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CHEMISTRY 115 Fall 2012 1 Why Study Chemistry? Think about current issues in the news. 2 Why Study Chemistry? Think about current issues in the news. Energy sources 3 Why Study Chemistry? Think about current issues in the news. Energy sources Greenhouse effect 4 Why Study Chemistry? Think about current issues in the news. Energy sources Greenhouse effect Pollution 5 Why Study Chemistry? Think about current issues in the news. Energy sources Greenhouse effect Pollution Ozone problem 6 Why Study Chemistry? Think about current issues in the news. Energy sources Greenhouse effect Pollution Ozone problem Food additives 7 Why Study Chemistry? Think about current issues in the news. Energy sources Greenhouse effect Pollution Ozone problem Food additives Drugs 8 INTRODUCTION CHEMISTRY: A science that deals with the composition, structure, and properties of substances and of the changes they undergo. There are six major subdivisions of Chemistry: 9 1. Organic chemistry: Covers the compounds of carbon and hydrogen (hydrocarbons). All compounds derived from hydrocarbons. 10 1. Organic chemistry: Covers the compounds of carbon and hydrogen (hydrocarbons). All compounds derived from hydrocarbons. 2. Inorganic chemistry: Covers all the elements, and all compounds, except the hydrocarbons and their derivatives. 11 1. Organic chemistry: Covers the compounds of carbon and hydrogen (hydrocarbons). All compounds derived from hydrocarbons. 2. Inorganic chemistry: Covers all the elements, and all compounds, except the hydrocarbons and their derivatives. 3. Physical chemistry: Measurement of physical properties. Interpretation of physical and chemical properties. 12 4. Biochemistry: Study of pure substances and chemical reactions in living systems. 13 4. Biochemistry: Study of pure substances and chemical reactions in living systems. 5. Analytical chemistry: Measurement of the amounts of substances. Measurement of chemical composition of materials. Separation of the components of mixtures 14 4. Biochemistry: Study of pure substances and chemical reactions in living systems. 5. Analytical chemistry: Measurement of the amounts of substances. Measurement of chemical composition of materials. Separation of the components of mixtures 6. Theoretical chemistry: Mathematical description of chemical structures and of chemical changes. 15 Scientific Method 16 Scientific Method A series of steps used to solve scientific problems. 17 Scientific Method A series of steps used to solve scientific problems. Objective: Some problem to be solved, e.g. how oxygen gas binds to the hemoglobin molecule in our blood. 18 Collection of data: Once the goal is defined, the next step involves making careful observations and collecting bits of information about the system. The bits of information are called data. The word system here means that part of the universe that is under investigation. 19 The information obtained may be both qualitative or quantitative. 20 The information obtained may be both qualitative or quantitative. Qualitative: general and non-mathematical. E.g. the object has a blue color. 21 The information obtained may be both qualitative or quantitative. Qualitative: general and non-mathematical. E.g. the object has a blue color. Quantitative : numerical – related to measurements. E.g. the density is 2.1 g/ml. 22 Law: After a large amount of data has been collected, it is often desirable to summarize the information in a concise way. This summarizing statement is called a law. 23 Law: After a large amount of data has been collected, it is often desirable to summarize the information in a concise way. This summarizing statement is called a law. A law is a concise verbal or mathematical statement of a relation between phenomena that is always the same under the same conditions. 24 Hypothesis: Once enough information has been gathered, a tentative explanation for the observations can be formulated – this is the hypothesis. 25 Hypothesis: Once enough information has been gathered, a tentative explanation for the observations can be formulated – this is the hypothesis. Further experiments are devised to test the validity of the hypothesis in as many ways as possible. 26 The hypothesis provides tentative explanations that must be tested by many experiments. If the hypothesis survives such tests, the hypothesis develops into a theory. 27 The hypothesis provides tentative explanations that must be tested by many experiments. If the hypothesis survives such tests, the hypothesis develops into a theory. Theory: A theory is a unifying principle that explains a body of facts and those laws that are based on them. Theories are constantly being tested. If a theory is proved incorrect by experiment, then it must be discarded or modified, so that it becomes consistent with experimental observations. 28 Scientific progress is made by modifying old laws and theories or replacing them with new ones. 29 Summary – the sequence 30 Summary – the sequence 1. Objective 31 Summary – the sequence 1. Objective 2. Data collection 32 Summary – the sequence 1. Objective 2. Data collection 3. Summarize data, generalization, formulation of law 33 Summary – the sequence 1. Objective 2. Data collection 3. Summarize data, generalization, formulation of law 4. Hypothesis (tentative explanation) 34 Summary – the sequence 1. Objective 2. Data collection 3. Summarize data, generalization, formulation of law 4. Hypothesis (tentative explanation) 5. Test hypothesis 35 Summary – the sequence 1. Objective 2. Data collection 3. Summarize data, generalization, formulation of law 4. Hypothesis (tentative explanation) 5. Test hypothesis 6. Formulate theory 36 Summary – the sequence 1. Objective 2. Data collection 3. Summarize data, generalization, formulation of law 4. Hypothesis (tentative explanation) 5. Test hypothesis 6. Formulate theory 7. Further testing 37 Summary – the sequence 1. Objective 2. Data collection 3. Summarize data, generalization, formulation of law 4. Hypothesis (tentative explanation) 5. Test hypothesis 6. Formulate theory 7. Further testing 8. Rejection or modification of theory as required to account for new observations. 38 Some basic definitions Matter: Anything that occupies space and possesses mass is called matter. 39 Some basic definitions Matter: Anything that occupies space and possesses mass is called matter. Mass: The mass of a body is a measure of the quantity of matter contained in that body. 40 Some basic definitions Matter: Anything that occupies space and possesses mass is called matter. Mass: The mass of a body is a measure of the quantity of matter contained in that body. Weight: Refers to the force which gravity exerts upon an object. Unfortunately, chemists very frequently use the word “weight” when they mean mass. 41 Substance: A substance is a form of matter that has a definite composition and distinct properties. Examples: gold, water, oxygen. 42 Substance: A substance is a form of matter that has a definite composition and distinct properties. Examples: gold, water, oxygen. Mixture: A combination of two or more substances in which the substances retain their identities. Examples: air, a solution of table sugar (sucrose) in water. 43 Substance: A substance is a form of matter that has a definite composition and distinct properties. Examples: gold, water, oxygen. Mixture: A combination of two or more substances in which the substances retain their identities. Examples: air, a solution of table sugar (sucrose) in water. Note: Mixtures do not have constant composition; samples of air collected in Los Angles will have different composition from samples collected in Eau Claire. 44 There are two types of mixtures: homogeneous and heterogeneous. 45 There are two types of mixtures: homogeneous and heterogeneous. Homogeneous mixture: The composition is the same throughout. Example: a small amount of sugar completely dissolved in water. 46 There are two types of mixtures: homogeneous and heterogeneous. Homogeneous mixture: The composition is the same throughout. Example: a small amount of sugar completely dissolved in water. Heterogeneous mixture: A mixture in which the individual components remain physically separate and can be seen as separate components. Example: a mixture of sugar and sand. 47 Any mixture, be it homogeneous or heterogeneous, can be put together and then separated into pure components without any change in the identity of the components, by physical means. 48 For example: sugar can be removed from a homogeneous sugar solution by evaporating off the solvent water. 49 For example: sugar can be removed from a homogeneous sugar solution by evaporating off the solvent water. A sugar/sand mixture could be separated by dissolving the sugar in water, drying the sand, and reclaiming the sugar by evaporation of the solution. 50 For example: sugar can be removed from a homogeneous sugar solution by evaporating off the solvent water. A sugar/sand mixture could be separated by dissolving the sugar in water, drying the sand, and reclaiming the sugar by evaporation of the solution. In the physical separation process, there has been no change in the composition of each substance making up the mixture. 51 Physical property: Any property of a substance that can be observed without permanently* changing the substance to form some other substance. Examples: color, density, melting point. * Some exceptions to this. E.g. some compounds decompose at their melting point. A physical property can be specified without reference to any other substance. 52 Chemical property: Any property of a substance that cannot be studied without resulting in a permanent change of the substance to form some other substance. 53 Chemical property: Any property of a substance that cannot be studied without resulting in a permanent change of the substance to form some other substance. Example: sodium metal is very reactive with water. 54 Chemical property: Any property of a substance that cannot be studied without resulting in a permanent change of the substance to form some other substance. Example: sodium metal is very reactive with water. Reactivity is a chemical property that refers to the tendency of a substance to undergo a particular chemical reaction. 55 Element: An element is a pure substance that cannot be separated into simpler substances by chemical means. 56 Element: An element is a pure substance that cannot be separated into simpler substances by chemical means. Compound: A pure substance composed of two or more elements chemically united in fixed proportions. 57 Element: An element is a pure substance that cannot be separated into simpler substances by chemical means. Compound: A pure substance composed of two or more elements chemically united in fixed proportions. Atom: The smallest particle of an element that retains the chemical nature of the element. 58 Element: An element is a pure substance that cannot be separated into simpler substances by chemical means. Compound: A pure substance composed of two or more elements chemically united in fixed proportions. Atom: The smallest particle of an element that retains the chemical nature of the element. Molecule: A structure consisting of two or more atoms that are chemically bound together and behave as an independent unit. 59 Energy: Is the capacity to do work or produce change. 60 Energy: Is the capacity to do work or produce change. Potential Energy: Energy available because of the position of an object. 61 Energy: Is the capacity to do work or produce change. Potential Energy: Energy available because of the position of an object. Kinetic Energy: Energy available because of the motion of an object. 62 Energy: Is the capacity to do work or produce change. Potential Energy: Energy available because of the position of an object. Kinetic Energy: Energy available because of the motion of an object. Chemical Energy: The energy stored by compounds. 63 CHEMICAL SYMBOLS 64 CHEMICAL SYMBOLS Chemical symbols are shorthand notation for the names of elements. There are 117 known elements. The last several have not been named. 65 If an element has a single letter to represent it – the letter must be capitalized. 66 If an element has a single letter to represent it – the letter must be capitalized. If an element is represented by two letters – the first letter must be capitalized and the second letter must be lower case. Examples: NO is not the symbol for Nobelium – the correct symbol is No. CO is not the symbol for cobalt – the correct symbol is Co. 67 CHEMICAL FORMULAS A chemical formula shows the chemical composition of a compound. Three types of chemical formulas will be of interest to us. 68 CHEMICAL FORMULAS A chemical formula shows the chemical composition of a compound. Three types of chemical formulas will be of interest to us. 1. Molecular formulas 69 CHEMICAL FORMULAS A chemical formula shows the chemical composition of a compound. Three types of chemical formulas will be of interest to us. 1. Molecular formulas 2. Empirical formulas 70 CHEMICAL FORMULAS A chemical formula shows the chemical composition of a compound. Three types of chemical formulas will be of interest to us. 1. Molecular formulas 2. Empirical formulas 3. Structural formulas 71 Molecular formulas An expression showing the exact numbers and types of elements present in a molecule. 72 Molecular formulas An expression showing the exact numbers and types of elements present in a molecule. Example: The molecular formula for water is H2O – this formula tells us that a molecule of water is formed from 2 atoms of hydrogen (symbol H) 1 atom of oxygen (symbol O) The subscript 2 tells how many atoms of the element on the left of the subscript are present. 73 When no subscripts are present, the implied number is one. When formulas like (CH3)2CO are encountered, the parenthesis followed by the subscript 2 mean that there are two CH3 groups, so for the given formula, we have: 3 atoms of carbon 6 atoms of hydrogen 1 atom of oxygen 74 The compound Ni(CO)4 has one atom of Ni, four atoms of carbon, and 4 atoms of oxygen. 75 Empirical formulas An expression showing types of elements present and the ratio of the different kinds of atoms. 76 Empirical formulas An expression showing types of elements present and the ratio of the different kinds of atoms. The empirical formula is useful information when the identity of an unknown compound is to be determined. 77 Empirical formulas An expression showing types of elements present and the ratio of the different kinds of atoms. The empirical formula is useful information when the identity of an unknown compound is to be determined. Note: Many different compounds may have the same empirical formula. 78 Examples: A molecule of glucose consists of 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms – and so its molecular formula is C6H12O6. 79 Examples: A molecule of glucose consists of 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms – and so its molecular formula is C6H12O6. From the formula we see that the proportion of atoms for carbon : hydrogen : oxygen is 6 : 12 : 6 or 1:2:1 The empirical formula of glucose is CH2O. 80 For many formulas, the empirical formula is the same as the molecular formula, e.g. H2O. 81 For many formulas, the empirical formula is the same as the molecular formula, e.g. H2O. Two examples with different molecular formulas but the same empirical formula: Benzene has the molecular formula C6H6 and its empirical formula is CH. Acetylene C2H2, has the empirical formula CH. 82 For ionic compounds, e.g. sodium chloride, the formula shows the ratio of elements that form the compound. 83 For ionic compounds, e.g. sodium chloride, the formula shows the ratio of elements that form the compound. Solid sodium chloride consists of a collection of positively charged sodium ions and negatively charged chloride ions in a three-dimensional structure. You cannot say which sodium ion is associated with any particular chloride ion. 84 For ionic compounds, e.g. sodium chloride, the formula shows the ratio of elements that form the compound. Solid sodium chloride consists of a collection of positively charged sodium ions and negatively charged chloride ions in a three-dimensional structure. You cannot say which sodium ion is associated with any particular chloride ion. The formula NaCl should be regarded as the empirical formula. 85 STRUCTURAL FORMULAS An expression showing the exact numbers and types of atoms present in a molecule, and information about how the atoms are chemically bonded to one another. Examples: O H water H 86 H O O hydrogen peroxide H 87 H O O hydrogen peroxide H 88 H H C H methane H 89 H H C H H C H O H ethanol (ethyl alcohol) 90 Naming inorganic compounds Two principal groups: 91 Naming inorganic compounds Two principal groups: (1) metal with nonmetal (or nonmetal group) use Stock system 92 Naming inorganic compounds Two principal groups: (1) metal with nonmetal (or nonmetal group) use Stock system (2) nonmetal with nonmetal use prefix system 93 Stock system 94 Stock system Oxidation number: is the charge that an atom in a compound would have if the electrons in each bond belonged entirely to the more electronegative atom. 95 Stock system Oxidation number: is the charge that an atom in a compound would have if the electrons in each bond belonged entirely to the more electronegative atom. Electronegativity: relative attraction of an atom for electrons. 96 Oxidation numbers: In many cases the oxidation number of an element in an ionic compound is the same as the formal charge on the ions present. Example: in the ionic compound KCl, which is composed of K+ ions and Cl- ions, the oxidation number of potassium is +1 and the oxidation number of the chlorine is -1. 97 For a neutral compound the sum of the oxidation numbers of the elements = 0. Example: for KMnO4 for which the oxidation numbers are K (+1), Mn (+7), and O (-2), the sum of the oxidation numbers = 1 + 7 + 4x(-2) = 0. 98 For a cation or an anion, the sum of the oxidation numbers of the elements = the charge on the species. Example: For MnO4the sum of the oxidation numbers = 7 + 4x(-2) = -1, where -1 is the charge on the anion. 99 1. The cation is named first. 100 1. The cation is named first. 2. The oxidation number is given next (using Roman numerals in parentheses) – if it is needed. 101 1. The cation is named first. 2. The oxidation number is given next (using Roman numerals in parentheses) – if it is needed. The oxidation number is given only if the element commonly has more than one oxidation state. 102 1. The cation is named first. 2. The oxidation number is given next (using Roman numerals in parentheses) – if it is needed. The oxidation number is given only if the element commonly has more than one oxidation state. 3. The anion is named second. 103 Examples: FeCl3 Iron forms two common cations, Fe2+ and Fe3+ , hence it will be necessary to specify the oxidation number of the Fe. The name is iron (III) chloride 104 Examples: FeCl3 Iron forms two common cations, Fe2+ and Fe3+ , hence it will be necessary to specify the oxidation number of the Fe. The name is iron (III) chloride CoPO4 Cobalt forms two common cations, Co2+ and Co3+ , hence it will be necessary to specify the oxidation number of the Co. The name is cobalt (III) phosphate 105 LiClO3 The cation has only one common oxidation state (which is +1), so it is not necessary to give this as part of the name. The name is lithium chlorate You need to be able to name in both directions: formula name name formula 106 Give the formula for the following: Iron (II) phosphate The two ions are Fe2+ and PO43- which can be put together Fe2+ PO43- Fe3(PO4)2 107 calcium sulfate The two ions are Ca2+ and SO42- which can be put together directly as CaSO4 (Note: there are no subscripts of 2 on the calcium and the sulfate ions in the final formula. Keep in mind that we are working with empirical formulas for ionic compounds. The one odd exception to this that you will encounter are the mercury (I) salts.) 108 To use the Stock system, you need to know the charges on the common cations and anions. 109 For the anions, pay attention to the endings. Within a given series, the name of the anion with more oxygen atoms usually ends with an “ate” ending and that with fewer oxygen atoms ends with an “ite” ending. Examples: ClO- hypochlorite ClO2- chlorite ClO3- chlorate ClO4- perchlorate 110 CHARGES OF SOME COMMON IONS IA II A IB II B III A IV A VA VI A VII A N-3 O-2 F- P-3 S-2 Cl- H+ Li+ Be+2 Na+ Mg+2 K+ Ca+2 Al+3 Cr+2 Cr+3 Rb+ Sr+2 Mn+2 Fe+2 Co+2 Fe+3 Co+3 Ni+2 Cu+1 Zn+2 Br- Cu+2 Ag+1 Cd+2 Sn+2 I- Sn+4 Cs+ Ba+2 Hg2+2 Pb+2 Bi+3 Hg+2 Group IA elements all form single positvely charged ions. Group IIA elements form doubly charged positive ions. Elements in the transition metal group are frequently able to form cations with different charges. Note the unusual formula for the mercury (I) ion. Caions with a +4 charge are very uncommon and are usually unstable in solution. The first four group VIA elements all form anions with a double negative charge. Group VIIA elements all form anions with a single negative charge. 111 (p.65) 112 (p.62) 113 Prefix system Use this for nonmetal nonmetal compounds. number prefix 1 2 3 4 5 6 7 8 9 10 mon di tri tetra penta hexa hepta octa nona deca 114 Examples: N2O dinitrogen monoxide (also called dinitrogen oxide) 115 Examples: N2O dinitrogen monoxide (also called dinitrogen oxide) CO carbon monoxide 116 Examples: N2O dinitrogen monoxide (also called dinitrogen oxide) CO carbon monoxide Cl2O7 dichlorine heptaoxide 117 Examples: N2O dinitrogen monoxide (also called dinitrogen oxide) CO carbon monoxide Cl2O7 dichlorine heptaoxide P4O10 tetraphosphorous decaoxide 118 Examples: N2O dinitrogen monoxide (also called dinitrogen oxide) CO carbon monoxide Cl2O7 dichlorine heptaoxide P4O10 tetraphosphorous decaoxide PF5 phosphorous pentafluoride 119 Examples: N2O dinitrogen monoxide (also called dinitrogen oxide) CO carbon monoxide Cl2O7 dichlorine heptaoxide P4O10 tetraphosphorous decaoxide PF5 phosphorous pentafluoride N2 dinitrogen 120 Chemical Reactions and Chemical Equations – An introduction Chemical reaction: The transformation of one or more chemicals into different compounds. 121 Examples: 2 Na + 2 H2O (sodium) (water) 2 NaOH + H2 (sodium dihydrogen hydroxide) 122 Examples: 2 Na + 2 H2O (sodium) (water) 2 NaOH + H2 (sodium dihydrogen hydroxide) Cu + 4 HNO3 Cu(NO3)2 + 2 NO2 (copper) (nitric acid) (copper (II) nitrate) (nitrogen dioxide) + 2H2O 123 Examples: 2 Na + 2 H2O (sodium) (water) 2 NaOH + H2 (sodium dihydrogen hydroxide) Cu + 4 HNO3 Cu(NO3)2 + 2 NO2 (copper) (nitric acid) (copper (II) nitrate) (nitrogen dioxide) + 2H2O Zn + 2HCl(aq) ZnCl2 + H2 124 Reactants: The starting substances in a chemical reaction. E.g. Na and H2O in the first reaction. 125 Reactants: The starting substances in a chemical reaction. E.g. Na and H2O in the first reaction. Products: Substances as a result of a chemical reaction. E.g. Cu(NO3)2, NO2, and H2O in the second reaction. 126 Reactants: The starting substances in a chemical reaction. E.g. Na and H2O in the first reaction. Products: Substances as a result of a chemical reaction. E.g. Cu(NO3)2, NO2, and H2O in the second reaction. Balanced equation: The number of atoms of each element (free or part of a compound) is the same on both sides of the equation. 127 The three equations given are all balanced. An example of a non-balanced equation is: H2 + O2 H2O The commonly employed definition of balanced equation is the smallest set of integers that leads to a balanced equation. 128 So, H2 + ½ O2 H2O and 4 H2 + 2 O2 4 H2O Are not balanced equations. The balanced equation is: 2 H2 + O2 2 H2O 129 The coefficients tell us how many molecules of each species are reacting. 130 Measurements and Units Most physical quantities we encounter in chemistry have units. Old system: The cgs system – centimeter, gram, second were some of the key units in use – hence the abbreviation cgs 131 The SI system: 132 The SI system: Basic Quantity Name of Unit Symbol 133 The SI system: Basic Quantity Name of Unit Symbol length meter m 134 The SI system: Basic Quantity Name of Unit Symbol length meter m mass kilogram kg 135 The SI system: Basic Quantity Name of Unit Symbol length meter m mass kilogram kg time second s 136 The SI system: Basic Quantity Name of Unit Symbol length meter m mass kilogram kg time second s temperature kelvin K 137 The SI system: Basic Quantity Name of Unit Symbol length meter m mass kilogram kg time second s temperature kelvin K amount of substance mole mol 138 Derived Units: These are obtained from the basic units just given. Examples: The SI unit of volume is derived as follows: volume = length3 (think about the volume of a cube). Since the unit of length is m, then the unit of volume is m3 139 Derived Units: These are obtained from the basic units just given. Examples: The SI unit of volume is derived as follows: volume = length3 (think about the volume of a cube). Since the unit of length is m, then the unit of volume is m3 The SI unit of density: density mass volume Hence the units of density are kg m-3 140 There are some non-SI units that are in common use in chemistry. A common unit of volume is the liter. 1 liter = 1 cubic decimeter (1 dm3) = (10 cm)3 = 1000 cm3 and 1 liter = 1000 ml, that is 1 ml = 1 cm3 141 The basic units are not always convenient for reporting measurements. Various prefixes are often employed. Examples: 1 kg = 1000 g 1 km = 1000 m 1 nm = 0.000 000 001 m 1 cm = 0.01 m 142 The factor-dimensional method of calculation (also called the factor-label method) 143 The factor-dimensional method of calculation (also called the factor-label method) Suppose we want to convert 45.6 m into centimeters. From the conversion factor 100 cm = 1m 144 The factor-dimensional method of calculation (also called the factor-label method) Suppose we want to convert 45.6 m into centimeters. From the conversion factor 100 cm = 1m we can write 1 1m 100 cm 1 or 100 cm 1m These are unit conversion factors. 145 Multiplication of a quantity by a unit conversion factor will change its units, and adjust the value of the quantity for the new unit system used. For the present problem we have: Number of cm centimeters = (45.6m) 100 1m = 4.56 x 103 cm 146 If we take the wrong conversion factor, the result would be: Number of centimeters = (45.6m) 1 m 100 cm = 4.56 x 10-1 m2 cm-1 147 If we take the wrong conversion factor, the result would be: Number of centimeters = (45.6m) 1 m 100 cm = 4.56 x 10-1 m2 cm-1 In this case we can see that the units on the right-hand side do not match with what we expect on the left-hand side of the expression. 148 Second example: convert 45.6 m to inches. Conversion factors are 1 m = 100 cm and 1 inch = 2.54 cm (exactly) 149 Second example: convert 45.6 m to inches. Conversion factors are 1 m = 100 cm and 1 inch = 2.54 cm (exactly) Number of inches = (45.6m) 100 cm 1 inch 1m 2.54cm = 1.80 x 103 inches 150 Second example: convert 45.6 m to inches. Conversion factors are 1 m = 100 cm and 1 inch = 2.54 cm (exactly) Number of inches = (45.6m) 100 cm 1 inch 1m 2.54cm = 1.80 x 103 inches In this factor label calculation, two conversion factors are strung together. You can multiply any number of conversion factors in a string. 151 Read the section in the book on: Atomic view of matter (section 2.2) Dalton’s atomic theory (section 2.3) Nuclear atom model (section 2.4) 152 Atomic number, Mass number, and Isotopes Atomic Number: The number of protons in the nucleus of each atom of an element. The standard symbol for the atomic number is Z. 153 Atomic number, Mass number, and Isotopes Atomic Number: The number of protons in the nucleus of each atom of an element. The standard symbol for the atomic number is Z. For a neutral atom, the atomic number also indicates the number of electrons. 154 Example: The atomic number of C is 6. Thus, carbon has 6 protons and 6 electrons. 155 Mass number: The total number of protons and neutrons in the nucleus of each atom of an element. The symbol employed is A. 156 Mass number: The total number of protons and neutrons in the nucleus of each atom of an element. The symbol employed is A. The number of neutrons present in the nucleus =A - Z 157 Mass number: The total number of protons and neutrons in the nucleus of each atom of an element. The symbol employed is A. The number of neutrons present in the nucleus =A - Z For example, the mass number of fluorine is 19 and the atomic number is 9, so the number of neutrons present in the fluorine nucleus is 19 - 9 = 10. 158 Isotopes: Atoms having the same atomic number, but different mass numbers. 159 Isotopes: Atoms having the same atomic number, but different mass numbers. The common symbol for denoting the atomic number and mass number of element X is A ZX 160 Examples of isotopes: hydrogen has three common isotopes 1H 1 hydrogen 2H 1 deuterium 3H 1 tritium 161 Examples of isotopes: hydrogen has three common isotopes 1H 1 hydrogen 2H 1 deuterium 3H 1 tritium Hydrogen is the only element for which special symbols are used for different isotopes. D for deuterium T for tritium 162 Key Point: The chemical properties of an element are primarily determined by the number of protons and electrons, not by the number of neutrons present. For this reason, isotopes of the same element are chemically similar. 163 Atomic mass scale By international agreement, the reference adopted for an atomic mass scale is the assignment of exactly a mass of 12 for the most abundant isotope of carbon, 12C 6 164 Atomic mass scale By international agreement, the reference adopted for an atomic mass scale is the assignment of exactly a mass of 12 for the most abundant isotope of carbon, 12C 6 The atomic mass unit (amu) is defined by 12C atom mass of one 6 1 amu = 12 165 Most naturally occurring elements contain more than one isotope – this means that when the atomic mass of a naturally occurring element is determined, it is an average quantity that is determined. 166 Most naturally occurring elements contain more than one isotope – this means that when the atomic mass of a naturally occurring element is determined, it is an average quantity that is determined. Example: the atomic masses of 12C and 13C 6 6 are 12.0000000 … amu (exact number) and 13.00335 amu and their natural abundances (relative amounts of the isotopes present) are 98.89% and 1.11% respectively. 167 The weighted average atomic mass of carbon 98.89 x 12 1.11 x 13.00335 1 00 1 00 = 12.011 amu 168 Molar mass Chemists frequently use the term molecular weight or for an atomic system, atomic weight. 169 Molar mass Chemists frequently use the term molecular weight or for an atomic system, atomic weight. The molar mass is the mass of 1 mol of a substance expressed in units of g/mol. 170 The mass of a single atom of carbon -12 is determined from: 12C molar mass of 6 mass of one 126C atom Avogadro's number -1 12 g mol 6.02x1023 mol-1 = 1.99 x 10-23 g 171 Molecules The simplest molecules consist of only two atoms – these are called diatomic molecules. 172 Molecules The simplest molecules consist of only two atoms – these are called diatomic molecules. Examples: O2, H2, and Cl2 These are dioxygen, dihydrogen, and dichlorine. Unfortunately, it is also very common to refer to these as oxygen, hydrogen, and chlorine. 173 Molecules containing more than two atoms are called polyatomic molecules. 174 Molecular mass The molecular mass of a molecule is the sum of the atomic masses of the atoms present in the molecule. Example: The molecular mass of H2O is given by 2 x(1.00794) + 15.9994 = 18.0153 amu 175 The molar mass (frequently termed the molecular weight) of a compound is the molecular mass expressed in grams/mole (units abbreviated as g/mol). 176 The molar mass (frequently termed the molecular weight) of a compound is the molecular mass expressed in grams/mole (units abbreviated as g/mol). Example: The molecular mass of H2O is 18.0153 amu and hence the molar mass of water is 18.0153 g/mol. 177 Laws of Chemical Combination Law of Definite Proportions: In a given compound, the elements are always combined in the same proportion by mass. 178 Laws of Chemical Combination Law of Definite Proportions: In a given compound, the elements are always combined in the same proportion by mass. Example: water contains hydrogen and oxygen in the proportion of 2.01588 g to 15.9994 g. 179 Law of Multiple Proportions: If two elements combine to form more than one compound, the amounts of one element that combine with a fixed amount of the other element are in the ratio of small whole numbers. 180 Law of Multiple Proportions: If two elements combine to form more than one compound, the amounts of one element that combine with a fixed amount of the other element are in the ratio of small whole numbers. Example: Nitrogen and oxygen combine to form several different oxides, of which three are NO, N2O, and NO2. From experiment it is known that 16 g of oxygen will combine with 14 g, 28 g, and 7 g of nitrogen, respectively, for the three compounds. These masses are in the ratio 14: 28: 7 or 2 : 4 : 1. 181 Percentage Composition Determining a chemical formula of an unknown compound begins with finding the elements present. 182 Percentage Composition Determining a chemical formula of an unknown compound begins with finding the elements present. The next step is to determine the percentage composition – which allows the empirical formula to be determined. 183 Percentage Composition Determining a chemical formula of an unknown compound begins with finding the elements present. The next step is to determine the percentage composition – which allows the empirical formula to be determined. Percentage composition: The percent by mass of each element present in a compound. 184 Example: For H2O2 the molar mass is 34.0147 g/mol. There are 2.01588 g of H and 31.9988 g of O. % H 2.01588 g x 100 5.92650% 34.0147g g x 100 94.0734% % O 31.9988 34.0147g 185 Example: For H2O2 the molar mass is 34.0147 g/mol. There are 2.01588 g of H and 31.9988 g of O. % H 2.01588 g x 100 5.92650% 34.0147g g x 100 94.0734% % O 31.9988 34.0147g The same percentage results would be obtained if we started with the empirical formula HO. 186 Example: Calculating % composition from mass data. 187 Example: Calculating % composition from mass data. A sample of a liquid with a mass of 8.657 g was decomposed into its elements to give 5.217 g of carbon, 0.9620 g of hydrogen, and 2.478 g of oxygen. What is the % by mass of each element in the liquid? 188 Example: Calculating % composition from mass data. A sample of a liquid with a mass of 8.657 g was decomposed into its elements to give 5.217 g of carbon, 0.9620 g of hydrogen, and 2.478 g of oxygen. What is the % by mass of each element in the liquid? Answer: Do a percentage calculation. 5.217 g x 100 60.26% C 8.657g 189 Example: Calculating % composition from mass data. A sample of a liquid with a mass of 8.657 g was decomposed into its elements to give 5.217 g of carbon, 0.9620 g of hydrogen, and 2.478 g of oxygen. What is the % by mass of each element in the liquid? Answer: Do a percentage calculation. 2.478 g x 100 28.62% O 5.217 g x 100 60.26% C 8.657g 8.657g 190 Example: Calculating % composition from mass data. A sample of a liquid with a mass of 8.657 g was decomposed into its elements to give 5.217 g of carbon, 0.9620 g of hydrogen, and 2.478 g of oxygen. What is the % by mass of each element in the liquid? Answer: Do a percentage calculation. 2.478 g x 100 28.62% O 5.217 g x 100 60.26% C 8.657g 8.657g 0.9620 g x 100 11.11%H 8.657g 191 Empirical Formulas To determine the empirical formula using percentage composition data – find the number of moles of each element present. 192 Empirical Formulas To determine the empirical formula using percentage composition data – find the number of moles of each element present. Example 1: Ascorbic acid (vitamin C) is composed of 40.92% C, 4.58% H and 54.50% O. Determine the empirical formula. 193 Since we have % data, it is convenient to start with 100 g of the substance. Therefore, in 100 g of ascorbic acid, there are 40.92 g of C, 4.58 g of H, and 54.50 g of O. Now calculate the number of moles: 194 Since we have % data, it is convenient to start with 100 g of the substance. Therefore, in 100 g of ascorbic acid, there are 40.92 g of C, 4.58 g of H, and 54.50 g of O. Now calculate the number of moles: moles of 1 mol C 3.406mol C 40.92 g C carbon = 12.011gC 195 Since we have % data, it is convenient to start with 100 g of the substance. Therefore, in 100 g of ascorbic acid, there are 40.92 g of C, 4.58 g of H, and 54.50 g of O. Now calculate the number of moles: moles of 1 mol C 3.406mol C 40.92 g C carbon = 12.011gC moles of 1 mol H 4.54 mol H 4.58 g H hydrogen = 1.008gH 196 Since we have % data, it is convenient to start with 100 g of the substance. Therefore, in 100 g of ascorbic acid, there are 40.92 g of C, 4.58 g of H, and 54.50 g of O. Now calculate the number of moles: moles of 1 mol C 3.406mol C 40.92 g C carbon = 12.011gC moles of 1 mol H 4.54 mol H 4.58 g H hydrogen = 1.008gH moles of 1 mol O 3.406mol O 54.50 g O oxygen = 15.9994gO 197 Hence, the empirical formula is obtained as follows: C3.406H4.54O3.406 198 Hence, the empirical formula is obtained as follows: C3.406H4.54O3.406 Now divide by the smallest number, so that C 3.406H 4.54 O 3.406 3.406 3.406 3.406 199 Hence, the empirical formula is obtained as follows: C3.406H4.54O3.406 Now divide by the smallest number, so that C 3.406H 4.54 O 3.406 3.406 3.406 3.406 that is, C1.000H1.33O1.000 = C1H4/3O1 200 Hence, the empirical formula is obtained as follows: C3.406H4.54O3.406 Now divide by the smallest number, so that C 3.406H 4.54 O 3.406 3.406 3.406 3.406 that is, C1.000H1.33O1.000 = C1H4/3O1 Now eliminate the fraction (multiply by 3) to obtain the empirical formula as C3H4O3. 201 Example: A major air pollutant in coal burning countries is a colorless pungent smelling gas containing sulfur and oxygen. Chemical analysis of a 1.078 g sample of this gas showed that it contained 0.540 g of sulfur and 0.538 g of oxygen. What is the empirical formula of this gas? 202 Example: A major air pollutant in coal burning countries is a colorless pungent smelling gas containing sulfur and oxygen. Chemical analysis of a 1.078 g sample of this gas showed that it contained 0.540 g of sulfur and 0.538 g of oxygen. What is the empirical formula of this gas? It is not necessary in this problem to calculate the % composition – just calculate the number of moles of each element present. 203 moles of sulfur = 0.540gS 1 mol S 0.0168mol S 32.064gS moles of oxygen = 0.538gO 1 mol O 0.0336mol O 15.999gO 204 moles of sulfur = 0.540gS 1 mol S 0.0168mol S 32.064gS moles of oxygen = 0.538gO 1 mol O 0.0336mol O 15.999gO Hence, S0.0168O0.0336 205 moles of sulfur = 0.540gS 1 mol S 0.0168mol S 32.064gS moles of oxygen = 0.538gO 1 mol O 0.0336mol O 15.999gO Hence, S0.0168O0.0336 Now divide by the smallest number to obtain, SO2 which is the empirical formula. 206 Molecular Formulas A molecular formula gives the actual composition of a molecule. Example: Calculating a molecular formula from an empirical formula and molar mass information. 207 Molecular Formulas A molecular formula gives the actual composition of a molecule. Example: Calculating a molecular formula from an empirical formula and molar mass information. An oxide of nitrogen gave the following analysis: 1.52 g of nitrogen combined with 3.47 g of oxygen. The molar mass of the compound was found to be 92.0 g/mol. Determine the molecular formula. 208 Approach 1: Determine the empirical formula and then determine the molecular formula. 209 Approach 1: Determine the empirical formula and then determine the molecular formula. moles of oxygen = 3.47g O 1 mol O 0.217mol O 15.999g O moles of nitrogen = 1.52g N 1 mol N 0.109mol N 14.007g N 210 Approach 1: Determine the empirical formula and then determine the molecular formula. moles of oxygen = 3.47g O 1 mol O 0.217mol O 15.999g O moles of nitrogen = 1.52g N 1 mol N 0.109mol N 14.007g N Empirical formula is N0.109O0.217 NO2 211 Approach 1: Determine the empirical formula and then determine the molecular formula. moles of oxygen = 3.47g O 1 mol O 0.217mol O 15.999g O moles of nitrogen = 1.52g N 1 mol N 0.109mol N 14.007g N Empirical formula is N0.109O0.217 NO2 The mass of the empirical formula unit is 14.0067 + 2 x 15.9994 = 46.0055 g 212 Approach 1: Determine the empirical formula and then determine the molecular formula. moles of oxygen = 3.47g O 1 mol O 0.217mol O 15.999g O moles of nitrogen = 1.52g N 1 mol N 0.109mol N 14.007g N Empirical formula is N0.109O0.217 NO2 The mass of the empirical formula unit is 14.0067 + 2 x 15.9994 = 46.0055 g Since the mass of 1 mol is 92.0 g, the number of empirical formula units in the compound is 213 equal to 92.0 g 2 46.0055g 214 equal to 92.0 g 2 46.0055g Hence, the molecular formula is (NO2)2, that is N2O4 (dinitrogen tetraoxide). 215 equal to 92.0 g 2 46.0055g Hence, the molecular formula is (NO2)2, that is N2O4 (dinitrogen tetraoxide). Example: A compound of carbon, hydrogen, and oxygen has the following % composition (by mass): C 42.10 %, H 6.479 %, O 51.421 % The molar mass of the compound is 342.3 g/mol. What is the molecular formula? 216 equal to 92.0 g 2 46.0055g Hence, the molecular formula is (NO2)2, that is N2O4 (dinitrogen tetraoxide). Example: A compound of carbon, hydrogen, and oxygen has the following % composition (by mass): C 42.10 %, H 6.479 %, O 51.421 % The molar mass of the compound is 342.3 g/mol. What is the molecular formula? Approach 2: Work directly with the given molar mass and find the moles of each element present. 217 moles of C = 42.10x 342.3g C 1 mol C 12.00mol C 100 12.01g C moles of H = 6.479x 342.3g H 1 mol H 22.00mol H 100 1.0079g H moles of O = 51.421x 342.3g O 1 mol O 11.00mol O 100 15.999g O Hence, the molecular formula is C12H22O11 218 Exercise: A compound of carbon, hydrogen, and oxygen has the following composition: C 40.00% and H 6.714 %. The molar mass of the compound is 180.16 g/mol. What is the molecular formula? Answer is C6H12O6 219