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CHAPTER 1 INTRODUCTION: ANALYTICAL CHEMISTRY SITI MAZLEENA MOHAMED SMM/FSG/UiTM NS

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INTRODUCTION ANALYTICAL CHEMISTRY

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  • CHAPTER 1INTRODUCTION: ANALYTICAL CHEMISTRYSITI MAZLEENA MOHAMEDSMM/FSG/UiTM NS

    SMM/FSG/UiTM NS

  • Review the type and steps in analysisReview the terms: Moles, molarity and concentration.Review other forms of expressing concentrationi) ppmii) ppbiii) pptiv) %(w/w)v) %(w/v)vi) %(v/v)CHAPTER OUTLINESMM/FSG/UiTM NS

    SMM/FSG/UiTM NS

  • A field that related to the measurements and characterization of a chemical species that contain in a sample

    Chemical analysis is more than just detecting or determining the general composition or a specific component of a sample.

    It is the resolution or interpretation of a given problem.What is chemical analysis?SMM/FSG/UiTM NS

    SMM/FSG/UiTM NS

  • Discipline of analytical chemistrySMM/FSG/UiTM NS

    QualitativeQuantitativeTo detect the presence of a substance in the sampleIs the determination of that substancesFinding out what substancesFinding the amount (quantity) of substances

    SMM/FSG/UiTM NS

  • Defining the problemSelecting the methodObtain a representative samplePrepare the sample for analysisPerforming the measurementsSteps in analysisConducting the experiment and gathering the dataSMM/FSG/UiTM NS

    SMM/FSG/UiTM NS

  • 1. Defining ProblemThe goal of every chemical analysis is to obtain the required information within a period of time acceptable to the customer.

    Many problem do not require complete identification and many cases require only a general classification.

    It is important for the analyst to determine the information required by the client.

    For example, in water analysis, only the total hardness is required rather than the concentration of individual Ca2+ and Mg2+ ion concentration is not necessary.

  • Should know the origin of the sample.

    Sample history include information on how, where, and when the sample were collected, transported and stored.

    Analyst will use the sample history to maximize advantage in solving the problem.

    2. Selecting the method

  • Literature search is essential to find an appropriate procedures.Patents or commercial literature usually help in determination of the composition of industrial materials.Others:- book- review articles/journal- standard organization : i) Association of official Analytical Chemists (AOAC)ii) Food and Drug Administration (FDA)- electronic media

  • Sampling: Process of selecting representative material from the lot and storing the sample. (Define the term sampling Apr 2007, Oct 2007)

    Chemical analysis usually performed on only a small portion of the material.

    If the amount of material is very small and it is not needed for future use, the entire sample may be used for analysis.

    The suitable sampling method differ from one substance to another depending on homogeneity.3. Obtain a representative sample

  • Homogeneous: Substance that has the same composition throughout the sample.

    Heterogenous: Substance that has different composition from one reagent to another reagent.

    Proper procedure must be used to store both samples and standards.

    All sample must be properly labeled and recorded.Continue

  • Step of sample preparation involved bulk materials:(Describe the steps involved in sampling bulk materials Apr 2007)

    1. Bulk sample must be reduced in size to obtain a laboratory sample of several grams from which a few grams to milligrams will betaken to be analyzed (analysis sample).

    2. The size reduction may require taking portions (eg, two quarters) and mixing, in several steps, as well as crushing and sieving to obtain a uniform powder for analysis.4. Prepare the sample for analysis

  • Why sample pretreatment is important?Laboratory samples are often subjected to physical or chemical pretreatment where it is converted to a form that is suitable for the measurement.

    During pretreatment:i)reduce and remove interferencesii)adjust analyte concentrations to a range suitable for measurementiii)produce species from analyte that have quantitatively measurable properties.

  • Analysis: Incorporates the measurement of the concentration of the analyte in replicates and comparing with standards.

    Replicate measurement: the practice of taking multiple readings. (What do you understand by the term replicate Nov 2005)

    The replicate measurements are necessary to obtain the measurement uncertainty.

    The uncertainty important as it indicates the reliability of the measurements.5. Performing the measurements

  • Factors:

    i)Calibration

    ii)Validation/Controls/Blank A blank contains the reagents and solvents use in analysis, but no analyte.

    6. Conducting the experiments

  • Statistical analysis (eg. Standard deviation)

    Deliver a clearly written, complete report and their limitations.

    Analyst should critically evaluate whether the results are reasonable and relate to the analytical problem.

    7. Gathering the result & report the data

  • Calculate the number of grams in one mole of CaSO4. 7H2O

    Calculate the molar mass of glucose?Calculate the molar mass of formaldehyde?In class revision (1)SMM/FSG/UiTM NS

    SMM/FSG/UiTM NS

  • MolesMoles = grams formula weight (g/mol)Milimoles = miligrams Formula weight (mg/mmol)SMM/FSG/UiTM NS

    SMM/FSG/UiTM NS

  • How many moles and milimoles of benzoic acid (molar mass = 122.1 g/mol) are contained in 2.00 g of the pure acid?

    How many grams of Na+ (23.00 g/mol) are contained in 25.0 g of Na2SO4 (142 g/mol)?

    Homework (1)SMM/FSG/UiTM NS

    SMM/FSG/UiTM NS

  • 3. Calculate the number of the moles in 500 mg Na2WO4 (Sodium Tungstarte)

    4. How many milligrams are in a 0.250 mmol Fe2O3 (ferric oxide)

    SMM/FSG/UiTM NS

    SMM/FSG/UiTM NS

  • Example:Calculate the number of moles in 500 mg Na2WO4Solution:Moles = grams1 g = 1000 mg formula weight=500 mg/1000 293.8= 0.0017 mol

    SMM/FSG/UiTM NS

    SMM/FSG/UiTM NS

  • Example:How many mg are in 0.250 mmol Fe2O3?Solution:Milimoles = miligrams formula weight (mg/mmol)Miligrams = milimoles x formula weight = 0.250 mmol x 159.7 mg/mmol = 39.9 mgSMM/FSG/UiTM NS

    SMM/FSG/UiTM NS

  • Molarity (M) : No of moles of the solute in 1L of solution.M = no.mol solute no.L solution

    Molality (m) : No of moles of the substance per kilograms of the solvent.m = no.mol solute kg solution

    How do we express concentrations of solution?Milimoles = molarity x mililitersSMM/FSG/UiTM NS

    SMM/FSG/UiTM NS

  • What is the molarity of a solution of 0.60 g NaCl in 100 mL of solution?Given, MW NaCl = 58.5 g/mol.Solution:No. of mol = 0.60 g 58.5 g/mol = 0.0102 mol

    Molarity = 0.0102 mol 0.1 L = 0.102 MExampleSMM/FSG/UiTM NS

    SMM/FSG/UiTM NS

  • How many grams Na2SO4 should be weighted out to prepare 500 mL of a 0.100 M solution?

    Ans: 7.1 gTry thisSMM/FSG/UiTM NS

    SMM/FSG/UiTM NS

  • Density Calculations How do we convert to molarity?The concentration of many fairly concentrated commercial acids and bases are usually given in terms of percent by weight.It is necessary to prepare solutions of a given approximate molarity from these substancesMust know the density in order to calculate molaritySMM/FSG/UiTM NS

    SMM/FSG/UiTM NS

  • Density Calculations How do we convert to molarity?Density - the mass of the solution per a unit of volume

    Density of a solution correlates with the concentration of the dissolved substance

    Molarity is the most common way to express the solution concentration that reports the number of moles of the dissolved compound (solute) in 1 liters (L) the solution.

    SMM/FSG/UiTM NS

    SMM/FSG/UiTM NS

  • ContinueExample:

    Calculate the molarity of a potassium hydroxide (KOH) solution having a density of 1.28 g/mL and containing 25 percent (by weight) of the dissolved compound.

    Solution:

    Formula weight of KOH = 39 + 16 + 1 = 56 g/mole.

    Multiply the solution density by 1000 to calculate the weight of 1000 ml (1 L) of the solution.

    1.28 g/ml x 1000 ml = 1280 g.

    SMM/FSG/UiTM NS

    SMM/FSG/UiTM NS

  • Multiply the mass from Step 2 by the mass percentage of the solute, and then divide by 100 to compute the mass of the dissolved compound in 1L of the solution. The mass of KOH is 1280 g x 25 = 320 g 100

    Divide the mass of the dissolved compound in 1L of the solution by its molecular mass to calculate the molarity of the solution.The molarity is 320 g = 5.71 M 56 (g/mole) ContinueSMM/FSG/UiTM NS

    SMM/FSG/UiTM NS

  • How many milliliters of concentrated sulfuric acid, 94.0% (g/100 g solution), density 1.831 g/cm3, are required to prepare 1 L of 0.100 M solution?

    Answer: 5.71 mLExample:SMM/FSG/UiTM NS

    SMM/FSG/UiTM NS

  • Dilute solution (with low concentration) can be prepared from a more concentrated solution.

    A known volume of the concentrated solution can be transferred into a new flask and diluted to the required volume or weight.DilutionSMM/FSG/UiTM NS

    SMM/FSG/UiTM NS

  • The laboratory grade concentrated HCl has a concentration of 12.1 M. What a volume of the concentrated acid is necessary to prepare 500 mL of 0.100 M HCl?

    Solution:M1V1 = M2V2(12.1 M) x V1 = (0.100 M) x 500 mLV1 = (50.0) / 12.1 = 4.13 mL.

    Procedures:Dilute 4.13 mL of concentrated acid in a 500-mL VFDilute with distilled water to the 500-mL markInvert the flask several times to ensure complete mixing.Example: To prepare 0.100 M HCl solutionSMM/FSG/UiTM NS

    SMM/FSG/UiTM NS

  • Solid sample

    Weight percent, %w/w = Weight solute (g) x 100 Weight of sample (g) ppm = weight solute (g) x 106 weight sample (g)

    ppb = weight solute (g) x 109 weight sample (g)

    ppt = weight solute (g) x 1012 weight sample (g)

    Expressions of Analytical ResultsSMM/FSG/UiTM NS

    SMM/FSG/UiTM NS

  • Liquid sample

    %w/v = Weight solute (g) x 100 Volume of sample (mL)

    ppm = weight solute (g) x 106 volume of sample (mL)

    ppb = weight solute (g) x 109 volume of sample (mL)

    ppt = weight solute (g) x 1012 volume of sample (mL)

    Expressions of Analytical ResultsSMM/FSG/UiTM NS

    SMM/FSG/UiTM NS

  • Common unit to express contentSMM/FSG/UiTM NS

    ExpressionAbbreviationUnitsw/ww/vv/vParts per millionppmg/gg/mLnL/mLmg/kgmg/LL/L

    Parts per billionppbng/gng/mLnL/Lg/kgg/L

    SMM/FSG/UiTM NS

  • Briefly explain how to prepare 1.0 L 10.50% (w/v) aqueous CH3CH2CH2OH

    Solution:

    %w/v = weight solute (g) x 100 volume sample (mL)

    10.50 = weight solute x 100 = 105 g 1000Example: Percent CompositionSMM/FSG/UiTM NS

    SMM/FSG/UiTM NS

  • Step of preparation:1.Weight a certain quantity of the solid reagent2.Dissolve it in a solvent in a volumetric flask.Preparation of solutionSMM/FSG/UiTM NS

    SMM/FSG/UiTM NS

  • As a student, you are given copper (II) sulphate pentahydrate (CuSO4.5H2O, MW 249.69 gmol-1). Describe how you can prepare a 500 mL standard solution of 10.00 mM Cu2+ from the reagent.Solution: 1 M = 1000 mMM x V = 10.00 x 10-3 x 0.500 L = 5.000 x 10-3 molWeight of reagent = 5.000 x 10-3 mol x 249.69 g/mol = 1.248 gProcedure: 1.Place solid CuSO4.5H2O (1.248 g) into 500-mL volumetric flask.2.Add some distilled water ~400-mL, stir to dissolve the solid reagent.3.Add distilled water to the 500-mL mark.4.Then, invert the flask several times (10x) to ensure complete mixing.Example: To prepare a solution of certain molarityM = mol LSMM/FSG/UiTM NS

    SMM/FSG/UiTM NS

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