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.

Lab 13: Molecular Geometry Lab

Lab 13

Molecular Geometry Lab

Shiva Senthil

7/27/17

Lab was done on paper.

Lab 12: Electron Configuration Battleship

Lab 12

Electron Configuration Battleship

Shiva Senthil

7/26/17

Introduction

The purpose of this lab was to practice electron configuration. The lab was playing the classic game of Battleship, but instead of with coordinates on a grid, electron configurations on the periodic table were used.

Conclusion

Through this game I got more practice and gained a better understanding of electron configurations and using the periodic table to find them. The biggest challenge was saying the electron configurations of the f-block elements.

Lab 11: Flame Test Lab

Lab 11

Flame Test Lab

Shiva Senthil

7/25/17

Introduction

The purpose of this lab was to observe how different atoms release different amounts of energy during chemical reactions. Nine known metallic ion solutions and two unknown solutions were heated to make the metal ions release energy and give off light.

Prelab Questions

1. What is the difference between ground state and an excited state?

The ground state is the lowest energy that an electron can have and excited state is when the electron has more energy than the ground state.

2. What does the word "emit" mean?

It means to "give off."

3. In this experiment, where are the atoms getting their excess energy from?

The atoms get energy from the heat from the flame.

4. Why do different atoms emit different colors of light?

Different atoms emit different colors of light because they release different amounts of energy during the reaction. The different energy levels mean different colors of light are emitted.

Data

Analysis Questions
2. What patterns do you notice in the groupings?
The color depends on the metal ion in the compound.

3. Predict the flame color for a substance called strontium sulfate.
The flame color would probably be red because the flame for strontium chloride was red.

4. What evidence do you have that atoms of certain elements produce flames of certain colors?
All of the compounds with the same metal burned at about the same color (e.g. all of the copper compounds burned green/green-blue, and both sodium compounds burned orange).

5. Can a flame test be used to identify a metal atom in a compound? Why or why not? What about a nonmetal atom?
Most of the time, a flame test can be used to identify the metal atom. This is because during a chemical reaction, the ion will always release the same color of light as the electron gains energy and moves closer to the electron. This is not true for a nonmetal ion because it is pushing the electron away from its nucleus, so it absorbs light rather than emitting it. Thus, nonmetal ions do not emit visible light and cannot be identified using a flame test.

6. Identify the two unknowns. What are they and how do you know?
Unknown 1 is a lithium compound and Unknown 2 is a potassium compound. The color of the flame in the first unknown was pink, the same as LiCl. The color of the flame in the second unknown was purple, the same as KCl. The nonmetal in the compounds cannot be identified, but the metals can.

7. What color flame would CuO produce?
Copper (II) oxide would produce a green or green-blue flame because all copper compounds produce the same colored flame. The three copper compounds in the lab all burned green or greenish blue.

The bright pink flame emitted when LiCl is heated.

Lab 10: Mole-Mass Relationships Lab

Lab 10

Mole-Mass Relationship Labs

Shiva Senthil

7/24/17

Introduction

The purpose of this lab was to use mass-mass stoichiometry to predict the theoretical yield of a chemical reaction. The lab consisted of executing a chemical reaction with sodium bicarbonate and hydrochloric acid, which yields sodium chloride and water. After the reaction, the product was heated to isolate the salt from the water. The salt was then measured and compared to the theoretical yield.

Data

The masses of the sodium bicarbonate and the resulting product were calculated by subtracting the mass of the dish from the mass of the dish with the substance inside. 

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

The limiting reactant is the sodium bicarbonate. A certain amount (~2.00g) of sodium bicarbonate was measured for the lab. The other reactant, hydrochloric acid, was continually added until all of the sodium bicarbonate was consumed in the reaction. The hydrochloric acid was in excess, and the sodium bicarbonate limited the amount of resulting product.


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

This is the overall balanced equation of the reaction in this lab.

A mass to mass conversion was used to find the theoretical yield of NaCl. The limiting reactant was NaHCO3, so that was used for the calculations.

There are 1.98g of sodium bicarbonate reacting with excess hydrochloric acid. To convert to moles, the 1.98g was divided by the molar mass of the substance, which is 84.007g. Since the mole ratio of NaHCO3 to NaCl is 1:1, that conversion was simply multiplying by 1/1. To convert from moles of NaCl to grams of NaCl, the number of moles was multiplied by the molar mass of NaCl, which is 58.44g. This calculation results in 1.38g of NaCl as the theoretical yield of product.

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

This calculation is simply subtracting the mass of the evaporating dish from the combined mass of the evaporating dish and the solid product after heating.

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

This calculation is dividing the experimental yield by the theoretical yield and converting to a percent. In this case, the percent yield is exactly 100%

Errors

Although the percent yield was 100%, there was mostly likely some error. The product may not have been evaporated for enough time to evaporate all of the water, and there may have been some water left. At the same time, the heat may have been too high, causing some of the product to pop and go out of the evaporating dish.

The product after heating.

Lab 9: Composition of a Copper Sulfate Hydrate Lab

Lab 9

Composition of a Copper Sulfate Hydrate Lab

Shiva Senthil

7/21/17

Introduction

The purpose of this lab was to be able to find the empirical formula of a hydrate. The lab consisted of heating a hydrate and measuring the change in mass from hydrate to anhydrous salt to find the amount of water and the empirical formula of the hydrate.

Data

The hydrate before heating

The anhydrous salt left behind after heating

This table shows the mass of the dish, the dish with the hydrate, and the dish with the anhydrate after heating.

To find the mass of the hydrate, subtract the mass of the dish from the mass of the dish and hydrate:

To find the mass of the water in the hydrate, subtract the mass of the dish and anhydrate from the mass of the dish and hydrate:

To find the percentage of water in the hydrate, divide the mass of the water by the mass of the hydrate:

This formula calculates the percent error. The correct percentage of water in the copper sulfate hydrate is 36.0%:

 There is an error of about 4.7%. This may be because of slightly inaccurate measurements of mass, or heating the hydrate too much. The error is not too high, but is still significant. If there was no error, the unrounded coefficient of the water would be lower, as the actual percent of water is less than the measured percent.

The next step is to find the number of moles of each molecule, which is done by dividing the number of grams by the molar mass of the molecule.

of water

of copper sulfate

 Finally, the ratio between the water and the copper sulfate is found by dividing the number of water moles by the number of copper sulfate moles. Because of the +4.7% error, the coefficient of water is most likely too high. However, the water coefficient is rounded to the nearest whole number because it is impossible to have part of a molecule; there must be a whole particle:

 Using the 5:1 ratio of water to copper sulfate molecules, the empirical formula of copper sulfate hydrate can be written:

This is the empirical formula of copper sulfate hydrate. 

Lab 8: Mole Baggie Lab

Lab 8

Mole Baggie Lab

Shiva Senthil

7/21/17

Introduction

The purpose of this lab was to identify compounds given either the number of moles or the number of particles in a bag of the substance. The compounds were weighed and that and the given measurements were used to calculate the molar mass and then identify the compound in the bag.

Data

The first step of the process was to find a formula to be able to use the mass and the given measurement to determine the molar mass (g/mol). The given measurement for each bag was different, so the formulas were slightly different. The next step was to measure the mass and subtract the given mass of the bag. The last step was to actually calculate the molar mass and identify the substance from one of five given compounds (sodium chloride [molar mass 58.44 g/mol], potassium sulfate [MM 174.3 g/mol], zinc oxide [MM 81.40 g/mol], sodium sulfate [MM 142.0 g/mol], and calcium carbonate [MM 100.1 g/mol]).

Bag A2:

Since the number of moles is given and the mass is measured, to find the molar mass, all that needs to be done is divide the number of grams by the number of moles:

 Plugging in the grams and the moles results in the molar mass:

The molar mass of the substance in bag A2 is about 173 g/mol. The closest to this number is the 174.3 g/mol of potassium sulfate. Thus, the substance in bag A2 is potassium sulfate.

Bag B6:

Since the number of representative particles is given and the mass measured, the number of particles must be converted to moles before dividing the grams by the moles:

Substituting this equation in for the mole in the basic molar mass equation results in:

After simplification:

 Then, the mass and number of representative particles is substituted in and calculated:

The molar mass is about 84.8 g/mol, which is closest to zinc oxide (81.41 g/mol). Therefore, the substance in bag B6 is zinc oxide.

Lab 7: Reactions Lab

Lab 7

Reactions Lab

Shiva Senthil

7/20/17

Introduction

The purpose of this lab was to make chemical reactions and to help strengthen the understanding of the different types of chemical reactions. The lab consisted of carrying out multiple reactions of many types (synthesis, decomposition, single replacement, double replacement, combustion), and writing balanced equations for each reaction.

Results

Magnesium and oxygen react and for a bright white light.

My favorite reaction was the magnesium and oxygen reaction. A magnesium strip is held over a Bunsen burner. Magnesium (Mg) reacts with oxygen gas (O2) to form magnesium oxide (MgO), emitting a bright white light in the process. The result, after the light fades, is a white, powdery solid.

Equations for each of the eight reactions.