CHM 211 Chapters 1-4
CHM 211 Chapters 1-4 CHM 211-114
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This 9 page Class Notes was uploaded by Alexia Combs on Friday October 2, 2015. The Class Notes belongs to CHM 211-114 at Marshall University taught by Laura R. McCunn-Jordan in Summer 2015. Since its upload, it has received 72 views. For similar materials see Principles of Chemistry I in Chemistry at Marshall University.
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I'm pretty sure these materials are like the Rosetta Stone of note taking. Thanks Alexia!!!
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Date Created: 10/02/15
Cha ter One Chemistry Study of matter and its transformations Matter 0 Descriptions 0 Measurements 0 Energy Matter Anything that has mass and volume The stuff of the universe books planets trees etc Composition Types and amounts of simpler substances that make up a sample of matter Properties Characteristics that give each substance a unique identity thsical Prooerties Can be observed while keeping the substance intact 0 ex Color mass shape volume boiling point Chemical Prooerties Can only be observed through the substance interacting with something else 0 ex Reactivity flammability ability to oxidize rust Extensive Prooerties Depend on how much of a substance you have 0 ex Mass volume lntensive Prooerties Do NOT depend on sample size 0 Ex Color density odor boiling point States of Matter 0 M Holds its shape Molecules particles packed closely together with limited motion 0 Will fill its container takes shape Pourable Particles close together but free to move 0 Completely fills container Particles are very spread out Have random motion Energy is the ability to do work 0 Potential Enerciv Energy due to the position of an object 0 Kinetic Energy Energy due to the movement of an object 0 Total Energy Potential Kinetic All measured quantities consist of a number and a unit Units are manipulated like numbers 0 ex 3ft 4ft 12ft2 o ex2 350mi7h 50mih or 50 mi h391 A conversion factor is a ratio of equivalent quantities used to express a quantity in different units The relationship 1 mi 5280 ft gives us the conversion factor 1 1mi5280 0 ex The height of Angel Falls is 3212 ft Express this quantity in miles if 1mi 5280ft I 3212ft 1mi5280ft 6083mi SI Base Units Mass Kilogram kg Length Meter m Time Second s 0 Temperature Kelvin K 0 Density 0 massvolume 0 At a given temperature and pressure the density of a substance is a characteristic physical property and has a specific value 0 Does depend on temperature 0 Temperature 0 Kelvin K Absolute temperature scale Can t go lower than 0 0 K is the absolute coldest o Celsius C Same increments as K scale but has negative temperatures I K 0 27315 0 Fahrenheit F WILL NOT USE 0 Significant Figures 0 Measurements include some uncertainty Rightmost digit is always estimated Recorded digits are called significant figures I RULES 0 1 Any number that is not zero 0 ex 35 2 sig figs 0 2 Zeroes in between sig figs are significant sandwiched 0 ex 305 3 sig figs 0 3 If at beginning NOT significant placeholders 0 ex 025 2 sig figs 0 4 If at the end ARE significant IF they are to the right of the decimal point 0 ex 0450 3 sig figs I CALCULATION RULES 0 1 For multiplication and division The answer contains same number of sig figs as there are in the measurement with the fewest sig figs 0 ex1 025444 111E11 o ex2 92cm 68cm 03744cm 23cm3 0 2 For addition and subtraction The answer has the same number of decimal places as there are in the measurement with the fewest places 0 ex 835mL 2328mL 10678 E 1068mL I Sid Fios in Calculations o ex1 32 109 35x 103 o ex2 00541 0000210 00539 c ex3 17 325 91 342 91 31 x104 0 Exact numbers have uncertainity associated with them 000 ex 1000mg 1g 60 min 1hr 254cm 1inch ex2 448 ft to inches I 448ft12in1ft 5376 E 538in Precision refers to how close the measurements in a series are to each other Accuracy refers to how close each measurement is to the actual value 0000 Cha ter 2 Element Type of matter that is completely uniform in its properties Consists of all one kind of atom Molecule Type of matter made of two or more atoms every molecule of the same identity should have same properties 0 ex H2 02 Compound More than one atom bound together not necessarily all the same elements 0 1 Covalent molecular o 2 lonic Mixture Physical intermingling of different types of matter atoms or molecules Law of Conservation of Mass 0 ex Calcium oxide carbon dioxide gt calcium carbonate I Ca0 CO2 gt CaCO3 I 5608g 4400g gt 10008g Law of Constant Composition Definite Proportions o In a chemical compound the relative amounts of the different elements remains constant I ex CaCO3 o 1 Ca 1 C 3 O I ex2 H20 is ALWAYS H20 Mass Percent of a Compound 0 ex Calcium carbonate I Analysis by Mass Mass Fraction Percent by Mass 80g Calcium 20 040 calcium 400 24g Carbon 012 carbon 12 96g Oxygen 048 oxygen 48 200g total 100 total 100 Law of Multiple Probortions o If elements A and B react to form two compounds the different masses of B that combine with a fixed mass ofA can be expressed as a ratio of small whole numbers I ex different oxides of carbon 0 carbon monoxide 571 oxygen and 429 carbon 0 carbon dioxide 727 oxygen and 273 carbon 0 Dalton s Atomic Theorv o Dalton postulated that I 1 Every element is made up of atoms the building blocks of matter I 2 Atoms of one element cannot be changed to another element I 3 In a given element all the atoms are identical I 4 Compounds can be made by combining different atoms 0 A compound will always have the same number and type of atoms 0 Atomic Structure 0 Protons Positive charge In nucleus 0 Neutrons Neutral ln nucleus 0 Electrons Negative charge Outside nucleus 0 Cathode Ray Tube Experiment 0 JJ Thomson 0 Led to masscharge ratio for electron 0 Oil Droo Experiment 0 Millikan 0 Similar to cathode ray experiment 0 Millikan s findings were used to calculate the mass on an electron 0 Gold Foil Experiment 0 Rutherford o or alpha particle is 2 protons and 2 neutrons 0 Atomic Number Mass Number and Atomic Symbol swamr a r set a r 7 sym w l r e simmer aquot I ex an atom of carbon 12 1260 I ex2 an atom of oxygen 1680 o Isotopes o Atoms of an element with the same number of protons but a different number of neutrons 0 Have same atomic number but a different mass number Atomic Mass Units amu o 1 amu 112 mass ofa carbon12 atom o 1 amu E 1 proton or1 neutron Molecular Masses 0 Sum of atomic masses 0 ex H20 molecule I molecular mass 2 atomic mass of H 1 atomic mass of O I 2 1008amu 1 1600amu I 1802amu lt need 4 sig figs or round to hundredths place 0 For ionic compounds we refer to a formula mass Ionic Compounds 0 Metal and nonmetal o Transferring electrons results in chaged species ions that are attached to each other 0 lons that have bigger charges attract more and the attraction increases as size decreases Covalent Compounds 0 Form when elements share electrons which usually occurs between nonmetals Molecule 0 Two or more atoms bonded together by atoms sharing electrons ln 0 Atom or molecule that has a positive or negative charge Diatomic Molecules 0 N2 02 H2 F2 Cl2 Br2 l2 Tetratomic Molecules 0 P4 Octatomic Molecules 0 S8 Se8 Chemical Formulas o Tells you type and numberratio of atoms I ex H20 0 2 H atoms 1 O atom I ex2 CH4 0 1 C atom 4 H atoms Naming Compounds 0 1 What kind of compound is it I covalent gt all nonmetals I ionic gtmetal nonmetal I acids gtcontain H bond to nonmetal o lonic Compounds I For all ionic compounds the name and formula lists the cation first and the anion second 0 ex NaCl 0 Sodium chloride I Naminci Cations 0 Name of cation is the same as the name of the metal element 0 ex CaBr2 I Calcium Bromide 0 Some metals can take on different charges Use a roman numeral to indicate the charge 0 ex FeCl2 I iron ll chloride 0 ex2 FeCl3 I iron lll chloride I Naminci Anions o Named by adding the suffix ide to the root of the nonmetal root 0 ex Cl39 I chloride 0 ex2N339 I nitride o ex3O239 I oxide Unless it s a polyatomic anion containing oxygen 0 exSO339 I sulfite o ex2SO4239 I sulfate I Namind Binarv Covalent Compounds 0 A binary covalent compound is typically formed by the combination of two nonmetals 0 ex 802 I sulfur dioxide 0 ex SF6 I su urhexa uonde 0 Some of these compounds are very common and have trivial names eg H20 water I Naming Acids 0 Always have hydrogen 0 ex HCI I hydrochloric acid also known as hydrogen chloride 0 ex2 HBr I hydrobromic acid 0 ex3 HF I hydrofluroic acid o Binary acid solutions form when certaain gaseous compounds dissolve in water 0 Oxoacid names are similar to those of the oxoanions except for two suffix changes 0 ate in the anion becomes ic in the acid 0 ite in the anion becomes ous in the acid 0 The oxoanion prefixes hypo and per are retained Thus BrO439 is perbromate and HBrO4 is perbromic acid IOZ39 is iodite and HIO2 is iodous acid Mixtures O O Heterogeneous Mixture Different parts of mixture are clearly distinguishable Homooeneous Mixture Looks completely uniform Chapter Three The Mol O O O 1 mol 602 x1023 M The number of carbon12 atoms in a twelvegram sample I 1 atom of 12C weighs 12 amu I 1 mol of 12C weighs 12 grams Molar Mass 0 The molar mass of a substance can be reported in gmol The periodic table tells us atomic weight in amu gramsmol I The molar mass of 02 2 1600gmol 3200gmol I The molar mass of SO2 3207 3200 6407 gmol o Avogadro s Number 602 x 1023 is often shortened to NA 0 I ex Convert 72g of H20 to moles o 72ng 1 mol1802g 040mol H20 I ex2 How many H20 molecules in 72g sample 0 040molHZO 602 x 1023molecules1 mol 24 x 1023 molecules Mass Percent Mass mass of element in compound mass of compound Determining ratio of atoms in a compound from mass of elements 000 1 Assume you have 100g of the compound 2 Use mass values to determines mass of each element 3 Convert each element s mass to moles 4 Divide by smallest mole value to find molar ratio I ex 0 104 C gt104g o 278 S gt 278g 617 CI gt 61 7g 0 617gCl 1mol3545g CI 174mol Cl 0866 201 0 27898 1mol3207g S 0867mol 80866 100 o 1049C 1mol12019 C 0866mol C 0866 100 I C1S1Cl2 CSCI2 lt empirical formula 0 Empirical and Molecular Formulas 0 Empirical Formula I Ratio of atoms in smallest whole numbers possible 0 Molecular Formula I Gives the actual number of atoms of each element in one molecule whole number multiple of empirical formula 0 Only applies to covalent compound 0 Determining a Molecular Formula from Combustion Analysis and Molar Mass 0 1 Combustion analysis determines amounts of C H and O in a compound 2 Can determine empirical formula 3 A separate experiment determines molecular weight of compound 4 Compare empirical formula mass to molecular mass to find molecular formula I ex 010059 menthol is burned and makess 028299 CO2 and 011599 H20 0 1 All of the C in CO2 came from original compound 0 Mass CO2 gt mols CO2 moles C o 2 All of the H in H20 came from original compound 0 Mass H20 gt mols H20 x2 moles H o 3 Subtract mass of C and H from original sample to find mass of 0 000 0 mass 0 gt moles O o 4 Find molar ratio of C H and O o 028299CO2 1molCO2 44019 0006428 mol C o 011599H20 1molH20 18029 0006432 mol H20 2mol1mol 001286 mol H 0006428mol C 12019 C 1mol C 0077209 C 001286mol H 1019 H mol H 0012999 H 01005 007720 001299 0010319 0 1molO 160090 00006444 mol 0 0 00006444 00006444 1 C 0006428 00006444 9975 E 10 H 0006432 0000644 1995 E 20 I C10H1oo o If the mol weight of menthol 1569mol what is molecular formula 0 CmHmO lt molecular formula I 10 1201 20 101 1600 1569mol 0 Writing and Balancing Chemical Equations 0 ex 2M9s 029 gt 2M90s o ex2 Aluminum metal reacts with hydrogen chloride gas to make aluminum chloride and hydrogen gas I 2Als 6HC9 gt 2AICI3s 3H2g O O O o Stoichiometric Calculations o The coefficients in a balanced chemical equation I represent the relative number of reactant and product particles I and the relative number of moles of each 0 ex 2Mgs 02g gt 2MgOs 0 Limiting Reactants o The limiting reactant will be completely used up in the reaction 0 The reactant that is not limiting is in excess some of this will be left over I ex 62mol Mg is allowed to react with 19mol 02 What is the limiting reactant and how much MgO is made 2Mgs 029 e 2MgOs o 62mol Mg 22 62 mol MgO o 19mol O2 2molMgO1mo02 38 mol MgO o 02 is the limiting reactant o 38 mol MgO 403044g1 mol 15315672 o 153gMgO lt theoretical yield 0 Reaction Yields 0 Yield actual yield theoretical yield x 100 0 Solution Stoichiometrv o Molarity moles solute moles of solution M mol L I ex How many moles of CESHQO6 are in 457mL of a 034M solution 0 M mol 06H1206 mL o 0034 mol 457 0 mol CESHQO6 0016 mol
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