THE CALCULATIONS OF GENERAL CHEMISTRY WITH DEFINITIONS, EXPLANATIONS, AND PROBLEMS
CONTENTS 1 ATOMIC STRUCTURE AND ISOTOPES 1 1 ATOMIC THEORY 1 2 THE STRUCTURE OF THE ATOM 1 3 ISOTOPES 1 4 RELATIVE ATOMIC MASS 1 5 PROBLEMS 2 FORMULA AND MOLECULAR MASS 2 1 FORMULA MASS 2 2 MOLECULAR MASS 2 3 MOLAR MASS 2 4 PROBLEMS 3 THE MOLE AND AVOGADRO’S NUMBER 3 1 THE MOLE AND AVOGADRO’S NUMBER (NA) 3 2 THE MOLE AND MOLAR MASS 3 3 CALCULATING THE NUMBER OF MOLES 3 4 PROBLEMS 4 FORMULAS OF COMPOUNDS AND PERCENT COMPOSITION 4 1 PERCENT COMPOSITION 4 2 TYPES OF CHEMICAL FORMULA 4 3 EMPIRICAL FORMULA FROM COMBUSTION ANALYSIS 4 4 MOLECULAR FORMULA 4 5 PROBLEMS 5 CHEMICAL FORMULAS AND NOMENCLATURE 5 1 GENERAL BACKGROUND 5 2 CHEMICAL FORMULA 5 3 OXIDATION NUMBERS 5 4 WRITING THE FORMULAS OF COMPOUNDS 5 5 NOMENCLATURE OF INORGANIC COMPOUNDS 5 6 PROBLEMS 6 CHEMICAL EQUATIONS 6 1 WRITING CHEMICAL EQUATIONS 6 2 BALANCING CHEMICAL EQUATIONS 6 3 TYPES OF CHEMICAL REACTIONS 6 4 PROBLEMS 7 STOICHIOMETRY 7 1 REACTION STOICHIOMETRY 7 2 INFORMATION FROM A BALANCED EQUATION 7 3 TYPES OF STOICHIOMETRIC PROBLEM 7 4 LIMITING REAGENTS 7 5 REACTION YIELDS: THEORETICAL, ACTUAL, AND PERCENT YIELDS 7 6 PROBLEMS 8 STRUCTURE OF THE ATOM 8 1 ELECTRONIC STRUCTURE OF THE ATOM 8 2 ELECTROMAGNETIC RADIATION 8 3 THE NATURE OF MATTER AND QUANTUM THEORY 8 4 THE HYDROGEN ATOM 8 5 THE QUANTUM-MECHANICAL DESCRIPTION OF THE HYDROGEN ATOM 8 6 QUANTUM MECHANICS AND ATOMIC ORBITALS 8 7 ELECTRONIC CONFIGURATION OF MULTIELECTRON ATOMS 8 8 PROBLEMS 9 CHEMICAL BONDING 1: BASIC CONCEPTS 9 1 INTRODUCTION: TYPES OF CHEMICAL BONDS 9 2 LEWIS DOT SYMBOLS 9 3 IONIC BONDING: FORMATION OF IONIC COMPOUNDS 9 4 COVALENT BONDING: LEWIS STRUCTURES FOR MOLECULES 9 5 COVALENT BONDING: WRITING LEWIS STRUCTURES 9 6 RESONANCE AND FORMAL CHARGE 9 7 EXCEPTIONS TO THE OCTET RULE 9 8 POLAR COVALENT BONDS: BOND POLARITY AND ELECTRONEGATIVITY 9 9 PROBLEMS 10 CHEMICAL BONDING 2: MODERN THEORIES OF CHEMICAL BONDING 10 1 VSPER THEORY: MOLECULAR GEOMETRY AND THE SHAPES OF MOLECULES 10 2 VSEPR THEORY: PREDICTING ELECTRON GROUP GEOMETRY AND MOLECULAR SHAPE WITH THE VSEPR MODEL 10 3 VSEPR THEORY: PREDICTING MOLECULAR SHAPE AND POLARITY 10 4 VALENCE BOND THEORY 10 5 VALENCE BOND THEORY: TYPES OF OVERLAP 10 6 HYBRIDIZATION 10 7 LIMITATIONS OF VALENCE BOND THEORY 10 8 MOLECULAR ORBITAL THEORY 10 9 PROBLEMS 11 GAS LAWS 11 1 STANDARD TEMPERATURE AND PRESSURE 11 2 BOYLE’S LAW: VOLUME VS PRESSURE 11 3 CHARLES’S LAW: VOLUME VS TEMPERATURE 11 4 THE COMBINED GAS LAW 11 5 GAY-LUSSAC’S LAW AND REACTIONS INVOLVING GASES 11 6 AVOGADRO’S LAW 11 7 THE IDEAL GAS LAW 11 8 DENSITY AND MOLECULAR MASS OF A GAS 11 9 MOLAR VOLUME OF AN IDEAL GAS 11 10 DALTON’S LAW OF PARTIAL PRESSURE 11 11 PARTIAL PRESSURE AND MOLE FRACTION 11 12 REAL GASES AND DEVIATION FROM THE GAS LAWS 11 13 GRAHAM’S LAW OF DIFFUSION 11 14 PROBLEMS 12 LIQUIDS AND SOLIDS 12 1 THE LIQUID STATE 12 2 VAPOR PRESSURE AND THE CLAUSIUS–CLAPEYRON EQUATION 12 3 THE SOLID STATE 12 4 THE CRYSTAL SYSTEM 12 5 CALCULATIONS INVOLVING UNIT CELL DIMENSIONS 12 6 IONIC CRYSTAL STRUCTURE 12 7 THE RADIUS RATIO RULE FOR IONIC COMPOUNDS 12 8 DETERMINATION OF CRYSTAL STRUCTURE BY X-RAY DIFFRACTION 12 9 PROBLEMS 13 SOLUTION CHEMISTRY 13 1 SOLUTION AND SOLUBILITY 13 2 CONCENTRATION OF SOLUTIONS 13 3 SOLVING SOLUBILITY PROBLEMS 13 4 EFFECT OF TEMPERATURE ON SOLUBILITY 13 5 SOLUBILITY CURVES 13 6 EFFECT OF PRESSURE ON SOLUBILITY 13 7 PROBLEMS 14 VOLUMETRIC ANALYSIS 14 1 INTRODUCTION 14 2 APPLICATIONS OF TITRATION 14 3 CALCULATIONS INVOLVING ACID-BASE TITRATION 14 4 BACK TITRATIONS 14 5 KJELDAHL NITROGEN DETERMINATION 14 6 PROBLEMS 15 IDEAL SOLUTIONS AND COLLIGATIVE PROPERTIES 15 1 COLLIGATIVE PROPERTIES 15 2 VAPOR PRESSURE AND RAOULT’S LAW 15 3 ELEVATION OF BOILING POINT 15 4 DEPRESSION OF FREEZING POINT 15 5 OSMOSIS AND OSMOTIC PRESSURE 15 6 PROBLEMS 16 CHEMICAL KINETICS 16 1 RATES OF REACTION 16 2 MEASUREMENT OF REACTION RATES 16 3 REACTION RATES AND STOICHIOMETRY 16 4 COLLISION THEORY OF REACTION RATES 16 5 RATE LAWS AND THE ORDER OF REACTIONS 16 6 EXPERIMENTAL DETERMINATION OF RATE LAW USING INITIAL RATES 16 7 THE INTEGRATED RATE EQUATION 16 8 HALF-LIFE OF A REACTION 16 9 REACTION RATES AND TEMPERATURE: THE ARRHENIUS EQUATION 16 10 PROBLEMS 17 CHEMICAL EQUILIBRIUM 17 1 REVERSIBLE AND IRREVERSIBLE REACTIONS 17 2 THE EQUILIBRIUM CONSTANT 17 3 THE REACTION QUOTIENT 17 4 PREDICTING THE DIRECTION OF REACTION 17 5 POSITION OF EQUILIBRIUM 17 6 HOMOGENEOUS VS HETEROGENEOUS EQUILIBRIA 17 7 CALCULATING EQUILIBRIUM CONSTANTS 17 8 CALCULATING EQUILIBRIUM CONCENTRATIONS FROM K 17 9 QUALITATIVE TREATMENT OF EQUILIBRIUM: LE CHATELIER’S PRINCIPLE 17 10 PROBLEMS 18 IONIC EQUILIBRIA AND PH 18 1 THE IONIZATION OF WATER 18 2 DEFINITION OF ACIDITY AND BASICITY 18 3 THE PH OF A SOLUTION 18 4 THE POH OF A SOLUTION 18 5 THE ACID IONIZATION CONSTANT, KA 18 6 CALCULATING PH AND EQUILIBRIUM CONCENTRATIONS IN SOLUTIONS OF WEAK ACIDS 18 7 PERCENT DISSOCIATION OF WEAK ACIDS 18 8 THE BASE DISSOCIATION CONSTANT, KB 18 9 RELATIONSHIP BETWEEN KA AND KB 18 10 SALT HYDROLYSIS: ACID–BASIS PROPERTIES OF SALTS 18 11 THE COMMON ION EFFECT 18 12 BUFFERS AND PH OF BUFFER SOLUTIONS 18 13 POLYPROTIC ACIDS AND BASES 18 14 MORE ACID–BASE TITRATION 18 15 PH TITRATION CURVES 18 16 PROBLEMS 19 SOLUBILITY AND COMPLEX-ION EQUILIBRIA 19 1 SOLUBILITY EQUILIBRIA 19 2 THE SOLUBILITY PRODUCT PRINCIPLE 19 3 DETERMINING KSP FROM MOLAR SOLUBILITY 19 4 CALCULATING MOLAR SOLUBILITY FROM KSP 19 5 KSP AND PRECIPITATION 19 6 COMPLEX-ION EQUILIBRIA 19 7 PROBLEMS 20 THERMOCHEMISTRY 20 1 INTRODUCTION 20 2 CALORIMETRY AND HEAT CAPACITY 20 3 ENTHALPY 20 4 HESS’S LAW OF HEAT SUMMATION 20 5 LATTICE ENERGY AND THE BORN–HABER CYCLE 20 6 BOND ENERGIES AND ENTHALPY 20 7 PROBLEMS 21 CHEMICAL THERMODYNAMICS 21 1 DEFINITION OF TERMS 21 2 THE FIRST LAW OF THERMODYNAMICS 21 3 EXPANSION WORK 21 4 ENTROPY 21 5 THE SECOND LAW OF THERMODYNAMICS 21 6 CALCULATION OF ENTROPY CHANGES IN CHEMICAL REACTIONS 21 7 FREE ENERGY 21 8 THE STANDARD FREE ENERGY CHANGE 21 9 ENTHALPY AND ENTROPY CHANGES DURING A PHASE CHANGE 21 10 FREE ENERGY AND THE EQUILIBRIUM CONSTANT 21 11 VARIATION OF G0 AND EQUILIBRIUM CONSTANT WITH TEMPERATURE 21 12 PROBLEMS 22 OXIDATION AND REDUCTION REACTIONS 22 1 INTRODUCTION 22 2 OXIDATION AND REDUCTION IN TERMS OF ELECTRON TRANSFER 22 3 OXIDATION NUMBERS (ON) 22 4 OXIDATION AND REDUCTION IN TERMS OF OXIDATION NUMBER 22 5 DISPROPORTIONATION REACTIONS 22 6 OXIDIZING AND REDUCING AGENTS 22 7 HALF-CELL REACTIONS 22 8 BALANCING REDOX EQUATIONS 22 9 OXIDATION-REDUCTION TITRATION 22 10 PROBLEMS 23 FUNDAMENTALS OF ELECTROCHEMISTRY 23 1 GALVANIC CELLS 23 2 THE CELL POTENTIAL 23 3 STANDARD ELECTRODE POTENTIAL 23 4 THE ELECTROCHEMICAL SERIES (ECS) 23 5 APPLICATIONS OF ELECTRODE POTENTIAL 23 6 CELL DIAGRAMS CELL FROM ELECTRODE POTENTIAL 23 8 RELATIONSHIP OF THE STANDARD ELECTRODE POTENTIAL, THE GIBBS FREE ENERGY, AND THE EQUILIBRIUM CONSTANT 23 9 DEPENDENCE OF CELL POTENTIAL ON CONCENTRATION (THE NERNST EQUATION) 23 10 ELECTROLYSIS 23 11 FARADAY’S LAWS OF ELECTROLYSIS 23 12 PROBLEMS 24 RADIOACTIVITY AND NUCLEAR REACTIONS 24 1 DEFINITIONS 24 2 RADIOACTIVE DECAY AND NUCLEAR EQUATIONS 24 3 NUCLEAR TRANSMUTATIONS 24 4 RATES OF RADIOACTIVE DECAY AND HALF-LIFE 24 5 ENERGY OF NUCLEAR REACTIONS 24 6 PROBLEMS ATOMIC STRUCTURE AND ISOTOPES 1 1 ATOMIC THEORY IN 1808, BASED ON EXPERIMENTAL DATA AND CHEMICAL LAWS KNOWN IN HIS DAY, DALTON PROPOSED HIS THEORY OF THE ATOM THE THEORY STATES THAT: 1 ALL CHEMICAL ELEMENTS ARE MADE UP OF TINY INDIVISIBLE PARTICLES CALLED ATOMS 2 ATOMS CANNOT BE CREATED OR DESTROYED CHEMICAL REACTIONS ONLY REARRANGE THE WAY THAT THE ATOMS ARE COMBINED 3 ATOMS OF THE SAME ELEMENT ARE IDENTICAL IN ALL RESPECTS AND HAVE THE SAME MASSES AND PHYSICAL AND CHEMICAL PROPERTIES ATOMS OF DIFFERENT ELEMENTS HAVE DIFFERENT MASSES AS WELL AS DIFFERENT PHYSICAL AND CHEMICAL PROPERTIES 4 A COMBINATION OF ELEMENTS TO FORM A COMPOUND OCCURS BETWEEN SMALL, WHOLE-NUMBER RATIOS OF ATOMS DALTON’S THEORY RESULTED IN THE FORMULATION OF THE LAW OF CONSERVATION OF MASS AND THE LAW OF MULTIPLE PROPORTIONS ALONG WITH THE LAW OF DEFINITE PROPORTION AND THE LAW OF RECIPROCAL PROPORTION, THESE FORM THE BASIS OF REACTION STOICHIOMETRY (SEE CHAPTER 7) 1 1 1 THE LAW OF CONSERVATION OF MASS IN A CHEMICAL REACTION, MATTER IS NEITHER CREATED NOR DESTROYED THE TOTAL MASS OF REACTANTS IS EQUAL TO THE TOTAL MASS OF THE PRODUCTS THIS LAW WAS FIRST STATED BY LAVOISIER, BASED ON HIS FINDINGS FROM COMBUSTION EXPERIMENTS 1 1 2 THE LAW OF DEFINITE PROPORTION THE LAW OF DEFINITE PROPORTIONS (ALSO KNOWN AS PROUST’S LAW OR THE LAW OF CONSTANT COMPOSITION) STATES THAT A GIVEN CHEMICAL COMPOUND IS ALWAYS COMPOSED OF THE SAME ELEMENTS IN THE SAME PROPORTIONS BY MASS WHEN ELEMENTS COMBINE TO FORM A PARTICULAR COMPOUND, THEY DO SO IN FIXED AND CONSTANT PROPORTIONS BY MASS, REGARDLESS OF THE SOURCE OF THE COMPOUND FOR EXAMPLE, A SAMPLE OF PURE WATER, REGARDLESS OF ITS SOURCE, ALWAYS CONTAINS 11 1% HYDROGEN AND 88 9% OXYGEN THE CALCULATIONS OF GENERAL CHEMISTRY WITH DEFINITIONS ATOMIC STRUCTURE AND ISOTOPES 1 2 2 MASS NUMBER (A) THE MASS NUMBER OF AN ELEMENT IS THE SUM OF PROTONS AND NEUTRONS IN THE NUCLEUS OF THE ATOM MASS NUMBER (A) = NO OF PROTONS (Z) + NO OF NEUTRONS (N) A = Z + N THE GENERAL SYMBOL FOR AN ELEMENT, SHOWING ITS MASS NUMBER AND ATOMIC NUMBER, IS: A ZE A = MASS NUMBER IN ATOMIC MASS UNITS (AMU) Z = ATOMIC NUMBER E = SYMBOL OF THE ELEMENT AS SHOWN ON THE PERIODIC TABLE FOR EXAMPLE, THE SYMBOLS FOR C AND MG, SHOWING THEIR MASS AND ATOMIC NUMBERS, ARE 6C AND DETERMINE THE NUMBER OF PROTONS, NEUTRONS, AND ELECTRONS IN (A) 11NA (B) 8O2? (C) 26FE3+ (D) 29CU SOLUTION THE SOLUTIONS ARE SUMMARIZED IN THE FOLLOWING TABLE SYMBOL PROTONS NEUTRONS ELECTRONS (A) 11NA (B) 8O2? (C) 26FE3+ (D) 29CU COMPLETE THE FOLLOWING TABLE PARTICLE ATOMIC MASS NUMBER OF NET CHARGE SYMBOL EN± NUMBER NUMBER PROTONS NEUTRONS ELECTRONS A MN B AL3+ C K G SOLUTION TO COMPLETE THIS TABLE, YOU NEED TO RECALL THE BASIC DEFINITIONS AND THE GENERAL EQUATION: MASS NUMBER (A) = NO OF PROTONS (Z) + NO OF NEUTRONS (N) USE THE PERIODIC TABLE TO IDENTIFY THE CORRESPONDING SYMBOLS ALSO REMEMBER THAT ATOMIC NUMBER IS EQUAL TO THE NUMBER OF PROTONS IN THE NUCLEUS OF AN ATOM IN A NEUTRAL ATOM (ZERO CHARGE), THE NUMBER OF ELECTRONS IS ALWAYS EQUAL TO THE NUMBER OF PROTONS THE DIFFERENCE BETWEEN THE NUMBER OF PROTONS AND THE NUMBER OF ELECTRONS IS THE CHARGE ON THE ATOMIC PARTICLE IF THE NUMBER OF PROTONS IS GREATER THAN THE NUMBER OF ELECTRONS, THE PARTICLE TAKES ON A NET POSITIVE CHARGE, AND VICE-VERSA THE COMPLETE TABLE IS PARTICLE ATOMIC MASS NUMBER OF NET CHARGE SYMBOL EN± NUMBER NUMBER PROTONS NEUTRONS ELECTRONS A BR? D P3? E PB2+ F F? H FE3+ 1 3 ISOTOPES ISOTOPES ARE ATOMS THAT HAVE THE SAME ATOMIC NUMBER, AND HENCE ARE THE SAME ELEMENT, BUT THAT HAVE DIFFERENT MASS NUMBERS (THAT IS, DIFFERENT NUMBERS OF NEUTRONS) THE DIFFERENCE IN MASS IS DUE TO THE VARIATION IN THE NEUTRON NUMBER ISOTOPES HAVE PRACTICALLY THE SAME CHEMICAL PROPERTIES BUT DIFFER SLIGHTLY IN PHYSICAL PROPERTIES MOST ELEMENTS EXHIBIT ISOTOPY FOR EXAMPLE, HYDROGEN HAS THREE ISOTOPES: HYDROGEN, DEUTERIUM, AND TRITIUM THE ATOMIC NUMBERS AND MASS NUMBERS ARE SHOWN IN TABLE 1 2 THE CALCULATIONS OF GENERAL CHEMISTRY WITH DEFINITIONS THE HYDROGEN ISOTOPES HYDROGEN (H) DEUTERIUM (D) TRITIUM (D) ATOMIC NUMBER MASS NUMBER WHICH OF THE FOLLOWING ARE ISOTOPES? A) 9 F AND 10NE B) 1H AND 1H C) 19K AND 20CA D) 12MG AND 12MG SOLUTION ISOTOPES ARE ATOMS WITH THE SAME ATOMIC NUMBER BUT DIFFERENT MASS NUMBER HERE ONLY (B) AND (D) ARE ISOTOPES 1 4 RELATIVE ATOMIC MASS THE RELATIVE MASS OF AN ATOM IS CALLED ATOMIC MASS (OLDER TEXTBOOKS CALL IT “ATOMIC WEIGHT”) AND IS CHARACTERISTIC OF EACH ELEMENT THE MASS OF AN INDIVIDUAL ATOM IS DIFFICULT TO MEASURE, BUT THE RELATIVE MASSES OF THE ATOMS OF DIFFERENT ELEMENTS CAN BE MEASURED, AND THIS DETERMINES THE ATOMIC MASS THE MASSES OF INDIVIDUAL ATOMS AND MOLECULES ARE EXCEEDINGLY SMALL AND CANNOT BE CON- VENIENTLY EXPRESSED IN GRAMS OR KILOGRAMS FOR EXAMPLE, ONE ATOM OF CA HAS A MASS OF 6 64×10?23G FOR THIS REASON, SCIENTISTS INSTEAD USE A RELATIVE ATOMIC MASS SCALE SCIENTISTS HAVE CHOSEN THE CARBON-12 ISOTOPE AS A REFERENCE, AND ITS MASS IS DEFINED AS EXACTLY TWELVE ATOMIC MASS UNITS (AMU) THEREFORE ONE ATOMIC MASS UNIT IS EQUAL TO 12 OF THE MASS OF THE CARBON-12 ATOM ONE AMU HAS BEEN DETERMINED EXPERIMENTALLY TO BE 1 66 × 10?24 G THE RELATIVE ATOMIC MASS OF AN ELEMENT IS DEFINED AS THE MASS OF AN ATOM OF THAT ELEMENT COMPARED WITH 12 OF THE MASS OF THE CARBON-12 ISOTOPE THE ATOMIC MASSES OF ELEMENTS GIVEN IN THE PERIODIC TABLE ARE NOT THE SAME AS MASS NUMBER IN NATURE, MOST ELEMENTS OCCUR AS A MIXTURE OF ISOTOPES THEREFORE, THE ATOMIC MASS OF AN ELEMENT IS AN AVERAGE MASS BASED ON THE ABUNDANCE OF THE ISOTOPES 1 4 1 CALCULATING ATOMIC MASSES THE INFORMATION NEEDED TO CALCULATE THE ATOMIC MASSES OF ELEMENTS INCLUDES: THE NUMBER OF ISOTOPES FOR THE ELEMENT, THE RELATIVE MASS OF EACH ISOTOPE ON THE CARBON-12 SCALE, AND THE PERCENT ABUNDANCE OF EACH ISOTOPE THE ATOMIC MASS OF A GIVEN ELEMENT IS THEN OBTAINED AS THE SUM OF THE PRODUCT OF THE EXACT MASS OF EACH ISOTOPE AND ITS PERCENT ABUNDANCE FOR EXAMPLE, NATURALLY OCCURRING CARBON CONSISTS OF THE CALCULATIONS OF GENERAL CHEMISTRY WITH DEFINITIONS X = 0 20 OR 20% AND 1 ?X = 0 80 OR 80% THE FRACTION AND % OF 10B IS 0 20 OR 20% ALSO, THE FRACTION OF 11B IS 0 80 OR 80% THE CALCULATIONS OF GENERAL CHEMISTRY WITH DEFINITIONS 1 5 PROBLEMS 1 HOW MANY PROTONS, NEUTRONS, AND ELECTRONS ARE IN EACH OF THE FOLLOWING? (A) 79AU (B) 5B (C) 20CA (D) 66DY 2 HOW MANY PROTONS, NEUTRONS, AND ELECTRONS ARE IN EACH OF THE FOLLOWING? (A) 36KR (B) 12MG (C) 31GA (D) 33AS 3 FIND THE NUMBER OF PROTONS, NEUTRONS, AND ELECTRONS IN THE FOLLOWING IONS: (A) 27CO2+ (B) 12MG2+ (C) 31GA3+ (D) 50SN2+ 4 IDENTIFY THE FOLLOWING ATOMS OR IONS (A) A HALOGEN (AN ELEMENT IN GROUP VII OF THE PERIODIC TABLE) ANION WITH THIRTY-SIX ELECTRONS AND THIRTY-FIVE PROTONS (B) AN ALKALI METAL (AN ELEMENT IN GROUP I OF THE PERIODIC TABLE) CATION WITH FIFTY-FIVE PROTONS AND FIFTY-FOUR ELECTRONS (C) A TRANSITION METAL CATION WITH TWENTY-FIVE PROTONS AND TWENTY-THREE ELECTRONS (NOTE: TRANSITION ELEMENTS ARE THOSE ELEMENTS EMBEDDED BETWEEN GROUPS II AND III IN THE PERIODIC TABLE) 5 COMPLETE THE FOLLOWING TABLE SYMBOL 20C A2+ 34S E2? 56B A2+ MASS NUMBER PROTONS NEUTRONS ELECTRONS 6 WITH THE AID OF A PERIODIC TABLE, IDENTIFY THE FOLLOWING ELEMENTS: (A) 26X; (B) 53 X; (C) 80 X; (D) 9 X 7 FILL IN THE BLANKS IN THE FOLLOWING TABLE: PARTICLE ATOMIC MASS NUMBER OF NET CHARGE SYMBOL NUMBER NUMBER PROTONS NEUTRONS ELECTRONS A P3? D ZR4+ THE CALCULATIONS OF GENERAL CHEMISTRY WITH DEFINITIONS CHEMISTRY IN QUANTITATIVE LANGUAGE 8 FIND THE NUMBER OF PROTONS, NEUTRONS, AND ELECTRONS IN THE FOLLOWING IONS: (A) 16S2? (B) 35BR? (C) 52TE2? (D) 7N3? 9 NATURALLY OCCURRING CHLORINE HAS TWO ISOTOPES, WHICH OCCUR IN THE FOLLOWING ABUNDANCE: 75 76% 17CL, WITH AN ISOTOPIC MASS OF 34 9689 AMU, AND 24 24% 17CL, WITH A ISOTOPIC MASS 36 9659 AMU CALCULATE THE ATOMIC MASS OF CHLORINE 10 BORON HAS TWO NATURALLY OCCURRING ISOTOPES: 80% OF 5 B, AND 20% OF 5B, WHICH HAS AN ISOTOPIC MASS OF 10 02 AMU IF THE ATOMIC MASS OF BORON IS 10 81, WHAT IS THE ISOTOPIC MASS OF 5 B? 11 NATURALLY OCCURRING OXYGEN GAS CONSIST OF THREE ISOTOPES USE THE INFORMATION IN THE TABLE TO CALCULATE THE ATOMIC MASS OF OXYGEN GAS ISOTOPE EXACT MASS ABUNDANCE 99 96% 0 037% 0 024% 12 NATURALLY OCCURRING GALLIUM (GA) EXISTS IN TWO ISOTOPIC FORMS, 69GA AND 71GA THE ATOMIC MASS OF GA IS 69 72 WHAT IS THE PERCENTAGE ABUNDANCE OF EACH ISOTOPE? (THE EXACT ISOTOPIC MASSES OF 69GA AND 71GA ARE 68 9259 AND 70 9249, RESPECTIVELY) THE CALCULATIONS OF GENERAL CHEMISTRY WITH DEFINITIONS 2 2 MOLECULAR MASS ONCE THE ACTUAL FORMULA OF A CHEMICAL SUBSTANCE IS KNOWN, THE MOLECULAR MASS CAN BE DETERMINED IN A MANNER SIMILAR TO THAT FOR CALCULATING THE FORMULA MASS THE MOLECULAR MASS OF A COMPOUND IS THE SUM OF THE ATOMIC MASSES OF ALL THE ATOMS IN ONE MOLECULE OF THE COMPOUND THE TERM APPLIES ONLY TO COMPOUNDS THAT EXIST AS MOLECULES SUCH AS H2O, SO2, AND GLUCOSE, C6H12O6 FOR EXAMPLE, THE MOLECULAR MASS OF ETHANOL, C2H5OH, IS (2 × C) + (6 × H) + (1 × O) = (2 × 12 0) + (6 × 1 0) + (1 × 16 0) = CALCULATE THE FORMULA MASS OF MGSO4 7H2O USING A TABLE OF ATOMIC MASSES SOLUTION 1 × AM OF MG = 24 3 AMU 1 × AM OF S = 32 1 AMU 4 × AM OF O = 64 0 AMU 7 × 2 × AM OF H = 14 0 AMU 7 × 1 × AM OF O = 112 0 AMU FM = 246 4 AMU THE FORMULA MASS OF MGSO4 7H2O IS 246 4 AMU 2 3 MOLAR MASS THE TERM MOLAR MASS IS NOW COMMONLY USED AS A GENERAL TERM FOR BOTH FORMULA MASS AND MOLECULAR MASS THE MOLAR MASS OF ANY SUBSTANCE IS THE MASS IN GRAMS OF ONE MOLE OF THE SUBSTANCE, AND IT IS NUMERICALLY EQUAL TO ITS FORMULA MASS (EXPRESSED IN AMU) NB: THE MOLE WILL BE DEFINED AND EXPLAINED IN CHAPTER 3 FOR EXAMPLE, THE FORMULA MASS OF GLUCOSE, C6H12O6, IS 180 0 AMU SO, THE MOLAR MASS OR THE MASS IN GRAMS OF 1 MOL OF GLUCOSE IS 180 0 G LET ME MENTION THAT, FOR MOST PURPOSES, YOU CAN ROUND ATOMIC AND MOLECULAR MASSES TO APPROXIMATELY ONE DECIMAL PLACE WHEN SOLVING PROBLEMS THE CALCULATIONS OF GENERAL CHEMISTRY WITH DEFINITIONS FORMULA AND MOLECULAR MASS CHEMISTRY IN QUANTITATIVE LANGUAGE 3 WHEN HYDROGEN SULFIDE GAS IS BUBBLED INTO AN AQUEOUS SOLUTION OF SBCL3, THE COMPOUND SB2S3 IS PRECIPITATED: 2 SBCL3 (AQ) + 3 H2S(G) ?SB2S3 (S) + 6 HCL(AQ) CALCULATE THE FORMULA MASS OF THE PRECIPITATED SB2S3 4 HYDROXYAPATITE, A CHEMICAL COMPOUND PRESENT IN TOOTH ENAMEL, HAS THE MOLECULAR FORMULA CA10(PO4)6(OH)2 CALCULATE ITS MOLAR MASS IN GRAMS 5 CALCULATE THE FORMULA MASS OF ASPIRIN, A MILD PAIN RELIEVER WITH THE FORMULA C9H8O4 6 PENICILLIN V IS AN IMPORTANT ANTIBIOTIC ITS CHEMICAL FORMULA IS C10H14N2 WHAT IS ITS FORMULA MASS? 7 THE CHEMICAL FORMULA OF TNT (TRINITROTOLUENE), AN INDUSTRIAL EXPLOSIVE, IS C7H5O6N3 CALCULATE ITS MOLECULAR MASS 8 THE MAJOR CHEMICAL COMPONENT OF EUCALYPTUS OIL, EXTRACTED FROM EUCALYPTUS LEAVES, IS EUCALYPTOL ITS CHEMICAL FORMULA IS C10H18O CALCULATE ITS FORMULA MASS 9 THE FORMULA MASS OF THE COMPOUND K2MCL6 IS 483 14 AMU WHAT IS THE ATOMIC MASS OF M? USE THE PERIODIC TABLE TO IDENTIFY M 10 THE FORMULA MASS OF THE INORGANIC ACID H4X2O7 IS 178 WHAT IS THE ATOMIC MASS OF X? USE THE PERIODIC TABLE TO IDENTIFY X THE CALCULATIONS OF GENERAL CHEMISTRY WITH DEFINITIONS THE MOLE AND AVOGADRO’S NUMBER 3 1 THE MOLE AND AVOGADRO’S NUMBER (NA) A MOLE IS DEFINED AS THE AMOUNT OF A GIVEN SUBSTANCE THAT CONTAINS THE SAME NUMBER OF ATOMS, MOLECULES, OR FORMULA UNITS AS THERE ARE ATOMS IN 12 G OF CARBON-12 FOR EXAMPLE, ONE MOLE OF GLUCOSE CONTAINS THE SAME NUMBER OF GLUCOSE MOLECULES AS THERE ARE CARBON ATOMS IN 12 G OF CARBON-12 THE NUMBER OF ATOMS IN EXACTLY 12 G OF CARBON-12 HAS BEEN DETERMINED TO BE 6 02 × 1023; THIS NUMBER IS CALLED AVOGADRO’S NUMBER (NA) THEREFORE, A MOLE IS THE AMOUNT OF A SUBSTANCE THAT CONTAINS AVOGADRO’S NUMBER OF ATOMS, IONS, MOLECULES, OR PARTICLES FOR EXAMPLE: 1 MOL HE ATOMS = 6 02 × 1023 ATOMS 1 MOL CH3OH MOLECULES = 6 02 × 1023 MOLECULES 1 MOL SO2? IONS = 6 02 × 1023 IONS 3 2 THE MOLE AND MOLAR MASS THE TERM MOLAR MASS IS NOW COMMONLY USED AS A GENERAL TERM FOR BOTH FORMULA MASS AND MOLECULAR MASS THE MOLAR MASS OF ANY SUBSTANCE IS THE MASS IN GRAMS OF ONE MOLE OF THE SUBSTANCE, AND IT IS NUMERICALLY EQUAL TO ITS FORMULA MASS (EXPRESSED IN AMU) FOR EXAMPLE, THE FORMULA MASS OF GLUCOSE, C6H12O6, IS 180 AMU SO, THE MOLAR MASS OR THE MASS IN GRAMS OF 1 MOL OF GLUCOSE IS 180 0 G 3 3 CALCULATING THE NUMBER OF MOLES IN TERMS OF CHEMICAL ARITHMETIC, THE MOLE IS THE MOST IMPORTANT NUMBER IN CHEMISTRY IT PROVIDES USEFUL STOICHIOMETRIC INFORMATION ABOUT REACTANTS AND PRODUCTS IN ANY GIVEN CHEMICAL REACTION THE QUANTITIES COMMONLY ENCOUNTERED IN CHEMICAL PROBLEMS INCLUDE THE NUMBER OF MOLES OF A SUBSTANCE; THE NUMBER OF ATOMS, MOLECULES, OR FORMULA UNITS OF A SUBSTANCE; AND THE MASS IN GRAMS THESE QUANTITIES ARE RELATED AND CAN BE READILY INTER-CONVERTED WITH THE AID OF THE MOLAR MASS AND AVOGADRO’S NUMBER CALCULATIONS BASED ON THE MOLE CAN BE CARRIED OUT BY USING CONVERSION FACTORS, OR WITH SIMPLE EQUATIONS BASED ON THOSE CONVERSION FACTORS C) NA2CO3 10 H2O SOLUTION FIRST DETERMINE THE MOLAR MASS OF EACH SUBSTANCE FROM THE INFORMATION GIVEN, DECIDE ON THE APPROPRIATE CONVERSION FACTOR OR FORMULA TO USE THIS IS A GRAMS-TO-MOLES CONVERSION THEREFORE, THE FOLLOWING CONVERSION FACTOR CAN BE USED: GRAMS TO MOLES: 1 MOLE OF THE SUBSTANCE 1 MOLAR MASS OF THE SUBSTANCE THE CALCULATIONS OF GENERAL CHEMISTRY WITH DEFINITIONS THE MOLE AND AVOGADRO’S NUMBER G AG × 1MOL AG FIND THE NUMBER OF MOLES IN EACH OF THE FOLLOWING SAMPLES: A) 0 027 G AL B) 1 ATOM NITROGEN C) 1 0 × 1015 HPO2? C) NO OF MOL HPO2? = NUMBER HPO2? IONS 6 02 × 1023 IONS/MOL = 1 0 × 1015 IONS 6 02 × 1023 IONS/MOL = 1 7 × 10?9 MOL HPO2? 8 ATOMS OF S WEIGH 8 × 32 = 4 ATOMS OF FE WEIGH 4 × 56 = 5 CALCULATE THE NUMBER OF MOLES PRESENT IN EACH OF THE FOLLOWING: (A) 5 0 × 1020 ATOMS OF MG (B) 2 5 × 1026 IONS OF FE3+ (C) 1 15 × 1023 MOLECULES OF H2O (D) 6 02 × 100 IONS OF SO2? 10 IRON (III) SULFATE (FE2(SO4)3) IONIZES ACCORDING TO THE FOLLOWING EQUATION: FE2(SO4)3 ?2 FE3+ + 3 SO2? CALCULATE THE NUMBER OF EACH ION PRODUCED FROM 2 00 MOLES OF FE2(SO4)3 11 A SUPERCONDUCTOR IS A MATERIAL THAT LOSES ALL RESISTANCE TO THE FLOW OF ELECTRICAL CURRENT BELOW A CHARACTERISTIC TEMPERATURE KNOWN AS THE SUPERCONDUCTING TRANSITION TEMPER- ATURE (TC) THE COPPER-CONTAINING OXIDE YBA2CU3O7, THE SO-CALLED 1-2-3 COMPOUND (1 YTTRIUM, 2 BARIUM, AND 3 COPPER), IS AN EXAMPLE OF A SUPERCONDUCTOR WITH A HIGH TC DETERMINE: (A) THE NUMBER OF MOLES OF Y, BA, AND CU IN 500 G OF THE OXIDE, AND (B) THE NUMBER OF ATOMS OF Y, BA, AND CU IN 500 G OF THE OXIDE 12 SILICON NITRIDE (SI3N4) IS A HIGH-TEMPERATURE CERAMIC THAT IS USED IN THE FABRICATION OF SOME ENGINE COMPONENTS HOW MANY ATOMS OF SI AND N ARE PRESENT IN 0 50 MOL OF SI3N4? THE CALCULATIONS OF GENERAL CHEMISTRY WITH DEFINITIONS FORMULAS OF COMPOUNDS AND PERCENT COMPOSITION 4 1 PERCENT COMPOSITION PERCENT MEANS ‘PARTS PER 100’ OR ‘PART DIVIDED BY WHOLE, MULTIPLIED BY 100’ IF THE FORMULA OF A COMPOUND IS KNOWN, ONE CAN EASILY CALCULATE THE MASS PERCENT OF EACH ELEMENT IN THE COMPOUND THE PERCENT COMPOSITION OF A COMPOUND IS THE MASS PERCENT OF EACH ELEMENT IN THE COMPOUND THE MOLAR MASS REPRESENTS 100% OF THE COMPOUND %ELEMENT = MASS OF ELEMENT FORMULA MASS × ASPIRIN, WHICH IS COMMONLY USED AS A PAIN RELIEVER, HAS THE CHEMICAL COMPOSITION C9H8O4 CALCULATE ITS PERCENT COMPOSITION SOLUTION FIRST, CALCULATE THE MOLAR MASS OF ASPIRIN 9 C = 9 × 12 0 = 108 0 G 8 H = 8 × 1 0 = 8 0 G 4 O = 4 × = 64 0 G MOLAR MASS = 180 0 G × 100 = CALCULATE THE PERCENTAGE OF WATER OF HYDRATION IN THE FOLLOWING HYDRATES: 1 CUSO4 5H2O 2 NA2CO3 10 H2O 3 ZN(NO3)2 6 H2O THE CALCULATIONS OF GENERAL CHEMISTRY WITH DEFINITIONS FORMULAS OF COMPOUNDS AND PERCENT COMPOSITION %H2O = 5 H2O CUSO4 5 H2O × 100 = 249 61 × 100 = 36 06% 2 NA2CO3 10 H2O = 2 × 23 + 12 + 3 × 16 + 10 (2 × 1 + 16) = %H2O = 10 H2O NA2CO3 10 H2O × 100 = %H2O = 6 H2O ZN(NO3)2 6 H2O × 100 = TITANIUM DISULFIDE IS A SEMICONDUCTOR WITH A LAYERED STRUCTURE THE LITHIATED FORM, LIXTIS2, IS USEFUL AS AN ELECTRODE IN RECHARGEABLE BATTERIES CALCULATE THE MASS PERCENT OF LITHIUM AND SULFUR PRESENT IN LIXTIS2 WHEN X = 0 25 SOLUTION FIRST, CALCULATE THE MOLAR MASS OF LI0 25TIS2 LI = 6 94 × 0 25 = 1 735 G TI = 47 9 × = 47 900 G S = 32 1 × THE BORON CONTENT OF A PYREX GLASS WAS REPORTED AS 17% B2O3 DETERMINE THE PERCENTAGE OF BORON IN THIS GLASS THE CALCULATIONS OF GENERAL CHEMISTRY WITH DEFINITIONS 0 75 × 100% = 85 3% 4 2 TYPES OF CHEMICAL FORMULA CHEMICAL FORMULAS ARE OFTEN DETERMINED FROM EXPERIMENTAL DATA GENERALLY, TWO TYPES OF FORMULA ARE POSSIBLE, DEPENDING ON THE AMOUNT OF EXPERIMENTAL INFORMATION AVAILABLE: THE EMPIRICAL FORMULA AND THE MOLECULAR FORMULA USUALLY, THE EMPIRICAL FORMULA IS DETERMINED FIRST, AND THE MOLECULAR FORMULA IS CALCULATED FROM THAT THE CALCULATIONS OF GENERAL CHEMISTRY WITH DEFINITIONS ANALYSIS OF THE PERCENTAGE COMPOSITION OF A NICOTINE PRODUCT YIELDED 74% C, 8 7% H, AND 17 3% N WHAT IS ITS EMPIRICAL FORMULA? SOLUTION FIRST ASSUME YOU HAVE 100 G OF THE SAMPLE, SO THAT THE GIVEN PERCENTAGE COMPOSITIONS WILL CORRESPOND TO MASSES IN GRAMS, I E 74 G C, 8 7 G H, AND 17 3 G N NOW, CONVERT THE RELATIVE MASSES OF EACH ELEMENT TO THE RELATIVE NUMBER OF MOLES OF EACH ELEMENT SINCE THIS IS A MASS-TO-MOLE CONVERSION, USE THE FOLLOWING CONVERSION FACTOR: GRAMS TO MOLE : 1 MOLE OF THE SUBSTANCE MOLAR MASS OF THE SUBSTANCE C : 74 G C × 1 MOLE C 12 0 G C = 6 17 MOL C H : 8 7 G H × 1 MOLE H 1 0 G H = 8 70 MOL H N : 17 3 G N × 1 MOLE N 14 0 G N = 1 24 MOL N SO, 100 G OF THE COMPOUND CONTAINS 6 17 MOL OF C, 8 70 MOL H, AND 1 24 MOL N CONVERT THESE TO WHOLE NUMBER RATIOS BY DIVIDING BY THE SMALLEST OF THE NUMBERS (I E 1 24) THE CALCULATIONS OF GENERAL CHEMISTRY WITH DEFINITIONS WE STILL DO NOT HAVE WHOLE NUMBER RATIOS THEREFORE, MULTIPLY BY THE LOWEST INTEGER THAT WILL CONVERT ALL TO WHOLE NUMBERS THE NUMBER 10 SEEMS TO DO THE TRICK C : 2 5 × 10 = H : 4 3 × 10 = N : 1 0 × 10 = MULTIPLY THE RATIOS BY 2 TO CLEAR THE DECIMALS THIS GIVES FE : O = 2:3 THE EMPIRICAL FORMULA OF THE OXIDE IS THUS FE2O3 4 3 EMPIRICAL FORMULA FROM COMBUSTION ANALYSIS COMBUSTION ANALYSIS IS ONE OF THE MOST COMMON METHODS USED TO DETERMINE THE EMPIRICAL FOR- MULA OF AN UNKNOWN COMPOUND CONTAINING CARBON AND HYDROGEN WHEN A COMPOUND CONTAINING CARBON AND HYDROGEN IS BURNED IN A COMBUSTION APPARATUS IN THE PRESENCE OF OXYGEN, ALL THE CARBON IS CONVERTED TO CO2, AND THE HYDROGEN TO H2O THE CO2 FORMED IS TRAPPED BY NAOH, AND THE WATER IS TRAPPED BY MAGNESIUM PERCHLORATE, MG(CLO4)2 THE MASS OF CO2 IS MEASURED AS THE INCREASE IN THE MASS OF THE CO2 TRAP, AND THE MASS OF H2O PRODUCED IS MEASURED AS THE INCREASE IN MASS OF THE WATER TRAP FROM THE MASSES OF CO2 AND H2O, THE MASSES AND MOLES OF CARBON AND HYDROGEN PRESENT IN THE COMPOUND CAN BE DETERMINED (IF NITROGEN IS PRESENT IN THE COMPOUND, IT MUST BE DETERMINED BY THE KJELDAHL METHOD IN A SEPARATE ANALYSIS, AS NH3 ) OXYGEN IN THE COMPOUND CAN THEN BE DETERMINED BY DIFFERENCE—THAT IS, THE PERCENTAGE REMAINING WHEN C, H, AND N (IF PRESENT) HAVE BEEN FOUND THE CALCULATIONS OF GENERAL CHEMISTRY WITH DEFINITIONS 100 × 38 7 MG 1 G N 1000 MG N = 0 0015 G OR 1 5 MG N NEXT, CALCULATE THE MASS OF O AS THE DIFFERENCE BETWEEN THE MASS OF THE SAMPLE AND THE MASSES OF C, H, AND N MASS OF O = 38 7 MG–26 7 MG–2 3 MG–1 5 MG = 8 2 MG O NOW CONVERT THE MASS OF EACH ELEMENT TO THE CORRESPONDING MOLES C : 0 0267 G C × 1 MOLE C 12 0 G C = 0 0022 MOL C H : 0 0023 G H × 1 MOLE H 1 01 G H = 0 0023 MOL H N : 0 0015 G N × 1 MOLE N 14 0 G N = 0 00011 MOL N O : 0 0082 G O × 1 MOLE O 16 0 G O = 0 00051 MOL O DIVIDE EACH MOLAR VALUE BY THE SMALLEST OF THEM THE CALCULATIONS OF GENERAL CHEMISTRY WITH DEFINITIONS C = 0 0022 MOL 0 00011 MOL = H = 0 0023 MOL 0 00011 MOL = N = 0 00011 MOL 0 00011 MOL = O = 0 00055 MOL 0 00011 MOL = THE EMPIRICAL FORMULA OF RESORCINOL, A COMMON CHEMICAL USED IN THE MANUFACTURE OF PLASTICS, DRUGS, AND PAPER PRODUCTS, IS C3H3O THE MOLECULAR MASS IS 110 0 WHAT IS THE MOLECULAR FORMULA? SOLUTION WE ARE GIVEN THE EMPIRICAL FORMULA AND MOLECULAR MASS OF THE COMPOUND THE UNKNOWN IS MOLECULAR FORMULA THE CALCULATIONS OF GENERAL CHEMISTRY WITH DEFINITIONS CHEMISTRY IN QUANTITATIVE LANGUAGE BEGIN BY CALCULATING THE FORMULA MASS OF C3H3O 3 × C + 3 × H + 1 × O = 3 × 12 + 1 × 3 + 1 × 16 = 55 AMU NEXT, CALCULATE THE MULTIPLICATION FACTOR, N, FROM THE EXPRESSION: MOLECULAR WEIGHT = N (EMPIRICAL FORMULA WEIGHT) OR N = MOLECULAR WEIGHT EMPIRICAL FORMULA WEIGHT N = 55 = EPINEPHRINE IS AN ADRENALINE HORMONE SECRETED INTO THE BLOODSTREAM IN TIMES OF DANGER OR STRESS ANALYSIS INDICATES IT HAS THE FOLLOWING COMPOSITION BY MASS: 59% C, 7 1% H, 26 2% O, AND 7 7% N THE MOLECULAR MASS IS 183 0 DETERMINE THE EMPIRICAL AND MOLECULAR FORMULA OF EPINEPHRINE SOLUTION BEGIN BY ASSUMING YOU HAVE 100 G OF SAMPLE WHICH CONTAINS 59 G C, 7 1 G H, 26 2 G O, AND 7 7 G N NEXT CONVERT EACH MASS OF AN ELEMENT INTO THE NUMBER OF MOLES OF THAT ELEMENT: C : 59 G C × 1 MOLE C 12 0 G C = 4 92 MOL C THE CALCULATIONS OF GENERAL CHEMISTRY WITH DEFINITIONS 4 CALCULATE THE PERCENT BY MASS OF MAGNESIUM, CHLORINE, HYDROGEN, AND OXYGEN IN HYDRATED MAGNESIUM CHLORIDE, MGCL2 6H2O 5 FIND THE MASS OF WATER OF CRYSTALLIZATION PRESENT IN 10 85 G OF HYDRATED SODIUM CARBONATE, NA2CO3 10H2O 6 CAFFEINE, A STIMULANT IN COFFEE AND TEA, HAS THE COMPOSITION 49 5% C, 5 19% H, 28 9% N, AND 16 5% O IT HAS THE MOLECULAR FORMULA C8H10N4O2 HOW MANY GRAMS OF CARBON AND NITROGEN ARE PRESENT IN 77 25 G OF CAFFEINE? 7 LARGE QUANTITIES OF HEMATITE, AN IMPORTANT ORE OF IRON, ARE FOUND IN AUSTRALIA, UKRAINE, AND USA A 100 G SAMPLE OF A CRUDE ORE CONTAINS 46 75 G OF FE2O3 WHAT IS THE PERCENTAGE OF IRON (FE) IN THE ORE? (ASSUME NO OTHER FE COMPOUND ) 8 THE ELEMENT LITHIUM IS COMMONLY OBTAINED FROM THE ORE SPODUMENE, LIALSI2O6, WHICH IS FOUND IN BRAZIL, CANADA, AND THE USA WHAT MASS OF LITHIUM IS PRESENT IN 17 85 G LIALSI2O6? THE CALCULATIONS OF GENERAL CHEMISTRY WITH DEFINITIONS 12 A COMPOUND THAT HAS THE EMPIRICAL FORMULA P2O5 IS FOUND TO HAVE A MOLECULAR MASS OF 283 9 WHAT