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Chemistry 211 Week 3 Notes

by: Lucas Kinsey

Chemistry 211 Week 3 Notes CHEM 211-003

Marketplace > George Mason University > Chemistry > CHEM 211-003 > Chemistry 211 Week 3 Notes
Lucas Kinsey
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These are the notes to Chemistry 211 week 3 at George Mason University
General Chemistry 1
Pritha G. Roy
Class Notes
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This 6 page Class Notes was uploaded by Lucas Kinsey on Sunday September 18, 2016. The Class Notes belongs to CHEM 211-003 at George Mason University taught by Pritha G. Roy in Summer 2016. Since its upload, it has received 21 views. For similar materials see General Chemistry 1 in Chemistry at George Mason University.


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Date Created: 09/18/16
Lucas Kinsey Chemistry 211 Week 3 Notes Chapter 3 Stoichiometry of Formulas and Equations 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 Avogadro’s Number: 1 mol = 6.022x 10 23 - 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_ O 2  _1_Mg O 2 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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