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LECTURE ThirteenCHM 151 ©slg
Topics:
1. Solution Stoichiometry: Molarity2. Titration Problems
GROUP WORK, BALANCE
C6H12O6 + O2 --> CO2 + H2O (sugar)
C5H11OH + O2 --> CO2 + H2O (alcohol)
a) Balance Cb) Balance Hc) Balance Od) Check
C6H12O6 + O2 --> CO2 + H2O
C6H12O6 + O2 --> 6 CO2 + H2O
C6H12O6 + O2 --> 6 CO2 + 6 H2O
C6H12O6 + 6 O2 --> 6 CO2 + 6 H2O
6C 12H 6 O + 12 O ---> 6 C 12 O + 12 H 6 O
Solution, Sugar
12 O + 6 O = 18 O
C5H11OH + O2 --> CO2 + H2OC5H11OH + O2 --> CO2 + H2O
C5H11OH + O2 --> 5 CO2 + H2O
C5H11OH + O2 --> 5 CO2 + 6 H2O
C5H11OH + 7.5 or 15/2 O2 --> 5 CO2 + 6 H2O 1 O 15 O 10 O 6 O
#1
#2
#3
C5H11OH + 7.5 or 15/2 O2 --> 5 CO2 + 6 H2O 10 O 6 O
X 2
2 C5H11OH + 15 O2 --> 10 CO2 + 12 H2O
[10C 24H 2 O] + 30 O --> [10C 20 O] + [24H 12 O]
Group work 2:
H2SO4 (aq) + Al(OH)3 (s) -----> ?
H3PO4 (aq) + KOH (aq) ---->
HCH3CO2 (aq) + Mg(OH)2 (s) ----->?
1. Form H2O and use leftover ions to form salt2. Write balanced equation3. Do total ionic (watch out for weak acids!)4. Do net ionic equation
Net Ionic Equations: What to “Break Up”:
A) all soluble salts “(aq)”
B) All soluble bases “(aq)”
C) All strong acids: HCl HBr HI; HNO3 H2SO4 HClO4
What not to “break up”:
A) all insoluble salts or bases “(s)”
B) H2O (and all molecules)
C) All weak acids: H3PO4, HCH3CO2, H2CO3.....
H2SO4 (aq) + Al(OH)3 (s) -----> ? ----> H2O + salt
Step One: Formula of salt:
Al3+ + SO42- -----> Al2(SO4)3
acid base
Always!
Step Two:
Complete Equation:
H2SO4 (aq) + Al(OH)3 (s) -----> H2O (l) + Al2(SO4)3 (aq)
Balance:
3 H2SO4 (aq) + 2 Al(OH)3 (s) -----> 6 H2O + Al2(SO4)3 (aq)
3 H2SO4 (aq) + 2 Al(OH)3 (s) -----> 6 H2O + Al2(SO4)3 (aq)
Step 3: Total Ionic Equation:
6 H+ (aq) + 3 SO42- (aq) + 2 Al(OH)3 (s) ----->
6 H2O(l) + 2 Al3+ (aq) + 3 SO42- (aq)
Strong acid,Break up!
Insoluble base,Don’t break up
Molecule,Don’t...
Soluble,Do...
Step 3: Total Ionic Equation:
6 H+ (aq) + 3 SO42- (aq) + 2 Al(OH)3 (s) ----->
6 H2O(l) + 2 Al3+ (aq) + 3 SO42- (aq)
Step Four: Net Ionic Equation
6 H+ (aq) + 2 Al(OH)3 (s) -----> 6 H2O (l) + 2 Al3+ (aq)
H3PO4 (aq) + KOH (aq) ----> ? -----> H2O + salt
Step One: salt formula
K+ + PO43- -----> K3PO4 (aq)
Step Two: Write Equation; Balance
H3PO4 (aq) + KOH (aq) ----> H2O (l) + K3PO4 (aq)
H3PO4 (aq) + 3 KOH (aq) ---->3 H2O (l) + K3PO4 (aq)
H3PO4 (aq) + 3 KOH (aq) ---->3 H2O (l) + K3PO4 (aq)
Step Three: Total Ionic:
H3PO4 (aq) + 3 K+ (aq) + 3 OH- (aq) -----> 3 H2O (l)
+ 3 K+ (aq) + PO43-(aq)
Step Four: Net Ionic:
H3PO4 (aq) + 3 OH- (aq) -----> 3 H2O (l) + PO43-(aq)
Weak acid, no! Yes! No! Yes!
HCH3CO2 (aq) + Mg(OH)2 (s) ----->? -----> H2O + salt
Step One: Salt Formula:
Mg2+ + CH3CO2- -----> Mg(CH3CO2)2 (aq)
Step Two: Write Equation, Balance:
HCH3CO2 (aq) + Mg(OH)2(s) -----> H2O + Mg(CH3CO2)2(aq)
2 HCH3CO2 (aq) + Mg(OH)2(s) -----> 2 H2O + Mg(CH3CO2)2(aq)
2 HCH3CO2 (aq) + Mg(OH)2(s) -----> 2 H2O + Mg(CH3CO2)2(aq)
Step Three: Total Ionic Equation:
2 HCH3CO2 (aq) + Mg(OH)2(s) -----> 2 H2O(l) + Mg2+ (aq)
+2 CH3CO2- (aq)
Step Four: Net Ionic Equation
SAME!
NO, weak acid! NO! NO! Yes!
Concentrations of Compounds in Aqueous Solutions
(Chapter 5, Section 5.8, p. 213)
Usually the reactions we run are done in aqueous solution, and therefore we need to add to our study of stoichiometry the concentration of compounds in aqueous solution.
We will utilize a very useful quantity known as “molarity,” the number of moles of solute per liter of solution.
Concentration(molarity) = # moles solute L solution
If we placed 1.00 mol NaCl (58.4 g) in a 1 L volumetric flask, dissolved it in water, swirled to dissolve and diluted the solution to the 1.00 liter mark you would have a solution that contains 1.00 mol NaCl per liter of solution. This may be represented several ways:
Concentration(molarity) = 1.00 mol NaCl / L soln = 1.00 M NaCl = [1.00] NaCl
Chemists call this a “1.00 molar solution”
Calculating Molar Amounts
TYPICAL PROBLEMS:
•What is the molarity of a solution made by dissolving25.0 g of BaCl2 in sufficient water to make up a solution of 500.0 mL?
•How many g of BaCl2 would be contained in 20.0 mL ofthis solution?
•How many mL of this solution would deliver 1.25 g ofBaCl2?
•How many mol of Cl- ions are contained in 10.00 ml ofthis solution?
What is the molarity of a solution made by dissolving25.0 g of BaCl2 in sufficient water to make up a solution of 500.0 mL?
25.0 g BaCl2 = ? mol/ L BaCl2 (= ? M BaCl2) 500.0 mL soln
1Ba = 1 X 137.33g = 137.332Cl = 2 X 35.45g = 70.90 208.23 g/mol
Molar mass, BaCl2:
25.0 g BaCl2
208.23 g BaCl2
1 mol BaCl2
500.0 mL
1000 mL
1 L= .240 mol / L BaCl2 = .240 M BaCl2
How many g of BaCl2 would be contained in 20.0 mL ofthis solution? (.240 M BaCl2)
Question: 20.0 mL soln = ? g BaCl2
Relationships: 1000 mL = 1 L 1 L soln = .240 mol BaCl2
1 mol BaCl2 = 208.23 g BaCl2
20.0 mL soln = ? g BaCl2
mL L mol g
20.0 mL soln
1000 mL
1 L
1L soln
.240 mol BaCl2
1 mol BaCl2
208.23 g BaCl2 = 1.00 g BaCl2
Molarity Molar Mass
•How many mL of this solution would deliver 1.25 g ofBaCl2?
1.25 g BaCl2 = ? mL soln.240 mol BaCl2 = 1 L soln208.23 g BaCl2 = 1 mol BaCl2
1.25 g BaCl2
208.23 g BaCl2
1 mol BaCl2
.240 mol BaCl2
1L soln
1 L soln
1000 mL= 25.0 mL
Molar Mass Molarity
•How many mol of Cl- ions are contained in 10.0 ml ofthis solution?
10.0 mL soln = ? Mol Cl- .240 mol BaCl2 = 1000 mL soln 1 mol BaCl2 = 2 mol Cl-
Note: BaCl2(aq) ---> Ba2+(aq) + 2 Cl- (aq)
10.0 mL soln
1000 mL soln
.240 mol BaCl2
1 mol BaCl2
2 mol Cl-
= .00480 mol Cl-
GROUP WORK:
If 35.00 g CuSO4 is dissolved in sufficient water to makeup 750. mL of an aqueous solution,
a) what is the molarity of the solution?
b) how many mL of the solution will deliver 10.0 g of CuSO4?
c) How many moles of sulfate ion (SO42-) will be
delivered in 10.0 mL of the solution?
1 Cu = 1 X 63.55 = 63.551 S = 1 X 32.07 = 32.074 O = 4 X 16.00 = 64.00 159.62 g/mol
35.00 g CuSO4
750 mL 159.62 g CuSO4
1 mol CuSO 4 1000 mL
1 L= .292 mol/L CuSO4
= .292 M CuSO4
10.0 g CuSO4
159.62 g CuSO4
1 mol CuSO 4
.292 mol CuSO4
1000 mL soln= 215 mL soln
10.0 mL soln
1000 mL soln
.292 mol CuSO4
1 mol CuSO4
1 mol SO42-
= .00292 mol SO42-
STOICHIOMETRY OF REACTIONS IN AQUEOUSSOLUTION: Chapter 5, Section 5.9
Let’s use our favorite reaction to add another dimension to calculating from balanced equations:
How many ml of 3.00 M HCl solution would be requiredto react with 13.67 g of Fe2O3 according to the followingbalanced equation:
Fe2O3(s) + 6HCl(aq) ---> 3H2O + 2FeCl3(aq)
159.70 g/mol 36.46 g/mol 18.02 g/mol 162.20 g/mol 3.00 M13.67 g ? mL
Fe2O3(s) + 6HCl(aq) ---> 3H2O + 2FeCl3(aq)
159.70 g/mol 3.00 M 13.67 g ? mL
Pathway: g Fe2O3 ---> mol Fe2O3---> mol HCl --- > mL soln
13.67 g Fe2O3 = ? mL soln
1000 mL soln = 3.00 mol HCl 6 mol HCl = 1 mol Fe2O3
159.70 g Fe2O3 = 1 mol Fe2O3
13.67 g Fe2O3
159.70 gFe2O3
1 mol Fe2O3
1 mol Fe2O3
6 mol HCl
3.00 mol HCl
1000 mL soln= 171 mL soln
MolarMass
BalancedEquation
Molarity
Group Work
How many g of Fe2O3 would react with 25.0 mL of 3.00 M HCl?
Fe2O3(s) + 6HCl(aq) ---> 3H2O + 2FeCl3(aq)
159.70 g/mol 3.00 M ? g 25.0 mL
mL L mol HCl mol Fe2O3 g Fe2O3
Fe2O3(s) + 6HCl(aq) ---> 3H2O + 2FeCl3(aq)
159.70 g/mol 3.00 M ? g 25.0 mL
25.0 mL soln 3.00 mol HCl 1 mol Fe2O3 159.70 g Fe2O3
1000 mL soln 6 mol HCl 1 mol Fe2O3
= 1.996 g Fe2O3 = 2.00 g Fe2O3
Molarity EquationMolar mass
Mass of product, grams
12
10
8
6
4
2
1 2 3 4
Mass of Fe, grams
Constant mass, Bromine, variable masses Fe
Ch 4,#60
(a) What mass of Br2 is used when the reaction consumes 2.0 g Fe?
Product 10.8 g; 10.8 - 2.0 = 8.8 g Br
(b) What is the mole ratio of Fe to Br in this reaction?
2.0 g Fe 1 mol Fe = .036 mol Fe .036 = 1 55.85 g Fe .036
8.8 g Br 1 mol Br = .11 mol Br .11 = 3 79.9 g Br .036
(c) What is the empirical formula of the product?
FeBr3
(d) Balanced chemical equation:
Fe + 3 Br2 -----> 2 FeBr3
(e) Name? Iron (III) Bromide
Which best describes graph:
1. When 1.00 g Fe is added, Fe is LR Yes
2. When 3.50 g of Fe is added, there is an excess of Br2
3. When 2.50 g of Fe is added to the Br2, both reactants are used up completely
4. When 2.00 g of Fe is added to the Br2, 10.0 g of product is formed. The percent yield must be 20%
TITRATION CALCULATIONS
(Titrations and makeup of standard solutions seen on CD ROM)
Titration: Procedure in which measured increments of one reactant are added to a known amount of a second reactantuntil some indicator signals that the reaction is complete. This point is called “the equivalence point.”
Indicators include many acid/base dyes, potentiometers, color change in one reagent...
Titrations are run with buret and Erlenmeyer flasks as seen on the CD ROM or demo just observed...
Two types of problems are encountered: •Standardizing an acid or a base solution(determining the exact molar concentration of the acid or base solution)
• Determining amount of acidic or basic material in a sample To be continued.....Happy Spring Break!