Successfully reported this slideshow.
We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. You can change your ad preferences anytime.

analytical chemistry: introduction

2,876 views

Published on

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

Published in: Education

analytical chemistry: introduction

  1. 1. Lecture 1 Introduction to Analytical Chemistry Azad H. Mohammed Azadalshatteri@Hotmail.com Chemistry Department University of Garmian
  2. 2. References 1. Fundamental analytical chemistry, 9th edition, Douglas A. Skoog, 2014.* 2. Modern Analytical Chemistry, 2nd edition, David Harvey, 2008.
  3. 3. This lecture ๏‚ง Introduction to analytical chemistry ๏‚ง Some important units of measurement ๏‚ง Moles and millimoles
  4. 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. 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. 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. 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. 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. 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. 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 (Avogadroโ€™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. 11. ๐‘€๐‘€ ๐ถ๐ป2 ๐‘‚ = 1 ๐‘š๐‘œ๐‘™ ๐ถ ๐‘š๐‘œ๐‘™ ๐ถ๐ป2 ๐‘‚ ๐‘ฅ 12.0 ๐‘” ๐‘š๐‘œ๐‘™ ๐ถ + 2 ๐‘š๐‘œ๐‘™ ๐ป ๐‘š๐‘œ๐‘™ ๐ถ๐ป2 ๐‘‚ ๐‘ฅ 1 ๐‘” ๐‘š๐‘œ๐‘™ ๐ป + 1 ๐‘š๐‘œ๐‘™ ๐‘‚ ๐‘š๐‘œ๐‘™ ๐ถ๐ป2 ๐‘‚ ๐‘ฅ 16 ๐‘” ๐‘š๐‘œ๐‘™ ๐ป = 30 ๐‘”/๐‘š๐‘œ๐‘™ ๐ถ๐ป2 ๐‘‚ or Molar mass of formaldehyde (๐ถ๐ป2 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 ๐ถ๐ป2 ๐‘‚ (g/mol) = โˆ‘ atomic weights = 12 + 2 + 16 = 30 g/mol
  12. 12. Millimole Sometimes it is more convenient to make calculations with millimoles (mmol) rather than moles. 1 ๐‘š๐‘š๐‘œ๐‘™ = 10โˆ’3 ๐‘š๐‘œ๐‘™, ๐‘Ž๐‘›๐‘‘ 103 ๐‘š๐‘š๐‘œ๐‘™ = 1 ๐‘š๐‘œ๐‘™ millimolar mass (g/mmol) is equal to 10-3 molar mass (g/mol, or mg/mmol).
  13. 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 ๐‘š๐‘œ๐‘™๐‘’ ๐ป๐ต๐‘ง = 2 ๐‘” ๐ป๐ต๐‘ง ๐‘ฅ 1 ๐‘š๐‘œ๐‘™ ๐ป๐ต๐‘ง 122.1 ๐‘” ๐ป๐ต๐‘ง = 0.0164 ๐‘š๐‘œ๐‘™ ๐ป๐ต๐‘ง
  14. 14. To obtain the number of millimoles, we divide by the millimolar mass (0.1221 g/mmol), that is, ๐‘š๐‘š๐‘œ๐‘™ ๐‘œ๐‘“ ๐ป๐ต๐‘ง = 2 ๐‘” ๐ป๐ต๐‘ง ๐‘ฅ 1 ๐‘š๐‘š๐‘œ๐‘™ ๐ป๐ต๐‘ง 0.1221 ๐‘” ๐ป๐ต๐‘ง = 16.38 ๐‘š๐‘š๐‘œ๐‘™ ๐ป๐ต๐‘ง
  15. 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 ๐‘š๐‘œ๐‘™๐‘’ ๐‘๐‘Ž+ = ๐‘š๐‘œ๐‘™ ๐‘๐‘Ž2 ๐‘†๐‘‚4 ๐‘ฅ 2 ๐‘š๐‘œ๐‘™ ๐‘๐‘Ž+ 1 ๐‘š๐‘œ๐‘™ ๐‘๐‘Ž2 ๐‘†๐‘‚4 To find the number of moles of Na2SO4 ๐‘š๐‘œ๐‘™๐‘’ ๐‘๐‘Ž2 ๐‘†๐‘‚4 = 25 ๐‘” ๐‘๐‘Ž2 ๐‘†๐‘‚4 ๐‘ฅ 1 ๐‘š๐‘œ๐‘™ ๐‘๐‘Ž2 ๐‘†๐‘‚4 142 ๐‘” ๐‘๐‘Ž2 ๐‘†๐‘‚4
  16. 16. Combining this equation with the first lead to ๐‘š๐‘œ๐‘™๐‘’ ๐‘๐‘Ž+ = 25 ๐‘” ๐‘๐‘Ž2 ๐‘†๐‘‚4 ๐‘ฅ 1 ๐‘š๐‘œ๐‘™ ๐‘๐‘Ž2 ๐‘†๐‘‚4 142 ๐‘” ๐‘๐‘Ž2 ๐‘†๐‘‚4 ๐‘ฅ 2 ๐‘š๐‘œ๐‘™ ๐‘๐‘Ž+ 1 ๐‘š๐‘œ๐‘™ ๐‘๐‘Ž2 ๐‘†๐‘‚4 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+ ๐‘š๐‘Ž๐‘ ๐‘  ๐‘๐‘Ž+ = ๐‘š๐‘œ๐‘™ ๐‘๐‘Ž+ ๐‘ฅ 22.99 ๐‘” ๐‘๐‘Ž+ ๐‘š๐‘œ๐‘™ ๐‘๐‘Ž+ 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. 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‘” ๐‘๐‘Ž2 ๐‘†๐‘‚4 ๐‘ฅ 1 ๐‘š๐‘œ๐‘™ ๐‘๐‘Ž2 ๐‘†๐‘‚4 142 ๐‘” ๐‘๐‘Ž2 ๐‘†๐‘‚4 ๐‘ฅ 2 ๐‘š๐‘œ๐‘™ ๐‘๐‘Ž+ 1 ๐‘š๐‘œ๐‘™ ๐‘๐‘Ž2 ๐‘†๐‘‚4 ๐‘ฅ 22.99 ๐‘” ๐‘๐‘Ž+ 1 ๐‘š๐‘œ๐‘™ ๐‘๐‘Ž+ = 8.09 ๐‘” ๐‘๐‘Ž+

ร—