preview

Copper Iron Stoichiometry Lab Report Essay

Better Essays

Copper-Iron Stoichiometry
Lab Report

10/3/12
Abstract:
The lab performed required the use of quantitative and analytical analysis along with limiting reagent analysis. The reaction of Copper (II) Sulfate, CuSO4, mass of 7.0015g with 2.0095g Fe or iron powder produced a solid precipitate of copper while the solution remained the blue color. Through this the appropriate reaction had to be determined out of the two possibilities. Through the use of a vacuum filtration system the mass of Cu was found to be 2.1726g which meant that through limiting reagent analysis Fe was determined to be the limiting reagent and the chemical reaction was determined to be as following:-
CuSO4(aq) + Fe(s) Cu(s) + FeSO4(aq)
Introduction:
There were …show more content…

Table 2: Time Measures Instance | Time (min) | Copper Sulfate solution cooled after heating | 10:00 min | Copper precipitate exposed on vacuum filtration setup | 10:11 min |

Table 3: Volume Measures Measures | Water added to anhydrous copper sulfate(±0.2ml) | Water required to wash copper solid(±0.2ml) | Acetone to wash the copper solid(±0.2ml) | HCl to wash crucible(±0.2ml) | 1 | 60.0ml | 15.0ml | 15.0ml | 45.0ml | 2 | - | - | 15.0ml | | 3 | - | - | 14.0ml | | 4 | - | - | 15.0ml | |

Table 4: Moles of CuSO4, Fe, Cu Substance | Moles | CuSO4 | 0.043866 mol | Fe | 0.035983 mol | Cu | 0.034189 mol |
Moles of Cu = 2.1726g × 1 mol Cu63.5463 g = 0.034189 mol [II]

Table 5: Limiting Reagent, Theoretical and Actual Yields, Percent Yields Equation | Limiting Reagent | Theoretical Yield Cu | Actual Yield Cu | Percent Yield Cu | [I] | Fe | 0.035983 mol | 0.034189 mol | 95.01% | [II] | CuSO4 | 0.043866 mol | | |

Limiting Reagent Determination-
If Fe Limiting Reagent:
Mol of Cu = 0.035983 mol Fe * 1 mol Cu1 mol Fe = 0.035983 mol [III]
If CuSO4 Limiting Reagent:
Mol of Cu = 0.043866 mol CuSO4 * 1 mol Cu1 mol CuSO4 = 0.043866 mol [IV]
Limiting Reagent = Fe
Theoretical Yield –
Mol of Cu = 0.035983 mol Fe * 1 mol Cu1 mol Fe = 0.035983 mol [V]
Percent Yield –
Percent Yield = Actual YieldTheoretical Yield* 100% = 0.034189mol0.035983mol*100% = 95.01% [VI]
Difference in theoretical yield and actual yield –
Difference = 0.035983 –

Get Access