Faculty of Allied Medical Sciences
General Chemistry
(MGGC-101)
Chemical stoichiometry
Supervision:
Prof.Dr.Shehata El-Sewedy
Dr.Fatma Ahmed
Outcomes
By the end of this lecture, the
students will be able to
1-Understand the Atomic mass
2-learn to Molecular mass
3-To differentiate between emprical and
molecular formula
4-Recognize mole of atoms
5-To know determination of empirical formula
6-learn to determination of molecular formula from
emprical forumla
Atomic mass:
Atomic mass of an atom is its mass in atomic mass
unit (amu)
Mass of 12C atom is exactly 12 amu
Average mass of F atom is 18.998 amu ( ~19amu)
Molecular mass:
It is mass of the molecule in atomic mass
unit, it the sum of atomic masses of all the
atoms in that molecule.
Example:
Calculate the molecular mass of CHCl3?
Molecular mass= 1X atomic mass of C+ 1X
atomic mass of H + 3X atomic mass of Cl
= 1X 12.01 + 1X1.008 + 3X35.5
= 119.37 amu
Example:
Calculate the molecular mass of KClO3?
molecular mass = 1X 39.098 + 1X 35.5 + 3X 15.999
= 122.548 amu
Mole of atoms:
Unit used for describing large number of atoms, ions and
molecules.
Mole of an object contains avogadro's No of this object.
e.g. 1 mol C = 6.023  1023 atom
1 mol H2O = 6.023  1023 molecule.
1 mol NO3 = 6.023  1023 ions.
*One mole of any substance contains
Avogadro’s number (6.02 x 1023) of
atoms or molecules.
Example :
Calculate the mass of 12C atom?
1 mol of 12C contains 6.022X 1023 atom
6.022X 1023 atom weights
1 atom
weights
12 gm
X gm
X gm = 12X 1/ 6.022 X 1023 gm
= 1.993 X 10-23 gm
Note:
No. of particles (atoms, molecules & ions) =
no. of moles X Avogadro`s no.
No. of moles = mass of substance/ its atomic mass
Weight/molecular Weight
Exercise:
Calculate the No of C-atoms in 0.35 mol of C6H12O6?
1 molecule C6H12O6  6 C-atoms
1 mol C6H12O6  6.023  1023 molecules
0.35 mol

X
molecules
No. of molecules of C6H12O6 = 0.35 x 6.023  1023
 No. of C atoms = 0.35 x 6.023  1023  6
= 1.26  1024 atom
Exercise:
What is the mass in grams of 1 mol glucose C6H12O6
Mwt = (6  12) + (12  1) + (6  16)
= 180 amu
1 mol= mass /180
Mass = 1 mol X 180 = 180 gm
 1 mol of C6H12O6 weight 180 grams.
Exercise:
How many Cl atoms are contained in 18.29 gm Cl?
1mol of Cl  35.5 gm of Cl
X mol of Cl  18.29 gm of Cl
X = 18.29/35.5= 0.5159 mol of Cl atom
No. of atoms= no. of moles X Avogadro`s no.
= 0.5159 X 6.023  1023
= 3.107  1023 Cl atoms.
Exercise:
Calculate the no. of grams found in 4.6 X10-4 mol of vitamin C
(C6H8O6) (molar mass= 176.1 gm)?
No. of moles = mass / mol. Mass
mol. Mass = gm/ mol
4.6 X10-4 mol = mass/ 176.1 gm.mol-1
Mass = 4.6 X10-4 mol X 176.1gm.mol-1
= 0.081 gm
Exercise:
How many moles of H –atoms are contained in 500 mg of
vitamin C (C6H8O6)?
500 mg = 0.5 gm
No. of moles of vitamin C in 500 mg = 0.5 gm/ 176.1 gm.
mol-1
= 2.84 X 10-3 mol
1mol of vitamin C  8 moles of H
2.84 X10-3 mol
 ?? moles of H
?? mol of H = 2.84 X10-3 X 8 = 2.27 X 10-2 mol
Determining Chemical Formulas
In order to determine a chemical formula, the mass
percent of each element in the compound must be
determined.
Mass percentage (percent composition)
The percent composition of a compound is the mass
percentage of each element in the compound. We define
the mass percentage of element “A” as the parts of
“A” per hundred parts of the total, by mass.
That is,
mass of " A" in whole
mass % " A" 
100%
mass of the whole
You can look at the mass percentage of A as the
number of grams of A in 100 g of the whole.
Example:
Calculate the percentage composition of C12H22O11?
Answer:
You can calculate %O by the same way or be
subtracting the %H and C from 100%:
% O = 100 – (42.1 +6.4) = 51.5 %
Example:
Formaldehyde, CH2O, is a toxic gas with a pungent odor.
Large quantities are consumed in the manufacture of
plastics, and a water solution of the compound is used to
preserve biological specimens. Calculate the mass
percentage of the elements of formaldehyde.
Answer:
% O = 100 - (40.0+6.7%) = 53.3
Types of Chemical Formula
1- Emprical Formula
Represent the relative number of each
atom in a molecule.
e.g. H2O2 has empirical formula HO
i.e. 1 H : 1O
e.g. Glucose C6H12O6  CH2O
1 C : 2 H : 1O
2- Molecular Formula
Represent the actual number of each
atom in a molecule.
e.g. C6H6 has 6 C-atoms and 6 H-atoms
Determination of Empirical Formula
Ex:
A compound composed of C, H, O with the weights
7.2, 1.2 and 9.69 grams respectively.
What is the empirical formula?
Answer:
C

Mass (grams) : 7.2
Moles (grams/at wt):
7.2/12
0.6

Divide by
(the least no.):
1
H
1.2
1.2/1
1.2
2
CH2O
O
9.69
9.69/16
0.6
1
Ex:
Compound composed of C H O with mass %
38.71, 6.44 and 51.58 respectively. What is the
empirical formula?
Answer:
C
H
O
 Assume 100 grams: 38.71
6.44
51.58

Moles:
3.22
6.44
3.22
 % by 3.22:
1
2
1
CH2O
Determination of molecular formula from emprical forumla
M.F = n  E.F
Where n =
M .F wt
E.F wt
Ex:
Compound composed of C, H, O has mass % of 26.7, 2.2 and
71.1 respectively and has molecular weight 90 g/mol. What is its
molecular formula?
Answer:
C
H
O
Assume 100 grams: 26.7
2.2
71.1 g
Moles:
2.22
2.2
4.4
% by 2.2:
1.01
1
2
CHO2
n=
M .F wt
E.F wt
=
M.F = 2  CHO2
= C2H2O4
90 = 2
45
Example:
Compound composed of Na, Cr, O with mass % 17.5, 39.7
and 42.8 respectively. What is the empirical formula?
Answer:
Na
Cr
O
 Assume 100 grams:
17.5
39.7
42.8
 Moles:
0.761
0.763
2.68
 divided by the least number, 0.761:
1
1
3.52
according to this, we get Na1Cr1O3.5, in this case the subscripts are not all
integers. They can be made into integers my multiplying each one by 2; then we get
the empirical formula Na2Cr2O7.
Quiz time
1-Calculate the molecular mass of Al2(SO4)3.
2-How many atoms or molecules present in
a)1.0 mol of Na
b)0.25 mol of hydrogen
c)2.781 mol of KNO3
3-How many grams in
a)0.25 mol H2O
b)2 mol H2
4-How many moles present in 9.8 gm of H2SO4
5-Compound composed of Na, Cr, O with mass %
17.5, 39.7 and 42.8 respectively. What is the
empirical formula?
6-Compound composed of C, H, O has mass % of
26.7, 2.2 and 71.1 respectively and has molecular
weight 90 g/mol. What is its molecular formula?
7-Formaldehyde, CH2O, is a toxic gas with a pungent odor.
Large quantities are consumed in the manufacture of
plastics, and a water solution of the compound is used to
preserve biological specimens. Calculate the mass
percentage of the elements of formaldehyde.
Student Question
* Compound composed of C, H, O has mass % of 26.7, 2.2 and 71.1
respectively and has molecular weight 90 g/mol. What is its
molecular formula?
* Compound composed of C H O with mass % 38.71, 6.44 and
51.58 respectively. What is the empirical formula?
* Calculate the percentage composition of C12H22O11?
* Calculate the No of C-atoms in 0.35 mol of C6H12O6?
* How many Cl atoms are contained in 18.29 gm Cl?
Assignments
Group A and Group
B
Emprical Formula
Samar Mostafa- Omnia MosaadEsraa Kamel- Reem khaledShrouk Essam
Molecular Formula
Ahmad Rada Kamel –
Ahmad Mohamad Abd El SalamAhmad El Said- Ahmad Alaa
Elemental analysis
Sara Lotfy- Shimaa Saied – Esraa
Sameh-Ahmad Shadra-Ibrahim ElHabet- Ibrahim El Said
Mass spectroscopy
Aya Ibrahim – Shreen Hassan –
Mohamad Ahmad Torky- Mohamad
Montaser- Doaa Sabri
RECOMMENDED TEXTBOOKS:
1-Raymond Chang. Chemistry. 10th ed.
2009
2-Zumdehl. International edition. 2009