Faculty of Biotechnology
General Chemistry
Lecture 4
STOICHIOMETRY
CHEMICAL CALCULATIONS
Dr. M. Abd-Elhakeem
1- Molecular weight
2- Mole
3- Avogadro’s number
4- Equivalent weight
5- Chemical equation stoichiometry
6- Gas’s laws
7-Element percent
8- Oxidation number
9- Solution stoichiometry
( Concentration – Titration)
Atomic weight:
The atomic weight of an atom is equal to the
number of protons in the nucleus of the atom plus
the number of neutrons in the nucleus of the atom
(one amu each). Therefore, a carbon atom with six
protons and six neutrons has an atomic weight of
12.
Molecular weight:
The weight of a molecule is the sum of the weights
of the atoms of which it is made.
Calculate the molecular weight of each of the
following Na2CO3 , HCl , NaOH
Na2CO3 = (23 X 2) + 12 + (16 X 3) = 106 g
Mole: Unit of amount of chemical substance
One mole is the molecular weight of a
substance in gram. It equals the summation
of atomic weight of the forming atoms.
Avogadro's number
Avogadro's number, also known as Avogadro's
constant, is defined as the quantity of atoms in
precisely 12 grams of 12C. The designation is a
recognition of Amedeo Avogadro, who was the
first to state that a gas' volume is proportional to
how many atoms it has. Avogadro's number is
given as 6.02214179 x 1023 mol-1.
How many moles in 56 g Na2CO3
Number of moles: weight in grams/ molecular weight
Then = 56/106 = 0.528 mole
F2 Mole contain …….fluorine atoms
= 2 X Avogadro’s number
• Keep in your mind
Mole -------Wight --------- number of atoms
Chemical equation
• is the symbolic representation of a chemical
reaction where the reactants are given on the left
hand side and the products on the right hand side.
Balancing chemical equations
NaOH + HCl
Na2CO3 +2 HCl
NaCl + H2O
2 NaCl + H2O +
CO2
Stoichiometry (more working with ratios)
Ratios are found within a chemical equation.
2HCl + 1Ba(OH)2  2H2O + 1 BaCl2
coefficients give MOLAR RATIOS
2 moles of HCl react with 1 mole of Ba(OH)2 to form 2 moles of H2O
and 1 mole of BaCl2
• How many moles of water are produced for each
mole of P4O10 (s) when this equation is balanced?
_
PH3 (g) + O2 (g) --> P4O10 (s) + H2O (l)
Mole – Mole Conversions
When N2O5 is heated, it decomposes:
2N2O5(g)  4NO2(g) + O2(g)
a. How many moles of NO2 can be produced from 4.3 moles of N2O5?
2N2O5(g)  4NO2(g) + O2(g)
4.3 mol
? mol
2 mol N2O5
4 mol NO2
4.3 mol N2O5
?? mol NO2
= 8.6
moles NO2
b. How many moles of O2 can be produced from 4.3 moles of N2O5?
2N2O5(g)  4NO2(g) + O2(g)
4.3 mol
? mol
=
2.2 mole O2
gram ↔ mole and gram ↔ gram conversions
When N2O5 is heated, it decomposes:
2N2O5(g)  4NO2(g) + O2(g)
a. How many moles of N2O5 were used if 210g of NO2 were produced?
2N2O5(g)  4NO2(g) + O2(g)
? moles
210g
2 moles N2O5
?? moles N2O5
(46 x4)g NO2 = 2.28
moles N2O5
210 g NO2
b. How many grams of N2O5 are needed to produce 75.0 grams of O2?
2N2O5(g)  4NO2(g) + O2(g)
75.0 g
? grams
= 506 grams N2O5
Concentration
Extensive quantities such as mass and volume are often used when
preparing solutions. We can talk about adding a teaspoon of
sugar when we make ice tea, for example, or filling a 2-L pitcher
with ice tea.
But when we want to compare solutions, we need intensive
quantities that tell us how much sugar has been added to a given
volume of lemonade or ice tea. We need to know the
concentration of the solution, which is the ratio of the amount of
solute to the amount of either solvent or solution.
Percent Concentration
Will be Studied in practical part
It is
w/w
concentration (g / g)
w/v
concentration (g / ml)
v/v
concentration (ml / ml)
Molarity
Molarity is a term used to express
concentration. The units of molarity are
moles per liter (It is abbreviated as a
capital M)
M = number of moles / Volume (L)
What is the number of moles in 50 ml
of 6 M HCl
Normality
Normality is a term used to express concentration. The units of normality
are number of equivalents per liter (It is abbreviated as a capital N)
N = Number of equivalent / Volume (L)
Equivalent weight
• equal the molecular weight / e
– No of H in acids
– No. of OH in base
– No. of acid molecules that react with basic salt,
and No. of base molecules that react with acidic
salt.
Compound
e
Compound
e
HCl
H2SO4
H3PO4
NaOH
Ca(OH)2
1
2
3
1
2
CaCl2
Na2HPO4
Na2SO4
KMnO4
H2O2
2
1, 2
2
1,3,5
2
Na2CO3
2
Calculate the equivalent weight of each of the
following
HCl, H2SO4, CaOH, Li2CO3
HCl = molecular wt/ 1 = (1+35.5)/1 = 36.5
You must learn how to prepare
1- percent solution
2- Molar solution
3- Normal solution
4- Transform between them
1- w/w % to Molarity or Normality
M = 10pd/M. wt. (Eq. wt.)
2- w/v % to Molarity or Normality
M = 10 p/M. wt. (Eq. wt.)
3- Molarity to Normality
M = N/e
Dilution
To dilute any concentrated solution use the relation
C X V (before dilution) = C X V (after dilution)
What is the needed volume of concentrated
HCl (11 M) to prepare 500 ml 1 M solution
Solution Stoichiometry
50.0 ml of 6.0 M H2SO4 were spilled and solid NaHCO3 is
to be used to neutralize the acid. How many grams of
NaHCO3 must be used?
H2SO4(aq) + 2NaHCO3  2H2O(l) + Na2SO4(aq) + 2CO2(g)
50.0 ml
? g Our Goal
6.0 M
=
6.0 mol
L
Look!
A conversion
factor!
Solution Stoichiometry
H2SO4(aq) + 2NaHCO3  2H2O(l) + Na2SO4(aq) + 2CO2(g)
50.0 ml
? g Our Goal
6.0 M
=
6.0 mol
L
6.0 mol H 2SO 4
H2SO4
50.0 ml
1000mL
H 2SO 4
NaHCO3
2 mol
1 mol
H2SO4
NaHCO3
84.0 g
= 50.4 g NaHCO3
mol
NaHCO3
Solution Stoichiometry:
Determine how many mL of 0.102 M NaOH solution are needed to
neutralize 35.0 mL of 0.125 M H2SO4 solution.
2
1 2SO4 
____NaOH
+ ____H
2 2O
____H
First write a balanced
Equation.
1 2SO4
+ ____Na
Solution Stoichiometry:
Determine how many mL of 0.102 M NaOH solution is needed to neutralize
35.0 mL of 0.125 M H2SO4 solution.
N X V (acid) = N X V (base)
N of NaOH = M X e = 0.102 X 1
N of H2SO4 = M X e = 0.125 X 2
0.102 X V = 0.250 X 35
V = 85.78 ml
Solution Stoichiometry
What volume of 0.40 M HCl solution is needed to
completely neutralize 47.1 mL of 0.75 M Ba(OH)2?
1st write out
a balanced chemical
equation
Solution Stoichiometry
2HCl(aq) +
Ba(OH)2(aq) 
0.40 M
47.1 mL
0.75 M
? mL
2H2O(l) + BaCl2
N X V (acid) = N X V (base)
Solution Stochiometry Problem:
48.0 mL of Ca(OH)2 solution was titrated with 19.2
mL of 0.385 M HNO3. Determine the molarity of
the Ca(OH)2 solution.
We must first
write a balanced
equation.
Solution Stochiometry Problem:
48.0 mL of Ca(OH)2 solution was titrated with 19.2
mL of 0.385 M HNO3. Determine the molarity of
the Ca(OH)2 solution.
Ca(OH)2(aq) + 2 HNO3(aq)  2 H2O(l) + Ca(NO3)2(aq)
48.0 mL
?M
19.2 mL
0.385 M
0.385 mol

L
Equivalent weight
NaOH + HCl
Na2CO3 +2 HCl
NaCl + H2O
2 NaCl + H2O +
CO2
Theoretical yield vs. Actual yield
Suppose the theoretical yield for an
experiment was calculated to be
19.5 grams, and the experiment was
performed, but only 12.3 grams of
product were recovered. Determine
the % yield.
Theoretical yield = 19.5 g based on limiting reactant
Actual yield = 12.3 g experimentally recovered
actual yield
% yield 
x 100
theoretica l yield
% yield 
12.3
x 100  63.1% yield
19.5
Try this problem (then check your answer):
Calculate the molarity of a solution prepared by dissolving 25.6 grams of
Al(NO3)3 in 455 ml of solution.
25.6 g mole
mol
 0.264
-3
213 g 455 x 10 L
L
Element’s percent
• If the bacterial medium additives listed below are
priced according to their nitrogen content, which
will be the least expensive per 50 kg. bag?
NH4Cl, (NH4)2SO4, NH2CONH2
Percent of element in its compounds: Simply to calculate
the percent of any element in a compound
Atomic weight of the element X number of element X 100
Molecular weight of the compound
Then to calculate the Nitrogen percent in
NH4Cl
1- M. wt of salt = 14 + 4+ 35.5 = 53.5 g
2= 14 X 1X100
= 26.1%
53.5
Gas’s Law
•
Boyle's Law: for a fixed amount of gas at
a constant temperature, the volume of the
gas varies inversely with its pressure
PV = constant
P1 V1 = P2 V2
•
Charles's Law: the volume of a fixed
amount of a gas at constant pressure is
directly proportional to its temperature
(Kelvin)
V/T = constant
V1/ T1= V2/ T2
•
Avogadro's Law: At fixed temperature
and pressure, the volume of a gas is
directly proportional to the amount of the
gas (either No. of moles or No. of
molecules)
V/n = constant
V1/ n1= V2/ n2
One mole of an ideal gas at STP occupies 22.4 liters.
The combined gas law( Ideal gas)
•
•
•
PV/nT = constant = R
P1V1/N1T1 = P2V2/N2T2
R = 0.082058 L atm /mol K
= 62.364 L Torr/ mol K
= 8.3145 j/mol K
Na2CO3 + H2SO4
Na2SO4 + CO2 + H2O
•How many grams of Na2CO3 are needed to neutralize 0.5
mole H2SO4
•How many moles of H2O are produced when we start
with 200 g Na2CO3
•How many liters of CO2 are produced when we start with
250 ml of 2N H2SO4 (at STP).
•How many moles of Na2CO3 are consumed when we need
to produce 200 L of CO2 at ( 20oC and 1 atm).
•What is the volume of 6 M H2SO4 was needed to
neutralize 106 g Na2CO3