1. INTRODUCTORY CHEMISTRY INTRODUCTORY CHEMISTRY Concepts and Critical Thinking Sixth Edition by Charles H. Corwin Chapter 3 1 © 2011 Pearson Education, Inc. Chapter 2 The Metric System

2. 2 Chapter 3 © 2011 Pearson Education, Inc. Metric System Basic Units

3. 3 Chapter 3 © 2011 Pearson Education, Inc. 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.

4. 4 Chapter 3 © 2011 Pearson Education, Inc. 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.

5. 5 Chapter 3 © 2011 Pearson Education, Inc. Metric System Prefixes The following table lists the common prefixes used in the metric system:

6. 6 Chapter 3 © 2011 Pearson Education, Inc. 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: –There are 1000 millimeters in 1 meter.

7. 7 Chapter 3 © 2011 Pearson Education, Inc. 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.

8. 8 Chapter 3 © 2011 Pearson Education, Inc. 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 1000 mm = 1 m.

9. 9 Chapter 3 © 2011 Pearson Education, Inc. 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 1000 mm = 1 m, we can write the following unit factors: 1000 mm or 1 m. 1 m 1000 mm

10. 10 Chapter 3 © 2011 Pearson Education, Inc. 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: 1.Write down the unit asked for in the answer. 2.Write down the given value related to the answer. 3.Apply unit factor(s) to convert the given unit to the units desired in the answer.

11. 11 Chapter 3 © 2011 Pearson Education, Inc. 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 (1000 mg = 1 g) and round to three significant figures. 325 mg x = 0.325 g 1000 mg 1 g

12. 12 Chapter 3 © 2011 Pearson Education, Inc. 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: 1 L and 1000 mL 10 dL 1 L

13. 13 Chapter 3 © 2011 Pearson Education, Inc. Two Metric–Metric Conversions, Continued Apply both unit factors, and round the answer to three significant digits. Notice that both dL and L units cancel, leaving us with units of mL. 125 dL x = 12,500 mL x 10 dL 1 L 1000 mL 1 L

14. 14 Chapter 3 © 2011 Pearson Education, Inc. 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 x 10 22 kg. Convert kilograms to grams, and then grams to megagrams. 7.35 x 10 22 kg × = 5.98 x 10 19 Mg x 1 kg 1000 g 1 Mg 1000000 g

15. 15 Chapter 3 © 2011 Pearson Education, Inc. 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.

16. 16 Chapter 3 © 2011 Pearson Education, Inc. 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 x = 110 m 1 yd 0.914 m

17. 17 Chapter 3 © 2011 Pearson Education, Inc. 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 x = 1,890 mL x 32 fl oz 1 qt 946 mL 1 qt

18. 18 Chapter 3 © 2011 Pearson Education, Inc. 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. 1.First, miles → meters 2.Next, hours → seconds

19. 19 Chapter 3 © 2011 Pearson Education, Inc. 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 x 1 km 1000 m 1 hr 3600 s 75 km hr x

20. 20 Chapter 3 © 2011 Pearson Education, Inc. 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 x w x 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).

21. 21 Chapter 3 © 2011 Pearson Education, Inc. 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.

22. 22 Chapter 3 © 2011 Pearson Education, Inc. Cubic and Liquid Volume Units 1 liter is equal to 1000 cubic centimeters. –10 cm x 10 cm x 10 cm = 1000 cm 3 1000 cm 3 = 1 L = 1000 mL. Therefore, 1 cm 3 = 1 mL.

23. 23 Chapter 3 © 2011 Pearson Education, Inc. 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 x 1 in 2.54 cm x 498 cm 3 x 1 in 2.54 cm 1 in 2.54 cm

24. 24 Chapter 3 © 2011 Pearson Education, Inc. 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.

25. 25 Chapter 3 © 2011 Pearson Education, Inc. 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.

26. 26 Chapter 3 © 2011 Pearson Education, Inc. 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.

27. 27 Chapter 3 © 2011 Pearson Education, Inc. 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

28. 28 Chapter 3 © 2011 Pearson Education, Inc. 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.

29. 29 Chapter 3 © 2011 Pearson Education, Inc. Densities of Common Substances

30. 30 Chapter 3 © 2011 Pearson Education, Inc. Estimating Density We can estimate the density of a substance by comparing it to another object. A solid object will float on top of 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.

31. Critical Thinking Question A glass cylinder contains four liquid layers: mercury (d = 13.6 g/mL), chloroform (d = 1.49 g/mL), water (d = 1.00 g/mL), and ether (d = 0.708 g/mL). If an ice cube (d = 0.92 g/mL) is dropped into the cylinder, where does it come to rest? A) on top of the ether layer B) on top of the water layer C) on top of the chloroform layer D) on top of the mercury layer E) on the bottom of the cylinder 31 Chapter 3 © 2011 Pearson Education, Inc.

32. 32 Chapter 3 © 2011 Pearson Education, Inc. 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. 33 Chapter 3 © 2011 Pearson Education, Inc. 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 x = 2350 g mL 1.84 g

34. 34 Chapter 3 © 2011 Pearson Education, Inc. 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)?

35. 35 Chapter 3 © 2011 Pearson Education, Inc. 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 x At 0 ºC: 1.00 gal x 730 g L 3.784 L 1 gal = 2700 g x At 25 ºC: 1.00 gal x 713 g L 3.784 L 1 gal

36. 36 Chapter 3 © 2011 Pearson Education, Inc. Temperature Temperature is a measure of the average kinetic energy of the individual particles in a sample. There are three temperature scales: 1.Celsius 2.Fahrenheit 3.Kelvin Kelvin is the absolute temperature scale.

37. 37 Chapter 3 © 2011 Pearson Education, Inc. 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 273 K and boils at 373 K on the Kelvin scale.

38. 38 Chapter 3 © 2011 Pearson Education, Inc. Temperature Conversions This is the equation for converting °C to °F. (°C * 9/5) + 32 = °F This is the equation for converting °F to °C. (°F – 32) * 5/9 = °C To convert from °C to K, add 273. °C + 273 = K

39. 39 Chapter 3 © 2011 Pearson Education, Inc. 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) x

40. 40 Chapter 3 © 2011 Pearson Education, Inc. 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).

41. 41 Chapter 3 © 2011 Pearson Education, Inc. Critical Thinking Question Which of the two beakers has higher heat? a)Beaker A b)Beaker B c)Both have the same amount of heat (100

42. 42 Chapter 2 © 2011 Pearson Education, Inc. The Percent Concept A percent, %, expresses the amount of a single quantity compared to an entire sample. A percent is a ratio of parts per 100 parts. The formula for calculating percent is shown below:

43. 43 Chapter 2 © 2011 Pearson Education, Inc. Calculating Percentages Sterling silver contains silver and copper. If a sterling silver chain contains 18.5 g of silver and 1.5 g of copper, what is the percent of silver in sterling silver?

44. 44 Chapter 2 © 2011 Pearson Education, Inc. Percent Unit Factors A percent can be expressed as parts per 100 parts. 25% can be expressed as 25/100 and 10% can be expressed as 10/100. We can use a percent expressed as a ratio as a unit factor. –A rock is 4.70% iron, so

45. 45 Chapter 2 © 2011 Pearson Education, Inc. Percent Unit Factor Calculation The Earth and Moon have a similar composition; each contains 4.70% iron. What is the mass of iron in a lunar sample that weighs 235 g? Step 1: We want g iron. Step 2: We write down the given: 235 g sample. Step 3: We apply a unit factor (4.70 g iron = 100 g sample) and round to three significant figures.

46. 46 Chapter 2 © 2011 Pearson Education, Inc. Chemistry Connection: Coins A nickel coin contains 75.0 % copper metal and 25.0 % nickel metal, and has a mass of 5.00 grams. What is the mass of nickel metal in a nickel coin?