Chem notes

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

2. Introduction to Chemistry and the Scientific Method ask a question make observations and collect data design an experiment analyze data draw a conclusion do research

5. 1.2 The Study of Chemistry Macroscopic Microscopic Chemists study the microscopic properties of matter, which in turn produce matter’s observable macroscopic properties – thus, we often switch back and forth between microscopic and macroscopic views of matter in this course.

6. The scientific method is a systematic approach to research. Although it is systematic, it is not a rigid series of steps that must be done in a particular order. ask a question make observations and collect data design an experiment analyze data draw a conclusion do research researcher’s hidden bias form a hypothesis

7. A theory is a unifying principle that explains a body of facts and/or those laws that are based on them. 1.3 Atomic Theory A hypothesis is a tentative explanation for a set of observations that can be tested . tested modified A law is a concise statement of a relationship between phenomena that is always the same under the same conditions. Force = mass x acceleration

8. Classification of Matter

9. Substances Matter is anything that has mass and occupies space. Matter that has a uniform and unchanging composition is called a (pure) substance. examples of pure substances include table salt, pure water, oxygen, gold, etc.

10. States of Matter Matter normally occupies one of three phases, or states. These are: P Solid P Liquid P Gas * Plasma is a 4 th state of matter in which the particles are at extremely high temperatures (over 1,000,000 °C).

11. As we shall see in more detail later, the phase (or state) of a substance is determined by the average kinetic energy of the particles that make up the substance, (i.e., temperature ) and the strength of the attractive forces holding the substance’s particles together. weak moderate strong States of Matter gas liquid solid

15. Technically, the word “gas” refers to a substance that is in the gas phase at room temperature. The word “vapor” refers to the gaseous state of a substance that is normally a solid or liquid at room temperature. States of Matter

17. A pure substance is a form of matter that has a definite composition and distinct properties. examples: gold, salt, iron, pure water, sugar A mixture is a combination of two or more substances in which each substance retains its own distinct identity. examples: salt water, oil & vinegar dressing, granite, air Classification of Matter

18. see page 13 Classification Summary

19. Heterogeneous mixture : the composition is not uniform throughout. You can visibly see the different components. Mixtures can be heterogeneous or homogeneous . examples: cement, iron filings in sand, granite, milk, oil and water, etc. Classification of Matter Mixtures

20. Homogenous mixture (also called a solution ): The composition of the mixture is the same throughout. Classification of Matter Solutions are made up of two components : (1) the solute which is dissolved in (2) the solvent . If the solvent is water , the solution is called an aqueous solution which is symbolized: ( aq ).

21. We often think of a solution as being a solid dissolved in a liquid. However… In a solution, both the solvent and solute can be in any phase – solid, liquid or gas. air (O 2 dissolved in nitrogen) ‏ gas gas alloys (brass, bronze, etc.) ‏ solid solid gasoline (a mix of liquids) ‏ liquid liquid salt dissolved in water solid liquid example solute solvent Classification of Matter

22. In the case of liquids , we use a special term: If two liquids completely dissolve in each other, they are said to be miscible . If they do not, they are immiscible. Classification of Matter If a substance dissolves in another substance, we say the first substance is soluble in the second. If they do not dissolve, they are said to be insoluble. example: carbon dioxide is soluble in air gold is in soluble in water example: alcohol and water are miscible gasoline and water are immiscible

23. A mixture can be separated into its pure components by simple physical methods. Classification of Matter Filtration is a means of separating a solids from liquids. For example, we can filter out the sand from a mix of sand and water. Magnetic substances can be separated using a magnet.

24. Fractional crystallization is a means of separating two solids by adding a solvent that will dissolve one of the solids but not the other; the mixture is then filtered to separate out the insoluble solid. Finally, the solvent is evaporated off to recover the remaining solid. Separation of a Mixture For example, we can separate salt from sand by adding hot water to dissolve the salt, then filter off the sand. The water is then evaporated off, leaving the salt behind.

25. Distillation is a means of separating two liquids based on differences in their boiling points. Separation of a Mixture This method is only effective for substances that are liquids at room temperature with significant differences in their boiling points. The substance with the lowest boiling point “boils off” and is then cooled and condensed back into a liquid. The liquid is collected in a receiver flask. distillation apparatus

26. Chromotography is the separation of a mixture based on solubility in a “mobile” solvent coupled with an adherence to a “stationary phase” medium, such as paper or silica gel, etc. Separation of a Mixture column chromotography is a common means of separating components from a mixture Thin Layer Chromotography can be used to separate the components of chlorophyll from a crushed plant leaf.

27. Other means of separation: Other techniques of separating a mixture include sublimation, extraction, and leaching, etc. If you had a jar containing both nails and marbles, the only way to separate them would be by hand Separation of a Mixture speak to the hand…

28. You are given a test tube which contains a mixture of water, methanol, aspirin, acetanilide and aluminum oxide. (Acetanilide, aluminum oxide and aspirin are all white, powdery solids at room temperature and are thus visibly indistinguishable from each other. Water and methanol are both colorless liquids at room temperature and are also visibly indistinguishable from each other.) Assume your only source of heat is a Bunsen burner which can produce a maximum temperature of 600  C . Using the following information, devise a method to separate this mixture. Be specific and complete in your answer. Example: dissolves only in hot (50  C) water or warm (25 °C) methanol 304  C 114  C acetanilide does not dissolve in either methanol or water at any temperature 2980  C 2072  C aluminum oxide dissolves in methanol or water (if above 10  C) decomposes at 140  C 135  C aspirin dissolves in cold or hot water 65  C – 97  C methanol dissolves in cold or hot methanol 100  C 0  C water what it dissolves/does not dissolve in boiling point melting point substance

30. see page 13 Classification Summary

32. Some symbols are based on the Latin name eg , iron is Fe (for ferrum ) and sodium is Na (for natrium ) ‏ Symbols for Elements Elements are identified by a one or two-letter symbol. The first letter, which is ALWAYS capitalized, is typically the first letter in the name of the element. eg, C = c arbon, H = h ydrogen The second letter (which is only used if other elements have the same first letter) is NEVER capitalized. eg, Cl = c h l orine, He = he lium.

33. http:// www.privatehand.com/flash/elements.html Symbols for Elements

34. A compound is a substance composed of atoms of two or more different elements chemically bonded in fixed proportions. Water (H 2 O) ‏ Sucrose (C 12 H 22 O 11 ) ‏ Classification of Matter table salt (NaCl) ‏ Compounds sugar As such, they can be chemically decomposed into their component elements.

35. The properties of a compound are different from the properties of its component elements Classification of Matter For example, table salt is composed of sodium and chlorine. Sodium is a soft, silver colored metal that reacts violently with water, and chlorine is a pale-green poisonous gas – yet when chemically combined, they form table salt, a white crystalline solid you put on your eggs in the morning! = +

36. Compounds can only be separated into their pure components (elements) by chemical means. Water can be separated into its elements, hydrogen and oxygen, by passing an electric current through it, a process called electrolysis . For example: Iron is separated from iron ore (Fe 2 O 3 ) by heating the ore in a blast furnace and reacting it with carbon monoxide and elemental carbon (in the form of “coke”). Compounds

37. There are TWO kinds of compounds, depending on the nature of the chemical bond holding the atoms together. Molecules form when two or more neutral atoms form bonds between them by sharing electrons Note that some elements exist as molecules . For example,the following elements occur in nature as molecular diatomic elements : H 2 , N 2 , O 2 , F 2 , Cl 2 , Br 2 and I 2 They are molecules, but they are NOT compounds , because they have only one kind of element present. Cl 2 O 2 Compounds H 2 N 2

39. PURE SUBSTANCE can it be separated by physical means? compound element YES YES NO NO YES NO Classification Summary MATTER heterogeneous mixture is the mixture uniform throughout? can the substance be chemically decomposed into simpler substances? solution MIXTURE YES

40. Properties of Matter

44. A physical change does not alter the composition or identity of a substance. Physical & Chemical Changes A chemical change (reaction) alters the identity or composition of the substance(s) involved. ice melting sugar dissolving in water hydrogen burns in air to form water

46. Measurement

47. Measurement The SI System of Measurement Scientists around the world use a unified system of measurement ( Le S ysteme I nternationale d’Unites, or SI for short). There are seven fundamental “quantities” that can be measured: Length Mass Time Temperature Electric Current Chemical quantity Luminous intensity

48. International System of Units (SI) page 16 Each base quantity is given a unit with a specific name and symbol

49. The SI units are based on metrics. Each power of ten change is given a special prefix used with the base unit. International System of Units (SI) You must know these prefixes see page 17

52. liquids —use a graduated cylinder. To read the scale correctly, read the volume at the lowest part of the meniscus - the curve of the liquid’s surface in a container. Measuring Volume meniscus Measurements with SI Units Your eye should be level with the meniscus when reading the volume regular solids : volume = length x width x height

53. Measurements with SI Units Measuring Volume continued irregular solids : volume is found by displacement. 4 6 2 Begin with a known volume of water. 4 6 2 Add the solid. volume of solid = volume displaced : 6.0 – 4.0 = 2.0 cm 3 The amount of water displaced is the volume of the solid.

54. Measurements with SI Units Mass (SI unit = kilogram): the amount of matter. The mass of a given object is constant . A kilogram is about 2.2 pounds -- this is too large a unit for most chemistry labs, so we will use grams instead. Note that mass and weight are two different things…

55. Weight is a measure of the force due to gravity acting on a mass. The weight of an object changes , depending on the gravitational force acting on it. http://www.exploratorium.edu/ronh/weight/ For example, on the moon you would weigh only 1/6th what you do on Earth, because the force of gravity on the moon is only 1/6th that of Earth . Measurements with SI Units

56. The Importance of Units On 9/23/99, the Mars Climate Orbiter entered Mar’s atmosphere 100 km (62 miles) lower than planned and was destroyed by heat because the engineers that designed the rocket calculated the force provided by the engines in pounds, but NASA engineers thought the force was given in the units of Newtons (N) when they determined when to fire the rockets… 1 lb = 1 N 1 lb = 4.45 N “ This is going to be the cautionary tale that will be embedded into introduction to the metric system in elementary school, high school, and college science courses till the end of time.”

57. Measurements with SI Units Measuring Mass We still use the term “weighing” even though we are finding the mass of an object, not its weight… Triple beam balance Electronic balance English/Metric equivalencies 1 kg = 2.203 lbs 1 paperclip  1 gram 1 lb = 453.6 grams

58. Temperature (SI unit = kelvin) is a measure of the average kinetic energy (energy due to motion) of the atoms and molecules that make up a substance. Measurements with SI Units There are three common temperature scales Fahrenheit ( o F) – English system, based on the freezing point of salt water. Centigrade ( o C) – metric system, based on the freezing and boiling points of pure water Kelvin (K) – SI unit , also called the “Absolute” scale; 0 K (Absolute Zero) is defined as the temperature at which all motion stops (kinetic energy = 0).

59. 32 o F = 0 o C 212 o F = 100 o C Temperature Conversions: K = o C + 273.15 273 K = 0 o C 373 K = 100 o C o C = ( o F – 32) 9 5 o F = ( o C) + 32 9 5

60. Temperature Examples A thermometer reads 12 o F. What would this be in o C ? The conversion formula from o F to o C is: o C = 5/9( o F – 32) Inserting the values gives: : o C = 5/9(12 o F – 32) o C = 5/9(-20) = -11.1 o C A thermometer reads 315.3 K. What would this be in o F ? The conversion formula from o C to o F is: o F = 9/5( o C) +32. Inserting the values gives: : o C = 9/5(42.15 o C) + 32 o C = (75.9) + 32 = 107.9 o F First convert K to o C: 315.3 K – 273.15 = 42.15 o C

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62. Measurements with SI Units Time (SI unit = second) . This is the only non-metric SI unit. We still use 1 day = 24 hours, 1 hour = 60 minutes, 1 minute = 60 seconds Chemical Quantity ( SI unit = mole). Since atoms are so tiny, it takes a LOT of them to make even one gram. In fact, you would have to put 602,200,000,000,000,000,000,000 atoms of carbon (that’s 6.022 X 10 23 ) on a balance to get just 12 grams of carbon! We do use metric fractions of time, however, such as milliseconds (1/1000th of a second), etc.

63. That huge number (6.022 X 10 23 ) is given a special name; it is called “ Avogadro’s Number, ” symbolized N A , after the Italian physicist, Lorenzo Romano Amedeo Avogadro who lived between 1776-1856. Just like 1 dozen = 12 things, we define: Measurements with SI Units The Mole continued Avogadro 1 mole = 6.022 X 10 23 things

65. 6.022 x 10 23 1.99 x 10 -23 N x 10 n N is a number between 1 and 10 n is a positive or negative integer Working with Numbers: Scientific Notation The number of atoms in 12 g of carbon: 602,200,000,000,000,000,000,000 The mass of a single carbon atom in grams: 0.0000000000000000000000199

66. Measurements with SI Units Derived Units Although we need only seven fundamental SI units, we can combine different units to obtain new units, called derived units. For example, speed is distance per unit time, so we must combine the unit for distance (m) and time (sec) to get the SI unit for speed: speed = meters per second (m/s) We will be working with many different derived units in this course. It is important to pay attention to the individual units that make up derived units!!

67. Derived Units SI derived unit for density is kg/m 3 . This is not a convenient unit in chemistry, so we usually use the units g/cm 3 or g/mL Density is the mass per unit volume of a substance. It is calculated using the equation: Every substance has a unique density . For example: You need to know the density of water. Any object that is more dense than water will sink in water; if it is less dense, it will float in water d = m V density = mass volume 0.70 g/cm 3 gasoline 11.35 g/cm 3 lead 2.70 g/cm 3 aluminum 1.00 g/cm 3 water density substance

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70. Dimensional Analysis: A problem solving technique given unit x = desired unit desired unit given unit

71. In algebra, we learn that: u x u = u 2 u u = 1 (the u’s cancel!) and If we let “u” = units, then every measured quantity is a number x a unit. We can solve problems by setting them up so that the unit we do NOT want gets cancelled out by dividing u/u in the problem. Thus, if a/u is a conversion (say 100 cm/1 m) then we can convert cm to meters etc. using this conversion factor so that the cm cancel… a u u x = a (2u) 3 = 8 u 3 and… The Mathematics of Units

73. Dimensional Analysis Method of Solving Problems Example: How many μ m are in 0.0063 inches? conversion factors needed: 0.0063 in = ? 1 inch = 2.54 cm 10 6 μ m = 1 m 1 m = 100 cm 0.0063 inch x x x = 160 μ m Begin with what units you have “in hand,” then make a list of all the conversions you will need. 2.54 cm 1 inch 1 m 10 2 cm 10 6 μ m 1 m

74. Example: The speed of sound in air is about 343 m/s. What is this speed in miles per hour? (1 mile = 1609 meters) 1 mi = 1609 m meters to miles seconds to hours 1.9 conversion units Dimensional Analysis Method of Solving Problems 1 min = 60 s 1 hour = 60 min 343 m s x 1 mi 1609 m 60 s 1 min x 60 min 1 hour x = 767 mi hour

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77. Uncertainty, Precision & Accuracy in Measurements

78. Measurements with SI Units Uncertainty, Precision and Accuracy in Measurements When you measure length using a meterstick, you often have to estimate to the nearest fraction of a line. The uncertainty in a measured value is partly due to how well you can estimate such fractional units. http:// www.mhhe.com/physsci/chemistry/chang7/esp/folder_structure/ch/m2/s2/chm2s2_2.htm The uncertainty also depends on how accurate the measuring device, itself, is.

79. Accuracy – how close a measurement is to the true or accepted value To determine if a measured value is accurate, you would have to know what the true or accepted value for that measurement is – this is rarely known! Precision and Accuracy Precision – how close a set of measurements are to each other; the scatter of repeated measurements about an average. We may not be able to say if a measured value is accurate, but we can make careful measurements and use good equipment to obtain good precision, or reproducibility.

80. accurate & precise precise but not accurate not accurate & not precise A target analogy is often used to compare accuracy and precision. Precision and Accuracy

81. example: which is more accurate: 0.0002 g or 2.0 g? answer: you cannot tell , since you don’t know what the accepted value is for the mass of whatever object this is that you are weighing! example: which is more precise: 0.0002 g or 2.0 g? answer: surprisingly, the most precise value is 2.0 g, not the 0.0002 g. The number of places behind the decimal is not what determines precision! If that were so, I could increase my precision by simply converting to a different metric prefix for the same measurement: Which is more precise: 2 cm or 0.00002 km? They are, in fact, identical! Precision and Accuracy

82. Precision and Accuracy Precision is a measure of the uncertainty in a measured value. Any measured value is composed of those digits of which you are certain, plus the first estimated digit. 1 2 3 4 1 The length of the object is at least 1.7 cm, and we might estimate the last digit to be half a unit, and say it is 1.75 cm long. Others might say 1.74 or possibly 1.76 – the last digit is an estimate, and so is uncertain.

83. We always assume an uncertainty of ±1 in the last digit. The percent error in a measured value is defined as: The smaller the percent error, the greater the precision – the smaller the % error, the more likely two measurements will be close together using that particular measuring instrument. Thus: 2.0 ± 0.1 has a % error of (0.1/2.0) x 100 = ±5% but 0.0002 has a % error of (0.0001/0.0002) x 100 = ± 50% Precision and Accuracy % error = ± uncertainty measured value x 100

84. Percent Difference When determining the accuracy of an experimentally determined value, it must be compared with the “accepted value.” One common method of reporting accuracy is called the percent difference (  %) – this gives how far off your value is, as a percent, from the accepted value: Percent difference: experimental value – accepted value accepted value x 100  % =

85. example: In an experiment, a student determines the density of copper to be 8.74 g/cm 3 . If the accepted value is 8.96 g/cm 3 , determine the student’s error as a percent difference. = − 2.46 % The (-) sign indicates the experimental value is 2.46% smaller than the accepted value; a (+)  % means the experimental value is larger than the accepted value. experimental value – accepted value accepted value x 100  % = x 100 8.74 – 8.96 8.96  % =

86. Precision and Accuracy We will be doing math operations involving measurements with uncertainties, so we need a method of tracking how the uncertainty will affect calculated values – in other words, how many places behind the decimal do we really get to keep the answer? The method requires us to keep track of significant digits. Significant digits (or significant figures) are all of the known digits, plus the first estimated or uncertain digit in a measured value.

88. How many significant figures are in each of the following measurements? 24 mL 2 significant figures 3001 g 4 significant figures 0.0320 m 3 3 significant figures 6.4 x 10 4 molecules 2 significant figures 560 kg You cannot tell!! Significant Figures: Rules

89. Significant Figures: Rules Suppose you wanted to estimate the number of jellybeans in a jar, and your best guess is around 400. Now – is the uncertainty in your estimate ±1 jellybean, or is it ±10 jellybeans, or maybe even ±100 jellybeans (if you weren’t very good at estimating jellybeans…) We need a way to write 400 and indicate in some way whether that was 400 ±1 vs 400±10 vs 400±100. The plain number “400” is ambiguous as to where the uncertain digit is. Use scientific notation to remove the ambiguity : 400 ± 1 = 4.00 x 10 2 = 3 sig figs 400 ± 10 = 4.0 x 10 2 = 2 sig figs 400 ± 100 = 4 x 10 2 = 1 sig fig

90. Rounding Numbers Given the number 6.82 and asked to round to 2 sig digits we would write 6.8. We write 6.8 because 6.82 is closer to 6.8 than it is to 6.9 Given the number 6.88 and asked to round to 2 sig digits, we would write 6.9. We write 6.9 because 6.88 is closer to 6.9 than it is to 6.8 You were taught this long ago. You were also probably taught that, given the number 6.85, and asked to round this to 2 sig digits, you would write 6.9. My question is, WHY did you round UP? 6.85 is JUST as close to 6.8 as it is to 6.9! Since it is in the middle, it could be rounded either way! And we should round it “either way.”

91. Rounding Numbers Since the rounding is “arbitrarily” up, this can introduce some round-off errors in chain calculations involving this number – the final value will be too large if you always round up when the next digit is exactly 5. Because rounding is “arbitrary” when the next digit is exactly 5, we introduce the following “ odd-even rounding rule : e.g. : 4.65 ≈ 4.6 but 4.75 ≈ 4.8 Note however, that 4.651 is closer to 4.7 than 4.6, so we round it to 4.7: only invoke the “odd-even rule” when the next digit is exactly 5. When the next digit is exactly 5, round up or down to make the number an even number .

92. Math Operations with Significant Digits We need a set of rules to determine how the uncertainty or error will “propagate” or move through a series of calculations and affect the precision of our final answer. There is one rule for addition and subtraction, and one rule for multiplication and division. Do not mix them and match them and confuse them!

93. Addition or Subtraction The answer cannot have more digits to the right of the decimal point than any of the original numbers. Significant Figures 89.332 1. 1 + 90 . 4 32 round off to 90.4 one digit after decimal point 3. 70 -2.9133 0. 78 67 two digits after decimal point round off to 0.79

94. Addition or Subtraction Significant Figures We often encounter two numbers that must be added that are in scientific notation. We cannot add them and determine the number of places “behind the decimal” unless they have the same power of 10 – we may have to convert! Example : What is the sum of 2.4 x 10 2 + 3.77 x 10 3 ? 3.77 x 10 3 0.24 x 10 3 4.01 x 10 3 Always convert the smaller power of 10 to the larger power of 10 The answer is good to 2 behind the decimal when written as x10 3 -- that is, the uncertain digit is in the “tens” place (± 10)

95. 10 n 10 n + + Decreasing the power of 10 means you must move the decimal to the RIGHT by one place for each power of 10 decrease. 10 n + Increasing the power of 10 means you must move the decimal to the LEFT one place for each power of 10 increase Moving the decimal determines both the magnitude and the +/- value of 10 n To determine the power of 10, visualize a see-saw when you move the decimal point: n n n n n n Significant Figures

96. Significant Figures Example: What is the answer to the following, to the correct number of significant digits? - 1.2 x 10 -4 - 0.012 x 10 -2 0 3.0268 x 10 -2 3.0148 x 10 -2 = 3.015 x 10 -2

97. Significant Figures Multiplication or Division The number of significant figures in the result is set by the original number that has the smallest number of significant figures 4.51 x 3.6666 = 16.5 36366 = 16.5 6.8 ÷ 112.04 = 0.060 6926 = 0.061 3 sig figs round to 3 sig figs 2 sig figs round to 2 sig figs

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99. Significant Figures 1.8 Exact Numbers Numbers from definitions or numbers of objects are considered to have an infinite number of significant figures Because 3 is an exact number the answer is not rounded to 7, but rather reported to be 6.67 cm (three sig figures). Find the average of three measured lengths: 6.64, 6.68 and 6.70 cm. These values each have 3 significant figures Example: 6.64 + 6.68 + 6.70 3 = 6.67333 = 6.67 = 7

100. Atoms, Molecules and Ions Chapter 2 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

101. Early Ideas Our understanding of the structure of matter has undergone profound changes in the past century. Nonetheless, what we know today did not arrive on a sudden inspiration. We can trace a fairly steady plodding towards our current understanding, starting as far back as 400 BCE…

102. Early Ideas The properties of matter could be explained by the shape and size of its atoms. Democritus (c.a. 400 BCE) All matter was composed of tiny, indivisible particles called atoms ( atomos = indivisible ) Each kind of matter had its own unique kind of atom – ie., there were water atoms, air atoms, fire atoms, bread atoms, etc. Fire atoms water atoms rolls & flows “ ouch!”

103. As a result, Democritus’ ideas were not very well receieved. It would be some 1200 years before the idea of atoms was revisited! Early Ideas Most importantly, Democritus believed atoms existed in a vacuum – that is, there was “nothing” in the spaces between atoms… Aristotle, among others, refused to believe in the existence of “nothingness” that still occupied space… Vacuum??

104. Early Ideas Aristotle Aristotle was the court philosopher to Alexander the Great . Because of this, Aristotle’s ideas were given a lot of weight . Aristotle believed that all matter was composed of four elements: earth, air, fire and water .

105. Heating WATER exchanged “hot” for “cold” which created “AIR” (which we see as steam…) WATER ( cold , wet ) AIR ( hot , wet ) These elements could be “inter-converted” into each other by exchanging the “properties” of hot, cold, dry and wet. example Early Ideas EARTH AIR WATER FIRE hot dry wet cold

106. Early Ideas This idea that one kind of element could be converted into another eventually led to the belief in Alchemy – that one could turn lead into gold by performing the right chemical reaction!

107. The “scientific method” of inquiry was developed during the 17 th and 18 th centuries. The invention of the balance and other instruments soon led to a new understanding about the nature of matter. Early Ideas The French chemist, Antoine-Laurent Lavoisier (1743-1794), presented two important ideas which would later help lead to a new, more developed atomic theory of matter…

109. Lavoiser experimenting with respiration

110. Joseph Proust, another 18 th century French scientist, proposed the Law of Definite Proportion , which states that the mass ratios of elements present in different samples of the same compound do not vary. Early Ideas For example, the percent by mass of the elements present in sugar are always found to be: 53.3% oxygen, 40.0% carbon and 6.7% hydrogen.

111. John Dalton (1766-1824) Dalton started out as an apothecary's assistant (today, we would call him a pharmacist). He was also interested in both meteorology and the study of gases. Dalton developed a new atomic theory of the nature of matter based on several postulates. His theory differed significantly from the early ideas of Democritus, but they both agreed that the simplest form of matter was the atom.

113. 2.1 Dalton’s Atomic Theory Law of Conservation of Matter and Definite Proportion Explained… + = 16 X 8 Y + 8 X 2 Y

114. Law of Multiple Proportions If Dalton’s ideas about atoms were correct, then he proposed that the mass of a compound containing different numbers of a given element (atom) would vary by the mass of that one whole atom – that is: If two elements can combine to form more than one compound, then the masses of one element that combine with a fixed mass of the other element are in ratios of small, whole numbers.

115. 16 12 = 1.33 = 2.67 2.67 / 1.33 = 2 Dalton’s Atomic Theory Consider the mass ratio of oxygen to carbon in the two compounds: CO and CO 2 32 12 Note that the mass of oxygen that combines with 12 g of carbon in carbon dioxide is 2 x greater than the mass of oxygen that combines with 12 g of carbon in carbon monoxide.

117. Cathode rays , discovered by William Crookes, are formed when a current is passed through an evacuated glass tube. Cathode rays are invisible, but a phosphor coating makes them visible.

118. J.J. Thomson The Electron is Discovered Sir Joseph John Thomson 1856-1940 J.J. Thomson helped show that cathode rays are made up of negatively charged particles (based on their deflection by magnetic and electric fields). N S

119. Thomson showed that all cathode rays are identical, and are produced regardless of the type of metals used for the cathode and anode in the cathode ray tube. Thomson was unable to determine either the actual electric charge or the mass of these cathode ray particles. He was, however, able to determine the ratio of the electric charge to the mass of the particles. J.J. Thomson

120. To do this, he passed cathode rays simultaneously through electric and magnetic fields in such a way that the forces acting on the cathode ray particles (now called electrons ) due to the fields cancelled out. The ratio of the electric field strength to the square of the magnetic field strength at this point was proportional to the charge to mass ratio of the electron. J.J. Thomson Magnetic field only Electric field only Both http://highered.mcgraw-hill.com/classware/ala.do?isbn=0072980605&alaid=ala_729122&showSelfStudyTree=true# _ +

121. The value he obtained, −1.76 x 10 8 C/g*, was always the same, regardless of the source of the cathode rays. This value was nearly 2000 times larger than the charge to mass ratio of a hydrogen ion (H + )! This indicated that either the charge of the electron was very large, or that the mass of the electron was very, very small – much smaller than the mass of a hydrogen atom, which was the lightest atom known. J.J. Thomson *the SI unit of electric charge is the Coulomb (C)

122. Thomson proposed that these electrons were not just very small particles, but were actually a sub-atomic particle present in all atoms . We thus credit Thomson with the “discovery” of the electron because of his work in determining their physical characteristics, and his rather bold hypothesis that they were present in all atoms (which was later shown to be true). J.J. Thomson

123. Since the atom is neutrally charged, if it has (-) charged electrons, there must also be a (+) part to the atom to cancel the negative electrons. This showed that Dalton’s idea that atoms were indivisible is NOT correct – instead, the atom is composed of TWO oppositely charged parts. Thomson thought the atom was a diffuse (+) charged object, with electrons stuck in it, like raisins in pudding (the plum pudding model ). The Plum Pudding Model

124. Thomson’s Plum Pudding Model of the Atom http://highered.mcgraw-hill.com/classware/ala.do?isbn=0072980605&alaid=ala_729122&showSelfStudyTree=true

125. Millikan’s Oil Drop Experiment Robert Millikan (1911) designed an experiment to determine the actual charge of an electron. He suspended charged oil drops in an electric field. The drops had become charged by picking up free electrons after passing through ionized air.

126. F ELEC = E x q F GRAVITY = m x g when the downward force of gravity on the drop was balanced by the upward force of the electric field, then: Millikan’s Oil Drop Experiment E x q = m x g or q = mg/E Knowing the mass (m) of the oil drop, and the strength of the electric field (E), he was able to find the charge (q) on the oil drop .

127. To find the charge of the electron, he found the smallest difference between the charges on any two oil drops. eg: Suppose you find three oil drops have the following charges: 12.4, 7.6, 10.8. The differences between the charges are: 12.4 – 10.8 = 1.6 10.8 – 7.6 = 3.2 12.4 – 7.6 = 4.8 4.8 – 3.2 = 1.6 You would conclude the charge of the electron was 1.6 charge units. Millikan’s Oil Drop Experiment

128. Millikan’s Oil Drop Experiment http://highered.mcgraw-hill.com/classware/ala.do?isbn=0072980605&alaid=ala_729122&showSelfStudyTree=true Using this technique, Millikan was able to determine the charge of an electron to be: Using Thomson’s charge to mass ratio and the charge for the electron, Millikan determined the mass of the electron to be 9.11 x 10 -31 kilogram. For his work, Millikan received the 1923 Nobel Prize in Physics. e = C 1.602 x 10 C 19 C

129. (Uranium compound) http://highered.mcgraw-hill.com/classware/ala.do?isbn=0072980605&alaid=ala_729122&showSelfStudyTree=true It was found that there are three distinct types of radiation: ( + ) alpha particles, ( - ) beta particles, and neutral gamma rays. Radioactivity was discovered in 1895

130. Rutherford’s Gold Foil Experiment Rutherford designed an experiment using these newly discovered alpha-particles to test if Thomson’s plum pudding model was correct. He fired (+) alpha particles at the gold foil. If the Thomson model was correct, most of the alpha particles would pass through the foil with little deflection. (1908 Nobel Prize in Chemistry)

131. R (+)  -particle Rutherford’s Experiment Expected Results of Rutherford’s Experiment θ The force of repulsion is directly proportional to the product of the charges of the alpha particle and nucleus and inversely proportional to the square of the distance between the center of the two charges. F = kQ 1 Q 2 /R 2 A large, diffuse positive charge is not able to repel a (+) alpha particle very strongly, because the alpha particle cannot make a close approach, so the angle of deflection, θ , would be fairly small.

132. However, some particles were deflected significantly, and perhaps one in 2000 were actually deflected nearly 180 degrees! Rutherford’s Gold Foil Experiment When Rutherford performed the experiment, nearly all the alpha particles passed through the foil without deflection, as expected…

133. Rutherford’s Experiment Rutherford was stunned. This would be like firing a machine gun at an apple, and having most of the bullets pass through -- but every once in a while one of the bullets would bounce off the apple and come back and hit you! Why would this happen??? something small and massive must be in there that deflects only those bullets that directly hit it… ?!? ? DUCK, ERNIE!

134. Only a positive charge with a very, very small radius would allow the alpha particle to approach close enough to experience a significant repulsion. Rutherford’s Experiment R Strong repulsion! nucleus  -particle By carefully measuring the angles of deflection, θ , Rutherford was able to determine the approximate size of this positive core to the atom. θ

135. Next, by measuring the kinetic energy of the alpha particle before and after the collision, Rutherford was able to apply conservation of momentum and determine the mass of the atom’s positive core. Putting it all together, he was able to conclude that all the positive charge -- and about 99.9% of the mass -- of an atom was concentrated in a very tiny area in the middle of the atom, which he called the nucleus. Rutherford’s Experiment

136. Only the very few (+) α -particles that passed very near this incredibly tiny (+) nucleus were strongly deflected; most α -particles never came near the nucleus and so were not deflected significantly. Rutherford’s Experiment *note carefully that the (+) α -particles never actually collide with the (+) nucleus – the repulsive force between the like charges is too great for that to occur!

137. atomic radius ~ 100 pm = 1 x 10 -10 m nuclear radius ~ 5 x 10 -3 pm = 5 x 10 -15 m Rutherford’s Model of the Atom “ If the atom is the Houston Astrodome, then the nucleus is a marble on the 50-yard line.” The estimated size of this nucleus was such a tiny fraction of the total volume of the atom, that at first Rutherford doubted his own conclusion.

138. As another size comparison, if the nucleus were the size of a basketball , placed at PHS, the atom would be over 20 km in diameter, reaching Martin to the North, and just missing the US 131 Business Loop exit to the South ! The basketball-sized nucleus would also mass about 70,000,000,000 tons! This is equivalent to about 100,000 cruise ship ocean liners!

141. mass n  mass p  1,840 x mass e -

142. Atomic number (Z) = number of protons in nucleus Mass number (A) = number of protons + number of neutrons also called the nucleon number = atomic number (Z) + number of neutrons Isotopes are atoms of the same element (X) with different numbers of neutrons in their nuclei Mass Number Atomic number, Mass number and Isotopes X A Z Element Symbol Atomic Number H 1 1 H (D) 2 1 H (T) 3 1 protium deuterium tritium U 235 92 Uranium-235 C 14 6 Carbon-14 examples nuclide

143. 6 protons, 8 (14 - 6) neutrons, 6 electrons 26 protons, 33 (59 - 26) neutrons, 26 electrons How many protons, neutrons, and electrons are in How many protons, neutrons, and electrons are in Fe 59 26 ? Atomic number, Mass number and Isotopes Examples: C 14 6 ?

144. see page 50

145. We now understand that the number of protons in the nucleus of the atom is what “defines” the element and gives each element its unique properties. The Periodic Table of the Elements

146. The Periodic Table of the Elements Transition metals period group

149. Elements Natural abundance of elements in the Earth’s crust Natural abundance of elements in the human body

150. Molecules and Ions

151. Molecules & Ions Note that some elements exist as molecules . For example,the following elements occur in nature as molecular diatomic elements : H 2 , N 2 , O 2 , F 2 , Cl 2 , Br 2 and I 2 They are molecules, but they are NOT compounds , because they have only one kind of element present. A molecule is an aggregate of two or more neutral atoms in a definite arrangement held together by chemical forces F 2 O 2 H 2 N 2

152. A polyatomic molecule contains more than two atoms O 3 , H 2 O, NH 3 , C 3 H 6 O An allotrope is one of two or more distinct molecular forms of an element, each having unique properties. For example, O 2 and O 3 are allotropes of oxygen; diamond, graphite and buckminster fullerene (C 60 ) are all different allotropes of carbon. Molecules & Ions

154. A monatomic ion contains only one atom Examples: Na + , Cl - , Ca 2+ , O 2- , Al 3+ , N 3- Examples: ClO 3 - , NO 2 - , CN - , SO 4 2 - Molecules and Ions A polyatomic ion contains more than one atom note that the convention is to indicate the magnitude of the charge first , and then the sign: e.g. , Ca 2+ , not Ca +2

155. 13 protons, so there are 13 – 3 =10 electrons 34 protons, so there are 34 + 2 = 36 electrons How many electrons are in ? How many electrons are in ? Molecules and Ions Examples Al 27 13 3+ Se 78 34 2-

156. Charges of common monatomic ions Note that some atoms, especially transition metals, have multiple charge states Note also that metals typically form (+) charged ions, nonmetals form (-) charged ions. s ee page 54

157. Also note the relation between the magnitude of the charge and the group number (1A, 5A, etc) for most elements. The charge of representative metals (group 1A, 2A and 3A) is equal to the group number The charge of representative nonmetals (group 4A-7A) is equal to: (the group number – 8)

158. Chemical Nomenclature Determining the names and formulas of chemical compounds IUPAC = International Union of Pure and Applied Chemists . This is the group that determines the official rules of nomenclature for all chemical elements and compounds

159. Chemical Formulas A chemical formula is a combination of element symbols and numbers that represents the composition of the compound. Subscripts following an element’s symbol indicate how many of that particular atom are present. If no subscripts are given, it is assumed that only one of that atom is present in the compound. NH 3 C 3 H 6 S P 4 O 10 1 N + 3 H atoms 3 C + 6 H + 1 S atoms 4 P + 10 O atoms

160. A molecular formula shows the exact number of atoms of each element in the smallest unit of a substance Chemical Formulas

161. An empirical formula shows the simplest whole-number ratio of the atoms in a substance H 2 O H 2 O molecular empirical C 6 H 12 O 6 CH 2 O N 2 H 4 NH 2 Chemical Formulas C 2 H 8 O 2 CH 2 O note that different molecular compounds may have the same empirical formula

162. For ionic compounds the formula is always the same as the empirical formula. The sum of the charges of the cation(s) and anion(s) in each formula unit must equal zero. Thus, the ratio of cations to anions can always be reduced to simple, whole number ratios. The ionic compound NaCl Ionic Formulas Na + 500 Cl - 500 = NaCl

163. Naming Binary Molecular Compounds

164. Naming Molecular Compounds We will only consider naming binary molecules . Binary molecular compounds typically form between two non-metals, or a non-metal and a metalloid. 1st element + root of 2nd element + “- ide ” Naming Molecules: e.g. : HCl = hydrogen chloride

166. HI hydrogen iodide NF 3 nitrogen tri fluoride SO 2 sulfur di oxide N 2 Cl 4 di nitrogen tetra chloride NO 2 nitrogen di oxide N 2 O Examples of naming molecules Naming Molecular Compounds di nitrogen mo noxide (laughing gas)

167. If the second element begins with a vowel, the terminal vowel of the prefix is allowed to be dropped. For example N 2 O 4 could be called dinitrogen tetroxide , rather than dinitrogen tetr a oxide. CO would be called carbon monoxide , not carbon mon oo xide Note, however, that the official IUPAC rule states that the vowel is only dropped for “compelling linguistic reasons.” Naming Molecular Compounds

168. Naming Compounds containing Hydrogen Compounds containing hydrogen can be named using the Greek prefixes, but most have common names that are accepted by IUPAC. The most common examples are: B 2 H 6 CH 4 SiH 4 NH 3 PH 3 H 2 O H 2 S diboron hexahydride diborane carbon tetrahydride silicon tetrahydride nitrogen trihydride phosphorus trihydride dihydrogen monoxide dihydrogen sulfide methane silane ammonia phosphine water hydrogen sulfide Naming Molecular Compounds

169. Determining the formula of molecules from the name The subscripts tell you the number of each type of element present, so naming molecules from the formula is straightforward. e.g. sulfur hexafluoride = SF 6 dichlorine heptoxide = Cl 2 O 7 Naming Molecular Compounds The order in which the atoms are listed in molecules is based on something called electronegativity. For now, we can predict the order using the chart on the next slide…

170. B Br Ge C N P As Sb O S Se Te F Cl Si I H Order of Elements in Writing Molecular Formulas Chemical Formulas

171. Organic chemistry is the branch of chemistry that deals with carbon compounds Carbon is unique among all the elements in its ability to catenate, or form long or branching chains of carbon atoms. C C C We usually write these chains as “condensed formulas” that assumes carbons are bonded to each other as follows: = CH 3 CH 2 CH 3 note that we could also write this as: C 3 H 8 H H H H H H H H

172. Organic molecules that contain only carbon and hydrogen are called hydrocarbons . The first 10 simple hydrocarbons Hydrocarbon compounds are named based on the number of carbon atoms in the “backbone” or chain of carbon atoms.

173. Naming Ionic Compounds

175. Naming Ionic Compounds BaCl 2 barium chloride K 2 O potassium oxide Binary ionic compounds are named: name of metal ion + root of non-metal + “-ide” e.g. Na 2 S Mg 3 N 2 Al 2 O 3 sodium sulfide magnesium nitride aluminum oxide

176. Determining the formula of ionic compounds from the name is a little more involved – unlike molecular compounds, the name does not give us the subscripts. These must be determined based on the charges of each ion. Remember that the total number of (+) and (-) charges in any ionic compound must sum to zero. Formula of Ionic Compounds

177. Formula of Ionic Compounds Al 2 O 3 2 x +3 = +6 3 x -2 = -6 Al 3+ O 2- CaBr 2 1 x +2 = +2 2 x -1 = -2 Ca 2+ Br - MgS 1 x +2 = +2 1 x -2 = -2 Mg 2+ S 2 - aluminum oxide calcium bromide magnesium sulfide

178. Formula of Ionic Compounds Note that if you take the magnitude of the charge of the cation, and make it the subscript on the anion, and take the magnitude of the anion’s charge and make it the subscript of the cation, the compound will always end up with a net neutral charge. Now, if possible, reduce the subscripts to a simpler ratio, and you have the correct formula for the compound! Al O Al 3+ O 2- Al 2 O 3 +3 -2 3 2

179. s ee page 58

180. Multivalent ions: The Non-Representative Atoms Cu W Mn Co Pb Fe

181. Most elements form only ions with one charge. However, most of the transition metals , as well as Pb and Sn, have more than one possible charge state. We say they are multi-valent. e.g. : copper can exist in either a +1 or +2 charge state: Cu + or Cu 2+ The formula or name of the compound must indicate which charge state the metal cation is in. Transition and other multi-valent metal ions

182. Transition and other multi-valent metal ions Cu + cuprous Fe 2+ ferrous Cu 2+ cupric Fe 3+ ferric Cr 2+ chromous Hg 2 2+ mercurous Cr 3+ chromic Hg 2+ mercuric Older method gives a common name for each valence state e.g. CuCl = cuprous chloride Hg 2 I 2 = mercurous iodide Fe 2 O 3 = ferric oxide PbO = plumbous oxide

183. Transition and other multi-valent metal ions To determine which charge state the cation is in, you must look at the anion, and calculate the charge of the cation… Fe 2 O 3 Subscript on O is the charge of the iron! Thus, Fe is +3 and this compound is ferric oxide. CuS S is always -2, and there is only one Cu to cancel this out, so copper must be +2. Thus, this is cupric sulfide .

185. Polyatomic Ions SO 4 2- C 2 O 4 2- C 2 H 3 O 2 2- NH 4 +

186. Naming Polyatomic Ions There are certain groups of neutral atoms that bond together, and then gain or lose one or more electrons from the group to form what is called a polyatomic ion. Most polyatomic ions are negatively charged anions. Examples: OH - = hydroxide ion CN - = cyanide ion NO 3 - = nitrate ion NH 4 + = ammonium ion SO 4 2- = sulfate ion SO 3 2- = sulfite ion

187. See page 60

188. Naming Polyatomic Ions Naming ionic compounds containing polyatomic ions is straightforward: Name the cation + name the (polyatomic) anion Examples: NaOH = sodium hydroxide K 2 SO 4 = potassium sulfate Fe(CN) 2 = iron (II) cyanide (NH 4 ) 2 CO 3 = ammonium carbonate

189. page 61

190. page 62

191. see page 64 Compound Summary

192. There are a different set of rules for naming acids. Some of the rules are based on a much older system of nomenclature, and so the rules are not as simple as they are for molecular and normal ionic compounds. NAMING ACIDS AND BASES

194. All acids have hydrogen as the first listed element in the chemical formula . For nomenclature purposes, there are two major types of acids: Oxoacids (also called oxyacids) = acids that contain oxygen. eg: H 2 SO 4 , HC 2 H 3 O 2 Non-oxo acids = acids that do not contain oxygen. eg: HCl (aq), H 2 S (aq) Acids

195. Acids Rules for naming non-oxoacids acid = “hydro-” + root of anion + “-ic acid” see page 65 *note that we add an extra syllable for acids with sulfur and phosphorus: it’s not hydrosulfic acid, but hydrosulf ur ic acid. Similarly, acids with phosphorus will end in phosph or ic, not phosphic acid. *

196. An oxoacid is an acid that contains hydrogen, oxygen, and another element – That is, oxoacids are the protonated form of those polyatomic ions that have oxygen in their formulas. HClO 3 chloric acid HNO 2 nitrous acid H 2 SO 4 sulfuric acid Acids examples:

197. If the name of the polyatomic anion ends in “ate,” drop the -ate and add “ic acid.” eg: SO 4 2- = sulfate anion H 2 SO 4 = sulfuric acid C 2 H 3 O 2 - = acetate anion HC 2 H 3 O 2 = acetic acid When naming oxoacids, NO “hydro” prefix is used. Instead, the acid name is the root of the name of the oxoanion + either “-ic” acid or “-ous” acid, as follows: If the name of the polyatomic anion ends in “ite,” drop the -ite and add “ous acid.” eg: SO 3 2- = sulfite anion H 2 SO 3 = sulfurous acid NO 2 - = nitrite anion HNO 2 = nitrous acid

198. Acids Naming Oxoacids and Oxoanions see page 66

199. Acids ic goes with ate because…. ”IC…I ATE it! ite goes with ous like……tonsil- ITE-OUS , senior-ITE-OUS As a mnemonic aid, I always use the following:

200. A base can be defined as a substance that yields hydroxide ions (OH - ) when dissolved in water. Bases NaOH sodium hydroxide KOH potassium hydroxide Ba(OH) 2 barium hydroxide

201. Hydrates are compounds that have a specific number of water molecules attached to them. BaCl 2 •2H 2 O LiCl•H 2 O MgSO 4 •7H 2 O Sr(NO 3 ) 2 •4H 2 O barium chloride dihydrate lithium chloride monohydrate magnesium sulfate heptahydrate strontium nitrate tetrahydrate CuSO 4 •5H 2 O cupric sulfate pentahydrate CuSO 4 anhydrous cupric sulfate Hydrates

202. Anhydrous: without water; this term describes hydrated compounds after “drying.” Hygroscopic: readily absorbs moisture directly from the air. Deliquescent: absorbs moisture from the air so readily, that these compounds can take on enough water to actually start to dissolve. Water of hydration: the water absorbed and incorporated into hygroscopic compounds Hydrates Other terms associated with hydrates

203. see page 68

204. Mass Relationships in Chemical Reactions Chapter 3 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

205. By definition : 1 atom 12 C “weighs” 12 amu On this scale 1 H = 1.008 amu and 16 O = 16.00 amu Atomic mass is the mass of an atom in atomic mass units ( amu ). This is a relative scale based on the mass of a 12 C atom. Micro World atoms & molecules Macro World grams Relative Masses of the Elements

206. How do we find the relative masses of the other elements? Imagine we have 66.00 grams of CO 2 . The compound is decomposed and yields 18.00 grams of C and 48 grams of O. Since there are two oxygen atoms for every 1 carbon atom, we can say that This means that the relative mass of each oxygen atom is 1.333 x the mass of a carbon atom (12.00 amu) , or… mass of oxygen = 1.333 x 12.00 amu = 16.00 amu Relative Masses of the Elements

207. 7.42% 6 Li (6.015 amu) 92.58% 7 Li (7.016 amu) The Average atomic mass of lithium would be: Average Atomic Mass The average atomic mass of an element is the weighted average mass of that element, reflecting the relative abundances of its isotopes. example: consider lithium (Li), which has two isotopes with the following relative percent abundances: 7.42 100 6.015 92.58 100 7.016 6.941               amu amu amu

208. 6.941 The masses reported at the bottom of the “box” for each element in the Periodic Table is the average atomic mass for that element, (in amu). Average Atomic Mass

209. see page 79 Average Atomic Mass

210. The mole (mol) is the SI unit for the amount of a substance that contains as many “things” as there are atoms in exactly 12.00 grams of 12 C. 1 mol = N A = 6.022 x 10 23 “things” The Mole & Avogadro’s Number This number, called Avogadro’s number ( N A ), has been experimentally determined to be approximately 6.0221367 X 10 23 things. We can have 1 mole of atoms, or molecules, or even dump trucks. The mole refers only to a number, like the term “dozen” means 12.

212. Molar mass is the mass, in grams , of exactly 1 mole of any object (atoms, molecules, etc.) Note that because of the way we defined the mole : 1 mole 12 C atoms = 6.022 x 10 23 atoms = 12.00 g 1 12 C atom = 12.00 amu For example: 1 mole 12 C atoms = 12.00 g 12 C 1 mole lithium atoms = 6.941 g of Li The Mole & Molar Mass Thus, for any element atomic mass (amu) = molar mass (grams)

213. One Mole of: C = 12.01 g S = 32.06 g Cu = 63.55 g Fe = 55.85 g Hg = 200.6 g The Mole & Molar Mass

214. The Mole & Molar Mass Solving Mole Problems We can now add the definitions of the mole, Avogadro’s number, and molar mass to our repertoire of conversion factors we can use in dimensional analysis problems. Thus, given the mass, we can use the molar mass to convert this to moles, and then use Avogadro’s number to convert moles to particles, and vice versa… N A = Avogadro’s number M = molar mass in g/mol

215. 0.551 g K How many atoms are in 0.551 g of potassium (K) ? 1 mol K = 39.10 g K 1 mol K = 6.022 x 10 23 atoms K = 8.49 x 10 21 atoms K Solving Mole Problems conversion factors x 6.022 x 10 23 atoms K 1 mol K 1 mol K 39.10 g K x

216. Solving Mole Problems see page 81

217. Solving Mole Problems see page 82

218. Solving Mole Problems see page 82

219. Molecular mass (or molecular weight ) is the sum of the atomic masses of the atoms in a molecule. 1 molecule SO 2 = 64.07 amu 1 mole SO 2 = 64.07 g SO 2 Molecular Mass As was the case for atoms, for any molecule Example: consider SO 2 SO 2 1S 32.07 amu 2O + 2 x 16.00 amu SO 2 64.07 amu molecular mass (amu) = molar mass (grams) SO 2 64.07 amu

220. Molecular Mass see page 83

221. Formula mass is the sum of the atomic masses (in amu) in a formula unit of an ionic compound. formula mass (amu) = molar mass (grams) 1 formula unit NaCl = 58.44 amu 1 mole NaCl = 58.44 g NaCl Formula Mass For any ionic compound 1Na 22.99 amu 1Cl + 35.45 amu NaCl 58.44 amu NaCl

222. What is the formula mass of Ca 3 (PO 4 ) 2 ? 1 formula unit of Ca 3 (PO 4 ) 2 310.18 amu Formula Mass Since the formula mass, in grams (per mole), is numerically equal to the molar mass, in amu, we find that the formula mass of Ca 3 (PO 4 ) 2 = 310.18 grams per mole of Ca 3 (PO 4 ) 2 . 3 Ca 3 x 40.08 2 P 2 x 30.97 8 O + 8 x 16.00

223. Example : How many H atoms are in 72.5 g of C 3 H 8 O ? 1 mol C 3 H 8 O = (3 x 12) + (8 x 1) + 16 = 60 g C 3 H 8 O 1 mol H = 6.022 x 10 23 atoms H = 5.82 x 10 24 atoms H 1 mol C 3 H 8 O molecules = 8 mol H atoms 72.5 g C 3 H 8 O Molecular/Formula Masses Using Molecular/Formula Masses in Dimensional Analysis Problems We can now add molecular & formula masses to our list of conversion factors. They are used similarly to the way we used the molar mass of the elements as conversion factors. conversion factors 1 mol C 3 H 8 O 60 g C 3 H 8 O x 8 mol H atoms 1 mol C 3 H 8 O x 6.022 x 10 23 H atoms 1 mol H atoms x

224. Solving Mole Problems see page 84

225. Solving Mole Problems see page 85

226. The Mass Spectrometer Atomic and molecular masses of unknown compounds are determined using a mass spectrometer . A gaseous sample of the unknown is bombarded with electrons in an electron beam. This knocks electrons loose from the unknown to produce cations. These cations are then accelerated through perpendicular electric and magnetic fields. The charge:mass ratio ( e/m ) of the unknown ions determines the degree to which the particles are deflected. The greater the charge:mass ratio, the smaller the angle through which the beam is deflected.

227. The Mass Spectrometer We know the angle that a given e/m produces, so we can identify the unknown ion when it registers on a special screen. high e/m low e/m Mass Spectrometer

228. Percent composition of an element in a compound is the percent, by mass, of that element in the compound. It can be calculated as follows: where n is the number of moles of the element in 1 mole of the compound Percent composition Knowing the percent composition, one can determine the purity of a substance, (are there contaminants present in the sample?) and you can even determine the empirical formula of an unknown compound. n x molar mass of element molar mass of compound x 100%

229. check: 52.14% + 13.13% + 34.73% = 100.0% Percent composition Example: What is the percent composition of ethanol, which has the formula, C 2 H 6 O ? First, we find the molecular mass of ethanol. This is found to be: 2(12.01) + 6(1.008) + 1(16.00) = 46.07 grams/mole. % Composition: C 2 H 6 O %C = 2 x (12.01 g) 46.07 g x 100% = 52.14% %H = 6 x (1.008 g) 46.07 g x 100% = 13.13% %O = 1 x (16.00 g) 46.07 g x 100% = 34.73%

230. We can also determine the % by mass of groups of atoms present in a compound in the same manner. Example: what is the percent water in epsom salts, which has the formula: MgSO 4 • 7 H 2 O ? % H 2 O = 24.31 + 32.07 + 4(16.00) + 7(18.02) 7(18.02) mass of water mass of compound x 100 = = 246.52 g cmpd 126.14 g H 2 O x 100 = 51.17% H 2 O Percent composition this is the molar mass of water

231. Example: How many grams of CaCl 2 • 2 H 2 O must be weighed out to obtain 12.20 grams of CaCl 2 ? There are two ways of solving this problem: Method 1 : First determine the % CaCl 2 in CaCl 2 • 2 H 2 O: ii. 75.49% of ( X grams) of CaCl 2 •2 H 2 O = 12.20 g of CaCl 2  0.7549( X ) = 12.20 or X = 12.20/0.7549 = 16.16 grams i. % CaCl 2 = 110.98 g CaCl 2 147.02 g CaCl 2 • 2 H 2 O x 100 = 75.49% Then we note that the 12.20 g of CaCl 2 desired must be 75.49% of the mass of the hydrate used: Percent composition

232. Example: How many grams of CaCl 2 • 2 H 2 O must be weighed out to obtain 12.20 grams of CaCl 2 ? There are two ways of solving this problem: 12.20 g CaCl 2 x = 16.16 g note that, math-wise, both methods involve the exact same calculations (i.e., the ratio of the molar mass of the hydrate to the molar mass of the anhydrous form had to be determined) . The only difference was the “logic” you followed which led you to that calculation! Percent composition Method 2 : Use dimensional analysis and molar masses: 110.98 g CaCl 2 1 mole CaCl 2 x 1 mole CaCl 2 1 mole CaCl 2 • 2 H 2 O 147.02 g CaCl 2 • 2 H 2 O x 1 mole CaCl 2 • 2 H 2 O

234. Percent Composition and Empirical Formulas 2. Next, knowing the mass of each element (in your 100 gram sample), determine the number of moles of that element in your sample, by dividing the mass by the molar mass of the element. The number of moles of Fe and O in our sample of the iron ore would be: 69.94 grams Fe x = 1.252 mol Fe 30.06 grams O x = 1.879 mol O 55.847 g 1 mol Fe 16.00 g 1 mol O

235. Percent Composition and Empirical Formulas 3. To find the simplest mole ratio of the elements, divide each by the smallest number: in our iron ore sample, we would have: = 1.501 mol O per mole of Fe 4. If this ratio is a whole number , then you are done – if the ratio is NOT a whole number, it must be converted to a whole number ratio (we cannot have fractions of an atom!) Fe 1.00 O 1.50 = Fe O = Fe 2 O 3 1.879 mol O 1.252 mol Fe 3 2 2 2

236. Percent Composition and Empirical Formulas The process is summarized in Figure 3.5 on page 89 in your textbook.

237. Percent Composition and Empirical Formulas Example : Determine the empirical formula of a compound that has the following percent composition by mass: K = 24.75%, Mn = 34.77%, and O = 40.51% n Mn = 34.77 g Mn x = 0.6329 mol Mn 1 mol Mn 54.94 g Mn n O = 40.51 g O x = 2.532 mol O 1 mol O 16.00 g O n K = 24.75 g K x = 0.6330 mol K 1 mol K 39.10 g K

238. n K = 0.6330, n Mn = 0.6329, n O = 2.532 Percent Composition and Empirical Formulas The empirical formula for the compound is: KMnO 4 K : ~ ~ 1.0 0.6330 0.6329 Mn : 0.6329 0.6329 = 1.0 O : ~ ~ 4.0 2.532 0.6329 divide each element by the smallest mole value

239. Combust 10.0 g compound Collect 24.078 g CO 2 and 11.088 g H 2 O Experimental determination of a molecular formula Determination of Molecular Formulas sample

240. Determination of Molecular Formulas C 4 H 9 O empirical formula = mass of each element moles of each element mole ratios of the elements empirical formula

241. Determination of Molecular Formulas The molecular weight of the compound was determined experimentally to be 146.2 g/mol. To determine the true molecular formula, divide the molecular weight by the formula weight. This ratio gives the number each subscript must be multiplied by to give the molecular formula. Formula weight of C 4 H 9 O = 73.1 g/mol Molecular weight of compound = 146.2 g/mol thus, the true molecular formula is (C 4 H 9 O) 2 = C 8 H 18 O 2

242. Working with Chemical Equations

243. For example, there are several ways of representing the reaction of H 2 with O 2 to form H 2 O A process in which one or more substances is changed into one or more new substances is a chemical reaction A chemical equation uses chemical symbols to show what happens during a chemical reaction Chemical Equations

244. How to “Read” Chemical Equations 2 Mg + O 2 2 MgO 2 atoms Mg + 1 molecule O 2 makes 2 formula units MgO 2 moles Mg + 1 mole O 2 makes 2 moles MgO 48.6 grams Mg + 32.0 grams O 2 makes 80.6 g MgO reactants form products IT DOES NOT IMPLY THAT 2 grams Mg + 1 gram O 2 makes 2 g MgO this is based on the molar masses of the species and the coefficients in the reaction…

249. example: balance the following equations: Fe + S -> Fe 2 S 3 K + H 2 O -> KOH + H 2 Al + S 8 -> Al 2 S 3 Pb(NO 3 ) 2 + KI -> KNO 3 + PbI 2 3 2 2 2 2 16 3 8 2 2

250. Stoichiometry One of the most important applications of balanced equations is in determining the amount of one reactant required to react completely with another, or in determining the theoretical amount of product that should be formed in a given reaction. These problems all follow the same set of “logic” steps – indeed, almost any problem involving balanced equations will always follow these same steps! Stoichiometry is the quantitative study of reactants and products in a chemical reaction.

252. Example : Methanol burns in air according to the equation: If 209 g of methanol are used up in the combustion, what mass of water is produced? 209 g CH 3 OH = 235 g H 2 O The sequence of steps we follow in solving the problem are: Stoichiometry 2CH 3 OH + 3O 2 2CO 2 + 4H 2 O grams CH 3 OH moles CH 3 OH grams H 2 O 1 mol CH 3 OH 32.0 g CH 3 OH x 4 mol H 2 O 2 mol CH 3 OH x 18.0 g H 2 O 1 mol H 2 O x moles H 2 O moles CH 3 OH

253. The molarity of a solution is the concentration of that solution, expressed as the moles of solute present in 1 liter of a solution Solution Stoichiometry It is often easier to work with solutions, rather than solids. This means we also need a means of quantitatively working with reactions in solution. read as, for example: 2M NaCl = 2 “molar” solution of NaCl see pages 142-150 in Chapter 4 M = molarity = moles of solute liters of solution

254. M KI MW KI 500. mL = 232 g KI x 2.80 mol KI 1 L soln x x Solution Stoichiometry example: what mass of KI is needed to prepare 500 mL of a 2.80 M solution of KI? Solution plan: convert volume to moles using molarity, then moles to mass using molar mass: volume of KI solution moles KI grams KI 166 g KI 1 mol KI 1 L 1000 mL

255. known mass of solute dissolve solute dilute to mark

256. Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. Solution Stoichiometry Dilution Add Solvent Moles of solute (concentrated) (c) Moles of solute after dilution (d) = M c V c M d V d =

257. example: How would you prepare 60.0 mL of 0.200 M HNO 3 from a stock solution of 4.00 M HNO 3 ? M c V c = M d V d M c = 4.00 M M d = 0.200 M V d = 0.0600 L V c = ? L = 0.003 L = 3 mL Thus, add 3 ml of acid to 57ml of water to form 60 ml of solution (dilute the 3 ml of acid to a total volume of 60 ml) Solution Stoichiometry V c = M d V d M c = 0.200 M x 0.0600 L 4.00 M

258. As with all stoichiometry problems, convert the starting unit to moles. Note that we now have THREE methods of converting to moles: Solving Solution Stoichiometry Problems 1. Use Avogadro’s number to convert particles to moles 2. Use the molar mass of the substance to convert grams to moles # particles x N A = moles grams x mole molar mass = moles

259. volume x = moles moles L sol’n 3. And now our third method is to use the molarity of the solution to convert volume* to moles. *the volumes must be in LITERS when converting to moles using the molarity of the solution. Solution Stoichiometry Note also that we can convert moles to volume by multiplying moles x 1/M

260. example : How many ml of 0.35 M Na 3 PO 4 are needed to react completely with 28.0 ml of a 0.42 M solution of Ba(NO 3 ) 2 , according to the balanced equation shown below: 3 Ba(NO 3 ) 2 + 2 Na 3 PO 4 -> Ba 3 (PO 4 ) 2 + 6 NaNO 3 0.0280 L x = 22.4 ml Solution plan: convert to moles, use mol : mol ratio from the balanced equation, convert moles to liters using molarity, then convert to mL. 0.42 mol Ba(NO 3 ) 2 Liter x 2 mol Na 3 PO 4 3 mol Ba(NO 3 ) 2 x 1 L 0.35 mol Na 3 PO 4 x 1000 mL I L

261. Gravimetric Analysis Gravimetric analysis is an analytical technique based on the measurement of the mass (usually of an ionic substance.) The substance of interest is typically reacted in solution and comes out as a precipitate. The precipitate is then filtered off, dried and weighed. Knowing the mass and chemical formula of the precipitate that formed, we can calculate the mass of a particular chemical component of the original sample.

263. example: A 0.5662 gram sample of an unknown ionic compound containing chloride ions is dissolved in water and treated with an excess of AgNO 3 . If 1.0882 grams of AgCl precipitated , what is the percent chlorine in the original sample? First determine the mass of Cl‾ ions in the AgCl ppt: 1.0882 g AgCl x 1 mol AgCl 143.4 g AgCl x x = 0.2690 g Now determine the % Cl in the original sample: 47.51 % 1 mol Cl‾ ions 1 mol AgCl 35.45 g Cl 1 mol Cl 0.2690 g Cl % Cl = 0.5662 g unknown x 100 =

264. Limiting Reagents The limiting reagent is the reactant that gets used up completely -- that is, the one not present in excess. It is very rare that you mix reactants together in the exact stoichiometric ratio needed for each to react completely with the other. Usually, you have a little “extra” of one of the reactants compared to the other one, that is, one reagent is present in excess .

265. The maximum amount of product that can be formed is thus limited by the amount of the limiting reagent present. Limiting Reagents When the reaction is completed, there will be no limiting reagent left over. But there will be some of the reagents in excess left over. The reagents present in quantities greater than the minimum amount necessary are called the reagents in excess.

266. Limiting Reagents NO is the limiting reagent O 2 is the excess reagent Consider the reaction between NO and O 2 to form NO 2 . If we start with the mix shown at the top right, and end with the mix shown at the bottom right, we see that oxygen was present in excess (some is left over) which means that NO was the limiting reagent. 2NO + 2O 2 2NO 2

267. example: In one process, 124 g of Al are reacted with 601 g of Fe 2 O 3 according to the rxn: 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe Calculate the mass of Al 2 O 3 formed. 124 g Al 367 g Fe 2 O 3 Have more Fe 2 O 3 (601 g) so Al is limiting reagent Limiting Reagents g Al mol Al mol Fe 2 O 3 needed g Fe 2 O 3 needed OR g Fe 2 O 3 mol Fe 2 O 3 mol Al needed g Al needed 1 mol Al 27.0 g Al x 1 mol Fe 2 O 3 2 mol Al x 160. g Fe 2 O 3 1 mol Fe 2 O 3 x = Start with 124 g Al need 367 g Fe 2 O 3

268. Use limiting reagent (Al) to calculate amount of product that can be formed. 124 g Al 234 g Al 2 O 3 3.9 Now… Limiting Reagents g Al mol Al mol Al 2 O 3 g Al 2 O 3 1 mol Al 27.0 g Al x 1 mol Al 2 O 3 2 mol Al x 102. g Al 2 O 3 1 mol Al 2 O 3 x = 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe

269. Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Actual Yield is the amount of product actually obtained from a reaction. Reaction Yields % Yield = Actual Yield Theoretical Yield x 100

270. Reactions in Aqueous Solution Chapter 4 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. AP Inorganic Chemistry

271. Symbols used in Equations (s) solid (l) liquid (g) gas ( aq ) aqueous = (dissolved in H 2 O) solid precipitate given off as a gas yields equilibrium  heated 20º C specified rxn temperature MnO 2 MnO 2 is a catalyst in the rxn arrows are only used for products!

272. examples Na 2 CO 3 (s) Na 2 O (s) + CO 2 (g)  HCl (g) + H 2 O (l) H 3 O + (aq) + Cl¯ (aq) We often write H 2 O over the yields arrows when we show that something is being dissolved in water. NaCl (s) + AgNO 3 (aq) AgCl ( ) + NaNO 3 (aq) 2 KClO 3 (s) 2 KCl (s) + 3 O 2 ( ) MnO 2  HCl (g) H 3 O + (aq) + Cl¯ (aq) H 2 O or

273. General Properties of Aqueous Solutions

274. A solution is a homogenous mixture of 2 or more substances The solute is the substance(s) present in the smaller amount. The solvent is the substance present in the larger amount Solutions

275. During the solution process, the solute is first surrounded by the solvent molecules, and the attractions between the solvent and the solute help to pull the solvent particles apart. The isolated solute particles are in turn surrounded by a “sphere” of solvent particles in a process called solvation – in the case of aqueous solutions, the term hydration is used. General Properties of Aqueous Solutions

276. Hydration (or solvation ) : the process in which a solute particle, such as an ion or a neutral molecule, is surrounded by water molecules arranged in a specific manner. Hydration “spheres” Solutions

277. An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity . A nonelectrolyte is a substance that, when dissolved, results in a solution that does not conduct electricity . Solutions nonelectrolyte weak electrolyte strong electrolyte

278. Strong Electrolyte – 100% dissociation Weak Electrolyte – not completely dissociated To conduct electricity, a solution must contain charged particles, that is, cations (+) and anions (-) Solutions (all three species are present at equilibrium) NaCl ( s ) Na + ( aq ) + Cl - ( aq ) H 2 O CH 3 COOH CH 3 COO - ( aq ) + H 3 O + ( aq ) H 2 O

279. Nonelectrolytes do not produce charged particles in solution Solutions does not dissociate into ions when in solution C 6 H 12 O 6 ( s ) C 6 H 12 O 6 ( aq ) H 2 O

280. Classification of Reactions PbI 2

283. Precipitation reactions involve the exchange of cations between two ionic compounds that results in the formation of an insoluble precipitate*: *we will learn to determine which substances are soluble and which are not later on in this chapter… Neutralization reactions involve an acid and a metal hydroxide; the acid’s H + ion is exchanged with the hydroxide’s metal cation to produce an ionic “salt” and water. Classes of Reactions (which is just H 2 O!) Ag NO 3 (aq) + Na Cl (aq) Ag Cl (s) + Na NO 3 (aq) eg: H Cl + K OH K Cl + H OH eg:

284. 2. Single displacement: A + B C B + A C Classes of Reactions Hydrogen Displacement Metal Displacement Halogen Displacement Single displacement is a reaction in which one element displaces another in a compound. The general pattern is: Single displacement reactions are typically one of three types: 2 Mg + Ti Cl 4 2 Mg Cl 2 + Ti Cl 2 + 2K Br 2K Cl + Br 2 Sr + 2 H 2 O Sr (OH) 2 + H 2

285. eg: 2 Al + 3 Br 2 2 AlBr 3 3. Synthesis: A + B AB Classes of Reactions H 2 O + SO 3 H 2 SO 4 A synthesis reaction is one in which two substances react and combine to form one substance. The general pattern of the reaction is: The reactants can be elements, compounds, or one of each.

286. eg: 2 KClO 3 2 KCl + 3 O 2 4. Decomposition: Decomposition is the opposite of synthesis: one substance decomposes (often by heating it) into two or more new substances. The general pattern is: 2 NaHCO 3 Na 2 O + 2 CO 2 + H 2 O Classes of Reactions AB A + B Δ Δ

287. Classes of Reactions 5. Combustion: eg: 2 Mg + O 2 2 MgO Technically, any reaction involving oxygen is a combustion reaction. Note that this reaction can also be classified as a synthesis reaction. A + O 2 A O x

288. C n H m + O 2 CO 2 + H 2 O Combustion reactions that involve hydrocarbon compounds (which may or may not contain O or N) reacting with oxygen gas will form CO 2 and H 2 O. The general reaction is: C 2 H 6 O + 3 O 2 2 CO 2 + 3 H 2 O Incomplete combustion forms CO (carbon monoxide). Classes of Reactions eg: CH 4 + 2 O 2 CO 2 + 2 H 2 O

290. We are now going to look at each class of reaction in more detail. We will begin with the sub-categories of the double displacement reaction

291. Precipitation Reactions

292. Solubility is the maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature. Precipitation Reactions If a substance dissolves in a solvent, it is said to be soluble ; if it does not, it is insoluble. Substances that are insoluble have a stronger attraction towards each other than they do towards the solvent (or sometimes, the solvent molecules have a stronger attraction towards themselves than towards the solute particles). Precipitate – insoluble solid that separates from solution One of the most common types of double displacement reaction is the precipitation reaction.

293. Precipitation Reactions PbI 2 precipitate of PbI 2 When aqueous solutions of Pb(NO 3 ) 2 and KI are mixed, a bright yellow precipitate of PbI 2 forms. Pb 2+ and I - form strong attractions Pb(NO 3 ) 2 ( aq ) + 2KI ( aq ) PbI 2 ( s ) + 2KNO 3 ( aq )

294. How do you know which substances are soluble and which form precipitates in aqueous solutions? There is a Table of general Solubility Rules on page 123 in your textbook. AP Chemistry students are required to memorize this list. (You need to know it for the AP Test!) Non-AP students are NOT required to memorize this list; it will be provided to you on tests and quizzes. Solubility Rules Precipitation Reactions

295. see page 123 Precipitation Reactions and acetates ( C 2 H 3 O 2 - )

296. While most double displacement reactions involve the formation of a precipitate, some double displacement reactions involve “dissolving” an insoluble compound by forming a soluble salt. These types of reactions most often involve the reaction of an insoluble ionic compound with an acid: AgCl (s) + HC 2 H 3 O 2 (aq) AgC 2 H 3 O 2 (aq) + HCl (aq) PbS (s) + 2 HNO 3 (aq) Pb(NO 3 ) 2 (aq) + H 2 S (g) Precipitation Reactions

297. Molecular and net ionic equations Precipitation Reactions The complete balanced equation, showing the formulas for each reactant and species is called the “molecular equation.” We can also show how each species in the reaction will dissociate or ionize when dissolved in water. Insoluble compounds do not dissociate, but soluble compounds will. We call this the “ionic equation.” molecular equation Pb(NO 3 ) 2 ( aq ) + 2NaI ( aq ) PbI 2 ( s ) + 2NaNO 3 ( aq ) ionic equation Pb 2+ + 2NO 3 - + 2Na + + 2I - PbI 2 ( s ) + 2Na + + 2NO 3 -

298. Molecular and net ionic equations Those species which appear unchanged on both sides of the yields sign did not “participate” in the reaction. They are said to be spectator ions. The net ionic equation shows only those species which actually participated in the reaction – all the spectator ions are cancelled out and not shown. net ionic equation In the above reaction, Na + and NO 3 - are the spectator ions – they do not participate in the net reaction. Pb 2+ ( aq ) + 2I − ( aq ) PbI 2 ( s ) be sure to include the phases, etc. in net ionic equations! Pb 2+ + 2NO 3 − + 2Na + + 2I − PbI 2 ( s ) + 2Na + + 2NO 3 −

300. example: Write the net ionic equation for the reaction of silver nitrate with sodium chloride. Net Ionic Equations molecular ionic net ionic AgNO 3 ( aq ) + NaCl ( aq ) AgCl ( s ) + NaNO 3 ( aq ) Ag + + NO 3 − + Na + + Cl − AgCl ( s ) + Na + + NO 3 − Ag + ( aq ) + Cl − ( aq ) AgCl ( s ) CuS (s) + 2 HC 2 H 3 O 2 ( aq ) Cu(C 2 H 3 O 2 ) 2 (aq) + H 2 S (g) CuS (s) + 2 H + (aq) Cu 2+ (aq) + H 2 S (g) CuS (s) + 2 H + + 2 C 2 H 3 O 2 − Cu 2+ + 2 C 2 H 3 O 2 − + H 2 S (g) example: Write the net ionic equation for the reaction of CuS with acetic acid molecular ionic net ionic

301. Note that if both reactants and products exist as solvated ions in solution, then NO REACTION has occurred – you began with a mix of hydrated ions, and you ended with the same mix of hydrated ions… K + + NO 3 − + Na + + Cl − K + + Cl − + Na + + NO 3 − They are ALL spectator ions! We write: Net Ionic Equations example KNO 3 ( aq ) + NaCl ( aq ) KCl ( aq ) + NaNO 3 ( aq ) KNO 3 ( aq ) + NaCl ( aq ) N.R.

302. Acid-Base Reactions H 3 O + = hydronium ion H H H O +

303. Arrhenius and Br  nsted-Lowry Definitions There are several “definitions” of acids or bases, from a chemical standpoint. The two most important definitions are those given by Svante Arrhenius in the late 19th century, and by J.N. Br  nsted and Thomas Lowry , who independently developed similar chemical descriptions of acids and bases in the 20th century. A very common type of double displacement reaction involves the neutralization of an acid with a base. Acid-Base Reactions

304. An Arrhenius acid is a substance that ionizes to produce H + ions in water An Arrhenius base is a substance that dissociates to produce OH - ions in water Acids and Bases i.e., Arrhenius bases are metal hydroxides that are soluble in water. H 2 O NaOH (s) Na + (aq) + OH ¯ (aq) H 2 O HCl (g) H + (aq) + Cl ¯ (aq) H 2 O

305. An H + ion is essentially a bare proton – this is an extremely reactive species! H + ions will instantly bond to a water molecule to form the polyatomic cation, H 3 O + , called the “hydronium ion.” Acids and Bases So, actually, an Arrhenius acid is a substance that produces H 3 O + ions in water .

306. A Br ø nsted-Lowry acid is a proton (H + ) donor A Br ø nsted-Lowry base is a proton (H + ) acceptor Acids and Bases Brønsted-Lowry made use of the fact that H + ions are essentially just a proton in their definition of acids and bases: B-L acid and base is a somewhat more “general” definition, since it does not require the presence of water as a solvent. However, one can certainly have an aqueous B-L acid or base!

307. acid base In the forward direction, water acts as the proton donor and NH 3 the acceptor… acid base Consider the reaction between NH 3 and H 2 O: … in the reverse direction, NH