Reaction Between Distilled Water, Iron, And Copper (II) Sulfate

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Ally Emerson Lab Report H. Chem. Period 3 1-11-15 How does mixing distilled water, iron, and copper (II) sulfate and then boiling the mixture help us determine the theoretical and percent yield of copper after the chemical reaction? Partner: Naomi Purpose: To determine the theoretical and percent yield of copper after a chemical reaction between distilled water, iron, and copper (II) sulfate. Hypothesis: There will be around 2 grams of copper left, which will equal a 89% yield. Materials: scale, filter paper, weighing paper, distilled water, copper (II) sulfate, iron, beaker, tongs, drip, and a funnel. Procedure: 1. Weigh the weighing paper. 2. Weigh the filter paper. 3. Collect 1 gram of iron onto weighing paper, weigh and record.…show more content…
Collect 6 grams of copper (II) sulfate on weighing paper. Weigh and record. 5. Collect 50 mL of distilled water into the beaker. 6. Combine the iron and copper (II) sulfate together into the distilled water. Heat to a slow boil for 10 minutes. 7. Remove from the heat and allow to cool until you can hold it. 8. Put filter paper into funnel and put the funnel into the drip. 9. Slowly pour the beaker contents into the drip paper. Make sure every piece is out of the beaker. 10. Remove the funnel from the drip and the filter paper from the funnel. Allow the filter paper to dry overnight. 11. Weigh the filter paper with the contents inside. Record the weight. Data: | Fe | CuSO4 | Cu | FeSO4 | Filter Paper | Filter Paper with contents | Weigh Paper | Before | 1.445 g | 6.185 g | -- | -- | 0.770 g | -- | 0.29 g | After | -- | -- | 1.850 g | (drained out) | -- | 2.620 g | -- | Evaluation: 1.445 g Fe x 1 mol Fe x 1 mol Cu x 63.3546 g Cu = 1.64 g Cu => Theoretical Yield 55.847 g Fe 1 mol Fe 1 mol Cu 1.850 g Cu = 112.8 % => Percent Yield 1.64 g

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