Lesson 11 and 12 Notes
Lesson 11 and 12 Notes CHM 111
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This 3 page Class Notes was uploaded by Kara Nichols on Friday June 17, 2016. The Class Notes belongs to CHM 111 at Calhoun Community College taught by Mona Chaudhary in Spring 2016. Since its upload, it has received 20 views.
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Date Created: 06/17/16
Lesson 11 Wednesday, June 15, 2016 5:29 PM • Atomic Weight / atomic mass ○ Decimal number on the periodic table ○ Weighted average of all the isotopes ○ We use it as grams ○ Weight and/or mass of one atom • Mole ○ Describes an amount of substance/compound ○ Mole is the SI unit used to represent 6.022 x 10^23 particles ○ 6.022 x 10^23= Avogadro's number ○ 1 mole of an atom - the weight (atomicmass) in grams ○ 1 mole of Cu = 63.54 grams 1 mole of Cu = 6.022 x 10^23atoms of Cu ○ 1 mole of O = 15.9994grams 1 mole of O = 6.022 x 10^23atoms of O ○ Molecular weight is used for molecules ○ Atomic weight is used for atoms • Formula / molecularweights ○ Formula weight (FW) = ionic compounds ○ Molecular Weight (MW) = covalent compounds ○ Calculation The atomic weights of all the atomspresent in the compound • Calculations of grams, moles, molecules,and atoms ○ If you start with moles: You will ALWAYS multiply To get to grams: □ Multiply by the molecular weight of the compound or molecule To get to the number of moleculesor atoms: □ Multiply by Avogadro's number (6.022 x 10^23) ○ If you start with grams: To get to moles: □ DIVIDE by the molecular weight To get to the number of moleculesor atoms: To get to the number of moleculesor atoms: □ Divide by the molecular weight □ Then multiply by Avogadro's number ○ If you start with the number of moleculesor atoms To get to moles: □ Divide by Avogadro's number To get to grams: □ Divide by Avogadro's number □ Then multiply by the molecular weight ○ To find the number of atoms in a molecule Multiply the number of moleculesby the subscript of the atom you are trying to find • Percent Composition ○ (Mass of the element/ mass of the compound ) x100 College Chemistry 1 – Mona Chaudhary 6/14/2016 Important Words or Definitions Important People Important Concepts Lesson 12 Formulas from element composition o Empirical formula Simplest formula Relative ration of atoms Smallest whole number ratio of atoms o Molecular (true) Formula True ratio of atoms Multiple of empirical formula ratios o Molecular formula = (empirical formula x n) Element Relative mass Relative # of atoms Divide by the Whole # of atoms smallest # ratio Symbol Change % Divide the relative mass by Divide by the Whole sign to Grams the atomic weight of the smallest relative number element value ratio Example: o A compound contains the following: 40% carbon, 6.7% hydrogen, 53.3% oxygen. What is the empirical formula? o What is the molecular formula if the compound has a molecular mass of 60.0 g/mol? o Element Relative mass of Relative # of Divide by the Whole # atoms atoms smallest # ratio C 40 g 40/12.01 3.33/3.33 1 H 6.7g 6.7/1.01 6.63/3.33 2 O 53.3g 53.3/16 3.33/3.33 1 Empirical formula = CH2O Molecular formula: Molecular weight of empirical formula = 12.01+2.02+16 = 30.03 60/30.03 = 1.998 (almost 2) Multiply all the subscripts by 2 C 2 4 2
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