Calorimetry Lab
1. Determining the concentration of KMnO4 from the solution created by the stockroom.
16H+ + 2MnO4- (aq) + 5C2O42- (aq) → 2Mn2+ (aq) + 10CO2 (g) + 8H2O (l)
Volume of potassium manganate (KMnO4) = 32.5 mL
Mass of Sodium Oxalate (NaC2O4): 0.104
[KMnO4] Calculation:
= (0.104g of NaC2O4)(1 mol NaC2O4 /134.0g)(2 mol KMnO4 / 5 mol NaC2O4)(1/32.5 mL)(1000 mL /1L)
= 0.00955 M KMnO4
2. Using the standardized concentration of KMnO4 calculated above to find the mass percentage of the oxalate ion
Equation: Volume of KMnO4 reacted with Oxalate ion = (Vol of KMnO4 used in titration - Vol of KMnO4 used in blank titration)
Titration Trial 1
Vf - Vi = VT
54.6 mL -28.1 mL = 26.5 mL
26.5 mL KMnO4 (1 L/ 1000 mL)(0.00955 M KMnO4 / 1L)( 5 mol C2O42- …show more content…
0.187 mol of K / 0.187 mol of K = (1) x 2 = 2
0.204 mol of Fe / 0.187 mol of K = (1.09) x 2 = 2
0.667 mol of C2O4 / 0.187 mol of K = (3.56) x 2 = 7
1.45 mol of H2O / 0.187 mol of K = (7.75) x 2 = 15
Therefore, the empirical formula for the Iron complex made is K2Fe2(C2O4)7*15H2O
7. Calculating whether or not the net ionic charge of the empirical formula is zero. If not, suggest a new molecule.
(2K)(1+) + (2Fe)(3+) + (7C2O4)(2-) = -6 charge …show more content…
Another means of modification include using an automatic titration device from Titralabs. The main issue with the volume used for the titrations is that the color change is highly subjective and may change by several mL between partners. These small variations can cause further changes in the calculations later on and could potentially change the results obtained. The automated titration devices use light sensors to notice the change in wavelength in the solution at a given temperature and will thus stop pouring the titrant into the solution when the equivalence point is achieved. In addition, an automated heating chamber could be used to maintain a specific temperature of 80 degrees Celsius when performing titrations because the hot plate will not maintain the temperature of the solution at 80 degrees at all times and therefore may cause you to add more or less volume to the titrated solution. Finally, over the course of the experiment, we noticed different accuracies in the scales we used throughout the lab. The differences between different scales for the same sample go differ by up to 0.100g which could certainly impact the amount of solute we believe we have
16H+ + 2MnO4- (aq) + 5C2O42- (aq) → 2Mn2+ (aq) + 10CO2 (g) + 8H2O (l)
Volume of potassium manganate (KMnO4) = 32.5 mL
Mass of Sodium Oxalate (NaC2O4): 0.104
[KMnO4] Calculation:
= (0.104g of NaC2O4)(1 mol NaC2O4 /134.0g)(2 mol KMnO4 / 5 mol NaC2O4)(1/32.5 mL)(1000 mL /1L)
= 0.00955 M KMnO4
2. Using the standardized concentration of KMnO4 calculated above to find the mass percentage of the oxalate ion
Equation: Volume of KMnO4 reacted with Oxalate ion = (Vol of KMnO4 used in titration - Vol of KMnO4 used in blank titration)
Titration Trial 1
Vf - Vi = VT
54.6 mL -28.1 mL = 26.5 mL
26.5 mL KMnO4 (1 L/ 1000 mL)(0.00955 M KMnO4 / 1L)( 5 mol C2O42- …show more content…
0.187 mol of K / 0.187 mol of K = (1) x 2 = 2
0.204 mol of Fe / 0.187 mol of K = (1.09) x 2 = 2
0.667 mol of C2O4 / 0.187 mol of K = (3.56) x 2 = 7
1.45 mol of H2O / 0.187 mol of K = (7.75) x 2 = 15
Therefore, the empirical formula for the Iron complex made is K2Fe2(C2O4)7*15H2O
7. Calculating whether or not the net ionic charge of the empirical formula is zero. If not, suggest a new molecule.
(2K)(1+) + (2Fe)(3+) + (7C2O4)(2-) = -6 charge …show more content…
Another means of modification include using an automatic titration device from Titralabs. The main issue with the volume used for the titrations is that the color change is highly subjective and may change by several mL between partners. These small variations can cause further changes in the calculations later on and could potentially change the results obtained. The automated titration devices use light sensors to notice the change in wavelength in the solution at a given temperature and will thus stop pouring the titrant into the solution when the equivalence point is achieved. In addition, an automated heating chamber could be used to maintain a specific temperature of 80 degrees Celsius when performing titrations because the hot plate will not maintain the temperature of the solution at 80 degrees at all times and therefore may cause you to add more or less volume to the titrated solution. Finally, over the course of the experiment, we noticed different accuracies in the scales we used throughout the lab. The differences between different scales for the same sample go differ by up to 0.100g which could certainly impact the amount of solute we believe we have