‫‪Chemical Formulas‬‬
‫‪and Composition‬‬
‫‪Stoichiometry‬‬
‫د‪ .‬كريم محمد الصاوى‬
‫استاذ مساعد الكيمياء الفيزيائية‬
‫قسم الكيمياء ‪ -‬جامعة القصيم‬
Stoichiometry
A word derived from the Greek stoicheion, which means “first principle or
element,” and metron, which means “measure.”
Stoichiometry describes the quantitative relationships
among elements in compounds (composition stoichiometry) and
among substances as they undergo chemical changes (reaction
stoichiometry).
The chemical formula for a substance shows its chemical composition.
This represents the elements present as well as the ratio in which the atoms
of the elements occur.
The chemical formula for a single atom is the
same as the symbol for the element. Thus, Na
can represent a single sodium atom.
A subscript following the symbol of an element
indicates the number of atoms in a molecule. For
instance, H2O is a molecule containing two
hydrogen atoms and one oxygen atom.
HO
2
Number of
hydrogen
atoms
Formula unit
(NaCl)n
The chemical formula
for a single atom is the
same as the symbol for
the element.
NaCl
A molecule is a definite
group of atoms that are
chemically bonded together
Formula unit: the lowest whole
number ratio of ions present in
an ionic compound
Composition is described
by the molecular formula
E.g. H2O
Composition is described by
the formula unit.
E.g. CaCl2
Subscript in the formula indicates the number of atoms
In a chemical reaction atoms are rearranged (Dalton’s Theory).
for example, six H2 molecules react with three
O2 molecules and rearrange to form six H2O
molecules
Equivalently, two H2 molecules react with one
O2 molecule rearrange to form two H2O
molecules
So, the reaction takes place if we have the
correct relative number of reactant
PARTICLES (e.g. atoms, molecules,…).
Atoms are so small to count (DIRECTLY), in
the lab it is much easier to WEIGH chemicals.
OK, If we get the correct relative weights of
reactants, does this mean that we get the
correct relative numbers of the reactant
particles (atoms)? NO
So. What!!!!!!!!!!!!!!!
The Mole
Chemists have adopted the mole concept as a convenient way to
deal with the enormous numbers of molecules or ions in the samples
they work with.
 The mole is the SI unit for amount of substance, it is abbreviated mol.
 The mole is defined as the amount of substance that contains as many
particles (atoms, molecules, ..) as there are atoms in exactly 12 g of
pure carbon-12 atoms.
 The number of particles in one mole is known as Avogadro’s
Number (NA)
 Avogadro's number = 6.022x1023
= 602200000000000000000000.
The mole is defined as the amount of substance that contains
as many particles (atoms, molecules, ..) as there are atoms in exactly 12 g of
pure carbon-12 atoms.
One mole of hydrogen (H) contains the same number of particles as 12 g of carbon-12
One mole of helium (He) contains the same number of particles as 12 g of carbon-12
One mole of neon
(Ne) contains the same number of particles as 12 g of carbon-12
The number of particle in one mole of hydrogen (H) =
The number of particle in one mole of helium (He) =
The number of particle in one mole of neon (Ne) =
.
.
The number of particle in one mole of any substance =
Avogadro's Number=6.022x1023
The mole is defined as the amount of substance that contains as many particles
(atoms, molecules, ..) as there are atoms in exactly 12 g of pure carbon-12
atoms.
=
12.000 g of carbon
(one mole of carbon)
6.022x1023
carbon atoms
This means we now have Avogadro's number of carbon
atoms (NA) = 6.022x1023 on the scales pan
The mole is defined as the amount of substance that contains as many particles
(atoms, molecules, ..) as there are atoms in exactly 12 g of pure carbon-12
atoms.
Then, by definition:
To get one mole of carbon all what we need to do is to weigh exactly 12 g
of carbon.
What if we need to get one mole of some other substance say Hydrogen,
Helium, Sodium, …?
How much do we need to weigh of these substance? (what is the molar
mass of these substance?)
It will be wrong to weigh 12 grams of each of these substances, 12 grams
will not contain the same number of particles since atoms of different
substances have different masses (Dalton’s Theory).
Then, to proceed, we need to know how heavy each element is relative to
carbon-12. In other words we need to know the relative atomic masses.
DETERMINATION OF RELATIVE ATOMIC MASSES (WEIGHTS)
Diagram of a simple mass spectrometer (top), showing the
separation of neon isotopes (left).
Neon gas enters an evacuated chamber, where neon atoms
form positive ions when they collide with electrons. Positively
charged neon atoms, Ne, are accelerated from this region by the
negative grid and pass between the poles of a magnet. The
beam of positively charged atoms is split into three beams by
the magnetic field according to the mass-to-charge ratios.
The three beams then travel to a detector at the end of the tube.
( The detector is shown here as a photographic plate)
Thousands of experiments on the composition of compounds have resulted
in the establishment of a scale of relative atomic masses (weights) based on
the atomic mass unit (amu).
The atomic mass unit (amu), is defined as exactly 1/12 of the mass of an
atom of carbon-12.
Carbon-12 isotope was arbitrarily assigned a mass of exactly 12 atomic
mass units
On this scale, the atomic weight (mass) of

hydrogen (H)
is
1.00794 amu

Helium (He)
is
4.0026 amu

sodium (Na)
is
22.9897 amu,

magnesium (Mg) is
24.3050 amu.
This tells us that Na atoms have nearly 23 times the mass of H atoms,
and Mg atoms are about 24 times heavier than H atoms.
Because of this relative scale, we can be sure that if we take 1:4
samples (by weight) of H and He, they will contain the same number
of particles.
The Molar Mass (M)
The molar mass of a substance is the mass of one mole of that substance.
Since one mole contains Avogadro's number of particles, then
The Molar mass = the total mass of Avogadro's number of particles.
Since we have established a relative atomic mass scale relative to a common
element (C-12), the following relationship is true.
The mass of one mole (Molar Mass) of atoms of a pure element in
grams is numerically equal to the atomic weight of that element in
atomic mass units.
The molar mass units are grams/mole, also written as g/mol or g mol-1.
 For instance, if you obtain a pure sample of the metallic element titanium
(Ti), whose atomic weight is 47.88 amu, and measure out 47.88 g of it,
you will have one mole, or 6.022 x1023 titanium atoms.
One mole of atoms of some common elements.
Back row (left to right): bromine, aluminium, mercury,
copper. Front row (left to right): sulfur, zinc, iron.
Each of these samples has a different mass but
contains the same number of atoms; a very large
number 6.022x1023 - Avogadro's number
How many moles of atoms does 136.9 g of iron (Fe) metal contain?
answer
We are given the total mass of the Fe sample (m) = 136.9 g
We look up the atomic mass of Fe = 55.85 amu
Then
The mass of one mole (molar mass M) of Fe= 55.85 g/mol
Then
number of moles (n) 
total mass
m

mass of one mole(molar mass) M
n
m
M
n
136.9 g
 2.451mol
55.85 g / mol
How many atoms are contained in 2.451 mol of iron?
answer
We are asked about the total number of atoms (N)
We are given the number of moles (n) =2.451 mol
We know that one mole of atoms of an element contains Avogadro’s
number (NA) of atoms, or 6.022 x1023 mol-1.
Then
total number of atoms( N )  number of moles(n)  avogadro' s number( N A )
N  n NA
N  2.451mol  6.022  10 mol  1.476  10
23
1
24
Calculate the average mass of one iron atom in grams.
answer
The atomic mass of Fe = 55.85 amu
Then
The mass of one mole (molar mass M) of Fe= 55.85 g mol-1
This is the mass of Avogadro’s number of atoms
Then
molar mass ( M )  mass of one atom Avogadro' s Number( N )
A
1
molar mass ( M )
55.85 g mol
mass of one atom 

 9.274  10 g
N
6.022  10 mol
 23
23
1
A
Thus, the average mass of one Fe atom is only 9.274 x10-23 g, that is,
0.00000000000000000000009274 g
FORMULA WEIGHTS, MOLECULAR WEIGHTS AND MOLES
So, we’ve learnt how to get the molar mass of atomic substances using
the relative atomic mass scale based on C-12
How about the molar mass of molecules and ionic substances?
The mass of one mole (molar mass) of a substance in grams is
numerically equal to the formula weight of that substance in atomic
mass units.
The formula weight = sum of the atomic masses (weights) of the elements in
the formula
Calculate the formula weight (mass) and molar mass of each of the following
a) chloroform, CHCl3; b). iron(III) sulfate, Fe2(SO4)3; c) water, H2O
Answer
a) chloroform, CHCl3
Formula weight  1 atomic mass of C 1 atomic mass of H  3  atomic mass of Cl
 112.0 amu
 1 1.008 amu
 3 35.45 amu
119.4 amu
Then the molar mass of CHCl3= 119.4 g mol-1
b) iron(III) sulfate, Fe2(SO4)3
Formula weight  2  atomic mass of Fe 3 ( atomic mass of S 4  atomic mass of O)
 2  55.8 amu
 3 (32.0 amu
 4  16.0 amu
)
 399.9 amu
Then the molar mass of Fe2(SO4)3= 399.9 g mol-1
c) water, H2O
Molecule, Mole and Molar Mass
Molar mass of H2O
=
mass of one mole of H2O =
Total mass of 6.022x1023
molecules of H2O
=
Formula weight in grams
of H2O
=
18 g mol-1
One
molecule of
H2O
One mole of H2O
(6.022x1023 molecules of H2O)
How many moles of substance are contained in each of the following samples?
(a) 18.3 g of NH3
(b) 5.32 g of ammonium bromide NH4Br
(c) 6.6 g of PCl5
Br 79.9 N 14.0 P 30.9 Cl 35.5
(d) 215 g of Sn
C 12.0 S 32.0 Sn 118.7
answer
number of moles (n) 
n
a)n
NH 3

m
M
NH 3
NH 3

total mass
m

mass of one mole(molar mass) M
m
M
18.3 g
18.3 g

 1.07 mol
14.0 g mol  3  1.0 g mol 17.0 g mol
1
1
1
What mass, in grams, should be weighed for an experiment that requires 1.54 mol
of (NH4)2HPO4?
answer
total mass
m

mass of one mole (molar mass) M
m
n
M
m
n

M
number of moles (n) 
( NH 4 ) 2 HPO4
( NH 4 ) 2 HPO4
( NH 4 ) 2 HPO4
m
( NH 4 ) 2 HPO4
M
n
( NH 4 ) 2 HPO4
( NH 4 ) 2 HPO4
 (14.0 g mol  4  1.0 g mol )  2 
1
( NH 4 ) 2 HPO4
M
1
1.0 g mol 
30.97 g mol 
4  16.0 g mol  132.1 g mol
1
1
1
m
( NH 4 ) 2 HPO4
n
( NH 4 ) 2 HPO4
M
1
( NH 4 ) 2 HPO4
 1.54 mol  132.1 g mol  203.4 g
1
How many (a) moles of O2, (b) molecules of O2 , and (c) O atoms are
contained in 40.0 g of oxygen gas (dioxygen)
a)
total mass
m

mass of one mole (molar mass) M
m
n
M
m(O )
40.0 g
n(O ) 

 1.25 mol
M (O ) (2  16.0) g / mol
number of moles (n) 
2
2
2
b)
total number of particles( N )  number of moles(n)  avogadro' s number( N A )
N  n NA
N (O )  n(O ) N A
2
2
N (O )  1.25 mol  6.022  10 mol  7.52  10
1
23
2
c)
Each oxygen molecule(O ) contains 2 oxygen atoms (O)
2
N (O)  2  N (O )
2
N (O)  2  7.52  10  1.50  10
22
23
22
Calculate the number of hydrogen atoms in 39.6 g of ammonium sulfate, (NH4)2SO4.
One molecule of (NH4)2SO4
contains
8
hydrogen atoms
One mole
of (NH4)2SO4
contains
8 x NA
hydrogen atoms
n
of (NH4)2SO4
contains
8 x NA x n hydrogen atoms
mole
number of hydrogen atoms  8  N  n
m
n
M
A
( NH 4 ) 2 SO4
( NH 4 ) 2 SO4
M
 (14.0 g mol  4  1.0 g mol )  2  32.0 g mol  4  16.0 g mol
1
( NH 4 ) 2 SO4
 132.0 g mol
1
1
1
39.6 g
number of hydrogen atoms  8  6.022  10 mol 
132.0 g mol
 1.44  10
23
24
1
1
1
How many hydrogen atoms are contained in 125 grams of butane, C4H10
Answer
One molecule of C4H10 contains
10
One mole
of C4H10 contains
10 x NA
n
of C4H10 contains
10 x NA x n hydrogen atoms
mole
hydrogen atoms
hydrogen atoms
number of hydrogen atoms  10  N  n
m
n
M
A
C 4 H10
C 4 H10
M
 4  12.0 g mol  10  1.0 g mol  58.0 g mol
1
C 4 H10
1
1
125 g
number of hydrogen atoms  10  6.022  10 mol 
58.0 g mol
 1.3  10
23
25
1
1
What mass of chromium is contained in 35.8 g of (NH4)2Cr2O7?
answer
Reading the given chemical formula tells us that
One mole of
(NH4)2Cr2O7 contains 2 moles of Cr
We can succinctly write this in terms of molar masses as:
→
→
One molar mass of (NH4)2Cr2O7
35.8 g
X
of (NH4)2Cr2O7
2 x Molar mass of Cr
X
2  molar mass of Cr  35.8 g
molar mass of ( NH ) Cr O
4
2
2
7
molar mass of ( NH ) Cr O  (14.0 g mol  4 1.0 g mol )  2
1
4
2
2
1
7
52.0 g mol  2  16.0 g mol  7
 252.0 g mol
molar mass of Cr  52.0 g mol
2  52.0 g mol  35.8 g
X
 14.8 g
252.0 g mol
1
1
1
1
1
1
What is the formula…?
When a chemist has discovered a new compound, the first
question to answer is, What is the formula?
To answer, you begin by analyzing the compound to determine
amounts of the elements for a given amount of compound.
This is conveniently expressed as percentage composition—
that is, as the mass percentages of each element in the
compound.
But, what do we really mean by a chemical formula?
The simplest, or empirical, formula for a compound is the smallest wholenumber ratio of atoms present.
The molecular formula indicates the actual numbers of atoms present in a
molecule of the compound. It may be the same as the simplest formula or
else some whole-number multiple of it.
E.g.
The molecular formula for hydrogen peroxide is
H2O2
This tells us that there are two hydrogen atoms and
two oxygen atoms in the molecule
The empirical formula for hydrogen peroxide is
HO
This tells us that the ratio between the hydrogen
and oxygen atoms in the molecule is 1 : 1
How to get these formula?
PERCENT COMPOSITION AND FORMULAS OF COMPOUNDS
Suppose that A is a part of something—that is, part of a whole. It
could be an element in a compound or one substance in a mixture.
We define the mass percentage of A as the parts of A per
hundred parts of the total, by mass. That is
If the formula of a compound is known, its chemical composition
can be expressed as the mass percent of each element in the
compound (percent composition).
E.g. Calculate the percent composition by mass of HNO3.
Answer
We are not given the mass of HNO3 but we can calculate its molar mass and then
derive its percent composition from it.
H
M
N
O
 1.0 g mol  14.0 g mol  3  16.0 g mol  63.0 g mol
1
HNO3
1
1
1
1
mass of H 1.0 g mol
%H 

100  1.6
M
63.0 g mol
1
HNO3
1
mass of N 14.0 g mol
%N 

100  22.2
M
63.0 g mol
1
HNO3
mass of O 3  16.0 g mol
%O 

100  76.2
M
63.0 g mol
1
1
HNO3
Nitric acid is 1.6% H, 22.2% N, and 76.2% O by mass.
N.B. All samples of pure HNO3 should have this composition, according to the
Law of Definite Proportions.
DERIVATION OF FORMULAS FROM ELEMENTAL COMPOSITION
To get a chemical formula for a compound we need the number (or the
relative number) of atoms in the compound. So if we have the mass of each
element we can proceed as follows:
mass of
each
element
Get the relative
number of
atoms for each
element
( mass/gram
atomic mass AW)
Divide by the
smallest
number
Convert to
whole
numbers
Chemical formula
corresponds to
the smallest
whole number
ratio of atoms
A 20.882-gram sample of an ionic compound is found to contain 6.072 grams
of Na, 8.474 grams of S, and 6.336 grams of O. What is its empirical (
simplest) formula?
answer
A 0.1014 g sample of purified glucose was burned in a C-H combustion train to
produce 0.1486 g of CO2 and 0.0609 g of H2O.
An elemental analysis showed that glucose contains only carbon, hydrogen, and
oxygen. Determine the masses of C, H, and O in the sample.
Answer
1 mole of CO2 → 1 mole of C
Molar mass of CO2= 12.0 g mol-1 + 2x16.0 g mol-1 = 44.0 g mol-1
Molar mass of C = 12.0 g mol-1
44 g mol-1 of CO2 → 12.0 g mol-1 of C
0.1486 g of CO2 → ?
g
C
0.1486 g 12.0 g mol
C
 0.04055g
44 g mol
1
1
1 mole of H2O → 2 mole of H
Molar mass of H2O= 2x1.0 g mol-1 + 16.0 g mol-1 = 18.0 g mol-1
Molar mass of H = 1.0 g mol-1
18 g mol-1 of H2O → 2.0 g mol-1 of H
0.0609 g of H2O → ?
g
H
0.0609 g  2.0 g mol
H
 0.00681g
18.0 g mol
O  mass of sample (mass of C  mass of H )
O  0.1014 g  (0.04055 g 0.00681g )  0.0540g
1
1
In the previous example a 0.1014 -gram sample of glucose was found to
contain 0.04055 grams of C, 0.00681 grams of H, and 0.0540 grams of O.
What is the empirical ( simplest) formula of glucose?
answer
mass of
each
element
Get the relative
number of
atoms for each
element
( mass/gram
atomic mass AM)
Divide by the
smallest
number
Convert to
whole
numbers
Chemical formula
corresponds to
the smallest
whole number
ratio of atoms
DETERMINATION OF MOLECULAR FORMULAS
As we have seen percent composition data yield only simplest (empirical)
formulas.
To determine the molecular formula for a molecular compound, both its
simplest formula and its molecular weight must be known.
Molecular formula = k x simplest formula
k = integer 1,2,3,….
k
molecular mass
simplist formula mass
If k=1, then the molecular formula and empirical formula are the same
E,g. H2O
In the previous example, the simplest formula for glucose was found to be
CH2O. Other experiments have shown that its molecular weight is
approximately 180 amu. Determine the molecular formula of glucose.
Answer
molecular formula  k  simplest formula
simplest formula  CH O
2
molecular formula  k  CH O
2
molecular mass
simplest formula mass
simplest formula mass  12.0 amu  2  1.0 amu  16.0 amu
 30.0 amu
180 amu
k
6
30.0 amu
molecular formula  6  CH O  C H O
k
2
6
12
6
Deduce the empirical formulas of the following molecular formulas
a) C6H12O6
b) H2O2 b) H2O
answer
We start from the definition:
Molecular formula = k x simplest formula
k = integer 1,2,3,….
Then we need to find k …k will be the largest common whole
number divisor of the numbers of atoms in the molecular
formula
a) C6H12O6
The molecular formula tells us that for each C6H12O6 molecule there are:
6
C atoms
12 H atoms
6
O atoms
Common divisors of 6,12 and 6 are 1,2,3 and 6
We choose the largest, k=6
The empirical formula is then C6/6 H12/6 O6/6 = CH2O