Experiment #1
SOLUTION PREP. AND BEER’S LAW
What is this experiment about?
This experiment has 2 parts to it.
They are as follows:
1.How to make solutions of required
concentration?
2. Using Beer’s law to determine
concentration of unknown solutions?
Experiment 1
Solution Preparation
and Beer’s Law
What is the game plan?
First: Briefly, review the concepts underlying
making solutions.
Second: Dilution of solutions.
Third: Deal with the BEER’S law……..
Solution Preparation
How do we make a solution?
Solute
+
(it dissolves
in the solvent)
solution
Solvent
(it dissolves
the solute)
Solution Preparation
The ratio of the amount of solute dissolved
in a certain volume of solution gives us the
concentration of the solute in that solution.
Units of concentration: g/mL, g/L, g/g, mL/mL
Molarity, Normality
g/g: concentration expressed as % by mass
mL/mL: concentration expressed as % by volume
Normality: Number of gram equivalents per liter
of solution
Molarity
Defined as the number of moles of solute
per liter of solution
moles of solute
Molarity 
Liters of solution
or
n
M
L
sol 'n
A Mole refers to a collection of 6.022 x 1023 items.
6.022 x 1023 is also called Avogadro’s number
Always remember!
Moles refers to a collection of particles.
Moles is a number
Moles is not a weight but Moles can be
calculated from the weight of the substance
So our goal is to know the moles of the solute
that is to be dissolved in a particular volume
of solvent, to make up a solution of required
molarity (a unit of concentration).
Why is preparing a solution correctly
so important?
How about these simple life situations?
1.Excess salt in your soup
2.Insufficient amount of sugar in the coffee
3.More or less alcohol in an alcoholic beverage
How about these life threatening situations?
1.Excess or insufficient amount of the drug in
a medication.
2.Excess chlorine in a swimming pool
A hypothetical example
Find the number of tennis balls inside the box without
opening the box, if each and every ball inside the box
weighs the same as the one outside the box? The weight
of empty box is 5.0 g. The Weight of the box with the
balls is 55.0 g. The weight of the tennis ball outside
the box is 2.0 g.
A box containing
tennis balls
tennis ball
outside the box
Solution:
Mass of Balls + box = 55.0 g
Mass of the box with no balls in it = 5.0 g
Mass of just the balls = 55.0 -5.0 = 50.0 g
Mass of the ball outside the box = 2.0 g
Since each ball inside the box weighs the same as the
ball outside the box, If we divide the mass of all the balls
by the weight of the single ball that is outside the box,
we can know how many balls are inside the box
(without opening the box).
Proceeding,
50.0 ÷ 2.0 = 25 balls.
Lessons learned from the example:
If we know the total mass of all the balls inside
the box and the mass of one ball, we can
determine the number of balls inside the box,
without opening it.
Important condition:
Mass of the ball outside the box = mass of each
and every ball inside the box.
How about we apply this example to our problem
of finding the number of solute particles that have
to be dissolved in our solution?
Think that all the particles that we are adding to make our
solution are inside the box that had the tennis balls. Let us say
just one of those particles is outside the box.
What are these particles that make up our solutions?
They are chemical substances. They are made up of atoms.
The atoms are combined in a certain way to form molecules.
For example: salt solution: salt + water
Salt = Sodium Chloride, NaCl
So now we can call these particles as molecules.
New lesson:
If we know the total mass of all the molecules
inside the box and the mass of a single
molecule outside the box, we can find
the number of molecules that we have inside
the box.
Important condition:
Mass of the molecule outside the box = mass of
each and every molecule inside the box
How do we find the following?
1. Total mass of all the molecules inside the box?
Weigh the substance on a balance.
2. The mass of a single molecule?
Mass of a single molecule is called its molecular weight.
Since a molecule is made up of atoms. We can find the
molecular weight by adding the mass of the individual
atoms (atomic mass) that make up the molecule.
3. Mass of a single atom?
Mass of a single atom can be obtained for each and every
atom from the atomic mass values given in a periodic table.
Atomic mass is given below the
symbol of the element
Units of Atomic Mass
Most common unit of mass: g, kg, lb
Mass of an atom or atomic mass has units: amu
1 a.m.u or atomic mass unit = 1.66 x 10-24 g
For example mass of a single atom of sodium or
the atomic mass of sodium = 22.99 amu
= 22.99 x 1.66 x 10-24 g
= 3.816 x 10-23 g
= 0.00000000000000000000003816 g
Since it is impossible to measure a single atom on
a common lab balance, we always measure a
collection of atoms or molecules.
How many are atoms/molecules are
in this collection?
the atomic mass of sodium = 22.99 amu
= 22.99 x 1.66 x 10-24 g
= 3.816 x 10-23 g
= 0.00000000000000000000003816 g
6.022 x 1023 Na atoms will be required make up 22.99 g of Na
= 22.99 x 1.66 x 10-24 g x 6.022 x 1023 Na atoms
= 22.98 g
This collection of 6.022 x 1023 Na atoms = 1 mole of Na atoms
So we can say that 1 mole of Na atoms weighs
22.99 g. Therefore, the weight off
1 Na atom = 22.99 g/mol
atomic mass has unit: amu or g/mol
Example Problem 1
What is the molarity of a 500 mL solution
that contains 10 g of sodium chloride (NaCl)?
Given:
Volume of the solution: 500 mL or 0.5 L (Remember! 1000 mL = 1.0 L)
Mass or weight of the solute: 10 g
Name and chemical formula of the compound: Sodium Chloride (NaCl)?
To be found:
Concentration or Molarity of the solution:?
Methodology:
moles of solute
Molarity 
Liters of solution
Example Problem 1
Weight of subs tan ce ( g )
Number of moles 
Molecular Weight ( g
)
mol
Calculation:
Number of moles 
Weight of NaCl ( g )
g
Molecular Weight of NaCl (
mol
Molecular Weight of NaCl= Atomic Weight of Na + Atomic Weight of Cl
From Periodic
Table
g
 22.99(
 58.44( g
mol
mol
g
)  35.45(
)
mol
)
)
Example Problem1
10 ( g )
g  mol 

Number of moles 
 0.17 

g
 1  g 
58.44 (
)
mol
10 ( g )
Number of moles 
 0.17 mol
g
58.44 (
)
mol
moles of solute 0.17 mol
Molarity 

 0.34 Molar
Liters of solution
0.5 L
Example Problem 2
How will you prepare 500 mL of 0.84 M solution
of glucose(C6H12O6)?
Given:
Volume of the solution to be prepared: 500 mL
Concentration or Molarity of the solution: 0.84 Molar
Name and chemical formula of the compound: glucose(C6H12O6)?
Information that needs to be found:
The mass or weight of glucose that is required to make up this 500 mL
0.84 M solution.
Methodology:
moles of solute
Molarity 
Liters of solution
Example Problem 2
Weight of subs tan ce ( g )
moles of solute 
Molecular Weight ( g
)
mol
Calculation:
moles of solute
Molarity 
Liters of solution


 Wt. of subs tan ce( g ) 


g
 Mol. Wt.(

)


mol

Liters of solution
(
)
1
Example Problem 2


1

 Wt. of subs tan ce( g )  
Molarity  



 Mol. Wt.( g
  Liters of So ln . 
)

mol 
Wt. of subst.( g )  Molarity  Mol. Wt.( g
mol
)  L of so ln
mol 

g
Wt. of subst.( g )  0.84
 180( mol )  0.5 L
 L 
Wt. of subst.( g )  75.6 g
When making solutions
500 mL mark
500mL
Volumetric
flask
Very accurate
500mL
Erlenmeyer
flask
less accurate
Beaker
less accurate
Summary I
moles of solute
1. Molarity 
Liters of solution
or
n
M
L
sol 'n
2. Moles refers to a collection of particles.
Moles is a number
Moles is not a weight but moles can be
calculated from the weight of the substance
3. Atomic mass has unit: amu or g/mol
4. Molecular mass or Molecular weight, which
is sum of the atomic mass of atoms in that
molecule, also has units amu or g/mol
Summary I Contd.
Weight of subs tan ce ( g )
5. Number of moles 
g
Molecular Weight (
)
mol
6. Mass of one mole of atoms or molecule is
called its molar mass. Unit of molar mass
is g. Ex. Molar mass of sodium = 22.99 g
Molar mass of NaCl= 58.44 g.
Dilution of solutions
Dilution means making a solution of
lower concentration from a solution
of higher concentration.
M1
More concentrated
(stock solution)
M2
Less concentrated
Need to take a certain volume from the more concentrated solution
And make it up to a certain volume of diluted solution
Example problem
How will you prepare 500 mL of 0.16 M solution
of glucose(C6H12O6) from a 500mL 0.84 M
glucose solution? Use water as the solvent.
Given:
Concentration of the stock solution: 0.84 M
Total volume of the stock solution: 500 mL or 0.5 L
Concentration of the dilute solution: 0.16 M
Volume of the dilute solution: 500 mL= 0.5 L
What makes a solution more or less concentrated is the number of solute molecules
per liter of solution.
To be found:
The volume of the stock solution that needs to be taken out of 500 mL stock so that
The dilution can be made
Example problem
Methodology and calculation:
Moles
Moles  Molarity (
)  Volume ( L)
L
Number of moles of glucose in the 500 mL stock solution:
Moles
Moles  0.84 (
)  0.5 ( L)
L
Moles  0.42 moles
Example problem
Number of moles of glucose in the 500 mL of 0.16 M solution
Moles
Moles  0.16 (
)  0.5 ( L)
L
Moles  0.08 moles
If we want to make a dilute (0.16 M, 500 mL) solution of glucose from the stock
Solution, the dilute solution dictates that we need to have only 0.08 moles
of glucose.
If we use all the 500 mL of 0.84 M stock to make the dilute solution, we will end up
With 0.42 moles of glucose which is much more than the 0.08 moles that we want.
So we will have to figure out what volume of stock will give us the required 0.08
moles of glucose.
Example problem
We can do this by trial and error by changing the volume of stock and figuring
Out which value of volume would give the required 0.08 moles.
Molarity of
Stock (M)
Volume of stock
(L)
Moles of stock (M
 L)
0.84
0.005 (5 mL)
0.0042
0.010 (10 mL)
0.0084
0.015 (15 mL)
0.0126
0.84
0.020 (20 mL)
0.0168
0.84
0.025 (25 mL)
0.0210
0.84
0.84
Or we can do this by solving a simple equation.
Example problem
Molarity of stock ( mole )
L
 Vol. of stock (L)

0.84 M
V
stock
mole
)
L
 Vol. of the dilute soln. (L)
Molarity of dilute soln. (
=
Vstock L
=
0.16 M
 0.5 L
 0.16 M  0.5 L 
( L)  
  0.095L
0.84 M


To make the solution:
Take 95 mL stock solution and water and fill a 500 mL
volumetric flask to the mark. Stopper the flask and shake
the flask few times so as to enable uniform mixing.
Lessons learned So far!!
1. From Example 1. We learnt how to make a
solution of specific concentration, given the molarity,
volume and the of course, the name of the compound
2. From Example 2. We learnt how to make a
dilution from a concentrated stock solution.
3. If solution need to be made accurately to a certain
concentration, they need to be made in a volumetric
flask and not in a beaker or a erlenmeyer flask.
Properties of light, Beer’s Law
and its application
Properties of light
Light exhibits both wave-like and particle
like properties
wave-like properties
Property
Greek symbol
Frequency
(nu)
Wave length
(lambda)
Speed
c (not a greek
Symbol)
Value/unit
( 1s )
nm or Å
1nm=10-9 m
1 Å = 10-10 m
3 x 108 m/s
wave-like properties
1
1
1 second
2
When wavelength increases, the frequency decreases
Wavelength is inversely proportional to frequency

1


c

  c
Particle-nature of light
Energy is directly proportional to number of
photons and their frequency
 n
= n h
photon
= nh
n – number of photons. It is not number of
moles of photons.
h – planck’s constant= 6.626 x 10-34 Js
– frequency of light, units 1 s
- Change in energy
Units of :
(no units)(Js)( 1 s ) = J
  c

c

Substituting the value of  from the above equation
into the equation for E (the equation below), we get
= nh
=
nhc

human eye can see only these colors
Interaction of light with matter
Definition of absorbance & transmittance
Iin
Iout
s
Absorbance
I out
A  log
I in
Transmittance
I in
%T 
100 %
I out
Absorbance (A) and transmittance (T) are unitless
Wavelength of maximum absorption, max
CuSO4
Graph #1
Abs
max=740 nm
400
500
600
700
800

KMnO4
Graph #2
Abs
max=530 nm
400
500
600

700
800
Beer-Lambert’s law
Absorbance Vs. Concentration (c ) Absorbance Vs. Path length (b)
Iin
Iout [1]
s
Iin
Iout [1]
s
Iin
Iout [2]
s
Iin
Iout [2]
s
Ac
This is not true at high
concentrations
Unit of concentration = M
Ab
Path length- the distance that light
travels in the sample
Unit of path length = cm
Beer-Lambert’s law
Ac
A  bc
Ab
A = abc
a – Molar absorptivity
Units of molar absorptivity:
unit of A  (unit of a)(unit of b)(unit of c)
unit of A
unit of a 
(unit of b)(unit of c)
unit of A
unit of a 
(unit of b)(unit of c)
unitless
1
unit of a 

 cm1M 1  M 1cm1
(cm)( M ) (cm)( M )
“A” is unitless but “a” has units M-1cm-1
Definition of molar absorptivity
A = abc
A
a
( b)(c)
a - is a measure of the amount of light
absorbed per unit concentration and pathlength
A compound with a high molar absorptivity is
very effective at absorbing light (of the appropriate
wavelength)
Finding unknown concentrations
of solution using Beer’s law
Diameter = d cm
Light source
Detector
Sample test tube
Diameter of the test tube = path length
Finding unknown concentrations
of solution using Beer’s law
Solutions of known concentration
and their absorbances
A1
A2
A3
A4
A5
C5,a5
Concentration
c1
c2
c3
c4
c5
Best-fit line
C4,a4
C2,a2
C3,a3
C1,a1
Absorbance
y
Concentration
Best-fit line
x c5,A5
n
cn
Equation of the best-fit line:
c4,A4
cm
c2,A2
m
x
Y=mX + z
m = slope =
c3,A3
c1,A1
x
Am
intercept
An
Absorbance
Y2  Y1
X 2  X1
Cn  Cm

An  Am
C= mA + Z
concentration
Absorbance
C= mA + Z
Cunknown= mAunkown + Z
If using Microsoft excel:
Please refer to section 3 of MATHCHEM 1.0
Obtain the equation of best-fit line
Sample Problem
Find the molar absorptivity of the a 0.5 M solution
whose absorbance is 0.77,When measured in a
tube of path length 2 cm?
Given:
Concentration of the solution = 0.5 M =c
Absorbance = 0.77= A
Path length = 2 cm = b
To be found:
Molar absorptivity of the solution= ?= a
Sample Problem
Methodology:
Beer’s law
A = abc
Calculation:
A
a
( b)(c)
0.77
a
 0.77cm1M 1  0.77M 1cm1
( 2 cm)(0.5M )
Summary
1. Beer’s Law: A  c and A  b and A = abc
2. Wave length is a property of light. It is the
distance between two consecutive crests or
troughs in a wave. The typical unit of wavelength
is nm or Ǻ.
3. Path length is NOT a property of light, It is the
distance that light travels in the sample. Typical
unit of path length is cm.
4. If we know the absorbance of solutions of known
concentration, we can use Beer’s law to find the
concentration of a solution, for which we know
only the value of absorbance.