Chemistry 101 - Book - Principles of Chemistry PDF

CHEM 101 - Book Principles of Chemistry: A Molecular Approach - Nivaldo Tro 1  CHEM 101- Syllabus Chapter 1: Matter, Measurement, and Problem Solving Chapter 2: Atoms and Elements Chapter 3: Molecules, Compounds & Chemical Equation...

CHEM 101 - Book Principles of Chemistry: A Molecular Approach - Nivaldo Tro 1  CHEM 101- Syllabus Chapter 1: Matter, Measurement, and Problem Solving Chapter 2: Atoms and Elements Chapter 3: Molecules, Compounds & Chemical Equation 2  Chapter 1 Matter, Measurement, and Problem Solving 1.1. Atoms and Molecules 1.2. The Scientific Approach to Knowledge 1.3. The Classification of Matter 1.4. Physical and Chemical Changes and Physical and Chemical properties 1.5. Energy: A Fundamental Part of Physical and Chemical Change 1.6. The Unit of Measurement 1.7. The Reliability of a Measurement 1.8. Solving Chemical Problems 3  Chemistry Matter  Anything that has mass and occupies space.  Properties of matter are determined by the properties of molecules and atoms. Chemistry (H2O) The science that helps to understand the behavior of matter by studying the behavior of atom and molecules. 4  Atoms and Molecules Atoms  Matter is composed of tiny particles called atoms.  Atom: smallest part of an element that is still that element. Molecules: Two or more atoms joined together by chemical bonds Chemical Bond  Scientific Method The scientific method is a systematic approach to research. A hypothesis is a tentative explanation for a set of observations. Theory (it is often called a model): Theory is a set of tested hypothesis that gives an overall explanation tested modified of some natural phenomenon. A theory is a possible explanation of why nature behaves in a particular way (it is an interpretation)  Observations  Two type of observation/experiments: 1. Qualitative: Describing how a process happens. 2. Quantitative: Measuring something about the process. Scientific Models: Law: Summarizes what happens. Hypothesis: A possible explanation for an observation. Theory (Model): (1) An attempt to explain why it happens, (2) Set of tested hypotheses that gives an overall explanation of some natural phenomenon. 7  Classification of Matter 3 states of Matter Solid Liquid Gas The particles The particles are The particles are packed closely packed, have complete close together but they have freedom from and are fixed in some ability to each other (No position. move around. fixed volume & (Fixed volume & (Fixed volume & shape) shape) indefinite Compressible 8  Classification of Solids Crystalline Solids: Amorphous Solids: Particles arranged in an Particles randomly orderly geometric pattern. distributed. - Salt, sugar, diamond - Plastic, glass, Charcoal Diamond and Charcoal are both composed of pure carbon!!! 9 Classify Matter by Composition? 10  Changes in Matter Chemical Physical Change Change Changes that alter the Changes that alter the state or appearance of composition of the the matter without matter. The atoms are altering the composition. rearranged into new  Example: Boiling of molecules water  Example: Burning CO2(g) gasoline + H2O(g) H2O (g) C3H8(g) H2O (l) 11  Common Physical Changes State changes  Evaporating / condensing  Melting / freezing  Subliming Gas Co ion nd Dissolving si t n t io e po nsa Bo ma De t io i l in bl i n g Su Melting Solid Liquid Freezing 12  Common Chemical Changes Processes that Rusting of metal release Example: Iron Iron lots of energy. oxide Example: Burning 13  Properties of Matter Physical Properties Chemical Properties Characteristics of Characteristics that matter that can be determine how the changed without composition of matter changing its changes. composition. Example:  Example: Flammability Color Acidity Smell Toxicity Attraction or repulsion Reactivity with to magnets. chemicals Density 14  Energy (A Fundamental Part of Physical and Chemical Change)  Energy is the capacity to do work (action of a force through a distance).  All matter possesses energy.  Physical and chemical changes usually come with an energy changes. Example:  Water evaporation (physical gy change): Absorb energy. E n er  Burn natural gas (chemical change): Release energy 15  Conversion of Energy  Energy can be converted from one form to another. Law of Conservation of Energy  Energy is neither created nor destroyed. (During conversion of energy from one form to another, the total amount of energy remains the same). 16  1.6 Units of Measurement  Measurement is a quantitative observation.  Measurement has a number and a unit (standard quantities used to specify measurements). Example: 25 g, 17 cm Most common unit systems  English system (USA) Units: inches, yards, pounds, etc.  Metric system (rest of the world) Units: centimeters, meters, kilogram, etc. 17  SI Units International system of Units (SI Units)  International System =Système International (SI)  The unit system used by scientists is called SI Units. Quantity SI Unit Symbol  It is based on metric system. Length Meter m Mass Kilogram kg Time Second s Temperature Kelvin K Amount of Mole Mol substance Electric current Ampere 18 A  Measure of Temperature Common Temperature scales: Fahrenheit scale, °F (USA) Celsius scale, °C (rest of the world) Kelvin scale, K  No negative temperature (absolute scale)  0 K = absolute zero 19  Converting between Temperature Scales Fahrenheit to Celsius: Celsius to Fahrenheit: °F = (°C x 1.8) + 32 Celsius to Kelvin: K = °C + 273.15 Kelvin to Celsius: °C = K – 273.15 Fahrenheit to Kelvin: °F °C K Kelvin to Fahrenheit: K °C °F 20 Convert 40 °C into K and °F Given: 40.00 °C Find: K and F Equation: K = °C + 273.15; °F = (°C X 1.8) + 32 Calculation: K = 40.00 + 273.15 °F = (40X 1.8) + 32 = 313.15 K = 72 + 32 = 104 °F 21  Prefix Multipliers in SI Units  Used to express very large or very small quantities in a compact manner (1000 meter = 1 kilometer) 22  Derived Units  Derived unit is a combination of other units. Formula SI Unit Speed distance/time m/s Volume Length x width x m3 height Density Mass/Volume kg/m3  Density of a substance is the ratio of its mass (m) to its volume (v). High Density  Common units for Density: Solid = g/cm3 Liquid = g/ml (note: 1 ml = 1 Low Density 23  Density  Density of a substance depends on its temperature (because heating an object generally causes it to expand).  Comparing density of a substance with its accepted value of density, we can determine the identity or purity. Osmium (Os) 22.59 Example: Densest naturally occurring element 24 Decide whether a ring with a mass of 3.15 g that displaces 0.233 cm3 of water is platinum Given: mass = 3.15 g; volume = 0.233 cm3 Find: density, g/cm3 Equation: Calculation: Compare with Density of platinum = 21.4 g/cm3; standard therefore, it is not platinum density: 25  1.7 Reliability of a Measurement In scientific measurements, Last digit: Estimated (uncertain) All other digits: Certain May be: Uncertainty is assumed to 5.212 be ±1 5.213 5.214 Written as: 5.213 ± 0.001 26  Measurement & Significant Figures  For instruments marked with a scale, you determine the last digit by estimating between the marks. 88 or 88.2 ? 88.2 ± 0.1 88 – Certain 0.2 – Estimated 27  Measurement & Significant Figures  The number of digits reported in a measurement depends on the measuring device. 28 Scientific Notation: 29 Scientific Notation: 30 Significant Figures 31 Significant Figures 32  Counting Significant Figures Significant Figures 0.00025 kg  The place-holding zeros are not Place-holding digits significant. 33  Counting Significant Figures 1) All nonzero digits are significant. 1.5 has 2 sig. figs. 2) Interior zeroes are significant. 1.05 has 3 sig. figs. 3) Leading zeroes are NOT significant. 0.001050 has 4 sig. figs. 1.050 x 10−3 4) Trailing zeroes may or may not be significant. a) Trailing zeroes after a decimal point are significant. 1.050 has 4 sig. figs. 34  Counting Significant Figures b) Trailing zeroes before a decimal point are significant if the decimal point is written. 150.0 has 4 sig. figs. c) Zeroes at the end of a number without a written decimal point are ambiguous and should be avoided by using scientific notation. 150 2 sig. fig.) 1.5 x 102 has 2 sig. figs. then 1.50 x 102 has 3 sig. figs. 5) Exact numbers have an unlimited number of significant figures (2 in following formula). Radius of circle = Diameter 35 /2  Counting Significant Figures How many significant figures are in each of the following? 0.04450 m 4 sig. figs.: the digits 4 and 5 and the trailing 0 5.0003 km 5 sig. figs.: the digits 5 and 3 and the interior 0’s 10 dm = 1 m Unlimited number of sig. figs.: exact number 1.000 × 105 s 4 sig. figs.: the digit 1 and the trailing 0’s 0.00002 mm 1 sig. fig.: the digit 2, not the leading 0’s 10000 m Generally assume 1 sig. fig. 36  Examples The radius of a carbon atom is approximately 0.000000000070 m. Express this number in scientific notation. 37 Rounding 38  Rounding Number Change to Answer 2.349865 2 Sig Fig 2.3 0.0234 2 Sig Fig 0.023 or 2.3 × 10−2 234 2 Sig Fig 230 or 2.3 × 102 39  Significant Figures in Calculation Multiplication or Division  Result has the same number of significant figures as the number with the fewest number of significant figures. Examples: 5.02 × 89.665 × 0.10 = 45.0118 = 45 3 sig. figs. 5 sig. figs. 2 sig. figs. 2 sig. figs. 5.892 ÷ 6.10 = 0.96590 = 0.966 4 sig. figs. 3 sig. figs. 40 3 sig. figs.  Significant Figures in Calculation Addition or Subtraction  Result has the same number of decimal places as the number with the fewest number of decimal places. Examples: 5.74 + 0.823 + 2.651 = 9.214 = 9.21 2 dec. pl. 3 dec. pl. 3 dec. pl. 2 dec. pl. 4.8 − 3.965 = 0.835 = 0.8 41  Significant Figures in Calculation Combined calculations (x or ÷ with + or -) 1. Do whatever is in parentheses first, and complete calculation. 2. Evaluate the significant figures or decimal places in the intermediate answer. 3. Then round-off the final answer (do not round intermediate steps). Example: 3.489 × (5.67 – 2.3) = 3.489 × 3.37 2 dp 1 dp 4 sf 2 sf = 11.75793 = 12 2 sf 42 Perform the following calculations and express the result to the correct number of significant figures. 43  1.8 Solving Chemical Problems Always write every number with its associated unit.  15 g , 2.7 ml Always include units in your calculations.  1 cm × 1 cm = 1 cm2  1 cm + 1 cm = 2 cm  1 cm – 1 cm = 0 cm  1 cm ÷ 1 cm = 1  Type of Chemical Problems: 1. Unit conversion problems 44  Converting from One Unit to Another  Using units as a guide  Example: 1 in=2.54 cm to problem solving is called dimensional analysis. Conversion factor Conversio n  A fractional quantity factor: with the unit we are ‘converting from’ on the bottom and the units we are Conversio n ‘converting to’ on the factor: top. 45 Convert 1.79 yd to centimeters (1 m = 1.094 yd, 0.01 m = 1 cm) Sort Given: 1.76 yd information Find: length, cm Strategize Conceptual yd m cm Plan: 1 m = 1.094 yd Relationships: 0.01 m = 1 cm Follow the Solution: conceptual plan to solve the problem Sig. figs. and Round: 160.8775 cm = 161 cm round Check Check: Units and magnitude make sense. 46 Convert 5.70 L to cubic inches Sort Information Given: 5.70 L Find: volume, in3 Strategize Conceptual L mL cm3 in3 Plan: 1 mL = 1 cm3, 1 mL = 10−3 L Relationships: 1 in = 2.54 cm Follow the Solution: conceptual plan to solve the problem. Sig. figs. and Round: 347.835 in3 = 348 in3 round Check Check: Units and magnitude make sense. 47  Density as a Conversion Factor  Density = Mass / Volume  Can use density as a conversion factor between mass and volume.  Density of Pb = 11.3 g/cm3  It means that: 11.3 g Pb = 1 cm3 Pb How much does 4.0 cm3 of lead weigh (Density=11.3g/cm3)? 48 What is the mass in k g of 173,231 L of jet fuel whose density is 0.768 g/mL? Sort Given: 173,231 L information density = 0.768 g/mL Find: mass, kg Strategize Conceptual L mL g kg Plan: 1 mL = 0.768 g, 1 mL = 10-3 L Relationships: 1 kg = 1000 g Follow the Solution: conceptual plan to solve the problem. Sig. figs. and Round: 1.33041 x 105 = 1.33 x 105 kg round Check Check: Units and magnitude make sense. 49 Homework – Chapter 1 See handout for selected Problem sets. Pages: 33 - 37 Exercises: 6, 8 20, 30, 40, 50, 56  Chapter 2 Atoms and Elements 2.4. The structure of the atom 2.5. Subatomic particles: protons, neutrons & electrons in atoms 2.6. Finding patterns: the periodic law and the periodic table 2.7. Atomic mass: the average mass of an element’s atoms 2.8. Molar mass: Counting atoms by weighing them Edited by: Dr. Ahmed Abu-Rayyan and Dr. Waad Alahmad,.  2.4 The Structure of Atom 1) The atom contains a tiny dense center called the nucleus with has the entire mass of the atom 2) The nucleus is positively charged 3) The electrons are dispersed in the empty space of the atom surrounding the nucleus. 4) Number of negatively charged electron are equal to number of positively charged particles (called protons).  So that the atom is electrically neutral. 52  2.5 Subatomic Particles: Protons, Neutrons, and Electrons in Atoms  Subatomic particles: 1. Protons 2. Neutrons 3. Electrons  Mass of atom expressed in: Atomic mass unit (amu)  1 amu = Mass of carbon atom/12 = 1.67 x 10-27 kg Symbol Mass Charge Location (amu) in atom Proton p or p+ 1.0072 +1 Nucleus 7 Neutron n or n0 1.0086 0 Nucleus 6 − 53  Elements Number of protons is the identity of an element. Atomic number (Z) = number of protons in nucleus Mass number (A) = number of protons + number of neutrons = atomic number (Z) + number of neutrons Mass Number A Atomic Number Z X Element Symbol Example Mass Number 23 Atomic Number 11 Na  Mass Number  Sum of the number of protons and neutrons is called mass number (symbol: A). Mass Number (A) = Number of Protons + Number of Neutrons Example: Nucleus of Carbon atom: 6 Protons and 6 Neutrons Mass Number (A) = 6 +6 = 12  Chemical Symbol  Chemical Symbol is a one or two-letter abbreviation of the name of the elements. Name of the Chemical element Symbol Oxygen O One capital letter Carbon C Calcium Ca Two letters: First letter is capital letter Chlorine Cl Second letter in lower case Sodium Na From the Latin name ‘Natrium’ Gold Au From the Latin name ‘Aurum’ 56  Periodic Table of Elements  Elements are arranged according to their atomic numbers.  Similar properties occur in the same column.  Each element is represented by its chemical symbol and atomic number. 57 57  Isotopes When the number of neutrons varies  Same element could have atoms with different masses, which are called isotopes.  All isotopes of an element have  same number of protons, same number of electrons, but different numbers of neutrons.  same chemical properties (undergo the same chemical reactions). Hydrogen Isotopes  Symbols for Isotopes  Isotopes are identified by their mass numbers.  Symbol of Isotopes:  Common symbol for Isotopes: Example: 21 20 10 Ne, Ne 10 or Ne-20, Ne-21 59  Neon isotopes Number of Percent Protons Number of A, Mass Natural Symbol (atomic number) Neutrons Number Abundance Ne-20 or 20 Ne 10 10 20 90.48% 10 21Ne Ne-21 or 10 10 11 21 0.27% 22 Ne Ne-22 or 10 10 12 22 9.25%  Number of protons = Atomic number  Number of neutrons = Mass number – Atomic number 60 How many protons, electrons, and neutrons are in an atom of ? Given: therefore, A = 52, Z = 24 Find: # p, # e, # n Formulas: Number of protons = Atomic number Number of electrons = Number of protons (in neutral atom) Number of neutron = Mass number – Atomic number Solution: Z = 24 # n= A – Z = 52 – 24 # e− = 24 = 28 Check: For most stable isotopes, n ≥ p 61  Number of p, e and n in Atoms Atomic Mass Atomic Protons Neutrons Electrons Number Number Symbol 6 7 6 6 13 42 54 42 42 96 13 14 13 13 27 55 78 55 55 133 62  2.6 The Periodic Law and the Periodic Table  Periodic Law: When the elements are arranged in order of increasing atomic number, certain sets of properties recur (repeat) periodically.  Elements with similar properties are arranged in the same column. 63  Modern Periodic Table  Elements are arranged in order of increasing atomic number.  3 Major types: Metals, Non-metals, Metalloids 64  Classification of Elements Metals: Good conductors of heat & electricity.  Lose electrons and form CATIONS during reactions.  Lower left and middle of the periodic table. Nonmetals: Poor conductors of heat & electricity.  Gain electrons to form ANIONS during reactions.  Upper right on the periodic table (except H). Metalloids: semiconductors  Show mixed properties. 65  Modern Periodic Table  Elements with similar chemical and physical properties are in the same column.  Columns are called Groups.  Rows are called Periods. Classification of Elements  Main Group Elements = “A” groups Properties are predictable based on the group #.  Transition Elements or Transition Metals = “B” groups 66 Group 1A: Alkali Metals, Group 2A: Alkaline Earth Metals  Properties of Elements  Group 1A: Alkali Metals  All are highly reactive.  Group 2A: Alkaline Earth Metals  All are fairly reactive (but not as reactive as Group 1A).  Group 8A: Noble gas  Non-metals in gas state at room temperature.  Mostly not reactive and chemically stable  Group 7A: Halogens  All are very reactive non-metals. 68  Ions and the Periodic Table An ion is an atom, or group of atoms, that has a net positive or negative charge. Cation – ion with a positive charge If a neutral atom loses one or more electrons it becomes a cation. Na 11 protons 11 protons 11 electrons Na+ 10 electrons Anion – ion with a negative charge If a neutral atom gains one or more electrons it becomes an anion. Cl 17 protons 17 protons 17 electrons Cl - 18 electrons 69  Number of p, e and n in Ions Atomic Electron Ion Ion Number Protons s Charge Symbol 2 16 16 18 2− S 2 12 12 10 2+ Mg 3 13 13 10 3+ Al 35 35 36 1− Br  Number of electrons (in ion)= (Number of Proton) - (Charge) 70 70  Ions and Compounds  Ions behave much differently than the neutral atom.  Sodium (Na) is highly reactive and quite unstable.  Sodium ions (Na+) very nonreactive and stable.  Since compounds like table salt (NaCl) are neutral, there must be equal amounts of charge from cations and anions in them. Compound: NaCl (charge: Neutral) Cation: Na+ (charge: +1) 71 71  Common Ions and the Periodic Table  Metals lose electrons to form cations (+ve ions).  For main group metals, Charge = Group number  Nonmetals gain electrons to form anions (-ve ions).  For nonmetals, Charge = group number − 8  2.7 Atomic Mass: The Average Mass of an Element’s Atoms  Isotopes of a given element have different masses.  Average atomic mass (also called Atomic mass) is mentioned in the periodic table.  It is calculated according to the natural abundance of each element. 73  Natural Abundance of Isotopes Percent Number of Number of A, Mass Natural Symbol Protons Neutrons Number Abundance Ne-20 or 20 Ne 10 10 20 90.48% 10 21Ne Ne-21 or 10 10 11 21 0.27% 22 Ne Ne-22 or 10 10 12 22 9.25% Average Atomic Mass of Ne = (20x90.48) + (21x0.27) + (22x9.25) 100 100 100 = 18.096 + 0.0567 + 2.035 = 20.1877 amu 74  2.8 Molar Mass Mole (mol)  Amount of substance containing 6.022 x 1023 particles (atoms or compounds)  like 1 dozen = 12 things 1 mole = 6.022 × 1023 things 6.022 x 1023 6.022 × 1023 = Avogadro’s Cu atoms number. 1 mole 75 Calculate the number of atoms in 2.45 mol of Cu Given: 2.45 mol Cu Find: atoms Cu Conceptual mol Cu atoms Cu Plan: Relationships: 1 mol = 6.022 × 1023 atoms Solution: 76 A silver ring contains 1.1 × 1022 silver atoms. How many moles of silver are in the ring? Given: 1.1 × 1022 atoms Ag Find: moles Ag Conceptual Plan: atoms Ag mol Ag Relationships: 1 mol = 6.022 × 1023 atoms Solution: 77  Mass and Amount (moles)  Mass in grams of 1 mole of atoms is called the molar mass. Example: Carbon (C)  Atomic mass = 12.011 amu (Mass of 1 atom)  Molar mass = 12.011 g/mol. (mass of 1 mole of atom)  1 mol C = 6.022 x 1023 C atoms = 12.011 g 1 amu = Mass of carbon atom/12 = 1.67 x 10-27 kg 78  Mole & Mass Relationships Element Number of Atoms Mass of Molar mass in 1 mole 1 mole H 6.022 x 1023 1.008 g 1.008 g/mol Na 6.022 x 1023 22.990 g 22.990 g/mol Cl 6.022 x 1023 35.45 g 35.45 g/mol 79 Calculate the moles of carbon in 0.0265 g of pencil lead Given: 0.0265 g C Find: mol C Conceptual gC mol C Plan: Relationships: 1 mol C = 12.01 g Solution: 80 How many copper atoms are in a coil weighing 3.10 g Given: 3.10 g Cu Find: atoms Cu Conceptual g Cu mol Cu atoms Cu Plan: Relationships: 1 mol Cu = 63.55 g, 1 mol = 6.022 × 1023 Solution: 81 Selected Exercises - Chapter 2 Pages: 68 - 71 Exercises: 26, 30, 34, 35 40, 45, 48 50c, 52, 56 82  Chapter 3 Molecules, Compounds, and Chemical Equations 3.2. Chemical Bonds 3. 3. Representing Compounds: Chemical Formulas and Molecular Models 3.4. An Atomic-Level View of Elements and Compounds 3.4. Ionic Compounds: Formulas and Names 3.5. Molecular Compounds: Formulas and Names 3.6. Formula Mass and the Mole Concept for Compounds 3.7. Composition of Compounds 3.8. Determining a Chemical Formula from Experimental Data 3.9. Writing and Balancing Chemical Equations 83  Elements and Compounds Elements Compounds Free atoms are rare in Elements combine with earth. each other to form Elements are substance compounds. that is made from only Example: one type of atom. H2O (water) Example: CH4 (methane) C (carbon) Properties of the H2 (hydrogen) compound are different O2 (oxygen) from the constituent elements. 84  3.2 Chemical Bonds Chemical bonds are forces that held the atoms together. Bonds are the result of interaction between the charged particles (electron and proton). Example: Hydrogen peroxide (H2O2) has 3 chemical bonds 85  Type of Bonds Covalent Bonds Ionic Bonds - Covalent bonds are - Ionic bonds are formed formed between two or between metal and non- more non-metals. metal. Example: Example: H-Cl Na-Cl - Covalent bond is formed - Ionic bond is formed by by sharing of electrons the transfer of electrons between two atoms. from Metal to Non-metal. 86  Ionic Bonds  When a metal reacts with a non-metal:  Metal lose electron to become cation (+ve charge)  Non-metals gain electron to become anion (-ve charge)  Opposite charged ions are attracted to form ionic bonds. 87  Covalent Bonds  When two non-metal reacts:  Electrons are shared between the two atoms  The shared electron interact with the nuclei of both atoms to form covalent bond.  Potential energy of bonded atoms is decreased due to the interaction between shared electron and nuclei. 88  3.3 Representing Compounds: Chemical Formulas and Molecular Models Representing Compounds Chemical Formulas Molecular Models Easiest way to represent Accurate and complete a compound (commonly way to represent a used) compound. 3 Types: 2 Types: 1. Empirical Formula 1. Ball and stick model 2. Molecular Formula 2. Space filling model 3.Structural Formula 89  Chemical Formulas  All chemical formulas indicate the elements present in the compound. 1. Molecular Formula  Give the actual number of atoms. 2. Empirical Formula  Give the relative number (ratio) of atoms. Empirical formula = Molecular formula Whole number 3. Structural Formula  Show how the atoms are connected together.  Use lines to represent a chemical bond. 90  Molecular Models Models show the 3D (three-dimensional) structure. 1. Ball and Stick models  Show the geometry of the molecule.  Represents atoms as balls.  Different atoms are in different color.  Chemical bonds are represented as sticks. 2. Space filling model  Give the relative size of atoms and how they are bonded together  Clearly represents the shape of the molecule 91 92  3.4 Elements and Compounds Single Multi-atom Made of Made of metal (cations) atoms molecules non-metals & non-metal (anion) 93  Molecular Elements Polyatomic elements: P4, S8, Se8 94 Classify the following as an atomic/molecular element or molecular/ionic compound. Aluminum, Al Atomic element Aluminum chloride, AlCl3 Ionic compound Chlorine, Cl2 Molecular element Acetone, C3H6O Molecular compound Carbon monoxide, CO Molecular compound Cobalt, Co Atomic element 95 3.5 Ionic Compounds : Formula and Names  In Ionic Compounds, cations and anions (+ve and –ve ions) interact to form a large lattice.  Charge of the ionic compound is neutral (Sum of +ve = Sum of –ve charge).  The formula unit reflects the smallest whole-number ratio of ions. Example: NaCl Na+ and Cl− ions in one to one ratio. 96 Formula Units for Ionic Compounds  In Ionic Compounds, cations and anions (+ve and –ve ions) interact to form a large lattice.  Charge of the ionic compound is neutral (Sum of +ve = Sum of –ve charge).  The formula unit reflects the smallest whole-number ratio of ions. Example: NaCl Na+ and Cl− ions in one to one ratio. 97  Charge of Metal Cations Cation name = Metal name Example: Na atom = Sodium Ca atom = Calcium Na+ ion = Sodium ion Ca2+ ion = Calcium ion Two types: Metals with invariant charge  Metals whose ions can have only one possible charge.  Example: Charge of sodium ion is always 1+ (Na+) Metal with variable charge  Metals whose ions can have more than one possible charge.  Example: Charge of iron is 2+ (Fe2+) in some compounds + 3+ Common Charges of Metal Cations 99  Nonmetal Anions: Anion name = Base name of non-metal + - ide Some common Anions: Non-metal Anion F Fluorine F− Fluoride Cl Chlorine Cl− Chloride Br Bromine Br− Bromide I Iodine I− Iodide O Oxygen O2− oxide S Sulfur S2− sulfide N Nitrogen N3− nitride P Phosphorous P4− phosphide 100  Naming Binary Ionic Compounds for Metals with Invariant Charge  Binary compounds contain only two different elements.  Ionic compounds contain metal cation and non-metal anion.  Metal listed first in formula and name. Naming ionic compounds (metal with fixed charge + non-metal) 101 Systemic name of Metals with Fixed Charge 1. Identify whether ionic or molecular compound K (Group 1A) = Metal Ionic compound Cl (Group 7A) = Non-metal 2. Name cation and anion. K (Potassium) = K+ = Potassium Cl (Chlorine) = Cl– = Chloride 4. Write the cation name first, then the anion name. Potassium chloride 102 Write the Systemic name of following 1.CaO Calcium oxide 2.MgBr2 Magnesium bromide 3.Al2S3 Aluminum sulfide 103  Naming Binary Ionic Compounds for Metals with Variable Charge  Some Metals (mostly transition metals) can have more than one charge state (example: Fe2+ or Fe3+).  Charge of the cation can be calculated form the charge of the anion. Charge of Fe in FeCl2 Charge of Au in Au2S3 Charge of Cl (group 7A) in Charge: FeCl2 = −1 2Au + 3S = 0 Charge of Cl2 in FeCl2 = −2 2Au + 3 (-2) = 0\ In ionic compounds, sum 2Au -6 = 0 of +ve and −ve charge is Charge of 2Au = +6 zero. Charge of Au = +6/2 Charge of Fe – 2 = 0 104 = +3  Naming Binary Ionic Compounds for Metals with Variable Charge  Name of cation is followed by a roman numeral indicating its charge. (I, II, III, IV, V in parentheses) Example: Fe2+ = Iron(II) 105  What is the name of CrBr3 ? 1. Identify whether ionic or molecular compound? Cr is transition metal then this compound is an ionic compound 2. Br is in group 7 then Br charge = -1, Charge of Cr = X 1 (X) + 3 (-1) = 0, X – 3 = 0, X=+3 3. Name cation and anion. Br (Bromine) then name of Br– is Bromide 3. Write the systemic name: Chromium(III) bromide 106 Determine Charge of the Metal and Name the Compound 1.CuO Copper(II) oxide Charge of Cu = X, Charge of O = -2 1(X) + 1(-2) = 0, X-2=0, X=+2 2.PbBr2 Lead(II) bromide Charge of Pb = X, Charge Br = -1 1(X) + 2(-1) = 0,X-2=0, X=+2 3.Fe2S3 Iron(III) sulfide Charge of Fe = X, Charge of S = -2 2(X) + 3(-2) = 0, 2X – 6 =0, 2X = 6, X = +3 107  Write Chemical Formula for Manganese(IV) sulfide 1. Type of compound: Binary ionic compound 2. Symbol for the cation: Mn4+ 3. Symbol for the anion: S2– 4. Charge (without sign) becomes subscript for other Mn4+ S2− ion. Mn2S4 5. Reduce subscripts to smallest whole number MnS2 ratio. 6. Check that the total charge Mn = (1) × (4+) = +4 of the cations cancels the total charge of the anions. S = (2) × (2–) = −4 108  Names of Polyatomic Ions Polyatomic ions are ions that contain more than one atom. Name and charge of polyatomic ion do not change. Example OH– : Hydroxide ion, NH4+ : Ammonium ion Ionic compound containing polyatomic ion: Polyatomic Cation Anion NH4Cl (NH4+, Cl–) Cation Polyatomic Anion NaOH (Na+, OH–) 109  Common Polyatomic Ions Name Formula Name Formula acetate C2H3O2– hypochlorite ClO– carbonate CO32– chlorite ClO2– hydrogen carbonate chlorate ClO3– (or bicarbonate) HCO 3 – perchlorate ClO4– hydroxide OH– sulfate SO42– nitrate NO3– sulfite SO32– nitrite NO2– hydrogen sulfate (or bisulfate) HSO4– chromate CrO42– dichromate Cr2O72– hydrogen sulfite (aka bisulfite) HSO3– ammonium NH4+ Name the Cation first and then the Anion. Example 1: Na2SO4 1. Identify the ions. Na = Na+ because in Group 1A SO4 = SO42− a polyatomic ion 2. Name the cation. Na+ = sodium (invariant charge) 3. Name the anion. SO42− = sulfate 4. Write the name of the cation followed by the name of the anion. Sodium sulfate 111 Example 2: Fe(NO3)3 1. Identify the ions. Fe = Cation (charge: Not known) NO3 = NO3− a polyatomic ion Fe = Fe3+ to balance the charge of the 3 NO3− 2. Name the cation. Fe3+ = iron(III), metal with variable charge 3. Name the anion. NO3− = nitrate 4. Write the name of the cation followed by the name of the anion. Iron(III) nitrate 112  Hydrates Hydrates are ionic compounds Prefix No. of containing a specific number of water Waters molecules in the formula unit. hemi ½ Example: mono 1 CoCl2∙6H2O di 2 Naming: tri 3 Name of ionic compound+hydrate tetra 4 penta 5 Example: hexa 6 CoCl2∙6H2O = Cobalt(II) chloride hexahydrate hepta 7 CaSO4∙½H2O = Calcium sulfate hemihydrate octa 8 113 Write the formula of a compound made from aluminum ions and oxide ions. 1. Check whether the atoms are Al: Metal, O: Nonmetal metals or non-metals. 2. Write the symbol for the Al+3 column 3A metal cation and its charge. 3. Write the symbol for the O2- column 6A nonmetal anion and its charge. 4. Charge (without sign) Al+3 O2- becomes subscript for other ion. Al2 O3 5. Reduce subscripts to smallest whole number ratio. 6. Check that the total charge of Al = (2) × (+3) = +6 the cations cancels the total O = (3) × (–2) = –6 charge of the anions. 114 3.6 Molecular Compounds: Formulas and Names 115  Molecular Compounds: Formulas and Names 1. Write the name of the 1st element in the formula. Superscript Prefix  Element furthest left and down on Periodic Table. 1 mono 2. Write the name of 2 element in the nd 2 di formula with -ide suffix.  Although these compounds do not contain ions! 3 tri 4 tetra 3. Use a prefix in front of each name to 5 penta indicate the number of atoms.  Never use the prefix mono- on first element. 6 hexa 7 hepta 8 Octa 9 nona 10 deca 116  Example—Naming Binary Molecular: BF3 1. Name the first element. boron 2. Name the second element with an -ide. fluorine  fluoride 3. Add a prefix to each name to indicate the subscript. monoboron, trifluoride 4. Write the first element with a prefix, then the second element with a prefix.  Drop the prefix mono from the first element. Boron trifluoride 117  Name the following NO2 Nitrogen dioxide PCl5 Phosphorus pentachloride I2F7 Diiodine heptafluoride 118  Acids: Formulas and Names 119  Acids Acids are molecular compounds that form H+ when dissolved in water. Formula generally starts with H.  e.g., HCl, H2SO4 HCl H2SO4 120  Naming Binary Acids Write a hydro prefix. Follow with the nonmetal name (change the end to –ic) Write the word acid at the end of the name. Example: HCl 1. Identify the anion. Cl = Cl−, chloride because Group 7A 2. Name the anion with an -ic suffix. Cl− = chloride  chloric 3. Add a hydro- prefix to the anion name. hydrochloric 4. Add the word acid to the end. Hydrochloric acid 121  Naming Oxyacids Oxyacids have H+ cation and polyatomic anion. If the polyatomic ion name ends in -ate, then change ending to -ic suffix. SO42− = sulfate (sulfur+oxygen)  sulfuric If the polyatomic ion name ends in -ite, then change ending to -ous suffix. SO32− = sulfite (sulfur+oxygen)  sulfurous Write word acid at the end of all names. 122  Naming Oxyacids - H2SO4 (aq) 1. Identify the anion. SO4 = SO42− = sulfate 2. If the anion has -ate suffix, change it to -ic. If the anion has -ite suffix, change it to -ous. SO42− = sulfate  sulfuric 3. Write the name of the anion followed by the word acid. Sulfuric acid 123  Naming Oxyacids - H2SO3 1. Identify the anion. SO3 = SO32− = sulfite 2. If the anion has -ate suffix, change it to -ic. If the anion has -ite suffix, change it to -ous. SO32− = sulfite  sulfurous 3. Write the name of the anion followed by the word acid. Sulfurous acid 124 Names and Formulas 125  3.7 Formula Mass and the Mole Concept for Compounds Formula Mass (molecular mass) The mass of an individual molecule or formula unit. Sum of the masses of the atoms in a single molecule or formula unit. Mass of 1 molecule of H2O = 2 (1.01 amu H) + 16.00 amu O = 18.02 amu Molar Mass The mass of 1 mole of molecule. Since 1 mole of H2O contains 2 moles of H and 1 mole of O. Molar Mass of 1 mole of H2O = 2 (1.01 g H) + 16.00 g O = 18.02 g/mol 126 How many oxygen atoms are present in 3.7 mol of PbO2? Given: 3.7 mol PbO2 Find: Number of ‘O’ Conceptual mol PbO2 mol O No. of O Plan: Relationships: 1 mol PbO2 has 2 mol ‘O’, 1 mol = 6.022x1023 Solution: Mole of oxygen = 2 x 3.7 mol = Number of oxygen = mol x 6.022 x 1023 = 127 How many moles are in 50 g of PbO2? Given: 50 g PbO2 Find: moles PbO2 Conceptual g PbO2 mol PbO2 Plan: Relationships: 1 mol PbO2 = 239.2 g Solution: Check: Since the given amount is less than 239.2 g, he moles being 46 g makes sense. 130  3.8 Composition of Compounds  Percent Composition: Percentage of each element in a compound by mass.  It can be determined from: 1. The formula of the compound. 2. The mass analysis of the compound. The percentages may not always total to 100% due to rounding. 131 Determine the mass percent composition of CaCl2 Given: CaCl2 Find: % composition of Ca and Cl in CaCl2 Conceptual Plan: Relationships: Molar mass (g/mol): Ca=40.08, Cl=35.45, CaCl2=110.98 Solution: Check: The total percentage is 99.99% 132 Find the mass of table salt NaCl containing 2.4 g of Na (molar mass: NaCl=58.44 g/mol, Na=22.99 g/mol). Given: 2.4 g Na, 39% Na Find: g NaCl Conceptual g Na g NaCl Plan: Relationships: 100. g NaCl : 39 g Na Solution: Check: Since the mass of NaCl is more than 2× the mass of Na, the number makes sense. 133  3.9 Determining a Chemical Formula from Experimental Data  Empirical Formula: Simplest, whole-number ratio of the atoms (of elements) in a compound.  It can be determined from elemental analysis.  masses of elements  percent composition. % composition Mass of elements Moles  Example of Empirical Forumla: CxHyOz 134 Moles (in whole  Finding an Empirical Formula 1) Convert the percentages to grams. a) assume you start with 100 g of the compound. b) skip if already grams. 2) Convert grams to moles. a) use molar mass of each element. 3) Write a pseudoformula using moles as subscripts. 4) Divide all by smallest number of moles. a) If result is within 0.1 of whole number, round to whole number. 5) Multiply all mole ratios by number to make all whole numbers. a) if ratio ?.5, multiply all by 2; if ratio ?.33 or ?.67, multiply all by 3; if ratio 0.25 or 0.75, multiply all by 4; etc. b) skip if already whole numbers. 135  Find the empirical formula of aspirin Laboratory analysis of aspirin determined the following mass percent composition. C = 60.00%, H = 4.48%, O = 35.53% Solution: In 100 g of aspirin there are 60.00 g C, 4.48 g H, and 35.53 g O. Conceptual Plan: whole ggCC molCC mol mole number pseudo- ratio empirical ratio empirical pseudo- ggHH molHH mol formula formula formula formula ggO O molO mol O 136  From Gram to Pseudo formula  Calculate the moles of each element: 1 mole C = 12.01 g, 1 mole H = 1.008 g, 1 mole O = 16.00 g  Write a pseudo formula: C4.996H4.44O2.220 137  Pseudo formula to Empirical formula  Find the mole ratio by dividing by the smallest number of moles:  Multiply subscripts by factor to give whole number: {C2.25H2O1} x 4 Empirical formula = C9H8O4 138  Molecular Formulas  The molecular formula is a multiple of the empirical formula.  It can be determined from:  empirical formula  molar mass of the compound (molecular formula) Molecular Formula = Empirical formula x n 139 Find the molecular formula of butanedione, its molar mass 86.09 g/mol and emp. formula C2H3O. Given: Empirical formula = C2H3O; MM = 86.09 g/mol Find: Molecular formula Conceptual Plan: and Relationships: Solution: Check: The molar mass of the calculated formula is in agreement with the given molar mass. 140  3.10 Writing and Balancing Chemical Equations Reactions involve chemical changes in matter resulting in new substances. Reactions involve rearrangement and exchange of atoms to produce new molecules.  Elements are not changed during a reaction. Reactants Products 141  Chemical Equations Shorthand way of describing a reaction. Provides information about the reaction:  formulas of reactants and products.  states of reactants and products.  relative numbers of reactant and product molecules that are required.  can be used to determine amount of reactants used and amount of products that is made. 142  Symbols Used in Equations Symbols used to indicate state after chemical  (g) = gas; (l) = liquid; (s) = solid  (aq) = aqueous = dissolved in water Energy symbols used above the arrow for decomposition reactions  D = heat  hn = light  shock = mechanical  elec = electrical 143  Combustion of Methane Methane gas burns with O2(g) to produce carbon dioxide gas and gaseous water. CH4(g) + O2(g) ® CO2(g) + H2O(g) O H H + + O C O O C H H H H O 1C+4H + 2O 1C+2O+2H+O 144  Combustion of Methane, Balanced The equation must be balanced to obey the Law of Conservation of Mass.  We adjust the numbers of molecules so there are equal numbers of atoms of each element on both sides of the arrow. CH4(g) + 2 O2(g) ® CO2(g) + 2 H2O(g) O O O O H H C + + C + H + H H H O O O O H H 1C + 4H + 4O 1C + 4H + 4O 145 Write a balanced equation for the combustion of butane, C4H10. Write a skeletal C4H10(l) + O2(g) CO2(g) + H2O(g) equation: Balance atoms in 4C1×4 complex substances C4H10(l) + O2(g) 4 CO2(g) + H2O(g) first: 10  H  2 × 5 C4H10(l) + O2(g) 4 CO2(g) + 5 H2O(g) Balance free elements 13/2 × 2  O  13 by adjusting coefficient C4H10(l) + 13/2 O2(g) 4 CO2(g) + 5 H2O(g) in front of free element: If fractional coefficients, {C4H10(l) + 13/2 O2(g) 4 CO2(g) + 5 H2O(g)} × 2 multiply thru by 2 C4H10(l) + 13 O2(g) 8 CO2(g) + 10 H2O(g) denominator: Check: 8  C  8; 20  H  20; 26  O  26 146 Homework – Chapter 3 Pages: 108 - 113 Exercises: 1, 7, 8 11, 12, 15 - 19 24, 35, 37, 39 41, 43, 59 147