Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Chapter 1 Chemistry.

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Chapter 1 Chemistry and Measurements 1.1 Chemistry and Chemicals 1

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. What Is Chemistry? Chemistry is the study of composition, structure, properties, and reactions of matter happens all around you everyday Antacid tablets undergo a chemical reaction when dropped in water. 2

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Chemistry Matter is another word for all substances that make up our world. Antacid tablets are matter. Water is matter. Glass is matter. Air is matter. 3

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Chemicals Used When Cooking 5 Many substances found in the kitchen are obtained using chemical reactions.

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Units of Measurement Scientists use the metric system of measurement and have adopted a modification of the metric system called the International System of Units as a worldwide standard. International System of Units (SI) is an official system of measurement used throughout the world for units in length, volume, mass, temperature, and time. 7

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Length, Meter (m) and Centimeter (cm) 9 1 m = 100 cm1 m = 1.09 yd 1 m = 39.4 in.2.54 cm = 1 in.

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Volume, Liter (L) and Milliliter (mL) 1 L = 1000 mL 1 L = 1.06 qt 946 mL = 1 qt 10 We use graduated cylinders to measure small volumes.

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Mass, Gram (g) and Kilogram (kg) 1 kg = 1000 g 1 kg = 2.20 lb 454 g = 1 lb The mass of a nickel is 5.01 g on an electronic scale. 11

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Temperature, Celsius ( o C) and Kelvin (K) Water freezes: 32 o F 0 o C The Kelvin scale for temperature begins at the lowest possible temperature, 0 K. A thermometer is used to measure temperature. 12

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Time, Second (s) The second is the correct metric and SI unit for time. The standard measure for 1 s is an atomic clock. A stopwatch is used to measure the time of a race. 13

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Chapter 1 Chemistry and Measurements 1.4 Scientific Notation People have an average of 1 x 10 5 hairs on their scalp. Each hair is about 8 x 10 −6 m wide. 14

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Writing a Number in Scientific Notation Numbers written in scientific notation have three parts: coefficient power of 10 unit To write 2400 m in correct scientific notation: the coefficient is 2.4 the power of 10 is 3 the unit is "m" 15

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Writing a Number in Scientific Notation 2400. m = 2.4 x 1000 = 2.4 x 10 3 m  3 places coefficient x power of 10 unit 0.00086 g = = 8.6 x 10 −4 g 4 places  coefficient x power of 10 unit 16

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Scientific Notation and Calculators Number to enter: 4 x 10 6 Enter:4 EXP (EE) 6 Display: 4 06 or 4 06 4 E06 Number to enter: 2.5 x 10 −4 Enter:2.5 EXP (EE) +/− 4 Display: 2.5 −04 or 2.5 −04 2.5 E−04 20

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Chapter 1 Chemistry and Measurements 1.5 Measured Numbers and Significant Figures 21

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Measured Numbers Measured numbers are the numbers obtained when you measure a quantity such as your height, weight, or temperature. 22

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Writing Measured Numbers To write a measured number, observe the numerical values of marked lines estimate value of number between marks the estimated number is the final number in your measured number 23

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Writing Measured Numbers for Length T he lengths of the objects are measured as (a) 4.5 cm (b) 4.55 cm (c) 3.0 cm 24

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. A Number Is Significant When A number is a significant figure (SF) if it is Example a. not a zero4.5 g 2 SF b. a zero between digits205 m 3 SF c. a zero at the end of a 50. L 2 SF decimal number d. in the coefficient of a4.8 x 10 5 m 2 SF number written in scientific notation 25

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. A Number Is NOT Significant When A number is not significant if it is Example a. at the beginning of a decimal number 0.0004 s 1 SF b. used as a placeholder in a large number without a decimal point 850 000 m 2 SF 26

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Learning Check Identify the significant and nonsignificant zeros in each of the following numbers: A. 0.002 650 m B. 43.026 g C. 1 044 000 L 27

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution Identify the significant and nonsignificant zeros in each of the following numbers: A. 0.002 650 m The zeros preceding 2 are not significant. The digits 2, 6, 5 are significant. The zero in last decimal place is significant. 4 SF B. 43.026 g The zeros between nonzero digits or at the end of decimal numbers are significant. 5 SF 28

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution Identify the significant and nonsignificant zeros in each of the following numbers: C. 1 044 000 L The zeros between nonzero digits are significant. The zeros at end of a number with no decimal are not significant. 4 SF 29

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Exact Numbers Exact numbers are numbers obtained by counting8 cookies in definitions that compare two units 6 eggs in the same measuring system1 qt = 4 cups 1 kg = 1000 g 30

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Chapter 1 Chemistry and Measurements 1.6 Significant Figures in Calculations 31

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Rules for Rounding Off 1. If the first digit to be dropped is 4 or less, then it and all the following digits are dropped from the number. 2. If the first digit to be dropped is 5 or greater, then the last retained digit of the number is increased by 1. 32

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Learning Check Select the correct value when 3.1457 g is rounded to: A. three significant figures B. two significant figures 34

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution Select the correct value when 3.1457 g is rounded to: A. To round 3.1457 to three significant figures, drop the final digits, 57 increase the last remaining digit by 1. The answer is 3.15 g. B. To round 3.1457 g to two significant figures, drop the final digits 457. do not increase the last number by 1 since the first of these digits is 4. The answer is 3.1 g. 35

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Rules for Multiplication and Division In multiplication or division, the final answer is written so it has the same number of significant figures as the measurement with the fewest significant figures (SFs). Example 1: Multiply the following measured numbers: 24.66 cm x 0.35 cm = 8.631 (calculator display) = 8.6 cm 2 (2 significant figures) Multiplying 4 SFs by 2 SFs gives us an answer with 2 SFs. 36

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Multiplication and Division with SFs Example 2: Multiply and divide the following measured numbers: 21.5 cm x 0.30 cm = 1.88 cm Put the following into your calculator: 21.5 x 0.30 ÷ 1.88 = 3.430851063 = 3.4 cm (2 significant figures) Multiplying 4 SFs by 2 SFs gives us an answer with 2 SFs. 37

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Multiplication and Division with SFs Example 3: Multiply and divide the following measured numbers: 6.0 g = 2.00 g Put the following into your calculator: 6.0 ÷ 2.00 = 3 (calculator display) = 3.0 g (2 significant figures) Add one zero to give 2 significant figures. 38

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Learning Check Perform the following calculation of measured numbers. Give the answer in the correct number of significant figures. 5.00 cm x 3.408 cm = 2.00 cm 39

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution Perform the following calculation of measured numbers. Give the answer in the correct number of significant figures. 5.00 cm x 3.408 cm = 2.00 cm (3 SF x 4 SF ÷ 3 SF ) = 8.52 cm calculator display and correct significant figures. 40

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Addition and Subtraction In addition or subtraction, the final answer is written so it has the same number of decimal places as the measurement with the fewest decimal places. Example 1: Add the following measured numbers: 2.012 g three decimal places 61.09 g two decimal places + 3.0 g one decimal place 66.102 g (calculator display) = 66.1 g answer rounded to one decimal place 41

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Addition and Subtraction with SFs Example 2: Subtract the following measured numbers: 65.09 g two decimal places − 3.0 g one decimal place 62.09 g (calculator display) = 62.1 g answer rounded to one decimal place 42

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Chapter 1 Chemistry and Measurements 1.7 Prefixes and Equalities 43

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Prefixes A special feature of the SI as well as the metric system is that a prefix can be placed in front of any unit to increase or decrease its size by some factor of ten. For example, the prefixes milli and micro are used to make the smaller units: milligram (mg) microgram (μg) 44

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Prefixes and Equalities The relationship of a prefix to a unit can be expressed by replacing the prefix with its numerical value. For example, when the prefix kilo in kilometer is replaced with its value of 1000, we find that a kilometer is equal to 1000 meters. kilometer=1000 meters kilogram=1000 grams 45

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Learning Check Fill in the blanks with the correct prefix: A. 1000 m = 1 ___m B. 1 x 10 −3 g = 1 ___g C. 0.01 m = 1 ___m 48

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution Fill in the blanks with the correct prefix: A. 1000 m = 1 ___m The prefix for 1000 is kilo; 1000 m = 1 km B. 1 x 10 −3 g = 1 ___g The prefix for 1 x 10 −3 is milli; 1 x 10 −3 g = 1 mg C. 0.01 m = 1 ___m The prefix for 0.01 is centi; 0.01 m = 1 cm 49

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Measuring Length Each of the following equalities describes the same length in a different unit. 1 m = 100 cm = 1 x 10 2 cm 1 m = 1000 mm = 1 x 10 3 mm 1 cm = 10 mm = 1 x 10 1 mm 50

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Measuring Length The metric length of 1 meter is the same as 10 dm, 100 cm, or 1000 mm. 51

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Measuring Volume Examples of Some Volume Equalities 1 L= 10 dL = 1 x 10 1 dL 1 L = 1000 mL = 1 x 10 3 mL 1 dL = 100 mL = 1 x 10 2 mL 52

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. The Cubic Centimeter The cubic centimeter (abbreviated as cm 3 or cc) is the volume of a cube whose dimensions are 1 cm on each side. A cubic centimeter has the same volume as a milliliter, and the units are often used interchangeably. 1 cm 3 = 1 cc = 1 mL 53

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. The Cubic Centimeter 1 cm 3 = 1 cc = 1 mL 10 cm x 10 cm x 10 cm = 1000 cm 3 = 1000 mL = 1 L 54

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Measuring Mass Examples of Some Mass Equalities 1 kg = 1000 g = 1 x 10 3 g 1 g = 1000 mg = 1 x 10 3 mg 1 g= 100 cg= 1 x 10 2 cg 1 mg = 1000 μg= 1 x 10 3 μg 55

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Learning Check Identify the larger unit in each of the following: A. mm or cm B. kilogram or centigram C. mL or μL 56

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution Identify the larger unit in each of the following: A. mm or cm A mm is 0.001 m, smaller than a cm, 0.01 m. B. kilogram or centigram A kilogram is 1000 g, larger than a centigram or 0.01 g. C. mL or μL A milliliter is 0.001 L, larger than a μL, 0.000 001 L. 57

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Chapter 1 Chemistry and Measurements 1.8 Writing Conversion Factors 58

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. 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 For example, 1 m = 1000 mm 1 lb = 16 oz 2.20 lb = 1 kg 59

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Exact and Measured Numbers in Equalities Equalities between units in the same system of measurement are definitions that use exact numbers different systems of measurement (metric and U.S.) use measured numbers that have significant figures Exception: The equality 1 in. = 2.54 cm has been defined as an exact relationship and therefore 2.54 is an exact number. 60

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Conversion Factors A conversion factor is obtained from an equality and written in the form of a fraction with a numerator and denominator Equality: 1 in. = 2.54 cm inverted to give two conversion factors for every equality 1 in. and 2.54 cm 2.54 cm 1 in. 62

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Learning Check Write conversion factors from the equality for each of the following: A. L and mL B. hours and minutes C. meters and kilometers 63

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution Write conversion factors from the equality for each of the following: A. 1 L = 1000 mL 1 L and 1000 mL 1000 mL 1 L B. 1 h = 60 min 1 h and 60 min 60 min 1 h C. 1 km = 1000 m 1 km and 1000 m 1000 m 1 km 64

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Conversion Factors in a Problem A conversion factor may be obtained from information in a word problem is written for that problem only Example 1: The price of one pound (1 lb) of red peppers is $2.39. 1 lb red peppers and $2.39 $2.39 1 lb red peppers Example 2: The cost of one gallon (1 gal) of gas is $2.89. 1 gal gas and $2.89 $2.89 1 gal gas 65

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Percent as a Conversion Factor A percent factor gives the ratio of the parts to the whole % = parts x 100 whole uses the same unit in the numerator and denominator uses the value of 100 can be written as two factors Example: A food contains 30% (by mass) fat: 30 g fatand 100 g food 100 g food 30 g fat 66

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Percent Factor in a Problem The thickness of the skin fold at the waist indicates 11% body fat. What factors can be written for percent body fat (in kg)? Percent factors using kg: 11 kg fat and 100 kg mass 100 kg mass 11 kg fat 67

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Given and Needed Units To solve a problem, identify the given unit identify the needed unit Example: A person has a height of 2.0 meters. What is that height in inches? The given unit is the initial unit of height. given unit = meters (m) The needed unit is the unit for the answer. needed unit = inches (in.) 69

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Study Tip: Problem Solving Using GPS The steps in the Guide to Problem Solving (GPS) are useful in setting up a problem with conversion factors. 70

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Setting Up a Problem How many minutes are in 2.5 hours? Solution: Step 1 State the given and needed quantities. Given unit: 2.5 hours Needed unit: min Step 2 Write a unit plan. Plan:hours min 71

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solving a Problem How many minutes are in 2.5 hours? Step 3 State equalities and conversion factors to cancel units. 60 min= 1 h60 min and 1 h 1 h 60 min Step 4 Set up problem to cancel units. Given Conversion Needed unit unit factor 2.5 h x 60 min = 150 min (2 SF) 1 h 72

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Learning Check A rattlesnake is 2.44 m long. How many centimeters long is the snake? 73

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution A rattlesnake is 2.44 m long. How many centimeters long is the snake? Step 1 State the given and needed quantities. Given unit: 2.44 m Needed unit: cm Step 2 Write a unit plan. Plan:meters centimeters 74

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution A rattlesnake is 2.44 m long. How many centimeters long is the snake? Step 3 State equalities and conversion factors to cancel units. 1 m = 10 2 cm10 2 cm and 1 m 1 m 10 2 cm Step 4 Set up problem to cancel units. Given Conversion Needed unit unit factor 2.44 m x 10 2 cm = 244 cm (3 SF) 1 m 75

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution How many minutes are in 1.4 days? Step 1 State the given and needed quantities. Given unit: 1.4 days Needed unit: minutes Step 2 Write a unit plan. Factor 1 Factor 2 Plan: days h min 77

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution How many minutes are in 1.4 days? Step 3 State equalities and conversion factors to cancel units. 1 day = 24 hours24 hours and 1 day 1 day 24 hours 1 hour = 60 minutes 60 min and 1 h 1 h 60 min 78

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution How many minutes are in 1.4 days? Step 4 Set up problem to cancel units. Given Conversion Conversion Needed unit unit factor factor 1.4 days x 24 h x 60 min = 2.0 x 10 3 min 1 day 1 h (rounded) 2 SF Exact Exact = 2 SF 79

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Percent Factor in a Problem If the thickness of the skin fold at the waist indicates 11% body fat, how much fat is in a person with a mass of 86 kg? Percent factor: 11 kg fat 100 kg mass percent factor 86 kg mass x 11 kg fat 100 kg mass = 9.5 kg of fat 80

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Chapter 1 Chemistry and Measurements 81 1.10 Density Objects that sink in water are more dense than water; objects that float in water are less dense.

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Density 82 Density compares the mass of an object to its volume is the mass of a substance divided by its volume Density Expression Density = mass = g or g or g/cm 3 volume mL cm 3 Note: 1 mL = 1 cm 3

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Learning Check 85 Osmium is a very dense metal. What is its density in g/cm 3 if 50.0 g of osmium has a volume of 2.22 cm 3 ? 1) 2.25 g/cm 3 2) 22.5 g/cm 3 3) 111 g/cm 3

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. 86 Step 1 State the given and needed quantities. Given: 50.0 g; 22.2 cm 3 Need: density, g/cm 3 Step 2 Write the density expression. D = mass volume Step 3 Express mass in grams and volume in mL or cm 3. Mass = 50.0 g Volume = 22.2 cm 3 Step 4 Substitute mass and volume into the density expression and calculate. D = 50.0 g = 22.522522 g/cm 3 2.22 cm 3 = 22.5 g/cm 3 (rounded to 3 SFs) Solution

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Volume by Displacement A solid completely submerged in water displaces its own volume of water has a volume calculated from the volume difference 45.0 mL − 35.5 mL = 9.5 mL = 9.5 cm 3 87

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Density Using Volume Displacement The density of the zinc object is calculated from its mass and volume. Density = mass = 68.60 g = 7.2 g/cm 3 volume 9.5 cm 3 88

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Learning Check 89 What is the density (g/cm 3 ) of a 48.0-g sample of a metal if the level of water in a graduated cylinder rises from 25.0 mL to 33.0 mL after the metal is added? 1) 0.17 g/cm 3 2) 6.0 g/cm 3 3) 380 g/cm 3 object 33.0 mL 25.0 mL

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution 90 Step 1 State the given and needed quantities. Given: 48.0 g Volume of water = 25.0 mL Volume of water + metal = 33.0 mL Need: Density Step 2 Write the density expression. Density = mass of metal volume of metal

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution 91 Step 3 Express mass in grams and volume in mL or cm 3. Mass = 48.0 g Volume of the metal is equal to the volume of water displaced. Volume of water + metal = 33.0 mL − Volume of water = 25.0 mL Volume of metal= 8.0 mL

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution 92 Step 4 Substitute mass and volume into the density expression and calculate the density. Density = 48.0 g = 6.0 g = 6.0 g/mL 8.0 mL 1 mL

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Sink or Float Ice floats in water because the density of ice is less than the density of water. Aluminum sinks because its density is greater than the density of water. 93

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Problem Solving Using Density 94 Density can be written as an equality. For a substance with a density of 3.8 g/mL, the equality is 3.8 g = 1 mL From this equality, two conversion factors can be written for density. Conversion 3.8 g and 1 mL factors1 mL 3.8 g

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Learning Check 96 The density of octane, a component of gasoline, is 0.702 g/mL. What is the mass, in kg, of 875 mL of octane? A. 0.614 kg B. 614 kg C. 1.25 kg

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution 97 The density of octane, a component of gasoline, is 0.702 g/mL. What is the mass, in kg, of 875 mL of octane? Step 1State the given and needed quantities. Given: Density of octane = 0.702 g/mL Volume = 875 mL Needed: Mass of octane Step 2 Write a plan to calculate the needed quantity. Density Conversion factor Plan: milliliters grams kilograms

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution 98 The density of octane, a component of gasoline, is 0.702 g/mL. What is the mass, in kg, of 875 mL of octane? Step 3 Write equalities and their conversion factors including density. density 0.702 g = 1 mL and 1 kg = 1000 g Step 4 Set up problem to calculate the needed quantity. 875 mL x 0.702 g x 1 kg = 0.614 kg 1 mL 1000 g Answer is A, 0.614 kg.

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Chapter 1 Chemistry.

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Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Chapter 1 Chemistry.

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