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Analytical Chemistry

National Technical Institute(NTI) Analytical Chemistry Second Year/ Gas and Petroleum Department Prof. Dr. Azad T. Faizullah. Analytical Chemistry Analytical Chemistry : The branch of chemistry that deals with the separation, identification and determination of components in a sample.

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Analytical Chemistry

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  1. National Technical Institute(NTI)Analytical ChemistrySecond Year/ Gas and Petroleum DepartmentProf. Dr. Azad T. Faizullah

  2. Analytical Chemistry • Analytical Chemistry: The branch of chemistry that deals with the separation, identification and determination of components in a sample. • It also traditionally includes coverage of chemical equilibrium and statistical treatment of data. • Qualitative analysis(شيكردنه وه ى جورى) • Quantitative analysis (شيكردنه وه ى برى مادده كه )

  3. 1- Qualitative analysis • Attempting to identify what materials are present in sample. Ag2CrO4

  4. 2- Quantitative analysis • Determining how much of material is present in a sample. The most reaction which is used in qualitative analysis can be used as a principle for quantitative analysis with some changes. • * The most important thing in quantitative analysis was purity…… • Always qualitative analysis must be done before quantitative analysis…. له زوربه ى شيكردنه وه كاندا دوو توخمى سه ره كى ههيه. • كيش (Weight) • قه باره (Volume)

  5. Analytical Chemistry Deals with development of new methods and techniques for the separation and determination of the analyte qualitatively and quantitatively Total analysis process consists of the following steps: 1-Sampling The most important conditions for sampling are a) The sample must be representative (reflect entire body from which it came). b) The sample must be homogeneous (having the composition everywhere) c) Transportation, from the field (sample place) to the laboratory without altering sample.

  6. 2-Method The analytical chemist must decide which of the numerous analytical methods available will be appropriate for the problem at hand. -Factors that the analyst may have to consider • Accuracy and precisionته واوى و وردى • Sensitivityهه ستيارى • Selectivityهه لبزاردنه يى • Speedخيرايى • Costنرخ • Legalityياسايى بوون

  7. 3- Separation • It involves the removing of the effect of interferences and increases the selectivity of the method toward the analyte with necessary accuracy and precision. 4- Quantitation • Involves quantititative measurement of the analyte 5- Evaluation • The statistical analysis is an important evaluative tool for analytical chemist. It can be used to determine the accuracy and precision required of the analytical technique.

  8. Quantitative Analysis • What type of information do you need? • Complete analysis: The goal is to determine the amount of each component in a sample. • Partial analysis: Determining one or a limited number of species in a sample (This is the most common approach). • Examples: • Iron in an ore sample. • Presence of lead in water sample • This type of analysis depends on measure the physical or chemical property of the substance If There is a mathematical relation between measuring this property with the amount of analyte.

  9. Chemical Methods This method depends on making a chemical reaction for the analyte and this reaction must be having some conditions. 1-Spontaneous and fast reaction 2- Irreversible reaction, and completely reacted 3- Equilibrated reaction 4- No side reaction 5- Clear equivalent point 6- Using indicators for measuring the equivalence point

  10. Solutions • Solutions are homogeneous mixtures. • The major component is called solvent(, and the minor components are called solute(. • If both components in a solution are 50%, the term solute can be assigned to either component. • When gas or solid material dissolve in a liquid, the gas or solid material is called the solute. • When two liquids dissolve in each other, the major component is called the solvent and the minor component is called the solute.

  11. Gravimetric Analysis • By definition, includes all methods of analysis in which the final stage of the analysis involves weighing. Volumetric AnalysisVolumetric analysis involves using volumes of liquids to analyse a concentration. To do this we need the following things: • A chemical of a known concentration that will react with our ‘unknown’ concentration chemical • An indicator that will tell us when all the chemical has been reacted • A number of pieces of equipment that we can use to measure volume accurately

  12. -Different methods for concentration expression • The mole: The mole (mol) is a fundamental unit describing the amount of chemical species. It is always associated with the chemical formula and represented one Avogadro's number (6.02 x 1023) of atom, ions, molecules or elements. • For example: chemical formula of Methane is CH4 ( 12+4x1=16 gm/mole) • For C6H12O6 ( 12x6+1x12+16x6 =180 gm/one mole) • Note: The mole is a big unit relatively, for this reason the millimole unit( mmole) unit is used mostly. mmole= 1/1000 mole. Number of mole = weight(gm)/ formula weight Number of mmole=weight(gm)/milliformula weight. no. of mole = gm. of species/M. Wt. no. of mmole=( gm. of species x M.wt.)/1000.

  13. Concentration expression 1- Molarity: Number of moles in one liter of solution, expressed as M : M = (no. of mole of solute) / (volume of solvent by liter). or M = (no. of mmole solute) / (volume of solvent by milliliter) M = (no. of moles) / (liter of solution) = [(grams of solute) / (M.Wt.) / liter of solution]. M= [(grams of solute) x 1000] / [(milliliter of solution) x M.Wt.] These two equations are very important in analytical chemistry especially in preparation of solutions………….

  14. 1- Preparation of solutions from Solid compounds. • Examples: Q/Prepare 2 liter of 0.1M Na2CO3 from the solid material. -What is needed to solve this example…..? • Request: how mush weighing from the solid material (Na2CO3)? • By another words : -How much gram we needed to weighing from Na2CO3 and dissolving in 2 liter of water to obtaining 0.1M Na2CO3?

  15. 0.1M = 0.1 mole Na2CO3 in one liter of solution = 0.1 x 2 = 0.2 mole Na2CO3 in 2liter of solution. no. of mole c= Wt. (g) / M.Wt. (g/mole). Wt. g = no. of mole x M.Wt. Wt. of Na2CO3 (g) = 0.2 x [(2x23) + (1x12) + (3x16)] = 21.2 g which is required to dissolved in 2 liter of water to obtain 0.1M Na2CO3 in this volume (2L).

  16. Q/ Prepare 0.1M of NaCl in 250ml volume? Question requirement was how many grams of solid NaCl was required to prepare a solution of this salt in 250ml volume. • 0.1M = 0.1 mole NaCl in 250ml of solution • = 0.1 x (250/1000) = 0.025 mole NaCl in 250ml of solution. • no. of mole c= Wt. (g) / M.Wt. (g/mole). • Wt. g = no. of mole x M.Wt. • Wt. of NaCl (g) = 0.025 x [(1x23) + (1x35.5)] = 1.4625 g NaCl is required to dissolved in 250ml of water to obtain 0.1M NaCl.

  17. Example, Explain how we can prepare each of the following solutions: 1- Prepare 0.2M BaCl2.2H2O in 500ml? 2- Prepare 0.03M CaCl2 in 1L. 3- Prepare 1x10-3M CuSO4.5H2O in 100ml.

  18. Q/Prepare 2 liter of 0.1M Na+ from Na2CO3 pure solid material. • What is needed to solve this example…..? • Request: Weight (in gm.) of (Na2CO3) to be dissolved in 2 liter of distilled water to obtain solution contains 0.1M Na+? • 0.1M Na+ = 0.1 mole Na+ in one liter of solution • Na2CO3 molecule contain 2 Na+ • M of Na2CO3 = 1/2 M Na+ • = 0.1 x 1/2 = 0.5 M • no. of grams Na2CO3 = 0.5 x 106 = 5.3 g must be dissolved in one liter. • no. of grams Na2CO3 in 2 liter = 5.3 x 2= 10.6 g of Na2CO3must be dissolved in two liter to obtain 0.1M Na+ solution.

  19. Q/ Prepare 0.25M Cl- in 250ml from 1- NaCl salt 2- BaCl2.2H2O 1- From NaCl • 0.25M Cl- = 0.25 mole Cl- in one liter of solution • NaCl molecule contain 1 Cl- • M of NaCl = M Cl-= 0.25M • no. of grams NaCl = M x M.wt=0.25 x 58.5 = 14.625 g must be dissolved in one liter. • no. of grams NaCl in 250ml = 14.625 x 0.25= 3.656 g of NaCl must be dissolved in 250ml to obtain 0.25M Cl- solution.

  20. 2- Preparation of solutions from Liquids The process will be made by two stepa; the first step to be calculated is Molar concentration of the concentrated solution( commercial) and the second step is the dilution process to prepare the required solution. To achieve that the following equations (steps) should be applied. M=( % x specific gravity( or density) x 1000 )/ molecular weight………….( 1) M1 x V1 ( conc.) = M2 x V2 ( diluted) M1= concentration calculated from step 1. V1= The volume of conc. Solution to be find out. M2= conc. Of the new diluted solution ( known.) V2= Volume of the diluted solution.( known ).

  21. First step: • calculate H3PO4 molarity in the bottle from the information. • Density of specific gravity = 1.696 g/ml • = 1.696 x 1000 = 1696 g/l this is when the percentage of H3PO4 was 100%. • But the percentage was 85%??? 1696 x (85/100) = 1441.6 g/l for this solution. • M = = = 14.7 M (molar concentration of the concentrated acid (H3PO4) in the bottle).

  22. There is another way to solve this example (fast way): • For the above example….. • = 14.7 M the same result.

  23. Second step: • Dilution (M1V1) concentrated = (M2V2) diluted • (14.7 x V1) = (6 x 500) • V1 = (6x500)/14.7 = 204.1 ml • This is meaning we take 204.1 ml from the concentrated solution (bottle) and diluted to 500 ml with distilled water to obtain 6M H3PO4.

  24. Q/ Prepare 250ml of 2M HNO3 from the traditional solution. The information on the bottle: 69%(w/w) HNO3 , specific gravity 1.42 g/ml, formula weight 63 g/mole. • First step: • Density= 1.42 x 1000 = 1420 g/l for 100% • But the percentage was 69%??? 1420 x (69/100) = 979.8 g/l for this solution. • M = = 97.98/63 =15.55 M (molar Con. acid) Or M= (% x Sp.g X 1000)/ M.wt =((69/100) x 1.42 x 1000)/63=15.55M

  25. Second step: • Dilution (M1V1) concentrated = (M2V2) diluted • (15.55 x V1) = (2 x 250) • V1 = (2x250)/15.55 = 32.2 ml • This is meaning we take 32.2 ml from the concentrated solution (bottle) and diluted to 250 ml with distilled water to obtain 2M HNO3. Example: A concentrated solution of aqueous ammonia is 28.0% w/w NH3 and has adensity of 0.899 g/mL and (F.wt=17.04g/mol). What is the molar concentration of NH3 in this solution, and how can you prepare 0.5M solution in 250m volume NH3 ?

  26. 2- Preparation of solutions from Liquids.

  27. Examples Q/ Describe a method for the preparation of 0.1M HCl in 250ml volume from the concentrated acid solution (2M). (M1V1)concentrated=(M2V2)diluted When M1=2M, M2=0.1M V2= 250ml V1= ? (2 x V1)=(0.1 x 250) V1=(0.1x250)/2 =12.5ml That is mean for the preparation of 0.1M HCl in 250ml we dilute 12.5ml concentrated acid to 250ml in the proper volumetric flask.

  28. Q/Describe the method for preparation of 500 ml of 0.03M K+ from the 0.7M K4Fe(CN)6 K4Fe(CN)6 contain 4K+ M K4Fe(CN)6 = (1/4)M K+ = (1/4)x 0.03 = 0.0075M (M1V1) concentrated = (M2V2) diluted (0.7 x V1) = (0.0075 x 500) V1 = (0.0075x500)/0.7 =5.357 ml Taking 5.357 ml from the 0.7M K4Fe(CN)6 then diluted to 500 ml with water to prepare 0.03M K+……….

  29. Q/Calculate the molar concentration of the solute species in (a) an aqueous solution that contain 2.30 g of ethanol (C2H5OH) (M.Wt. = 46.1g/mol), in 3.50L. (b) an aqueous solution that contains 285 mg of trichloroacetic acid (Cl3CCOOH) (M.Wt.= 163g/mol), in 10 ml (the acid is 73% ionized in water).

  30. b)

  31. 2- Formality: • Both molarity and formality express concentration as moles of solute per liter of solution. The only differences between them are: • Molarity (M)is the concentration of a particular chemical species in solution (mol/L). • Formality (F)on the other hand, The number of moles or formula of solute, regardless of chemical form, per liter of solution (mol/L). • There is no difference between a substance’s molarity and formality if it dissolves without dissociating into ions.

  32. The molar concentration of a solution of glucose, for example, is the same as its formality, Since it is not dissociated in the solution to other form. • 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. While, the formality of NaCl, however, is 0.1 F because it represents the total amount of NaCl in solution.

  33. 3- Normality: • Normality makes use of the chemical equivalent, which is the amount of one chemical species reacting stoichiometrically with another chemical species. -Normality (N) :The number of equivalents of solute per liter of solution, or the number of milliequivalents in ml of solution.

  34. The Definitions of Equivalent and Milliequivalent • The amount of a substance contained in one equivalent can vary from reaction to reaction. Consequently, the weight of one equivalent of a compound can never be computed without reference to a chemical reaction in which that compound is, directly or indirectly, a participant. • The number of equivalents, n, is based on a reaction unit, which is that part of a chemical species involved in a reaction.

  35. Equivalent weight for different reaction types: 1- Acid-base reaction (neutralization reaction) Reactive species (n) in acid base reactions were H+ and OH-, respectively. Accordingly number of equivalent (n) in this case will depend on the number of H+ or OH- that participate in the reaction.

  36. 2- Precipitation reaction Reactive species (n)in this case was cation or anion charge 3- Oxidation-reduction (Redox) reaction Reactive species (n) was the number of electrons released by the reducing agent or accepted by the oxidizing agent

  37. 4- Complex formation reduction The reactive species (n) is the number of electron pairs that can be accepted by the metal or donated by the ligand.

  38. Example Q/Calculate the number of gram millequivalent for Oxalic acid (H2C2O4.2H2O , M.Wt. = 126.1) in 0.5 g pure material, we supposed that the two hydrogen ions displaced. • Eq. Wt. =126.1/2 = 63.05 gm • Meq. Wt. = 63.05/1000= 0.063305 gm • no. of gram milliequivalent for H2C2O4.2H2O = 0.5/0.063305= 7.93 gm.

  39. Q/ What is the volume of 0.1N HCl which is prepared from dilution of 150ml of 1.24N HCl. No. milleq. for acid before dilution = No. milleq for acid after dilution (N1V1) before dilution = (N2V2) after dilution (1.24 x 150) = (0.1 x V2) • V2 = 1860 ml. • i.e, we can prepare 0.1N HCl by diluting 150ml of 1.24N of HCl to final volume of 1860ml.

  40. 4- Percentage concentrations: a) Weight -weight percent (% w/w):Grams (g) of solute per 100 g of solution. b) Volume-volume percent (% v/v): Milliliters (mL) of solute per 100 mL of solution. c) Weight-to-volume percent (% w/v):Grams (g) of solute per 100 mL of solution.

  41. ppm ppm c) parts per million (ppm): • Micrograms of solute per gram of solution; for aqueous solutions the units are often expressed as milligrams of solute per liter of solution.

  42. Example: • 35% (w/w) of HCl 35 g in 100 g of water, or 35 g in 100 ml of water (density of water = 1). • 98% (w/v) of NaCl 98 g NaCl dissolved in 100 ml of water. • How can you prepare 200ml of a solution of 5%(v/v) acetic acid (CH3COOH). It can be prepared by addition of 10ml of the acid to a little amount of distilled water then dilute the solution to 200ml final volume.

  43. Q/ Describe the methods for preparation of anhydrous ethanol. 1) 500 ml of 6% (w/v) 2) 500 ml of 6% (w/w) 3) 500 ml of 6% (v/v) • 6% (w/v) meaning dissolving 6 gm of ethanol in 100 ml of water this is meaning (6 x 5) = 30 gm of ethanol in 500 ml of water. • 2) 6% (w/w) meaning dissolving 6 gm of ethanol in 100 gm of water this is meaning (6 x 5) = 30 gm of ethanol in 500 gm of water. • 3) 6% (v/v) meaning mixing 6 ml of ethanol in 94 ml of water this is meaning (6 x 5) = 30 ml of ethanol in (94 x 5) = 470 ml of water

  44. 0.2% (w/v) K2Cr2O7 = 0.2 gm in 100 ml of water = 2 gm in 1000 ml of water. M= [(grams of solute) / (M.wt.)] / liter of solution M = [2 gm / 294.19 gm/mole]/ L = 0.00674 M = 6.7 x10-3 M (M1V1) before dilution = (M2V2) after dilution (0.137 x V1) = (6.7 x 10-3 x 250) V1 = 8.759 ml. Taking 8.579 ml from the 0.2% (w/v) K2Cr2O7 solution and diluted to 250 ml with water. Q/ Prepare 250 ml of 0.2% (w/v) K2Cr2O7 form 0.137 M K2Cr2O7.

  45. Q/The maximum allowed concentration of chloride in a municipal drinking water supply is 2.5x102 ppm Cl–. When the supply of water exceeds this limit, it often has a distinctive salty taste. What is this concentration in moles Cl–/Liter (M)? M Cl-= 2.5x102 mg/L/ ( 35.453g/mol) x 1000 = 7.05x10-3 mol Cl-/ L = 7.05x10-3M

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