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CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES

CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

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Page 1: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

CHE 111 - Module 4

CHAPTER 4 & 5

LECTURE NOTES

Page 2: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Stoichiometry & Balancing Equations

• Remember we stated in the previous chapter that stoichiometry is the study of the quantitative relationships between the amounts of reactants and products in chemical reactions.

• We use BALANCED equations to understand stoichiometric relationships of the elements and compounds within a chemical reaction.

Page 3: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

The Balanced Equation

2Al(s) + 3Br2(l) Al2Br6(s)

2mol of Al : 3mol of Br2 : 1mol of Al2Br6

2 atoms of Al = 2 atoms of Al

6 atoms of Br = 6 atoms of Br

The number of the same atom of each element must

be equal on each side of the equation.

Page 4: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

A Closer Look at the Equation

2Al(s) + 3Br2(l) Al2Br6(s)

• The chemicals on the left are the reactants and the right are the products.

• The coefficient in front of the chemical denotes the stoichiometric relationship.

• The numerical subscript represents the number of atoms present in the molecule.

• The letter subscripted denotes the phase of matter.

Page 5: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Balancing Equations

For example the following is balanced.CH4 + 2O2 CO2 + 2H2O

Try to balance the following:• Fe2S3 + O2 Fe2O3 + S

• Al + H2SO4 Al2(SO4)3 + H2

• Ca + Al2Br6 CaBr2 + Al

Page 6: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Balanced Equations

Check your answer from the previous slide:

• 2Fe2S3 + 3O2 2Fe2O3 + 6S

• 2Al + 3H2SO4 Al2(SO4)3 + 3H2

• 3Ca + Al2Br6 3CaBr2 + 2Al

Page 7: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Types of Reactions

• Combination Reactions • Decomposition Reactions • Displacement (Single-Replacement)

Reactions • Metathesis (Double-Replacement)

Reactions• Combustion Reactions

Page 8: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Combination Reactions

• A combination reaction is a reaction where two substances chemically combine to form another substance.

A + B AB

2Na(s) + Cl2(g) 2NaCl(s)

P4(s) + 6Cl2(g) 4PCl3(s)

Page 9: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Decomposition Reaction

• A decomposition reaction is when a single compound decomposes into two or more other substances.

AB A + B

2KClO3(s) 2KCl(s) + 3O2(g)

2NO2(g) 2NO(g) + O2(g)

Page 10: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Displacement Reaction

• A displacement (single replacement) reaction is a reaction where one element displaces another element.

A + BC B + AC

Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

Cu(s) + 2AgNO3(aq) Cu(NO3)2(aq) + 2Ag(s)

Page 11: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Metathesis Reaction• A metathesis (double replacement)

reaction is a reaction where two compounds switch cations to form two new compounds.

A+B- + C+D- A+D- + C+B-

CaCl2(aq) + Na2CO3(aq) CaCO3(s) + 2NaCl(aq)

AgNO3(aq) + KCl(aq) ???

Page 12: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

SOLUBILITY

• Solubility – the amount of a substance that can be dissolved in a given quantity of solvent (like water) at a specific temperature

• Unsaturated – amount of substance less than saturated

• Saturated – the exact amount at solubility• Supersaturated – excess amount of substance

Page 13: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

How Solubility Influences Rxn

• When a substance is soluble in water, it will appear with a subscript of (aq) meaning that the substance is broken up into it’s ions incorporated into the water lattice.

• When a substance is insoluble in water, it will be written with a subscript of (s), (l), or (g) and will precipitate out of solution.

Page 14: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Solubility of Ionic Compounds in Water

Sodium, potassium, and ammonium compounds

Acetates and nitrates

Halides (chlorides, bromides, Lead(II), silver, and mercury (I) and iodides) halides are insoluble

Sulfates Calcium, strontium, barium, and Lead(II) sulfates are insoluble

Soluble Compounds Exceptions

Page 15: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Insolubility of Ionic Compounds in Water

Carbonates and phosphates Sodium, potassium, and ammonium compounds are soluble.

Hydroxides Sodium, potassium, calcium, strontium, and barium compounds are soluble

Sulfide Sodium, potassium, calcium,and ammonium compoundsare soluble

Insoluble Compounds Exceptions

Page 16: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

A Look at Metathesis Again

• Looking back at slide 10 to the first reaction: when the cations rearranged, the CaCO3 being insoluble by our definition is recorded as CaCO3 (s). The CaCO3 would precipitate out of solution as a solid.

• Looking at AgNO3(aq) + KCl(aq) ?, we can rearrange the cations and conclude that the AgCl is a solid and will precipitate out of solution

Page 17: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Types of Metathesis Reactions

• Three classifications of metathesis reactions– Precipitation reaction - formation of a solid

Pb(NO3)2(aq) + Na2CO3(aq) PbCO3(s) + 2NaNO3(aq)

– Neutralization reaction - formation of water

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

– Gas formation reaction - CO2, H2S, SOx, & NOx are typically formed

Page 18: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Precipitation Reactions

• A solid precipitate is produced in the rearrangement of cations as follows:

Pb(NO3)2(aq) + Na2CO3(aq) PbCO3(s) + 2NaNO3(aq)

• The Ionic Equation is expressed as: Pb+2 + 2NO3

- + 2Na+ + CO3-2 PbCO3(s) + 2Na+ + 2NO3

-

• After neglecting the spectator ions, the net ionic equation will look like:

Pb+2(aq) + CO3

-2(aq) PbCO3(s)

Page 19: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Reviewing Ionic Compounds

Ca+2 + 2Cl- CaCl2

Each ion comes together based on charge to form an overall neutral ionic compound.

3Ca+2 + 2PO4-3 Ca3(PO4)2

The cation and the polyatomic ion come together based on charge to form an overall neutral ionic compound.

Page 20: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Net Ionic Equations (NIE)

• If you were given the reactants Ca(NO3)2 and Na3PO4 you should be able to predict the precipitate and write a balance equation, the ionic equation, and the net ionic equation (NIE) for this reaction.

• The NIE for these reactants is as follows:

3Ca+2(aq) + 2PO4

-3(aq) Ca3(PO4)2(s)

Page 21: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Common Polyatomic Ions

• carbonate ion CO3-2

• sulfate ion SO4-2

• sulfite ion SO3-2

• hydroxide OH-

• phosphate PO4-3

• permanganate MnO4-

• chromate CrO4-2

• dichromate Cr2O7-2

• ammonium NH4+

• oxalate C2O4-2

• bicarbonate HCO3-

• cyanide ion CN-

• acetate C2H3O3

-

Page 22: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Neutralization Reaction

• A neutralization reaction is a reaction that occurs between an acid and a base with the production of a salt and water.

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

acid base salt water

Page 23: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Gas Formation Reaction

• A gas formation reaction is a metathesis reaction that generates a gas as a product.

– Metal carbonates or bicarbonates + acid– Metal sulfides + acid– Metal sulfites + acid– Ammonium salts and strong base

Page 24: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Metal Carbonates

• Metal carbonates or bicarbonates when combined with an acid form salt, water and carbon dioxide gas.

Na2CO3(aq)+ 2HCl(aq) 2NaCl(aq)+ H2O(l)+ CO2(g)

– Where CO2 gas is given off

Page 25: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Metal Sulfides

• Metal sulfides when combined with an acid form salt and hydrogen sulfide gas.

Na2S(aq)+ 2HCl(aq) 2NaCl(aq)+ H2S(g)

– Where H2S gas is given off

Page 26: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Metal Sulfites

• Metal sulfites when combined with an acid form salt, water, and sulfur dioxide gas.

Na2SO3(aq)+ 2HCl(aq) 2NaCl(aq)+ H2O(l)+ SO2(g)

– Where SO2 gas is given off

Page 27: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Ammonium Salts

• Ammonium salts when combined with a base produce salt, water and ammonia.

NH4Cl (aq)+ NaOH(aq) NaCl(aq)+ H2O(l)+ NH3(g)

– Where ammonia gas is given off

Page 28: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Combustion Reactions

• A combustion reaction is a reaction with molecular oxygen to form products in which all elements are combined with oxygen.

CH4 + 2O2 CO2 + 2H2O

Page 29: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Limiting Reactants

• One of the reactants is in limited supply and thus restricts the amount of product formed.

• Think of it as: If you wanted to bake a batch of peanut butter cookies and the recipe calls for 1 cup of peanut butter and all you have is ½ a cup, even though you have all the other ingredients, you can at most make ½ a batch of cookies.

Page 30: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Limiting Reactants (cont.)

• Consider the combustion reaction:

CH4 + 2O2 CO2 + 2H2O

• How much CO2 can be produced if you have 0.13g of methane and 0.45g of O2?

Page 31: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Percent Yield

• The maximum amount of product that can be obtained from a chemical reaction is the theoretical yield.

• The actual amount produced in a chemical process is the actual yield.

• The percent yield is equal to the actual yield divided by the theoretical yield times 100%.

Page 32: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Redox Reactions

• Oxidation of an element takes place when electrons are lost from the valence shell of the element.

• Reduction of an element takes place when electrons are added to the valence shell of the element.

• Redox reactions show the transfer of electrons that takes place during oxidation and reduction.

Page 33: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Redox Reactions (cont.)

• All oxidation and reduction reactions involve transfer of electrons between substances.

• View CD-ROM screen 5.12

• Ag+ accepts electrons for Cu and is reduced to Ag and Cu loses electrons to Ag+ and is oxidized to Cu+2 in the following redox rxn:

2Ag+(aq) + Cu(s) 2Ag(s) + Cu+2

(aq)

Page 34: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Redox Reactions (cont.)

• The oxidation half reaction is :

Cu(s) – 2e- Cu+2(aq)

• The reduction half reaction is:

2Ag+(aq) + 2e- 2Ag(s)

• Cu is called the reducing agent because it caused Ag+ to be reduced; and Ag+ is called the oxidizing agent because it caused Cu to be oxidized.

Page 35: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Determining Oxidation Numbers• Each atom in a pure element has an oxidation

number of zero.• For monoatomic ions, the ox. number is equal to

it’s ionic charge.• F is always –1, other halogens are –1 as well

except with oxygen or fluorine.• The ox. number for H is +1 except with hydrides

(CaH2).and O is –2 except with peroxides (H2O2).

• The ox.# must = 0 for a compound or = to the overall charge of polyatomic ion being considered.

Page 36: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Balancing Redox Reactions

• We can use the balance of electrons transferred in a redox reaction to help us balance the overall equation.

• Consider the unbalanced equation:

Zn(s) + HCl(aq) ZnCl2(aq) + H2(g)

• The balanced equation takes into consideration the oxidation of the Zn and the reduction of the H+.

Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

Page 37: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Molarity

Molarity = Moles of Solute Liters of Total

Solution

Symbol for molarity is M

Units are moles/Liter

Page 38: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Solution Preparation• To prepare a 1.0M solution of NaCl, you would

determine how many grams of NaCl is contained in 1.0 moles of NaCl and then dissolve that amount in a 1.0L volumetric flask. You would then qs with distilled H2O.

• 1.0M NaCl = 1 mole (or 58.44g) NaCl 1.0L of solution• How much NaCl would you use to make a 0.1M

solution of NaCl?

As 1/10 of a mole = 5.844g NaCl, you would dissolve 0.1mole (5.844g) of NaCl in 1.0L of solution.

Page 39: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Acids

An acid is defined as follows:• Arrhenius – releases H+ when dissolved in

H2O• Bronsted-Lowrey – a substance that can

donate a proton to another substance• Lewis – a substance that can accept a pair

of electrons from another atom to form a new bond

Page 40: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Bases

A base is defined as follows:• Arrhenius – releases OH- when dissolved in

H2O• Bronsted-Lowrey – a substance that can

accept a proton from another substance• Lewis – a substance that can donate a pair

of electrons to another atom to form a new bond

Page 41: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

pH and Concentrations of Acids and Bases

pH = -log [H+]

1 – acidic – 7 – basic – 14

When dealing with [H+] less than 0.1M (pH=1), we use activity coefficients instead of pH.

Page 42: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

pH of Household Items

• pH of vinegar = 2.80

• pH of soda = 2.90

• pH of orange juice = 3.80

• pH of pure water = 7.00

• pH of blood = 7.40

• pH of ammonia = 11.00

• pH of oven cleaner = 11.7

Page 43: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the

Titration

• A method for quantitative analysis of a substance by essentially complete reaction in solution with a measured quantity of a reagent of known concentration.

• Often used in redox reactions – Many redox reactions go rapidly to completion in

aqueous media to determine the equivalency point.

• Typically used for neutralization reactions.– Acid is titrated with a base using an indicator to

determine the equivalency point of the neutralization reactions.

Page 44: CHE 111 - Module 4 CHAPTER 4 & 5 LECTURE NOTES. Stoichiometry & Balancing Equations Remember we stated in the previous chapter that stoichiometry is the