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.
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.
3.7
2C2H6 NOT C4H12
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
3.7
start with C2H6
start with C or H but not O
Balancing Chemical Equations
4. Balance those elements that appear in only once on each side of the arrow first.
C2H6 + O2 CO2 + H2O
3.7
2 carbonon left
1 carbonon right
multiply CO2 by 2
C2H6 + O2 2CO2 + H2O
6 hydrogenon left
2 hydrogenon right
multiply H2O by 3
C2H6 + O2 2CO2 + 3H2O
start with C or H but not O
Balancing Chemical Equations
5. Balance those elements that appear more than once on a side. Balance free elements last.
3.7
2 oxygenon left
4 oxygen(2x2)
C2H6 + O2 2CO2 + 3H2O
+ 3 oxygen(3x1)
multiply O2 by 72
= 7 oxygenon right
C2H6 + O2 2CO2 + 3H2O72
remove fractionmultiply both sides by 2
2C2H6 + 7O2 4CO2 + 6H2O
The smallest whole-number coefficients are preferred!
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) 4 C
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
3.7
2C2H6 + 7O2 4CO2 + 6H2O
Reactants Products
4 C12 H14 O
4 C12 H14 O
6. Check to make sure that you have the same number of each type of atom on both sides of the equation.
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 carbon-
Hydrogen-Oxygen Compound. Liquid triethylene glycol, C6H14O4,
is used as a solvent and plasticizer for vinyl and polyurethane pla
stics. Write a balanced chemical equation for its complete combu
stion.
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
Solution
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
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 form
ulas, and the chemical equation.
The coefficients in the chemical equation
2H2(g) + O2(g) 2H2O(l)
Mean that
2x molecules H2 + 1x molecules O2 2x molecules H2O
Suppose that we let x=6.02214×1023 .
Then the chemical equation also means that
2 mol H2 + 1 mol O2 2 mol H2O
3-3 Equations on a Macroscopic Scale
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 t
wo substances involved in a chemical reaction; thus a stoichiom
etric factor is a mole ratio.
The mole is the key to quantitative relationships between
substances involved in chemical changes on a practical scale.
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)
3-4 Mass Relationship in Chemical Reaction
Grams of H2 GIVENUse inverse of molar mass
as conversion factor:1 mol H2/2.016 g H2
Moles of H2
Use coefficients in balanced chemical Equation to find mole
ratio: 2 mol H2O/2 mol H2
Moles of H2O= (2/2) ×mol H2
Use molar mass as conversion factor: 18.02 g H2O/1 mol H2
Grams of H2O FOUND
( g H2 mol H2 mol H2O g H2O )1 2 3
1 mol H2
2.016 g H2
? g H2O =4.16 g H2 × × × 2 mol H2
2 mol H2O 18.02 g H2O1 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 = 2H2 + O2
Moles 2 2 1
Formula Mass 18.0 2.02 32.0
Mass 36.0 4.03 32.0
When all the reactants are completely and simultaneously co
nsumed in a chemical reaction, the reactants are said to be in stoi
chiometric proportions--in the mole ratios dictated by the coeffic
ients in the balanced equation. At other times, the reactants are n
ot 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.
3-5 Limiting Reactants
Solution
1. The first step in a stoichiometric calculation is to writeA balanced equation for the reaction. If the equation
Is not given, you must supply your own.
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?
Example
Phosphorus trichloride
? Mol Cl2 = 323g Cl2 × = 4.56 mol Cl2
? Mol P4 = 125g P4 × = 1.01 mol P4
1mol Cl2
70.91 g Cl2
1 mol P4
123.9 g P4
P4(s) + 6 Cl2(g) 4 PCl3(l)
125 g 323 g ?g
2. Note what’s given and asked for the above reaction.
3. Write the formula masses needed below the reaction.
formula masses, u 123.9 70.91 137.3
4. Determine which reactant in limiting. Convert grams to moles.
? g PCl3 = 323 gCl2 × × ×
= 417 g PCl3
1 mol Cl270.91 g Cl2
4 mol PCl3
6 mol Cl2
137.3 g PCl3
1 mol PCl3
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.
If calculated mole tatio <6/1 chlorine is limiting
If calculated mole tatio >6/1 phosphorus is limiting
Grams of P4
Use inverse of molarMass as conversion factor
: 1 mol P4/123.9g P4
Moles of P4
Grams of Cl2
Use inverse of molarMass as conversion factor
: 1 mol Cl2/70.91g Cl2
Moles of Cl2
Calculate mole ratio of Cl2 to P4
Mole ratio = moles of Cl2/moles of P4
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 :
?g P4 = 323 g Cl2× 1 mol Cl270.91 g Cl2
1 mol P4
6 mol Cl2× ×
123.9 g P4
1 mol P4
= 94.1 g P4
Or on the mass of PCl3 produced.
?g P4 = 417 g PCl3× 1 mol PCl3137.3gPCl3
1 mol P4
4 mol PCl3× ×
123.9 g P4
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
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%
3-6 Theoretical Yield, Actual Yield, and Percent Yield
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?
60.1 g CO(NH2)2
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× ×
= 60.1g CO(NH2)2
1 mol CO(NH2)2
1 mol CO2 1 mol CO(NH2)2
(b) Actual yield = 47.7 g CO(NH2)2
(c) %yield = × 100% = 79.4%47.7 g CO(NH2)2
60.1 g 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.
100sample of mass total
samplein A masssamplein A of massby percent
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
• Molecular formulas
– give the actual numbers and types of atoms in a
molecule.
– Examples: H2O, CO2, CO, CH4, H2O2, O2, O3, and
C2H4.
3.9 Molecular and Structural Formulas
• Most molecular substances that we will study in this class contain only nonmetals.
Space-filling models
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-and-
stick model, or space-filling model is used.
Representing Structure in Molecules
Accurately representsthe angles at which molecules are attached.
Class Practice ExerciseThe structural formula of propane and butane is
What is the chemical and empirical formula for thesemolecules?
C C C
H H
H
H
H
HH
H C C C
H H
H
H
H
HH
H C
H
H