Introductory Chemistry, 3rd Edition Nivaldo Tro Chapter 2 Measurement and Problem Solving Graph of global Temperature rise in 20th Century. Cover page Opposite page 11. Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2009, Prentice Hall Outline 2.1 Measuring Global Temperature 2.2 Scientific Notation: Writing Large and Small Numbers 2.3 Significant Figures: Writing Numbers to Reflect Precision 2.4 Significant Figures in Calculations 2.5 The Basic Units of Measure 2.6 Converting One Unit to Another 2.7 Solving Multistep Conversions 2.8 Units Raised to a Power 2.9 Density 2.10 Numerical Problem Solving Strategies and the Solution Map Tro's "Introductory Chemistry", Chapter 2 2 2.1 Measuring Global Temperature Tro's "Introductory Chemistry", Chapter 2 3 Scientists have Measured the Average Global Temperature Rise over the Past Century to be 0.6 °C • 0.6 – a number, quantitative: between 0.5 and 0.7 °C. • °C – unit of a standardized scale. • 0.60 °C - indicates a number between 0.59 and 0.61°C • Number + unit = a measurement Tro's "Introductory Chemistry", Chapter 2 4 • • • • • Chapter 2 -Measurement and Problem Solving Topics in this Chapter. Measurements Precision Basic Units Converting Measurements Derived Units Tro's "Introductory Chemistry", Chapter 2 5 Scientific Notation A way of writing large and small numbers. A small number : Radius of a hydrogen nucleus is about 0.00000000000005 m (13 zeros between decimal and 5) A large number: Distance to nearest star other than the sun is 24,500,000,000,000 miles 24.5 trillion miles (245 + 11 zeroes) Tro's Introductory Chemistry, Chapter 2 6 Scientific Notation • Expresses a number as a number between 1 and 10 multiplied by a power of 10. • Example: 3.89 x 105 = 389,000 • More compact way of writing large or small numbers • Easier to compare numbers • Easier to determine precision Tro's "Introductory Chemistry", Chapter 2 7 Exponents or Powers of Ten • Positive exponent = larger number • Negative exponent = smaller number • Write out powers 10 = 101 1/10 = 10-1 100 =102 1/100 = 10-2 1,000= 103 1/1,000 = 10-3 10,000 = 104 1/10,000 = 10-4 1 = 100 Tro's "Introductory Chemistry", Chapter 2 8 2.2 Scientific Notation: Writing Large and Small Numbers Tro's "Introductory Chemistry", Chapter 2 9 Scientific Notation • Parts of a number in scientific notation Figure from the center of page 12 showing the parts Of a number written in scientific notation exponent 1.35 x 10-8 exponential part (10) decimal point of number between 1 and 10 Tro's "Introductory Chemistry", Chapter 2 10 Writing Numbers in Scientific Notation for Numbers Greater than 1. 1. Move the decimal to give a number between 1 and 10. Only one digit to the left of the decimal. 2. Decimal from right to left, the number is now smaller than the original number so you must multiply by a number greater than 1. This means increase in the exponent. 3. Example 34,780 move decimal 4 places to left () 4. 34,780 = 3.4780 x 10,000 = 3.4780 x 104 5. Decimal to left, increase exponent, the number of places you moved the decimal point. Tro's "Introductory Chemistry", Chapter 2 11 Our Examples (1) • • • • • A large number: Distance to nearest star other than the sun is 24,500,000,000,000 miles 24.5 trillion miles (245 + 11 zeroes) • To get a number between 1 and 10 the decimal is moved - (left) by 13 places. 2,45 is smaller than the actual number so we need a positive exponent. • 2.45. x 10+13 miles Tro's "Introductory Chemistry", Chapter 2 12 Writing Numbers in Scientific Notation for Numbers Less than 1. 1. Move the decimal to give a number between 1 and 10. Only one digit will be to the left of the decimal. 2. Decimal from left to right, the number is now larger than the original number so you must multiply by a number less than 1. This means decrease in the exponent. 3. Example 0.0067 move decimal 3 places to right () 4. 0.0067 = 6.7 x 1/1,000 = 6.7 x 10-3 5. Decimal to right, decrease exponent, the number of places you moved the decimal point. Tro's "Introductory Chemistry", Chapter 2 13 Our Examples (2) • A small number : • Radius of a hydrogen nucleus is about • 0.00000000000005 m • (13 zeros between decimal and 5) • To get a number between 1 and 10 the decimal is moved - (right) by 14 places. 5. is bigger than the actual number so we need a negative exponent. • 5. x 10-14 m Tro's "Introductory Chemistry", Chapter 2 14 Writing a Number in Scientific Notation, Other Examples Continued • The U.S. population in 2007 was estimated to be 301,786,000 people. Express this number in scientific notation. • 301,786,000 people = 3.01786 x 108 people Tro's "Introductory Chemistry", Chapter 2 15 Example with a Number that Already has a Power of Ten • 458.98 x 105 • Move decimal 4.5898 This is left () 2 places, so exponent increase by 2 (5 +2 = 7) • 4.5898 x 107 Tro's "Introductory Chemistry", Chapter 2 16 Inputting Scientific Notation into a Calculator • Learn how your calculator handles scientific notation. • Usually through a key marked EXP Tro's "Introductory Chemistry", Chapter 2 17 Reporting Measurements • Measurements depend upon the equipment being used. • All the digits except the last are certain. • The last is an estimate. 24 25 cm 24.7? cm ? Is a guess 24.72cm 3 certain one digits + guess = four significant digits Range is 24.71 to 24.73 cm Tro's "Introductory Chemistry", Chapter 2 18 2.3 Significant Figures: Writing Numbers to Reflect Precision Tro's "Introductory Chemistry", Chapter 2 19 Significant Figures Writing numbers to reflect precision. Why do we need significant figures? All measures have some degree of uncertainty. Tro's "Introductory Chemistry", Chapter 2 20 Precision Increased by Dividing? Consider traveling 237 miles in 3.7 hours. 237 mi 3.7 h = 64.05405405405 mi/h Tro's "Introductory Chemistry", Chapter 2 21 Significant Figures in a Single Measurement • Nonzero numbers are significant (23.89 cm) • Zeros Interior zeros – always significant (1.5067 in) Trailing zeros after a decimal – always significant (45.900 L) Leading zeros – never significant (0.0000456 m) Trailing zeros before a decimal – ambiguous, avoid by using scientific notation. Tro's "Introductory Chemistry", Chapter 2 22 Exact Numbers have Unlimited Significant figures • Counted numbers (3 trials of an experiment) • Defined numbers (1 in = 2.54 cm) • Numbers as part of a formula (2pr = C) Tro's "Introductory Chemistry", Chapter 2 23 Examples of —Determining the Number of Significant Figures in a Number Tro's "Introductory Chemistry", Chapter 2 24 2.4 Significant Figures in Calculations Tro's "Introductory Chemistry", Chapter 2 25 Multiplication and Division with Significant Figures • The overall answer is limited by the measurement with the fewest number of significant figures. 5.02 × 89,665 × 0.10 = 45.0118 = 45 3 sig. figs. 5.892 4 sig. figs. 5 sig. figs. ÷ 2 sig. figs. 2 sig. figs. 6.10 = 0.96590 = 0.966 3 sig. figs. Tro's "Introductory Chemistry", Chapter 2 3 sig. figs. 26 Rounding • Find the number that is the last significant figure. Look at the number to the right. 1. For 0 to 4, keep the last significant figure the same . 34.8244 to four significant figures 34.82 1. For 5 to 9, add 1 the last significant figure the same. 45.967346 to four significant figures 45.97 Tro's "Introductory Chemistry", Chapter 2 27 Examples of Multiplication and Division Tro's "Introductory Chemistry", Chapter 2 28 Addition and Subtraction with Significant Figures • The overall answer is limited by the measurement with the fewest number of decimal places. 5.74 2 decimal places +0.823 3 decimal places +2.651 3 decimal places 9.214 2 decimal places rounds to 9.21 Tro's "Introductory Chemistry", Chapter 2 29 Both Multiplication/Division and Addition/Subtraction with Significant Figures . • Follow the standard order of operations. Please Excuse My Dear Aunt Sally. n - Keep track of number of significant figures at each step, but round at the end. Example next page. Tro's "Introductory Chemistry", Chapter 2 30 Example 1.6—Perform the Following Calculations to the Correct Number of Significant Figures a) 1.10 0.5120 4.0015 3.4555 b) For addition and subtraction, it is useful to draw a line To help determine the correct decimal place. 0.355 105.1 100.5820 Tro's "Introductory Chemistry", Chapter 2 31 Addition or Subtraction and Multiplication and Division Together Do the steps in parentheses and apply the rules to determine the Correct number of significant figures or decimal p[laces in the Intermediate answer. Keep track of significant figures and only round at the last step. Tro's "Introductory Chemistry", Chapter 2 32 Both Multiplication/Division and Addition/Subtraction with Significant Figures (cont.) 12.95 × (12.96 – 11.346) 2 right 3 right decimal dcecimal 12.95 × (1.614) 2 right decimal (by subt. rules) 12.95 × (1.614) = 20.9013 20.9 four three three sig. fig sig. fig. sig. fig (by multip. Rules) 33 Example 1.6—Perform the Following Calculations to the Correct Number of Significant Figures, Continued c) 4.562 3.99870 452.6755 452.33 52.79904 53 d) 14.84 0.55 8.02 0.142 0.1 Tro's "Introductory Chemistry", Chapter 2 34 2.5 The Basic Units of Measure Tro's "Introductory Chemistry", Chapter 2 35 Measurement = Number + Unit • Units - agreed upon standardized quantities • Scientists use the SI system which is a form of the metric system • SI = Système International • Base units are the units for fundamental quantities Tro's "Introductory Chemistry", Chapter 2 36 Base Units in the SI System Quantity Length Mass Time Temperature Unit meter kilogram second kelvin Tro's "Introductory Chemistry", Chapter 2 Symbol m kg s K 37 Length • Measure of a 1-D distance • SI unit = meter = 39.37 in, 3.37 in longer than a yard • Commonly used centimeters (cm) = 1/100 of a meter. 1 inch = 2.54 cm (exactly, by definition) A figure showing a meterstick slightly longer than a yardstick 38 Mass • Measure of the amount of matter present in an object. • SI unit = kilogram (kg) = 2.2 lb • A replica of the standard mass is shown • Commonly measure in the lab mass in grams (g) or milligrams (mg). • 1 kg = 1,000 g = 1,000,000 mg • 1 oz = 23.35 g • 1 lb = 453.6 g • A nickel (5 cent piece) weighs about 5 g Tro's "Introductory Chemistry", Chapter 2 Standard Mass Figure 2.6 Top loading Balance Page 23 39 Time • Measure of the duration of an event. • SI units = second (s) • 1 s is as a certain number of repeats of a Cs atom in a so called atomic clock. Tro's "Introductory Chemistry", Chapter 2 40 Temperature • Measure of the average energy of motion of molecules in a sample. • Is a result of molecular motions • Faster molecular motions = higher temperature • Heat flows from warmer objects to cooler objects through molecular collisions until they reach the same temperature. • Unit is Kelvin (K). Water boils at 373.15 K, and freezes at 272.15Tro's K"Introductory Chemistry", Chapter 2 Three figures Student in 22 °C Room temp. Ice at 0 °C Death Valley at 40 ° C 41 Nonbase Units in the SI System • Units in SI: Prefix (a multiplier) + base unit Example centimeter = centi + meter = centimeter 0.01 x meter = 0.01 meter Tro's "Introductory Chemistry", Chapter 2 42 Common Prefixes in the SI System (see the stuff to memorize on the web page) Prefix Symbol Decimal Equivalent Power of 10 1,000,000 Base x 106 1,000 Base x 103 mega- M kilo- k deci- d 0.1 Base x 10-1 centi- c 0.01 Base x 10-2 milli- m 0.001 Base x 10-3 micro- m or mc 0.000 001 Base x 10-6 nano- n 0.000 000 001 Base x 10-9 Tro's "Introductory Chemistry", Chapter 2 43 Prefixes Indicate What Power of 10 the Base Unit has been Multipied by • Kilometer = kilo (1000) + meter = 1000 x meter (km) • Nanosecond = nano (10-9) + second = 10-9 x second (ns) • Milligram = milli (10-3) + gram = 10-3 x gram (mg) • If the multiplier is greater than 1 the new unit will be larger than the base unit. • If the multiplier is less than 1 the new unit will be smaller than the base unit. Tro's "Introductory Chemistry", Chapter 2 44 Volume, an Example of a Derived Unit • Derived unit. A combination of the base units. • This may be a multiplication, division, addition, subtraction or other mathimatical combination A cube 10 cm On a side which Is the volume = 1 Liter • Examples Area = length2 Volume = length3 • SI system = Area = square meter (m2) Volume = cubic meter (m3) Graduated Cylinder Tro's "Introductory Chemistry", Chapter 2 45 More Convenient Volume Units • Liter (L) is defined as a cubic decimeter (dm3). • A dm = 10 cm so a Liter is a cube with 10 cm sides. • 10 cm x 10 cm x 10 cm = 1000 cm3 • So a Liter also equals 1000 cm3 • Milliliter = 1/1000 L = 1 cm3 • REMEMBER: 1 mL = 1 cm3 • Solids usually use cm3 and liquids usually use mL Tro's "Introductory Chemistry", Chapter 2 46 Table 2.3 (page 24) Common Units and Their Equivalents Length 1 kilometer (km) 1 meter (m) 1 meter (m) 1 foot (ft) 1 inch (in.) = = = = = 0.6214 mile (mi) 39.37 inches (in.) 1.094 yards (yd) 30.48 centimeters (cm) 2.54 centimeters (cm) exactly Tro's "Introductory Chemistry", Chapter 2 47 Table 2.3 (cont.) Common Units and Their Equivalents, Continued Mass 1 kilogram (km) = 2.205 pounds (lb) 1 pound (lb) = 453.59 grams (g) 1 ounce (oz) = 28.35 (g) Volume 1 liter (L) 1 liter (L) 1 liter (L) 1 U.S. gallon (gal) = = = = 1000 milliliters (mL) 1000 cubic centimeters (cm3) 1.057 quarts (qt) 3.785 liters (L) Tro's "Introductory Chemistry", Chapter 2 48 2.6 Converting One Unit to Another •Metric metric (easy, a power of 10) •Metric English (a little harder, a conversion factor) •Arithmetic operations apply to numbers and units •Keep track of units by dimensional analysis Tro's "Introductory Chemistry", Chapter 2 49 Dimensional Analysis • Using units to help solve problems is called dimensional analysis • Units may be operated on mathematically like numbers Tro's "Introductory Chemistry", Chapter 2 50 SI to SI Conversions • This involves a change in the unit that differs by a power of 10. The number will change by moving the decimal. • A way to guide you through the problem is a plan that your text calls solution map. • A solution map lists the steps one mus go through to solve the problem Tro's "Introductory Chemistry", Chapter 2 51 Example SI to SI • How many mm are there in 34.9 m? • Solution map: m mm • How do we make the change? Using a conversion factor Tro's "Introductory Chemistry", Chapter 2 52 Conversion Factor • A conversion factor is derived from an equation. It is a fraction = 1 • Multiplying by a conversion factor does not change a quantities size, but does change the unit used to measure it. • Example from the table 1 mm = 0.001 m • 1mm 0.001 m 0.001 m 1mm Tro's "Introductory Chemistry", Chapter 2 53 Conversion Factors (cont) • Notice the equation gives two conversion factors. • The one to choose is determined by which unit is supposed top cancel out • Back to our problem 34.9 m is how many mm? Tro's "Introductory Chemistry", Chapter 2 54 Conversions Convert an Old Unit to a New Unit new unit • old unit × = new unit old unit • In the solution map • Solution map: m mm • m = old unit and mm = new unit 55 Carrying out the Solution 1 mm • 34.9 m × = 34,900 mm 0.001 m 3.49 x 104 mm • Writing in scientific notation to give 3 sig. fig. Tro's "Introductory Chemistry", Chapter 2 56 Problem to Work in Class • How many kg are there in 4,392 mm? • Solution map: mm m km • Could be done in one step, but it is easy to convert to the base unit. Tro's "Introductory Chemistry", Chapter 2 57 SI to Other Systems of Measure • A piece of notebook paper is 8.50 in wide, what is that length in cm? • Solution map: in cm • Need a conversion factor from the definition 1 in = 2.54 cm • 1in or 2.54 cm 2.54 cm 1in Tro's "Introductory Chemistry", Chapter 2 58 Solution • 8.50 in × 2.54 cm 1 in = 21.59 cm 21.6 cm ( 3 sig. fig. ) Tro's "Introductory Chemistry", Chapter 2 59 2.7 Solving Multistep Conversions Tro's "Introductory Chemistry", Chapter 2 60 Some Conversions Require Two or More Conversion Factors • Given that 2 Tablespoons (tbsp) = 1 oz, 8 oz = 1 cup, 4 cup = 1 qt, and 1 L = 1.097 qt • How many mL are there in 1.00 tbsp? • Solution map: tbsp oz cupqtLmL • Each step represents a conversion. 5 steps. (Solution next slide) Tro's "Introductory Chemistry", Chapter 2 61 • 1.00 tbsp × • • • • 1 oz = 0.500 oz 2 tbsp 0.500 oz × 1 cup = 0.0625 cup 8 oz 0.0625 cup × 1 qt = 0.015625 qt 4 in 0.015625 qt × 1 L = 0.0142339 L 1.097 in 0.0142339 L × 1000 mL = 14.2339 mL 1 L 14.2 mL Tro's "Introductory Chemistry", Chapter 2 62 2.8 Units Raised to a Power Tro's "Introductory Chemistry", Chapter 2 63 Units Raised to a Power • An injection of a drug has a volume of 4.52 cm3. What is the volume in cubic inches (in3)? • Solution map: cm3 in3 • 4.52 cm3 × 1 in = 1.78 in3 (NO!!!!) 2.54 cm • What does cm3 really mean Tro's "Introductory Chemistry", Chapter 2 64 Units Raised to a Power • 4.52 cm3 = 4.53 cm × cm × cm • For squared or cubed units, one must also square or cube the entire conversion factor • 4.52 cm3 × 1 in × 1 in × 1 in = 2.54 cm 2.54 cm 2.54 cm or 3 • 4.52 cm3 × 1 in = 4.52 cm3× 13 in3 = 2.54cm (2.54) 3cm3 Tro's "Introductory Chemistry", Chapter 2 65 Units Raised to a Power • 4.52 cm3 × 1 in3 = 16.387 cm3 Tro's "Introductory Chemistry", Chapter 2 0.276 in3 66 2.9 Density Tro's "Introductory Chemistry", Chapter 2 67 Mass and Volume • Two main characteristics of matter. • Mass or volume alone cannot be used to identify what type of matter something is. If you are given a large glass containing 100 g of a clear, colorless liquid and a small glass containing 25 g of a clear, colorless liquid, are both liquids the same stuff? • Even though mass and volume are individual properties, for a given type of matter they are related to each other! Tro's "Introductory Chemistry", Chapter 2 68 Mass vs. Volume of Brass Mass grams Volume cm3 20 2.4 32 3.8 40 4.8 50 6.0 100 11.9 150 17.9 Tro's "Introductory Chemistry", Chapter 2 69 This slide was a graph of the mass as y and the volume as x resulting in a straight line this indicates that the ratio of mass to volume is a constant = the density of a substance For these brass samples d = 8.3 g/mL Tro's "Introductory Chemistry", Chapter 2 70 Density • Ratio of mass:volume. • Its value depends on the kind of material, not the amount. • Solids = g/cm3 1 cm3 = 1 mL Mass Density Volume • Liquids = g/mL • Gases = g/L • Volume of a solid can be determined by water displacement—Archimedes Principle. Explain • Density : solids > liquids > gases Except ice is less dense than liquid water! Tro's "Introductory Chemistry", Chapter 2 71 For a Table of Sample Densities see Table 2.4 on page 33 Tro's "Introductory Chemistry", Chapter 2 72 Volume by Water Displacement • • • • Graduated cylinder – water is added = Vinitial A solid object is added to the cylinder. Volume of water rises = Vfinal Vobject = Vfinal - Vinitial Tro's "Introductory Chemistry", Chapter 2 73 Using Density in Calculations Three Possible Formulas Solution Maps: Mass Density Volume m, V D Mass Volume Density m, D V V, D m Mass Density Volume Tro's "Introductory Chemistry", Chapter 2 74 Density Problems •Think of density as a conversion factor between mass and volume of a given substance. •Problems from text for class 2.96, 2.100, Tro's "Introductory Chemistry", Chapter 2 75 Problem 2.96 An aluminum engine block has a volume of 4.77 L and a mass of 12.88 kg. What is the density of the aluminum in grams per cubic centimeter? Solution map Volume, L Volume, mL=cm3 mass Density = Mass, kg mass, g Tro's "Introductory Chemistry", Chapter 2 volume 76 Problem 2.96 (cont.) Volume to cm3: 4.77 L × 1000 mL 1L Mass to g: 12.88 kg × = 4,770 mL = 4,770 cm3 1000 g 1 kg = 12,880 g Density using the formula: 12,880 g = 2.7002096 g/cm3 2.70 g/cm3 4,770 cm3 3 sig. fig. Tro's "Introductory Chemistry", Chapter 2 77 Problem 2.100 Acetone (fingernail polish remover) has a density of 0.7857 g/cm3. (a) What is the mass in grams of 28.56 mL of acetone? (b) What is the volume in milliliters of 6.54 g of acetone? There are two basic ways to work this problem. 1)Plug the variables into the formula for density and solve for the unknown. Or 2) Treat the density as a conversion factor between mass and volume of a substance. We will use # 2 Solution maps (a) ml g (b) g mL Tro's "Introductory Chemistry", Chapter 2 78 Problem 2.100 (cont.) A density of 0.7857 g/cm3 means that 1 cm3 or mL of the substance Has a mass of 0.7857 g. (i.e. 1 mL = 0.7857 g of acetone) From this equation we can derive conversion factors. (a) The conversion is from mL to g so we want mL to cancel 28.56 mL × 0.7857g = 22.44 g (4 sig. fig.) 1 mL (a) The conversion is from g to mL so we want g to cancel 6.54 g × 1 mL = 8.32 g (3 sig. fig.) 0.7857g Tro's "Introductory Chemistry", Chapter 2 79
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