Part B
1) Weigh 6.16g of MgSO4.7H20 to the nearest 0.01g on a filter paper, using the digital balance.
2) Weigh 41.85g of water to the nearest 0.01g into a polystyrene cup using the balance.
3) Measure the initial temperature of the measured amount of water using the thermometer and record this value.
4) Dissolve the MgSO4.7H20 in the water and record the temperature change associated with this process.
Results:
Discussion:
*First the enthalpy change of the reaction in part A is calculated:
(MgSO4(s) + 100H2O(l) MgSO4(aq,100H2O)
* Q = m x c x delta T
* M= 45.1 + 3.02
= 48.1g
* C = 4.18 J/g.K
* Delta T = 21 – 18
= 3.00 K
* Q = 603J
* Number of moles of water = mass / molar mass
= 45.1 / 18.0
= 2.50 moles
*Number of moles of MgSO4(s) = Mass/ Molar mass
= 3.02 / (24.31 + 32.06 + (4 x 16.00))
= 3.02 / 120
= 0.0251 moles
* Ratio of moles = 1 : 100
Thus both reactants are limiting
* Enthalpy change = Q / number of moles
= 603 / 0.0251
= - 24023.90 (rounded to 3 significant figures) = -24000 J/mol
(Exothermic reaction)
Enthalpy change for part B:
(MgSO4.7H20(s) + 93H2O(l) MgSO4(aq,100H2O))
* Q = m x c delta T
* M = 6.19 + 41.9
= 48.09 (rounded to 3 significant figures) = 48.1 g
* C = 4.18 J/g.K
* Delta T = 16 – 18
= - 2 K (place only the absolute value)
= 2.00
Q = 402.116 (rounded to 3 significant figures) = 402 J
* Number of moles of water = mass / molar mass
= 41.9 / 18.0
= 2.33 moles
*number of moles of MgSO4.7H2O = mass / molar mass
= 6.19 / (24.31 + 32.06 + (4 x 16.00) + (7x 18.02))
= 93.0 moles
* Ratio of moles = 2.33 : 0.0251
= 93.0 : 1.00
* Both reactants are limiting
* Enthalpy change = Q / number of moles
= 402/ 0.0251
= 16015.93 = 16000 J/mole (endothermic reaction)
* Enthalpy change for the reaction of:
(MgSO4(s) + 7H2O MgSO4.7H2O(s) )
= -24000 - 16000
= - 40000 J/ mole
= - 40 Kj / mole
Uncertainties:
Total percentage uncertainties = 0.166 + 0.0808 + 0.0111 + 0.0119 + 0.278 + 0.238 + 0.278 + 0.313 = 1.3768 = 1.38 %
The enthalpy change = - 40 ± ((1.38/100) x 40)
= - 0 ± 0.552 Kj/ mole
Conclusion:
Percentage error: difference between calculated value and literature value divided by literature value times 100
Percentage error = ((40 – 104) / - 104) x 100
= (- 64 / 104) x 100
= 61.5 % (take absolute value)
Sources of error:
1) Heat lost to the surrounding
2) The specific heat capacity used in calculations was that of water, not the mixture itself
3) Parallax error
4) Position of thermometer
5) Stirring was not constant
Improvements:
1) Use a well insulated vessel with lid
2) Find the specific heat capacity of the mixture itself
3) Use digital thermometer connected to a computer
4) Ensure that the line of sight is perpendicular to the reading of the thermometer and measuring cylinder
5) Use magnetic stirrer for uniform stirring
6) Repeat the experiment several times, taking an average value for the heat emitted or gained by the mixture