Chapter 8 Hydrates.

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Presentation transcript:

Chapter 8 Hydrates

Hydrates A hydrate is a solid crystalline substance that has a certain number of water molecules as part of its structure. Example: CoCl2 • 6H2O

Hydrates A hydrate is a solid crystalline substance that has a certain number of water molecules as part of its structure. Example: CoCl2 • 6H2O An anhydrous salt is the hydrate minus the water. Example: CoCl2

Determination of a hydrate formula: CoCl2 • 6H2O Although all problems for hydrates may sound a little different they all require that you use the information in the problem to find the: Grams of water Grams of anhydrous substance.

Determination of a hydrate formula Calculate the mass of water and convert it moles. Calculate the mass of the anhydrous salt (dried salt) and convert it to moles. Divide your moles of water by your moles of anhydrous salt. Round your answer to the nearest whole number to determine the number of water molecules in the hydrate formula.

4. 32g of the hydrate FeSO4  xH2O is heated to drive off the water 4.32g of the hydrate FeSO4  xH2O is heated to drive off the water. After heating we find that we have 2.36g of anhydrous salt. What is the hydrate formula? Example problems are in the note outline.

4. 32g of the hydrate FeSO4  xH2O is heated to drive off the water 4.32g of the hydrate FeSO4  xH2O is heated to drive off the water. After heating we find that we have 2.36g of anhydrous salt. What is the hydrate formula? 4.32g FeSO4  xH2O – 2.36g FeSO4 = 1.96g H2O

4. 32g of the hydrate FeSO4  xH2O is heated to drive off the water 4.32g of the hydrate FeSO4  xH2O is heated to drive off the water. After heating we find that we have 2.36g of anhydrous salt. What is the hydrate formula? 4.32g FeSO4  xH2O – 2.36g FeSO4 = 1.96g H2O 1.96g H2O mol = 0.109 mol H2O 18.0 g 2.36g FeSO4 mol = 0.0155 mol FeSO4 151.9g

4. 32g of the hydrate FeSO4  xH2O is heated to drive off the water 4.32g of the hydrate FeSO4  xH2O is heated to drive off the water. After heating we find that we have 2.36g of anhydrous salt. What is the hydrate formula? 4.32g FeSO4  xH2O – 2.36g FeSO4 = 1.96g H2O 1.96g H2O mol = 0.109 mol H2O 18.0 g 2.36g FeSO4 mol = 0.0155 mol FeSO4 151.9g 0.109 mol H2O = 7.03 0.0155 mol FeSO4

4. 32g of the hydrate FeSO4  xH2O is heated to drive off the water 4.32g of the hydrate FeSO4  xH2O is heated to drive off the water. After heating we find that we have 2.36g of anhydrous salt. What is the hydrate formula? 4.32g FeSO4  xH2O – 2.36g FeSO4 = 1.96g H2O 1.96g H2O mol = 0.109 mol H2O 18.0 g 2.36g FeSO4 mol = 0.0155 mol FeSO4 151.9g 0.109 mol H2O = 7.03 ≈ 7 0.0155 mol FeSO4

4. 32g of the hydrate FeSO4  xH2O is heated to drive off the water 4.32g of the hydrate FeSO4  xH2O is heated to drive off the water. After heating we find that we have 2.36g of anhydrous salt. What is the hydrate formula? 4.32g FeSO4  xH2O – 2.36g FeSO4 = 1.96g H2O 1.96g H2O mol = 0.109 mol H2O 18.0 g 2.36g FeSO4 mol = 0.0155 mol FeSO4 151.9g 0.109 mol H2O FeSO4 • 7H2O = 7.03 ≈ 7 0.0155 mol FeSO4

A student uses a balance to measure the mass of an evaporating dish to be 19.82g. She then pours some of the hydrate CuSO4  xH2O into the evaporating dish and measures the mass again to obtain a reading of 21.54g. She heats the hydrate gradually at first and then more rapidly to vaporize the water from the hydrate. After vaporizing the water she removes the evaporating dish from the hotplate and allows it to cool. She then measures the mass for the last time to be 20.94g. What is the formula of the hydrate?

A student uses a balance to measure the mass of an evaporating dish to be 19.82g. She then pours some of the hydrate CuSO4  xH2O into the evaporating dish and measures the mass again to obtain a reading of 21.54g. She heats the hydrate gradually at first and then more rapidly to vaporize the water from the hydrate. After vaporizing the water she removes the evaporating dish from the hotplate and allows it to cool. She then measures the mass for the last time to be 20.94g. What is the formula of the hydrate? Find the grams of water and the grams of anhydrous salt before attempting to solve the problem.

A student uses a balance to measure the mass of an evaporating dish to be 19.82g. She then pours some of the hydrate CuSO4  xH2O into the evaporating dish and measures the mass again to obtain a reading of 21.54g. She heats the hydrate gradually at first and then more rapidly to vaporize the water from the hydrate. After vaporizing the water she removes the evaporating dish from the hotplate and allows it to cool. She then measures the mass for the last time to be 20.94g. What is the formula of the hydrate? 21.54g – 19.82g = 1.72g hydrate 21.54g – 20.94g = 0.60g H2O 1.72g – 0.60g = 1.12g CuSO4

CuSO4 • 5H2O mol 18.0 g 0.60g H2O 1.12g CuSO4 = 0.033 mol H2O mol The mass of an evaporating dish is 19.82g. When the hydrate CuSO4  xH2O is added to the evaporating dish the mass is 21.54g. After heating for the last time the mass is 20.94g. What is the formula of the hydrate? mol 18.0 g 0.60g H2O 1.12g CuSO4 = 0.033 mol H2O mol 159.6g = 0.00702 mol CuSO4 0.033 mol H2O 0.00702 mol CuSO4 = 4.7 ≈ 5 CuSO4 • 5H2O

Homework Hydrate Worksheet (due tomorrow). Do the lab summary for the Lab: “Determination of a Hydrate Formula” (due tomorrow). Study Guide Chapter 8 (due in two days).