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Analytical Chemistry Definition: the science of extraction, identification, and quantitation of an unknown sample. Example Applications: Human Genome Project.

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Presentation on theme: "Analytical Chemistry Definition: the science of extraction, identification, and quantitation of an unknown sample. Example Applications: Human Genome Project."— Presentation transcript:

1 Analytical Chemistry Definition: the science of extraction, identification, and quantitation of an unknown sample. Example Applications: Human Genome Project Lab-on-a-Chip (microfluidics) and Nanotechnology Environmental Analysis Forensic Science

2 Course Philosophy develop good lab habits and technique background in classical “ wet chemical ” methods (titrations, gravimetric analysis, electrochemical techniques) Quantitation using instrumentation (UV-Vis, AAS, GC)

3 Analyses you will perform Basic statistical exercises %purity of an acidic sample %purity of iron ore %Cl in seawater Water hardness determination UV-Vis: Amount of caffeine and sodium benzoate in a soft drink AAS: Composition of a metal alloy GC: Gas phase quantitation titrations

4 Chapter 1: Chemical Measurements

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7 Chemical Concentrations

8 Dilution Equation Concentrated HCl is 12.1 M. How many milliliters should be diluted to 500 mL to make 0.100 M HCl? M 1 V 1 = M 2 V 2 (12.1 M)(x mL) = (0.100 M)(500 mL) x = 4.13 M

9 Chapter 3: Math Toolkit accuracy = closeness to the true or accepted value precision = reproducibility of the measurement

10 Significant Figures Digits in a measurement which are known with certainty, plus a last digit which is estimated beakergraduated cylinderburet

11 Rules for Determining How Many Significant Figures There are in a Number  All nonzero digits are significant (4.006, 12.012, 10.070)  Interior zeros are significant (4.006, 12.012, 10.070)  Trailing zeros FOLLOWING a decimal point are significant (10.070)  Trailing zeros PRECEEDING an assumed decimal point may or may not be significant  Leading zeros are not significant. They simply locate the decimal point (0.00002)

12 Reporting the Correct # of Sig Fig ’ s Multiplication/Division 12.154 5.23 Rule: Round off to the fewest number of sig figs originally present 36462 24308 60770 63.56542 ans = 63.5

13 Reporting the Correct # of Sig Fig ’ s Addition/Subtraction 15.02 9,986.0 3.518 Rule: Round off to the least certain decimal place 10004.538

14 Rounding Off Rules digit to be dropped > 5, round UP 158.7 = 159 digit to be dropped < 5, round DOWN 158.4 = 158 digit to be dropped = 5, round UP if result is EVEN 158.5 = 158 157.5 = 157

15 Wait until the END of a calculation in order to avoid a “rounding error” (1.235 - 1.02) x 15.239 = 2.923438 = 1.12 1.12 1.235 -1.02 1.235 -1.02 0.215 = 0.22 0.215 = 0.22 ? sig figs 5 sig figs 3 sig figs

16 Propagation of Errors A way to keep track of the error in a calculation based on the errors of the variables used in the calculation error in variable x 1 = e 1 = "standard deviation" (see Ch 4) e.g. 43.27  0.12 mL percent relative error = %e 1 = e 1 *100 x 1 e.g. 0.12*100/43.27 = 0.28%

17 Addition & Subtraction Suppose you're adding three volumes together and you want to know what the total error (e t ) is: 43.27  0.12 42.98  0.22 43.06  0.15 129.31  e t

18 Multplication & Division

19 Combined Example

20 Chapter 4: Statistics

21 Gaussian Distribution: Fig 4.2

22 Standard Deviation – measure of the spread of the data (reproducibility) Infinite populationFinite population Mean – measure of the central tendency or average of the data (accuracy) Infinite population Finite population N  

23 Standard Deviation and Probability

24 Confidence Intervals

25 Confidence Interval of the Mean The range that the true mean lies within at a given confidence interval x True mean “  ” lies within this range

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27 Example - Calculating Confidence Intervals In replicate analyses, the carbohydrate content of a glycoprotein is found to be 12.6, 11.9, 13.0, 12.7, and 12.5 g of carbohydrate per 100 g of protein. Find the 95% confidence interval of the mean. ave = 12.55, std dev = 0.465 N= 5, t = 2.776 (N-1)  = 12.55 ± (0.465)(2.776)/sqrt(5) = 12.55 ± 0.58

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29 Rejection of Data - the "Q" Test A way to reject data which is outside the parent population. Compare to Q crit from a table at a given confidence interval. Reject if Q exp > Q crit

30 Example: Analysis of a calcite sample yielded CaO percentages of 55.95, 56.00, 56.04, 56.08, and 56.23. Can the last value be rejected at a confidence interval of 90%?

31 Linear Least Squares - finding the best fit to a straight line

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