1. Stoichiometry is the quantitative relationship between reactants and products in a balanced chemical equation. It allows calculations of mass, moles, particles, and volume from a balanced equation under specified conditions. 2. Calculations include mass-mass, mass-mole, mass-volume, and mole-mole relationships between substances. Conversions use mole ratios as conversion factors. 3. Limiting reagent refers to the reactant that is completely used up first and thus limits the amount of product that can be formed, analogous to the relationship between kababs and bread slices in making sandwiches.

1. STOICHIOMETRY
Aga Khan Board

2. SLO#1.1.1 Discuss
the significance of
Chemistry as a
quantitative
science in daily
life.
QUANTITATIVE
SCIENCE:- Quantitative Science in chemistry is
applied throughout industry, medicine,
and all the sciences.
The concentrations of oxygen and of
carbon dioxide are determined in
millions of blood samples every day and
used to diagnose and treat illnesses.
Quantities of hydrocarbons, nitrogen
oxides, and carbon monoxide present in
automobile exhaust gases are measured
to determine the effectiveness of
emission-control devices.

3. 4. Quantitative measurements of ionized calcium in blood serum help diagnose parathyroid disease in
humans.
5. Quantitative determination of nitrogen in foods establishes their protein content and thus their nutritional
value.
6. Analysis of steel during its production permits adjustment in the concentrations of such elements as
carbon, nickel, and chromium to achieve a desired strength, hardness, corrosion resistance, and ductility.
7. The mercaptan content of household gas supplies is monitored continually to ensure that the gas has a
sufficiently obnoxious odor to warn of dangerous leaks.
8. Farmers tailor fertilization and irrigation schedules to meet changing plant needs during the growing
season, gauging these needs from quantitative analyses of plants and soil.

4. SLO#1.2.1 Define ‘mole’ and
‘Avogadro’s number.
• MOLE:-
• “The amount of substance that contains as many
elementary particle(atoms,molecules,ions or other)
as there are atoms in exactly 12 grams of carbon-
12”
• “The mass of substance containing the same same
number of fundamental units as there are atoms in
exactly 12.0000 g of C-12. Fundamental units may
be atoms, molecules or formula units, depending on
the substance concerned.”

6. • Avogadros’s Number:-
• “Number of units in one mole of any substance
(defined as its molecular weight in grams) , equal to
6.02 x 10^23. The units may be electrons , atoms ,
ions or molecules, depending on the nature of the
sunbstance and character of the reaction.”

7. SLO# Relate the
concept of mole
with Avogadro’s
number:

8. To understand Avogadro's number let us consider the following
quantities of substances.
1.008 g of hydrogen=1mole of hydrogen = 6.02 x 1023
atoms of H
23 g of sodium = 1 mole of Na = 6.02 x1O23 atoms
of Na
238 g of uranium = 1 mole of U = 6.02 x 1023
atoms of U

9. Another Example Let us consider:
18 g of H2O =1 mole of water
=6.02 x 1023 molecules of water
180 g of glucose = 1 mole of glucose
= 6.02 x 1023 molecules of glucose
342 g of sucrose = 1 mole of sucrose
= 6.02 x 1023 molecules of sucrose

10. When we take into consideration the ions,
then
96 g of SO4^2- = 1 mole of SO4^2- = 6.02 x
10^23 ions of SO4^2-
62 g of NO^3- = 1 mole of NO^3- = 6.02 x
10^23 ions of NO^3-

12. S.L.O#1.2.3 Calculate the
number of following
chemical species/ particles,
i.e.
(a). atoms (b). molecules
(c). moles (d). ions (e).
protons (f). neutrons (g).
electrons

13. • The mathematical equation, N = n ×
NA, can also be used to find the
number of atoms of each element in
a known amount (in moles) of a
compound.
• For a compound with the molecular
formula XaYb:
• 1 molecule of compound
XaYb containsa atoms of element X
• b atoms of element Y
• 1 mole of compound
XaYb containsa moles of atoms of
element X
• b moles of atoms of element Y

14. •n moles of compound XaYb contains
(n × a) moles of atoms of element X
(n × b) moles of atoms of element Y
• n moles of compound XaYb contains
(n × a) × NA atoms of element X
(n × b) × NA atoms of element Y
• n moles of compound XaYb contains
(n × a) × 6.022 × 1023 atoms of
element X
(n × b) × 6.022 × 1023 atoms of
element Y
Consider n moles of each of these
compounds with the general formula
XY2.
The table below gives the moles of
each element present in the
compound, and also shows us how to
calculate the number of atoms of each
element present:

15. Example:-

16. Question
1: Calculate the
number of ammonia,
NH3, molecules in
3.5 moles of
ammonia.
No of Molecule= no
of mole x Av. No

17. Question 1: Calculate
the number of
ammonia, NH3,
molecules in
3.5 moles of
ammonia.
No of Molecule= no of mole
x Av. No
No of Molecule= 3.5 x 6.02
x10^23

18. Question 1: Calculate
the number of
ammonia, NH3,
molecules in
3.5 moles of
ammonia.
No of Molecule= no of mole
x Av. No
No of Molecule= 3.5 x 6.02
x10^23
No of Molecule=21.07 x10
^23

19. S.L.O#1.2.4 calculate, using a balanced chemical equation, the a.
interacting moles b. representative particles c. masses and volume of
gases at STP (22.4 L) and RTP (24 L);
STOICHIOMETRY:-
“Stoichiometry is a branch
of chemistry which tells us
the quantitative
relationship between
reactants and products in a
balanced chemical
equation.”

20. CONDITIONS:-
when stoichiometeric
calculations are performed,
we have to assume the
following conditions.
1. All the reactants are
completely converted into the
products.
2. No side reaction occurs.

21. The following type of
relationships can be
studied with the help
of a balanced chemical
equation.
1) Mass-mass Relationship
“If we are given the mass
of one substance, we can
calculate the mass of the
other substances involved
in the chemical reaction.”

22. 2) Mass-mole Relationship or
Mole-mass Relationship
•“If we are given
the mass of one
substance, we
can calculate the
moles of other
substance and
viceversa.”

23. 3) Mass-volume
Relationship:-
• “If we are given
the mass of one
substance, we can
calculate the
volume of the
other substances
and vice-
versa.Similarly,
mole-mole
calculations can
also be
performed.”

24. SLO#1.2.5 solve
problems based on
stoichiometry using
mole ratios as
conversion factor:

25. SLO#1.3.1 calculate
percentage (by
mass) of :(a).
elements in
compounds

26. b. water of
crystallisation
in hydrated
salts:

27. SLO# 1.3.2:
Deduce empirical
and molecular
formula of
compounds:
Empirical Formula:-
“ It is the simplest formula that gives the small
whole number ratio between the atoms of
diferent elements present in a compound.”
Example:-
The empirical formula of glucose (C6H12O6) is
CH2O and that of benzene (C6H6) is CH.

28. Molecular Formula:-
That formula of a
substance which is
based on the actual
molecule is called
molecular formula.
Example:-
Molecular formula
of benzene is C6H6
while that of
glucose is C6H12O6.

29. SLO#1.4.1:
Deduce the
limiting reagent
in chemical
reactions:
LIMITIG REAGENT:-
“The limiting reactant is a
reactant that controls the
amount of the product formed
in a chemical reaction due to
its smaller amount.”

30. Example:-
The concept of limiting reactant is analogous to the
relationship between the number of “kababs” and
the “slices” to prepare “sandwiches”. If we have 30
“kababs” and ive breads “having 58 slices”, then we
can only prepare 29 “sandwiches”. One “kabab” will
be extra (excess reactant) and “slices” will be the
limiting reactant. It is a practical problem that we
can not purchase exactly sixty “slices” for 30
“kababs” to prepare 30 “sandwiches”.