Stoichiometry

1
Stoichiometry
In solving a problem of this sort, the grand
thing is to be able to reason backward. This is a
very useful accomplishment, and a very easy one,
but people do not practice it much.
Sherlock Holmes, in Sir Arthur Conan Doyles A
Study in Scarlet
2
The Mole
1 dozen
12
1 gross
144
1 ream
500
1 mole
6.022 x 1023
There are exactly 12 grams of carbon-12 in one
mole of carbon-12.
3
Avogadros Number
6.022 x 1023 is called Avogadros Number in
honor of the Italian chemist Amadeo Avogadro
(1776-1855).
I didnt discover it. Its just named after me!
Amadeo Avogadro
4
Calculations with MolesConverting moles to grams
How many grams of lithium are in 3.50 moles of
lithium?
3.50 mol Li
6.94 g Li
g Li
45.1
1 mol Li
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Calculations with MolesConverting grams to moles
How many moles of lithium are in 18.2 grams of
lithium?
18.2 g Li
1 mol Li
mol Li
2.62
6.94 g Li
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Calculations with MolesUsing Avogadros Number
How many atoms of lithium are in 3.50 moles of
lithium?
3.50 mol Li
6.022 x 1023 atoms Li
atoms Li
2.11 x 1024
1 mol Li
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Calculations with MolesUsing Avogadros Number
How many atoms of lithium are in 18.2 g of
lithium?
18.2 g Li
1 mol Li
6.022 x 1023 atoms Li
6.94 g Li
1 mol Li
atoms Li
1.58 x 1024
(18.2)(6.022 x 1023)/6.94
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Calculating Formula Mass
Calculate the formula mass of magnesium
carbonate, MgCO3.
24.31 g 12.01 g 3(16.00 g)
84.32 g
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Calculating Percentage Composition
Calculate the percentage composition of magnesium
carbonate, MgCO3.
From previous slide 24.31 g 12.01 g
3(16.00 g) 84.32 g
100.00
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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

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Formulas (continued)
Formulas for ionic compounds are ALWAYS empirical
(lowest whole number ratio).
Examples
NaCl
MgCl2
Al2(SO4)3
K2CO3
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Formulas (continued)
Formulas for molecular compounds MIGHT be
empirical (lowest whole number ratio).
Molecular
C6H12O6
C12H22O11
H2O
Empirical
H2O
CH2O
C12H22O11
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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.

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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?
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Empirical Formula Determination(part 2)
Divide each value of moles by the smallest of the
values.
Carbon
Hydrogen
Oxygen
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Empirical Formula Determination(part 3)
Multiply each number by an integer to obtain all
whole numbers.
Carbon 1.50
Hydrogen 2.50
Oxygen 1.00
x 2
x 2
x 2
3
5
2
C3H5O2
Empirical formula
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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
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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
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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
(C3H5O2) x 2
C6H10O4
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Review 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
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Solving a Stoichiometry Problem

1. Balance the equation.
2. Convert masses to moles.
3. Determine which reactant is limiting.
4. Use moles of limiting reactant and mole ratios to
   find moles of desired product.
5. Convert from moles to grams.

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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.
Al

O2
4
3
Al2O3
2
a. Every reaction needs a yield sign!
b. What are the reactants?
c. What are the products?
d. What are the balanced coefficients?
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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 3 O2 ? 2Al2O3
6.50 g Al
1 mol Al
2 mol Al2O3
101.96 g Al2O3

? g Al2O3
4 mol Al
1 mol Al2O3
26.98 g Al
6.50 x 2 x 101.96 26.98 4
12.3 g Al2O3
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Limiting Reactant

• The limiting reactant is the reactant
• that is consumed first, limiting the amounts of
  products formed.