Transcript Stoichiometry and the mole

Stoichiometry and the
mole
Chapter 8
What is stoichiometry?
Quantitative aspects of chemistry
 Stoicheon Greek root (element)
 Metron Greek root( to measure)
 Calculate how much of a reactant
needed to produce a product or how
much product could be expected

Stoichiometry
How much do I want?
 How much do I have?
 How much will I get?

The Mole
•Defined as the number of carbon
atoms in exactly 12 grams of carbon12.
•1 mole is 6.02 x 1023 particles.
•Treat it like a very large dozen
•6.02 x 1023 is called Avogadro's
number.
The Mole

One mole = 6.022 x 1023 (Avogadro’s
number)
Avogadro’s hypothesis
When all of the following are the
same:
 Volumes of containers
 Temperature of the gases
 The pressure exerted by and on each
gas
Then, the number of molecules in each
container will be the same, too.
However, the masses are not equal.
N2(g) + 3H2(g)2NH3(g)
1
3
2
1 dozen
(12)
1 gross
(144)
1mole
3 dozen
(36)
3 gross
(432)
3 moles
2 dozen
(24)
2 gross
(288)
2 moles
(6.02x1023) (18.06x1023)
(12.04 x1023)
Representative particles
•The smallest pieces of a substance.
•For an element it is an atom.
–Unless it is diatomic
•For a molecular compound it is a molecule.
•For an ionic compound it is a formula unit.
Molar Mass
A substance’s molar mass (molecular
weight) is the mass in grams of one mole
of the compound.
CO2 = 44.01 grams per mole
H2O = 18.02 grams per mole
Ca(OH)2 = 74.10 grams per mole
Chemical Equations
Chemical change involves a reorganization of
the atoms in one or more substances.
C2H5OH + 3O2  2CO2 + 3H2O
reactants
products
When the equation is balanced it has quantitative
significance:
1 mole of ethanol reacts with 3 moles of oxygen
to produce 2 moles of carbon dioxide and 3 moles
of water
Mole Relations
Calculating Masses of Reactants and
Products
1.
2.
3.
4.
Balance the equation.
Convert mass to moles.
Set up mole ratios.
Use mole ratios to calculate moles of
desired substance.
5. Convert moles to grams, if necessary.
Working a Stoichiometry Problem
6.50 grams of aluminum reacts with an excess of
oxygen. How many grams of aluminum oxide are
formed.
1. Identify reactants and products and write
the balanced equation.
4 Al
+ 3 O2
2 Al2O3
a. Every reaction needs a yield sign!
b. What are the reactants?
c. What are the products?
d. What are the balanced coefficients?
Working a Stoichiometry Problem
6.50 grams of aluminum reacts with an excess of
oxygen. How many grams of aluminum oxide are
formed?
4 Al
6.50 g Al
+
3 O2  2Al2O3
1 mol Al
2 mol Al2O3 101.96 g Al2O3
26.98 g Al
4 mol Al
1 mol Al2O3
= ? g Al2O3
6.50 x 2 x 101.96 ÷ 26.98 ÷ 4 = 12.3 g Al2O3
Percent yield
The % yield is calculated from:
(actual yield / theoretical yield) x
100
 The theoretical yield is how much
product is predicted from balanced
chemical equation.
 The actual yield is how much is
recovered when actual experiment is
conducted.

Limiting Reactant
The limiting reactant is the reactant
that is consumed first, limiting the amounts of
products formed.
How to solve stoichiometry
problems?
Write the balanced chemical equation
 Convert to moles information on
reactants and products
 Use mole ratios from equation to
determine number of moles of
unknown
 Convert moles to unit desired
 g xmoles xmoles yg y

Limiting reactant

Reactant present in short supply
Excess reactant
•Reactant in excess relative to limiting
reactant
Formulas
Empirical formula: the lowest whole number
ratio of atoms in a compound.
Molecular formula: the true number of
atoms of each element in the formula of a
compound.
 molecular formula = (empirical
formula)n [n = integer]
 molecular formula = C6H6 = (CH)6
 empirical formula = CH
Formulas
(continued)
Formulas for ionic compounds are ALWAYS
empirical (lowest whole number ratio).
Examples:
NaCl
MgCl2
Al2(SO4)3
K2CO3
Formulas
(continued)
Formulas for molecular compounds MIGHT
be empirical (lowest whole number ratio).
Molecular:
H2O
C6H12O6
C12H22O11
Empirical:
H2O
CH2O
C12H22O11
Empirical Formula Determination
1. Base calculation on 100 grams of compound.
2. Determine moles of each element in 100
grams of compound.
3. Divide each value of moles by the smallest of
the values.
4. Multiply each number by an integer to obtain
all whole numbers.
Empirical Formula Determination
Adipic acid contains 49.32% C, 43.84% O, and
6.85% H by mass. What is the empirical formula
of adipic acid?
 49.32 g C 1 mol C   4.107 mol C
12.01 g C 
 6.85g H 1 mol H   6.78 mol H
1.01 g H 
 43.84 g O 1 mol O   2.74 mol O
16.00 g O 
Empirical Formula Determination
(part 2)
Divide each value of moles by the smallest of the
values.
4.107
mol
C
Carbon:
 1.50
2.74 mol O
6.78 mol H
Hydrogen:
 2.47
2.74 mol O
2.74 mol O
Oxygen:
 1.00
2.74 mol O
Empirical Formula Determination
(part 3)
Multiply each number by an integer to obtain all
whole numbers.
Carbon: 1.50
x 2
3
Hydrogen: 2.50
x 2
5
Oxygen: 1.00
x 2
2
Empirical formula: C3H5O2
Finding the Molecular Formula
The empirical formula for adipic acid is
C3H5O2. The molecular mass of adipic acid
is 146 g/mol. What is the molecular
formula of adipic acid?
1. Find the formula mass of C3H5O2
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
Finding the Molecular Formula
The empirical formula for adipic acid is
C3H5O2. The molecular mass of adipic acid
is 146 g/mol. What is the molecular
formula of adipic acid?
2. Divide the molecular mass by the
mass given by the emipirical formula.
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
146
2
73
Finding the Molecular Formula
The empirical formula for adipic acid is
C3H5O2. The molecular mass of adipic acid
is 146 g/mol. What is the molecular
formula of adipic acid?
3. Multiply the empirical formula by this
number to get the molecular formula.
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
146
2
73
(C3H5O2) x 2
=
C6H10O4
Combustion Analysis
Technique that requires burning of an unknown substance and trap
the gases from burning.
Used to determine molecular
formula of an unknown
 Combustion= burning
 Using oxygen

Points about combustion






Element that makes unknown almost always contain carbon
and hydrogen. Oxygen is often involved and nitrogen is
involved sometimes.
Must know mass of the unknown substance before burning
it
Unknown will be burnt in pure oxygen, present in excess
Carbon dioxide and water are the products
All the carbon winds up as carbon dioxide and all the
hydrogen winds up as water
The end result will be to determine the empirical formula of
the substance