Chapter 4

Chapter 4

 The combustion of fossil fuels (or any combustion reaction) releases _____&_____

 So how much carbon dioxide? How much is released during one “cycle” or one complete balanced combustion reaction?

 The balanced chemical reaction relates the reactants and products to one another

 The combustion of fossils fuels (octane) can be represented by the following:

??

WHAT IS OCTANE??

 The balanced chemical reaction relates the reactants and products to one another

 The combustion of fossils fuels (octane) can be represented by the following:

___ C

8

H

18(l)

+ ___ O

2(g)

 ___ CO

2(g)

+ ___H

2

O

(g)

 The balanced chemical reaction relates the reactants and products to one another

 The combustion of fossils fuels (octane) can be represented by the following:

2 C

8

H

18(l)

+ 25 O

2(g)

 16CO

2(g)

+ 18H

2

O

(g)

2 C

8

H

18(l)

+ 25 O

2(g)

 16CO

2(g)

+ 18H

2

O

(g)

 This statement says that for every 2 mols of octane to burn completely you need 25 mols of oxygen gas… in this “completion” reaction

(no reactants left over) you will produce 16 mol carbon dioxide and 18 mol water vapor.

 Stoichiometry is the relationships between reactants and products in a chemical reaction

 The reaction must be balanced before making these connections

 The coefficients out front (your balanced numbers) are relationships or conversion factors between components in the reaction

 In the balanced chemical reaction, we have a

“recipe” for how the reactants combine to form the products

 From our octane reaction, we have a mole to mole relationship of

2 mol octane : 16 mol carbon dioxide

 Let’s pretend that 22.0 mols of C

8

H

18(g) are put into this reaction. How many mols of

CO

2(g) should be released?

22.0 mol C

8

H

18

16 mol CO

2

2 mol C

8

H

18

= 176 mol CO2

 What if we are given grams? Can we go into the “heart of the reaction”?

 41.5 grams of octane are burned. How many mols of CO2 will be liberated?

 108 grams of O2 is consumed. What mass of

CO2 is produced?

 4.22x10

23 molecules of octane are burned.

How many water molecules will be released?

In photosynthesis, plants convert carbon dioxide and water into glucose.

Suppose you determine a particular plant consumes 37.8 g of CO

2 in one week. Assuming there is more than enough water (excess), how much glucose will be made?

6CO

2(g)

+ 6H

2

O

(l)

6O

2(g)

+ 1C

6

H

12

O

6(aq)

37.8 g CO

2

1 mol CO

2

44.0098 g CO

2

1 mol C

6

H

12

O

6

6 mol CO

2

180.15768g C

6

H

12

O

6

1 mol C

6

H

12

O

6

= ??

How about if you were given excess carbon dioxide and 82.22 grams of water? What mass of oxygen gas would be released?

 Sulfuric acid is a component of acid rain that forms when sulfur dioxide, a pollutant, reacts with oxygen and water.

 If a company is allowed to release 25 kg of

SO

2 per year, what mass of sulfuric acid will be produced (in kg)?

 Conversions can be made from any component to any component in a chemical reaction assuming you have balanced it correctly

 However, you must be in mol to be able to make the conversion (go through the heart)

 If given any unit OTHER THAN MOLES be sure to find a way to first convert to moles

 Because, remember the molar rations (the stoichiometric coefficients) are in moles!!

 Particles (molecules, atoms, ions, compounds)  moles (use Avogadro’s #)

Grams  moles (use a molar mass)

Kilograms to moles (go to grams first)

 Each bicycle needs 2 tires, one seat, and 2 handlebars.

 How many complete bikes can be made from 6 tires, 7 seats, and 5 handlebars?

 Which one runs out first???

This is the LIMITING REACTANT

 Just like the bike example, a reaction will run out of a reactant before all parts are consumed.

 A limiting reactant is the substance that controls the quantity of product that can form in a chemical reaction.

 The handlebars were limiting in the bike problem

 The “other” reactant must be in excess if it is not limiting.

 The excess reactant is the substance that is not used up completely in a reaction.

 Some remains still.

 The tires and seats were in excess in the bike example.

 Perhaps you do not have “excess” of one reactant (like if you are given information on both reactants).

 Meaning, you may run out of one or the other before completion

 You must determine which one will “control” the outcome… the one the runs out, or is limited, will control the reaction

 If you are given quantities of all the reactants involved, it is a Limiting Reactant problem.

 You must first determine which reactant is limiting before you can calculate the theoretical amount, or theoretical yield, of the products.

1.

Rules for making L.R. calculations:

2.

3.

4.

5.

6.

Write out and balance the equation.

Determine if the problem is an L.R.

(quantities of all reactants are given)

Convert all quantities to moles

Divide each value of moles by its own corresponding stoichiometric coefficient.

Whichever number is smallest is the L.R.

Calculate the products using that reactant.

Identify the limiting reactant and the theoretical yield of phosphorous acid, H

3

PO

3

, if

225 g of PCl

3

___PCl

3 is mixed with 125 g of H

2

O.

+ ___H

2

O  ___H

3

PO

3

+ ___HCl

___CuO

(s)

+ ___H

2(g)

 ___Cu

(s)

+ ___H

2

O

(g)

What is the limiting reactant when 19.9 g CuO react with 2.02 g H

2

?

 Equations tell you what should happen in a reaction, not always what does happen

 The “theoretical yield” is what SHOULD be produced. What is calculated stoichiometrically.

 The “actual yield” is what is ACTUALLY produced in lab by a reaction.

Percentage Yield is the measured amount of a product of a reaction.

 In lab, we may only get 85% of the desired amount. Not everything is perfect in this world 

 The percent yield is a ratio of the ACTUAL yield and the THEORETICAL yield.

% Yield = Actual Yield x100 = %

Theoretical Yield

What is the percent yield if stoichiometry tells us we should get 14.7 g of product but actually only get 10.85 grams?

Determine the limiting reactant, the theoretical yield, and the percentage yield if

14.0 g N

2

NH

3 are mixed with 9.0 g H are recovered.

2

, and 16.1 g

___N

2

+ ___H

2

 ___NH

3

 Even though the actual yield can only be determined experimentally, we can estimate the amount if we know the percentage yield.

 Say a reaction usually yields about 80.5% of the theoretical yield. We have enough to make 200 grams (theoretically). We can assume that we will get around 80.5% of

200 g or roughly 161 grams.

How many grams of C

7

H

8

O

2 should form if

4808 g are theoretically possible and the percent yield for the reaction is 83.2%?

The percent yield of NH

3 from the following reaction is 85.0%. What actual yield is expected from the reaction of 1.00 kg N

2 with 225 g H

2

?

 Concentration – the amount of particular substance in a given quantity of a solution.

 Concentration can be expressed in many forms:

 ppm – parts per million

 molarity – moles solute over liters solution

 molality – moles solute over kg of solvent

 We can make these calculations if we know the quantity of solute and volume (quantity) of solvent.

 We often use molarity when dealing with concentrations.

 Molarity – a concentrations unit of a solution expressed as moles of solute dissolved per liter of solution.

 How MUCH solute is in how MUCH solvent

M = mol / L

 Molarity describes a solution in terms of volume

of solution not volume of solvent.

 If you need 1 Liter of solution, you are NOT adding 1 Liter of solvent. You must add “enough” solvent to make the total volume 1 Liter.

 The solute will take up some of the volume.

 You can determine the number of moles of solute if you know the mass used.

 What is the molarity of a potassium chloride solution that has a volume of 400.0 mL and contains 85.0 g KCl?

 Vinegar contains 5.0 g acetic acid, CH in 100.0 mL of solution. Calculate the

3

COOH, molarity of the acetic acid in the vinegar.

 What mass of KCl is present in 25 mL of a 0.85

M solution of potassium chloride?

 Solutions of known concentrations are often used in the lab.

 When reactions involving these solutions occur, we can use stoichiometry to calculate the mass of products we should obtain.

 We could then use this amount of product formed along with the total volume of the system to determine the concentration of product formed.

 An excess of zinc is added to 125 mL of 0.100

M HCl solution. What mass of zinc chloride is formed?

___ Zn + ___ HCl  ___ ZnCl

2

+ ___H

2

 What volume (in milliliters) of a 0.500 M solution of copper (II) sulfate is needed to react with an excess of aluminum to provide

11.0 grams of copper?

 Hint: Step 1: Write out and balance your reaction

 Ionic compounds do not dissolve

 Ionic (a metal and a nonmetal) will “dissociate”

(ionize) or come apart into its cation and anion

 Covalent molecules DO dissolve

 If there are ions in solution (ionic compounds dissociate) then the solution will be able to carry a charge… this is said to be

“electrolytic”

 An electrolyte is a substance that will come apart in water to form solutions that will carry a charge

 If a compound COMPLETELY ionize, then it will be a strong electrolyte (make a strong electrolytic solution)

 If it does not dissociate into its ions (like covalent, i.e. sugar) it will not carry a charge

 We call this nonelectrolytic

 What do you know about acids?

 What are the 6 strong acids?

 What does this mean???

 Your strong acids will dissociate completely!!

 They are great conductors of electricity (strong electrolytes) – Strong Acids!!

 If your solubility rules tell you an ionic compound is soluble then it will dissociate and be electrolytic

 Think about the rules for hydroxide

 All group 1 hydroxides are soluble, therefore all of those will be strong electrolytes (dissociate completely) – Strong Base!!

 Weak acids DO NOT dissociate completely

 They usually only dissociate less than 1%

 This fact makes them poor conductors of electricity. But, there are still ions in solution so they do conduct a little bit

 Makes them weak electrolytes and will make weak electrolytic solutions

 The more a compound, acid, base, etc. dissociates, the stronger its ability to conduct electricity will be.

 The less it comes apart, the weaker it will be

 If it does not at all…. It will not conduct at all

(nonelectrolytic)

 Precipitation reactions

 Makes an insoluble product (remember solubility rules)

 T.I.E.s and N.I.E.s

 Spectator ions

 In an acid-base reaction, the acid (H + ) will react with the base (OH ) to make water and a salt

 This neutralization reaction is a special type of double displacement

 Acid – substance that produces H + ions in aqueous solutions

 Base – substance that produces OH ions in aqueous solution

 Think about the ionic equation for a strong acid

HCl

(aq)

 H +

(aq)

+ Cl -

(aq)

 The complete dissociate releases 100% hydrogen ions in solution (another characteristic of a strong acid)

 The same can be considered with a strong base (using solubility rules, what is a strong base?)

NaOH

(aq)

 Na +

(aq)

+ OH -

(aq)

 The dissociation of a SA and SB (100%) does not go back and forth… meaning it is a completion reaction

 WA, WB, and other weak electrolytes will be reversible… these reactions will go back and forth and will be a mixture of both reactants and products

 Another sign of weakness… why is this signify weak electrolytes?

 In a titration, a substance in a solution of known concentration is reacted with another substance in a solution of unknown concentration

 A titration reacts ALL of the acid (H+) with

ALL of the base (OH-)

 This point is known as the equivalence point

Equivalence point – the point in the titration when the number of moles hydroxide equals the number of moles of hydrogen

(hydronium) in solution

 At the equivalence point, the moles of A = the moles of B

 Using stoich, the concentration of the unknown can be determined.

 To see this equivalence, we use an indicator solution that will exhibit different colors in a base than in an acid

 Phenolphthalein changes from clear (acid) to deep pink (base) when you reach the equivalence point

 The color change (that you SEE) is called the end point

Oxidation-Reduction Reactions or REDOX

reactions are reactions in which electrons are transferred from one reactant to the other

 Examples:

 Rusting of a car

 The bleaching of hair

 The production of electricity in batteries

 Many redox reactions involve the reaction of a substance with oxygen

 However, redox reaction do not need to involve oxygen

2Na

(s)

+ 1Cl

2(g)

 2NaCl

(s)

4Na

(s)

+ 1O

2(s)

 2Na

2

O

(s)

 In both of these cases, a metal (which typically loses electrons) reacts with a nonmetal (which has the tendency to gain electrons)

 In both, the metal loses and the nonmetal gains

 A fundamental definition:

Oxidation is the loss of electrons

Reduction is the gain of electron

 OIL RIG

 Oxidation is losing, reduction is gaining

 Identifying a reaction between a metal and a nonmetal is fairly easy due to the ion formation

 However, how can we identify REDOX reactions involving two nonmetals?

 We need to “track” the electrons through the reaction to see who is losing and who is gaining them (from reactants to products)

 An oxidation state or an oxidation number is assigned to each element

 This is kind of its “charge”

1.

2.

3.

4.

5.

The oxidation state of an atoms in a free element is 0

The oxidation state of a monoatomic ion is equal to its charge

The sum of the oxidation states of all atoms in

▪ A neutral molecule is 0

▪ An ion is equal to the charge of the ion

In their compounds, metals have positive oxidation states

Group 1A always have a +1

Group 2 A always have a +2

Use the table for nonmetals (17 has a -1, 16 has a -2, and so on)

3.

4.

5.

1.

2.

Examples

Cu

0 ox state

Cl

2

Ca +2 Cl -1

+2 ox -1 ox

H

2

NO

NaCl, the Na has a

+1 ox state

Na

O, overall is 0

2

3

-1 , overall is -1

O, the O has a -2 ox state

 Assign an oxidation state to each atom in each of the following compounds

Cl

2

Na +

KF

CO

2

SO

4

-2

K

2

O

2

170

 Oxidation states can be used to identify

REDOX reactions

 Remember, OIL RIG

 Look for this in a reaction

Ex: 1C + 2S  1CS2

Which element is reduced? Which is oxidized?

 Oxidation is an increase in oxidation state

 Reduction is a decrease in oxidation state

 OIL RIG  oxidation is losing electrons and reduction is gaining electrons

 Use oxidation states to identify the element that is being oxidized and the elements that is being reduced in the following REDOX reaction

1Mg

(s)

+ 2H

2

O

(l)

 1Mg(OH)

2(aq)

+ 1H

2(g)

0 +1 -2 +2 -2 +1 0

 What is being “oxidized” must be doing the reducing and can be called the “reducing

agent

 And… Just the opposite:

 What is being “reduced” must be doing the oxidizing and can be called the “oxidizing

agent

 Which of the following is a redox reaction?

For the redox rxns, identify the oxidizing agent and the reducing agent

2Li

(s)

+ 1Cl

2(g)

2LiCl

2(s)

2Al

(s)

+ 3Sn 2+

(aq)

2Al 3+

(aq)

+ 3Sn

(s)

1C

(s)

+ O

2(g)

CO

2(g)

END OF CHAPTER 4

 Practice Problems assigned from book are excellent review / study / test questions