lecture 1 chem.pdf

Chemistry The Molecular Nature of Matter and Change Tenth Edition Martin S. Silberberg and Patricia G. Amateis  Chapter 1: Keys Studying Chemistry 1.1 Some Fundamental Definitions 1.2 The Scientific Approach: Developing a Model 1.3 Units of Measurement 1.4 Uncertainty in Measurement: Significan...

Chemistry The Molecular Nature of Matter and Change Tenth Edition Martin S. Silberberg and Patricia G. Amateis  Chapter 1: Keys Studying Chemistry 1.1 Some Fundamental Definitions 1.2 The Scientific Approach: Developing a Model 1.3 Units of Measurement 1.4 Uncertainty in Measurement: Significant Figures 1.5 Units and Conversion Factors in Calculations 2  Chemistry Chemistry is the study of matter, its properties, the changes that matter undergoes, and the energy associated with these changes. Mitigate climate change by reducing green-house gases emission Saving marine ecosystems by tackling ocean acidification Maintain food supply by producing drought- and disease resistant crops Produce and promote renewable energies by solar cells Diagnose and treatments of diseases like cancer, AIDS, Alzheimer, etc 3  Definitions Matter: anything that has both 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. 4  The States of Matter A solid has a fixed shape and volume. Solids may be hard or soft, rigid or flexible. A liquid has a varying shape that conforms to the shape of the container, but a fixed volume. A liquid has an upper surface. A gas has no fixed shape or volume and therefore does not have a surface. 5 The Physical States of Matter 6  Physical and Chemical Properties Physical Properties: Properties a substance shows by itself without interacting with another substance. Color, melting point, boiling point, density. Chemical Properties: Properties a substance shows as it interacts with, or transforms into, other substances. Flammability, corrosiveness. 7  The Distinction Between Physical and Chemical Change 𝑤𝑎𝑡𝑒𝑟 𝑠𝑜𝑙𝑖𝑑 𝑠𝑡𝑎𝑡𝑒 → 𝑤𝑎𝑡𝑒𝑟(𝑙𝑖𝑞𝑢𝑖𝑑 𝑠𝑡𝑎𝑡𝑒) 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑤𝑎𝑡𝑒𝑟 ℎ𝑦𝑑𝑟𝑜𝑔𝑒𝑛 + 𝑜𝑥𝑦𝑔𝑒𝑛 8  Sample Problem 1.1 – Problem and Plan Visualizing Change on the Atomic Scale PROBLEM: The scenes below represent an atomic-scale view of substance A undergoing two different changes. Decide whether each scene shows a physical or a chemical change. PLAN: We need to determine what change is taking place. The numbers and colors of the little spheres that represent each particle tell its “composition.” If the composition does not change, the change is physical, whereas a chemical change results in a change of composition. 9  Sample Problem 1.1 – Solution 1 SOLUTION: Each particle of substance A is composed of one blue and two red spheres. Sample B is composed of two different types of particles—some have two red spheres, while some have one red and one blue. As A changes to B, the chemical composition has changed. A → B is a chemical change. 10  Sample Problem 1.1 – Solution 2 SOLUTION: Each particle of C is still composed of one blue and two red spheres, but the particles are closer together and are arranged in a regular pattern. The composition remains unchanged, but the physical form is different. A → C is a physical change. 11  Problems Identify the chemical and physical change from two atomic scale view described below: A) B) 12  Temperature and Change of State A change of state is a physical change. Physical form changes, composition does not. Changes in physical state are reversible. By changing the temperature. A chemical change cannot simply be reversed by a change in temperature. heating heating Ice liquid water water vapor cooling cooling 13 Some Characteristic Properties of Copper Table 1.1 Some Characteristic Properties of Copper Source: (copper) Stephen Frisch/McGraw Hill; (copper wire) dgstudiodg/iStockphoto/Getty Images; (candlestick) Willard/iStock/Getty Images; (copper carbonate, copper reacting with acid, copper and ammonia) Stephen Frisch/McGraw Hill 14  Sample Problem 1.2 – Problem and Plan Distinguishing Between Physical and Chemical Change PROBLEM: Decide whether each of the following processes is primarily a physical or a chemical change, and explain briefly: (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) A match ignites to form ash and a mixture of gases. (d) Perspiration evaporates when you relax after jogging. (e) A silver fork tarnishes slowly in air. PLAN: “Does the substance change composition or just change form?” 15  Sample Problem 1.2 – Solution (a) Frost forms as the temperature drops on a humid winter night— physical change: water vapor changes to ice crystals. (b) A cornstalk grows from a seed that is watered and fertilized— chemical change: the seed undergoes changes in composition. (c) A match ignites to form ash and a mixture of gases—chemical change: substances in the match head are converted into other substances. (d) Perspiration evaporates when you relax after jogging—physical change: liquid water changes to gaseous water. (e) A silver fork tarnishes slowly in air—chemical change: silver changes to silver sulfide by reacting with sulfur-containing substances in the air. 16  Energy in Chemistry Energy is the ability to do work. Potential energy is energy due to the position of an object. Kinetic energy is energy due to the movement of an object. Total Energy = Potential Energy + Kinetic Energy 17  Energy Changes Lower energy states are more stable and are favoured over higher energy states. Energy is neither created nor destroyed. It is conserved. It can be converted from one form to another. 18  Potential Energy Is Converted to Kinetic Energy 1 A gravitational system. The potential energy gained when a weight is lifted is converted to kinetic energy as the weight falls. A lower energy state is more stable. 19  Potential Energy Is Converted to Kinetic Energy 2 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. Energy is conserved when it is transformed. 20  Potential Energy Is Converted to Kinetic Energy 3 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. 21  Potential Energy Is Converted to Kinetic Energy 4 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. 22  The Scientific Approach: Developing a Model In the beginning our ancestors used trial and error to solve any problems. But we use quantitative rather qualitative theories to understand chemistry of any materials. This approach is called Scientific Method. 23 The Scientific Approach: Developing a Model 1. Observation An observation may be qualitative and as simple as indicating some phenomenon observed by the naked eye. An observation may also be quantitative in nature, requiring tools or instruments to be quantified, recorded, and understood. 2. Formulation of a Hypothesis - a tentative or speculative explanation for observations. This becomes the basis for future experiments. 1. A hypothesis must be testable and falsifiable. 3. Experimentation - Having formulated a hypothesis, it can now be used to make predictions about experimental outcomes. Experiments are now designed and executed in order to test these hypothetical predictions. 4. Conclusions - After reviewing the experimental data, if they support or fail to support the hypothesis, it can be concluded that the hypothesis is either valid or invalid. If it is validated, it becomes the basis for further experimental testing and refinement. 24 A Scientific Law - is a summary statement A Scientific Theory - is a model which (or mathematical equation) which describes the underlying explanations of describes a set of observations and can be observations. used to make predictions about the outcome of future events or experiments. 25  Scientific Method (pgs. 3-5 in text) A Scientific Theory - is a model which describes the underlying explanations of observations. Theories are the height of scientific knowledge. They are models of how the world works, which are supported by large bodies of experimental data and can be used to predict entirely new observations across a wide range of phenomena. 26  Scientific Notation In common usage, large numbers often carry commas every three spaces to make their interpretation easier. As chemists however, we will often encounter numbers that are so much larger (or smaller) than 150 million that even this convention is ineffective. For example, diamond is a form of elemental carbon, and in a 1-carat diamond there are 10,027,000,000,000,000,000,000 atoms of carbon (we will learn more about atoms later). It is in these instances that we will need a different method for quickly and easily reading and writing such numbers. That method is scientific notation. A number written in scientific notation will consist of 2 parts: A coefficient – number greater than or equal to 1 but less than 10. An exponential part – 10 raised to an appropriate power (an exponent) if the decimal is moved to the left, the resulting exponent will be positive, and if the decimal must be moved to the right, the exponent will be negative. 27 Table 1.3 Common Decimal Prefixes Used with SI Units* Standard As a Power of Ten Scientific Notation Notation 1,000,000 1 × 10 10 10 10 10 10 = 1 × 106 100,000 1 × 10 10 10 10 10 = 1 × 105 10,000 1 × 10 10 10 10 = 1 × 104 1,000 1 × 10 10 10 = 1 × 103 100 1 × 10 10 = 1 × 102 10 1 × 10 = 1 × 101 1 1= 1 × 100 1 0.1 1× = 1 × 10−1 10 1 1 0.01 1× = 1 × 10−2 10 10 1 1 1 0.001 1× = 1 × 10−3 10 10 10 1 1 1 1 0.0001 1× = 1 × 10−4 10 10 10 10 1 1 1 1 1 0.00001 1× = 1 × 10−5 10 10 10 10 10 1 1 1 1 1 1 0.000001 1× = 1 × 10−6 10 10 10 10 10 10 28 Convert the following to Scientific Notation: a) 2700 b.) 0.002700 Solutions: a. Step 1. Move the decimal to ensure a numeric value ≥ 1 and < 10. In this case this will require moving the decimal 3 spaces to the left. 2700 → 2.700 Step 2. Account for moving the decimal 3 spaces to the “left” by multiplying by 10 raised to the 3rd power. 𝟐. 𝟕𝟎𝟎 × 𝟏𝟎𝟑 29 Convert the following to Scientific Notation: a) 2700 b.) 0.002700 Solutions: b) Step 1. Again, move the decimal place to ensure a numeric value ≥ 1 and < 10. In this case it will require moving the decimal 3 spaces to the right. 0.002700 → 2.700 Step 2. Account for moving the decimal 3 spaces to the “right” by multiplying by 10 raised to the negative 3rd power. 𝟐. 𝟕𝟎𝟎 × 𝟏𝟎−𝟑 30 Convert the following to standard notation: a) 4.93 x 105 b) 7.421 x 10-8 Solutions: a. Since exponential part I 10 raised to the 5 th power, move the decimal 5 spaces to the right and drop the exponential term. 4.93 × 105 → 𝟒𝟗𝟑, 𝟎𝟎𝟎 b. Since exponential part I 10 raised to the negative 8th power, move the decimal 8 spaces to the left and drop the exponential term. 7.421 × 10−8 → 𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟕𝟒𝟐𝟏 31  SI Base Units Table 1.2 SI Base Units Physical Quantity (Dimension) Unit Name Unit Abbreviation Mass kilogram kg Length meter m Time second s Temperature kelvin K Amount of substance mole mol Electric current ampere A Luminous intensity candela cd 32 Common SI–English Equivalent Quantities Table 1.4 Common SI–English Equivalent Quantities Quantity SI Units SI Equivalents English Equivalents English to SI Equivalent Length 1 kilometer (km) 1000 (10³) meters 0.62137 mile (mi) 1 mile = 1.6093 km 1 meter (m) 100 (10²) centimeters 1.094 yards (yd) 1 yard = 0.9144 m 1 meter (m) 1000 millimeters (mm) 39.370 inches (in) 1 foot (ft) = 0.3048 m 1 centimeter (cm) 0.01 (10⁻²) meter 0.39370 inch 1 inch = 2.54 cm (exactly) Volume 1 cubic meter (m³) 1,000,000 (10⁶) cubic centimeters 35.31 cubic feet (ft³) 1 cubic foot = 0.02832 m³ 1 cubic decimeter (dm³) 1000 cubic centimeters 0.2642 gallon (gal) 1 gallon = 3.7854 dm³ 1 cubic decimeter (dm³) 1000 cubic centimeters 1.0569 quarts (qt) 1 quart = 0.9464 dm³ 1 quart = 946.4 cm³ 1 cubic centimeter (cm³) 0.001 dm³ 0.03381 fluid ounce 1 fluid ounce = 29.574 cm³ Mass 1 kilogram (kg) 1000 grams (g) 2.2046 pounds (lb) 1 pound = 0.4536 kg 33  Some Volume Relationships in SI 1 Figure 1.5 34  Common Laboratory Volumetric Glassware Figure 1.6 Source: (A) Stephen Frisch/McGraw Hill; (B) Matthew Jones/18percentgrey/123RF 35  Quantities of Length (A), Volume (B), and Mass (C) Figure 1.7 36  Temperature Temperature is a measure of how hot or cold one object is relative to another. Heat is the energy that flows from an object with a higher temperature to an object with a lower temperature. 37  Temperature Scales Celsius (°C) – The Celsius scale is based on the freezing and boiling points of water. This is the temperature scale used most commonly around the world. Kelvin (K) – The “absolute temperature scale” begins at absolute zero and has only positive values. The Celsius and Kelvin scales use the same size degree although their starting points differ. Note that the kelvin is not used with the degree sign (°). Fahrenheit (°F) – The Fahrenheit scale is commonly used in the U.S. The Fahrenheit scale has a different degree size and different zero points than both the Celsius and Kelvin scales. 38  Some Interesting Temperatures Figure 1.8 39  Freezing and Boiling Points of Water Figure 1.9 40 Temperature Conversions T ( in K ) = T ( in °C ) + 273.15 T ( in °C ) = T ( in K ) − 273.15 9 T ( in °F ) = T ( in °C ) + 32 5 5 T ( in °C ) = T ( in °F ) − 32  9 41  The Three Temperature Scales Table 1.5 The Three Temperature Scales Scale Unit Size of Degree Freezing Boiling T at Conversion (Relative to K) Point Point Absolute of H2O of H2O Zero Kelvin kelvin (K) — 273.15 K 373.15 K 0K to °C (Equation (absolute) 1.1) Celsius Celsius 1 0°C 100°C −273.15°C to K (Equation degree (°C) 1.2) to °F (Equation 1.3) Fahrenheit Fahrenheit 5over 9 32°F 212°F −459.67°F to °C (Equation degree (°F) 1.4) 42  Sample Problem 1.3 – Problem, Plan and Solution Converting Units of Temperature PROBLEM: A child has a body temperature of 38.7°C, and normal body temperature is 98.6°F. Does the child have a fever? What is the child’s temperature in kelvins? PLAN: We have to convert °C to °F to find out if the child has a fever. We can then use the °C to Kelvin relationship to find the temperature in Kelvin. SOLUTION: 9 Converting from °C to °F: ( 38.7 °C ) + 32 = 101.7F 5 Yes, the child has a fever. Converting from °C to K 38.7℃ + 273.15 = 311.8 K 43  Significant Figures Every measurement includes some uncertainty. The rightmost digit of any quantity is always estimated. The recorded digits, both certain and uncertain, are called significant figures. The greater the number of significant figures in a quantity, the greater its certainty. 44  The Number of Significant Figures in a Measurement Figure 1.10 45  Determining Which Digits Are Significant All digits in a measurement are significant. Except zeros used only to position the decimal point. 0.0055 has 2 significant figures. Zeros that end a number are significant. Whether they occur before or after the decimal point. As long as a decimal point is present. 1.030 mL has 4 significant figures. 5300. L has 4 significant figures. If no decimal point is present. Zeros at the end of the number are not significant. 5300 L has only 2 significant figures. 46  Sample Problem 1.4 – Problem and Plan Determining 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 (e) to (g), express each in exponential notation first. (a) 0.0030 L (b) 0.1044 g (c) 53.069 L (d) 3040 kg (e) 0.00004715 m (f) 57,600. s (g) 0.0000007160 cm3 PLAN: We determine the number of significant figures by counting digits, paying particular attention to the position of zeros in relation to the decimal point, and underline zeros that are significant. 47  Sample Problem 1.4 – Solution SOLUTION: (a) 0.0030 L has 2 sf (b) 0.1044 g has 4 sf (c) 53.069 mL has 5 sf (d) 3040 kg has 3 sf (e) 0.00004715 m = 4.715  10−5 m has 4 sf (f) 57,600. s = 5.7600  104 s has 5 sf (g) 0.0000007160 cm3 = 7.160  10−7 cm 3 has 4 sf 48  Rules for Significant Figures in Calculations 1 1. For multiplication and division. The answer contains the same number of significant figures as there are in the measurement with the fewest significant figures. Multiply the following numbers: 9.2 cm  6.8 cm  0.3744 cm = 23.4225 cm 3 = 23 cm 3 Two significant figures are allowed in the answer because two of the values in the calculation have only two significant figures. 49  Rules for Significant Figures in Calculations 2 2. For addition and subtraction. The answer has the same number of decimal places as there are in the measurement with the fewest decimal places. Example: adding two volumes 83.5 mL + 23.28 mL = 106.78 mL = 106.8 mL The answer has only one digit to the right of the decimal as does 83.5. Example: subtracting two volumes 865.9 mL − 2.8121 mL = 863.0879 mL = 863.1 mL The answer has only one digit to the right of the decimal as does 865.9. 50  Rules for Rounding Off Numbers 1 1. If the digit removed is more than 5, the preceding number increases by 1. 5.379 rounds to 5.38 if 3 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 3 significant figures are retained. 3. If the digit removed is 5 followed by zeros or with no following digits, 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 by other nonzero digits, rule 1 is followed: 17.6500 rounds to 17.6, but 17.6513 rounds to 17.7. 51  Rules for Rounding Off Numbers 2 4. Be sure to carry two or more additional significant figures through a multistep calculation and round off the final answer only. 52  Significant Figures in the Lab The measuring device used determines the number of significant digits possible. Figure 1.11 Source: (both) Stephen Frisch/McGraw Hill 53  Exact Numbers Exact numbers have no uncertainty associated with them. Numbers may be exact by definition: 1000 mg = 1 g. 60 min = 1 hr. 2.54 cm = 1 in. Numbers may be exact by count: Exactly 26 letters in the alphabet. Exact numbers do not limit the number of significant digits in a calculation. 54  Sample Problem 1.5 – Problem and Plan Significant Figures and Rounding PROBLEM: Perform the following calculations and round each answer to the correct number of significant figures: 16.205 cm 2 − 1.5 cm 2 (a) 7.081 cm  1g  4.80  10 mg  4   1000 mg  (b) 11.55 cm3 PLAN: We use the rules for rounding presented in the text: (a) We subtract before we divide. (b) We note that the unit conversion involves an exact number. 55  Sample Problem 1.5 – Solution SOLUTION: 16.205 cm 2 − 1.5 cm 2 14.7 cm 2 (a) = = 2.08 cm 7.081 cm 7.081 cm The answer to the subtraction (14.705) is rounded to 14.7 with one decimal place because 1.5 has only one decimal place.  1g  4.80  10 mg   1000 mg  4 (b)   = 48.0 g = 4.16 g / cm 3 11.55 cm3 11.55 cm3 56  Precision, Accuracy, and Error Precision refers to how close the measurements in a series are to each other. Accuracy refers to how close each measurement is to the actual value. Systematic error produces values that are either all higher or all lower than the actual value. This error is part of the experimental system. Random error produces values that are both higher and lower than the actual value. Random error always occurs. 57  Precision and Accuracy in a Laboratory Calibration Figure 1.12 58  Chemical Problem Solving All measured quantities consist of a number and a unit. Units are manipulated like numbers: 3 ft  4 ft = 12 ft 2 350 mi 50 mi = or 50 mi  h −1 7h 1h 59  Conversion Factors A conversion factor is a ratio of equivalent quantities used to express a quantity in different units. The relationship 1 mi = 5280 ft gives us the conversion factor: 1 mi 5280 ft = =1 5280 ft 5280 ft 60  Conversion Factor Problem A conversion factor is chosen and set up so that all units cancel except those required for the answer. PROBLEM: The height of the Empire State building is 1454 ft. Express this quantity in miles (m i) if 1 mi = 5280 ft. PLAN: Set up the conversion factor so that ft will cancel and the answer will be in mi. 1 mi SOLUTION: 1454 ft  = 0.2754 mi 5280 ft 61  Systematic Approach to Solving Chemistry Problems State Problem. Plan. Clarify the known and unknown. Suggest steps from known to unknown. Prepare a visual summary of steps that includes conversion factors, equations, known variables. Solution. Check. Comment. Follow-up Problem. 62  Sample Problem 1.6 – Problem and Plan Converting Units of Length PROBLEM: To wire your stereo equipment, you need 325 centimeters (cm) of speaker wire that sells for $0.15/ft. What is the price of the wire? PLAN: We know the length (in cm) of wire and cost per length ($/ft). We have to convert cm to inches and inches to feet. Then we can find the cost for the length in feet. 63  Sample Problem 1.6 – Solution SOLUTION: Length (in) = length (cm) × conversion factor 1 in = 325 cm  = 128 in 2.54 cm Length (ft) = length (in) × conversion factor 1 ft = 128 in  = 10.7 ft 12 in Price ($) = length (ft) × conversion factor $0.15 = 10.7 ft  = $1.60 1 ft 64  Sample Problem 1.7 – Problem and Plan Converting Units of Volume PROBLEM: A graduated cylinder contains 19.9 mL of water. When a small piece of galena, an ore of lead, is added, it sinks and the volume increases to 24.5 mL. What is the volume of the piece of galena in cm³ and in L? PLAN: The volume of the galena is equal to the difference in the volume of the water before and after the addition. 65  Sample Problem 1.7 – Solution SOLUTION: (24.5 − 19.9) mL = volume of galena = 4.6 mL 1 cm3 4.6 mL  = 4.6 cm 3 1 mL 10−3 L 4.6 mL  = 4.6  10−3 L 1 mL 66  Sample Problem 1.8 – Problem Converting Units of Mass PROBLEM: Many international computer communications are carried out by optical fibers in cables laid along the ocean floor. If one strand of optical fiber weighs 1.19  10−3 lb/m, what is the mass (in kg) of a cable made of six strands of optical fiber, each long enough to link New York and Paris ( a distance of 8.94  103 km ) ? 67  Sample Problem 1.8 – Plan and Solution PLAN: We convert the length of one fiber from km to m and then find its mass (lb) by using the conversion factor 1.19  10−3 lb = 1 m. Then we multiply the fiber mass by 6 to get the cable mass and convert lb to kg. SOLUTION: Converting length of fiber from km to m: 103 m 8.84  10 km  3 = 8.84  106 m 1 km Converting length of fiber (m) to mass (lb): −3 1.19  10 lb 8.84  106 m  = 1.05  10 lb 1 m 68  Sample Problem 1.8 – Solution Finding the mass of the cable: 1.05  104 lb 6 fibers lb  = 6.30  10 4 1 fiber 1 cable cable Converting mass of cable from lb to kg: 6.30  104 lb 1 kg  = 2.86  104 kg / cable 1 cable 2.205 lb 69  Sample Problem 1.9 – Problem and Plan Converting Units Raised to a Power PROBLEM: A furniture factory needs 31.5 ft² of fabric to upholster one chair. Its Dutch supplier sends the fabric in bolts that hold exactly 200 m². How many chairs can be upholstered with three bolts of fabric? PLAN: We know the amount of fabric in one bolt in m²; multiplying the m² of fabric by the number of bolts gives the total amount of fabric available in m². We convert the amount of fabric from m² to ft² and use the conversion factor 31.5 ft² of fabric = 1 chair to find the number of chairs (see the road map). 70  Sample Problem 1.9 – Solution SOLUTION: Converting from number of bolts to amount of fabric in m²: Amount ( m ) 200 m 2 2 of fabric = 3 bolts  = 600 m 2 1 bolt Converting the amount of fabric from m² to ft² : Since 0.3048 m = 1 ft , we have ( 0.3048 ) m = (1) ft 2 ,so 2 2 2 Amount ( ft ) 1 ft 2 2 of fabric = 600 m  2 = 6460ft 2 ( 0.3048) m 2 2 Finding the number of chairs: 1 chair Number of chairs = 6460 ft 2  2 = 205 chairs 31.5 ft 71  Density Density is the mass of a sample divided by its volume: mass density = volume At a given temperature and pressure, the density of a substance is a characteristic physical property and has a specific value. 72  Densities of Some Common Substances Table 1.7 Densities of Some Common Substances* Substance Physical State Density Hydrogen gas 0.0000899 Oxygen gas 0.00133 Grain alcohol liquid 0.789 Water liquid 0.998 Table salt solid 2.16 Aluminium solid 2.70 Lead solid 11.3 Gold solid 19.3 *At room temperature (20°C) and normal atmospheric pressure (1 atm). 73  Sample Problem 1.10 – Problem and Plan Calculating Density from Mass and Volume PROBLEM: (a) Lithium, a soft, gray solid with the lowest density of any metal, is a key component of advanced batteries. A slab of lithium weighs 1.49  103 mg and has sides that are 20.9 mm by 11.1 mm by 11.9 mm. Find the density of lithium in g∕cm³. (b) What is the volume (in mL) of a piece of lithium with a mass of 895 mg? PLAN: (a) Density is expressed in g∕cm³ so we need to convert the mass from mg to g. We convert the lengths of the three sides from mm to cm and multiply them to obtain the volume in cm³. (b) We convert the mass of sample from mg to g and use its density as a conversion factor between mass and volume in cm³. Finally, we convert cm³ to mL. 74 Sample Problem 1.10 –Plan 75  Sample Problem 1.10 – Solution 1 SOLUTION: −3 10 g (a) 1.49  10 3 mg  = 1.49 g 1 mg 1 cm 20.9 mm  = 2.09 cm 10 mm Similarly the other sides will be 1.11 cm and 1.19 cm, respectively. Volume = 2.09 cm  1.11 cm  1.19 cm = 2.76 cm3 1.49 g Density of Li = = 0.540 g / cm 3 2.76 cm3 76  Sample Problem 1.10 – Solution 2 SOLUTION: 10−3 g (b) 895 mg  = 0.895 g 1 mg 1 cm3 1 mL 0.895 g   3 = 1.66 mL 0.540 g 1 cm 77  Extensive and Intensive Properties Extensive properties are dependent on the amount of substance present; mass and volume, for example, are extensive properties. Intensive properties are independent of the amount of substance; density is an intensive property. Figure 1.13 78