Department of Industrial Chemistry
Chemistry for Nursing (Slides)
by
Dr. Faady Siouri
Fall 2022
Department of Industrial Chemistry
Chemistry for Nursing (Slides)
by
Dr. Faady Siouri
Chapter 1: Introduction
Fall 2022
Dr. Faady Siouri
Sec. 1.3: The International System of Units (SI)
• The SI specifies a set of seven base units shown in Table 1.2
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Derived Units
There are many quantities that did not appear in the previous table.
Units for such quantities are obtained by appropriate combinations of
the base units are called derived units.
Example:
 Area of a rectangle = length × width
m × m = m2
Distance meter m
=
=
 Speed =
Time
second s
Dr. Faady Siouri
Working With Larger and Smaller Units
• Larger and smaller units are obtained by modifying the basic units with decimal
factors and prefixes.
Example
Here is another way of writing
the exact same thing.
1 m = 1 x 10-12 terameter (Tm)
1 m = 1 x 10-9 gigameter (Gm)
1 m = 1 x 10-6 megameter (Mm)
1 m = 1 x 10-3 kilometer (km)
1 m = 10 decimeter (dm)
1 m = 100 centimeter (cm)
1 m = 1000 millimeter (mm)
1 m = 1 x 106 micrometer (μm)
1 m = 1 x 109 nanometer (nm)
1 m = 1 x 1012 picometer (pm)
Dr. Faady Siouri
Conversion Among SI Units
Example 1.1: A desk is found to be 1437 mm wide, what is this width expressed in meters?
Conversion problems can be solved in different ways:
Directly (using information from table 1.3)
We know from table 1.3 that 1mm = 0.001 m = 1 x 10-3 m
1) Cross multiplication
1 mm
= 1 × 10-3 m
1437 mm = x (?)
2) Bracket multiplication
 1103 m 
1437 mm 
  1.437 m
 1 mm 
OR


1m
1437 mm 
  1.437 m
3
 110 mm 
x = 1.437 m
Either one works; chose
the one you are more
comfortable with
Dr. Faady Siouri
Indirectly: When the table is not provided or it does not give a direct conversion, use your general
information to do the conversion. WHEN IN DOUBT, TAKE THE EXTRA STEP.
We know that 10 mm = 1 cm
and
100 cm = 1 m
1) Cross multiplication
10 mm
= 1 cm
1437 mm = y (?)
y = 143.7 cm
100 cm
= 1m
143.7 cm = x (?)
x = 1.437 m
2) Bracket multiplication
 1 cm   1 m 
1437 mm 

  1.437 m
 10 mm   100 cm 
**I find this to be the best and safest way but
again, it is your choice.
Dr. Faady Siouri
Example 1.2: A certain person is 172 cm tall. Express this height in
decimeters.
Table 1.3 does not give a direct conversion between cm and dm.
We know that 1 cm = 0.01 m and 1 dm = 0.1 m
Therefore, it is better to convert indirectly.
 0.01 m   1 dm 
172 cm 

  17.2 dm
 1 cm   0.1 m 
OR:
 1 m   10 dm 
172 cm 

  17.2 dm
 100 cm   1 m 
Dr. Faady Siouri
Example 1.3: Calculate the number of cubic centimeters in 0.225 dm3.
We know that 1 cm = 0.01 m
3
 0.1 m 
0.225 dm 

 1 dm 
3
3

10
m
0.225 dm3 
3
1
d
m

3
and
1 dm = 0.1 m
3
 1 cm 


 0.01 m 
  1 cm3

6
3
10
m



3

2
2
5
c
m


OR:
3
 1m 
0.225 dm 

10
dm


3

1
m
0.225 dm3  3
3
1
0
d
m

3
3
 100 cm 


1
m


  106 cm3

3
1
m



3

2
2
5
c
m


Dr. Faady Siouri
Units for Laboratory Measurements
Mass: gram
1 kg = 1000 g
Length: meter
1 m = 100 cm
1 m = 1000 mm
1 cm = 10 mm
Volume: liter
1 L = 1 dm3 = 1000 cm3 = 1000 mL
1 cm3 = 1 mL
Dr. Faady Siouri
Practice Question: A piece of gold weighs 2500 mg, what is this mass
expressed in kg?
Dr. Faady Siouri
Temperature
Temperature is commonly measured with thermometer.
Two reference temperatures are chosen to make the markings on the scale of a
thermometer.
The reference temperature used for defining common scales are freezing point and
boiling point of H2O.
• In Celsius scale, the normal freezing point and boiling point of water are 0°C and 100°C
respectively.
• In Fahrenheit scale, the normal freezing point and boiling point of water are 32°F and
212°F respectively.
• In Kelvin scale (SI base unit), the normal freezing point and boiling point of water are
273.15 K and 373.15 K respectively.
Dr. Faady Siouri
Comparison of the Three Temperature Scales
Dr. Faady Siouri
Temperature Conversion Formulas
o

5
C 
o
o
o
C  ( F - 32 F )   o 
9 F 
If you solve for oF:
o

9 F 
o
o
o
F   o   ( C )  32 F
5 C 
Memorize
 1K 
K  ( C  273.15 C )  o 
1 C 
o
o
** Note: The red symbols are the variables **
Dr. Faady Siouri
Practice Question: Assume the temperature in this room is 300 K, what
is this temperature expressed in Fahrenheit and Celsius?
Dr. Faady Siouri
English vs Metric System
• Table 1.4 on page 12 shows a comparison of the English and metric systems.
 Length Measurements:
1 inch (in) = 2.54 cm
0.39 inch = 1 cm
1 foot (ft) = 30.48 cm
3.28 feet = 1 meter
1 yard (yd) = 0.91 meter
1.09 yards = 1 meter
1 mile (mi) = 1.61 km
0.62 miles = 1 km
No Need to
Memorize
 Weight Measurements:
1 ounce (oz) = 28.35 grams
0.035 ounces = 1 grams
1 pound (lb) = 0.45 kg
2.21 pounds = 1 kg
Dr. Faady Siouri
Example 1.4: Calculate the number of meters in 0.200 mile using the
information given below.
Hint: 1 mile = 5280 ft
1 ft = 12 inches
1 m = 39.37 inches
 5280 ft  12 in  1 m 
0.200 mi 


  322 m
 1 mi  1 ft  39.37 in 
Dr. Faady Siouri
Practice Question: How many cubic millimeters are there in a room
measuring 5m x 10m x 2m?
Dr. Faady Siouri
Matter and its Properties
• Matter: Anything that takes up space and has mass.
• Mass: The amount of matter in an object.
• Weight: A measure of the force with which an object of a given mass is
attracted by gravity.
• Weight is measured in units of Newton.
Mass is constant and does not depend on where the object is but the weight
does.
Mass is measured in units of grams (or kilograms).
Dr. Faady Siouri
Example 1.7: An aluminum bar was weighed and found to have a mass of
14.2 g. Its volume was measured to be 5.26 cm3. What is the density of
aluminum?
Mass
14.2 g
g
Density 

 2.70 3
Volume 5.26 mL
cm
Example 1.8: A certain copper coin has a mass of 3.14 g. The density of
copper is 8.96 g/cm3. What is the volume of the coin?
** The density tells us that 8.96 g of copper is equivalent to 1.00 cm3 of
copper.
Mass
g
3.14 g
Density 
 8.96 3 
Volume
cm
V

V  0.350 cm3
Dr. Faady Siouri
Physical and Chemical Properties
 Physical property is the one that can be observed or measured without changing
the chemical makeup of a substance.
Example: Color, density, melting point, and boiling point.
• Chemical property is the tendency of a substance to undergo a particular chemical
reaction.
1
2
Example: Na + H2O  NaOH + H2
• A chemical property of H2O is that it reacts violently with Na.
• H2O and Na undergo a change called chemical change.
Dr. Faady Siouri
Sec. 1.6: Elements, Compound, and Mixtures
Dr. Faady Siouri
Sec. 1.6: Elements, Compound, and Mixtures
• Elements: Substances that cannot be decomposed into simpler
substances by chemical reactions.
• Compound: Substance that is composed of two or more elements. The
elements in the compound are always present in the same proportion
by mass (elements are always combined in the same, fixed proportions
by mass).
Example: H2O  1 : 8
Dr. Faady Siouri
Mixture Types
• Homogeneous mixture: Called solution and has uniform properties, consists of a
single phase.
* Phase: Part of a system which has uniform properties and composition.
Example: NaCl in H2O, powdered drinks.
• Heterogeneous mixture: Non-uniform, many consist of two phases or more.
Example: Oil in H2O, pizza, salad.
* To distinguish between a pure substance (element, compound)
and a mixture, we measure melting point. A pure substance
has a constant melting point, but if the temperature changes
the sample is impure and classifies as a mixture.
Dr. Faady Siouri
Sec. 1.6: Symbols, Formulas, and Equations
• Each element has been assigned a chemical symbol, which consists of one
or two letters.
Example: Carbon (C), Calcium (Ca), Sodium (Na), Iron (Fe).
• A chemical compound is represented symbolically by its chemical formula.
Example: Water (H2O), Methane (CH4), Carbon dioxide (CO2).
• Chemical formulas show the quantitative composition of substances.
• The subscripts in the formula give the relative number of atoms of each
element in the compound.
Dr. Faady Siouri
• If two or more atoms are combined, they give a molecule.
• The molecule may be:
1) Molecule of an element
Example: H2, N2, P4, S8
2) Molecule of a compound
Example: CO, NO, H2O, CO2, (NH4)2SO4
•
Some substances form crystals that contain water molecules, they are called
hydrates.
Example: CuSO4.5H2O (blue crystals). If the blue crystals are heated, water can
be driven off to leave pure, almost white, CuSO4.
Dr. Faady Siouri
Equations
• A chemical equation shows the chemical changes that occur during a
chemical reaction.
Zn + S
Reactants
ZnS
React to
yield
Products
• The equation shows the physical state of substances.
CaCO3 (aq) +
H2O (l) + CO2 (g)
Ca(HCO3)2 (aq)
• The chemical equation must be balanced!
Dr. Faady Siouri
Example 1.10: The combustion of butane follows the chemical equation
2C4H10 + 13O2
8CO2 + 10H2O
• How many oxygen atoms are included among the reactant molecules?
• How many oxygen atoms are included among the product molecules?
• Is the equation balanced?
Answer: There are 26 oxygen atoms in both reactants and products.
Yes, the equation is balanced.
Dr. Faady Siouri
Department of Industrial Chemistry
Chemistry for Nursing (Slides)
by
Dr. Faady Siouri
Chapter 2: Atoms, Molecules, and Moles
Fall 2022
Dr. Faady Siouri
Sec. 2.3: The Mole Concept
• Stoichiometry: The term we use in describing the quantitative aspects of
chemical composition and reaction.
• Mole: Avogadro’s number of objects (6.022 x 1023)
• The mole is a unit that is used to quantitatively measure the amount of
substance. The mole designates an extremely large number,
6.02214179 x 1023, which is the number of atoms determined
experimentally to be found 12 grams of C12.
Dr. Faady Siouri
• Avogadro proposed that equal volumes of gases under the same conditions
contain the same number of molecules.
• The number of atoms or other particles in a mole is the same for all
substances. The mole is related to the mass of an element in the following
way:
One mole of C12 has 6.022 x 1023 atoms and a mass of
12 grams. In comparison, one mole of oxygen consists
of the same number of atoms as C12 but it has a mass of
16 grams. Therefore, O has a greater mass than C12.
Dr. Faady Siouri
• The ratio by which moles of substances react is the same as the ratio by which
atoms and molecules react. To clarify this, let us take carbon monoxide (CO) as an
example:
• To make one molecule of CO, we need 1 C atom and 1 O atom so the ratio of C:O
is 1:1
1 atom C + 1 atom O  1 molecule CO
(small scale)
1 dozen C atoms + 1 dozen O atoms  1 dozen CO molecules
(larger scale)
1 mole C atoms + 1 mole O atoms  1 mole CO molecules.
(largest scale)
• Regardless of the scale (units), the ratio has always remained constant.
Dr. Faady Siouri
Example 2.2: What mole ratio of C to Cl must be chosen to prepare hexachloroethane
C2Cl6?
 To make one C2Cl6:
1 Molecule C2Cl6
2 C Atom
Similarly 
6 Cl Atom
1 Mole C2Cl6
2 C Moles
6 Cl Moles
• Atomic ratio: 2 C atoms : 6 Cl atoms
• Mole ratio: 2 mole C: 6 mole Cl  1 mole C : 3 mole Cl
Example 2.3: How many moles of carbon atoms are needed to combine with 4.87
mole Cl to form the substance C2Cl6 ?
We know that the mole ratio of C:Cl is 2:6 or 1:3 so can perform cross or bracket
2 mole C
6 mole Cl
OR
multiplication. The mole ratio can be written as
6 mole Cl
2 mole C
2 mole C
4.87 mole Cl 
 1.62 mole C
6 mole Cl
Dr. Faady Siouri
Example 2.4: How many moles of carbon are in 2.65 mole C2Cl6?
We know that the mole ratio of C2Cl6:C is 1:2 . We can perform cross or
bracket multiplication.
1 Molecule C2Cl6
2 C Atom
6 Cl Atom
Similarly 
1 Mole C2Cl6
2 C Moles
2 mole C
The mole ratio can be written as
1 mole C2Cl6
6 Cl Moles
1 mole C2Cl6
OR
2 mole C
2 mole C
2.65 mole C2Cl6 
 5.30 mole C
1 mole C2Cl6
Dr. Faady Siouri
Sec. 2.4: Measuring Moles of Atoms
Mass of Avogadro’s number of atoms of an element equals to that element’s
atomic mass.
There are three ways to convert from mass to moles or moles to mass:
1. Formulas
Mass ( g )
m ( g)
Atomic Mass or Molecular Mass ( g / mole ) 
 M ( g / mole ) 
Moles
n ( mole )
Mass ( g )
Moles 
Atomic Mass or Molecular Mass ( g / mole )
Mass ( g )  Atomic Mass or Molecular Mass ( g / mole )  Moles
2. Cross multiplication
3. Bracket multiplication
Dr. Faady Siouri
Example 2.5: How many moles of silicon (Si) are in 30.5 g of Si?
Given: 1 mole of Si = 28.1 g.
1. Formula:
M = Molar mass in g/mol
Mass ( g )
Moles 
Atomic Mass ( g / mole)
NA = Avogadro’s number

30.5 g
 1.09 moles Si
28.1 g / mole
Dr. Faady Siouri
2. Cross Multiplication:
1 mole Si  28.1 g Si
x
 30.5 g Si
(30.5 g Si ) (1 mole Si )
x
 1.09 moles Si
28.1 g Si
3. Bracket Multiplication:
1 mole Si
28.1 g Si
1 mole Si  28.1 g , which can be written as
OR
28.1 g Si
1 mole Si
 1 mole Si 
30.5 g Si  
  1.09 moles Si
 28.1 g Si 
Dr. Faady Siouri
Example 2.6: How many grams of copper (Cu) are there in 2.55 moles
of Cu?
Given: 1 mole of Cu = 63.5 g.
1. Formula:
M = Molar mass in g/mol
NA = Avogadro’s number
Mass ( g )  Atomic Mass  Moles  63.5 ( g / mol )  2.55 mol  162 g Cu
Dr. Faady Siouri
2. Cross Multiplication:
1 mole Cu
 63.5 g Cu
2.55 mole Cu 
x
(63.5 g Cu ) (2.55 mole Cu )
x
 162 g Cu
1 mole Cu
3. Bracket Multiplication:
1 moleCu
63.5 g Cu
1 mole Cu  63.5 g , which can be written as
OR
63.5 g Cu
1 mole Cu
 63.5 g Cu 
2.55 mole Cu  
  162 g Cu
 1 moleCu 
Dr. Faady Siouri
Example 2.7: How many moles of Ca are required to react with 2.50
mole of Cl to produce CaCl2?
1 Mole CaCl2
We know that the mole ratio Ca:Cl is 1:2
1 Ca Mole
1 mole Ca
The mole ratio can be written as
2 mole Cl
1 mole Ca
2.50 mole Cl 
 1.25 mole Ca
2 mole Cl
2 Cl Moles
2 mole Cl
OR
1 mole Ca
Dr. Faady Siouri
Example 2.8: How many grams of Ca must react with 41.5 g of Cl to
produce CaCl2?
Given :1 mole Cl  35.5 g
And 1 mole Ca  40.1 g
1 mole Cl

35.5 g Cl
1 mole Ca

40.1 g Ca
We know that the mole ratio of Ca:Cl is 1:2
35.5 g Cl
OR
1 mole Cl
40.1 g Ca
OR
1 mole Ca
1 Mole CaCl2
1 Ca Mole
2 Cl Moles
 1 mole Cl   1 mole Ca   40.1 g Ca 
41.5 g Cl 


  23.4 g Ca
 35.5 g Cl   2 mole Cl   1 mole Ca 
Convert mass of Cl
into moles of Cl
Convert moles of Cl
into moles of Ca
Convert moles of Ca
into mass of Ca
Dr. Faady Siouri
Example 2.9: What is the mass of one atom of calcium?
We know 1 mole Ca = 6.022 x 1023 atoms Ca which can be written as
1 mole Ca
6.022  10 atoms Ca
OR
23
6.022  10 atoms Ca
1 mole Ca
23
From the periodic table: 1 mole Ca = 40.1 g Ca which can be written as
1 mole Ca
40.1 g Ca
OR
40.1 g Ca
1 mole Ca

  40.1 g Ca 
1 mole Ca
23
1 atom Ca 

6.
6
6

10
g


23
 6.022  10 atoms Ca  1 mole Ca 
Dr. Faady Siouri
Sec. 2.5: Measuring Moles of Compounds
• Molecular mass (molecular weight): The sum of atomic masses of
elements in the molecule.
• The molecular mass of CO2 =
(1 x C atomic mass) + (2 x O atomic mass)
= (1 x 12.0 u) + (2 x 16.0 u) = 44.0 u
• The weight of one mole of a substance is simply the molecular mass
followed by the units, grams. Thus,
1 mole CO2 = 44.0 g
Dr. Faady Siouri
• Ions: Atoms or groups of atoms that have acquired an electrical charge.
(Na+, Cl-, NH4+, SO4-2).
• Ionic compounds are chemical compounds composed of ions and held
together by ionic bonds. Ionic compounds are not molecules.
• Formula unit: specifies the number of ions in the formula.
NaCl : 2 ions
CaCl2: 3 ions
• Formula mass (formula weight): The sum of atomic masses of elements
present in one formula unit.
For NaCl, this is 22.99 + 35.45 = 58.44
1 mole of NaCl = 58.44 g
Negative
Ion
Positive
Ion
Dr. Faady Siouri
Example 2.10:
a) How many grams does 0.25 mole Na2CO3 weigh?
b) How many moles of Na2CO3 are in 132 g Na2CO3?
Molar Mass of Na
Molar Mass of C
Molar Mass of O
1 mole Na2CO3 = (2 x 23.0) + (1 x 12.0) + (3 x 16.0) = 106.0 g
a)
b)
 106.0 g Na2CO3 
0.25 mole Na2CO3  
  26.5 g Na2CO3
 1 mole Na2CO3 
 1 mole Na2CO3 
132 g Na2CO3  
  1.25 mole Na2CO3
 106.0 g Na2CO3 
Dr. Faady Siouri
Sec. 2.6: Percent Composition
• Percent composition: The total mass contributed by each element.
Weight of part
In general , % by weight 
 100
Weight of whole
OR
• Percent composition is the percent by mass of each element in a
compound.
grams of element
% Composition 
 100
grams of compound
Dr. Faady Siouri
Example 2.11: What is the percent composition of CHCl3?
1 mole of CHCl3 = (1 x 12.01) + (1 x 1.008) + (3 x 35.45) = 119.37 g
12.01 g C
% of C 
 100  10.06 %
119.37 g CHCl3
1.008 g H
% of H 
 100  0.844 %
119.37 g CHCl3
3  35.45 g Cl
% of Cl 
 100  89.09 %
119.37 g CHCl3
Total = 10.06% + 0.844% + 89.09% = 100%
Dr. Faady Siouri
Example 2.12: Calculate the mass of iron Fe in a 10.0 g sample of iron oxide Fe2O3?
 The ratio of Fe : Fe2O3 = 2 Fe : 1 Fe2O3
1 Mole Fe2O3
2 Fe Moles
3 O Moles
 The mass of 1 mole of Fe = 55.85 g so the mass of 2 moles of Fe = 2 (55.85) = 111.7 g
 The mass of 1 mole of Fe2O3 = (2 x 55.85) + (3 x 16.00) = 159.7 g
111.7 g
The mass ratio of Fe in Fe2O3 
 0.699  OR 69.9 %
159.7 g
 However, the question is not asking for mass ratio or percent composition, it is
asking for the mass of Fe if we have 10.0 g sample of Fe2O3
 111.7 g Fe 
10.0 g Fe2O3  
  6.99 g Fe
 159.7 g Fe2O3 
Dr. Faady Siouri
Sec. 2.7: Chemical Formulas
The chemical formula can be written in 3 ways:
1) Simplest Formula (Empirical formula): This formula uses the smallest
set of a whole-number subscripts to specify the relative number of atoms
of each element present in a formula unit. This formula does not
necessary show the exact number of atoms but rather the simplest ratio of
the atoms in the compound.
Example: CH2, H2O, Glucose empirical formula is CH2O, Butane’s (C4H10)
empirical formula is C2H5
Dr. Faady Siouri
2) Molecular formula: This formula shows the actual number of each kind of
atom found in a molecule.
Example: C2H4, H2O, Glucose (C6H12O6); the ratio of C:H:O = 6:12:6
3) Structural formula: This formula gives information about the way in
which atoms in a molecule are linked together and provides information to
write molecular and empirical formulas. This graphical representation of the
molecule’s structure helps us picture how atoms are arranged in the molecule.
Example:
C2H4O2
CH2O
Dr. Faady Siouri
Sec. 2.8: Empirical Formulas and Molecular
Formulas
How to calculate the empirical formula of a compound:
1. We need to know the mass of each of the elements in a given mass of
the compound.
2. Convert masses to moles.
3. Find the smallest whole-number ratio of moles = subscripts in the
formula = atom ratio.
Dr. Faady Siouri
Example 2.13: A sample of a compound contains 2.34 g N and 5.34 g O. What is the
simplest formula of the compound?
1 mole N = 14.0 g N
1 mole O = 16.0 g O
 1 mole N 
2.34 g N 
  0.167 mole N
 14.0 g N 
 1 mole O 
5.34 g O 
  0.334 mole O
 16.0 g O 
0.167
N
0.167
0.334
O
0.167
N1 : O2  NO2
Dr. Faady Siouri
Example 2.14: What is the empirical formula of a compound composed of 43.7% P
and 56.3% O by weight?
1 mole P = 31.0 g P
1 mole O = 16.0 g O
Let’s assume we have 100 g sample = 43.7 g P + 56.3 g O
 1 mole P 
43.7 g P 
  1.41 mole P
 31.0 g P 
 1 mole O 
56.3 g O 
  3.52 mole O
 16.0 g O 
1.41
P
1.41
3.52
O
1.41
P1 : O2.5  P2O5
Dr. Faady Siouri
Example 2.15: A 1.025 g sample of a compound that contains only C and H was
burned in O2 to give CO2 and H2O. It was found that 3.007 g of CO2 and 1.845 g
H2O were formed. What is the empirical formula of the compound?
1 mole H2O = 18.02 g H2O
1 mole CO2 = 44.01 g CO2
CxHy + O2  CO2 + H2O
x, y =??
3.007 g
1.845 g
 We are not given the mass of C but it can be determined from the mass of CO2
 Similarly, we are not given the mass of H but it can be determined from the
mass of H2O .
Dr. Faady Siouri
 1 mole CO2  1 mole C 
3.007 g CO2 

  0.06833 mole C
 44.01 g CO2  1 mole CO2 
 1 mole H 2O  2 mole H 
1.845 g H 2O 

  0.2048 mole H
 18.02 g H 2O  1 mole H 2O 
1 Mole CO2
1 C Mole
2 O Moles
1 Mole H2O
2 H Moles
1 O Mole
** Why didn’t we convert the number of moles of C and H into grams?
0.06833
C
0.06833
0.2048
H
0.06833
C1 : H 2.977  CH3
Dr. Faady Siouri
Example 2.16: A 0.100 g sample of ethyl alcohol contains C, H, and O was burned
in oxygen to give 0.1910 g CO2 and 0.1172 g H2O. What is the empirical formula of
ethyl alcohol?
CxHyOz + O2  CO2 + H2O
x, y, z = ??
0.100g
0.1910 g
0.1172 g
 We are not given the mass of C but it can be determined from the mass of CO2
 Similarly, we are not given the mass of H but it can be determined from the
mass of H2O .
 The mass of O can be determined by subtracting the mass of C and H from the
total mass.
Dr. Faady Siouri
 1 mole CO2  1 mole C 
0.1910 g CO2 

  0.004340 mole C
 44.01 g CO2  1 mole CO2 
 1 mole H 2O  2 mole H 
0.1172 g H 2O 

  0.0130 mole H
 18.02 g H 2O  1 mole H 2O 
Mass of O = Mass of sample – (mass of C + mass of H)
= 0.1000g – (0.05212 g + 0.01311 g) = 0.0348 g O
 1 mole O 
0.0348 g O 
  0.00218 mole O
 16.00 g O 
0.004340
0.0130
0.00218
C
H
O
0.00218
0.00218
0.00218
C 2 : H 6 : O 1  C2H6O
1 Mole CO2
1 C Mole
2 O Moles
1 Mole H2O
2 H Moles
1 O Mole
** Convert moles of
C and H into grams
so you can subtract
them from the total
mass of the sample.
Dr. Faady Siouri
Determining Molecular Formula of a Compound
Example 2.17: A liquid whose empirical formula is NO2, has a molecular mass of
92.0 . What is its molecular formula?
The formula mass of NO2 = 14.0 + (2 x 16.0) = 46.0 g
The number of times the empirical formula is included in the compound (molecular
formula) =
Molecular Mass
Empirical Molar Mass

92.0
2
46.0
So, the molecular formula = 2 x empirical formula
= 2 x NO2
= N2O4
Dr. Faady Siouri
Example: A sample of a compound containing boron (B) and hydrogen (H) contains
6.444 g of B and 1.803 g of H. The molar mass of the compound is about 30 g.
What is its molecular formula? What is its exact molar mass?
 1 mol B 
nB  6.444 g B 
  0.5961 mol B
 10.81 g B 
 1 mol H 
nH  1.803 g H 
So the formula at this point is B0.5961H1.789
  1.789 mol H
 1.008 g H 
Now simplify to whole number ratio by dividing over the smallest subscript.
0.5961
1.789
B:
1
H:
 3  The empirical formula is BH3
0.5961
0.5961
Empirical molar mass = 10.81 g + 3(1.008 g) = 13.834 g
molar mass
30 g

=2.17  2
empirical molar mass 13.834 g
 molecular formula = (empirical formula) 2 = (BH 3 ) 2  B2 H 6
The exact molar mass  2 (empirical molar mass)  2 (13.834 g)  27.668 g
Dr. Faady Siouri
Department of Industrial Chemistry
Chemistry for Nursing Slides
by
Dr. Faady Siouri
Chapter 3: Chemical Reactions and the Mole Concept
Fall 2022
Dr. Faady Siouri
Sec. 3.1: Chemical Reactions and Chemical Equations:
Balancing Chemical Equations
Writing a balanced chemical equation should always be viewed as a
two-step process.
1) Write an unbalanced chemical equation, being careful to write the
correct formula for each substance involved.
2) Balance the equation by adjusting the coefficients that precede the
formulas of the reactants and products so that the same number of atoms
of each kind on both sides of the arrow.
Dr. Faady Siouri
Hydrochloric acid + sodium carbonate  sodium chloride + carbon
dioxide + water.
HCl + Na2CO3  NaCl + CO2 + H2O
HCl + Na2CO3  2NaCl + CO2 + H2O
2HCl + Na2CO3  2NaCl + CO2 + H2O
Dr. Faady Siouri
Example 3.1: Balance the equation for the combustion of octane.
C8H18 + O2  CO2 + H2O
C8H18 + O2  8CO2 + H2O
C8H18 + O2  8CO2 + 9H2O
C8H18 + 12.5O2  8CO2 + 9H2O
I prefer that we multiply any fractions in order to get an integer.
We multiply the equation by 2
2C8H18 + 25O2  16CO2 + 18H2O
Dr. Faady Siouri
Sec. 3.2: Calculations Based on Chemical Equations
C2H5OH + 3O2  2CO2 + 3H2O
1 Molecule C2H5OH + 3 Molecules O2  2 Molecules CO2 + 3 Molecules H2O
1 Mole C2H5OH + 3 Moles O2  2 Moles CO2 + 3 Moles H2O
Dr. Faady Siouri
Mass Changes in Chemical Reactions
Mass (g) of
compound A
?
Use molar
mass (g/mol)
of compound A
Moles of
compound A
Mass (g) of
compound B
Use molar
mass (g/mol)
of compound B
Use mole ratio
of A and B
from balanced
equation
Moles of
compound B
1. Convert quantities of known substances into moles
2. Write balanced chemical equation
3. Use coefficients in balanced equation to calculate the number of moles of the sought
quantity
4. Convert moles of sought quantity into desired units
Dr. Faady Siouri
Example 3.2: How many moles of O2 are needed to burn 1.80 mole
C2H5OH according to the balanced equation
C2H5OH + 3O2  2CO2 + 3H2O
moles C2H5OH
moles O2
Mole ratio of C2H5OH to O2 is 1 C2H5OH : 3 O2
 3 mole O2
1.80 mole C2 H 5OH 
 1 mole C2 H 5OH

  5.40 mole O2

Dr. Faady Siouri
Example 3.4: How many moles of water will form when 3.66 mole CO2
are produced during the combustion of C2H5OH ?
C2H5OH + 3O2  2CO2 + 3H2O
moles CO2
moles H2O
Mole ratio of CO2 to H2O is 2 CO2 : 3 H2O
 3 mole H 2O 
3.66 mole CO2 
  5.49 mole H 2O
 2 mole CO2 
Dr. Faady Siouri
Example 3.5: How many grams of O2 are required to react with 0.300 mole Al according to the
equation
Al + O2  Al2O3
We are given the number of moles of Al and are asked to calculate the mass in grams of O2 needed.
WAIT……before you start, is the equation balanced?
NO, it is not !!! …….so balance it first.
4Al + 3O2  2Al2O3
Mole ratio of Al to O2 is 4 Al : 3 O2
moles Al
moles O2
grams O2
1 mole O2 = 32.0 g O2
 3 mole O2   32.0 g O2 
0.300 mole Al 
  7.20 g O2

 4 mole Al   1 mole O2 
Dr. Faady Siouri
Example 3.6: Calculate the number of grams of Al2O3 that could be formed if 12.5 g
O2 react completely with Al.
4Al + 3O2  2Al2O3
We are given the number of grams of O2 and are asked to calculate the mass in
grams of Al2O3 produced.
Mole ratio of O2 to Al2O3 is 3 O2 : 2 Al2O3
grams O2
moles O2
moles Al2O3
mass Al2O3
1 mole O2 = 32.0 g O2
1 mole Al2O3 = 102 g Al2O3
 1 mole O2  2 mole Al2O3   102 g Al2O3 
12.5 g O2 
  26.6 g Al2O3


 32 g O2  3 mole O2  1 mole Al2O3 
Dr. Faady Siouri
Example: How many grams of Li are needed to produce 9.89 g of H2
according to the reaction?
2 Li(s) + 2 H2O(l)
grams H2
moles H2
2LiOH(aq) + H2(g)
moles Li
mass Li
 1 mol H 2   2 mol Li   6.941 g Li 
9.89 g H 2 


 =68.1 g Li
 2.016 g H 2   1 mol H 2   1 mol Li 
Dr. Faady Siouri
Practice Question: If 856 g of C6H12O6 is consumed by a person over a
certain period, what is the mass of CO2 produced ?
C6H12O6 + O2 → CO2 + H2O
Dr. Faady Siouri
Sec. 3.3: Limiting Reactant Calculations
• Limiting reactant is the reactant that when it is completely consumed,
no further reaction can occur and no further product can be formed.
2
+10
2
The car without tires limited how many usable cars we can make;
therefore, it is the limiting reactant.
We are left with 2 extra tires that were not used.
Dr. Faady Siouri
Dr. Faady Siouri
Dr. Faady Siouri
Example 3.7: Zinc and sulfur react to form zinc sulfide. The equation for the
reaction is:
Zn + S  ZnS
In a particular experiment, 12.0 g of Zn are mixed with 6.50 g of S and allowed to
react.
a) Which is the limiting reactant?
b) How many grams of ZnS can be formed?
c) How many grams of which reactant will remain unreacted in this experiment?
Given: 1 mole Zn = 65.4 g Zn
1 mole S = 32.1 g S
** Before you solve the problem, you need to make sure the equation is balanced **
Dr. Faady Siouri
Method 1: Relationship between the reactants and the product
Which is the L.R? How much product is produced?
Start with Zn:
12.0 g Zn
1 mol Zn
1 mol ZnS
97.5 g ZnS
65.4 g Zn
1 mol Zn
1 mol ZnS
1 mol S
1 mol ZnS
97.5 g ZnS
=
17.8 g ZnS
=
19.7 g ZnS
Limiting Reactant
Now S:
6.50 g S
32.1 g S
1 mol S
1 mol ZnS
Excess
Dr. Faady Siouri
How many grams of which reactant will remain unreacted in this
experiment?
S is the excess reactant in this reaction. When Zn reacts completely, it
produces 17.8 g ZnS (0.183 mole ZnS). So, the question is how many
grams of S remains unreacted when ZnS is completely produced.
 1 mol S   32.1g S 
c) 0.183 mol ZnS 

 = 5.87 g S reacted
 1 mol ZnS   1 mol S 
6.5 g - 5.87 g = 0.63 g S
excess/remaing/unreacted
OR
Dr. Faady Siouri
How many grams of which reactant will remain unreacted in this
experiment?
S is the excess reactant in this reaction. Zn reacts completely with a
certain amount of S. So, the question is how many grams of S remains
unreacted when Zn is completely consumed.
 1 mol Zn   1 mol S  32.1 g S 
12 g Zn 


 = 5.89 g S reacted
 65.4 g Zn   1 mol Zn  1 mol S 
6.5 g - 5.89 g = 0.61 g S
excess/remaing/unreacted
**Your answer here might be slightly different from your previous answer due to
rounding**
Dr. Faady Siouri
Method 2: The common method
Which reactant is limiting?
Zn +
Convert grams of
Zn into moles
Divide the
moles of Zn
by its
coefficient
12 g
S

ZnS
6.50 g
0.183 mole 0.202 mole
0.183
0.202
= 0.183 mole
= 0.202 mole
1
1
Convert grams of
S into moles
Divide the
moles of S
by its
coefficient
 The number of moles of Zn is smaller so it is the limiting reactant.
Dr. Faady Siouri
How much product is produced?
Since Zn is the limiting reactant, it limits how much product is
produced. When Zn is completely consumed, the reaction stops and no
further product could be formed.
 1 mol Zn   1 mol ZnS  97.5 g ZnS 
12 g Zn 


 = 17.8 g ZnS
 65.4 g Zn   1 mol Zn  1 mol ZnS 
You could start with the number of
moles of Zn which was previously
calculated when determining the
Limiting Reactant
Dr. Faady Siouri
Method 3: Relationship between the reactants
Which reactant is limiting?
 1 mol Zn 
12 g Zn 
 = 0.183 mol Zn
 65.4 g Zn 
 1 mol S 
6.5 g S 
 = 0.202 mol S
 32.1 g S 
Before we choose the reactant with less value as the limiting reactant, we
need to take the mole ratio into consideration.
 1 mol S 
0.183 mol Zn 
 = 0.183 mol S
 1 mol Zn 
This means that “Zn” reacts completely with 0.183 moles of “S” leaving us with no Zn but
some S leftover.  Zn is the limiting reactant.
Dr. Faady Siouri
How much product is produced?
Since Zn is the limiting reactant, it limits how much product is
produced.
 1 mol Zn   1 mol ZnS  97.5 g ZnS 
12 g Zn 


 = 17.8 g ZnS
 65.4 g Zn   1 mol Zn  1 mol ZnS 
You could start with the number of
moles of Zn which was previously
calculated when determining the
Limiting Reactant
Dr. Faady Siouri
Example: 10.0 g of aluminum reacts with 35.0 g of chlorine gas to produce
aluminum chloride. Which reactant is limiting? How much product is
produced? How many grams of which reactant will remain unreacted in this
experiment?
Method 1: Relationship between the reactants and the product
Which is the L.R? How much product is produced?
2Al + 3Cl2  2AlCl3
Start with Al:
Now Cl2:
10.0 g Al
1 mol Al
27.0 g Al
35.0g Cl2 1 mol Cl2
71.0 g Cl2
Limiting Reactant
2 mol AlCl3 133.5 g AlCl3
2 mol Al
1 mol AlCl3
2 mol AlCl3 133.5 g AlCl3
3 mol Cl2
1 mol AlCl3
= 49.4 g AlCl3
= 43.9 g AlCl3
Dr. Faady Siouri
How many grams of which reactant will remain unreacted in this
experiment?
Al is the excess reactant in this reaction. When Cl2 reacts completely, it
produces 43.9 g AlCl3 (0.33 mole AlCl3). So, the question is how many
grams of Al remains unreacted when AlCl3 is completely produced.
 2 mol Al   27g Al 
c) 0.33 mol AlCl3 

 = 8.91 g Al reacted
 2 mol AlCl3   1 mol Al 
10 g - 8.91 g = 1.09 g Al
excess/remaing/unreacted
OR
Dr. Faady Siouri
How many grams of which reactant will remain unreacted in this
experiment?
Al is the excess reactant in this reaction. Cl2 reacts completely with a
certain amount of Al. So, the question is how many grams of Al remains
unreacted when Cl2 is completely consumed.
 1 mol Cl2   2 mol Al   27 g Al 
35 g Cl2 


 = 8.87 g Al reacted
 71 g Cl2   3 mol Cl2   1 mol Al 
10 g - 8.87 g = 1.13 g Al
excess/remaing/unreacted
**Your answer here might be slightly different from your previous answer due to
rounding**
Dr. Faady Siouri
Method 2: The common method
Which reactant is limiting?
2Al +
3Cl2
10 g
35 g
0.37 mole
0.49 mole
0.37
= 0.185 mole
2

2AlCl3
0.49
= 0.163 mole
3
 Cl2 is the limiting reactant.
Dr. Faady Siouri
How much product is produced?
Since Cl2 is the limiting reactant, it limits how much product is
produced. When Cl2 is completely consumed, the reaction stops and no
further product could be formed.
 1 mol Cl2   2 mol AlCl3   133.5 g AlCl3 
35 g Cl2 
 = 43.9 g AlCl3


 71 g Cl2   3 mol Cl2  1 mol AlCl3 
You could start with the number of
moles of Cl2 which was previously
calculated when determining the
L.R.
Dr. Faady Siouri
Method 3: Relationship between the reactants
Which reactant is limiting?
 1 mol Cl2 
35 g Cl2 
 = 0.49 mol Cl 2
 71 g Cl2 
 1 mol Al 
10 g Al 
 = 0.37 mol Al
 27 g Al 
Before we choose the reactant with less value as the limiting reactant, we need to take
the mole ratio into consideration.
 3 mol Cl2 
0.37 mol Al 
 = 0.555 mol Cl2
 2 mol Al 
For Al to react completely, it requires 0.555 mole Cl2 , which is not available. This means
that Cl2 will be consumed completely leaving us with no Cl2, but with Al leftover.
Dr. Faady Siouri
How much product is produced?
Since Cl2 is the limiting reactant, it limits how much product is
produced.
 1 mol Cl2   2 mol AlCl3   133.5 g AlCl3 
35 g Cl2 
 = 43.9 g AlCl3


 71 g Cl2   3 mol Cl2  1 mol AlCl3 
You could start with the number of
moles of Cl2 which was previously
calculated when determining the
L.R.
Dr. Faady Siouri
Practice Question: Consider the reaction:
C6H12O6 + O2  CO2 + H2O + Energy
What mass of carbon dioxide forms in the reaction of 25 grams of
glucose with 40 grams of oxygen? How many grams of which reactant
will remain unreacted in this experiment?
Dr. Faady Siouri
Sec. 3.4: Theoretical Yield, Actual Yield, and
Percent Yield
• Theoretical yield: The maximum yield of a given product that could
be obtained if reactants gave only that product, with no side reactions.
It is a calculated quantity that is obtained by using balanced equations.
• Actual yield: The amount of product actually obtained in a given
experiment. It is obtained by isolating the product and weighing it.
• Percent yield: Measures the efficiency of the reaction.
Actual Yield
Percent Yield 
 100
Theoretical Yield
Dr. Faady Siouri
Example 3.8: Use the information calculated in example 3.7 to solve
this problem. If in the given experiment, 14.6 g of ZnS is obtained, what
is the percent yield of ZnS?
The theoretical value
from example 3.7
Actual Yield of ZnS
Percent Yield 
 100
Theoretical Yield of ZnS
14.6 g ZnS

 100  85.3%
17.8 g ZnS
Dr. Faady Siouri
Sec. 3.5: Reactions in Solution
• If we mix crystals of sodium chloride (NaCl) with crystals of silver nitrate
(AgNO3), nothing much appears to happen, only their outer surfaces can come in
contact.
• However, when these substances are dissolved in water, the particles of each
reactant are free. When the solutions are combined, a rapid reaction
takes place and a white precipitate of silver chloride (AgCl) is formed.
AgNO3 (aq) + NaCl (aq)  AgCl (s) + NaNO3 (aq)
Dr. Faady Siouri
• Another example:
Zn (s) + H2SO4 (aq)  ZnSO4 (aq) + H2 (g)
• Terminology applied to solutions:
Solvent: The component whose physical state doesn’t change when the
solution is formed.
Solute: The component that dissolved in the solvent.
Concentrated solution: is the one with a relatively large concentration of
solute.
Diluted solution: has relatively small concentration of solute.
Dr. Faady Siouri
Molar Concentration
Moles of Solute
Molarity ( M ) 
Volume of Solution ( Liters )
Example 3.9: A 2.00 g sample of NaOH was dissolved in water to give a
solution with a volume of 200 mL. What is the molarity of this NaOH
solution?
First, we convert the number of grams of NaOH to number of moles.
1 mole NaOH = 40.0 g NaOH
 1 mol NaOH 
2.00 g NaOH 
 = 0.0500 mol NaOH
 40.0 g NaOH 
Dr. Faady Siouri
Next, we convert 200 mL to Liters.
We know that 1000 mL = 1 L
 1L 
200 mL 
 = 0.200 L
 1000 mL 
Finally, we plug in into the molarity equation
Moles of Solute
Molarity ( M ) 
Volume of Solution ( Liters )
0.0500 mole NaOH
mole

 0.250
NaOH
0.200 L Solution
L
 0.250 M NaOH
Dr. Faady Siouri
Example 3.10: How many milliliters of 0.250 M NaOH solution are needed to provide
0.0200 mole of NaOH?
1 mole NaOH = 40.0 g NaOH
Moles of Solute
Molarity ( M ) 
Volume of Solution ( Liters )
0.0200 mole NaOH
0.250 M 
Volume of Solution ( Liters )
Volume of solution = 0.080 L
and because the question is asking for the answer in milliliters, then the volume of
solution = 80.0 mL
Dr. Faady Siouri
Example 3.11: How many grams of NaOH are in 50.0 mL of 0.400 M
NaOH solution?
1 mole NaOH = 40.0 g NaOH
Moles of Solute
Molarity ( M ) 
Volume of Solution ( Liters )
Make sure you
convert mL to L
Moles of NaOH
0.400 M 
0.05 L Solution
Moles of NaOH = 0.0200 mole NaOH
 40.0 g NaOH 
0.0200 mole NaOH 
 = 0.800 g NaOH
 1 mole NaOH 
Dr. Faady Siouri
Practice Question: How would you prepare 50.0 mL of 0.400 M NaOH?
Answer this question using numbers and words.
Dr. Faady Siouri
Sec. 3.5: Preparing Solutions by Dilution
• The process of dilution involves mixing a concentrated solution with
additional solvent to give some larger final volume.
• In this process, the number of moles remains constant and only the
volume increases.
Moles of solute
before dilution (i)
=
Moles of solute
after dilution (f)
MiVi
=
MfVf
Dr. Faady Siouri
Example 3.13: How many milliliters of concentrated H2SO4 (18.0 M) are
required to prepare 750 mL of 3.00 M H2SO4 solution?
We are given the initial concentration, final concentration, and final volume.
Mi = 18.0 M
;
Mf = 3.00 M
;
Vf = 750 mL
Mi Vi = Mf Vf
Vi 
M f Vf
Mi
3.00 M  750 mL
Vi 
 125 mL
18.0 M
750 mL
18 M H2SO4
Dr. Faady Siouri
Lab Safety
• When concentrated chemicals are diluted, a large amount of heat is
sometimes liberated. This is especially true for sulfuric acid.
• To absorb this heat safely, you must always add the concentrated acid
to the water, never the reverse.
• If water is added to the concentrated acid,
so much heat is liberated that it can cause the
water to boil suddenly, spattering the acid.
Dr. Faady Siouri
Example 3.14: How much water must be added to 25.0 mL of 0.500 M KOH
solution to produce a solution of 0.350 M?
We are given the initial concentration, initial volume, and final concentration
Mi = 0.500 M
M i Vi
Vf 
Mf
;
Vi = 25.0 mL
;
Mf = 0.350 M
Mi Vi = Mf Vf
0.500 M  25.0 mL
Vf 
 35.7 mL
0.350 M
Be Careful!!!!
This is the final volume
of the solution and not
the amount of water
that must be added.
35.7 mL – 25.0 mL = 10.7 mL water must be added.
Dr. Faady Siouri
Example 3.15: If 200 mL of water were added to 300 mL solution of 0.600 M
HNO3. What will be the new concentration of the solute in the solution?
We are given the initial concentration, initial volume, and final volume.
Mi = 0.600 M
;
Vi = 300 mL
;
Vf = (200 mL + 300 mL) = 500 mL
Mi Vi = Mf Vf
M i Vi
Mf 
Vf
0.600 M  300 mL
Mf 
 0.360 M
500 mL
Dr. Faady Siouri
Sec. 3.7: The Stoichiometry of Reactions in
Solution
This is almost the same type of problems we practiced in chapters 2 and
3 with a little twist.
Example 3.16: How many milliliters of 0.200 M NaOH solution are
needed to completely react with 3.50 g Al2(SO4)3 ?
Al2(SO4)3 (aq) + 6NaOH (aq)  2Al(OH)3 (s) + 3Na2SO4 (aq)
We are given the number of grams of Al2(SO4)3 and asked to find how
many milliliters of 0.200 M NaOH is required to react completely with
the given amount of Al2(SO4)3 .
Dr. Faady Siouri
Moles of Solute
Molarity ( M ) 
Volume of Solution ( Liters )
NaOH molarity : 0.200 M
Moles of solute (moles of NaOH) : Can be calculated from the
information given in the question
Volume of solution : This is what the question is asking for!
We convert the number of grams of Al2(SO4)3 into moles then we use
the mole ratio between Al2(SO4)3 and NaOH to calculate the number of
moles of NaOH.
Dr. Faady Siouri
1 mole Al2(SO4)3 = 342.2 g Al2(SO4)3
The mole ratio of Al2(SO4)3 to NaOH is 1:6
 1 mole Al2 ( SO4 )3   6 mole NaOH 
2
3.50 g Al2 ( SO4 )3 
=
6.12

10
mole NaOH


 342.2 g Al2 ( SO4 )3   1 mole Al2 ( SO4 )3 
6.12 102 mole NaOH
Volume of Solution 
 0.306 L Solution
0.200 M NaOH
= 306 mL Solution
Dr. Faady Siouri
Example 3.17: How many milliliters of 0.250 M CaCl2 are needed to
react completely with 50.0 mL of 0.150 M Na2CO3 solution?
CaCl2 (aq) + Na2CO3 (aq)  CaCO3 (s) + 2NaCl (aq)
It is solved the same way like the previous question except that here
they are not giving us the number of grams of Na2CO3. Instead, they
gave us the molarity and volume which we could use to calculate the
number of moles of Na2CO3.
Moles of Solute
Molarity (Na 2CO 3 ) 
Volume of Solution ( Liters )
Dr. Faady Siouri
Moles of Na2CO3 = (0.150 M Na2CO3 ) (0.050 L solution)
= 7.50 x 10-3 mole Na2CO3
From the balanced equation, the mole ratio of Na2CO3 to CaCl2 is 1:1
 1 mole CaCl2 
3
7.50 10 mole Na2CO3 
 = 7.50 10 mole CaCl2
 1 mole Na2CO3 
3
7.50 103 mole CaCl2
Volume of Solution 
 0.0300 L Solution
0.250 M CaCl2
= 30.00 mL solution
Dr. Faady Siouri
Example 3.18: How may grams of solid AgBr will be formed if 50.0 mL of
0.180 M AgNO3 are mixed with 60.0 mL of 0.0850 M CaBr2?
2AgNO3 (aq) + CaBr2 (aq)  2AgBr (s) + Ca(NO3)2 (aq)
From the molarity equation, find the number of moles of both AgNO3 and
CaBr2 in order to determine which one is the limiting reactant.
Moles of AgNO3 = (0.180 M Na2CO3 ) (0.050 L solution)
= 9.00 x 10-3 mole AgNO3
Moles of CaBr2 = (0.0850 M Na2CO3 ) (0.060 L solution)
= 5.10 x 10-3 mole CaBr2
Dr. Faady Siouri
Before we pick the small number (CaBr2) as the limiting reactant, we
need to take the mole ratio into consideration.
 2 mole AgNO3 
3
5.10 10 mole CaBr2 
=
10.20

10
mole AgNO3

 1 mole CaBr2 
3
10.20 x 10-3 mole AgNO3 is greater than 9.00 x 10-3 mole AgNO3 so
AgNO3 is the limiting reactant.
 2 mole AgBr 
3
9.00 10 mole AgNO3 
=
9.00

10
mole AgBr

 2 mole AgNO3 
 187.8 g AgBr 
3
9.00 10 mole AgBr 
 = 1.69 g AgBr
 1 mole AgBr 
3
Dr. Faady Siouri
Department of Industrial Chemistry
Chemistry for Nursing Slides
by
Dr. Faady Siouri
Chapter 5: Chemical Reactions in Aqueous Solution
Fall 2022
Dr. Faady Siouri
Sec. 5.8: Chemical Analysis and Titrations
Chemical Analysis: Determination of the chemical composition of a
substance or a mixture of substances.
Example 5.16: A 1.244 g of a white powder contains NaCl and Na2SO4 was
dissolved in water and a solution of Ba(NO3)2 was added until the
precipitation of BaSO4 was complete. The weight of the dry precipitate was
found to be 0.851 g. What was the percentage by mass of Na2SO4 in the
sample?
Given: 1 mole BaSO4 = 233.4 g
1 mole Na2SO4 = 142.0 g
Dr. Faady Siouri
Na2SO4 + Ba(NO3)2  BaSO4 + 2NaNO3
The addition of NaCl has no effect on the chemical equation so the
equation can be written as
NaCl + Na2SO4 + Ba(NO3)2  BaSO4 + 2NaNO3 + NaCl
Total Mass: 1.244 g
Mass: 0.851 g
We are given the mass of BaSO4 so we can use it to determine the
number of moles (hence the number of grams) of Na2SO4 from the
balanced chemical equation. Then we calculate the percent by mass of
Na2SO4 in the sample that contains NaCl and Na2SO4.
Dr. Faady Siouri
Convert the
mass of BaSO4
into moles
Determine the
moles of Na2SO4
from the moles of
BaSO4 using the
balanced equation
Convert the
moles of Na2SO4
into grams
 1 mole BaSO4 
3
0.851 g BaSO4 
  3.65 10 mole BaSO4
 233.4 g BaSO4 
 1 mole Na2 SO4 
3
3.65 10 mole BaSO4 
  3.65 10 mole Na2 SO4
 1 mole BaSO4 
3
 142.0 g Na2 SO4 
3.65 10 mole Na2 SO4 
  0.518 g Na2 SO4
 1 mole Na2 SO4 
3
Divide the mass of
Na2SO4 by the total
mass of the sample
Mass Na2 SO4
% Na2 SO4 
 100
Mass of Sample
0.518 g Na2 SO4

 100  41.6%
1.244 g Sample
Dr. Faady Siouri
• Titration: Analytical procedure used to measure the amount of one
solution needed to react exactly with the contents of another solution.
• In titration we measure volumes (volumetric analysis)
• The indicator detects the end point and changes
color according to the medium acidic or basic.
Litmus: pink in acidic solution and blue in basic.
Phenolphthalein: colorless in acidic solution and
pink in basic.
Dr. Faady Siouri
Example 5.17: A solution of NaOH was prepared with 0.1 M concentration.
In the presence of phenolphthalein, 20.00 mL portion of 0.1000 M solution
of HCl required 18.47 mL of NaOH solution to reach the end point. What is
the exact molarity of NaOH solution?
NaOH + HCl  NaCl + H2O
(Balanced)
At the End Point:
Moles of HCl = Moles of NaOH
Moles of Solute
Molarity ( M ) 
Volume of Solution ( Liters )
Dr. Faady Siouri
Moles of HCl Solution   0.0200 L HCl  0.1000 M HCl   2.000 103 mole HCl
 1 mole NaOH 
3
2.000 10 mole HCl 

2.000

10
mole NaOH

 1 mole HCl 
3
18.47 mL NaOH Solution  0.01847 L NaOH Solution
 2.000 103 mole NaOH 
Molarity of NaOH  
  0.1083 M NaOH
 0.01847 L NaOH Solution 
OR
[ M  V ] Acid  [ M  V ]Base
[0.1000  20.00]HCl  [ M 18.47]NaOH
M NaOH  0.1083 M NaOH
Dr. Faady Siouri
Example 5.18: A 0.500 g sample of a drug containing aspirin, monoprotic acid,
HC9H7O4, required 21.50 mL of 0.100 M NaOH for complete neutralization.
What percent by mass of the drug was aspirin?
HC9H7O4 + NaOH  NaC9H7O4 + H2O
(Balanced)
Given: 1 mole HC9H7O4 = 180.2 g
At the End Point:
Moles of HC9H7O4= Moles of NaOH
Moles of Solute
Molarity ( M ) 
Volume of Solution ( Liters )
Dr. Faady Siouri
Moles of NaOH So ln   0.0215 L NaOH  0.100 M NaOH   0.00215 mole NaOH
 1 mole HC9 H 7O4 
0.00215 mole NaOH 
  0.00215 mole HC9 H 7O4
 1 mole NaOH 
 180.2 g HC9 H 7O4 
0.00215 mole HC9 H 7O4 
  0.387 g HC9 H 7O4
 1 mole HC9 H 7O4 
Mass of HC9 H 7O4
% HC9 H 7O4 
100
Mass of Drug Sample
0.387 g HC9 H 7O4

100  77.4%
0.500 g Drug Sample
Dr. Faady Siouri
Practice Question: In the presence of phenolphthalein indicator, 25 mL
portion of 0.5 M NaOH required 50 mL solution of HCl solution to
reach the end point. What is the concentration of the HCl solution?
Dr. Faady Siouri
Department of Industrial Chemistry
Chemistry for Nursing Slides
by
Dr. Faady Siouri
Chapter 16: Acid-Base Equilibrium in Aqueous Solutions
Fall 2022
Dr. Faady Siouri
Sec. 16.1: The Ionization of Water
• Water is very weak electrolyte.
• Electrolytes are chemicals that break into ions in water.
• Strong electrolytes include the strong acids,
strong bases, and salts. These chemicals
completely dissociate into ions in aqueous solution.
Dr. Faady Siouri
• The equilibrium expression for water autoionization (self-ionization)
reaction is:
[ H 3O  ][OH  ]
2


KC 

K
[
H
O
]

[
H
O
][
OH
]
C
2
3
2
[ H 2O ]
[ H 2O ]  Constant 
KW  [ H 3O  ][OH  ]
• Kw is called ion product constant, ionization constant or dissociation
constant of water. Its value varies with temperature, at 25°C it has a
value of (1.0×10-14).
Dr. Faady Siouri
1.0 10
14
Because


 2
 2
 [ H 3O ][OH ]  [ H 3O ]  [OH ]
[ H 3O  ]  [OH  ]
1.0 107  [ H 3O  ]  [OH  ]
Solutions in which
[ H 3O  ]  [OH  ] are Neutral
[ H 3O  ]  [OH  ] are Acidic
[ H 3O  ]  [OH  ] are Basic
Dr. Faady Siouri
Sec. 16.2: Solutions of Strong Acids and Bases
• Strong acids and also strong bases are 100% dissociated in aqueous
solution.
Acid:
HNO3

H 2O  H 3O 

NO3
• This means that [H3O+] of a strong acid is equal to the concentration of the

acid because all the acid dissociates  [ HNO3 ]  [ H 3O ]
• [H3O+] from autoionization of H2O is negligible, because the autoionization
is shifted to the left.
Base:
NaOH ( s )
Ca (OH ) 2 ( s )
 Na 
 Ca 2
 OH 
 2OH 
• [OH-] from autoionization of H2O is negligible.
Dr. Faady Siouri
Example 16.1:
a) What is [OH-] in 0.0010 M HCl solution?
b) What is the H+ concentration derived from the dissociation of the
solvent?
a)
HCl
 H
 Cl 
Initial (M)
0.0010 M
0.0 M
0.0 M
Final (M)
0.0 M
0.0010 M
0.0010 M
[ H  ]  0.0010 M
Dr. Faady Siouri
1.0  10
14


 [ H 3O ][OH ]
1.0 1014  (0.0010 M )[OH  ]

[OH ]  1.0  10
11
M
b) [OH-] derived from H2O = [H+] derived from H2O = 1.0×10-11 M.
Dr. Faady Siouri
Example 16.1: What are the concentrations of H+ and OH- in a 0.0040 M
solution of Ca(OH)2? What part of the [OH-] in the solution comes from
the ionization of water?
Ca (OH ) 2
 Ca 2 
 2OH 
Initial (M)
0.0040 M
0.0 M
0.0 M
Final (M)
0.0 M
0.0040 M
0.0080 M
[OH  ]  0.0080 M
1.0  1014  [ H 3O  ][OH  ]
1.0 1014  [ H  ](0.0080 M )
[ H  ]  1.2  10 12 M  [OH  ] from H 2O
Dr. Faady Siouri
Hydronium Ion, Hydrated Proton, H3O+
• The size of proton is very small (10–15 m) compared to (10–10 m), the
diameter of an average atom or ion.
• Such exceedingly small charged particle can not exist as separate entity in
aqueous solution due to the strong attraction for the negative pole (the O
atom).
• Thus proton exists in the hydrated form as hydronium ion (H3O+)
Dr. Faady Siouri
Sec. 16.3: The pH Concept
pH   log[ H  ]

pOH   log[OH ]
• We use pH to simplify the expression of H+ concentration which may
be very small or of high value.

pH   log[ H ]
2
  log 1.5  10  1.8
  log 1.6  10
12
 11.8
Dr. Faady Siouri


KW  [ H 3O ][OH ]
log KW  log[ H  ]  log[OH  ]
 log KW   log[ H  ]  ( log[OH  ])
pKW  pH  pOH
14  pH  pOH
at 25 oC
Solution is
Neutral
Acidic
Basic
pH = 7
pH < 7
pH > 7
Dr. Faady Siouri
Measuring pH: Acid – Base Indicators and
pH Meters
• Acid – base indicators are organic compounds whose color depends on the
pH of the solution in which they are dissolved.
Example:
 Litmus paper is pink in acidic solution and blue in basic solutions.
• Other pH test papers contain mixture of indicator dyes and used to estimate
an approximate value of pH.
 pH meters: electronic devices used to measure pH with a high degree of
precision and accuracy.
Dr. Faady Siouri
pH and pOH Calculations
Example 16.3: What is the pH of a 0.0020 M solution of HCl?
HCl  H 
 Cl 
Initial (M)
0.0020 M
0.0 M
0.0 M
Final (M)
0.0 M
0.0020 M
0.0020 M
pH   log[ H  ]
  log0.0020  2.7
Dr. Faady Siouri
Example 16.4: What is the pH of a 5.00 x 10-4 M solution of NaOH at
25°C?
NaOH
 OH
0.0 M
0.0 M
 Na
Initial (M) 5.00x10-4 M
Final (M)

5.00x10-4 M
0.0 M

5.00x10-4 M
pOH   log[OH  ]
  log (5.0  10 4 )  3.30
14  pH  pOH
14  pH  3.30

pH  10.70
Dr. Faady Siouri
Example 16.5: A sample of orange juice was found to have a pH of
3.80. What were the H+ and OH- concentrations in the juice?
pH   log[ H  ]


 pH  log[ H ]  [ H ]  10
 pH
 10
3.8
4
 1.6 10 M
14
KW 1.0 10
11
[OH ]   
 6.3 10 M
4
[ H ] 1.6 10

Dr. Faady Siouri
Sec. 16.4: Conjugate Acid-Base Systems in
Aqueous Solutions
 Bronsted – Lowry Concept of Acids and Bases:
Acid: a proton H+ donor
Base: a proton H+ acceptor
conjugate pair
H 2 O + H 2O
acid base
H3O+ + OHacid base
conjugate pair
conjugate pair
NH4+ + H2O
acid base
NH3 + H3O+
base acid
conjugate pair
Dr. Faady Siouri
Equilibrium Constant Expressions for acids
in water
• If we represent an acid by the general formula HA, then it reacts with
water according to the equation: HA  H 2O
A  H 3O 
[ H 3O  ][ A ]
KC 
[ HA][ H 2O]
• The concentration of H2O in dilute aqueous solutions is constant and
can be incorporated into Kc to give Ka.
K C [ H 2O ]  K a
[ H 3O  ][ A ]
 Ka 
[ HA]
Dr. Faady Siouri
Equilibrium Constant Expressions for bases
in water
• If we use the symbol B to stand for a base, then it reacts with water according to
B  H 2O
the equation:
BH   OH 
The equilibrium constant expression is:
[ BH  ][OH  ]
KC 
[ B][ H 2O]
• The concentration of H2O in dilute aqueous solutions is constant and can be
incorporated into Kc to give Kb.
K C [ H 2O ]  K b
[ BH  ][OH  ]
 Kb 
[ B]
The larger the value of Ka or Kb the stronger is the acid or the base.
Dr. Faady Siouri
The relation between Ka and Kb for an
acid – base conjugate pair
Example:
NH 3  H 2 O
NH 4   H 2 O
NH 4   OH 
NH 3  H 3O 
[ NH 4  ][OH  ]
Kb 
[ NH 3 ]
[ H 3O  ][ NH 3 ]
Ka 
[ NH 4  ]
[ NH 4  ][OH  ] [ H 3O  ][ NH 3 ]


Kb  K a 
x

[
H
O
][
OH
]
3

[ NH 3 ]
[ NH 4 ]
K b  K a  KW
• From this relation, as Ka gets larger, Kb gets smaller. As a result, the stronger is the
acid the weaker is its conjugate base.
• The value of Ka and Kb can be determined experimentally by measuring pH of a
solution of a known concentration.
Dr. Faady Siouri
Example 16.6: A 0.10 M acetic acid solution has a pH value of 2.88. Calculate
Ka for acetic acid. What percentage of the acetic acid dissociated?



CH 3COOH
CH 3COO
Initial (M)
0.10
0.00
0.00
Change (M)
-x
+x
+x
0.10 - x
x
x
Equilibrium (M)


2
[ H ][CH 3COO ]
x
Ka 

[CH 3COOH ]
0.10  x
( aq )
H
( aq )
Notice: x is the concentration of H+.
• Since “x” is the same as the [H+] at equilibrium and the pH of the solution
is known (pH is given to be 2.88), “x” is determined as follows:
Dr. Faady Siouri
pH   log[ H  ]
 pH  log[ H  ]  [ H  ]  10 pH  102.88  1.3 103 M  x
x2
(1.3  103 ) 2
5
Ka 


1.7

10
0.10  x (0.10  1.3  103 )
 Now, what percentage of the acetic acid dissociated?
concentration of acid or base dissociated
Percent Dissociation 
 100
concentration of acid or base available
1.3  103 M
Percent Dissociation 
 100  1.3%
0.1 M
Dr. Faady Siouri
Example 16.7: A 0.010 M NH3 solution was prepared and it was
determined that the NH3 had undergo 4.2% ionization. Calculate Kb for
NH3.
NH 3  H 2 O
NH 4

 OH

concentration of acid or base dissociated
Percent Dissociation 
 100
concentration of acid or base available
x
4.2% 
 100
0.010 M

x  4.2  104 M
x  4.2  104 M  [OH  ] at equilibrium
Dr. Faady Siouri
NH 3  H 2 O
Initial (M)
0.010
Change (M)
- 4.1 x 10-4 M
Equilibrium (M)
0.010 - 4.1 x 10-4 M
NH 4

0.00
 OH

0.00
+ 4.1 x 10-4 M + 4.1 x 10-4 M
4.1 x 10-4 M
4.1 x 10-4 M
[ NH 4  ][OH  ] (4.1 104 )(4.1 104 )
Kb 

[ NH 3 ]
(0.010  4.1 104 )
K b  1.8  105
Dr. Faady Siouri
Example 16.8: What are the concentrations of all the species present in a 0.50 M
solution of acetic acid. For acetic acid, CH3COOH, Ka=1.8x10-5
CH 3COOH
CH 3COO  ( aq )

H  ( aq )
Initial (M)
0.50
0.00
0.00
Change (M)
-x
+x
+x
0.50 - x
x
x
Equilibrium (M)


2
2
[
H
][
CH
COO
]
x
x
3
1.8  105 


[CH 3COOH ]
0.50  x 0.50
x 2  0.9  105
x  3.0  103 M
Make this assumption.
Then check if your
assumption was valid
Dr. Faady Siouri


3
x  [ H ]  [CH 3COO ]  3.0  10 M
 Check your assumption (Later we show how to properly check your
assumption)
[CH 3COOH ]  0.5  x  0.50 M
14
K
1.0

10
12
W
[OH  ] 


3.3

10
[ H 3O  ]
3.0 103
[OH  ]  3.3 1012 M
Dr. Faady Siouri
Calculating the pH of a Solution That Contains
a Weak Acid and a Strong Acid
Example 16.9: What is the pH of a solution that contains 0.10 M HCl
and 0.10 M CH3COOH? For acetic acid CH3COOH, Ka = 1.8×10-5.
CH 3COOH
Initial (M)
Change (M)
Equilibrium (M)
CH 3COO  ( aq )

H  ( aq )
0.10
0.00
0.00
-x
+x
+x
0.10 - x
x
x
HCl  H

 Cl

Initial (M)
0.10 M
0.0 M
0.0 M
Final (M)
0.0 M
0.10 M
0.10 M
Dr. Faady Siouri
[ H  ][CH 3COO  ]
Ka 
[CH 3COOH ]
(0.10  x) x 0.10 x
1.8  10 

(0.10  x)
0.10
5
x  1.8  105
Make this assumption.
Then check if your
assumption was valid
So, [ H  ]  0.10  1.8 105  0.10 M
pH   log[ H  ]   log 0.10  1
 Therefore, the pH is determined from H+ coming from the strong acid.
Dr. Faady Siouri
Calculating the pH of a Solution of a
Weak Base
Example 16.10: What is the pH of a 0.010 M solution of the weak base
diethylamine,(C2H5)2NH, for which Kb = 9.6×10-4 ?
(C2 H 5 )2 NH ( aq )  H 2O
Initial (M)
Change (M)
Equilibrium (M)
(C2 H 5 )2 NH 2  ( aq )  OH  ( aq )
0.010
0.00
0.00
-x
+x
+x
0.010 - x
x
x
[(C2 H 5 ) 2 NH 2  ][OH  ]
Kb 
[(C2 H 5 ) 2 NH]
Dr. Faady Siouri
9.6  104
( x)( x)
x2


(0.010  x) 0.010
x  3.1 103 M  [OH  ]
pOH   log 3.1 103  2.51
pH  14  pOH
Make this assumption.
Then check if your
assumption was valid
 14  2.51  11.49
 If ([B]/Kb) ≥ 100, neglect the value of x.
0.010
 10.41
4
9.6 10
 If x is less than 5% of the initial concentration we can neglect it.
In this example x is about 30% of the initial concentration 0.010 M, so it is not
negligible.
Dr. Faady Siouri
3.1103 M
Percent dissociation 
 100  33.1%
0.010 M
5%
Therefore, solve for x exactly using the quadratic equation
[(9.6 104 )  0.010  (9.6 104 ) x]  x 2
x 2  9.6  104 x  9.6  106  0
b  b  4ac
x

2a
2
4
9.6 10 
 9.6 10 
4
2 1
2

 4 1 9.6 106

x  2.7 103 ,  3.6 103
[OH  ]  2.7 103 M
pH  14  pOH

pOH   log 2.7 103  2.57
 14  2.57  11.43
Dr. Faady Siouri
Sec. 16.6: Buffers: The Control of pH
• Buffers usually consist of a weak acid and its conjugate base like
CH3COOH/CH3COONa. The pH of a buffer solution changes very little
when a small amount of strong acid or base is added to it.
• If a small amount of strong acid is added to this buffer, its H3O+ can react
with the conjugate base.
H3O+ + CH3COO-
CH3COOH + H2O
• If a small amount of strong base is added to this buffer, its OH- can react
with the acid to give the conjugate base.
OH- + CH3COOH
CH3COO- + H2O
Dr. Faady Siouri
 Also Buffers may consist of a weak base and its conjugate acid like
NH3/NH4Cl.
• If a small amount of strong acid is added to this buffer, its H3O+ can
react with the base.
H3O+ + NH3
NH4+ + H2O
• If a small amount of strong base is added to this buffer, its OH- can
react with the conjugate acid to give the base.
OH- + NH4+
NH3 + H2O
Dr. Faady Siouri
Buffer Calculations
Example 16.11: What is the pH of a buffer solution prepared by
dissolving 0.10 mole of CH3COONa and 0.20 mol of CH3COOH in
enough water to give 1.00 L of solution? (Ka for CH3COOH = 1.8×10-5)
CH 3COOH
Initial (M)
Change (M)
Equilibrium (M)
CH 3COO  ( aq )
H  ( aq )

0.20
0.00
0.00
-x
+x
+x
0.20 - x
x
x
CH 3COONa ( aq )
 CH 3COO  ( aq )

Na  ( aq )
Initial (M)
0.10 M
0.0 M
0.0 M
Final (M)
0.0 M
0.10 M
0.10 M
Dr. Faady Siouri
[ H  ][CH 3COO  ]
Ka 
[CH 3COOH ]
(0.10  x) x 0.10 x
1.8  10 

(0.20  x)
0.20
5
x  3.6 105 M  [ H  ]
pH   log[ H  ]
  log 3.6 105  4.44
Dr. Faady Siouri
Example 16.12: What ratio of acetic acid to sodium acetate concentration is
needed to form a buffer whose pH is 5.70? (Ka for CH3COOH = 1.8×10-5) ?
CH 3COO  ( aq )
CH 3COOH
CH 3COONa ( aq )
 CH 3COO

( aq )
H  ( aq )


Na

( aq )
pH   log[ H  ]
 pH  log[ H  ]  [ H  ]  10 pH  105.70  2.0 106 M
[ H  ][CH 3COO  ]
Ka 
[CH 3COOH ]
[ H  ] [CH 3COOH ] 2.0  106 1




5
Ka
[CH 3COO ] 1.8  10
9
Dr. Faady Siouri
OR

[CH 3COO ]
pH  pK a  log
[CH 3COOH ]
[CH 3COO  ]
5.70  4.74  log
[CH 3COOH ]

[CH 3COO ]
[CH 3COOH ] 1
 9.12 


[CH 3COOH ]
[CH 3COO ] 9
Dr. Faady Siouri
How Effective a Buffer is at Holding the pH Nearly Constant?
Suppose we have CH3COOH/CH3COONa buffer and the concentration of each is 1.0 M, so the pH of this
buffer is 4.74. What happens to pH when we add 0.20 mole HCl to a liter of this buffer? Ka = 1.8 x 10-5
CH 3COOH
CH 3COO  ( aq )

H  ( aq )
Initial (M)
1.0
0.00
0.00
Change (M)
-x
+x
+x
1.0 - x
x
x
Equilibrium (M)
CH 3COONa ( aq )  CH 3COO  ( aq )  Na  ( aq )
Initial (M)
Final (M)
1.0 M
0.0 M
0.0 M
0.0 M
1.0 M
1.0 M
HCl  H 
 Cl 
Initial (M)
0.20 M
0.0 M
0.0 M
Final (M)
0.0 M
0.20 M
0.20 M
Dr. Faady Siouri
[CH 3COOH ]
[H ]  Ka
[CH 3COO  ]

[1.0  x  0.20]
[ H ]  1.8  10
[1.0  x  0.20]

5
Assumption
[1.0  0.20]
[ H ]  1.8  10
[1.0  0.20]

5
[ H  ]  2.7  105 M
pH   log 2.7  105  4.57
Dr. Faady Siouri
 If we add 0.20 mole of OH- to a the 1 L of our original buffer:
OH 
 CH 3COOH
 CH 3COO 
 H 2O
[CH 3COOH ]
[H ]  Ka
[CH 3COO  ]

[1.0  x  0.20]
[ H ]  1.8  10
[1.0  x  0.20]

5 [1.0  0.20]
[ H ]  1.8  10
[1.0  0.20]

5
Assumption
[ H  ]  1.2  105 M
5
pH   log1.2  10  4.92
Dr. Faady Siouri
Example 16.13: A buffer was prepared by mixing 200 mL of a 0.60 M
NH3 solution and 300 mL of a 0.30 M NH4Cl solution.
a) What is the pH of this buffer, if we assume a final volume of
500 mL? For NH3, Kb = 1.8×10-5.
b) What will be the pH after 0.020 mol of HCl is added?
a) M1V1 = M2V2
(0.60 M )(200 mL)  M 2 (500 mL)  M 2  0.24 M NH 3 Solution
(0.30 M )(300 mL)  M 2 (500 mL)  M 2  0.18 M NH 4Cl Solution
Dr. Faady Siouri
NH 3  H 2 O
NH 4

 OH
Initial (M)
0.24
0.00
0.00
Change (M)
-x
+x
+x
0.24 - x
x
x
Equilibrium (M)
NH 4Cl  NH 4

 Cl


Initial (M)
0.18 M
0.0 M
0.0 M
Final (M)
0.00 M
0.18 M
0.18 M
[ NH 4  ][OH  ]
( x  0.18) x
5
Kb 
 1.8  10 
[ NH 3 ]
(0.24  x)
x  [OH ]  2.4  10


0.18 x
1.8  10 
0.24
5
5
pOH   log 2.4  105  4.62
pH  14  4.62  9.38
Dr. Faady Siouri
0.020 mole
b) Molarity ( HCl ) 
 0.040 M
0.500 L
HCl  H 
 Cl 
Initial (M)
0.040 M
0.0 M
0.0 M
Final (M)
0.0 M
0.040 M
0.040 M
NH 3  H 3O   NH 4   H 2O
[ NH 4  ][OH  ]
( x  0.18  0.040) x
5
Kb 
 1.8  10 

[ NH 3 ]
(0.24  x  0.040)
(0.18  0.040) x
1.8  10 
0.24  0.040
5
x  [OH  ]  1.6  105
pOH   log1.6  105  4.80
pH  14  4.80  9.20
Dr. Faady Siouri
Sec. 16.10: Acid-Base Titrations
Titration of a Strong Acid and a Strong Base
• If we titrate 25.00 mL of 0.10 M HCl with 0.10 M NaOH gradually we can determine
the pH throughout the titration by calculating H+ concentration. The plot of pH versus
the volume of NaOH added should give a graph with the equivalence point at pH = 7
because neither of the ions of the salt left in solution (NaCl) undergo hydrolysis.
• The pH of 0.1 M HCl solution before starting the titration is equal to 1.
 Calculate the pH after adding 10 mL of 0.10 M NaOH solution.
Dr. Faady Siouri
• The neutralization reaction is
NaOH

HCl  NaCl
 H 2O
Mole HCl  (0.1 mole) (0.025L)  2.5  103 mole HCl
Mole NaOH  (0.1 mole) (0.010 L)  1.0  103 mole NaOH
• The number of moles of HCl remaining =
(2.5  103 )  (1.0  103 )  1.5  103 mole HCl
• The molar concentration of HCl is now =
3
1.5 10 mol
2
[ HCl ] 
 4.3 10 M
0.035 L
2
pH   log 4.3  10  1.37

 [H ]
Total Volume
Dr. Faady Siouri
• The concentrations of HCl after further additions of NaOH have occurred are summarized in the table below
• The table shows that the pH increases slowly at first, then rises rapidly near the equivalence point, and finally
levels off gradually after the equivalence point is reached.
Volume of
HCl (mL)
Volume of
NaOH (mL)
Total volume
(mL)
Moles of HCl
Moles of
NaOH
Molarity of
substance in
excess
pH
25.00
0.00
25.00
2.5×10-3
0
0.10 (H+)
1.00
25.00
10.00
35.00
2.5×10-3
1.0×10-3
4.3×10-2 (H+)
1.37
25.00
24.99
49.99
2.5×10-3
2.499×10-3
2.0×10-5 (H+)
4.70
25.00
25.00
50.00
2.5×10-3
2.5×10-3
0
7.00
25.00
25.01
50.01
2.5×10-3
2.501×10-3
2.0×10-5 (OH-)
9.30
25.00
26.00
51.00
2.5×10-3
2.60×10-3
2.0×10-3 (OH-)
11.30
25.00
50.00
75.00
2.5×10-3
5.0×10-3
3.3×10-2 (OH-)
12.52
Dr. Faady Siouri
Dr. Faady Siouri
Department of Industrial Chemistry
Chemistry for Nursing Slides
by
Dr. Faady Siouri
Chapter 4: The Periodic Table and Some Properties of Elements
Fall 2022
Dr. Faady Siouri
Sec. 4.4: Atomic Numbers and the Modern
Periodic Table
In the periodic table used today, elements are arranged in order of
increasing atomic numbers.
Dr. Faady Siouri
• The table consists of a number of rows called periods and are
identified by Arabic numbers (1, 2, 3, …)
• The vertical columns are called groups and are identified by Roman
numeral and a letter, A or B.
• Elements in groups with
letter A (IA to VIIA) and
group 0 are called
representative elements.
Dr. Faady Siouri
• Elements in groups with letter B (IB to VIIIB) are called transition
elements.
• The two rows below the table are called inner transition elements. The first
row follows the element lanthanum La and are called lanthanides (rare
earth elements). The second row follows actinium Ac and called actinides.
Group IA: Alkali metals
Group IIA: Alkaline earth metals
Group VIIA: Halogens
Group 0: Noble gases (inert gases)
Dr. Faady Siouri
Sec. 4.9: Naming Chemical Compounds
Binary Compounds of a Metal and a Nonmetal :
Binary Compound: Compound composed of atoms of only two different
elements.
• If the two elements are a metal and a nonmetal, the metal is named first,
followed by the nonmetal.
 If the metal has one oxidation state, we use the English name for the metal
followed by the name of the nonmetal + ide.
NaCl: sodium chloride
SrO: strontium oxide
Al2S3: aluminum sulfide
Mg3P2: magnesium phosphide
Dr. Faady Siouri
 For metals having more than one oxidation state, two methods are used for naming:
The Old method: - ous
- ic
Cr
chromium
for lower oxidation state
for higher oxidation state
Cr3+
CrCl3
chromic chloride
Cr2+
CrCl2
chromous chloride
• If metal has a symbol derived from the Latin name, its Latin name is generally used.
Fe
iron
Fe3+
FeBr3
ferric bromide
Fe2+
FeBr2
ferrous bromide
For more examples see Table (4.8)
Dr. Faady Siouri
The preferred method for naming ions like this is the Stock method. In this
method a Roman numeral equals to the metal's oxidation state is placed in
parentheses following the English name of the metal.
Formula
Old Method
Stock Method
Fe2+
 FeCl2
ferrous chloride
iron(II) chloride
Fe3+
 FeCl3
ferric chloride
iron(III) chloride
Cu1+ 
Cu2O
cuprous oxide
copper(I) oxide
Cu2+ 
CuO
cupric oxide
copper(II) oxide
Dr. Faady Siouri
Common metals that have more than one oxidation state:
Ion
Name (Stock Method)
Name (Old Method)
Fe3+
iron(III)
ferric
Fe2+
iron(II)
ferrous
Cu2+
copper(II)
cupric
Cu+
copper(I)
cuprous
Co3+
cobalt(III)
cobaltic
Co2+
cobalt(II)
cobaltous
Cr3+
chromium(III)
chromic
Cr2+
chromium(II)
chromous
Pb4+
lead(IV)
plumbic
Pb2+
lead(II)
plumbous
Sn4+
tin(IV)
stannic
Sn2+
tin(II)
stannous
Dr. Faady Siouri
Name, symbol, and charge for some of the most commonly used elements :
Symbol /
Charge
Name
Symbol /
Charge
Name
Symbol /
Charge
Name
Li +
Lithium
Ba 2+
Barium
O 2-
Oxygen
Na +
Sodium
B 3+
Boron
S 2-
Sulfur
K+
Potassium
Al 3+
Aluminum
F-
Fluorine
Mg 2+
Magnesium
C 4+,4-
Carbon
Cl -
Chlorine
Ca 2+
Calcium
N 3-
Nitrogen
Br -
Bromine
Sr 2+
Strontium
P 3-
Phosphorus
I-
Iodine
Non-metal examples: H, C, N, O, P, S, F, Cl, Br, I, He, Ne, Ar
Semi-metals examples: B, Si
Binary Compounds of Nonmetals:
• For naming, Greek prefixes are used to indicate the number of atoms
of each kind in one molecule of the substance.
• In the name, the first element in the formula is given its English name.
The second element is specified by adding the suffix – ide to the root of
the element's name.
Dr. Faady Siouri
P4O10
tetraphosphorus decoxide
NO2
nitrogen dioxide
N2O4
dinitrogen tetroxide
N2O5
dinitrogen pentoxide
CO
CO2
carbon monoxide
carbon dioxide
Table 4.7: Names of monatomic anions derived from nonmetals
Dr. Faady Siouri
Compounds Containing Polyatomic Ions:
Na2CO3
Ba(OH)2
Ca(C2H3O2)2
(NH4)2SO4
Formula
CuSO4
Fe2(C2O4)3
sodium carbonate
barium hydroxide
calcium acetate
ammonium sulfate
Stock Method
Old Method
copper(II) sulfate
iron(III) oxalate
cupric sulfate
ferric oxalate
See Table (4.4) p. 99 : some common polyatomic ions
- ide ending is added to monatomic anions. Hydroxide (OH-) and cyanide (CN-) are
exceptions.
Dr. Faady Siouri
Some Common Polyatomic Ions:
NH4+
ammonium
SO42-
sulfate
CO32-
carbonate
SO32-
sulfite
HCO3-
bicarbonate
NO3-
nitrate
ClO3-
chlorate
NO2-
nitrite
Cr2O72-
dichromate
SCN-
thiocyanate
CrO42-
chromate
OH-
hydroxide
C2H3O2-
acetate
C2O42-
oxalate
PO33-
phosphite
PO43-
phosphate
ClO4-
perchlorate
ClO-
hypochlorite
ClO2-
chlorite
BrO3-
bromate
Dr. Faady Siouri
Dr. Faady Siouri
Binary Acids:
Acids: substances that release H+ ions when they are dissolved in water.
Binary acid (hydro acid): water solution of a binary compound of
hydrogen and a nonmetal.
HF(aq)
hydrofluoric acid
HCl(aq)
HBr(aq)
HI(aq)
H2S(aq)
HCl(g)
hydrochloric acid
hydrobromic acid
hydroiodic acid
hydrosulfuric acid
hydrogen chloride
Dr. Faady Siouri
• Neutralization Reaction: Reaction between an acid and a base having
hydroxide ion, an ionic compound is formed (salt).
• Salt: is used for ionic compounds not containing oxide ion or hydroxide ion.
Salts formed from hydro … ic acids contain anions end in – ide.
• Oxoacids: Are acids that contain hydrogen, oxygen and at least one other
element (usually a nonmetal).
H2SO4
sulfuric acid
S is +6
H2SO3
sulfurous acid
S is +4
The prefix hydro is not used here.
Dr. Faady Siouri
Anion derived from – ic acid has a name ends in – ate.
Anion derived from – ous acid has a name ends in – ite.
H2SO4
sulfuric acid
SO42-
sulfate
H2SO3
HNO3
HNO2
HClO3
HClO2
HClO
HClO4
sulfurous acid
nitric acid
nitrous acid
chloric acid
chlorous acid
hypochlorous acid
perchloric acid
SO32NO3NO2ClO3ClO2ClOClO4-
sulfite
nitrate
nitrite
chlorate
chlorite
hypochlorite
perchlorate
Dr. Faady Siouri
Dr. Faady Siouri
Department of Industrial Chemistry
Chemistry for Nursing Slides
by
Dr. Faady Siouri
Chapter 7: Electronic Structure and the Periodic Table
Fall 2022
Dr. Faady Siouri
Sec. 7.3: Quantum Numbers
• Quantum numbers are used to describe the distribution of electrons in
atomic orbitals in hydrogen and other atoms.
• The four quantum number:
1) The principle quantum number (n)
2) The azimuthal quantum number (l)
3) The magnetic quantum numbers (ml)
4) The electron spin quantum number (ms)
Dr. Faady Siouri
 The principle quantum number (n): Determines the size of orbitals in
the energy levels.
• The size of the orbital and the energy of the energy level increases with
increasing n.
• The value of n (n = 1, 2, 3,…,∞ ) can be a positive integer starting at 1,
where n=1 designates the first principle
shell (ground state or lowest energy state).
Dr. Faady Siouri
 The azimuthal quantum number (l): This is also called the orbital angular
momentum quantum number.
• This quantum number tells us the shape of the orbitals.
• Each shell is composed of one or more subshells specified by l. For a given shell
(n), l takes the values of: 0, 1, …, (n-1).
For n = 1, l = 0
n = 3, l = 0, 1, 2
n = 2, l = 0, 1
n = 4, l = 0, 1, 2, 3
n = n, l = 0, 1, 2, 3, …, n-1
Dr. Faady Siouri
l = 0 (s orbitals)
l = 1 (p orbitals)
l = 2 (d orbitals)
7.6
Dr. Faady Siouri
 The magnetic quantum numbers (ml): Describes the orientation of the orbital in space
relative to other orbitals.
• Each subshell is composed of one or more orbitals specified by its value of ml.
• For each value of l, there are (2l+1) values of ml range between +l and –l.
[-l, (-l+1), …., 0, …, (+l-1), +l]
• The number of ml values for a given value of l = the number of orbitals in the subshell.
For
l = 0, ml = 0,
l = 1, ml = -1, 0, 1
l = 2, ml = -2, -1, 0, 1, 2
l = 3, ml = -3, -2, -1, 0, 1, 2, 3
s subshell, one orbital
p subshell, three orbitals
d subshell, five orbitals
f subshell, seven orbitals
Dr. Faady Siouri
Dr. Faady Siouri
Example: List the values of n, l, and ml for orbitals in the 4d subshell.
For 4d: n = 4 and l = 2
ml values range between –l, (–l + 1), …. , 0, ….. , (+l – 1), +l
ml values for 4d subshell are: –2, –1, 0, 1, 2
Example: Give the values of the quantum numbers associated with the orbitals
in the 3p subshell.
For 3p: n = 3 and l = 1
ml values rages between –l, (–l + 1), …., 0, ….., (+l – 1), +l
ml values for 3p subshell are: –1, 0, 1
Dr. Faady Siouri
Example: What is the total number of orbitals associated with the principal
quantum number n = 3?
l value ranges between 0 and (l – 1)
For n = 3 and l = 0, 1, 2
ml values rages between –l, (–l + 1), …., 0, ….., (+l – 1), +l
ml values for l = 0 is 0
ml values for l = 1 is –1, 0, +1
ml values for l = 2 is –2, –1, 0, +1, +2
(one 3s orbital)
(three 3p orbitals)
(five 3d orbitals)
The total number if orbitals is 1 + 3 + 5 = 9
The total number of orbitals in the shell n is n2. So we have 32 =9
Dr. Faady Siouri
Energy Level Diagram for Multi Electron
Atom/Ion
4f
6s
5p
4d
5s
4p
3d
4s
Energy
3p
3s
2p
2s
 The average energies of the shells increases with increasing
the value of n.
 As n becomes larger, the spacing between successive shells
becomes less, so we observe overlap among the subshells.
1s
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f
Dr. Faady Siouri
Sec. 7.4: Electron Spin and the Pauli
Exclusion Principle
• Another quantum number determining how the electron is spinning is
the electron spin quantum number (ms).
• The electron spin quantum number value is either +1/2 or -1/2.
Dr. Faady Siouri
• Pauli exclusion principle: No two electrons in any one atom can have all four
quantum numbers the same. This limits the number of electrons in any given
orbital to two.
• Two electrons in the same atomic orbital have the same n, l, and ml but differ in
their spins ms .
2He:
1s2
He
He
1s2
He
1s2
1s2
• Maximum number of electrons in any shell (n) = 2n2.
Dr. Faady Siouri
Sec. 7.5: The Electron Configuration of Elements
• Electron configuration is how the electrons are distributed among the
various atomic orbitals in an atom.
• The four quantum numbers n, l, ml and ms are considered as the
address of an electron in an atom.
• For a 2s electron:
n = 2, l = 0, ml = 0 and ms = +1/2 or -1/2.
(2, 0, 0, +1/2) or (2, 0, 0, -1/2)
Dr. Faady Siouri
• For H atom in the ground state, the electron resides in the 1s orbital as
shown in the diagram below.
Number of electrons
in the orbital or subshell
1s1
Angular momentum
quantum number (l)
Principal quantum
number (n)
Orbital diagram
H
1s1
Dr. Faady Siouri
Order of Orbitals (Filling) in Multi-electron Atom
 Arrangement of subshells in order of increasing energy
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d
Dr. Faady Siouri
• When we distribute the electrons in orbitals, they will be found in the
lowest energy levels available.
Called orbital diagram
1
H:
1s
1
Spins or electrons are paired
2
He:
1s
2
22s1
Li:
1s
3
[He]2s1
[He] noble gas core
4Be:
1 s 2 2 s2
[He]2s2
Dr. Faady Siouri
• We use noble gas for abbreviation because we focus our attention on
the outer most shell with high value of n responsible for chemical
changes in chemical reactions.
• Core electrons: electrons in shells below the outer one.
22s22p1
B:
1s
5
22s22p2
C:
1s
6
 Question: How are the 2 electrons in 2p2
are distributed among the 3 p orbitals??
Dr. Faady Siouri
6C:
Paired in the same orbital
Paired in the different orbital
Spins are in the same direction (unpaired)
Lowest energy (most stable)
Hundʹs rule: Electrons entering a subshell containing more than one orbital
will be spread out over the available orbitals with their spins in the same
direction.
22p3
N:
[He]
2s
7
He
22p4
O:
[He]
2s
8
He
Dr. Faady Siouri
9F:
[He] 2s22p5
He
22p6
Ne:
[He]
2s
10
He
1
Na:
[Ne]
3s
11
Ne
24Cr: [Ar]
4s23d4 = [Ar] 4s13d5
23d9 = [Ar] 4s13d10
Cu:
[Ar]
4s
29
21Sc:
Ar
Ar
[Ar] 4s23d1 = 1s22s22p63s23p64s23d1
Ar
= 1s22s22p63s23p63d14s2
The most stable configuration for the nd subshell is nd10 and nd5.
Filled and half-filled orbitals are more stable
Dr. Faady Siouri
 Arrangement of subshells in order of increasing energy
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s
< 5f < 6d
Example 7.6: What is the electron configuration of antimony (51Sb)?
51Sb:
1s22s22p63s23p64s23d104p65s24d105p3
1s22s22p63s23p63d104s24p64d105s25p3
Example 7.7: What is the electron configuration of lead (82Pb)?
22s22p63s23p64s23d104p65s24d105p66s24f145d106p2
Pb:
1s
82
1s22s22p63s23p63d104s24p64d104f145s25p65d106s26p2
Dr. Faady Siouri
• Electrons in the outer shell are called valence shell electrons. The word
valence relates to the number of chemical bonds an atom is able to form.
Example: Tin (50Sn), from period 5 and group IVA (Latin for 4). What is the
outer shell electron configuration?
50Sn:
[Kr] 4d105s25p2
Outer shell electrons 5s25p2
Example 7.8: What is the outer-shell configuration of silicon (14Si)?
Silicon: period 3 and group IVA
14Si:
[Ne] 3s23p2
Outer-shell configuration = 3s23p2
Dr. Faady Siouri
Example 7.9: Use the periodic table to construct the orbital diagram of
the valence shell of tellurium (52Te).
Te: period 5 and group VIA (Latin for 6)
Te: [Kr] 4d105s25p4
Valence shell: 5s25p4
Kr
Orbital diagram of the valence shell of Te
Dr. Faady Siouri
Dr. Faady Siouri
 Diamagnetic material:
 No attraction for another magnet
 They have the same number of electrons of each spin
 Examples: Copper, silver and gold.
 Paramagnetic material:
 Weakly attracted to magnetic field
 More electrons of one spin than the other
 Examples: Magnesium and lithium.
 Ferromagnetic material:
 Strongly attracted to the magnet due to interactions among paramagnetic atoms in the
solid state.
 Examples: Iron, cobalt, and nickel.
Dr. Faady Siouri
Paramagnetism vs Diamagnetism
Paramagnetic
Unpaired electrons
2p2
Diamagnetic
All electrons paired
2p6
Dr. Faady Siouri
• If the two electrons have the same spin
, their net magnetic
field would reinforce each other and the substance is paramagnetic.
• If the two electrons are paired (antiparallel)
, the magnetic effect
cancel out and the substance is diamagnetic.
22s1
Li:
1s
3
One unpaired electron  Paramagnetic
22s2
Be:
1s
4
Paired electrons  Diamagnetic
22s22p1
B:
1s
5
Paramagnetic
Dr. Faady Siouri