1. 1 Chapter 2 Measurements 2.7 Problem Solving Basic Chemistry Copyright © 2011 Pearson Education, Inc. A health professional obtains a measured volume from a vial for an injection.

2. 2 To solve a problem identify the given unit identify the needed unit Problem: A person has a height of 2.0 m. 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.) Given and Needed Units Basic Chemistry Copyright © 2011 Pearson Education, Inc.

3. 3 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 in this problem. Given unit= _______ Needed unit = _______ Basic Chemistry Copyright © 2011 Pearson Education, Inc.

4. 4 Solution An injured person loses 0.30 pints of blood. How many milliliters of blood would that be? Identify the given and needed units in this problem Given unit=pints Needed unit =milliliters Basic Chemistry Copyright © 2011 Pearson Education, Inc.

5. 5 STEP 1 State the given and needed quantities. STEP 2 Write a plan to convert the given unit to the needed unit. STEP 3 State the equalities and conversion factors needed to cancel units. STEP 4 Set up problem to cancel units and calculate answer. Unit 1 x Unit 2 = Unit 2 Unit 1 Given Conversion Needed unit factor unit Guide to Problem Solving Using Conversion Factors Basic Chemistry Copyright © 2011 Pearson Education, Inc.

6. Guide to Problem Solving 6 Basic Chemistry Copyright © 2011 Pearson Education, Inc. The steps in the Guides to Problem Solving Using Conversion Factors are useful in setting up a problem with conversion factors.

7. 7 Setting up a Problem How many minutes are 2.5 hours? STEP 1 Given 2.5 h Need min STEP 2 Plan hours minutes STEP 3 Equalities 1 h = 60 min STEP 4 Set up problem 2.5 h x 60 min = 150 min 1 h (2 SFs) given conversion needed unit factor unit Basic Chemistry Copyright © 2011 Pearson Education, Inc.

8. 8 A rattlesnake is 2.44 m long. How long is the snake in centimeters? 1) 2440 cm 2) 244 cm 3) 24.4 cm Learning Check Basic Chemistry Copyright © 2011 Pearson Education, Inc.

9. 9 A rattlesnake is 2.44 m long. How long is the snake in centimeters? STEP 1 Given 2.44 m Need centimeters STEP 2 Plan meters centimeters STEP 3 Equality 1 m = 100 cm Factors 1 m and 100 cm 100 cm 1 m STEP 4 Set up problem 2.44 m x 100 cm = 244 cm (answer 2) 1 m Solution Basic Chemistry Copyright © 2011 Pearson Education, Inc.

10. 10 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 Using Two or More Factors Basic Chemistry Copyright © 2011 Pearson Education, Inc.

11. 11 How many minutes are in 1.4 days? STEP 1 Given 1.4 days Need minutes STEP 2 Plan days hours minutes STEP 3 Equalties: 1 day = 24 h 1 h = 60 min STEP 4 Set up problem: 1.4 days x 24 h x 60 min = 2.0 x 10 3 min 1 day 1 h 2 SFs Exact Exact = 2 SFs Example: Problem Solving Basic Chemistry Copyright © 2011 Pearson Education, Inc.

12. 12 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. Example: What is wrong with the following setup? 1.4 day x 1 day x 1 h 24 h 60 min Units = day 2 /min, which is not the unit needed. Units do not cancel properly; the setup is wrong. Check the Unit Cancellation Basic Chemistry Copyright © 2011 Pearson Education, Inc.

13. 13 Using the GPS What is 165 lb in kilograms? STEP 1 Given 165 lb Need kg STEP 2 Plan lb kg STEP 3 Equalities/conversion 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 = 74.8 kg (3 SFs) 2.20 lb Basic Chemistry Copyright © 2011 Pearson Education, Inc.

14. 14 A bucket contains 4.65 L water. How many gallons of water is that? Learning Check Basic Chemistry Copyright © 2011 Pearson Education, Inc.

15. 15 STEP 1 Given 4.65 L Need gallons STEP 2 Plan liters quarts gallons STEP 3 Equalities 1 L = 1.057 qt1 gal = 4 qt STEP 4 Set Up Problem 4.65 L x 1.057 qt x 1 gal = 1.23 gal 1 L 4 qt 3 SFs 4 SFs Exact 3 SFs Solution Basic Chemistry Copyright © 2011 Pearson Education, Inc.

16. 16 If a ski pole is 3.0 feet in length, how long is the ski pole in mm? Learning Check Basic Chemistry Copyright © 2011 Pearson Education, Inc.

17. 17 STEP 1 Given 3.0 ft Need mm STEP 2 Plan ft in. cm mm STEP 3 Equalities 1 ft = 12 in., 2.54 cm = 1 in., 1 cm = 10 mm STEP 4 Set up problem 3.0 ft x 12 in. x 2.54 cm x 10 mm = 910 mm (2SFs) 1 ft 1 in. 1 cm Check initial unit:ft Check factor setup: units cancel properly Check needed unit:mm Solution Basic Chemistry Copyright © 2011 Pearson Education, Inc.

18. 18 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? Learning Check Basic Chemistry Copyright © 2011 Pearson Education, Inc.

19. 19 STEP 1 Given 7500 ft; 65 m/min Need min STEP 2 Plan ft in. cm m min STEP 3 Equalities 1 ft = 12 in. 1 in. = 2.54 cm 1 m = 100 cm 1 min = 65 m (walking pace) STEP 4 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 Answer rounded off (2 SFs) Solution Basic Chemistry Copyright © 2011 Pearson Education, Inc.

20. 20 Percent Factor in a Problem If the thickness of the skin fold at the waist indicates 11% of body fat, how much fat, in kg, does a person have with a body mass of 86 kg? STEP 1 Given 86 kg; 11% fat Need kg fat STEP 2 Plan kg body mass kg fat STEP 3 Equalities/Conversion factors 100 kg of body mass = 11 kg of fat 100 kg body mass and 11 kg fat 11 kg fat 100 kg body mass STEP 4 Set Up Problem 86 kg mass x 11 kg fat = 9.5 kg of fat 100 kg mass Basic Chemistry Copyright © 2011 Pearson Education, Inc.

21. 21 How many lb of sugar are in 120 g of candy if the candy is 25% (by mass) sugar? Learning Check Basic Chemistry Copyright © 2011 Pearson Education, Inc.

22. 22 STEP 1 Given 120 g of candy Need lb of sugar STEP 2 Plan g lb of candy lb of sugar STEP 3 Equalities 1 lb = 453.6 g 100 g of candy = 25 g of sugar STEP 4 Set Up Problem % factor 120 g candy x 1 lb candy x 25 lb sugar 453.6 g candy 100 lb candy = 0.066 lb of sugar Solution Basic Chemistry Copyright © 2011 Pearson Education, Inc.