Lab 14: Titration Lab

Looking happy with our titration setup even though the lab is very stressful

In this lab, my partners Meghana, Maya, Katie, and I titrated acetic acid with a strong base, sodium hydroxide, in order to determine the percent ionization of the acid in vinegar. We started off by cleaning and then filling a burette with 50 mL of NaOH (base) that had a molarity of 0.25M. We then filled a flask with 7.25mL vinegar, which is made up of a large amount of acetic acid, and added four drop of phenolphthalein. The phenolphthalein we added was an indicator that changed color from clear to pink when it turned from an acidic form to a basic form. As we added drops of base to the acid, we saw the color of the acid turn from clear to pink, and then back to clear as the acid/base solution was stirred. As the solution took longer to turn back to clear after drops of base were added, we knew it was nearing the equivalence point between H+ ions and OH- ions. We added fewer drops at a time until the solution became a light pink and there were a higher amount of OH- ions. This meant that pH increased over 7 and the solution became a base.

• Trial One:
• Molarity of Acid: ?
• Molarity of Base: 0.25 M
• Volume of Acid: 0.00725 L
• Volume of Base: 0.0250 L
• Calculations:
• M1V1=M2V2
• M(.00725) = .25(.025)
• M = 0.86
• Trial Two:
• Molarity of Acid: ?
• Molarity of Base: 0.25 M
• Volume of Acid: 0.00720 L
• Volume of Base: 0.0255 L
• Calculations:
• M1V1=M2V2
• M(.0072) = .25(.0255)
• M = 0.89

Average concentration of the vinegar (acetic acid): 0.875 M

[H3O+] = 10^-pH = 10^-2.4 = 0.0040 M

Percent ionization: 0.0040 M / 0.875 M * 100 = 0.46%

Why is this such a low number?

We know vinegar is a weak acid, which means that not all of the ions break apart/ionize,  meaning there are not many H+ and OH- ions in the solution. Therefore, the percent ionization is very low.

Lab 13: Solubility: A Guided Inquiry Lab

Introduction

In this lab, my lab partner Meghana and I used the "Solubility of 3 Salts" graph in order to design our own procedure to find the identity of the solid we were given. The solubility curve graph clearly illustrates the curve in which each solid (NaNO3, KNO3, and NaCl) will dissolve or not dissolve in water. The curve follows the largest mass in grams of each solid at degrees Celsius in which the solute will completely dissolve. Because of this, any amount of solid (in grams) below the line will make an unsaturated solution and completely dissolve, and any amount of solid above the line will be a saturated solution and not completely dissolve. Therefore, by making the solution one temperature and seeing if each mass dissolved or not, we were able to find which solid we had.

Procedure

1. We chose to raise the temperature of the solution to 44 degrees Celsius because at that temperature, there were large gaps in between the mass that was on the solubility line.
2. We measured out 5 grams of the solid in order to test if the solid was NaCl, as it was 3.7g at the solubility line, and if it dissolved, we knew that the solution would be unsaturated, and therefore not NaCl since it was above the solubility line.
3. We filled a larger beaker that would act as a hot bath, and filled the smaller beaker to 10 mL. Since we filled it to 10 mL, we divided all the values on the graph by 10 since the graph was based on 100g/100mL of H2O.
4. We heated the smaller beaker to 44 degrees Celsius by placing it in the larger beaker and placing that on the hot plate.
5. Once the smaller beaker was 44 degrees C (measured with a thermometer) we took it out of the larger beaker and checked if the solid was dissolved. After stirring, it dissolved completely, so we knew it couldn't be NaCl.
6. We added 4 grams to the small beaker, making the total 9g, as KNO3's solubility line value was 7g at 44 degrees C.
7. We again heated the smaller beaker to 44 degrees C and stirred the solid to see if it would dissolve. The solid dissolved, so we knew the solution was unsaturated, and therefore couldn't be KNO3, as it was above the solubility line.
8. Since the solid was dissolved for both tries and therefore was not NaCl or KNO3, we knew that it had to be NaNO3. NaNO3 was the only solid whose solubility line was above 9 grams at 44 degrees C, and the only solid in which 9 grams would make a unsaturated solution in 10 mL of water.

Data

• Trial One:
• Temperature: 44 degrees Celsius
• Mass of Solute: 5 grams
• Mass of Solvent: 10 grams
• Dissolved/Didn't Dissolve: Dissolved

• Trial Two:
• Temperature: 44 degrees Celsius
• Mass of Solute: 9 grams
• Mass of Solvent: 10 grams
• Dissolved/Didn't Dissolve: Dissolved

Conclusion

The unknown solid we had was NaNO3. We know this because in the first trial, we measured 5 grams of solute into 10 grams of solvent, and after being heated to 44 degrees, the solute dissolved. This meant that the solid could either be KNO3 or NaNO3, since if it was NaCl, it would be a saturated solution. Then in the second trial, we measured 9 grams of solute into 10 grams of solvent, and after being heated to 44 degrees, the solute dissolved again. This meant that the solid had to be NaNO3, since if it was KNO3, it would be a saturated solution and wouldn't dissolve. The relationship between temperature and solubility of a solid is that as the temperature of the solution increases, the solubility of the solid increases. For example, in NaNO3, the largest amount of grams that could be dissolved completely is about 10g at 36 degrees Celsius, and at 100 degrees Celcius, the number is 18g.

9 grams of our solid dissolved at 44 degrees Celsius

Lab 12: Alka Seltzer and the Ideal Gas Law

In this lab, my lab partner Meghana and I practiced using the ideal gas law to find the mass of CO2 in a balloon after the reaction between sodium bicarbonate and citric acid occurred within the balloon and a test tube below it. Since dissolving the two substances (both in an alka seltzer tablet) in water lets them react, we placed a balloon filled with the crushed tablets over a test tube of water. When we flipped the balloon over and let the contents into the water, they reacted and CO2 was released into the balloon. The circumference of the balloon and the amount of water that filled the balloon creating the same circumference were used to find the mass of CO2 in the balloon.

1. Discuss an area in this lab where experimental error may have occurred.

Experimental error may have occurred when transferring the crushed tablet powder from the weigh boat to the balloon; some of the powder could have missed the balloon, decreasing the amount of sodium bicarbonate and citric acid that reacted. Another area could also have been while pouring the water from the balloon into the graduated cylinder, some of the water missed, resulting in less water being measured as the volume of the balloon. The measurement of the circumference could have also been wrong, since the balloon wasn't spherical and was very misshapen, and it was difficult to find the actual circumference.

2. Choose one error from above and discuss if it would make "n" too big or too small.

If the measurement of the circumference is too low, that would mean that while filling the balloon with water, we wouldn't have filled it with enough, which would make the volume too low, which would then result in n being too small.

3. Calculate the volume of the balloon mathematically.

2(3.14)r = 31.20 cm

r = 4.97 cm

v = (4/3)(3.14)r^3

v = (4/3)(3.14)(4.97)^3

v = 514 cm^3 = 514 mL

4. Are the two volumes close? Which do you feel is more accurate and why?

705.9 mL - 514 mL = 192 mL greater

I believe that our volume is more accurate because the CO2 released from the reaction filled the balloon to a non-spherical shape, as did the water we filled the balloon with. We measured the circumference in the same area each time, so it was more accurate than a calculated volume, since neither shapes of the balloon were spherical, as the calculations would assume.

5. Give two differences between a real gas and an ideal gas.

Two of the ideal gas laws are that in an ideal gas, there are no interactive forces, meaning none of the particles of gas repulse or attract each other, and that there is no net loss of energy from collisions between the particles and the walls of the container the gas is in. Real gases don't follow these rules perfectly, and in a real gas, some of the particles could repulse/attract each other, and there could be a net loss of energy from the collisions.

6. Would the CO2 you collected in this lab be considered ideal?

The CO2 collected in this lab would not be considered ideal because all non-theoretical gases are real gases.

Advanced Questions:

1. Calculate the mass of CO2 that should be collected for two tablets.

Each tablet had:

1000 mg C6H8O7

1916 mg NaHCO3

We used two tablets, and after finding the theoretical yields, we found that the C6H8O7 would theoretically produce 1.37439g CO2, and the NaHCO3 would theoretically produce 2.00748g CO2. Therefore, the theoretical yield of CO2 would be 1.37g CO2.

2. What percent is the percent yield for the CO2 collected in your sample?

actual/theoretical = 1.26g / 1.37g = 92.0% yield

3. What effect does the fact that CO2 is water soluble have on your calculated "n" value?

The fact that CO2 is water soluble could decrease the n value. This is because the CO2 could have dissolved in the water while the alka seltzer powder was poured into the water in the test tube. If the CO2 dissolved in the water and not all of it gathered in the balloon, the circumference of the balloon could have been smaller than it would have been without dissolving. This would make the amount of water we filled the balloon with smaller, therefore decreasing the volume, and overall decreasing the n value.

Lab 11B: Calories in Food Lab

In this lab, my lab partner Meghana and I found the amount of calories in various food items (a cashew, pecan, and cheese-puff) by burning them and keeping track of the heat released. We built a makeshift calorimeter with a paper clip to hold the food item, a tin can to insulate the reaction and prevent heat loss, and a water-filled flask in which the heat lost by the food item was gained by the water. Each food item burned, which released heat, and the water absorbed the heat, causing the temperature to change, and this change in temperature was used to calculate the amount of energy in the food, and then the calories in the food.

1. Did you measure a temperature change in the food sample or the water?

We measured a temperature change in the water. Since the water absorbs all the heat released by the food sample, we knew that the heat released would be equal to the heat gained. Therefore, we were able to call the change in temperature of the water the change in temperature of the food sample as well. 

2. Did you measure the energy released by the food sample or the energy gained by the water?

We measured the energy released by the food sample because in the calculation, we plugged in the food's mass into q = mc(change in)T. However, since the energy released by the food sample was gained by the water, one could say we also calculated the energy gained by the water.

3. What happens to the small amount of energy that is not absorbed by the water?

The small amount of energy that was not absorbed by the water could have not been released by the food sample or could have escaped through the vent holes in the tin can.

4. Were you surprised by any of the results? Explain.

I was surprised that the cheese-puff had less Calories per gram than the nuts do. I always thought that cheese-puffs had a lot of Calories. Now, I can buy cheese-puffs and eat them more often than I eat nuts without feeling bad about my health.

Lab 10: Evaporation and Intermolecular Attractions

PRELAB: 

DATA:

2. Explain the differences in the difference in temperature of these substances as they evaporated. Explain your results in terms of intermolecular forces.

We know that when the minimum temperature is lower, the substance evaporated more because more energy from the surroundings was put in in order to evaporate it, making the surroundings colder. The substances in order of change from lowest to highest difference are glycerin, n-Butanol, water, ethanol, and methanol. This can be explained by the intermolecular forces in each substance. Glycerin not only has three strong hydrogen bonds, but it is also has a high mass, supporting why it had the lowest change in temperature/evaporated the least/had the strongest intermolecular forces. The temperature change for glycerin is negative because since its bonds are so strong and it is such a large molecule, it actually solidifies instead of evaporating and gives off heat to its surroundings. N-Butanol is also a large molecule, and taking its mass and dipole-dipole bonds into consideration shows why its forces kept it from evaporating much of its substance. Water also had a low difference in maximum and minimum temperatures. Since water only has three atoms and both bonds are strong hydrogen bonds, they take a lot of energy to break. Ethanol had a larger difference in temperature than the previous three substances because it only had one hydrogen bond, and didn't have as much mass as n-Butanol. Methanol had the largest difference in temperatures. This is because it only has one hydrogen bond, and is not as heavy of a molecule as ethanol. In conclusion, much of the difference in temperature depended not only on the amount of hydrogen bonds, which increase intermolecular force, but also on how massive the molecule was, as mass also increases intermolecular forces.

3. Explain the difference in evaporation of methanol, ethanol, and n-Butanol. Explain your results in terms of intermolecular forces.

Since all three substances have the same amount of hydrogen bonds, the difference in evaporation can be explained by the effect of mass on the strength of intermolecular forces. The higher the mass of the molecule, the stronger the bonds, and then the strength of the intermolecular forces increase. N-Butonal is the most massive, explaining its low amount of evaporation (higher temperature = lower evaporation, as explained in the above question). Ethanol has the second highest mass of the three, which is why it evaporated more than N-Butonal but less than methanol. Methanol has the lowest mass of the three, which makes its intermolecular forces less strong, explaining why it evaporated the most.

4. Explain how the number of -OH groups in the substances tested affects the ability of the tested compounds to evaporate. Explain your results in terms of intermolecular forces. 

Since the -OH bonds are very strong hydrogen bonds, they contribute to the strength of intermolecular forces by bonding with oxygens in other molecules, therefore making the intermolecular forces stronger. Making the intermolecular forces stronger means that more energy would be required to break these bonds, which slowed the evaporation rates. Therefore, substances such as glycerin and water with a larger amount of -OH groups would have a smaller change in temperature/evaporate less.

Me and my lab parter with our LabQuest graph-data-collector

Lab 7: Flame Test Lab

In the flame test lab, we used wooden popsicle sticks soaked in various compounds made of a metal + chloride and placed them over a flame in order to see the color of light emitted. The loosely held electrons of the metals were excited by the flame, which then caused the electrons to release a certain color of light when they returned to their ground state. We also had to test two unknown compounds, see what colors were emitted, and identify the compounds.

1. What is the difference between ground state and an excited state?

The ground state is when the electrons are in their normal configuration. The excited state is when some electrons absorb energy and jump to a higher energy level.

2. What does the word "emit" mean?

The word emit means to produce or release. Electrons emit light in the form of electromagnetic radiation when they fall back to the ground state after being excited.

3. In this experiment, where are the atoms getting their excess energy from?

The atoms are getting excess energy from the flame that the metals are held over. Without the flame, the electrons would not get excited, and the light released would be much weaker.

4. Why do different atoms emit different colors of light?

Different atoms emit different colors of light because each atom is quantized, meaning each atom will only produce one certain color, and none in between. Different atoms also means that they have different numbers of electrons, which leads to different energy levels. The electrons of one atom could be able to jump and then fall from the fourth level to the second, emitting one color, while another atom's electrons might jump then fall from the third level to the second, emitting a different color.

5. Why is it necessary to clean the nichrome wires between each flame test?

It is necessary to clean the wires/use different wooden sticks between each flame test because otherwise, the flame will react with both compounds. This would confuse the viewer as to which color corresponds to which compound.

Above are pictures of the color produced by NaCl, a selfie of me and my lab partner holding one of the wooden sticks, and the color produced by LiCl

Unknown #1: LiCl - We know Unknown #1 is LiCl because it produced a bright fuchsia/red color that looked exactly like the color produced by LiCl.

Unknown #2: KCl - We know Unknown #2 is KCl because it produced a light purple color that looked exactly like the color produced by KCl.

Lab 8: Electron Configuration Battleship

My board with an angry looking Meghana (my opponent)

Looking happy even though we want to defeat each other

The biggest challenge I had while playing was getting the hang of saying the electron configurations without needing to look at the reference sheet, and only looking at the periodic table. I was also extremely slow at saying the configurations at first, but gradually became better as I remembered that you can use the noble gases. One thing I learned while playing was that the noble gases can be used as short cuts, and that it is really annoying to say the electron configurations of elements in the 6 or 7p rows.