1. How many hydrogen atoms are in 1.00 g of hydrogen?

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1 MOLES AND CALCULATIONS USING THE MOLE CONCEPT INTRODUCTORY TERMS A. What is an amu? 1.66 x g B. We need a conversion to the macroscopic world. 1. How many hydrogen atoms are in 1.00 g of hydrogen? 1 H atom 1.00 g H x = 6.02 x H atoms 1.66 x g H Avogadro s Number

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3 2. Consider carbon-12 (the most abundant isotope of C) What is the mass of one carbon-12 atom? amu g g 12 x 1.66 x = 1.99 x C atom amu C atom 3. What is the mass of Avogadro s number of C atoms? g 6.02 x C atoms x 1.99 x = g C atom

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5 C. Definition of a mole 1. A mole is the amount of any substance that contains as many elementary entities as there are atoms in exactly 1.00 g of hydrogen A mole is the amount... in exactly g of carbon x of anything 4. It is important to state the entities involved: atoms, molecules, ions, electrons, etc. 5. How large a number is this?

6 How Large a Quantity is Avogadro s Number??

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8 D. Definition of a molar mass (g/mol) 1. Molar mass of an element - The mass in grams that is numerically equal to the atomic weight of the element. C, g/mol Na, g/mol Xe, g/mol 2. Molar mass of a molecular form of an element - May be different N2, g/mol Cl2, g/mol (O2, I2, F2, Br2) 3. Molar mass of a compound - The mass in grams that is numerically equal to the formula weight of the compound.(the sum of the atomic weights involved) H2O, g/mol ( ) BaCl2, g/mol ( ) CO2, g/mol ( )

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10 4. Sample Calculations of Molar Mass a. Na2HPO4 Na2 2 x = H 1 x = P 1 x = O4 4 x = g/mol b. Ca3(PO4)2 Ca3 3 x = (P )2 2 x = ( O4)2 8 x = g/mol c. C15H22ClNO2 C15 15 x = (Demerol) H22 22 x = Cl 1 x = N 1 x = O2 2 x = g/mol

11 THE ABOVE CONCEPTS LEAD TO NEW "CONVERSION FACTORS" 1 mole x 1023 objects or 6.02 x 1023 objects mole How many atoms are represented by 3.00 moles of calcium (Ca)? 3.00 mol Ca X 6.02 x 1023 atoms 1.00 mol Ca = 1.81 x 1024 atoms of Ca

12 How many moles are represented by 1,000,000 molecules of carbon monoxide (CO)? 1.0 x 106 molecules CO X 1.00 mole CO 6.02 x 1023 molecules CO =1.66 x mol CO MOLES OF SUBSTANCE PARTICLES OF SUBSTANCE Avogadro s Number

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15 ANOTHER PAIR OF "CONVERSION FACTORS" g of CO mole of CO2 or 1 mole of CO g of CO2 How many grams are contained in 4.25 moles of carbon dioxide (CO2)? 4.25 moles CO 2 X g of CO mole of CO2 = 187 g CO2

16 3.77 g of CO2 is equal to how many moles of CO2? 3.77 g CO2 X 1.00 mole of CO g of CO2 =.0856 mol CO 2 MOLES OF SUBSTANCE GRAMS OF SUBSTANCE Molar Mass

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19 How many atoms are in 7.67 g of cobalt (Co)? 7.67 g Co X 1 mol of Co X 6.02 x atoms g of Co 1.00 mol Co = 7.84 x 1022 atoms of Co What is the mass of 1.77 x molecules of CO2? 1.77 x 1030 molecules of CO 2 X 1.00 mol CO x 1023 molecules CO 2 X g of CO mole of CO2 = 1.29 x 108 g CO2

20 GRAMS OF SUBSTANCE MOLES OF SUBSTANCE PARTICLES OF SUBSTANCE

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24 Information Available from A Balanced Equation

25 Conversion Factors Relevant to Stoichiometry Use Avogadro s number as a conversion factor. Moles of A Particles of A Use molar mass as a conversion factor. Moles of A Grams of A Use mole ratio as a conversion factor. Moles of A Moles of B

26 Solution Map for Stoichiometry Grams of A Grams of B Molar Mass Molar Mass Moles of A Coefficients Moles of B Avogadro s Number Avogadro s Number Particles of A Particles of B

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28 Limiting Reactant, Theoretical Yield, and Percent Yield Limiting Reactant The reactant that is completely consumed in a chemical reaction Theoretical Yield The amount of product that can be made in a chemical reaction based on the amount of limiting reactant Actual Yield The amount of product actually produced by a chemical reaction Actual Yield Percent Yield = x 100% Theoretical Yield

29 Consider the Reaction: Ti (s) + 2 Cl2 (g) > TiCl4 (s) If we begin the reaction with 1.8 mol of titanium and 3.2 mol of chlorine, what is the limiting reactant and theoretical yield of TiCl4 in moles? (theoretical yield) If only 1.5 mol of TiCl4 is actually isolated from the reaction, what is the percent yield for the process? Actual Yield 1.5 Percent Yield = x 100% = x 100% = 94% Theoretical Yield 1.6

30 Consider the Reaction: 2 Al (s) + 3 Cl2 (g) > 2AlCl3 (s) If we begin the reaction with mol of aluminum and mol of chlorine, what is the limiting reactant and theoretical yield of AlCl3 in moles? (theoretical yield) If only mol of AlCl3 is actually isolated from the reaction, what is the percent yield for the process? Actual Yield.448 Percent Yield = x 100% = x 100% = 81.1% Theoretical Yield.552

31 Consider the Reaction: 2 NO (g) + 5 H2 (g) > 2 NH3 (g) + 2 H2O (g) What is the maximum amount of ammonia in grams that can be synthesized from 45.8 g of NO and 12.4 g of H2? (theoretical yield) If only 16.4 g of NH3 is actually isolated from the reaction, what is the percent yield for the process? Actual Yield 16.4 Percent Yield = x 100% = x 100% = 63.1% Theoretical Yield 26.0

32 Consider the Reaction: 2 Na (s) + Cl2 (g) > 2 NaCl (s) If we begin the reaction with 53.2 g of sodium and 65.8 g of chlorine, what is the limiting reactant and theoretical yield of NaCl in grams? (theoretical yield) If only 86.4 g of NaCl is actually isolated from the reaction, what is the percent yield for the process? Actual Yield 86.4 Percent Yield = x 100% = x 100% = 80.0% Theoretical Yield 108

33 Consider the Reaction: 2 Fe (s) + 3 S (l) > Fe 2 S 3 (s) When 10.4 g of Fe are allowed to react with 11.8 g of S, 14.2 g of Fe2S3 are obtained. Find the limiting reactant, theoretical yield and percent yield? (theoretical yield) Actual Yield 14.2 Percent Yield = x 100% = x 100% = 73.2% Theoretical Yield 19.4

34 Consider the Reaction: Cu2O (s) + C (g) > 2 Cu (s) + CO (g) When 11.5 g of C are allowed to react with g of Cu2O, 87.4 g of Cu are obtained. Find the limiting reactant, theoretical yield and percent yield? (theoretical yield) Actual Yield 87.4 Percent Yield = x 100% = x 100% = 85.9% Theoretical Yield 101.7

35 EMPIRICAL FORMULA A chemical formula that indicates the relative proportions of the elements in a molecule rather than the actual number of atoms of the elements. (An empirical formula may be obtained from percentage composition of elements in a compound.) MOLECULAR FORMULA A chemical formula that indicates the actual number of atoms of the elements in a molecule. (Information in addition to percentage composition of elements is needed to determine a molecular formula.)

36 Molecular Formulas May Differ from Empirical Formulas Benzene Empirical Formula, CH Molecular Formula, C6H6 Acetylene Empirical Formula, CH Molecular Formula, C2H2

37 Molecular Formulas May Differ from Empirical Formulas Glucose Empirical Formula, CH2O Molecular Formula, C6H12O6 Fructose Empirical Formula, CH2O Molecular Formula, C6H12O6

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39 Solutions of Emprical Formula Problems

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