Chapter 8 Hydrates. A hydrate is a solid crystalline substance that has a certain number of water molecules as part of its structure. –Example: CoCl 2.

Chapter 8Hydrates

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:

CoClCoCl22 • 6H • 6H22OOAlthough 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 waterGrams of water

• Grams of anhydrous substance.Grams of anhydrous substance.

Determination of a hydrate formula

1. Calculate the mass of water and convert it moles.

2. Calculate the mass of the anhydrous salt (dried salt) and convert it to moles.

3. 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 FeSO4.32g of the hydrate FeSO44 xH xH22O is heated to drive off O is heated to drive off

the water. After heating we find that we have 2.36g of the water. After heating we find that we have 2.36g of anhydrous salt. What is the hydrate formula?anhydrous salt. What is the hydrate formula?

Example Example problems are problems are in the note in the note outline.outline.

4.32g of the hydrate FeSO4.32g of the hydrate FeSO44 xH xH22O is heated to drive off O is heated to drive off

the water. After heating we find that we have 2.36g of the water. After heating we find that we have 2.36g of anhydrous salt. What is the hydrate formula?anhydrous salt. What is the hydrate formula?

4.32g FeSO4.32g FeSO44 xH xH22O – 2.36g FeSOO – 2.36g FeSO44 = 1.96g H= 1.96g H22OO

4.32g of the hydrate FeSO4.32g of the hydrate FeSO44 xH xH22O is heated to drive off O is heated to drive off

the water. After heating we find that we have 2.36g of the water. After heating we find that we have 2.36g of anhydrous salt. What is the hydrate formula?anhydrous salt. What is the hydrate formula?

4.32g FeSO4.32g FeSO44 xH xH22O – 2.36g FeSOO – 2.36g FeSO44 = 1.96g H= 1.96g H22OO

1.96g H1.96g H22OO

18.0 g18.0 g

molmol= 0.109 mol H= 0.109 mol H22OO

2.36g FeSO2.36g FeSO44

151.9g151.9g

molmol= 0.0155 mol FeSO= 0.0155 mol FeSO44

4.32g of the hydrate FeSO4.32g of the hydrate FeSO44 xH xH22O is heated to drive off O is heated to drive off

the water. After heating we find that we have 2.36g of the water. After heating we find that we have 2.36g of anhydrous salt. What is the hydrate formula?anhydrous salt. What is the hydrate formula?

4.32g FeSO4.32g FeSO44 xH xH22O – 2.36g FeSOO – 2.36g FeSO44 = 1.96g H= 1.96g H22OO

1.96g H1.96g H22OO

18.0 g18.0 g

molmol= 0.109 mol H= 0.109 mol H22OO

2.36g FeSO2.36g FeSO44

151.9g151.9g

molmol= 0.0155 mol FeSO= 0.0155 mol FeSO44

0.109 mol H0.109 mol H22OO

0.0155 mol FeSO0.0155 mol FeSO44

= 7.03= 7.03

4.32g of the hydrate FeSO4.32g of the hydrate FeSO44 xH xH22O is heated to drive off O is heated to drive off

the water. After heating we find that we have 2.36g of the water. After heating we find that we have 2.36g of anhydrous salt. What is the hydrate formula?anhydrous salt. What is the hydrate formula?

4.32g FeSO4.32g FeSO44 xH xH22O – 2.36g FeSOO – 2.36g FeSO44 = 1.96g H= 1.96g H22OO

1.96g H1.96g H22OO

18.0 g18.0 g

molmol= 0.109 mol H= 0.109 mol H22OO

2.36g FeSO2.36g FeSO44

151.9g151.9g

molmol= 0.0155 mol FeSO= 0.0155 mol FeSO44

0.109 mol H0.109 mol H22OO

0.0155 mol FeSO0.0155 mol FeSO44

= 7.03 ≈ 7= 7.03 ≈ 7

4.32g of the hydrate FeSO4.32g of the hydrate FeSO44 xH xH22O is heated to drive off O is heated to drive off

the water. After heating we find that we have 2.36g of the water. After heating we find that we have 2.36g of anhydrous salt. What is the hydrate formula?anhydrous salt. What is the hydrate formula?

4.32g FeSO4.32g FeSO44 xH xH22O – 2.36g FeSOO – 2.36g FeSO44 = 1.96g H= 1.96g H22OO

1.96g H1.96g H22OO

18.0 g18.0 g

molmol= 0.109 mol H= 0.109 mol H22OO

2.36g FeSO2.36g FeSO44

151.9g151.9g

molmol= 0.0155 mol FeSO= 0.0155 mol FeSO44

0.109 mol H0.109 mol H22OO

0.0155 mol FeSO0.0155 mol FeSO44

= 7.03 ≈ 7= 7.03 ≈ 7 FeSOFeSO44 • 7H • 7H22OO

• 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 Find the grams of water and the grams of anhydrous salt before attempting to solve the problem.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

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?

0.60g H0.60g H22OO

1.12g CuSO1.12g CuSO44

molmol

18.0 g18.0 g

molmol

159.6g159.6g

= 0.033 mol H= 0.033 mol H22OO

= 0.00702 mol CuSO= 0.00702 mol CuSO44

0.033 mol H0.033 mol H22OO

0.00702 mol CuSO0.00702 mol CuSO44

= 4.7= 4.7 ≈ ≈ 55

CuSOCuSO44 • 5H • 5H22OO

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).