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Stoichiometry
Chapter 3
Chemical Stoichiometry
•
•
Stoichiometry: The study of quantities of materials
consumed and produced in chemical reactions.
In macroworld, we can count objects by weighing
assuming that
– Objects behave as though they were all identical
– We know the average mass of the objects
•
•
•
In microworld, atoms are too small to count directly. But
If we know the average mass of atoms, we can count
them.
12C is the standard for atomic mass, with a mass of
exactly 12 atomic mass units (u).
The masses of all other atoms are given relative to this
standard.
2
Atomic Mass
•
•
Atomic mass is the mass of an atom in atomic mass
units (amu or u)
One atomic mass unit is defined as a mass exactly
equal to one-twelfth the mass of one carbon-12 atom
mass of one carbon  12 atom
1 amu 
12
By definition:
1 12C (6 1p + 6 1n) atom “weighs” 12 u
On this scale and on average
1H = 1.008 u
16O = 16.00 u
56Fe = 55.93 u
3
Average Atomic Mass
• Elements occur in nature as mixtures of isotopes
• The atomic mass of an element is the average mass of a
representative sample of atoms of the element (also call
the weighted average atomic mass)
Isotope
12C
Natural
abundance
98.89%
13C
1.11%
14C
<0.01%
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atomic number
C
12.01
atomic mass
The weighted average is the average calculated by
considering the relative amounts!
average atomic mass
of natural carbon
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Calculation of Average Atomic Mass
The weighted average is the average calculated by
considering the relative amounts…
2
2
3
5
Calculate the average atomic mass of chlorine:
Isotope
Mass relative to C-12
Percentage natural abundance
Chlorine-35
34.96885 amu
75.77
Chlorine-37
36.96590 amu
24.23
5
The Mole (mol)
•
•
•
•
A unit to count numbers of particles.
Defined as the amount of a substance that contains as
many elementary entities (atoms, molecules, or other
particles) as there are atoms in exactly 12 g of the
carbon-12 isotope.
The actual number of atoms in 12 g of carbon-12 is
called Avogadro’s Number (NA), in honor of the Italian
chemist Amadeo Avogadro (1776-1855).
The currently accepted value is
– NA = 6.0221367  1023
– NA = 6.022  1023 (4 significant figures)
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The Molar Mass (M)
1 mole of anything = 6.022 × 1023 units of that thing
Molar mass
• Is the mass (g or kg) of 1 mole of units of a substance
• Has units of g/mol
• Is numerically equal to the atomic mass in amu
Thus, for any element,
atomic mass (amu) = molar mass (grams)
1 carbon-12 atom = 12.00 amu
1 mole carbon-12 atoms = 12.00 g = 6.022  1023 atoms
1 mole carbon atoms = 12.01 g of C
1 mole lithium atoms = 6.941 g of Li
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One Mole of
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Q1. Which of the following is closest to the average
mass of one atom of copper?
a.
b.
c.
d.
e.
.
63.55 g
52.00 g
58.93 g
65.38 g
1.055 x 10-22 g
Q2. Which of the following 100.0 g samples contains
the greatest number of atoms?
a.
b.
c.
d.
e.
Magnesium
Silicon
Zinc
Silver
Cesium
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Q3. Rank the following according to number of atoms
(greatest to least):
a.
b.
c.
d.
e.
.
107.9 g of silver
70.0 g of zinc
21.0 g of magnesium
39.0 g of potassium
53.0 g of chromium
Q4. Calculate the number of copper atoms in a 63.55 g
sample of copper.
a.
b.
c.
d.
e.
6.022×1021 Cu atoms
6.022×1022 Cu atoms
6.022×1023 Cu atoms
6.022×1024 Cu atoms
6.022×1025 Cu atoms
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Molecular Mass
• A molecular mass (or molecular weight, MW) is
the sum of the atomic weights of the atoms in a
molecule.
• For the molecule sulfur dioxide, SO2, the
molecular weight would be
S:
O:
11
The Molar Mass
• Is the mass (g or kg) of 1 mole of units of a
substance
• Has units of g/mol
• Is numerically equal to the molecular mass
• Thus, for a molecule,
– molecular mass (u) = molar mass (grams)
Cl has an AW of 35.45 u  1 mol Cl has a mass of 35.45 g
H2O has an MW of 18.02 u  1 mol H2O has a mass of 18.02 g
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Formula Mass
• A formula mass (or formula weight, FW) is the
sum of the atomic weights for the atoms in a
formula unit of an ionic compound.
• So, the formula weight of calcium chloride,
CaCl2, would be
• The molar mass is also numerically equal to the
molecular mass.
– NaCl has an FW of 58.44 u  1 mol NaCl has a
mass of 58.44 g
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EX. 1 Consider separate 100.0 gram samples of each of the
following:
H2O, CO2, C3H6O2, N2O4
Rank them from greatest to least number of oxygen atoms.
You need to calculate the number of oxygen atoms by first
dividing the mass of each sample by its molar mass to obtain
the moles, then multiplying the moles by the number of oxygen
in the formula, and finally multiplying the Avogadro’s number
14
Percent Composition of Compounds
•
•
The percent composition by mass is the percent by
mass of each element in a compound.
You can use the following formula to find the percent by
mass of each element in a compound:
% composition of element 
Example: C2H5OH
Check:
no. of atomes  atomic weight
 100%
formula weight of substance
%C =
 100% = 52.14%
%H =
 100% = 13.13%
%O =
 100% = 34.73%
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EX. 2 Consider separate 100.0 gram samples of each of the
following:
H2O, CO2, C3H6O2, N2O4
Rank them from highest to lowest percent oxygen by mass.
Using the molecular weight obtained in EX. 1, we can calculate
the percent oxygen by mass as follows:
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Analyzing for Carbon and Hydrogen
A schematic diagram of the combustion device used to analyze substances
for carbon and hydrogen. The sample is burned in the presence of excess
oxygen, which converts all its carbon to carbon dioxide and all its hydrogen
to water. These products are collected by absorption using appropriate
materials, and their amounts are determined by measuring the increase in
masses of the absorbents.
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Review: Formulas
• Empirical formula: the simplest whole-number
ratio of atoms in a compound.
• Molecular formula: the exact number of atoms
of each element in the formula of a compound.
• Molecular formula = (empirical formula)n [n =
integer]
• Empirical formula of dinitrogen tetroxide:
– molecular formula: N2O4 =
• Empirical formula of benzene:
– Molecular formula: C6H6 =
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Empirical Formula Determination
1. Convert the percentages to grams
a. assume you start with 100 g of the compound
b. skip if already grams
2. Convert grams to moles
a. use molar mass of each element
3. Divide all by smallest number of moles
a. if result is within 0.1 of whole number, round to whole
number
4. Multiply all mole ratios by number to make all
whole numbers
a. if ratio ?.5, multiply all by 2; if ratio ?.33 or ?.67, multiply
all by 3; if ratio 0.25 or 0.75, multiply all by 4, …
b. skip if already whole numbers
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Molecular Formula Determination
1. Obtain the empirical formula (see previous slide)
2. Compute the empirical formula mass or molar
mass
3. Calculate the ratio:
Molarmass
Empiricalformulamass
4. The integer from Step 3 represents the no. of
empirical formula units in one molecule. When
the empirical formula subscripts are multiplied
by this integer, the molecular formula results.
molarmass
Molecularformula empiricalformula
empiricalformulamass
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Example 3.10 Determine the empirical and molecular formulas
for a compound that gives the following percentages on analysis
(in mass percent): 71.65% Cl
24.27%C
4.07% H
The molar mass is known to be 98.96 g/mol.
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Example 3.10 Determine the empirical and molecular formulas
for a compound that gives the following percentages on analysis
(in mass percent): 71.65% Cl
24.27%C
4.07% H
The molar mass is known to be 98.96 g/mol.
22
Chemical Rxn and Chemical Eqn
•
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
•
When the methane (CH4) in natural gas combines with
oxygen (O2) in the air and burns, carbon dioxide (CO2)
and water (H2O) are formed. This process is
represented by an unbalanced chemical reaction:
CH4 + O2

reactants 
•
CO2 + H2O
products
Reactants are only placed on the left side of the arrow,
products are only placed on the right side of the arrow.
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How to “Read” Chemical Equations?
•
•
•
An unbalanced chemical equation gives qualitative
information:
CH4 + O2
CO2 + H2O
A balanced chemical equation gives both qualitative
and quantitative information.
CH4 + 2O2
CO2 + 2H2O
Use the coefficients in the balanced equation to decide
the amount of each reactant that is used, and the
amount of each product that is formed.
1 CH4 molecule + 2 O2 molecules makes 1 CO2 molecule & 2 H2O molecules
1 mole CH4 + 2 moles O2 makes 1 mole CO2 and 2 moles H2O
16 g CH4 + 64 g O2 makes 44 g CO2 and 36 g H2O
But NOT 1 g CH4 + 2 g O2 makes 1 g CO2 and 2 g H2O
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Writing and Balancing the Equation for
a Chemical Reaction
1. Determine what reaction is occurring. What are
the reactants, the products, and the physical
states involved?
2. Write the unbalanced equation that
summarizes the reaction described in Step 1.
3. Balance the equation by inspection, starting
with the most complicated molecule(s). The
same number of each type of atom needs to
appear on both reactant and product sides. Do
NOT change the formulas of any of the
reactants or products.
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Notice for Balancing Chem Eqns
•
•
•
•
The number of atoms of each type of element
must be the same on both sides of a balanced
equation.
Subscripts must not be changed to balance an
equation.
A balanced equation tells us the ratio of the
number of molecules which react and are
produced in a chemical reaction.
Coefficients can be fractions, although they are
usually given as lowest integer multiples.
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Balancing Chemical Equations
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Balancing Chemical Equations
1. Write the correct formula(s) for the reactants on
the left side and the correct formula(s) for the
product(s) on the right side of the equation.
Ethane reacts with oxygen to form carbon dioxide & water:
C2H6 + O2
CO2 + H2O (unbalanced)
2. Change the numbers in front of the formulas
(stoichiometric coefficients) to make the number
of atoms of each element the same on both sides
of the equation. Do not change the subscripts.
2C2H4
NOT
C4H8
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Balancing Chem Eqns (cont’d)
3. Start by balancing those elements that occur in
the fewest chemical formulas.
C2H6 + O2
carbon
on left
C2H6 + O2
CO2 + H2O
carbon
on right
CO2 + H2O
hydrogen
on left
C2H6 + O2
start with C or H but not O
hydrogen
on right
CO2 +
H2O
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Balancing Chem Eqns (cont’d)
4. Balance those elements that appear in two or
more reactants or products.
C2H6 + O2
CO2 +
H2O
oxygen
on left
C2H6 +
O2
CO2 +
H2O
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Balancing Chem Eqns (cont’d)
5. Check to make sure that you have the same
number of each type of atom on both sides of the
equation.
C2H6 +
O2
CO2 +
H2O
Reactants
Products
4C
12 H
14 O
4C
12 H
14 O
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Stoichiometric Calculations
• Chemical equations can be used to relate the masses of
reacting chemicals.
• Mole method: treat the stoichiometric coefficients in a
balanced chemical equation as the number of moles of
each substance.
C2H5OH + 3O2
2CO2 + 3H2O
1 molecule
3 molecules
1 mol
3 mol
2 molecules
3 molecules
2 mol
3 mol
1 mole of ethanol reacts with 3 moles of oxygen to form 2 moles
of carbon dioxide and 3 moles of water.
1 mol C2H5OH =
3 mol O2 =
3 mol O2
3 mol H2O
1 mol C2H5OH
1 mol C2H5OH
=
=
2 mol CO2
3 mol H2O
The equal sign (=) here means “stoichiometrically equivalent
to” or simply “equivalent to”
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Calculating Masses of Rxts & Prdts in Rxns
1. Balance the equation for the reaction.
2. Convert the known mass of the reactant or
product to moles of that substance.
3. Use the balanced equation to set up the
appropriate mole ratios.
4. Use the appropriate mole ratios to calculate the
number of moles of the desired reactant or
product.
5. Convert from moles back to grams if required
by the problem.
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Calculating Masses of Rxts & Prdts in Rxns
34
EX. 3 Methanol burns in air according to the equation
2CH3OH + 3O2
 2CO2 + 4H2O
If 209 g of methanol are used up in the combustion,
(a) what mass of water is produced?
grams CH3OH  moles CH3OH  moles H2O  grams H2O
= 235 g H2O
(b) what mass of oxygen is required?
= 314 g O2
35
Stoichiometric Mixture
•
•
Consider the reaction: N2(g) + 3H2(g)  2NH3(g)
If we start with a stoichiometric mixture, i.e. the molar
ratio of N2:H2 is 1:3, all reactants will be consumed to
form products.
– 5 N2 molecules + 15 H2 molecules  10 NH3
molecules with no reactants left over.
36
Non-Stoichiometric Mixture
•
•
Consider the reaction: N2(g) + 3H2(g)  2NH3(g)
If we start with 5 N2 molecules and 9 H2 molecules, after
the reaction is complete, not all reactants will be
consumed to form products.
– 5 N2 molecules + 9 H2 molecules  6 NH3 molecules
with 2 N2 molecules left over.
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Limiting Reactants
•
•
Consider the reaction: N2(g) + 3H2(g)  2NH3(g)
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 (H2).
•
In the example (start with 5 N2 and 9 H2), the N2 would
be the excess reactant.
38
Limiting Reactants
39
Q5. Which of the following reaction mixtures could
produce the greatest amount of product? Each
involves the reaction symbolized by the equation:
2H2 + O2 2H2O
a.
b.
c.
d.
e.
2 moles of H2 and 2 moles of O2
2 moles of H2 and 3 moles of O2
2 moles of H2 and 1 mole of O2
3 moles of H2 and 1 mole of O2
Each produce the same amount of product.
40
Reaction Yield
• Reaction yield is an important indicator of the
efficiency of a particular laboratory or industrial
reaction.
• Theoretical Yield is the amount of product that
would result if all the limiting reactant reacted.
• Actual Yield is the amount of product actually
obtained from a reaction.
• Percent Yield is the proportion of the actual
yield to the theoretical yield.
Actual Yield
Percent Yield 
 100%
Theoretical Yield
41
EX. 4 Consider the reaction, 2NH3 + 5F2 → N2F4 + 6HF. If
25.0 g of NH3 are reacted with 150. g of F2, (a) What is the
limiting reactant? (b) Calculate the theoretical yield of N2F4 in
grams. (c) Calculate the percent yield if 56.8 g of N2F4 are
actually obtained. (d) Calculate the actual yield of N2F4 in grams
if the percent yield is 90%.
(a) What is the limiting reactant?
Start with 25.0 g NH3, calculate g of F2 needed.
= 140. g F2
42
EX. 4 Consider the reaction, 2NH3 + 5F2 → N2F4 + 6HF. If
25.0 g of NH3 are reacted with 150. g of F2, (a) What is the
limiting reactant? (b) Calculate the theoretical yield of N2F4 in
grams. (c) Calculate the percent yield if 56.8 g of N2F4 are
actually obtained. (d) Calculate the actual yield of N2F4 in grams
if the percent yield is 90%.
(b) Calculate the theoretical yield of N2F4 in grams.
Start with 25.0 g NH3, calculate g of N2F4 produced.
= 76.5 g N2 F4
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EX. 4 Consider the reaction, 2NH3 + 5F2 → N2F4 + 6HF. If
25.0 g of NH3 are reacted with 150. g of F2, (a) What is the
limiting reactant? (b) Calculate the theoretical yield of N2F4 in
grams. (c) Calculate the percent yield if 56.8 g of N2F4 are
actually obtained. (d) Calculate the actual yield of N2F4 in grams
if the percent yield is 90%.
(c) Calculate the percent yield if 56.8 g of N2F4 are actually
obtained.
= 74.2%
(d) Calculate the actual yield of N2F4 in grams if the percent
yield is 90%.
= 68.9 g
44