Lab 20: Solubility Inquiry Lab

Lab 20

Solubility Inquiry Lab

Shiva Senthil

8/2/17

Introduction

The purpose of this lab was to create a procedure to identify an unknown solid using given solubility data. The lab consisted of adding solute to the solvent, water, at certain temperatures and using the solubility data given to identify it. Important terms in this lab are solute, solvent, and solubility. The solute is the substance that is dissolved. The solvent is the substance that dissolves the solute. Solubility is the measurement of how much solute will dissolve in water, and is usually defined in grams dissolved per 100 grams of water. The solubility for solid solutes generally increases with increasing temperature.

Procedure

The first step in the procedure was to measure about 6 grams of the unknown solid and place it into 10 grams of water. The water was at room temperature (about 21.5°C). The reason we chose room temperature is because the two of the possible solids (KNO3 and NaCl) had similar solubilities at that temperature and the other solid (NaNO3) had a much higher solubility. The solubility of KNO3 and NaCl is about 37 grams per 100 grams of water. Scaling it down to per 10 grams of water, the solubility is 3.7g/10gH2O. The solubility of NaNO3 at room temperature is about 8.7g/10gH2O. A value in between those two (~6) was chosen as the amount of solute to put into 10 grams of 21.5°C water. If the solid was NaNO3, the solute would completely dissolve, and the solid would be identified. If the solid was KNO3 or NaCl, some of the solute would not dissolve and would settle at the bottom. Then, to distinguish between KNO3 and NaCl, 10 grams of water would be heated to about 60°C and again about 6 grams of the solid would be added. At 60°C, the solubility of KNO3 is about 11g/100gH2O and the solubility of NaCl is about 3.8g/100gH2O. If the 6 grams dissolved, the solid would be KNO3, and if it didn't dissolve completely it would be NaCl.

Data

The procedure needed either one or two trials depending on the results of the first trial. In this case, the second trial was not necessary as the first trial delivered the results needed to identify the solid.

Conclusion

The solid was identified as NaNO3. Because of the planned procedure, only one trial was necessary to prove that the solid was indeed NaNO3. At 21.5°C, NaNO3 was the only one of the three solids that was soluble enough for 10mL of water to dissolve 6.04 grams.
When the 6.04 grams were placed into the 10mL of water, all of the solute was dissolved. Thus NaNO3 was a viable option as its solubility allows for more than 6.04 grams to be dissolved.

Had the solute been NaCl or KNO3, some of the solute would not have dissolved as they are not soluble enough. If that occurred, a second trial would have been necessary to distinguish between NaCl and KNO3.
This second trial would increase the temperature. In solids, as temperature increases, solubility increases. KNO3's solubility increases much more than NaCl's, so another test at a higher temperature with the same 6 grams of solute would have identified the solid.

Set up with 10mL water and 6.04g solute.

Lab 19: Concentration and Molarity Lab

Lab 19

Concentration and Molarity

Shiva Senthil

8/2/17

Done on paper.

Lab 18: Alka Seltzer and the Ideal Gas Law Lab

Lab 18

Alka Seltzer and the Ideal Gas Law

Shiva Senthil

8/1/17

Introduction

The purpose of this lab was to calculate how much gas was produced in a chemical reaction. Alka seltzer powder was mixed with water to produce carbon dioxide, which was collected in a balloon.

Data

Calculations

Questions
1. Discuss an area in this lab where experimental error may have occurred.
Errors could have occurred in measuring the alka seltzer's mass, the measuring of the circumference of the balloon, and filling the balloon to the same circumference with water.

2. Choose one error from above and discuss if it would make "n" the number of moles of CO2 too big or too small.
While measuring the mass of the alka seltzer, the whole alka seltzer powder was measured. It was nearly impossible to scoop all of the alka seltzer into the balloon. Since there would be less alka seltzer than you would expect, the moles of CO2 would be too small.

3. Using the value for the circumference of the balloon in cm, calculate the volume of the balloon mathematically.
Using 2πr for circumference and 4/3πr^3 for volume of the sphere, we can solve for the volume using the measured circumference of 31.5cm. 31.5 = 2πr → r = 5.01cm. V = 4/3πr^3 → V = 527cm^3

4. Compare your answer to #3 to the volume obtained by filling the balloon with water. Is it close? Which is more accurate and why?
The mathematically calculated value is much lower. Filling the balloon with water is probably more accurate because the shape of the balloon is not exactly a sphere. The circumference measured was shorter than it would have been if the balloon had been a perfect sphere (the measurement was made where the balloon was thinner). Thus, a mathematical calculation would be smaller than the actual value, and filling the balloon up with water until it is about the same size is a more accurate representation.

5. The ideal gas law technically applies to ideal gases. Give two differences between a real gas and an ideal gas.
Real gases actually have a volume, while ideal gases are assumed to have particles infinitesimally small. Real gases also do not have elastic collisions like ideal gases do.

6. Would the CO2 you collected in this lab be considered ideal? Why or why not?
The CO2 would not be ideal because gases will behave ideally only when the temperature is very high and the pressure is very low (this allows the particles to move without being restricted and are less likely to be affected by intermolecular forces, which can cause collisions to transfer energy).


Advanced Questions
1. Using info from the label, calculate the mass of CO2 that should be collected per tablet.

2. What is the percent yield?

3. CO2 is water soluble. The solubility around room temperature is around 90mL/100mL of water. What effect does this have on your calculated n value?
This means that not all of the CO2 went in to the balloon as some was dissolved in water. Thus the n value was smaller than expected.

Balloon with CO2 inside.

Lab 17: Calories in Food Lab

Lab 17

Calories in Food

Shiva Senthil

8/1/17

Didn't get to do this lab

:(

Lab 16: Specific Heat of a Metal

Lab 16

Specific Heat of a Metal

Shiva Senthil

7/31/17

Introduction

The purpose of this lab was to measure the specific heat of a metal and identify the metal. The lab consisted of heating up a piece of metal, then letting it cool in water and observing differences in temperature and using specific heat of water to solve for the specific heat of the metal.

Data

Calculations

The measurements needed to calculate the specific heat are q, the amount of energy transferred; m, the mass of the metal; and ΔT, the change in temperature of the metal. To find q, we must first find the energy the water gained.

With the data from the lab, we can calculate m and ΔT for the water. C is constant for water, and is defined at 4.18J/g°C. We then use the formula:

We can plug in the values on the right side, as they have all been found:

Solving for q:

We get that 1800 joules transferred from the hot metal to the water.

Thus, we use -1800 joules as q for the metal. The mass and change in temperature can be calculated from the data.

To solve for c, the specific heat, we use this formula derived from the previous formula:

Then, we plug in the values:

Solving for c, then converting to kilojoules:

We get the specific heat of the metal is 356J/kg°C.

According to the given values, this is closest to brass, which has a specific heat of 380J/kg°C.

Thus the metal was identified as brass.

Errors

There were many opportunities for error in this lab. The thermometer was only accurate to the nearest degree, so the number of significant figures was only three. Also, the environment was not completely closed, as the hole for the thermometer allows for some vapor to escape. Many of the metals had similar specific heats, so small errors could have changed the identity of the metal.

Lab 15: Evaporation and Intermolecular Attractions Lab

Lab 15

Evaporation and Intermolecular Attractions Lab

Shiva Senthil

7/30/17

Introduction

The purpose of this lab was to observe how intermolecular attractions affect the evaporation of different substances. The lab consisted of placing temperature probes in a substance and then pulling it out and observing how the temperature changed over a 240 second period.

Data

Table with Lewis structure, molar mass, intermolecular forces

Table with temperature readings 240 seconds apart

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

The two factors that affected the difference in temperature were the strength of the intermolecular forces and the molar mass of the substance; however, the intermolecular force was the more important factor. Although all five tested substances had hydrogen bonding as one of their intermolecular forces, the strengths of the hydrogen bonding varied. The stronger the hydrogen bonding, the less likely the substance evaporated, which meant the temperature went up or only went down a little bit. This is because the substance was more likely to absorb energy from the surrounding air rather than evaporate and release the energy as the bonds were so strong. Glycerin, which had the strongest hydrogen bonds, was the only substance that increased in temperature. Methanol, ethanol, and n-Butanol, which had the weakest hydrogen bonds, saw significant decreases in temperature. The other factor, which only applied if the hydrogen bonds were the same strength, was molar mass. A higher molar mass meant that more energy was needed to cause the substance to evaporate. Of the three substances with the weakest molar masses, n-Butanol had the highest molar mass and thus had the smallest temperature decrease. Methanol had the lowest and saw the largest temperature decrease.

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

If two compounds had similar molar masses, the compound with the greater intermolecular forces would evaporate less because more energy would be needed to break the forces. Glycerin and n-Butanol had somewhat similar molar masses, but glycerin had stronger hydrogen bonds. Thus, glycerin evaporated less, and absorbed more energy, with an increase in temperature. N-Butanol evaporated more because its bonds were easier to break, resulting in a significant drop in temperature.

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.

A greater number of OH groups means the hydrogen bonds are stronger. There is more opportunity for hydrogen bonding to occur as there are more very positive hydrogen atoms and very negative oxygen atoms. Glycerin had most OH groups (three) and thus the strongest hydrogen bonds. As hydrogen bonds are the greatest intermolecular force, glycerin had the greatest intermolecular forces overall. Glycerin was the only substance whose temperature actually increased. Water had two OH groups, and had the smallest decrease in temperature. Methanol, ethanol, and n-Butanol all had only one OH group. These substances had temperature changes correlating to their molar masses.

Lab 14: Ester Synthesis Lab

Lab 14

Ester Synthesis Lab

Shiva Senthil

7/27/17

Introduction

The purpose of this lab was to observe how molecules change during chemical reactions. The lab consisted of synthesizing ester from different acids and alcohols and adding heat. The mixtures were smelled before and after heating and were compared.

Data

The smells created in the lab

Questions

1. Compare the odors of the three mixtures before and after heating.

The mixtures all smell foul before heating, but they smell sweet after heating.

2. Based on the smell of the mixtures after heating, what functional group must be present in the final molecules that were produced? Draw it.

The functional group is ester as the smells were sweet.

3. Were the new compounds easily identified as a specific fragrance like apple or banana? In the case where a specific fragrance was detected, how does the odor compare to the natural fragrance?

All three scents were hard to identify, but the easiest test tube to identify was test tube M, which was identified as mint. Compared to their natural fragrances, these synthetic compounds smelled much more artificial, as you would see in an artificially flavored candy.

Challenge. Draw the structural formula of water and isopentyl acetate.

The structural formula of water and isopentyl acetate. The ester group is seen in the middle.