1. 1 Chapter 2 Measurements 2.6 Writing Conversion Factors Basic Chemistry Copyright © 2011 Pearson Education, Inc. The mass of a packaged product is listed in both U.S. and metric units.

2. 2 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 Examples: 1 m = 1000 mm 1 lb = 16 oz 2.205 lb = 1 kg Equalities Basic Chemistry Copyright © 2011 Pearson Education, Inc.

3. 3 Exact and Measured Numbers in Equalities Equalities between units of the same system are definitions and are exact numbers different systems (metric and U.S.) use measured numbers and count as significant figures Basic Chemistry Copyright © 2011 Pearson Education, Inc.

4. 4 Some Common Equalities Basic Chemistry Copyright © 2011 Pearson Education, Inc.

5. 5 Equalities on Food Labels The contents of packaged foods are listed as both metric and U.S. units indicate the same amount of a substance in two different units Basic Chemistry Copyright © 2011 Pearson Education, Inc. The mass of a packaged product is listed in both U.S. and metric units.

6. 6 A conversion factor is a fraction obtained from an equality Equality: 1 in. = 2.54 cm written as a ratio with a numerator and denominator is inverted to give two conversion factors for every equality 1 in. and 2.54 cm 2.54 cm 1 in. Conversion Factors Basic Chemistry Copyright © 2011 Pearson Education, Inc.

7. 7 Write conversion factors for each pair of units: A. liters and mL B. hours and minutes C. meters and kilometers Learning Check Basic Chemistry Copyright © 2011 Pearson Education, Inc.

8. 8 Write conversion factors for each pair of units: A. liters and mL (1 L = 1000 mL) 1 L and 1000 mL 1000 mL 1 L B. hours and minutes (1 h = 60 min) 1 h and 60 min 60 min 1 h C. meters and kilometers (1 km = 1000 m) 1 km and 1000 m 1000 m 1 km Solution Basic Chemistry Copyright © 2011 Pearson Education, Inc.

9. 9 Factors with Powers A conversion factor can be squared or cubed on both sides of the equality. Equality 1 in. = 2.54 cm 1 in. and 2.54 cm 2.54 cm 1 in. Equality squared (1 in.) 2 = (2.54 cm) 2 (1 in.) 2 and (2.54 cm) 2 (2.54 cm) 2 (1 in.) 2 Equality cubed (1 in.) 3 = (2.54 cm) 3 (1 in.) 3 and (2.54 cm) 3 (2.54 cm) 3 (1 in.) 3 Basic Chemistry Copyright © 2011 Pearson Education, Inc.

10. 10 An equality and conversion factors may be obtained from information in a word problem are for that problem only Example: The price of one pound (1 lb) of red peppers is $2.39. Equality: 1 lb peppers = $2.39 Conversion factors: 1 lb red peppers and$2.39 $2.391 lb red peppers Conversion Factors in a Problem Basic Chemistry Copyright © 2011 Pearson Education, Inc.

11. 11 A percent factor gives the ratio of the part to the whole. Percent(%) = part x 100 whole uses matching units to express the percent uses the value 100 and a unit for the whole Example: A food contains 30% (by mass) fat. Equality: 100 g food = 30 g fat Conversion factors: 30 g fat and100 g food 100 g food30 g fat Percent as a Conversion Factor Basic Chemistry Copyright © 2011 Pearson Education, Inc.

12. 12 Percent Factor in a Problem The thickness of the skin fold at the waist indicates 11% body fat. What equality and percent conversion factors can be written for body fat in kg? Equality (kg): 100 kg of body mass = 11 kg of fat Percent conversion factors (kg) 11 kg fat and 100 kg mass 100 kg mass 11 kg fat Basic Chemistry Copyright © 2011 Pearson Education, Inc. The thickness of the skin fold at the waist is measured in millimeters (mm) to determine the percent of body fat.

13. ppm and ppb Relationships of small percent values are ppm (parts per million or mg/kg) ppb (parts per billion or  g/kg) Example: A soil sample contains 2 ppm of lead. Equality: 2 mg of lead = 1 kg of soil Conversion factors: 2 mg lead and 1 kg soil 1 kg soil 2 mg lead 13 Basic Chemistry Copyright © 2011 Pearson Education, Inc.

14. 14 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 costing $3.29 D. a water sample with 55 ppb of chromium (Cr) Basic Chemistry Copyright © 2011 Pearson Education, Inc.

15. 15 Solution A. 1 m 2 = (100 cm) 2 (1m) 2 and (100 cm) 2 (100 cm) 2 (1m) 2 B. 100 g of jewelry = 18 g of gold 18 g gold and 100 g jewelry 100 g jewelry 18 g gold C. 1 gal of gas = $3.29 1 gal and $3.29 $3.29 1 gal D. 1 kg of water = 55  g of chromium 1 kg water and 55  g chromium 55  g chromium 1 kg water Basic Chemistry Copyright © 2011 Pearson Education, Inc.