Morgan Gallagher

CHEM 101H-202

September 30, 2000

 

- Exp. 5 - 

Determination of the 

Molar Volume of a Gas and 

Atomic Weight of a Metal

 
    Unknown Number:____

Abstract

        The experiment will explore Hess's Law, specific heat, heat capacity and other thermodynamic principles. A calorimeter will be calibrated. The specific heat and atomic weight of a metal will be calculated. The heat of solution will be determined and the heat of reaction using magnesium, magnesium oxide and hydrochloric acid. The experiment will be considered a success if the atomic weight of the metal, heat of solution, and enthalpy of reaction are within 10% of the accepted values.

 

Experimental

Stoichiometric Equations:

        Mg(s) + 2 H⁺(aq)  ^---> Mg²⁺(aq) + H₂(g)          DH_{rxn 1}
        H₂(g) + ½O₂(g)  ^---> H₂O(l)                           DH_{rxn 2}
        Mg²⁺(aq) + H₂O(l)  ^---> MgO(s) + 2H⁺(aq)     DH_{rxn 3}
        Mg(s) + ½O₂(g) ^---> MgO(s)                         DH_f  = DH_{rxn 1} + DH_{rxn 2} + DH_{rxn 3}

Mathematical Formulas:

        DH = H_final - H_inital
        q = m_*s_*DT
        C = m_*s
        -q_metal = q_water +q_calorimeter = q_total
        -m_m*s_m*DT = m_w*s_{w *}DT + m_c*s_c*DT = q_total
        DH_solution = U + DH_h = q_(water) + q_(salt) + q_{(calorimeter)}
        DH_sol = m_w*s_w*DT +C_c*DT
        DH^o =S_{(bonds in reactants)} -S_{(bonds in products)}
        DH_{f (MgO)} = DH_{rxn 1} +DH _{rxn 2} - DH _rxn = DH_{rxn 1} + DH_{rxn 2} + DH_{rxn 3}
        DT_c = T_max - T_i
        DT_metal = T_max - T_hot
        

Reagents:

data from Hazardous Chemical Database, The University of Akron,
Department of Chemistry http://ull.chemistry.uakron.edu/erd/

Materials:

        -2 Styrofoam cups                                -lid
        -stir bar                                                -2 thermometer
        -100 mL beaker                                   -400 mL beaker
        -magnetic stirrer                                   -ring stand
        -ring                                                    -Bunsen burner
        -test tube                                             -stopper
        -sand paper

Experimental Apparatus

 

Procedure:

Part A:

1. The calorimeter was constructed by placing one Styrofoam cup inside the other.
2. The stir bar and thermometer-stopper assembly were inserted through the lid and placed on the top cup.
3. The assembly was weighed and 40 mL of water added.
4. The temperature was observed and the assembly was reweighed.
5. The assembly was placed inside a 400 mL beaker atop a magnetic stirrer, held by an iron ring.
6. The thermometer was clamped so that the thermometer was suspended just above the bottom of the calorimeter.
7. A 100 mL beaker was weighed, 40 mL of water added and reweighed.
8. The second thermometer was placed inside it and the beaker was heated.
9. With constant and gentle stirring, the temperature of the water in the calorimeter was recorded every 30 seconds for three minutes.
10. At the end of the third minute the temperature of the hot water was recorded and it was transferred into the calorimeter.
11. Stirring continued and at the fourth minute started recording the temperature again every 30 seconds for an additional 6 to 8 minutes.(T increased and then decreased)
12. The calorimeter was emptied, cleaned and dried.
13. The calorimeter was weighed and 40 mL of water added and reweighed.
14. The temperature of the water was recorded.
15. 6 to 10 g of an unknown metal was obtained.
16. 5 to 8 g of the metal was weighed in a tared test tube.
17. A stopper was placed in the test tube and it was put into the hot water (the metal should be below the water level, but no water should get into the test tube-allow to heat for 15 minutes).
18. The temperature of the calorimeter was recorded every 30 seconds for three minutes.
19. At the end of the third minute the temperature of the hot water bath was recorded and the test tube was removed, quickly dried and the metal pieces were placed in the calorimeter.
20. The calorimeter was stirred gently and its temperature record from the fourth minute until the 10th or 12th minute every thirty seconds.
21. Repeat the experiment once if time permits.

Part B:

1. The calorimeter was emptied, dried and weighed.
2. 30.0 mL of water was added and it was reweighed.
3. The lid, thermometer, and stirrer were put on.
4. The assembly was allowed to sit for 4 minutes for thermal equilibrium to be obtained.
5. The temperature was recorded every 30 seconds for three minutes.
6. A solid salt was obtained from the instructor and 5 g of it was weighed and transferred into the calorimeter.
7. The temperature was recorded every 30 seconds for 10 more minutes.
8. If time permits perform a second run.

Part C:

1. The calorimeter was emptied, dried and reweighed.
2. 25.00 mL of 1.00 M HCl was added and it was reweighed.
3. The lid, thermometer and stirrer were put back on and the assembly was placed on a sand bath and allowed to stand for 4 minutes.
4. The temperature was recorded and then again every 30 seconds for the next 3 minutes.

Reaction of Magnesium

5. 100 mg of sanded magnesium ribbon was weighed, coiled and added to the acid solution.
6. The lid was immediately replaced and the magnesium reacted with the acid to form Mg²⁺ & H₂ gas.
7. The mixture was continually swirled as the temperature was recorded every 30 seconds for at least 12 minutes.
8. The calorimeter was emptied, rinsed, and dried and a second run performed if time permitted.

Reaction of Magnesium Oxide

9. The calorimeter was weighed, 25.00 mL of 1.00 M HCl was added to it and it was reweighed.
10. After 4 minutes, the temperature of the calorimeter was recorded every 30 seconds for 3 minutes.
11. 200 mg of MgO was weighed on weighing paper, and transferred into the solution.
12. The lid was immediately replaced and made sure that all the MgO was in solution.
13. While the solution was stirred, the temperature was recorded every 30 seconds until it began to fall; the temperature was recorded for another 12 minutes.
14. The weighing paper was reweighed and the mass of the MgO calculated.
15. A second run was performed if time permitted.

Results

Reduced Data Table:

Omitted in this example report.  Data section will consist of a table of reduced data with proper units, labels, and significant figures.

Example Calculations:

Omitted in this example report.  Calculations will consist of ONE example taken from each set of different calculations.

Graphs:

None applicable in this experiment

Discussion

        Specific heat is the amount of heat it takes to raise one gram of a substance by one Kelvin. The specific heat of metal is much less than that of water since it is not able to store heat as well. As it takes much less heat to raise one gram of metal (Dulong and Petit in 1819 proposed that any metal will absorb 25 J mol^-1 K^-1) by one Kelvin than it does to raise one gram of water (75.9 J mol^-1 K^-1 -- value calculated by multiplying the value given in the text on page 128 for 288 K by the MW of water) metals will conduct heat sooner and not store as much heat. Since the metal's heat capacity is less than that of water the metal cannot transfer as much heat to the cold water in the calorimeter and consequently to the calorimeter, or closed system.

Heat of solution is the heat generated or absorbed when an amount of solute is dissolved in an amount of solvent. The heat of solution of the unknown salt 509 should be positive because positive enthalpy corresponds to an endothermic reaction. An endothermic reaction takes heat from the surroundings. Thus, when the salt was added to the water in the calorimeter the temperature decreased, since heat was being taken from the water. The calorimeter is a closed system; the Styrofoam insulated the reaction, preventing it from taking heat from the surroundings. Because of this we are able to calculate how much change in heat occurs since there is no unaccounted source for it. For exothermic reactions enthalpy is negative, corresponding to the reaction giving off heat. Temperature of the water would have increased if this had been the case.

        Hess's Law states that a reaction enthalpy is the sum of the enthalpies of any sequence of reactions (at the same temperature and conditions) into which the overall reaction can be divided. By manipulating this law the enthalpy of formation of magnesium oxide can be calculated using the experimentally determined reaction enthalpy for magnesium and magnesium oxide.  The amount of magnesium and magnesium oxide added in the experiments to determine their respective reaction enthalpies was kept small to simplify the calculations.  The heat of solution is comprised of three components in these experiments; the heat gained or lost by the water, the  calorimeter, and the metal or salt.  The small additions simplified the equations by allowing us to neglect the heat lost or gained by the salt.  As a result, as in the previous parts of the experiment, the heat gained or lost by the components of the sytem under experimental control could be found by using the general equation:

q = (m)(s)(DT)                 (1)

Where q is the heat gained or lost, m is the measured mass , s is the specific heat, and DT is the measured change in temperature of the component.  By applying the Law of Conservation of Energy the sum of the known terms in the system where used to determine the value of the unknown term.  In the first part of the experiment the unkown term was the mass times the specific heat of all the parts of the calorimeter, the closed system, except the water.  The known terms were the heat (energy) gained by the water in the calorimeter, and the heat lost by the hot water added to the system.
        In the second part of the experiment, determination of the specific heat and atomic weight of a metal, the known terms were the heat gained by the water in the calorimeter and that gained by the rest of the colrimeter.  By determining the value of the remaining term, the heat lost by the metal, as described, the value for the specific of the metal could be determined using equation 1.  Known values for equation 1 were the mass of the metal and the change in temperature.  Knowledge of the specific heat and the use of Delong and Petit's proposal allowed the determintaion of the AW of the metal.
        The third part, heat of solution of a salt, of the experiment was performed in a similar manner substituting a salt for the metal and again using the above described procedure of summing known experimentally determined values to find value of the unknown term as that left over from the total energy in the closed system.  The general relationship of heat of solution to heat of reaction allowed us to find the value for the one unknown term in our Hess's Law equation for the net reaction Mg_(s) + 1/2 O_2(g) ^---> MgO_(s).  The individual reactions we used to sum for this net reaction were those for the ionization of metallic magnesium to divalent magnesium, the formation of water, and the subsequent reaction of water and Mg²⁺  to form MgO and aqueous protons.  the value for the formation of water was taken from the literature.  Good advantage was made of equation 1 and the Law of Conservation of Energy in all parts of the experiment, limiting the unknown value of any part or subpart to one.

Conclusion

        The experimentally determined calorimeter constant was 32.918 J/K, with a %RSD of 11.745. The experimentally determined specific heat of the unknown metal was 0.300 J/(g K). Its atomic weight was 105.415 g/mole with a %error of 11.184. The metal was tin. The heat of solution was -10.499 J/g with a %RSD of 576.691. The enthalpy of formation of MgO was -286.857 kJ/mole with a %RSD of 0.028. Precision and accuracy were fair to poor.

 

Appendices

Attach notebook pages and any appropriate pages of information here.

 

Prelaboratory Questions

1. Define the following:

A. Specific heat:  the amount of heat, q, necessary to raise the temperature of 1 g of a substance by 1 K. (s)
                                                               s = q/(m)(DT)

B. Heat capacity: the amount of heat, q, required to raise the temperature of a given quantity of any given substance by 1 K. (C)

                                                                  C = (m)(s)
C. Heat of solution: the heat generated (or absorbed) when a certain quantity of  solute dissolves in a certain quantity of solvent

D. Heat of formation: (enthalpy of formation-ÊH_f) is the standard reaction enthalpy for the formation of a substance from its elements in their most stable form.

E. Molar enthalpy: the enthalpy (heat content at constant pressure) per mole of a substance.

2.  A student uses 100.0 mg of Mg and 300.0 mg of MgO in this experiment. She adds 50.0 mL of 1.50 M HCl to each sample separately. Write the balanced equations for the reactions. Calculate the number of moles of Mg and MgO and the limiting reagent for each reaction.

                                      Mg_(s) + 2HCl_(aq)t MgCl₂ + H_2(g)

                                      MgO_(s) + 2HCl_(aq) t MgCl₂ + H_2(g)

       _{(0.1 g Mg)( 1 mole Mg )  / ( 1 )(24.305 g Mg) = 0.004114 moles Mg}

          (0.3 g MgO)( 1 mole MgO ) / ( 1 )(40.304 g MgO)  = 0.00744 moles MgO

   Mg:   (0.004114 moles Mg)( 1 mole MgCl₂)/( 1 mole Mg ) = 0.004114 moles MgCl₂
   MgO:   (0.00744 moles MgO)(1mole MgCl₂)/( 1 mole MgO) = 0.00744 moles MgCl₂
   HCl:   (0.05 L HCl)(1.5 mole HCl)(1 mole MgCl₂)/( 1 L HCl )( 2 mole HCl ) = 0.0375 moles MgCl₂
                     

Mg and MgO are the limiting reagents in the reactions.

 

Post laboratory Questions

1. If a metal container is used instead of a Styrofoam cup as a calorimeter, will Cc be larger or smaller? Explain your answer.

        Cc would be less for the metal than for Styrofoam since the metal is a poorer insulator. Metal conducts heat readily but does not store it well.

2. Using the standard heat of formation of the species involved in the reaction,

                Mg(s) + 2HCl(aq) ^---> MgCl₂ (aq) + H₂ (g)

        Calculate the change in heat of formation for the reaction.

        DH_rxn = [0 + 796 kJ/mole] - [0 + 2*(167.2 kJ/mole)] = 461.6 kJ/mole

             -endothermic

3. Given that

             H₂O(l) ^---> H₂0(g)                               DH = 44 kJ/mole

             C₂H₅OH (l) ^---> C₂H₅OH (g)              DH = 42.2 kJ/mole

        Explain why you feel a cooling effect when

        A.) You come out of the shower

              Since the change in enthalpy for the evaporation of water is positive the reaction is endothermic. This means heat will be taken in from the surroundings. When you get out of the shower with liquid water on your skin you feel the cooling effect since heat is taken from your body to change it from liquid to gas.

        B.) You rub ethanol on your hands

        Since the change in enthalpy for the evaporation of ethanol is positive thereaction is endothermic. This means heat will be taken from the surroundings. When you rub ethanol on your hands heat is taken from your body as the reaction takes place, converting it from liquid to gas.