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STOICHIOMETRY

Chapter 4

§4.1 Stoichiometry

Stoichiometry - calculating the quantities of reactants and products in a chemical reaction.

Stoichiometry is entirely based on the Law of Conservation of Mass.

§4.2 Stoichiometric Calculations

The amounts of any other substance in a chemical reaction can be determined from the amount of just one substance:

(where A and B are in the same reaction)

§4.2 Stoichiometry

Sulfuric acid dissolves aluminum metal according to the following reaction. If you had 15.2 g aluminum, what amount of sulfuric acid would you need to complete the reaction?

3 H2SO4 (aq) + 2 Al(s) → Al2(SO4)3 (aq) + 3 H2(g)

15.2 g Al 1 mol Al26.98 g Al

3 mol H2SO4

2 mol Al

98.09 g H2SO4

1 mol H2SO4

= 82.9 g H2SO

4

§4.2 Stoichiometry

Pearls and baby teeth dissolve in vinegar according to this reaction.

CaCO3 (s) + 2 HC2H3O2 (aq) → Ca(C2H3O2)2 (aq) + H2CO3

(aq)

10 g CaCO3 1 mol CaCO3

100 g CaCO3

2 mol HC2H3O2

1 mol CaCO3

60 g HC2H3O2

1 mol HC2H3O2

= 12 g HC2H3O2

§4.3 Limiting Reactant

Sometimes all reactants are completely consumed; sometimes one or more is in excess.

A limiting reactant is present in lesser molar amounts and is consumed first, limiting the amount of product that can form.

Always determine which reactant is limiting when performing stoichiometry calculations!


N2(g) + 3H2(g) → 2NH3(g)

in excess limiting reagent


Reaction of ammonia and oxygen gives nitrogen monoxide and water. Determine the mass of nitrogen monoxide produced from 10.0 g ammonia and 10.0 g oxygen.

NH3(g) + O2(g) NO(g) + H2O(l)

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(l)

1. Balance the equation

mol NH3

17.034 g NH3

mol O2

32.00 g O2

2. Limiting reagent?

10.0 g NH3

10.0 g O2

4 mol NO4 mol

NH3

4 mol NO5 mol O2

= 0.587 mol NO

= 0.250 mol NO

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(l)

Limiting Reagent

30.01 g NO1 mol NO

30.01 g NO1 mol NO

= 17.7 g NO

= 7.50 g NO

§4.3 Percent Yield

Theoretical Yield (calculated) – the maximum amount of product that can form during a reaction

Actual Yield (measured) – the amount of pure product collected from a reaction

(actual yield) ≤ (theoretical yield)

Percent Yield = Actual Yield (100%) Theoretical Yield

§4.3 Percent Yield

Reaction of titanium (IV) oxide with graphite produces titanium metal and carbon monoxide. When 28.6 kg of C is allowed to react with 88.2 kg of titanium (IV) oxide, 42.8 kg of Ti is produced. Find the percent yield of Ti from this reaction.

TiO2(s) + C(s) Ti(s) + CO(g)

TiO2(s) + 2C(s) Ti(s) + 2CO(g)

1. Balance the equationmol TiO2

79.88 g TiO2

mol C12.01 g C

2. Limiting reagent?

88.2 x 103 g TiO2

28.6 x 103 g C

1 mol Ti1 mol TiO2

1 mol Ti2 mol C

= 1104 mol Ti

= 1191 mol Ti

TiO2(s) + 2C(s) Ti(s) + 2CO(g)

Limiting Reagent

47.88 g Ti mol Ti

42.8 kg Ti (100%)

42.8 kg Ti (100%)= 52.9 kg Ti

= 80.9%= 5.29 x 104 g Ti

§4.4 Aqueous Solutions

Water is one of the most important substances on earth.

All chemistry that makes life possible occurs in water.

Stoichiometry also applies to chemistry in aqueous solutions.

§4.4 Some Definitions

Solution – a homogenous mixtureSolvent – the majority solution component, retains its stateSolute – the minority solution component, becomes the

state of the solventSalt – 1) any ionic compound; 2) NaCl

§4.3 Solution Concentration

Molarity (M) = moles of solute liters of solution

0.5 M CuSO4

0.3 M CuSO4

0.1 M CuSO4

§4.4 Molarity Calculations

Determine the molarity of a solution where 45.5 mg of NaCl is dissolved in 154.4 mL of solution.

How many moles of Cl- ion are contained in 114 mL of a 1.85 M CaCl2 solution?

45.5 mg NaCl

154.4 mL soln

1000 mL soln1 L soln

1 g NaCl1000 mg

NaCl

1 mol NaCl58.44 g NaCl

= 5.04 x 10-3 M NaCl

CaCl2(s) -> Ca2+(aq) +

2Cl-(aq)

114 mL soln

1.85 mol CaCl2

1 L soln

1 L soln1000 mL

soln

2 mol Cl-

1 mol CaCl2

= 0.422 mol Cl-

§4.5 Electrolytes

An electrolyte is a substance that when dissolved in water produces a solution that conducts electricity.

Nonelectrolytes dissolve in water but don't ionize.

§4.5 Electrolytes

Electrical conductivity is a qualitative measure of the degree to which a solute ionizes.

weakelectrolyte

non-electrolyte

strongelectrolyte

§4.5 Strong Electrolytes

Strong electrolytes completely ionize in water. There are three types:

1) Soluble salts – NaCl, KBr, LiNO3, etc.

2) Strong acids – HCl, HNO3, H2SO4, HClO4

3) Strong bases – NaOH, KOH

Example: NaOH(s) → Na+(aq) + OH-

(aq)

100% ionized

H2

O

§4.5 Weak Electrolytes

Weak electrolytes slightly ionize in water.Examples include weak acids such as

HC2H3O2 and weak bases such as NH3.For instance, acetic acid is ~1% ionized in

water:

HC2H3O2(l) H+(aq) + C2H3O2

-(aq)

In contrast, strong electrolytes are 100% ionized.

→←

99% 1%

§4.5 Solubility

Dissolution occurs if interactions between solvent and solute particles overcome interactions of the solute to itself.

§4.5 Solubility

An ionic compound dissolving in water:

Just because a compound is ionic doesn’t automatically mean it dissolves in water. (Chapter 16)

§4.5 Solubility

A nonelectrolyte (sucrose) dissolved in water:

§4.3 Percent Yield

If 18.20 g of NH3 (mm 17.03 g/mol) are produced by a reaction mixture that initially contains 6.00 g of H2 and an excess of N2, what is the percent yield of the reaction?

N2(g) + 3H2(g) → 2NH3(g)

Calculate theoretical yield of NH3 :

1 mol H2

2.016 g H2

6.00 g H2

2 mol NH3

3 mol H2

17.03 g NH3

1 mol NH3

Percent yield: 18.20 g NH3

33.8 g NH3

(100%) = 53.8%

N2(g) + 3H2(g) → 2NH3(g)

= 33.8 g NH3

= 33.8 g NH3

§4.6 Three Types of Reactions

1. Precipitation reactions2. Acid-base reactions3. Oxidation-reduction reactions

We will not be covering gas-evolution reactions (§4.6).

§4.5 Solubility RulesLearn these

.

§4.6 Precipitation Reactions

How to predict a precipitation reaction:

1) Swap the anions and cations.

K2SO4(aq) + BaI2(aq) ??? K2SO4(aq) + BaI2(aq) BaSO4(?) + 2KI(?)

2) Consult the solubility rules. If any of the “products” are insoluble in water, a precipitation reaction has occurred. Otherwise, there is no reaction.

K2SO4(aq) + BaI2(aq) BaSO4(s) + 2KI(aq)

§4.6 Demo – Precipitation Reactions

Pb(NO3)2 + 2KI --> PbI2 + 2KNO3

soluble

• NO3- salts are soluble.

soluble

• Li+, Na+, and K+ salts are soluble.

soluble

insoluble

• Cl-, Br- and I- salts are soluble unless the cation is Ag+, Pb2+ or Hg2

2+.

§4.6 Demo – Precipitation Reactions

AgNO3 + NaCl --> NaNO3 + AgCl

soluble

• NO3- salts are soluble.

soluble

• Li+, Na+, and K+ salts are soluble.

soluble

insoluble

3. Cl-, Br- and I- salts are soluble unless the cation is Ag+, Pb2+ and Hg2

2+.

§4.6 Water-soluble or -insoluble?

Sr3(PO4)2 insoluble

Na3PO4 soluble

CaSO4 insoluble

RbCl solubleBaS soluble

(NH4)2CrO4 soluble

Ag2SO4 insoluble

CdBr2 soluble

NH4OH soluble

NiCO3 insoluble

§4.6 Precipitation Reactions

A 1.000 g sample of a metal dichloride was dissolved in water and treated with excess AgNO3(aq) to give 1.286 g of a white solid. What is the metal?

1.000 g MCl2

2 mol AgCl 1 mol MCl2

mol MCl2

x g MCl2

= 1.286 g AgCl

MCl2(aq) + 2AgNO3(aq) 2AgCl(s) + M(NO3)2(aq)

MCl2(aq) + 2AgNO3(aq) 2AgCl(s) + M(NO3)2(aq)

143.32 g AgCl

1 mol AgClx = 222.89 g/mol MCl2 M + 2(35.45 g/mol) = 222.89 g/mol

M = 152.0 g/mol M = europium

§4.4 Dilution

It is very common to make dilute solutions from more concentrated ones (stock solutions).

For dilutions: more solvent is added but the amount of solute remains constant. Hence:

M1V1 = M2V2

Molarity = mol solute

L solution(Molarity) (L solution) = mol

solute

M1 = molarity of initial solutionV1 = volume of initial solutionM2 = molarity of final solutionV2 = volume of final solution

mol solute before dilution = mol solute after dilution

§4.6 Solubility

100.0 mL of a Pb(NO3)2 solution was concentrated to 80.0 mL. To 2.00 mL of the concentrated solution was added an excess of NaCl(aq), and 3.407 g of a solid was obtained. What was the molarity of the original Pb(NO3)2 solution?

+M2

2 mL

excessNaCl(s)

concentration

M1

V1 V2

M2

M1V1 = M2V2

§4.6 Solubility

100.0 mL of a Pb(NO3)2 solution was concentrated to 80.0 mL. To 2.00 mL of the concentrated solution was added an excess of NaCl(aq), and 3.407 g of a solid was obtained. What was the molarity of the original Pb(NO3)2 solution?

Pb(NO3)2(aq) + NaCl(aq) PbCl2(s) + NaNO3(aq)

Pb(NO3)2(aq) + 2NaCl(aq) PbCl2(s) + 2NaNO3(aq)

V1M1 = V2M2

Use the same equation for concentration as for dilution:

2.00 x 10-3 L LN soln

1 mol PbCl2

1 mol LN

x mol LN1 L soln

= 3.407 g PbCl2

278.10 g PbCl2 1 mol PbCl2

x = 6.125 M Pb(NO3)2

LN = Pb(NO3)2

(100 mL) · M1 = (80.0 mL)(6.125 M)

M1 = 4.90 M

A 2.25 g sample of scandium metal is reacted with excess hydrochloric acid to produce 0.1502 g hydrogen gas. What is the formula of the scandium chloride produced in the reaction?

2 Sc (s) + HCl (aq) → ScClx (aq) + H2 (g)Sc (s) + x HCl (aq) → ScClx (aq) + H2 (g)2 Sc (s) + 2x HCl (aq) → 2 ScClx (aq) + x H2 (g)

0.1502 g H2 1 mol H2

2.018 g H2

= 0.1487 mol Cl-2x mol HClx mol H2

1 mol Cl-1 mol HCl

= 0.0500 mol Sc2.25 g Sc 1 mol Sc44.96 g Sc

2 Sc (s) + x HCl (aq) → ScClx (aq) + x/2 H2

(g)

§4.6 Stoichiometry

A 2.25 g sample of scandium metal is reacted with excess hydrochloric acid to produce 0.1502 g hydrogen gas. What is the formula of the scandium chloride produced in the reaction?

2 Sc (s) + 2x HCl (aq) → 2 ScClx (aq) + x H2 (g)

= 0.1487 mol Cl-

= 0.0500 mol Sc

≈ 3.0

ScCl3

§4.6 Stoichiometry

§4.7 Writing Chemical Equations for Solutions

Molecular Equation – reactants and products are written as compounds.

Complete Ionic Equation – all strong electrolytes are written as ions.

Net Ionic Equation – species that are the same on both sides of the equation are removed.

Cs2S(aq) + HgBr2(aq) 2CsBr(aq) + HgS(s)

2Cs+(aq) + S2-

(aq) + Hg2+(aq) + 2Br -

(aq) 2Cs+(aq) + 2Br -

(aq) + HgS(s)

Hg2+(aq) + S2-

(aq) HgS(s)

In this reaction, Cs+ and Br- are calledspectator ions.

Write the net ionic equation (if any) for these reactions.

FeSO4(aq) + KCl(aq)

Al(NO3)3(aq) + LiOH(aq)

No precipitate = No reaction

Al3+(aq) + 3 OH-

(aq) Al(OH)3(s)

FeCl2(?) + K2SO4(?) 2 FeCl2(aq) + K2SO4(aq)

Al(OH)3(?) + 3LiNO3(?)Al(OH)3(s) + 3LiNO3(aq)3

§4.8 Acid-Base Reactions

Acid – produces H+ in waterBase – produces -OH in waterAcid-base reactions often have the format:

acid + base → water + saltExample:

HBr(aq) + LiOH(aq) → H2O(l) + LiBr(aq)

The net ionic equation is:H+

(aq) + OH-(aq) → H2O(l)

§4.8 Acid-Base Titrations

Titration – analysis of the amount of a substance by reaction with a solution of known concentration.

Equivalence point –the reaction is complete with neither acid nor base in excess.

For titration stoichiometry: mole ratios are mols acid/mols base or vice versa.

§4.8 Acid-Base Reactions

Titration of a 20.0 mL solution of H2SO4 (a diprotic acid) required 22.87 mL of 0.158 M KOH solution. What is the concentration of the H2SO4 solution?

H2SO4(aq) + 2KOH(aq) 2H2O(l) + K2SO4(aq)

0.02287 L KOH soln0.158 mol KOH1 L KOH soln

1 mol H2SO4

2 mol KOH= 1.81 x 10-3 mol H2SO41.81x 10-3 mol H2SO4

0.0200 L H2SO4 soln= 0.0903 M H2SO4

H2SO4(aq) + 2KOH(aq) 2H2O(l) + K2SO4(aq)

Oxidation – loss of electronsReduction – gain of electronsOxidizing agent – gains electrons, is itself reducedReducing agent – loses electrons, is itself oxidized

Oxidation and reduction always go together. One species transfers electrons to another.

§4.9 Oxidation-Reduction Reactions


“LEO says GER” Loss of Electrons:

OxidationGain of Electron: Reduction

“OIL RIG” Oxidation Is LossReduction Is Gain

Oxidation used to mean reaction with oxygen.Anything that reacts with O2 (except F2) is

oxidized (loses electrons), but many redox reactions don’t involve oxygen at all:

2 K(s) + F2(g) → 2 KF(s)


Reactions of metals and nonmetals to give ionic compounds are typical redox reactions.

4 Na(s) + O2(g) → 2 Na2O(s)

2 Na(s) + Cl2(g) → 2 NaCl(s)

The metal is oxidized, the nonmetal is reduced.


Assigning elements in a reaction an oxidation state allows us to trace electron flow during the reaction.

Except for monoatomic ions, oxidation states are not ionic charges.

Oxidation states are assigned by “giving” electrons in a bond (for counting purposes, not in reality) to the more electronegative element.

§4.9 Oxidation State

1. Free elements have an oxidation state of 0. Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g)

2. Monoatomic ions have an oxidation state equal to their charge. For NaCl, Na = +1 and Cl = -1

3. a. The sum of the oxidation states for all the atoms in a compound is 0. For NaCl, (+1) + (-1) = 0

b. The sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ion. For NO3

–, N = +5 and O = -2, (+5) + 3(-2) = -1

§4.9 Oxidation State Rules

4. a. Group I metals have an oxidation state of +1 in all their compounds. Na = +1 in NaCl

b. Group II metals have an oxidation state of +2 in all their compounds. Mg = +2 in MgCl2


5. In their compounds, nonmetals have oxidation states in the table below. Elements higher on the table take priority.

Nonmetal Oxidation State Example

F -1 CF4

H +1 CH4

O -2 CO2


§4.9 Oxidation States

What are the oxidation states of the elements in the following species?F2 0

Mg2P2O7 +2, +5, -2

Hg2Cl2 +1, -1

S2O32- +2, -2

Consider the following reaction:

2 F2 + 2 NaOH → OF2 + 2 NaF + H2O

§4.9 Oxidation State

What is reduced? F

What is oxidized? OWhat is the reducing agent? NaOHWhat is the oxidizing agent? F2

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

§4.6 Precipitation Stoichiometry

When 159 mL of a 0.423 M Al(NO3)3 solution of is added to 378 mL of a 0.296 M Ba(OH)2 solution, a solid forms. What mass of solid forms and what is the concentration of each ion in solution after the reaction?2Al(NO3)3(aq) + 3Ba(OH)2(aq) 2Al(OH)3(s) +

3Ba(NO3)2(aq) Al(NO3)3(aq) + Ba(OH)2(aq) Al(OH)3(s) +

Ba(NO3)2(aq) 0.159 L Al(NO3)3 soln 0.423 mol Al(NO3)3

1 L Al(NO3)3 soln2 mol Al(OH)3

2 mol Al(NO3)3

= 6.73 x 10-2 mol Al(OH)3

0.378 L Ba(OH)2 soln 0.296 mol Ba(OH)2

1 L Ba(OH)2 soln2 mol Al(OH)3

3 mol Ba(OH)2

= 7.46 x 10-2 mol Al(OH)3

2Al(NO3)3(aq) + 3Ba(OH)2(aq) 2Al(OH)3(s) + 3Ba(NO3)2(aq)

Limiting reagent

78.004 g Al(OH)3

1 mol Al(OH)3

= 5.25 g Al(OH)3= 6.73 x 10-2 mol Al(OH)3

0.159 L Al(NO3)3 soln(0.159 L + 0.378 L)

0.423 mol Al(NO3)3

1 L Al(NO3)3 soln3 mol NO3

-

1 mol Al(NO3)3

0.378 L Ba(OH)2 soln(0.159 L + 0.378

L)

0.296 mol Ba(OH)2

1 L Ba(OH)2 soln1 mol Ba2+

1 mol Ba(OH)2


Limiting reagent

= 0.376 M NO3-

= 0.208 M Ba2+

[Al3+] = ?[Al3+] = 0 M[NO3

-]

[Ba+]

When 159 mL of a 0.423 M Al(NO3)3 solution of is added to 378 mL of a 0.296 M Ba(OH)2 solution, a solid forms. What mass of solid forms and what is the concentration of each ion in solution after the reaction?



Limiting reagent

0.378 L Ba(OH)2 soln0.296 mol Ba(OH)2

1 L Ba(OH)2 soln 2 mol OH- =1 mol Ba(OH)2

Initial mol OH-

mol OH- consumed

0.223 mol OH-

0.159 L Al(NO3)3 soln0.423 mol Al(NO3)3

1 L Al(NO3)3 soln2 mol Al(OH)3

2 mol Al(NO3)3

3 mol OH-

1 mol Al(OH)3

0.202 mol OH-

( - ) / (0.378 L + 0.159 L) =

=

= 3.91 x 10-2 M OH-


When 159 mL of a 0.423 M Al(NO3)3 solution of is added to 378 mL of a 0.296 M Ba(OH)2 solution, a solid forms. What mass of solid forms and what is the concentration of each ion in solution after the reaction?

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