Titration Lab

Post a picture of your titration setup, and a photo of your analyte at the equivalence point. Write a short summary of your procedure. Calculate the percent ionization of vinegar and describe why it is such a low number.

Setup:

Following the directions on the lab sheet, we filled up the burette to the zero point with the 0.25 M NaOH base. We added vinegar to the flask along with distilled water and four drops of phenolphthalein. After securing the burette to the stand and placing the flask on the hot place, we started the magnetic stir of the mixture. Slowly, we opened the valve of the burette to release the NaOH base. Once the drops started to make more noticeably pink streaks in the mixture and it took longer to return to the clear color, we knew the equivalence point was approaching. At that point, we switched to a drop by drop method so that the mixture would not pass the equivalence point and become bright pink. After the drop that caused the mixture to turn light pink, we took the tip of the burette out of the opening of the flask and took the photos below. We also repeated this procedure for a second trial, knowing the relative amount of NaOH base that would be needed.

Trial one:

Ms. Kapinos looking pleased with our titration success:

Trial two:

Trial one:

Volume of NaOH used = 25.0 mL
pH of vinegar = 2.4
Molarity of acid = 0.86 M
Volume of acid = 0.00725 L
Molarity of base = 0.25 M
Volume of base = 0.0250 L

Trial two:

Volume of NaOH used = 25.5 mL
pH of vinegar = 2.4
Molarity of acid = 0.98 M
Volume of acid = 0.00720 L
Molarity of base = 0.25 M
Volume of base = 0.0255 L


Average concentration of vinegar = 0.875 M

[H3O+] = 10^-2.4 = 0.0040

Percent ionization = 0.46%

The percent ionization is such a low number because it does not have the properties that would allow it to break into ions. Strong bases have a 100% ionization because the acid and base completely ionize. Since vinegar is a weak acid, it does not completely break apart into ions.

pic creds to Ms. Kapinos #candid:

Solubility: A Guided Inquiry Lab

Introduction:

The purpose of this lab is to identify a solid by making use of its solubility curve. Solubility is the maximum amount of solute that with be dissolved in a specific solvent. Any solution under the curve is considered unsaturated, as more solute can be added. Any solution on the curve (or above) is considered saturated, as more solute can not be dissolved. My lab partner, Alison, and I made our own procedure that involved heating the solution of the salt and water in a bath and then stirring the solution to see whether the salt dissolved or not.

Procedure:

1) First, we measured the 5g of the salt using a weigh boat and scale.

2) Next, we filled a quarter of a beaker with faucet water. We also measured 10mL of distilled water in a graduated cylinder and then transferred that into a smaller beaker.

3) We added the salt into the smaller beaker with the 10mL of distilled water. Then we put the larger beaker on the hot plate to heat it up, and placed the smaller beaker into the larger one. We held the smaller beaker in place with a pair of tongs to be make sure no extra water from the larger beaker entered.

4) We placed a thermometer into the solution with the salt, and waited until the temperature reached 44 degrees Celsius.

5) Once the solution reached 44 degrees, we removed the solution to stir it. 

6) Since the solute dissolved, we repeated the process, this time adding 4g into the solution, so it had a total of 9g of salt and the same 10mL of water.

7) Since the solute dissolved again, we knew which salt it had to be, and cleaned up.

Data:

Trial one:

Mass of solute = 5g

Mass of solvent = 10g

Temperature of solution = 44 degrees Celsius

Trial two:

Mass of solute = 9g

Mass of solvent = 10g

Temperature of solution = 44 degrees Celsius

Solubility (from graph):

NaCl at 44 degrees Celsius = 3.7g
KNO3 at 44 degrees Celsius = 7g
NaNO3 at 44 degrees Celsius = 10.7g

After stirring the solution in the first trial, the solute dissolved completely. After stirring the solution in the second trial, the solute again dissolved completely.

Discussion/Conclusion:

The solute we had was sodium nitrate (NaNO3). Our claim can be justified when taking the solubility graph into consideration. If the salt was sodium chloride (NaCl), 5g of it in 44 degrees Celsius would cause the solution to be saturated, meaning not all the solute would have dissolved. If the salt was potassium nitrate (KNO3), 9g of it in 44 degrees Celsius would cause the solution to be saturated. However, after 9g of the salt were added, all the solute still dissolved. The only other salt it could have been was sodium nitrate, because its solubility curve was above those of the other two. 

As the temperature of a solution (with a solid as the solute) increases, the solubility of the solid also increases.

Alka Seltzer and the Ideal Gas Law

In this lab, we found the mass of carbon dioxide that filled up a balloon. When Alka Seltzer tablets react with water, carbon dioxide gas is produced. We filled up a test tube with water and put crushed Alka Seltzer tablets (powder form) into a balloon. Next, we stretched the opening of the balloon around the rim of the test tube and let the powder mix with the water, which caused the solution in the tube to fizz. Once the fizzing stopped (when the maximum amount of carbon dioxide was given off), we measured the circumference. Then, we found the volume of the space inside the balloon by filling it up to the same circumference with water, and measuring that amount. Using all these measurements, we were able to calculate the mass of the carbon dioxide.

Data:

Calculations:

Before powder and water mixed:

While powder and water mixed:

Jason lookin' $weg in the background:

Analysis Questions:

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

There were several areas for errors to have occurred in this experiment. For example, some of the crushed tablet powder fell outside of the balloon when we poured it in. Also, we may have measured the circumference of the inflated balloon inaccurately, as it was not a perfect sphere but rather a three dimensional oval shape. In addition, some of the water from the balloon fell out as we poured it into the graduated cylinder.

2) Choose one error from above and discuss if it would make "n" the number of moles of CO2 too big or too small.

If the balloon circumference that we measured is too low, the volume used in the ideal gas law equation is too low. As volume is directly related to the number of moles, this would cause the number of moles to also be too low.

3) Filling the balloon with water may be one place where error could have occurred. Using the value for the circumference of the balloon in cm, calculate the volume of the balloon mathematically.

r = 4.97 cm

V = 514.23 cm^3 = 514 mL

4) Compare your answer to #3 to the volume obtained by filling the balloon with water. Is it close? Which do you feel is more accurate and why?

No, the measured volume and the calculated volume are not close. The volume obtained by filling the balloon with water is 192 mL greater, and is more accurate. This is due to the fact that the balloon was not filled to a perfect sphere by the carbon dioxide gas (as seen above) nor the water. In both instances it was a three dimensional oval shape.

5) The ideal gas law technically applies to ideal gases. Give two differences between a real gas and an ideal gas. You may use your computer or book to research.

An ideal gas doesn't have attractions or repulsions among its particles whereas a real gas could have these interactive forces. An ideal gas also doesn't have a net loss of energy caused by its particles colliding, but a real gas could.

6) Would the CO2 you collected in this lab be considered ideal? Why or why not?

No, the carbon dioxide gas would not be considered ideal. This is because real gases aren't ideal, and ideal gases don't exist.

Advanced Questions:

1) Using the information provided on the label, and stoichiometry, calculate the mass of CO2 that should be collected. 

(We used two tablets instead of three)

2g citric acid ==> 1.374g CO2

3.832g sodium bicarbonate ==> 2.007g CO2

In this case, citric acid is limiting, so the theoretical yield of carbon dioxide is 1.37g CO2.

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

1.26g/1.37g x 100 = 92.0%

3) CO2 is water soluble. The solubility around room temperature is around 90mL/100mL of water. What effect does that have on your calculated "n" value?

As carbon dioxide is water soluble, some of the gas could have dissolved in the water when we shook the test tube-and-balloon system. This would cause less carbon dioxide to fill the balloon, and the circumference and volume of the balloon would be too low. Therefore, our n value would also be lower.

Calories in Food Lab

In this lab, we figured out how much energy (in calories) was contained in three different foods: a cashew, pecan, and cheesepuff. We did so by placing each piece of food on a paper clip and lighting it on fire. On top of the burning food we placed a tin can, and on top of the can, a flask of water with a thermometer inside. Since the water reached its maximum temperature after the piece of food burned entirely, we recorded the thermometer reading after the flame went out. Using the mass of the water samples, the specific heat of water, and each change in temperature of the water to figure out the calories contained in the foods.

Before setting this poor cashew on fire:

While this cashew is on fire, burning slowly and peacefully:

Data:

 Burnt food:

 Alison and I:

Questions:

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

As there is no way to directly measure the temperature of a food sample, we measured the change in the water instead. This served the purpose of finding the change in temperature of the food.

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

The energy released by the food sample is actually the same as the energy absorbed or gained by the water. By measuring the change in temperature of the the water, we were also measuring the energy released by the food. 

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

The energy that was not absorbed by the water could've escaped through the holes in the tin can. In this expense of energy, there wasn't a completely effective transfer, as some of the heat released by the food was not absorbed by the water.

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

I was not severely surprised by any of the results, but I found it interesting that the nuts contained more energy than the cheesepuff. It seems more intuitive to think that a cheesepuff has more Calories, or the "unhealthy stuff" as we know them to be, than healthy nuts.

Evaporation and Intermolecular Attractions

Pre-lab:

Data:

Questions:

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

Temperature change is directly related to the evaporation rate. This is because a substance gets energy from its surroundings to evaporate in the form of heat, so its surroundings get colder. The difference in temperature of the substances increases from lowest to highest in the following order: glycerin, n-Butanol, water, ethanal, methanol. All the substances have hydrogen bonds, but have different masses and different hydrogen bonding ratios, which cause these differences in temperature. Higher mass and more hydrogen bonds increase intermolecular force. Glycerin has the highest mass and contains many hydrogen bonds, and its change in temperature is lowest. (Also, Glycerin's temperature increases because it actually solidifies, due to its many hydrogen bonds and large mass, and discharges energy to its surroundings.) N-Butanol has a mass less than glycerin but greater than the other substances, and therefore has the next highest change in temperature. Although water has a low mass, both bonds between its atoms are hydrogen bonds, which are the strongest. Thus, its low mass and high hydrogen bond ratio balanced each other out, causing water to have an intermediate change in temperature. Ethanol has a higher mass than the previous substances, so a higher change in temperature. Finally, methanol has the highest mass and also the highest change in temperature.

3) Explain the difference in evaporation of any two compounds that have similar molar masses. Explain your results in terms of intermolecular forces.

When methanol, ethanol, and n-Butanol are compared, their individual masses need to be taken into consideration. When particles of a substance have a high masses,  the intermolecular forces increase and the substance takes a longer time to evaporate. Methanol has a high evaporation rate due to its low mass. Ethanol has the next highest evaporation rate due to its mass being in between the other two substances. N-Butanol has the lowest evaporation rate because it has the highest mass.

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.

The number of —OH groups in the substances slows the evaporation rates of the substances. This is due to the fact that hydrogen bonding is relatively strong. Hydrogen bonding increases intermolecular forces by keeping the molecules of the substances close together.

Flame Test Lab

NaCl + flame:

Alison and me in our natural habitat (because were hot like this flame):

The purpose of this lab was to explore the fact that energy levels of elements are quantized. To demonstrate this we conducted a series of flame tests by placing wooden sticks that had solutions of different compounds over the flame of a bunsen burner. Different colors of flame were produced due to the different metals in the compounds.

Questions:

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

Ground state is the lowest energy level that an atom or ion of an element can be, and has the normal, most stable electron configuration. An excited state is said to be unstable, when electrons of an atom or ion absorb energy and jump to a higher energy level.

2) What does the word "emit" mean?

"Emit" means to produce or discharge. For example, if a compound emits a red color when heated, it produces a visibly red color when heated.

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

The atoms are getting their excess energy from the heat of the flame of a bunsen burner.

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

The energy of each atom is quantized and each atom has a certain number of electrons, therefore only a certain color will be emitted by a compound.

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

The flame will react with all metal elements present, so not cleaning the wires could produce unwanted results. Therefore, the wires need to be cleaned to remove metal from previous trial(s).


Identity of Unknown #1: LiCl

Identity of Unknown #2: KCl


We found the identities of these unknowns by comparing the colors of the flames they emitted to the other solutions. Unknown #1 produced a vivid red color, like the LiCl. Unknown #2 produced a purple color with specks of red, like the KCl.

Electron Configuration Battleship

Alison and me playing Battleship:

Game face:

Since I'm #1:

Since she thinks she's #1:

My setup:

Her setup:

When our selfie game strong:

Playing Electron Configuration Battleship helped me become more fluent with the electron orbitals and configurations. Before this game, I had very little understanding of how electrons and orbitals worked, but now I feel much more confident identifying atoms with their electron configurations. My biggest challenge while playing this game was in remembering to subtract one for the d-block levels and two for the f-block levels, so Alison would correct me. However, I eventually got better at catching my own mistakes and it all worked out!

Mole-Mass Relationships Lab

The purpose of today's lab is to look into the relationship between moles and mass in chemical reactions, and to find the percent yield of a product.

Equation of chemical reaction: NaHCO3 (s) + HCl (aq) ==> NaCl (aq) + CO2 (g) + H2O (g)

Using a known mass of sodium hydrogen carbonate, my lab partner, Orest, and I  added hydrochloric acid and found the relationships within the equation above. Below are some questions regarding the lab with my answers.

Questions:

1) Which reactant is limiting? How do you know?

The limiting reactant in this lab is the sodium hydrogen carbonate. I know this because the hydrochloric acid was added until the substance stopped fizzing. As there was no limit to the HCl added until there was no NaHCO3 left for it to react with, the latter is limiting. 

2) Find the theoretical yield of NaCl based on your limiting reactant. Show your work below.

(2.05g NaHCO3 / 1) x (1 mol NaHCO3 / 84.0g NaHCO3) x (1 mol NaCl / 1 mol NaHCO3) x 

(58.4g NaCl / 1 mol NaCl) = 1.425 g

1.43g NaCl

3) Find the mass for the remaining solid product after the evaporation of water based on your experimental data. (Show work.)

45.48g - 44.23g = 1.25g NaCl

4) Find the percent yield for this experiment for the solid product. (Show work.)

(1.25g / 1.43g) x 100 = 87.4%

NaCl ultimately left in the dish:

There could be various reasons as to why our percent yield is 87.4% rather than 100%. Firstly, we noticed that some salt (NaCl) ended up on the lab desk, instead of in the evaporating dish. This could have been caused by our not holding the watch glass close enough to the rim of the dish, allowing some NaCl particles to fly onto the desk. Additionally, we may have kept the evaporating dish on the hot plate for too long on our last heating. It is faint in the photo above but in the lab, Orest and I noticed that there is some brown coloring in the center ring of the NaCl. We concluded that the salt may have burned. Therefore, the salt mass ended up being less than the theoretical yield. 

Composition of a Copper Sulfate Hydrate Lab

Before heating:

During heating:

After heating:

Above are the photos of the hydrate before, during, and after heating the evaporating dish on the hot plate. Below I have included some calculations to find the empirical formula of the compound.

1) Calculate the mass of the hydrate used.

45.83g - 45.06g = 0.77g

2) Calculate the mass of the water lost.

45.83g - 45.58g = 0.25g

3) Calculate the percentage of water in the hydrate.

0.25g/0.77g = 32.5%

4) The accepted value for the percentage of water in this hydrate is 36.0%. Find your percent error and provide a possible explanation for your error.

percent error = [(experimental value - accepted value)/accepted value] x 100

= [(0.325 - 0.360)/0.360] x 100

= -9.27%

= 9.27%

There are various possible reasons for our error. For example, the scale fluctuated every time we checked the mass of the evaporating dish. Therefore, our calculations for the values of the substance could be inaccurate. Also, we may have lost a few grams of the hydrate when stirring the solid in the dish. I tried to break the larger crystals down so that more of the hydrate would be exposed to the heat. However, if I recall correctly, a piece of a hydrate crystal flew out of the evaporating disk.

5) Find the exact formula of the hydrate from your experimental data.

(a) Moles of water evaporated: 0.014 mol H20

(b) Moles of CuSO4 (anhydrate) that remain in evaporating dish: 0.0033 mol CuSO4

(c) Find the ratio of moles of CuSO4 to moles of H2O.

     1 CuSO4 : 4.24 H2O

(d) What is the empirical formula of the hydrate: 1 CuSO4 * 4 H2O

Our percent of error was 9.27%, which is not a huge one, but the formula above may be incorrect. For the percent of water in the hydrate to be closer to the accepted value of 36%, more water would have to be lost. We did not let the water evaporate from the hydrate enough. If more water had evaporated, the moles of water would increase. Subsequently, the ratio of moles of water to moles of copper sulfate hydrate would be greater than 4.24:1. Thus, if the coefficient 4 for H2O in the empirical formula is incorrect, I would predict the actual value to be slightly greater.

At the end of the lab, we added water to the hydrate, and found that it reverts to look like the original:

Mole Baggie Lab

The purpose of this lab is to use molar mass to identify mystery substances. We were given two baggies with two different substances, and with the mass of the empty baggies and either the number of representative particles OR the number of moles, we figured out which substances the baggies held. Since molar mass = # of grams/# of moles, we set up equations to find the number of grams per mole of each substance, and matched those with the molar masses of the given possible compounds.

Identifications:

A1: Sodium chloride

B6: Zinc oxide

Double Replacement Reaction Lab

Our well plate:

Balanced chemical reaction equations:

Net ionic equations:

This lab was very difficult in the beginning, having literally just learned how to balance chemical reactions. However, my lab partner, Alison, and I worked hard and eventually figured out the proper notation for balancing chemical reaction equations. It surprised me that our predictions were actually all correct. I was pretty sure that we must have messed something up, either in balancing or identifying precipitates. After completing the experiment, we found out that our predictions were right!

Nomenclature Puzzle Activity

The goal of this activity was to familiarize ourselves with naming compounds by matching chemical formulas with their nominal equivalents to make a square puzzle. Today we learned how to properly name a compound, given its chemical formula, and this activity helped us exercise this new skill!

Here's my group's completed puzzle:

The biggest difficulty Grace, Reshma, and I had was keeping track of the puzzle pieces while flipping through all our pages on the periodic table, ionic compounds, polyatomic ions, and the nomenclature flowchart. We first separated the pieces into piles, depending on if they contained a certain element, such as potassium, iron, and barium. Next, we each took a pile and started putting pieces together. My pile was potassium, and although it took some time to look up names and formulas, I eventually figured it out. I feel that this was my greatest contribution to the group effort. Finally, we put our pieces together, looking for matches, sides, and corners. It required a great amount of focus and diligence, and we did not win the race, but we took our time and finished strong!

Atomic Mass of Candium

The purpose of this lab is to find the average atomic mass of a newly discovered element called candium. Regular M&Ms, peanut M&Ms, and pretzel M&Ms (shown below) are the three different isotopes of this element. Using a random sample of candium, we are to calculate the average atomic mass to assign to our new YUM-my element!

The average atomic mass that my lab partner, Megan, and I calculated was 1.43g. Below are a few conclusion questions and answers about this lab:

Ask a group nearby what their average atomic mass was. Why would your average atomic mass be different than theirs?

Katie and Nick's average atomic mass was 1.52g, which is 0.09g larger than ours. This difference is due to the different amounts of each isotope that our two groups have. Different amounts of each isotope would alter the overall average. Although we used random samples of candium, there weren't enough total M&Ms to get even closer averages.

If larger samples of candium were used, for example if I gave you a whole backpack filled with candium, would the differences between your average atomic mass and others' average atomic masses be bigger or smaller? Defend you answer.

With larger samples of candium, the difference between all the average atomic masses calculated would be smaller. As there's a larger pool of atoms, there is less room for an unpredictable occurrence of a certain isotope. The percent abundances of each isotope would be more similar and the average mass of each isotope would be more similar. Thus, the average atomic mass of candium would be more similar.

If you took any piece of candium from your sample and placed it on the balance, would it have the exact average atomic mass that you calculated? Why or why not?

No, a random piece of candium would not have the exact average atomic mass that we calculated. This is because the average atomic mass is an average of all the isotope masses, not that of a specific isotope.

You have been chosen to design the square that will be placed on the periodic table for candium. Draw a rough sketch of what your square would look like. Include an atomic symbol for candium (that doesn't already represent another element!) as well as the average atomic mass for candium.

Chromatography

Before:

After:

Our chromatograms:

Above is a photo of my lab partner, Megan, and I with our favorite chromatograms that we made together. Below, I answer a few questions about this colorful lab!


Q & A:

Why is it important that only the wick and not the filter paper circle be in contact with the water in the cup?

The purpose of this lab is for the black ink to spread itself across the filter paper. If the paper is in direct contact with the water, the paper will get saturated quickly. By having the wick in between, the water gradually reaches the paper and the ink slowly spreads.

What are some of the variables that will affect the pattern of colors produced on the filter paper?

The components of the ink used and the pattern that is initially made with the ink on the filter paper affect the pattern of colors produced.

Why does each ink separate into different pigment bands?

Components of the ink that are not easily adsorbed onto the paper will spend more time in the solution and will move up the paper at a faster rate. Therefore, the different components end up in different bands on the paper based on their color.

Choose one color that is present in more than one type of ink. Is the pigment that gives this color always the same? Do any of the pens appear to contain common pigments? Explain.

Looking at the color blue on both chromatograms, we see that it's on the outer band. The two chromatograms have the same blue pigment. Though the chromatogram on the left (above) looks like it has a more concentrated band of blue, that whole chromatogram itself looks more concentrated. The two pens used to create these pigments appear to contain common pigments, as both chromatograms go from yellow to red to blue respectively from the inner band to the outer band.

Why are only water-soluble markers or pens used in this activity? How could the experiment be modified to separate the pigments in "permanent" markers or pens?

Water-soluble markers and pens were used in this activity because water was the substance used to separate the components of the ink. If we used other types of markers or pens, water would not be sufficient to separate the pigments. The experiment could be modified to use a different solution, such as rubbing alcohol (instead of water), to catalyze the pigment separation for permanent markers or pens.