Stoichiometry and mole concept

Mole concept
 Ctd
• The mole concept is a convenient method of
expressing the amount of a substance.
• Any measurement can be broken down into
two parts – the numerical magnitude and the
units that the magnitude is expressed in.
• For example, when the mass of a ball is
measured to be 2 kilograms, the magnitude is
‘2’ and the unit is ‘kilogram’.
 Ctd
• When dealing with particles at an atomic (or
molecular) level, even one gram of a pure
element is known to contain a huge number
of atoms.
• This is where the mole concept is widely
used. It primarily focuses on the unit known as
a ‘mole’, which is a count of a very large
number of particles.
 Ctd
• What is a Mole?
• In the field of chemistry, a mole is defined as the amount
of a substance that contains exactly 6.02214076 * 1023
‘elementary entities’ of the given substance.
• The number 6.02214076*1023 is popularly known as the
Avogadro constant and is often denoted by the symbol ‘NA’.
• The elementary entities that can be represented in moles
can be atoms, molecules, monoatomic/polyatomic ions,
and other particles (such as electrons).
 Ctd
• For example, one mole of a pure carbon-12
(12C) sample will have a mass of exactly 12
grams and will contain 6.02214076*1023 (NA)
number of 12C atoms.
• The number of moles of a substance in a given
pure sample can be represented by the
following formula:
• n = N/NA
 Ctd
• Quantities Related to mole concept and their
Formulae
• 1. Atomic and Molecular Mass
• 2. Molar Mass
• 3. Number of Moles = (Mass of the
Sample)/(Molar Mass)
 Calculations
• 1. How many C atoms are there in a carbon
rod weighing 8 g (C=12
Answer:
• 12 g (RAM of carbon in g or 1 mole) of carbon
have 6.0 x 1023 atoms.
8 g of carbon has,
8g × 6.0 x 1023 / 12 = 4.0 x 1023 atoms
 Ctd
• 2. How many grams of copper (Cu=64) contain
4.5 x 1023 atoms?
• Answer:
6.0 x 1023 atoms are in 1 mole of copper(64 g)
• 4.5 x 1023 atoms are in,
• 4.5 x 1023 x 64/ 6.0 x 1023 = 48 g
 Ctd
• 3. How many ions are there in 13.5 g of copper (II)
chloride, CuCl2?
• Answer:
• Mass of 1 mole of CuCl2 = 64 + (2 x 35.5)= 135g
• There are 3 ions (one copper and two chloride ions) in
CuCl2 molecule.
• i.e. 1 mole or 135 g of CuCl2 contain 3 x 6.0 x 1023 ions
13.5 g of CuCl2 contain.
13.5 x 3 ions x NA/135 = 1.8 x 1023 ions
 Ctd
4. Calculate the mole of each element present in
18.4 g of iron (III) sulphate(Fe2(SO4)3)
 Ctd
Answer:
Molar mass of Fe2(SO4)3 = 400 g
No. of moles of Fe2(SO4)3 = 18.5/400 =
0.046
No. of moles of iron = 0.046 x 2 = 0.092 moles.
No. of moles of sulphur = 0.046 x 3 = 0.138 moles
No. of moles of oxygen = 0.046 x 12 = 0.552 moles
 Ctd
stoichiometry,
• In chemistry, the determination of the
proportions in which elements or compounds
react with one another.
• The rules followed in the determination of
stoichiometric relationships are based on the
laws of conservation of mass and energy and
the law of combining weights or volumes.
 Ctd
• Stoichiometry is a section of chemistry that involves using
relationships between reactants and/or products in a
chemical reaction to determine desired quantitative data.
• In Greek, stoikhein means element and metron means
measure, so stoichiometry literally translated means the
measure of elements.
• In order to use stoichiometry to run calculations about
chemical reactions, it is important to first understand the
relationships that exist between products and reactants
and why they exist, which require understanding how to
balance reactions.
 Ctd
• Balancing
• To balance an equation, it is necessary that
there are the same number of atoms on the
left side of the equation as the right. One can
do this by raising the coefficients.
• The following equation demonstrates the
typical format of a chemical equation:
• 2Na(s)+2HCl(aq)→2NaCl(aq)+H2(g)
 Ctd
• In the above equation, the elements present in
the reaction are represented by their chemical
symbols.
• Based on the Law of Conservation of Mass,
which states that matter is neither created nor
destroyed in a chemical reaction, every chemical
reaction has the same elements in its reactants
and products, though the elements they are
paired up with often change in a reaction.
 Ctd
• Stoichiometric Coefficients
• In a balanced reaction, both sides of the equation
have the same number of elements.
• The stoichiometric coefficient is the number written
in front of atoms, ion and molecules in a chemical
reaction to balance the number of each element on
both the reactant and product sides of the equation.
• Though the stoichiometric coefficients can be
fractions, whole numbers are frequently used and
often preferred.
 Ctd
• This stoichiometric coefficients are useful since they
establish the mole ratio between reactants and products. In
the balanced equation:
• 2Na(s)+2HCl(aq)→2NaCl(aq)+H 2(g)
• we can determine that 2 moles of HCl will react with 2 moles
of Na(s) to form 2 moles of NaCl(aq) and 1 mole of H2(g).
• If we know how many moles of Na reacted, we can use the
ratio of 2 moles of NaCl to 2 moles of Na to determine how
many moles of NaCl were produced or we can use the ratio
of 1 mole of H2 to 2 moles of Na to convert to NaCl. This is
known as the coefficient factor.
 Ctd
• The balanced equation makes it possible to
convert information about the change in one
reactant or product to quantitative data about
another reactant or product. Understanding
this is essential to solving stoichiometric
problems.
 Ctd
• Stoichiometry and Balanced Equations
Calculations
1.Propane (C3H8) burns in this reaction:
C3H8+5O2→4H2O+3CO2
If 200 g of propane is burned, how many g of
H2O
is produced?
 Ctd
• Answer:
Step 1: 200 g C3H8 is equal to 4.54 mol C3H8.
Step 2: Since there is a ratio of 4:1 H2O to C3H8,
for every 4.54 mol C3H8 there are 18.18 mol H2O.
• Step 3: Convert 18.18 mol H2O to g H2O. 18.18
mol H2O is equal to 327.27 g H2O.
 Ctd
• Concentration
• Concentration is a general measurement unit
that reports the amount of solute present in a
known amount of solution
 Ctd
Unit Definition Application
Used for quantitative
reactions in solution and
moles of solute/liter of titrations; mass and
Molarity(M)
solution (mol/L) molecular mass of solute and
volume of solution are
known.
Used for partial pressures of
gases and vapor pressures of
moles of solute/total moles
mole fraction (X) some solutions; mass and
present (mol/mol)
molecular mass of each
component are known.
Used in determining how
colligative properties vary
moles of solute/kg of solvent
molality (m) with solute concentration;
(mol/kg)
masses and molecular mass
of solute are known.
 Ctd
Useful when masses are
[mass of solute (g)/mass of
mass percentage (%) known but molecular
solution (g)] × 100
masses are unknown.
Used in the health sciences,
[mass of solute/mass of
ratio solutions are typically
parts per thousand (ppt) solution] × 103 (g solute/kg
expressed as a proportion,
solution)
such as 1:1000.
Used for trace quantities;
[mass of solute/mass of
masses are known but
parts per million (ppm) solution] × 106 (mg solute/kg
molecular masses may be
solution)
unknown.
Used for trace quantities;
[mass of solute/mass of
masses are known but
parts per billion (ppb) solution] × 109 (µg solute/kg
molecular masses may be
solution)
unknown.
• Calculation
• 1. A solution is prepared by mixing 100.0 mL of
toluene with 300.0 mL of benzene. The densities of
toluene and benzene are 0.867 g/mL and 0.874
g/mL, respectively. Assume that the volume of the
solution is the sum of the volumes of the
components. Calculate the following for toluene.
• mass percentage
• mole fraction
• molarity
 Ctd
• Answer:
• mass percentage toluene = 24.8%
• Xtoluene = 0.219
• 2.35 M toluene
 Preparation of solution
• Solutions of known concentration can be
prepared either by dissolving a known mass
of solute in a solvent and diluting to a desired
final volume or by diluting the appropriate
volume of a more concentrated solution (a
stock solution) to the desired final volume.
• 1. From solid:
 Ctd
1. Starting with a solid
• Determine the mass in grams of one mole of solute, the molar
mass, MMs.
• Decide volume of solution required, in liters, V.
• Decide molarity of solution required, M.
• Calculate grams of solute (gs) required using equation
gs = MMs x M x V
• • Example: Prepare 800 mL of 2 M sodium chloride.
• (MMNaCl = 58.45 g/mol)
• gNaCl = 58.45 g/mol x 2 mol/L x 0.8 L
• gNaCl = 93.52 g NaCl
• Dissolve 93.52 g of NaCl in about 400 mL of distilled water,
• then add more water until final volume is 800 mL
 Ctd
2.Starting with a solution or liquid reagent:
• hen diluting more concentrated solutions,
decide what
• volume (V2) and molarity (M2) the final solution
should be.Volume can be expressed in liters or
milliliters.
• Determine molarity (M1) of starting, more
concentrated solu
• tion.
• • Calculate volume of starting solution (V 1) required using
equation Note: V1 must be in the same units as V2.
eq M1V1 = M2V2
• Example: Prepare 100 mL of 1.0 M hydrochloric acid from
• concentrated (12.1 M) hydrochloric acid.
• M 1V 1 = M 2V 2
• (12.1 M)(V1) = (1.0 M)(100 mL)
V 1 = 8.26 mL conc. HCl
• Add 8.26 mL of concentrated HCl to about 50 mL of distilled
water, stir, then add water up to 100 mL.
 Ctd

3.Starting from a commercial reagent where % and density is provided

• Determine the mass of solution by multiplying the volume of
the solution by the density of the solution.
mass = volume x density
• Determine concentration in percent by mass of the solute in
solution.
• Calculate the molar mass of the compound, MM.
• Multiply mass by mass % and divide by
• molecular mass to find the number of moles present in
• the whole solution.
• 5. Divide the number of moles by the volume in liters of
• the solution to find the molarity of the solution.
 Ctd
• Example: Determine molarity of 37.2% hydrochloric
acid(density 1.19 g/mL).
1. Mass of solution = 1,000 mL x 1.19 g/mL = 1,190 g
2. Mass % = 37.2 % = 0.372
3. Molar mass of hydrochloric acid = 36.4 g/mol
4. mass x mass % /MM = 1,190 g x 0.372/36.4
= 12.1 mol
• 5. Molarity = moles/liters = 12.1 moles/1 liter = 12.1
M
 Ctd
• Or use the formula,
• % purity × density × 1000/ MM × 100
• 37 ×1.19 × 1000/100 × 36.4 =12.1 M
 Titrimetry
• Definition
• Titration also known as titrimetry and volumetric analysis is a
common laboratory method of quantitative chemical analysis
to determine the concentration of an identified analyte (a
substance to be analyzed).
• A reagent, termed the titrant or titrator,is prepared as a
standard solution of known concentration .
• The titrant reacts with a solution of analyte (which may also
be termed the titrand) to determine the analyte's
concentration.
• The volume of titrant that reacted with the analyte is termed
the titration volume.
 Ctd
• Procedure
• A typical titration begins with a beaker or
Erlenmeyer flask containing a very precise amount of
the analyte and a small amount of indicator (such as
phenolphthalein) placed underneath a calibrated
burette or chemistry pipetting syringe containing the
titrant.
• Small volumes of the titrant are then added to the
analyte and indicator until the indicator changes color in
reaction to the titrant saturation threshold, representing
arrival at the endpoint of the titration,
 Ctd
• Meaning the amount of titrant balances the
amount of analyte present, according to the
reaction between the two. Depending on the
endpoint desired, single drops or less than a
single drop of the titrant can make the
difference between a permanent and
temporary change in the indicator
 Ctd
Preparation techniques
• Typical titrations require titrant and analyte to
be in a liquid (solution) form.
• Solids are usually dissolved into an aqueous
solution, other solvents such as
glacial acetic acid or ethanol are used for
special purposes (as in petrochemistry,)
• Concentrated analytes are often diluted to
improve accuracy.
 Ctd
• Many non-acid–base titrations require a
constant pH during the reaction. Therefore, a
buffer solution may be added to the titration
chamber to maintain the pH.
• In instances where two reactants in a sample
may react with the titrant and only one is the
desired analyte, a separate masking solution
may be added to the reaction chamber which
eliminates the effect of the unwanted ion
 Ctd
• Example: consider a solution of Mg2+ and Fe3+
to be titrated by EDTA. Since Fe3+ interferes
with the titration, one can mask it by adding a
cyanide compound to the solution to form
Fe(CN6​)3−. This complex does not react with
EDTA, which allows for the selective titration
of Mg2+ without interference from Fe3+.
 Ctd
• Some reduction-oxidation (redox) reactions
may require heating the sample solution and
titrating while the solution is still hot to
increase the reaction rate. For instance, the
oxidation of some oxalate solutions requires
heating to 60 °C (140 °F) to maintain a
reasonable rate of reaction.
 Ctd
• Titration curves
• A titration curve is a curve in graph the x-
coordinate of which represents the volume of
titrant added since the beginning of the
titration, and the y-coordinate of which
represents the concentration of the analyte at
the corresponding stage of the titration (in an
acid–base titration, the y-coordinate usually
represents the pH of the solution).
Titration curve
 Ctd
Types of titrations
• Titration are classified according to reaction
involved in the process of titration
1. Acid-base
2. Redox
3. Precipitation
4. Complexometric
- Redox and acid base are the most common
 Acid-base
• An acid–base titration is a method of quantitative
analysis for determining the concentration of an
acid or base by exactly neutralizing it with a
standard solution of base or acid having known
concentration.
• Hundreds of compounds both organic and
inorganic can be determined by a titration based on
their acidic or basic properties. Acid is titrated with
a base and base is titrated with an acid. The
endpoint is usually detected by adding an indicator.
 Ctd
• Theory
• An acid-base titration involves strong or weak
acids or bases. Specifically, an acid-base titration
can be used to figure out the following.
• The concentration of an acid or base
• Whether an unknown acid or base is strong or
weak.
• pKa of an unknown acid or pKb of the unknown
base.
 Ctd
• Key Terms
• Titrant – A solution of known strength of concentration used in
the titration.
• Titrand or titrate – The titrand is any solution to which the titrant
is added and which contains the ion or species being determined.
• Titration curve – A plot of pH Vs millilitres of titrant showing the
manner in which pH changes Vs millilitres of titrant during an
acid-base titration.
• Equivalence point – The point at which just an adequate reagent
is added to react completely with a substance.
• Buffer solution – A solution that resists changes in pH even when
a strong acid or base is added or when it is diluted with water
 Types of acid and bases
S.No Types Examples

Hydrochloric acid and

1. Strong acid-strong base
sodium hydroxide

Ethanoic acid and sodium

2. Weak acid-strong base
hydroxide

Hydrochloric acid and

3. Strong acid-weak base
ammonia

4. Weak acid-weak base Ethanoic and ammonia

 Titration Curve & Equivalence Point

• In a titration, the equivalence point is the

point at which exactly the same number of
moles of hydroxide ions have been added as
there are moles of hydrogen ions. In a
titration, if the base is added from the burette
and the acid has been accurately measured
into a flask. The shape of each titration curve
is typical for the type of acid-base titration.
 Ctd
• Determination of the concentration of a given
solution:
• The concentration of an acid solution can be
determined by titration with a strong base.
• First, calculate the number of moles of strong base
required to reach the equivalence point of the
titration.
• Then, using the mole ratio from the balanced
neutralization equation, convert from moles of
strong base to moles of acid.
 Ctd
• E.g:
• Suppose 25.66 mL (or 0.02566 L) of 0.1078 M
HCl was used to titrate an unknown sample of
NaOH. What mass of NaOH was in the
sample?
 Ctd
• Answer:
1.The number of moles of HCl reacted:
(0.02566 L)(0.1078 M) = 0.002766 mol HCl
2. the balanced chemical reaction between HCl and NaOH:
HCl+NaOH→NaCl+H2O
3. Conversion factor to convert to number of moles of
NaOH reacted:
0.002766molHCl×1molNaOH/1molHCl=0.002766molNaOH
4. Convert mole to mass
0.002766 × 40g of NaOH = 0.110 g
 Ctd
Activity 1.
• What mass of Ca(OH)2 is present in a sample if
it is titrated to its equivalence point with 44.02
mL of 0.0885 M HNO3? The balanced chemical
equation is as follows:
• 2HNO3+Ca(OH)2→Ca(NO3)2+2H2O
Answer:
 Ctd
• Answer:
- 44.02 ml × 0.0885 M/1000 = 0.00390
molHNO3
- 0.00390 molHNO3 × 1 mol /2 mol = 0.00195
mol of Ca(OH)2

- 0.00195 × 74.1 g = 0.144g of Ca(OH)2

 Ctd
• Activity 2
• What mass of H2C2O4 is present in a sample if it
is titrated to its equivalence point with 18.09
mL of 0.2235 M NaOH?
• Answer:
• H2C2O4+2NaOH→Na2C2O4+2H2O
• 18.09 × 0.2235/1000 = 0.00404 mol NaOH
• 0.00404 /2 = 0.00201 mol of H2C2O4
• 0.00808 × 90.03 = 0.182g
 Ctd
• Activity 3
• If 25.00 mL of HCl solution with a
concentration of 0.1234 M is neutralized by
23.45 mL of NaOH, what is the concentration
of the base?
• Answer
25.00× 0.1235/ 23.45 = 0.132 M
 Ctd
• Activity 4
• A 20.0 mL solution of strontium hydroxide, Sr(OH) 2,
is placed in a flask and a drop of indicator is added.
The solution turns color after 25.0 mL of a standard
0.0500 M HCl solution is added. What was the
original concentration of the Sr(OH)2 solution?
• Answer
• 2HCl + Sr(OH)2 → SrCl2 + H2O
• 25.0 × 0.0500/2 = 0.625 mol of Sr(OH) 2
• 0.625/20 = 0.0312 M