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1 LECTURE Eight CHM 151 ©slg Formulas of “Hydrates” Unit Two: Balancing Chemical Equations Formulas of “Hydrates” Unit Two: Balancing Chemical Equations TOPICS:

2 Hydrated Compounds Crystalline ionic solids (salts!) are frequently found in nature or are produced from aqueous solutions with a specific number of water molecules associated with each set of formula ions: CuSO 4. 5H 2 O NiCl 2.6H 2 O CaSO 4.2H 2 O Frequently the color of the salt depends on the presence of these “waters of hydration.”

3 The number of water molecules associated with a particular salt is characteristic but not easy to predict: therefore the value is determined experimentally... The hydrated salt is weighed, heated carefully to drive off the water, and reweighed. The mass of the water driven off is calculated, converted to moles and compared to moles of the parent, anhydrous salt to determine the formula of the hydrate...

4 Naturally occurring hydrated copper(II) chloride is called eriochalcite. If 0.235 g of CuCl 2.xH 2 O is heated to drive off the water, 0.185 g residue remains. What is the value of x? 0.235 g hydrate - 0.185 g parent salt = 0.050 g H 2 O 1Cu= 1 X 63.55 = 63.55 2Cl = 2 X 35.45 =70.90 134.45 g/mol CuCl 2 2H= 2 x 1.008= 2.016 1O= 1 x 16.00= 16.00 18.02 g/mol H 2 O

5 grams Molar mass Moles Simplest mole ratio CuCl 2 0.185 g 134.45g/mol.00138.00138/.00138= 1 H 2 O 0.050 g 18.02 g/mol.00278.00278/.00138= 2 Formula of eriochalcite: CuCl 2. 2H 2 O Note: The correct answer will always be a whole number, with one mole of parent compound and (usually) several moles of water....

6 GROUP WORK If 1.023 g of a hydrated compound, CuSO 4. xH 2 O shows a mass of 0.654 g when dehydrated, what is the formula of the compound? (CuSO 4, 159.6 g/mol; H 2 O, 18.02 g/mol).

7 End, Unit One material Solution: grams Molar mass Moles Simplest mole ratio CuSO 4 0.654 g 159.6 g/mol.00410.00410/.00410= 1 H 2 O 0.369 g 18.02 g/mol.0205.0205/.00410= 5 1.023 g CuSO 4. xH 2 O - 0.654 g CuSO 4 0.369 g H 2 O CuSO 4. 5 H 2 O

8 UNIT TWO Overall Topics: Kotz, Chapter Four: Chemical Equations and Stoichiometry (Calculating from Balanced Equations) Kotz, Chapter Five: Reactions in Aqueous Solutions (Prediction of Products, Net Ionic Equations)

9 TOPIC ONE “The chemical equation: what it represents and how to balance...”

10 When a substance reacts with another substance or interacts with energy to form one or more different substances, a chemical reaction occurs. The changes that have taken place are represented by the chemical equation, which keeps track of all the atoms involved in the reactants and the products. The chemical equation must be carefully balanced to insure that it adequately describes the reaction without loosing or gaining any atoms in the process!

11 One cannot create or destroy matter in a chemical reaction, so all reactant atoms must show up somewhere in the product(s)... To completely describe the reaction that has taken place, the equation should include not only the formula or symbol for each reactant and product, but also their physical state, and the energy process involved. We will indicate physical state by subscripts (s) solid, (l)liquid, (g) gas, (aq) aqueous, in water solution.

12 We will not at this point add the energy component to our chemical equations but we must keep in mind that all reactions involve energy in some form: Either: The reaction requires some form of heat and is termed “endothermic” or The reaction gives off heat and is described as “exothermic”

13 Typical chemical reactions include predictable types: Combination: A + B --> C Decomposition: AB ---> A + B Double Replacement in Aqueous solution: AB + CD --> AD + CB Combustion: C x H y + O 2 --> CO 2 + H 2 O (We’ll consider another type, oxidation/reduction later in the unit!)

14 As we review balancing equations, let’s describe them by type: Combination or “Synthesis”: A + B --> C H 2(g) + O 2(g) --> H 2 O (g) 2 atoms O 1 atom O UNBALANCED!

15 2 H 2(g) + O 2(g) --> 2 H 2 O (g) We have now accounted for all atoms 4 atoms H, left and right 2 atoms O, left and right Equation is balanced... H 2(g) + O 2(g) --> 2 H 2 O (g) Rebalance: To balance O’s, place coefficient before formula: Now 4 atoms H

16 NaCl (l) --> Na (l) + Cl 2(g) 2NaCl (l) --> Na (l) + Cl 2(g) 2NaCl (l) --> 2Na (l) + Cl 2(g) Balancing process, Decomposition Reactions: energy AB ----------> A + B Al 2 (CO 3 ) 3(s) --> Al 2 O 3(s) + CO 2(g) Al 2 (CO 3 ) 3(s) --> Al 2 O 3(s) + 3CO 2(g) Balance Cl Balance Na Balance C, then O ok

17 Balancing Combustion Reactions: (C,H) + O 2 --> CO 2 + H 2 O + heat (C,H,O) + O 2 --> CO 2 + H 2 O + heat C 5 H 12 + O 2 --> CO 2 + H 2 O Methodology: a) Balance C b) Balance H c) Balance O Organic molecules: hydrocarbons, carbohydrates, alcohols: “fuels”

18 c) Balance O: C 5 H 12 + 8 O 2 --> 5 CO 2 + 6 H 2 O a) Balance C: C 5 H 12 + O 2 --> 5 CO 2 + H 2 O b) Balance H: C 5 H 12 + O 2 --> 5 CO 2 + 6 H 2 O 10 O 6 O

19 Now, a more interesting one: C 8 H 18 + O 2 --> CO 2 + H 2 O a) C 8 H 18 + O 2 --> 8 CO 2 + H 2 O b) C 8 H 18 + O 2 --> 8 CO 2 + 9 H 2 O c) C 8 H 18 + 12.5 O 2 --> 8 CO 2 + 9 H 2 O 16 O + 9 O = 25 O 25 O 16 O 9 O

20 At this point we have: C 8 H 18 + 25/2 or 12.5 O 2 --> 8 CO 2 + 9 H 2 O Should have whole number coefficients, so MULTIPLY ALL SPECIES BY 2: 2 C 8 H 18 + 25 O 2 --> 16 CO 2 + 18 H 2 O Checking for balance: 16 C 36 H 50 O -->; 16 C 36 H; 32 O +18 O

21 GROUP WORK, BALANCE C 6 H 12 O 6 + O 2 --> CO 2 + H 2 O (sugar) C 5 H 11 OH + O 2 --> CO 2 + H 2 O (alcohol) a) Balance C b) Balance H c) Balance O d) Check

22 C 6 H 12 O 6 + O 2 --> CO 2 + H 2 O C 6 H 12 O 6 + O 2 --> 6 CO 2 + H 2 O C 6 H 12 O 6 + O 2 --> 6 CO 2 + 6 H 2 O C 6 H 12 O 6 + 6 O 2 --> 6 CO 2 + 6 H 2 O 6C 12H 6 O + 12 O ---> 6 C 12 O + 12 H 6 O Solution, Sugar 12 O + 6 O = 18 O

23 C 5 H 11 OH + O 2 --> CO 2 + H 2 O C 5 H 11 OH + O 2 --> 5 CO 2 + H 2 O C 5 H 11 OH + O 2 --> 5 CO 2 + 6 H 2 O C 5 H 11 OH + 7.5 or 15/2 O 2 --> 5 CO 2 + 6 H 2 O 1 O 15 O 10 O 6 O #1 #2 #3

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