Lab 10 Marlena Slade

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Dec 6, 2023

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Hydrated Lab: Heating Copper(II) Sulfate pentahydrate Lab Introduction: Hydrates are crystalline solids that contain a fixed number of water molecules as an integral part of their crystalline structure. The number of water molecules bonded per metal ion is often characteristic of that particular metal ion. One of the more common hydrates is copper(II) sulfate pentahydrate, which contains 5 moles of water per 1 mole of copper(II) sulfate, written as CuSO 4 ·5H 2 O. It is used as a catalytic precursor, fungicide, and as a source of copper in chemical manufacturing processes. Epsom salt is magnesium sulfate heptahydrate, MgSO 4 ·7H 2 O. Epsom salt is used to reduce inflammation when applied externally. Many hydrates can be transformed to the anhydrous (without water) compound when heated strongly. For example, copper(II) sulfate pentahydrate can be converted into anhydrous copper(II) sulfate by losing its water. This change can be followed visually. The blue crystalline copper(II) sulfate pentahydrate is converted when heated to a white, powdery, anhydrous salt, according to the following endothermic equation. Heat + CuSO 4 ·5 H 2 O(s) -----------± CuSO 4 (s) + 5 H 2 O(g) Blue Crystals white hydrated salt anhydrous salt + water vapor Copper(II) sulfate pentahydrate Copper(II) sulfate It is also possible to reverse the above process, as shown in the equation below: CuSO 4 + 5 H 2 O → CuSO 4 ·5 H 2 O white blue If water is added to the white anhydrous copper(II) sulfate, a blue color of hydrated compound is obtained indicating that the blue pentahydrate is regenerated. Objective : 1. To observe the changes that occur to hydrated copper(II) sulfate when it is heated with a Bunsen Burner and 2. Calculate the Empirical Formula of the hydrated compound.

Materials : 1. Bunsen Burner 2. Crucible 3. Copper(II) Sulfate pentahydrate 4. Clay Triangle 5. Stand/Clamp 6. Tongs Directions: : 1. Watch the following videos a. https://www.youtube.com/watch?v=5fZlLEBAfEU b. https://www.youtube.com/watch?v=RPWkndwt2FA 2. Record the data in the lab sheet as you watch the first video. 3. As you watch the 2 nd video, follow the instruction for calculating the % of water in the hydrated compound (theoretically). Procedure: 1. Set up the stand, clamp and ceramic triangle 2. Weigh the crucible. 3. Place 1cm depth of copper(II) sulfate pentahydrate in the bottom of the crucible and weigh it. 4. Place the crucible on the clay triangle. 5. Light the Bunsen burner and turn it to a roaring blue flame. 6. Adjust the height of the crucible, so that the bottom of the crucible is at the top of the blue cone in the flame. 7. After 5 minutes, remove the Bunsen burner and record your observations. 8. Light the Bunsen burner again and heat for another 5 minutes. 9. Turn off the Bunsen burner and allow everything to cool for 10 minutes. 10. When the crucible is cool, weigh the crucible with the anhydrous.

Data from the first video: 1. Mass of empty crucible= 12.51g 2. Mass of crucible with the hydrated compound=12.70g 3. Mass of crucible with the anhydrous salt (after heating)=12.62g Observations: What did you observe during the heating of copper(II) sulfate pentahydrate? Explain. The copper(II) sulfate pentahydrate turned from blue color to white color during the heating process. The water dissipated and got removed from the copper (II) sulfate due to the heat introduced. Calculations: Watch the 2 nd video to help you with the calculation. Show your work for credit. Use proper units & SF. 1. Mass of hydrated compound=249.72.g 2. Mass of anhydrous compound=159.62g 3. Mass of water loss=90.10g 4. Calculate the theoretical % (by mass) of water in copper(II) sulfate pentahydrate. % water = (90.10 g/249.72 g) x 100 = 36.08 5. Calculate the experimental % of water (by mass) in the hydrated compound using your data. % water = [(12.70 g - 12.62 g) / (12.70 g - 12.51 g)] x 100 = 42.11 6. Use your data and calculate the experimental % (by mass) of anhydrous compound used in the lab. % anhydrous = [(12.62 g- 12.51 g) / (12.70 g - 12.51 g)] x 100 = 57.89 7. Calculate the percent error for water. % error = (|42.11 - 36.08|/ 36.08) x 100 = 16.71

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