LECTURE 2 CHM 151 ©slg TOPICS: SF's IN CALCULATIONS SI, METRIC, ENGLISH UNITS DIMENSIONAL ANALYSIS UNIT CONVERSIONS DENSITY

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.

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.

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 off 23.45231 to 4 SF’s: 23.45231 = 23.45 Less than five

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

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 = 23.46 23.44500 = 23.44 Note: 23.455001 = 23.46 23.445001 = 23.45

SF’s in Calculations, Samples: 1.30 in. X .20 in. X 2960. in. = 769.60 in.3 = 7.7 X 10 2 in.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.

Group Work 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 =?

Group Work #1, Answers: 24.569 g - .0056 g = ? 24.569 g - .0055 g “ No more digits after decimal than value with least number of digits after decimal”

32.35 cm X 21.9 cm X 0.76 cm = ? 32.35 cm X 21.9 cm X 0.76 cm = “538.4334” cm3 allowed SF’s: 2 538.4334 = 5.384334 X 102 = 5.4 X 102 cm3 You may also use 540 cm3 which is also 2 SF’s

54.01 lb + .6789 lb + 1,312.0 lb =? 54.01 lb .6489 lb + 1,312.0 lb 1,366.6589 lb = 1,366.7 lb

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”

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

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

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, new unit 135 lb = ? kg

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

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 1 kg 103g

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!

 SF’s All defined conversions within a system are exact; The “1” in all conversion relationships is exact 4 SF 3 SF Exact 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!)

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

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: 109 nm = 1 m 1 m = 102 cm Pathway: nm  m  cm

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

Group Work 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?

A football field is 100. yd long. What length is this in meters? (1 m = 39.37 in) 1. State question: 100. yd = ? m 2. State relationships: 1 yd = 3 ft 1 ft = 12 in 39.37 in = 1 m Pathway: yd  ft  in  m

Pathway: yd  ft  in  m 3. Setup and Solve: 100. yd X 3 ft X 12 in X 1 m = 91.44018 m 1 yd 1 ft 39.37 in = 91.4 m

In water, H2O, the bond length between each H and O is 94 pm. What is this length in millimeters? 1. State the question: 94 pm = ? mm 2. State the needed relationships: 1012 pm = 1 m 1 m = 103 mm Pathway: pm  m  mm

Pathway: pm  m  mm 3. Setup and solve: 94 pm X 1 m X 103 mm = 94 X 103 - 12 mm 1012 pm 1 m = 94 X 10-9 mm = 9.4 X 10-8 mm

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

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

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 103 cm3 = 1.9 X 103 cm3

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: 1.9 X 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

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

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”

Common Density Values 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 Magnesium: 1.74 g/cm3 Aluminum: 2.70 g/cm3 Silver: 10.5 g/cm3 Gold: 19.3 g/cm3

SOLVING DENSITY PROBLEMS 1. When the Density factor itself is desired: Density = mass of object, etc (g) volume occupied (mL, cm3, L) 2. In all other 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

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

3. Substitute values and solve: D = mass = 205 g = .86497 g = .865 g 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

.865 g = ? oz avoir mL in3 2. State relationships: 453.6 g = 1 lb 1 lb = 16 oz avoir 1 mL = 1 cm3 2.540 cm = 1 in ( 2.540 cm)3 = (1 in)3 16.39 cm3 = 1 in3 Pathway: g  lb  oz avoir; mL  cm3  in3

.865 g = ? oz avoir mL in3 Pathway: g  lb  oz avoir; mL  cm3  in3 3. Setup and Solve: .865 g X 1 lb X 16 oz avoir X 1 mL X 16.39 cm3 1 mL 453.6 g 1 lb 1 cm3 1 in3 = .50008 oz avoir in3 = .500 oz avoir

Density as a Conversion Factor: Peanut oil has a density of .92 g/mL. If the recipe calls for 1 cup of oil (1 cup = 237 mL), what mass of oil, in lb, are you going to use? State question: 237 mL oil =? Lb oil State relationship: 1 mL oil = .92 g oil 453.6 g = 1 lb Pathway: mL  g  lb SETUP AND SOLVE AS GROUP WORK!

Solution: 237 mL oil =? Lb oil Pathway: mL  g  lb 237 ml oil X .92 g oil X 1 lb = .48068 lb oil 1 mL oil 453.6 g = .48 lb oil