Experiment 17 and 18 - note: not everything is right but I know I did get higher than 80/100 on all PDF

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note: not everything is right but I know I did get higher than 80/100 on all my labs! My grade ranged from 85-99 for all my labs, keep that in mind when referencing. ...

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Cindy Huynh Matthew Depouli November 23rd, 2020

Experiment 17: S tealing Joules ~ Introduction: Thermodynamic quantities, also known as entropy, enthalpy, and free energy are an important factor in chemistry. They are used to determine the energy of heat flows, chemicals, and fuels in a comparable way. In this experiment, we are going to test the relationships between the thermodynamics quantities using a calorimeter (insulated container that prevents heat from entering or leaving). Using a calorimeter helps us control the conditions required (standard conditions) to accurately conclude our data. This will help us understand how surroundings obtain heat and how the system distributes heat (release and absorb) in a lab setting.

~ Materials: ● Lab Manual: follow procedures given ● Microsoft excel: to create graphs ● 2 styrofoam cups: needed to make calorimeter ● Lid: to close cup so heat doesn’t escape ● Thermometer: to measure temperature ● 50.0mL of 2.0M HCl: acid ● 50.0Ml of 2.0M NaOH: base ● Timer: used to track time (30 seconds) ● Electronic balance: used to measure mass ● Graduated cylinder: used to measure and store solutions ● 50.0mL of cold water: initial solution in calorimeter ● 200mL of cold water: boiling it ● 400mL beaker: holds water ● Heating plate: heat up water ● Unknown metal: finding heat capacity it ● 50.0mL of deionized water: initial solution in calorimeter ● 2g NH4Cl ● Heat capacity table: to solve for q (table at the end of this lab section) *calorimeter = styrofoam cups + lid + thermometer + stirrer + first solution.

~ Observation and Experimentals: Part 1: Heat of Neutralization

Time in seconds

Temp ( ° C )

0

24.8

30

37.2

60

37.1

90

37

120

36.9

150

36.8

180

36.7

210

36.6

240

36.5

270

36.4

300

36.3

330

36.2

360

36.1

Figure 1. This graph illustrates the relationship between the heat of neutralization and the time in seconds.

mL of 2M HCl

mL of 2M NaOH

G of 2M HCl

G of solution

G of 2M NaOH

50

50

53.2

106.64

53.44

Calculations: ▵T = y-intercept - initial temperature ▵T = 33.866 - 24.8 T = 9.066 q = m C▵T q = 106.64 x 3.89 x 9.066 q = 3760.845154 ≈ 3760.85 Nrxn = volume (L) x concentration (M) Nrxn = 0.1L x 1M = 0.1 q rxn (J )

△H = n rxn (mole)

T (°C)

9.066

nrxn

0.1

H neutralization (J/mol)

37608.5

△H =

3760.85 0.1

H = 37608.5 J/mol | 37.6085 kJ/mol

Part 2: Specific Heat Capacity Time in seconds

Temp ( ° C )

0

20

30

26.5

60

27.6

90

27.6

120

27.6

150

27.5

180

27.4

210

27.4

240

27.40

270

27.3

300

27.3

330

27.3

360

27.3

mass - water + calorimeter (g)

58.83

390

27.2

mass water (g)

52.6

420

27.2

450

27.2

480

27.1

q water, J

1345.12

510

27.1

mass metal (g)

57.52

540

27.1

mass metal (g)

-73.888

570

27.1

C metal (J/g

600

27

Calculations:

Figure 2. This graph illustrates the relationship between the temperature and time in seconds of a certain unknown metal.

6.112

T (°C) Specific heat of water (J/g

)

)

4.184

0.316

▵T = y-intercept - initial temperature ▵T = 26.112 - 20 T = 6.112 q = m C▵T q = 52.6 x 4.184 x 6.112 q = 1345.119181 ≈ 1345.12 qwater = -qmetal 1345.12 = -1345.12 ▵T = y-intercept - initial temperature (after placing in boiling water) ▵T = 26.112 - 100 T = -73.888 q = m C▵T - 1345.12 = 57.52 x C x (-73.888) - 1345.12 = - 4250.03776 x C Divide… C = 0.3164960116 ≈ 0.316

Part 3: Molar Heat of Solution of a Salt Time in Seconds

Temp ( )

0

23.3

30

18.8

60

20.2

90

20.4

120

20.5

150

20.5

180

20.5

210

20.6

240

20.6

270

20.6

300

20.7

Figure 3. This graph illustrates the relationship between the temperature of salt and the time.

330

20.7

360

20.7

mass NH4Cl g

2.00

390

20.7

n NH4Cl

0.037

420

20.8

mass in g (water)

54.11

450

20.8

Mass in g (solution)

56.55

480

20.8

T initial (°C)

23.3

510

20.8

y -intercept (°C)

20.603

540

20.8

ΔT (°C)

- 2.697

570

20.8

600

20.9

ΔH solution kJ/mol

16.330552

Calculations: molar mass of NH4Cl = 53.491 g/mol n (moles) = mass / molar mass n (moles) = 2 / 53.491 n (moles) = 0.0373894674 ≈ 0.037 ▵T = y-intercept - initial temperature ▵T = 20.603 - 23.3 T = - 2.697 q = m C▵T q = 54.11 x 4.184 x (-2.697) q = - 610.59065928 ≈ - 610.59 -qsystem = qsurroundings - 610.59 = 610.59 △H =

q rxn (J ) nrxn (mole)

△H =

610.59 0.0037

H ≈ 16330.552 J/mol | 16.330552 kJ/mol

~ Discussion and Conclusion: In this experiment, we were able to see how a calorimeter is used to find thermal energy transfer. We were able to calculate the heat of neutralization using

the table provided to create a time vs. temperature graph in part 1 and in part 2 we were given an unknown metal and were asked to calculate the specific heat capacity as well as determining what the unknown metal was (copper). Finally, in part 3, we calculated the molar heat of the salt in the calorimeter. Overall, we were able to copy our own isolated system and used it to find values asked for like enthalpy etc. There was room for error if we were to do the lab in person as well as calculation error that might have arisen.

~ References: ● Smeureanu, G. & Geggier, S. (2019). General Chemistry Laboratory .  New York, NY Zumdahl, S. (2014). Chemistry. 9 th e d. Belmont, California: Cengage Learning ● Smeureanu, Gabriela. “Lab 17 Part 1.” YouTube , 21 Mar. 2020, www.youtube.com/watch?v=oJWyV-wPNqQ. Accessed 23 Nov. 2020. ● Smeureanu,  Gabriela. “Lab 17 Part 2.” YouTube , 21 Mar. 2020, www.youtube.com/watch?v=qJBqIKpFDAI. Accessed 23 Nov. 2020. ● Smeureanu, Gabriela. “Lab 17 Part 3.” YouTube , 21 Mar. 2020, www.youtube.com/watch?v=FfcLIDwxSD4. Accessed 23 Nov. 2020. ● Smeureanu,  Gabriela. “Lab 17 Part 4.” YouTube , 21 Mar. 2020, www.youtube.com/watch?v=6qGmEunLvlY. Accessed 23 Nov. 2020. ● “Chapter 9.5: Enthalpies of Solution - Chemistry LibreTexts.” Libretexts.org , 22 Nov. 2014, chem.libretexts.org/Courses/Howard_University/General_Chemistry%3A_An_Atoms_Fi rst_Approach/Unit_4%3A__Thermochemistry/09%3A_Thermochemistry/Chapter_9.05 %3A_Enthalpies_of_Solution. Accessed 4 Dec. 2020.

~ Focus Questions: 1. What is the heat of neutralization for a strong acid/ strong base reaction? The heat of neutralization for a strong acid/ strong base reaction is the heat needed or heat released when neutralization takes place. For part 1, our heat of neutralization is 37.6085 kJ/mol (37608.5 J/mol). 2. What is the identity of your unknown metal? The identity of the unknown metal is copper as it has the closest specific heat capacity that I calculated. Calculated heat capacity: 0.316 From chart: copper, 0.385. 3. What is the molar heat of solution of ammonium chloride salt? The molar heat of solution of ammonium chloride salt is 16.33 kJ/mol (16330.552 J/mol).

~ Post Lab Questions:

1. Compare your results for the molar heat of your salt (kJ/mol) to the literature value. What is your percent error? Try to give three reasons why your value (kJ/mol) may be different from the tabulated values. My results for the molar heat of the salt is 16.330552 kJ/mol compared to the literature value of 14.8 kJ/mol. There is a percent error of approximately 10.34%. Some reasons as to why my value differs from the tabulated values could be because of a calculation error, maybe there was a loss of heat/energy into the surroundings if someone did not close the lid fast enough or did not create the calorimeter properly, or any sort of miscalculation including rounding and plugging wrong numbers into the formula. 2. In 1819, french physicists Dulong and Petit found experimentally that for many solids at room temperature, c ≈  3R (R = gas constant in SI units). Show the Dulong-Petit’s rule is in agreement with the specific heat capacities of metals in the table. Explain why the specific heat capacity of magnesium is roughly twice as large as the one of titanium, and why the specific heat capacities of lead and gold are nearly identical. The specific heat capacity of magnesium is roughly twice as large as titanium but close to lead and gold because of their size. Aluminum and gold have similar masses so they have similar specific heat capacities. 3. Write the net ionic equation for the reaction between hydrochloric acid and aqueous sodium hydroxide reaction. How would the temperature change and the calculated heat of neutralization vary if the concentration of the acid and the base used in this experiment will double? Explain HCl + NaOH → NaCl + H2O If you double the concentration of acid and base, the temperature would double but the specific heat capacity would stay the same. 4. In a coffee cup calorimeter, 2.80g of NH4NO3 is mixed with 83.0g of water at an initial temperature of 22.57°C. After dissolution of the salt, the final temperature of the calorimeter contents is 20.36°C. assuming the solution has a heat capacity of 4.184 J/g x °C and assuming no heat loss to the calorimeter, calculate the enthalpy change for the dissolution of NH4NO3 in units of kJ/mol. Molar mass: 80.043 Specific Heat Capacity: 4.184 J/g T = 20.36 - 22.57 = - 2.21 qrxn = - qsolution and q = m C T - (83 x 4.184 x ( -2.21)) = 767.47 J Moles of NH4NO3= 2.08 / 80.05 = 0.026 moles ΔH = (767.47 / 0.026) ≈ 29533.94 J/mol | 29.533 kJ/mol

Specific Heat Capacity Table: Substance

Specific Heat Capacity (J/g ° C)

Water at 20 ° C

4.184

1M NaCl (aq)

3.89

Magnesium

1.023

Aluminum

0.897

Titanium

0.524

Iron

0.449

Copper

0.385

Silver

0.235

Lead

0.130

Gold

0.129

Cindy Huynh Matthew Depouli

November 23rd, 2020

Experiment 18: B  urning Food. Where are my carbs? ~ Introduction: Eating is essential for us to survive and is our fuel to perform various tasks throughout the day. The chemical energy from food is converted to different forms of energy like heat to maintain our constant body temperature along with mechanical energy and electrical energy. The total amount of energy that is released from this is called the calorie content and is expressed in calories. The total amount of energy that is released is called the calorie content and is expressed in calories. In this experiment, we will be using the calorimeter to determine the actual release of energy from food and identifying any patterns that may arise.

~ Materials: ● Lab manual: has lab procedures to follow ● Calculator: calculate values asked

~ Observation and Experimentals: Part 1: Determination of the Calorimeter Constant C6H12O6 (s) + 6O2(g) → 6CO2(g) + 6H2O (g) + heat Glucose (Reactant) = -1,275 CO2 = -393.5 H2O (Products) = -242 Moles of glucose 3g / 180.156 = 0.01665 mole of glucose ΔH °rxn = Σnp ΔH °f (products) − Σnr Δ°f (reactants) ΔH °rxn = [ 6 (-393.5) + 6 (-242) ] - [ 6 (0) + (-1275) ] = -2,538 kJ/mol ΔH °rxn = 0.01665 x -2,538 = - 42.26 kJ q = -q → 42.3846 △T = final - initial → 29.2 - 24.9 = 4.3℃ c (cal constant) = 42.26 / 4.3 ≈ 9.83 kJ/

Part 2: Energy Content of Various Snacks Food Sample

Mass (g)

Initial T( )

Final T ( )

ΔT ( )

Energy content on label (per serving)

q cal kJ

q comb kJ

Energy content kJ/g

Energy content cal/oz

Label cal/ oz

% error

Oreo Cookies

2.41

26.12

30.92

4.8

160 cal (34g)

47.184

- 47.184

19.578

132.739

133.41

- 0.50

Popcorn

2.29

25.76

29.95

4.19

130 cal (30g)

41.188

- 41.188

17.986

121.945

122.85

- 0.74

Cheetos

2.21

25.13

30.49

5.36

160 cal (1 oz)

52.689

- 52.689

23.841

161.642

160

1.03

Beef Jerky

2.23

25.89

29.54

3.65

80 cal (1 oz)

35.880

- 35.880

16.090

109.090

80

36.36

Oil Roasted Salted Peanuts

2.45

25.64

31.79

6.15

170 cal (1 oz)

60.455

- 60.455

24.671

167.269

170

- 1.59

Formulas: 1 ounce = 28.35 grams 1 kJ/g = 6.78 kilocalories/oz △T = final - initial q cal = C calΔT q comb = − q cal q cal e nergy content kJ g = g e nergy content cal oz l abel cal oz Percent error =

calculated energy content (cal/oz ) − label energy content (cal/oz ) label energy content (cal/oz )

x 100

Example calculation for oreo cookies: q cal = 9.83 x 4.8 = 47.184 → to find q comb = − q cal → - 47.184 = 47.184 q 47.184 = cal e nergy content kJ g → 2.41 ≈ 19.578 g e nergy content cal oz = 19.578 x 6.78 ≈ 132.739 cal l abel oz = 160 cal (34g). convert 34g to oz which is approximately 1.20. 160/ 1.20 = 133.33 − 133.41 Percent error = 132.739 x 100 = -0.50 133.41 Ranking Snacks from lowest to highest energy content. 1. Beef jerky 2. Popcorn 3. Oreo cookie 4. Cheetos 5. Oil roasted salted peanuts Explain what the snack with the highest energy content (in cal/oz) has more calories than the one with the lowest energy content. My thoughts on why the peanut has the highest energy content would be because it contains a lot of oil while beef jerky is more of a preservative type snack meaning it has low oil content since it’s probably just protein. Plus the higher energy content is all carbohydrates while beef jerky is just dried meat.

~ Discussion and Conclusion: In this experiment, we found the calories of different snacks and compared their levels of energy. We concluded that peanuts had the highest energy constant even though it had one of the highest amounts of proteins. However, this is due to the fact that oil was added into them, causing them to have a higher energy constant. Beef jerky had the least energy constant because it only had protein which contains low amounts of calories

compared to oil. We were able to conclude this by finding the amount of heat released for the different snacks as well as any patterns that occurred. In conclusion, the calorie content is not based on the food itself but all the added fats and oils that are added onto the snack causing an increase in calories. Using this knowledge, we are now able to figure out which foods fit our body’s calorie intake.

~ References: ● Smeureanu, G. & Geggier, S. (2019). General Chemistry Laboratory .  New York, NY Zumdahl, S. (2014). Chemistry. 9 th e d. Belmont, California: Cengage Learning

~ Focus Questions: 1. How much energy is released when food burns in the body? The amount of energy released from the body when food burns depends on the food consumed since different types of food have different calorific values. If the body consumes more than the required amount of calories, there will be excess calories being stored as fat to be used later on. It also depends on the person (weight height etc). 2. How is the caloric content of food determined? The caloric content of food is determined by a bomb calorimeter.

~ Post Lab Questions: 1. Which food sample had the highest number of calories per gram? Are you surprised by these findings? The oiled roasted salted peanuts had the highest number of calories per gram. I am surprised because peanuts are usually very healthy and have high amounts of protein but I am guessing it’s because they added oil and how it was prepared made the calories skyrocket. 2. Which of the foods above stores the most energy? Again, the oil roasted salted peanuts store the most energy. 3. Why is it important to calculate calories per gram rather than calories per burned? It is important to calculate calories per gram rather than calories per burned because it gives a more accurate measure of how much energy gets released when combustion happens and will tell us which snack has the most energy instead of how much energy will be needed for the product. 4. 3.550g of sucrose goes through combustion in a bomb calorimeter. If the temperature rose from 23.42°C to 27.74℃ and the calorimeter constant was 4.70 kJ/℃, determine the heat of combustion of sucrose in kJ/mol. Q = c(cal) T → 4.7 x (27.74-23.42) = 20.304 Qcomb = -qcal = -20.3 3.55/342.4 = 0.0104

-20.204/0.0104 = -1957.96 kJ/mol...