Chemistry 211 Midterm 1 Study Guide
Chemistry 211 Midterm 1 Study Guide CHEM 211-003
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This 11 page Study Guide was uploaded by Lucas Kinsey on Sunday September 25, 2016. The Study Guide belongs to CHEM 211-003 at George Mason University taught by Pritha G. Roy in Summer 2016. Since its upload, it has received 64 views. For similar materials see General Chemistry 1 in Chemistry at George Mason University.
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Date Created: 09/25/16
Lucas Kinsey Chemistry 211 Chemistry 211 Study Guide Chapters 1-3 Matter: anything that has mass and volume Composition: the types and amounts of simpler substances that make it up Properties: characteristics that give each substance its unique identity Chemical reaction: when one substance is changed into another substance Physical properties: characteristics that show without interaction with other particles - Can change back between two states of matter Chemical properties: characteristics that show during interaction with other substances - Cant change back and forth between different states Extensive Properties: A property that changes when the size of the sample changes (ex: mass, volume, length, charge, ect..) Intensive properties: A property that doesn’t change when you take away the sample (ex: temperature, hardness, color, melting point, boiling point, ect..) Derived Unit: An SI unit of measurement comprised of a combination of the seven base units (ex: Force = ma, Density = m/v, ect…) Energy: the ability to do work - Potential energy + Kenetic energy = Total Energy Scientific Approach: Observation Hypothesis Hypothesis is revised if experiment results do not support it ExTheorynt Further Experiments Model is altered if predicted events do not support it Unit Conversion Factors in Calculations - Conversion Factors: ratios used to express a quantity in different units o Uses equivalent quantities Ex:) 1 Mile = 5280 Feet - 150 Miles = 79200 Feet - The conversion factor you use must cancel all units except those you want to answer - To get from kg to L: kg g L o 1 × kg× g = L Sig Figs - The more Significant Figures there are, the more certainty there is in the measurement. Every measurement has some degree of uncertainty - For Multiplication and division: o Answer contains the same number of sig figs as there are in the measurement with the fewest sig figs 1,400 x 223 = 312,200 310,000 - For Addition and Subtraction: o The answer has the same number of decimal places as there are in the measurement with the fewest decimal places 13.13 + 0.0093 = 13.14 - Precision: Reproducibility, the amount of times a measurement can be consistently obtained - Accuracy: How close the measurement is to the actual value - Systematic error produces values that are either all higher or all lower than the actual value. (This issue can easily be solved by calibrating measuring equipment) - Random error produces values that are higher and lower than the actual value sporadically throughout the data (This issue is usually due to human error) PREFIX SYMBOL WORD CONVENTIONA SCIENTIFIC L NOTATION NOTATION TERA T Trillion 1,000,000,000,0 1x10^12 00 GIGA G Billion 1,000,000,000 1x10^9 MEGA M Million 1,000,000 1x10^6 KILO K Thousand 1,000 1x10^3 HECTOR h Hundred 100 1x10^2 DEKA da Ten 10 1x10^1 - - One 1 1x10^0 DECI d Tenth 0.1 1x10^-1 CENTI c Hundredth 0.01 1x10^-2 MILLI m Thousandth 1x10^-3 MICRO Millionth 0.001 1x10^-6 NANO n Billionth 0.000001 1x10^-9 PICO Trillionth 0.000000001 1x10^-12 FEMTO f Quadrilliont0.00000000000 1x10^-15 0001 PHYSICAL QUANTITY UNIT NAME UNIT ABBREVIATION MASS Kilogram kg LENGTH Meter M TIME Second S TEMPERATURE Kelvin K ELECTRICAL Ampere A CURRENT AMOUNT OF Mole mol SUBSTANCE LUMINOUS Candela cd INTENSITY Law of Mass Conservation: - The total mass of a substance does not change during a chemical reaction - Mainly applies to chemical reactions - The number of substances is subject to change but the total amount never changes Law of Definite Composition: No matter what its source, a particular compound is composed of the same elements in the same parts (fractions) by mass MAssof Element x∈Compound A Mass Fraction = Massof Compound A Mass Percent = Mass Fraction x 100 Mass of Element = Mass of Compound x Mass Fraction Mass of Element in Sample = Mass of Compound in Sample x Mass Fraction Law of Multiple Proportions: - If elements A and B react to form 2 compounds, the different masses of B that combine with a fixed mass of A can be expressed as a ratio of small whole numbers - Example: - Carbon Oxide I (CO) is 57.1% Oxygen and 42% Carbon Carbon Dioxide II (C0_2)is 72.7% Oxygen and 27.3% Carbon - We are looking for the proportion of Oxygen in both substances, so first we look for the ratio between the two elements in each substance, and we make it a 100g sample so the percentages will be easier to convert to grams - Carbon Oxide I = O = 51.7g = a ratio of 1.33g Dalton’s Atomic Theory: C 42.9g O 2 72.7g 1. All matte- cCarbon Dioxide II =hich ca= ne27.3g = a ratio ofor destroyed C 2. An atom of one element cannot be converted into an atom of another 3. Atoms of one element are all alike in their properties, but different from any other atom of any other element 4. Compound result from chemical combinations of different elements in a certain ratio Isotopes: atoms of a certain element that have different numbers of neutrons and therefore different mass numbers - Mass # - Atomic # = # of Neutrons Isotopic Composition: Specifies the proportional abundance of each isotope - Atomic Mass Unit (amu) = Dalton (Da) o All relative to carbon (1C ) which has 12 amu exactly ❑ Mass Spectrometry: a method for measuring the relative masses and abundances of atomic- scale particles precisely Isotopic Mass: the relative mass of an isotope - Isotopic mass of ❑Si = Measured Mass Ratio x Mass of ❑C = 2.33 x 12 amu = 27.97 amu Calculating Atomic Mass of an Isotope: - Atomic Mass = (Frequency of 1 Isotope (%) x Mass (amu) of 1st Isotope) + (Frequency of 2dIsotope (%) x Mass (amu) of 2dIsotope) Common Monatomic Ions: Charge Name Formula Cations 1+ Hydrogen H+ Lithium* Li+ Sodium* Na+ Potassium* K+ Cesium Cs+ Silver* Ag+ 2+ Magnesium* Mg2+ Calcium* Ca2+ Stronium Sr2+ Barium* Ba2+ Zinc* Zn2+ Cadmium Cd2+ 3+ Aluminum Al3+ Anions 1- Hydride H- Flouride* F- Chloride* Cl- Bromide* Br- Iodide* I- 2- Oxide* O2- Sulfide* S2- 3- Nitride N3- *= most common Common Metals That form more than One Monatomic Ion: Element Ion Formula Systemic Name Common Name Chromium Cr2+ Chromium (II) Chromous Cr3+ Chromium (III) Chromic Cobalt Co2+ Cobalt (II) Co3+ Cobalt (III) Copper Cu+ Copper (I) Cuprous Cu2+ Copper(II) Cupric Iron Fe2+ Iron (II) Ferrous Fe3+ Iron (III) Ferric Lead Pb2+ Lead (II) Pb4+ Lead (IV) Mercury Hg_2 2+ Mercury (I) Mercurous Hg2+ Mercury (II) Mercuric Tin Sn2+ Tin (II) Stannous Sn4+ Tin(IV) Stannic Stoichiometry: the study of the quantitative aspects of formulas and reactions Mole (mol): the amount of a substance that contains the same number of entities as the number of atoms in 12 g of Carbon-12 # of atoms in 12g of Carbon 12 = Avogadro’s Number 23 Avogadro’s Number: 1 mol = 6.022x 10 - Moles relate the number of entities to the mass of the sample of entities - Moles maintain the same numerical relationship between amu and mass in grams on the macroscopic scale Molar Mass ( μ ): mass per mole (g/mol) Ways to find molar mass: 1. With Elements: find atomic mass on periodic table, figure out if its monatomic or molecular a. If monatomic: Molar mass = g/mol b. If molecular: Look at the formula to determine molar mass (# of atoms = # of entities) 2. With Compounds: Molar mass is the sum of the molar masses of the atoms in the formula EX: 1 mol of S O2 molecules = 1 mol of S and 2 mols of O Info Contained in the Chemical Formula of Glucose C 6 12 6 ( μ = 180.16 g/mol): Carbon (C) Hydrogen (H) Oxygen (O) Atoms/Molecule 6 atoms 12 atoms 6 atoms Mass/ Moles of 6 (6.022x 1023 ) 12 (6.022x 1023 6 (6.022x 1023 ) Compound F.U. ) F.U. F.U. Atoms/Moles of 6 (12.01 amu) 12 (1.008 amu) 6 (16.00 amu) compound Moles of 6 mol of atoms 12 mol of atoms 6 mol of atoms atoms/moles of compound Mass/ Moles of 72.06 g 12.10 g 96.00 g compound Road Map for Conversions: Mass (g) Molar Mass (g/mol) Chemical Formula Moles (mol) Moles in Compound (mol) Avogadro’s Number Atoms or Molecules (F.U.) Mass Percent: - Each element contributes a fraction of a compound’s mass - For a molecule (F.U.) use molecular mass and formula to find the mass percent: o Mass % of element x in compound= Atoms of x in formula * atomic mass of x *100 Molecular mass of compound - For a mole of compound, use molar mass and formula to find the mass percent of each element: o Mass % of element x = Moles of x in formula * molar mass of x *100 Mass(g) of 1 mol of compound - An element always constitutes the same fraction of mass of a given compound o Mass of element = (Mass of compound x)* mass of element in 1 mol of compound mass of 1 mol of compound 3 Common Types of Formula: 1. Empirical Formula - Shows the lowest whole number of moles, relative # of atoms of each element in compound 2. Molecular Formula – Shows actual number of atoms of each element in a molecule 3. Structural Formula – Shows the relative placement and connections of the atoms How to Find Empirical Formula: 1. Determine mass (g) of each component element 2. Convert each mass to amount (mol), and write a preliminary formula 3. Convert the amounts (mol) mathematically to whole number subscripts a. Divide each subscript by the smallest subscript b. If necessary, multiply through by the smallest integer that turns all subscripts into whole number integers If we know molar mass of compound we can use empirical formula to obtain molecular formula When Given the Mass Percent: 1. Assume 100g of compound to express each mass percent directly as mass 2. Convert each mass (g) to mole (mol) 3. Derive empirical formula 4. Divide molar mass of compound by empirical formula mass to find the whole-number multiple and the molecular formula Combustion Analysis: Used to measure the amounts of carbon and hydrogen in a combustible organic compound Compound is burned in pure oxygen, C and H react with O to get carbon dioxide and water, after that we measure carbon dioxide and water, then carbon and hydrogen, and from that the empirical formula is found **Empirical and molecular formulas tell us nothing about the structure of a molecule Isomers: Compounds with the same molecular formula and thus, molar mass, yet have different properties When a chemical equation is balanced it means there are the same number of each type of atoms in both the reactants and the products Reactants: Substances before a chemical reaction occurs Products: Substances after a chemical reaction has occurred Steps for Balancing an Equation: 1. Creation of the Equation – make a skelato equation of the reactants and products EX: __Mg + __ O2 __Mg O 2 2. Balancing Atoms – We match the numbers of each type of atom on the left and right side of the yield arrow -Start with most complex molecule and end with the least complicated, such as the element on its own -Balancing Coefficient: a numerical multiplier of all the atoms in the formula that follows is _1_ Mg + _1/2_ O2 _1_Mg O2 3. Adjusting the Coefficients – Multiply the balancing coefficients all by the same number until you reach the lowest common multiple for each balancing coefficient (must be whole number in final answer) 4. Checking – In order to make sure it is balanced, make sure there are the same amount of each type of atom in the reactants as there are in the product 5. Specify the state of matter – make sure you label what state of matter each substance is a. g = gas b. l = liquid c. s = solid d. aq = aqueous solution **Balancing Coefficients refer to both individual chemical entities and moles of entities In balanced equations, the amounts (mol) of substances are stoichiometrically equivalent to each other - Means reactions occur when that specifi amoun of each substance is present For Stoichiometric purposes, when the same substance forms in one reaction and reacts in the next reaction in sequence, we eliminate it in an overall (net) equations - Steps for writing overall equations: 1. Write the sequence of balanced equations 2. Adjust the equations arithmetically to cancel the common substances 3. Add the adjusted equations together to obtain the overall balanced equation Metabolic Pathways: Multistep reaction sequences in a biological system Limiting Reactant: The reactant in a chemical reaction that limits the amount of product that can be formed. The reaction will stop when all of the limiting reactant is consumed - The reactant that is not limiting is present in excess, which means the amount that doesn’t react is left over To determine which is the limiting reactant, we use the molar mass ratios in th balanced equation to perform a series of calculations to see which reactant forms less product Theoretical Yield: The amount of product calculated from the molar ratio in the balanced Equation - Theoretical yield is never obtained in a limiting reaction because: o Side reactions take away from substance o Many reactions don’t use up all of the limiting reactant o Physical losses occur in every step of the process, its never “perfect” Actual Yield: The amount of the product actually obtained through a reaction Percent Yield: the actual yield expressed as a percentage of the theoretical yield - % Yield = actual yield *100 theoretical yield Atom Economy: the proportion of reactant atoms that end up in the desired product % atom economy = # of moles * molar mass of desired product *100 sum of (# of moles *molar mass) for all products
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