2) Quantitative analysis: It is aimed to determine the quantities of the substances

Analytical Chemistry is a branch of chemical science that deals with identification, separation and quantification of the components of a sample.

Analysis could be divided two part.

1) Qualitative analysis: It is aimed to determine the chemical components of a

sample.

2) Quantitative analysis: It is aimed to determine the quantities of the substances

present in the sample.

Separation methods are included in both qualitative and quantitative analysis.

Analytical methods are classified according to the measurement of some

quantities proportional to the quantity of analyte. Classical Methods and

Instrumental Methods.

The classical methods are divided in to two part.

1) Gravimetric methods: The product formed as a result of a chemical reaction is based on the weight measurement, such as precipitation, electrodeposition and evaporation.

2) Volumetric methods: These are based on measuring the volume of a phase proportional to the quantity of the component to be determined.

Titrimetric methods are example to volumetric methods.

These methods are fairly accurate and precise methods, but their operation takes a long time and requires a large amount of manual skill as long as the amount of material to be assigned is high.

In instrumental analysis, signals are analyzed proportinal to type and quantity of a compound using some analytical instruments. Instrumental analytical methods are divided some parts.

1) Spectroscopic methods: This techniques is based on analysis of spectroscopic characteristics of a compound. For this aim, UV-Visible spectrophotometry, spectrofluorimetry, IR spectrophotometry, atomic spectroscopy, X-ray spectroscopy can be used.

2) Electroanalytical methods: This techniques is based on analysis of electrochemical characteristics of a compound. Voltammetry, polarography, coulometry, conductometry and potentiometry.

3) Chromatographic methods: This techniques is based on separation of

compounds related with structure of these compounds. Liquid chromtography,

gas chromatography, thin layer chromatography, super critic fluid chromatograpy

are exampled to this technique.

Beside these, there are capillary electrophoresis, thermal analytical methods and combined methods such as LC-MS, GC-MS. These are also classified in instrumental methods.

CONCENTRATIONS UNITS OF SOLUTIONS

Solutions are mixtures which are formed by homogeneous distribution of two or more phases in one of them. In solutions, higher amount part is solvent, lower amount part is solute. The amount of solubility is determined in various forms in chemical analysis.

1) Atom weight: The total weight of an atom as many as the avogadro number (6.02 x 10

) of an element.

2) Molecular weight: It is the sum of the atomic weights of the atoms that bring a molecule to the square. For example: Mn: 55 g, Zn: 65 g.

3) Mole number = weight / molecular weight

What is mole of 20.0 g NaOH? (Na:23, H:1, O:16) 20 / 40 = 0.5 mol

1 mol = 1000 mmol 1 mmol = 10

mol

Percent solubles: Solubles in solution are expressed as percentages.

Weight / weight (w / w), weight / volume (w / v) and volume / volume (v / v). If there isnt any knowledge about this, units will be excepted as w / v.

For example, % 3 boric acid solution means that 3 g boric acid in 100 mL

water. In commercial acids like HNO

, HCl etc, this unit is w/w. For liquid –

liquid mixture, this unit is v/v. For example, % 20 methanol solution means

that 20 mL methanol in 100 mL solution (80 mL of this solution is water).

Molarity (M): Mole number of a solute or analyte in 1 L solutions. the number of moles

of dissolved substance in 1 liter of solution.

If 5.20 g of KMnO

are dissolved in 300 mL of the solution, what is the molarity of this solution? (K:39, Mn:55, O:16)

In 300 mL 5.2 g KMnO

In 1000 mL 17.33 g KMnO

mol of KMnO

= 17.33 / 158 = 0.1097 mol This solution is 0.1097 M

How to prepare the solution of 100 mL 0.2 M KNO

solution? K:39, N:14, O:16 1 L 1 M KNO

is 1 mole = 101 g

0.1 L 0.2 M is 2.02 g

2.02 g KNO

is weighted and solve in 100 mL water.

Normality (N): Equivalent gram number of an analyte in 1 L solutions.

Equivalent gram = weight / equivalent weight

Equivalent weight = Molecular weight / effect value Effect value:

1) For acids: It is the number of releasing H + to media. HCl is 1, H

SO

is 2, H

PO

is 3, but H

BO

is 1. Because H

BO

is a very weak acid and only release 1 H +, under special conditions (After complexing with polyols).

2) For bases: It is the number of releasing OH- to media. which the base can give when it is dissolved. NaOH is 1, Ca(OH)

is 2.

3) For salt: The total value of the cathion (or the total value of the cation that replaces the protons in the acid forming the salt). KCl is 1, CaCO

is 2

4) For oxydation and reduction reactions: It is the number of electrons taken or given per molecule.

Zn

+ 2e

 Zn Effect value is 2.

IO

+ 6H

+ 6e

→ I

+ 3H

O Effect value is 6.

What is the normality of a solution containing 2.42 g HCl in 420 mL? (H:1, Cl:35.5)

Equivalent weight of HCl = 36.5 / 1 = 36.50 g In 420 mL 2.42 g HCl

In 1000 mL 5.76 g HCl

5.76 / 36.5 = 0.16 equivalent gram HCl

For this reason normality of HCl solutions is 0.16 N N = M x Effect value

For example: 0.1 M H

SO

Normality of this solution: 0.1 x 2 = 0.2 N There are other concentrations units:

ppm = mg/mL

ppb = ng/mL

Purity of HNO

is % 70 and density is 1.25 g/mL. What are the normality and molarity of this solution? (H:1, N:14, O:16)

In 1 mL solution 1.25 g HNO

In 1000 mL solution 1250 g HNO

After that we have to calculate pure amount of HNO

in 1000 mL solution.

1250 x 70/100 = 875 g pure HNO

Molarity of this solution is : 875 / 63 = 13.89 M

Normality of this solution is : 875 / (63/1) = 13.89 N

How do you prepare 0.1050 M, 300 mL H

PO

solution from % 85 pure, d = 1.88 g/mL H

PO

? (H:1, P:31, O:16)

In 1 mL H

PO

solution 1.88 g H

PO

In 1000 mL H

PO

solution 1880 g H

PO

Pure amount of H

PO

is;

1880 x 85/100 = 1598 g

1 L 1M H

PO

98 g 0.1050 M 0.3 L H

PO

3.087 g

In 1000 mL solution pure amount of H

PO

is 1598 g x 3.087 g

x = 1.932 mL

First of all, an amount of water should be added to flask and then 1.932 mL H

PO

solution should be added. After that the volume is completed to 300 mL with

water.