Lab Report 18 - Calorimetry and Combustion \"Burning Food. Where are my Carbs?\" PDF

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Calorimetry and Combustion “Burning Food. Where are my carbs?” 1.0

Introduction I this lab, we primarily calculated the calorimeter constant of glucose, by finding the q of combustion of glucose. Then, we used this to calculate the energy content of various snacks in Cal/oz. Lastly, we ranked these snacks from lowest energy content to highest energy content.

2.0 Materials -

Lab manual

-

Calculator

3.0 Observation and Experimental Part One: C6 H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2 O(g) + Heat

kJ kJ 𝐤𝐉 kJ ) + 6 (−393.5 ) − 6(O) − (−1275 ) = −𝟐𝟓𝟑𝟖 mol mol mol 𝐦𝐨𝐥 1 mole nC6 H12O6 = 3.00g ∗ = 0.0167 mol of C6 H12O6 180.156 g

∆HRXN = 6 (−242

∆T = 29.20 ℃ − 24.90 ℃ = 4.3 ℃

qcombustion = 0.0167 mol ∗ −2538

kJ

mol qcalorimeter = 42.26 kJ

Ccal =

= −42.26 kJ

kJ 1000 J 𝐉 42.26 kJ = 9.83 ∗ = 𝟗𝟖𝟐𝟖. 𝟔𝟗 4.3 ℃ ℃ 1 kJ ℃

Part Two: Sample Calculations: (Oreo Cookies) ∆T = 30.92 ℃ − 26.12 ℃ = 4.8 ℃ qcalorimeter = 9.83

kJ ∗ 4.8 ℃ = 47.184 kJ ℃

qcombustion = −47.184 kJ Experimental Engery Content =

−(−47.184 kJ) 28.3495 g 1 Cal Cal ∗ ∗ = 132.78 2.41 g 1 oz 4.18 kJ oz

Theoretical Energy Content =

160 Cal 28.3495 g Cal ∗ = 133.41 34 g 1 oz oz

% 𝐸𝑟𝑟𝑜𝑟 =

|132.78 − 133.41| ∗ 100 % = 0.472 % 133.41

Data: Food Sample

mass (g)

Ti (℃)

Tf (℃)

Oreo Cookies

2.41

26.12

30.92

Microwave popcorn

2.29

25.76

Crunchy Cheetos

2.21

Beef Jerky Oil-roasted, salted peanuts

∆T (℃)

qcal (kJ)

qcomb (kJ)

4.8

47.184

47.184

29.95

4.19

41.188

41.188

25.13

30.49

5.36

52.689

52.689

2.23

25.89

29.54

3.65

35.88

-35.88

2.45

25.64

31.79

6.15

60.455

60.455

Energy Experimental Theroetical % Content on (Cal/oz) (Cal/oz) Error label 160 Cal/serving 132.78 133.41 0.472 (34 g) 130 Cal/serving 121.98 122.85 0.708 (30 g) 160 Cal/serving (1 161.69 160 1.06 oz) 80 Cal/serving (1 oz) 109.12 80 36.4 170 Cal/serving (1 oz) 167.35 170 1.56

Ranking: (Low to High energy content) 1. Beef Jerky 2. Microwave Popcorn 3. Oreo Cookies 4. Crunchy Cheetos 5. Oil-roasted, salted peanuts

•

Explain why the snack with the highest energy content has more calories than the one with the lowest energy content o The snack with the highest energy content has more calories because it contains more glucose and fats, which add up to more calories.

4.0 Discussion and Conclusion In the first part of this lab, we found that the calorimeter constant of glucose is 9.83 kJ/mol. Then we found that the energy content of Oreo cookies is 132.78 Cal/oz, of Microwave

popcorn is 121.98 Cal/oz, of Crunchy Cheetos is 161.69 Cal/oz, of Beef Jerky is 109.12 Cal/oz, and of Oil-roasted, salted peanuts is 167.35 Cal/oz. Sources of error in this lab could occur with the many calculations. 5.0 Focus Questions 1. How much energy is released when food burns in the body? It depends on the number of calories in the food. The general amount of energy released per calories is 4.18 J. 2. How is the caloric content of food determined? Caloric content of food can be determined using a bomb calorimeter. In a bomb calorimeter you can find out how much energy is released by the combustion of the food, by measuring the temperature change of the water in this type of calorimeter. 6.0 References Smeureanu, Gabriela. “Experiment 18 Calorimetry and Combustion ‘Burning Food. Where are my carbs?’” Chemistry 106 General Chemistry Laboratory, edited by Stephanie Geggier, Custom Publishing, Inc., 2019, pp. 139-145. 7.0 Post Lab Questions 1. Which food sample had the highest number of calories per gram? Are you surprised by these findings? The peanuts had the highest number of calories per gram, which was 6 Cal/gram. This is not surprising as we found that they have the highest energy content, and the largest amount of glucose and fat in them. 2. Which of the foods above stores the most energy? The peanuts also store the most energy because they have the most calories. 3. Why is it important to calculate calories per gram rather than calories burned? Calories per gram is a more accurate and universal unit that allows us to compare all foods with each other. 4. 3.550 g of sucrose goes through combustion in a bomb calorimeter. If the temperature rose from 23.42°C to 27.64°C and the calorimeter constant was 4.90 kJ/°C, determine the heat of combustion of sucrose in kJ/mol. ∆T = 27.64 ℃ − 23.42 ℃ = 4.22 ℃

q calorimeter = 4.90

kJ ∗ 4.22 ℃ = 20.678 kJ ℃

qcombustion = −20.678 kJ Sucrose: C12H22O11

Molar Mass = 12(12.01) + 22(1.008) + 11(16) = 342.296 mol 3.550 g ∗

g

1 mole = 0.01037 mol sucrose 342.296 g

heat of combustion =

−20.678 kJ 𝐤𝐉 = −𝟏𝟗𝟗𝟑. 𝟖 0.01037 mol 𝐦𝐨𝐥...