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LECTURE TOPIC I: Problem Solving in Chemistry

LECTURE TOPIC I: Problem Solving in Chemistry. Significant Figures and Scientific Notation Allowed Digits and “Rounding Off” Dimensional Analysis and SI Unit Conversions Density and Percent Problems. (Kotz & Treichel text: 1.6-1.8). Handling Numbers in Chemistry: The Necessary

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LECTURE TOPIC I: Problem Solving in Chemistry

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  1. LECTURE TOPIC I: Problem Solving in Chemistry • Significant Figures and Scientific Notation • Allowed Digits and “Rounding Off” • Dimensional Analysis and SI Unit Conversions • Density and Percent Problems (Kotz & Treichel text: 1.6-1.8)

  2. Handling Numbers in Chemistry: The Necessary Skills: Correct use of significant figures (“sig figs”) Handling scientific notation Rounding off computational values Knowledge of SI, Metric, and English Units Use of dimensional analysis as problem solving tool

  3. Correct use of significant figures (“SF’s”) All measured values are limited in scope by the accuracy of the instrument used. When correctly expressed, all measured values contain all the digits which can be read directly plus one estimated digit. The digits in measured values are described as “significant figures” when they are presented in this fashion.

  4. TEST YOUR MEMORY, Counting SF’s:

  5. Correct Number of SF’s:

  6. RECALL THE RULES, COUNT ONLY: 1. All non zero digits: 2.35 cm, 3 SF 4.689 g, 4 SF 2. All Internal zeros: 25.01 oz, 4 SF 3. All ending zeros when the decimal point is expressed: 2.00 lb, 3 SF 2.0 in, 2 SF 200. ft, 3 SF

  7. DO NOT COUNT AS SIG FIGS: 1. Ending zeros, no decimal point 2000 g, 1 SF 2. All beginning zeros .0005 g, 1 SF

  8. Group Work (Test Yourself!) These “Group Work Exercises” will occur 2-4 times during each two hour lecture session... Your “table-mates” or nearest neighbors will make up your group... minimum 2, maximum 4, ideal: 3. Pick a “secretary” to write down your solutions; all members sign name on top right hand corner of 1st page; hand in at end of lecture, front desk... The group work counts 5 points/lecture session!

  9. I will present the solution key in class following each activity, but it will not be in the Topics slides. The Group Work Solution Keys will be found on-line at groups.yahoo.com along with the homework and resources for each topic.... On-line students: “group work activities” are to be used as “test yourself” exercises as you proceed through each topic; do the activity and then check out the answers before proceeding...

  10. GROUP WORK 1.1: How many SF’s?

  11. SCIENTIFIC NOTATION Scientific Notation is an alternate method of expressing numerical values in which the original value is multiplied or dividedby ten until there is only ONE digit to the left of the decimal. The resulting number is multiplied by 10 raised to the appropriate power to restore the worth of the original value.

  12. Scientific Notation is used to express very large and very small values, and to facilitate expression of some value to correct number of SF’s. .0000270 = 2.70 X 10-5= 2.70 10x10x10x10x10 27,000 = 2.70 X 104 = 2.70 x10x10x10x10 1,000,000 = 1.00 X 106 (3 SF) or 1 X 106 (1 SF)

  13. The Methodology: 1. Recall: 10-1 = .1 100 = 1 101 = 10 102 =100 Equivalent Values: 27.0 = 27.0 x 1 = 27.0 x 100 2. Small Numbers: subtract one from the power of ten for each right move of decimal: .000027= .000027 x 100 = 2.7 x10 0 - 5 = 2.7 X 10-5 3. Large Numbers: add one to the power of ten for each left move of decimal: 27,000 = 27,000 x 100 = 2.7 x10 0 + 4 = 2.7 X 104

  14. In a nutshell: Add +1 to power of ten for each LEFT MOVE Add -1 to power of ten for each RIGHT MOVE

  15. GROUP WORK 1.2 Place into Scientific Notation: 95,000 (4 SF’s) .00593 X 10-4 .0000008090 0.02030 X 10+5 4578.2 346.00 X 10 +4

  16. SF’s IN CALCULATIONS When doing calculations involving measured values (always the case in science!), you must limit the number of digits in your results to reflect the degree of uncertainty introduced by these values . You must be familiar with the rules for number of SF’s or digits allowed and also with the rules for rounding values down to the allowed number of digits.

  17. Calculations Involving SF’s Multiplication and Division: The final answer in a computation involving these operations should have no more SF’s than the value in the original problem with the least number of SF’s. Addition and Subtraction: The sum of these operations is allowed no more digits after the decimal than the original value with the least number of digits after the decimal.

  18. Rounding off Answers to Correct # SF’s If first digit to be dropped is <5, drop it and all following digits, leaving rest of number unchanged. Round off23.45231 to 4 SF’s: 23.45231 = 23.45 Less than five

  19. If the first digit to be dropped is >5, drop it and all following digits, but increase the last “retained digit” by one: Round off 23.45678 to 4 SF’s: 23.45678 = 23.46 >5

  20. If the first digit to be dropped is exactly five, no non zero digits following, “even up” the resulting rounded-off value: Increase the last retained digit to make it even if it is odd only. Round off to 4 SF’s: 23.45500 = 23.4623.44500 = 23.44 Note: 23.455001 = 23.4623.445001 = 23.45

  21. SF’s in Calculations, Samples: 1.30 in. X .20 in. X 2960. in. = 769.60 in.3 = 7.7X 10 2in.3 3 SF 2 SF 4 SF 2 SF allowed 1.30 in. Since one value has .20 in. no digits after decimal, + 2960. in. none are allowed! 2961.50 in. = 2962 in.

  22. Group Work 1.3 Calculate and round off to correct # SF’s: 24.569 g - .0055 g = ? 32.35 cm X 21.9 cm X 0.76 cm = ? 54.01 lb + .6489 lb + 1,312.0 lb =?

  23. METRIC AND SI UNITS Prefixes you should know: M, 1,000,000 (106) X basic unit “mega” k, 1,000 (103) X basic unit “kilo” d, 1/10 (10-1) X basic unit “deci” c, 1/100 (10-2) X basic unit “centi” m, 1/1000 (10-3) X basic unit “milli” , 1/1,000,000 (10-6) X basic unit “micro” n, 1/1,000,000,000 (10-9) X basic unit “nano” p, 1/1,000,000,000,000 (10-12) X basic unit “pico”

  24. Common SI/Metric/ English Conversions (Know!) 1. MASS 1000 g = 1 kg 1000 mg = 1 g (1 g = 10-3 kg ) (1 mg = 10-3 g) 1 lb = 453.6 g 16 oz avoir = 1 lb SI / English gateway

  25. 2. LENGTH: 1 m = 100 cm = 103 mm 1000 m = 1 km (1 cm = 10-2 m) (1 m = 10-3 km) (1 mm = 10-3 m) 1 m = 109 nm = 1012 pm 2.540 cm = 1 inch SI/ ENG gate 1 yd = 3 ft = 12 in. 1760 yd = 1 mi

  26. 3. VOLUME: 1 L = 1000 mL = 1000 cm3 1 qt = .9463 L= 946.3 mL SI / ENG GATE 1 qt = 2 pt 1 gal = 4 qt 1pt = 16 fl oz

  27. Unit Conversion Using Dimensional Analysis Three Steps Involved: a) Recognizing and stating the question b) Recognizing and stating relationships c) Multiplication of Initial value by appropriate conversion factors to get desired value

  28. Dimensional Analysis and Mass Conversion Problems A typical goal weight for a 5’ 6” female might be 135 pounds. What is this weight in kilograms (kg) ? 1. State the Question: (Use following format): Original value, old unit = ? Value,newunit 135 lb = ? kg

  29. Eng / SI conversion 135 lb = ? kg 2. State the relationship(s) between the old unit and thenew unit: 1 lb = 453.6 g 103 g = 1 kg Eng / SI mass gateway Problem pathway: lb g  kg

  30. 135 lb = ? kg Problem pathway: lb g  kg 3. Multiply the old value, unit by the appropriate conversion factor(s) to arrive at the new value, unit: 1 lb = 453.6 g 1 lb = 1 = 453.6 g 453.6 g 1 lb 103 g = 1 kg 103 g = 1 = 1 kg 1kg 103g

  31. 135 lb = ? kg Problem pathway: lb g  kg 3. Setup and Solve: Old value, unit X factor 1 X factor 2 = new value, unit 135 lb X 453.6 g X 1 kg = 61.236 kg 1 lb 103 g Calculator answer Not done yet!

  32. All defined conversions within a system are exact; The “1” in all conversion relationships is exact 4 SF Exact 3 SF 135 lb X 453.6 g X 1 kg = 61.236 kg 1 lb 103 g Exact Exact,  SF’s # SF’s allowed, 3 61.236 kg = 61.2 kg (Answer!)

  33. 2. LENGTH: 1 m = 100 cm = 103 mm 1000 m = 1 km (1 cm = 10-2 m) (1 m = 10-3 km) (1 mm = 10-3 m) 1 m = 109 nm = 1012 pm 2.540 cm = 1 inch SI/ ENG gate 1 yd = 3 ft = 36 in. 1760 yd = 1 mi

  34. Dimensional Analysis and Length Conversion Problems Orange light has a wavelength of 625 nm. What would this length be in centimeters? 1. State the question: 625 nm = ? cm 2. State the needed relationships: 109nm = 1 m 1 m = 102cm Pathway: nm m  cm

  35. 3. Setup and solve: 625 nm X 1 m X 102cm = 625 X 102 - 9cm 109nm 1 m = 625 X 10-7cm = 6.25 X 10-5 cm

  36. Group Work 1.4 A football field is 100. yd long. What length is this in meters? (1 m = 39.37 in) In water, H2O, the bond length between each H and O is 94 pm. What is this length in millimeters?

  37. 3. VOLUME: 1 L = 1000 mL = 1000 cm3 1 qt = .9463 L= 946.3 mL SI / ENG GATE 1 qt = 2 pt 1 gal = 4 qt 1pt = 16 fl oz

  38. Dimensional Analysis and Volume Conversion Problems What is the volume in cm3 occupied by one half gallon ( .50 gal) of Sunkist Orange Juice? What is this value in in3? 1. State First Question: .50 gal = ? cm3 2. Relationships: 1 gal = 4 qt 1 qt = 946.3 mL 1 mL = 1 cm3 Pathway: gal  qt  mL  cm3

  39. Pathway: gal  qt  mL  cm3 3. Setup and Solve: .50 gal X 4 qt X 946.3 mL X 1 cm3 = 1892.6 cm3 1 gal 1 qt 1 mL = 1.8926 X 103cm3 = 1.9 X 103 cm3

  40. What is the volume in cm3 occupied by one half gallon ( .50 gal) of Sunkist Orange Juice? What is this value in in3? ( .50 gal = 1.9 X 103 cm3) 1. State the Question: 1.9X 103 cm3 =? in3 2. State the relationships: 2.540 cm = 1 in ( 2.540 cm)3 = ( 1 in)3 16.39 cm3= 1 in3 Pathway: cm3 in3

  41. Pathway: cm3 in3 3. Setup and solve: 1.9 X 103cm3 X 1 in3 = 115.9 in3 16.39 cm3 = 1.159 X 102in3 = 1.2 X 102 in3 Calculator: enter 1.9, then “E” button, then “3”

  42. Dimensional Analysis and Density Conversions “Density” is a handy conversion factor which relates the mass of any object or solution or gas to the amount of volume it occupies. It is a physical property of matter, and can be obtained for all elements, most compounds and common solutions in a reference handbook (or on line). Denser matter “feels heavier” and accounts for the properties of “floating” and “sinking”

  43. Common Density Values gases Dry air: 1.2 g/L at 25 oC, 1 atm pressure Water: .917 g/mL at 0 oC 1.00 g/ mL at 4.0 oC .997 g/mL at 25 oC Sea Water: 1.025 g/mL at 15 oC Antifreeze: 1.1135 g/mL at 20oC liquids Magnesium: 1.74 g/cm3 Aluminum: 2.70 g/cm3 Silver: 10.5 g/cm3 Gold: 19.3 g/cm3 solids

  44. SOLVING DENSITY PROBLEMS 1. Density itself can be calculated from experimental values: Density = mass of object, solution, substance volume occupied (mL, cm3, L) Volume can be determined in several ways: a) direct measurement, liquid b) liquid displacement, solid c) measurement of dimensions, calculation

  45. Volume by Displacement: A liquid in which the solid does not dissolve is suitable for this technique

  46. Calculation, Volume: V = l X w X ht (rectangle) V = e3 (cube) V =  r2 ht (cylinder) radius = diameter / 2; all dimensions in same units w diameter height ht l

  47. SOLVING DENSITY PROBLEMS 2. In allother cases, where conversion of mass to volume or volume to mass is desired, density is used as a “conversion factor”: D, Al = 2.70 g /cm3 2.70 g Al = 1 cm3 Al 2.70 g Al = 1 = 1 cm3 Al 1 cm3 Al 2.70 g Al “conversion factors”

  48. Sample, Obtaining Density Value A “cup” is a volume used by cooks in US. One cup is equivalent to 237 mL. If one cup of olive oil has a mass of 205 g, what is the density of the oil in g/mL and in oz avoir / in3? 1. State the question: D, olive oil = ? g/ mL = ? oz avoir / in3 2. State formula: Density = mass, g volume, mL

  49. 3. Substitute values and solve: D = mass = 205 g = .86497 g = .865g volume 237 mL mL mL Second question: What is this density factor expressed as oz avoir/ in3? 1. State Question: .865 g = ? oz avoir mL in3

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