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!
§4.3 Limiting Reactant
N2(g) + 3H2(g) → 2NH3(g)
in excess limiting reagent
§4.3 Limiting Reactant
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
§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)
§4.9 Oxidation-Reduction Reactions
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.
§4.9 Oxidation-Reduction Reactions
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
§4.9 Oxidation State Rules
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 State Rules
§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
2Al(NO3)3(aq) + 3Ba(OH)2(aq) 2Al(OH)3(s) + 3Ba(NO3)2(aq)
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?
§4.6 Precipitation Stoichiometry
2Al(NO3)3(aq) + 3Ba(OH)2(aq) 2Al(OH)3(s) + 3Ba(NO3)2(aq)
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-
§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?
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