chapter 4 - chemical quantities

CHAPTER 4CHEMICAL QUANTITIES AND AQUEOUS REACTIONS

Global Warming• Scientists have measured an average

0.6 °C rise in atmospheric temperature since 1860

• During the same period atmospheric CO2 levels have risen 25%

• Are the two trends causal?

3

The Sources of Increased CO2• One source of CO2 is the combustion reactions of fossil fuels we use to get energy

• Another source of CO2 is volcanic action• How can we judge whether the increase in CO2 is natural or due to our use of fossil fuels?

Tro: Chemistry: A Molecular Approach, 2/e

4

Quantities in Chemical Reactions• The amount of every substance used and made in a

chemical reaction is related to the amounts of all the other substances in the reaction• Law of Conservation of Mass• Balancing equations by balancing atoms

• The study of the numerical relationship between chemical quantities in a chemical reaction is called stoichiometry.

© 2009, Prentice-Hall, Inc.

Stoichiometric Calculations

The coefficients in the balanced equation give the ratio of moles of reactants and products.

Stoichiometry

• stoichiometry: a numerical relationship between chemical quantities in a balanced chemical equation• allows one to predict the amounts of products that form in a

chemical reaction based on the amounts of reactants

3 H2(g) + N2(g) 2 NH3(g)

For every molecule of N2 consumed in the reaction there are 3 molecules of H2 and 2 molecules of NH3.

Equivalence for Chemical Reactions• Since chemists don’t usually think of things in terms of

molecules, you can transform these relationships into mole to mole values.

3 H2(g) + N2(g) 2 NH3(g)

you can determine several equivalent statements~ (mole ratios)• For every 3 moles of H2 you can produce 2 moles of NH3.• For every 1 mole of N2 you can produce 2 moles of NH3.• For every 3 moles of H2 you will need 1 mole of N2 to perform

this reaction.

Mole – Mole Relationships• putting stoichiometric relationships to work…

3 H2(g) + N2(g) 2 NH3(g)

• If we have 3 moles of N2 and more than enough H2, how much NH3 can we make?

1 N2 ≡ 2 NH3 conversion factor 2 moles NH3

1 mole N2

moles N2 moles NH3

3 moles N2 x 2 moles NH3 = 6 moles NH3

1 mole N2

Practice Problems (moles to moles)3 H2(g) + N2(g) 2 NH3(g)

• How many moles of N2 are needed to completely react with 6.75 moles of H2?

• How many moles of H2 are required to produce 4.50 moles of NH3?

Practice Problems (grams to grams)

3 H2(g) + N2(g) 2 NH3(g)

• How many grams of N2 are needed to completely react with 45.30 grams of H2?

• How many grams of NH3 form when 16.90 grams of N2 react?

Estimate the mass of CO2 produced in 2007 by the combustion of 3.5 x 1015 g gasolne

• Assuming that gasoline is octane, C8H18, the equation for the reaction is

2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g)

11


Which Produces More CO2; Volcanoes or Fossil Fuel Combustion?

• Our calculation just showed that the world produced 1.1 x 1016 g of CO2 just from petroleum combustion in 2007• 1.1 x 1013 kg CO2

• Estimates of volcanic CO2 production are 2 x 1011 kg/year

• This means that volcanoes produce less than 2% of the CO2 added to the air annually

12

%.%.

.81100

1011

1002

yrkg13

yrkg11



How Many Cookies Can I Make?

• You can make cookies until you run out of one of the ingredients.

• Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat).


How Many Cookies Can I Make?

• In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make.


Limiting Reactants

• The limiting reactant is the reactant present in the smallest stoichiometric amount.• In other words, it’s the reactant you’ll run out of first (in this

case, the H2).

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

Limiting Reactant Concept• Reactants will not always be present in the exact

amounts necessary for all reactants to be converted completely into products.

• limiting reactant (limiting reagent): the reactant that controls the amount of products that can be produced• a limiting reactant is used up before the other reactants; all

other reactants are said to be “in excess”


Limiting ReactantsIn the example below, the O2 would be the excess reagent.

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

Determining the Limiting Reactant1. Calculate the amount (moles, mass, volume) of

product formed when using the amount of each reactant given

2. Whichever reactant produces the smallest amount (smallest mass) of product is your limiting reactant

3. All other reactants are “in excess”

Sample Problem2 Al (s) + 6 HCl (aq) 2 AlCl3 (aq) + 3 H2 (g)

• How many moles of H2 are produced when 5.00 moles Al react with 5.00 moles HCl?• Limiting reactant …• Reactant in excess …• Moles H2 produced …

Theoretical Yield• The theoretical yield is the maximum amount of product

that can be made.• In other words it’s the amount of product possible as calculated

through the stoichiometry problem.

• This is different from the actual yield, which is the amount one actually produces and measures when performing the experiment.

Sample Problem2 Al (s) + 6 HCl (aq) 2 AlCl3 (aq) + 3 H2 (g)

• How many grams of H2 are produced when 4.55 grams of Al reacts with 6.05 grams of HCl?

Percent YieldOne finds the percent yield by comparing the amount actually obtained (actual yield) to the amount it was possible to make (theoretical yield).

Actual YieldTheoretical YieldPercent Yield = x 100

Percent Yield2 Al (s) + 6 HCl (aq) 2 AlCl3 (aq) + 3 H2 (g)

• How many grams of H2 are produced when 4.55 grams of Al reacts with 6.05 grams of HCl?

• For the problem above, what would the percent yield be if only 0.100 grams of hydrogen gas were obtained from the reaction? ©

2009,

Prentice-

Hall,

Inc.

Sample Problems• The reaction below is performed using 1.56

grams of copper(II) nitrate and 1.00 grams of sodium carbonate. • What is the theoretical yield for the experiment?

• What would the percent yield be, if only 0.875 grams of copper(II) carbonate was obtained?

Cu(NO3)2(aq) + Na2CO3(aq) → CuCO3(s) + 2 NaNO3(aq)

Sample Problem• 32.7 g Zn reacted with 18.23 g HCl. If 0.252 g H2 was

captured, what is the percent yield of H2 obtained?

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


SolutionsSolutions

• Solutions are defined as homogeneous mixtures of two or more pure substances.

• The solvent is present in greatest abundance.

• All other substances are solutes.

Describing Solutions• Because solutions are mixtures, the composition can vary from

one sample to another• pure substances have constant composition• saltwater samples from different seas or lakes have different amounts of

salt

• So to describe solutions accurately, we must describe how much of each component is present• we saw that with pure substances, we can describe them with a single

name because all samples are identical

27


Solution Concentration• Qualitatively, solutions are often

described as dilute or concentrated

• Dilute solutions have a small amount of solute compared to solvent

• Concentrated solutions have a large amount of solute compared to solvent

28


Concentrations—Quantitative Descriptions of Solutions

• A more precise method for describing a solution is to quantify the amount of solute in a given amount of solution

• Concentration = amount of solute in a given amount of solution• occasionally stated in terms of the amount of solvent

29


Solution ConcentrationMolarity

• Moles of solute per 1 liter of solution• Used because it describes how many molecules of solute in

each liter of solution

30


moles of solute

volume of solution in litersMolarity (M) =

Preparing 1 L of a 1.00 M NaCl Solution

31


What is the molarity of a solution that has 25.5 g KBr dissolved in 1.75 L of solution?


What Is the molarity of a solution containing 3.4 g of NH3 in 200.0 mL of solution?

Using Molarity in Calculations• Molarity shows the relationship between the moles of

solute and liters of solution• If a sugar solution concentration is 2.0 M, then 1 liter of

solution contains 2.0 moles of sugar • 2 liters = 4.0 moles sugar • 0.5 liters = 1.0 mole sugar

• 1 L solution : 2 moles sugar

33


or

What volume of 0.125 M NaOH (in Liters) contains a total of 0.255 mol of NaOH?


Determine the mass of CaCl2 in 1.75 L of 1.50 M CaCl2 solution.

What mass of copper(II) sulfate pentahydrate would be required to prepare 250.0 mL of a 1.00 M solution of CuSO45 H2O?


How would you prepare 250.0 mL of 0.150 M CaCl2?

Dilution• Often, solutions are stored as concentrated stock solutions

• To make solutions of lower concentrations from these stock solutions, more solvent is added• the amount of solute doesn’t change, just the volume of solution

moles solute in solution 1 = moles solute in solution 2

• The concentrations and volumes of the stock and new solutions are inversely proportional

M1∙V1 = M2∙V2

36

To what volume should you dilute 0.200 L of 15.0 M NaOH to make a 3.00 M NaOH solution?

37


What is the concentration of a solution prepared by diluting 45.0 mL of 8.25 M HNO3 to 135.0 mL?

How would you prepare 200.0 mL of 0.25 M NaCl solution from a 2.0 M solution?

Solution Stoichiometry• Because molarity relates the moles of solute to the liters

of solution, it can be used to convert between amount of reactants and/or products in a chemical reaction

38


What volume of 0.150 M KCl is required to completely react with 0.150 L of 0.175 M Pb(NO3)2 in the reaction below?

2 KCl(aq) + Pb(NO3)2(aq) PbCl2(s) + 2 KNO3(aq)

39


If 43.8 mL of 0.107 M HCl is needed to neutralize 37.6 mL of Ba(OH)2 solution. What is the molarity of the barium hydroxide solution?

2 HCl + Ba(OH)2 BaCl2 + 2 H2O

What Happens When a Solute Dissolves?• There are attractive forces between the solute particles holding them together

• There are also attractive forces between the solvent molecules

• When we mix the solute with the solvent, there are attractive forces between the solute particles and the solvent molecules

• If the attractions between solute and solvent are strong enough, the solute will dissolve

40



DissociationDissociation

• When an ionic substance dissolves in water, the solvent pulls the individual ions from the crystal and solvates them.

• This process is called dissociation.


ElectrolytesElectrolytes• The result is a solution with

free moving charged particles able to conduct electricity

• An electrolyte is a substance that dissociates into ions when dissolved in water.

• A nonelectrolyte may dissolve in water, but it does not dissociate into ions when it does so.

Electrolytes and Nonelectrolytes

• Materials that dissolve in water to form a solution that will conduct electricity are called electrolytes

• Materials that dissolve in water to form a solution that will not conduct electricity are called nonelectrolytes

43


Molecular View of Electrolytes and Nonelectrolytes• To conduct electricity, a material must have charged particles that are able to flow

• Electrolyte solutions all contain ions dissolved in the water• ionic compounds are electrolytes because they dissociate into

their ions when they dissolve

• Nonelectrolyte solutions contain whole molecules dissolved in the water• generally, molecular compounds do not ionize when they

dissolve in water• the notable exception being molecular acids

44


Salt vs. Sugar Dissolved in Water45


ionic compounds dissociate into ions when

they dissolve

molecular compounds do not dissociate when

they dissolve

Acids• Acids are molecular compounds that ionize when they dissolve in water• the molecules are pulled apart by their attraction for the water• when acids ionize, they form H+ cations and also anions

• The percentage of molecules that ionize varies from one acid to another

• Acids that ionize virtually 100% are called strong acidsHCl(aq) H+

(aq) + Cl−(aq)

• Acids that only ionize a small percentage are called weak acids

HF(aq) H+(aq) + F−

(aq)

46


©

2009,

Prentic

e-Hall,

Inc.

Electrolytes• A strong electrolyte

dissociates completely when dissolved in water.• ionic compounds and

strong acids

• A weak electrolyte only dissociates partially when dissolved in water.• weak acids

When compounds containing a polyatomic ion dissolve, the polyatomic ion stays together

HC2H3O2(aq) H+(aq) + C2H3O2−(aq)

Practice – Write the equation for the process that occurs when the following strong electrolytes dissolve in water

48

CaCl2

HNO3

(NH4)2CO3

CaCl2(aq) Ca2+(aq) + 2 Cl−(aq)

HNO3(aq) H+(aq) + NO3−(aq)

(NH4)2CO3(aq) 2 NH4+(aq) + CO3

2−(aq)


Solubility of Ionic Compounds• Some ionic compounds, such as NaCl, dissolve very well in water at room temperature

• Other ionic compounds, such as AgCl, dissolve hardly at all in water at room temperature• Compounds that dissolve in a solvent are said to be soluble• Compounds that do not are said to be insoluble

NaCl is soluble in water, AgCl is insoluble in water• the degree of solubility depends on the temperature• even insoluble compounds dissolve, just not enough to be

meaningful

49


When Will a Salt Dissolve?• Predicting whether a compound will dissolve in water is

not easy• The best way to do it is to do some experiments to test

whether a compound will dissolve in water, then develop some rules based on those experimental results• we call this method the empirical method

50


Practice – Determine if each of the following is soluble in water

52

KOH

AgBr

CaCl2

Pb(NO3)2

PbSO4

KOH is soluble because it contains K+

AgBr is insoluble; most bromides are soluble, but AgBr is an exception CaCl2 is soluble; most chlorides are soluble, and CaCl2 is not an exception

Pb(NO3)2 is soluble because it contains NO3−

PbSO4 is insoluble; most sulfates are soluble, but PbSO4 is an exception


Precipitation Reactions• Precipitation reactions are reactions

in which a solid forms when we mix two solutions• the solid (insoluble product) is called a

precipitate

• When one mixes ions that form compounds that are insoluble (as could be predicted by the solubility guidelines), a precipitate is formed.

53


No Precipitate Formation = No Reaction

55

KI(aq) + NaCl(aq) KCl(aq) + NaI(aq)all ions still present, no reaction


56


Predict the products and balance the equation for each interaction below:

KCl(aq) + AgNO3(aq)

Na2S(aq) + CaCl2(aq)

Write the balanced chemical equation for the precipitation reaction between an aqueous solution of potassium carbonate and an aqueous solution of nickel(II) chloride.

Write an equation for the reaction that takes place when an aqueous solution of (NH4)2SO4 is mixed with an aqueous solution of Pb(C2H3O2)2.

57


©

2009,

Prentic

e-Hall,

Inc.

Molecular Equation

The molecular equation lists the reactants and products in their molecular form.

AgNO3 (aq) + KCl (aq) AgCl (s) + KNO3 (aq)

©

2009,

Prentic

e-Hall,

Inc.

Complete Ionic Equation• In the complete ionic equation all strong electrolytes

(strong acids, strong bases, and soluble ionic salts) are dissociated into their ions.

• This more accurately reflects the species that are found in the reaction mixture.

Ag+ (aq) + NO3- (aq) + K+ (aq) + Cl- (aq)

AgCl (s) + K+ (aq) + NO3- (aq)

©

2009,

Prentic

e-Hall,

Inc.

Net Ionic Equation• To form the net ionic equation, cross out anything that

does not change from the left side of the equation to the right.

Ag+(aq) + NO3-(aq) + K+(aq) + Cl-(aq)

AgCl (s) + K+(aq) + NO3-(aq)

©

2009,

Prentic

e-Hall,

Inc.



• The only things left in the equation are those things that change (i.e., react) during the course of the reaction.

Ag+(aq) + Cl-(aq) AgCl (s)

©

2009,

Prentic

e-Hall,

Inc.



• The only things left in the equation are those things that change (i.e., react) during the course of the reaction.

• Those things that didn’t change (and were deleted from the net ionic equation) are called spectator ions.

Ag+(aq) + NO3-(aq) + K+(aq) + Cl-(aq)

AgCl (s) + K+(aq) + NO3-(aq)

©

2009,

Prentic

e-Hall,

Inc.

Writing Net Ionic Equations1. Write a balanced molecular equation.2. Dissociate all strong electrolytes.3. Cross out anything that remains unchanged from

the left side to the right side of the equation.4. Write the net ionic equation with the species that

remain.

Write the complete ionic and net ionic equation for each

64

K2SO4(aq) + 2 AgNO3(aq) 2 KNO3(aq) + Ag2SO4(s)

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


For each of the following, write the molecular equation, complete ionic equation, and net ionic equation.

• An aqueous solution of barium chloride is added to an aqueous solution of sodium sulfate.

• An aqueous solution of calcium hydroxide is added to an aqueous solution of magnesium bromide.

Acid-Base Reactions• Also called neutralization reactions because the acid and

base neutralize each other’s properties

2 HNO3(aq) + Ca(OH)2(aq) Ca(NO3)2(aq) + 2 H2O(l)

• The net ionic equation for an acid-base reaction isH+(aq) + OH(aq) H2O(l)

• as long as the salt that forms is soluble in water

66


Acids and Bases in Solution• Acids ionize in water to form H+ ions

• more precisely, the H from the acid molecule is donated to a water molecule to form the hydronium ion, H3O+

• most chemists use H+ and H3O+ interchangeably

• Bases dissociate in water to form OH ions• bases, such as NH3, that do not contain OH ions, produce

OH by pulling H off water molecules

acid + base salt + water

• the H+ (acid) combines with the OH (base) to make water

• The cation from the base combines with the anion from the acid to make a salt

67


Common Acids 68


69

Common Bases


HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)70


For each of the following, write the molecular equation, complete ionic equation, and net ionic equation.

71


HCl(aq) + Ba(OH)2(aq)

H2SO4(aq) + Sr(OH)2(aq)

Titration• Often in the lab, a solution’s concentration is determined by reacting it with another material and using stoichiometry – this process is called titration

• In the titration, a known volume of a solution with a known concentration is reacted with a known volume of a solution until the reaction is just completed. • At this point, called the endpoint, the reactants

are in their stoichiometric ratio.• the known solution is added slowly from an

instrument called a buret

72


73


an indicator is added that changes color when the solution undergoes large changes in acidity/alkalinity•At the endpoint of an acid-base titration, the number of moles of H+ equals the number of moles of OH (equivalence point)

Titration

The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?

74


What is the concentration of NaOH solution that requires 27.5 mL to titrate 50.0 mL of 0.1015 M H2SO4?

Gas-Evolving Reactions• Some reactions form a gas directly from the ion exchange

K2S(aq) + H2SO4(aq) K2SO4(aq) + H2S(g)

• Other reactions form a gas by the decomposition of one of the ion exchange products into a gas and water

K2SO3(aq) + H2SO4(aq) K2SO4(aq) + H2SO3(aq)

H2SO3(aq) H2O(l) + SO2(g)

75


NaHCO3(aq) + HCl(aq) NaCl(aq) + CO2(g) + H2O(l)

76


Compounds that UndergoGas-Evolving Reactions

77


Write a balanced chemical equation for the following gas evolving reactions.

78


HCl(aq) + Na2SO3(aq)

H2SO4(aq) + CaS(aq)

Other Patterns in Reactions• The precipitation, acid-base, and gas-evolving reactions all

involve exchanging the ions in the solution• Other kinds of reactions involve transferring electrons from

one atom to another – these are called oxidation-reduction reactions

• also known as redox reactions

• many involve the reaction of a substance with O2(g)

4 Fe(s) + 3 O2(g) 2 Fe2O3(s)

79


Combustion as Redox2 H2(g) + O2(g) 2 H2O(g)

80


Redox without Combustion2 Na(s) + Cl2(g) 2 NaCl(s)

81

2 Na 2 Na+ + 2 e

Cl2 + 2 e 2 Cl Tro: Chemistry: A Molecular Approach, 2/e

Reactions of Metals with Nonmetals

• Consider the following reactions:4 Na(s) + O2(g) → 2 Na2O(s)

2 Na(s) + Cl2(g) → 2 NaCl(s)

• The reactions involve a metal reacting with a nonmetal

• In addition, both reactions involve the conversion of free elements into ions

Na2O → 2 Na+ & O2–

NaCl → Na+ & Cl–

82


© 2009,

Prentice-

Hall,

Inc.

Oxidation-Reduction Reactions

• An oxidation occurs when an atom or ion loses electrons.

• A reduction occurs when an atom or ion gains electrons.

• One cannot occur without the other.

© 2009,

Prentice-

Hall,

Inc.

Oxidation NumbersTo determine if an oxidation-reduction reaction has occurred, we assign an oxidation number to each element in a neutral compound or charged entity.

Oxidation number (oxidation state) is a hypothetical charge assigned to the atom.

© 2009,

Prentice-

Hall,

Inc.

Oxidation Numbers

• Elements in their elemental form have an oxidation number of 0.

• The oxidation number of a monatomic ion is the same as its charge.

© 2009,

Prentice-

Hall,

Inc.

Oxidation Numbers

• Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions.Oxygen has an oxidation number of −2, except in the peroxide

ion in which it has an oxidation number of −1.Hydrogen is −1 when bonded to a metal, +1 when bonded to a

nonmetal.

© 2009,

Prentice-

Hall,

Inc.

Oxidation Numbers

• Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions.Fluorine always has an oxidation number of −1.The other halogens have an oxidation number of −1 in most

binary compounds; they can have positive oxidation numbers, however, most notably in oxyanions.

© 2009,

Prentice-

Hall,

Inc.

Oxidation Numbers

• The sum of the oxidation numbers in a neutral compound is 0.

• The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.

Practice Problems• Determine the ox. #’s of each element for the following:

• MgCrO4

• ClO4-

• N2

• FeN

• MnO4-

Assign an oxidation state to each element in the following

90

• Br2

• K+

• LiF

• CO2

• SO42−

• Na2O2


Oxidation and ReductionAnother Definition

• Oxidation occurs when an atom’s oxidation state increases during a reaction

• Reduction occurs when an atom’s oxidation state decreases during a reaction

91


Oxidation–Reduction• Oxidation and reduction must occur simultaneously

• if an atom loses electrons another atom must take them • The reactant that reduces an element in another reactant is called the reducing agent• the reducing agent contains the element that is oxidized

• The reactant that oxidizes an element in another reactant is called the oxidizing agent• the oxidizing agent contains the element that is reduced

92

2 Na(s) + Cl2(g) → 2 Na+Cl–(s)Na is oxidized, Cl is reduced

Na is the reducing agent, Cl2 is the oxidizing agentTro: Chemistry: A Molecular Approach, 2/e

Assign oxidation states, then determine the element oxidized and reduced, and determine the oxidizing agent and reducing agent in the following reaction:

93

Fe + MnO4− + 4 H+ → Fe3+ + MnO2 + 2 H2O


Sn4+ + Ca → Sn2+ + Ca2+

F2 + S → SF4

chapter 4 - chemical quantities

Engineering