LECTURE EightCHM 151 ©slg

Formulas of “Hydrates”

Unit Two:

Balancing Chemical Equations

Formulas of “Hydrates”

Unit Two:

Balancing Chemical Equations

TOPICS:

Hydrated Compounds

Crystalline ionic solids (salts!) are frequently found innature or are produced from aqueous solutions with a specific number of water molecules associated with each set of formula ions:

CuSO4. 5H2O NiCl2.6H2O CaSO4.2H2O

Frequently the color of the salt depends on the presence of these “waters of hydration.”

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 theformula of the hydrate...

Naturally occurring hydrated copper(II) chlorideis called eriochalcite. If 0.235 g of CuCl2.xH2O is heatedto 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 H2O

1Cu= 1 X 63.55 = 63.552Cl = 2 X 35.45 =70.90 134.45 g/mol CuCl2

2H= 2 x 1.008= 2.0161O= 1 x 16.00= 16.00 18.02 g/mol H2O

grams Molar mass Moles Simplest mole ratio

CuCl2 0.185 g 134.45g/mol .00138 .00138/.00138= 1H2O 0.050 g 18.02 g/mol .00278 .00278/.00138= 2

Formula of eriochalcite: CuCl2. 2H2O

Note: The correct answer will always be a whole number, with one mole of parent compound and(usually) several moles of water....

GROUP WORK

If 1.023 g of a hydrated compound, CuSO4. xH2Oshows a mass of 0.654 g when dehydrated, what is the formula of the compound? (CuSO4, 159.6 g/mol;H2O, 18.02 g/mol).

End, Unit One material

Solution:

grams Molar mass Moles Simplest mole ratio

CuSO4 0.654 g 159.6 g/mol .00410 .00410/.00410= 1H2O 0.369 g 18.02 g/mol .0205 .0205/.00410= 5

1.023 g CuSO4. xH2O- 0.654 g CuSO4

0.369 g H2O

CuSO4 . 5 H2O

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)

TOPIC ONE

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

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!

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.

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”

orThe reaction gives off heat and is described as “exothermic”

Typical chemical reactions include predictable types:

Combination: A + B --> C

Decomposition: AB ---> A + B

Double Replacement in Aqueous solution: AB + CD --> AD + CB

Combustion: CxHy + O2 --> CO2 + H2O

(We’ll consider another type, oxidation/reduction later in the unit!)

As we review balancing equations, let’s describe them by type:

Combination or “Synthesis”: A + B --> C

H2(g) + O2(g) --> H2O(g)

2 atoms O

1 atom O

UNBALANCED!

2 H2(g) + O2(g) --> 2 H2O(g)

We have now accounted for all atoms4 atoms H, left and right2 atoms O, left and rightEquation is balanced...

H2(g) + O2(g) --> 2 H2O(g)

Rebalance:

To balance O’s, place coefficient before formula:

Now 4 atoms H

NaCl(l) --> Na(l) + Cl2(g)

2NaCl(l) --> Na(l) + Cl2(g)

2NaCl(l) --> 2Na(l) + Cl2(g)

Balancing process, Decomposition Reactions: energy AB ----------> A + B

Al2(CO3)3(s) --> Al2O3(s) + CO2(g)

Al2(CO3)3(s) --> Al2O3(s) + 3CO2(g)

Balance Cl

Balance Na

Balance C,then O ok

Balancing Combustion Reactions: (C,H) + O2 --> CO2 + H2O + heat (C,H,O) + O2 --> CO2 + H2O + heat

C5H12 + O2 --> CO2 + H2O

Methodology: a) Balance C b) Balance H c) Balance O

Organic molecules:hydrocarbons, carbohydrates, alcohols: “fuels”

c) Balance O:

C5H12 + 8O2 --> 5 CO2 + 6 H2O

a) Balance C: C5H12 + O2 --> 5 CO2 + H2O

b) Balance H:

C5H12 + O2 --> 5 CO2 + 6 H2O

10 O 6 O

Now, a more interesting one: C8H18 + O2 --> CO2 + H2O

a) C8H18 + O2 --> 8CO2 + H2O

b) C8H18 + O2 --> 8CO2 + 9 H2O

c) C8H18 + 12.5 O2 --> 8 CO2 + 9 H2O

16 O + 9 O = 25 O

25 O 16 O 9 O

At this point we have:

C8H18 + 25/2 or 12.5 O2 --> 8CO2 + 9 H2O

Should have whole number coefficients, so MULTIPLY ALL SPECIES BY 2:

2 C8H18 + 25 O2 --> 16CO2 + 18 H2O

Checking for balance:

16 C 36 H 50 O -->; 16 C 36 H; 32 O +18 O

GROUP WORK, BALANCE

C6H12O6 + O2 --> CO2 + H2O (sugar)

C5H11OH + O2 --> CO2 + H2O (alcohol)

a) Balance Cb) Balance Hc) Balance Od) Check

C6H12O6 + O2 --> CO2 + H2O

C6H12O6 + O2 --> 6 CO2 + H2O

C6H12O6 + O2 --> 6 CO2 + 6 H2O

C6H12O6 + 6 O2 --> 6 CO2 + 6 H2O

6C 12H 6 O + 12 O ---> 6 C 12 O + 12 H 6 O

Solution, Sugar

12 O + 6 O = 18 O

C5H11OH + O2 --> CO2 + H2OC5H11OH + O2 --> CO2 + H2O

C5H11OH + O2 --> 5 CO2 + H2O

C5H11OH + O2 --> 5 CO2 + 6 H2O

C5H11OH + 7.5 or 15/2 O2 --> 5 CO2 + 6 H2O 1 O 15 O 10 O 6 O

#1

#2

#3

C5H11OH + 7.5 or 15/2 O2 --> 5 CO2 + 6 H2O 10 O 6 O

X 2

2 C5H11OH + 15 O2 --> 10 CO2 + 12 H2O

[10C 24H 2 O] + 30 O --> [10C 20 O] + [24H 12 O]

Balancing Double Replacement Reactions: AB + CD --> AD + CB

General directions: save H for next to last, O for last:

a) balance P:

2 H3PO4 + Ca(OH)2 --> Ca3(PO4)2 + H2O

b) balance Ca:

2H3PO4 + 3 Ca(OH)2 --> Ca3(PO4)2 + H2O

H3PO4 + Ca(OH)2 --> Ca3(PO4)2 + H2O

c) balance H

2 H3PO4 + 3 Ca(OH)2 --> Ca3(PO4)2 + H2O 6H 6H 2H

2 H3PO4 + 3 Ca(OH)2 --> Ca3(PO4)2 + 6 H2O

6H 6H 12H

d) checkout O’s:

8O + 6O ---> 8O + 6O

BALANCED!