Lab 14: Titration Lab

 Summary: In this lab we performed a titration in which we found the molarity of an acid by the amount of base that it took to neutralize the acid. It was a very frustrating lab since a little too much base would cause the experiment of completely go awry. It was an interesting lab and I liked the stirring thing.

Why is the percent ionization such a low number?
The percent ionization is such a low number because acetic acid is a weak acid and therefore does not fully ionize. Therefore most of the H+ protons stay connected and do not bond with water to form hydronium. The percent ionization was 0.46%.

Data

	Trial 1	Trial 2
Volume of NaOh used	24.4 mL	23.7 mL
Volume of of Acid used	7 mL	7 mL
pH of Acid	2.4	2.4
Molarity of Acid	0.871 M	0.846 M
Molarity of Base	0.250 M	0.250 M

Lab 13: Solubility

Introduction: Solubility is the property of one substance dissolving in another. In this lab we were given a solute and were asked to figure out what the solute was. This was based on the solubility of the salt in 10 mL of water at 50 degrees Celsius. By dissolving a certain amount of solute in the water and looking if there is any solute left undissolved we can figure out what solute it based on the graph given. Our solute was Potassium Nitrate.

Procedure:

1. Measure out 13 grams of solute between three weigh boats in 5 grams, 5 grams, and 3 grams.
2. Measure out 10 mL of distilled water into a graduated cylinder and then pour into a beaker
3. Plug in hot plate and turn it on
4. Fill up a large beaker with water and place on hot plate, check the temperature with a thermometer until the water reaches of 50 degrees Celsius.
5. Place small beaker in large beaker filled with water and heat indirectly until small beaker has a temperature of 50 degrees Celsius (take care that no water from the large beaker enters the small beaker)
6. Pour 5 grams of solute in and stir with glass stirrer keeping the water at 50 degrees Celsius (this may require taking off the heat, just check the temperature with the thermometer every so often)
7. If the solute dissolves repeat step 6 pouring in the other 5 grams of solute
8. If solute dissolves repeat step 6 again pouring in the final 3 grams

Observations (Qualitative Data):  I observed that when the first amount of solute was poured in it started as cloudy but then the solution became clear and no particles of solute were left at the bottom. When the second amount of solute was added it remained cloudy when stirred, and when the solution settled there was particles of solute at the bottom of the beaker. That meant that the solution was saturated.

Quantitative Data:

Grams of Water	10 grams
Grams of Solute (Trial 1)	5 grams
Grams of Solute (Trial 2)	10 grams
Grams of Solute (Trial 3)	13 grams
Temperature of Water	50 °C
Trial in which Water was Saturated	Trial 2

Conclusion: Since the water was saturated with 10 grams of solute of at 50 °C that means according to the graph we were given the solute was Potassium Nitrate. We knew that the solution was saturated with 10 g of solute at 50 °C because there were particle of the solute at the bottom of the beaker, meaning it had not dissolved all the way making it saturated. That differed from when 5 g of solute were dissolved in the solution at 50 °C because then the solution had now particles floating around and was completely clear, so the solute was dissolved. So, the solute could not be Sodium Chloride because sodium Chloride would have saturated the solution at 5 g and it could not be Sodium Nitrate because the solute would have dissolved all the way if there was 10 g of solute and the solute was Sodium Nitrate. This means that the solute could only be Potassium Nitrate. Certain amounts of solute dissolve in solutions at certain temperatures because the transfer of energy from breaks the intermolecular forces, and the strength of the forces depend on the molecule. For Ions the more heat the more solute can be dissolved for gases the opposite is true.

Lab 12: Alka Seltzer and the Ideal Gas Law

Summary: Today in class we learned about gasses and how the move and act in certain environments. In this experiment we found the data needed to find the ideal amount of gas in the balloon and the actual gas. This showed the ideal gas law is ideal in that all gas occupy's a certain area completely.

Analysis Questions:
1. Some mistakes that could have occurred are not having an accurate volume of how much space the carbon dioxide took up, as well as not all carbon dioxide going into the balloon.

2. The amount of carbon dioxide in the water could affect the amount of atoms of carbon dioxide and therefore the amount of moles or n making n too small.

3. The volume of the balloon is 1072.67 mL.

4. The difference between the volume obtained by filling the balloon minus the answer from #3 is 6.33 mL. Which is fairly close, however I think that the volume obtained by filling the balloon was more accurate because the balloon is not a perfect sphere, so measuring it by by filling it with water you get a more accurate volume because the water takes up the actual space inside the balloon.

5. The difference between ideal gas and real gas is ideal gas occupies one space alone, and none of it would dissolve in water or somehow be released, but real gas would.

6. No, because not all of the gas was present in the balloon when we found out the grams in the balloon.

Advanced Questions:
1. The amount of gas that should be collected in total from 3 tablets of Alka Seltzer is 2.06 grams of carbon dioxide.

2. The percent yield of the sample of carbon dioxide collected was 93.7%.

3. Some of the carbon dioxide would have dissolved in the water making the actual amount of moles slightly less than the theoretical amount of moles.

Data

Mass of Alka Seltzer Powder	9.14 grams
Circumference of Balloon	39.9 cm
Volume of Water that fits in Balloon	1079 mL
Room Temperature	20 °C
Barometric Pressure	759.2 mmHg
CO2   Pressure	741.2 mmHg
Moles of CO2 in Balloon	0.0437 moles of CO2
Mass of CO2	1.93 grams

Lab 11B: Calories in Food Lab

Summary: The purpose of this the lab was to use what we used about energy in class today to measure the calories in food through the energy they transferred in the form of heat into water.

Lab 10: Evaporation and Intermolecular Attractions

Data

Substance	Maximum or Initial Temp (°C)	Minimum or Final Temp. (°C)	∆ Temp. (°C) (- and + indicates whether temps. increased or decreased)
Methanol	18.9°C	3.2°C	-15.7°C
Ethanol	19.6°C	10.6°C	-9.0°C
n-Butanol	20.3°C	18.2°C	-2.1°C
Glycerin	20.3°C	23.3°C	+3.0°C
Water	19.9°C	14.8°C	-5.1

Pre-Lab Questions:

2. The substances that we tested had a vast range of the difference in temperature, in order from the substances that had the biggest decrease in temperature to the ones which had the smallest decrease in temperature it goes Methanol with a decrease of 15.7°C,  Ethanol with a decrease of 9.0°C, water with a decrease of 5.1°C, n-Butanol with a decrease of  2.1°C, and Glycerin which actually gained 3.0°C. The temperature increase or decrease were due to intermolecular forces. Each substance posses hydrogen bonds, which are the attraction between the partially positive positive hydrogen and partially negative atoms, which in this case is the oxygen in all these substances. More hydrogen bonds the more energy it takes to evaporate, and a less acute drop in temperature. So, the substance with the most hydrogen bonds was Glycerin with 3 potential places for hydrogen bonds per molecule, and Glycerin was the only substance to increase in temperature during the lab.

3. Hydrogen bonds only explain part of the difference in temperature seeing as three substances, Methanol, Ethanol, and n-Butanol, had the same amount of potential places for hydrogen bonds and yet had differences in the decreases in temperature. This is due to London Dispersion Forces. London Dispersion Forces create a temporary poles on molecules called a momentary dipole-dipole. This is due to the fact that electrons move around the nucleus and when more electrons move to one side that side becomes more negative and the other becomes more positive, this also forces the molecules around the molecule with the momentary dipole-dipole also to become momentary dipole-dipole. It takes more energy (and therefore a smaller decrease in temperature during evaporation) in molecules that have more molar mass. This is due to the fact more molar mass means more electrons, which mean a stronger momentary dipole-dipole. That explain why n-Butanol, which has the biggest mass had the smallest decrease in temperature.

4. The number of -OH’s correlated to evaporation because of that is where hydrogen bonds will form. This is expected because hydrogen bonds usually form between atoms with higher electronegativity such as oxygen, which is often partially negative in many molecules, and hydrogen. 1 -OH means one potential hydrogen bond could be formed, because there is one partially negative atom and one hydrogen which will for the hydrogen bond. This is why water has two places for bonding because it has two hydrogens which will form bonds, and more bonds means more energy needed for evaporation, and a less severe temperature decrease.

Lab 7: Flame Test

Unkown #1: We put unknown #1 as Strontium Nitrate. We did this because in both the Strontium Nitrate and the unknown #1 when burned turned the flame a vivid solid red.  Lithium Chloride also has a reddish pink hue, but when we burned the Lithium Chloride it burned inconstantly as well as flickered. So, we thought the unknown substance compared better with the striking scarlet Strontium Nitrate rather than the coral Lithium Chloride.

Unknown #2: We thought unknown #2 was Potassium Chloride because compound turned the flame when burned a shade of fuchsia and violet that the only compound close to it was the Potassium chloride which also possessed a lavandar flame when burned..

Use Link To see the Video

https://plus.google.com/112388463111199653366/posts/bvZ2mhhem2D?pid=6177013114767068210&oid=112388463111199653366

Lab 8: Electron Configuration Battleship

Challenge:  The biggest challenge of this game was using the periodic table to try to find the electron configuration. It was hard to go back and forth from the s, p, d, and f block. My biggest help was listening to the last orbital to find the element.

What I learned:  What I was able to better lean is where the elements fall based off their last orbital and to use the periodic table to find the last orbitals by separating the s, p, d, and f blocks, and then counting the electrons in them.

Lab 6: Mole-Mass Relationships Lab

Summary: The purpose of this experiment was to use what we learned from Stoichimetry to find the theoretical mass of the salt using the limiting reactant and determine the percent yield based on our actual result.
Error in Percent Yield: The error of 5% was probably due to the fact that while we were boiling the water out of the salt, some of the salt escaped from the evaporating dish changing the overall mass of the actual data.

Lab 5B: Composition of a Copper Sulfate Hydrate Lab

Prediction: I predicted that we did not let the water in the copper sulfate evaporate enough, so we our empirical formula for the hydrate had a smaller coefficient than then actual formula for the Hydrate. So, there would be more water in the actual equation than our equation, because we did not let all the water evaporate.
Data: Mass of Evaporating dish: 42.22g
Mass of evaporating dish + hydrate : 43.16g
Mass of evaporating dish + anhydrous salt : 43.01
Mas of anhydrous salt: 0.94g
Mass of evaporated water: 0.15g
Percent of water: 16%
Percent error: 58%
Equation: CuSO4 ●2H2O
Actual Equation: CuSO4 ●5H2O

Lab 5A: Mole Baggie Lab

1. In bag A2 there was Potassium Sulfate. We got the answer of Potassium Sulfate by weighing the bag, subtracting the empty bag's mass from the full bag's mass, and then dividing by the number of moles to find molar mass. We wrote out all the compounds formulas as well as the elemental mass. We added the mass of each element in each of the compounds together to find the weight of 1 mole of the substance and then found the compound with the same molar mass as the molar mass in the bag.

2. In bag B1 there was  Sodium sulfate. What we did in this one was we divided the number of atoms by Avogadro's number to find the amount of moles in the bag. Then by repeating the process we used to find the compound in bag A2.

Lab 4: Double Replacement Reaction Lab

Challenge: I think the biggest challenge of this lab was writing out an equation (neatly) and balancing it. This was the first time I had balanced equations in a while so, sometimes I found it difficult. The other hard part was following the rules for solids and liquids, because there is much to remember as well as check, which add to my growing list of Chemistry rules. What surprised me about the lab was how little was actual lab work, and how much of the Lab was writing down equations, balancing equations, and finding the solids.

Lab 3: Nomenclature Puzzle

Goal: The goal of this activity was to see if we could use what we learned about Nomenclature to pair binary ions as well as polyatomic ions formulas together with the name of each compound.

Challenge: The biggest challenge for me was finding the matching piece of the puzzle because there were so many pieces as well as having to think of the formula or name that would be a match to the compound.

Contribution: I thought my biggest contribution was helping separate and match pieces together as well as matching a few bigger segments of the puzzle together. I think that we worked really well together to complete the puzzle in a sh

 The Completed puzzle

Lab 2B: Atomic Mass of Candium

Purpose: The purpose of the lab was to see if we could use what we learned in class, isotopes, as well as the formulas for finding the atomic mass,  to determine the relative atomic mass based of the cadium isotopes that we were given.

1. The Atomic mass of the group nearby us was 1.25 while ours was 1.49. Their atomic mass was different then our atomic mass because they different amounts of the different isotopes of candium, so their calculations varied from ours, making ours different due to differnce in decimal abundence and average mass of an isotope.

2. The differences  of our atomic mass would be smaller if we had closer ratios of each isotope in each bag, if that happened then the calculations would be closer as well the atomic mass, because there would be closer decimal abundance as well as closer average mass. However if the backpacks had huge differences of the ratios of isotopes then the difference would be larger because the  difference in decimal abundance.

3. If a random piece of candy was taken and placed on a scale it would be profoundly unlikely that  it would have the exact average atomic mass because it is an average based off three isotopes so each atomic mass would not be necessarily close to any of the individuals isotopes averages' mass. It is possible that there is a chance that it could weigh the average atomic mass because each atom of candium has a slightly differently mass, so it could possibly happen but it is slim chance that it would happen.

4.

The Periodic Table Symbol for Candium 

The Different Isotopes of Candium

Lab 2A: Chromatography

1.The water needs to be absorbed slowly into the filter paper for the ink to separate to for a chromatogram. The inks separate due to their attraction to the filter paper. If the water is absorbed quickly not all the ink will form the rings of color.

2. Some variables that could affect the different pigment bands is what type of marker was use, the what figures are drawn on the filter paper, how far away the dots are from the center,  and how long the chromatogram is absorbing the water. The  different types of makers could have different pigments which leave different color bands. The ink of the figures could spread more or less depending on the size of the figure. The distance between the center and the ink could change the spread if the filter paper has been saturated in water and the ink has not fully saturated. Finally if the chromatogram has only been absorbing water for a short amount of time the water might have parted the various inks.

3. The ink separates into different ink bands because different inks make up the black ink. Each are attracted to the paper in some ways, but some inks are much more strongly adsorbed than other inks as the water is absorbed up the paper, this creates rings of the different colored inks contained in black ink.

4. I chose the color blue which was present in what I could observe two types of ink. However they were not the same shade of blue. One was a dark blue that was toward the middle of a chromatogram, and the other was a lighter blue that was on the outside of a chromatogram. I will talk about the lighter blue. This blue from what I could  discern was almost exclusively on the edge of the chromatograph. So, the pigment appears to always give the color in the same way. The pigment for the darker blue is most likely different because it was expressed by a different pen that from what I could see had no light blue ink. The marker that did have the light blue also had a ring of dark blue just below it, so they have to be different pigments because they are in different rings. The light blue pigment only appeared in one of the four markers we tested so it seems it is not a commonly used pigment in ink.

5. Water-soluble pens were only used in this activity because since it is water soluble. This means the black ink comes of in water, meaning the inks separate in water. "Permanent" markers which are more hydrophobic would have a harder time  being separated by the water they are supposed to at least somewhat repel. I would modify the experiment by adding  soap to the  solute because "permanent" markers are much more soluble to soap and water than to water, so it would be easier to the pigment rings.
Links to research of ingredients of some ingredients in Sharpies
Diacetone alcohol  http://www.inchem.org/documents/sids/sids/123422.pdf
Propyl alcohol http://www.inchem.org/documents/icsc/icsc/eics0553.htm
Butyl alcohol http://www.inchem.org/documents/icsc/icsc/eics0111.htm

Look at the blues in the ink rings

Lab 1B: Aluminum Foil Lab

Procedure:

1. Obtain a piece of Aluminum from your instructor
2. Weigh piece of Aluminum on a scale (in grams) to find the mass
3. Observe and Record the mass of Aluminum
4. Fill a graduated cylinder with 100 ml of water
5. Gently tip graduated cylinder and slide piece of Aluminum into the cylinder without splashing water out of the cylinder
6. Record the the amount in the graduated cylinder
7. Subtract 100 ml from the amount of ml in the graduated cylinder (this is water displacements, which is a way to find volume) 
8. Observe and record Volume of Aluminum
9. Use Density formula to find density  of Aluminum(SEE Lab 1A for Equation)
10. Observe and Record Density of Aluminum
11. Obtain a rectangle of Aluminum from Instructor
12. Measure the width and length of the rectangle using the cm side of the ruler
13. Weigh the rectangle of Aluminum on a scale in grams
14. Using data found set up this equation D x V = L x W x H (H is the variable, use density found with piece of aluminum)
15. Solve for H
16. Record Data

Data Type	Data Amount
Density	2.73 g/cm3
Mass	0.42 g
Volume	1.1 cm3
Length	11.05 cm
Width	11.55 cm
Height (Thickness)	0.089 mm

Find The Thickness of the foil