Analytical chemistry
Second lecture
Measurements in analytical chemistry
• Analytical chemistry is a quantitative science.
Whether determining the concentration of a
species, evaluating an equilibrium constant,
measuring a reaction rate, or drawing a
correlation between a compound’s structure and
its reactivity, In this section we briefly review the
use of units and significant figures in analytical
chemistry.
Units of Measurement
• A measurement usually consists of a unit and a
number expressing the quantity of that unit. We
may express the same physical measurement
with different units, which can create confusion.
For example, the mass of a sample weighing 1.5
g also may be written as 0.0033 lb or 0.053 oz.
To ensure consistency, and to avoid confusion,
scientists use a common set of fundamental
units, several of which are listed in the Table
below; These units are called SI units after the
Système International d’Unités.
Mass/Amount of Substance:
• Mass
• Volume
• Amount of Substance
kilogram (kg) ⇒ gram (g)
liters (L)⇒ milliliters (mL)
mole (mol)
• 1 mole = 6.022 x 1023 particles (e.g. atoms, molecules,
ions)
• Atomic Mass = number of grams containing Avogadro’s
number (6.022 x 1023) of atoms
• Molecular Mass = number of grams containing
Avogadro’s number (6.022 x 1023) of molecules; sum of
atomic masses of elements in a molecule
Example:
• Atomic mass = the sum of proton and neutron in an
atom.
• Atomic number = number of protons in an atom.
• No. of neutrons = atomic mass – atomic no.
• No. of protons = no. of electrons;
• No. of neutrons = atomic mass – no. protons.
• No. of neutrons = atomic mass – no. of electrons.
• According to that, Atomic mass can be defined as:
It is the mass of an atom or particle in atomic mass unit
(amu).
On that scale,
1 atom C12 weighs 12 amu
1 atom H1 =1.00794 amu
1 atom O16 = 15.9994 amu
as defined earlier, moles are the number of atoms, ions or
molecules in a substance. We can calculate the weight of a
substance if we know the formula,
For any element:
Atomic mass = molar mass (grams)
1 mole of C12 atom= 12.00 grams of C12 .
1mole of H1 = 1.00794 g of H.
• Molecular mass ( molecular weight) :
the sum of the atomic mass of each constituent atom
multiplied by the number of atoms of that element in the
molecular formula.
Ex:
The molecular mass of H2O = 18 amu
As atomic mass of H=1 and O=16
Molecular mass = (2x1)+ 16 = 18
For any element:
Molecular mass (amu) = molar mass (grams)
So:
1 mole of H2O = 18 grams of H2O/mole
Examples:
• Calculate the molecular mass of H2SO4?
If H=1, S=32 ,O =16 ,
Molecular mass = (2x1)+32+(4x16) = 98
1 mole of H2SO4 = 98 gram/mole
• CO2
C= 12, O=16,
Molecular mass = 12+ (2x16) = 44
1 mole of CO2= 44 gram/mole
How to calculate number of moles in a
compound:
• From the equation below:
No. of moles = weight (grams) / molecular weight (gram/mole)
The unit is: mole
No. of millimoles = weight (mg) / molecular weight (mg/mmole)
Unit is :mmole
From the equation above we can caculate the weight in grams or mg:
Weight(gram)= no. of moles x molecular weight (g/mole)
Weight (mg) = no. of mmole x molecular weight (mg/mmole)
Ex:
Calculate the no. of moles in 5 grams of Fe2O3 (ferric oxide)?
Fe= 55.8 ,O=16
Solution:
No. of moles = weight (gram)/ molecular weight
Weight = 5 grams
Molecular weight = (2x55.8)+ (3x16) = 159.6 gram /mole
No. of moles = 5/ 159.6 = 0.0312 moles
• Ex2:
Calculate the number of moles in 500 mg of Na2SO4?
Na=23, S=32, O=16?
No. of mmole= wieght(mg) / molecular weight (mg/mmole)
= 500 / 142 = 3.521 mmole
No. of moles= 3.521 / 1000 = .003521 mole
Concentration:
• Concentration: is a general measurement unit stating the
amount of solute present in a known amount of solution
Concentration= amount of solute/ amount of solvent
And it can be expressed as :
1.
2.
3.
4.
Molarity
Molality
Normality
Percentages.
1. Molarity:
• It is the number of moles of solutes in a liter of solvent.
• Molarity= moles of solute / liter of solvent.
• The unit is molar (M).
Example:
calculate the molarity of a solution of Nacl prepared by
dissolving 0.735 mol of it in water where the final volume
is 650 ml?
Molarity = no. of moles/ volume (L)
No. of moles = 0.735 mole
Volume=650 ml /1000 ml = 0.65 L
Molarity= 0.735/0.65 = 1.13 molar
2. molality:
• it is the number of moles of solute /kilogram of solvent.
• Molality= no. of solute/ kg solvent
• The unit is molal (m)
Example:
• Determine the molality of a solution prepared by
dissolving 75.0g Ba(NO3)2 (s) in to 374.00g of water at
250C.
Molality= no. moles/kg solvent
• No. of moles= weight(g)/molar mass of Ba(NO3)2 =
75.0 g/261.32 g/mole = 0.28700 mole
molality = 0.28700 mole/ 0.37400 kg = 0.767 m
3.
Normality:
• Defined as the number of equivalents dissolved in liter of the
solvent.
N= no. of eq. / volume(L)
No. of eq. can be calculated as :
No. of eq.= weight (g) / equivalent weight
N = weight (g) / equivalent weight x volume (L)
And the eq. weight can be calculated as:
Molecular weight/ n
Where n= no. of reacting units (‫)التكافؤ‬
N varies depending on the reaction.
• In acid –base reaction :
n = no. of hydrogen if its an acid.
Or the no. of hydroxyl if it’s a base.
Ex: (acids)
HCl ;
H2SO4 ;
H3PO4 ;
Ex: ( base)
NaOH ;
Ca(OH)2 ;
Al(OH)3 ;
n=1
n=2
n=3
n=1
n=2
n= 3
• In redox reactions:
n= no. of electrons (take on or supply).
If we know the no. of equivalents (n), then we can calculate N
by:
N = weight (g) x n x 1000 / molecular weight x volume (mL)
or simply:
N = Molarity x n
As molarity = no. of moles / volume (L)
And no. of moles = weight (g) / molecular weight
3. percentages:
• They are a set of concentrations expressions based on the%
representation and are of three types:
i.
Weight per volume percentage (w/v )%
Can be defined as the weight of solutes in grams dissolved in mL
of the solution and can be expressed as (g/mL) %.
w/v % = weight solute (g)/ volume solution(mL).
Ex:
5 grams of NaOH dissolved in 500 ml of solution, calculate w/v %?
w/v %= 5/ 500 x100 = 1% (g/mL)
ii.
Weight per weight percentage (w/w)%:
Can be defined as the number of grams of solutes in number of
grams of solution.
And its expressed as (g/g)% .
w/w % = weight solute(g) / weight of solution (g) x100
Ex:
Find the weight percentage (w/w)% of HCl, if 20 grams of it was
dissolved in 1 kg of solution?
w/w %= 20 (g) / 1000 (g) x 100 = 2 % (g/g)
iii. Volume per volume percentage (v/v)% :
can be defined as the volume of solutes (mL) dissolved in (mL)
of solution.
And can be expressed as (mL/mL)%
v/v % = volume solute/ volume solution x 100
Ex:
What is the volume of ethyl alcohol dissolved in in water if the
v/v% is 15% and volume of solution 350 mL?
15% = volume of alcohol/ 350 mL x100
Volume of alcohol = 15 x 350 /100 = 52.50 mL
Preparing solution:
• Now as you know how to calculate M, n, V , what
does that mean?
• that means you will be able to prepare your own
solutions.
What is a solution?
• Solution: a homogeneous mixture of two or
more substances.
• Solute: a substance in a solution that is present
in the smallest amount.
• Solvent: a substance in a solution that is present
in the largest amount.
• In an aqueous solution, the solute is a liquid or
solid and the solvent is always water.
• The majority of chemical reactions occur in
Aqueous solution.
• All solutes that dissolve in water fit into one of
two categories: electrolyte or non-electrolyte.
• Electrolyte: a substance that when dissolved in
water conducts electricity.
• Non-electrolyte: a substance that when
dissolved in water does not conduct electricity.
• Electrolyte can be divided into; strong or weak
electrolyte.
• Strong electrolyte: conduct current very efficiently by
Completely ionized or dissociate when dissolved in water
( cations (+) and anions (-) ).
• Ex: strong acids, strong bases and soluble salts.
NaCl(s) → Na+(aq) + Cl–(aq)
HCl(s) → H+(aq) + Cl–(aq)
• Weak electrolyte: conduct only a small current
By Slightly ionized in solution.
Ex: weak acids, weak bases.
CH3COOH(aq)
↔
CH3COO–(aq) + H+(aq)
• Non electrolyte does not conduct any current as they do
not dissociate in the aqueous solution.
• No cations (+) and anions (-) in the solution.
• Ex:
• Sugars( glucose, sucrose), alcohol (ethanol, methanol)
• C6H12O6 (s) → C6H12O6 (aq)
Classification of solutes in aqueous
solution:
What are the steps for preparing solution?
•Preparing solution of known concentration is perhaps the
most common activity in any analytical lab.
•Pipets and volumetric flasks are used when a solution’s
concentration must be exact; graduated cylinders, beakers
when concentrations need only be approximate.
•Two methods for preparing solutions are used; preparing
stock solution, preparing solution by dilution.
Preparing stock solution:
• A Stock Solution is a concentrated solution that will be
diluted to some lower concentrated for actual use. Stock
solutions are used to save preparation time, conserve
materials, reduce storage space, and improve the
accuracy with which working lower concentration
solutions are prepared.
• A stock solution is prepared by weighing out an
appropriate portion of a pure solid or by measuring out
an appropriate volume of a pure liquid and diluting to a
known volume.
For example, to prepare a solution with a desired molarity
you weigh out an appropriate mass of the reagent, dissolve
it in a portion of solvent, and bring to the desired volume.
• How many grams of Potassium Dichromate, K2Cr2O7, are
required to prepare a 250mL solution with a
concentration of 2.16M?
250mL x 1L/ 1000mL = 0.250L
M= n/v
n= M x v
n= 2.16M x .250L
n= 0.54 mol But in the lab we weigh grams not
moles, so …
No.moles = weight (g)/MW
Weight (g) = no.moles x MW
Weight (g)= 0.54 mol K2Cr2O7 x 294.2 g K2Cr2O7
Weight (g)= 159 grams
159 grams of K2Cr2O7 are needed to prepare the
requested solution
• Explain the process of making 1L of 3.0M KCl.
M = n/volume (L)
n = M x volume (L)
n = 3.0M x 1L
n = 4.0 mol of KCl needed
No. of moles= weight (g)/MW
weight (g)= moles x MW
weight (g)= 4.0 mol KCl x 36.0g KCl
weight (g)= 144g KCl
Weigh out 144g of KCl. Put in a 1L flask. Add enough
distilled H20 to dissolve KCl. Fill flask to 1L.
Preparing solution by dilution:
• Solutions are often prepared by diluting a more
concentrated stock solution. A known volume of the
stock solution is transferred to a new container and
brought to a new volume.
• So we can define Dilution as the procedure for preparing
a less concentrated solution from a more concentrated
one.
• A given volume of a stock solution contains a specific
number of moles of solute.
e.g.: 25 mL of 6.0 M HCl contains 0.15 mol HCl
(How do you know this???)
Solvent
Ex.(water)
+
25 mL
0.15 mol
=
25 mL
50 mL
0.15 mol
• If 25 mL of 6.0 M HCl is diluted with 25 mL of water, the
number of moles of HCl present does not change.
Still contains 0.15 mol HCl
no. moles solute
before dilution
=
no. moles solute
after dilution
• Although the number of moles of solute does not
change, the volume of solution does change.
• The concentration of the solution will change since:
Molarity = no .of moles / volume (L)
Dilution calculations:
• When a solution is diluted, the concentration of the new
solution can be found using the formula below:
Mc x Vc = Md x Vd
Where,
Mc= initial concentration (mol/L)= more concentrated
Vc = initial volume of more conc. solution
Md = final concentration (mol/L) in dilution
Vd = final volume of diluted solution
Practice:
What is the concentration of a solution prepared
by diluting 25.0 mL of 6.00 M HCl to a total volume
of 50.0 mL?
Given:
Vc = 25.0 mL
Mc = 6.00 M
Vd = 50.0 mL
Find:
Md ??
• By using the formula:
Mc x Vc = Md x Vd
Md = Mc x Vc / Vd
Md = 6.00 x 0.025/ 0.05 = 3 mol/L
• How many mL of a 14 M stock solution must be used to
make 250 mL of a 1.75 M solution?
Mc x Vc = Md x Vd
Mc = 14 M,
Vc = ?,
Md = 1.75 M,
Vd = 250 mL
Vc = Md Vd / Mc = (1.75 M) x (0.250 L) / (14 M)
Vc = 0.03125 L = 31.25 mL
• 100 mL of 6.0 M CuSO4 must be diluted to what
final volume so that the resulting solution is 1.5
M?
Mc = 6 M,
Vc = 100 mL,
Md = 1.5 M,
Vd = ?
Vd = McVc / Md = (6 M) x (0.100 L) / (1.5 M)
Vd = 0.4 L or 400 mL
Summary:
• Expressing Concentration
• Solutions
• Preparing stock solutions.
• Preparing diluted solutions.
Exercises:
• How many grams of nitric acid are present in 1.0
L of a 1.0 M HNO3 solution?
63 g
• Give the molarity of a solution containing 10 g of
H2SO4 solute in 2.5 L of solution?
0.041 mol/L
• Water is added to 4 L of 6 M antifreeze until it is
1.5 M. What is the total volume of the new
solution?
Vd = 16 L