chem 112 Lab 1 - 3.55g = 34.00g 30.45g %Yield = Weight of...

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02/25/08 Chemistry 112 Synthesis of an Iron Coordination Complex Introduction : The purpose of this laboratory is to prepare potassium tris (oxalato) ferrate(III) trihydrate (K 3 [Fe(C 2 O 4 ) 3 ]*3H 2 O). The compound is formed by K + , Fe 3+ , C 2 O 4 2 , and H2O. Results: At the end of the experiment 3.55g of the complex was yielded. The expected amount of K 3 [Fe(C 2 O 4 ) 3 ]*3H 2 O is .013 moles or 6.38g. This was determined by calculating the moles in the 5.198g of Fe(NH 4 ) 2 (SO 4 ) 2 *6H 2 O which was .013 using the molecular weight given in the lab. It is determined that the mole ratio of K 3 [Fe(C 2 O 4 ) 3 ]*3H 2 O : Fe(NH 4 ) 2 (SO 4 ) 2 *6H 2 O is 1:1. Data Collection: Weight of Empty Beaker = 30.45g Weight of Beaker + Product = 34.00g Weight of K 3 [Fe(C 2 O 4 ) 3 ]*3H 2 O = 3.55g Calculations: Weight of Product = Weight of Beaker+Product – Weight of Empty Beaker
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Unformatted text preview: 3.55g = 34.00g 30.45g %Yield = Weight of Product Obtained * 100 Weight of Product Expected 55% = (3.55g/6.38g) * 100 Weight of Product Expected = (Moles of Fe(NH 4 ) 2 (SO 4 ) 2 *6H 2 O)* (1 Mole K 3 [Fe(C 2 O 4 ) 3 ]*3H 2 O ) *(MW of K 3 [Fe(C 2 O 4 ) 3 ]*3H 2 O) (1 Mole Fe(NH 4 ) 2 (SO 4 ) 2 *6H 2 O) 6.38g = (.013mol Fe(NH 4 ) 2 (SO 4 ) 2 *6H 2 O)*(1)*(492.1g/mol) Discussion: The obtained amount of the complex (3.55g) almost half of the expected value (6.38g). Possible reasons for this may be that some of the product was lost throughout the course of the experiment. This could have happened as it was being handled and emptied from one container to another. Ideally the % yield would be more than the 55% obtained....
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This note was uploaded on 04/17/2008 for the course CHEM 112 taught by Professor Hardy during the Spring '08 term at UMass (Amherst).

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