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.

Lab 6: Mole-Mass Relationships Lab

The purpose of this lab was to practice calculating both theoretical and percent yields, and look at the relationship between moles and mass. The reaction we investigated was NaHCO3 + HCl => NaCl + CO2 + H2O. My lab partner Leila and I predicted how much NaCl would remain at the end of the lab by using conversions to see how much NaCl would result as a product of two grams of the reactant NaHCO3. We had to convert two grams of NaHCO3 to moles, then used the mole ratio in the equation (1 NaHCO3 : 1 NaCl), and then converted to grams of NaCl. This helped us practice theoretical yield calculations. We then did the lab, boiling down the H2O in our evaporating dish until it was mostly evaporated and only the NaCl remained. After weighing the dish and the NaCl, we found the mass of the NaCl and used that and our theoretical yield to calculate our percent yield. 

Above is the product, NaCl, after the H2O was evaporated out of the dish. The yellow-green color on the top could have resulted due to contamination from the tongs used to lift the dish.

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

NaHCO3 is the limiting reactant. This is because we added as much of the other reactant (HCl) as needed to a measured amount of NaHCO3. We also didn't need to use of the 15 mL of HCl that we measured, showing that it wasn't limiting.

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

(2.00g NaHCO3 / 1) * (1 mol NaCO3 / 84.006g NaCO3) * (1 mol NaCl / 1 mol NaHCO3) * (58.443g NaCl / 1 mol NaCl) = 1.3914 g = 1.39g NaCl

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

46.30g (mass of dish + remaining solid product) - 44.91g (mass of dish) = 1.39g

4. Find the percent yield for this experiment for the solid product.

(1.39g (actual) / 1.39g (theoretical)) * 100 = 100%

Since it was probably very unlikely to actually achieve a 100% yield, there are possible sources that would allow this to happen. While the NaCl began to pop after boiling, some of the product could have popped out of the dish, lowering the mass. However, we could have not boiled down all the H2O, which would increase the mass. These two factors could even out to produce a 100% yield. The scale also fluctuated, which could have lead to a reading that was not completely accurate. 

Lab 5B: Composition of a Copper Sulfate Hydrate Lab

Post a photo of your hydrate before heating and after heating. Include calculations for #1-#4 in your lab manual. For #5, report your data, including the empirical formula you obtained for the hydrate. If your percent error is high, include a prediction for whether or not your predicted coefficient on water is higher or lower than the actual value.

DATA: 

Mass of Evaporating Dish: 45.06 g

Mass of Dish + Hydrate: 45.83 g

First Heating Mass of Dish + Anhydrous Salt: 45.57 g 

Second Heating Mass of Dish + Anhydrous Salt: 45.59 g

1. Calculate mass of hydrate used.

45.83 g - 45.06 g = 0.77 g CuSO4

2. Calculate mass of water lost.

45.83 g - 45.58 g = 0.25 g H2O

3. Calculate percentage of water in hydrate.

0.25 g / 0.77 g = 32.4675% = 32.5%

4. Find the percent error and provide a possible explanation for your error.

Percent error = (( | Experimental Value - Accepted Value | ) / Accepted Value ) * 100

(( | 0.325 - 0.360 | ) / 0.360 ) * 100 = 9.72% error

Possible explanations for this percent error include how the scale fluctuated and the possible loss of anhydrous salt while stirring. Because of the fluctuating scale while finding the mass of the dish, then of the dish and hydrate, and then after the first and second heatings, my lab partner and I might not have gotten the most accurate readings, as we had to estimate the masses as closely as possible. While stirring the anhydrous salt during the heating, we also accidentally flicked a small piece out of the evaporating dish when we tried to break a large blue crystal down.

5. Moles of water evaporated: 0.014 mol H2O

Moles of CuSO4 that remain in evaporating dish: 0.0033 mol CuSO4

Ratio: 4.24 H2O : 1 CuSO4

Empirical formula: 1CuSO4 * 4H2O

Although our percent error wasn't too high and three could possibly be the correct coefficient, if it is wrong, I predict that our coefficient would be too low. Since getting closer to 36% would mean more water would have to be lost than what happened in our lab, the moles of H2O evaporated would have to increase, meaning the ratio would have to increase, which would then lead to the coefficient increasing.

Lab 5A: Mole Baggie Lab

The purpose of this Mole Baggie Lab was to practice calculating molar mass. In Set A, the mass of the bag and the number of moles of substance was known. To calculate the molar mass, my lab partner Meghana and I weighed the bag with the substance inside. We then subtracted the bag mass by the total mass to find the mass of the substance in grams. After dividing the mass by the amount of moles, we found the molar mass, which was 60.4 g/mol. In Set B, the mass of the bag and the number of atoms was known. We again found the mass of the substance in the bag by weighing the bag and subtracting the mass of just the ziploc from it,. We then used a conversion to find the molar mass; we multiplied the number of atoms in a mol/1 mol by 1.96 g (mass of the substance)/1.51*10^22 atoms to cancel out the representative particles and leave g/mol. The molar mass turned out to be 78.1 g/mol. We then found the molar masses of the possible compounds and matched them up.

A1: sodium chloride

B6: zinc oxide

Lab 4A: Double Replacement Reaction Lab

Balanced chemical reactions 1-5 from Lab 4A

Balanced chemical reactions 6-10 from Lab 4A

Net ionic equations 2-5 from Lab 4A

Net ionic equations 6-7 from Lab 4A

What challenged me the most this lab was getting the hang of writing the first couple of reactions. I didn't fully understand that no matter what, a double replacement occurs and the two outside and two inside elements pair up. I was also overwhelmed with the amount of rules to figure out whether the product would be a solid or an aqueous solution. However, after understanding that I had to do the double replacement, then make sure the products were neutrally charged, then change the coefficients to balance the equation, I was surprised that I finally understood how write the balanced chemical reactions, and was able to do it efficiently.

Lab 3: Nomenclature Puzzle

The goal of this activity was to familiarize ourselves with the names and corresponding formulas of various binary and polyatomic ions. The biggest challenge while completing this activity was recognizing and searching for symbols and polyatomic ion formulas I didn't recognize. I'm also very unfamiliar with certain elements in which the symbols don't seem to relate to the name at all - for example, iron and Fe, so I had to familiarize myself with those by repeating them in my head and searching them on the periodic table. Some of the polyatomic ion formulas were also extremely long and difficult to memorize, but I eventually got the hang of it. I believe my biggest contribution to the group was organizing who would start out with which element (I started with barium), because creating the pairs was the basis of forming the complete square. I also was very intent on setting goals for the group and calling out what name/formula we should have searched for next.

Lab 2B: Atomic Mass of Candium

The purpose of this lab was to determine the average atomic mass of the so fresh, so new, recently discovered element candium. The average atomic mass of an element is the weighted average of the masses of each isotope of the element. An element could have multiple isotopes, meaning each atom has a different amount of neutrons, so the average atomic mass is used on the periodic table to show the most common atomic mass of the element. 

We found the average atomic mass of candium to be 1.25 grams. We counted 95 regular candium isotopes, 11 peanut candium isotopes, and 22 pretzel candium isotopes. We found the average masses to be 0.85g, 2.4g, and 2.4g, respectively. We then calculated the average atomic mass using the equation (decimal abundance*mass) + (2nd decimal abundance*2nd mass) + etc. 

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

Another group's average atomic mass was 1.49g. Our average atomic masses could be different because each group received candium isotopes from a large bag of the three isotopes combined. This means that we would all possibly have different amounts of each isotope, which would alter our average atomic masses after plugging the numbers into the equation.

2. If larger samples of candium were used, would the differences between your average atomic mass and others' be bigger or smaller?

If larger samples were used, the differences between our average atomic mass and others' would be smaller, because of the larger amount of different isotopes in the assortment of candium. If there were more isotopes, an additional peanut candium wouldn't make as much of a difference in the average atomic mass as compared to if there were only four peanut candiums in the sample. Therefore, even if the groups had varying amounts of each isotope, these differences wouldn't make as much of a difference in the final average atomic masses.

3. 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?

If you took any piece of candium from the sample and placed it on the balance, it would not have the exact average atomic mass that was calculated. This is because the average atomic mass is the average of every isotope of candium - therefore, it doesn't only represent one isotope, but represents the average of all three.

4. Draw a rough sketch of what candium's square would look like on the periodic table. 

Lab 2A: Chromatography

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

It is important than only wick be in contact with the water in the cup because you don't want to saturate the entire filter paper with water. The way chromatography works is that the components/pigments will be absorbed into the paper and travel at different rates due to their different physical properties and tendency to attract to the paper/water. As the water slowly moves out from the wick, pigments will be picked up and then absorbed and be allowed to travel and end at different rates. Without this gradual spread of water, the pigments wouldn't spread, and would stay black.

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

Some variables that affect the pattern of colors produced on the paper include the type of pen being used and the pigment that makes up the pen's color, how long the wick touched both the water and the filter paper, the patterns that were drawn on the filter paper, and the size of the hole in the filter paper the wick was placed in.

3. Why does each ink separate into different pigment bands?

Since the black ink is made up of different pigments and each pigment has different properties that allow them to be absorbed more/less quickly into the filter paper, each pigment will stop at around the same area and different will travel farther.

4. Choose one color (e.g. yellow, red, or blue) that is present in more than one type of ink. Is the pigment (compound) that gives this color always the same? Do any of the pens appear to contain common pigments? Explain.

A color present in more than one type of ink is red. The pigment/compound that gives this color is always the same, because when comparing chromatograms made by the same pen, this red pigment seems to be around the same area of the filter paper. Another example is a bright blue seen around the edges of many of the chromatograms; this shows that the pigment stopped at the same place, meaning it has the same properties. This common area that the red and bright blue show up in on different filter papers also proves that multiple pens contained common pigments.

5. 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?

Only water soluble markers or pens were used in this activity because in order for the pigment to spread, it needed to be able to dissolve/become a homogeneous mixture with the water. If it was not water-soluble, meaning it could become a solution with water, the water would be unable to spread the pigment. Using rubbing alcohol in place of water could work with this experiment if using permanent markers or pens, since permanent ink is soluble in alcohol. 

Lab 1B: Aluminum Foil Lab

The purpose of the Aluminum Foil Lab was to create a procedure that could be used to determine the thickness/height of the aluminum foil in millimeters. In order to find the thickness, which is a key part in the volume (volume = l*w*h), the known value of density of aluminum was used, along with the measured mass. The length and width were estimated using a ruler. 

Procedure: 

1. In order to find the mass, we used the scale, and found that the mass of the aluminum foil was 0.42 grams. 
2. To find the length and width, we measured two sides of the aluminum foil and found the length to be 11.05 cm and the width to be 11.55 cm.
3. We labeled the variable for height "h" and multiplied the known length and width in order to find a value for volume: 11.05cm*11.55cm*h = 127.628h => 127.6h (four significant figures). 
4. Since we found the density of aluminum to be 2.73 g/cm^3 in a previous lab, we plugged in our density, volume, and mass into the equation density = mass/volume. 
5. 2.73 g/cm^3 = 0.42g/127.6h
6. We solved for h and got that the height, or thickness, of the aluminum foil was .001176 cm, or .011755 mm, which was rounded to .012 mm because of the two significant figures in the equation.

Data:

• Aluminum Foil
• mass: 0.42 g
• density: 2.73 g/cm^3
• volume
• length: 11.05 cm
• width: 11.55 cm 
• height/thickness: .012 cm

Lab 1A: Density Block Lab

In this lab, the mass of a plastic block was found using its known density and measured volume. The mass of the block, determined by the equation mass = density * volume, is how much matter is in an object. The density, determined by the equation density = mass/volume, is how much matter is in a unit of space in an object. To find the volume of the block, or amount of space in it, the equation length*width*height can be used. After calculating the experimental mass, the actual mass was found using a scale, and the percent error was calculated. In this lab, we aimed for a <2% error.

Procedure:

1. We knew that the density of the plastic block to be 1.42 g/cm^3. The goal of the lab was to calculate the mass, so we needed to find the volume in order to solve the equation mass = density * volume. 
2. In order to find the volume of the block, we used a ruler to measure its length, width, and height. We estimated the length to be 7.50 cm, the width to be 3.25 cm, and the height to be 2.45 cm. 
3. We then plugged these values into the formula volume = length*width*height. 7.50*3.25*2.45 = 59.7188, but since the other values had three significant figures, we rounded to 59.7 cm. 
4. We then plugged this number into the equation as well as the known density, giving us the equation 1.42 g/cm^3 = mass/59.7 cm. 
5. To solve for mass, we multiplied 59.7 * 1.42 = 84.774 = mass, but since the values had three significant figures, we rounded to 84.8 grams.
6. We then brought the block to Ms. Kapinos and she measured its mass using the scale, finding the actual mass to be 85.1 grams.
7. In order to calculate the percent error, we used the equation percent error = ((actual-experimental)/actual) * 100, and plugged in the experimental mass as 84.8 g and the actual mass as 85.1 g in order to find the percent error of .353%.

Data:

• Block One
• density: 1.42 g/cm^3
• volume: 59.7 cm^3
• length: 7.50 cm
• width: 3.25 cm
• height: 2.45 cm
• experimental mass: 84.8 g
• actual mass: 85.1 g

Conclusion:

In conclusion, we found our experimental mass of the plastic block to be 0.3 g off of the actual mass, resulting in a .353% error. Since the purpose of the lab was to calculate the volume manually using a ruler and our knowledge of significant figures and then use that volume to calculate the mass of the block and get within a 2% error of the actual mass, we did fulfill the purpose of this lab. Possible error could be within the initial measurement of the sides of the plastic block. If the estimation was a few micrometers off, the volume could be affected, and therefore the mass could be affected. In the future, I would look more closely at the ruler and think about my estimation more.