Lab 14: Titration Lab

Last lab!

In this lab we determined the percent ionization of acetic acid in vinegar.  We filled a burette with 50 mL of NaOH (the base) that had a molarity of 0.25 M and slowly added drops to a flask with 7.6 mL vinegar (the acid) and water while stirring.  We added 5 drops of phenolphthalein to the solution as an indicator that would turn from clear to pink when the solution changed from an acidic form to a basic form.  The drops of NaOH created a momentary pink color in the solution, and once the solution reached a very light pink, we knew that it was the equivalence point between H+ ions and OH- ions.  We then recorded the volume of NaOH used by looking at the volume on the burette.

Titration setup

Analyte at equivalence point

Trial 1:

Molarity of acid: 0.86 M
Volume of acid: 7.60 mL

Molarity of base: 0.25 M
Volume of base: 26.0 mL

Trial 2:

Molarity of acid: 0.87 M
Volume of acid: 7.40 mL

Molarity of base: 0.25 mL
Volume of base: 25.7 mL

Average concentration of acetic acid: 0.87 M
[H3O+] = 10^(-pH) = 10^(-2.4) = 0.0040
Percent ionization: [H3O+] / molarity of acetic acid x 100 = 0.0040 / 0.87 = 0.46%

The percent ionization of acetic acid in vinegar is 0.46%.  This is such a low number because acetic acid is weak, which means that not all of the ions break apart and ionize, so there are not many H+ and OH- ions in the solution.

Lab 13: Solubility (A Very Salty Lab)

Introduction:

The goal of this inquiry lab was to create and implement our own procedure to identify an unknown solid using experimental data and known solubility data.  We used a graph of the solubility curves of three solids (NaCl, NaNO3, and KNO3) to help identify.  We were only allowed to use the given solid, 10 mL of water, and other equipment such as hot plates, balances, thermometers, flasks, and stirring rods.

Important terms:

-solute: the substance that is dissolved
-solvent: the substance that the solute is dissolving in
-solubility: the measure of the amount of solute that will dissolve in a given amount of solvent

Procedure:

1.  We chose 45°C as our testing temperature because the solubility varied greatly at that temperature, so it was easier to differentiate between the solids based on the amount of sample dissolved.

2.  We measured 10 mL of water and poured that into a small beaker.  We knew that that was 10 g because 1 gram of water is equal to 1 mL of water.

3.  Next, we filled a larger beaker about halfway with water to act as the hot bath for the small beaker, and placed that on a hot plate.

4.  While that was heating up, we used the solubility graph to find the solubility in grams for each salt at 45°C.  We divided that mass by ten because the mass given on the graph is per 100g of water, and we only had 10g water.

5.  We measured out 4.8 grams of NaCl, which was one gram more than the mass on the solubility line, so we would know for sure if the substance was above the solubility line and unsaturated or saturated.

6.  When the thermometer in bath on the hot plate reached 45°C, we added that mass of substance to the small beaker and placed that into the bath (bigger beaker).  After stirring with a stirring rod for a couple of minutes, we checked to see if the substance had dissolved, which it had, meaning it was an unsaturated solution.

7.  We then moved on to the next salt, KNO3, and added 3.5 more g to the beaker to obtain 8.3 total g of substance, which was one more gram than the mass on the solubility line for KNO3.

8.  Because the substance dissolved for both NaCl and KNO3, we knew that the mystery substance was NaNO3.  11.7 grams of NaNO3 did not dissolve in the 10 g of water, making it a saturated solution.

Data:

Trial 1 (NaCl test):

Temperature - 45°C
Mass of solute - 4.8 grams
Mass of solvent (water) - 10 grams
Dissolved

Trial 2 (KNO3):

Temperature - 45°C
Mass of solute - 8.3 grams
Mass of solvent (water) - 10 grams
Dissolved

Trial 3 (NaNO3):

Temperature - 45°C
Mass of solute - 11.7 grams
Mass of solvent (water) - 10 grams
Not dissolved

Conclusion:

The mystery substance we had was NaNO3.  We found this out because in the first trial, we measured 4.8 grams of solute into 10 grams of solvent at 45°C and the solute dissolved.  In the second trial, we measured 8.3 grams of solvent into the same amount of solvent at the same temperature and the solute again dissolved.  This left NaNO3 as the identity, because the solution would be saturated if it was NaCl or KNO3.  The relationship between the solubility of a substance and the temperature is that as solubility increases, temperature also increases.  For NaNO3, the saturation point at 45°C is 38 grams, and the saturation point at 100°C is 242 grams.  A challenge I encountered during the trials was making sure the temperature was right at 45°C when we started adding and stirring the substance, or else the results would be off.

Lab 12: Alka Seltzer and the Ideal Gas Law

In today's lab, we dissolved three Alka Seltzer tablets and collected the gas (carbon dioxide) given off from the citric acid and sodium bicarbonate. We placed a balloon filled with crushed Alka Seltzer on top of a test tube containing water and turned the balloon over to start the reaction.  After the fizzing ceased, we measured the circumference of the balloon, now filled with gas, and then used the ideal gas law learned in class today to determine the mass of the gas produced.

Analysis Questions:

1.  Discuss an area in this lab where experimental error may have occurred.

Experimental error may have occurred during the transfer of Alka Seltzer to the balloon, since a little bit of powder remained in the weigh boat, and some powder was spilled.  Also, when we filled the balloon with water to measure the circumference, we may have been off and the size of the balloon might not have been exactly the same as it was filled with CO2.  Error could have also taken place when we mixed the water in the test tube with the Alka Seltzer if we didn't wait long enough for the fizzing to stop when we measured the balloon.

2.  Choose one error from above and discuss if it would make "n" the number of moles of CO2 too big or too small.

Since some Alka Seltzer spilled, it would make the number of moles of CO2 too small, because not enough of the reaction would have occurred between the Alka Seltzer and the water, and therefore not enough carbon dioxide would have been collected in the balloon.

3.  Calculate the volume of the balloon mathematically.

C = 2πr

38 = 2πr

r = 19/π

V = (4/3)π * (19/π)^3 = 926.6 cm^3= 926.6 mL

4.  Are the two volumes close?  Which do you feel is more accurate and why?

Yes, the two volumes are close.  The volume found during the experiment is more accurate than the calculated volume, because the shape of the balloon isn't exactly a sphere, so using the formula for the volume of a sphere isn't precise.  However, the volume we found during the experiment isn't 100% accurate either because of the human error possibilities listed in the previous questions.

5.  Give two differences between a real gas and an ideal gas.

A real gas molecule interacts with other particles through repulsions and attractions.  An ideal gas molecule (imaginary) reacts independently of others and there are no interactive forces.  Another difference is that when real gas molecules collide between particles or the wall of the container, there is a net loss of energy, where as in ideal gases, there is no loss of energy during collisions.

6.  Would the CO2 you collected in this lab be considered ideal?  Why or why not?

No, the CO2 would not be considered ideal, because it lost energy from particle collisions and had interactive forces between its molecules.  It also occupied space and mass.

Advanced Questions:

1.  Calculate the mass of CO2 that should be collected per tablet.

Each tablet had 1000 mg C6H8O7 and 1916 mg NaHCO3.  We used three tablets and found that the citric acid produced 2.062 g CO2, and the sodium bicarbonate produced 3.011 g CO2.  Due to citric acid being the limiting reactant, the theoretical yield is 2.062 g CO2.

2.  What percent is the percent yield for the CO2 collected in your sample?

We divided the experimental mass of CO2, 1.86 g, by the calculated mass, 2.062 g, and obtained 90.2% as our percent yield.

3.  What effect does the fact that CO2 is water soluble have on your calculated "n" value?

The fact that CO2 is water soluble may have decreased our n value, since some of the CO2 may have dissolved in the water and so was not measured in the balloon.  This is why our actual results were smaller than the theoretical yield.

Lab 11B: Calories in Food

In this lab, we indirectly measured the amount of Calories per gram of three sample food items: a pecan, cashew, and cheese puff.  To do this, we used a calorimeter made of an Erlenmeyer flask filled with water on top of a metal can (which acted as a stove).  We ignited the food item, which was underneath the metal can, and measured the change in temperature of the water in the flask above.  By doing this, we were then able to calculate the amount of energy in the food tested because the heat gained by the water equals the heat lost by the food item.

A pecan burning in the calorimeter

Data Table and Calculations:

Questions:

1.  We measured a temperature change in the water.  The water absorbed the heat given off of the burning food, so we knew that the heat gained by the water would be equal to the heat released by the food sample.  Therefore, the temperature increase in the water is also the temperature change in the sample.

2.  We measured the energy released by the food sample, which is the same as the energy gained by the water, since the energy released heated the water.

3.  The small amount of energy that was not absorbed by the water could have escaped through the holes in the metal can as smoke and been released into the surrounding air.

4.  I was expecting the nuts to have more Calories per gram than the cheese puff because of previous knowledge, but it turned out to be the other way around.  This is most likely because our cheese puff didn't burn thoroughly and the mass of food burned was too small (0.04 g compared to 1.25 g and 1.38 g of nut burned).

Lab 10: Evaporation and Intermolecular Attractions

Pre-Lab Table:

Data Table:

Questions #2-4:

2.  Explain the differences in the difference in temperature of these substances as they evaporated.  Explain your results in terms of intermolecular forces.

The differences in temperature change of the five substances varied due to the difference in the strengths of intermolecular forces.  Methanol had the highest change and highest vapor pressure because it had the weakest intermolecular force, so it evaporated and cooled faster.  On the other hand, glycerin had the smallest temperature change (the initial temperature actually increased) and the slowest evaporation rate because its intermolecular force was the strongest --- it had three -OH groups (three hydrogen bonds).

3.  Explain the difference in evaporation of any two compounds that have similar molar masses.  Explain in terms of intermolecular forces.

Two compounds that had similar molar masses were methanol (32.042 g) and ethanol (46.068 g).  The change in temperature for methanol (15.9°C) was higher than that of ethanol (10.1°C).  This is because even though they both have one hydrogen bond, the molar mass of methanol is smaller, so the bond is weaker.  A weaker bond causes faster evaporation and also lower vapor pressure.  Ethanol's molar mass is larger, so its bond is stronger, causing slower evaporation and thus lower temperature change.

4.  Explain how the number of -OH groups in the substances tested affects the ability of the tested compounds to evaporate.  Explain in terms of intermolecular forces.

Methanol and ethanol both have one -OH group (hydrogen bond), but because methanol's molar mass is smaller, it's temperature change is greater and its evaporation rate is quicker.  Ethanol had the second highest temperature change.  Water has two hydrogen bonds, but it has the smallest molar mass, so it had the third quickest evaporation rate and third highest temperature change.  n-Butanol has one hydrogen bond, but because its molar mass is so high, the bonds are stronger and thus the temperature change is smaller and the evaporation rate is slower.  Lastly, glycerin has three -OH groups, or the strongest intermolecular force, causing an increase in temperature (smallest difference in temperature) and the slowest evaporation rate.  This is because the rate glycerin was absorbing heat from the condensation in the room was faster than the evaporation rate.

Lab 7: Flame Test Lab

In this lab, we had to burn different compounds over a flame with a wooden stick and observe the color of light emitted.  We then used those recordings to identify two mystery chlorine compounds in the end.

Pre-Lab Questions:

1.  The ground state is the normal electron configuration of atoms or ions of an element, whereas the excited state is when atoms or ions in the ground state are heated to high temperatures and some electrons absorb energy and "jump" to a higher energy level.

2.  The word "emit" means to release.  Electrons emit light in the form of electromagnetic radiation as they return to ground state.

3.  In this experiment, the atoms are getting their excess energy from the bunsen burner flame, since the atoms are being heated at high temperatures, causing the electrons to absorb energy and jump to higher energy levels.

4.  Different atoms emit different colors of light because each atom will produce only one color (each atom is quantized).  Different atoms also have different electrons, and those electrons could jump to different energy levels, creating different colors.

5.  It's necessary to clean the nichrome wires (or, in this experiment, use a different stick each time) between each flame test to ensure that different compounds don't mix.

Unknown #1 was lithium chloride (LiCl) and unknown #2 was potassium chloride (KCl).  We found this because we compared the color burned to the other eight recorded color observations.  Unknown #1 burned a magenta color, shown above, which matched LiCl, and unknown #2 burned a light purplish color, which matched KCl.

Lab 8: Electron Configuration Battleship

The aftermath of Battleship with my partner Rachel

The biggest challenge I had while playing was naming the right electron configurations, especially in the f block and elements with exceptions.  As we continued playing, this became easier, and I felt more comfortable with naming the electron configurations using the noble gas shortcut in the end.

Lab 6: Mole-Mass Relationships

The purpose of this lab was to practice calculating theoretical yield and percent yield using the experimental data of a reaction of sodium bicarbonate and hydrochloric acid.  We also had to find the limiting reactant using the reaction NaHCO3 + HCl --> NaCl + CO2 + H2O by looking at the relationship of the reactants and products (how much product each reactant yielded).

Questions 1-4:

Our percent yield is lower than 100% most likely because some salt popped out during boiling or we didn't wait long enough for all the water to evaporate before weighing the dish.

The remaining solid in the evaporating dish after boiling.  The salt is a bluish color because the same dish
 was used in the copper sulfate hydrate lab, and the dish was probably not cleaned very well after use. It was
also interesting that the tongs left a yellow-greenish mark on the salt.

Lab 5B: Composition of a Copper Sulfate Hydrate Lab

Hydrate before heating:

Hydrate after heating:

Calculations for Questions 1-4

Question 5: The empirical formula we calculated for the hydrate was 1 CuSO₄ · 4 H₂O.  We predict that the coefficient for H2O will be slightly smaller, if not equal to the actual value since our percent error was so small (8.3%).

Lab 5A: Mole Baggie Lab

The purpose of this lab was to identify a mystery substance in a plastic baggie given only the mass of the empty bag and the number of moles/particles.  We determined the identity of the substance by first weighing the bag on the scale to find the total mass, and then subtracted the mass of the empty bag from that, which gave us the substance mass.  Afterwards, we calculated the molar mass by dividing the substance mass (in grams) by the number of moles in the substance.  For Set B, we were given the number of particles instead of moles, so we just converted it into moles and plugged that into the molar mass equation.  Finally, we matched the calculated answer with one of the given possible compounds using the periodic table.

Bag A4 contained calcium carbonate, and bag B3 contained potassium sulfate.

Lab 4A: Double Replacement Reaction Lab

Well plates after the reactions.  Plates 2-7 show
chemical reactions in which a solid precipitate formed.

Balanced chemical reactions #1-5 with net ionic equations

Balanced chemical reactions #6-10 with net ionic equations

The most surprising part of this lab was how easy writing the net ionic equations turned out to be.  I expected it to be more complicated, but taking the shortcut presented in class instead of writing out the complete ionic equations was a lot quicker and more convenient.  The most challenging part was looking up if certain compounds were aqueous or solid using the solubility rules.

Lab 3: Nomenclature Puzzle

The goal of this activity was to solve a binary and polyatomic ions puzzle by matching the ion formula to their name, using our newly learned knowledge of chemistry nomenclature!  The biggest challenge we encountered was combining the pairs of triangles to form bigger squares and chains.  This puzzle wouldn't have been possible without organization and teamwork, so I think my biggest contributions were helping separate the squares into categories of certain elements, combining single squares at the very beginning, and also looking up certain unknown ions in the lab book.

Our finished puzzle!

Lab 2B: Atomic Mass of Candium

Candium's three isotopes - M&Ms!

The purpose of this lab was to plan and implement a procedure to determine the average atomic mass of the element candium, given a random sample of three different isotopes of the element: regular M&Ms, peanut M&Ms, and pretzel M&Ms.

Average atomic mass: 1.43 g

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

Another group's calculated atomic mass was 1.52 g.  Ours is different because each group received different amounts of isotopes (type of M&M) and different total amount and weight.  The sample sizes were also not very big, leading to more variation.

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

The differences would be smaller, because the larger the sample, the closer the calculated masses will be to the average value as a result of less variation.

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?

No, because the calculated atomic mass is just an average and isn't necessarily the same value, but should be close.  It would be extremely rare for a random candium sample to be exactly the same as the average atomic mass.

4.  Periodic table square for candium!

Lab 2A: Chromatography Lab

Before:

After:

My partner Meghana and I with our two favorite chromatograms!

Questions:

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

It's important that the filter paper isn't entirely saturated at first so the water can seep through the wick and slowly onto the paper, which then gives the ink time to spread out from the center and separate into the different pigments.

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

Some variables include the type of pen used (since different inks are composed of different colors), the pattern drawn using the pens, the distance the pattern was drawn from the center, the amount of ink used, and the length of the wick.

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

Each ink has different mixtures and will travel up the paper at different rates depending on their characteristic physical properties.  Some components in the mixture are more strongly adsorbed onto the paper than others, and those will move up the paper more slowly than the solvent.  Components that are not strongly adsorbed onto the paper will move up the paper more slowly than the solvent.  This "partitioning" of the components of the mixture between the paper and the solvent separates the components and creates different pigment bands.

4)  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.

Blue is one color that is present in both chromatograms, and the pigment that gives this color is always the same, since blue is used in the ink of many types of pens and markers.  Many of the pens appear to contain common pigments besides blue, such as yellow, orange, pink, and a bit of violet.  The order that the colors show up from the circle is also similar on both papers -- yellow, orange, and pink (warmer colors) are near the center, and blue, the cooler color, is on the outer side.

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 pens were used so that water could cause the ink to spread across the filter paper.  The experiment could be modified to separate the pigments in permanent markers by using a solvent, for example, rubbing alcohol, that is able to "remove" or separate permanent markers.

Lab 1B: Aluminum Foil Lab

Part II: Determining the Thickness of Aluminum Foil

Procedure:

To find the thickness of a piece of aluminum foil, we first used the formula for volume: V = L x H x W.  We measured the length (15.50 cm) and width (8.13 cm) of the foil, leaving the height H (thickness) as the variable to solve for.  For the volume of the foil, we used the expression 15.50 x 8.13 x H (126 x H).  The next step was to find the mass and density to plug into the formula D=M/V.  We calculated the mass, which was 0.48 grams, by weighing the foil on the scale.  We then determined the density by referring to the given density of aluminum from Part I of the lab, which was 2.8 g/cm³.  By plugging these three values into D=M/V, we obtained:

Finally, we solved for H and rounded to two significant figures, which was 0.0014 cm.  We converted that to millimeters by multiplying it by ten and obtained our final answer for the thickness of aluminum foil, 0.014 mm.

Lab 1A: Density Block Lab

Introduction:

The purpose of this lab is to determine the mass of a plastic block using its density and volume.  My partner and I calculated the mass of the block using the calculated volume and the given density of the block.  Afterwards, we found the actual mass by weighing the block on a scale, and calculated the percent error using the percent error formula.  Important terms include density (mass per unit volume of an object), mass (the amount of matter in an object), and volume (the amount of space an object occupies).

Procedure:  

First, we used a ruler to carefully measure each side of the plastic block --- length, width, and height.  We then calculated the volume by multiplying the three numbers.  Afterwards, we plugged the volume and density of the block (which was already given to us) into the D=M/V equation to find the mass.  To determine whether our calculation was accurate or not, we weighed the block on a scale. Finally, using the percent error formula, ((actual - experiment)/actual) x 100%, we found the percent error.

Data:  

Block:

• height - 1.25 cm
• length - 9.33 cm
• width - 6.82 cm

Volume: 82.1 cm^3

On our first try, we calculated the mass to be 44.4 grams using the D=M/V formula.  82.1 cm^3 was the volume, and the given density was 0.541 g/cm^3.  The block was actually 44.5 grams, leaving a percent error of 0.225%.

Conclusion: 

We successfully fulfilled the purpose of the lab on our first try.  We learned that we had to be very precise when measuring objects with a ruler, and some sources of error include inaccuracy with measuring and rounding to significant figures.  In the future, to prevent the same errors and improve the accuracy of our results, we would need to pay closer attention to significant figures and rounding.