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Chemical Stoichiometry Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1 Atomic Masses Elements occur in nature as mixtures of isotopes Carbon = 98.89% 12C 1.11% 13C <0.01% 14C Carbon atomic mass = 12.01 amu Copyright©2000 by Houghton Mifflin Company. All rights reserved. 2 Atomic Mass Atoms are so small, it is difficult to discuss how much they weigh in grams Use atomic mass units. an atomic mass unit (amu) is one twelfth the mass of a carbon-12 atom This gives us a basis for comparison The decimal numbers on the table are atomic masses in amu They are not whole numbers Because they are based on averages of atoms and of isotopes. can figure out the average atomic mass from the mass of the isotopes and their relative abundance. add up the percent as decimals times the masses of the isotopes. The Mole • • • • The mole is a number a very large number, but still, just a number 6.022 x 1023 of anything is a mole a large dozen The Mole The number equal to the number of carbon atoms in exactly 12 grams of pure 12C. 1 mole of anything = 6.022 1023 units of that thing Copyright©2000 by Houghton Mifflin Company. All rights reserved. 6 Avogadro’s number equals 23 6.022 10 units Copyright©2000 by Houghton Mifflin Company. All rights reserved. 7 Molar Mass A substance’s molar mass (molecular weight) is the mass in grams of one mole of the compound. CO2 = 44.01 grams per mole Copyright©2000 by Houghton Mifflin Company. All rights reserved. 8 Percent Composition Mass percent of an element: mass of element in compound mass % 100% mass of compound For iron in iron (III) oxide, (Fe2O3) 111.69 mass % Fe 100% 69.94% 159.69 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 9 Formulas molecular formula = (empirical formula)n [n = integer] molecular formula = C6H6 = (CH)6 empirical formula = CH Copyright©2000 by Houghton Mifflin Company. All rights reserved. 10 Empirical Formula Determination 1. Base calculation on 100 grams of compound. 2. Determine moles of each element in 100 grams of compound. 3. Divide each value of moles by the smallest of the values. 4. Multiply each number by an integer to obtain all whole numbers. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 11 Working backwards From percent composition, you can determine the empirical formula. Empirical Formula the lowest ratio of atoms in a molecule Based on mole ratios A sample is 63.68% C, 9.80 %H, 12.38%N, and 14.14%O what is its empirical formula. (question #70; p.119_ Example: Question 79, p.119 Cumene is a compound containing only carbon and hydrogen that is used in the production of acetone and phenol. Combustion of 47.6 mg cumene produces some CO2 and 42.8 mg water. The molar mass of cumene is between 115 and 125 g/mol. What is the empirical formula and the molecular formula of cumene? Pure O2 in Sample is burned completely to form CO2 and H2O CO2 is absorbed H2O is absorbed Empirical To Molecular Formulas • • • • Empirical is lowest ratio Molecular is actual molecule Need Molar mass Ratio of empirical to molar mass will tell you the molecular formula • Must be a whole number because... Chemical Equations Chemical change involves a reorganization of the atoms in one or more substances. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 16 Chemical Equations • Are sentences. • Describe what happens in a chemical reaction. • Reactants Products • Equations should be balanced • Have the same number of each kind of atoms on both sides because ... Chemical Equation A representation of a chemical reaction: C2H5OH + 3O2 2CO2 + 3H2O reactants products Copyright©2000 by Houghton Mifflin Company. All rights reserved. 18 Abbreviations • (s) , for product) • (g) , for product) • (aq) • heat • D • catalyst Meaning • A balanced equation can be used to describe a reaction in molecules and atoms. • Not grams. • Chemical reactions happen molecules at a time • or dozens of molecules at a time • or moles of molecules. Chemical Equation C2H5OH + 3O2 2CO2 + 3H2O The equation is balanced. 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water Copyright©2000 by Houghton Mifflin Company. All rights reserved. 21 Calculating Masses of Reactants and Products 1. 2. 3. 4. Balance the equation. Convert mass to moles. Set up mole ratios. Use mole ratios to calculate moles of desired substituent. 5. Convert moles to grams, if necessary. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 22 Limiting Reactant The limiting reactant is the reactant that is consumed first, limiting the amounts of products formed. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 23 Limiting Reagent • Reactant that determines the amount of product formed. • The one you run out of first. • Makes the least product. • Book shows you a ratio method. • It works. • So does mine Limiting reagent • To determine the limiting reagent requires that you do two stoichiometry problems. • Figure out how much product each reactant makes. • The one that makes the least is the limiting reagent. Solving a Limiting Reactant Problem 1. 2. 3. 4. Balance the equation. Convert masses to moles. Determine which reactant is limiting. Use moles of limiting reactant and mole ratios to find moles of desired product. 5. Convert from moles to grams. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 26 Percent Yield % yield = Actual Theoretical x 100% % yield = what you got x 100% what you could have got Copyright©2000 by Houghton Mifflin Company. All rights reserved. 27