CHE 101 BASIC ANALYTICAL CHEMISTRY Basic Analytical Chemistry 27/12/2018 03:08 1 2 Basic Analytical Chemistry 27/12/2018 03:08 Introduction to Analytical Chemistry Analytical Chemistry deals with; identification and determination of components/chemical species/analytes in a sample. 3 Basic Analytical Chemistry Introduction…….continues 27/12/2018 03:08 • Analytical Chemistry intends to gather and interpret chemical information. • The gathered information is of value to society in a wide range of contexts. 4 Basic Analytical Chemistry 27/12/2018 03:08 Introduction…….continues Chemical analysis can be divided into two categories; Qualitative analysis Quantitative analysis 5 The role of Analytical Chemistry Basic Analytical Chemistry 27/12/2018 03:08 Analytical Chemistry is applied in the following areas; Ensure raw materials meet the required specifications Industry Quality control – Ensure manufactured products contains essential components within predetermined range of composition. Monitoring of pollutants from industrial processes 6 The role of Analytical Chemistry Basic Analytical Chemistry 27/12/2018 03:08 Health Diagnosis of illness Monitoring the patients’ conditions Monitoring levels of chemical species in foodstuffs, pharmaceuticals and water 7 The role of Analytical Chemistry Basic Analytical Chemistry Agriculture Determine the nature and amount of fertilizers to be used in a given soil and for a particular crop. Monitoring of pesticides and herbicides levels in the environment. 27/12/2018 03:08 Geology and mining Determination of composition of the numerous rock and soil samples collected in the field. Routine determination of quantity of chemical species found in soil samples 8 Basic Analytical Chemistry 27/12/2018 03:08 Analytical Methods 9 Basic Analytical Chemistry 27/12/2018 03:08 Analytical methods The methods or techniques are based on; • Suitable chemical reactions and measuring either the amount of the reagent needed to complete reaction or the amount of the product obtained. • Measurement of electrical quantities (I,R,V,Q) • Measurement optical properties (e.g absorption spectra) • Combination of optical or electrical measurements and quantitative chemical reaction (potentiometric titration, spectrophotometric determination) 10 Basic Analytical Chemistry Analytical Methods 27/12/2018 03:08 Classical methods of analysis include; • Gravimetric analysis –in which mass is the final measurement. • Titrimetry analysis –involves measurement of volume standard solutionneeded for complete reaction. The common reactions are Neutralization (acid-base) reactions Complex-forming reactions Precipitation reaction Reduction oxidation reaction • Volumetry- involves measuring the volume of gas evolved or absorbed in a chemical reaction 11 Basic Analytical Chemistry 27/12/2018 03:08 Analytical Methods Instrumental methods of analysis include; • Electrical methods –measurement of I,V, R, Q in relation to the concentration of certain species in solution. The techniques include; Voltametry – measurement of current Coulometry – measurement of current and time Potentiometry - measurement of potential of an electrode Conductometry – measurement of electrical conductivity of solution. 12 Basic Analytical Chemistry 27/12/2018 03:08 Analytical Methods Instrumental methods of analysis include; • Optical methods – depend on measurement of amount of radiant energy either absorbed or emission by a sample at a specific wavelength. ▫ Absorption methods are classified according to the wavelength used. eg. Visible spectrophotometry (colorimetry), Ultraviolet spectrophotometry, Infrared spectrophotometry ▫ Emission methods involve the measurement the intensity of the emitted energy. eg. Emission spectroscopy, Flame photometry 13 Analytical Methods Basic Analytical Chemistry Instrumental methods of analysis include; • Chromatography – used in separation as well as in qualitative and quantitative analysis of species in a sample. • X-ray methods, • Radioactivity, • Mass spectrometry, • Kinetic methods, • Thermal methods, • Optical rotation, and • Refractometry 27/12/2018 03:08 14 Basic Analytical Chemistry 27/12/2018 ERRORS IN CHEMICAL ANALYSIS 15 Measurement errors Basic Analytical Chemistry 27/12/2018 • Error is any departure of a result from a true or accepted value. • Every experiment is accompanied by many uncertainties which combine to produce a scatter of results. • Since the measurement uncertainties can not be completely eliminated, a true value for any quantity is always unknown. • However it is possible to evaluate a probable magnitude of error in a measurement 16 Basic Analytical Chemistry Types of errors • Systematic (determinate) errors • Random (indeterminate) errors • Gross errors 27/12/2018 17 Systematic errors Basic Analytical Chemistry 27/12/2018 Systematic errors show consistent deviation of analytical results from the true value, i.e are of the same magnitude for replicate measurement. These are classified into; Instrumental errors Caused by faults in measuring instruments and instability in power supply. Method errors Arise from non ideal chemical or physical conditions of reagents and reactions on which the analytical method is based upon. Personal errors Originating from the individual analyst and are not related to the method or procedure; result from carelessness, personal limitations and ignorance. 18 Systematic errors Basic Analytical Chemistry 27/12/2018 Systematic errors can either be constant or proportional Constant errors ▫ Magnitude is independent of the size of quantity measured ▫ Become more serious as the size of the quantity measured decreases. ▫ An effective way of minimizing constant error is to use large sample sizes Proportional errors ▫ Magnitude increases or decreases in proportion to the size of the sample 19 Correction of systematic errors Basic Analytical Chemistry 27/12/2018 • Personalerrors Instrumental errors • Detected and corrected by calibration ▫ Periodic calibration is important because instruments performance change over time due to corrosion, misuse and wear. • Care and self discipline • Careful select an analytical method that you are competent in • Get as much information or knowledge as possible on the method as well as reagents being used. Method errors These can be detected by analyzing standard reference materialsmaterials containing exactly known concentration levels of one or more analytes 20 Random errors Basic Analytical Chemistry 27/12/2018 • Arise from random fluctuations in measured quantities, which always occur even under closely controlled conditions. • It is impossible to eliminate them entirely, but they can be minimized by careful experimental design and control. • Experimental factors such as temperature, pressure and humidity, and electrical properties such as current, voltage and resistance are susceptible to small continuous and random variations. 21 Basic Analytical Chemistry 27/12/2018 Gross errors • Cause the result to be either too high or too low i.e they lead to outlier. • They are often caused by real blunders e.g spilling of a solution required in a certain step or accidentally skipping a step in a procedure. 22 Reporting the experimental data Basic Analytical Chemistry 27/12/2018 • A scientifically accepted manner of expressing a quantity is; Where q is the actual measurement and y is the uncertainty or error associated with a particular measurement. • If y is not given, it is taken as ½ the unit of the last significant figure. e.g if m = 1.2g, then m=1.2±0.05g i.e the true value may lies between 1.15 to 1.25g. • Similarly m = 1.00g, then m=1.00±0.005g and m lies between 0.995 and 1.005g. 23 Basic Analytical Chemistry 27/12/2018 Accuracy and Precision 24 Basic Analytical Chemistry 27/12/2018 Accuracy • Accuracy is the closeness of an experimental measurement or result to the true or accepted value. • It is often more difficult to determine because the true value is usually unknown. • It is expressed in terms of either absolute error or relative error. 25 Basic Analytical Chemistry 27/12/2018 Absolute error • Absolute error, E is the difference between the measured quantity xi and the true value xt. E= xi - xt Example If the true value of volume of a gas is 42.5cm3 and the measured values are 41.8cm3 and 43.7cm3, then their absolute errors will be -0.7 and +1.2 cm3 respectively. 26 Relative error Basic Analytical Chemistry 27/12/2018 • Relative error is the absolute error divided by the true value. It can be expressed in percent, parts per thousand or parts per million. − = × 100% Thus, the relative errors for the gas volumes are; 41.8 − 42.5 = × 100% = −1.65% 42.5 43.7 − 42.5 = × 100% = 2.82% 42.5 27 Basic Analytical Chemistry 27/12/2018 Precision • Precision is the closeness or agreement between replicated measurement or results obtained under the same prescribed conditions. • It is expressed by; standard deviation, variance and coefficient of variance. y 28 Accuracy • Measures the agreement between the result and its true value. • Can never be exactly determined since the true value can not be known. 27/12/2018 Precision • Describes the agreement among several results which have been measured by the same procedure. • Can be determined by replicating a measurement. 29 Measures of central tendency Basic Analytical Chemistry 27/12/2018 • Mean, x is a quantity obtained by dividing the sum of the measurements of data, xi by the number of measurements, N. 1 = =1 • Median – middle value in ranked data. • Mode – most frequent datum or value. 30 Measures of spread/scatter Basic Analytical Chemistry 27/12/2018 a) Spread/range, w is the difference between the largest value and the smallest value. b) Deviation from the mean, d is the deviation of the individual result xi from the mean. = − c) Variance, V is the sum of the deviations from the mean, divided by N-1 2 = = −1 − −1 2 31 Measures of spread/scatter Basic Analytical Chemistry 27/12/2018 d) Standard deviation, s is the square root of the variance. = − −1 2 e) Relative standard deviation (RSD) = ( %) 32 Significant figures in numerical computation Basic Analytical Chemistry 27/12/2018 • Significant figures in a number are all of the certain digits plus the first uncertain digit. • Care is to be taken to determine the appropriate number of significant figures to be retained in a result of computation of two or more numbers. 33 Significant figures Basic Analytical Chemistry 27/12/2018 • Addition and subtraction ▫ The result should contain the same number of decimal places as the number with the smallest number of decimal places. Example: 3.4+0.020+7.31=10.730. This is rounded to 10.7. ▫ When adding and subtracting numbers in scientific notation, express the numbers to the same power of 10. Example 7.234 10 6 1.215 10 4 7.234 10 6 0.01215 10 6 7.24615 10 6 This is rounded to 7.246 10 6 34 Significant figures Basic Analytical Chemistry 27/12/2018 • Multiplication and division It is suggested to round off the answer to the same number of significant figures as the original number with the smallest number of significant figures. • Logarithm and antilogarithm The number of significant figures in the mantisa is the same as the number of significant figures in the original number. Example; log( 9.57 10 ) 4.981 4 35 Basic Analytical Chemistry 27/12/2018 Error propagation • It is necessary to estimate the standard deviation of a result that has been calculated from two or more experimental data points, each of which has a known sample standard deviation. • Given; a±sa, b±sb, and c±sc ,the estimation of the standard deviation of the value y is as shown in the table in the next slide. 36 Basic Analytical Chemistry Type of calculation Example Addition or subtraction =++ Multiplication or division Standard deviation of y × = Exponentiation = 27/12/2018 2 + 2 + 2 = = 2 + = 2 + Logarithm = 10 Antilogarithm = 10 = 0.434 = 2.303 2 37 Basic Analytical Chemistry 27/12/2018 Example Consider the following 4.10( 0.02) 0.0050( 0.0001) 0.010406( ?) 1.97( 0.04) compute the standard deviation of the result where the numbers in parenthesis are absolute standard deviation. 38 Basic Analytical Chemistry 27/12/2018 • The standard deviation is calculated as follows; 2 2 2 0.02 0.0001 0.04 0.0289 y 4.10 0.0050 1.97 sy The calculation is completed by calculating the absolute standard deviation s y y (0.0289) 0.010406 (0.0289) 0.000301 We can write the answer with its uncertainty as 0.0104(±0.0003) 39 Basic Analytical Chemistry 27/12/2018 Rounding results after error propagation • First,, round the error term (sy) to one significant figure. • Second, round the answer (y) to the same number of decimal places as the error term. 40 Basic Analytical Chemistry 27/12/2018 Thank you for listening CH 101; BASIC ANALYTICAL CHEMISTRY EVALUATION OF ANALYTICAL DATA Introduction • Experimental results seldom agree exactly with those predicted from a theoretical model. • Scientists must judge whether a numerical difference is a result of the random errors or a result of systematic errors. • The judgment is done by using the significance tests such as Q-test, F-test and t- test. SIGNIFICANCE TESTS • Significance testing involves a comparison between a calculated experimental factor and a tabulated factor determined by the number of values in a set(s) of experimental data or the degree of freedom and a selected probability level. OUTLIERS • Outliers are measurements in a set of replicate measurements or results that appear to be considerably higher or lower than the remainder and are suspected to be outside the expected range. Q-test • It is devised to test suspected outliers in a set of replicate results. • It involves the calculation of a ratio, Qcalcu (calculated value of Q), defined as the absolute difference between the suspected value and the value closest to it divided by the spread of all values in the set. − = Cont… • The calculated value of Q (Qcalcu) for the questionable datum or value is compared with a critical value Qcritical from the table of critical values. (see table 1). • The result or datum is rejected if Qcalcu exceeds Qcritical Table 1:Critical values of Q at 95% confidence level Sample size Critical value 4 5 6 7 8 0.831 0.717 0.621 0.570 0.524 Example Four replicate values were obtained for the determination of pesticide in river water, 0.403, 0.410, 0.410, 0.380 µgdm-3. ▫ Inspect the obtained results; ▫ Is there a possible outlier? Identify it. ▫ Should it be rejected or retained? to answer this you need to calculate the value of Q Cont… 0.380 µgdm-3 is a suspect value, the value of Q is calculated as follows; 0.380 − 0.403 = = 0.767 0.410 − 0.380 Qcritical = 0.831 for four value at the 95% probability level. As Qcalcu is less than Qcritical, 0.380 µgdm-3 is not an outlier at 95% level and it should be retained. F-test • This is used to compare the precision of; ▫ two sets of data which may originate from two analysts in the same laboratory. ▫ two different methods of analysis for the same analyte. ▫ results from two different laboratories. • Statistical F is defined as the ratio of either the population variance, = 12 22 • Or the sample variance of the two sets of data = 12 22 • The large variance is always taken as numerator. • When Fcalcu exceeds the critical value (see table 2), there is a significant difference between the two variances and hence between the precisions of the two sets of data Table 2:critical values of F at 95% confidence level Degrees of Degrees of freedom of the numerator freedom of the denominator 5 6 7 5 7.146 5.285 4.484 6 6.853 4.995 4.197 7 6.681 4.823 4.026 Example • A proposed new method for the determination of sulphate in an industrial waste effluent ios compared wih an existing method, giving the results in the table below. Is there a significanmt difference between the the precisions of the two methods? Method Mean No. of No. of s (mgdm-3) (gdm-3) replicate degrees of freedom Existing 72 8 7 3.38 New 72 8 7 1.06 Fcalcu 2 sexisting 2 snew (3.38) 2 5.08 2 (1.50) • The tabular value for F with 7 degrees of freedom for both the numerator and the denominator is Fcritical = 4.026 at 95% probability level • As Fcalcu is greater than Fcritical, the two methods are giving significantly different precisions t-test It is used to compare; • experimental means of two sets of data. • the mean of one set of data with a known reference value. Comparison of two means • When two experimental means are to be compared, the value of t is obtained by the following equation, with (M+N-2) degree of freedom − = +−2 1 2 • Where x is the mean of M determinations, y the mean of N determinations and spooled the pooled standard deviation. Spooled, pooled standard deviation • The pooled standard deviation (spooled) for sets A and B is given by the following equation; = − 1 2 + − 1 2 +−2 • Where sA and sB are the the standard deviations for sets A and B respectively. Comparison of the mean with known • When the set of data is compared to an accepted value, t is value computed from the following equation; − 1 2 = Where x is the mean of the experimental set of data, µ is the accepted mean, s is the experimental standard deviation and N the number of results • When tcalcu exceeds the critical value (see table 3) for an appropriate number of degree of freedom the difference between the means is said to be significant • When texp exceeds the critical value (see table 3) for an appropriate number of degree of freedom the difference between the means is said to be significant Table 3: Critical values of t at 95% confidence level Number of degrees of freedom 2 t values 5 2.57 10 2.23 18 2.10 4.03 •Example A method of determination of mercury by atomic absorption spectrometry gave values of 400, 385 and 382 ppm for a standard known to contain 400 ppm. Does the mean value differ significantly from the true value or is there any evidence of systematic error(bias)? Solution • Mean =389 ppm • s = 9.64ppm • µ = 400 ppm x 12 (389 400) 12 tcalcu N 3 1 . 98 s 9.=1.98 64 • Ignore the negative sign, tcalcu • For 2 degrees of freedom tcritical is 4.30 at 95% probability level. As tcalcu is less than tcritical, the mean is not significantly different from the true value. Therefore there is no evidence of a systematic error or bias Some practical problems with relevant statistical tests Practical problems Relevant tests One result in a replicate set differs Examine for widely from the rest. Is it a significant gross error. result Apply Q-test Two operators analysing the same Examine the sample by the same method obtain data for results with different spreads. Is unreliable there a significant difference in the results. Apply precisions between the results? F-test Practical problems Relevant tests A new method of analysis is Examine the data for being tested by the analysis unreliable results. of a standard sample with an Apply t-test accurately known composition. Is the difference between the experimental value and the accepted value significant? SAMPLING AND SAMPLE PREPARATION 66 MWANKUNA C.J. 12/27/2018 INTRODUCTION • Sampling is a process of obtaining a small amount of a material that accurately represent the bulk of the material being analysed. • Or sampling is the selection of some parts of an aggregate or totality on the bases of which judgment or inference about the aggregate is made 67 MWANKUNA C.J. 12/27/2018 • Both sample collection and sample preparation play an important role in chemical analysis. • If the samples collected are wrong, even the best method of laboratory analysis will not be able to correctly answer the questions inspiring the analysis. 68 MWANKUNA C.J. 12/27/2018 SAMPLE COLLECTION • Choosing appropriate samples can be difficult especially where the material (population) is heterogeneous. • For example, consider a segregated or stratified sample population. Should the whole sample be taken or separate samples of each layer? 69 Choosing appropriate samples can be difficult especially where the materia MWANKUNA C.J. 12/27/2018 is heterogeneous. For example, consider a segregated or stratified sample p Should the whole sample be taken or separate samples of each layer? steps involves the following Sampling involves • Sampling the following steps Identify population Collect a gross or bulk sample Reduce the gross sample to a laboratory sample A gross sample is a miniature replica of the entire mass of the material to b 70 MWANKUNA C.J. 12/27/2018 • A gross sample is a miniature replica of the entire mass of the material to be analysed. It corresponds to the bulk material in chemical composition and in particle size distribution if composed of particles. • A laboratory sample is a reduced gross sample which is homogeneous. 71 MWANKUNA C.J. 12/27/2018 SAMPLE STORAGE • Often samples are collected from places remote from the analytical laboratory, therefore several days or weeks elapse before they are received and analysed. • Moreover the workload of many laboratories is such that the incoming samples are stored for a period of time prior to analysis. 72 MWANKUNA C.J. 12/27/2018 • For this case the sample containers as well as the storage conditions (e.g temperature, humidity, and level of light) must be controlled such that no significant changes occur that could affect the validity of the analytical data. 73 MWANKUNA C.J. 12/27/2018 • It should be ensured that the containers do not allow contamination of samples. For instance sodium potassium, boron, and silicates can be leached from glass into sample solutions. Therefore plastic containers should be used for such samples. 74 MWANKUNA C.J. 12/27/2018 • Sample solution containing organic solvents and other organic liquids should be stored in glass containers because the base plastic may be leached from the walls of plastic containers. 75 MWANKUNA C.J. 12/27/2018 SAMPLE PREPARATION • Sample preparation refers to the ways in which a sample is treated prior to its analysis. Preparation is a very important step in most analytical techniques, because the techniques are often not responsive to the analyte in its insitu form, or the results are distorted by interfering species. 76 MWANKUNA C.J. 12/27/2018 • Sample preparation may involve dissolution, reaction with some chemical species, masking filtering, dilution , sub-sampling or many other techniques. 77 Basic Analytical Chemistry 27/12/2018 03:08 General Chemistry: An Integrated Approach Hill, Petrucci, 4th Edition Chapter 3 Stoichiometry: Chemical Calculations Mark P. Heitz State University of New York at Brockport © 2005, Prentice Hall, Inc. 79 Chapter 3: Stoichiometry Molecular Mass Molecular mass is the sum of the masses of the atoms represented in a molecular formula. 1 Oxygen atom Example: water - H2O 2(1.0079 u) + 15.9994 u = 18.0152 u 2 Hydrogen atoms EOS 80 Molecular Mass Chapter 3: Stoichiometry Sulfur dioxide - SO2 = 32.066 u + 2(15.9994 = 64.065 u Glucose - C6H12O6 = 6(12.0107 u) + 12(1.0079 u) + 6(15.9994 u) = 180.1154 u EOS 81 Chapter 3: Stoichiometry Formula Mass Formula mass is the sum of the masses of the atoms or ions present in a formula unit. ClNa+ Na+ Cl- Na+ Cl- ClNa+ Crystal of sodium chloride One Na+ and one Cl– make a formula unit for sodium chloride The mass of one formula unit is: = 22.9898 u + 35.4527 u = 58.4425 u EOS 82 Chapter 3: Stoichiometry Avogadro’s Number, NA The number of elementary entities in one mole of anything NA = 6.02214199 x 1023 entities mol–1 -or- 1 mole 6.022 x 1023 entities A mole (mol) is an amount of substance that contains as many elementary entities as there are atoms in exactly 12 g of the carbon-12 isotope. EOS 83 Chapter 3: Stoichiometry Examples of NA 1 mole of stars in the universe 1 mole of pennies 1 mole of tennis balls 1 mole of glucose molecules 1 mole of helium atoms 1 mole of potassium ions (K+) = 6.022 x 1023 stars = 6.022 x 1023 pennies (beats the lottery!) = 6.022 x 1023 tennis balls = 6.022 x 1023 molecules = 6.022 x 1023 atoms = 6.022 x 1023 ions Clearly, NA is most practical for counting microscopic entities. EOS 84 Chapter 3: Stoichiometry NA and Reactions Example: consider the formation of carbon dioxide At the molecular level ... Problem: how does one mass out a single carbon atom? Note that the mass in grams is ~2.00 x –23 g! 10 Answer: one doesn’t! EOS 85 Chapter 3: Stoichiometry The Solution ... Use a measurable amount – molar quantities For carbon, mass out: 2.0 × 10–23 g atom–1 × 6.0 × 1023 atoms mol–1 = 12 g C EOS 86 Making CO2 Chapter 3: Stoichiometry same number different entities macroscopic amounts EOS 87 Chapter 3: Stoichiometry Molar Mass The molar mass of a substance is the mass of one mole of that substance. numerically equal to the atomic mass, molecular mass, or formula mass. The units of molar mass are grams per mole (g mol–1) Examples: 1 mol Na = 22.99 g mol–1 1 mol CO2 = 44.01 g mol–1 EOS 88 Chapter 3: Stoichiometry Calculations and Molar Mass Molar mass is an equality that can be expressed in fractional form grams of substance 1mol or 1mol grams of substance These are used as conversion factors when problem solving … EOS 89 Chapter 3: Stoichiometry Sample Calculation Calculate the mass, in grams, of 0.250 moles of ammonia (NH3) Molar mass = 14.00 g mol–1 + 3(1.008 g mol–1) = 17.024 g mol–1 17.024 g NH 3 0.250 mol NH 3 x 1 mol NH 3 = 4.256 g NH3 EOS 90 Chapter 3: Stoichiometry Percent Composition A simple example: calculate the percent composition, by color, of the collection of blocks below Fraction of colors 4 orange blocks 6 blocks 0.67 or 67% 2 green blocks 6 blocks 0.33or 33% EOS 91 Chapter 3: Stoichiometry Mass Percent Composition The ratio of the total mass of a given element in a compound to the mass of the compound EOS 92 Chapter 3: Stoichiometry Mass Percent Composition EOS 93 Chapter 3: Stoichiometry Molecular and Empirical Formulas Empirical formula: the simplest whole number ratio of elements in a compound Example: Molecular formula of glucose – C6H12O6 The elemental ratio C:H:O is 1:2:1, so the empirical formula is CH2O. EOS 94 Chapter 3: Stoichiometry Determining Empirical Formulas A compound is comprised of 40.01% carbon, 6.72% hydrogen, and 53.27% oxygen. Calculate the empirical formula of the compound. 1) Given the percent composition, assume a mass of sample 100.00 g for convenience 6.72% H 40.01% C - use 6.72 g H 53.27% O 40.01 g C 53.27 g O EOS 95 Chapter 3: Stoichiometry Determining Empirical Formulas 2) Convert masses to mole mol C 40.01 g C x 1 mol C 12.01 g C 3.331 mol C mol H 6.72 g H x 1 mol H 1.008 g H 6.667 mol H 1 mol O mol O 53.27 g O x 15.99 g O 3.331 mol O 3) Divide by the least number of moles 3.331 C 1 3.331 6.667 H 2 3.331 3.331 O 1 3.331 = CH2O EOS 96 Chapter 3: Stoichiometry Stoichiometry of Chemical Reactions For any chemical conversion of reactants, Reactants Products stoichiometry expresses the mathematical connections between all reactants and products involved in the reaction EOS 97 Chapter 3: Stoichiometry Writing Chemical Equations • • • • • A chemical equation is a shorthand description of a chemical reaction, using symbols and formulas to represent the elements and compounds involved. Starting substances are called reactants. The substances formed from the reaction are called products. A plus sign (+) is used between each reactant and between each product. An arrow () points from the reactants to the products. EOS 98 Chapter 3: Stoichiometry Equation Notation • Extra notation added to each substance: – – – – (g) = gas (l) = liquid (s) = solid (aq) = aqueous (water) solution • Extra notation added above the arrow: – = heat – sometimes the actual temperature is used instead EOS 99 Chapter 3: Stoichiometry Chemical Equations Example: consider the formation of water H2(g) + O2(g) H2O(g) Law of Conservation of Mass must be obeyed … therefore, equations must be balanced. EOS 100 Chapter 3: Stoichiometry Balancing Equations Chemical “bookkeeping” of atoms involved in the reaction H2(g) + O2(g) H–2 O–2 Reactants H2O(g) H–2 O–1 Products Note the imbalance in oxygen atoms Whole units of reactants or products must be added to equate numbers of reactant atoms with numbers of product atoms EOS 101 Chapter 3: Stoichiometry Balancing Equations EOS 102 Chapter 3: Stoichiometry Stoichiometric Equivalents Coefficients from a balanced chemical equation show molar equivalents of reactants and products In the formation of water: ==> form conversion 2 mol H2 = 1 mol O2 2 mol H2 = 2 mol H2O 1 mol O2 = 1 mol H2O factors 2 mol H 2 1 mol O2 or 1 mol O2 2 mol H 2 2 mol H 2 2 mol H 2O or 2 mol H 2O 2 mol H 2 1 mol O2 2 mol H 2O or 2 mol H 2O 1 mol O2 EOS 103 Chapter 3: Stoichiometry Limiting Reagents Chemical reactant that is completely consumed in a reaction and therefore limits the quantity of product formed. **Depends on stoichiometry of reaction EOS 104 Chapter 3: Stoichiometry Limiting Reagents (cont.) EOS 105 Chapter 3: Stoichiometry Limiting Reagents (cont.) How many meals can be made from 105 sandwiches, 202 cookies, and 107 oranges? 105 sandwiches x 1 meal 105 meals 1 sandwich 202 cookies x 1 meal 101 meals 2 cookies 107 oranges x 1 meal 107 meals 1orange Cookies limit the total number of whole meals with excess sandwiches and oranges VideoClip EOS 106 Reaction Yields Chapter 3: Stoichiometry Theoretical yield – predicted amount of product formed from the limiting reagent, based only on the stoichiometry of the reaction 2H2(g) + O2(g) 2 H2O(g) If all worked perfectly ... Example: 1 mol H2 will produce 1 mol of water Actual yield – amount of product produced In practice, actual < theoretical: errors, poor technique, etc. ... EOS 107 Chapter 3: Stoichiometry Reaction Yields (cont.) Reaction yields are expressed as a ratio in the form of a percentage actual percent yield x 100 theoretical EOS 108 Chapter 3: Stoichiometry Solution Components Solute – the substance that is dissolved; usually present in lesser quantities than the solvent Solvent - the substance that does the dissolving (examples: water, hexane, methanol) Making Solutions NaCl NaCl NaCl EOS 109 Chapter 3: Stoichiometry Concentrations • Substances enter into chemical reactions according to certain molar ratios. • Volumes of solutions are more convenient to measure than their masses. Molar concentration expresses the amount of solute in one liter of molarity ( M ) solution mol solute L of solution EOS 110 Chapter 3: Stoichiometry Examples of Concentration Dilute solution NaCl Concentrated solution NaCl NaCl NaCl NaCl NaCl e.g., 0.1 M NaCl solution e.g., a 5 M NaCl solution (0.05 mol NaCl dissolved in 500 mL of water) (5 mol NaCl dissolved in 1.0 L of water) EOS 111 Chapter 3: Stoichiometry Dilutions • Dilution is the process of preparing a more dilute solution by adding solvent to a more concentrated one. • Addition of solvent does not change the amount of solute in a solution but does change the solution concentration. Dilution Illustration EOS 112 Chapter 3: Stoichiometry Same number of solute molecules after dilution Added solvent volume, therefore less concentrated Concentrated solution Dilute solution EOS 113 Chapter 3: Stoichiometry Summary of Concepts • • • • • Molecular and formula masses relate to the masses of molecules and formula units. A mole is an amount of a substance containing Avogadro’s number of elementary entities. Avogadro’s number (NA) = 6.022 × 1023. The molar mass of a substance is the mass in grams of one mole of that substance. Formulas and molar masses can be used to calculate mass percent composition. EOS 114 Chapter 3: Stoichiometry Summary (cont.) • • • • A chemical equation uses symbols and formulas for the elements and/or compounds involved in a reaction. Calculations involving reactions use stoichiometric factors based on stoichiometric coefficients in the balanced equation. The limiting reactant determines the amounts of products in a reaction. The molarity of a solution is the number of moles of a solute per liter of solution. EOS 115 Basic Analytical Chemistry 27/12/2018 03:08 CONCENTRATIONS OF SOLUTIONS CONCENTRATIONS OF SOLUTIONS An intimate homogeneous mixture of two or more substances Components of Solutions • Solvent- a component which its physical state is retained (or substance present in a greater amount) • Solute – Substances being dissolved (with lesser amount). Solution expression. The concentration of the Solution is given by a measure of the amount of solute in specific quantity of solvent/solution. • Mass percent, Volume percent and mass/volume percent – Percent by mass (%w/w) is widely used in industrial chemical applications. CONCENTRATIONS OF SOLUTIONS • Calculate the mass of nickel(II) sulfate, NiSO4 contained in 200. g of a 6.00% solution of NiSO4 . Solution • The percentage information tells us that the solution contains 6.00 grams of NiSO4 per 100. grams of solution. The desired information is the mass of NiSO4 in 200. grams of solution. • A unit factor is constructed by placing 6.00 grams NiSO4 over 100. grams of solution. • Multiplication of the mass of the solution, 200. grams, by this unit factor gives the mass of NiSO4 in the solution. CONCENTRATIONS OF SOLUTIONS Also the percentage concentration can be expressed as • percent by volume (%v/v) • Mass/Volume percent (%m/v). This is commonly used in medicines and Pharmacy CONCENTRATIONS OF SOLUTIONS; MOLARITY • Molarity: the ratio between the moles of dissolved substance (solute) and the volume of the solution (in liters or cubic decimeters) Ammount of a Solute in moles Molarity Volume of Solution in litres Example: If 0.455mol. of Urea (CO(NH2)2 is dissolved in enough water to make exactly 1 L of solution, what is the molarity of that solution? No. of Moles of Urea 0.455 mol. Molarity 0.455 M .Urea. Volume of Solutionof urea 1.00 L of a So ln . The SI unit for Molarity, M is mol. Per Litre. CONCENTRATIONS OF SOLUTIONS; MOLARITY Contd. Sample Problem: 1 Calculate the molarity of the following • 145 g (NH4)2C4H4O6 in 500 mL of solution • 13.2 g MnSeO4 in 500 mL of solution • 45.1 g cobalt (II) sulfate in 250 mL of solution • 41.3 g iron (II) nitrate in 100 mL solution 2 Consider the following reaction CaCO3(s) + 2HCl(aq) → Ca2+(aq) + 2Cl1-(aq) + H2O(l) + CO2(g). ▫ Calculate the molarity of C a2+ and Cl1- if 10 dm3 of CO2 was produced at STP,. ▫ What is the mass of CaCO3 required to produce 10 Litres of CO2 gas at STP in excess amount of Acid? What will be minimum amount (Moles) of HCl required? ▫ What will be the volume of CO2 produced if • the concentration of Ca2+ is 0.15 mol./litre? • The concentration of Cl1- is 15mol./litre? CONCENTRATIONS OF SOLUTIONS; Mole Fraction. • Mole Fraction (xi)/Mole percent ammount of component i in moles xi Total ammount of all component in moles CONCENTRATIONS OF SOLUTIONS; Molality Molality (m) • Used when the experiment requires high precision. • If the temperature of a solution is changed, the amount of a solute in moles/weight remains constant. • However the volume of the solution changes. • Therefore there is a slight change of molarity of the solution. Amount of solute in moles molality (m) mass of Solvent in kg CONCENTRATIONS OF SOLUTIONS; Normality • Normality is the number of gram equivalents of a substance dissolved in one litre of a solution. Number of gram equivalents Normality Number of litres CONCENTRATIONS OF SOLUTIONS • Concentration of very dilute solutions. • This is expressed in ppm, ppb, ppt etc. CONCENTRATIONS OF SOLUTIONS • Examples 1. A typical shot of hard liquor drink is 40% by mass of C2H5OH. Assuming the rest is water, how much alcohol would we consume if the drink has a mass of 30g? Calculate the molality, molarity and the mole fraction of the solution. Solution • % by mass of C2H3OH = 40%, mass of drink = 30g % compositionof etOH Mass of Solution Mass of EtOH 100% 40% 30 g of So ln 12 g etOH 100% CONCENTRATIONS OF SOLUTIONS • Molality of ethanol • First of all calculate the mass of the solvent in kg = total mass – mass of Ethanol • 30g of solution – 12g of ethanol = 18g = 0.018kg. • Then calculate the number of moles of Ethanol nethanol nethanol mass of dissolved ethanol 12 g 0.26mol. molar mass of ethanol 46.1g Mole Fraction of Ethanol From above massof water = 18g = 1 mole of H2O. ammount of component i in moles 0.26mol EtOH xi 0.21 Total ammount of all component in moles 0.26mol EtOH 1molH 2 O CONCENTRATIONS OF SOLUTIONS • Example 2. A 2.00 M aqueous NaCl solution has a density of 1.08 g/mL. What is the molality of the solution? Solution • First of all calculate the mass of the solution in using the density given 1.08 g Mass desity Volume 1000mL 1080 g 1mL Then calculate the mass of NaCl. Mass of solute Molarity ( M ) Volume of solution Molar mass of solute 2M 1L 58.449g/mo le 117 gNaCl. Then calculate the mass of solvent in kg Mass of solvent =total mass – mass of a solute = 1080g117g = 963g = 0.963kg of H2O CONCENTRATIONS OF SOLUTIONS • Finally calculate molality nsolute 2mol NaCl m 2.08 molal NaCl msolvent in kg 0.963kg H 2 O CONCENTRATIONS OF SOLUTIONS; dilution • Dilution is the process of preparing a more dilute solution by adding solvent to a more concentrated one. • Is the process of reducing the concentration of a solute in solution, usually simply by adding more solvent. • Addition of solvent does not change the amount of solute in a solution but does change the solution concentration. • the volume times molarity (ViCi) before dilution is equal to the volume times molarity (VfCf) after dilution. • This gives the following general equation; ViCi= VfCf. • The concentrations can be in molarities, molalities, percentages etc. • We must use the same concentration units thus the general equation becomes ViCi= VfCf. Same number of solute molecules after dilution Added solvent volume, therefore less concentrated Concentrated solution Dilute solution CONCENTRATIONS OF SOLUTIONS; dilution • Example: How many milliliters of 18.0 M H2SO4 are required to prepare 1.00 L of a 0.900 M solution of H2SO4? Solution • Vf =1.00 L, Cf=0.900 M and Ci=18.0M • Therefore, the relationship ViCi=VfCf can be used,with subscript i for the initial acid solution and subscript f for the dilute solution. • We solve ViCi=VfCf for Vi Vi Vf C f Ci 1.00 L 0.900M 0.050 L 50.0mL 18.0M USING SOLUTIONS IN CHEMICAL REACTIONS • If we plan to carry out a reaction in a solution, we must calculate the amounts of solutions that we need. • If we know the molarity of a solution, we can calculate the amount of solute contained in a specified volume of that solution. • Example: Calculate the volume in liters and in milliliters of a 0.324 M solution of sulfuric acid required to react completely with 2.792 grams of Na2CO3 according to the equation, H2SO4+Na2CO3→Na2SO4+CO2+H2O • Solution: From the balanced equation the mole ratio for all species is one to one • We convert g Na2CO3 to mol Na2CO3 to mol H2SO4 to L H2SO4 soln ? L . H 2 SO4 2.792 g . Na2CO3 1.00mol.Na2CO3 1.00mol.H 2 SO4 1.00 L.H 2 SO4 106.0 g .Na2CO3 1.00mol.Na2CO3 0.324 mol.H 2 SO4 0.0818 L.H 2 SO4 So ln 81.3mL.H 2 SO4 so ln 134 03:08 Gravimetric methods 135 Gravimetry or gravimetric methods 03:08 • Quantitative methods that involve measurement of mass of a pure compound to which the analyte is chemically related Types: • volatilization methods • precipitation methods • electro-analytical methods • extraction and chromatographic methods 136 03:08 •1.1 Involves collecting andmethod weighing of a volatilized Volatilization product OR Determining mass of product from loss in mass of a sample, following volatilization of an analyte or its decomposition product. • e.g. A sample containing a bicarbonate may be treated with sulphuric acid to evolve CO2 ie NaHCO3 (aq) + H2SO4 (aq) CO2(g) + H2O(l)+NaHSO4(aq) • The liberated CO2 be dried to remove H2O and finally retained in weighed tube that contains an adsorbent [NaOH (to trap CO2)]. Volatization method 137 03:08 • The difference in mass of the tube before and after adsorption can be used to calculate the amount of carbonate in the sample. 138 1.2 Precipitation methods 03:08 • The analyte is converted to a sparingly soluble precipitate filtered, washed free of impurities and converted to a product of known composition by suitable heat treatment. • The product is finally weighed to determine its mass: e.g. A sample known to contain Cl¯ as analyte is dissolved in water and the Cl¯ is converted to a precipitate of AgCl by adding excess AgNO3. 139 03:08 • Ag+ (aq) + Cl(aq) AgCl(s) • The precipitate is washed with water in glass crucible, and dried in an oven. It is cooled and weighed to obtain the weight of AgCl. • Thus chloride content in a sample can be established. 140 Gravimetry or gravimetric methods 03:08 1.2.1 Advantages of gravimetric analysis • Accurate and precise when using modern analytical balances • Determination can be carried out using relatively inexpensive apparatus (except for muffle furnace and platinum crucibles) • Possible sources of errors are readily checked. Eg. Completeness of precipitation can be readily tested. 1.3 Gravimetric Calculations 141 03:08 • Results of gravimetric analysis are computed from mass of product of known composition and mass of sample mass of x 100% % of x mass of sample • Example: 0.3516 g of sample of commercial phosphate detergent was ignited to destroy all the organic matter. The residue was then dissolved in hot HCl which converted the P to H3PO4. The phosphate was precipitated as MgNH4PO4.6H2O by addition of Mg2+ followed by ammonia. After being filtered and washed, the precipitate was converted to Mg2P2O7 (222.57 g/mol) by ignition at 1000oC. This residue weighed 0.2161 g. Calculate the percentage P (30.974 g/mol.) in the sample. 142 1.3 Gravimetric Calculation 03:08 • Solution mass of P %P 100%; mass of sample but mass of P 0.2161gMg 2 P2O7 2mol.P 30.974 gP / mol. 1mol.Mg 2 P2O7 222.57 gMg 2 P2O7 / mol Therefore; 0.2161gMg 2 P2 O7 %P 2mol.P 30.974 gP / mol. 1mol.Mg 2 P2 O7 222.57 gMg 2 P2 O7 / mol 100% 17.11% 0.3516 g sample 143 Example 2 Magnetite is a mineral with the formula Fe2O4. A 1.1324g sample of magnetite ore was dissolved in concentrated HCl to give a solution that contained a mixture of Fe3+ and Fe2+. Nitric acid was added and solution boiled for a few minutes to convert all Fe2+ to Fe3+. The Fe3+ was then precipitated as Fe2O3 x H2O by the addition of NH3. After filtration, the residue was ignited at high temperature to give 0.5394g of pure Fe2O3 (159.69g/mol). Calculate (a) percent Fe (55.847 g/mol), (b) percent Fe3O4 (231.54 f/mol). 03:08 144 1.3 Gravimetric Calculation 03:08 Solution, (a) % Fe mass of Fe 100%; mass of sample 1molFe2O3 2molFe 55.847 gFe but mass of Fe gFe2O3 159.69 gFe2O3 1molFe2O3 1molFe 2mol.Fe 55.847 gFe 0.5394 gFe2O3 159.69 gFe2O3 2mol.Fe 55.847 gFe 0.5394 gFe2O3 159.69 gFe2O3 Percent Fe 100% 33.32% 1.1324 g magnetite ore 145 1.3 Gravimetric Calculation 03:08 (b) In this calculation, we assume that 3moles of Fe2O3 are formed from 2 moles of Fe3O4. i.e. 2Fe3O4+0.5O2→ 3Fe2O3., Then; 1molFe2O3 2molFe3O4 231.54 gFe3O4 mass of Fe3O4 gFe2O3 159.69 gFe2O3 3molFe2O3 1molFe3O4 2mol.Fe3O4 55.847 gFe3O4 0.5394 gFe2O3 3molFe2O3 159.69 gFe2O3 Therefore; 2mol.Fe3 O4 55.847 gFe3O4 3molFe2 O3 159.69 gFe2 O3 100% 46.04% 1.1324 gmagnetite ore 0.5394 gFe2 O3 Percent Fe3 O4 2.0.Precipitates and precipitating reagents 146 03:08 • An ideal precipitating agent used in gravimetric analysis should react either selectively or specifically with the analyte of interest. • For example, dimethylglyoxime is a specific reagent for Ni2+ under alkaline conditions; AgNO3 is a selective reagent which forms precipitates with Cl¯, l¯ Br¯, SCN¯. • In addition the product formed must have the following properties: (i) Readily filtered and washed free of contaminants (ii) Sufficiently low solubility to minimize loses during filtration and washing (iii) Inert to constituents of the atmosphere (iv) Known composition after filtration or ignition. 2.0 Precipitates and precipitating reagents 147 03:08 2.1 Particle size and filterability of precipitates • For gravimetric analysis, large particles are more desirable than the finely divided precipitates because they are easy to filter and wash free of contaminations. • Particles formed in liquids vary enormously in size from colloidal suspensions to crystalline suspension. • Colloid is a solid consisting of particles with diameter less than 10-4 cm. 148 • Thus colloidal suspension consists of tiny particles which are invisible to human eye. • Such solutions are difficult to filter and particles remain permanently suspended in solution. • However, colloidal particles can be made to stick together to give filterable mass. 03:08 149 2.0 Precipitates and precipitating reagents 03:08 • Crystalline suspension is made up of particles which are at least 0.1 mm; • Crystalline suspension tend to settle in the solution and can be readily filtered. • Particles size (Crystalline or colloidal) is influenced by experimental variables: temperature, precipitate solubility, reactant reactions and rate at which reactants are mixed. • The net effect of these variables can be quantitatively accounted by a single property of a system known as relative supersaturation. 150 • supersaturated solution is an unstable solution which contains more solute than it can hold at a particular temperature. • Eventually, supersaturation is relieved by precipitation of the excess solute. • According to Von Weiman, supersaturation plays an important part in determining the particle size of the precipitate. 03:08 QS Von Weiman Equation Re lative sup ersaturation • Where Q = concentration ofSspecies at any time; S = equilibrium solubility. 2.0 Precipitates and precipitating reagents 151 03:08 • Q-S will denote the super saturation at the moment precipitation commences. • Note: The expression applies only when Q is larger than S. • When (Q-S)/S is large, a colloidal precipitate forms, otherwise crystalline precipitant is Formed. • Many salts with low solubility, S, eg. Oxides of Fe3+, Cr3+ form colloidal precipitates. 2.0 Precipitates and precipitating reagents 152 03:08 2.1.1Mechanism of precipitation • Precipitation occurs by two parallel mechanisms: nucleation and particle growth. • The size of the particles formed is determined by a faster mechanism: • If nucleation predominates, colloidal precipitates is formed; if particle growth is faster, then crystalline precipitate is formed 153 03:08 • Nucleation is a process where a minimum number of atoms, ions or molecules come together to form a stable solid. • Nuclei may form on the surface of the solid contaminant e.g. dust particles. • Further precipitation is a competition between additional nucleation and growth on existing nuclei. 2.0 Precipitates and precipitating reagents 2.1.2 154 03:08 Experimental control of particles size • Crystalline ppt formation depends on any experimental variables which minimize supersaturation. • They include; (i)Elevated temperature (increase S of pt). (ii)Dilute solution (Q is kept low), (iii)Slow addition of precipitating reagent with good stirring (minimize Q). (iv)Control of pH can be used to obtain large particles provided that solubility of the precipitate depends upon pH. 2.0 Precipitates and precipitating reagents 155 03:08 2.2 Colloidal precipitates • Colloidal precipitates cannot be analyzed effectively by gravimetry, because the particles are difficult to filter. • Heating, stirring, and/or adding electrolyte binds the individual colloidal particles together to give amorphous mass that settles out of the solution and is filterable. • A process of converting colloidal suspension into filterable mass is known as coagulation or agglomeration. 2.0 Precipitates and precipitating reagents 156 03:08 2.2.1 Coagulation of Colloids • Colloidal suspensions are stable because they are made up of similarly charged particles • The charge originates from cations or anions which are bound on the surface of the particles. • This process is known as adsorption, a process by which ions are retained on the surface of a solid. • For Example, when Sodium chloride is first added to a solution containing silver Nitrate, the colloidal particle of silver chloride formed are positively charged as shown in the fig below. • This charge is due to adsorption of some of the excess silver ion in the medium. • The charge on the particle becomes negative, however when enough sodium chloride has been added to provide excess of Chloride ion. 157 03:08 2.0 Precipitates and precipitating reagents 158 03:08 Coagulation may be brought about by: (i)Short periods of heating accompanied by stirring; since it decreases the number of adsorbed particles, making it possible for particles to approach one another. In addition, heating has an effect of increasing the kinetic energy to overcome barrier which keeps the particles apart. 159 03:08 (i) Increasing electrolyte concentration of a solution; adding the right electrolyte increases the concentration of counter ions around the particle. Consequently, the volume of solution containing the counter ions decreases. 2.0 Precipitates and precipitating reagents 160 03:08 2.2.2 Peptization • Process by which a coagulated colloidal reverts to its original dispersed state. • Washing coagulated colloid results in leaching of the electrolyte responsible for coagulation from the internal liquid in contact with the solid particles. 161 03:08 • Removing the electrolyte responsible for coagulation increases the volume of layer surrounding the particle. • The repulsive forces which were responsible for the original colloidal state is reestablished and particles detach themselves from the coagulated mass. 2.0 Precipitates and precipitating reagents 162 03:08 2.2.3 Treatment of colloidal ppt • Colloids are best pppted from hot, stirred solution containing enough electrolytes for coagulation. • Colloidal ppt can be more easily filtered if digested (process by which a ppt is heated for at least 1 h in mother liquor). 2.0 Precipitates and precipitating reagents 163 03:08 2.3 Crystalline particles • Crystalline particles are generally more easily filtered and purified than are coagulated colloidal. • Also the particles size and their filterability can be improved. 2.0 Precipitates and precipitating reagents 164 03:08 2.4 Co precipitation • Phenomenon, in which soluble compounds are removed from solution by a precipitate. • Note: solution is not saturated with the coprecipitated species. 4 types of coprecipitation • Surface adsorption (equilibrium process) • Mixed crystal formation (equilibrium process) • Occlusion (kinetics of crystal growth) • Mechanical entrapment (kinetics of crystal growth) 2.0 Precipitates and precipitating reagents 165 03:08 2.4.1 Surface adsorption • Source of coprecipitation which occurs to a greater extent with coagulated colloids than with crystalline solids and is therefore a common source of contamination with large specific surface area (i.e. colloids). • The net effect of surface adsorption is carrying down of an otherwise soluble compound as surface contaminant. 166 03:08 • E.g. coagulated AgCl formed in gravimetric determination of Cl¯ is contaminated with adsorbed Ag+, NO3¯. • Consequently, silver nitrate, a soluble compound is coprecipitated with AgCl. 2.0 Precipitates and precipitating reagents 03:08 167 Adsorbed impurities can be minimized by (i)Digestion– water is expelled by this process to give a denser mass of smaller specific surface area for adsorption. (ii)Washing the coagulated colloid with a solution containing volatile electrolyte e.g. volatile HCl be used to wash AgCl ppt. (iii)Reprecipitation – • Is a drastic but an effective way to minimize effects adsorption. • In reprecipitation the filtered solids are redisolved and reprecipitated. 168 03:08 • It significantly reduces contamination originally present in the ppt but substantially increase the analysis time. • E.g. Hydrous oxides of Fe3+ and Al3+ have tendencies to adsorb the hydroxide Zn, Cd and can be removed by reprecipitation. 2.0 Precipitates and precipitating reagents 03:08 169 2.4.2 Mixed crystal formation • Coprecipitation where a contaminat ion replaces an ion in the lattice of the crystal. • E.g. Pb2+ or Sr2+ may replace Barium ions in the BaSO4 crystals. • For exchange to occur the ions must have similar charge and size. • Occurs in both colloid and crystalline ppts Control • Remove the contaminant ion before pptn • Use an alternative ppting reagent which does not give rise to mixed crystal. 170 2.0 Precipitates and precipitating reagents 03:08 2.4.3 Occlusion and mechanical entrapment • Occlusion – coprecipitation whereby a compound is trapped within a pocket formed during rapid crystal growth. • The amount of occluded material is greatest in the part of the crystal that forms first. • Mechanical entrapment-Occurs when crystals lie close together during growth, thereby trapping a portion of the solution in tiny pocket. 171 03:08 • Both Occlusion and mechanical entrapment are at a minimum when the rate of ppt formation is low • Digestion is often remarkably helpful in reducing this type of coprecipitation. • Rapid solution and reprecipitation, which occur at elevated temperature of digestion open pockets and frees impurities into solution. 2.0 Precipitates and precipitating reagents 172 03:08 2.4.4 Errors due to coprecipitation • Coprecipitation can cause either positive or negative errors. 2.0 Precipitates and precipitating reagents 03:08 173 2.5 precipitation from Homogeneous solution • is a process in which a precipitate is formed by slow generation a precipitating reagent homogeneously throughout a solution. • Local reagent excess do not occur because the precipitating agent appear gradually and homogeneously throughout the solution and reacts immediately with the analyte. 174 • As the result, the relative supersaturation is kept low during the entire precipitation. • Homogeneously formed precipitates, both colloidal and crystalline, are better suited to analysis than those formed by direct addition of a precipitating agent. 03:08 • just below 100oC, Urea is used to homogeneous generation of OH¯ as follows; • (NH2)2CO+3H2O→CO2+2NH4++2OH¯. 2.0 Precipitates and precipitating reagents 175 03:08 • For example, anhydrous oxides of iron and aluminium are formed by adding a base in a solution containing iron and aluminium. • direct addition of a base (OH¯) results to formation of gelatinous masses that are heavily contaminated and difficult to filter. • In contrast when same products are produced by homogeneous generation of OH¯, they are dense readily filtered and have considerable high purity. 3.0 Drying and ignition of the ppt 176 03:08 • After filtration, a gravimetric ppt is subjected to heat until constant mass is obtained. • Heating removes solvents and any volatile species which may be carried down with ppt. • In some cases the ppt are ignited to a solid compound of known composition. • E.g. CaC2O4 may be heated to 200oC to expel both unbounded and bonded water and further heated to 900oC to convert it to CaO. 4.0 Application of gravimetric methods 177 03:08 • Determination of most inorganic cations and anions as well as neutral species e.g. H2O, SO2, CO2 and l2 • Determination of organic compounds – lactose, salicylates nicotine, cholesterol, etc. 4.1 Inorganic precipitating agents • These are inorganic reagents which form slightly soluble salts or anhydrous oxides with the analyte. • Most of them are not selective. 178 4.0 Application of gravimetric methods 03:08 4.2 Reducing precipitating reagents • These are the reagents that convert an analyte to its elemental form for weighing. Table: Reducing agents employed in gravimetric methods. Reducing agent SO2 H2C2O4 H2 SnCl2 HCOOH NaNO2 SnCl2 Electrolytic reduction Analyte Se, Au Au Re, Ir Hg Pt Au Hg Co, Ni, Cu, Zn, Ag, In, Sn, Sb, Cd, Re, Bi 179 4.0 Application of gravimetric methods 4.3 Organic precipitating reagents 03:08 • Form intensely coloured chelating complexes with certain species • They are more soluble in organic solvents than in aqueous solution 8-hydroxyquionoline. 8-Hydroxyquinoline • It forms sparingly soluble compounds with most metal ions except the alkali ions. • The solubility of the different compounds depends strongly on the acidity of the solution. • Other metal ions that form sparingly soluble quinolates in a near neutral solution (Co2+, Ni2+ etc.) should be absent. • Ions that require a basic solution to be precipitated as quinolates (e.g. Mg2+) will not disturb. 180 4.0 Application of gravimetric methods 03:08 • Aluminium ions are precipitated quantitatively by 8hydroxyquoinoline from a near neutral solution. • Al3+ + 3C9H7NO + 3 H2O → Al(C9H6NO)3 + H3O+ 181 03:08 OH N C HC HC C N C CH C CH O O Al C N N O 8-Hydroxoquinoline Aluminiumm quinolate 4.0 Application of gravimetric methods182 03:08 • Dimethylgyloxime DMG • Only Nickel ions are precipitated quantitatively by dimethylglyoxime from a weakly acidic or ammoniacal solution. • Ni2+ + 2 C4H8N2O2 + 2 H2O → Ni(C4H7N2O2)2 + 2 H3O+ 183 03:08 O H3C H N O H3C C H O N N C Ni C H3C C N O H H3C N H Dimethylglyoxime C C N O CH3 O Nickel dimethylglyoximate CH3 QUALITATIVE ANALYSIS ◦ Shows the chemical identity of the component/chemical species/analytes in a sample. ◦ May cover confirming presence or absence of species ANALYTICAL CHEMISTRY QUANTITATIVE ANALYSIS ◦ Establishes the amount (absolute, relative) of the analyte contained in a sample.