03_lecture-akins021511 - Chemistry 10301, Spring 2011...

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1 3 Stoichiometry: Ratios of Combination Chapter Chemistry 10301, Spring 2011 Instructor: Dr. Daniel L. Akins
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Stoichiometry: Ratios of Combination 3 3.1 Molecular and Formula Masses 3.2 Percent Composition of Compounds 3.3 Chemical Equations Interpreting and Writing Chemical Equations Balancing Chemical Equations 3.4 The Mole and Molar Mass The Mole Determining Molar Mass Interconverting Mass, Moles, and Numbers of Particles Empirical Formula from Percent Composition 3.5 Combustion Analysis Determination of Empirical Formula Determination of Molecular Formula 3.6 Calculations with Balanced Chemical Equations Moles of Reactants and Products Mass of Reactants and Products 3.7 Limiting Reactants Determining the Limiting Reactant Reaction Yield
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Molecular and Formula Masses The molecular mass is the mass in atomic mass units (amu) of an individual molecule. To calculate molecular mass, multiply the atomic mass for each element in a molecule by the number of atoms of that element and then total the masses Molecular mass of H 2 O = 2(atomic mass of H) + atomic mass of O = 2(1.008 amu) + 16.00 amu = 18.02 amu 3.1
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Molecular and Formula Masses Calculate the molecular mass of ibuprofen, C 13 H 18 O 2 . Note that the isotopes of an element occur in the same proportion in a compound as they do in the free element, and as a result, the calculated molecular mass is usually an average value. Solution: Molecular mass = 13(12.01 amu) + 18(1.008 amu) + 2(16.00 amu) = 206.27 amu Calculate the molecular mass of glycerol,C 3 H 8 O 3 . Solution: Molecular mass = 3(12.01 amu) + 8(1.008 amu) + 3(16.00 amu) = 92.09 amu
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A list of the percent by mass of each element in a compound is known as the compound’s percent composition by mass . where n is the number of atoms of the element in a molecule or formula unit of the compound. atomic mass of element percent mass of an element = 100% molecular or formula mass of compound n Percent Composition of Compounds 3.2
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For a molecule of H 2 O 2 : 22 2 1.008 amu H %H = 100% = 5.926% 34.02 amu H O 2 16.00 amu O %O = 100% = 94.06% 34.02 amu H O Percent Composition of Compounds
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Percent Composition of Compounds Determine the percent composition by mass of each element in acetaminophen (C 8 H 9 NO 2 ). Solution: Step 1: First determine the molecular mass: MM = 8(12.01 amu) + 9(1.008 amu) + 1(14.01 amu) + 2(16.00 amu) = 151.16 amu Step 2: Calculate the percent by mass of each element: 8 9 2 8 12.01 amu H %C = 100% = 63.56% 151.16 amu C H NO 8 9 2 9 1.008 amu H %H = 100% = 6.002% 151.16 amu C H NO 8 9 2 1 14.01 amu H %N = 100% = 9.268% 151.16 amu C H NO 8 9 2 2 16.00 amu H %O = 100% = 21.17% 151.16 amu C H NO
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Chemical Equations A chemical equation uses chemical symbols to denote what occurs in a chemical reaction. NH 3 + HCl → NH 4 Cl Ammonia and hydrogen chloride react to produce ammonium chloride. Each chemical species that appears to the left of the arrow is called a reactant . NH 3 and HCl Each species that appears to the right of the arrow is called a product.
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03_lecture-akins021511 - Chemistry 10301, Spring 2011...

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