Hydrated Compound Lab U n i t 2 : M o l e C o n c e p t Pre-lab Problem Cobalt (II) Chloride is a hydrated crystal in its solid form. In the lab, you want to determine the formula of this hydrated compound (ie. How many water molecules attach to a formula unit of CoCl2? CoCl2 • X H2O?) In the lab, you find the mass of the test-tube to be 12.443g, and the mass of the compound and the test-tube to be 17.122g. After heating and mixing multiple times to drive off the water, you find the mass of the ionic compound (also known as the anhydrous compound) and the test-tube to be 14.992g. The hydrated compounds is always a 1: X ratio. Write a balanced equation: CoCl2 • 6 H2O(s) --> CoCl2 (s) + 6 H2O(g) Create a data table to organize the data: Object/substance Calculations Mass test-tube -- 12.443g compound and test-tube -- 17.122g compound (CoCl2 • X H2O) 17.122g - 12.443g = 4.679g 4.679g ionic compound and testtube -- 14.992g ionic compound (without water) 14.992g-12.443g = 2.549g 2.549g water 4.679g - 2.549g = 2.130g What is the mass of the ionic compound without water (also called the anhydrous compound)? How many moles is this? mass of ionic compound without water (anhydrous compound) = 2.549g What is the mass of the water heated away? How many moles is this? mass of water heated away = 2.130g What is the mole ratio of the ionic compound to the water? 1.963 x 10-2 mol CoCl2 / 1.963 x 10-2 mol CoCl2 = 1 1.182 x 10-1 mol H2O / 1.963 x 10-2 mol CoCl2 = 6.021 ≈ 6 1 CoCl2 : 6 H2O What is the formula of the reactant hydrate compound? CoCl2 • 6 H2O What is the name of this compound? Cobalt (II) Chloride hexahydrate Part 2: Lab Date: October 28, 2013 Title: Hydrated Compound Lab Purpose: To determine the formula of Copper (II) Sulfate by measuring the mass of the hydrated compound before and after it is heated. Procedure: 1. Mass the empty crucible, then mass the crucible with Copper (II) Sulfate in it. 2. Put the crucible with hydrated salt in it on the clay triangle, which is placed the bunsen burner, and heat it until the blue salt turns into an ash color. 3. If time permits, heat it a second time to ensure all the water (H2O) has heated and evaporated. 4. Using tongs to remove the crucible from the ring stand. Mass the crucible with the dehydrated salt. 5. Record all mass in the data table. Materials: - Tongs - Spatula - Crucible - Scale - Copper (II) Sulfate (CuSO4) - Ring stand - Clay Triangle - Bunsen burner Data Table: Object/Substance Calculations Mass crucible (empty) -- 18.877g crucible with hydrated CuSO4 -- 22.011g crucible with dehydrated CuSO4 (water evaporated) -- 20.830g hydrated CuSO4 (with H2O) 22.011g - 18.877g = 3.134g 3.134g dehydrated CuSO4 (without H2O) 20.830g - 18.877g = 1.953g 1.953g water (H2O) 3.134g - 1.953g = 1.181g 1.181g Calculations: mass of anhydrous compound (in grams) into moles: mass of water (in grams) into moles: ratio of 1 mol CuSO4 : X H2O: 1.224 x 10-2 mol CuSO4 / 1.224 x 10-2 mol CuSO4 = 1 6.554 x 10-2 mol H2O / 1.224 x 10-2 mol CuSO4 = 5.355 1 CuSO4 : 5.355 H2O rounded ratio: (1 CuSO4 : 5.355 H2O) x 7 = 7 CuSO4 : 37.50 H2O (7 CuSO4 : 37.50 H2O) x 2 = 14 CuSO4 : 75.00 H2O the correct answer: Copper (II) Sulfate Pentahydrate (CuSO4·5H2O) http://en.wikipedia.org/wiki/Copper(II)_sulfate Percentage error: Discussion Questions: Make sure to address the following questions in your discussion section: Would the error cause your ratio to increase or decrease? " 1. What was the actual ratio you obtained (with all significant digits)? What was your rounded ratio to whole numbers? Is your actual ratio higher or lower than the rounded ratio? Research what the actual ratio should be. Is your answer too high or too low? What is the percent error? The actual ratio we obtained is 1 CuSO4 : 5.355 H2O. Our rounded ratio to whole numbers 14 CuSO4 : 75 H2O. Our actual ratio is higher the rounded ratio. The actual ratio for Copper (II) Sulfate is 1 CuSO4 : 5 H2O. Our answer is a little lower. The percent error is 7.100%. 2. How would your ratio change if you did not dehydrate the compound fully? Yes, the ratio would change. If the water isn’t fully dehydrated, then the mass of the dehydrated compound would be higher, which would make the mass of water lower than it is supposed to be. Then when you calculate the ratio of 1 salt to x molecules of water, the number you use to divide, which is the moles of the salt, will be larger than it is supposed to be, and therefore the quotient will be smaller. So the ratio will become larger. 3. Suppose some of the compound fell out of the test tube before heating but after massing. How would that change your ratio? The ratio would become much smaller. When you mass the anhydrous compound after compound, you would assume that the loss of mass was all from the water that evaporated. So your mass for water would be much higher than the actual mass, which would increase the number for x amount of water molecules to 1 salt. Therefore, it would the ratio would become smaller. 4. How would your ratio change if the test tube was wet when the hydrated compound was added to it at the beginning of the experiment? If the test tube was wet when the hydrated compound was added to it, then the ratio would become smaller. The original mass of the hydrated compound would be higher than the actual mass. When all the water evaporated, the mass of the dehydrated compound would be lower than the actual mass, and the mass of water would be higher. When calculating the ratio, there would be more water molecules to every anhydrous salt, which would make the ratio smaller. 5. How would your ratio relate to another group’s who used the same compound, if you used 2.0 grams and they used 2.5grams? The ratio should be the same thing if both of the two groups didn’t make any error during the experiment because we are all using the same compound. Even though the other group’s mass for the hydrated salt, anhydrous compound, and evaporated water would be higher than ours, but the ratio should be the same. 6. Suggest what might happen if you left the anhydrous compound to sit overnight before weighing it for your data? The anhydrous compound can revert back to hydrated compound if water is added to it. So if the anhydrous compound is left overnight, some water molecules could go back into the compound, which would increase the mass of the anhydrous compound. This would then decrease the mass of the water, and the ratio would be larger. 7. If you could improve this lab, what you do differently and why? Discuss at least three feasible improvements and why with reference to the other discussion questions and errors. (Note: relate this to #1 and the answers for the other questions). 1) One way to improve this lab would be to heat the compound multiple times instead of one. As answered in question 1, the compound could be not fully dehydrated, and it would affect the data. 2) Another improvement would be to do several trials. The experiment should have at least 3 trials, so the averaged data would be more accurate. 3) It would be better if we used different amounts of compound for each trial, so the data would be more accurate and we could prove that the ratio doesn’t change even though the amount does. Conclusion Statement: In conclusion, we found out the formula for Copper (II) Sulfate by evaporating water from the hydrated compound into the anhydrous compound. Our result, 1 CuSO4 : 5.355 H2O, is pretty close with the actual result, 1 CuSO4 : 5 H2O, with a percent error of 7.100%. Here is the link to the Google doc: (my partner is Corinna) https://docs.google.com/a/hkis.edu.hk/document/d/12JnjNUYflo0OABJCRYadXcG_G_d0mFmaAHBU4zUSS8/edit
