Download Chemistry 211

Chapter 3
Chemical Equations
Stoichiometry: Calculations
with Chemical Formulas and
Equations
Reactants
Products
C(s) + O2(g)
CO2(g)
Combination Reaction (Combustion)
The rusting of iron
Fe + O2
Ways of presenting molecular composition
Fe2O3
•The mass of each element
Need to balance
4 Fe + 3O2
•A molecular formula
2 Fe2O3
•Percent composition
The first thing you always do is
examine to see whether the equation
is balanced.
Stoichiometry
Stoichiometry
Quantitative relationship
between amounts of
reactants and products.
aA + bB
cC + dD
Answers questions:
•How much product is formed from some
given amount of reactant?
(Theoretical Yield)
•How much of one reactant is required to
bring a reaction to completion given
some amount of another reactant?
(Limiting Reagent)
The Concept of the mole:
We talk here about the chemist’s
way for counting the amount of “stuff”
as compared to some standard.
One mole of “stuff” contains the
same number of “particles” as there
are atoms in one mole of carbon-12
# moles of an element =
mass of element/ molar mass
You have 53.0 grams of Na metal.
How many moles will you have?
53.0 g/(22.99 g/mol) = 2.31 mol
MOLE: Important Concepts
# moles of an element =
mass of element/ molar mass
1 mol contains Avogadro’s Number
•One mole of carbon-12 will weigh
12.0000 grams and contain
6.02*1023 atoms (Avogadro’s number)
of carbon-12.
•One mole of oxygen-16 will weigh
16.0000 grams and contain
6.02*1023 atoms (Avogadro’s number)
of oxygen-16.
•One mole of nitrogen-14 will weigh
14.0000 grams and contain
6.02*1023 atoms (Avogadro’s number)
of nitrogen-14.
# moles of an element =
mass of element/ molar mass
How many grams of Fe are found
in 0.25 mole?
(0.25 mol) X (55.87 g/mol) = 14 g
Problem
How many atoms of Hg (mercury) will
be contained in 15 mL (cm3) of Hg? The
density of Hg is 13.534 g/mL.
Molecules, Compounds and the Mole
Avogadro’s number: (N) 6.02 x 1023 “things”
per mol of “things”
# Things = # mol x N
Stoichiometry: concerned with
mass/mass relationships in
chemical reactions
2
A
+
B
C
#mol = mass/Molar mass
(Molar mass (Molecular Weight )) =
Sum Individual atomic masses
Stoichiometry: concerned with
mass/mass relationships in
chemical reactions
Mass B
Mass A
2
A
+
Mass C
B
C
MM C
MM A MM B
Moles A
Moles B
REAL!
In theory
2 Molecules
+ 1 Molecule
1 Molecule
From any balanced equation:
If you know masses and
molar masses you can
calculate any missing
quantity.
# mol = mass / molar mass
mass = # mol x molar mass
Moles C
Structural Formula
Percent Composition
H2
C
H 2C
EXAMPLE: What is the
Percent composition of the
elements in the sex attractant
of the Asian Civet cat?
CH2
CH
H 3C
C8H14O3
CH
O
CO2H
C
H2
Molar mass = 8 C *12.01 g/mol C +
14 H *1.008 g/mol H +
3 O *15.994 g/mol C = 158.174 g/mol
C8H14O3
What percent of this mass is due to oxygen?
Empirical Formulas
%C = mass C/Molar mass (*100) =
(96.08 g/mol) / (158.2 g/mol) * 100 = 60.73%
The simplest formula of a compound
in small whole numbers.
%H = mass of H/Molar mass(*100) =
(14.11 g/mol) / (158.2 g/mol) * 100 = 8.92%
%O = mass of O/ Molar Mass (*100) =
(47.98 g/mol) / (158.2 g/mol) * 100 = 30.33%
You have a sample of the venom of
the fire ant. You send it for analysis
and the company sends back a
printout telling you that your
sample contains:
Method:
1. ASSUME 100 grams of sample
Thus your sample contains
80.4 grams C, 13.8 grams H and 5.8 grams N
2. What is mole ratio of the constituents?
# mol = mass/molar mass of atoms
80.4 %C, 13.8 %H, and 5.8 %N
#mol C = 80.4 g / (12.01 g/mol) = 6.69 mol
#mol H = 13.8 g / (1.001 g/mol) = 13.8 mol
#mol N = 5.8 g / (14.00 g/mol) = .414 mol
C6.69H13.8N.414
3. Find a common denominator
(divide by smallest)
H2
C
H 2C
6.69 mol /.414 mol = 16.17
13.8 mol /.414 mol = 33.3
.414 mol /.414mol = 1.0
CH2
CH
H 3C
C
H2
C
H2
C16H33N
H2
C
H2
C
H2
C
H2
C
CH
N
H
C
H2
C
H2
CH3
C
H2
Determining an Empirical Formula
C,H,O
Burn in O2
Grams
All the carbon
in carbon dioxide
comes from here
CO2
H2O
Grams
Grams
Moles
Moles
#moles H2O = 0.7425g/(2.011g/mol + 15.994g/mol)
= 0.04024 mol
But, 1 mole of H2O has 2 moles of H
Grams
0.523
Moles =
CO2
H2O
Grams
Grams
1.612
0.03664
0.7425
Moles(C) =
0.03664
Moles(H) =
Burn in O2
0.04024
0.08248
CO2
H2O
Grams
Grams
All the hydrogen
in water
comes from here
Moles
#moles CO2 = 1.612g/(12.011g/mol + 31.988g/mol)
= 0.03664 mol
C,H
Grams
0.523
Burn in O2
Moles =
Moles(C) =
What is the formula?
Burn in O2
C,H,O
Grams
Exercise
0.523 g of CxHy was burned to
give 1.612 g of carbon dioxide
and 0.7425 g of water. A
separate mass spectrum showed
the molar mass = 114 g/mole.
C,H
Determining an Empirical Formula
CO2
H2O
Grams
Grams
1.612
0.03664
0.7425
0.03664
Find a common denominator
(divide by smallest)
.08248 mol /.03664 mol = 2.251
.03664 mol /.03664 mol = 1.000
Another example
C4H9
Empirical Formula
Molar Mass = (4 x 12g/mol) + (9 x 1g/mol) = 57g/mol
Analysis gives the following data:
%C = 74.0
%H = 8.64
%N = 17.3
What is the empirical formula?
Given: Molar Mass = 114g/mol
Thus,
Molecular Formula is
Tobacco contains a number of alkaloids . .
the most important is nicotine
C8H18
If you know the molar mass = 162,
what is the molecular formula?
Using formulas(Balancing Reactions)
CH4 + O2
CO2 + H2O
Methane + oxygen burn to give
carbon dioxide and water
We have to conserve atoms:
Have to have same number of each
type of atom on both sides of arrow
CH4 + O2
C =1
H=4
CO2
+ 2H2O
C=1
O=2
We need
2 more O’s
in reactant
H=4
O=4
Put a 2 as
coefficient to
oxygen.
CH4 + O2
CO2 + H2O
C =1
H=4
C=1
H=2
O=3
O=2
We need
2 more H’s
in product
CH4 +2 O 2
C =1
H=4
Put a 2 as
coefficient to
water.
CO2
+ 2H2O
C=1
O=4
H=4
O=4
We have a balanced equation.
It says “ 1 mole of methane +
2 moles of oxygen will produce 1 mole
of carbon dioxide + 2 moles of water.”
24 grams
CH4 +2 O 2
CO2
+ 2H2O
If we have 24 grams of methane
(and plenty of air!) how much water
will be formed? (i.e. methane is the
Limiting Reagent)
CH4 +2
1 mole of
methane
? grams
CO2
O2
+ 2H2O
2 moles
of water
gives
Molar mass of methane = 16.0 g/mol
Molar mass of water=18 g/mol
Moles Methane = 24 g /16 g/mol = 1.5 mol
The process: We need to work in moles.
How many moles of methane do
we REALLY have?
Thus, REALLY will have:
1.5 mol CH4 * (2 mol H2O mol CH4)
= 3 moles of H2O
3 mol H2O *18 g/mol = 54 grams H 2O
Combustion of Carbon
Compounds
A sugar:
Glucose: C6H12O6
Our metabolism is a combustion
process:
Carbon compounds in . . .
carbon dioxide and water out!
A fat:
Glyceryl stearate: C37H110O6
Lets compare fat with sugar.
C6H12O6 +O2
CO2 +
Balance:
C6H12O6 + 6O2
6 CO2 + 6H2O
H2O
Molar Mass =72.07g C +12.096g H + 95.96g O
=180.13 g/mol
Thus, 1.00 g =
1.00 g / (180.13 g/mol) = 5.53 x 10-3 mol
Lets examine the amount of oxygen
needed to “burn” 1.00 gram
of each. We’ll also see how much
water and carbon dioxide we
produce from each.
C6H12O6 + 6O2
6 CO2 +6H2O
5.53 x 10 -3 mol C6H12O6
We would expect to use
5.53 x 10-3 mol C6H12O6 * (6 mol of O2 / mol C6H12O6) =
3.33 x 10 -2 mol O2
This is equal to
3.33 x 10 -2 mol O2 * 31.98 g/mol =
1.07 grams O2
C37H110O6 + O2
CO2 + H2O
Balance:
C37H110O6 + 123/2 O2
37 CO2 +55 H2O
Molar Mass =444.4g C +110.9g H + 95.96g O
=651.3 g/mol
Thus, 1.00 g =
1.00 g / (651.3 g/mol) = 1.54 x 10-3 mol
Limiting Reagent Problems:
AgNO3 + NaCl -----> AgCl +NaNO 3
How much silver nitrate?
How much sodium chloride?
How much silver chloride will
be formed?
Stoichiometry Involves
• Balancing the equations
– This gives theory mole ratios
• Assigning molar masses
• Calculating “real” moles
– This gives real mole ratios
• Converting moles to mass
C37 H110O6 +(123/2)O2
37 CO2 +55 H2O
1.54 x 10 -3 mol C37H110O6
We would expect to use
1.54 x 10-3 mol C37H110O6 *
(123 mol of O 2 / 2 mol C6H12O6) =
9.47 x 10 -2 mol O2
This is equal to
9.47 x 10 -2 mol O2 * 31.98 g/mol =
3.03 grams O2
Determining Percent Yield
% Yield = Actual/Theory