The uploaded slides are covering the whole chapter of "Stoichiometry" especially related to KPK boards.

1. STOICHIOMETRY
Presented by
SHABAB HUSSAIN
M. Phil “Chemistry”
CIIT, Abbottabad 1

2. Contents
 Stoichiometry
 Atomic Mass
 One Gram-Atom
 Mole
 Avogadro's Number (NA)
 Mole Mass Calculations
 Mole-Particle Calculations
 The Mole and Chemical Equations
 Writing a Chemical Equation
 Balanced Chemical Equation
 Law of Conservation of Mass/Matter
 Interpretation of Chemical Equation
 Mole-Mole Conversion
 Mole-Mass Conversion
 Mass-Mass Conversion
 Stoichiometric Calculations Involving
Gases
 Percentage Composition
 Excess and Limiting Reagents
 Theoretical Yield and Actual Yield as
Percentage
 Why actual yield is always less than
theoretical yield?
 Percentage Yield (% yield)
2

3. Stoichiometry: -
The branch of Chemistry which deals with
the study of quantitative calculations
involved in the interconversion of matter
during any physical or chemical change is
known as stoichiometry.
OR
The study of the relationship between the
amounts of reactants and products formed, by
a balanced chemical equation is also called
stoichiometry. 3

4. Atomic Mass: -
 Atomic mass of an element is defined as; the mass of
one atom of the element compared with one-twelfth
the mass of one atom of carbon 12
6C.
 The unit of atomic mass is “atomic mass unit (amu)”.
𝟏𝐚𝐦𝐮 =
𝐦𝐚𝐬𝐬 𝐨𝐟 𝐨𝐧𝐞 𝐚𝐭𝐨𝐦 𝐨𝐟 𝐂𝐚𝐫𝐛𝐨𝐧 − 𝟏𝟐
𝟏𝟐
 So amu is defined as; mass exactly equal to one-
twelfth the mass of one carbon-12 atom.
4

5. One Gram-Atom: -
 One gram-atom of any element is the relative atomic mass of
the element expressed in grams.
 For example, relative atomic masses of Chlorine (Cl) is
35.5amu while that of Sulphur (S) is 32.1amu. So one gram-
atom of Chlorine and Sulphur would be 35.5g and 32.1g
respectively.
 The concept of gram-atom is useful because it is impossible to
see or weigh individual atoms or even a large number of
atoms. We can actually see or weigh one gram-atom of an
element.
 One gram molecule and formula unit means molecular mass
and formula mass when expressed in grams respectively.5

6. Mole: -
 The atomic masses of atoms, molecular masses of molecules
and formula masses of formula units when expressed in gram
is called a mole of that substance.
OR
 The gram atomic mass of atoms, gram molecular mass of
molecules and gram formula mass of formula units is also
known as mole of that substance.
 Examples: - H = 1.008amu= 1.008g = 1Mol
H2 = 2.016amu = 2.016g= 1Mol
H2O =18amu =18g = 1Mol
NaCl = 58.5amu = 58.5g = 1Mol 6

7. Avogadro's Number (NA): -
• Avogadro's Number is defined as; the number of atoms,
molecules or ions in one mole correspond to atomic mass,
molecular mass or formula mass of a substance is known as
Avogadro number.
• The number was determined by an Italian scientist, Amadeo
Avogadro, represented by NA, which is equal to 6.023x1023.
 Examples: - H = 1.008g = 1Mol = 6.023x1023 atoms.
H2 = 2.016g = 1Mol = 6.023x1023 molecules.
H2O = 18g = 1Mol = 6.023x1023 molecules.
NaCl =58.5g = 1Mol = 6.023x1023 formula units.
7

8. Mole Mass Calculations: -
 The relationship between a mole and a molar mass
can be treated as a conversion factor. Based on this
relationship the following equation will help us to
convert any mass into moles and any number of
moles into mass.
Number of mole (n) =
Mass (in gram)
Molar mass
8

9. Mole-Particle Calculations: -
 As we know that there exists a connection between mole of a substance and the
Avogadro's number (NA). This means that another quantitative relationship is
available to us:
1 mole = 6.023x1023 particles
 Based on above fundamental relationship, we are able to convert any number
of moles into particles (atoms, molecules or formula units) and vice versa,
using the following mathematical equations:
No. of moles (n) =
No. of particles (atoms, molecules or formula units)
Particles (atoms, molecules or formula units) per mole
No. of moles (n) =
No. of particles (atoms, molecules or formula units)
Avogadro′s number (NA) {6.023x1023}
No. of particles (atoms, molecules or formula units) = No. of moles x NA
9

10. “The Mole and Chemical Equations”
Chemical Equations: -
• A chemical change is called a chemical reaction.
• The representation of chemical reaction via chemical
symbols is called chemical equation.
OR
• The symbolic representation of any chemical reaction is
also known as chemical equation.
10
H2 (g) + O2 (g) H2O

11. Writing a Chemical Equation: -
 Hydrogen gas (H2) reacts with Oxygen gas (O2) to yield “Water”. The reaction can be
represented by the following chemical equation:
 The sign (+) means “reacts with” and the means “to yield”. Thus, this symbolic
expression can be read: “Molecular Hydrogen reacts with molecular Oxygen to yield
water. The reaction is assumed to proceed from left to right as the arrow indicates.
 We refer to H2 and O2 in above Equation as reactants, which are the starting materials in
a chemical reaction. Water is the product, which is the substance formed as a result of a
chemical reaction.
 A chemical equation, then, can be thought of as the chemist's shorthand description of a
reaction.
 In a chemical equation the reactants are conventionally written on the left and the
product on the right side of arrow:
Reactants Products
 In a chemical equations, chemists often indicate the physical states of the reactants and
products by using the abbreviations g, l, and s in parentheses to denote the gas, liquid,
and solid state, respectively.
11
H2 (g) + O2 (g) H2O

12. Balanced Chemical Equation: -
 A type of chemical equation which follows the “Law of
Conservation of Mass/Matter” is known as a balanced
chemical equation.
OR
 In other words, when the amount (mass) of products are equal to
the amount (mass) of reactants is also known as balanced
chemical equation.
 We must have as many atoms after the reaction ends as we had
before it started.
12
2H2O2H2 + O2

13. Law of Conservation of Mass/Matter:-
 Lavoisier, a French chemist, performed a number of
experiments on combustion reaction and finally proved
that the total mass of the product was always equal to
the total mass of reactants.
 According to this Lavoisier in 1885 concluded that:
“mass is neither created nor destroyed during a
chemical reaction”.
 This is called law of conservation of mass or matter.
 According to this law a chemical reaction must,
therefore, be balanced.
13

14. Interpretation of Chemical Equation:-
 The above chemical equation can be interpreted(read) as:
14
2H2O2H2 + O2
2H2O2H2 + O2
Two molecules + one molecule
2 moles + 1 mole
2(2.02) = 4.04g + 32.00g
two molecules
2 moles
2(18.02)=36.04g
36.04g reactants 36.04g products
2H2 O2
+
Two Hydrogen
Molecules
One Oxygen
Molecule
Two Water
Molecules
+
+
H H
H H
O O
O
H
H
O
H H

15. Mole-Mole Conversion: -
 Consider the following balanced chemical equation:
 The equation can be read/interpreted in several ways:
 2g-atoms of Mg + 1g-molecules of O2 2g-formula units of
MgO.
 2x6.023x1023 atoms of Mg + 6.023x1023 molecules of O2
2x6.023x1023 formula units of MgO.
 2 mol of Mg + 1 mol of O2 2 mol of MgO.
 2x24g of Mg + 1x32g of O2 2(24+16)g of MgO.
 All the above statements are true because they maintain the
same ratio of coefficients for all the species involved:
Mg : O2: MgO
15
2Mg(s) + O2(g) 2MgO(s)

16. Mole-Mole Conversion: -
• The mole relationships, expressed by the coefficients of a
chemical equation, can be used to calculate the number of
moles of the reactants consumed or that of the products
formed.
• In the given example, the equation says that 2mol of Mg
requires 1mol of O2 to produce 2mol of MgO. We can also
say that 2mol of Mg are equivalent to 1mol of O2
(i.e., 2mol of Mg ≈1mol of O2).
• The equation can also be stated as:
2mol of Mg ≈ 2mol of MgO
and 1mol of O2 ≈ 2mol of MgO.
• Each of these equivalencies can be used as conversion factor.
The relationship of “Mole” and “Avogadro's number” in the
form of equations help us in calculating the number of moles
of a particular reactant consumed or of a product formed.16

17. Mole-Mass Conversion: -
 Actually these types of stoichiometric problems are
extension of mole-mole problem.
 Being both substances (reactants & products) in mole,
one of them is given/asked in mass units (g).
 According to mole equation i.e.,
Mole =
Mass
Molar Mass
 Any mole of a substance either of reactants or products
can be interconverted to mole and mass easily.
17

18. Mass-Mass Conversion: -
 In case of substances (reactants or products) in mass units, we
might be asked to give answer in mass units, however, possible to
do it.
 In such a case we have to do two additional steps i.e., mol-mol
conversion one at the beginning and the other at the end of
calculation.
Mass(reactants) Mole(reactants) Mole(products) Mass(product)18
mass of
reactants
Mole of Reactants Mole of Products
mass of
products
up
down
use ratio from
the equation
across

19. Stoichiometric Calculations Involving Gases:-
 As compared to solids and liquids, weight of gases is quite
difficult in gaseous reactions. Instead of weight of gases, finding
volume is easy way in the stoichiometric calculations.
 According to Avogadro's Law “equal volume of all gases, at the
same temperature and pressure, contain equal number of particles
(atoms, molecules or formula units).
Molar Volume (volume of one Mole) of Gas at STP: -
 It has been found experimentally that one mole of any gas at
standard temperature (0oC or 273K) and pressure (1atmosphere
‘atm’ or 760mmHg or 760 torr) {STP}, occupies a volume of
22.4dm3.
 The above mentioned law is applicable to all gases and based on
this law one can convert volume of gas into moles and vice versa
using the following relationship:
Moles(n)=
Volume (dm3)
Molar Volume (22.4dm3/mol)19

20. Percentage Composition: -
 The percent mass (number of grams) of an element in a 100g
of a compound is called its percentage composition.
 Elements in a compound having molecular mass or formula
mass, the percentage composition can be found as:
𝐏𝐞𝐫𝐜𝐞𝐧𝐭𝐚𝐠𝐞 𝐨𝐟 𝐞𝐥𝐞𝐦𝐞𝐧𝐭 =
𝐓𝐨𝐭𝐚𝐥 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐞𝐥𝐞𝐦𝐞𝐧 𝐢𝐧 𝐜𝐨𝐦𝐩𝐨𝐮𝐧𝐝
𝐓𝐨𝐭𝐚𝐥 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐂𝐨𝐦𝐩𝐨𝐮𝐧𝐝
𝐱 𝟏𝟎0
𝐏𝐞𝐫𝐜𝐞𝐧𝐭𝐚𝐠𝐞 𝐨𝐟 𝐞𝐥𝐞𝐦𝐞𝐧𝐭 =
𝐀𝐭𝐨𝐦𝐢𝐜 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐞𝐥𝐞𝐦𝐞𝐧𝐭 𝐱 𝐍𝟎. 𝐨𝐟 𝐚𝐭𝐨𝐦𝐬 𝐢𝐧 𝐜𝐨𝐦𝐩𝐨𝐮𝐧𝐝
𝐌𝐨𝐥𝐚𝐫 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐂𝐨𝐦𝐩𝐨𝐮𝐧𝐝
𝐱 𝟏𝟎𝟎
 Sum of individual percentages of all elements must be equal to
100.
20

21. Excess and Limiting Reagents: -
 In a chemical reaction when the reactants are not exactly to
stoichiometric amounts as indicated by balanced chemical
equation, the reactants used up first in a reaction is called limiting
reactant.
 When this reactant is used up, no more product can be formed.
 Other reactants, present in quantities greater than those needed to
react with the quantity of the limiting reagent present, are called
excess reactants.
 In short, those reactants which take part in a chemical reaction and
vanish early because they formed the products thus stops the
chemical reaction are called limiting reagents. Limiting reagents
limit the reaction. On the other hand, those reagents which are in
excess not taking part in reaction (not able to form products) are
called excess reagents.
21

22. Excess and Limiting Reagents: -
 Sulfur burns in atmosphere with Fluorine forming Sulfur hexafluoride, which
is colorless, odorless and extremely sable compound.
 The above balanced equation tells us that 1mol of S reacts with 3moles of F2
to produce 1mol of SF6.
 Suppose in certain preparation 4moles of S are added to 20moles of F2. Since
1mol of S reacts with 3moles of F2, the number of moles of F2 needed to react
with 4moles of S is;
 Since 4moles of S will react with 12moles of F2, and limit the reaction because
no more S is present so ‘S’ is limiting reagent. While 12 out of 20moles (F2)
will take part in formation of SF6 and remaining 8moles (20-12) will be in
excess, viz not taking part in reaction hence called excess reagent.
22
S(l) + 3F2(g) SF6(g)
4 mol S x
3 mol F2
1 mol S
=12 mol F2

23. Theoretical Yield and Actual Yield as Percentage: -
 Theoretical Yield:- The amount of products formed as
predicted by balanced chemical equation when all the
limiting reactant has reacted is called theoretical yield.
 The theoretical yield then, is the maximum obtainable
yield because it is the calculated quantity of the product
that we should get, using the starting amounts of the
reactants.
 Actual Yield:- In actual practice, the amount of product
obtained, is less as theoretical amount which is known
as actual yield, experimental yield or practical yield.
23

24. Why actual yield is always less than
theoretical yield?
The reasons due to which actual yield is always less than theoretical yield are
as follow:
 Due to reversibility of a chemical reaction, the reaction may not gone to
completion. Means all of the reactants are not converted into products.
 Due to side, parallel or chain reaction, some amount of reactants may converted
to some other unknown products that we do not need or desire.
 Due to mechanical loss i.e., from a reaction medium, the complete recovery of
products is impossible. As a result, some of the products will be lost and
consequently effect the amount of products negatively.
 Due to disturbance (up or down) in reaction conditions like temperature,
pressure, concentration, pH etc actual yield might be disturbed.
24

25. Percentage Yield (% yield): -
 The comparison/proportion of actual yield to the
theoretical yield is called percentage yield. It is defined
as follows:
(The units in the ratio should be the same so that they will cancel)
 Percent yields may range from a fraction of less than 1
percent to 100 percent.
 Reactions that give yield above 90% are considered as
excellent.
 Percentage yield expresses the efficiency of a chemical
reaction. 25
% yield =
actual yield
theoretical yield
x100

26. Thanks
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