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Chapter 3 – Stoichiometry of Formulas and Equations
This chapter addresses two basic concepts: 1) the masses of molecules and salts at both the
atomic and macroscopic levels and 2) the number relationship between reacting species.
3.1 The Mole
You were introduced briefly to the atomic mass unit in the previous chapter as a convenient
way of discussing the masses of atoms and subatomic particles. As you have already surmised, on
the macroscopic scale the amu is as inconvenient as the gram is at the atomic scale, so we use the
gram for “normal” scale situations. However, it would be convenient to be able to move easily
between these scales. The conversion factor 6.022 x 1023 allows for such easy movement.
As your book observes, even small amounts of material contain unimaginably large numbers of
atoms. For example, one tablespoon of water holds 1023 molecules. If you can see it, a sample of
anything contains at least a million, million, million molecules. For that reason, a collective term
was developed. The mole contains 6.022 x 1023 items (Avogadro’s number). Memorize this value.
The mole is similar to the dozen (12 items) and gross (144 items) in that it is a counting term.
Technically, a mole contains the number of atoms in exactly 12 g of 12C. Remember that “exactly”
means there is no error in this number. Using the g/amu conversion factor means if one carbon-12
atom weighs exactly 12 amu, then one mole of 12C weighs exactly 12 g. Therefore, one chooses the
mass units based on the scale one works on. This is why the periodic table does not have units next
to the masses.
Before we go further, there is a language issue we must address. Your book refers to molar
mass, which, not surprisingly, is the mass of one mole of a substance. The following terms are used
interchangeably with it: molecular mass, molecular weight, formula mass, formula weight, etc.
2
Molecular weight (MW) is probably the most common of the terms, even if the words are least
accurate.
All can be thought of as the sum of the masses of the atoms that compose a compound. Thus,
for glucose, C6H12O6:
MWglucose =

15.9994 amu 
1.0079 amu
12.011 amu
 (6 atomsC )(
)  / moleculeglu cos e
)  (6 atomsO )(
)  (12 atomsH )(
atomO
atomH
atomC


= 180.157 amu
Or

15.9994 g 
1.0079 g
12.011 g
)  / molglu cos e
)  (6 molO )(
)  (12 molH )(
molO 
molH
molC
MWglucose =  (6 molC )(

=180.157 g/mol
Note that I don’t round off to the tenths place like the book does. You shouldn’t either for two
reasons: 1) with calculators it isn’t worth the trouble and 2) in some cases, you may make a
significant figure error.
Example: How many moles of glucose, C6H12O6, are in 1.00 g? How many molecules? How
many carbon atoms?
MWC6H12O6 = 180.157 amu/molecule  MWC6H12O6 = 180.157 g/mole
 1 mol 
 = 5.55 x 10-3 mol
180.157
g


molC6H12O6 = ( 1.00 g ) 
 1 mol  6.022 x 10 23 molecules 


mol

 180.157 g 
moleculesC6H12O6 = 1.00 g 
3
= 3.34 x 1021 molecules

1 mol  6.022 x 10 23 molecules  6 atoms C 



mol
 molecule 
 180.157 g 
atomsC = 1.00 g 
= 2.00 x 1022 atomsC
When discussing materials, be careful to think about what is really being asked for in the
text. In particular, if nitrogen is listed as a product of reaction, what is typically meant is
molecular nitrogen (N2), not atomic nitrogen (N). There is similar ambiguity in other materials (e.g.
oxygen and chlorine) and you should learn to anticipate what is expected in these situations.
Mass percent (sometimes called percent composition) is a concept related to molecular weight.
It is a common and very useful calculation in chemistry that provides the elemental composition as
the percent of total molecular weight. When a new compound is prepared, its elemental composition
is usually determined. The following calculation provides the expected composition based on
molecular formula. For glucose:
 12.011 amu
( 6 atoms C ) 
 atom C
%C =
180.157 amu
%H =
%O =



 x 100%
 1.0079 amu 

( 12 atoms H ) 
 atom H 
180.157 amu
 15.9994 amu 

( 6 atomO ) 
 atomO 
180.157 amu
= 40.00%
x 100% =
x 100%
6.71%
= 53.28%
99.99%
There are three points worth noting here: (1) Substituting grams for amu gives exactly the same
answers. Use whichever unit the problem requires. (2) You should add the percentages together and
make sure they total 100%  0.02%. There will frequently be a very small deviation from 100%
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because of rounding, but it should never be much larger than a couple hundredths of a percent. (3)
By convention, percent composition is almost always reported to a maximum accuracy of hundredths
of a percent, even though the calculation may be more accurate. This is because percent composition
calculations are usually used in the determination of the identity of unknown materials. It is quite
difficult to obtain materials more than 99.9% pure and, combined with experimental error in making
the measurements, hundredths of a percent does quite well for almost all purposes.
3.2 Determining the Formula of an Unknown Compound
As it turns out, mass percent isn’t a very useful concept as is, but using it in the reverse direction
to determine a molecular formula is very important to chemists. Before we see how, it may have
occurred to you that there is a bit of a problem with mass percent. Two compounds with the same
ratio of atoms, will give the same mass percent.
Formulas that contain the actual numbers of each type of atom present are called molecular
formulas. Those with the smallest whole number ratio of atoms are called empirical formulas.
Sometimes molecular formulas and empirical formulas are the same (e.g. H2O) and sometimes they
differ (e.g. for glucose molecular formula = C6H12O6 and empirical formula = CH2O). There is
something to be careful about here: Sometimes the empirical formula for one compound is the same
as the molecular formula for a second. For example, the molecular formula for formaldehyde, CH2O,
is the same as the empirical formula for glucose.
This calculation is done in the reverse direction as often as the forward direction. That is, we
often start with the percentages of each element and try to determine the relative number of atoms
present. For example, we can burn a sample containing C & H and collect and weigh the CO2 &
5
H2O produced. From the weights, the amount of C and H in the original sample can be determined.
Similar experiments are performed for other elements.
The calculation follows the following steps:
1) Confirm the percentages total to 100%.
2) Assume there are 100 g of compound present. (Then the amount of each element present in
grams will equal the percentage of that element present.)
3) Calculate the number of moles present of each element. (Make sure to use the element in
elemental form (e.g. oxygen as 15.9994 g/mol not 31.9998 g/mol).
4) Calculate the molar ratios, dividing by the element with the smallest number of moles
present. (Retain units throughout the process.)
Example: An unknown compound is found to be composed of 40.0% carbon, 6.7% hydrogen, and
53.3% oxygen. Its molecular weight was measured as 180 g/mol. What are its
empirical and molecular formulas?


molC = (40.0 gC)  1 mol C  = 3.33 molC
 12.011 gC 
 1 mol H
 1.0079 g H
molH = (6.7 gH) 

 = 6.65 molH



molO = (53.3 gO)  1 mol O  = 3.33 molO
 15.9994 gO 
molar ratios:
C:
H:
3.33 mol C
= 1.00
3.33 mol C
6.65 mol H
= 2.00 molH per molC
3.33 mol C
6
O:
3.33 mol O
= 1.00 molO per molC
3.33 mol C
The empirical formula is CH2O.
Note that with one exception, all of the numbers in this example have units. Be careful in
rounding the molar ratios. You should round numbers very close to integers (e.g. 0.97 rounds to 1),
but numbers close to 0.25, 0.33, and 0.5 should be treated differently. In these cases, all ratios are
multiplied by a factor to yield whole number values. E.g. a compound has the following ratios: Fe
= 1.00, O = 1.33 oxygen atoms per iron atom. Multiply through by 3 to get Fe3O4.
If you also have the molecular weight of the material, you can determine the molecular formula.
When going from empirical formula to molecular formula, all of the subscripts are multiplied by a
whole number. This is the only time it is permissible to change the subscripts on a formula.
Example: In the previous problem you were told the molecular weight was about 180 g/mol, you
calculate the molecular formula as follows:
The empirical weight of CH2O is 30 g/mol.
MW 180 g/mol
=
= 6.0
EW 30 g/mol
Thus, there are 6 empirical units in each molecule  C6H12O6
3.3 Writing and Balancing Chemical Equations
By far and away, the most convenient way of describing chemical reactions is the chemical
equation. The generic form is given as:
aA + bB  cC + dD
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where a, b, c, and d are whole number, balancing coefficients (“1” is not used), A & B are reactants,
while C & D are products. Sometimes the symbols  (heat) or h (light) are placed above the arrow
to indicate if outside sources of energy were used to facilitate the reaction. Also, sometimes the
solvent (section 4.1) is placed below the arrow. Usually the physical state of each species will be
indicated by placing one of the following symbols after each reactant and product: (g) = gas, () =
liquid, (s) = solid, (aq) = aqueous (dissolved in water).
A chemical equation does not tell you:
1) Reaction rate (how fast the reaction goes)
2) Reaction pathway. (Most reactions go through several steps on the way to completion.)
You will learn about both of these topics in CHM 212.
How does one balance a chemical equation (determine the coefficients)? We begin by assuming
you know the reactants and products in the reaction mixture. The following method is different
from the book’s and, in my experience, works well. You may use any method that works, though.
The steps are:
1)
Write out all reactants and products on the correct side of the arrow.
2)
Count out how many atoms of each element (or group) appear on either side of the arrow.
3)
Choose the element (or group) that appears in the fewest number of places and has different
numbers on either side of the arrow. Balance using coefficients.
4)
Repeat with the remaining atoms/groups.
5)
Make all coefficients whole numbers by multiplying through by the least common denominator
of the existing coefficients.
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Examples: (The blue atoms are being balanced.)
__ P + __ Cl2  __ PCl3
__ C7H14 + __ O2  __ CO2 + __ H2O
P + 3 Cl2  2 PCl3
C7H14 + O2  7 CO2 + H2O
2 P + 3 Cl2  2 PCl3
C7H14 + O2  7 CO2 + 7 H2O
C7H14 +
21/
2
O2  7 CO2 + 7 H2O
2 C7H14 + 21O2  14 CO2 + 14 H2O
There are a couple of points worth mentioning: (1) Never change subscripts in formulas. Doing
so changes the identity of the substance. For example, H2O is water, but H2O2 is hydrogen peroxide.
(2) Coefficients should not have a common divisor. That is, reaction between hydrogen and oxygen
to form water is not properly balanced as: 4 H2 + 2 O2  4 H2O because each coefficient is an
even number. To balance it correctly you must divide by 2 (point 5 above).
One common mistake when writing out consecutive reactions is to write the first reaction, then
add a reactant to the product, draw an arrow, then the final reaction. For example, let’s say that
you burn phosphorus in oxygen to yield diphosphorus pentaoxide, then react it with water to produce
phosphoric acid. It is incorrect to write:
4 P(s) + 5 O2 (g)  2 P2O5 (s) + 6 H2O()  4 H3PO4 (aq)
The reason for this is that it implies that the first reaction produces the water, which then further
reacts with the P2O5. The correct way to write this pair of reactions is on two lines:
4 P(s) + 5 O2 (g)  2 P2O5 (s)
P2O5 (s) + 3 H2O ()  2 H3PO4 (aq)
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3.4 Calculating Quantities of Reactant and Product
Once you have a balanced equation, you can then employ it in useful ways. A term you should
know is stoichiometry, which refers to the quantity relationship between reacting chemical species.
For example, when hydrogen and oxygen gases react to form water, two hydrogen molecules react
with one oxygen molecule to form two water molecules.
O
H
H
+
H
H
H
O
O
O
H
H
H
2 H2
+
O2

2 H2O
The 2 to 1 to 2 relationship is the stoichiometry of the reaction. Stoichiometry also allows the number
of atoms or molecules involved in a reaction to be related to the measured quantities: mass and
volume (for gases or liquids) of the substances. Thus, 1 gram of hydrogen reacts with 8 grams of
oxygen to yield 9 grams of water.
In the real world, when we want to know how much of a substance we have, we usually
measure its mass or volume. We never count the number of molecules present. On the other hand
when we discuss a chemical reaction, we consider the reaction of individual particles. How do we
reconcile these views? In a balanced equation, the coefficients tell us the relative numbers of atoms
or molecules reacting. Because the mole is a number term, the coefficients also tell us the relative
number of moles of atoms/molecules involved in a reaction. Finally, molecular weights allow us to
relate numbers of species to their masses. This allows the amount of all substances produced or
consumed in a reaction to be calculated from only one reaction species. Two frequently asked
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questions in a chemistry lab are: (1) if I have x grams of reactant A, what mass of reactant B will I
need to react with it? and (2) if I have x grams of reactant A, what mass of product C will be
generated by the reaction?
The general solution to these problems is:
1) Write out the balanced equation.
2) Convert all quantities to moles.
3) Using the balanced equation, determine the molar ratio of desired to known species.
4) Convert the molar amount of desired species into the appropriate units.
5) Check for reasonability. This is very important.
Example: Consider the reaction: CS2 + 3 O2  CO2 + 2 SO2. What mass of O2 is required to react
completely with 9.34 g carbon disulfide?


massO2 = (9.34 gCS2)  1 mol CS2  3 mol O2
 32.00 gO2 



 76.13 gCS2  1 mol CS2  1 mol O2 
= 11.8 gO2
In a similar manner, the masses of CO2 and SO2 produced can be calculated. (5.40 gCO2
and 15.7 gSO2)
Combustion reactions are a particular class of reactions defined as those producing a flame.
(The book’s definition is incorrect.) Oxygen is usually a reactant, although sometimes another
substance such as nitrogen in the magnesium example below can also participate.
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CH4 + 2 O2  CO2 + 2 H2O
(burning of natural gas)
There are other important, common reactions too. Combination reactions occur when 2 or more
substances react to form one product. The reactions that occurred in an old-fashioned flash bulb
were combination reactions (and also combustion reactions).
3 Mg + N2  Mg3N2
&
2 Mg + O2  2 MgO
Decomposition reactions occur when one substance breaks up into more than one substance.
They are the reverse of combination reactions. The following reaction shows how limestone
decomposes to lime.
CaCO3  CaO + CO2 (with heating)
Reactions that Involve a Limiting Reagent
In practice, reactions are rarely conducted with exact stoichiometric amounts of each reagent
for three reasons. First, if one of the reactants is a gas, it may be difficult to measure out that
reactant. Second, one reactant is nearly always more expensive than the others. In this case, one
wants to use up all of that reagent. Adding small excesses of the others usually accomplishes this.
Finally, some reactions can proceed through multiple paths and using excess amounts of some
reagents can maximize the yield of a desired product.
When some reagents are used in excess, the completely consumed reagent is called the limiting
reagent (or less commonly: limiting reactant). When doing a limiting reagent calculation, first
convert the masses of all reactants to moles, which then must be converted to a common scale using
stoichiometric coefficients.
Example: Reaction of tungsten with chlorine gas yields tungsten(VI) chloride. Find the mass of
unreacted starting material when 12.6 g of tungsten is treated with 13.6 g of chlorine
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gas. How much tungsten(VI) chloride is formed?
We begin by generating a balanced chemical equation.
W(s) + 3 Cl2 (g)  WCl6 (s)
 1 mol W
 183.9 g W
a) molW = 12.6 gW 

 = 0.0685 molW



molCl2 = 13.6 gCl2  1 mol Cl2  = 0.192 molCl2
 70.90 g Cl2 
Now calculate how much chlorine is needed to completely react with tungsten
 3 mol Cl2 
 = 0.206 molCl2 but you only have 0.192 mol of chlorine, so it is

 1 mol W 
(0.0685 molW) 
the limiting reagent.
Calculate the mass of tungsten consumed:
 1 mol Cl2
 70.90 gCl2
massW = 13.6 gCl2 
 1 mol W

 3 mol
Cl2

 183.9 g W

 mol
W


 = 11.8 gW

massW(excess) = 12.6 g - 11.8 g = 0.8 gW
 1 mol Cl2
 70.90 g Cl2
b) massWCl6 = 13.6 gCl2 
 1 mol WCl6

 3 mol
Cl2

 396.6 g WCl6

 mol
WCl6


 = 25.4 gWCl6


Theoretical Yields, Actual, and Percent Yields
In real life, reactions rarely generate all of the product you would expect from the reaction
stoichiometry because there are frequently side reactions that use some of the starting materials to
make other products. Equilibria can also occur. Also, sometimes the products or reactants aren’t
extremely stable under the reaction conditions. The theoretical yield for a reaction is the amount of
product expected based on stoichiometry. Part (b) of the previous example was a theoretical yield
calculation.
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The amount of product obtained is called the actual yield. The ratio of actual to theoretical
yields is called the percent yield.
percent yield =
actual yield
x 100%
theoretical yield
Using the previous example, what would the percent yield be if 17.6 g of tungsten(VI)
chloride actually formed?
 17.6 g 
 (100%) = 69.3 %
 25.4 g 
percent yield = 
In this case, the side reactions might be production of WCl2 or WCl4 resulting from incomplete
reaction of the starting materials.
3.5
Fundamentals of Solution Stoichiometry
In nearly all reactions, at least one of the components is in liquid solution. If you think about
it, this is actually quite reasonable. For two things to react, they typically must come into contact
with each other and that involves mixing. Doing that with solids is very difficult so reactions
typically occur in liquids or gases, or when a liquid or gas comes into contact with a solid (e.g.
rusting). For this reason, it is useful to be able to describe the amounts of substances dissolved in a
liquid.
The composition of a solution is most frequently described in terms of concentration, the amount
of one material dissolved in a second material. The most common and important unit for describing
the concentration of a solution is molarity: the number of moles of solute dissolved in a liter of
solution (abbreviated “M”). Placing the dissolved species in square brackets also means molar
concentration (see below). Note that the volume of solution is used,
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molar concentration =
moles of solute
volume of solution in liters
not just the solvent added. If large amounts of solute are added to pure solvent the total volume
will increase, but if small amounts are added either an increase or decrease may occur. (A small
solute species might draw solvent molecules towards itself, thus causing a small shrinkage.)
Molarity allows convenient comparison of reactants in different solutions through balanced
equations.
Example: What is the molarity of a solution that contains 17.4 g NaCl per 112 mL solution?
 1 mol  1  1000 mL 
mol

= 2.66 M

 = 2.66
L
 58.44 g  112 mL  L 
[NaCl] = (17.4 g) 
Likewise, we can find the mass of a solute or volume of solvent if 2 of the 3 (molarity, mass, volume)
are known.
When an ionic solid dissolves, it exists entirely as ions. For convenience we write 0.1 M NaCl,
but in reality no NaCl actually exists in solution. The solution consists of 0.1 M Na+ and 0.1 M Cl-.
For 0.1 M Na2CO3:
 0.1 mol Na2CO3  2 mol Na  
= 0.2 M

L

 1 mol Na2CO3 
Example: [Na+] = 
 0.1 mol Na2CO3  1 mol (CO3)2 - 
= 0.1 M

L

 1 mol Na2CO3 
[CO32-] = 
(This calculation is shown in your book in Section 1 of Chapter 4.)
Dilution
Dilution is the process of lowering the concentration of a solution, almost always by adding
more solvent to the solution. Here the total amount of solute remains the same. The following
15
equality allows you to calculate the concentration of a solution after more solvent has been added,
or more usefully, to calculate how much solvent must be added to an existing solution to dilute to a
new concentration.
MiVi = MdVd
Example: What is the concentration of a solution after 25.0 mL of the previous sodium chloride is
added to 125 mL of water? (Assume the volumes are additive.)
(2.66 M)(25.0 mL) = Md (1.50 x 102 mL)
Md = 0.443 M
How much water must be added to 50.0 mL of the original NaCl solution to make it
1.00 M?
(2.66 M)(50.0 mL) = (1.00 M)Vd
Vd = 133 mL
Volume added = 133 mL - 50 mL = 83 mL
One thing you might have noticed is that the volume term in molarity is L and the volumes used in
the examples are mL. We can get away with this because of the equality. In each case, either the
concentration or volume units will cancel.
One very common error in solution problems involving reactions is to use the dilution equation.
In problems where the reagents are in a one-to-one molar ratio it will give the correct numerical
answer, but for the wrong reason. Thus, using the dilution equation in stoichiometry/titration
(Section 4.6) problems is always wrong.
August 24, 2015