1. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-1 Keys to the Study of Chemistry Chapter 1

2. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-2 1.1 Some Fundamental Definitions 1.2 The Scientific Approach: Developing a Model 1.3 Chemical Problem Solving 1.4 Measurement in Scientific Study 1.5 Uncertainty in Measurement: Significant Figures Chapter 1 : Keys to the Study of Chemistry

3. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-3 Is the study of matter, its properties, the changes that matter undergoes, and the energy associated with these changes.

4. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-4 Definitions Chemical Properties those which the substance shows as it interacts with, or transforms into, other substances such as flammability, corrosiveness Matter anything that has mass and volume -the “stuff” of the universe: books, planets, trees, professors, students Composition the types and amounts of simpler substances that make up a sample of matter Properties the characteristics that give each substance a unique identity Physical Properties those which the substance shows by itself without interacting with another substance such as color, melting point, boiling point, density

5. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-5 Figure 1.1 A Physical change B Chemical change The distinction between physical and chemical change.

6. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-6 Figure 1.2 The physical states of matter.

7. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-7 Sample Problem 1.1Distinguishing Between Physical and Chemical Change PROBLEM:Decide whether each of the following process is primarily a physical or a chemical change, and explain briefly: PLAN:“Does the substance change composition or just change form?” SOLUTION: (a) Frost forms as the temperature drops on a humid winter night. (b) A cornstalk grows from a seed that is watered and fertilized. (c) Dynamite explodes to form a mixture of gases. (d) Perspiration evaporates when you relax after jogging. (e) A silver fork tarnishes slowly in air. (a) physical change(b) chemical change(c) chemical change (d) physical change(e) chemical change

8. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-8 energy due to the position of the object or energy from a chemical reaction Potential Energy Kinetic Energy energy due to the motion of the object Energy Energy is the capacity to do work. Potential and kinetic energy can be interconverted.

9. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-9 Energy Energy is the capacity to do work. Figure 1.3A less stable more stable change in potential energy EQUALS kinetic energy A gravitational system. The potential energy gained when a lifted weight is converted to kinetic energy as the weight falls.

10. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-10 Energy Energy is the capacity to do work. Figure 1.3B less stable more stable change in potential energy EQUALS kinetic energy A system of two balls attached by a spring. The potential energy gained by a stretched spring is converted to kinetic energy when the moving balls are released.

11. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-11 Energy Energy is the capacity to do work. Figure 1.3C less stable more stable change in potential energy EQUALS kinetic energy A system of oppositely charged particles. The potential energy gained when the charges are separated is converted to kinetic energy as the attraction pulls these charges together.

12. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-12 Energy Energy is the capacity to do work. Figure 1.3D less stable more stable change in potential energy EQUALS kinetic energy A system of fuel and exhaust. A fuel is higher in chemical potential energy than the exhaust. As the fuel burns, some of its potential energy is converted to the kinetic energy of the moving car.

13. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-13 Scientific Approach: Developing a Model Observations : Natural phenomena and measured events; universally consistent ones can be stated as a natural law. Hypothesis: Tentative proposal that explains observations. Experiment: Procedure to test hypothesis; measures one variable at a time. Model (Theory): Set of conceptual assumptions that explains data from accumulated experiments; predicts related phenomena. Further Experiment: Tests predictions based on model. revised if experiments do not support it altered if predictions do not support it

14. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-14 A Systematic Approach to Solving Chemistry Problems Problem statement Plan Clarify the known and unknown. Suggest steps from known to unknown. Prepare a visual summary of steps. Solution Check Comment and Follow-up Problem

15. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-15 Sample Problem 1.2Converting Units of Length PROBLEM:To wire your stereo equipment, you need 325 centimeters (cm) of speaker wire that sells for $0.15/ft. What is the price of the wire? PLAN:Known - length (in cm) of wire and cost per length ($/ft) We have to convert cm to inches and inches to ft followed by finding the cost for the length in ft. SOLUTION: length (cm) of wire length (ft) of wire length (in) of wire Price ($) of wire 2.54 cm = 1 in 12 in = 1 ft 1 ft = $0.15 Length (in) = length (cm) x conversion factor = 325 cm xin 2.54 cm = 128 in Length (ft) = length (in) x conversion factor = 128 in xft 12 in = 10.7 ft Price ($) = length (ft) x conversion factor = 10.7 ft x$0.15 ft = $1.60

16. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-16 Table 1. 1 SI Base Units Physical Quantity (Dimension) Unit Name Unit Abbreviation mass meter kg length kilogram m timeseconds temperaturekelvinK electric currentampereA amount of substancemolemol luminous intensitycandelacd

17. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-17 Common Decimal Prefixes Used with SI Units Table 1.2

18. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-18 English Equivalent Length 1 kilometer(km) 1000(10 3 )m 0.62mi Common SI-English Equivalent Quantities Table 1.3 QuantitySI Unit SI Equivalent English to SI Equivalent 1 kilometer(km) 1000(10 3 )m 0.62miles(mi) 1 mi = 1.61km 1 meter(m) 100(10 2 )cm 1.094yards(yd) 1000(10 3 )mm 39.37inches(in) 1 yd = 0.9144m 1 foot (ft) = 0.3048m 1 centimeter(cm) 0.01(10 -2 )m 0.3937in 1 in = 2.54cm (exactly!)

19. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-19 Volume 1,000,000(10 6 ) cubic centimeters 35.2cubic feet (ft 3 ) 1 cubic meter(m 3 ) 1 ft 3 = 0.0283m 3 1 cubic decimeter(dm 3 ) 1000cm 3 0.2642 gallon (gal) 1.057 quarts (qt) 1 gal = 3.785 dm 3 1 qt = 0.9464 dm 3 1 cubic centimeter (cm 3 ) 0.001 dm 3 0.0338 fluid ounce 1 qt = 946.4 cm 3 1 fluid ounce = 29.6 cm 3 English Equivalent QuantitySI Unit SI Equivalent English to SI Equivalent Common SI-English Equivalent Quantities Table 1.3

20. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-20 English Equivalent QuantitySI Unit SI Equivalent English to SI Equivalent Mass 1 kilogram (kg) 1000 grams 2,205 pounds (lb) 1 (lb) = 0.4536 kg 1 gram (g) 1000 milligrams 0.03527 ounce(oz) 1 lb = 453.6 g 1 ounce = 28.35 g Common SI-English Equivalent Quantities Table 1.3

21. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-21 Sample Problem 1.3Converting Units of Volume PROBLEM:The volume of an irregularly shaped solid can be determined from the volume of water it displaces. A graduated cylinder contains 19.9mL of water. When a small piece of galena, an ore of lead, is submerged in the water, the volume increases to 24.5mL. What is the volume of the piece of galena in cm 3 and in L? PLAN:The volume of galena is equal to the change in the water volume before and after submerging the solid. volume (mL) before and after addition volume (mL) of galena volume (cm 3 ) of galena subtract volume (L) of galena 1 mL = 1 cm 3 1 mL = 10 -3 L SOLUTION: (24.5 - 19.9)mL = volume of galena 4.6 mL x mL 1 cm 3 = 4.6 cm 3 4.6 mL x mL 10 -3 L = 4.6x10 -3 L

22. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-22 Sample Problem 1.4Converting Units of Mass PROBLEM:International computer communications are often carried by optical fibers in cables laid along the ocean floor. If one strand of optical fiber weighs 1.19 x 10 -3 lbs/m, what is the total mass (in kg) of a cable made of six strands of optical fiber, each long enough to link New York and Paris (8.84 x 10 3 km)? PLAN:The sequence of steps may vary but essentially you have to find the length of the entire cable and convert it to mass. length (km) of fiber mass (lb) of fiber length (m) of fiber 1 km = 10 3 m mass (kg) of cablemass (lb) of cable 6 fibers = 1 cable1 m = 1.19x10 -3 lb 2.205 lb = 1 kg SOLUTION: 8.84 x 10 3 km x km 10 3 m 8.84 x 10 6 m x m 1.19 x 10 -3 lbs 1.05 x 10 4 lb x cable 6 fibers = 1.05 x 10 4 lb = 8.84 x 10 6 m 2.205 lb 1kg x 6.30x 10 4 lb cable 6.30x 10 4 lb = cable 2.86x10 4 kg cable =

23. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-23 Sample Problem 1.5Calculating Density from Mass and Length PROBLEM:Lithium (Li) is a soft, gray solid that has the lowest density of any metal. If a slab of Li weighs 1.49 x 10 3 mg and has sides that measure 20.9 mm by 11.1 mm by 11.9 mm, what is the density of Li in g/cm 3 ? PLAN:Density is expressed in g/cm 3 so we need the mass in grams and the volume in cm 3. mass (mg) of Li lengths (mm) of sides mass (g) of Li density (g/cm 3 ) of Li 10 3 mg = 1 g 10 mm = 1 cm lengths (cm) of sides volume (cm 3 ) multiply lengths SOLUTION: 20.9mm x 1mg 10 -3 g = 1.49g1.49x10 3 mg x 10mm 1cm = 2.09cm Similarly the other sides will be 1.11 cm and 1.19 cm, respectively. 2.09 x 1.11 x 1.20 = 2.76cm 3 density of Li = 1.49g 2.76 cm 3 = 0.540 g/cm 3

24. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-24 Figure 1.8 The freezing and boiling points of water.

25. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-25 Temperature Scales and Interconversions Kelvin ( K ) - The “Absolute temperature scale” begins at absolute zero and only has positive values. Celsius ( o C ) - The temperature scale used by science, formally called centigrade, most commonly used scale around the world; water freezes at 0 o C, and boils at 100 o C. Fahrenheit ( o F ) - Commonly used scale in the U.S. for our weather reports; water freezes at 32 o F and boils at 212 o F. T (in K) = T (in o C) + 273.15 T (in o C) = T (in K) - 273.15 T (in o F) = 9/5 T (in o C) + 32 T (in o C) = [ T (in o F) - 32 ] 5/9

26. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-26 Sample Problem 1.6Converting Units of Temperature PROBLEM:A child has a body temperature of 38.7 0 C. PLAN:We have to convert 0 C to 0 F to find out if the child has a fever and we use the 0 C to kelvin relationship to find the temperature in kelvins. (a) If normal body temperature is 98.6 0 F, does the child have a fever? (b) What is the child’s temperature in kelvins? SOLUTION: (a) Converting from 0 C to 0 F 9 5 (38.7 0 C) + 32 = 101.7 0 F (b) Converting from 0 C to K38.7 0 C + 273.15 = 311.8K

27. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-27 The number of significant figures in a measurement depends upon the measuring device. Figure 1.9A 32.3 0 C32.33 0 C

28. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-28 Rules for Determining Which Digits are Significant All digits are significant Make sure that the measured quantity has a decimal point. Start at the left of the number and move right until you reach the first nonzero digit. Count that digit and every digit to it’s right as significant. Numbers such as 5300 L are assumed to only have 2 significant figures. A terminal decimal point is often used to clarify the situation, but scientific notation is the best! except zeros that are used only to position the decimal point. Zeros that end a number and lie either after or before the decimal point are significant; thus 1.030 ml has four significant figures, and 5300. L has four significant figures also.

29. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-29 Sample Problem 1.7Determining the Number of Significant Figures PROBLEM:For each of the following quantities, underline the zeros that are significant figures(sf), and determine the number of significant figures in each quantity. For (d) to (f), express each in exponential notation first. PLAN:Determine the number of sf by counting digits and paying attention to the placement of zeros. SOLUTION: (b) 0.1044 g(a) 0.0030 L(c) 53,069 mL (e) 57,600. s(d) 0.00004715 m (f) 0.0000007160 cm 3 (b) 0.1044 g(a) 0.0030 L(c) 53.069 mL (e) 57,600. s(d) 0.00004715 m (f) 0.0000007160 cm 3 2sf4sf5sf (d) 4.715x10 -5 m4sf(e) 5.7600x10 4 s5sf (f) 7.160x10 -7 cm 3 4sf

30. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-30 Rules for Significant Figures in Answers 1. For addition and subtraction. The answer has the same number of decimal places as there are in the measurement with the fewest decimal places. 106.78 mL = 106.8 mL Example: subtracting two volumes 863.0879 mL = 863.1 mL 865.9 mL - 2.8121 mL Example: adding two volumes83.5 mL + 23.28 mL

31. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-31 = 23.4225 cm 3 = 23 cm 3 9.2 cm x 6.8 cm x 0.3744 cm 2. For multiplication and division. The number with the least certainty limits the certainty of the result. Therefore, the answer contains the same number of significant figures as there are in the measurement with the fewest significant figures. Rules for Significant Figures in Answers Multiply the following numbers:

32. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-32 Rules for Rounding Off Numbers 1. If the digit removed is more than 5, the preceding number increases by 1. 5.379 rounds to 5.38 if three significant figures are retained and to 5.4 if two significant figures are retained. 2. If the digit removed is less than 5, the preceding number is unchanged. 0.2413 rounds to 0.241 if three significant figures are retained and to 0.24 if two significant figures are retained. 3.If the digit removed is 5, the preceding number increases by 1 if it is odd and remains unchanged if it is even. 17.75 rounds to 17.8, but 17.65 rounds to 17.6. If the 5 is followed only by zeros, rule 3 is followed; if the 5 is followed by nonzeros, rule 1 is followed: 17.6500 rounds to 17.6, but 17.6513 rounds to 17.7 4. Be sure to carry two or more additional significant figures through a multistep calculation and round off only the final answer.

33. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-33 Issues Concerning Significant Figures graduated cylinder < buret ≤ pipet numbers with no uncertainty 1000 mg = 1 g 60 min = 1 hr These have as many significant digits as the calculation requires. be sure to correlate with the problem FIX function on some calculators Electronic Calculators Choice of Measuring Device Exact Numbers

34. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-34 Sample Problem 1.8Significant Figures and Rounding PROBLEM:Perform the following calculations and round the answer to the correct number of significant figures: PLAN:In (a) we subtract before we divide; for (b) we are using an exact number. SOLUTION: 7.085 cm 16.3521 cm 2 - 1.448 cm 2 (a) 11.55 cm 3 4.80x10 4 mg (b) 1 g 1000 mg 7.085 cm 16.3521 cm 2 - 1.448 cm 2 (a) = 7.085 cm 14.904 cm 2 = 2.104 cm 11.55 cm 3 4.80x10 4 mg (b) 1 g 1000 mg = 48.0 g 11.55 cm 3 = 4.16 g/ cm 3

35. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-35 Precision and Accuracy Errors in Scientific Measurements Random Error - In the absence of systematic error, some values that are higher and some that are lower than the actual value. Precision - Refers to reproducibility or how close the measurements are to each other. Accuracy - Refers to how close a measurement is to the real value. Systematic Error - Values that are either all higher or all lower than the actual value.

36. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-36 Figure 1.10 precise and accurate precise but not accurate Precision and accuracy in the laboratory.

37. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-37 systematic error random error Precision and accuracy in the laboratory. Figure 1.10 continued