Transcript Chapter 3 Stoichiometry: Calculations with Chemical

Chapter 3
Stoichiometry:
Calculations with Chemical
Formulas and Equations
Reaction
Types
Combination Reactions
• Two or more substances react to form one
product
• Examples:
N2 (g) + 3 H2 (g)  2 NH3 (g)
C3H6 (g) + Br2 (l)  C3H6Br2 (l)
2 Mg (s) + O2 (g)  2 MgO (s)
2 Mg (s) + O2 (g)  2 MgO (s)
Link to Video
Decomposition Reactions
• One substance breaks down into two or
more substances
• Examples:
CaCO3 (s)  CaO (s) + CO2 (g)
2 KClO3 (s)  2 KCl (s) + O2 (g)
2 NaN3 (s)  2 Na (s) + 3 N2 (g)
Link to Video
Combustion Reactions
• Rapid reactions that produce a flame
• Most often involve hydrocarbons
reacting with oxygen in the air
• Examples:
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)
Balancing
Reactions
Law of Conservation of Mass
“We may lay it down as an
incontestable axiom that, in all
the operations of art and nature,
nothing is created; an equal
amount of matter exists both
before and after the experiment.
Upon this principle, the whole
art of performing chemical
experiments depends.”
--Antoine Lavoisier, 1789
Dalton’s Postulates
Atoms of an element are not changed into
atoms of a different element by chemical
reactions; atoms are neither created nor
destroyed in chemical reactions. They just
rearrange!
A process in which one or more
substances is changed into one or more
new substances is a chemical reaction.
A chemical equation uses chemical symbols
to show what happens during a chemical
reaction.
reactants
products
3.7
How to “Read” Chemical Equations
2 Mg + O2
2 MgO
2 atoms Mg + 1 molecule O2 makes 2 formula units MgO
2 moles Mg + 1 mole O2 makes 2 moles MgO
48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO
IS NOT
2 grams Mg + 1 gram O2 makes 2 g MgO
3.7
Balancing Chemical Equations
1. Write the correct formula(s) for the reactants and
the product(s).
Ethane reacts with oxygen to form carbon dioxide and water
C2H6 + O2
CO2 + H2O
2. Change the numbers in front of the formulas
(coefficients) to make the number of atoms of
each element the same on both sides of the
equation. Do not change the subscripts.
2C2H6
NOT
C4H12
3.7
Balancing Chemical Equations
3. Start by balancing those elements that appear in
only one reactant and one product.
C2H6 + O2
2 carbon
on left
C2H6 + O2
6 hydrogen
on left
C2H6 + O2
CO2 + H2O
start with C or H but not O
1 carbon
on right
multiply CO2 by 2
2CO2 + H2O
2 hydrogen
on right
2CO2 + 3H2O
multiply H2O by 3
3.7
Balancing Chemical Equations
4. Balance those elements that appear in two or
more reactants or products.
C2H6 + O2
2 oxygen
on left
2CO2 + 3H2O
multiply O2 by 7
2
4 oxygen + 3 oxygen = 7 oxygen
(3x1)
on right
(2x2)
C2H6 + 7 O2
2
2CO2 + 3H2O
2C2H6 + 7O2
4CO2 + 6H2O
remove fraction
multiply both sides by 2
3.7
Balancing Chemical Equations
5. Check to make sure that you have the same
number of each type of atom on both sides of the
equation.
2C2H6 + 7O2
4CO2 + 6H2O
12
4 CH
(2
6)
14
O(2
(7xx2)
2)
4(6
C
2)6)
1412
OH
(4
x 2x +
Reactants
4C
12 H
14 O
Products
4C
12 H
14 O
3.7
Remember
2 Mg + O2
2 MgO
IS NOT
2 grams Mg + 1 gram O2 makes 2 g MgO
So How do we relate masses of reactants
and products? That’s what we can observe!
3.7
Formula
Weights
Formula Weight (FW)
• Sum of the average atomic weights for the
atoms in a chemical formula
• So, the formula weight of calcium chloride,
CaCl2, would be
Ca: 1(40.1 amu)
+ Cl: 2(35.5 amu)
111.1 amu
• These are generally reported for ionic
compounds
Molecular Weight (MW)
• Sum of the average atomic weights of the
atoms in a molecule
• For the molecule ethane, C2H6, the
molecular weight would be
C: 2(12.0 amu)
+ H: 6(1.0 amu)
30.0 amu
Percent Composition
One can find the percentage of the mass of a
compound that comes from each of the
elements in the compound by using this
equation:
(number of atoms)(atomic weight)
% element =
(FW of the compound)
x 100
Percent Composition
So the percentage of carbon in ethane is…
(2)(12.0 amu)
%C =
(30.0 amu)
24.0 amu
x 100
=
30.0 amu
= 80.0%
Moles
Using Moles
Moles provide a bridge from the molecular scale to
the real-world scale
Avogadro’s Number
• 6.02 x 1023
• 1 mole of 12C has a
mass of 12 g
The mole (mol) is the amount of a substance that
contains as many elementary entities as there
are atoms in exactly 12.00 grams of 12C
1 mol = NA = 6.0221367 x 1023
Avogadro’s number (NA)
3.2
eggs
Molar mass is the mass of 1 mole of shoes in grams
marbles
atoms
1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g
1 12C atom = 12.00 amu
1 mole 12C atoms = 12.00 g 12C
1 mole lithium atoms = 6.941 g of Li
For any element
atomic mass (amu) = molar mass (grams/mol)
3.2
Percent composition revisited
n x molar mass of element
x 100%
molar mass of compound
n is the subscript number … Assume one mole of
substance and the math goes like this.
2 x (12.01 g)
x 100% = 52.14%
46.07 g
6 x (1.008 g)
%H =
x 100% = 13.13%
46.07 g
1 x (16.00 g)
%O =
x 100% = 34.73%
46.07 g
%C =
C2H6O
52.14% + 13.13% + 34.73% = 100.0%
3.5
Question
• What is the mass on one atom of carbon 12?
• What is the mass in grams of one amu?
Mole Relationships
• One mole of atoms, ions, or molecules contains
Avogadro’s number of those particles
• One mole of molecules or formula units contains
Avogadro’s number times the number of atoms or ions
of each element in the compound
Do You Understand Molar Mass?
How many atoms are in 0.551 g of potassium (K) ?
3.2
Do You Understand Molar Mass?
How many atoms are in 0.551 g of potassium (K) ?
1 mol K = 39.10 g K
1 mol K
0.551 g K x
39.10 g K
= 0.0141 moles K
3.2
Molecular mass (or molecular weight) is the sum of
the atomic masses (in amu) in a molecule.
1S
SO2
2O
SO2
32.07 amu
+ 2 x 16.00 amu
64.07 amu
For any molecule
molecular mass (amu) = molar mass (grams)
1 molecule SO2 = 64.07 amu
1 mole SO2 = 64.07 g SO2
3.3
Do You Understand Molecular Mass?
How many H atoms are in 72.5 g of C3H8O ?
3.3
Do You Understand Molecular Mass?
How many H atoms are in 72.5 g of C3H8O ?
1 mol C3H8O = (3 x 12.01) + (8 x 1.008) + 16.00 =
60.09 g C3H8O
1 mol C3H8O molecules = 8 mol H atoms
1 mol H = 6.022 x 1023 atoms H
72.5 g C3H8O x
1 mol C3H8O
8 mol H atoms
x
60.09 g C3H8O 1 mol C3H8O
= 9.65 moles H
or 5.82 x 1024 atoms H
3.3
Mass Changes in Chemical Reactions
1. Write balanced chemical equation
2. Convert quantities of known substances into moles
3. Use coefficients in balanced equation to calculate the
number of moles of the sought quantity
4. Convert moles of sought quantity into desired units
3.8
Methanol burns in air according to the equation
2CH3OH + 3O2
2CO2 + 4H2O
If 209 g of methanol are used up in the combustion,
what mass of water is produced?
grams CH3OH
moles CH3OH
molar mass
CH3OH
209 g CH3OH x
moles H2O
grams H2O
molar mass
coefficients
H2O
chemical equation
4 mol H2O
18.0 g H2O
1 mol CH3OH
=
x
x
32.0 g CH3OH
2 mol CH3OH
1 mol H2O
235 g H2O
3.8
Finding
Empirical
Formulas
Calculating Empirical Formulas
One can calculate the empirical formula from the
percent composition
Calculating Empirical Formulas
The compound para-aminobenzoic acid (you may have
seen it listed as PABA on your bottle of sunscreen) is
composed of carbon (61.31%), hydrogen (5.14%),
nitrogen (10.21%), and oxygen (23.33%). Find the
empirical formula of PABA.
Calculating Empirical Formulas
Assuming 100.00 g of para-aminobenzoic acid,
C:
H:
N:
O:
1 mol
12.01 g
1 mol
5.14 g x
1.01 g
1 mol
10.21 g x
14.01 g
1 mol
23.33 g x
16.00 g
61.31 g x
= 5.105 mol C
= 5.09 mol H
= 0.7288 mol N
= 1.456 mol O
Calculating Empirical Formulas
Calculate the mole ratio by dividing by the smallest number
of moles:
C:
5.105 mol
0.7288 mol
= 7.005  7
H:
5.09 mol
0.7288 mol
= 6.984  7
N:
0.7288 mol
0.7288 mol
= 1.000
O:
1.458 mol
0.7288 mol
= 2.001  2
Calculating Empirical Formulas
These are the subscripts for the empirical formula:
C7H7NO2
Combustion Analysis
• Compounds containing C, H and O are routinely analyzed
through combustion in a chamber like this
– C is determined from the mass of CO2 produced
– H is determined from the mass of H2O produced
– O is determined by difference after the C and H have been
determined
Elemental Analyses
Compounds
containing other
elements are analyzed
using methods
analogous to those
used for C, H and O
Stoichiometric Calculations
The coefficients in the balanced equation give the
ratio of moles of reactants and products
Stoichiometric
Calculations
From the mass of
Substance A you can
use the ratio of the
coefficients of A and B
to calculate the mass of
Substance B formed (if
it’s a product) or used
(if it’s a reactant)
Stoichiometric
Calculations
C H O + 6 O  6 CO + 6 H O
6
12
6
2
2
2
Starting with 1.00 g of C6H12O6…
we calculate the moles of C6H12O6…
use the coefficients to find the moles of H2O…
and then turn the moles of water to grams
Real Life Analysis
• What is the percent composition of an
organic substance called ethanol?
Combust 11.5 g ethanol
Collect 22.0 g CO2 and 13.5 g H2O
3.6
Combust 11.5 g ethanol
Collect 22.0 g CO2 and 13.5 g H2O
g CO2
mol CO2
mol C
gC
6.0 g C = 0.5 mol C
g H2O
mol H2O
mol H
gH
1.5 g H = 1.5 mol H
g of O = g of sample – (g of C + g of H)
4.0 g O = 0.25 mol O
Empirical formula C0.5H1.5O0.25
Divide by smallest subscript (0.25)
Empirical formula C2H6O
3.6
Limiting
Reactants
How Many Cookies Can I Make?
• You can make cookies
until you run out of one
of the ingredients
• Once this family runs out
of sugar, they will stop
making cookies (at least
any cookies you would
want to eat)
How Many Cookies Can I Make?
• In this example the sugar
would be the limiting
reactant, because it will
limit the amount of
cookies you can make
Limiting Reactants
The limiting reactant
is the reactant present
in the smallest
stoichiometric amount
Limiting Reactants
• The limiting reactant is the reactant present in the
smallest stoichiometric amount
– In other words, it’s the reactant you’ll run out of first (in
this case, the H2)
Limiting Reactants
In the example below, the O2 would be the excess
reagent
Do You Understand Limiting Reagents?
In one process, 124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O3
Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.
g Al
mol Al
g Fe2O3
124 g Al x
mol Fe2O3 needed
OR
mol Al needed
mol Fe2O3
1 mol Al
27.0 g Al
x
g Fe2O3 needed
1 mol Fe2O3
2 mol Al
Start with 124 g Al
160. g Fe2O3
=
x
1 mol Fe2O3
g Al needed
367 g Fe2O3
need 367 g Fe2O3
Have more Fe2O3 (601 g) so Al is limiting reagent
3.9
Use limiting reagent (Al) to calculate amount of product that
can be formed.
g Al
mol Al
mol Al2O3
2Al + Fe2O3
124 g Al x
1 mol Al
27.0 g Al
x
1 mol Al2O3
2 mol Al
g Al2O3
Al2O3 + 2Fe
102. g Al2O3
=
x
1 mol Al2O3
234 g Al2O3
3.9
Theoretical Yield
• The theoretical yield is the amount of
product that can be made
– In other words it’s the amount of product
possible as calculated through the stoichiometry
problem
• This is different from the actual yield, the
amount one actually produces and measures
Percent Yield
A comparison of the amount actually
obtained to the amount it was possible to
make
Actual Yield
Percent Yield =
Theoretical Yield
x 100