Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Successfully reported this slideshow.

Like this presentation? Why not share!

- Mbr โฌย ย ู ูู โฌย ย ุฑูู 5 by Dr. Ashraf Refaat 789 views
- Sokalanยฎ RO 3500 by BASF 2906 views
- Quality control & Assurance in Anal... by Lavakusa Banavatu 7333 views
- B.sc. frst yr fundamental concepts by Ram Darash Pandey 912 views
- Chemistry Basics by I Wonder Why Science 2201 views
- FREE SIMPLE POWER POINT TEMPLATE by calonmayat 684 views

2,876 views

Published on

analytical chemistry by Azad H. M. Alshatteri, MSc. at UoM

Published in:
Education

No Downloads

Total views

2,876

On SlideShare

0

From Embeds

0

Number of Embeds

26

Shares

0

Downloads

170

Comments

0

Likes

20

No embeds

No notes for slide

- 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 ๏ง Introduction to analytical chemistry ๏ง Some important units of measurement ๏ง 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 (ยตg) are used. Volumes of liquids are measured in units of liters (L), milliliters (mL), or microliters (ยตL). 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- ยต 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 (Avogadroโs 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. ๐๐ ๐ถ๐ป2 ๐ = 1 ๐๐๐ ๐ถ ๐๐๐ ๐ถ๐ป2 ๐ ๐ฅ 12.0 ๐ ๐๐๐ ๐ถ + 2 ๐๐๐ ๐ป ๐๐๐ ๐ถ๐ป2 ๐ ๐ฅ 1 ๐ ๐๐๐ ๐ป + 1 ๐๐๐ ๐ ๐๐๐ ๐ถ๐ป2 ๐ ๐ฅ 16 ๐ ๐๐๐ ๐ป = 30 ๐/๐๐๐ ๐ถ๐ป2 ๐ or Molar mass of formaldehyde (๐ถ๐ป2 ๐) 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 ๐ถ๐ป2 ๐ (g/mol) = โ 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 ๐๐๐๐ = 10โ3 ๐๐๐, ๐๐๐ 103 ๐๐๐๐ = 1 ๐๐๐ 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 ๐๐๐๐ ๐ป๐ต๐ง = 2 ๐ ๐ป๐ต๐ง ๐ฅ 1 ๐๐๐ ๐ป๐ต๐ง 122.1 ๐ ๐ป๐ต๐ง = 0.0164 ๐๐๐ ๐ป๐ต๐ง
- 14. To obtain the number of millimoles, we divide by the millimolar mass (0.1221 g/mmol), that is, ๐๐๐๐ ๐๐ ๐ป๐ต๐ง = 2 ๐ ๐ป๐ต๐ง ๐ฅ 1 ๐๐๐๐ ๐ป๐ต๐ง 0.1221 ๐ ๐ป๐ต๐ง = 16.38 ๐๐๐๐ ๐ป๐ต๐ง
- 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 ๐๐๐๐ ๐๐+ = ๐๐๐ ๐๐2 ๐๐4 ๐ฅ 2 ๐๐๐ ๐๐+ 1 ๐๐๐ ๐๐2 ๐๐4 To find the number of moles of Na2SO4 ๐๐๐๐ ๐๐2 ๐๐4 = 25 ๐ ๐๐2 ๐๐4 ๐ฅ 1 ๐๐๐ ๐๐2 ๐๐4 142 ๐ ๐๐2 ๐๐4
- 16. Combining this equation with the first lead to ๐๐๐๐ ๐๐+ = 25 ๐ ๐๐2 ๐๐4 ๐ฅ 1 ๐๐๐ ๐๐2 ๐๐4 142 ๐ ๐๐2 ๐๐4 ๐ฅ 2 ๐๐๐ ๐๐+ 1 ๐๐๐ ๐๐2 ๐๐4 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+ ๐๐๐ ๐ ๐๐+ = ๐๐๐ ๐๐+ ๐ฅ 22.99 ๐ ๐๐+ ๐๐๐ ๐๐+ 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 ๐ ๐๐2 ๐๐4 ๐ฅ 1 ๐๐๐ ๐๐2 ๐๐4 142 ๐ ๐๐2 ๐๐4 ๐ฅ 2 ๐๐๐ ๐๐+ 1 ๐๐๐ ๐๐2 ๐๐4 ๐ฅ 22.99 ๐ ๐๐+ 1 ๐๐๐ ๐๐+ = 8.09 ๐ ๐๐+

No public clipboards found for this slide

Be the first to comment