1 / 49

Unit 1 – Introduction to Chemistry

Unit 1 – Introduction to Chemistry. Part 2: Scientific Measurement. Uncertainty in Measurement. The # associated with a measurement is obtained using a measuring device Last digit has to be estimated Estimated digits are “uncertain” A measurement always has some degree of uncertainty.

edupree
Download Presentation

Unit 1 – Introduction to Chemistry

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Unit 1 – Introduction to Chemistry Part 2: Scientific Measurement

  2. Uncertainty in Measurement • The # associated with a measurement is obtained using a measuring device • Last digit has to be estimated • Estimated digits are “uncertain” • A measurement always has some degree of uncertainty. • Communicating our uncertainty in a measurement is equally important as the measurement itself!

  3. Uncertainty in Measurement

  4. Uncertainty with Reading Volumes Smallest div. = 10 mL 1 mL Smallest div. = 0.1 mL Smallest div. = Measurement = 47 mL Measurement = 36.7 mL Meas. = 20.38 mL • A rule of thumb: read the volume to 1/10 or 0.1 of the smallest division. • Read at the bottom of the meniscus AT EYE LEVEL!

  5. You Try It! What is the length of the wooden stick? 1) 4.5 cm 2) 4.54 cm 3) 4.547 cm

  6. Accuracy vs. Precision • Accuracy - how close a measurement is to the accepted value • Precision - how close a series of measurements are to each other ACCURATE = CORRECT PRECISE = CONSISTENT

  7. Accuracy vs. Precision

  8. Error  • Definition = The difference between an accepted value and an experimental value • Accepted- what you are “supposed” to get • Experimental- what you “actually” get • Take the absolute value • No negative values • |experimental – accepted| = error

  9. your value Percent Error • Indicates accuracy of a measurement • Ratio of an error to an accepted value

  10. % error = 3% Percent Error Example • A student determines the density of a substance to be 1.40 g/mL. Find the % error if the accepted value of the density is 1.36 g/mL.

  11. 2.35 cm Significant Figures • Indicate precision (uncertainty) of a measurement. • Recording Sig Figs • Sig figs in a measurement include the known digits plus a final estimated digit • The more digits reported, the more precise the measurement

  12. 4 4 6 6 3 3 5 5 Implied range of uncertainty in a measurement reported as 5 cm. Implied range of uncertainty in a measurement reported as 5.0 cm. 4 6 3 5 Implied range of uncertainty in a measurement reported as 5.00 cm. Implied Range of Uncertainty Dorin, Demmin, Gabel, Chemistry The Study of Matter 3rd Edition, page 32

  13. Significant Figures • Counting Sig Figs • Count all numbers EXCEPT: • Leading zeros -- 0.0025 • Trailing zeros without a decimal point -- 2,500

  14. Atlantic Pacific Rule • If a decimal point is Present, start counting from the Pacific (left) side at the 1st non-zero number. • If the decimal point is Absent, start counting from the Atlantic (right) side at the 1st non-zero number. • Count until you run out of numbers!

  15. Significant Figures Practice Counting Sig Figs Practice Problems 1. 23.50 cm 1.23.50 4 sig figs 3 sig figs 2. 402 g 2.402 3. 5,280 g 3.5,280 3 sig figs 2 sig figs 4. 0.080 L 4. 0.080

  16. 3 SF Significant Figures • Calculating with Sig Figs • Multiplication/Division • The measurement with the fewest sig figs determines the # of sig figs in the answer. (13.91g/cm3)(23.3cm3) = 324.103g 4 SF 3 SF 324g

  17. Significant Figures Calculating with Sig Figs (cont.) Addition/Subtraction The measurement with the fewest decimal places determines how many decimal places can be in the answer. 224 g + 130 g 354 g 224 g + 130 g 354 g 3.75 mL + 4.1 mL 7.85 mL  350 g  7.9 mL

  18. Significant Figures • Calculating with Sig Figs (cont.) • Exact Numbers do not limit the # of sig figs in the answer. • Counting numbers: 12 students • Exact conversions: 1 m = 100 cm • “1” in any unit conversion: 1 in = 2.54 cm

  19. 1. (15.30 g) ÷ (6.4 mL)  2.4 g/mL 2 SF Significant Figures Practice Calculating with Sig Figs Practice Problems 4 SF 2 SF = 2.390625 g/mL 2. 18.9 g - 0.84 g  18.1 g 18.06 g 1 DECIMAL PLACE

  20. PRACTICE! • Accuracy, Precision & Sig Figs WS

  21. Scientific Notation Way of expressing really big numbers or really small numbers How to write scientific notation Product of 2 factors a number with only one digit in front of the decimal power of 10 # x 10 x

  22. Scientific Notation Converting into Sci. Notation: Move decimal until there’s 1 digit to its left. Places moved = exponent. Large # (>1)  positive exponentSmall # (<1)  negative exponent Only include sig figs! 65,000 kg  ? 6.5 × 104 kg

  23. PRACTICE: Change to scientific notation. 12,340 = 0.369 = 0.008 = 1,000,000,000 = 1.234 x 104 3.69 x 10–1 8 x 10–3 1 x 109

  24. Scientific Notation Converting out of Sci. Notation: Simply move the decimal point to the right for positive exponent of 10. Move the decimal point to the left for negative exponent of 10. Use zeros for placeholders. 5.4 x 10-3 kg  ? 0.0054 kg

  25. PRACTICE: Change to standard notation. 1.87 x 10–5 = 3.7 x 108 = 7.88 x 101 = 2.164 x 10–2 = 0.0000187 370,000,000 78.8 0.02164

  26. 1. 2,400,000 cg 2. 0.00256 kg 3. 7  10-5 km 4. 6.2  104 mm Converting into and out of Scientific Notation Practice 2.4  106 cg 2.56  10-3 kg 0.00007 km 62,000 mm

  27. Rule for MultiplicationCalculating with Numbers Written in Scientific Notation When multiplying numbers in scientific notation, multiply the first factors andadd the exponents. Sample Problem: Multiply 3.2 x 10-7 by 2.1 x 105 (3.2) x (2.1) = 6.7 6.7 x 10-2 (-7) + (5) = -2 or 10-2 Exercise: 1.46 x 108 x 1.5 x 104 2.2 x 1012

  28. Rule for DivisionCalculating with Numbers Written in Scientific Notation When dividing numbers in scientific notation, divide the firstfactor in the numerator by the first factor in the denominator. Then subtract the exponent in the denominator from the exponent in the numerator. Sample Problem: Divide 6.4 x 106 by 1.7 x 102 (6.4)  (1.7) = 3.8 3.8 x 104 (6) - (2) = 4 or 104 7.7 x 10-22 Exercise: 2.4 x 10-7 3.1 x 1014

  29. Rule for Addition and SubtractionCalculating with Numbers Written in Scientific Notation In order to add or subtract numbers written in scientificnotation, you must express them with thesame power of 10. Sample Problem: Add 5.8 x 103 and 2.16 x 104 (5.8x103) + (21.6x103) = 27.4 x 103 2.74 x 104 Exercise: 8.32 x 10-7 + 1.2 x 10-5 1.3 x 10-5

  30. Calculating with Scientific Notation Practice =-65.25 x 10-10 report -6.5 x 10-9 (2 sig. figs.) =5.3505 x 103 report 5.35 x 103 (3 sig. figs.) = 0.5842 x 10-12 report 5.84 x 10-13 (3 sig. figs.) 0.88 x 1012 = 4.18 x 1012 report 4.2 x 1012 (1 dec. place) =4.665 x 10-8 0.435 x 10-8 report4.7 x 10-8 (1 dec. place)

  31. EE EXP Using the Exponent Keyon a Calculator for Scientific Notation

  32. 6 6 6 6 1 1 0 0 0 0 0 0 x x EE EE y x y x 2 2 2 2 2 2 2 2 3 3 3 3 . . . . How to type out 6.02 x 1023: EE or EXP means “times 10 to the…” Don’t do it like this… WRONG! WRONG! …or like this… …or like this: TOO MUCH WORK.

  33. EXE EXP EXP ENTER EE EE Scientific Notation Calculator Practice • Solve the following problem using your calculator. (5.44 × 107 g) ÷ (8.1 × 104 mol) = Type on your calculator: 5.44 7 8.1 4 ÷ 6.7 × 102 g/mol = 670 g/mol = = 671.6049383

  34. Units of Measurement • Quantity = number + unit UNITS MATTER!!

  35. SI Units of Measurement Quantity Symbol Unit Abbrev. Length l meter m Mass m kilogram kg Time t second s Temp T Kelvin K Amount n mole mol

  36. mega- 106 M --- deci- 10-1 BASE UNIT d 100 kilo- k 103 centi- c 10-2 milli- m 10-3 micro- 10-6  nano- 10-9 n Metric Prefixes Symbol Power of 10 Prefix

  37. Giga- Mega- Kilo- base milli- micro- nano- pico- femto- atomo- 1024 g 1021 g Earth’s atmosphere to 2500 km 1018 g 1015 g 1012 g Ocean liner 109 g Indian elephant 106 g Average human 103 g 1.0 liter of water 100 g 10-3 g 10-6 g Grain of table salt 10-9 g 10-12 g 10-15 g 10-18 g Typical protein 10-21 g Uranium atom 10-24 g Water molecule Kelter, Carr, Scott, Chemistry A Wolrd of Choices 1999, page 25

  38. To the left or right? Metric Conversions 1.Find the difference between the exponents of the two prefixes. 2. Move the decimal that many places.

  39. Metric Conversions

  40. NUMBER NUMBER = UNIT UNIT Metric Conversions 0.532 532 m = ______ km

  41. Metric Conversions Practice 0.2 • 20 cm = ________ m • 0.032 L = ________ mL • 80,500 cm = ________ km 32 0.805

  42. Density How much mass takes up a certain amount of volume The density of a substance will NOT change If the mass changes, then the volume will change also So, what kind of property is density? M V D =

  43. Density M V D = M M = D x V ass D V M D V = ensity olume

  44. Density via Volume Displacement • To find the volume of irregularly shaped objects, we must use Archimedes’ principle. • an object, when placed in a liquid, will displace a volume of liquid equal to its own volume

  45. Thread Vfinal = 98.5 cm3 - Vinitial = 44.5 cm3 Vfishingsinker = 54.0 cm3 98.5 cm3 44.5 cm3 Fishing sinker Water Before immersion After immersion Volume Displacement

  46. Density via Graphing Mass (g) Volume (cm3)

  47. Common Units of Density grams/ cm3 grams/ mL Important Conversion to Know: 1 mL = 1 cm3

  48. Density Practice Problem #1 An object has a volume of 825 cm3 and a density of 13.6 g/cm3. Find its mass. GIVEN: V = 825 cm3 D = 13.6 g/cm3 M = ? WORK: M = DV M = (13.6 g/cm3)(825cm3) M = 11,200 g How many kg is this???

  49. Density Practice Problem #2 A liquid has a density of 0.87 g/mL. What volume is occupied by 25 g of the liquid? V = 25 g 0.87 g/mL GIVEN: D = 0.87 g/mL V = ? M = 25 g WORK: V = M D V = 29 mL

More Related