Experimental Analytical Chemistry 227                             Spring 2011

 

Chapter 0  THE ANALYTICAL PROCESS                                                            Lectures 1 & 2

 

SAMPLING:  How does one get a true representative sample of the material he/she wishes to study?  How do we get the "analyte" in a form that we can analyze quantitavely?  How do we make sure that the sampel is uniform?  Seawater at the surface of the sea is very different in composition than that even 200 ft down, and also different again at 1000 ft down..  It is heterogeneous in composition.

 

SAMPLE PREPARATION:  Sometimes, dissolution of a solid material is all that is necessary.  In other cases, liquids may be analyzed directly.  Gases may be anaylyzed directly, or bubbled into a liquid in order to cause a reaction to occur.  At times, many steps are required to be able to analyze certain components: dissolution, extraction with various solvents, drying, etc.

 

THE CHEMICAL ANALYSIS:  Difference between qualitative analysis (what it is) and quantitative analysis (how much of a specific analyte is present).  There are many techniques that are available for these determinations:  HPLC, IR, UV-Vis spectrometry, Mass spectrometry, titrations, Atomic Absorption spectroscopy, electroanalytical techniques, column chromatography, nmr.  The analyst must decide on which technique will give him/her the "best" and most reliable analysis.

 

CALIBRATION CURVES:  Standard materials are usually run to prepare calibration curves.  Then the "unknown" is run in exactly the same way, and from the calibration curve, the quantity of the unknown can be determined with some degree of accurace and precision.

 

INTERPRETING THE RESULTS:  Once the analysis is done, then it is time to interpret the results:  usually, one does not make a single measurement and is not satisfied that he/she has done such a magnificent job that this is the "true" answer!  Usually, analyses are done at least in triplicate, and the average gives a much better feeling than would only a single measurement.  Also, we can calculate the standard deviation of the results; the standard deviation is a measure of the reproducibility of the measurements.  When the standard deviation is 0, then we have the same result over  and over again; when the standard deviation is very large, then we do not have much confidence that the results are very reproducible.  Now, we have to ask if we have done everything that needs be done to answer some of the questions that were initially asked: is this the only sample, does the answer we got make any sense, etc.???

 

QUALITY ASSURANCE:  We have to make sure that the analyses are correct.  One way is to use "quality assurance" techniques.  One of these methods involves "spiking" the unknown with the specific analyte of interest, re-analyzing the sample, and then finding the original amount of analyte has been increased by exactly the amount of added spike.  That means that you should have great confidence in the whole method and that your measurements were done accurately.

 

 

 

Chapter 1  CHEMICAL MEASUREMENTS

 

1-1    SI Units and Prefixes:  SI stands for "Systčme International d"Unités" = Fundamental Units from which all others are derived.  Length (m), mass (kg), time (s), electric current (A), temperature (K), amount of substance (mole).  There are other derived units:  frequency (Hz) = 1/s, force (N) = m.kg/s2; pressure (Pa) = N/m2; energy (J) = kgm2/s2; power (W) = J/s; qunatity of electricity (C) = A.s; electric potential (V) = W/A; electric resistance (ohm) = V/A

1-2    Conversion Between Units:  ALWAYS use units!!!!!     1 L = 1000 mL; 1 mL = 1 cc = 1 cm3; 1 in. = 2.54 cm; 1 mm = 1 x 10-3 m; 1 µm = 1 x 10-6 m;1 nm = 1 x 10-9 m; 1 pm = 1 x 10-12 m

            Ex.:  Convert 0.0123 m into nm:        0.0123 m (1 nm / 1 x 10-9 m) = 12300000 nm.  However, we need to look at the correct number of digits to be reported in the answer.  The correct answer = 1.23 x 107 nm

1-3    Chemical Concentrations

Molarity (M) and Molality (m):  M = moles solute / L solution

                                                 m = moles solute / kg of solvent

These are very different!!  Will be the same numerical value when the density of the solvent is 1.00 g/mL = H2O (because 1000 g  = 1 kg = 1000 mL).  BUT, when the density is not 1.00 (all the otehr solvents) , then these values are very different!

 

Percent Composition:  % composition = [# g of analyte / sample wt (g)] x 100

Parts per Million and Parts per Billion: 

               1 ppm = 1 mg of solute /1000 g solvent = 1 mg of solute / 1 L solvent (usually water)

               1 ppm = 1 μg of solute / 1 g solvent  =  1 μg of solute / 1 mL of water

               1 ppb = 1 μg of solute / 1000 g solvent  = 1 μg of solute / 1 L solvent (usually water)

               1 ppb = 1 ng of solute / 1 g solvent  =  1 ng of solute / 1 mL of water

Ex.:  Find the M of concentrated H2SO4 (MW = 98.02 g/mol).  The density of this H2SO4 is 1.800 g/mL and it is 98.00% pure.

        The easiest way to do this problem is:     Take 1L of H2SO4.  =  This is 1000 mL.      1000 mL H2SO4  (1.800 g H2SO4 / mL H2SO4) (0.9800)(1 mol H2SO / 98.02 g H2SO4)  =  18.00 M H2SO4


Ex.:  Find the M of 42.5 g H2SO4 (MW = 98.02 g/mol) dissolved in enough water to make 500.0 mL of solution.  The density of H2SO4 is 1.800 g/mL and it is 98.00% pure.
        M = 42.5 g H2SO(1 mol H2SO / 98.02 g H2SO4) / 0.500 L = 0.8672 M H2SO

 

1-4    Preparing Solutions:  NEVER dissolve any solid by putting the solid in a volumetric flask and then adding solvent to dissolve it in the flask.  You should dissolve the solid in the solvent in a BEAKER.  After it is dissolved, then trnasfer thiw solution quantitatively into th eflask and bring it up to the mark (BEFORE putting on the cap and shaking it to make it uniform!!)

1-5    The Equilibrium Constant:

                      Manipulating Equilibrium Constants

            Le Chatelier’s Principle

 

 

 

Chapter 2  TOOLS OF THE TRADE

 

2-1      SAFETY, waste Disposal, and Green Chemistry:  eye protection, gloves, ALWAYS use the waste bottles to get rid of chemicals; do not pour used chemicals down the sink (see TAs for help).

 

2-2      Your Lab Notebook:  IMPORTANT!!!!

 

2-3      The Analytical Balance:  Make sure the balance is level!  Reset the “zero” on the balance.  Do NOT use weighing paper unless you are directed to do so.  DO NOT ADD ANY CHEMICALS INTO THE WIEGHING CHAMBER, EVER!!!!  Take the empty weighed flask or beaker out of the balance, put sample into it, and then re-weigh.

 

2-4      Burets:  Make sure that the buret is CLEAN!  Rinse the buret with ~ 5 mL of the solution, making sure to run it through the stopcock.  Do this twice.  Then fill the buret BEYOND the 0 mark, open the stopcock full-bore open to push out any entrained air in the stopcock, then bring the level of the solution BELOW the 0.00 mL mark and take the initial reading.

 

2-5      Volumetric Flasks:  Never try to dissolve a solid in a volumetric flask!!   First, dissolve the solid in a small amount of distilled water and then quantitatively transfer this to the volumetric flask.  After the quantitative transfer, then fill the flask to the mark with distilled water and finally, put the ground glass stopper on the flask and invert it and shake it.

 

2-6      Pipets and Syringes:  Always use a bulb for pipetting solutions.  Used syringes CANNOT be thrown in the basket; they MUST be handled as though they were medical waste.

 

2-7   Filtration


2-8   Drying:  Both standard and unknown solid samples need to be dried to drive off the moisture that adheres to many materials that have been ground into fine powders.  Some materials are necessarily much more hygroscopic than others, and therefore
         have to be handled accordingly.  Also, once a sample is dry, then it becomes necessary to keep the sample dry by storing it in a desiccator over calcium chloride (why this material?), only taking the material out of the desiccator to weigh a sample and then
         immediately putting it back in to keep it dry.

 

2-9    Calibration of Volumetric Glassware

 

2-10      Methods of Sample Preparation:  gases, liquids, solids (ex.:  how sample an oil spill in the ground?), biological fluids (ex.: when blood was collected, how was it kept? = on ice in a cooler?)

  Handling the sample:  minimize contamination, loss, decomposition, matrix change.

  Have to be wary of contamination by:  container, atmosphere, light.  Ex.:  SO2 often reacts with container materials.

  How measure the sample? 

1)      Weighing

2)      Volume (aliquot = definition)

                    Analyte = definition

 

 

Chapter 3  MATH TOOLKIT

 

3-1    Significant Figures:  All of the known numbers in a measured quantity + the last one that has the error.  Ex.:  the number 12.3456 has 6 significant digits = the 1st five of them are absolutely known quantities, and the last digit (the 6) is the one with the error.  Usually, the error is taken as  ±1; if it is not that value, it will be given.

 

3-2     Significant Figures in Arithmetic

Addition and Subtraction:  The answer CANNOT have more decimal places than the number in the problem with the LEAST number of decimal places.

Ex.:  12.1234 + 1.089 – 0.12 = 13.0924 is the INCORRECT answer   =>  13.09 is the CORRECT answer, because it has 2 decimal places just like 0.12.

 

Another ex.:  13 + 1.2345 + 123.21 + 4.23 = 141.6745 is the INCORRECT answer   =>  142 is the CORRECT answer, because it has NO decimal places just like the 13.

When do you round up or down?  Always round up if the digit being dropped is larger than 5.  If the digit being dropped is 5, look at the next digit to see if you should round the 5 up or down.

 

Multiplication and Division:  The answer CANNOT have more significant figures than the number in the problem with the LEAST number of significant figures.

Ex.:  12.123 x 17.23 / 134.6521 = 1.551251633 is the INCORRECT answer   =>  1.551 is the CORRECT answer, because it has 4 significant figures just like 17.23 has 4 significant figures.

 

Another ex.:  315.2 mg x 0.9995 / 42.11mL x 74.55 mg/mmol = 0.100354331 M  is the INCORRECT answer   =>  0.1004 M is the CORRECT answer, because it has 4 significant figures just like all of the numbers in the problem.

 

Logarithms and Antilogarithms:  Only use the mantissa part of the log (the decimal part) when you consider sig. figs.  The characteristic is the integer part of the log.

 

Ex.:  log 3.462 = 0.539327064 is the INCORRECT answer   =>  0.539 is the CORRECT answer, because it has 3 decimal places just like 0.462.

 

Ex.:  The antilog of 0.1064 = 1.277614884 is the INCORRECT answer   =>  1.2776 is the CORRECT answer, because it has 4 decimal places just like 0.1064.

 

3-3     Types of Error

Gross Error:  negligence = spill material on the counter, pick it up and put it back in the flask; spill material outside of the flask when you are weighing, splash titrant outside of the Erlenmeyer flask while titrating.

 

Systematic Error (determinate error):  repeatable if you do everything the same way over and over again.  Can be discovered and corrected, usually by someone watching what you are doing and finding out that you are doing something incorrectly over and over again.

 

Random Error (indeterminate error):  errors inherent in the physical measurements, including the type of instrumentation being used.  Ex.:  The analytical balance has an inherent error of  ±0.0002 g for each reading.  Therefore, if you weigh an empty flask = 102.2345 g and then weigh some material in it and it now weighs 102.6781 g, the material inside the flask weighs =  0.4436 gHowever, the error associated with this weight = ±0.0003 g.  This may be shown as: 

0.4436 ±0.0003 g  or   0.4436(3) g.  This latter is read as is the previous value where the number in parentheses is known to be in the last digit.

 

Precision and Accuracy:  Precision is the reproducibility of a result.  Accuracy is how close a measured value is to the “true” value.  Show targets.

 

Absolute and Relative Uncertainty:  or Absolute and Relative Error:

 

ABSOLUTE ERROR:  difference between the true value and a measured value

        Ex.:  Known % Cu in a sample = 6.48%.  We measure 6.63%. 

                    Therefore, absolute error = 6.48% - 6.63% = 0.15%

 

RELATIVE ERROR:  the absolute error divided by the true value, expressed in %, ppt, ppm, ppb

       Relative error in the above measurement = [(6.48% – 6.63%) / 6.48%] x 100 = 2.47% = 2.47 pph

            OR = 24.7 ppt = [(6.48% – 6.63%) / 6.48%] x 1000

 

Ex.:  Measure the N of NaOH solution:  0.1023, 0.1026, 0.1034, 0.1027

                    Average = Xbar = (0.1023 + 0.1026 + 0.1034 + 0.1027) / 4 = 0.1027 N

                    Relative error (in ppt) = [(high – low) / Xbar] x 1000 = [(0.1034 – 0.1023) / 0.1027] x 1000 = (0.0011 / 0.1027) x 1000 = 11 ppt

 

 

3-4    Propagation of Uncertainty:  after all systematic errors have been eliminated, there still are random errors to be considered.

 

Addition and Subtraction:  the absolute error of the answer is found from the absolute errors of each of the values:    e(answer)  =  (e12 + e22 + e32 + ...)1/2

0.125 (±0.003) + 1.232(±0.001) – 0.5821(±0.0008) = 0.775 is the answer.

But, what is the error associated with this answer?  The absolute error   e = (e12 + e22 + e32 + ...)1/2

Therefore:  e  =  [(0.003)2  + (0.001)2  + (0.0008)2]1/2    =  (0.00001064) 1/2    = 0.003261901

                 e = ±0.003     Therefore, the answer with its error is:  0.775 ± 0.003

This is called absolute error or absolute uncertainty.

To find the relative error = (absolute error / value) x 100 =  (0.003/0.775)x 100 = 0.4%     (only 1 sig. fig. because the error only has 1 sig. fig).     Therefore:  0.775 ± 0.4 %

 

Multiplication and Division:  the relative error of the answer is found by the relative errors of each of the values:  %e(answer)  =  (%e12 + %e22 + %e32 + ...)1/2

3.26(±0.03) x 1.893(±0.003) / 0.591(±0.003) x 13.21(±0.02)  = 0.790 is the answer

But, what is the relative error associated with this answer? 

The relative error     %e(answer)  =  (%e12 + %e22 + %e32 + ...)1/2

 

Therefore:  %e  =  {[(0.03/3.26)x100]2  + [(0.003/1.893)x100)]2  + [(0.003/0.591)x100)]2] + [(0.02/13.21)x100)]2]} 1/2    =  (5.129639) 1/2    =  ±2 %

                   %e = ±2 %     Therefore, the answer with its error is:  0.790 ± 2 %

This is called relative error or relative uncertainty.

To find the absolute error = (relative error x value) / 100 =  0.790 x (2 / 100) = 0.0158  =>  0.016    Therefore:  0.790 ± 0.016 (the error must have same number of decimals as the answer).

 

Mixed Operations:   Let’s do this calculation:  

        [1.783(±0.002)  + 1.22(±0.02) – 0.5943(±0.0008)] / 0.678(±0.002) x 1.22(±0.02)  =2.4087 / 0.82716 = 2.912012         =>   real answer = 2.91

 

     For the numerator:   answer = 2.4087   =>  2.41 with the absolute error: 

               e = (e12 + e22 + e32 + ...)1/2  =  [(0.002)2  + (0.02)2  + (0.0008)2]1/2    =  (0.00040464) 1/2   =  ± 0.02012   =>  ± 0.02 

          

     Now, for the whole problem:    2.41(± 0.02) / 0.678(±0.002) x 1.22(±0.02) =  2.91358  =>  2.91 is the real answer.

 

                 For the relative error:    %e(answer)  =  (%e12 + %e22 + %e32 + ...)1/2

                                         =  {[(0.02/2.41)x100]2  + [(0.002/0.678)x100)]2  + [(0.02/1.22)x100)]2]}1/2  =  (3.463159) 1/2  =  1.8610    =>   ±2 %  

                           

                And, therefore the absolute error = 2.91 x (2 / 100) = 0.0482   =>  0.05

 

The answer with its absolute error is:   2.91 ± 0.05 

The answer with its relative error is:    2.91 ± 2 % 

 

 

The Real Rule for Significant Figures:

                        Ex.:  27.52 (± 0.03) / 29.01 (± 0.02) = 0.948638401  =>  0.949 (± 0.1%) RELATIVE ERROR  = 0.949 (± 0.001)  ABSOLUTE ERROR        and NOT = 0.9486 (± 0.0009)

      

                        Ex.:  27.58 (± 0.03) / 25.01 (± 0.02) = 1.102759  =>  1.103 (± 0.1%) RELATIVE ERROR      = 1.103 (± 0.001)  ABSOLUTE ERROR


                        Ex.:   0.1971 (± 0.0005) / 2.007 (± 0.004) = 0.098206  =>  0.0982 (± ?)  OR 0.09821 (± ?)

                                     =  0.0982 (± 0.20%)  RELATIVE ERROR

                                     =  0.0982 (± 0.0002)  ABSOLUTE ERROR

 

3-5    Introducing Spreadsheets:  Learn how to use XL spreadsheets.  This will simplify many of the routine calculations thta you do in the lab, allow you to do plotting and doing calculations such as finding the average of a series of numbers, the standard deviation, etc.


3-6    Graphing in Excel:  We have to learn how to use XL for graphing.

 

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