Chapter 04: Stoichiometry
1. Insight
2. Fundamentals of Stoichiometry
3. Limiting Reactants
4. Theoretical and Percentage Yields
5. Solution Stoichiometry
CONCENTRATION
Molarity
(M)
%v/v
%w/w
Number of moles/ volume
%w/v
Composition of Solutions: qualitative
• Mixtures have variable composition?
• Need a way of indicating the amounts in the composition
• Relatively speaking, solutions are often described as dilute or concentrated
• Dilute solutions: low amount of solute compared to solvent
• Concentrated solutions: large amount of solute compared to solvent
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Solution concentration
• Molarity: moles of solute per liter of solution
• Describes #molecules per liter of solution
Molarity (M) =
moles of solute
volume of solution in liters
Total volume
not just the volume of solution used
Preparing 1 L of a 1.00 M NaCl Solution
DI water (Distilled,
deionized water)
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Practice — Find the molarity of a solution that has
25.5 g KBr dissolved in 1.75 L of solution
moles solute
M=
Total Volume ( L )
Step 1:
g KBr
mol KBr
M
Step 2:
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L sol’n
Practice — Determine the mass of CaCl2
(MM = 110.98) in 1.75 L of 1.50 M solution
L sol’n
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mol CaCl2
g CaCl2
Molarity Calculator
• http://www.tocris.com/molarityCalculator.php
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Given a concentrated stock solution, prepare a less concentrated (more dilute) working solution.
Glacial acetic acid (99+%) m.p. 16.7 °C Flash point 40 °C
The Dilution processes of concentrated acid and base are exothermic.
Dilution
• Given a concentrated stock solution, prepare a less
concentrated (more dilute) working solution
# moles of solute remains constant
Concentration (M) X volume (V) = # moles
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Dilution
• # moles of solute remains constant
McVc = MdVd
where “c” stands for the concentrated solution & “d” for the
diluted solution.
(In the following exams)
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Dilution Exam ple
Add 28.7 ml
of 17.4M
concentrated
stock
A student added 28.7 ml of 17.4M concentrated stock solution to a 500.0 ml
volumetric flask, after adding D.I. water to the curve, after well mixing, what
is the final molar concentration (molarity) of this solution?
Fill to the mark
with water
M=
( 28.7ml ) (17.4 M ) = 1.0M
500.ml
1.00M
500 ml
volumetric
flask
Figure 4-12 p153
Chapter 04: Stoichiometry
1. Insight
2. Fundamentals of Stoichiometry
3. Limiting Reactants
4. Theoretical and Percentage Yields
5. Solution Stoichiometry
6. Insight
STOICHIOMETRY
CONCENTRATION
MOLARITY (M)
Solution Stoichiometry
• Molarity
• A function of the # moles of solute  #molecules
• Can be used to convert between amount of reactants and/or products in a chemical
reaction
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Solution vs. Solid Stoichiometry
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Example: 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 2 KCl(aq) + Pb(NO3)2(aq)  PbCl2(s) + 2 KNO3(aq)?
Method 2
Given:
0.150 M KCl, 0.150 L of 0.175 M Pb(NO3)2
Find:
L KCl
Conceptual Plan:
L Pb(NO3)2
mol Pb(NO3)2
mol KCl
L KCl
Relationships:
1 L Pb(NO3)2 = 0.175 mol, 1 L KCl = 0.150 mol, 1 mol Pb(NO3)2 : 2 mol KCl
Solution:
Check: because you need 2x moles of KCl as Pb(NO3)2, and the molarity of
Pb(NO3)2 > KCl, the volume of KCl should be more than 2x the
volume of Pb(NO3)2
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Acids & bases
Electrolytes & nonelectrolytes
Strong Acids
• completely dissociate (100%)
• HCl(aq)  H+(aq) + Cl−(aq)
• H2SO4(aq)  2 H+(aq) + SO42−(aq)
Form ula N am e
HCl
Hy drochloric acid
HBr
Hy drob ro mic acid
HI
Hy droio dic acid
HNO3
N itric acid
H2 SO4
Su lfu ric acid
HClO4
Perch loric acid
notice that the polyvalent
ion stays together
Form ula N am e
LiOH
Lith iu m h yd ro xide
NaOH
So diu m hy dro xide
KOH
Po tass iu m h yd ro xide
Ba( OH) 2 Barium hy droxide
Weak Acids
• only partially dissociate
• HF(aq)  H+(aq) + F−(aq)
• Acetic acid: only 4 out every 1000 molecules (0.4%) are
converted to acetate ions
CH3 COOH(aq) + H 2 O( l)
Acetic acid
-
CH3 COO ( aq) + H 3 O+ ( aq)
A cetate ion
Common Acids
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Common Bases
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Practice – Write the equation for the process that occurs when the following
strong electrolytes dissolve in water
CaCl2
CaCl2(aq)  Ca2+(aq) + 2 Cl−(aq)
HNO3
HNO3(aq)  H+(aq) + NO3−(aq)
(NH4)2CO3 (NH4)2CO3(aq)  2 NH4+(aq) + CO32−(aq)
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Acid-Base Systems
① An acid MUST react with a base
② A base always reacts with an acid
 They always come in pairs
Acid – Base Definitions
• Acid: proton (H+) donor
• Base: hydroxide (OH-) donor
𝐻 + 𝑂𝐻 − → 𝐻2 𝑂
If you put the Base in water……..
NH3(aq) + H+(aq)

NH4+(aq) proton acceptor
H+(aq) + OH-(aq)
+ H2O 
NH3(aq)+ H2O
 NH4+(aq) + OH-(aq)
OH- donor
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Acids & Bases
• Acid: proton donor
HCl(aq)  H+(aq) + Cl-(aq)
• Base: proton acceptor:
NH3(aq)+ H+(aq) NH4+(aq)
+
HCl(aq) + NH3(aq)  NH4+(aq) + Cl-(aq)
Bases are anti-acids (antacids)
• What happens when you add a base to an acid?
HCl(g)  H+(aq) + Cl-(aq)
NaOH(s)  Na+(aq) + OH-(aq)
HCl(g) + NaOH(s)  H+(aq) + OH-(aq)+ Cl-(aq) + Na+(aq)
HCl(g) + NaOH(s)  H2O(l) + Cl-(aq) + Na+(aq)
You get water!
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Acid – Base reactions: look to form water and exothermic
• For the acid base reaction: HNO3(aq) + Ca(OH)2(aq) 
• Dissociate the acid and the base into ions (inverse criss-cross rule):
(H+ + NO3−) + (Ca2+ + 2OH−) 
• Rearrange: H+ combines with OH− to make H2O(l)
H+ + 2OH− + Ca2+ + NO3−
• – notice there are 2OH- ions  need 2 H+ ions
2H+ + 2OH− + Ca2+ + 2NO3−
• Now form water
2H2O + Ca2+ + 2NO3−
• write the formula of the salt
Ca2+ + NO3−  Ca(NO3)2
• Combine and make sure it’s balanced
HNO3(aq) + Ca(OH)2(aq)  2H2O + Ca(NO3)2
Practice(Diacid/binary acid/base) – Predict the products
and balance the equation
HCl(aq) + Ba(OH)2(aq) 
(H+ + Cl−) + (Ba2+ + OH−) → (H+ + OH−) + (Ba2+ + Cl−)
HCl(aq) + Ba(OH)2(aq) → H2O(l) + BaCl2
2 HCl(aq) + Ba(OH)2(aq)  2 H2O(l) + BaCl2(aq)
H2SO4(aq) + Sr(OH)2(aq) 
(H+ + SO42−) + (Sr2+ + OH−) → (H+ + OH−) + (Sr2+ + SO42−)
H2SO4(aq) + Sr(OH)2(aq) → H2O(l) + SrSO4
H2SO4(aq) + Sr(OH)2(aq) → 2 H2O(l) + SrSO4
H2SO4(aq) + Sr(OH)2(aq)  2 H2O(l) + SrSO4(s)
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Acid-Base Titrations
• Neutralization reaction
• [acid] or [base] known
• Other is unknown
• Stoichiometry of neutralization known
• Volumes of acid (VH) & base (VOH) needed to reach neutralization known
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Acid-Base Titration Reactions (Exp 8 inFor
Chem301)
Titration exp 8
Analytical procedure involving acids & bases
in Chem301
1) Goal: determine [unknown reactant]
2) Materials: acid solution, base solution, indicator
a) [1 reactant] known
b) [2nd reactant] unknown (to be determined)
3) Stoichiometry of reaction known
4) Procedure: Slowly add 1 reactant into another until neutrality
is reached (molesH = molesOH)
a)Result: volumes of acid & base used
5) Analysis: use MHVH=MOHVOH or solution stoichiometry to
determine [unknown reactant]
Conclusion: [unknown reactant]
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Titration
Here the titrant is the base
solution in the burette.
As the titrant is added to
the flask, the H+ reacts
with the OH– to form
water.
Once the acid is
neutralized (endpoint),
the phenolphthalein will
show a stable light pink
color for 30 secs
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Example:
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?
① Start with a balanced eqn
② Convert given  moles
HCl + NaOH  H2O + NaCl
æ 0.100moles_ NaOH ö
-3
0.01254L ç
÷ =1.254x10 moles_ NaOH
è
ø
L
③ Convert moles given  desired (balanced eqn)
æ 1HCl ö
-3
1.254x10 moles_ NaOH ç
÷ =1.254x10 moles_ HCl
è 1NaOH ø
-3
④ Convert desired moles  Molarity using V &
definition of M
1.254x10-3 moles_ HCl
= 0.1254M _ HCl
-2
1.000x10 L
Let’s go to the Lab.
Please do not forget to
1. Sign-in
2. Keep your PPEs on.
3. Checkout with my signature on your datasheet before you
leave
Practice −
What is the concentration of NaOH solution that requires 27.5 mL to titrate 50.0 mL
of 0.1015 M H2SO4?
① Start with a balanced eqn
② Convert given  moles
H2SO4 + 2NaOH  2H2O + Na2SO4
æ 0.1015moles_ H 2 SO4 ö
-3
0.050L ç
÷ = 5.075x10 moles_ H 2 SO4
è
ø
L
③ Convert moles given  desired (balanced eqn)
æ 2NaOH ö
-2
5.075x10 moles_ H 2 SO4 ç
÷ =1.015x10 moles_ NaOH
è 1H 2 SO4 ø
-3
④ Convert desired moles  Molarity using V &
definition of M
1.015x10-2 moles_ NaOH
= 0.369M _ NaOH
-2
2.75x10 L
Practice — Solution stoichiometry
• 43.8 mL of 0.107 M HCl is needed to neutralize 37.6 mL of Ba(OH)2 solution. The
unbalanced equation is HCl + Ba(OH)2  BaCl2 + H2O What is the molarity of the
base?
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Practice 43.8 mL of 0.107 M HCl is needed to neutralize 37.6 mL of Ba(OH)2 solution. The unbalanced equation is
HCl + Ba(OH)2  BaCl2 + H2O What is the molarity of the base?
Given:
Find:
0.0438 L of 0.107 M HCl, 0.0376 L Ba(OH)2
M Ba(OH)2
Conceptual Plan: Balance the equation
L HCl
mol HCl
mol Ba(OH)2
M Ba(OH)2
Relationships:
L Ba(OH)2
1 mL= 0.001 L, 1 L HCl = 0.107 mol, 1 mol Ba(OH)2 : 2 mol HCl
Solution:
2HCl + Ba(OH)2  BaCl2 + 2H2O