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.