Chapter 1 Chemical Foundations 1.6 Dimensional Analysis

Transcription

1 Chapter 1 Chemical Foundations 1.6 Dimensional Analysis Copyright 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings 1

2 Equalities Equalities use two different units to describe the same measured amount. are written for relationships between units of the metric system, U.S. units, or between metric and U.S. units. For example, 1 m = 1000 mm 1 lb = 16 oz lb = 1 kg 2

3 Exact and Measured Numbers in Equalities Equalities between units of the same system are definitions and use exact numbers. different systems (metric and U.S.) use measured numbers and count as significant figures. 3

4 Some Common Equalities 4

5 Equalities on Food Labels The contents of packaged foods in the U.S. are listed as both metric and U.S. units. indicate the same amount of a substance in two different units. Copyright 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings 5

6 Conversion Factors A conversion factor is a fraction obtained from an equality. Equality: 1 in. = 2.54 cm is written as a ratio with a numerator and denominator. can be inverted to give two conversion factors for every equality. 1 in. and 2.54 cm 2.54 cm 1 in. 6

7 Learning Check Write conversion factors for each pair of units. A. liters and ml B. hours and minutes C. meters and kilometers 7

8 Solution Write conversion factors for each pair of units. A. liters and ml Equality: 1 L = 1000 ml 1 L and 1000 ml 1000 ml 1 L B. hours and minutes Equality: 1 hr = 60 min 1 hr and 60 min 60 min 1 hr C. meters and kilometers Equality: 1 km = 1000 m 1 km and 1000 m 1000 m 1 km 8

9 Conversion Factors in a Problem A conversion factor may be obtained from information in a word problem. is written for that problem only. Example 1: The price of one pound (1 lb) of red peppers is $ lb red peppers and $2.39 $ lb red peppers Example 2: The cost of one gallon (1 gal) of gas is $ gallon of gas and $2.34 $ gallon of gas 9

10 Percent as a Conversion Factor A percent factor gives the ratio of the parts to the whole. % = Parts x 100 Whole uses the same unit to express the percent. uses the value 100 and a unit for the whole. can be written as two factors. Example: A food contains 30% (by mass) fat. 30 g fat and 100 g food 100 g food 30 g fat 10

11 Percent Factor in a Problem The thickness of the skin fold at the waist indicates 11% body fat. What percent factors can be written for body fat in kg? Percent factors using kg: 11 kg fat and 100 kg mass 100 kg mass 11 kg fat Copyright 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings 11

12 Learning Check Write the equality and conversion factors for each of the following. A. square meters and square centimeters B. jewelry that contains 18% gold C. One gallon of gas is $

13 Solution A. square meters and square centimeters (1m) 2 and (100 cm) 2 (100 cm) 2 (1m) 2 B. jewelry that contains 18% gold 18 g gold and 100 g jewelry 100 g jewelry 18 g gold C. One gallon of gas is $ gal and $2.27 $ gal 13

14 Given and Needed Units To solve a problem Identify the given unit Identify the needed unit. Example: A person has a height of 2.0 meters. What is that height in inches? The given unit is the initial unit of height. given unit = meters (m) The needed unit is the unit for the answer. needed unit = inches (in.) 14

15 Learning Check An injured person loses 0.30 pints of blood. How many milliliters of blood would that be? Identify the given and needed units given in this problem. Given unit = Needed unit = 15

16 Solution An injured person loses 0.30 pints of blood. How many milliliters of blood would that be? Identify the given and needed units given in this problem. Given unit = pints Needed unit = milliliters 16

17 Problem Setup Write the given and needed units. Write a unit plan to convert the given unit to the needed unit. Write equalities and conversion factors that connect the units. Use conversion factors to cancel the given unit and provide the needed unit. Unit 1 x Unit 2 = Unit 2 Unit 1 Given x Conversion = Needed unit factor unit 17

18 Guide to Problem Solving (GPS) The steps in the Guide to Problem Solving are useful in setting up a problem with conversion factors. 18

19 Setting up a Problem How many minutes are 2.5 hours? Given unit = 2.5 hr Needed unit = min Unit Plan = hr min Setup problem to cancel hours (hr). Given Conversion Needed unit factor unit 2.5 hr x 60 min = 150 min (2 SF) 1 hr Copyright 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings 19

20 Learning Check A rattlesnake is 2.44 m long. How many centimeters long is the snake? 1) 2440 cm 2) 244 cm 3) 24.4 cm 20

21 Solution A rattlesnake is 2.44 m long. How many centimeters long is the snake? 2) 244 cm Given Conversion Needed unit factor unit 2.44 m x 100 cm = 244 cm 1 m 21

22 Using Two or More Factors Often, two or more conversion factors are required to obtain the unit needed for the answer. Unit 1 Unit 2 Unit 3 Additional conversion factors are placed in the setup to cancel each preceding unit Given unit x factor 1 x factor 2 = needed unit Unit 1 x Unit 2 x Unit 3 = Unit 3 Unit 1 Unit 2 22

23 Example: Problem Solving How many minutes are in 1.4 days? Given unit: 1.4 days Plan: Factor 1 Factor 2 days hr min Set up problem: 1.4 days x 24 hr x 60 min = 2.0 x 10 3 min 1 day 1 hr 2 SF Exact Exact = 2 SF 23

24 Check the Unit Cancellation Be sure to check your unit cancellation in the setup. The units in the conversion factors must cancel to give the correct unit for the answer. What is wrong with the following setup? 1.4 day x 1 day x 1 hr 24 hr 60 min Units = day 2 /min is not the unit needed Units don t cancel properly. 24

25 Using the GPS What is 165 lb in kg? STEP 1 Given 165 lb Need kg STEP 2 Plan STEP 3 Equalities/Factors 1 kg = 2.20 lb 2.20 lb and 1 kg 1 kg 2.20 lb STEP 4 Set Up Problem 165 lb x 1 kg = 75.0 kg 2.20 lb 25

26 Learning Check A bucket contains 4.65 L of water. How many gallons of water is that? Unit plan: L qt gallon Equalities: 1.06 qt = 1 L 1 gal = 4 qt Set up Problem: 26

27 Solution Given: 4.65 L Needed: gallons Plan: L qt gallon Equalities: 1.06 qt = 1 L; 1 gal = 4 qt Set Up Problem: 4.65 L x x 1.06 qt x 1 gal = 1.23 gal 1 L 4 qt 3 SF 3 SF exact 3 SF 27

28 Learning Check If a ski pole is 3.0 feet in length, how long is the ski pole in mm? 28

29 Solution 3.0 ft x 12 in x 2.54 cm x 10 mm = 1 ft 1 in. 1 cm Calculator answer: mm Needed answer: 910 mm (2 SF rounded) Check factor setup: Check needed unit: Units cancel properly mm 29

30 Learning Check If your pace on a treadmill is 65 meters per minute, how many minutes will it take for you to walk a distance of 7500 feet? 30

31 Solution Given: 7500 ft 65 m/min Need: min Plan: ft in. cm m min Equalities: 1 ft = 12 in. 1 in. = 2.54 cm 1 m = 100 cm 1 min = 65 m (walking pace) Set Up Problem: 7500 ft x 12 in. x 2.54 cm x 1m x 1 min 1 ft 1 in. 100 cm 65 m = 35 min final answer (2 SF) 31

32 Percent Factor in a Problem If the thickness of the skin fold at the waist indicates an 11% body fat, how much fat is in a person with a mass of 86 kg? percent factor 86 kg mass x 11 kg fat 100 kg mass = 9.5 kg fat Copyright 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings 32

33 Learning Check How many lb of sugar are in 120 g of candy if the candy is 25% (by mass) sugar? 33

34 Solution How many lb of sugar are in 120 g of candy if the candy is 25%(by mass) sugar? percent factor 120 g candy x 1 lb candy x 25 lb sugar 454 g candy 100 lb candy = lb sugar Ref: Timberlake, Chemistry, Pearson/Benjamin Cummings, 2006, 9 th Ed. 34