Lab course on general chemistry 1 (2)

Experiment 25 Prelaboratory Assignment Calorimetry A 20.94g sample of a metal is heated to 99.4 oC in a hot water bath until thermal equilibrium is reached The metal sample is quickly transferred to 100.0 mL of water at 22.0oC contained in a calorimeter The thermal equilibrium temperature of the metal sample plus water mixture is 24.6oC What is the speci c heat of the metal? Express the speci cheat with the correct number of signi can’t gures To find the specific heat of the metal, let's calculate the heat absorbed by the water first: mwater is the mass of water (100 g) cwater is the specific heat of water (4.18 J/g·°C) q = m × c × ΔT qwater = 100.0 × (4.18) × (24.6 - 22.0) qwater = 836 (J) Since the heat released by the metal is equal in magnitude but opposite in sign to qwater: qmetal = - qwater qmetal = - 836 (J) Now we can substitute the values into the equation to find the specific heat of the metal: c= −q (water ) m(metal) × ΔT ¿ (metal ) c= −(−836) 20.94 ×(24.6−99.4) c ≈ -1.17 (J/g·°C) a Experimental Procedure, Part A.1 What is the procedure for heating a metal to an exact but measured temperature? Step 1: Set up a hot water bath by heating water in a container Step 2: Measure the initial temperature of the hot water bath Step 3: Weigh the metal sample to determine its mass Step 4: Immerse the metal sample in the hot water bath and stir gently Step 5: Monitor and adjust the heat source to maintain a constant temperature Step 6: Wait until the metal sample reaches thermal equilibrium with the hot water bath Step 7: Remove the metal sample from the water bath and transfer it to a separate container Step 8: Measure and record the final temperature of the metal sample using a thermometer b Experimental Procedure, Part A.1 How can bumping be avoided when heating water in a beaker? To avoid bumping when heating water in a beaker: Use a boiling chip or boiling stone Gently stir the water while heating Choose a larger beaker to allow for more space Heat the water gradually Avoid overheating and superheating Place a heat-resistant mat or wire gauze between the beaker and the heat source Experimental Procedure, Parts A.4, a When a metal at a higher temperature is transferred to water at a lower temperature, heat is inevitably lost to the calorimeter (Figure 25.4) Will this unmeasured heat loss increase or decrease the calculated value of the specific cheat of the metal? Explain See equation 25.5 When a metal at a higher temperature is transferred to water at a lower temperature, heat is indeed lost to the calorimeter This unmeasured heat loss will decrease the calculated value of the specific heat of the metal The specific heat of the metal is determined based on the heat exchanged between the metal and the water However, if there is heat loss to the calorimeter or the surroundings that is not accounted for, the calculated value of the specific heat will be lower than the actual value The unmeasured heat loss reduces the amount of heat that is transferred from the metal to the water As a result, the calculated specific heat will be lower because it is based on a smaller amount of heat transfer The heat loss acts as an additional heat sink, leading to an underestimation of the metal's specific heat To obtain a more accurate value, it is important to minimize heat loss during the experimental setup Insulating the calorimeter and ensuring efficient thermal contact between the metal and water can help reduce unmeasured heat loss b Explain why the extrapolated temperature is used to determine the maximum temperature of the mixture rather than the highest recorded temperature in the experiment See Figure 25.5 Lag time: In Figure 25.5, you can see that there is a delay between the heat transfer and the response of the thermometer This delay is known as lag time The highest recorded temperature in the experiment may not accurately represent the true maximum temperature because it might not have reached equilibrium The extrapolated temperature takes into account the lag time and provides a more accurate estimation of the maximum temperature reached by the mixture Temperature equilibrium: The extrapolated temperature considers the trend of the temperature change and estimates the temperature the mixture would have reached if there were no lag time It takes into account the approaching equilibrium between the metal sample and the water This equilibrium temperature is a more reliable indicator of the maximum temperature achieved by the mixture By using the extrapolated temperature, researchers can compensate for the lag time and obtain a more accurate estimation of the maximum temperature reached during the experiment, providing a better understanding of the heat transfer and the behavior of the system Experimental Procedure, Part B Three student chemists measured 50.0 mL of 1.00 M NaOH in separate Styrofoam coffee cup calorimeters (Part B) Brett added 50.0 mL of 1.10 M HCl to his solution of NaOH; Dale added 45.5 mL of 1.10 M HCl (equal moles) to his NaOH solution Lyndsay added 50.0 mL of 1.00 M HCl to her NaOH solution Each student recorded the temperature change and calculated the enthalpy of neutralization Identify the student who observes a temperature change that will be different from that observed by the other two chemists Explain why and how (higher or lower) the temperature will be different Based on the information provided, the student who will observe a different temperature change compared to the other two chemists is Dale Dale added 45.5 mL of 1.10 M HCl to his NaOH solution, which is equal in moles to the other students' additions However, the volume of the acid added by Dale is different from the volumes added by Brett and Lyndsay, who both added 50.0 mL of acid Since the volumes of acid and base used in the reaction are not the same for Dale, the resulting mixture will have a different concentration compared to the mixtures prepared by Brett and Lyndsay This difference in concentration will affect the heat generated during the reaction and, consequently, the observed temperature change If the concentration of the acid or base is different, the heat evolved or absorbed during the reaction will also be different In this case, since Dale used a smaller volume of acid, his mixture will have a higher concentration of acid compared to the other two students This higher concentration of acid will result in a more exothermic reaction, generating more heat As a result, the temperature change observed by Dale will be higher than that observed by Brett and Lyndsay 5 Experimental Procedure, Part C Angelina observes a temperature increase when her salt dissolves in water a The lattice energy for the salt is greater than the hydration energy for the salt Lattice energy refers to the energy required to separate the ions of a solid salt and convert them into isolated gaseous ions It is an endothermic process that requires energy input On the other hand, hydration energy refers to the energy released when the gaseous ions combine with water molecules to form hydrated ions in solution Hydration energy is an exothermic process that releases energy In this case, since Angelina observes a temperature increase when the salt dissolves in water, it suggests that the dissolution process is absorbing energy from the surroundings This indicates that the lattice energy, which represents the energy required to break the ionic lattice of the salt, is greater than the hydration energy, which represents the energy released when the salt dissolves The higher energy requirement for breaking the ionic lattice implies that more energy is absorbed during the dissolution process, leading to a temperature increase b The solubility of the salt will typically increase with temperature increases In general, as temperature increases, the solubility of most salts tends to increase This is because an increase in temperature provides more thermal energy to the solvent molecules, allowing them to move more rapidly and collide with the solute particles more frequently and with greater energy These energetic collisions can overcome the intermolecular forces holding the solute together and facilitate its dissolution However, it is worth noting that this relationship does not apply universally to all salts Some salts may exhibit different solubility behavior with temperature changes, such as decreased solubility with temperature increases Therefore, it is important to consider the specific solute and solvent involved when determining the effect of temperature on solubility A 5.00-g sample of KBr at 25.0oC dissolves in 25.0 mL of water also at 25.0 oC The nal equilibrium temperature of the resulting solution is 18.1oC What is the enthalpy of solution, ∆Hs, of KBr expressed in kilojoules per mole? See equation 25.12 m is the mass of water (25.0 mL is approximately 25.0 g) C is the specific heat capacity of water (4.18 J/g·°C) ΔT is the change in temperature (final temperature - initial temperature) q = (25.0 g) × (4.18 J/g·°C) × (18.1°C - 25.0°C) q = -585.57 J (negative sign indicates heat released) Next, we need to calculate the number of moles of KBr We can use the molar mass of KBr to convert the mass of the sample to moles: molar mass of KBr = 39.10 g/mol + 79.90 g/mol = 119.0 g/mol n = mass / molar mass n = 5.00 g / 119.0 g/mol n = 0.0420 mol Now we can substitute the values into the equation to find the enthalpy of solution: ΔHs = q / n ΔHs = (-585.57 J) / (0.0420 mol) ΔHs ≈ -1.39 × 10^4 J/mol Finally, we can convert the enthalpy of solution from joules to kilojoules by dividing by 1000: ΔH ≈ -1.39 × 10^4 J/mol / 1000 ΔH ≈ -13.9 kJ/mol Therefore, the enthalpy of solution (ΔH) of KBr is approximately -13.9 kJ/mol Experiment 25 Report Sheet Calorimetry Laboratory Questions Part A.1 The 200mm test tube also contained some water (besides the metal) that was subsequently added to thecalorimeter (in Part A.4) Considering a higher speci c heat for water, will the temperature change in the calorimeter be higher, lower, or unaffected by this technique error? Explain If the 200mm test tube contained water in addition to the metal, and some of that water was subsequently added to the calorimeter, the temperature change in the calorimeter would be higher This is because water has a higher specific heat capacity compared to the metal, meaning it can absorb and retain more heat energy The presence of water increases the overall heat capacity of the system, requiring more heat energy to produce a given temperature change in the calorimeter Part A.4 When a student chemist transferred the metal to the calorimeter, some water splashed out of the calorimeter Will this technique error result in the specific heat of the metal being reported as too high or too low? Explain The technique error of water splashing out of the calorimeter during the transfer of the metal would result in the specific heat of the metal being reported as too high This is because the decreased mass of the system would lead to a higher calculated specific heat capacity, as the heat energy is distributed over a smaller mass, resulting in a larger temperature change for the same amount of heat transferred Part B The enthalpy of neutralization for all strong acid–strong base reactions should be the same within experimental error Explain Will that also be the case for all weak acid–strong base reactions? Explain In Part B, the statement is made that the enthalpy of neutralization for all strong acid-strong base reactions should be the same within experimental error This statement is generally true for strong acid-strong base reactions due to the nature of these compounds Strong acids completely dissociate in water, releasing a high concentration of hydrogen ions (H+) into the solution Similarly, strong bases completely dissociate in water, producing a high concentration of hydroxide ions (OH-) When a strong acid reacts with a strong base, the hydrogen ions from the acid combine with the hydroxide ions from the base to form water This is known as a neutralization reaction The enthalpy of neutralization is the heat released or absorbed during a neutralization reaction In the case of strong acid-strong base reactions, the formation of water is highly exothermic, meaning it releases a significant amount of heat Since strong acids and strong bases completely dissociate, the reaction proceeds to completion, resulting in the same number of moles of water formed per mole of acid and base reacted As a result, the enthalpy of neutralization for strong acid-strong base reactions tends to be consistent within experimental error However, it's important to note that some variations in experimental conditions or specific acid-base combinations may introduce minor deviations, but these are usually within an acceptable range On the other hand, for weak acid-strong base reactions, the enthalpy of neutralization may not be the same as for strong acid-strong base reactions Weak acids not fully ionize in water, meaning that not all acid molecules donate hydrogen ions Consequently, the neutralization reaction with a strong base may not proceed to completion, and some of the acid may remain unreacted This incomplete reaction can result in a different enthalpy of neutralization compared to strong acid-strong base reactions In summary, while the enthalpy of neutralization is generally consistent for strong acid-strong base reactions, it may vary for weak acid-strong base reactions due to incomplete dissociation of weak acids Part B Heat is lost to the Styrofoam calorimeter Assuming a 6.22 oC temperature change for the reaction of HCl(aq) with NaOH(aq), calculate the heat loss to the inner 2.35-g Styrofoam cup The specific heat of Styrofoam is 1.34 J/g•oC Q is the heat energy (in joules), m is the mass of the Styrofoam cup (in grams), c is the specific heat capacity of Styrofoam (in J/g°C), ΔT is the temperature change (in °C) Given: m = 2.35 g c = 1.34 J/g°C ΔT = 6.22°C Plugging in the values, we have: Q = mcΔT Q = (2.35 g) × (1.34 J/g°C) × (6.22°C) Q ≈ 19.95 (J) Therefore, the heat loss to the inner Styrofoam cup is approximately 19.95 Joules Part B.3 Jacob carelessly added only 40.0 mL (instead of the recommended 50.0 mL) of 1.1 M HCl to the 50.0 mL of 1.0 M NaOH Explain the consequence of the error By adding only 40.0 mL of 1.1 M HCl instead of the recommended 50.0 mL to the 50.0 mL of 1.0 M NaOH, Jacob has deviated from the balanced stoichiometry of the neutralization reaction The consequence of this error is that there will be an excess of NaOH remaining in the solution after the reaction is complete This means that not all of the HCl will react with the NaOH, leading to an incomplete neutralization The resulting solution will be basic because there is excess base (NaOH) present Part B.3 The chemist used a thermometer that was miscalibrated by oC over the entire thermometer scale Will this factory error cause the reported energy of neutralization, ∆Hn, to be higher, lower, or unaffected? Explain The miscalibration of the thermometer by 2°C over the entire scale will cause the reported energy of neutralization (ΔHn) to be unaffected The reason is that the miscalibration affects the absolute temperature measurement, but the change in temperature (ΔT) is what matters for the calculation of ΔHn Since ΔT is the difference between the initial and final temperatures, the miscalibration of the thermometer will have the same effect on both the initial and final temperatures, canceling out in the calculation of ΔT Therefore, the miscalibrated thermometer will not impact the reported energy of neutralization 7 Part C.3 If some of the salt remains adhered to the weighing paper (and therefore is not transferred to the calorimeter), will the enthalpy of solution for the salt be reported too high or too low? Explain If some of the salt remains adhered to the weighing paper and is not transferred to the calorimeter during the measurement of the enthalpy of solution, the enthalpy of solution for the salt will be reported as too low This is because the weighing paper acts as an insulator, preventing heat transfer between the salt and the calorimeter As a result, the heat released or absorbed during the dissolution of the salt will not be fully captured and measured in the calorimeter, leading to an underestimated enthalpy of solution Part C The dissolution of ammonium nitrate, NH 4NO3, in water is an endothermic process Since the calorimeter is not a perfect insulator, will the enthalpy of solution, ∆Hs, for ammonium nitrate be reported as too high or too low if this heat change is ignored? Explain The dissolution of ammonium nitrate (NH4NO3) in water is an endothermic process, meaning it absorbs heat from the surroundings If this heat change is ignored due to the calorimeter not being a perfect insulator, the enthalpy of solution (ΔHs) for ammonium nitrate will be reported as too low This is because the heat absorbed during the dissolution process is lost to the surroundings through the imperfect insulation of the calorimeter As a result, the measured heat change will be lower than the actual value, leading to an underestimated enthalpy of solution Experiment 12 Prelaboratory Assignment Molar Mass of a Volatile Liquid The following data were recorded in determining the molar mass of a volatitle liquid following the Experimental Procedure for this experiment Mass of dry ask, foil, and rubber band (g) 74.722 g Temperature of boiling water (oC, K) 98.7 ° C Mass of dry ask, foil, rubber band, and vapor (g) 74.921 g Volume of 125-mL ask (L) 0.152 L Atmospheric pressure (torr, atm) 752 torr / 0.989 atm a How many moles of vapor are present? n( vapor )= 0.989 atm ×0.125 L ≈ 0 04927(mol) atm 0.08206 ∙ ×37 1.85 K mol b What is the molar mass of the vapor? M ( compound )= 0.199 g =40.39 ¿) 0.004927 mol a If the atmospheric pressure of the ask is assumed to be 760 torr in question 1, what is the reported molar mass of the vapor? Moles of Vapor= Molar Mass= atm ×0.152 L ≈ 0.004981(moles) atm 0.08206 L∙ × 371.85 K mol mass of vapor 0.199 g = =39.95(g /mol) moles of vapor 0.004981 moles b What is the percent error caused by the error in the recording of the pressure of the vapor? % error= M difference × 100 Mactual 40.39 g/mol−39 95 g /mol ¿ × 100=1.09 % 40.39 g /mol The ideal gas law equation (equation 12.1) is an equation used for analyzing ideal gases According to the kinetic molecular theory that de nes an ideal gas, no ideal gases exist in nature, only real gases Van der Waals’ equation is an attempt to make corrections to real gases that not exhibit ideal behavior Describe the type of gaseous molecules that are most susceptible to nonideal behavior The deviations from ideal behavior are more pronounced in gases that have: High pressure: At high pressures, the volume occupied by gas molecules becomes significant compared to the total volume of the gas This leads to intermolecular interactions, such as attractive forces between molecules, becoming more significant and causing deviations from ideal behavior Low temperature: At low temperatures, the kinetic energy of gas molecules decreases, reducing the tendency to overcome intermolecular forces As a result, molecules are more likely to interact with each other, leading to deviations from ideal behavior Large molecular size: Gases composed of molecules with large sizes have a greater likelihood of experiencing intermolecular interactions These interactions can result in deviations from ideal behavior due to increased molecular volume and potential attractive forces between molecules Polar molecules: Gases composed of polar molecules have an uneven distribution of electron density, leading to dipole-dipole interactions between molecules These intermolecular forces can cause nonideal behavior, particularly at high pressures and low temperatures These factors contribute to deviations from ideal behavior and require corrections in the ideal gas law Van der Waals' equation is one such attempt to account for these deviations by introducing correction terms for molecular size and intermolecular forces, providing a more accurate description of the behavior of real gases 4 a How is the pressure of the vaporized liquid determined in this experiment? A barometer is used b How is the volume of the vaporized liquid determined in this experiment? Filling a 125 mL Erlenmeyer flask with water, volume is measured by transferring water to a graduated a cylinder c How is the temperature of the vaporized liquid determined in this experiment? After the water reaches a gentle boil the temperature is recorded with a thermometer d How is the mass of the vaporized liquid determined in this experiment? Allow the gas to cool to room temperature then dry the outside of the flask a measure with a balance The molar mass of a compound is measured to be 30.7, 29.6, 31.1, and 32.0 g/mol in four trials a What is the average molar mass of the compound? To calculate the average molar mass of the compound, you need to sum up the measured values and divide by the number of trials Let's the calculation: Average molar mass = (30.7 + 29.6 + 31.1 + 32.0) / = 123.4 / = 30.85 (g/mol) Therefore, the average molar mass of the compound is 30.85 (g/mol) b Calculate the standard deviation and the relative standard deviation (as %RSD) (see Appendix B) for the determination of the molar mass The deviation of each measurement from the mean Square each deviation Deviation1 = 30.7 - 30.85 = -0.15 (g/mol) 0.0225 (g2/mol2) Deviation2 = 29.6 - 30.85 = -1.25 (g/mol) 1.5625 (g2/mol2) Deviation3 = 31.1 - 30.85 = 0.25 (g/mol) 0.0625 (g2/mol2) Deviation4 = 32.0 - 30.85 = 1.15 (g/mol) 1.3225 (g2/mol2) Sum of squared deviations = 0.0225 + 1.5625 + 0.0625 + 1.3225 = 1.97 (g2/mol2) Calculate the variance by dividing the sum of squared deviations by (n-1), where n is the number of measurements: Variance = 1.97 / (4-1) = 0.6567 (g2/mol2) Take the square root of the variance to find the standard deviation: Standard deviation = √ 0.6567 = 0.8101 (g/mol) Calculate the relative standard deviation (%RSD) by dividing the standard deviation by the mean and multiplying by 100: %RSD = (0.8101 / 30.85) * 100 = 2.63% Experiment 12 Report Sheet Trial Trial Trial A Preparing the Sample Mass of dry ask, foil, and rubber band (g) 85.2672 B Vaporize the Sample Temperature of boiling water (° C, K) 368.15 K 367.65 K 368.74 K Mass of dry ask, foil, rubber band, and vapor (g) 85.8673 86.1092 86.1258 C Determine the Volume and Pressure of the Vapor Volume of 125-mL ask (L) 0.144 0.08 L + 0.02 L + 0.014 = total volume Atmospheric pressure (torr, atm) 0.997 atm D Calculations Moles of vapor, nvapor (mol) 4.80E-03 4.80E-03 4.80E-03 Mass of vapor, mvapor (g) 6.00E-01 8.42E-01 7.36E-01 Molar mass of compound (g/mol) 125.0208* 175.4167 154.6391 Average molar mass (g/mol) 150.2188 Standard deviation of molar mass 35.635 Appemdix B Relative standard deviation of molecular mass (%RSD) 23.72 Appemdix B Molar Mass of a Volatile Liquid E Calculations (van der Waals’ equation) Trial 1* Trial Trial Moles of vapor, nvapor (mol) 0.0048 0.0048 0.0047 Mass of vapor, mvapor (g) 0.6001 0.8420 0.8586 125.0208 175.4167 182.6809 Molar mass of compound (g/mol) Average molar mass (g/mol) 161.0395 Laboratory Questions Part A.1 The mass of the ask (before the sample in placed into the ask) is measured when the outside of the ask is wet However, in Part B.3, the outside of the ask is dried before its mass is measured a Will the mass of vapor in the ask be reported as too high or too low, or will it be unaffected? Explain The mass of vapor in the ask will be reported as too high if the outside of the ask is wet during the measurement When the outside of the ask is wet, it includes the mass of the water or moisture present on the surface Therefore, when the mass of the ask is measured, it will include the mass of the water vapor, leading to an overestimation of the vapor mass b Will the molar mass of vapor in the ask be reported as too high or too low, or will it be unaffected? Explain The molar mass of the vapor in the ask will be unaffected by whether the outside of the ask is wet or dry during the measurement The molar mass is a property of the gas and is independent of any external conditions Therefore, even if the mass of the vapor is overestimated due to the wet surface of the ask, the molar mass will remain the same It is determined by the composition of the vapor and the molar masses of its constituent elements or compounds, and it does not change with variations in the measurement procedure or the presence of moisture on the outside surface of the ask