Chemistry Lab Report 1 Nicole H

Chemistry Lab Report 1 Nicole H. Healey
(Experiment 1 and 2) October 7, 2014

Data Collection:
Table 1: (First Titration)
C2O42- Analysis
Sample 1
Sample 2
Molarity of KMnO4
0.02m
0.02m
Weight of Sample
0.237g
0.225g
Final Buret Reading
28.5ml
26.3ml
Initial Buret Reading
0ml
0ml
Volume of KMnO4 dispensed
28.5ml
26.3ml
Moles of KMnO4
5.7E-4
5.26E-4
Moles of C2O42-
1.425E-3
1.315E-3

Table 2: (Second Titration)
C2O42- Analysis
Sample 1
Sample 2
Molarity of KMnO4
0.02m
0.02m
Weight of Sample
0.237g
0.225g
Final Buret Reading
4.2ml
4.5ml
Initial Buret Reading
0ml
0ml
Volume of KMnO4 dispensed
4.2ml
4.5ml
Moles of KMnO4
8.4E-5
9.0E-5
Moles of C2O42-
2.1E-4
2.25E-4

Calculations:
1.) 2 KMnO4 + 5 K2C2O4 + 8 H2SO4 = 2 MnSO4 + 10 CO2 + 8 H2O + 6 K2SO4
Find moles of C2O42-
Trial 1:
KMnO4 = 5.7E-4 /1mol KMnO4 x 5 moles K2C2O4 /2moles KMnO4 = .001425 moles C2O42-
KMnO4 =5.26E-4 /1mol KMnO4 x 5moles K2C2O4 /2moles KMnO4 = .001315 moles C2O42-

Trial 2:
KMnO4 = 8.4E-5 /1mol KMnO4 x 5 moles K2C2O4 /2moles KMnO4 = .00021 moles C2O42-
KMnO4 = 9.0E-5 /1mol KMnO4 x 5 moles K2C2O4 /2moles KMnO4=.000225 moles C2O42-

2.) 5 Fe2+ + MnO4- + 8 H+ =Fe3+ + Mn2+ + 4H2O

Find Moles of Fe3+

Trial 1:
KMnO4 = 5.7E-4 / 1mol KMnO4 x 5 molesFe3+ / 1 mole KMnO4 = .00285 moles Fe3+
KMnO4 = 5.26E-4/ 1mol KMnO4 x 5 molesFe3+ / 1 mole KMnO4 = .00263 moles Fe3+

Trial 2:
KMnO4 = 8.4E-5/ 1mol KMnO4 x 5 molesFe3+ / 1 mole KMnO4 = .00042moles Fe3+
KMnO4 = 9.0E-5/ 1mol KMnO4 x 5 molesFe3+ / 1 mole KMnO4 = .00045moles Fe3+

3.) Mole Ratio of C2O42- / Fe3+

Ratio = .001425 moles C2O42 / .00285 moles Fe3+ =.5 = 1:2 Ratio = .001315 moles C2O42 / .00263 moles Fe3+ =.5 = 1:2

Discussion:
My mole ratio was not 3:1, because there was either too little titration or my concentration was off in experiment one.