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Chemistry 111 for
Engineering students
Dr. Ayman H. Kamel
Office: 33
Chapter 3
Stoichiometry: Chemical Calculations
Objectives.
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Define Atomic masses.
Molecular Masses and Formula Masses.
Define the mole.
Calculate the number of moles.
Calculate the molar mass of any compound.
Determine the percent composition of
compounds.
Determine the formula of compounds.
Balance chemical equations.
Solutions and solution stoichiometry
3
3.1.Molecular masses [ Molecular Weight]
Molecular mass is the sum of the masses of the atoms
represented in a molecular formula.
 Ex.1.
M. Wt. of O2= 2x atomic mass of O
= 2 x 15.9994 u = 31.9988u.
Ex. 2.

Calculate the molecular mass of glycerol CH2(OH)CH(OH)CH2OH
3 x atomic mass of C = 3 x 12.011u = 36.033u.
3 x atomic mass of O = 3 x 15.9994u = 47.9982u.
8 x atomic mass of H = 8 x 1.00794u = 8.06352u.
Molecular Mass of C3H8O3 = 92.095u.
4
3.2.The Mole
A mole is an amount of substance that contains
as many elementary entities as there are atoms
in exactly 12 g of C12.
Avogadro’s Number:

is the number of carbon atoms in exactly 12 grams of
pure C12. This No. was found to be 6.022 x 1023.
SO,
 ONE MOLE OF SOMETHING CONSISTS OF
AVOGADRO’S NUMBER OF THAT SUBSTANCE i.e.
6.022 x1023 units of that substance.

The mass of 1 mole of an element is equal to its atomic
mass in grams.
5
Example 1
Aluminum (Al) is a metal with a high strength to mass
ratio and high resistance to corrosion. Compute both
the number of moles of atoms and the number of atoms
in a 10 g sample of aluminum.
Solution
The mass of 1 mole (6.022x1023 atoms) of aluminum is
26.98 g.
1 mole--------------------------------------------26.98 9
? mole--------------------------------------------10.0 g
10.0 x 1/ 26.98 = 0.371 mol Al atoms
1 mole-------------------------------------------6.022x1023atoms
0.37 mole Al-----------------------------------? atoms
The no. of atoms in 10.0 g (0.37 mol) of aluminum is
0.371 x 6.022x1023/1 =2.23x1023

6
Example 2
Cobalt Co is metal that is added to steel to improve its
resistance to corrosion. Calculate both the number of
moles in a sample of cobalt containing 5.00x1020 atoms
and the mass of the sample.
Solution
1 mole Co-------------------------------------6.022x1023 atoms Co
? mole------------------------------------------5.00x1020 atoms Co
No of mole of Co is = 5.00x1020 x1/6.022x1023 = 8.30x10-4

Since the mass of 1 mole of cobalt is 58.93
1 mole Co------------------------------------------------------58.93 g
8.30x10-4 mole Co-------------------------------------------?g
The mass of Co is = 58.93 x 8.30x10-4/1 = 4.89x10-2 g Co.
7
Example 3
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A silicon chip used in an integrated circuit of a
microcomputer has a mass of 5.68 mg. How many
silicon (Si) atoms are present in the chip?
Solution
First we have to convert weights to g , so the weight of
silicon chip is 5.68x10-3 g.
1 mol Si --------------------------------------------28.09 g Si
? Mol------------------------------------------------5.68x10-3 g Si.
No of moles of Si is 5.68x10-3 x1/28.09 = 2.02x10-4 mol Si.
1mol Si---------------------------------------------6.022x1023 atoms
2.02x10-4 mol Si-------------------------------- ? Atoms
The no of silicon atoms is
2.02x10-4 x6.022x1023/1 = 1.22x1023 atoms.
8
3.3.Molar Mass
A chemical compound is a collection of atoms,
for example , methane consists of molecules that
each contain one carbon and four hydrogen atoms
(CH4).
How to calculate the mass of 1 mole of methane?
 Mass of 1 mol C =
12.01 g
 Mass of 4 mol H = 4x 1.008 g
 Mass of 1 mol CH4 = 16.04 g
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Thus the molar mass of a substance is the
mass in grams of one mole of the compound
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Example 1
For the natural dye, Jug lone, C10H6O3,Calculate the molar
mass of jug lone. How many moles of jug lone are
found in a 1.56x10-2 g sample of pure dye?
Solution
The molar mass can be obtained by summing the masses
of the component atoms, so, we have
10 C: 10 x 12.01 g = 120.1g
6 H: 6 x 1.008 g = 6.048 g
3 O: 3 x 16.00 g = 48.00 g
Mass of I mol C10H6O3 = 174.1 g.
1 mol jug lone--------------------------------------174.1 g
? mol------------------------------------------------- 1.56x10-2 g juglone
 No. of moles of jug lone is 1.56x10-2 x 1/174.1
= 8.96x10-5 mol juglone.
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Example 2
Calculate the molar mass of calcium carbonate CaCO3
(calcite). For a certain sample of calcite containing
4.86 moles, what is the mass in grams of this sample?
And what is the mass of carbonate ions present.
Solution
The molar mass Ca(40) + C (12) + O (3x16 =48) = 100.09 g
1 mol--------------------------------------100.09 g
4.86 mol--------------------------------------? g
The no. of moles of Calcium carbonate is 4.86 x1/100.09 = 486 gm
CaCO3
2
To find the mass of CO3 present in this sample, we must realize that
4.86 moles of CaCO3 contains 4.86 moles of Ca2+ ions and 4.86
moles of carbonate ions.
The mass of 1 mole carbonate ion is C (1x 12.01) + O (3x16.00 =48.00)
= 60.01 g
 1 mol carbonate ion--------------------------------------------------60.01 g
 4.86 mol -----------------------------------------------------------------?
 The mass of 4.86 moles is 4.86 x 60.01/1 = 292 g
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3.4.Percent composition of compounds
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There are two common ways of describing the
composition of a compound. In terms of the
number of its constituent atoms and in terms
of the percentage (by mass ) of its elements.
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The mass percents of the elements can be
obtained by comparing the mass of each
element present in 1 mole of the compound to
the total mass of 1 mole of the compound.
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Example 1
Carvone is a substance with the molecular formula
(C10H14O), compute the mass percent of each element in
carvone.
Solution
Mass of C = 10 x 12.01 = 120.01 g
Mass of H = 14 x 1.008 = 14.11 g
Mass of O = 1 x 16.00 = 16.00 g
 Mass of 1 mol (C10H14O) = 120.01 + 14.11 + 16.00 = 150.2g
So the mass percent of each component will be:
Mass percent of C= 120.1/150.2 x 100% = 79.96%
Mass percent of H = 14.11/150.2 x 100% = 9.394%
Mass percent of O = 16.00/150.2 x 100% = 10.65%
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Example 2
Penicillin, has the formula C14H20N2SO4. Compute the mass percent of
each element.
Solution
The molar mass of penicillin is computed as follows,
C: 14 x 12.01 = 168.1 g
H: 20 x 1.008 = 20.16 g
N: 2 x 14.01 = 28.02 g
S: 1 x 32.07 = 32.07 g
O: 4 x 16.00 = 64.00 g
 Mass of 1 mol C14H20N2SO4 = 312.4 g , then,
Mass percent of C: 168.1/312.4 x 100% = 53.81%
Mass percent of H: 20.16/312.4 x 100% = 6.543%
Mass percent of N: 28.02/312.4 x 100% = 8.969%
Mass percent of S: 32.07/312.4 x 100% = 10.27%
Mass percent of O: 64.00/312.4 x 100% = 20.49%
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3.5.Determining the formula of a compound
The formula of any substance can either be represented
by,
Empirical formula,
which is the simplest whole number ratio of atoms in a
compound (CnHnOn..)
or
Molecular formula,
which is the exact formula of a molecule, giving the types
of atoms and the exact number of each type.
(CnHnOn….)n
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Example 1
Determine the empirical and molecular formulas for a compound that
gives the following analysis in mass percents: 71.65% Cl, 24.27%C,
4.07%H, The molar mass is known to be 98.96 g/mol.
Solution
From the above given values we can say that in a 100 g sample of the
compound we have 71.65 g Cl, 24.27 g C and 4.07 g H, so to calculate
the no of moles of each component we should divide each by its atomic
mass So,
No of moles of Cl is = 71.65/35.45 = 2.021 mol Cl
No of moles of C is = 24.27/ 12.01 = 2.021 mol C
No of moles of H is = 4.07/1.008 = 4.04 mol H
To get the empirical formula we should divide each of the above
values by the smallest number of moles that is (2.021)
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Example 1 (cont.)
2.021/2.021 CL :2.021/2.021 C : 4.04/2.021H
1 Cl: 1C: 2H
(notice that you should approximate the values to the nearest whole
numbers, i.e. no decimal values should be written the formula)
So the empirical formula can be represented as ClCH2.
To determine the molecular formula, we must compare
the empirical formula mass to the molar mass.
The empirical formula mass is 49.48 (35.45+12.01+2(1.008)).,
Molar mass/empirical formula mass = 98.96/49.48 = 2
 Molecular formula is (ClCH2)2 = Cl2C2H4
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Example 2
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Caffeine, is a stimulant found in coffee, tea and
chocolate, contains 49.48% carbon, 5.51% hydrogen,
28.87% nitrogen and 16.49% oxygen by mass and has a
molar mass of 194.2 g/mol. Determine the molecular
formula of caffeine.
Another way of solving molecular formula problems:
We will first determine the mass of each element in 1 mole (194.2 g) of
caffeine,
Mass of C = 49.48/100 x 194.2 = 96.09 g/mol C
Mass of H = 5.15/100 x 194.2 = 10.0 g/mol H
Mass of N = 28.87/100 x 194.2 = 56.07 g/mol N
Mass of O = 16.49/100 x 194.2 = 32.02 g/mol O
Now we will convert to moles,
No of moles of C = 96.09/12.01 = 8.001 mol C/mol caffeine.
No of moles of H = 10.0/1.008 = 9.92 mol H/ mol caffeine.
No of moles N = 56.07/14.01 = 4.002 mol N / mol caffeine.
No of moles of O = 32.02/16.00 = 2.001 mol O/ mol Caffeine.
the molecular formula of caffeine can be given as: C8H10N4O2

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3.7.Writing and balancing Chemical equations
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Chemical reactions are represented by chemical
equations, which identify reactants and products.
Formulas of reactants appear on the left side of the
equation and those of products are written on the
right.
In a balanced chemical equation, there are the
same number of atoms of a given element on
both sides.
You can’t write an equation unless you know what
happens in the reaction that it represents. So all
the reactants and products should be identified.
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Example 1
At 1000 °C, ammonia gas, NH3 , reacts with oxygen gas to form gaseous nitric
oxide, NO and water vapour. Balance the equation for this reaction.
Solution
The unbalanced equation for the reaction is
NH3 + O2
NO + H2O
To start balancing, we can first chose hydrogen, as follows
2NH3 + O2
NO + 3H20
The nitrogen can be balanced by a coefficient of 2 for NO
2NH3 + O2
2NO + 3H2O
Finally note that there are two atoms of oxygen on the left and 5 on the right so
oxygen can be balanced with a coefficient 5/2 for O2
2NH3 + 5/2O2
2NO + 3H2O
However, the usual custom is to have whole number coefficients, so we multiply the
entire equation by 2
4NH3 + 5O2
4NO + 6H2O
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3.8. Reaction Stoichiometry
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The principal reason for writing balanced equations is to make
it possible to relate the masses of reactants and products,
taking into consideration that: “the coefficient of a balanced
equation represent numbers of moles of reactants and
products”.
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For the reaction,
2N2H4 + N2O4
3N2 + 4H2O
2 molecules N2H4 + 1 molecule N2O4 gives 3 molecules N2 +
4 molecules H2O.
A balanced equation remains valid even if each
coefficient is multiplied by the same number including
Avogadro’s number:
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2AN molecules N2H4 + 1AN molecule N2O4 gives 3AN
molecules N2 + 4AN molecules H2O.
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Example 1
Ammonia can be prepared by the following reaction
N2 + 3H2
2NH3
Determine
The number of moles of ammonia formed when 1.34 mol of N2 reacts.
The mass in grams of N2 required to form 1.00x103 g NH3.
The number of molecules of ammonia formed when 1.34 g of H2
reacts.
Solution
The coefficients of the balanced equation shows that each
mol of N2 produces 2 moles of ammonia thus 1.34 mol N2
should produce 2.68 mol NH3.
I mole N2-------------------------------------------------2moles NH3
(2 x 14.01) = 28.02g----------------------------------2(14.01 +3 x1.008) = 34.06g
? g--------------------------------------------------- 1.00x103 g NH3
Mass of N2 = 1.00x103 x 28.02/ 34.06 = 823 g N2
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Example 1(Cont.)
3 mol H2-------------------------------------------2 mol NH3
3( 2 x 1.008) = 6.048 g------------------------ 2 (14.01 + 3 x 1.008) =34.02 g
1.34 g H2------------------------------------------? g NH3
Mass of NH3 = 34.02 x 1.34/6.048 = 7.537 g
To get the number of molecules,
1 mol [17 gm]------------------------------------------------------6.022 x 1023 molecule
7.537-gm-----------------------------------------------------?
The number of molecules is =
7.537 x 6.022x1023 = 2.67 x 1023 molecule NH3
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Example 2
Solid lithium hydroxide is used in space vehicles to remove
exhaled carbon dioxide by forming lithium carbonate and water.
What mass of carbon dioxide can be absorbed by 1kg of lithium
hydroxide?
Solution
LiOH + CO2
Li2CO3 + H2O
We should first balance the equation
2LiOH + CO2
Li2CO3 + H2O
1 mol LiOH-------------------------------------23.95 g LiOH
? mol-------------------------------------------------1000g
Mass of LiOH is 1000 x 1/23.95= 41.8 mol LiOH

From the equation 1 mol of CO2 can be produced from 2 moles LiOH
1 mol CO2--------------------------------------------------2 mol LiOH
(44.0)---------------------------------------------------------(2 x 23.95)
?g--------------------------------------------------------------41.8 LiOH
The mass of CO2= 44.0 x 41.8/2x23.95 = 920 g CO2
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3.11 Solutions and solution Stoichiometry
Molarity, M:
i. The Molecular Weight, MW:
It is the summation of the atomic weights
of all the atoms composing the molecule
● Example:
MW of NaOH = 23 + 16 + 1 = 40
Where: at wt of Na = 23, O = 16, and H = 1
iv. Molarity:
It is the Number of Moles of solute per
one liter of solution
wt / MW
No of moles of solute
M=
=
VL
sol.n Volume in Liters
 M x VL = wt of solute / its MW

We want to prepare a 6.68 molar solution of NaOH ( 6.68 M NaOH).
A- How many moles of NaOH are required to prepare 0.500 L of 6.68 M
NaOH?
B- How many liters of 6.68 M NaOH can we prepare with 2.35 kg NaOH?
Solution
A. moles of NaOH = 0.500 L x 6.68 mol NaOH/ 1 L solution = 3.34 mol
NaOH
B. gm NaOH/ Molar mass of NaOH = M x Volume (L)
2.35 x 1000 gm / 40 = 6.68 x VL
VL = 8.79 L
Exercise :
How many grams of 1- butanol, CH3CH2CH2OH, are required to
prepare 725 mL of a 0.350 M aqueous solution of this solute?
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 Dilution:
By diluting a solution, the number of moles
doesn’t change as we don’t increase or decrease
the amount of solute
 no of moles of solute before dilution = no of
moles of solute after dilution
,∵ no of moles = wt(g)/MW = M x VL
 (M x V)Before dilution = (M’ x V’)After dilution
 Dilution:
● Example:
Calculate the molarity of a solution prepared by
diluting 10 ml of 0.1 M with 90 ml of water
- Answer:
∵ (M x V)Before dilution = (M’ x V’)After dilution
 (0.1 x 10) = (M’ x (90+10))
 M’ = 1/100 = 0.01 moles/L