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3 Stoichiometry
Contents
3-1 Law of Conservation of Matter
3-2 Balancing Equation
3-3 Equations on a Macroscopic Scale
3-4 Mass Relationship in Chemical Reactions
3-5 Limiting Reactants
3-6 Theoretical Yield, Actual Yield, and Percent Yield
3.7 Quantitative analysis
3.8 Empirical Formulas from Percent Composition
3.9 Molecular and Structural Formulas
3.10 Percent Composition from Formulas
Stoichiometry is the study of quantitative relationships
between substances involved in chemical changes.
It is a very important subject from the point of view of
both theory and practice.
Analytical chemistry is based on Stoichiometry.
3-1 Law of Conservation of Matter
The Quantity of matter is not changed by chemical reactions;
Matter is neither created nor destroyed by chemical reactions.
The number of atoms of each kind must be the same after
reaction as it was before reaction.
In chemical reactions, the quantity of matter does not
change. The total mass of the products equals the total mass
of the reactants.
Law of Conservation of Matter provided the foundation for
modern chemistry.
H2 + N2
3H2 + N2
NH3
=
2NH3
3-2 Balancing Equation
A chemical reaction is a process in which one set of
substances called reactants is converted to a new set of
substances called products. In other words, a chemical reaction
is the process by which a chemical change occurs.
We need evidence before we can say that a reaction has
occurred. Some of the types of physical evidence to look for
are shown here.
• a color change
• formation of a solid (precipitate) with a clear solution
• evolution of a gas
• evolution or absorption of heat
When none of these signs of a chemical reaction appears,
we need chemical evidence. This requires a detailed chemical
analysis of the reaction mixture to discover whether any new
substances are present.
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 and water
C2H6 + O2
CO2 + H2O
2. Change the numbers in front of the formulas
(coefficients) to make the number of atoms of each
element the same on both sides of the equation. Do
not change the subscripts.
2C2H6
NOT
C4H12
3.7
Balancing Chemical Equations
3. Begin with the compound that has the most atoms
or the most kinds of atoms and use one of these
atoms as a starting point.
C2H6 + O2
CO2 + H2O
start with C2H6
start with C or H but not O
3.7
Balancing Chemical Equations
4. Balance those elements that appear in only once
on each side of the arrow first.
C2H6 + O2
2 carbon
on left
C2H6 + O2
6 hydrogen
on left
C2H6 + O2
CO2 + H2O
1 carbon
on right
start with C or H but not O
multiply CO2 by 2
2CO2 + H2O
2 hydrogen
on right
2CO2 + 3H2O
multiply H2O by 3
3.7
Balancing Chemical Equations
5. Balance those elements that appear more than
once on a side. Balance free elements last.
C2H6 + O2
2 oxygen
on left
2CO2 + 3H2O
multiply O2 by 7
2
4 oxygen + 3 oxygen = 7 oxygen
(3x1)
on right
(2x2)
C2H6 + 7 O2
2
2CO2 + 3H2O
2C2H6 + 7O2
4CO2 + 6H2O
remove fraction
multiply both sides by 2
The smallest whole-number coefficients are preferred!
3.7
Balancing Chemical Equations
6. Check to make sure that you have the same
number of each type of atom on both sides
of the equation.
2C2H6 + 7O2
4CO2 + 6H2O
4 C (2 x 2)
4C
12 H (2 x 6)
12 H (6 x 2)
14 O (7 x 2)
14 O (4 x 2 + 6)
3.7
Balancing Chemical Equations
6. Check to make sure that you have the same
number of each type of atom on both sides
of the equation.
2C2H6 + 7O2
4CO2 + 6H2O
Reactants
4C
12 H
14 O
Products
4C
12 H
14 O
3.7
The system to balance an equation
•
Begin with the compound that has the most atoms or
the most kinds of atoms and use one of these atoms as
a starting point.
•
Balance elements that appear only once on each side
of the arrow first.
•
Then balance elements that appear more than once
on a side.
•
Balance free elements last.
Here are some useful strategies for balancing equations
If an element occurs in only one compound on each side of
the equation, try balancing this element first.
When one of the reactants or products exists as the free
element, balance this element last.
In some reactions, certain groups of atoms(for example,
polyatomic ions) remain unchanged. In such cases, balance
these groups as a unit.
It is permissible to use fractional as well as integral numbers
as coefficients. At times, an equation can be balanced most easily
by using one or more fractional coefficients and then, if desired,
clearing the fractions by multiplying all coefficients by a common
multiplier.
Example
Writing and balancing an equation: The combustion of a carbonHydrogen-Oxygen Compound. Liquid triethylene glycol, C6H14O4,
is used as a solvent and plasticizer for vinyl and polyurethane
plastics. Write a balanced chemical equation for its complete
combustion.
Solution
Carbon- hydrogen-oxygen compounds, like hydrocarbons,
yield carbon dioxide and water when burned in oxygen gas.
Starting expression : C6H14O4 + O2
CO2 + H2O
Balance C : C6H14O4 + O2
6CO2 + H2O
Balance H : C6H14O4 + O2
6CO2 + 7H2O
At this point, the right side of the expression has 19 O atoms(12 in
six CO2 molecules and 7 in seven H2O molecules). To get 19
atoms on the left, we start with 4 in a molecule of C6H14O4 and
need 15 more. This requires a fractional coefficient of 15/2 for O2.
Balance O : C6H14O4 + (15/2) O2
6CO2 + 7H2O (balanced)
To remove the fractional coefficient, multiply all coefficients by 2,
the denominator of the fractional coefficient, 15/2.
2 C6H14O4 + 15 O2
12CO2 +14H2O (balanced)
Check:
Left : (2×6) =12 C;
(2 ×14) = 28 H; [(2×4) + (15×2)] = 38 O
Right : (12×1) =12 C;
(14 ×2) = 28 H;
[(12×2) + (14×1)] = 38 O
3-3 Equations on a Macroscopic Scale
The term stoichiometry means, literally, to measure the elements,
but from a more practical standpoint, it includes all the quantitative
relationships involving atomic and formula masses, chemical
formulas, and the chemical equation.
The coefficients in the chemical equation
2H2(g) + O2(g)
2H2O(l)
Mean that
2x molecules H2 + 1x molecules O2
Suppose that we let x=6.02214×1023 .
Then the chemical equation also means that
2 mol H2 + 1 mol O2
2 mol H2O
2x molecules H2O
The coefficients in the chemical equation allow us to make
statements, such as:
Two moles of H2O are produced for every two moles of H2
consumed.
Two moles of H2O are produced for every one mole of O2
consumed.
Two moles of H2 are consumed for every one mole of O2
consumed.
A stoichiometric factor relates the amounts a mole basis, of any
two substances involved in a chemical reaction; thus a
stoichiometric factor is a mole ratio.
The mole is the key to quantitative relationships between
substances involved in chemical changes on a practical scale.
3-4 Mass Relationship in Chemical Reaction
Example
Relating the Mass of a reactant and a product. What mass of
H2O is formed in the reaction of 4.16g H2 with an excess of O2?
Solution
The general strategy for reaction stoichiometry problems, outlined
earlier and illustrated in below, suggests these three steps.
1. Convert the quantity of H2 from grams to moles. (use the
inverse of molar mass of H2)
2. From the number of moles of H2, calculate the number of
moles of H2O formed.
3. Convert the quantity of H2O from moles to grams. (use the
inverse of molar mass of H2O)
Grams of H2 GIVEN
Moles of H2
Moles of H2O
= (2/2) ×mol H2
Use inverse of molar mass
as conversion factor:
1 mol H2/2.016 g H2
Use coefficients in balanced
chemical Equation to find mole
ratio: 2 mol H2O/2 mol H2
Use molar mass as conversion
factor: 18.02 g H2O/1 mol H2
Grams of H2O FOUND
( g H2
1
2
mol H2
3
mol H2O
g H 2O )
18.02 g H2O
1 mol H2
2 mol H2O
? g H2O =4.16 g H2 ×
×
×
2.016 g H2
2 mol H2
1 mol H2O
= 37.2 g H2O
Always begin solving problem about quantitative relationships
in reaction by writing an equation for the reaction. The
coefficients in the equation describe the relations between moles
of products and moles of reactions. For example:
2H2O
Moles
=
2H2
+
O2
2
2
1
Formula Mass
18.0
2.02
32.0
Mass
36.0
4.03
32.0
3-5 Limiting Reactants
When all the reactants are completely and simultaneously
consumed in a chemical reaction, the reactants are said to be in
stoichiometric proportions--in the mole ratios dictated by the
coefficients in the balanced equation. At other times, the reactants
are not usually present in the proportions shown by the equation.
The quantity of one reactant controls the amount of products that
can be formed. The reactant that is completely consumed: the
limiting reactant--determines the quantities of products formed.
Example
Determining the Limiting Reactant in a Reaction. Phosphorus
trichloride, PCI3, is a commercially important compound used in the
manufacture of pesticides, gasoline additives, and a number of other
products. It is made by the direct combination of phosphorus and
chlorine.
P4(s) + 6 Cl2(g)
4 PCl3(l)
What mass of PCl3(l) forms in the reaction of 125 g P4 with 323 g Cl2?
Solution
1. The first step in a stoichiometric calculation is to write
A balanced equation for the reaction. If the equation
Is not given, you must supply your own.
2. Note what’s given and asked for the above
reaction.
3. Write the formula masses needed below the
reaction.
125 g
Phosphorus trichloride
formula masses, u
323 g
P4(s) + 6 Cl2(g)
4 PCl3(l)
123.9
137.3
70.91
4. Determine which reactant in limiting. Convert grams to moles.
? Mol Cl2 = 323g Cl2 ×
? Mol P4 = 125g P4 ×
1mol Cl2
70.91 g Cl2
1 mol P4
123.9 g P4
?g
= 4.56 mol Cl2
= 1.01 mol P4
We see rather clearly that there is less than 6 mol
Cl2 per mole of P4- Chlorine is the limiting reactant.
The remainder of the calculation is to determine the
mass of PCl3 formed in the reaction of 323 g Cl2 with
an excess of P4.
4. Calculation and check.
? g PCl3 = 323 gCl2 ×
= 417 g PCl3
1 mol Cl2
70.91 g Cl2
×
4 mol PCl3
6 mol Cl2
×
137.3 g PCl3
1 mol PCl3
Grams of P4
Grams of Cl2
Use inverse of molar
Mass as conversion factor
: 1 mol P4/123.9g P4
Use inverse of molar
Mass as conversion factor
: 1 mol Cl2/70.91g Cl2
Moles of P4
Moles of Cl2
Calculate mole ratio of Cl2 to P4
Mole ratio = moles of Cl2/moles of P4
If calculated mole tatio
<6/1 chlorine is limiting
If calculated mole tatio
>6/1 phosphorus is limiting
Example
Determining the quantity of excess reactant(s) remaining
after a reaction. What mass of P4 remains in excess
following the reaction in example above?
Solution
The key to this problem is to calculate the mass of P4 that is
consumed, and we can base this calculation either on the
mass of Cl2 consumed：
1 mol Cl2
1 mol P4
×
?g P4 = 323 g Cl2×
70.91 g Cl2
6 mol Cl2
= 94.1 g P4
×
123.9 g P4
1 mol P4
Or on the mass of PCl3 produced.
?g P4 = 417 g PCl3×
1 mol PCl3
1 mol P4
123.9 g P4
×
×
137.3gPCl3 4 mol PCl3
1 mol P4
= 94.1 g P4
The mass of P4 remaining after the reaction is simply the
difference between what was originally present and what
was consumed; that is,
125 gP4 initially – 94.1 g P4 consumed = 31 gP4 remaining
3-6 Theoretical Yield, Actual Yield, and Percent Yield
Theoretical Yield, Actual Yield, and Percent Yield
The theoretical yield of a reaction is the amount of
product calculated to be formed by the reaction. The
amount of product that is actually produced is called
the actual yield. The percent yield is defined as:
percent yield = actual yield / theoretical yield
× 100%
In few reactions the actual yield almost exactly equals the theoretical
yield, and the reactions are said to be quantitative. On the other hand,
in some reactions the actual yield is less than the theoretical yield,
and the percent yield is less than 100%. The yield may be less than
100% for many reasons:
(1) The product of a reaction rarely appears in a pure form, and in the
necessary purification steps, some product may be lost through handling.
This reduces the yield.
(2) In many cases the reactants may participate in reactions other than
the one of central interest. These are called side reactions, and the
unintended products are called by-products. To the extent that side
reactions occur, the yield of the main product is reduced.
(3) Finally, if a reverse reaction occurs, some of the expected product
may react to re-form the reactants, and again the yield is less than
expected.
Example
Determining Theoretical, Actual, and Percent Yields. Billions
of pounds of urea, CO(NH2)2, are produced annually for use as
a fertilizer. The reaction used is
2 NH3 + CO2
CO(NH2)2 + H2O
The typical starting reaction mixture has a 3: 1 mole ratio of NH3
to CO2. If 47.7 g urea forms per mole of CO2 that reacts, what is
the (a) theoretical yield; (b) actual yield; and (c) percent yield in
this reaction?
Solution
(a) The stoichiometric proportions are 2 mol NH3: 1 mol CO2.
Because the mole ratio of NH3 to CO2 used is 3: 1, NH3 is in
excess and CO2 is the limiting reactant. Because the quantity of
urea is given per mole of CO2, we should base the calculation
on 1.00 mol CO2.
Theoretical yield = 1.00mol CO2×
1 mol CO(NH2)2
1 mol CO2
= 60.1g CO(NH2)2
(b) Actual yield = 47.7 g CO(NH2)2
47.7 g CO(NH2)2
(c) %yield =
60.1 g CO(NH2)2
× 100% = 79.4%
×
60.1 g CO(NH2)2
1 mol CO(NH2)2
3.7
Quantitative analysis
Quantitative analysis is finding out how much of a given
substance is present in a sample.
Qualitative analysis is finding out what substances are
present in a sample.
mass A in sample
percent by mass of A in sample 
100
total mass of sample
3.8
Empirical Formulas from Percent Composition
The empirical formulas for a compound is the simplest
formula that shows the ratios of the numbers of atoms of each
kind in the compound.
•Empirical formulas
–give the relative numbers and types of atoms in a
molecule.
–That is, they give the lowest whole number ratio
of atoms in a molecule.
–Examples: H2O, CO2, CO, CH4, HO, CH2.
Example
The composition of a compound is 36.4% Mn, 21.2% S, and
42.4% O. All percents are mass percent. What is the empirical
formulas of the compound?
Solution: Suppose there is 100 g sample, then:
Mn: 36.4/54.9=0.663mol; S: 21.1/32.1=0.660mol;
O: 42.4/16.0=2.65mol
0.663:0.660:2.65≈1: 1: 4
MnSO4
3.9 Molecular and Structural Formulas
• Molecular formulas
– give the actual numbers and types of atoms in a
molecule.
– Examples: H2O, CO2, CO, CH4, H2O2, O2, O3, and
C2H4.
• Most molecular substances that we will study in this
class contain only nonmetals.
Space-filling models
• Molecular and empirical formulas do not show how
atoms are arranged when bonded together.
Picturing Molecules
• Molecules occupy three dimensional space.
• However, we often represent them in two dimensions.
• The structural formula gives the connectivity between
individual atoms in the molecule.
• The structural formula may or may not be used to
show the three dimensional shape of the molecule.
• If the structural formula does show the shape of the
molecule, then either a perspective drawing, ball-andstick model, or space-filling model is used.
Representing Structure in Molecules
Accurately represents
the angles at which
molecules are attached.
Class Practice Exercise
The structural formula of propane and butane is
H
H
H
H
C
C
C
H
H
H
H
H
H
H
H
H
C
C
C
C
H
H
H
H
H
What is the chemical and empirical formula for these
molecules?
3.10
Percent Composition from Formulas
Example
What is the percent by mass of iron (III) to one decimal place
of FeCl3?
Solution:
55.8
percent by mass of iron in FeCl 
100  34.4%
55.8  35.5  3
3