Chapter 2 Scientific Measurements Chemistry The Molecular Nature

Chapter 2: Scientific Measurements Chemistry: The Molecular Nature of Matter, 6 E Jespersen/Brady/Hyslop

Properties Characteristics used to classify matter Physical properties Can be observed without changing chemical makeup of substance Ex. Gold metal is yellow in color Sometimes observing physical property causes physical change in substance Ex. Melting point of water is 0 °C Measuring melting temperature at which solid turns to liquid Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 2

Solids: States of Matter Fixed shape & volume Particles are close together Have restricted motion Liquids: Fixed volume, but take container shape Particles are close together Are able to flow Gases: Expand to fill entire container Particles separated by lots of space Ex. Ice, water, steam Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 3

States of Matter Physical Change from 1 state to another Physical States Important in chemical equations Ex. 2 C 4 H 10(g) + 13 O 2(g) 8 CO 2(g) + 10 H 2 O(g) Indicate after each substance with abbreviation in parentheses Solids = (s) Liquids = (ℓ) Gases = (g) Aqueous solutions = (aq) Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 4

Chemical Properties Chemical change or reaction that substance undergoes Chemicals interact to form entirely different substances with different chemical & physical properties Describes behavior of matter that leads to formation of new substance “Reactivity" of substance Ex. Iron rusting Iron interacts with oxygen to form new substance Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 5

Learning Check: Chemical or Physical Property? Chemical Physical X Magnesium metal is grey Magnesium metal tarnishes in air X X Magnesium metal melts at 922 K Magnesium reacts violently with hydrochloric acid Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E X 6

Your Turn! Which one of the following represents a physical change? A. when treated with bleach, some dyed fabrics change color B. grape juice left in an open unrefrigerated container turns sour C. when heated strongly, sugar turns dark brown D. in cold weather, water condenses on the inside surface of single pane windows E. when ignited with a match in open air, paper burns Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 7

Intensive vs. Extensive Properties Intensive properties Independent of sample size Used to identify substances Ex. Color Density Boiling point Melting point Chemical reactivity Extensive properties Depend on sample size Ex. volume & mass Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 8

Identification of Substances by Their Properties Ex. Flask of clear liquid in lab. Do you drink it? What could it be? What can we measure to determine if it is safe to drink? Density Melting Point Boiling Point Electrical conductivity Jespersen/Brady/Hyslop 1. 00 g/m. L 0. 0 °C 100 °C None Chemistry: The Molecular Nature of Matter, 6 E 9

Gold or “Fool’s Gold? ” Can test by heating in flame A. Real gold Nothing happens B. Pyrite (Fool’s Gold) Sputters Smokes Releases foul-smelling fumes Due to chemical ability to react chemically with oxygen when heated Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 10

Your Turn! Which of the following is an extensive property? A. Density B. Melting point C. Color D. Temperature E. Mass Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 11

Observations Fall into 2 categories 1. Quantitative observations Numeric data Measure with instrument Ex. Melting point, boiling point, volume, mass 2. Qualitative observations Do not involve numerical information Ex. Color, rapid boiling, white solid forms Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 12

Measurements Include Units!! 1. Measurements involve comparison Always measure relative to reference Ex. Foot, meter, kilogram Measurement = number + unit Ex. Distance between 2 points = 25 What unit? inches, feet, yards, miles Meaningless without units!!! 2. Measurements are inexact Measuring involves estimation Always have uncertainty The observer & instrument have inherent physical limitations Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 13

International System of Units (SI) Standard system of units used in scientific & engineering measurements Metric 7 Base Units Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 14

SI Units Focus on 1 st six in this book All physical quantities will have units derived from these 7 SI base units Ex. Area Derived from SI units based on definition of area length × width = area meter × meter = area m × m = m 2 SI unit for area = square meters = m 2 Note: Units undergo same kinds of mathematical operations that numbers do! Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 15

Learning Check What is the SI unit for velocity? What is the SI unit for volume of a cube? Volume (V) = length × width × height V = meter × meter V = m 3 Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 16

Your Turn! The SI unit of length is the A. B. C. D. E. millimeter yard centimeter foot Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 17

Table 2. 2 Some Non-SI Metric Units Commonly Used in Chemistry Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 18

Table 2. 3 Some Useful Conversions Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 19

Decimal Multipliers Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 20

Using Decimal Multipliers Use prefixes on SI base units when number is too large or too small for convenient usage Only commonly used are listed here For more complete list see Table 2. 4 in textbook Numerical values of multipliers can be interchanged with prefixes Ex. 1 m. L = 10– 3 L 1 km = 1000 m 1 ng = 10– 9 g 1, 130, 000 s = 1. 13 × 109 s = 1. 13 Gs Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 21

Laboratory Measurements 4 common 1. 2. 3. 4. Length Volume Mass Temperature Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 22

Laboratory Measurements 1. Length SI Unit is meter (m) Meter too large for most laboratory measurements Commonly use Centimeter (cm) 1 cm = 10– 2 m = 0. 01 m Millimeter (mm) 1 mm = 10– 3 m = 0. 001 m Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 23

2. Volume (V) Dimensions of (length)3 SI unit for Volume = m 3 Most laboratory measurements use V in liters (L) 1 L = 1 dm 3 (exactly) Chemistry glassware marked in L or m. L 1 L = 1000 m. L What is a m. L? 1 m. L = 1 cm 3 Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 24

3. Mass SI unit is kilogram (kg) Frequently use grams (g) in laboratory as more realistic size 1 kg = 1000 g 1 g = 0. 1000 kg = g Mass is measured by comparing weight of sample with weights of known standard masses Instrument used = balance Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 25

4. Temperature Measured with thermometer 3 common scales A. Fahrenheit scale Common in US Water freezes at 32 °F and boils at 212 °F 180 degree units between melting & boiling points of water Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 26

4. Temperature B. Celsius scale Rest of world (aside from U. S. ) uses Most common for use in science Water freezes at 0 °C Water boils at 100 °C 100 degree units between melting & boiling points of water Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 27

4. Temperature C. Kelvin scale SI unit of temperature is kelvin (K) Note: No degree symbol in front of K Water freezes at 273. 15 K & boils at 373. 15 K 100 degree units between melting & boiling points Only difference between Kelvin & Celsius scale is zero point Absolute Zero point on Kelvin scale Corresponds to nature’s lowest possible temperature Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 28

Temperature Conversions How to convert between °F and °C? Ex. 100 °C = ? °F t. F = 212 °F Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 29

Temperature Conversions Common laboratory thermometers are marked in Celsius scale Must convert to Kelvin scale Amounts to adding 273. 15 to Celsius temperature Ex. What is the Kelvin temperature of a solution at 25 °C? = 298 K Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 30

Learning Check: T Conversions 1. Convert 100. °F to the Celsius scale. = 38 °C 2. Convert 100. °F to the Kelvin scale. We already have in °C so… TK = 311 K Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 31

Learning Check: T Conversions 3. Convert 77 K to the Celsius scale. = – 196 °C 4. Convert 77 K to the Fahrenheit scale. We already have in °C so = – 321 °F Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 32

Your Turn! In a recent accident some drums of uranium hexafluoride were lost in the English Channel. The melting point of uranium hexafluouride is 64. 53 °C. What is the melting point of uranium hexafluoride on the Fahrenheit scale? A. B. C. D. E. 67. 85 °F 96. 53 °F 116. 2 °F 337. 5 °F 148. 2 °F Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 33

Uncertainties in Measurements all inexact Contain uncertainties or errors Sources of errors Limitations of reading instrument Ways to minimize errors Take series of measurements Data clusters around central value Calculate average or mean values Report average value Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 34

Limits in Reading Instruments Consider 2 Celsius thermometers Left thermometer has markings every 1 °C T between 24 °C & 25 °C About 3/10 of way between marks Can estimate to 0. 1 °C = uncertainty T = 24. 3 0. 1 °C Right thermometer has markings every 0. 1 °C T reading between 24. 3 °C & 24. 4 °C Can estimate 0. 01 °C T = 24. 32 0. 01 °C Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 35

Limits in Reading Instruments Finer graduations in markings Means smaller uncertainties in measurements Reliability of data Indicated by number of digits used to represent it What about digital displays? Mass of beaker = 65. 23 g on digital balance Still has uncertainty Assume ½ in last readable digit Record as 65. 230 0. 005 g Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 36

Significant Figures Scientific convention: All digits in measurement up to & including 1 st estimated digit are significant. Number of certain digits plus 1 st uncertain digit Digits in measurement from 1 st non-zero number on left to 1 st estimated digit on right Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 37

Rules for Significant Figures 1. All non-zero numbers are significant. Ex. 3. 456 has 4 sig. figs. 2. Zeros between non-zero numbers are significant. Ex. 20, 089 or 2. 0089 × 104 has 5 sig. figs 3. Trailing zeros always count as significant if number has decimal point Ex. 500. or 5. 00 × 102 has 3 sig. figs Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 38

Rules for Significant Figures 4. Final zeros on number without decimal point are NOT significant Ex. 104, 956, 000 or 1. 04956 × 108 has 6 sig. figs. 5. Final zeros to right of decimal point are significant Ex. 3. 00 has 3 sig. figs. 6. Leading zeros, to left of 1 st nonzero digit, are never counted as significant Ex. 0. 00012 or 1. 2 × 10– 4 has 2 sig. figs. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 39

Learning Check How many significant figures does each of the following numbers have? scientific notation # of Sig. Figs. 1. 2. 3. 4. 5. 413. 97 0. 0006 5. 120063 161, 000 3600. Jespersen/Brady/Hyslop 4. 1397 × 102 6 × 10– 4 5 1 5. 120063 1. 61 × 105 7 3 3. 6 × 103 2 Chemistry: The Molecular Nature of Matter, 6 E 40

Your Turn! How many significant figures are in 19. 0000? A. B. C. D. E. 2 3 4 5 6 Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 41

Rounding to Correct Digit 1. If digit to be dropped is greater than 5, last remaining digit is rounded up. Ex. 3. 677 is rounded up to 3. 68 2. If number to be dropped is less than 5, last remaining digit stays the same. Ex. 6. 632 is rounded to 6. 63 3. If number to be dropped is 5, then if digit to left of 5 is a. Even, it remains the same. Ex. 6. 65 is rounded to 6. 6 b. Odd, it rounds up Ex. 6. 35 is rounded to 6. 4 Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 42

Scientific Notation Clearest way to present number of significant figures unambiguously Report number between 1 & 10 followed by correct power of 10 Indicates only significant digits Ex. 75, 000 people attend a concert If rough estimate? Uncertainty 1000 people 7. 5 × 104 Number estimated from aerial photograph Uncertainty 100 people 7. 50 × 104 Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 43

Learning Check Round each of the following to 3 significant figures. Use scientific notation where needed. 1. 37. 459 37. 5 or 3. 75 × 101 2. 5431978 5. 43 × 106 3. 132. 7789003 133 or 1. 33 × 102 4. 0. 00087564 8. 77 × 10– 4 5. 7. 665 7. 66 Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 44

Accuracy & Precision Accuracy How close measurement is to true or accepted true value Measuring device must be calibrated with standard reference to give correct value Precision How well set of repeated measurements of same quantity agree with each other More significant figures = more precise measurement Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 45

Significant Figures in Calculations Multiplication and Division Number of significant figures in answer = number of significant figures in least precise measurement Ex. 10. 54 × 31. 4 × 16. 987 = 5620 = 5. 62× 103 4 sig. figs. × 3 sig. figs. × 5 sig. figs = 3 sig. figs. Ex. 5. 896 ÷ 0. 008 = 700 = 7× 102 4 sig. figs. ÷ 1 sig. fig. = 1 sig. fig. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 46

Your Turn! Give the value of the following calculation to the correct number of significant figures. A. 1. 21213 B. 1. 212 C. 1. 212132774 D. 1. 2 E. 1 Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 47

Significant Figures in Calculations Addition and Subtraction Answer has same number of decimal places as quantity with fewest number of decimal places. Ex. 12. 9753 319. 5 + 4. 398 336. 9 4 1 397 – 273. 15 124 0 decimal places 2 decimal places 0 decimal place Jespersen/Brady/Hyslop decimal places decimal place Chemistry: The Molecular Nature of Matter, 6 E 48

Your Turn! When the expression, 412. 272 + 0. 00031 – 1. 00797 + 0. 000024 + 12. 8 is evaluated, the result should be expressed as: A. 424. 06 B. 424. 064364 C. 424. 1 D. 424. 064 E. 424 Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 49

Exact Numbers Number that come from definitions 12 in. = 1 ft 60 s = 1 min Numbers that come from direct count Number of people in small room Have no uncertainty Assume they have infinite number of significant figures. Do not affect number of significant figures in multiplication or division Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 50

Learning Check For each calculation, give the answer to the correct number of significant figures. 1. 10. 0 g + 1. 03 g + 0. 243 g = 11. 3 g or 1. 13 × 101 g 2. 19. 556 °C – 19. 552 °C = 3. 327. 5 m × 4. 52 m = 1. 48 × 103 m 4. 15. 985 g ÷ 24. 12 m. L = Jespersen/Brady/Hyslop 0. 004 °C or 4 × 10– 3 °C 0. 6627 g/m. L or 6. 627 g/m. L Chemistry: The Molecular Nature of Matter, 6 E 51

Learning Check For the following calculation, give the answer to the correct number of significant figures. 1. = 2 × 10– 4 m/s 2 2. = 0. 87 cm 3/s Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 52

Your Turn! For the following calculation, give the answer to the correct number of significant figures. A. 179 cm 2 B. 1. 18 cm C. 151. 2 cm D. 151 cm E. 178. 843 cm 2 Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 53

Dimensional Analysis Factor-Label Method Not all calculations use specific equation Use units (dimensions) to analyze problem Conversion Factor Fraction formed from valid equality or equivalence between units Used to switch from one system of measurement & units to another Given × Conversion Quantity Factor Jespersen/Brady/Hyslop = Desired Chemistry: The Molecular Nature of Matter, 6 E Quantity 54

Conversion Factors Ex. How to convert a person’s height from 68. 0 in to cm? Start with fact 2. 54 cm = 1 in. Dividing both sides by 1 in. or 2. 54 cm gives 1 =1 =1 Cancel units Leave ratio that equals 1 Use fact that units behave as numbers do in mathematical operations Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 55

Dimensional Analysis Now multiply original number by conversion factor that cancels old units & leaves new Given × Conversion = Desired Quantity Factor Quantity = 173 cm Dimensional analysis can tell us when we have done wrong arithmetic = 26. 8 in 2/cm Units not correct Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 56

Using Dimensional Analysis Ex. Convert 0. 097 m to mm. Relationship is 1 mm = 1 × 10– 3 m Can make 2 conversion factors Since going from m to mm use on left. = 173 cm Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 57

Learning Check Ex. Convert 3. 5 m 3 to cm 3. Start with basic equality Now cube both sides Units & numbers (1 cm)3 = (0. 01 m)3 1 cm 3 = 1 × 10– 6 m 3 1 cm = 0. 01 m Can make 2 conversion factors or 3. 5 × 106 Cm 3 Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 58

Non-metric to Metric Units Convert speed of light from 3. 00× 108 m/s to mi/hr Use dimensional analysis 1 min = 60 s 60 min = 1 hr 1 km = 1000 m 1 mi = 1. 609 km 1. 08 × 1012 m/hr 6. 71 × 108 mi/hr Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 59

Your Turn! The Honda Insight hybrid electric car has a gas mileage rating of 56 miles to the gallon. What is this rating expressed in units of kilometers per liter? 1 gal = 3. 784 L A. 1. 3 × 102 km L– 1 B. 24 km L– 1 C. 15 km L– 1 D. 3. 4 × 102 km L– 1 E. 9. 2 km L– 1 Jespersen/Brady/Hyslop 1 mile = 1. 609 km Chemistry: The Molecular Nature of Matter, 6 E 60

Law of Multiple Proportions If 2 elements form more than 1 compound they combine in different ratios by mass Same mass of 1 element combines with different masses of 2 nd element in different compounds Experimentally hard to get exactly same mass of 1 element in 2 or more experiments Can use dimensional analysis to calculate Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 61

Applying Law of Multiple Proportions Titanium forms 2 different compounds with bromine. In compound A we find that 4. 787 g of Ti are combined with 15. 98 g of bromine. In compound B we find that 6. 000 g of Ti are combined with 40. 06 g of bromine. Determine whether these data support the law of multiple proportions. Analysis Need same mass of 1 element & compare masses of 2 nd element 6. 000 g Ti for each How much Br? Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 62

Applying Law of Multiple Proportions Know: In compound A: In compound B: 4. 787 g Ti ⇔ 15. 98 g Br 6. 000 g Ti ⇔ 40. 06 g Br Must find: 6. 000 g Ti ⇔ ? g Br (compound A) Solution: = 20. 03 g Br Compare Ratio of small whole numbers Supports law of multiple proportions Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 63

Density Ratio of object’s mass to its volume Intensive property (size independent) Determined by taking ratio of 2 extensive properties (size dependent) Frequently ratio of 2 size dependent properties leads to size independent property Sample size cancels Units g/m. L or g/cm 3 Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 64

Learning Check A student weighs a piece of gold that has a volume of 11. 02 cm 3 of gold. She finds the mass to be 212 g. What is the density of gold? 19. 3 g/cm 3 Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 65

Density Most substances expand slightly when heated Same mass Larger volume Less dense Density slightly as T Liquids & Solids Change is very small Can ignore except in very precise calculations Density useful to transfer between mass & volume of substance Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 66

Learning Check 1. Glass has a density of 2. 2 g/cm 3. What is the volume occupied by 22 g of glass? 10. g/cm 3 2. What is the mass of 400 cm 3 of glass? 880 g Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 67

Your Turn! Titanium is a metal used to make artificial joints. It has a density of 4. 54 g/cm 3. What volume will a titanium hip joint occupy if its mass is 205 g? A. 9. 31 × 102 cm 3 B. 4. 51 × 101 cm 3 C. 2. 21 × 10– 2 cm 3 D. 1. 07 × 10– 3 cm 3 E. 2. 20 × 10– 1 cm 3 Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 68

Your Turn! A sample of zinc metal (density = 7. 14 g cm-3) was submerged in a graduated cylinder containing water. The water level rose from 162. 5 cm 3 to 186. 0 cm 3 when the sample was submerged. How many grams did the sample weigh? A. 1. 16 × 103 g B. 1. 33 × 103 g C. 23. 5 g D. 1. 68 × 102 g E. 3. 29 g Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 69

Specific Gravity Ratio of density of substance to density of water Unitless Way to avoid having to tabulate densities in all sorts of different units Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 70

Learning Check Concentrated sulfuric acid is sold in bottles with a label that states that the specific gravity at 25 °C is 1. 84. The density of water at 25 °C is 0. 995 g cm– 3. How many cubic centimeters of sulfuric acid will weigh 5. 55 kilograms? Analysis: 5. 55 kg sulfuric acid = ? cm 3 sulfuric acid Solution: density sulfuric acid = specific gravity × density water dsulfuric acid = 1. 84 × 0. 995 g/cm 3 = 1. 83 5. 58 cm 3 Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 71

Your Turn! Liquid hydrogen has a specific gravity of 7. 08 × 10– 2. If the density of water is 1. 05 g/cm 3 at the same temperature, what is the mass of hydrogen in a tank having a volume of 36. 9 m 3? A. 7. 43 × 10– 2 g B. 2. 74 g C. 274 g D. 2. 74 × 106 g E. 2. 61 × 106 g Jespersen/Brady/Hyslop = 7. 43 × 10– 2 g/cm 3 Chemistry: The Molecular Nature of Matter, 6 E 72

Importance of Reliable Measurements To trust conclusions drawn from measurements Must know they are reliable Must be sure they are accurate Measured values must be close to true values Otherwise can’t trust results Can’t make conclusions based on those results Must have sufficient precision to be meaningful So confident that 2 measurements are same for 2 samples Difference in values must be close to uncertainty in measurement Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 73

Learning Check You have a ring? Is it made of 24 K gold? Calculate density & compare to known Density of 24 K gold = 19. 3 g/m. L Use inaccurate glassware Volume of ring = 1. 0 m. L Use kitchen balance Mass of ring = 18 1 g Anywhere between 17 & 19 g Density range is 17 – 19 g/m. L Could be 24 k gold or could be as low as 18 K gold (density = 16. 9 g/m. L) Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 74

Learning Check (cont) Use more precise laboratory balance Mass of ring = 18. 153 0. 001 g Use more precise glassware Volume of ring = 1. 03 m. L Density of ring = 18. 153 g/1. 03 m. L = 17. 6 g/m. L Calculate difference between d 24 K gold & dring 19. 3 g/m. L – 17. 6 g/m. L = 1. 7 g/m. L Larger than experimental error in density ~ 0. 1 g/m. L Conclude: ring NOT 24 K gold! Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6 E 75