suggest that NH4NO3 would be endothermic as when it is dissolved this

compound absorbs energy from its surrounding making them cooler at the rate of 34

degrees Celsius per mole. The first two results are unusual and do not make sense as they

suggest that the calorimeter in the presence of heat would cool and that sand which is

known to vary tremendously with temperature has a higher specific heat capacity or is

more stable than water. Different forms of human and experimental error have been

attributed to these results.


This lab’s purpose has three goals. The first goal was to determine the heat capacity of a calorimeter

by mixing hot and cold water together and deducing from the heat (q) consumed by both of these how

much heat was absorbed the calorimeter and therefore what its heat capacity was. The second goal was

to find the specific heat capacity of sand by combining hot sand with cold water inside a calorimeter

OPEN ACCESS

CHEM 111 A09 FALL 2016, LAB 10, 46-50 47

and measuring the relative change of temperature of both of them. The final goal was to determine

whether ammonium nitrates dissociation is an endothermic or an exothermic reaction.

2. Procedure

The experiment involved using a homemade calorimeter to determine changes in temperature of

reactions. The calorimeter was made by combining two coffee cups and covering it with a cardboard

cover that allowed a thermometer and temperature probe (Logger Pro) to analyze the reaction being

undergone inside. In each case the final temperature was found by measuring the highest temperature

the reaction attained as it was assumed after this point the energy would be unreflective of total energy

as some would have been lost. The first experiment involved the mixing of hot water with cold water

in a calorimeter. This was performed by boiling hot water and adding it to the calorimeter with the

intention of finding the heat loss or inefficiency attributed to the calorimeter and thus being able to

solve for the heat capacity. The second part of the experiment mirrored the first half but this time the

variable being found was the specific heat capacity of sand. This was done by uniformly heating sand

in a test tube that was surrounded by a water bath. Following this, the sand was added to the cold water

in a calorimeter and the peak temperature was noted. The final experiment dealt with determining if

Ammonium nitrate was a heat producing or heat absorbing when it dissociated. To determine this

massed amount of this compound was added to a calorimeter with cold water and the final temperature

was recorded as a means to determine the change in temperature per mole of Ammonium nitrate.

3. Results and Calculations

Determination of the Heat Capacity of a Calorimeter

0

5

10

15

20

25

30

35

40

0 20 40 60 80 100 120

Tem

per

atu

re (℃

)

Time (seconds)

Time vs. Temperature of Calorimeter

Mass of Water (g)

Initial

Temperature (ºC)

Final Temperature

(ºC)

Change in

Temperature (ºC)

Cold Water

(Calorimeter)

48.875 25.7 36.2 10.5

Warm Water

(Beaker)

49.283 27.1 47.1 20.0

CHEM 111 A09 FALL 2016, LAB 10, 46-50 48

The capacity of the calorimeter was calculated using the final and initial temperatures of a sample of

warm and cold water before and after mixing. The graph above shows the water’s temperature 100

seconds after being mixed. It demonstrates that the temperature sharply increased and reached its peak

of 36.2 degrees Celsius after 25 seconds. From this point onward, it steadily declined to a steady

temperature of 36 degrees. The heat capacity of the calorimeter could then be calculated using equation

“8” shown above. In this particular sample tested, Ccal was found to be 9.56 J/ ºC mean that it takes

about 9.56 joules of energy to raise the calorimeter’s temperature by one degree. The class average

was found to be significantly different at -13.123 J/ ºC with a deviation of 18.713 J/ ºC. This means

that when heat is added to the system, it actually lowers or cools the calorimeter’s temperature by this

rate.

Determination of the Specific Heat of Sand

Mass of Water (g)

Initial

Temperature (ºC)

Final Temperature

(ºC)

Change in

Temperature (ºC)

Cold Water

(Calorimeter)

98.555 24.8 27.4 2.6

Hot Sand

(Beaker)

6.000 60.0 27.4 -32.6

(-13.123) (27.4-24.8) + (6.000)(Cs)(27.4-60.0)+98.555(4.184)(27.4-24.8)=0

24.5

25

25.5

26

26.5

27

27.5

28

0 20 40 60 80 100 120

Tem

per

atu

re (

ºC)

Time (seconds)

Time vs. Temperature of Calorimeter

CHEM 111 A09 FALL 2016, LAB 10, 46-50 49

Cs= 5.31 J/ gºC

The specific heat of sand was calculated using the final and initial temperatures of a sample of hot

sand and cold water before and after mixing. The graph above shows the water’s temperature 100

seconds after being mixed. It demonstrates that the temperature sharply increased and reached its peak

of 27.4 degrees Celsius after 35 seconds. From this point onward, it steadily declined to a steady

temperature of 27.3 degrees. The specific heat could then be calculated using equation “6” shown

above and solving for Cs of sand. In this particular sample tested, was found to be 5.31 J/ ºC mean that

it takes about 5.31 joules of energy to raise one gram of the sand by one degree Celsius.

Dissolution of Salt

NH4NO3(aq) → H+ (aq) + NO3

- (aq) + NH3(g)

Salt Mass of Salt (g) Moles of Salt Initial Temperature

(ºC)

Final Temperature

(ºC)

Change in

temperature/moles

(ºC/mol)

NH4NO3 4.007 .05006 26.6 24.9 -34.0

NH4NO3 would be endothermic as when it is dissolved this compound absorbs energy from its

surrounding making them cooler at the rate of 34 degrees Celsius per mole. This can be determined by

the negative change in temperature per mole calculated.

Error Analysis

Error was noted in both the calculation of the heat capacity of the calorimeter and the determination

of the specific heat of the sand. The problem with the heat capacity of the calorimeter was that it is a

large negative number. This poses two problems first of all the number should not realistically be

negative. This result suggests the improbability than when heat is added to the system, it actually

lowers or cools the calorimeter’s temperature by this rate. Moreover, the calorimeters heat capacity

would likely have been slightly inefficient or above zero but not by such a degree. This means that the

calorimeter as opposed to being an insulator which is not supposed to absorb heat is actually capable of

absorbing it. Even the large standard deviation of +/- 18.713 J/ ºC for a mean of -13.123 J/ ºC suggests

something is very wrong. This means that the variation could more than double the conclusion in the

negative direction or make the heat capacity a small positive number. The most likely conclusion for

this is problem in calculating it or problems in actually using the calorimeter to attain the numbers.

The specific heat seems to have been calculated in error as its value of 5.31 joules per gram per

degree Celsius is higher than that of water. This does not make sense in either the case of the

experiment or in real life. The sand in the experiment changed temperature by a significant margin of

30+ degrees when it was mixed in the water but the cold water in the calorimeter itself changed less

than 5 degrees. This would suggest that the specific heat capacity would have to be lower. This also

goes against natural phenomenon as water is known to retain heat longer as is the case with the ocean

in the late summer months’ versus the relatively cool sand. The flaws in the experiment could have

been due to human errors such as a mistake in the calculations, problem with the heat capacity

calculated earlier on, problems mixing the sand within the calorimeter, or problems in the structure of

the calorimeter which was found to have a small hole.

CHEM 111 A09 FALL 2016, LAB 10, 46-50 50


There were three main goals in this experiment. The first goal was to determine the heat capacity of

a calorimeter by mixing hot and cold water together and deducing from the heat (q) consumed by both

of these how much heat was absorbed the calorimeter and therefore what its heat capacity was. The

second goal was to find the specific heat capacity of sand by combining hot sand with cold water

inside a calorimeter and measuring the relative change of temperature of both of them. The final goal

was to determine whether ammonium nitrates dissociation is an endothermic or an exothermic

reaction. The class average was found to be -13.123 J/ ºC with a deviation of 18.713 J/ ºC. Moreover, it

was found that the specific heat capacity of sand is 5.31 J/ ºC mean that it takes about 5.31 joules of

energy to raise one gram of the sand by one degree Celsius. The results also suggest that NH4NO3

would be endothermic as when it is dissolved this compound absorbs energy from its surrounding

making them cooler at the rate of 34 degrees Celsius per mole.

The calorimeter was not very good as there were gaps between the lid and there was a hole at its

bottom. Also, in terms of insulation it would allow for a ready transfer of heat. To this end, the class

average does not tell much about the quality of calorimeter as it suggests that the calorimeter did not

get hot but rather got cold at the rate of -13.123 J/ ºC when heat was added to the system. This

therefore, cannot be used to accurately reflect how good or bad the calorimeter is. Design changes

would include making the calorimeter air tight, a better insulator, or submerging it in water as is the

case with bomb calorimeters.

The sand had a higher specific heat than that of water according to our experiment. This does not

make sense and in actuality water is believed to have a higher capacity to hold heat. For example, in

natural phenomena such as the beach or deserts, the land of these locations are known to gain or lose

heat really quickly. This explains why deserts like El’ Azizia vary from such high to low temperatures

as it does not take much heat from the sun for the sand to either rapidly gain or lose heat. In contrast,

the water of areas of bodies of water have warm temperatures long into the summer as the water

retains heat and even radiates. Tripoli likewise had moderate temperatures as the water is an

intermediary force that in cold months retains the heat and in warm months would not be extremely

affected or absorb the heat in large quanities. This retainment of heat is attributed to a higher specific

heat capacity so it doesn’t make sense for sand to have had a higher specific heat capacity when it does

the opposite. The results were therefore most likely due to human error and inaccuracy. This is also

chemically supported as water molecules are bonded by strong hydrogen bonds that take a lot more

energy to break and would therefore take a lot of energy to fluctuate in terms of heat. In contrast

covalent bonds such as those in sand are relatively easily broken.

Acknowledgments

The author gratefully acknowledges his lab partner Brad O’Hara for his assistance in performing the

experiment and all other aspects of the process, including data collection, statistical calculations, and

lab clean-up to enumerate a few.


CHEM 111 A09 FALL 2016, LAB 11, 51-55

CHEM 111 A09

GENERAL CHEMISTRY



Creation of Hot Pack Using Calcium Chloride

Mohammed Ali




Abstract: A hot pack was created that through salt dissolution that would increase the

temperature of a calorimeter by three degrees Celsius. The salt used was calcium chloride

due to it being more environmentally friendly and conducive to larger sized hot packs. The

formula Ccal∆T+qdissolution+(mCs∆T)water was used to determine the exact number of grams

of calcium chloride required. This was found to theoretically be 0.375 grams of calcium

chloride. Unfortunately, when this was experimentally tested the temperature rose to only

about 1.5 degrees. This resulted in a 50% error. This error was most likely due to a

calorimeter’s design flaws but could also be caused by error in the amount of reactant

added or procedural inconsistencies. If these were to be marketed based on the current

results about double the predicted amount of calcium chloride would likely need to be

added to achieve a desired change in temperature.


This lab’s purpose is to create a hot pack through salt dissolution that would increase the

temperature of a calorimeter by three degrees Celsius. This is caused by an exothermic reaction in

which chemical energy is released into its surroundings. The focus of this experiment is to also to

ensure that the salt used in its design makes the hot pack safe, cost effective, and efficient. This

experiment will be performed by using the molar enthalpy of the reaction with regards to the formula

Ccal∆T+qdissolution+ (mCs∆T)water to predict how much reactant will be needed to generate the required

temperature change. Hot packs play an important role in the treatment of both acute and chronic

injuries. They increase blood flow and help restore movement to injured tissue. Warmth can also

effectively reduce joint stiffness, pain, and muscle spasms.

OPEN ACCESS

CHEM 111 A09 FALL 2016, LAB 11, 51-55 52

The scope of this paper is to create an operative hot pack capable of increasing the overall

temperature by three degrees Celsius. This paper will show the methodology performed to come to

both conclusions by focusing on the materials and procedures, data and analysis, and the error analysis

and overall conclusions that can be drawn from the data.

2. Experimental Procedure for Creation of Hot Pack

This experiment required 1.000g Calcium Chloride, 50 mL distilled water,1-100mL graduated

cylinder, 2-styrofoam cups with cardboard cover, weighing boat, scoopula, temperature probe,

thermometer, electronic pan balance, LoggerPro, and computer. This lab was performed by using

enthalpies of formation to calculate the total molar enthalpy of reaction of calcium chloride. This

amount was then inputted into the formula Ccal∆T+qdissolution+(mCs∆T)water where the heat of

dissolution is equal to molar enthalpy times number of moles, ∆H(n). With all other variables known,

“n” or the number of moles of the reactant is solved for and this can be converted to find the exact

number of grams of calcium chloride needed to raise the temperature by three degrees Celsius (see

“Results and Calculations” section below). This amount was then massed and added to 25 mL of

distilled water within a homemade calorimeter. The initial and final temperature were recorded to

determine how much the temperature had increased due to the salt’s dissolution. This was performed a

total of two times.

This lab used a homemade calorimeter to determine changes in temperature of the reaction. The

calorimeter was made by combining two coffee cups and covering it with a cardboard cover. A

thermometer and a temperature probe (Logger Pro) were placed inside the calorimeter to analyze the

reaction being undergone and heat change within. In each case, the initial temperature was recorded

before the calcium chloride was added to the calorimeter and only the water was present. The final

temperature was found by measuring the highest temperature the reaction attained. It was assumed

after this point the energy would be unreflective of total energy released as some would have been lost.

3. Results and Calculations

Dissolution Reaction:

Calculation of Molar Enthalpy:

=

Calculation of Amount of Reactant:

0= Ccal∆T+∆H(n)+ (mCs∆T)water

0=

0=

0=

CHEM 111 A09 FALL 2016, LAB 11, 51-55 53

The amount of calcium chloride that needed to be dissociated to change the temperature by 3

degrees Celsius was found by using the above equation in which all the variables are known. The value

of Ccal or the calorimeter’s heat capacity in a prior experiment was found to be -13.124 joules per

degree Celsius. This was determined by mixing hot and cold water within a calorimeter and finding the

total heat loss or inefficiency attributed to the calorimeter. ∆T or the change in temperature is a

positive 3.0 degrees Celsius as the intended result is for the temperature to increase by three degrees.

∆H is the molar enthalpy of calcium chloride that was calculated from the individual enthalpies of

formation to be -81.2 kJ per mole or -81,200 joules per mole. The mass of the water added was 25.0

grams and the specific heat capacity of water is 4.184 joules per gram per degree Celsius. These

numbers were algebraically rearranged to solve for “n” or the number of moles of calcium chloride

needed. Finally, “n” was multiplied by calcium chloride’s molar mass to determine how many grams

of calcium chloride are needed to raise the temperature by three degrees Celsius.

24.5

25

25.5

26

26.5

0 20 40 60 80 100 120

Tem

per

atu

re (

)

Time (seconds)

Trial #1

24

24.5

25

25.5

26

26.5

0 20 40 60 80 100 120

Tem

per

atu

re (

)

Time (seconds)

Trial #2

Trial #1

Time

(seconds)

Temperature

0 25.1

20 25.0

40 25.8

60 26.0

68 26.3

80 26.2

100 26.2

Trial #2

Time

(seconds)

Temperature

0 24.2

20 24.2

40 24.5

60 25.6

78 26.0

80 26.0

100 26.0

Trial # Mass of Calcium Chloride

(g)

Temperature Change

Trial#1 0.376 1.2

Trial#2 0.376 1.8

Avg. +/- Std. 1.5 (+/-) 0.4

CHEM 111 A09 FALL 2016, LAB 11, 51-55 54

Two trials were performed in which 0.376 grams of calcium chloride was dissociated to

theoretically increase the calorimeter’s temperature by 3.0 degrees Celsius. In the first trial the initial

temperature was measured at 25.1 with the final temperature reaching 26.3 . The total change in

temperature was 1.2 overall. In the second trial the initial temperature was measured at 24.2 with

the final temperature reaching 26.0 . The total change in temperature was 1.8 overall. The average

temperature change based on these two trials was found to be 1.5 with a standard deviation of plus

or minus 0.4 . The temperature change graphs for both trials show a relatively smooth increase in

temperature with an eventual leveling out after 100 seconds. The maximum temperature was reached

around 60-80 seconds after the calcium chloride was added.

Error Analysis

There is a 50% error from the predicted and actual value of temperature change. The predicted

change in temperature was 3.0 but the actual value for ∆T was found to be 1.5 . Due to the fact

that the calculated value for temperature change differs by such a significantly large margin from the

actual value this means that a substantial amount of error has occurred. Also, the positive sign for

percent error means that the predicted amount was actually smaller than what it should be. This error is

most likely due to problems in the calorimeters design. It is possible that heat escaped from the loose

cardboard lid especially through the large holes designated for the probe and thermometer. Moreover,

an error in the heat capacity of the calorimeter that was previously calculated would influence the

results. This is presumed because the heat capacity is negative which suggests the unlikely scenario

that a heat producing reaction would actually lower the calorimeter’s temperature. The implications of

this erroneous value for Ccal is that the calculations would suggest a lower amount of calcium chloride

required to change the temperature by three degrees than what is actually necessary. Inconsistencies in

procedures could also have influenced the results. For example, in the second trial the calcium chloride

was shaken vigorously inside the calorimeter while in the first trial the calorimeter was not shaken at

all. This could potentially explain why the temperature increase of the second trial was 60% larger than

that of the first trial. It is also a possibility that no one major error caused the calculated temperature to

be lower than the actual value but it could have been a culmination of these factors, namely; design

issues with the calorimeter, error in the calculation of how much calcium chloride to use, and

procedural inconsistencies.


This lab’s purpose was to create a hot pack through salt dissolution that would increase the

temperature of a calorimeter by three degrees Celsius. The salt used was calcium chloride. To

determine the amount of reactant needed, the formula Ccal∆T+qdissolution+(mCs∆T)water was used in

which the heat of dissolution is equal to molar enthalpy times number of moles, ∆H(n). Notably, ∆H of

calcium chloride was calculated by subtracting the sum of the degrees of freedom of products minus

that of the reactants. It is -81.2kJ/mol which is important because this means that the dissolution of

CaCl2 releases heat and would therefore be appropriate for use within a hot pack. With all other

variables known, “n” was solved for and this was used to determine the exact number of grams of

calcium chloride needed to raise the temperature by three degrees. This was found to theoretically be

CHEM 111 A09 FALL 2016, LAB 11, 51-55 55

0.375 grams of calcium chloride. When this was experimentally tested though, the temperature was

raised only to about 1.5 degrees. This means that between the predicted and actual values of

temperature change, a significant error of 50% was present. This error was most likely due to a

calorimeter’s design flaws but could also be caused by error in the amount of reactant added or

procedural inconsistencies. If these were to be marketed based on the current results about double the

predicted amount of calcium chloride would likely need to be added to achieve a desired change in

temperature.

When considering the construction of the hot pack both magnesium chloride and calcium chloride

were considered as viable options. In regards to price and overall heat released the difference between

the two were very similar. For example, magnesium chloride costs about $0.018 to raise the

temperature of calorimeter three degrees while calcium chloride costs a very comparable $0.022 to do

the same thing. Assuming that a hot pack was desired to raise the temperature from room temperature

(about 20 ) to 50 ,the total cost of the chemicals in a magnesium chloride hot pack would be 18

cents while those made of calcium chloride would be 22 cents. Due to such a small difference in cost

the choice to use calcium chloride came down to a matter of size and convenience. The calcium

chloride hot pack would weigh 37.5 grams while the magnesium chloride hot pack would weigh 16.34

grams. It seems that the extra weight of the calcium chloride would make it more desirable than the

relatively lighter and less substantial feeling of the magnesium hot pack. Moreover, because most hot

packs are disposable and one-time-use products it is important to note that calcium chloride has been

found to be more environmental friendly. In term of safety both chemicals are considered relatively

safe if used with common sense. They are rated as having low toxicity and the safety warnings for both

include standard precautions. Namely that these chemicals can be stored in a typical lab setting and

that they should not be ingested, inhaled, or applied to the bare skin for extended periods of time.

The majority of sports hot packs actually contain calcium chloride as their active ingredient. If a hot

pack was to be designed it would be designed in a similar manner to current hot packs. It would

contain a heavy-duty plastic bag filled with water or gel with a tiny bag containing calcium chloride.

The outer bag would have pressure or a twisting force applied to it to combine the dry chemical with

the water-based solution. After this is done the contents of the pack should be shaken to combine the

two and further the dissolution of the salt. As far as safety precautions go to biggest risk is that the bag

will be punctured and remain on the skin for an extended period of time of that the contents would be

some ingested within the body. To prevent this big warning labels should be present on the hot pack

and a thick, tear resistant outer bag should be used in the hot pack’s construction.

Acknowledgments

The author gratefully acknowledges his lab partner Brad O’Hara for his assistance in performing the

experiment and all other aspects of the process, including data collection, statistical calculations, and

lab clean-up to enumerate a few.


Chemistry Department

975 Oakland Avenue

143 Weyandt Hall


Indiana, PA 15705

www.iup.edu/chemistry

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