Chapter 3 - Calculations with Chemical Formulas & Eqns

1. Chapter 3 - Calculations with Chemical Formulas & Eqns Online HW 3 due: ……………….. Equation Practice HW Available at any time. Online Stoichiometry Lab due before 5/2/2014 - Exam #2 will cover chapters 3 & 4 on………….

2. End-of-ChapterSuggested Problems, Chapter 3, pp 116-125 3, 5, 6, 7, 14, 15, 16, 25, 26, 27, 33 37, 43, 47, 51, 57d, 61, 65, 67, 71 75, 79, 81, 83, 89, 91, 95, 97

3. I. Mole A. Introduction • Chemists devised a new term called a “mole” or “mol.” • A mole has two definitions: (know these two) Definition #1: 6.02x1023 number of that object; so, 1.00 mole = 6.02x1023 # Definition #2:The formula weight (FW) in grams; so 1.00 mole = FW in g (from periodic chart) • The above two definitions work because 6.02x1023 protons or neutrons weights 1.00 gram. • Following is true for chemicals: 6.02x1023 # = FW in g (Note: 6x1023 ≈ number of stars in the universe)

4. I. Mole A. Introduction

5. I. Mole A. Introduction • Mole Conversion Factors to know A) 1.00 mole = Formula Weight in Grams (from Periodic Chart) B) 1.00 mole = 6.02x1023 # C) 6.02x1023 # = FW in g • A few examples of conversion factors using 1.0 mol H2O 1.00 mole H2O 18.0 g H2O 1.00 mole H2O 18.0 g H2O 1.00 mole H2O 6.02x1023 molec. H2O 1.0 mole H2O 1.0 mole H2O 1.00 mole H2O 12x1023 H atoms 18x1023 total atoms 2.02 g H

6. I.Mole B. Calculations 1. How many grams are in 1.0 mole of H2SO4. 2x1 + 32 + 4x16 = 98 g H2SO4 2. How many grams are in 1.00 mole of Ca(NO2)2. 40 + 2x14 + 4x16 = 132 g Ca(NO2)2 3. How many moles are in 0.36 g Be. 0.36g Be 1.0 mole Be = 0.040 mole Be 9.0g Be 4. How many moles are in 36 g of H2O. 36g H2O 1.0 mole H2O = 2.0 moles H2O 18 g H2O

7. I. Mole B. Calculations continued 5. How many g present in 1.5x10-3 moles CaCO3? 1.5x10-3 moles CaCO3 x 100. g CaCO3 = 0.15 g CaCO3 1.00 mole CaCO3 6. How many molecules are in 0.20 moles H20 0.20 mole H2O x 6.0x1023 molecule = 1.2x1023 molecules 1.0 mole H2O 7. How many atoms are in 0.20 moles H2O. 0.20 mol H2O 6.0x1023 molec H2O 3 atoms = 3.6x1023 atom 1.0 mol H2O1.0 molec H2O

8. I.Mole B. Calculations continued 8. How many moles (m) of O are in 2.0 moles of H3PO4? 2.0 m H3PO4 x 4 m O = 8.0 mole of O 1 m H3PO4 9. How many grams of O atoms are in 0.60 moles of H3PO4 ? 0.60 m H3PO4 4 m O 16 g O = 38 g O 1 m H3PO4 1 m O

9. I.Mole C. Mass % Composition • Note: % = Part x 100. % - Know Total • % composition = g element x 100. %. total g compound • Example 1: Calculate % composition of C in C4H10 to 2 SF. - Pick mass of 1 mole of compound for the TOTAL. Why? • Total g (1 mole) of C4H10 = 48 g C + 10 g H = 58 g (4 mol C x 12 g C/mol = 48 g C & 10 mol H x 1.0 g H/mol H = 10 g H) % C = 48 g C x 100 % = 83 % C 58 g C4H10

10. I. Mole C. Mass % Composition • Example 2: A compound contains 53.3 (W/W) % S. How many g of S are in 71 g of the compound? (53.3 g S = 100. g compd = conversion factor from 53.3 w/w % S) 71 g compd x 53.3 g S = 38 g S 100. g compd • Example 3: 0.056 g of a hydrocarbon gave 0.082 g of CO2 upon burning. Calculate w/w % C.CxHy + O2 -------) CO2 + H2O (unbalanced) % C = [ g C / g compound ] x 100 % = [ g C / 0.056 g ] x 100 % 0.082g CO2 1m CO2 1m C 12.1g C = 0.022g C 44.0g CO2 1m CO2 1 m C [ 0.022 g C / 0.056 g compd ] x 100 % = 39 % C

11. II. Empirical Formulas A. Definition - The empirical formula (EF) is the simplest whole number ratio of atoms (or mols) in the unit. Examples:Unit EF H2O2 HO C6H6 CH N2O4 NO2 P4S10 P2S5 - Obtain the EF from experimental % or from gram (g) data. • Procedure: % g mols simplest whole # ratio

12. II. Empirical Formulas B. Calculation • Example: Calculation from grams or from w/w %. (Pick 100 g of compound, then get mols of each atom). A compound was found to be 50.0 % S and 50.0 % O. Calculate EF. Formula = SxOy 50.0 g S x (1 mol S / 32.0 g S) = 1.56 mol S 50.0 g O x (1 mol O / 16.0 g O) = 3.12 mol O Convert to simplest whole # ratio by dividing both by the smaller number of mols: S1.56 O3.12 EF = S1.56/1.56O3.12/1.56 = S1O2 or SO2 - Note: if get something like S1.00O1.33 then what do you do?

13. II. Empirical Formulas C. Molecular Weight - If one has the empirical formula and the molecular weight (MW), then one can calculate the molecular formula (MF). MW / EW = # ; Then multiply EF by # to get the MF. 1) The MW of a compound is 284 g & the EF is P2O5 (142 g). Calculate the MF. 284g / 142 g = 2.00 MF = 2 x P2O5 = P4O10 2) The MW = 78 g for CxHy& the EF = C1H1. Obtain the MF. 78 g / 13 g = 6.0 MF = 6 x C1H1 = C6H6

14. III. Stoichiometry A. Introduction Mole Conversion Factors from Equations - The balancing coefficients in a chemical equation can be assigned units of moles & used as conversion factors. - Example:2 H2 + 1 O2 ---) 2 H2O (2 mol of H2 will react with 1 mol of O2 to give 2 mol of H2O) a) How many moles of H2 will react with 5.5 mol of O2? 5.5 m O2 2 m H2 = 11 moles of H2 1 m O2 b) How many moles of H2 are needed to produce 13 mol H2O? 13 mol H2O 2 m H2 = 13 mol H2 2 m H2O

15. III. Stoichiometry B. Methodology - The coefficients in a balanced chemical reaction can be used as conversion factors in calculating the amount of any other chemical in the reaction (RXN). - The problem solving scheme is: a b c g known -----) moles known -----) moles unknown -----) g unknown - Conversion factors for a & c are from Periodic Chart (1 mole = FW in grams); conversion factor for step b is from the balanced chemical equation.

16. III. Stoichiometry B. Regular Problems - In regular problems, you are given the amount of ONE reagent & you assume there is enough of the other reagents for complete reaction. Examples: Given: 2 Al(OH)3 + 3 H2SO4 -----) 6 H2O + 1 Al2(SO4)3 1.How many moles of H2SO4 are needed to produce 8.0 moles of Al2(SO4)3 8.0 moles Al2(SO4)3x 3.0 moles H2SO4 = 24 moles H2SO4 1.0 mole Al2(SO4)3 2. How many moles of H2O will be produced from 156 g of Al(OH)3? 156 g Al(OH)31 mole Al(OH)36 mole H2O = 6.00 mole H2O 78.0 g Al(OH)32 mole Al(OH)3

17. III. Stoichiometry B. Regular Problems Continued Given: 2 Al(OH)3 + 3 H2SO4 -----) 6 H2O + 1 Al2(SO4)3 Example: 3.How many g of Al(OH)3 will react with 59 g of H2SO4? 59 g H2SO41 mol H2SO42mol Al(OH)378 g Al(OH)3= 31 g Al(OH)3 98 g H2SO4 3 mol H2SO41 mol Al(OH)3

18. III. Stoichiometry C. Limiting Reagent Problems • You are given TWO starting materials & one will limit the amount of product. Either a) determine the limiting reagent ahead of time or b) calculate the result for each reagent & use the one which gives the smallest mols or g of product. 1.How many g of HgS will be produced from 20. g of Hg and 64 g of S ? 1 Hg + 1 S -----) 1 HgS 64 g S 1 mol S 1 mol HgS 232 g HgS = 464 g HgS 32 g S 1 mole S 1 mol HgS 20. g Hg 1 mol Hg 1 mol HgS 232 g HgS =23 g HgS 200. g Hg 1 mole Hg 1 mole HgS

19. III. Stoichiometry C. Limiting Reagent Problems 2.How many g of CS2 can be made from 20. g C and 30. g SO2? 5 C + 2 SO2 -----) 1 CS2 + 4 CO 20.g C 1 mol C 1 mol CS2 76 g CS2 = 25 g CS2 12 g C 5 mol C 1 mol CS2 30.g SO2 1mol SO2 1 mol CS2 76g CS2 = 18 g CS2(Correct) 64g SO2 2 mol SO2 1 mol CS2 3.11 moles of Hg are mixed with 9 moles of S. How many moles of excess reagent are left over? 1 Hg + 1 S -----) 1 HgS 11 m Hg gives 11 m HgS & 9 m S gives 9 m HgS; so, S is LR & excess Hg 11 m Hg initially – m Hg that reacts = m Hg left over (calculate m Hg that react) 9 mol S x 1 mol Hg = 9 mols of Hg react. 1 mol S 11 mols Hg - 9 mols Hg = 2 mols Hg Left Over

20. III. Stoichiometry C. % Yield Example: 48 mol of O2 reacts with excess Mg; 39 moles of MgO are obtained in the lab. Calculate the % Yield. 2 Mg + 1O2 -----) 2 MgO - % Yield = Actual Yield x 100 % = 39 moles MgO x 100 % Theoretical YieldTheoretical Yield - Now calculate the theoretical yield. 48. mol O2 x 2 mol MgO = 96 mol MgO (Theoretical Yield in moles) 1 mol O2 % Yield = 39. mol MgO x 100% = 41 % 96 mol MgO - Note: Can use either g or moles in % problems; get same answer.