Download Chapter 3 Stoichiometry STOICHIOMETRY: The chemical arithmetic

Chapter 3
Stoichiometry
STOICHIOMETRY:
The chemical
arithmetic used to relate the amount of
products and reactants to each other
1st Write Chemical Equation
2nd Balance Equation
3rd Interpret Equation
1st Write Chemical Equation
REACTANTS PRODUCTS
A2 + B2 A2B
nd
2
Balance Equation
A balanced chemical equation has the
same type and number of ______ in
the reactants as in the products.
A2 + B2 =A=
=B=
A2 B
3rd Interpret Equation
2 A2 + B2 2 A2B
Two units of A2 _
React with One unit of B2 __
Forming Two units of A2B__
Magnesium reacts with “air”
forming Magnesium Oxide
1st Write Chemical Equation
Mg(s) + O2(g) → MgO(s)
2nd Balance Equation
2 Mg + O2
→
2 MgO
2 = Mg = 2
2 = O = 2
3rd Interpret Equation
2 Mg + O2 → 2 MgO
Some Balanced Chemical Reactions
Combination & Decomposition reactions
In a combination reaction two or more
substances form a single compound.
1 2(g) + __H
3 2(g) → __NH
2
__N
3(g)
In a decomposition reaction, a single
compound forms two or more new
substances.
2
2
3 2(g)
__KClO
+ __O
3(s) → __KCl(s)
CHEMICAL REACTIONS
(YOU SHOULD KNOW)
1. COMBUSTION
(Of a Hydrocarbon)
2. NEUTRALIZATION
(Acid + Base)
3. ACID + ACTIVE METAL
4. FORMATION Reaction
CHEMICAL REACTIONS
(YOU SHOULD KNOW)
1. COMBUSTION (Of a Hydrocarbon)
The combustion of a Hydrocarbon
produces CO2 and H2O :
CH4 + O2 → CO2 + H2O
Combustion of Butane
C4H10 + 6 ½ O2
→ 4 CO2 + 5 H2O
Is This Reaction Balanced ?
With Fractions ?
CHEMICAL REACTIONS
(YOU SHOULD KNOW)
2. NEUTRALIZATION
(ACID+ BASE)
HCl (aq) + NaOH → H2O + A SALT
In this case the salt is NaCl Sodium Chloride
CHEMICAL REACTIONS
(YOU SHOULD KNOW)
3. ACID + ACTIVE METAL
HCl (aq) + Zn (s) → H2 (g) + A SALT
In this case the salt is
CHEMICAL REACTIONS
(YOU SHOULD KNOW)
4. FORMATION Reaction
Zn (s) + Cl2 (g) → ZnCl2 (s)
Reactants are in their “natural” state
Which of the following are
Balanced ?
C4H10 + 6 ½ O2
→ 4 CO2 + 5 H2O
4 C4H10 + 26 O2 → 16 CO2 + 20 H2O
Uniquely balanced equation
C4H10 + 6 ½ O2
→ 4 CO2 + 5 H2O
4 C4H10 + 26 O2 → 16 CO2 + 20 H2O
The ONLY Uniquely balanced equation is:
2 C4H10 + 13 O2 → 8 CO2 + 10 H2O
(MOLECULAR) WEIGHTS
The FORMULA WEIGHT of a
substance is the sum of the atomic
weights of each atom in its chemical
formula.
Calculate the Formula Weight of
Hydrogen Sulfide
Nickel II Carbonate
Magnesium Acetate
Ammonium Sulfate
Potassium Phosphate
Iron III Oxide
Diphosphorus pentasulfide
1st you must know
the formula
Calculate the Formula Weight of
H2S
NiCO3
Mg(C2H3O2)2
(NH4)2SO4
K3PO4
Fe2O3
P2S5
Avogadro’s Number & The Mole
Avogadro’s number is chosen so that
1 mole of 12C atoms has a mass of exactly
12 grams.
1 mole
12C
atoms = 6.02x1023
6.02x1023
12C
atoms = 12 g
12C
atoms
Mascots
Lamar University
University of Texas
LSU
Rice
Oregon
Nederland
CARDINALS
LONGHORNS
TIGERS
OWLS
DUCKS
BULLDOGS
Biology
Department
Mascot
Chemistry
Department
Mascot
Rabbits (& other animals)
If you had a mole of
rabbits, how many
rabbits would you
have ?
How many rabbit ears
would you have?
How many rabbit feet?
Interpreting Chemical Formulas
If you had a mole of water ,
how many molecules of
water would you have ?
How many Hydrogen atoms
would you have?
How many Oxygen atoms?
• The mole is just a conversion factor
that allows us to accurately work
with atoms and molecules.
• One mole of any substance contains
6.02 x 1023 units of that substance.
Grams, Moles & Avogadro
Conversion Factors
Use Molecular Weight
To Convert
To Convert
Grams to Moles
or
Moles to Grams
Grams to Moles
3.2 Grams of Oxygen = ? Moles
1 Mole
? Grams x
= ? Moles
? Grams
1 Mole
3.2 Grams x
= 0.10 Moles
32 Grams
Moles to Grams
1.5 Moles of Methane = ? Grams
? Grams
? Moles x
= ? Grams
1 Mole
16 Grams
1.5 Moles x
= 24 Grams
1 Mole
Use AVOGARDRO’S Number
• For NUMBER of Atoms
• Or NUMBER of Molecules
• 32 Grams of Oxygen = ? Molecules
• 32 Grams of Oxygen = ? Atoms
• 1 Molecule of Oxygen = ? Grams
Interpretation of Chemical Reactions
Using Stoichiometry
Example 1
Nitrogen monoxide reacts with oxygen to
produce Nitrogen dioxide
1st Write the reaction
2nd balanced reaction
3rd Interpret Reaction
How many moles of oxygen gas are required
to react completely with 2.0 mole NO?
2 NO (g) + O2 (g) 2 NO2 (g)
(a) 0.5 mol O2
(b) 1.0 mol O2
(c) 1.5 mol O2
(d) 2.0 mol O2
How many moles of oxygen gas are required
to react completely with 1.0 mole NO?
2 NO (g) +
O2 (g)
(a) 0.5 mol O2
(b) 1.0 mol O2
(c) 1.5 mol O2
(d) 2.0 mol O2
(e) 2.5 mol O2
2 NO2 (g)
How many moles of oxygen gas are required
to react completely with 2.50 moles NO?
2 NO (g) +
O2 (g)
(a) 0.5 mol O2
(b) 1.00 mol O2
(c) 1.25 mol O2
(d) 1.50 mol O2
(e) 2.00mol O2
2 NO2 (g)
How many moles of oxygen gas are required
to react completely with 10 moles NO?
2 NO (g) +
O2 (g)
(a) 5 mol O2
(b) 10 mol O2
(c) 15 mol O2
(d) 20 mol O2
(e) 25 mol O2
2 NO2 (g)
If 6.0 moles of NO are reacted with 3.0 mole O2,
how many moles NO2 are produced?
2 NO (g)
(a)
(b)
(c)
(d)
(e)
+
2.0 mol NO2
6.0 mol NO2
10.0 mol NO2
16.0 mol NO2
32.0 mol NO2
O2 (g)
2 NO2 (g)
If 10.0 moles of NO are reacted with 5.0 mole
O2, how many moles NO2 are produced?
2 NO (g)
(a)
(b)
(c)
(d)
(e)
+
2.0 mol NO2
6.0 mol NO2
10.0 mol NO2
16.0 mol NO2
32.0 mol NO2
O2 (g)
2 NO2 (g)
If 10.0 moles of NO are reacted with 6.0 moles
O2, how many moles NO2 are produced?
2 NO (g)
(a)
(b)
(c)
(d)
(e)
+
2.0 mol NO2
6.0 mol NO2
10.0 mol NO2
16.0 mol NO2
32.0 mol NO2
O2 (g)
2 NO2 (g)
• 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.
.
If the actual amount of product formed in a
reaction is less than the theoretical amount
we can calculate a percentage yield.
Actual product yield
% yield =
× 100%
Theoretica l product yield
The Limiting Reactant
A reaction stops when one reactant
is totally consumed.
This is the limiting reactant.
The other reactants are excess reactants.
The Limiting Reactant
• How many bikes can be made
from 10 frames and 16 wheels ?
1 frame + 2 wheels → 1 bike
What is the limiting “reagent” ?
Excess “reagent” ?
If 10.0 moles of NO are reacted with 6.0 moles
O2, how many moles NO2 are produced?
2 NO (g)
+
O2 (g)
(a)What LIMITS
the reaction?
(b)What is in
excess ?
2 NO2 (g)
If 10.0 moles of NO are reacted with 6.0
moles O2, how many moles of the excess
reagent remain?
2 NO (g)
•
•
•
•
+
1.0 mol O2
5.0 mol O2
4.0 mol NO
8.0 mol NO
O2 (g)
2 NO2 (g)
Example 2:
Hydrogen reacts with Nitrogen
to form Ammonia
1st Write Reaction
Next Balance Equation
Interpret Balanced Equation
Interpretation of a Chemical Reaction
3 H2 +
Three moles
of Hydrogen
1 N2
→
One mole
of Nitrogen
2 NH3
Two moles
of Ammonia
====================================================
Mole Ratio MUST Always be 3 : 1 : 2
3 H2 +
3/3 H2 +
3/2 H2 +
1 N2
→
1/3 N2
→
1/2 N2
→
2 NH3
2/3 NH3
1 NH3
3/4 H2 +
1/4 N2
→
2/4 NH3
3/8 H2 +
1/8 N2
→
2/8 NH3
All are
3 :1 : 2
How many moles of Nitrogen are needed to
react with 3 moles of Hydrogen ?
3 H2 + 1 N2 → 2 NH3
(a) ½ (b) 1 (c)1 ½ (d) 2 (e) 2 ½ (f) 3
How many moles of Nitrogen are needed to
react with 1 ½ moles of Hydrogen ?
3 H2 + 1 N2 → 2 NH3
(a) ½ (b) 1 (c)1 ½ (d) 2 (e) 2 ½ (f) 3
Chemical Reactions are
Interpreted on the
Mole basis but chemicals are weighed
in the laboratory in Grams
Use
Molecular Weight
To Convert
To Convert
Grams to Moles
or
Moles to Grams
Units, Units, Units
___Grams
1 moles
x
? Grams
= ___moles
Units, Units, Units
___moles
? Grams
x
1 moles
= ___Grams
How many moles of Nitrogen are needed to
react with Three grams of Hydrogen ?
3 g x 1 mole = 1 ½ moles of H2
2 g
3 H2 + 1 N2 → 2 NH3
(a) ½ (b) 1 (c)1 ½ (d) 2 (e) 2 ½ (f) 3
3 H2 + 1 N2 → 2 NH3
How many moles of Ammonia ( NH3 ) are
produced from
(a) 3 grams of H2 and ½ mole of N2?
1 mole = (1 mole)x(17 g/mole) grams of NH3
(b) 3 grams of H2 and 28 grams of N2?
1 mole NH3
with 14 g of Nitrogen in excess
3 H2 + 1 N2 → 2 NH3
With a 50 % Yield, How many moles of NH3
are produced from
(a) 3 grams of H2 and ½ mole of N2?
½ mole = (½ mole)x(17 g/mole) grams of NH3
(b) 3 grams of H2 and 28 grams of N2?
½ mole NH3
with 14 g of Nitrogen in excess
DESCRIBING COMPOUNDS
Percent Composition
Empirical Formulas
Molecular Formulas
Two Approaches to Chemical Formulas
1. Given Formula
Determine Percent Composition of
Each Element in Compound
2. Given Percent Composition of Each
Element in Compound
Determine Formula
A student obtains the following data
Mass of crucible and cover = 28.288 gms
Mass crucible, cover & K = 28.709 gms
Mass crucible, cover &
= 28.793 gms
potassium oxide
Determine the EMPIRICAL formula
First determine % by Weight
Weight K = 28.709 – 28.288 = 0.421
Weight O = 28.793 – 28.709 = 0.084
Total =
0.505
0.421
% by Wt K =
x 100 = 83.36633
0.505
83.4 % by wt K and 16.6 % by wt O
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
= 2