CH104chapter5.pdf CH 104
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This 2 page Class Notes was uploaded by Regan Dougherty on Thursday October 15, 2015. The Class Notes belongs to CH 104 at University of Alabama - Tuscaloosa taught by Stephen Woski in Summer 2015. Since its upload, it has received 27 views. For similar materials see Introductory Chemistry in Chemistry at University of Alabama - Tuscaloosa.
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Date Created: 10/15/15
Thursday October 8 2015 CH 104 Chapter 5 Chemical Reactions The difference between a chemical change and a physical change In a chemical reaction you are changing things at a molecular level bonds being broken and bonds being formed You must have the same amount of each atom on the reactant side of the equation as the product side balancing coefficients if there is no number in front the coefficient is one mole coefficient conversion factor In a mole of methane CH4 there are 4 moles of hydrogen atoms and 1 mole of carbon OxidationReduction oxidation lose an electron charge increases less negativemore positive Reduction gain electron charge is reduced less positivemore negative reducing agent undergoes an oxidation reaction causes the other thing to be reduced oxidizing agent is reduced causes the other thing to be oxidized Avogadro s Number as a Conversion Factor Avogadro s number of a substance is a mole of that substance Thursday October 8 2015 1mol 602 x1023OR 602 x10231mol 015 mol of C How many atoms 015 mol x 602 x 1023 atomsmol 903 x 1022 atoms The decimal number under each element on the periodic table is the atomic weight Atomic weight how much a mole of that atom weighs in grams atomic mass units 454 g Au x 1 mol1970 g 230 mol Au atomic weight of gold 1970 230 mol x 602 x 10231 mol 138 x 1024 atoms in 454 g Au Calculate the Formula Weight of a Compound Write the correct formula and determine the number of atoms of each element from the subscripts Multiply the number of atoms of each element by the atomic weight and add the results Atomic Weight of FeSO4 15192 amu 7 g FeSO4 How many moles 7g x 1 mol15192g MATH HELP grams of A can be related to moles of A by the formula weight of A moles of A can be related to moles of B by the coefficient of a balanced reaction moles of B can be related to grams of B by the formula weight of B Actual Yield amount of a substance that is actually produced in a chemical reaction Percent Yield actual yield divided by theoretical yield x 100 Limiting Reagent the substance on the reactant side that controls how much of a product is a produced To determine the limiting reagent look at quantities in moles