Chapter 3
Stoichiometry
1
Preview
the contents of this chapter will introduce
you to the following topics:
 Atomic mass, Mole concept, and Molar mass
(average atomic mass).
 Number of atoms per amount of element.
 Percent composition and Empirical formula
of molecules.
 Chemical equations, Balancing this equate
and Stoichiometric calculations including
limiting reactants.
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• in 1961, 12C-isotope has been instituted as the standard
for determining the atomic masses
Heavy
2. The isotopic composition of a natural element
Light
Mass spectrometer
will be used to determine:
3.1 Atomic
Masses
1. Accurate mass value for individual atoms.
"12C is assigned a mass of exactly 12 atomic mass unit (amu)"
KE = 1/2 x m x v2
v = (2 x KE/m)1/2
F=qxvxB
Heavy
Light
• The most accurate method currently available for
comparing the masses of atoms involves the use of
Mass Spectrometer (figure 3.1)
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3.1 Atomic Masses
 All the mass in the periodic table are related to the 12C-mass.
Sometimes, they are called the atomic weight for each element.
 For carbon element the mass in the table is "12.01" and not exact
12.00??
 Natural carbon found on earth is a mixture of the isotopes C-12, C13 and C-14 isotopes, and the mass in table is an average value
reflecting the average of the isotopes composing it. The average
masses are calculating using the following equation:
A.A.M = ∑ (atomic mass X fractions of its abundance)
Fraction = z%
⁄ 100
 Average atomic mass is traditionally has been called atomic weight
of element.
 the average atomic mass will enable us to count atoms by weighing
a sample of any element.
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3.1 Atomic Masses
Example 3.1:
When a sample of natural copper is vaporized and injected into
a mass spectrometer we got two isotopes: 63.29Cu (69.09%) and
65.29Cu (30.91%). The mass value for 63Cu & 65Cu are 62.93 and
64.93 amu, respectively.
Use these data to compute the average mass of natural copper.
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3.2 The Mole
The mole is “the number equal to number of carbon
atoms in exactly 12 grams of pure 12C”
Avogadro has determined this number to be
NA = 6.02214 X 1023 Particle/mole
1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g
1 12C atom = 12.00 amu
Molar mass is the mass of 1 mole of
eggs
shoes in grams
marbles
atoms
For any element
atomic mass (amu) = molar mass (grams)
6.022 X 1023 amu = 1 g exact
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3.2 The Mole
Example 3.3:
Aluminum (Al) is high resistance to corrosion. Compute both
the number of moles of atoms and the number of atoms in a
10.0-g sample of aluminum.
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3.3 Molar Mass of Compounds
Compounds are a collection of atoms, and Molar Mass is the
mass in grams of 1 Mole of the compound (traditionally is called
Molecular weight)
Molar mass for molecule or compound = ∑atomic masses in g
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
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3.3 Molar Mass of Compounds
Example 3.7:
Calcium carbonate (Ca CO3), also called calcite, is the principle
found in limestone, marble, pearls…
1. Calculate the molar mass of calcium carbonate.
2. A certain sample of calcium carbonate contains 4.86 moles.
What is the mass in grams of this sample?
3. What is the mass of the CO32- ion present?
Example 3.8:
Isopentylacetate (C7H14O2), the compound responsible for the
scent of bananas interestingly, bees release about 1 mg of this
compound when they sting. the resulting scent attracts other
bees to join the attack.
1. How many molecules of isopentyl acetate are releases in a
typical bee sting?
2. How many atoms of carbon are present?
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3.4 percent Composition of Compounds
The percent composition of elements in any compound
con be determined using the comparing ratio equation:
n x molar mass of element
x 100%
molar mass of compound
n is the number of moles of the element in 1 mole of the compound
%C =
%H =
C2H6O
%O =
2 x (12.01 g)
46.07 g
6 x (1.008 g)
46.07 g
1 x (16.00 g)
46.07 g
x 100% = 52.14%
x 100% = 13.13%
x 100% = 34.73%
52.14% + 13.13% + 34.73% = 100.0%
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3.4 percent Composition of Compounds
Example 3.10:
Penicillin, the first of a now large number of antibiotic was
discovered accidentally by the Scottish bacteriologist Alexander
Aeming in 1928, but he was never able to isolate it as a pure
component. This and similar antibiotics have saved millions of
lives that might have been lost due to infections. Penicillin-F
has the formula C14 H20 N2 SO4. Compute the mass percent of
each element.
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3.5 Determining the Formula of a Compound
For any new compound the 1st item of interest is the
determination of its chemical formula. This performed by
decomposing know sample mass into its component or
reacting it with Oxygen to produce its oxides e.g.
O2
Organic Compound
CO2, H2O, N3
complete reaction
figure 3.5: is a schematic diagram for combustion device:
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3.5 Determining the Formula of a Compound
Masses of CO2, H2O and other gases will be used to determine:
 Empirical Formula and then Molecular Formula if the Molar Mass
is known
Exercise: (Determining Empirical Formula)
Combust 11.5 g ethanol
Produce 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 H2 O
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
Empirical formula is the simplest chemical formula for any compound
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3.5 Determining the Formula of a Compound
molecular formula = (empirical formula)n
[n = integer]
molecular formula = C6H6 = (CH)6
empirical formula = CH
If the Molar mass for any compound is known one can get the
chemical or molecular formula by the following:
a. calculate empirical mass = ∑ atomic mass
b. find the multiplication factor = MM/EM
c. molecular formula = multiplicity x EF
Note: if MM = EM
molecular formula = empirical formula
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3.5 Determining the Formula of a Compound
Example 3.13:
Caffeine a stimulant found in coffee, tea, and chocolate,
contain 49.48 % C, 5.159% H, 28.87 % N, and 16.49 % O by
mass has a molar mass 194.2 g/mol. Determine the
molecular formula of caffeine.
Solution:
1.
2.
3.
4.
Assume 100g of caffeine
mC = 49.48 g
mc
mH = 5.15 g
mH
mN = 28.87 g
mN
mO = 16.49 g
mO
= 49.48/12
=5.15/1
= 28.87/14
= 16.49/16
=4
=5
=2
=1
empirical formula ---(C4 H5 N2 O)n
E.M.
= (12x4) + (5x1) + (14x2) + (16x1)
= 48 + 5 + 28 + 16
= 97
E.F x n = 194.2 ----- n = 194.2/ 97 ~ 2
Molecular Formula ----> C8 H10 N4 O2
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3.6 Chemical Equations:
A process in which one or more substances is changed
into one or more new substances is a chemical reaction
Note:
In chemical reaction atoms are neither created nor destroyed, i.e., all
atoms present in reactants must be accounted for among the
products. this is the called balancing process.
For example:
The combustion (oxidation) of methane (CH4) produces
(yields) carbon dioxide (CO2) and dihydrogen oxide (H2O)
[common name, water]
CH4 + O2  CO2 + H2O
Reactants
Products
(this equation is not balanced)
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3.6 Chemical Equations:
How to “Read” Chemical Equations
Chemical Equation for a reaction gives two important types of
information:
– The nature of reactants and products
(including the physical state: "s, l, g, aq“)
– the relative number of each of the components
(including coefficient moles, and number of species)
A representation of a chemical reaction
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
IS NOT: 2 grams Mg + 1 gram O2 makes 2 g MgO
48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO
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3.7 Balancing Chemical Equation:
Most chemical equations can be balanced by inspection, that is, by
trial and error, but you have to follow some procedure:
 Write the correct formula(s) for the reactants and the
product(s)
e.g. Ethane reacts with oxygen to form carbon dioxide and water

C2H6 + O2
CO2 + H2O
C2H6 + O2
CO2 + H2O start with C or H but not O
Start by balancing those elements that appear in only one
reactant and one product or the most complicated one.
2 carbon
on left
C 2 H 6 + O2
6 hydrogen
on left
1 carbon
on right
multiply CO2 by 2
2CO2 + H2O
2 hydrogen
on right
multiply H2O by 3
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3.7 Balancing Chemical Equation:
C2H6 + O2
2CO2 + 3H2O
 Balance those elements that appear in two or more reactants or products.
C2H6 + O2
2 oxygen
on left
2CO2 + 3H2O
4 oxygen+ 3 oxygen = 7 oxygen
(2x2)
(3x1)
on right
C2H6 + 7 O2
2CO2 + 3H2O
2C2H6 + 7O2
4CO2 + 6H2O
2
remove fraction
multiply both sides by 2
 Check to make sure that you have the same number of each type of atom
on both sides of the equation.
Reactants
Products
4C
4C
12 H
12 H
14 O
14 O
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3.8 Stoichiometric Calcalations:
Amount of reactants and Products
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
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3.8 Stoichiometric Calcalations:
Amount of reactants and Products
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
moles H2O
coefficients
chemical equation
grams H2O
molar mass
H2O
4 mol H2O
18.0 g H2O
1 mol CH3OH
=
x
x
209 g CH3OH x
32.0 g CH3OH
2 mol CH3OH
1 mol H2O
235 g H2O
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3.9 Calculations Involving a Limiting Reactant
When chemical reaction undergo, the reactants are
often Mixed in Stoichiometric quantities, but if not what
happen:
The limiting reactant is the reactant that is consumed
first, limiting the amounts of products formed.
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3.9 Calculations Involving a Limiting Reactant
1.
2.
3.
4.
5.
Balance the equation.
Convert masses to moles.
Determine which reactant is limiting.
Use moles of limiting reactant and mole ratios to
find moles of desired product.
Convert from moles to grams.
Theoretical Yield is the amount of product that would
result if all the limiting reagent reacted.
Actual Yield is the amount of product actually
obtained from a reaction.
% Yield =
Actual Yield
x 100
Theoretical Yield
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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
mol Fe2O3 needed
g Fe2O3 needed
OR
g Fe2O3
124 g Al
mol Fe2O3
x
1 mol Al
27.0 g Al
x
mol Al needed
1 mol Fe2O3
2 mol Al
Start with 124 g Al
x
160. g Fe2O3
1 mol Fe2O3
g Al needed
=
367 g Fe2O3
need 367 g Fe2O3
Have more Fe2O3 (601 g) so Al is limiting reagent
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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
x
102. g Al2O3
=
1 mol Al2O3
234 g Al2O3
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