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Lecture PowerPoint Chemistry The Molecular Nature of Matter and Change Sixth Edition Martin S. Silberberg 1-1 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 1: Keys to the Study of Chemistry 1.1 Some Fundamental Definitions 1.2 Chemical Arts and the Origins of Modern Chemistry 1.3 The Scientific Approach: Developing a Model 1.4 Chemical Problem Solving 1.5 Measurement in Scientific Study 1.6 Uncertainty in Measurement: Significant Figures 1-2 Chapter 1 Keys to the Study of Chemistry 1-3 Chemistry Chemistry is the study of matter, its properties, the changes that matter undergoes, and the energy associated with these changes. 1-4 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 1-5 the characteristics that give each substance a unique identity 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 1-6 Figure 1.1 The distinction between physical and chemical change. 1-7 Sample Problem 1.1 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: 1-8 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. Sample Problem 1.1 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. 1-9 Sample Problem 1.1 Each particle of C is still composed of one blue and two red spheres, but the particles are closer together and are more organized. The composition remains unchanged, but the physical form is different. A → C is a physical change. 1-10 Table 1.1 Some Characteristic Properties of Copper 1-11 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. 1-12 Figure 1.2 1-13 The physical states of matter. 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. 1-14 Sample Problem 1.2 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: 1-15 “Does the substance change composition or just change form?” Sample Problem 1.2 SOLUTION: (a) Frost forms as the temperature drops on a humid winter night. physical change (b) A cornstalk grows from a seed that is watered and fertilized. chemical change (c) A match ignites to form ash and a mixture of gases. chemical change (d) Perspiration evaporates when you relax after jogging. physical change (e) A silver fork tarnishes slowly in air. chemical change 1-16 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 1-17 Energy Changes Lower energy states are more stable and are favored over higher energy states. Energy is neither created nor destroyed – it is conserved – and can be converted from one form to another. 1-18 Figure 1.3A Potential energy is converted to kinetic energy. A gravitational system. The potential energy gained when a lifted weight is converted to kinetic energy as the weight falls. A lower energy state is more stable. 1-19 Figure 1.3B Potential energy is converted to 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. Energy is conserved when it is transformed. 1-20 Figure 1.3C Potential energy is converted to 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. 1-21 Figure 1.3D Potential energy is converted to 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. 1-22 Figure 1.6 The scientific approach to understanding nature. Observations Hypothesis is revised if experimental results do not support it. Hypothesis Tentative proposal that explains observations. Experiment Procedure to test hypothesis; measures one variable at a time. Model (Theory) Model is altered if predicted events do not support it. 1-23 Natural phenomena and measured events; can be stated as a natural law if universally consistent. Further Experiment Set of conceptual assumptions that explains data from accumulated experiments; predicts related phenomena. Tests predictions based on model Chemical Problem Solving • All measured quantities consist of – a number and a unit. • Units are manipulated like numbers: – 3 ft x 4 ft = 12 ft2 – 350 mi = 7h 1-24 50 mi 1h or 50 mi.h-1 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 1-25 = 5280 ft 5280 ft =1 A conversion factor is chosen and set up so that all units cancel except those required for the answer. PROBLEM: The height of the Angel Falls is 3212 ft. Express this quantity in miles (mi) if 1 mi = 5280 ft. PLAN: Set up the conversion factor so that ft will cancel and the answer will be in mi. SOLUTION: 1-26 3212 ft x 1 mi 5280 ft = 0.6083 mi Systematic Approach to Solving Chemistry Problems • State Problem Clarify the known and unknown. • Plan 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 1-27 Sample Problem 1.3 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. length (cm) of wire 2.54 cm = 1 in length (in) of wire 12 in = 1 ft length (ft) of wire 1 ft = $0.15 Price ($) of wire 1-28 Sample Problem 1.3 SOLUTION: Length (in) = length (cm) x conversion factor 1 in = 325 cm x = 128 in 2.54 cm Length (ft) = length (in) x conversion factor = 128 in x 1 ft = 10.7 ft 12 in Price ($) = length (ft) x conversion factor = 10.7 ft x 1-29 $ 0.15 1 ft = $ 1.60 Table 1. 2 SI Base Units Physical Quantity (Dimension) Unit Name Mass kilogram kg Length meter m Time second s Temperature kelvin K Electric Current ampere A Amount of substance mole Luminous intensity candela 1-30 Unit Abbreviation mol cd Table 1.3 1-31 Common Decimal Prefixes Used with SI Units Table 1.4 Common SI-English Equivalent Quantities Quantity SI to English Equivalent English to SI Equivalent Length 1 km = 0.6214 mile 1 m = 1.094 yard 1 m = 39.37 inches 1 cm = 0.3937 inch 1 mi = 1.609 km 1 yd = 0.9144 m 1 ft = 0.3048 m 1 in = 2.54 cm Volume 1 cubic meter (m3) = 35.31 ft3 1 dm3 = 0.2642 gal 1 dm3 = 1.057 qt 1 cm3 = 0.03381 fluid ounce 1 ft3 = 0.02832 m3 1 gal = 3.785 dm3 1 qt = 0.9464 dm3 1 qt = 946.4 cm3 1 fluid ounce = 29.57 cm3 Mass 1 kg = 2.205 lb 1 g = 0.03527 ounce (oz) 1 lb = 0.4536 kg 1 oz = 28.35 g 1-32 Figure 1.7 Some volume relationships in SI. Some volume equivalents: 1 m3 = 1000 dm3 1 dm3 = 1000 cm3 = 1 L = 1000 mL 3 1 cm = 1000 mm3 = 1 mL = 100= μL 1 mm3 = 1 μL 1-33 Figure 1.8 1-34 Common laboratory volumetric glassware. Sample Problem 1.4 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. volume (mL) before and after subtract volume (mL) of galena 1 mL = 1 cm3 volume (cm3) of galena 1-35 1 mL = 10-3 L volume (L) of galena Sample Problem 1.4 SOLUTION: (24.5 - 19.9) mL = volume of galena = 4.6 mL 1-36 1 cm3 4.6 mL x 1 mL = 4.6 cm3 10-3 L 4.6 mL x 1 mL = 4.6 x 10-3 L Sample Problem 1.5 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 x 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 (8.94 x 103 km)? PLAN: The sequence of steps may vary but essentially we need to find the length of the entire cable and convert it to mass. length (km) of fiber 1 km = 103 m length (m) of fiber 1 m = 1.19 x 10-3 lb mass (lb) of fiber 6 fibers = 1 cable mass (lb) of cable 1-37 2.205 lb = 1 kg Mass (kg) of cable Sample Problem 1.5 SOLUTION: 8.84 x 103 km 103 m x 1 km = 8.84 x 106 m -3 lb 1.19 x 10 8.84 x 106 m x = 1.05 x 104 lb 1m 1.05 x 104 lb 6 fibers x 1 fiber 1 cable 6.30 x 104 lb 1 kg x 1 cable 2.205 lb 1-38 = 6.30 x 104 lb/cable = 2.86 x 104 kg/cable Figure 1.9 Some interesting quantities of length (A), volume (B), and mass (C). 1-39 Density mass density = volume At a given temperature and pressure, the density of a substance is a characteristic physical property and has a specific value. 1-40 Table 1.5 Densities of Some Common Substances* Substance Physical State Density (g/cm3) Hydrogen gas 0.0000899 Oxygen gas 0.00133 Grain alcohol liquid 0.789 Water liquid 0.998 Table salt solid 2.16 Aluminum solid 2.70 Lead solid 11.3 Gold solid 19.3 *At 1-41 room temperature (20°C) and normal atmospheric pressure (1atm). Sample Problem 1.6 Calculating Density from Mass and Length PROBLEM: Lithium, a soft, gray solid with the lowest density of any metal, is a key component of advanced batteries. A slab of lithium weighs 1.49x103 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. PLAN: Density is expressed in g/cm3 so we need the mass in g and the volume in cm3. lengths (mm) of sides 10 mm = 1 cm mass (mg) of Li lengths (cm) of sides 103 mg = 1 g multiply lengths mass (g) of Li volume (cm3) divide mass by volume density (g/cm3) of Li 1-42 Sample Problem 1.6 SOLUTION: 1.49x103 mg x 20.9 mm x 1g = 1.49 g 3 10 mg 1 cm = 2.09 cm 10 mm Similarly the other sides will be 1.11 cm and 1.19 cm, respectively. Volume = 2.09 x 1.11 x 1.19 = 2.76 cm3 density of Li = 1-43 1.49 g 2.76 cm3 = 0.540 g/cm3 Figure 1.10 Some interesting temperatures. 1-44 Figure 1.11 Freezing and boiling points of water in the Celsius, Kelvin (absolute) and Fahrenheit scales. 1-45 Table 1.6 The Three Temperature Scales 1-46 Temperature Scales Kelvin ( K ) - The “absolute temperature scale” begins at absolute zero and has only positive values. Note that the kelvin is not used with the degree sign (°). Celsius ( oC ) - The Celsius scale is based on the freezing and boiling points of water. This is the temperature scale used most commonly around the world. The Celsius and Kelvin scales use the same size degree although their starting points differ. Fahrenheit ( oF ) – The Fahrenheit scale is commonly used in the US. The Fahrenheit scale has a different degree size and different zero points than both the Celsius and Kelvin scales. 1-47 Temperature Conversions T (in K) = T (in oC) + 273.15 T (in oC) = T (in K) - 273.15 T (in °F) = 9 T (in °C) + 32 5 T (in °C) = [T (in °F) – 32] 5 9 1-48 Sample Problem 1.7 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: Converting from °C to °F 9 (38.7 °C) + 32 = 101.7 °F 5 Yes, the child has a fever. Converting from °C to K 1-49 38.7 °C + 273.15 = 311.8 K 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. 1-50 Figure 1.12 1-51 The number of significant figures in a measurement. Determining Which Digits are Significant All digits are significant - except zeros that are used only to position the decimal point. • Make sure the measured quantity has a decimal point. • Start at the left and move right until you reach the first nonzero digit. • Count that digit and every digit to its right as significant. 1-52 • 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. 1-53 Sample Problem 1.8 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 (d) to (f), express each in exponential notation first. (a) 0.0030 L (b) 0.1044 g (c) 53,069 mL (d) 0.00004715 m PLAN: 1-54 (e) 57,600. s (f) 0.0000007160 cm3 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. Sample Problem 1.8 SOLUTION: (a) 0.0030 L has 2 sf (b) 0.1044 g has 4 sf (c) 53,069 mL has 5 sf (d) 0.00004715 m = 4.715x10-5 m has 4 sf (e) 57,600. s = 5.7600x104 s has 5 sf (f) 0.0000007160 cm3 = 7.160x10-7 cm3 has 4 sf 1-55 Rules for Significant Figures in Calculations 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 x 6.8 cm x 0.3744 cm = 23.4225 cm3 = 23 cm3 1-56 Rules for Significant Figures in Calculations 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 Example: subtracting two volumes 865.9 mL - 2.8121 mL 863.0879 mL = 863.1 mL 1-57 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 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. 1-58 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 4. Be sure to carry two or more additional significant figures through a multistep calculation and round off the final answer only. 1-59 Figure 1.13 Significant figures and measuring devices. The measuring device used determines the number of significant digits possible. 1-60 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. 1-61 Sample Problem 1.9 Significant Figures and Rounding PROBLEM: Perform the following calculations and round each answer to the correct number of significant figures: (a) 16.3521 cm2 - 1.448 cm2 7.085 cm (b) 4.80x104 mg 1g 1000 mg 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. 1-62 Sample Problem 1.9 SOLUTION: (a) 16.3521 cm2 - 1.448 cm2 = 7.085 cm (b) 4.80x104 mg 1g 1000 mg 11.55 cm3 1-63 14.904 cm2 7.085 cm = = 2.104 cm 48.0 g 11.55 cm3 = 4.16 g/ cm3 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. 1-64 Figure 1.14 Precision and accuracy in a laboratory calibration. precise and accurate precise but not accurate 1-65 Figure 1.14 Precision and accuracy in the laboratory. continued random error systematic error 1-66
Lecture PowerPoint Chemistry The Molecular Nature of Matter and Change Sixth Edition Martin S. Silberberg 2-1 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 2 The Components of Matter 2-2 Chapter 2: The Components of Matter 2.1 Elements, Compounds, and Mixtures: An Atomic Overview 2.2 The Observations That Led to an Atomic View of Matter 2.3 Dalton’s Atomic Theory 2.4 The Observations That Led to the Nuclear Atom Model 2.5 The Atomic Theory Today 2-3 Chapter 2: The Components of Matter 2.6 Elements: A First Look at the Periodic Table 2.7 Compounds: Introduction to Bonding 2.8 Formulas, Names, and Masses of Compounds 2.9 Mixtures: Classification and Separation 2-4 Definitions for Components of Matter Element - the simplest type of substance with unique physical and chemical properties. An element consists of only one type of atom. It cannot be broken down into any simpler substances by physical or chemical means. Molecule - a structure that consists of two or more atoms that are chemically bound together and thus behaves as an independent unit. Figure 2.1 2-5 Definitions for Components of Matter Compound - a substance composed of two or more elements which are chemically combined. Figure 2.1 Mixture - a group of two or more elements and/or compounds that are physically intermingled. 2-6 Table 2.1 2-7 Some Properties of Sodium, Chlorine, and Sodium Chloride. Property Sodium Melting point + Chlorine Sodium Chloride 97.8°C -101°C 801°C Boiling point 881.4°C -34°C 1413°C Color Silvery Yellow-green Colorless (white) Density 0.97 g/cm3 0.0032 g/cm3 2.16 g/cm3 Behavior in water Reacts Dissolves slightly Dissolves freely Sample Problem 2.1 PROBLEM: PLAN: 2-8 Distinguishing Elements, Compounds, and Mixtures at the Atomic Scale The following scenes represent an atomic-scale view of three samples of matter. Describe each sample as an element, compound, or mixture. A sample that contains only one type of particle is either an element or a compound. The particles of an element consist of only one type of atom whereas the particles of a compound have two or more types of atom bonded together. Sample Problem 2.1 SOLUTION: Sample (a) contains three different types of particles and is therefore a mixture. Sample (b) contains only one type of particle and each particle has only one atom. This is an element. Sample (c) contains only one type of particle, each of which contains two different types of atoms. This is a compound. 2-9 Figure 2.2 The law of mass conservation. The total mass of substances does not change during a chemical reaction. 2-10 Law of Mass Conservation The total mass of substances present does not change during a chemical reaction. reactant 1 + total mass calcium oxide + carbon dioxide 2-11 product reactant 2 = total mass calcium carbonate CaO + CO2 CaCO3 56.08 g + 44.00 g 100.08 g Law of Definite (or Constant) Composition No matter the source, a particular compound is composed of the same elements in the same parts (fractions) by mass. Figure 2.3 2-12 Calcium carbonate Analysis by Mass (grams/20.0 g) 8.0 g calcium 2.4 g carbon 9.6 g oxygen 20.0 g 2-13 Mass Fraction (parts/1.00 part) Percent by Mass (parts/100 parts) 0.40 calcium 0.12 carbon 0.48 oxygen 40% calcium 12% carbon 48% oxygen 1.00 part by mass 100% by mass Sample Problem 2.2 Calculating the Mass of an Element in a Compound PROBLEM: Analysis of 84.2 g of the uranium containing compound pitchblende shows it is composed of 71.4 g of uranium, with oxygen as the only other element. How many grams of uranium can be obtained from 102 kg of pitchblende? PLAN: The mass ratio of uranium/pitchblende is the same no matter the source. We can use the ratio to find the answer. mass (kg) of pitchblende mass ratio of U in pitchblende mass (kg) of uranium 1 kg = 1000 g mass (g) of uranium 2-14 Sample Problem 2.2 SOLUTION: mass (kg) of uranium = mass (kg) pitchblende x mass (kg) uranium in pitchblende mass (kg) pitchblende = 102 kg pitchblende x 86.5 kg uranium x 2-15 71.4 kg uranium = 86.5 kg uranium 84.2 kg pitchblende 1000 g 1 kg = 8.65 x 104 g uranium Law of Multiple Proportions If elements A and B react to form two compounds, the different masses of B that combine with a fixed mass of A can be expressed as a ratio of small whole numbers. Example: Carbon Oxides A & B Carbon Oxide I : 57.1% oxygen and 42.9% carbon Carbon Oxide II : 72.7% oxygen and 27.3% carbon 2-16 Assume that you have 100 g of each compound. In 100 g of each compound: g O = 57.1 g for oxide I & 72.7 g for oxide g C = 42.9 g for oxide I & 27.3 g for oxide II For oxide I: gO 57.1 = gC 42.9 = 1.33 For oxide II: gO gC = 2.66 = 72.7 27.3 2.66 g O/g C in II 1.33 g O/g C in I 2-17 = 2 1 Dalton’s Atomic Theory Dalton postulated that: 1. 2. 3. 4. 2-18 All matter consists of atoms; tiny indivisible particles of an element that cannot be created or destroyed. Atoms of one element cannot be converted into atoms of another element. Atoms of an element are identical in mass and other properties and are different from the atoms of any other element. Compounds result from the chemical combination of a specific ratio of atoms of different elements. Dalton’s Atomic Theory explains the mass laws Mass conservation Atoms cannot be created or destroyed postulate 1 or converted into other types of atoms. postulate 2 Since every atom has a fixed mass, postulate 3 during a chemical reaction the same atoms are present but in different combinations; therefore there is no mass change overall. 2-19 Dalton’s Atomic Theory explains the mass laws Definite composition Atoms are combined in compounds in specific ratios postulate 3 and each atom has a specific mass. postulate 4 Each element constitutes a fixed fraction of the total mass in a compound. 2-20 Dalton’s Atomic Theory explains the mass laws Multiple proportions Atoms of an element have the same mass postulate 3 and atoms are indivisible. postulate 1 When different numbers of atoms of elements combine, they must do so in ratios of small, whole numbers. 2-21 Sample Problem 2.3 Visualizing the Mass Laws PROBLEM: The following scene represents an atomic-scale view of a chemical reaction. Which of the mass laws (mass conservation, definite composition, or multiple proportions) is (are) illustrated? PLAN: Note the numbers, types and combination of atoms before and after the reaction. 2-22 Sample Problem 2.3 SOLUTION: There are 7 purple and 9 green atoms both before and after the reaction. Mass is therefore conserved. After the reaction some purple atoms remain unreacted, but some have combined with green atoms to form a compound. Each particle of this compound contains 1 purple and 2 green atoms – the composition is constant, illustrating the law of definite composition. The ratio of the elements in the compound is a small, whole number. The ratio of their masses will also be a small, whole number. This illustrates the law of multiple proportions. 2-23 Figure 2.4 Observations that established the properties of cathode rays. Observation Ray bends in magnetic field. Conclusion Ray consists of charged particles. Ray bends toward positive plate Ray consists of negative particles. in electric field. Ray is identical for any cathode. These particles are found in ALL matter. 2-24 Figure 2.5 2-25 Millikan’s oil-drop experiment for measuring an electron’s charge. Millikan’s findings were used to calculate the mass on an electron. determined by J.J. Thomson and others mass of electron = mass charge x charge = (-5.686x10-12 kg/C) x (-1.602x10-19 C) = 9.109x10-31 kg = 9.109x10-28 g 2-26 Figure 2.6 2-27 Rutherford’s a-scattering experiment and discovery of the atomic nucleus. Figure 2.7 General features of the atom. The atom is an electrically neutral, spherical entity composed of a positively charged central nucleus surrounded by one or more negatively charged electrons. The atomic nucleus consists of protons and neutrons. 2-28 Table 2.2 Properties of the Three Key Subatomic Particles Charge Name Relative Absolute (C)* (Symbol) Mass Relative (amu)† Absolute (g) Location in Atom Proton (p+) 1+ +1.60218x10-19 1.00727 1.67262x10-24 Nucleus Neutron (n0) 0 0 1.67493x10-24 Nucleus Electron (e-) 1- -1.60218x10-19 0.00054858 9.10939x10-24 Outside nucleus * The † 2-29 1.00866 coulomb (C) is the SI unit of charge. The atomic mass unit (amu) equals 1.66054x10-24 g. Atomic Symbol, Number and Mass Figure 2.8 X = Atomic symbol of the element A = mass number; A = Z + N Z = atomic number (the number of protons in the nucleus) N = number of neutrons in the nucleus 2-30 Isotopes Isotopes are atoms of an element with the same number of protons, but a different number of neutrons. Isotopes have the same atomic number, but a different mass number. Figure 2.8 2-31 Sample Problem 2.4 Determining the Number of Subatomic Particles in the Isotopes of an Element PROBLEM: Silicon (Si) has three naturally occurring isotopes: 28Si, 29Si, and 30Si. Determine the number of protons, neutrons, and electrons in each silicon isotope. PLAN: The mass number (A) is given for each isotope and is equal to the number of protons + neutrons. The atomic number Z, found on the periodic table, equals the number of protons. The number of neutrons = A – Z, and the number of electrons equals the number of protons for a neutral atom. SOLUTION: 2-32 The atomic number of silicon is 14; therefore 28Si has 14p+, 14e- and 14n0 (28-14) 29Si has 14p+, 14e- and 15n0 (29-14) 30Si has 14p+, 14e- and 16n0 (30-14) Tools of the Laboratory Figure B2.1 2-33 Formation of a positively charged neon particle (Ne+). Tools of the Laboratory Figure B2.2 2-34 The mass spectrometer and its data. Sample Problem 2.5 Calculating the Atomic Mass of an Element PROBLEM: Silver (Ag, Z = 47) has two naturally occurring isotopes, 107Ag and 109Ag. From the mass spectrometric data provided, calculate the atomic mass of Ag. Isotope 107Ag 109Ag Mass (amu) Abundance (%) 106.90509 108.90476 51.84 48.16 PLAN: Find the weighted average of the isotopic masses. mass (g) of each isotope multiply by fractional abundance of each isotope portion of atomic mass from each isotope add isotopic portions atomic mass 2-35 Sample Problem 2.5 SOLUTION: mass portion from 107Ag = 106.90509 amu x 0.5184 = 55.42 amu mass portion from 109Ag = 108.90476amu x 0.4816 = 52.45amu atomic mass of Ag = 55.42amu + 52.45amu = 107.87amu 2-36 Figure 2.9 2-37 The modern periodic table. Figure 2.10 Copper Some metals, metalloids, and nonmetals. Cadmium Chromium Silicon Lead Bismuth Arsenic Antimony Chlorine Bromine Sulfur Iodine Boron 2-38 Tellurium Carbon (graphite) Figure 2.11 The formation of an ionic compound. Transferring electrons from the atoms of one element to those of another results in an ionic compound. 2-39 Figure 2.12 2-40 Factors that influence the strength of ionic bonding. Figure 2.13 2-41 The relationship between ions formed and the nearest noble gas. Sample Problem 2.6 Predicting the Ion an Element Forms PROBLEM: Predict the monoatomic ion formed by each of the following elements: (a) Iodine (Z = 53) (b) Calcium (Z = 20) (c) Aluminum (Z = 13) PLAN: Use Z to find the element on the periodic table and see where it lies relative to its nearest noble gas. SOLUTION: (a) Iodine is a nonmetal in Group 7A(17). It gains one electron to have the same number of electrons as 54Xe. The ion is I(b) Calcium is a metal in Group 2A(2). It loses two electrons to have the same number of electrons as 18Ar. The ion is Ca2+ (c) Aluminum is a metal in Group 3A(13). It loses three electrons to have the same number of electrons as 10Ne. The ion is Al3+ 2-42 Figure 2.14 Formation of a covalent bond between two H atoms. Covalent bonds form when elements share electrons, which usually occurs between nonmetals. 2-43 Molecules and Ions Molecule – the basic unit of an element or covalent compound, consisting of two or more atoms bonded by the sharing of electrons. Most covalent substances consist of molecules. Ion – a single atom or covalently bonded group of atoms that has an overall electrical charge. There are no molecules in an ionic compound. 2-44 Figure 2.15 2-45 Elements that occur as molecules. Figure 2.16 The carbonate ion in calcium carbonate. A polyatomic ion consists of two of more atoms covalently bonded together and has an overall charge. In many reactions the polyatomic ion will remain together as a unit. 2-46 Chemical Formulas • A chemical formula consists of – element symbols with – numerical subscripts. • The chemical formula indicates the – type and number of each atom present – in the smallest unit of a substance. 2-47 Naming Binary Ionic Compounds For all ionic compounds, the name and formula lists the cation first and the anion second. In a binary ionic compound, both the cation and the anion are monatomic. The name of the cation is the same as the name of the metal. Many metal names end in -ium. The anion is named by adding the suffix -ide to the root of the nonmetal name. Calcium and bromine form calcium bromide. 2-48 Common Monatomic Ions* Table 2.3 Cations Formula Name Charge +1 H+ Li+ Na+ K+ Cs+ Ag+ hydrogen lithium sodium potassium cesium silver -1 HFClBrI- hydride fluoride chloride bromide iodide +2 Mg2+ Ca2+ Sr2+ Ba2+ Zn2+ Cd2+ magnesium calcium strontium barium zinc cadmium -2 O2S2- oxide sulfide +3 Al3+ aluminum -3 N3- nitride Charge *Listed 2-49 Anions Formula Name by charge; those in boldface are most common. Figure 2.17 Some common monatomic ions of the elements. Most main-group elements form one monatomic ion. Most transition elements form two monatomic ions. 2-50 Sample Problem 2.7 Naming Binary Ionic Compounds PROBLEM: Name the ionic compound formed from each of the following pairs of elements: (a) magnesium and nitrogen (b) iodine and cadmium (c) strontium and fluorine (d) sulfur and cesium PLAN: Use the periodic table to decide which element is the metal and which the nonmetal. The metal (cation) is named first and the suffix-ide is added to the root of the non-metal name. SOLUTION: 2-51 (a) magnesium nitride (b) cadmium iodide (c) strontium fluoride (d) cesium sulfide Sample Problem 2.8 Determining Formulas of Binary Ionic Compounds PROBLEM: Write empirical formulas for each of the compounds named in Sample Problem 2.7. (a) magnesium nitride (b) cadmium iodide (c) strontium fluoride (d) cesium sulfide PLAN: A compound is neutral. We find the smallest number of each ion that will produce a neutral formula. These numbers appear as right subscripts to the relevant element symbol. SOLUTION: (a) Mg2+ and N3-; three Mg2+(6+) and two N3-(6-); Mg3N2 (b) Cd2+ and I-; one Cd2+(2+) and two I-(2-); CdI2 (c) Sr2+ and F-; one Sr2+(2+) and two F-(2-); SrF2 (d) Cs+ and S2-; two Cs+(2+) and one S2- (2-); Cs2S 2-52 Some Metals That Form More Than One Monatomic Ion* Table 2.4 Element Ion Formula Systematic Name Common Name Chromium Cr2+ Cr3+ Co2+ Co3+ Cu+ Cu2+ Fe2+ Fe3+ Pb2+ Pb4+ Hg22+ Hg2+ Sn2+ Sn4+ chromium(II) chromium(III) cobalt(II) cobalt(III) copper(I) copper(II) iron(II) iron(III) lead(II) lead(IV) mercury (I) mercury (II) tin(II) tin(IV) chromous chromic Cobalt Copper Iron Lead Mercury Tin *Listed 2-53 cuprous cupric ferrous ferric mercurous mercuric stannous stannic alphabetically by metal name; the ions in boldface are most common. Sample Problem 2.9 Determining Names and Formulas of Ionic Compounds of Elements That Form More Than One Ion PROBLEM: Give the systematic name for each formula or the formula for each name for the following compounds: (a) tin(II) fluoride PLAN: (b) CrI3 (c) ferric oxide (d) CoS Find the smallest number of each ion that will produce a neutral formula. SOLUTION: (a) Tin(II) is Sn2+; fluoride is F-; so the formula is SnF2. (b) The anion I- is iodide; 3I- means that Cr (chromium) is +3. CrI3 is chromium(III) iodide. (c) Ferric is a common name for Fe3+; oxide is O2-; therefore the formula is Fe2O3. (d) Co is cobalt; the anion S2- is sulfide; the compound is cobalt(II) sulfide. 2-54 Table 2.5 Formula Some Common Polyatomic Ions* Name Formula Name Cations NH4+ H 3 O+ ammonium hydronium Common Anions CH3COOCNOHClOClO2ClO3NO2NO3MnO4*Bold 2-55 acetate cyanide hydroxide hypochlorite chlorite chlorate nitrite nitrate permanganate face ions are most common. CO32HCO3CrO42Cr2O72O22PO43HPO42SO32SO42- carbonate bicarbonate chromate dichromate peroxide phosphate hydrogen phosphate sulfite sulfate (partial table) Naming oxoanions Figure 2.18 No. of O atoms Prefix per hypo 2-56 Root Suffix Example root ate ClO4- perchlorate root ate ClO3- chlorate root ite ClO2- chlorite root ite ClO- hypochlorite Table 2.6 Numerical Prefixes for Hydrates and Binary Covalent Compounds Number Prefix Number Prefix Number Prefix 1 mono- 4 tetra- 8 octa- 2 di- 5 penta- 9 nona- 3 tri- 6 hexa- 10 deca- 7 hepta- 2-57 Sample Problem 2.10 Determining Names and Formulas of Ionic Compounds Containing Polyatomic Ions PROBLEM: Give the systematic name for each formula or the formula for each name for the following compounds: (a) Fe(ClO4)2 (b) sodium sulfite (c) Ba(OH)2·8H2O PLAN: Remember to use parentheses when more than one unit of a particular polyatomic ion is present in the compound. SOLUTION: (a) ClO4- is perchlorate; Fe must have a 2+ charge since there are 2 ClO4- ions. This is iron(II) perchlorate. (b) The anion sulfite is SO32-; therefore you need 2 Na+ for each sulfite. The formula is Na2SO3. (c) The ionic compound is barium hydroxide. When water is included in the formula, we use the term “hydrate” and a prefix that indicates the number of molecules of H2O. This compound is barium hydroxide octahydrate. 2-58 Sample Problem 2.11 Recognizing Incorrect Names and Formulas of Ionic Compounds PROBLEM: There is an error in the second part of each statement. Provide the correct name or formula in each case. (a) Ba(C2H3O2)2 is called barium diacetate. (b) Sodium sulfide has the formula (Na)2SO3. (c) Iron(II) sulfate has the formula Fe2(SO4)3. (d) Cesium carbonate has the formula Cs2(CO3). SOLUTION: (a) The charge of Ba2+ must be balanced by two C2H3O2- ions. The prefix “di” is not required and is not used in this way when naming ionic compounds. The correct name is simply barium acetate. (b) An ion of a single element does not need parentheses, and sulfide is S2-, not SO32-. The correct formula is Na2S. 2-59 Sample Problem 2.11 (c) Sulfate or SO42- has a 2- charge, and only one Fe2+ is needed to form a neutral compound. The formula should be FeSO4. (d) The parentheses are unnecessary, since only one CO32- ion is present. The correct formula is Cs2CO3. 2-60 Naming Acids 1) Binary acid solutions form when certain gaseous compounds dissolve in water. For example, when gaseous hydrogen chloride (HCl) dissolves in water, it forms a solution called hydrochloric acid. Prefix hydro- + anion nonmetal root + suffix -ic + the word acid hydro + chlor + ic + acid hydrochloric acid 2) Oxoacid names are similar to those of the oxoanions, except for two suffix changes: -ate in the anion becomes –ic in the acid -ite in the anion becomes –ous in the acid The oxoanion prefixes hypo- and per- are retained. Thus, BrO4- is perbromate, and HBrO4 is perbromic acid; IO2- is iodite, and HIO2 is iodous acid. 2-61 Sample Problem 2.12 Determining Names and Formulas of Anions and Acids PROBLEM: Name the following anions and give the name and formula of the acid derived from each: (a) Br (b) IO3 (c) CN (d) SO4 2(e) NO2 SOLUTION: (a) The anion is bromide; the acid is hydrobromic acid, HBr. (b) The anion is iodate; the acid is iodic acid, HIO3. (c) The anion is cyanide; the acid is hydrocyanic acid, HCN. (d) The anion is sulfate; the acid is sulfuric acid, H2SO4. (e) The anion is nitrite; the acid is nitrous acid, HNO2. 2-62 Naming Binary Covalent Compounds A binary covalent compound is typically formed by the combination of two non-metals. Some of these compounds are very common and have trivial names, eg., H2O is water. For a binary covalent compound, the element with the lower group number in the periodic table is first in the name and formula. Its name remains unchanged. The element that is second is named using the root with the suffix –ide. Numerical prefixes indicate the number of atoms of each element present. 2-63 Sample Problem 2.13 Determining Names and Formulas of Binary Covalent Compounds PROBLEM: (a) What is the formula of carbon disulfide? (b) What is the name of PCl5? (c) Give the name and formula of the compound whose molecules each consist of two N atoms and four O atoms. SOLUTION: (a) Carbon is C, sulfide is sulfur S and di-means two; the formula is CS2. (b) P is phosphorous, Cl is chloride, the prefix for 5 is penta-. This is phosphorous pentachloride. (c) N is nitrogen and is in a lower group number than O (oxygen). The compound formula is N2O4 and the name is dinitrogen tetraoxide. 2-64 Sample Problem 2.14 PROBLEM: Recognizing Incorrect Names and Formulas of Binary Covalent Compounds Explain what is wrong with the name of formula in the second part of each statement and correct it: (a) SF4 is monosulfur pentafluoride. (b) Dichlorine heptaoxide is Cl2O6. (c) N2O3 is dinitrotrioxide. SOLUTION: (a) The prefix mono- is not needed if there is only one atom of the first element, and the prefix for four is tetra-. So the name is sulfur tetrafluoride. (b) Hepta- means 7; the formula should be Cl2O7. (c) The first element is given its elemental name so this is dinitrogen trioxide. 2-65 Naming Straight-Chain Alkanes Hydrocarbons are compounds that contain only carbon and hydrogen atoms. Alkanes are the simplest type of hydrocarbon. Alkanes are named using a root name followed by the suffix –ane. 2-66 Table 2.7 The First 10 Straight-Chain Alkanes 2-67 Molecular Masses from Chemical Formulas Molecular mass = sum of atomic masses For the H2O molecule: molecular mass = (2 x atomic mass of H) + (1 x atomic mass of O) = (2 x 1.008 amu) + (1 x 16.00 amu) = 18.02 amu By convention, we read masses off the periodic table to 4 significant figures. For ionic compounds we refer to a formula mass since ionic compounds do not consist of molecules. 2-68 Sample Problem 2.15 Calculating the Molecular Mass of a Compound PROBLEM: Using the periodic table, calculate the molecular (or formula) mass of: (a) tetraphosphorous trisulfide (b) ammonium nitrate PLAN: Write the formula and then multiply the number of atoms by the respective atomic masses. Add the masses for each compound. SOLUTION: (a) P4S3 molecular mass = (4 x atomic mass of P) + (3 x atomic mass of S) = (4 x 30.97 amu) + (3 x 32.07 amu) = 220.09 amu (b) NH4NO3 formula mass = (2 x atomic mass of N) + (4 x atomic mass of H) + (3 x atomic mass of O) = (2 x 14.01 amu) + (4 x 1.008 amu) + (3 x 16.00 amu) = 80.05 amu 2-69 Sample Problem 2.16 Using Molecular Depictions to determine Formula, Name, and Mass for a compound PROBLEM: Each scene represents a binary compound. Determine its formula, name, and molecular (formula) mass. PLAN: Each compound contains only two elements. Find the simplest whole number ratio of atoms in each compound and use this formula to determine the name and the formula mass. 2-70 Sample Problem 2.16 SOLUTION: (a) There is 1 brown Na+ for every green F-, so the formula is NaF, an ionic compound, which is named sodium fluoride. Formula mass = (1 x atomic mass of Na) + (1 x atomic mass of F) = 22.99 amu + 10.00 amu = 41.99 amu (b) There are 3 green F for every blue N, so the formula is NF3, a covalent compound, which is named nitrogen trifluoride. Molecular mass = (1 x atomic mass of N) + (3 x atomic mass of F) = 14.01 amu + (3 x 19.00) = 71.01 amu 2-71 Representing Molecules with Formulas and Models H2O Molecular formula for water. Structural formulas for water. Ball-and-stick model for water. Space-filling model for water. 2-72 Representing Molecules with Formulas and Models 2-73 Figure 2.19 The distinction between mixtures and compounds. S2- Fe2+ A physical mixture of Fe and S8 can be separated using a magnet. 2-74 Fe and S have reacted chemically to form the compound FeS. The elements cannot be separated by physical means. Mixtures A heterogeneous mixture has one or more visible boundaries between the components. A homogeneous mixture has no visible boundaries because the components are mixed as individual atoms, ions, and molecules. A homogeneous mixture is also called a solution. Solutions in water are called aqueous solutions. 2-75 Tools of the Laboratory Basic Separation Techniques Filtration: Separates components of a mixture based upon differences in particle size. Filtration usually involves separating a precipitate from solution. Crystallization: Separation is based upon differences in solubility of components in a mixture. Distillation: separation is based upon differences in volatility. Extraction: Separation is based upon differences in solubility in different solvents (major material). Chromatography: Separation is based upon differences in solubility in a solvent versus a stationary phase. 2-76 Tools of the Laboratory Figure B2.3 2-77 Distillation Tools of the Laboratory Figure B2.4 2-78 Procedure for column chromatography Tools of the Laboratory Figure B2.5 2-79 Principle of gas-liquid chromatography (GLC).
Lecture PowerPoint Chemistry The Molecular Nature of Matter and Change Sixth Edition Martin S. Silberberg 3-1 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 3 Stoichiometry of Formulas and Equations 3-2 Mole - Mass Relationships in Chemical Systems 3.1 The Mole 3.2 Determining the Formula of an Unknown Compound 3.3 Writing and Balancing Chemical Equations 3.4 Calculating Quantities of Reactant and Product 3.5 Fundamentals of Solution Stoichiometry 3-3 The Mole The mole (mol) is the amount of a substance that contains the same number of entities as there are atoms in exactly 12 g of carbon-12. The term “entities” refers to atoms, ions, molecules, formula units, or electrons – in fact, any type of particle. One mole (1 mol) contains 6.022x1023 entities (to four significant figures). This number is called Avogadro’s number and is abbreviated as N. 3-4 Figure 3.1 3-5 One mole (6.022x1023 entities) of some familiar substances. Molar Mass The molar mass (M) of a substance is the mass per mole of its entites (atoms, molecules or formula units). For monatomic elements, the molar mass is the same as the atomic mass in grams per mole. The atomic mass is simply read from the Periodic Table. The molar mass of Ne = 20.18 g/mol. 3-6 For molecular elements and for compounds, the formula is needed to determine the molar mass. The molar mass of O2 = 2 x M of O = 2 x 16.00 = 32.00 g/mol The molar mass of SO2 = 1 x M of S + 2 x M of O = 32.00 + 2(16.00) = 64.00 g/mol 3-7 Table 3.1 Information Contained in the Chemical Formula of Glucose C6H12O6 ( M = 180.16 g/mol) Atoms/molecule of compound Carbon (C) Hydrogen (H) Oxygen (O) 6 atoms 12 atoms 6 atoms 12 mol of atoms 6 mol of atoms Moles of atoms/mole 6 mol of atoms of compound Atoms/mole of compound 6(6.022x1023) atoms 12(6.022x1023) atoms 6(6.022x1023) atoms Mass/molecule of compound 6(12.01 amu) = 72.06 amu 12(1.008 amu) = 12.10 amu 6(16.00 amu) = 96.00 amu Mass/mole of compound 72.06 g 12.10 g 96.00 g 3-8 Interconverting Moles, Mass, and Number of Chemical Entities Mass (g) = no. of moles x no. of grams g 1 mol 1 mol No. of moles = mass (g) x M no. of grams No. of entities = no. of moles x 6.022x1023 entities 1 mol No. of moles = no. of entities x 3-9 1 mol 6.022x1023 entities Figure 3.2 3-10 Mass-mole-number relationships for elements. Sample Problem 3.1 PROBLEM: Calculating the Mass of a Given Amount of an Element Silver (Ag) is used in jewelry and tableware but no longer in U.S. coins. How many grams of Ag are in 0.0342 mol of Ag? PLAN: To convert mol of Ag to mass of Ag in g we need the molar mass of Ag. amount (mol) of Ag multiply by M of Ag (107.9 g/mol) mass (g) of Ag SOLUTION: 0.0342 mol Ag x 3-11 107.9 g Ag 1 mol Ag = 3.69 g Ag Sample Problem 3.2 Calculating the Number of Entities in a Given Amount of an Element PROBLEM: Gallium (Ga) is a key element in solar panels, calculators and other light-sensitive electronic devices. How many Ga atoms are in 2.85 x 10-3 mol of gallium? PLAN: To convert mol of Ga to number of Ga atoms we need to use Avogadro’s number. mol of Ga multiply by 6.022x1023 atoms/mol atoms of Ga 3-12 Sample Problem 3.2 SOLUTION: 2.85 x 10-3 mol Ga atoms x 6.022x1023 Ga atoms 1 mol Ga atoms = 1.72 x 1021 Ga atoms 3-13 Sample Problem 3.3 Calculating the Number of Entities in a Given Mass of an Element PROBLEM: Iron (Fe) is the main component of steel and is therefore the most important metal in society; it is also essential in the body. How many Fe atoms are in 95.8 g of Fe? PLAN: The number of atoms cannot be calculated directly from the mass. We must first determine the number of moles of Fe atoms in the sample and then use Avogadro’s number. mass (g) of Fe divide by M of Fe (55.85 g/mol) amount (mol) of Fe multiply by 6.022x1023 atoms/mol atoms of Fe 3-14 Sample Problem 3.3 SOLUTION: 95.8 g Fe x 1 mol Fe = 1.72 mol Fe 55.85 g Fe 23 1.72 mol Fe x 6.022x10 atoms Fe 1 mol Fe = 1.04 x 1024 atoms Fe 3-15 Figure 3.3 3-16 Amount-mass-number relationships for compounds. Sample Problem 3.4 PROBLEM: PLAN: Calculating the Number of Chemical Entities in a Given Mass of a Compound I Nitrogen dioxide is a component of urban smog that forms from the gases in car exhausts. How many molecules are in 8.92 g of nitrogen dioxide? Write the formula for the compound and calculate its molar mass. Use the given mass to calculate first the number of moles and then the number of molecules. mass (g) of NO2 divide by M amount (mol) of NO2 multiply by 6.022 x 1023 formula units/mol number of NO2 molecules 3-17 Sample Problem 3.4 SOLUTION: NO2 is the formula for nitrogen dioxide. M = (1 x M of N) + (2 x M of O) = 14.01 g/mol + 2(16.00 g/mol) = 46.01 g/mol 8.92 g NO2 x 1 mol NO2 46.01 g NO2 = 0.194 mol NO2 23 0.194 mol NO2 x 6.022x10 molecules NO2 1 mol NO2 = 1.17 x 1023 molecules NO2 3-18 Sample Problem 3.5 PROBLEM: Calculating the Number of Chemical Entities in a Given Mass of a Compound II Ammonium carbonate, a white solid that decomposes on warming, is an component of baking powder. a) How many formula units are in 41.6 g of ammonium carbonate? b) How many O atoms are in this sample? PLAN: Write the formula for the compound and calculate its molar mass. Use the given mass to calculate first the number of moles and then the number of formula units. The number of O atoms can be determined using the formula and the number of formula units. 3-19 Sample Problem 3.5 mass (g) of (NH4)2CO3 divide by M amount (mol) of (NH4)2CO3 multiply by 6.022 x 1023 formula units/mol number of (NH4)2CO3 formula units 3 O atoms per formula unit of (NH4)2CO3 number of O atoms SOLUTION: (NH4)2CO3 is the formula for ammonium carbonate. M = (2 x M of N) + (8 x M of H) + (1 x M of C) + (3 x M of O) = (2 x 14.01 g/mol) + (8 x 1.008 g/mol) + (12.01 g/mol) + (3 x 16.00 g/mol) = 96.09 g/mol 3-20 Sample Problem 3.5 41.6 g (NH4)2CO3 x 1 mol (NH4)2CO3 96.09 g (NH4)2CO3 = 0.433 mol (NH4)2CO3 23 0.433 mol (NH4)2CO3 x 6.022x10 formula units (NH4)2CO3 1 mol (NH4)2CO3 = 2.61x1023 formula units (NH4)2CO3 2.61x1023 formula units (NH4)2CO3 x 3 O atoms 1 formula unit of (NH4)2CO3 = 7.83 x 1023 O atoms 3-21 Mass Percent from the Chemical Formula Mass % of element X = atoms of X in formula x atomic mass of X (amu) x 100 molecular (or formula) mass of compound (amu) Mass % of element X = moles of X in formula x molar mass of X (g/mol) mass (g) of 1 mol of compound 3-22 x 100 Sample Problem 3.6 Calculating the Mass Percent of Each Element in a Compound from the Formula PROBLEM: Glucose (C6H12O6) is a key nutrient for generating chemical potential energy in biological systems. What is the mass percent of each element in glucose? PLAN: Find the molar mass of glucose, which is the mass of 1 mole of glucose. Find the mass of each element in 1 mole of glucose, using the molecular formula. The mass % for each element is calculated by dividing the mass of that element in 1 mole of glucose by the total mass of 1 mole of glucose, multiplied by 100. 3-23 Sample Problem 3.6 PLAN: amount (mol) of element X in 1 mol compound multiply by M (g/mol) of X mass (g) of X in 1 mol of compound divide by mass (g) of 1 mol of compound mass fraction of X multiply by 100 mass % X in compound 3-24 Sample Problem 3.6 SOLUTION: In 1 mole of glucose there are 6 moles of C, 12 moles H, and 6 moles O. 6 mol C x 12.01 g C = 72.06 g C 1 mol C 6 mol O x 16.00 g O = 96.00 g O 1 mol O mass percent of C = mass percent of H = 1.008 g H = 12.096 g H 1 mol H M = 180.16 g/mol 72.06 g C = 0.3999 x 100 = 39.99 mass %C 180.16 g glucose 12.096 g H 180.16 g glucose mass percent of O = 3-25 12 mol H x = 0.06714 x 100 = 6.714 mass %H 96.00 g O = 0.5329 x 100 = 53.29 mass %O 180.16 g glucose Mass Percent and the Mass of an Element Mass percent can also be used to calculate the mass of a particular element in any mass of a compound. Mass of element X present in sample = mass of compound x mass of element in 1 mol of compound mass of 1 mol of compound 3-26 Sample Problem 3.7 Calculating the Mass of an Element in a Compound PROBLEM: Use the information from Sample Problem 3.6 to determine the mass (g) of carbon in 16.55 g of glucose. PLAN: The mass percent of carbon in glucose gives us the relative mass of carbon in 1 mole of glucose. We can use this information to find the mass of carbon in any sample of glucose. mass of glucose sample multiply by mass percent of C in glucose mass of C in sample 3-27 Sample Problem 3.7 SOLUTION: Each mol of glucose contains 6 mol of C, or 72.06 g of C. Mass (g) of C = mass (g) of glucose x 6 mol x M of C (g/mol) mass (g) of 1 mol of glucose = 16.55 g glucose x 3-28 72.06 g C 180.16 g glucose = 6.620 g C Empirical and Molecular Formulas The empirical formula is the simplest formula for a compound that agrees with the elemental analysis. It shows the lowest whole number of moles and gives the relative number of atoms of each element present. The empirical formula for hydrogen peroxide is HO. The molecular formula shows the actual number of atoms of each element in a molecule of the compound. The molecular formula for hydrogen peroxide is H2O2. 3-29 Sample Problem 3.8 Determining an Empirical Formula from Amounts of Elements PROBLEM: A sample of an unknown compound contains 0.21 mol of zinc, 0.14 mol of phosphorus, and 0.56 mol of oxygen. What is its empirical formula? PLAN: Find the relative number of moles of each element. Divide by the lowest mol amount to find the relative mol ratios (empirical formula). amount (mol) of each element use # of moles as subscripts preliminary formula change to integer subscripts empirical formula 3-30 Sample Problem 3.8 SOLUTION: Using the numbers of moles of each element given, we write the preliminary formula Zn0.21P0.14O0.56 Next we divide each fraction by the smallest one; in this case 0.14: 0.21 = 1.5 0.14 0.14 = 1.0 0.14 0.56 = 4.0 0.14 This gives Zn1.5P1.0O4.0 We convert to whole numbers by multiplying by the smallest integer that gives whole numbers; in this case 2: 1.5 x 2 = 3 1.0 x 2 = 2 4.0 x 2 = 8 This gives us the empirical formula Zn3P2O8 3-31 Sample Problem 3.9 Determining an Empirical Formula from Masses of Elements PROBLEM: Analysis of a sample of an ionic compound yields 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What is the empirical formula and the name of the compound? PLAN: Find the relative number of moles of each element. Divide by the lowest mol amount to find the relative mol ratios (empirical formula). mass (g) of each element divide by M (g/mol) amount (mol) of each element use # of moles as subscripts preliminary formula change to integer subscripts empirical formula 3-32 Sample Problem 3.9 SOLUTION: 2.82 g Na x 1 mol Na = 0.123 mol Na 22.99 g Na 4.35 g Cl x 1 mol Cl = 0.123 mol Cl 35.45 g Cl 7.83 g O x 1 mol O = 0.489 mol O 16.00 g O 0.489 0.123 = 1 and O = = 3.98 Na and Cl = 0.123 0.123 The empirical formula is Na1Cl1O3.98 or NaClO4; this compound is named sodium perchlorate. 3-33 Determining the Molecular Formula The molecular formula gives the actual numbers of moles of each element present in 1 mol of compound. The molecular formula is a whole-number multiple of the empirical formula. molar mass (g/mol) empirical formula mass (g/mol) 3-34 = whole-number multiple Sample Problem 3.10 Determining a Molecular Formula from Elemental Analysis and Molar Mass PROBLEM: Elemental analysis of lactic acid (M = 90.08 g/mol) shows it contains 40.0 mass % C, 6.71 mass % H, and 53.3 mass % O. Determine the empirical formula and the molecular formula for lactic acid. PLAN: assume 100 g lactic acid; then mass % = mass in grams divide each mass by M amount (mol) of each element use # mols as subscripts; convert to integers empirical formula divide M by the molar mass for the empirical formula; multiply empirical formula by this number molecular formula 3-35 Sample Problem 3.10 SOLUTION: 40.0 g C x Assuming there are 100. g of lactic acid; 1 mol C 12.01 g C 6.71 g H x 1 mol H 1.008 g H = 3.33 mol C C3.33 H6.66 3.33 3.33 = 6.66 mol H O3.33 3.33 molar mass of lactate mass of CH2O 53.3 g O x = 3.33 mol O CH2O empirical formula 90.08 g/mol 30.03 g/mol C3H6O3 is the molecular formula 3-36 1 mol O 16.00 g O =3 Figure 3.4 Combustion apparatus for determining formulas of organic compounds. m m CnHm + (n+ ) O2 = n CO(g) + H O(g) 2 2 2 3-37 Sample Problem 3.11 Determining a Molecular Formula from Combustion Analysis PROBLEM: When a 1.000 g sample of vitamin C (M = 176.12 g/mol) is placed in a combustion chamber and burned, the following data are obtained: mass of CO2 absorber after combustion = 85.35 g mass of CO2 absorber before combustion = 83.85 g mass of H2O absorber after combustion = 37.96 g mass of H2O absorber before combustion = 37.55 g What is the molecular formula of vitamin C? PLAN: The masses of CO2 and H2O produced will give us the masses of C and H present in the original sample. From this we can determine the mass of O. 3-38 Sample Problem 3.11 (mass after combustion – mass before) for each absorber = mass of compound in each absorber mass of each compound x mass % of oxidized element mass of each oxidized element mass of vitamin C – (mass of C + H) mass of O divide each mass by M mol of C, H, and O use # mols as subscripts; convert to integers empirical formula 3-39 molecular formula Sample Problem 3.11 SOLUTION: For CO2: 85.35 g - 83.85 g = 1.50 g 1.50 g CO2 x 12.01 g C = 0.409 g C 44.01 g CO2 For H2O: 37.96 g - 37.55 g = 0.41 g 0.41 g H2O x 2.016 g H = 0.046 g H 18.02 g H2O mass of O = mass of vitamin C – (mass of C + mass of H) = 1.000 g - (0.409 + 0.046) g = 0.545 g O 3-40 Sample Problem 3.11 Convert mass to moles: 0.409 g C = 0.0341 mol C 12.01 g/mol C 0.545 g O 0.046 g H = 0.0456 mol H 1.008 g/mol H = 0.0341 mol O 16.00 g/mol O Divide by smallest to get the preliminary formula: 0.0456 0.0341 = 1.34 =1 H O C 0.0341 0.0341 C1H1.34O1 = C3H4.01O3 0.0341 =1 0.0341 C 3 H 4 O3 Divide molar mass by mass of empirical formula: 176.12 g/mol 88.06 g 3-41 = 2.000 mol C 6 H 8 O6 Table 3.2 Some Compounds with Empirical Formula CH2O (Composition by Mass: 40.0% C, 6.71% H, 53.3% O) M Molecular Whole-Number Formula (g/mol) Multiple Name Use or Function formaldehyde CH2O 1 30.03 acetic acid C2H4O2 2 60.05 disinfectant; biological preservative lactic acid C3H6O3 3 90.09 acetate polymers; vinegar (5% soln) erythrose C4H8O4 4 120.10 sour milk; forms in exercising muscle ribose C5H10O5 5 150.13 part of sugar metabolism glucose C6H12O6 6 180.16 component of nucleic acids and B2 major energy source of the cell CH2O 3-42 C2H4O2 C3H6O3 C4H8O4 C5H10O5 C6H12O6 Table 3.3 Two Pairs of Constitutional Isomers C4H10 Property Butane C2H6O 2-Methylpropane Ethanol Dimethyl Ether M (g/mol) 58.12 58.12 46.07 46.07 Boiling Point -0.50C -11.060C 78.50C -250C Density at 200C 0.579 g/mL 0.549 g/mL (gas) (gas) Structural formula H H H H H H C C C C H H H H H H C H C 3-43 H C H H C H H H H C C H H H H H H H Space-filling model 0.789 g/mL 0.00195 g/mL (liquid) (gas) OH H C H O C H H Chemical Equations A chemical equation uses formulas to express the identities and quantities of substances involved in a physical or chemical change. Figure 3.6 The formation of HF gas on the macroscopic and molecular levels. 3-44 Figure 3.7 3-45 A three-level view of the reaction between magnesium and oxygen. Features of Chemical Equations A yield arrow points from reactants to products. Mg + O2 MgO Reactants are written on the left. Products are written on the right. The equation must be balanced; the same number and type of each atom must appear on both sides. 3-46 Balancing a Chemical Equation translate the statement magnesium and oxygen gas react to give magnesium oxide: Mg + O2 → MgO balance the atoms using coefficients; formulas cannot be changed 2Mg + O2 → 2MgO adjust coefficients if necessary check that all atoms balance specify states of matter 2Mg (s) + O2 (g) → 2MgO (s) 3-47 Sample Problem 3.12 Balancing Chemical Equations PROBLEM: Within the cylinders of a car’s engine, the hydrocarbon octane (C8H18), one of many components of gasoline, mixes with oxygen from the air and burns to form carbon dioxide and water vapor. Write a balanced equation for this reaction. PLAN: SOLUTION: translate the statement balance the atoms C8H18 + C8H18 + 25 2 O2 O2 CO2 + 8 CO2 + H2 O 9 H2 O adjust the coefficients 2C8H18 + 25O2 16CO2 + 18H2O check the atoms balance 2C8H18 + 25O2 16CO2 + 18H2O specify states of matter 3-48 2C8H18(l) + 25O2 (g) 16CO2 (g) + 18H2O (g) Molecular Scene Combustion of Octane 3-49 Sample Problem 3.13 Balancing an Equation from a Molecular Scene PROBLEM: The following molecular scenes depict an important reaction in nitrogen chemistry. The blue spheres represent nitrogen while the red spheres represent oxygen. Write a balanced equation for this reaction. PLAN: Determine the formulas of the reactants and products from their composition. Arrange this information in the correct equation format and balance correctly, including the states of matter. 3-50 Sample Problem 3.13 SOLUTION: The reactant circle shows only one type of molecule, composed of 2 N and 5 O atoms. The formula is thus N2O5. There are 4 N2O5 molecules depicted. The product circle shows two types of molecule; one has 1 N and 2 O atoms while the other has 2 O atoms. The products are NO2 and O2. There are 8 NO2 molecules and 2 O2 molecules shown. The reaction depicted is 4 N2O5 → 8 NO2 + 2 O2. Writing the equation with the smallest whole-number coefficients and states of matter included; 2 N2O5 (g) → 4 NO2 (g) + O2 (g) 3-51 Stoichiometric Calculations • The coefficients in a balanced chemical equation – represent the relative number of reactant and product particles – and the relative number of moles of each. • Since moles are related to mass – the equation can be used to calculate masses of reactants and/or products for a given reaction. • The mole ratios from the balanced equation are used as conversion factors. 3-52 Table 3.4 Information Contained in a Balanced Equation Viewed in Terms of Reactants C3H8(g) + 5 O2(g) Molecules 1 molecule C3H8 + 5 molecules O2 Amount (mol) 1 mol C3H8 + 5 mol O2 Mass (amu) 44.09 amu C3H8 + 160.00 amu O2 Mass (g) Total Mass (g) 3-53 44.09 g C3H8 + 160.00 g O2 204.09 g Products 3 CO2(g) + 4 H2O(g) 3 molecules CO2 + 4 molecules H2O 3 mol CO2 + 4 mol H2O 132.03 amu CO2 + 72.06 amu H2O 132.03 g CO2 + 72.06 g H2O 204.09 g Figure 3.8 3-54 Summary of amount-mass-number relationships in a chemical equation. Sample Problem 3.14 PROBLEM: Calculating Quantities of Reactants and Products: Amount (mol) to Amount (mol) Copper is obtained from copper(I) sulfide by roasting it in the presence of oxygen gas) to form powdered copper(I) oxide and gaseous sulfur dioxide. How many moles of oxygen are required to roast 10.0 mol of copper(I) sulfide? PLAN: write and balance the equation use the mole ratio as a conversion factor moles of oxygen SOLUTION: 2 Cu2S (s) + 3 O2 (g) → 2 Cu2O (s) + 2 SO2 (g) 10.0 mol Cu2S x 3 mol O2 2 mol Cu2S 3-55 = 15.0 mol O2 Sample Problem 3.15 PROBLEM: PLAN: Calculating Quantities of Reactants and Products: Amount (mol) to Mass (g) During the process of roasting copper(I) sulfide, how many grams of sulfur dioxide form when 10.0 mol of copper(I) sulfide reacts? Using the balanced equation from the previous problem, we again use the mole ratio as a conversion factor. mol of copper(I) sulfide use the mole ratio as a conversion factor mol of sulfur dioxide multiply by M of sulfur dioxide mass of sulfur dioxide 3-56 Sample Problem 3.15 SOLUTION: 2 Cu2S (s) + 3 O2 (g) → 2 Cu2O (s) + 2 SO2 (g) 10.0 mol Cu2S x 2 mol SO2 x 64.07 g SO2 2 mol Cu2S 3-57 1 mol SO2 = 641 g SO2 Sample Problem 3.16 PROBLEM: PLAN: Calculating Quantities of Reactants and Products: Mass to Mass During the roasting of copper(I) sulfide, how many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide? mass of oxygen divide by M of oxygen mol of oxygen use mole ratio as conversion factor mol of copper(I) oxide multiply by M of copper(I) oxide mass of copper(I) oxide 3-58 Sample Problem 3.16 SOLUTION: 2 Cu2S (s) + 3 O2 (g) → 2 Cu2O (s) + 2 SO2 (g) 3 2.86 kg Cu2O x 10 g 1 kg 20.0 mol Cu2O x 3 mol O2 2 mol Cu2O 3-59 x 1 mol Cu2O 143.10 g Cu2O x 32.00 g O2 1 mol O2 = 20.0 mol Cu2O x 1 kg 103 g = 0.959 kg O2 Reactions in Sequence • Reactions often occur in sequence. • The product of one reaction becomes a reactant in the next. • An overall reaction is written by combining the reactions; – any substance that forms in one reaction and reacts in the next can be eliminated. 3-60 Sample Problem 3.17 Writing an Overall Equation for a Reaction Sequence PROBLEM: Roasting is the first step in extracting copper from chalcocite, the ore used in the previous problem. In the next step, copper(I) oxide reacts with powdered carbon to yield copper metal and carbon monoxide gas. Write a balanced overall equation for the two-step process. PLAN: Write individual balanced equations for each step. Adjust the coefficients so that any common substances can be canceled. Add the adjusted equations together to obtain the overall equation. 3-61 Sample Problem 3.17 SOLUTION: Write individual balanced equations for each step: 2Cu2S (s) + 3O2 (g) → 2Cu2O (s) + 2SO2 (g) Cu2O (s) + C (s) → 2Cu (s) + CO (g) Adjust the coefficients so that the 2 moles of Cu2O formed in reaction 1 are used up in reaction 2: 2Cu2S (s) + 3O2 (g) → 2Cu2O (s) + 2SO2 (g) 2Cu2O (s) + 2C (s) → 4Cu (s) + 2CO (g) Add the equations together: 2Cu2S (s) + 3O2 (g) + 2C (s) → 2SO2 (g) + 4Cu (s) + 2CO (g) 3-62 Limiting Reactants • So far we have assumed that reactants are present in the correct amounts to react completely. • In reality, one reactant may limit the amount of product that can form. • The limiting reactant will be completely used up in the reaction. • The reactant that is not limiting is in excess – some of this reactant will be left over. 3-63 Figure 3.10 3-64 An ice cream sundae analogy for limiting reactions. Sample Problem 3.18 Using Molecular Depictions in a LimitingReactant Problem PROBLEM: Chlorine trifluoride, an extremely reactive substance, is formed as a gas by the reaction of elemental chlorine and fluorine. The molecular scene shows a representative portion of the reaction mixture before the reaction starts. (Chlorine is green, and fluorine is yellow.) (a) Find the limiting reactant. (b) Write a reaction table for the process. (c) Draw a representative portion of the mixture after the reaction is complete. (Hint: The ClF3 molecule has 1 Cl atom bonded to 3 individual F atoms). 3-65 Sample Problem 3.18 PLAN: Write a balanced chemical equation. To determine the limiting reactant, find the number of molecules of product that would form from the given numbers of molecules of each reactant. Use these numbers to write a reaction table and use the reaction table to draw the final reaction scene. SOLUTION: The balanced equation is Cl2 (g) + 3F2 (g) → 2ClF3 (g) There are 3 molecules of Cl2 and 6 molecules of F2 depicted: 3 molecules Cl2 x 2 molecules ClF3 1 molecule Cl2 6 molecules F2 x 2 molecules ClF3 3 molecule Cl2 = 6 molecules ClF3 = 4 molecules ClF3 Since the given amount of F2 can form less product, it is the limiting reactant. 3-66 Sample Problem 3.18 We use the amount of F2 to determine the “change” in the reaction table, since F2 is the limiting reactant: Molecules Cl2 (g) + 3F2 (g) → 2ClF3 (g) Initial Change 3 -2 6 -6 0 +4 Final 1 0 4 The final reaction scene shows that all the F2 has reacted and that there is Cl2 left over. 4 molecules of ClF2 have formed: 3-67 Calculating Quantities in a LimitingReactant Problem: Amount to Amount PROBLEM: In another preparation of ClF3, 0.750 mol of Cl2 reacts with 3.00 mol of F2. Sample Problem 3.19 (a) Find the limiting reactant. (b) Write a reaction table. PLAN: Find the limiting reactant by calculating the amount (mol) of ClF3 that can be formed from each given amount of reactant. Use this information to construct a reaction table. SOLUTION: The balanced equation is Cl2 (g) + 3F2 (g) → 2ClF3 (g) 0.750 mol Cl2 x 2 mol ClF3 1 mol Cl2 3.00 mol F2 x 2 mol ClF3 3 mol F2 3-68 = 1.50 mol ClF3 = 2.00 mol ClF3 Cl2 is limiting, because it yields less ClF3. Sample Problem 3.19 All the Cl2 reacts since this is the limiting reactant. For every 1 Cl2 that reacts, 3 F2 will react, so 3(0.750) or 2.25 moles of F2 reacts. 3-69 + 3F2 (g) → 2ClF3 (g) Moles Cl2 (g) Initial Change 0.750 -0.750 3.00 - 2.25 0 +1.50 Final 0 0.75 1.50 Sample Problem 3.20 Calculating Quantities in a LimitingReactant Problem: Mass to Mass PROBLEM: A fuel mixture used in the early days of rocketry consisted of two liquids, hydrazine (N2H4) and dinitrogen tetraoxide (N2O4), which ignite on contact to form nitrogen gas and water vapor. (a) How many grams of nitrogen gas form when 1.00 x 102 g of N2H4 and 2.00 x 102 g of N2O4 are mixed? (b) Write a reaction table for this process. PLAN: 3-70 Find the limiting reactant by calculating the amount (mol) of ClF3 that can be formed from each given mass of reactant. Use this information to construct a reaction table. Sample Problem 3.20 mass (g) of N2H4 mass (g) of N2O4 divide by M divide by M mol of N2O4 mol of N2H4 mole ratio mole ratio mol of N2 mol of N2 select lower number of moles of N2 multiply by M mass of N2 3-71 Sample Problem 3.20 SOLUTION: For N2H4: 2N2H4 (l) + N2O (l) → 3N2 (g) + 4H2O (g) 1.00x 102 g N2H4 x 1 mol N2H4 = 3.12 mol N2H4 32.05 g N2H4 3.12 mol N2H4 x 3 mol N2 2 mol N2H4 For N2O4: 2.00x 102 g N2O4 x 1 mol N2O4 92.02 g N2O4 2.17 mol N2O4 x 3 mol N2 1 mol N2O4 = 4.68 mol N2 = 2.17 mol N2 = 6.51 mol N2 N2H4 is limiting and only 4.68 mol of N2 can be produced: 4.68 mol N2 x 28.02 g N2 1 mol N2 3-72 = 131 g N2 Sample Problem 3.20 All the N2H4 reacts since it is the limiting reactant. For every 2 moles of N2H4 that react 1 mol of N2O4 reacts and 3 mol of N2 form: 3.12 mol N2H4 x 1 mol N2O4 2 mol N2H4 3-73 = 1.56 mol N2O4 reacts Moles 2N2H4 (l) + N2O4 (l) → 3N2 (g) + 4H2O (g) Initial Change 3.12 -3.12 Final 0 2.17 - 1.56 0 +4.68 0 +6.24 0.61 4.68 6.24 Reaction Yields The theoretical yield is the amount of product calculated using the molar ratios from the balanced equation. The actual yield is the amount of product actually obtained. The actual yield is usually less than the theoretical yield. % yield = actual yield theoretical yield 3-74 x 100 Figure 3.11 3-75 The effect of side reactions on the yield of the main product. Sample Problem 3.21 Calculating Percent Yield PROBLEM: Silicon carbide (SiC) is made by reacting sand(silicon dioxide, SiO2) with powdered carbon at high temperature. Carbon monoxide is also formed. What is the percent yield if 51.4 kg of SiC is recovered from processing 100.0 kg of sand? PLAN: write balanced equation find mol reactant find mol product find g product predicted 3-76 percent yield Sample Problem 3.21 SOLUTION: SiO2(s) + 3C(s) → SiC(s) + 2CO(g) 3 100.0 kg SiO2 x 10 g x 1 mol SiO2 = 1664 mol SiO 2 1 kg 60.09 g SiO2 mol SiO2 = mol SiC = 1664 mol SiC 1664 mol SiC x 40.10 g SiC x 1 kg = 66.73 kg 1 mol SiC 103g 51.4 kg x 100 66.73 kg 3-77 = 77.0% Solution Stoichiometry • Many reactions occur in solution. • A solution consists of one or more solutes dissolved in a solvent. • The concentration of a solution is given by the quantity of solute present in a given quantity of solution. • Molarity (M) is often used to express concentration. Molarity = moles solute liters of solution 3-78 M= mol solute L soln Sample Problem 3.22 Calculating the Molarity of a Solution PROBLEM: What is the molarity of an aqueous solution that contains 0.715 mol of glycine (H2NCH2COOH) in 495 mL? PLAN: SOLUTION: Molarity is the number of moles of solute per liter of solution. mol of glycine divide by volume concentration in mol/mL 103 mL = 1 L molarity of glycine 3-79 0.715 mol glycine x 1000 mL 495 mL soln 1L = 1.44 M glycine Figure 3.13 3-80 Summary of mass-mole-number-volume relationships in solution. Sample Problem 3.23 Calculating Mass of Solute in a Given Volume of Solution PROBLEM: How many grams of solute are in 1.75 L of 0.460 M sodium monohydrogen phosphate buffer solution? PLAN: Calculate the moles of solute using the given molarity and volume. Convert moles to mass using the molar mass of the solute. volume of solution multiply by M moles of solute multiply by M SOLUTION: grams of solute 1.75 L x 0.460 moles = 0.805 mol Na2HPO4 1L 0.805 mol Na2HPO4 x 141.96 g Na2HPO4 1 mol Na2HPO4 3-81 = 114 g Na2HPO4 Figure 3.14 Laboratory preparation of molar solutions. A •Weigh the solid needed. •Transfer the solid to a volumetric flask that contains about half the final volume of solvent. 3-82 C Add solvent until the solution reaches its final volume. B Dissolve the solid thoroughly by swirling. Figure 3.15 Converting a concentrated solution to a dilute solution. 3-83 Sample Problem 3.24 Preparing a Dilute Solution from a Concentrated Solution PROBLEM: “Isotonic saline” is a 0.15 M aqueous solution of NaCl. How would you prepare 0.80 L of isotonic saline from a 6.0 M stock solution? PLAN: To dilute a concentrated solution, we add only solvent, so the moles of solute are the same in both solutions. The volume and molarity of the dilute solution gives us the moles of solute. Then we calculate the volume of concentrated solution that contains the same number of moles. volume of dilute soln multiply by M of dilute soln moles of NaCl in dilute soln = mol NaCl in concentrated soln divide by M of concentrated soln L of concentrated soln 3-84 Sample Problem 3.24 Mdil x Vdil = # mol solute = Mconc x Vconc SOLUTION: Using the volume and molarity for the dilute solution: 0.80 L soln x 0.15 mol NaCl 1 L soln = 0.12 mol NaCl Using the moles of solute and molarity for the concentrated solution: 0.12 mol NaCl x 1 L soln 6.0 mol NaCl = 0.020 L soln A 0.020 L portion of the concentrated solution must be diluted to a final volume of 0.80 L. 3-85 Sample Problem 3.25 Visualizing Changes in Concentration PROBLEM: The beaker and circle represents a unit volume of solution. Draw the solution after each of these changes: (a) For every 1 mL of solution, 1 mL of solvent is added. (b) One third of the volume of the solution is boiled off. PLAN: 3-86 Only the volume of the solution changes; the total number of moles of solute remains the same. Find the new volume and calculate the number of moles of solute per unit volume. Sample Problem 3.25 SOLUTION: (a) Ndil x Vdil = Nconc x Vconc where N is the number of particles. Ndil = Nconc x Vconc = 8 particles x 1 mL = 4 particles Vdil 2 mL Vdil (b) N = N x = 8 particles x 1 mL = 12 particles conc dil Vconc 2 mL 3 (a) 3-87 (b) Sample Problem 3.26 Calculating Quantities of Reactants and Products for a Reaction in Solution PROBLEM: A 0.10 M HCl solution is used to simulate the acid concentration of the stomach. How many liters of “stomach acid” react with a tablet containing 0.10 g of magnesium hydroxide? PLAN: Write a balanced equation and convert the mass of Mg(OH)2 to moles. Use the mole ratio to determine the moles of HCl, then convert to volume using molarity. mass Mg(OH)2 divide by M mol Mg(OH)2 use mole ratio mol HCl L HCl divide by M 3-88 Sample Problem 3.26 SOLUTION: Mg(OH)2 (s) + 2HCl (aq) → MgCl2 (aq) + 2H2O (l) 0.10 g Mg(OH)2 x 1 mol Mg(OH)2 = 1.7x10-3 mol Mg(OH)2 58.33 g Mg(OH)2 = 1.7x10-3 mol Mg(OH)2 x 2 mol HCl 1 mol Mg(OH)2 3.4x10-3 mol HCl x 1L HCl soln 0.10 mol HCl 3-89 = 3.4x10-3 mol HCl = 3.4x10-2 L HCl Sample Problem 3.27 Solving Limiting-Reactant Problems for Reactions in Solution PROBLEM: In a simulation mercury removal from industrial wastewater, 0.050 L of 0.010 M mercury(II) nitrate reacts with 0.020 L of 0.10 M sodium sulfide. How many grams of mercury(II) sulfide form? Write a reaction table for this process. PLAN: Write a balanced chemical reaction. Determine limiting reactant. Calculate the grams of mercury(II) sulfide product. volume of Hg(NO3)2 soln volume of Na2S soln multiply by M multiply by M mol of Hg(NO3)2 mol of Na2S mole ratio mole ratio mol of HgS mol of HgS select lower number of moles of HgS multiply by M mass of HgS 3-90 Sample Problem 3.27 SOLUTION: Hg(NO3)2 (aq) + Na2S (aq) → HgS (s) + 2NaNO3 (aq) 1 mol HgS 0.050 L Hg(NO3)2 x 0.010 mol Hg(NO3)2 x 1 L Hg(NO3)2 1 mol Hg(NO3)2 = 5.0x10-4 mol HgS 0.020 L Na2S x 0. 10 mol Na2S x 1 mol HgS 1 L Na2S 1 mol Na2S = 2.0x10-3 mol HgS Hg(NO3)2 is the limiting reactant because it yields less HgS. 5.0 x 10-4 mol HgS x 232.7 g HgS 1 mol HgS 3-91 = 0.12 g HgS Sample Problem 3.27 The reaction table is constructed using the amount of Hg(NO3)2 to determine the changes, since it is the limiting reactant. Amount Hg(NO3)2 (aq) + Na2S (aq) → HgS (s) + 2NaNO3 (aq) Initial Change 5.0 x 10-4 -5.0 x 10-4 0 +5.0 x 10-4 0 +1.0 x 10-3 Final 0 5.0 x 10-4 +1.0 x 10-3 3-92 2.0 x 10-3 -5.0 x 10-4 1.5 x 10-3 Figure 3.16 3-93 An overview of amount-mass-number stoichiometric relationships.
Lecture PowerPoint Chemistry The Molecular Nature of Matter and Change Sixth Edition Martin S. Silberberg 4-1 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 4 Three Major Classes of Chemical Reactions 4-2 The Major Classes of Chemical Reactions 4.1 The Role of Water as a Solvent 4.2 Writing Equations for Aqueous Ionic Reactions 4.3 Precipitation Reactions 4.4 Acid-Base Reactions 4.5 Oxidation-Reduction (Redox) Reactions 4.6 Elements in Redox Reactions 4.7 Reaction Reversibility and the Equilibrium State 4-3 Water as a Solvent • Water is a polar molecule – since it has uneven electron distribution – and a bent molecular shape. • Water readily dissolves a variety of substances. • Water interacts strongly with its solutes and often plays an active role in aqueous reactions. 4-4 Figure 4.1 Electron distribution in molecules of H2 and H2O. A. Electron charge distribution in H2 is symmetrical. B. Electron charge distribution in H2O is asymmetrical. C. Each bond in H2O is polar. D. The whole H2O molecule is polar. 4-5 Figure 4.2 4-6 An ionic compound dissolving in water. Figure 4.3 4-7 The electrical conductivity of ionic solutions. Sample Problem 4.1 Using Molecular Scenes to Depict an Ionic Compound in Aqueous Solution PROBLEM: The beakers shown below contain aqueous solutions of the strong electrolyte potassium sulfate. (a) Which beaker best represents the compound in solution? (H2O molecules are not shown). (b) If each particle represents 0.10 mol, what is the total number of particles in solution? 4-8 Sample Problem 4.1 PLAN: (a) Determine the formula and write and equation for the dissociation of 1 mol of compound. Potassium sulfate is a strong electrolyte; it therefore dissociates completely in solution. Remember that polyatomic ions remain intact in solution. (b) Count the number of separate particles in the relevant beaker, then multiply by 0.1 mol and by Avogadro’s number. SOLUTION: (a) The formula is K2SO4, so the equation for dissociation is: K2SO4 (s) → 2K+ (aq) + SO42- (aq) 4-9 Sample Problem 4.1 There should be 2 cations for every 1 anion; beaker C represents this correctly. (b) Beaker C contains 9 particles, 6 K+ ions and 3 SO42- ions. 6.022x1023 particles 9 x 0.1 mol x 1 mol 4-10 = 5.420x1023 particles Sample Problem 4.2 PROBLEM: (a) (b) (c) (d) PLAN: 4-11 Determining Amount (mol) of Ions in Solution What amount (mol) of each ion is in each solution? 5.0 mol of ammonium sulfate dissolved in water 78.5 g of cesium bromide dissolved in water 7.42x1022 formula units of copper(II) nitrate dissolved in water 35 mL of 0.84 M zinc chloride Write an equation for the dissociation of 1 mol of each compound. Use this information to calculate the actual number of moles represented by the given quantity of substance in each case. Sample Problem 4.2 SOLUTION: (a) The formula is (NH4)2SO4 so the equation for dissociation is: (NH4)2SO4 (s) → 2NH4+ (aq) + SO42- (aq) 5.0 mol (NH4)2SO4 x 2 mol NH4+ 1 mol (NH4)2SO4 5.0 mol (NH4)2SO4 x 1 mol SO42- 1 mol (NH4)2SO4 4-12 = 10. mol NH4+ = 5.0 mol NH4+ Sample Problem 4.2 SOLUTION: (b) The formula is CsBr so the equation for dissociation is: CsBr (s) → Cs+ (aq) + Br- (aq) + 78.5 g CsBr x 1 mol CsBr x 1 mol Cs 212.8 g CsBr 1 mol CsBr = 0.369 mol Cs+ There is one Cs+ ion for every Br- ion, so the number of moles of Br- is also equation to 0.369 mol. 4-13 Sample Problem 4.2 SOLUTION: (c) The formula is Cu(NO3)2 so the formula for dissociation is: Cu(NO3)2 (s) → Cu2+ (aq) + 2NO3- (aq) 7.42x1022 formula units Cu(NO3)2 x 1 mol 6.022x1023 formula units = 0.123 mol Cu(NO3)2 0.123 mol Cu(NO3)2 x 1 mol Cu2+ 1 mol Cu(NO3)2 = 0.123 mol Cu2+ ions There are 2 NO3- ions for every 1 Cu2+ ion, so there are 0.246 mol NO3- ions. 4-14 Sample Problem 4.2 SOLUTION: (d) The formula is ZnCl2 so the formula for dissociation is: ZnCl2 (s) → Zn2+ (aq) + 2Cl- (aq) 35 mL soln x 1 L x 103 mL 2.9x10-2 mol ZnCl2 x 0.84 mol ZnCl2 = 2.9x10-2 mol ZnCl2 1 L soln 2 mol Cl- = 5.8x10-2 mol Cl- 1 mol ZnCl2 There is 1 mol of Zn2+ ions for every 1 mol of ZnCl2, so there are 2.9 x 10-2 mol Zn2+ ions. 4-15 Writing Equations for Aqueous Ionic Reactions The molecular equation shows all reactants and products as if they were intact, undissociated compounds. This gives the least information about the species in solution. 2AgNO3 (aq) + Na2CrO4 (aq) → Ag2CrO4 (s) + 2NaNO3 (aq) When solutions of silver nitrate and sodium chromate mix, a brick-red precipitate of silver chromate forms. 4-16 The total ionic equation shows all soluble ionic substances dissociated into ions. This gives the most accurate information about species in solution. 2Ag+ (aq) + 2NO3- (aq) + 2Na+ (aq) + CrO42- (aq) → Ag2CrO4 (s) + 2Na+ (aq) + NO3- (aq) Spectator ions are ions that are not involved in the actual chemical change. Spectator ions appear unchanged on both sides of the total ionic equation. 2Ag+ (aq) + 2NO3- (aq) + 2Na+ (aq) + CrO42- (aq) 4-17 → Ag2CrO4 (s) + 2Na+ (aq) + 2NO3- (aq) The net ionic equation eliminates the spectator ions and shows only the actual chemical change. 2Ag+ (aq) + CrO42- (aq) 4-18 → Ag2CrO4 (s) Figure 4.4 4-19 An aqueous ionic reaction and the three types of equations. Precipitation Reactions • In a precipitation reaction two soluble ionic compounds react to give an insoluble products, called a precipitate. • The precipitate forms through the net removal of ions from solution. • It is possible for more than one precipitate to form in such a reaction. 4-20 Figure 4.5 The precipitation of calcium fluoride. 2NaF (aq) + CaCl2 (aq) → CaF2 (s) + 2NaCl (aq) 2 Na+ (aq) + 2 F- (aq) + Ca2+ (aq) + 2 Cl- (aq) 2 NaF(aq) + CaCl2 (aq) 4-21 → → CaF2(s) + 2 Na+ (aq) + 2 Cl- (aq) CaF2(s) + 2 NaCl (aq) Figure 4.6 The precipitation of PbI2, a metathesis reaction. 2NaI (aq) + Pb(NO3)2 (aq) → PbI2 (s) + NaNO3 (aq) 2Na+(aq) + 2I- (aq) + Pb2+ (aq) + 2NO3- (aq) → PbI2 (s) + 2Na+ (aq) + 2NO3-(aq) Pb2+ (aq) + 2I- (aq) → PbI2 (s) Precipitation reactions are also called double displacement reactions or metathesis. 2NaI (aq) + Pb (NO3)2 (aq) → PbI2 (s) + 2NaNO3 (aq) Ions exchange partners and a precipitate forms, so there is an exchange of bonds between reacting species. 4-22 Predicting Whether a Precipitate Will Form • Note the ions present in the reactants. • Consider all possible cation-anion combinations. • Use the solubility rules to decide whether any of the ion combinations is insoluble. – An insoluble combination identifies the precipitate that will form. 4-23 Table 4.1 Solubility Rules for Ionic Compounds in Water Soluble Ionic Compounds 1. All common compounds of Group 1A(1) ions (Li+, Na+, K+, etc.) and ammonium ion (NH4+) are soluble. 2. All common nitrates (NO3-), acetates (CH3COO- or C2H3O2-) and most perchlorates (ClO4-) are soluble. 3. All common chlorides (Cl-), bromides (Br-) and iodides (I-) are soluble, except those of Ag+, Pb2+, Cu+, and Hg22+. All common fluorides (F-) are soluble except those of Pb2+ and Group 2A(2). 4. All common sulfates (SO22-) are soluble, except those of Ca2+, Sr2+, Ba2+, Ag+, and Pb2+. Insoluble Ionic Compounds 1. All common metal hydroxides are insoluble, except those of Group 1A(1) and the larger members of Group 2A(2)(beginning with Ca2+). 2. All common carbonates (CO32-) and phosphates (PO43-) are insoluble, except those of Group 1A(1) and NH4+. 3. All common sulfides are insoluble except those of Group 1A(1), Group 2A(2) and NH4+. 4-24 Sample Problem 4.3 Predicting Whether a Precipitation Reaction Occurs; Writing Ionic Equations PROBLEM: Predict whether or not a reaction occurs when each of the following pairs of solutions are mixed. If a reaction does occur, write balanced molecular, total ionic, and net ionic equations, and identify the spectator ions. (a) potassium fluoride (aq) + strontium nitrate (aq) → (b) ammonium perchlorate (aq) + sodium bromide (aq) → PLAN: 4-25 Note reactant ions, write the possible cation-anion combinations, and use Table 4.1 to decide if the combinations are insoluble. Write the appropriate equations for the process. Sample Problem 4.3 SOLUTION: (a) The reactants are KF and Sr(NO3)2. The possible products are KNO3 and SrF2. KNO3 is soluble, but SrF2 is an insoluble combination. Molecular equation: 2KF (aq) + Sr(NO3)2 (aq) → 2 KNO3 (aq) + SrF2 (s) Total ionic equation: 2K+ (aq) + 2F- (aq) + Sr2+ (aq) + 2NO3- (aq) → 2K+ (aq) + 2NO3- (aq) + SrF2 (s) K+ and NO3- are spectator ions Net ionic equation: Sr2+ (aq) + 2F- (aq) → SrF2 (s) 4-26 Sample Problem 4.3 SOLUTION: (b) The reactants are NH4ClO4 and NaBr. The possible products are NH4Br and NaClO4. Both are soluble, so no precipitate forms. Molecular equation: NH4ClO4 (aq) + NaBr (aq) → NH4Br (aq) + NaClO4 (aq) Total ionic equation: NH4+ (aq) + ClO4- (aq) + Na+ (aq) + Br- (aq) → NH4+ (aq) + Br- (aq) + Na+ (aq) + ClO4- (aq) All ions are spectator ions and there is no net ionic equation. 4-27 Sample Problem 4.4 Using Molecular Depictions in Precipitation Reactions PROBLEM: The following molecular views show reactant solutions for a precipitation reaction (with H2O molecules omitted for clarity). (a) Which compound is dissolved in beaker A: KCl, Na2SO4, MgBr2, or Ag2SO4? (b) Which compound is dissolved in beaker B: NH4NO3, MgSO4, Ba(NO3)2, or CaF2? 4-28 Sample Problem 4.4 PLAN: Note the number and charge of each kind of ion and use Table 4.1 to determine the ion combinations that are soluble. SOLUTION: (a) Beaker A contains two 1+ ion for each 2- ion. Of the choices given, only Na2SO4 and Ag2SO4 are possible. Na2SO4 is soluble while Ag2SO4 is not. Beaker A therefore contains Na2SO4. (b) Beaker B contains two 1- ions for each 2+ ion. Of the choices given, only CaF2 and Ba(NO3)2 match this description. CaF2 is not soluble while Ba(NO3)2 is soluble. Beaker B therefore contains Ba(NO3)2. 4-29 Sample Problem 4.4 PROBLEM: (c) Name the precipitate and spectator ions when solutions A and B are mixed, and write balanced molecular, total ionic, and net ionic equations for this process. (d) If each particle represents 0.010 mol of ions, what is the maximum mass (g) of precipitate that can form (assuming complete reaction)? PLAN: (c) Consider the cation-anion combinations from the two solutions and use Table 4.1 to decide if either of these is insoluble. SOLUTION: The reactants are Ba(NO3)2 and Na2SO4. The possible products are BaSO4 and NaNO3. BaSO4 is insoluble while NaNO3 is soluble. 4-30 Sample Problem 4.4 Molecular equation: Ba(NO3)2 (aq) + Na2SO4 (aq) → 2NaNO3 (aq) + BaSO4 (s) Total ionic equation: Ba2+ (aq) + 2NO3- (aq) + 2Na+ (aq) + SO42- (aq) → 2Na+ (aq) + 2NO3- (aq) + BaSO4 (s) Na+ and NO3- are spectator ions Net ionic equation: Ba2+ (aq) + SO42- (aq) → BaSO4 (s) 4-31 Sample Problem 4.4 PLAN: (d) Count the number of each kind of ion that combines to form the solid. Multiply the number of each reactant ion by 0.010 mol and calculate the mol of product formed from each. Decide which ion is the limiting reactant and use this information to calculate the mass of product formed. SOLUTION: There are 4 Ba2+ particles and 5 SO42- particles depicted. 2+ 4 Ba2+ particles x 0.010 mol Ba x 1 mol BaSO4 = 0.040 mol BaSO 4 1 mol Ba2+ 1 particle 24 SO42- particles x 0.010 mol SO4 x 1 mol BaSO4 = 0.050 mol BaSO4 1 mol SO421 particle 4-32 Sample Problem 4.4 Ba2+ ion is the limiting reactant, since it yields less BaSO4. 0.040 mol BaSO4 x 233.4 g BaSO4 = 9.3 g BaSO 4 1 mol BaSO4 4-33 Acid-Base Reactions An acid is a substance that produces H+ ions when dissolved in H2O. HX H2O → H+ (aq) + X- (aq) A base is a substance that produces OH- ions when dissolved in H2O. MOH H2O → M+ (aq) + OH- (aq) An acid-base reaction is also called a neutralization reaction. 4-34 Figure 4.7 The H+ ion as a solvated hydronium ion. H+ interacts strongly with H2O, forming H3O+ in aqueous solution. 4-35 Table 4.2 Selected Acids and Bases Acids Bases Strong Strong hydrochloric acid, HCl sodium hydroxide, NaOH hydrobromic acid, HBr potassium hydroxide, KOH hydriodic acid, HI calcium hydroxide, Ca(OH)2 nitric acid, HNO3 strontium hydroxide, Sr(OH)2 sulfuric acid, H2SO4 barium hydroxide, Ba(OH)2 perchloric acid, HClO4 Weak hydrofluoric acid, HF phosphoric acid, H3PO4 acetic acid, CH3COOH (or HC2H3O2) 4-36 Weak ammonia, NH3 Figure 4.8 Acids and bases as electrolytes. Strong acids and strong bases dissociate completely into ions in aqueous solution. They are strong electrolytes and conduct well in solution. 4-37 Figure 4.8 Acids and bases as electrolytes. Weak acids and weak bases dissociate very little into ions in aqueous solution. They are weak electrolytes and conduct poorly in solution. 4-38 Determining the Number of H+ (or OH-) Ions in Solution Sample Problem 4.5 PROBLEM: How many H+(aq) ions are in 25.3 mL of 1.4 M nitric acid? PLAN: Use the volume and molarity to determine the mol of acid present. Since HNO3 is a strong acid, moles acid = moles H+. volume of HNO3 convert mL to L and multiply by M mol of HNO3 mole of H+ = mol of HNO3 mol of H+ multiply by Avogadro’s number number of H+ ions 4-39 Sample Problem 4.5 SOLUTION: 35.3 mL soln x 1 L x 1.4 mol HNO3 = 0.035 mol HNO3 103 mL 1 L soln One mole of H+(aq) is released per mole of nitric acid (HNO3). H2O HNO3 (aq) = 0.035 mol HNO3 x 4-40 → H+ (aq) + NO3- (aq) 1 mol H+ 23 x 6.022x10 ions 1 mol HNO3 1 mol = 2.1x1022 H+ ions Sample Problem 4.6 Writing Ionic Equations for Acid-Base Reactions PROBLEM: Write balanced molecular, total ionic, and net ionic equations for the following acid-base reactions and identify the spectator ions. (a) hydrochloric acid (aq) + potassium hydroxide (aq) → (b) strontium hydroxide (aq) + perchloric acid (aq) → (c) barium hydroxide (aq) + sulfuric acid (aq) → PLAN: All reactants are strong acids and bases (see Table 4.2). The product in each case is H2O and an ionic salt. Write the molecular reaction in each case and use the solubility rules to determine if the product is soluble or not. 4-41 Sample Problem 4.6 SOLUTION: (a) hydrochloric acid (aq) + potassium hydroxide (aq) → Molecular equation: HCl (aq) + KOH (aq) → KCl (aq) + H2O (l) Total ionic equation: H+ (aq) + Cl- (aq) + K+ (aq) + OH- (aq) → K+ (aq) + Cl- (aq) + H2O (l) Net ionic equation: H+ (aq) + OH- (aq) → H2O (l) Spectator ions are K+ and Cl- 4-42 Sample Problem 4.6 SOLUTION: (b) strontium hydroxide (aq) + perchloric acid (aq) → Molecular equation: Sr(OH)2 (aq) + 2HClO4 (aq) → Sr(ClO4)2 (aq) + 2H2O (l) Total ionic equation: Sr2+ (aq) + 2OH- (aq) + 2H+ (aq) + 2ClO4- (aq) → Sr2+ (aq) + 2ClO4- (aq) + 2H2O (l) Net ionic equation: 2H+ (aq) + 2OH- (aq) → 2H2O (l) or H+ (aq) + OH- (aq) → H2O (l) Spectator ions are Sr2+ and ClO4- 4-43 Sample Problem 4.6 SOLUTION: (c) barium hydroxide (aq) + sulfuric acid (aq) → Molecular equation: Ba(OH)2 (aq) + H2SO4 (aq) → BaSO4 (s) + 2H2O (l) Total ionic equation: Ba2+ (aq) + 2OH- (aq) + 2H+ (aq) + SO42- (aq) → BaSO4 (s) + H2O (l) The net ionic equation is the same as the total ionic equation since there are no spectator ions. This reaction is both a neutralization reaction and a precipitation reaction. 4-44 Figure 4.9 4-45 An aqueous strong acid-strong base reaction as a proton-transfer process. Figure 4.11 A gas-forming reaction with a weak acid. Molecular equation NaHCO3 (aq) + CH3COOH(aq) → CH3COONa (aq) + CO2 (g) + H2O (l) Total ionic equation Na+ (aq)+ HCO3- (aq) + CH3COOH (aq) → CH3COO- (aq) + Na+ (aq) + CO2 (g) + H2O (l) Net ionic equation HCO3-(aq) + CH3COOH (aq) → CH3COO- (aq) + CO2 (g) + H2O (l) 4-46 Sample Problem 4.7 Writing Proton-Transfer Equations for Acid-Base Reactions PROBLEM: Write balanced total and net ionic equations for the following reactions and use curved arrows to show how the proton transfer occurs. (a) hydriodic acid (aq) + calcium hydroxide (aq) → Give the name and formula of the salt present when the water evaporates. (b) potassium hydroxide (aq) + propionic acid (aq) → Note that propionic acid is a weak acid. Be sure to identify the spectator ions in this reaction. 4-47 Sample Problem 4.7 PLAN: In (a) the reactants are a strong acid and a strong base. The acidic species is therefore H3O+, which transfers a proton to the OH- from the base. SOLUTION: Total Ionic Equation: H+ transferred to OH- 2H3O+ (aq) + 2I- (aq) + Ca2+ (aq) + 2OH- (aq) → 2I- (aq) + Ca2+ (aq) + 4H2O (l) Net Ionic Equation: H3O+ (aq) + OH- (aq) → + H2O (l) When the water evaporates, the salt remaining is CaI2, calcium iodide. 4-48 Sample Problem 4.7 PLAN: In (b) the acid is weak; therefore it does not dissociate much and largely exists as intact molecules in solution. SOLUTION: Total Ionic Equation: H+ transferred to OH- K+ (aq) + OH- (aq) + CH3CH2COOH (aq) → K+ (aq) + H2O (l) + CH3CH2COO- (aq) Net Ionic Equation: CH3CH2COOH (aq) + OH- (aq) → CH3CH2COO- (aq) + H2O (l) K+ is the only spectator ion in the reaction. 4-49 Acid-Base Titrations • In a titration, the concentration of one solution is used to determine the concentration of another. • In an acid-base titration, a standard solution of base is usually added to a sample of acid of unknown molarity. • An acid-base indicator has different colors in acid and base, and is used to monitor the reaction progress. • At the equivalence point, the mol of H+ from the acid equals the mol of OH- ion produced by the base. – Amount of H+ ion in flask = amount of OH- ion added • The end point occurs when there is a slight excess of base and the indicator changes color permanently. 4-50 Figure 4.11 4-51 An acid-base titration. Sample Problem 4.8 Finding the Concentration of Acid from a Titration PROBLEM: A 50.00 mL sample of HCl is titrated with 0.1524 M NaOH. The buret reads 0.55 mL at the start and 33.87 mL at the end point. Find the molarity of the HCl solution. PLAN: Write a balanced equation for the reaction. Use the volume of base to find mol OH-, then mol H+ and finally M for the acid. volume of base (difference in buret readings) multiply by M of base mol of OHuse mole ratio as conversion factor mol of H+ and acid divide by volume (L) of acid molarity (M) of acid 4-52 Sample Problem 4.8 SOLUTION: NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (l) volume of base = 33.87 mL – 0.55 mL = 33.32 mL 33.32 mL soln x 1L x 0.1524 mol NaOH = 5.078x10-3 mol NaOH 103 mL 1 L soln Since 1 mol of HCl reacts with 1 mol NaOH, the amount of HCl = 5.078x10-3 mol. 5.078x10-3 mol HCl x 103 mL 1L 50.00 mL 4-53 = 0.1016 M HCl Oxidation-Reduction (Redox) Reactions Oxidation is the loss of electrons. The reducing agent loses electrons and is oxidized. Reduction is the gain of electrons. The oxidizing agent gains electrons and is reduced. A redox reaction involves electron transfer Oxidation and reduction occur together. 4-54 Figure 4.12 4-55 The redox process in compound formation. Table 4.3 Rules for Assigning an Oxidation Number (O.N.) General rules 1. For an atom in its elemental form (Na, O2, Cl2, etc.): O.N. = 0 2. For a monoatomic ion: O.N. = ion charge 3. The sum of O.N. values for the atoms in a compound equals zero. The sum of O.N. values for the atoms in a polyatomic ion equals the ion’s charge. Rules for specific atoms or periodic table groups 1. For Group 1A(1): O.N. = +1 in all compounds 2. For Group 2A(2): O.N. = +2 in all compounds 3. For hydrogen: O.N. = +1 in combination with nonmetals 4. For fluorine: O.N. = -1 in combination with metals and boron 5. For oxygen: O.N. = -1 in peroxides O.N. = -2 in all other compounds(except with F) O.N. = -1 in combination with metals, nonmetals (except O), and other halogens lower in the group 6. For Group 7A(17): 4-56 Sample Problem 4.9 Determining the Oxidation Number of Each Element in a Compound (or Ion) PROBLEM: Determine the oxidation number (O.N.) of each element in these species: (a) zinc chloride (b) sulfur trioxide (c) nitric acid PLAN: The O.N.s of the ions in a polyatomic ion add up to the charge of the ion and the O.N.s of the ions in the compound add up to zero. SOLUTION: (a) ZnCl2. The O.N. for zinc is +2 and that for chloride is -1. (b) SO3. Each oxygen is an oxide with an O.N. of -2. The O.N. of sulfur must therefore be +6. (c) HNO3. H has an O.N. of +1 and each oxygen is -2. The N must therefore have an O.N. of +5. 4-57 Sample Problem 4.10 Identifying Redox Reactions PROBLEM: Use oxidation numbers to decide whether each of the following is a redox reaction or not. (a) CaO (s) + CO2 (g) → CaCO3 (s) (b) 4 KNO3 (s) → 2 K2O(s) + 2 N2(g) + 5 O2(g) (c) NaHSO4 (aq) + NaOH (aq) → Na2SO4 (aq) + H2O (l) PLAN: Use Table 4.3 to assign an O.N. to each atom. A change in O.N. for any atom indicates electron transfer. SOLUTION: (a) CaO(s) + CO2(g) → CaCO3(s) +2 -2 +2 -2 +2 -2 +2 This is not a redox reaction, since no species change O.N. 4-58 Sample Problem 4.10 (b) 4 KNO3 (s) → 2 K2O(s) + 2 N2(g) + 5 O2(g) +1 -2 +5 +1 -2 0 0 This is a redox reaction. N changes O.N. from +5 to 0 and is reduced. O changes O.N. from -2 to 0 and is oxidized. 4-59 Sample Problem 4.10 (c) NaHSO4 (aq) + NaOH (aq) → Na2SO4 (aq) + H2O (l) +1 +1 +6 -2 +1 +1 -2 -2 +6 +1 -2 +1 This is not a redox reaction since no species change O.N. 4-60 Figure 4.13 4-61 A summary of terminology for redox reactions. Sample Problem 4.11 Identifying Oxidizing and Reducing Agents PROBLEM: Identify the oxidizing agent and reducing agent in each of the following reactions: (a) 2Al (s) + 3H2SO4 (aq) → Al2(SO4)3 (aq) + 3H2(g) (b) PbO (s) + CO (g) → Pb (s) + CO2 (g) (c) 2H2 (g) + O2 (g) → 2H2O (g) PLAN: Assign an O.N. to each atom and look for those that change during the reaction. The reducing agent contains an atom that is oxidized (increases in O.N.) while the oxidizing agent contains an atom that is reduced (decreases in O.N.). 4-62 Sample Problem 4.11 SOLUTION: (a) 2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g) 0 +1 +6 -2 +3 -2 +1 +6 -2 0 Al changes O.N. from 0 to +3 and is oxidized. Al is the reducing agent. H changes O.N. from +1 to 0 and is reduced. H2SO4 is the oxidizing agent. 4-63 Sample Problem 4.11 SOLUTION: (b) PbO (s) + CO (g) → Pb (s) + CO2 (g) +2 -2 +2 0 -2 +4 -2 Pb changes O.N. from +2 to 0 and is reduced. PbO is the oxidizing agent. C changes O.N. from +2 to +4 and is oxidized. CO is the reducing agent. 4-64 Sample Problem 4.11 SOLUTION: (c) 2H2 (g) + O2 (g) → 2H2O (g) 0 0 +1 -2 H2 changes O.N. from 0 to +1 and is oxidized. H2 is the reducing agent. O changes O.N. from 0 to -2 and is reduced. O2 is the oxidizing agent. 4-65 Balancing Redox Equations (oxidation number method) 1. Assign O.N.s to all atoms. 2. Identify the reactants that are oxidized and reduced. 3. Compute the numbers of electrons transferred, and draw tie-lines from each reactant atom to the product atom to show the change. 4. Multiply the numbers of electrons by factor(s) that make the electrons lost equal to the electrons gained. 5. Use the factor(s) as balancing coefficients. 6. Complete the balancing by inspection and add states of matter. 4-66 Sample Problem 4.12 Balancing Redox Equations by the Oxidation Number Method PROBLEM: Use the oxidation number method to balance the following equations: (a) Cu (s) + HNO3 (aq) → Cu(NO3)2 (aq) + NO2 (g) + H2O (l) SOLUTION: Assign oxidation numbers and identify oxidized and reduced species: (a) Cu (s) + HNO3 (aq) → Cu(NO3)2 (aq) + NO2 (g) + H2O (l) 0 4-67 +1 +5 -2 +2 +5 -2 +4 -2 +1 -2 Sample Problem 4.12 loses 2e-; oxidation Cu(s) + HNO3(aq) → Cu(NO3)2(aq) + NO2(g) + H2O(l) gains 1e-; reduction Multiply to make e- lost = e- gained: Cu (s) + 2HNO3 (aq) → Cu(NO3)2 (aq) + 2NO2 (g) + H2O (l) Balance other atoms by inspection: Cu (s) + 4HNO3 (aq) → Cu(NO3)2 (aq) + 2NO2 (g) + 2H2O (l) 4-68 Sample Problem 4.12 (b) PbS (s) + O2 (g) → PbO (s) + SO2 (g) SOLUTION: Assign oxidation numbers and identify oxidized and reduced species: (b) PbS (s) + O2 (g) → PbO (s) + SO2 (g) 0 +2 -2 4-69 +2 -2 +4 +5 -2 Sample Problem 4.12 loses 6e-; oxidation PbS (s) + O2 (g) → PbO (s) + SO2 (g) gains 2e- per O; reduction Multiply to make e- lost = e- gained: PbS (s) + 3 O (g) → PbO (s) + SO2 (g) 2 2 Balance other atoms by inspection, and multiply to give whole-number coefficients: 2PbS (s) + 3O2 (g) → 2PbO (s) + 2SO2 (g) 4-70 Figure 4.14 4-71 The redox titration of C2O42- with MnO4- Sample Problem 4.13 Finding the Amount of Reducing Agent by Titration PROBLEM: To measure the Ca2+ concentration in human blood, 1.00 mL of blood was treated with Na2C2O4 solution to precipitate the Ca2+ as CaC2O4. The precipitate was filtered and dissolved in dilute H2SO4 to release C2O42-, which was titrated with KMnO4 solution. The solution required 2.05mL of 4.88x10-4 M KMnO4 to reach the end point. The balanced equation is 2 KMnO4 (aq) + 5 CaC2O4 (s) + 8 H2SO4 (aq) → 2 MnSO4 (aq) + K2SO4 (aq) + 5 CaSO4 (s) + 10 CO2 (g) + 8 H2O (l) Calculate the amount (mol) of Ca2+ in 1.00 mL of blood. 4-72 Sample Problem 4.13 PLAN: Calculate the mol of KMnO4 from the volume and molarity of the solution. Use this to calculate the mol of C2O42- and hence the mol of Ca2+ ion in the blood sample. volume of KMnO4 soln convert mL to L and multiply by M mol of KMnO4 molar ratio mol of CaC2O4 ratio of elements in formula mol of Ca2+ 4-73 Sample Problem 4.13 SOLUTION: -4 mol KMnO 4.88x10 4 2.05 mL soln x x 103 mL 1L soln 1L 1.00x10-6 mol KMnO4 x 2.50x10-6 mol CaC2O4 x 4-74 = 1.00x10-6 mol KMnO4 5 mol CaC2O4 2 mol KMnO4 = 2.50x10-6 mol CaC2O4 1 mol Ca2+ 1 mol CaC2O4 = 2.50x10-6 mol Ca+2 Elements in Redox Reactions Types of Reaction • Combination Reactions – Two or more reactants combine to form a new compound: – X+Y→Z • Decomposition Reactions – A single compound decomposes to form two or more products: – Z→X+Y • Displacement Reactions – double diplacement: AB + CD → AC + BD – single displacement: X + YZ → XZ + Y • Combustion – the process of combining with O2 4-75 Figure 4.15 Combining elements to form an ionic compound. 4-76 Figure 4.16 4-77 Decomposition of the compound mercury(II) oxide to its elements. Figure 4.17 4-78 The active metal lithium displaces H2 from water. Figure 4.18 The displacement of H2 from acid by nickel. O.N. increasing O.N. decreasing oxidation occurring reduction occurring reducing agent oxidizing agent 0 +1 +2 0 Ni (s) + 2H+ (aq) → Ni2+ (aq) + H2 (g) 4-79 Figure 4.19 4-80 A more reactive metal (Cu) displacing the ion of a less reactive metal (Ag+) from solution. Figure 4.20 4-81 The activity series of the metals. Sample Problem 4.14 Identifying the Type of Redox Reaction PROBLEM: Classify each of the following redox reactions as a combination, decomposition, or displacement reaction. Write a balanced molecular equation for each, as well as total and net ionic equations for part (c), and identify the oxidizing and reducing agents: (a) magnesium (s) + nitrogen (g) → magnesium nitride (aq) (b) hydrogen peroxide (l) → water (l) + oxygen gas (c) aluminum (s) + lead(II) nitrate (aq) → aluminum nitrate (aq) + lead (s) PLAN: Combination reactions combine reactants, decomposition reactions involve more products than reactants and displacement reactions have the same number of reactants and products. 4-82 Sample Problem 4.14 SOLUTION: (a) This is a combination reaction, since Mg and N2 combine: 3Mg (s) + N2 (g) 0 0 → Mg3N2 (s) +2 -3 Mg is the reducing agent; N2 is the oxidizing agent. 4-83 Sample Problem 4.14 (b) This is a decomposition reaction, since H2O2 breaks down: 2 H2O2 (l) → + 2H2O (l) + O2 (g) +1 -2 +1 -2 0 H2O2 is both the reducing and the oxidizing agent. 4-84 Sample Problem 4.14 (c) This is a displacement reaction, since Al displaces Pb2+ from solution. 2Al (s) + 3Pb(NO3)2 (aq) → 2Al(NO3)3 (aq) + 3Pb (s) 0 +2 +3 +5 -2 +5 -2 0 Al is the reducing agent; Pb(NO3)2 is the oxidizing agent. The total ionic equation is: 2Al (s) + 3Pb2+ (aq) + 2NO3- (aq) → 2Al3+ (aq) + 3NO3- (aq) + 3Pb (s) The net ionic equation is: 2Al (s) + 3Pb2+ (aq) → 2Al3+ (aq) + 3Pb (s) 4-85 Figure 4.21 4-86 The equilibrium state.
Lecture PowerPoint Chemistry The Molecular Nature of Matter and Change Sixth Edition Martin S. Silberberg 5-1 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 5 Gases and the Kinetic-Molecular Theory 5-2 Gases and the Kinetic Molecular Theory 5.1 An Overview of the Physical States of Matter 5.2 Gas Pressure and Its Measurement 5.3 The Gas Laws and Their Experimental Foundations 5.4 Rearrangements of the Ideal Gas Law 5.5 The Kinetic-Molecular Theory: A Model for Gas Behavior 5.6 Real Gases: Deviations from Ideal Behavior 5-3 An Overview of the Physical States of Matter Distinguishing gases from liquids and solids. • Gas volume changes significantly with pressure. – Solid and liquid volumes are not greatly affected by pressure. • Gas volume changes significantly with temperature. – Gases expand when heated and shrink when cooled. – The volume change is 50 to 100 times greater for gases than for liquids and solids. • Gases flow very freely. • Gases have relatively low densities. • Gases form a solution in any proportions. – Gases are freely miscible with each other. 5-4 Figure 5.1 5-5 The three states of matter. Gas Pressure and its Measurement force Pressure = area Atmospheric pressure arises from the force exerted by atmospheric gases on the earth’s surface. Atmospheric pressure decreases with altitude. 5-6 Figure 5.2 5-7 Effect of atmospheric pressure on a familiar object. Figure 5.3 5-8 A mercury barometer. Figure 5.4 A Closed-end manometer The Hg levels are equal because both arms of the U tube are evacuated. 5-9 A gas in the flask pushes the Hg level down in the left arm. The difference in levels, Dh, equals the gas pressure, Pgas. Figure 5.4 B Open-end manometer When Pgas is less than Patm, subtract Dh from Patm. Pgas < Patm Pgas = Patm - Dh 5-10 When Pgas is greater than Patm, add Dh to Patm. Pgas > Patm Pgas = Patm + Dh Table 5.1 Common Units of Pressure 5-11 Sample Problem 5.1 Converting Units of Pressure PROBLEM: A geochemist heats a limestone (CaCO3) sample and collects the CO2 released in an evacuated flask attached to a closed-end manometer. After the system comes to room temperature, Δh = 291.4 mm Hg. Calculate the CO2 pressure in torrs, atmospheres, and kilopascals. PLAN: Construct conversion factors to find the other units of pressure. SOLUTION: 291.4 mmHg x 1torr 1 mmHg 291.4 torr x 1 atm 760 torr = 0.3834 atm 0.3834 atm x 101.325 kPa 1 atm 5-12 = 291.4 torr = 38.85 kPa The Gas Laws • The gas laws describe the physical behavior of gases in terms of 4 variables: – – – – pressure (P) temperature (T) volume (V) amount (number of moles, n) • An ideal gas is a gas that exhibits linear relationships among these variables. • No ideal gas actually exists, but most simple gases behave nearly ideally at ordinary temperatures and pressures. 5-13 Figure 5.5 5-14 Boyle’s law, the relationship between the volume and pressure of a gas. Boyle’s Law At constant temperature, the volume occupied by a fixed amount of gas is inversely proportional to the external pressure. V 1 P or PV = constant At fixed T and n, P decreases as V increases P increases as V decreases 5-15 Figure 5.6 AB 5-16 Charles’s law, the relationship between the volume and temperature of a gas. Figure 5.6 C 5-17 Absolute zero (0 K) is the temperature at which an ideal gas would have a zero volume. Charles’s Law At constant pressure, the volume occupied by a fixed amount of gas is directly proportional to its absolute (Kelvin) temperature. VT V = constant T At fixed T and n, P decreases as V increases P increases as V decreases 5-18 Figure 5.7 The relationship between the volume and amount of a gas. At fixed temperature and pressure, the volume occupied by a gas is directly proportional to the amount of gas. Avogadro’s Law: at fixed temperature and pressure, equal volumes of any ideal gas contain equal numbers of particles (or moles). 5-19 Figure 5.8 5-20 The process of breathing applies the gas laws. Gas Behavior at Standard Conditions STP or standard temperature and pressure specifies a pressure of 1 atm (760 torr) and a temperature of 0°C ( 273.15 K). The standard molar volume is the volume of 1 mol of an ideal gas at STP. Standard molar volume = 22.4141 L or 22.4 L 5-21 Figure 5.9 5-22 Standard molar volume. Figure 5.10 5-23 The volume of 1 mol (22.4 L) of an ideal gas and of some familiar objects: 1 gal of milk (3.79 L), a basketball (7.50 L) and 2.00 L of a carbonated drink. The Ideal Gas Law pV = nRT PV R= nT = 1 atm x 22.414 L 1 mol x 273.15 K = 0.0821 atm·L mol·K R is the universal gas constant; the numerical value of R depends on the units used. The ideal gas law can also be expressed by the combined equation: P1V1 P2V2 = T1 T2 5-24 Figure 5.11 5-25 The individual gas laws as special cases of the ideal gas law. Sample Problem 5.2 Applying the Volume-Pressure Relationship PROBLEM: Boyle’s apprentice finds that the air trapped in a J tube occupies 24.8 cm3 at 1.12 atm. By adding mercury to the tube, he increases the pressure on the trapped air to 2.64 atm. Assuming constant temperature, what is the new volume of air (in L)? PLAN: The temperature and amount of gas are fixed, so this problem involves a change in pressure and volume only. V1 (cm3) unit conversions V1 (L) multiply by P1/P2 V2 (L) 5-26 Sample Problem 5.2 SOLUTION: P1 = 1.12 atm V1 = 24.8 cm3 P2 = 2.64 atm V2 = unknown 24.8 cm3 x 1 mL 1 cm3 P1V1 n1T1 = P2V2 L 103 mL = 0.0248 L P1V1 = P2V2 n2T2 V2 = V1 x P1 P2 5-27 x n and T are constant = 0.0248 L x 1.12 atm 2.46 atm = 0.0105 L Sample Problem 5.3 Applying the Pressure-Temperature Relationship PROBLEM: A steel tank used for fuel delivery is fitted with a safety valve that opens when the internal pressure exceeds 1.00x103 torr. It is filled with methane at 23°C and 0.991 atm and placed in boiling water at exactly 100°C. Will the safety valve open? PLAN: We must determine if the pressure will exceed 1.00x103 torr at the new temperature. Since the gas is in a steel tank, the volume remains constant. P1 (atm) T1 and T2 (°C) 1 atm = 760 torr T1 and T2 (K) P1 (torr) multiply by T2/T1 P2 (torr) 5-28 K = °C + 273.15 Sample Problem 5.3 SOLUTION: P1 = 0.991 atm T1 = 23°C P2 = unknown T2 = 100.°C 0.991 atm x 760 torr 1 atm P1V1 n1T1 = P2 = P1 x T2 T1 n and V are constant T1 = 23 + 273.15 = 296 K T2 = 100. + 273.15 = 373 K = 753 torr P2V2 P1 n2T2 T1 = 753 torr x 373 K 296 K = P2 T2 = 949 torr The safety valve will not open, since P2 is less than 1.00 x 103 torr. 5-29 Sample Problem 5.4 Applying the Volume-Amount Relationship PROBLEM: A scale model of a blimp rises when it is filled with helium to a volume of 55.0 dm3. When 1.10 mol of He is added to the blimp, the volume is 26.2 dm3. How many more grams of He must be added to make it rise? Assume constant T and P. PLAN: The initial amount of helium (n1) is given, as well as the initial volume (V1) and the volume needed to make it rise (V2). We need to calculate n2 and hence the mass of He to be added. n1 (mol) of He multiply by V2 /V1 n2 (mol) of He subtract n1 mol to be added multiply by M g to be added 5-30 Sample Problem 5.4 SOLUTION: n1 = 1.10 mol V1 = 26.2 dm3 P1V1 n1T1 n2 = unknown V2 = 55.0 dm3 = n2 = n1 x V2 V1 T and P are constant P2V2 V1 n2T2 n1 = 3 = 1.10 mol x 55.0 dm 26.2 dm3 V2 n2 = 2.31 mol He Additional amount of He needed = 2.31 mol – 1.10 mol = 1.21 mol He 1.21 mol He x 5-31 4.003 g He 1 mol He = 4.84 g He Sample Problem 5.5 PROBLEM: PLAN: Solving for an Unknown Gas Variable at Fixed Conditions A steel tank has a volume of 438 L and is filled with 0.885 kg of O2. Calculate the pressure of O2 at 21oC. We are given V, T and mass, which can be converted to moles (n). Use the ideal gas law to find P. SOLUTION: V = 438 L n = 0.885 kg O2 (convert to mol) 3 0.885 kg O2 x 10 g x 1 kg P= 5-32 nRT = V 1 mol O2 32.00 g O2 27.7 mol x 0.0821 438 L T = 21°C = 294 K P is unknown = 27.7 mol O2 atm·L x 294.15 K mol·K = 1.53 atm Sample Problem 5.6 Using Gas Laws to Determine a Balanced Equation PROBLEM: The piston-cylinders is depicted before and after a gaseous reaction that is carried out at constant pressure. The temperature is 150 K before the reaction and 300 K after the reaction. (Assume the cylinder is insulated.) Which of the following balanced equations describes the reaction? 5-33 (1) A2(g) + B2(g) → 2AB(g) (2) 2AB(g) + B2(g) → 2AB2(g) (3) A(g) + B2(g) → AB2(g) (4) 2AB2(g) + A2(g) + 2B2(g) Sample Problem 5.6 PLAN: We are told that P is constant for this system, and the depiction shows that V does not change either. Since T changes, the volume could not remain the same unless the amount of gas in the system also changes. SOLUTION: n1T1 = n2T2 n2 = n1 T1 T2 = 150 K =½ 300 K Since T doubles, the total number of moles of gas must halve – i.e., the moles of product must be half the moles of reactant. This relationship is shown by equation (3). A(g) + B2(g) → AB2(g) 5-34 The Ideal Gas Law and Gas Density The density of a gas is - directly proportional to its molar mass and - inversely proportional to its temperature. m density = V PV = m V 5-35 and m M =d= moles = RT M xP RT m M Sample Problem 5.7 PROBLEM: Calculating Gas Density Find the density (in g/L) of CO2 (g) and the number of molecules per liter (a) at STP and (b) at room conditions (20.°C and 1.00 atm). PLAN: We can use the molar mass of CO2 to find its density from the ideal gas equation. SOLUTION: d= (a) At STP, or 273 K and 1.00 atm: M xP RT 44.01 g/mol x 1.00 atm = 0.0821 atm·L x 273 K = 1.96 g/L mol·K 1.96 g CO2 x 1 mol CO2 x 6.022 x 1023 molecules 1L 1 mol 44.01 g CO2 = 2.68 x 1022 molecules CO2/L 5-36 Sample Problem 5.7 SOLUTION: (b) At 20.°C and 1.00 atm: T = 20.°C + 273.15 = 293 K d= M xP RT 44.01 g/mol x 1.00 atm = 0.0821 atm·L x 293 K = 1.83 g/L mol·K 1.83 g CO2 x 1 mol CO2 x 6.022 x 1023 molecules 1L 1 mol 44.01 g CO2 = 2.50 x 1022 molecules CO2/L 5-37 Molar Mass from the Ideal Gas Law n= m M M= 5-38 PV = RT mRT PV Sample Problem 5.8 PROBLEM: Finding the Molar Mass of a Volatile Liquid An organic chemist isolates a colorless liquid from a petroleum sample. She places the liquid in a preweighed flask and puts the flask in boiling water, causing the liquid to vaporize and fill the flask with gas. She closes the flask and reweighs it. She obtains the following data: Volume (V) of flask = 213 mL mass of flask + gas = 78.416 g T = 100.0°C P = 754 torr mass of flask = 77.834 g Calculate the molar mass of the liquid. PLAN: The variables V, T and P are given. We find the mass of the gas by subtracting the mass of the flask from the mass of the flask with the gas in it, and use this information to calculate M. 5-39 Sample Problem 5.8 SOLUTION: m of gas = (78.416 - 77.834) = 0.582 g V = 213 mL x 1 L = 0.213 L 103 mL T = 100.0°C + 273.15 = 373.2 K P = 754 torr x 1 atm = 0.992 atm 760 torr M= 5-40 mRT = PV atm·L x 373 K mol·K 0.213 L x 0.992 atm 0.582 g x 0.0821 = 84.4 g/mol Mixtures of Gases • Gases mix homogeneously in any proportions. – Each gas in a mixture behaves as if it were the only gas present. • The pressure exerted by each gas in a mixture is called its partial pressure. • Dalton’s Law of partial pressures states that the total pressure in a mixture is the sum of the partial pressures of the component gases. • The partial pressure of a gas is proportional to its mole fraction: PA = XA x Ptotal 5-41 XA = nA ntotal Sample Problem 5.9 Applying Dalton’s Law of Partial Pressures PROBLEM: In a study of O2 uptake by muscle at high altitude, a physiologist prepares an atmosphere consisting of 79 mole % N2, 17 mole % 16O2, and 4.0 mole % 18O2. (The isotope 18O will be measured to determine the O2 uptake.) The pressure of the mixture is 0.75 atm to simulate high altitude. Calculate the mole fraction and partial pressure of 18O2 in the mixture. PLAN: Find X 18 O2 and P18 O2 from Ptotal and mol % 18O2. mole % 18O2 divide by 100 mole fraction, X 18 O2 multiply by Ptotal partial pressure P18 5-42 O2 Sample Problem 5.9 SOLUTION: X 18 P18 5-43 O2 O2 = = X 18 O2 4.0 mol % 18O2 100 = 0.040 x Ptotal = 0.040 x 0.75 atm = 0.030 atm Table 5.2 Vapor Pressure of Water (P 5-44 H2O T(0C) P (torr) H2O T(0C) 0 5 10 12 14 16 18 20 22 24 26 28 30 35 4.6 6.5 9.2 10.5 12.0 13.6 15.5 17.5 19.8 22.4 25.2 28.3 31.8 42.2 40 45 50 55 60 65 70 75 80 85 90 95 100 ) + at Different T P H2O (torr) 55.3 71.9 92.5 118.0 149.4 187.5 233.7 289.1 355.1 433.6 525.8 633.9 760.0 Figure 5.12 5-45 Collecting a water-insoluble gaseous product and determining its pressure. Sample Problem 5.10 Calculating the Amount of Gas Collected over Water PROBLEM: Acetylene (C2H2) is produced in the laboratory when calcium carbide (CaC2) reacts with water: CaC2(s) + 2H2O(l) → C2H2(g) + Ca(OH)2(aq) A collected sample of acetylene has a total gas pressure of 738 torr and a volume of 523 mL. At the temperature of the gas (23oC), the vapor pressure of water is 21 torr. How many grams of acetylene are collected? PLAN: The difference in pressures will give P for the C2H2. The number of moles (n) is calculated from the ideal gas law and converted to mass using the molar mass. 5-46 Sample Problem 5.10 PLAN: SOLUTION: Ptotal subtract P for H2O P of C2H2 P C2H2 = (738 - 21) torr = 717 torr P = 717 torr x 1 atm = 0.943 atm 760 torr V = 523 mL x 1L 103 mL use ideal gas law n of C2H2 multiply by M mass of C2H2 5-47 = 0.523 L T = 23°C + 273.15 K = 296 K Sample Problem 5.10 SOLUTION: n C 2H2 = PV = RT 0.0203 mol x 0.943 atm x atm·L 0.0821 mol·K 26.04 g C2H2 1 mol C2H2 5-48 0.523 L x 296 K = 0.529 g C2H2 = 0.0203 mol The Ideal Gas Law and Stoichiometry P, V, T of gas A P, V, T of gas B Amount (mol) of gas A Amount (mol) of gas B Figure 15.13 The relationships among the amount (mol, n) of gaseous reactant (or product) and the gas pressure (P), volume (V), and temperature (T). 5-49 Sample Problem 5.11 Using Gas Variables to Find Amounts of Reactants and Products I PROBLEM: What volume of H2 gas at 765 torr and 225°C is needed to reduce 35.5 g of copper(II) oxide to form pure copper and water? PLAN: Write a balanced equation. Convert the mass of copper (II) oxide to moles and find the moles of H2, using the mole ratio from the balanced equation. Calculate the corresponding volume of H2 using the ideal gas law. mass (g) of CuO divide by M mol CuO use mole ratio mol H2 volume of H2 ideal gas law 5-50 Sample Problem 5.11 SOLUTION: CuO(s) + H2(g) → Cu(s) + H2O(g) 35.5 g CuO x 1 mol CuO x 1 mol H2 = 0.446 mol H 2 79.55 g CuO 1 mol CuO P = 765 torr x V= 5-51 nRT = P 1 atm = 1.01 atm 760 torr T = 225°C + 273.15 K = 498 K 0.446 mol H2 x 0.0821 1.01 atm atm·L x 498 K mol·K = 18.1 L H2 Sample Problem 5.12 PROBLEM: PLAN: Using Gas Variables to Find Amounts of Reactants and Products II What mass of potassium chloride forms when 5.25 L of chlorine gas at 0.950 atm and 293 K reacts with 17.0 g of potassium metal? First we must write a balanced equation. Since the quantities of both reactants are given, we must next determine which reactant is limiting. We will use the ideal gas law to calculate the moles of Cl2 present. SOLUTION: The balanced equation is: Cl2(g) + 2K(s) → 2KCl(s) For Cl2: P = 0.950 atm V = 5.25 L T = 293 K n = unknown 5-52 Sample Problem 5.12 n Cl2 = PV = RT 0.50 atm 0.0821 x atm·L mol·K 5.253 L x 293 K = 0.207 mol Cl2 2 mol KCl = 0.435 mol KCl 1 mol Cl2 For Cl2: 0.207 mol Cl2 x For K: 17.0 g K x 1 mol K x 2 mol KCl = 0.414 KCl 39.10 g K 2 mol K Cl2 is the limiting reactant. 0.435 mol KCl x 74.55 g KCl 1 mol KCl 5-53 = 30.9 g KCl The Kinetic-Molecular Theory: A Model for Gas Behavior Postulate 1: Gas particles are tiny with large spaces between them. The volume of each particle is so small compared to the total volume of the gas that it is assumed to be zero. Postulate 2: Gas particles are in constant, random, straight-line motion except when they collide with each other or with the container walls. Postulate 3: Collisions are elastic, meaning that colliding particles exchange energy but do not lose any energy due to friction. Their total kinetic energy is constant. Between collisions the particles do not influence each other by attractive or repulsive forces. 5-54 Figure 5.14 Distribution of molecular speeds for N2 at three temperatures. 5-55 Figure 5.15 5-56 Pressure arise from countless collisions between gas particles and walls. Figure 5.16 A molecular view of Boyle’s law. Pext increases, T and n fixed At any T, Pgas = Pext as particles hit the walls from an average distance, d1. 5-57 Higher Pext causes lower V, which results in more collisions, because particles hit the walls from a shorter average distance (d2 < d1). As a result, Pgas = Pext again. Figure 5.17 5-58 A molecular view of Dalton’s law Figure 5.18 At T1, Pgas = Patm. 5-59 A molecular view of Charles’s law Higher T increases collision frequency, so Pgas > Patm. Thus, V increases until Pgas = Patm at T2. Figure 5.19 For a given amount, n1, of gas, Pgas = Patm. 5-60 A molecular view of Avogadro’s law When gas is added to As a result, V increases reach n2 the collision until Pgas = Patm again. frequency of the particles increases, so Pgas > Patm. Kinetic Energy and Gas Behavior At a given T, all gases in a sample have the same average kinetic energy. 1 Ek = mass x speed2 2 Kinetic energy depends on both the mass and the speed of a particle. At the same T, a heavier gas particle moves more slowly than a lighter one. 5-61 Figure 5.20 The relationship between molar mass and molecular speed. 5-62 Graham’s Law of Effusion Effusion is the process by which a gas escapes through a small hole in its container into an evacuated space. Graham’s law of effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. A lighter gas moves more quickly and therefore has a higher rate of effusion than a heavier gas at the same T. Rate of effusion  5-63 1 √M Figure 5.21 5-64 Effusion. Lighter (black) particles effuse faster than heavier (red) particles. Sample Problem 5.13 Applying Graham’s Law of Effusion PROBLEM: A mixture of helium (He) and methane (CH4) is placed in an effusion apparatus. Calculate the ratio of their effusion rates. PLAN: The effusion rate is inversely proportional √M for each gas, so we find the molar mass for each substance using its formula and take the square root. The ratio of the effusion rates is the inverse of the ratio of these square roots. SOLUTION: rate rate 5-65 M of CH4 = 16.04 g/mol He CH4 = √ 4.003 16.04 M of He = 4.003 g/mol = 2.002 Figure 5.22 5-66 Diffusion of gases Chemical Connections Figure B5.1 Variations in pressure and temperature with altitude in Earth’s atmosphere Variations in pressure, temperature, and composition of the Earth’s atmosphere. 5-67 Chemical Connections 5-68 Real Gases: Deviations from Ideal Behavior • The kinetic-molecular model describes the behavior of ideal gases. Real gases deviate from this behavior. • Real gases have real volume. – Gas particles are not points of mass, but have volumes determined by the sizes of their atoms and the bonds between them. • Real gases do experience attractive and repulsive forces between their particles. • Real gases deviate most from ideal behavior at low temperature and high pressure. 5-69 Table 5.3 Molar Volume of Some Common Gases at STP (0°C and 1 atm) Gas He H2 Ne Ideal gas Ar N2 O2 CO Cl2 NH3 5-70 Molar Volume (L/mol) 22.435 22.432 22.422 22.414 22.397 22.396 22.390 22.388 22.184 22.079 Boiling Point (oC) -268.9 -252.8 -246.1 ‒ -185.9 -195.8 -183.0 -191.5 -34.0 -33.4 Figure 5.23 5-71 Deviations from ideal behavior with increasing external pressure Figure 5.24 5-72 The effect of interparticle attractions on measured gas pressure. Figure 5.25 5-73 The effect of particle volume on measured gas volume. The van der Waals equation • The van der Waals equation adjusts the ideal gas law to take into account – the real volume of the gas particles and – the effect of interparticle attractions. Van der Waals equation for n moles of a real gas n2a (P + 2 )(V − nb) = nRT V The constant a relates to factors that influence the attraction between particles. 5-74 The constant b relates to particle volume. Table 5.4 Van der Waals Constants for Some Common Gases a Gas He Ne Ar Kr Xe H2 N2 O2 Cl2 CH4 CO CO2 NH3 H2O 5-75 atm∙L2 b L mol2 mol 0.034 0.211 1.35 2.32 4.19 0.244 1.39 1.36 6.49 2.25 1.45 3.59 4.17 5.46 0.0237 0.0171 0.0322 0.0398 0.0511 0.0266 0.0391 0.0318 0.0562 0.0428 0.0395 0.0427 0.0371 0.0305

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