1. Christopher G. Hamaker, Illinois State University, Normal IL © 2008, Prentice Hall Chapter 3 The Metric System INTRODUCTORY CHEMISTRY INTRODUCTORY CHEMISTRY Concepts & Connections Fifth Edition by Charles H. Corwin

2. Chapter 3 2 The Metric System The English system was used primarily in the British Empire and wasn’t very standardized. The French organized a committee to devise a universal measuring system. After about 10 years, the committee designed and agreed on the metric system. The metric system offers simplicity with a single base unit for each measurement.

3. Chapter 3 3 Metric System Basic Units

4. Chapter 3 4 Original Metric Unit Definitions A meter was defined as 1/10,000,000 of the distance from the North Pole to the equator. A kilogram (1000 grams) was equal to the mass of a cube of water measuring 0.1 m on each side. A liter was set equal to the volume of one kilogram of water at 4  C.

5. Chapter 3 5 Metric System Advantage Another advantage of the metric system is that it is a decimal system. It uses prefixes to enlarge or reduce the basic units. For example: –A kilometer is 1000 meters. –A millimeter is 1/1000 of a meter.

6. Chapter 3 6 Metric System Prefixes The following table lists the common prefixes used in the metric system:

7. Chapter 3 7 Metric Prefixes, continued For example, the prefix kilo- increases a base unit by 1000: –1 kilogram is 1000 grams The prefix milli- decreases a base unit by a factor of 1000: –1 millimeter is 0.001 meters

8. Chapter 3 8 Metric Symbols The names of metric units are abbreviated using symbols. Use the prefix symbol followed by the symbol for the base unit, so: –nanometer is abbreviated nm –microgram is abbreviated  g –deciliter is abbreviated dL –gigasecond is abbreviated Gs

9. Chapter 3 9 Metric Equivalents We can write unit equations for the conversion between different metric units. The prefix kilo- means 1000 basic units, so 1 kilometer is 1000 meters. The unit equation is 1 km = 1000 m. Similarly, a millimeter is 1/1000 of a meter, so the unit equation is 1 mm = 0.001 m.

10. Chapter 3 10 Metric Unit Factors Since 1000 m = 1 km, we can write the following unit factors for converting between meters and kilometers: 1 km or 1000 m 1000 m 1 km Since 1 m = 0.001 mm, we can write the following unit factors. 1 mm or 0.001 m 0.001 m 1 mm

11. Chapter 3 11 Metric-Metric Conversions We will use the unit analysis method we learned in Chapter 2 to do metric-metric conversion problems. Remember, there are three steps: –Write down the unit asked for in the answer. –Write down the given value related to the answer. –Apply unit factor(s) to convert the given unit to the units desired in the answer.

12. Chapter 3 12 Metric-Metric Conversion Problem What is the mass in grams of a 325 mg aspirin tablet? Step 1: We want grams. Step 2: We write down the given: 325 mg. Step 3: We apply a unit factor (1 mg = 0.001 g) and round to three significant figures. 325 mg ×= 0.325 g 1 mg 0.001 g

13. Chapter 3 13 Two Metric-Metric Conversions A hospital has 125 deciliters of blood plasma. What is the volume in milliliters? Step 1: We want the answer in mL. Step 2: We have 125 dL. Step 3: We need to first convert dL to L and then convert L to mL: 0.1 L and 0.001 L 1 dL 1 mL.

14. Chapter 3 14 Problem, continued Apply both unit factors, and round the answer to 3 significant digits. Notice that both dL and L units cancel, leaving us with units of mL. 125 dL ×= 12,500 mL× 1 dL 0.1 L 1 mL 0.001 L

15. Chapter 3 15 Another Example The mass of the Earth’s moon is 7.35 × 10 22 kg. What is the mass expressed in megagrams, Mg? We want Mg; we have 7.35 × 10 22 kg. Convert kilograms to grams, and then grams to megagrams. 7.35 × 10 22 kg ×= 7.35 × 10 19 Mg× 1 kg 1000 g 1 Mg 1000000 g

16. Chapter 3 16 Metric and English Units The English system is still very common in the United States. We often have to convert between English and metric units.

17. Chapter 3 17 Metric-English Conversion The length of an American football field, including the end zones, is 120 yards. What is the length in meters? Convert 120 yd to meters given that 1 yd = 0.914 m. 120 yd ×= 110 m 1 yd 0.914 m

18. Chapter 3 18 English-Metric Conversion A half-gallon carton contains 64.0 fl oz of milk. How many milliliters of milk are in a carton? We want mL; we have 64.0 fl oz. Use 1 qt = 32 fl oz, and 1 qt = 946 mL. 64.0 fl oz ×= 1,890 mL× 32 fl oz 1 qt 946 mL 1 qt

19. Chapter 3 19 Compound Units Some measurements have a ratio of units. For example, the speed limit on many highways is 55 miles per hour. How would you convert this to meters per second? Convert one unit at a time using unit factors. –first, miles → meters –second, hours → seconds

20. Chapter 3 20 Compound Unit Problem A motorcycle is traveling at 75 km/hour. What is the speed in meters per second? We have km/h; we want m/s. Use 1 km = 1000 m and 1 h = 3600 s. = 21 m/s× 1 km 1000 m 1 hr 3600 s 75 km hr ×

21. Chapter 3 21 Volume by Calculation The volume of an object is calculated by multiplying the length (l) by the width (w) by the thickness (t). volume = l × w × t All three measurements must be in the same units. If an object measures 3 cm by 2 cm by 1 cm, the volume is 6 cm 3 (cm 3 is cubic centimeters).

22. Chapter 3 22 Cubic Volume and Liquid Volume The liter (L) is the basic unit of volume in the metric system. One liter is defined as the volume occupied by a cube that is 10 cm on each side.

23. Chapter 3 23 Cubic & Liquid Volume Units 1 liter is equal to 1000 cubic centimeters –10 cm × 10 cm × 10 cm = 1000 cm 3 1000 cm 3 = 1 L = 1000 mL Therefore, 1 cm 3 = 1 mL.

24. Chapter 3 24 Cubic-Liquid Volume Conversion An automobile engine displaces a volume of 498 cm 3 in each cylinder. What is the displacement of a cylinder in cubic inches, in 3 ? We want in 3 ; we have 498 cm 3. Use 1 in = 2.54 cm three times. = 30.4 in 3 × 1 in 2.54 cm ×498 cm 3 × 1 in 2.54 cm 1 in 2.54 cm

25. Chapter 3 25 Volume by Displacement If a solid has an irregular shape, its volume cannot be determined by measuring its dimensions. You can determine its volume indirectly by measuring the amount of water it displaces. This technique is called volume by displacement. Volume by displacement can also be used to determine the volume of a gas.

26. Chapter 3 26 Solid Volume by Displacement You want to measure the volume of an irregularly shaped piece of jade. Partially fill a volumetric flask with water and measure the volume of the water. Add the jade, and measure the difference in volume. The volume of the jade is 10.5 mL.

27. Chapter 3 27 Gas Volume by Displacement You want to measure the volume of gas given off in a chemical reaction. The gas produced displaces the water in the flask into the beaker. The volume of water displaced is equal to the volume of gas.

28. Chapter 3 28 The Density Concept The density of an object is a measure of its concentration of mass. Density is defined as the mass of an object divided by the volume of the object. = density volume mass

29. Chapter 3 29 Density Density is expressed in different units. It is usually grams per milliliter (g/mL) for liquids, grams per cubic centimeter (g/cm 3 ) for solids, and grams per liter (g/L) for gases.

30. Chapter 3 30 Densities of Common Substances

31. Chapter 3 31 Estimating Density We can estimate the density of a substance by comparing it to another object. A solid object will float on top a liquid with a higher density. Object S 1 has a density less than that of water, but larger than that of L 1. Object S 2 has a density less than that of L 2, but larger than that of water.

32. Chapter 3 32 Calculating Density What is the density of a platinum nugget that has a mass of 224.50 g and a volume of 10.0 cm 3 ? Recall, density is mass/volume. = 22.5 g/cm 3 10.0 cm 3 224.50 g

33. Chapter 3 33 Density as a Unit Factor We can use density as a unit factor for conversions between mass and volume. An automobile battery contains 1275 mL of acid. If the density of battery acid is 1.84 g/mL, how many grams of acid are in an automobile battery? We have 1275 mL; we want grams: 1275 mL ×= 2350 g mL 1.84 g

34. Chapter 3 34 Critical Thinking: Gasoline The density of gasoline is 730 g/L at 0 ºC (32 ºF) and 713 g/L at 25 ºC (77 ºF). What is the mass difference of 1.00 gallon of gasoline at these two temperatures (1 gal = 3.784L)? The difference is about 60 grams (about 2 %). = 2760 g× At 0 ºC: 1.00 gal × 730 g L 3.784 L 1 gal = 2700 g×At 25 ºC: 1.00 gal × 713 g L 3.784 L 1 gal

35. Chapter 3 35 Temperature Temperature is a measure of the average kinetic energy of the individual particles in a sample. There are three temperature scales: –Celsius –Fahrenheit –Kelvin Kelvin is the absolute temperature scale.

36. Chapter 3 36 Temperature Scales On the Fahrenheit scale, water freezes at 32 °F and boils at 212 °F. On the Celsius scale, water freezes at 0 °C and boils at 100 °C. These are the reference points for the Celsius scale. Water freezes at 273K and boils at 373K on the Kelvin scale.

37. Chapter 3 37 This is the equation for converting °C to °F. This is the equation for converting °F to °C. To convert from °C to K, add 273. °C + 273 = K Temperature Conversions = °F °C × 100°C 180°F ( ) + 32 ( ) 180°F 100°C = °C (°F - 32°F) ×

38. Chapter 3 38 Fahrenheit-Celsius Conversions Body temperature is 98.6 °F. What is body temperature in degrees Celsius? In Kelvin? K = °C + 273 = 37.0 °C + 273 = 310 K ( ) 180°F 100°C = 37.0°C (98.6°F - 32°F) ×

39. Chapter 3 39 Heat Heat is the flow of energy from an object of higher temperature to an object of lower temperature. Heat measures the total energy of a system. Temperature measures the average energy of particles in a system. Heat is often expressed in terms of joules (J) or calories (cal).

40. Chapter 3 40 Heat vs. Temperature Although both beakers below have the same temperature (100 ºC), the beaker on the right has twice the amount of heat, because it has twice the amount of water.

41. Chapter 3 41 Specific Heat The specific heat of a substance is the amount of heat required to bring about a change in temperature. It is expressed with units of calories per gram per degree Celsius. The larger the specific heat, the more heat is required to raise the temperature of the substance.

42. Chapter 3 42 Chapter Summary The basic units in the metric system are grams for mass, liters for volume, and meters for distance. The base units are modified using prefixes to reduce or enlarge the base units by factors of 10. We can use unit factors to convert between metric units. We can convert between metric and English units using unit factors.

43. Chapter 3 43 Chapter Summary, continued Volume is defined as length × width × thickness. Volume can also be determined by displacement of water. Density is mass divided by volume.

44. Chapter 3 44 Chapter Summary, continued Temperature is a measure of the average energy of the particles in a sample. Heat is a measure of the total energy of a substance. Specific heat is a measure of how much heat is required to raise the temperature of a substance.