STOICHIOMETRY
Dr. Tanweer Ahmad
Department of Chemistry
School of Mathematics and Natural Science
The Copperbelt University
Kitwe, Zambia
Email: tanweer.ahmad@cbu.ac.zm
INTRODUCTION

Describes the quantitative relationships that exist
between substances undergoing chemical changes.
COUNTING BY WEIGHING
Suppose a wholesaler is selling a large number of toffee
sweets that weigh 5 g each.
 If a customer asks for 1000 sweets, the mass of sweets
required is 1000 sweets 5g or 5000 g (5 kg).
 Since weighing 5 kg of sweets takes a shorter time than
counting 1000 sweets, weighing is a preferred method of
“counting” the 1000 sweets for the customer.
 It is to be noted that since in real life sweets do not have
identical masses, the key piece of information needed to
count sweets by weighing is the average mass of the toffee
sweets.
 To relate mass and number of atoms, the average
atomic mass is required.

ATOMIC MASSES

Mass of an atom depends on the number of its nucleons
(protons and neutrons given as A = N + Z).

The modern system of atomic masses, instituted in
1961, is based on 12C (“carbon twelve”) as the standard.

In this system, 12C is assigned a mass of exactly 12 atomic
mass units (amu), and the masses of all other atoms are
given relative to this standard.

The most accurate method currently available for
comparing masses of atoms involves the use of an
instrument called the mass spectrometer.
ATOMIC MASS UNIT (AMU)
Atomic mass (or atomic weight) is the mass of an atom in
atomic mass units (amu).
 One atomic mass unit is defined exactly equal to onetwelfth of the mass of one carbon-12 (or 12C) atom.
 Thus, the mass of one carbon-12 atom is 12 amu.

Average Atomic Mass (Aav) of an Element
 This is the weighted average mass of the naturally
occurring mixture of isotopes of an element given by the
equation
A av 
Fi  Ai

i1 100

n

n
f
i1 i
 Ai
CONTD…
where Fi is the % abundance of the ith isotope, Ai is the
atomic mass in amu of the ith isotope, is the fractional
abundance of the ith isotope and n is the number of
isotopes for the element.
Example:

MOLE

It is the amount of a substance that contains as many
elementary entities (atoms, molecules, or other
particles) as there are atoms in exactly 12 g (or 0.012
kg) of carbon-12 isotope.

The actual number of atoms in 12 g of carbon-12 is an
experimentally determined number called Avogadro’s
number (NA) whose accepted value is NA = 6.022 1023.

Thus, 1 mole of carbon-12 atoms has a mass of 12 g and
6.022 1023 atoms.

This mass of carbon-12 is its molar mass defined as mass
(in grams or kilograms) of 1 mole of units (such as
atoms or molecules) of a substance.
CONTD…
The relationships between mass in grams of an
element (m) and number of moles of an element (n) as
well as between number of moles (n) of an element
and number of atoms of element (N).
 M in the figure is the molar mass (g/mol) of an element
and NA is Avogadro’s number.

CONTD…
Example 1:
 Helium (He) is valuable gas used in industry, low temperature
research, deep-sea diving tanks and balloons. How many
moles of He atoms are in 6.46 g of He?
Solution:
To convert grams to moles we need molar mass. From the
periodic table we see that the molar mass of He is 4.003 g, that
is, 1 mol of He = 4.003 g. We write 6.46 g He.

No. of moles in 6.46 g He = 6.46 g x 1 mol/4.003 g = 1.61mol
Thus, there are 1.61 moles of He atoms in 6.46 g of He.
CONTD…
Example 2:
 Zinc (Zn) is a silvery metal that is used in making
brass (with copper) and in plating iron to prevent
corrosion. How many grams of Zn are in 0.356 mole of
Zinc?
Solution:
 To convert moles to grams we remember that
molar mass of Zn
1
1 mol Zn

Since molar mass of Zn is 65.39g, the mass of 0.356
mol is 0.0356 molZn .  65 .39 g Zn  23 .3 g Zn
1 mol Zn
CONTD…
Example 3:
 Sulphur is a non-metallic element. Its presence in coal gives
rise to acid rain phenomenon. How many atoms are in 16.3
g of S?
Solution:
 Solving this problem has 2 steps. We first get the number of
moles of S in 16.3 g of S. We thereafter calculate the number
of atoms.
 n, the number of moles of S in 16.3 g is given by the relation
n

m
16 .3 g

M 32.07 g / mol
The number of atoms (N) is given by the relation
16.3
N  n  NA 
 6.022  10 23  3.06  10 23
32.07
S atoms.
MOLECULAR AND FORMULA MASS

The molecular mass (sometimes called the molecular
weight) is the sum of the atomic mass (in amu) in the
molecule. For example, the molecular mass of H2O is
2 x (atomic mass of H) + atomic mass of O or


2  1.008 amu  16.00 amu  18.02 amu

In general, we need to multiply the atomic mass of each
element by the number of atoms of the element present in
the molecule and sum over all the elements.
CONTD…
Example:
 Calculate the molecular mass of the following compounds
(a) Sulphur dioxide (SO2) and (b) caffeine (C8H10N4O2).
Reasoning and Solution: To calculate molecular mass we
need to count the number of each type of atom in the
molecule and look up its atomic mass in the periodic table.
 There is one S atom and two O atoms is sulphur dioxide, so
that molecular mass of SO2 = 32.07 amu+2 (16.00) amu
= 64.07 amu.
 There are eight C atoms, ten H atoms, four N atoms, and two
O atoms in caffeine, so that the molecular mass of
C8H10N4O2 is given by 8 (12.01 amu)+10 (1.008 amu)+4
(14.01 amu)+2 (16.00 amu) =194.20 amu.
MOLAR MASS OF MOLECULE OR COMPOUND

From the molecular mass we can determine the molar mass
of a molecule or compound.

The molar mass of a compound (in grams) is numerically
equal to its molecular mass (in amu).

For example, the molecular mass of water is 18.02 amu, so
that its molar mass is 18.02 g.

Note that 1 mole of water weighs 18.02 g and contains
6.022x1023 H2O molecules, just as 1 mole of elemental
carbon contains 6.022x1023 carbon atoms.
CONTD…
Example:
 Methane (CH4) is the principal component of natural gas.
How many moles of CH4 are present in 6.07 g of CH4?
Reasoning and Solution:
 Therefore the first step is to calculate the molar mass of
CH4.
 Molar mass of CH4 = 12.01 g + 4 (1.008) g = 16.04g.
 We calculate the number of moles of n by dividing the
molar mass (Mr) into the mass of CH4 given (m) as follows
n
m
6.07 g

 0.378 mol CH 4
M r 16.04 g / mol
CONTD…
Example:
 Given that the molar mass of urea is 60.6 g, determine how
many hydrogen atoms are present in 25.6 g of urea whose
formula is (NH2)2CO?
Reasoning and Solution:
 We first calculate the number of molecules present in 25.6 g
of urea. Next we note that there are 4 hydrogen atoms in
every urea molecule.
 Number of molecules of urea (N) is given by

Number of H atoms is given as
4 N 
25.6
 6.022  10 23  4  1.03  10 24 atoms
60.06
PERCENT COMPOSITION


OF COMPOUNDS
The percent composition by mass is the percent by mass of
each element in compound.
% composition of an element =
n  molar mass of element
 100 %
molar mass of compound

where n is the number of moles of the element in 1 mole of
the compound.
CONTD…
For example:
 1 mole of hydrogen peroxide (H2O2) has 2 moles of H atoms
and 2 moles of O atoms.
 The molar masses of H2O2, H and O are 32.02 g, 1.008 g and
16.00 g, respectively.
 % composition of H2O2 is
2  1.008 g
%H 
 100  5.93 %
34 .02 g
2  16 .00 g
%O 
 100  94 .06 %
34 .02 g
CONTD…
Example:
 Phosphoric acid (H3PO4) is a colourless, syrupy liquid used
in detergents, fertilizers, toothpastes, and in carbonated
beverages for a “tangy” flavour. Calculate the percent
composition by mass of H, P, and O in this compound.
Reasoning and Solution:
 The molar mass of H3PO4 is 97.99 g. The percent by mass of
each of the elements in H3PO4 is calculated as follows:
%H 


3  1.008 g H
 100 %  3.086 % %P  30 .97 g P  100 %  31 .61 %
97 .99 g H 3PO 4
97 .99 g H 3PO 4


4  16 .00 g O
%O 
 100 %  65 .31 %
97 .99 g H 3PO 4
DETERMINING THE FORMULA OF A COMPOUND
CONTD…
Mass percent
Convert to grams and
divide by molar mass
Moles of each element
Divide by the smallest
number of moles
Mole ratios of elements
Change to integer
subscripts
Empirical formula
EXAMPLE
CHEMICAL REACTION AND CHEMICAL EQUATION
A chemical reaction is a process in which a substance (or
substances) is changed into one or more new substances.
 Chemical equations use chemical symbols to show what
happens during a chemical reaction.
 “Molecular hydrogen reacts with molecular oxygen to yield
water”.

H 2  O 2  H 2O

Chemical equations must conform to the Law of
Conservation of Mass that requires that there must be the
same number of each type of atom on both sides of the
arrow.
2H 2  O 2  2H 2O
CONTD…

Chemists often indicate the physical states of the reactants
and products using the letters g, l, and s in brackets to
denote gas, liquid, and solid, respectively. These letters are
called state symbols. For example,
2CO(g)  O2(g)  2CO2(g)
2HgO (s)  2Hg (l)  O2(g)

To represent what happens when sodium chloride (NaCl) is
added to water we write
BALANCING CHEMICAL EQUATION
Example:
 Consider the combustion of the natural gas component
ethane (C2H6) in oxygen or air to yield CO2 and water. Write
a balanced equation for this.
Solution:
 The unbalanced equation of reactants and products is:
C2H6  O2  CO2  H2O
To balance C, place 2 in front of CO2 to get the
equation
C2 H6  O2  2CO2  H2O
 To balance H, place 3 in front of H2O

C2 H6  O2  2CO2  3H2O
CONTD…

C and H are balanced but the O atoms are not; there are 7
O atoms on the RHS and 2 O atoms on the left hand side
(LHS). This can be eliminated by writing in front of O2 on
the LHS as shown below
C2H6 

7
2
O2  2CO2  3H2O
However, it is normally preferred to express
coefficients as whole numbers rather than fractions.
2C2 H6  7O2  4CO2  6H2O
AMOUNTS OF REACTANT AND PRODUCT

To interpret a reaction quantitatively, we need to apply our
knowledge of molar masses and the mole concept.
Stoichiometry is the quantitative study of reactants and
products in a chemical reaction.

Whether the units given for reactants (or products) are
moles, grams, litres (for gases) or some other units, we use
moles to calculate the amount of product formed in the
reaction.

This approach is called is the mole method which simply
means that the stoichiometric coefficients in a chemical
equation can be interpreted as the number of moles of each
substance.
CONTD…
The mole method consists of the following steps:
 Write correct formulae for all reactants and products, and
balance the resulting equation.
 Convert the quantities of some or all given or known
substances (usually reactants) into moles.
 Use the coefficients in the balanced equations to calculate
the number of moles of the sought or unknown quantities
(usually products) in the problem.
 Using the calculated numbers of moles and the molar mass,
convert the unknown quantities to whatever units are
required (typically grams).
 Check that your answer is reasonable in physical terms.
CONTD…
Example:
 All alkali metals react with water to produce hydrogen gas and the
corresponding alkali metal hydroxide. A typical reaction is that between
lithium and water
2Li (s)  2H 2O(l)  2LiOH (aq)  H 2(g)
How many moles of H2 will be formed by the complete reaction of 6.23
moles of Li with water?
 How many grams of H2 will be formed by the complete reaction of
80.57g of Li with water?
Reasoning and Solution: (a)
 The balanced equation is given in the problem.
 No conversion is needed because the amount of starting material,
Li, is given in moles.
 Since 2 moles of Li produce 1 mole of hydrogen molecules, we
calculate moles of H2 produced as follows:

moles of H 2  6.23 mol Li 
1 mol H 2
 3.12 mol H 2
2 mol Li
CONTD…
(b) We follow the steps used in (a) above
 The balanced equation is the same as in (a).
 The number of moles of Li, is given by the equation
moles of Li  n Li 


mLi
80.57 g

 11.61 mol Li
M Li
6.941 g/mol
Because 2 moles of Li produce 1 mole of H2, we calculate the
number of moles of H2 as follows:
1 mol H 2
n H  11.61 mol Li 
 5.805 mol H 2
2 mol Li
From the molar mass of H2 (2.016 g), we calculate the mass
of H2 produced (mH) as
m H  n H  M H  5.805  2.016 g  11.70 g H 2

Because the molar mass of H2 is smaller than that of Li and
two moles of Li is needed to produce one mole of H2, we
expect the answer to be smaller than 80.57 g.
THE CONCEPT LIMITING REACTANTS




Suppose you have a part-time job in a sandwich shop. One
very popular sandwich is always made as follows:
Assume that you come to work one day and find the
following quantities of ingredients:
8 slices bread; 9 slices meat; 5 slices cheese
How many sandwiches can you make? What will be
left over?
CONTD…

To solve this problem, let’s see how many sandwiches we
can make with each component:

How many sandwiches can you make? The answer is three.
When you run out of meat, you must stop making
sandwiches. The meat is the limiting ingredient.
CONTD…

What do you have left over? Making three sandwiches
requires six pieces of bread. You started with eight slices, so
you have two slices of bread left.

You also used three pieces of cheese for the three
sandwiches, so you have two pieces of cheese left.

In this example, the ingredient present in the largest
number (the meat) was actually the component that limited
the number of sandwiches you could make.

This situation arose because each sandwich required three
slices of meat—more than the quantity required of any
other ingredient.
CONTD…
Example:
 Urea [(NH2)2CO] is prepared by reacting ammonia with carbon dioxide
2NH3(g)  CO2(g)  (NH 2 ) 2 CO(aq)  H 2O (l)
In one process, 637.2 g of NH3 are allowed to react with 1142 g of CO2.
 Which of the two reactants is the limiting reagent?
 Calculate the mass of (NH2)2CO formed.
 How much of the excess reagent (in grams) is left at the end of the
reaction?
Reasoning and solution
 Because we can’t tell by inspection which of the two reactants is the
limiting reagent, we have to proceed by converting their masses into
numbers of moles. The molar mass of NH3 and CO2 are 17.03 g and
44.01 g, respectively. Thus
m NH3
637.2 g
Moles of NH3 = n NH 

 37.42 mol
3
M NH3
17.03 g/mol
CONTD…
Moles of CO2 = n CO 
2




mCO2
MCO2

1142 g
 25.95 mol
44.01 g/mol
From the balanced equation, 2 moles of NH3 react with 1 mole of CO2.
Therefore, the number of moles of NH3 needed to react with 25.95 moles
of CO2 is 25.95 x 2 = 51.90 moles of NH3.
Since there is only 37.42 mol of NH3 present, not enough to completely
react with the CO2, NH3 must be the limiting reagent and CO2 the
excess reagent.
The amount of (NH2)2CO produced is determined by the amount of
limiting reagent present. We know that 2 moles of NH3 produces 1
mole of (NH2)2CO and the molar mass of (NH2)2CO is 60.06 g. Thus,
we calculate the mass (NH2)2CO produced according to the equation
m( NH 2 )2 CO  n( NH 2 )2 CO  M ( NH 2 )2 CO
n (NH2 )2 CO  37.42 mol NH3 
1 mol (NH2 ) 2 CO
 18.71 mol (NH2 ) 2 CO
2 mol NH3
CONTD…
Therefore,
m (NH2 )2 CO  n (NH2 )2 CO  M (NH2 )2 CO  18.71 mol  60.06 g/mol  1124 g ( NH 2) 2 CO

The number of moles of the excess reagent left is determined
from the fact that the number of CO2 used is equal to the
number of (NH2)2CO produced, that is,
n (NH2 )2 CO  n CO2 used  18.71 mol

Therefore moles of CO2 left is 25.95 - 18.71 = 7.24 mol so that
the mass of CO2 left is calculated from the equation
m CO 2  n CO 2  M CO 2  7.24 mol  44.01g/mol  319 g
REACTION YIELD





The amount of limiting reagent present at the start of a reaction
determines the theoretical yield of the reaction.
Theoretical yield is the amount of product that would result if all the
limiting reagent reacted.
The theoretical yield is the maximum obtainable yield predicted by the
balanced equation.
In practice, the actual yield, or the amount of the product actually
obtained from a reaction, is almost always less than the theoretical
yield.
The percent yield, which describes the proportion of the actual yield to
the theoretical yield. It is calculated as follows:
actual yield
% yield 
100%
theoretical yield

Its value is a fraction < 100 %
CONTD…
Example:
 Titanium is prepared by the reaction of titanium (IV) chloride with molten
magnesium between 950oC and 1150oC.
TiCl 4(g)  2Mg (l)  Ti (s)  2MgCl 2(l)
In a certain industrial operation 3.54 x 107g of TiCl4 are reacted with 1.13
x 107g of Mg.
 Calculate the theoretical yield of Ti in gram.
 Calculate the percent yield if 7.91 x 106 g of Ti are actually obtained.
 How much of the excess reagent (in grams) is left at the end of the
reaction?
Reasoning and solution
 We find out which of the two reactants is the limiting reagent. This
knowledge will enable us to calculate the theoretical yield. The percent
yield can be obtained by applying the % yield equation.
 First we calculate the number of moles of TiCl4 and Mg initially present.
7
m
3
.
54

10
g
TiCl
4
Moles of TiCl4 = n


 1.87 105 mol

TiCl 4
M TiCl 4
189.7 g/mol
CONTD…




Moles of Mg = n Mg 
m Mg
M Mg
1.13 107 g

 4.65 105 mol
24.31 g/mol
Next we must determine which of these substances is the limiting
reagent. From the balanced equation we see that 1 mol of TiCl4used
2 mol of Mg. Therefore, the number of moles of Mg needed to react
with 1.87 x 105moles of TiCl4 is
1.87  10 5 2  3.74  105
Since 4.65 x 105 moles of Mg are present, TiCl4 is the limiting
reagent.
The equation shows that 1 mol of TiCl4 gives 1 mol of Ti, thus, the
theoretical yield is obtained from the equation
m Ti  n Ti  M Ti  1.87 105 mol  47.88 g/mol  8.93 106 g Ti
6
7
.
91

10
g
actual yield
%
yield

100%  88.6%
% yield 
100%
6
8.93 10 g
theoretical yield
THANKS