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.

Aluminum Foil Lab

Procedure:

1. First, my lab partner, Zoe, and I weighed our piece of aluminum foil by placing it on a scale. We recorded all the digits of this measurement.
2. Next, we found the volume of the piece by using the density of aluminum with the density equation: density = mass/volume.
3. Using a ruler, we measured the length and width of the piece of foil (as shown below).
4. Finally, we calculated the height of the aluminum foil using the equation for volume: volume = length x width x height. Our last step was converting our calculated thickness into millimeters.

Calculations (a deeper look):

density of aluminum (ours) = 2.8 g/cm^3

mass of aluminum foil = 0.43 g

To find the volume of the aluminum foil, we divided the mass by the density.

volume = mass/density = 0.43/2.8 = 0.15 cm^3

length of aluminum foil = 11.52 cm
width of aluminum foil = 9.21 cm

To find the height, or thickness of the aluminum foil, we divided the volume by the product of the length and width.

height = volume/(length x width) = 0.15/(11.52 x 9.21) = .0014 cm

.0014 cm = .014 mm

Thickness of aluminum foil = 0.014 mm

Density Block Lab

In this lab, we explore the relationship between mass, volume, and density. Mass is the amount of matter in an object, and is measured in grams. Volume, measured in liters, is the amount of space an object occupies. Density is the amount of matter in a certain amount of space, or the mass per unit volume. Therefore, we see that there is a connection between these three terms.

density = mass/volume

Today, we are trying to find the mass of a plastic block, given its density. Our only tool is a ruler, which we can use to find the volume. Then we should be able to obtain the mass of the block using its density and volume.

Procedure:

1. After obtaining a plastic block and a ruler, my partner, Zoe, and I first measured the dimensions of the block. We did this by placing the block and ruler on the lab desk. Lining up a bottom corner of the block with the "0 cm" mark on the ruler (in photo below), we found measurements for all the sides and recorded them, taking significant figures into account.
2. Next, we multiplied the length, width, and height in order to find the volume of the block, again following the rules of significant figures.
3. Using the equation for density, we calculated the mass of the block.
4. Then, we asked our RTA, Rachel, to weigh the block for us.
5. Finally we calculated the percent of error. To find the percent of error, we subtracted our calculated (experimental) mass from the actual mass and divided the difference by the actual mass. Since our percent of error was less than 2%, we did not repeat this procedure.

Data:

density = 0.985 g/mL

length of block = 8.75 cm

width of block = 2.45 cm

height of block = 3.75 cm

volume = (l)(w)(h) = 80.4 mL

calculated mass = (80.4)(0.985) = 79.2 g

Conclusion:

This lab shows that the mass of a block can be estimated, using its density and volume. Although the actual mass of the block was not equal to our calculated mass, it was within 2% of error. The cause of this inaccuracy would most likely be the estimation of the block's dimensions. However, it is very difficult to record entirely accurate measurements, considering the significant figures permitted when using the rulers. In the future, if possible, perhaps the ruler should be even more specific in order to calculate a mass closer to the actual mass. Overall, this lab fulfilled its purpose, as we learned how to find the volume of a block, and use its density to ultimately calculate its mass.