Ruth Zalalem (1) - Hydrates and Thermal Decomposition - Checked on 9_8

docx

School

Georgia Gwinnett College *

*We aren’t endorsed by this school

Course

Subject

Chemistry

Date

Feb 20, 2024

Type

docx

Pages

Uploaded by PrivateCheetah8480

Hydrates and Thermal Decomposition Objectives ● To identify properties of hydrates ● To determine the number of waters of hydration in an unknown analyte ● Design a plan in order to collect data on the synthesis or decomposition of a compound to confirm the conservation of matter and the law of definite proportions Pre-Lab Read the following background information on hydrates and answer the pre-lab questions. Hydrates are chemical substances of definite composition formed from ionic compounds and integer or half-integer numbers of water molecules. In this experiment you will investigate some of the properties of hydrates and you will verify the number of water molecules present in a formula unit of one particular hydrate compound. A common hydrate is bright blue copper sulfate pentahydrate, formula CuSO 4 •5H 2 O. The dot (•) indicates that five water molecules are bound to the copper and sulfate ions. Water bound in this manner is called a ligand. Note that the dot is not a multiplication symbol. When you calculate the molar mass of a hydrate, the dot actually functions as a “plus” sign: the mass of five moles of water is added to the mass of one mole of copper (II) sulfate. One of the most important hydrates is calcium sulfate dihydrate, CaSO 4 •2H 2 O. The mineral gypsum is a hard, dense form of this compound. When it is heated and the two molecules of water are driven off, the anhydrous (water-free) calcium sulfate that remains is known as plaster of Paris. The addition of water to plaster of Paris to form the dihydrate once again is quite exothermic; producing plaster. The plaster ‘sets’ rapidly to a highly interlocked crystalline form that has water molecules arrayed between layers of calcium and sulfate ions. Before the invention of cement, plaster of Paris was a popular construction component. It can be found as the mortar between bricks in ancient structures still standing in Pompeii. Anhydrous calcium chloride, CaCl 2 , aggressively picks up water, ultimately forming a hexahydrate, CaCl 2 •6H 2 O. The reaction is exothermic and the resulting solution has a very low freezing point, so the primary use of the compound is for deicing roads, which it is capable of doing down to temperatures as low as -51°C. Anhydrous calcium chloride is commonly used in the laboratory as a desiccant; it will remove moisture from air (in a closed vessel called a dessicator) or from gases that flow through particles of the solid. It also has the property of removing water from organic liquids such as diethyl ether, which can dissolve about 8% of its weight in water. It can be useful in the summer, controlling dust in highway construction by removing moisture from the air to form a damp solid. Copper sulfate pentahydrate is blue, while the anhydrous salt is white. Cobalt chloride in moist air is a hexahydrate, CoCl 2 •6H 2 O, which is maroon-red, while in dry, cool air it is the blue anhydrous salt. At moderate humidity levels it is present as the purple dihydrate, CoCl 2 •2H 2 O, so it can serve as an indicator of relative humidity. It is incorporated, for example, with the desiccant silica gel found in the packaging of electronic equipment; when the desiccant turns pink it needs to be replaced or reactivated by heating. Some hydrates lose water at room temperature and normal humidity; they are said to effloresce (or, they are efflorescent). Other compounds, such as calcium chloride and sodium hydroxide, pick up so much moisture from the air that they dissolve in the resulting liquid. These substances deliquesce; they are deliquescent. Sodium hydroxide is deliquescent. A very large class of organic compounds is called carbohydrates. This might seem to imply that they are hydrates of carbon; that misconception arose in the middle of the 18th century when only the empirical formulas, [C(H 2 O)n], of many sugars were known. The empirical formula for glucose, C 6 H 12 O 6 , seems as if it could be written C 6 (H 2 O) 6 , but this sugar is not carbon hexahydrate. True hydrate formation is completely reversible, which is not the case with carbohydrates. They can be

partially dehydrated; that is, some of the material can be driven off as water as a result of heating, but when a sugar is heated it will lose the elements of water, but is decomposed irreversibly in the process. Partial dehydration of sucrose, C 12 H 22 O 11 , leads to taffy, caramel, and finally to the dark brown substance from which peanut brittle is made. Pre-Lab Questions 1. Define the term hydrate (in your own words) and explain the function of the dot in the formulas of hydrates. How is the dot treated when you are calculating the molar mass of a hydrate? A hydrate is an ionic compound that contains water. The dot in formulas represents how many water molecules are bonded. When calculated molar mass you add 5 of the molar mass. 2. What is the meaning of the term anhydrous? Containing no water 3. Determine the theoretical mass percent of water in CuSO 4 •5H 2 O. (90.8 g)/ (249. 691 g) (100)= 36.0766% Materials (list all materials below) -crucible -bunsen burner -clay triangle -ring stand -iron ring - CuSO4 - weighing boat -electric balance - scoopula - ring stand - boyces fan - pencil - instructions/ lab directions Procedure 1. Set up an empty crucible as shown in the figure to the right but without the lid. Heat it strongly with a hot burner flame for 2-3 minutes, to drive off moisture and any impurities present Allow the crucible to cool completely before you proceed. 2. Obtain a sample of copper (II) sulfate pentahydrate. Carefully weigh the cooled crucible with its cover, then add about 1.5-2.0 g of CuSO 4 •5H 2 O to the crucible. Weigh the crucible, cover, and contents (“the system”) again and record the mass. 3. Place the cover on the crucible and transfer the system to a clay triangle supported by an iron ring. The bottom of the crucible should be just high enough that it will be in the hottest part of a strong burner flame. Set the cover very slightly ajar, just enough so that steam can escape. Heat the crucible gently, holding the burner in your hand and moving it back and forth for about 5-6 minutes. Much of the water of hydration will be driven off during this time. Gentle heating is necessary at first to avoid loss of materials due to spattering. 4. Increase the intensity of the flame by opening the air vents at the base of the Bunsen burner. Heat for 10 minutes or longer, but keep an eye on the bottom surface of the crucible - if it starts to glow orange or red, back off on the heating for a moment. Copper (II) sulfate is stable up to 650°C but will start to

decompose above that temperature. If you notice even a hint of a foul, choking odor,reduce the heating immediately. 5. At the conclusion of heating, remove the crucible and place on a heat-resistant pad and turn off the burner. Allow the crucible and contents to cool for at least 10 minutes, then weigh the system. Determine the mass of water lost. 6. Return the crucible to the clay triangle and heat strongly for an additional 5 minutes. Turn off the burner and allow the crucible and contents to cool fully on a heat-resistant pad and weigh the system again. Determine the mass change from the previous weighing. The difference in mass should be less than 10 mg; this is referred to as heating to constant mass , and is one way to be certain that the dehydration was complete. If dehydration is not complete, and if time permits, heat for an additional 5-minute period, cool, and reweigh. The final mass will be used in the calculations. If class time is close to being over, reserve this heating for your lunch period or after school, but extend it from 5 minutes to 10 minutes. 7. Dispose of the dehydrated copper (II) sulfate in the waste bucket in the fume hood. Clean with soap and water and dry all glassware, the crucible, and cover. Data Cooled crucible with cover (g) 50.52 g Mass of CuSO 4 •5H 2 O (g) 2.00 g Mass of crucible, cover, and CuSO 4 •5H 2 O (g) 52.52 g Mass after 1st heating (g) 51.70 g Mass after 2nd heating (g) Mass after 3rd heating (g) - if needed Qualitative observations - Blue CuSO 4 after heated became a grayish white substance - There was no odor after or during heating Analysis 1. Determine the total mass of water lost during all heating periods. 52.52 g- 51. 7g =0.82 g of water lost 2. Determine the percentage of water (by mass) in the original hydrated sample. (.82g H20/ 2.00 CuSO 4 •5H 2 O )100= 41 % of H20 3. Compare the actual percentage of water to the theoretical percentage of water and determine the percent error. 159.62 g CuSO 4. + 18.016 g H20= 249.72 g/mol CuSO 4 •5H 2 O (18.016 g H20) (5) = 90.1 g 5H20 [(90.1 g 5H20) / ( 249.72 g/mol CuSO 4 •5H 2 O)](100)= 36.08% 36.08% theoretical percentage of water (.82 g of water/ 2.00 g CuSO 4 •5H 2 O) (100) = 41%

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

41.00% actual percentage (Actual- theoretical)/ ( actual) (41- 36.08) / 41= 12% 12.00% error 4. a. Determine the moles of anhydrous copper (II) sulfate present in your crucible at the conclusion of the procedure. 2.00 g CuSO 4 •5H 2 O-.82 g H20= 1.18g copper (II) sulfate 1.18g copper (II) sulfate / 159.6 g CuSO 4 = .00739 mol CuSO 4 b. Determine the number of moles of water that were lost during heating. .82 g H20/ 18.016 g H20= .04529 mol H20 c. Determine the ratio of moles of water per mole of anhydrous solid. 0.04529 mol H20 / .00739 mol CuSO 4 = 6.13 moles of water per mole of anhydrous solid d. Write the formula of the hydrate in the form, CuSO 4 • Y H2O, where Y is the number of moles calculated in part c. CuSO 4 • 6 H2O Conclusion 1. Suppose a student forgets to cover the crucible while they are heating the copper (II) sulfate, and that as a result, some of the material is lost to spattering. How would this affect the calculated value for the mass percent of water in the hydrate? The student would not be able to calculate the mass because of the splattering. This is because the student would not know how much material is there. 2. Examine your results for the percentage of water in copper (II) sulfate pentahydrate. Discuss the accuracy of your result. If you were more than 5% off in your determination, suggest possible explanations for the difference and ways that you would change your procedure if you were to repeat the experiment. The percentage of water was 41%, this is somewhat accurate but still deviates from the theoretical value of 36.08%. I believe this occurred because we overheated the CuSO 4 , this evaporated all the water and some of the CuSO 4 . If we were to repeat this experiment we would have decreased the amount of gas during the first heating and watch to make sure it doesnt burn. 3. In the procedure, you are directed to heat gently at first, since heating too strongly might cause some loss of material, due to spattering. Why would too-rapid heating give rise to spattering? Because too rapid heating would increase the kinetic energy of the ionic compound as well as the water.If the kinetic energy of the CuSO 4 is too high the crucible can not contain it, allowing it to splatter 4. Although copper metal is unreactive toward hydrochloric acid, the same is not true for copper (II) oxide. In fact, in cases in which it is difficult to remove the solid oxide from a crucible, one technique is to add 6 M HCl to the contents. This results in the formation of a deep blue solution. Write the formula equation, the complete ionic equation, and the net ionic equation for the reaction between hydrochloric acid and copper (II) oxide. 2HCl(aq) + CuO(s) → CuCl 2 (aq) + H 2 O(l) 2H⁺ + 2Cl⁻ + CuO → Cu²⁺ + 2Cl⁻ + H 2 O 2H⁺ + CuO → Cu²⁺ + H 2 O 5. Consider this same experiment with copper (II) carbonate. Write the balanced formula equation for the decomposition of copper (II) carbonate. If the copper (II) carbonate system is heated for too short of a time, will the value for the mass percent of copper be too large or too small? Justify your answer. CuCO3 (s) →Δ CuO (s) + CO2 (g) The value of the copper will be too large, this is because there will still be water inside it.

6. Identify two potential sources of error in this experiment. Also, indicate how these errors could affect the results. - To rapid heating which can cause splattering - Having the crucible cover closed or having a small space open where the water can not evaporate