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Chapter 3:
Chemical Stoichiometry
Chemical REACTIONS you should know
• Chemical Equations – (Write,
Write, Balance, Interpret)
Interpret
• Reactions You Should Know
• Formula Weights – (Must
Must know chemical formula
COMBUSTION (Of a Hydrocarbon)
C6H5OH + O2 CO2 + H2O
•
•
•
•
•
•
NEUTRALIZATION (Acid + Base)
HC2H3O2(aq) + NaOH(aq) H2O + NaC2H3O2(aq)
Avogadro’s number and the Mole
Limiting Reactants
Per Cent Yield
Empirical Formulas
Solubility Rules
Oxidation Rules
Chemical Reactions
Combustion of Ethanol
1st Write Reaction
C2H6OH +
O2 CO2 + H2O
2nd Balance Equation
1 C2H6OH + ? O2 2 CO2 + 7/2 H2O
must start somewhere
why not here
then
and
What number for oxygen?
Calculate Formula Weights
Must Know Formulas
Oxygen
O2 : 2 x 16 = 32 grams per unit
Ozone
O3 : 3 x 16 = 48 grams per mole
Methane
CH4 : 12 + 4 x 1 = 16 gms / mole
Ammonia
NH3 :14 + 3 x 1 = 17 g / mole
C6H12O6
(6 x 12) + (12 x 1) + (6 x 16) = 180
PRECIPITATION
BaCl2(aq) + Na2SO4(aq) BaSO4(s) + NaCl(aq)
A balanced chemical equation has the same number of
atoms in the reactants as in the products.
4 C2H6OH + 13 O2 8 CO2 +14 H2O
8=C=8
28=H=28
30=O=30
3rd Interpret Reaction
The reactants must be in the ratio of 4 : 13
The balanced equation has
4 + 13 + 8 + 14 = 39 total moles
Use Formula Weight to Convert
Grams to Moles & Moles to Grams
How many moles in 2.4 grams of Ozone
2.4 Grams x
1 moles
= ___moles
? Grams
Must know formula for Ozone !
2.4 Grams x
1 moles
= 0.050 moles
48 Grams
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Use Formula Weight to Convert
Grams to Moles & Moles to Grams
How many grams in 1½ moles of Methane
Use AVOGARDRO’S Number to Convert
Molecules to Moles & Moles to Molecules
___Molecules x
? Grams
= ___Grams
1 moles
1.5 moles x
Units, Units, Units
Formula for Methane ?
1.5 moles x
16 Grams
= 24 Grams
1 moles
Use Chemical Formula to Convert
Molecules to Atoms &
Atoms to Molecules
1
moles
= ___moles
6.02x10 23 Molecules
___moles x
6.02x10 23 Molecules
= ___Molecules
1
moles
Interpretation of Chemical Reactions
Using Stoichiometry
How many Hydrogen atoms in 1 mole of Methane
___Molecules x
? Atoms
= ___Atoms
1 Molecules
Formula for methane?
6.02 x 1023 Molecules x
4 hydrogen atoms
= 2.41 x 1024 hydrogen atoms
1
Molecules
Yields of Chemical Reactions
• Chemical Reactions do not always go
the way we expect them to
• Using stoichiometry we can calculate
the theoretical (Maximum) amount of
product formed in a reaction.
• The chemical arithmetic necessary to relate the
moles of reactants & products
• Chemical Reactions are Interpreted on the
Mole basis but chemicals are weighed in the
laboratory in Grams
If 10.0 moles of NO are reacted with 5.0 mole
O2, how many moles NO2 are produced?
2 NO (g)
+
(a) 2.0 mol NO2
(b) 6.0 mol NO2
(c) 10.0 mol NO2
(d) 16.0 mol NO2
(e) 32.0 mol NO2
O2 (g)
2 NO2 (g)
(c) 10.0 mol NO2
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The Limiting Reactant
If 10.0 moles of NO are reacted with 6.0 moles
O2, how many moles NO2 are produced?
A reaction stops when one reactant
is totally consumed.
The other reactants are excess reactants.
If 10.0 moles of NO are reacted with 6.0
moles O2, how many moles of the excess
reagent remain?
•
•
•
•
+
1.0 mol O2
5.0 mol O2
4.0 mol NO
8.0 mol NO
O2 (g)
+
O2 (g)
(a)What LIMITS
the reaction?
This is the limiting reactant.
2 NO (g)
2 NO (g)
2 NO2 (g)
1.0 mol O2
With a 50 % Yield, How many moles of NH3 are
produced from 1 ½ moles of H2 and 1 mole of N2?
1st Write and Balance Reaction
3 H2 + 1 N2 → 2 NH3
or
1 ½ H2 + ½ N2 → 1 NH3
So ½ mole NH3 formed
with ½ mole of Nitrogen in excess
(b)What is in
excess ?
2 NO2 (g)
The amount of NO
The amount of Oxygen
Yields of Chemical Reactions
• Using stoichiometry the theoretical (Maximum)
amount of product formed is calculated.
• If the actual amount of product formed is less
than the theoretical amount, the percentage yield
is calculated.
% yield =
Actual product yield
× 100%
Theoretica l product yield
If 0.720 grams Ozone reacts with 0.600 grams of
nitrogen monoxide, to produce nitrogen dioxide
1. How many grams of product will be formed ?
2. What is the limiting reagent?
3. How much excess reagent will remain ?
1st Write & Balance Reaction :
1 O3 + 3 NO → 3 NO2
6=O=6
3=N=3
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0.720 grams O3 & 0.600 grams of NO
For O 3 : 0.720 grams ×
1 moles
= 0.015 moles O 3
48 grams
For NO have 0.600 grams ×
1 mole
= 0.0200moles
30.0 grams
1 O3 + 3 NO → 3 NO2
1 (0.015) O3 + 3 (0.015) NO → 3 (0.015) NO2
0.015 O3 + 0.045 NO → 0.045 NO2
Have 0.0200 mol NO
Need 0.045 mole NO
Therefore Nitrogen monoxide is limiting reagent
moles O3 needed =
1
× 0.0200 moles = 0.00667
3
grams O 3 needed = 0.00667moles ×
0.015 O3 + 0.045 NO → 0.045 NO2
48 grams
= 0.320
1 mole
Grams O3 excess = 0.720 – 0.320 = 0.400
Two Approaches to Chemical Formulas
1. Given Formula Determine % Composition of
Each Element in Compound.
What is the % H & O in hydrogen peroxide?
2. Determine Formula Given % Composition of
Each Element in Compound
What is the Formula of the compound that is
5.88% H & 94.12 %O?
Find Empirical Formula
Weight K = 0.421
Weight O = 0.084
Total Weight = 0.505
First determine % by Weight
% by Wt K =
0.421
x 100 = 83.36633
0.505
83.4 % by wt K and 16.6 % by wt O
Chapter 4
Choose any total weight 100 grams is convenient
Construct the following work sheet
Element
%
Wt
moles
ratio
K
83.4
83.4 g
83.4 g / 39 = 2.14
O
16.6
16.6 g
16.6 g / 16 = 1.04 = 1
Therefore formula is K2O
= 2
Reactions in
Aqueous Solutions
&
Solution Stoichiometry
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WHAT IS A SOLUTION ?
A
HOMOGENEOUS
MIXTURE OF
TWO OR MORE SUBSTANCES
Solvent + Solute = Solution
Concentrations in Solutions
Molarity
• ELECTROLYTE – A substance that
dissolves in water to produce IONS
Example: HCl(aq), NaOH(aq), NaCl(aq)
• NON ELECTROLYTE A substance that
DOES NOT produce IONS, remain as
molecules, when dissolved in water.
Example: sugar(aq) , Ethylene glycol
(aq)
What do the following compounds do when
mixed with water ? {are they water soluble?)
WHY DOES A REACTION OCCUR ?
1. Precipitation
Reaction
An INSOLUBLE
Solid is Formed
2. Neutralization
Reaction
Acid-Base slightly
ionized Substance
water Formed
3. Oxidation-Reduction
Reaction
Change in Oxidation
State
Magnesium Iodide
MgI2 Mg2+(aq) + 2 I-(aq)
Aluminum Nitrate
Al(NO3)3 Al3+(aq) + 3NO3-(aq)
Ammonium Sulfate
(NH4)2SO4 2NH4+(aq) + SO42-(aq)
Perchloric Acid
HClO4 H+(aq) + ClO4-(aq)
Silver Chloride
AgCl Is Insoluble
Will precipitation occur when the following solutions
are mixed?
If so, write net ionic equation for reaction
(a) Na2CO3 and Ag C2H3O2 …….. Yes
2 Ag+(aq) + CO32-(aq) Ag2CO3 (s)
(b) NaNO3 and NiSO4
………….
No reaction
(c) FeSO4 and Pb(ClO4)2 ………….Yes
Pb2+(aq) + SO42- (aq) PbSO4 (s)
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Equations for Silver Perchlorate + Sodium Nitrate
CHEMICAL EQUATIONS
• MOLECULAR - Uses The Full
Formula Of Reactants and Products
• IONIC – ONLY Ions Are Shown
• NET IONIC – SPECTATOR Ions Are
Removed From Ionic Equation
The Molecular Equation is :
AgClO4 (aq) + NaNO3 (aq) → AgNO3 (aq) + NaClO4 (aq)
The Ionic Equation is :
Ag+(aq) + ClO4-(aq) + Na+(aq) + NO3-(aq) →
Ag+(aq) + NO3-(aq) + Na+(aq) + ClO4-(aq)
Spectator Ions: all are spectator ions
There is no Net Ionic Equation
because there is NO REACTION !!!!!!
• SPECTATOR Ions – Ions that do not
take part in the chemical reaction
Write (1) Molecular (2) Ionic &
(3) Net Ionic Equations for
Equations for Sulfuric Acid + Potassium Hydroxide
The Molecular Equation is :
H2SO4(aq) + KOH(aq) H2SO4(aq) + 2 KOH(aq) H2O(liq) + K2 SO4(aq)
?????
The Ionic Equation is :
H2O(liq) + K2 SO4(aq)
2
H+(aq)
+
SO42-(aq)
2 H +(aq) + SO42-(aq) + K+ (aq) + OH-(aq) H2O(liq) +
K+ (aq)
H2O(liq) + 2 K+(aq) + SO42-(aq)
+ SO4
2-(aq)
Spectator Ions
K+ &
Write (1) Molecular (2) Ionic &
(3) Net Ionic Equations for
???????
NaNO3(aq) + HgI2(S)
Hg +2(aq) + NO3-(aq) + Na+ (aq) + I-(aq) Na+(aq)
+ NO3
-(aq)
Hg+2(aq) + 2 I-(aq) HgI2(S)
-2
SO4
Net Ionic Equation is :
2 H +(aq) + OH-(aq) H2O(liq)
Hg(NO3)2(aq) + NaI(aq) + 2 K+(aq) + OH-(aq) 2 H+ (aq)
Equations for
+ OH-(aq) H2O (liq)
Mercury II Nitrate + Sodium Iodide
The Molecular Equation is :
Hg(NO3)2 (aq) + NaI (aq) NaNO3 (aq) + HgI2 (s)
The Ionic Equation is :
Hg+2(aq) + NO3-(aq) + Na+(aq) + I-(aq) Na+(aq) + NO3-(aq) + HgI2(s)
Spectator Ions Na+ & NO3-
+ HgI2(S)
Net Ionic Equation is :
Hg+2 (aq) + 2 I- (aq) HgI2(s)
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How do you recognize an
Oxidation Reduction Reaction ?
H2 (g) + O2 (g) H2O (g)
On “one side” of element all by its self
While on the “other side” element in combination
Another example
Zn (s) + 2 HCl (aq) → H2 (g) + ZnCl2 (aq)
Assigning Oxidation Numbers
Must Know
Oxidation“RULES”
See Text / Lab Manual
For example
Sodium ......................Na
Atomic Number …….11
1. EACH ATOM in a
PURE ELEMENT has an
Oxidation Number of ZERO
H2 , O2 , N2 , F2 ,
Na , Fe , Zn etc
Number of Protons … 11
Number of Electrons ..11
Same number of protons
and electrons
and
2. Alkali Metals always = +1 {Li , Na, K, Rb, Cs
3. Alkaline Earth Metals always = +2 {Mg , Ca , Sr, Ba
4. The oxidation number of H in its compounds = + 1
Most of the time: H2 O ; Na OH ; HCl ; H C2H 3O 2
But Not all the time Ca H 2 {H = -1 in Hydrides
5. Oxidation number of F in compounds always = - 1
6. Oxidation number of O in its compounds = – 2
Most of the time: H2 O ; Na OH ; H C2H 3O 2
But Not all the time H 2 O2 {O = -1 in Peroxides
OXIDATION REDUCTION
• What is The OXIDATION NUMBER
For EACH Element Present ?
• What is OXIDIZED ?
• What is REDUCED ?
• What is The OXIDIZING AGENT ?
• What is The REDUCING AGENT ?
“Types” of Equations
(a) Molecular (b) Ionic & (c) Net Ionic
(a) Molecular: the formula for each substance
are written in the molecular form
Zn (s) + 2 HBr (aq)→ H2 (g)+ ZnBr2 (aq)
(b) Ionic: shows dissolved species in ionic form
Zn(s) + 2H+(aq) + 2Br-(aq)→ H2(g) + Zn2+(aq) + 2Br-(aq)
only shows the ionic species that
actually take part in the reaction
Zn (s) + 2 H+ (aq) → H2 (g) + Zn2+ (aq)
(c) Net Ionic:
Redox Reactions
Zn(s) + HCl(aq) → H2(g) + ZnCl2(aq)
Oxidation Numbers
Zn(s) & H2(g) = 0
Zn in ZnCl2(aq) = +2
Cl in HCl and ZnCl2 = -1
What is OXIDIZED
Zn 0 → Zn 2+
What is REDUCED
H + → H2O
Oxidizing AGENT ?
HCl(aq)
Reducing AGENT ?
Zn metal
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Putting Concepts Together
A sample of 70.5 mg of potassium phosphate is added
to 15.0 mL of 0.050 M silver nitrate,
Potassium phosphate’s chemical formula is K3PO4
Silver nitrate’s chemical formula is AgNO3
Molecular Equation
(b) Write ionic and net ionic equations
K3PO4(aq) + AgNO3(aq) Ag3PO4(?) + KNO3(?)
According to solubility guidelines Ag3PO4 will precipitate
(c) What is the limiting reactant in the reaction?
while K3PO4, AgNO3, & KNO3 are soluble
(a) Write the molecular equation for the reaction.
(d) Calculate the theoretical yield, in grams, of the
precipitate that forms.
ionic equation
K+(aq) + PO43– (aq) + Ag+(aq) + NO3– (aq) Ag3PO4(s) + K+(aq) + NO3-1(aq)
(b)
Net Ionic Equation
Ag+(aq) + PO43– (aq) Ag3PO4(s)
Spectator ions: K+ & NO3– ions
(c) To determine the limiting reactant, must
determine the number of moles of each reactant.
Molarity (M)
Memorize the definition
Molarity =
moles of solute
Liters of solution
or
Moles Solute = Molarity x Liters of solution
moles = M x V
Complete ionic equation
K+(aq) + PO43– (aq) + Ag+(aq), + NO3– (aq) Ag3PO4(s) + K+(aq) + NO3-1(aq)
According to the balanced equation,
1 K3PO4(aq) + 3AgNO3(aq) Ag3PO4 + 3 KNO3(aq)
1 mol K3PO4 requires 3 mol of AgNO3.
There are 3.32 × 10–4 moles of K3PO4
and 7.5 × 10–4 moles of AgNO3
Rebalance equation
3.32 K3PO4 + 9.96AgNO3 3.32Ag3PO4 + 9.96 KNO3
Not enough AgNO3 (have 7.5x 10-4 moles) for the
K3PO4 therefore AgNO3 is the limiting reactant.
How are Moles determined from Molarity?
Moles of Solute = Molarity x (Volume in Liters)
------------------------------------Calculate the number of moles of HCl
in 50.0 mL of 2.00 M HCl(aq)
Moles = M x V = (0.0500)x(2.00) = 0.100
-----------------------------------Calculate the number of moles of NaOH
in 51.0 mL of 2.00 M NaOH
Moles = M x V = (0.0510)x(2.00) = 0.102
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Write & Balance Reaction
1 HCl (aq) + 1 NaOH (aq) 1 H2O + 1 NaCl (aq)
Preparing solutions By Dilution
or
0.1HCl(aq) + 0.102 NaOH(aq) 0.1H2O + 0.1NaCl(aq)
What is the Limiting
Reagent ?
--------------------------
What is the Theoretical
Yield of water?
----------------------
5.85 grams of NaCl
formed, % Yield is
HCl(aq)
-----------------------0.1 mole = 1.8 grams
-----------------------5.85 g NaCl = 0.1 mole
Start with 50.0 mL of 2.00 M HCl(aq)
Add water until have 100.0 mL of solution
What is the Molarity of the new solution?
(1) M1 V1 = M2 V2 = moles of Solute
(2) (2.00)(50.0) = ( M2)(100.0) = ???
(3) M2 = 1.00
Therefore 100% yield
Chap 5 THERMO chemistry Review
HEAT CHANGES that take place during
1. Physical Changes (Change of Phase)
&
2. Chemical Changes (Chemical Reactions)
Energy cannot be created or destroyed
BUT it can be transformed from one form to another
(potential & kinetic)
or transferred between system and surroundings
One of the basic assumptions of thermodynamics
is the idea that the Universe can be divided into
a System and its Surroundings
The 1st Law of Thermodynamics
states that the energy of the universe is constant
∆Uuniverse = ∆Usystem - ∆Usurrounding = 0
Energy can be transferred
from system to surroundings (or vice versa)
The study of energy and its transformation is known as
THERMODYNAMICS
THERMOCHEMISTRY is the portion of
thermodynamics that pertains to chemical reactions
The energy possessed by a system is called its
internal energy
internal energy of a system is the sum of all kinetic
and potential energies of all components of the system
Note: Most books use U for internal energy.
This book uses the older nomenclature E
The change in the internal energy of a system is equal to the
sum of the heat ….(gained or lost by the system)
system and the
work …………...(done by or on the system)
system
∆Usystem = q + w
For q
+ means system GAINS heat (endothermic)
- means system LOSES heat (exothermic)
For w
+ means work done ON system
- means work done BY system
For ∆Usys
+ means system GAINS energy
- means system LOSES energy
---------------------------------------------------------------------------------------------------------
but it can't be created or destroyed
∆Usystem = ∆Usurrounding
--------------------------------------------------------------------------------------------------
A more useful form of the first law:
∆Usystem = q + w
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Changes involving energy can either be
PHYSICAL or CHEMICAL
PHYSICAL Changes
1. Change in state (No temperature change)
2. Change within a state (Change in temp)
CHEMICAL Changes (Reactions
1. Formation
2. Combustion
3. Neutralization
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