This document provides an introduction and overview of analytical chemistry concepts: - Analytical chemistry involves separating, identifying, and determining the relative amounts of compounds in a sample. It has two main parts: qualitative analysis to identify what is present, and quantitative analysis to determine how much is present. - Important units of measurement in analytical chemistry include grams, milligrams, microliters, and the mole. A mole represents 6.02x1023 units of a substance and is used to relate amounts of substances. - Calculations involving moles and millimoles allow determining the amount of a substance in a sample based on its mass, molar mass, and the chemical formula. Factor-label and dimensional analysis

1. Lecture 1
Introduction to Analytical Chemistry
Azad H. Mohammed
Azadalshatteri@Hotmail.com
Chemistry Department
University of Garmian

2. References
1. Fundamental analytical chemistry, 9th edition, Douglas A. Skoog, 2014.*
2. Modern Analytical Chemistry, 2nd edition, David Harvey, 2008.

3. This lecture
 Introduction to analytical chemistry
 Some important units of measurement
 Moles and millimoles

4. Introduction to analytical chemistry
Analytical Chemistry: it involves separating, identifying, and determining the
relative amounts of the compounds in a sample of matter. Analytical chemistry
is divided into two parts include: qualitative analysis; and quantitative analysis.
Qualitative analysis (what is present): it is to find out the sample components,
it is therefore the identification process.
Quantitative analysis (how much is present): it is the analysis that concerns
with the percentages content of the matter or unknown samples. Therefore, it
is the determination process of sample components.

5. A solution is a homogenous mixture of the solute and the solvent. When
we used water as a solvent this solution called aqueous solution. When
the solvent does not water this solution is called nonaqueous solution.
Diluted solution: it is a solution that contains only a small amount of
solute.
Concentrated solution: it is a solution that a large amount of solute.

6. Classifying solutions of electrolytes:
An electrolyte is a substance that dissociates into ions in solution. In
general, electrolytes are more dissociated in water than in other
solvents. We refer to a compound that is mostly dissociated into ions as
a strong electrolyte (example: strong acids, strong bases, and salts).
One that is partially dissociated is called a weak electrolyte (example:
weak acids, and weak bases).
A non-electrolyte is a substance that does not dissociate into ions in
solution, sugar is a good example of a nonelectrolyte.

7. Some important units of measurement
1. SI units
Scientists throughout the world have adopted a standardized system of
units known as the international system of units (SI). This system is
based on the seven fundamental base units shown in Table 1.1.

8. In analytical chemistry, we often determine the amount of chemical
species from mass measurements. For such measurements, kilograms (kg),
grams (g), milligrams (mg), or micrograms (µg) are used. Volumes of liquids
are measured in units of liters (L), milliliters (mL), or microliters (µL). The SI
units of volume is liter and it is defined as exactly 10-3 m3. The milliliter is
defined as 10-6 m3, 1 cm3 or 10-2 dL.
To express small and large measured quantities in terms of a few simple
digits, prefixes are used with these base units and their derived units. As
shown in Table 1.2.

9. Table 1.2: prefixes for base units
Prefix Abbreviation Multiplier Prefix Abbreviation Multiplier
yotta- Y 1024 deci- d 10-1
zetta- Z 1021 centi- c 10-2
exa- E 1018 milli- m 10-3
peta- P 1015 micro- µ 10-6
tera- T 1012 nano- n 10-9
giga- G 109 pico- p 10-12
mega- M 106 femto- f 10-15
kilo- k 103 atto- a 10-18
hecto- h 102 zepto- z 10-21
deca- da 101 yocto- y 10-24

10. Mole
Mole (abbreviated mol) is the SI units for the amount of a chemical
substance. A mole of a chemical species is 6.022 x 1023 (Avogadro’s
number) atoms, molecules, ions, electrons, ion pairs.
Molar mass (abbreviated MM), (g/mol) of a substance is the mass in
grams of 1 mole of that substance. We calculate molar masses by
summing atomic masses of all atoms appearing in a chemical formula.
For example, the molar mass of formaldehyde CH2O is

11. 𝑀𝑀 𝐶𝐻2 𝑂 =
1 𝑚𝑜𝑙 𝐶
𝑚𝑜𝑙 𝐶𝐻2 𝑂
𝑥
12.0 𝑔
𝑚𝑜𝑙 𝐶
+
2 𝑚𝑜𝑙 𝐻
𝑚𝑜𝑙 𝐶𝐻2 𝑂
𝑥
1 𝑔
𝑚𝑜𝑙 𝐻
+
1 𝑚𝑜𝑙 𝑂
𝑚𝑜𝑙 𝐶𝐻2 𝑂
𝑥
16 𝑔
𝑚𝑜𝑙 𝐻
= 30 𝑔/𝑚𝑜𝑙 𝐶𝐻2 𝑂
or
Molar mass of formaldehyde (𝐶𝐻2 𝑂) is
1 C 1 x atomic weight of C = 1 x 12 = 12
2 H 2 x atomic weight of H = 2 x 1 = 2
1 O 1 x atomic weight of O = 1 x 16 = 16
Molar mass of 𝐶𝐻2 𝑂 (g/mol) = ∑ atomic weights = 12 + 2 + 16 = 30 g/mol

12. Millimole
Sometimes it is more convenient to make calculations with millimoles (mmol)
rather than moles.
1 𝑚𝑚𝑜𝑙 = 10−3 𝑚𝑜𝑙, 𝑎𝑛𝑑 103 𝑚𝑚𝑜𝑙 = 1 𝑚𝑜𝑙
millimolar mass (g/mmol) is equal to 10-3 molar mass (g/mol, or mg/mmol).

13. Calculations the amount of a substance in moles or millimoles
Example 1.1
Find the number of moles and millimoles of benzoic acid (MM = 122.1
g/mol) that are contained in 2 g of the pure acid.
Solution
If we use HBz to represent benzoic acid, we can write that 1 mole of HBz has
a mass of 122.1 g. Therefore
𝑚𝑜𝑙𝑒 𝐻𝐵𝑧 = 2 𝑔 𝐻𝐵𝑧 𝑥
1 𝑚𝑜𝑙 𝐻𝐵𝑧
122.1 𝑔 𝐻𝐵𝑧
= 0.0164 𝑚𝑜𝑙 𝐻𝐵𝑧

14. To obtain the number of millimoles, we divide by the millimolar mass (0.1221
g/mmol), that is,
𝑚𝑚𝑜𝑙 𝑜𝑓 𝐻𝐵𝑧 = 2 𝑔 𝐻𝐵𝑧 𝑥
1 𝑚𝑚𝑜𝑙 𝐻𝐵𝑧
0.1221 𝑔 𝐻𝐵𝑧
= 16.38 𝑚𝑚𝑜𝑙 𝐻𝐵𝑧

15. Example 1.2
What is the mass in grams of Na+ (A. wt. = 22.99 g/mol) in 25 g of Na2SO4 (142
g/mol)?
Solution
The chemical formula tells us that 1 mole of Na2SO4 contains 2 moles of Na+,
that is
𝑚𝑜𝑙𝑒 𝑁𝑎+
= 𝑚𝑜𝑙 𝑁𝑎2 𝑆𝑂4 𝑥
2 𝑚𝑜𝑙 𝑁𝑎+
1 𝑚𝑜𝑙 𝑁𝑎2 𝑆𝑂4
To find the number of moles of Na2SO4
𝑚𝑜𝑙𝑒 𝑁𝑎2 𝑆𝑂4 = 25 𝑔 𝑁𝑎2 𝑆𝑂4 𝑥
1 𝑚𝑜𝑙 𝑁𝑎2 𝑆𝑂4
142 𝑔 𝑁𝑎2 𝑆𝑂4

16. Combining this equation with the first lead to
𝑚𝑜𝑙𝑒 𝑁𝑎+ = 25 𝑔 𝑁𝑎2 𝑆𝑂4 𝑥
1 𝑚𝑜𝑙 𝑁𝑎2 𝑆𝑂4
142 𝑔 𝑁𝑎2 𝑆𝑂4
𝑥
2 𝑚𝑜𝑙 𝑁𝑎+
1 𝑚𝑜𝑙 𝑁𝑎2 𝑆𝑂4
To obtain the mass of sodium in 25 g of Na2SO4, we multiply the number of moles of Na+
by the molar mass of Na+
𝑚𝑎𝑠𝑠 𝑁𝑎+
= 𝑚𝑜𝑙 𝑁𝑎+
𝑥
22.99 𝑔 𝑁𝑎+
𝑚𝑜𝑙 𝑁𝑎+
Substituting the previous equation gives the mass in grams of Na+:
mass Na+ = 25 g Na2SO4 x
1 mol Na2SO4
142 g Na2SO4
x
2 mol Na+
1 mol Na2SO4
x
22.99 g Na+
1 mol Na+
= 8.11 g Na+

17. The factor-label approach to above example
Another method can be used to solve this kind of example which has been
referred to as the factor-label method, or dimensional analysis. For instance, in
example 1.2, the units of the answer are g Na+, and the units given are g
Na2SO4. Thus, we can write
25 𝑔 𝑁𝑎2 𝑆𝑂4 𝑥
1 𝑚𝑜𝑙 𝑁𝑎2 𝑆𝑂4
142 𝑔 𝑁𝑎2 𝑆𝑂4
𝑥
2 𝑚𝑜𝑙 𝑁𝑎+
1 𝑚𝑜𝑙 𝑁𝑎2 𝑆𝑂4
𝑥
22.99 𝑔 𝑁𝑎+
1 𝑚𝑜𝑙 𝑁𝑎+
= 8.09 𝑔 𝑁𝑎+