FACULTY OF PRE-MEDICAL SCIENCES
CHEMISTRY DEPARTMENT
CH 2021: ANALYTICAL CHEMISTRY
Lecturer: Mr.J.Sichinga
What is Analytical Chemistry?
It is the “Science of Chemical Measurements”
Analytical Chemistry provides the methods and tools needed for insight into our material
world…for answering four basic questions about a material sample.
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What?
Where?
How much?
What arrangement, structure or form?
Areas of Chemical Analysis and Questions that analysts answer.
 Quantitation:
How much of substance X is in the sample?
 Detection:
Does the sample contain substance X?
Identification:
What is the identity of the substance in the sample?
Separation:
How can the species of interest be separated from the sample matrix for better
quantitation and identification?
Qualitative analysis
An analysis to ascertain what is contained in the sample.
Quantitative analysis
An analysis to ascertain how much is contained in the sample
Expressing concentration of solution
Concentration is a general measurement unit stating the amount of solute present in a
known amount of solution.
Although the terms “solute” and “solution” are often associated with liquid samples, they
can be extended to gas-phase and solid-phase samples as well. The actual units for
reporting concentration depend on how the amounts of solute and solution are measured.
Molality (m)
Molality, m, tells us the number of moles of solute dissolved in exactly one
kilogram of solvent. It is represented by a lower case m. Molality is
independent of temperature as it does not change when temperature
changes.
We need two pieces of information to calculate the molality of a solute in a
solution:
The moles of solute present in the solution.
The mass of solvent (in kilograms) in the solution.
To calculate molality we use the equation:
Example
If you have 10.0 grams of Br2 and dissolve it in 1.00 L of cyclohexane, what
is the molality of the solution? The density of cyclohexane is 0.779 kg/L at
room temperature.
Solution
First, work out the number of moles of bromine. Br2 has a molecular
weight of 159.8 g/mole, so we have
10 g / (159.8 g/mole) = 0.063 moles Br2
Next, convert the volume of solvent to the weight of solvent using the
density.
1.00L × 0.779 kg/L= 0.779 kg
Now just divide the two to get the molality
0.063 moles Br2/ 0.779 kg cyclohexane = 0.081 mol/kg or 0.081 molal or 0.081 m
Normality (N):
It is the number of equivalents of solute dissolved in one liter of solution. The
units, therefore are equivalents per liter or simply (N). Specifically it's
equivalents of solute per liter of solution.
Normality =
No. of equivalents of solute
liter of solution
Weight (g)
No. of equivalents = --------------------------Equivalent Weight (g/eq)
Eq.Wt =
M.Wt
-----------n
Where,
n = No. of (H) atoms for acids
For HCl
n=1
n = No of OH groups for bases
For NaOH
n=1
n = No of Cation atoms (M+) for salts
For Na2CO3
n= 2
n = No. of gained or lost electrons for oxidants and reductants.
For KMnO4
n= 7
Relationship between Molarity and Normality
N=nxM
Where, N = Normality, n = number of reacting units and M =Molarity
Example
What is the normality of 0.1 mol /L of Na2SO4?
Solution
N = 2 eq/mol × 0.1mol/L = 0.2 eq / L or 0.2 N
Normality and Molarity of Concentrated Reagents
Specific Gravity x Percentage (%) x 1000
Normality = ------------------------------------------------------------Equivalent Weight (g/ eq)
Specific Gravity x Percentage (%) x 1000
Molarity =
------------------------------------------------------------Molecular Weight (g/mol)
Formality (F)
Formality is a substance’s total concentration in solution without regard to its
specific chemical form. It is the number of moles of solute, regardless of
chemical form, per liter of solution .The symbol for Formality is F.
There is no difference between a substance’s Molarity and Formality if it
dissolves without dissociating into ions. The Molar concentration of a solution
of glucose, for example, is the same as its Formality. For substances that
ionize in solution, such as NaCl, Molarity and Formality are different. For
example, dissolving 0.1 mol of NaCl in 1 L of water gives a solution
containing 0.1 mol of Na+ and 0.1 mol of Cl-.The Molarity of NaCl, therefore,
is zero since there is essentially no undissociated NaCl in solution. The
solution, instead, is 0.1 M in Na+ and 0.1 M in Cl–. The Formality of NaCl,
however, is 0.1F because it represents the total amount of NaCl in solution.
When we state that a solution is 0.1 M NaCl we understand it to consist of Na+
and Cl– ions. The unit of Formality is used only when it provides a clearer
description of solution chemistry. Molar concentrations are used so frequently
that a symbolic notation is often used to simplify its expression in equations
and writing. The use of square brackets around a species indicates that we
are referring to that species’ Molar concentration. Thus, [Na+] is read as the
“Molar concentration of sodium ions.”
Concentration: ppm and ppb
Parts per million (ppm) and parts per billion (ppb) are examples of
expressing concentrations by mass. These units turn out to be
convenient when the solute concentrations are very small (almost
trace amounts). For example, if a solution has 1 ppm solute this
would mean that 1 g of solution would have one "millionth" gram of
solute. Equivalently, 1 kg of this solution will have 1 mg of solute etc.
By definition we have:
Parts per million (ppm) =
1
106
Parts per billion (ppb) =
1
109
1 Ppm = 1000 Ppb
For example, suppose a 155.3 g sample of pond water is found to
have 1.7x10-4g of phosphates. What is the concentration of
phosphates in ppm?
A similar procedure would be followed to calculate ppb. In the above
example the pond water would be 1,100 ppb.
Now suppose we have 400 g sample of pond water and it has a
concentration of 3.5 ppm dissolved nitrates. What is the mass of
dissolved nitrates in this sample?
Relationship between Ppm, Molarity and Normality.
Ppm= M x M.Wt x 1000
Ppm = N x Eq.Wt x 1000
Converting weight/volume (w/v) concentrations to ppm
Ppm = 1g/m3 = 1mg/L = 1μg/mL
Example 1
A solution has a concentration of 1.25 g/L. What is its concentration in ppm?
Solution
a. Convert the mass in grams to a mass in milligrams:
1.25g = 1.25 x 1000mg = 1250 mg
b. Re-write the concentration in mg/L = 1250mg/L = 1250 ppm
Example 2
A solution has a concentration of 0.5 mg/mL. What is its concentration
in ppm?
Solution
a. Convert the volume to litres:
volume = 1mL = 1mL ÷ 1000mL/L = 0.001L
b. Re-write the concentration in mg/L = 0.5mg/0.001L = 500 mg/L =
500 ppm
Converting weight/weight (w/w) concentrations to ppm
1ppm = 1mg/kg = 1μg/g
Example 1
A solution has a concentration of 0.033 g/kg. What is its concentration
in ppm?
Solution
a. Convert mass in grams to mass in milligrams:
0.033g = 0.033g x 1000mg/g = 33mg
b. Re-write the concentration in mg/kg = 33mg/kg = 33 ppm
Example 2
A solution has a concentration of 2250 μg/kg. What is its concentration
in ppm?
Solution
a. Convert mass in μg to mass in mg:
2250μg = 2250μg ÷ 1000μg/mg = 2.25mg
b. Re-write the concentration in mg/kg = 2.25mg/kg = 2.25 ppm
Parts per Million (ppm) Concentration Calculations
Example 1
150mL of an aqueous sodium chloride solution contains 0.0045g NaCl.
Calculate the concentration of NaCl in parts per million (ppm).
Solution
a. ppm = mass solute (mg) ÷ volume solution (L)
b. mass NaCl = 0.0045g = 0.0045 x 1000mg = 4.5mg
volume solution = 150mL = 150 ÷ 1000 = 0.150L
c. concentration of NaCl = 4.5mg ÷ 0.150L = 30mg/L = 30 ppm
Example 2
What mass in milligrams of potassium nitrate is present in 0.25kg of a
500ppm KNO3 (aq)?
Solution
a. ppm = mass solute (mg) ÷ mass solution (kg)
b. Re-arrange this equation to find the mass of solute:
mass solute (mg) = ppm x mass solution (kg)
c. Substitute in the values:
Mass KNO3 = 500ppm x 0.25kg = 125 mg
Example 3
A student is provided with 500mL of 600ppm solution of sucrose.
What volume of this solution in millilitres contains 0.15 g of
sucrose?
Solution
a. ppm = mass solute (mg) ÷ volume solution (L)
b. Re-arrange this equation to find volume of
solution: volume solution (L) = mass solute (mg) ÷
ppm
c. Substitute in the values:
volume solution (L) = (0.15g x 1000mg/g) ÷ 600 = 0.25L
d. Convert litres to millilitres: volume solution = 0.25L x 1000mL/L =
250 mL
DILUTIONS
Whenever you need to go from a more concentrated solution [“stock”] to a
less concentrated one, you add solvent [usually water] to “dilute” the
solution. No matter what the units of concentration are, you can always
use this formula.
C1 V1 = C2 V2
[Concentration of the stock] x [Volume of the stock] = [Concentration of the
final solution] x Volume of the final solution]
N1 V1 = N2 V2
M1 V1 = M2 V2