Introduction: Analytical Chemistry NURSAKINAH MAT HAZIR
OUTLINE OF CHAPTER Types & Steps in analysis Review the terms: Moles, molarity and concentration. Review other forms of expressing concentration i) ppm ii) ppt iii) ppb iv) %(w/w) v) %(w/v) vi) %(v/v)
What is Analytical Chemistry? “Analytical Chemistry is what analytical chemists do.” --- C. N. Reilley 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 Analytical Chemistry? Chemical analysis determination of the chemical composition or chemical make up as well as the quantity of each composition presence in a sample E.g: The chemical analysis of a blood sample involves the determination of its iron content, its alcohol content, or perhaps its drug content as well as how much each of the composition presence in the blood sample. E.g: The chemical analysis of a water sample involves the determination of its mineral content, its pollutant content, and its dissolved oxygen content and the percentage of each constituents.
What is actually involve in CHEMICAL ANALYSIS? Chemical analysis procedures involve two areas; the GOAL of the analysis and the METHOD used in the analysis. The GOAL of the analysis can be classified into QUALITATIVE or QUANTITATIVE analysis: QUALITATIVE: Analysis conducted to identify what are the constituents present in the sample (identification of the sample component) QUANTITATIVE: Analysis to determine how much of each constituent present in the sample.
Qualitative Analysis determination of chemical identity of the species Quantitative Analysis determination of the amount of species or analytes, in numerical terms. Hence, Math is heavily involved. In order to perform quantitative analysis, typically one needs to complete qualitative analysis. One needs to know what it is and then select the means to determine the amount
Qualitative analysis is what. Quantitative analysis is how much. ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley)
WHAT HAPPEN TO OUR RIVER?
Why it is important? Analytical chemists must use a systematic approach to obtain the required information from samples. This approach consists of a series of steps and processes. Let we see the step in analytical analysis.
Step in chemical analysis…. NOV 2005 Understanding and defining the problem Familiarize with background of problem and history of sample Literature search (selecting the analytical procedures) Sampling (obtain and store sample) Preparation and pretreatment of sample
Analysis (perform measurement and compare results with standards) Apply required stastitical tecniques Verify results Reporting
1. Understanding and Defining Problem The 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.
2. Sample History and Background of the Problem 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.
3. Literature Search 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 - standard organization : i) Association of official Analytical Chemists (AOAC) ii) Food and Drug Administration (FDA) - electronic media
4. Sampling APR 2007 OCT 2007 Definition: Process of selecting representative material from the lot and storing the sample. The suitable sampling method differ from one substance to another depending on homogeneity. 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.
5. Sample Preparation and Pretreatment APR 2007 Step of sample preparation involved bulk materials: 1. Bulk sample must be reduced in size to obtain a laboratory sample of several grams. 2. From which a few grams to milligrams will be taken to be analyzed (analysis sample). 3. 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.
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 interferences ii) adjust analyte concentrations to a range suitable for measurement iii) produce species from analyte that have quantitatively measurable properties.
6. Analysis Analysis: Incorporates the measurement of the concentration of the analyte in replicates and comparing with standards. Replicate measurement: the practice of taking multiple readings. The replicate measurements are necessary to obtain the measurement uncertainty. The uncertainty important as it indicates the reliability of the measurements.
ii) EXTERNAL STANDARDIZATION iii) INTERNAL STANDARDIZATION Suitable standardization methods should be selected to ensure the precision and accuracy of results. i) CALIBRATION CURVE – A graph of detector response as a function of analyte concentration. ii) EXTERNAL STANDARDIZATION iii) INTERNAL STANDARDIZATION iv) STANDARD ADDITION - Addition of a known quantity of analyte to a real sample undergoing analysis. v) BLANKS – A blank contains the reagents and solvents use in analysis, but no analyte. vi) CONTROL CHARTS APR 2006
7. Reporting and Interpretation Deliver a clearly written, complete report and their limitations.
HOMEWORK Turn to page 17 (Analytical Chemistry, 5th Ed. Gary D. Christian, Wiley, New York, 1994.) and please answer all questions. This is your tutorial question !!! It not required you to Submit this homework to me…. But if I ask you to explain Later on.. I do hope you can… thank you.
PREFIX NOTATION Main measures such as gram and meter are given prefix number for the power of ten. Exponential Prefix name Symbol 1012 tera T 109 giga G 106 mega M 103 kilo k 10 - 10-1 deci d 10-2 centi c 10-3 milli m 10-6 micro µ 10-9 nano n 10-12 pico p
CONCENTRATION UNITS Concentration : the quantity of solute in a specified volume or mass of solution or solvent. The concentration of solutes can be expressed in various ways. Name Component measured by Total sample measured by Abbreviation Units used Molarity Number Volume M Mol L-1 Molality Weight m Mol kg-1 Percent composition Weight-to-weight w/w %(w/w) Weight-to-volume w/v %(w/v) Volume-to-volume v/v %(v/v)
DEFINITION 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
Continued…. Analytical molarity Species molarity There are 2 versions of molarity Analytical Molarity: The total number of moles of the solute in 1L of solution. For example, a HCl solution that has an analytical concentration of 1.0 M can be prepared by dissolving 1.0 mol or 36.5 g of HCl in water and diluting to exactly 1.0 L. Species Molarity (Equilibrium molarity) : The number of moles of a particular species in 1L of a solution at equilibrium. The concentration unit is usually written in bracket, [H+],means the concentration of H+. Species molarity
Continued…. Species Molarity For example, the species molarity of HCl in a solution with an analytical molarity of 1.0M is 0.0 M because HCl is entirely ionized into H+ and Cl- ions and essentially no HCl molecules as such present in the solution. CHCl = 1.0 M [HCl] = 0.0 M [H+ ] = 1.0 M [Cl-] = 1.0 M
You should calculate this……. What is the molarity of a solution of 0.60 g NaCl in 100 mL of solution? Given, MW NaCl = 58.5g/mol.
Example: Saline water concentration Typical seawaters contain sodium chloride, NaCl as much as 2.7 g per 100 mL. (a) What is the molarity of NaCl in the saline water ? (b) The MgCl2 content of the saline water is 0.054 M. Determine the weight (grams) of MgCl2 in 50 mL of the saline water ?
Solution: (a) What is the molarity of NaCl in the saline water ? MW of NaCl = 22.99 + 35.54 = 58.44 g/mol No.of mol of NaCl in 100 mL saline water = 2.7 g 58.44 g/mol = 0.046 mol Molarity of saline water = no.mol = 0.046 mol = 0.46 M L (100/1000) L
Continued………. (b) The MgCl2 content of the saline water is 0.054 M. Determine the weight (grams) of MgCl2 in 50 mL of the saline water ? MW of MgCl2 = 24.30 + 2(35.45) = 95.20 g/mol Weight of in 50 mL of saline water = (M x V) x MW = 0.054 M x (50/1000) x 95.20 g/mol = 0.26 g
DEFINITION Weight percent, %w/w = mass of solute x 100 weight of solution Volume percent, %v/v = volume of solute x 100 volume of solution Volume percent, %w/v = mass of solute x 100 volume of solution
DEFINITION Parts per million (ppm) is grams of solute per million grams of total solution. ppm = mass of solute x 106 mass of solution Parts per billion (ppb) is grams of solute per billion grams of total solution. Parts per trillion (ppt) is grams of solute per trillion grams of total solution.
Common unit to express content Expression Abbreviation Units w/w w/v v/v Parts per million ppm µg/g µg/mL nL/mL mg/kg mg/L µL/L Parts per billion ppb ng/g ng/mL nL/L µg/kg µg/L
Example: Percent Composition A solution contains 118.5 KI per liter solution. Calculate the concentration in % (w/v) % (w/w) Given the density of the solution at 25oC is 1.078 g/mL.
Solution…… (a) Volume percent, % (w/v) = mass of solute x 100 volume of solution = 118.5 g x 100 1000 mL = 11.85% (w/v) (b) weight percent, % (w/w) = mass of solute x 100 weight of solution = 118.5 g x 1mL x 100 1000 mL 1.078 g = 10.99% (w/w)
Example: Conversion of ppb to molarity An aqueous solution contains 56 ppm SO2. Calculate the molarity of the solution. Solution: MW of SO2 = 32 + 2(16) = 64 g/mol 56 ppm = 56 mg/L = (56 x 10-3 g / 64 gmol-1)/L solution = 8.75 x 10-4 M