Lab 14: Titration Lab

Summary

The purpose of this lab was to find the percent ionization of vinegar. This was done tritating the acetic acid present with sodium hydroxide. First we cleaned the burette to make sure that only sodium hydroxide was in the burette. Then, we measured out 8 mL of vinegar into a flask, and added 5 drops of phenolphtalein to the flask. Then, we slowly added sodium hydroxide to it. We did this until we got a very light pink color. We used 26.5 mL of sodium hydroxide for our first trial and 2.4 pH of vinegar. For our second trial we used 27.4 mL of sodium hydroxide and 2.4 pH of vinegar. We also knew that the molarity of sodium hydroxide was 0.25 M. Then we used the equation M(Molarity of acid) x V(volume of acid) equals M(Molarity of base) x V(Volume of base). Sodium hydroxide was our base so we plugged in 26.5 mL and 0.25 M on the base side of the equation. Since vinegar was the acid we were left with 8 mL on the left side. By solving for the unknown M on the acid side of the equation, this will give us the molarity of the vinegar during the first trial. After conducting our calculations, we got 0.828 M for the first trial. We repeated the same process for the second trial, except with a different value for our base, sodium hydroxide (27.4 mL of sodium hydroxide, 0.25 M of NaOH, and 8 mL of vinegar). For our second trial we got, 0.856 M. Hence, the most reliable molarity we got was from our first trial, 0.828 M. We were close on the second trial, but one drop slipped of NaOH slipped from the burette. Therefore, to calculate the percent ionization we divided the H3O+ concentration by the concentration of NaOH. Hence, to determine the concentrations, we calculated the average and divided by the concentration of NaOH. Then, we divided the molarity of H3O+ by the molarity of CH3COOH, and then we multiplied that result by 100 to get the percent ionization.

Percent Ionization

 Our percent ionization turned out to be 0.46%. The percent of ionization is a really low number because vinegar is a weak acid. Therefore, it will not completely disassociate in the solution, leaving behind many acid molecules.

Pictures

This is a picture of our best trial

               This is a picture of our setup

Lab 12: Gases

For case 1(volume vs. pressure), we set the constant as the temperature and raised or lowered the volume. For case 2(volume vs. temperature), we set the pressure as the constant and pumped in the volume. The  volume adjusted by itself. For case 3(pressure vs. temperature), we set the volume as the constant and adjusted the temperature. The temperature remained at a constant 100 degrees Celsius or 200 K. From this, we measured the pressure. For case 4(number of particles vs. volume), we set the pressure as the constant. Then, we added the number of particles. We increased the number of particles by 20 for each trial.

Questions

3. 

4. a. Bicycle tires seem more flat in the winter because of Charles Law. Charles Law states that as the temperature decreases, the volume also decreases. Hence, since the temperature decreases during the winter, the volume of the gas particles in the tires will decrease making them seem more flat during the winter.

    b. A soda can will explode due to Gay-Lussac's Law. Gay-Lussac's Law states that as the temperature increases, the pressure also increases. Hence, when the soda can is left under the hot sun, the pressure increases which would make it explode.

   c. The pressure will decrease as the temperature decreases due to Gay-Lussac's Law. The rigid container will have a constant volume as a rigid container does not get bigger nor smaller.

   d. No, it would make it worse since the pressure is one of the main factors which is causing the pain. due to Gay-Lussac's Law, as the temperature increases, the pressure will also increase, making it more painful.

Lab 13B: Solubility Inquiry

 Introduction
The purpose of this lab was to identify an unknown solid by using a given solubility curve for three solids, NaCl, NaNO3, and KNO3. We were given a solid from which it could be NaCl, NaNO3, or KNO3. NaCl is sodium chloride, NaNO3 is sodium nitrate, and  From the data given, we were supposed to create and execute a procedure which helped us find the identity of the unknown solid. Solubility curves are curves of different types of substances, which help us find the solubility in 100 grams of water at a given temperature. Solubility is a measure of the amount of solute that will dissolve in a given amount of solvent. A solute is the minor component in a solution. A solute is dissolved in the solvent. A solvent is the major component in a solution which dissolves another liquid, gas, or solid (solute). Together, a solvent and a solute form a solution.

 Procedure

Materials

1. Big and small beaker
2. Pencil
3. Paper
4. Hot pan
5. Plastic tray
6. Thermometer
7. Test tubes
8. Erlenmeyer flasks
9. Stirring rods.

Procedure

1. Collect unknown solute

2. Fill the small water with 10 mL water. The water should be enough to submerge at least half of the small beaker in. We are using 10 mL of water because we were using a proportional amount of the solute and solvent in the solution.

3. Fill the big beaker with enough water so that half of the small beaker can get submerged into the water. This step is necessary so that the water from the small beaker does not evaporate when the temperature is being raised.

4. Measure the mass of a plastic tray. This has to be done first because the plastic tray's mass is not relevant in the lab. Hence, we will have to subtract the plastic tray's mass from the tray and solute mass to get the mass of the solute.

5. Put 4 grams of solute into the plastic tray.

6. Put the solute in the small beaker and stir for 1-2 minutes.

7. Heat the big beaker with the small beaker submerged halfway into the water in the big beaker until the temperature reaches 45 degrees Celsius.

8. After the small beaker reaches 45 degrees Celsius, stir for at least 1 minute. This is to dissolve as many particles as possible.

9. After stirring for 1-2 minutes, take the small beaker out and observe the mixture. If there is a solid, it must be NaCl. Since there is a solid, more solute must be added in the 10 mL.

10. Since the compound was NaCl, it should become saturated. Since our compound was NaCl, it did become saturated.

Data

Mass of solute- 4.00 grams
Volume of water- 10 mL
Temperature of heated water- 45 degrees Celsius
Temperature of room temperature water- 20 degrees Celsius
Mass of plastic tray- 2.04 g
Mass of plastic tray + solute= 6.04 g
Solute- NaCl

Conclusion

We identified the unknown solid as NaCl. One piece of evidence was how it became saturated when added to water at 45 degrees Celsius. This meant that our compound had a low solubility, which meant that it was NaCl as on the solubility curve, NaCl has the lowest solubility out of all of the compounds.When the temperature of the solute was 45 degrees Celsius and the mass was 4 grams, the solubility curve "told" us that our substance was NaCl. From the solubility curve, since the point of 45 degrees and 4 grams was near the graph for NaCl, this led to our prediction that our solute was NaCl. Also when the solute was not dissolved after step 7, the solute was saturated. Hence, the solute was NaCl. When the solubility increases the temperature also increases. Therefore, the relationship between solubility and temperature is a direct relationship.

Lab 11B: Calories in Food Lab

Summary

 The purpose of this lab was to find the Calories per gram when burning different food items while using the concepts of specific heat. The calculations done to find the Calories per gram was to find the mass of the food burned, which was the initial-final mass of food. Then, we figured out the change in temperature. Next, we used the equation q=mass x C x change in temperature to figure out the number of calories. From this we divided by 1000 calories to get Calories with the upper case C as the difference between calories and Calories is that Calories have 1000 of calories. Finally, we divided the number of Calories by the mass of the food burned to get our Calories per gram. 

Questions

1. We measured the temperature change in the water.

2. We measured the energy released by the food sample.

3. The small amounts of energy escape through the sides of the can.

4. I was surprised by the results for the pecan. This was because the energy released by the pecan was very significant, but we did not get a large Calorie per gram. I was also surprised by the cashew's results as it did not burn that long. 

Data and Calculations

Lab 11A: Specific Heat of a Metal

Summary

The purpose of this lab was to find the identity of an element by calculating the specific heat. I found the specific heat of the element by calculating the amount of energy by the change in room temperature and the water heated by the metal block. Then, I divided the amount of energy by the change in room temperature and the water heated by the metal block values multiplied. These calculations can be connected to the equation Q=m*change in temperature*c. We got the needed information to perform our calculations from boiling water with the metal block. We got 1.2 J/gC for our specific heat. My partner and I concluded that Metal A was Lead from our calculations. 

Data

We only figured out the identity of one metal, hence the data below is for Metal A.

Mass of Styrofoam Cup- 7.65 g

Mass of cup + water- 301.7 g

Mass of water- 294.05 g

Initial temperature of H2O in a cup- 23.1 C

Initial temperature of metal in boiling water- 97.0 C

Final temperature of metal and water- 24.8 C

Mass of metal alone- 250.2 g

Lab 10: Evaporation and Intermolecular Attractions

Pre-Lab

Data

Calculations and Results

1. Shown in data table

2. From my data, Methanol had the greatest difference in temperature and glycerin had the least difference in temperature. Methanol is a small molecule which has weak hydrogen bonding. Hence, it evaporated very fast. Ethanol is the same as methanol but it is larger, which means that there will be more London forces. Water is also a small molecule, but it has strong hydrogen bonds. n-Butanol is a larger molecule which means that it will have more London forces. Glycerin is also a large molecule which means that it has greater London forces. Glycerin has three -OH groups which gives increased hydrogen bonding. Therefore, glycerin has more interactions which would make it least volatile when compared to the other substances.

3. Methanol, ethanol, and n-Butanol are similar in the way that they are alcohols. As molar mass increases, London forces increase. Therefore, the n-Butanol has more London forces than methanol and ethanol since it has a greater molar mass.

4. All of the compounds had a single -OH group with the exception of glycerin. Glycerin had three -OH groups which allowed more interactions to happen which made it less volatile. 

Lab 7: Flame Tests

 Pre-Lab

1. Ground state is when the electrons are in the lowest energy levels available. Excited state is when atoms absorb energy and "jump" to a higher energy level.

2. In this case, emit means the discharge of energy when the atom transitions back from its excited state to the grounded state.

3. The atoms are getting their excess energy from the heat.

4. Atoms emit light at different wavelengths. These different wavelengths represent different colors. Also, different atoms have different numbers of levels and electrons.

5. It is necessary to clean the nichrome wires between each flame test because the chemicals from the prior test can affect your results by altering the actual color of the chemicals under the flame test.

Summary

 In this lab we had popsicle sticks dipped into different chemicals and solids. Then, we put the popsicle stick over the flame and the flame changed color. We recorded this change of color next to the chemical name. Overall, I believe the purpose of the lab was to show how the different elements' excited atoms emit their varying energies. I also believe that this lab emphasized that different colors are unique to one substance as shown by the identifying the unknown substances exercise part of the lab.

Flame Tests

 LiCl- Red/pink
 NaCl- Orange
 SrCl₂- Red/orange
 CuCl₂- Blue/green (cyan)
 KCl- Violet
 FeCl₃- Sparks
 CaCl₂- Orange
 ZnCl₂- Orange


Unknowns

 The first unknown was Lithium Chloride and the second unknown was Potassium Chloride. We know what the unknowns are because when atoms get heated, they get into an excited state. Then they emit different wavelengths which are converted into different colors of light. So, when we put the unknown substances under the flame test, they emitted the same colors as Lithium Chloride(red/pink) and Potassium Chloride(violet) which is how we knew what the unknown substances were.

This is a picture of the flame test for CuCl₂

Lab 8: Electron Configuration Battleship

 The biggest challenge I had while playing was finding the electron configuration of the elements. Through playing electron battleship, I got a better understanding of how to do electron configuration of elements through the periodic table and electron configurations in general.

This is a picture of my setup

   This is a picture of my partner Elie's setup

Lab 6: Mole-Mass Relationships Lab

Purpose

The purpose of this lab was a representation of mass to mass relationships that exist in a chemical reaction. This lab also helped us practice to calculate and understand limiting reactants, theoretical yields, and percent yields.

Data

Mass of evaporating dish-28.98 g
Mass of evaporating dish + NaHCO ₃ - 31.01 g
Mass of NaHCO ₃ - 2.03 g
Volume of 6 M HCl- 15 mL
Mass of evaporating dish + remaining (solid) product- 30.31 g
Mass of remaining solid product- 1.33 g

Possible sources of Error

The possible sources of error included popping as this would reduce the mass and, hence percent yield. Also, the shape of the beaker was another source of error. The beaker was cylindrical while the evaporating was more like a bowl. Therefore, the cylindrical beaker would be a downfall because the water would condense on the sides and fall back in.

This is a picture of my work and answers for numbers 1-4. To clarify number 1: Because it has a higher molar mass which means that there would be less moles than the others.

Lab 5B: Mole-Mass Relationship Lab

Data

Mass of evaporating dish- 1.32 g

Mass of evaporating dish + Hydrate- 2.16 g

Mass of evaporating dish + anhydrous salt- 1.96 g (1st heating)

                                                                      1.84 g (2nd heating)

                                                                      1.79 g (3rd heating)

Calculations

1. 2.16g - 1.32g = 0.84g (mass of hydrate used)

2. 1.79g - 1.32g = 0.47g

    0.84g - 0.47g = 0.37g (mass of water lost)

3. (0.37g/0.84g) * 100 = 44% (percentage of water in the hydrate)

4. Percent Error = |44-36|/36 x 100

    Percent Error = 22% error

5. CuSO₄* 7H₂O

Since our percent error was very high(22%), the empirical formula above is not correct. Since the percentage of water in the hydrate was higher than the accepted value(36%), the ratio would most likely be lower than 7.

Hydrate prior to heating

Hydrate after heating

Lab 5A: Mole Baggie Lab

 The purpose of the lab was to figure out what type of compound the substance in the ziplock bag was. This lab made us practice our skills in finding the molar mass, number of moles, and identifying the type of compound just by knowing a few pieces of information. Our first bag's label was A2 and we identified it as Potassium Sulfate. Our second bag's label was B6 and we identified it as Zinc oxide.

Lab 4A: Double Replacement Lab

 Writing the net ionic equations challenged me a lot. I was also confused at first when I was trying to write the balanced chemical reactions, but after writing some of the reactions, it became easy.

This is a picture of my balanced chemical reactions for all of my reactions. NR beneath a reaction means no reaction.

This is a picture of my net ionic equations.

This is a picture of my micro-well plate.

Lab 3: Nomenclature Puzzle

 The goal of this activity was to put together a puzzle by forming formulas with compounds and matching the compounds with their respective formulas. The biggest challenge of this puzzle was the organization part. First, my group tried to divide everyone up by element. This plan backfired because many of the pieces overlapped. Therefore, we had to find the pieces we were looking for. At some times, this took us a long time to do as all of the pieces were scattered. I believe I contributed to my group the most by finding pieces and help to form a portion of the puzzle.

Lab 2B: Atomic Mass of Candium

Purpose
 The overall purpose of this lab was to find the average atomic mass of the element Candium, To do this we had to find the masses of each of the three isotopes, decimal abundances of the isotopes, and the percent abundance of the three isotopes. From these three calculations, we were able to plug them into the average atomic mass formula and get the average atomic mass.
 Conclusions
1. The group we asked had an average atomic mass of 1.58 g while ours was 1.423 g. This difference might be because of a difference in the number of isotopes(M&Ms). The other group also could have had a number of different sized isotopes. There were also confusions between the pretzel and peanut isotopes.

2. The difference would be smaller. This is because if a bigger data set was used, more accurate results would show as the chance of error would decrease and there would be less differences between the decimal abundances.

3. No, but it would be similar. This is because the average atomic mass is a general value, calculating a value of which all of the isotopes' atomic mass is around. Hence, one Candium would be close or similar to the atomic mass and highly unlikely to have the same mass as there are many other isotopes included in the calculation.

4.

Lab 2A: Chromatography Lab

Questions

1. It would get soaked and the water would not spread out like we were wanting it to. The water has to start from the wick/center and spread to the ends of the filter circle. The filter paper should not be in contact with the cup because the ink spots will just dissolve in the water.

2. The marker/type of ink and the designs put on the filter paper circle.

3. Physical properties of the pigment makes some colors travel at different rates(faster than others). Components that are not strongly absorbed onto the paper will spend more time in the solution and will move up the paper faster. This "partitioning" of the components of a mixture between the paper and solvent separates the components and gives rise to different bands. If the components of the mixture are colored, then these bands are easy to see.

4. The blue pigment does appear commonly along the edges of the chromatograph and is always the same. Therefore, the pens do appear to contain common pigments. To achieve a single color for the pens or markers, multiple pigments are used to get the right color or shade.

5. Only water-soluble pens are used in this lab because water-soluble pens' ink is able to travel/dissolve in water. If a different type of ink/pen would have been used, then the colors would not separate. Therefore, a different substance would be needed instead of water to make the colors separate.

Lab 1B: Aluminum Foil thickness

Procedure

 First, we had to figure out the mass of the aluminum which we received. Then, we calculated the mass of the aluminum by calculated the water displacement in a graduated cylinder. From this we divided the mass and the volume to get the density. Then, we measured the length and width of the aluminum foil in centimeters. In this case, height was the thickness of the aluminum foil. We also had to find the mass of the aluminum foil, which we weighed on the weighing scale. Since density=mass/volume we then plugged in the values we knew and broke the volume part of the equation apart. Hence, the volume part became length x width x height. Therefore, we only had one unknown(the height(thickness)). Then we multiplied the volume by the density and divided both sides by the coefficient of the expression(27839.84mmH=0.46g(h stand for height/thickness)). This in turn gave us the variable H, which was the thickness. In the end, we had to convert from millimeters to centimeters.  The important materials needed for this lab was an aluminum block, aluminum foil, ruler, pen, paper, computer, calculator, graduated cylinder, and water.

Data

 We found out that the mass of the aluminum block was 80.3 g. From this we figured the volume of it by using the water displacement method which 30 mL. From this, we divided the mass by the volume and found the density, which was 2.68 g/cm^3. Then, we figured out the length and width of the aluminum foil which was 10.6 cm and 9.8 cm, respectively. From plugging this into the density equation, we found that the thickness was 0.0165 mm.

Lab 1A: Density Block Lab

Introduction
My partner, Luis, and I calculated the mass and compared it to the mass of the actual mass of the block. From this we calculated the percent error((actual-experiment)/actual x 100%)). Percent error shows you how precise your result is compared with the actual "result". The purpose of this lab was to determine the mass of a plastic block using its density and volume. Density is the result of the mass of an object(amount of matter in an object) divided by the volume of the object (how much space it takes up).

Procedure
My partner and I selected different density blocks and then measured the length, width, and height to find the volume. After figuring out the volume, we multiplied it by the density of the block (which was given on the density block) to figure out the mass of the block. Then, we weighed the block to figure out the actual mass and then calculated the percent error. We went through 5 trials to get our lowest percent error. The important materials needed for this lab was a plastic block, weighing scale, ruler, pen, and paper.

Data
Our lowest percent error was 2.84%. For this block we found that the length of the block was 7.3 cm, the width being 3.3 cm, and the height being 2.4 cm. From this the volume was 57.816 cm^3. The density, given on the block, was 1.42 g/cm^3. Hence, our calculated mass was 82.1 g and the actual mass was 84.5 g. The percent error for this trial was 2.84%.

Conclusion
We partially fulfilled our purpose of the lab as our calculated mass  We determined the mass of a plastic block by figuring out the volume and using the density given. The reason why our percent errors were very high on many of our trials was maybe because we were a little off when measuring the length, width, and height of the block. I learned that accuracy plays a very big part in measurements, as even if we were a little off in our measurements, we had a magnificent difference between the actual and calculated mass of the block. In the future, I will make sure to make more accurate measurements.