Ch3.ppt
- 1. Copyright McGraw-Hill 2009 1 Chapter 3 Stoichiometry: Ratios of Combination Insert picture from First page of chapter
- 2. Copyright McGraw-Hill 2009 2 3.1 Molecular and Formula Masses • Molecular mass - (molecular weight) – The mass in amu’s of the individual molecule – Multiply the atomic mass for each element in a molecule by the number of atoms of that element and then total the masses • Formula mass (formula weight)- – The mass in amu’s of an ionic compound
- 3. Copyright McGraw-Hill 2009 3 Calculating Molar Mass • Calculate the molar mass for carbon dioxide, CO2 • Write down each element; multiply by atomic mass – C = 1 x 12.01 = 12.01 amu – O = 2 x 16.00 = 32.00 amu – Total: 12.01 + 32.00 = 44.01 amu
- 4. Copyright McGraw-Hill 2009 4 Your Turn! • Calculate the molar mass for each of the following: – Sulfur trioxide – Barium phosphate – Silver nitrate – Acetic acid
- 5. Copyright McGraw-Hill 2009 5 3.2 Percent Composition of Compounds • Calculate by dividing the total mass of each element in a compound by the molecular mass of the compound and multiplying by 100 • % composition allows verification of purity of a sample
- 6. Copyright McGraw-Hill 2009 6 % Composition
- 7. Copyright McGraw-Hill 2009 7 % Composition • Calculate the percent composition of iron in a sample of iron (III) oxide • Formula: Fe2O3 • Calculate formula mass – Fe = 2 x 55.85 = 111.70 amu – O = 3 x 16.00 = 48.00 amu – Total mass: 111.70 + 48.00 = 159.70 amu
- 8. Copyright McGraw-Hill 2009 8 % Composition % by mass = 111.70 159.70 100 = 69.9% Fe What is the % oxygen in this sample? (hint :100%)
- 9. Copyright McGraw-Hill 2009 9 Your Turn! • Calculate the percent oxygen in a sample of potassium chlorate
- 10. Copyright McGraw-Hill 2009 10 3.3 Chemical Equations • Chemical equations represent chemical “sentences” • Read the following equation as a sentence – NH3 + HCl NH4Cl – “ammonia reacts with hydrochloric acid to produce ammonium chloride”
- 11. Copyright McGraw-Hill 2009 11 Chemical Equations • Reactant: any species to the left of the arrow (consumed) • Product: any species to the right of the arrow (formed) • State symbols: – (s) solid (l) liquid (g) gas – (aq) water solution
- 12. Copyright McGraw-Hill 2009 12 Balancing Equations • Balanced: same number and kind of atoms on each side of the equation
- 13. Copyright McGraw-Hill 2009 13 Balancing Equations • Steps for successful balancing 1. Change coefficients for compounds before changing coefficients for elements.(never change subscripts!) 2. Treat polyatomic ions as units rather than individual elements. 3. Count carefully, being sure to recount after each coefficient change.
- 14. Copyright McGraw-Hill 2009 14 Balancing Equations • Balance the equation representing the combustion of hexane __C6H14(l) +__O2(g) __CO2(g) +__H2O(l) (Hint: Make a list of all elements and count to keep track)
- 15. Copyright McGraw-Hill 2009 15 Balancing Equations • Balance the equation representing the combustion of hexane C6H14(l) +19/2O2(g) 6CO2(g) + 7H2O(l) Or…multiply through the entire equation to eliminate fractions 2C6H14(l) +19O2(g) 12CO2(g) + 14H2O(l)
- 16. Copyright McGraw-Hill 2009 16 Chemical Equations Equations can represent physical changes: KClO3(s) KClO3(l) Or chemical changes: • Note the symbol for heat above the arrow 2 KClO3(s) 2 KCl(s) + 3 O2(g)
- 17. Copyright McGraw-Hill 2009 17 3.4 The Mole and Molar Masses • Balanced equations tell us what is reacting and in what relative proportions on the molecular level. • However, chemists must work with the chemical reactions on a macroscopic level.
- 18. Copyright McGraw-Hill 2009 18 The Mole • The unit of measurement used by chemists in the laboratory • 1 mole = 6.022 x 1023 – (Avogadro’s number represents the number of atoms that exist in exactly 12 grams of carbon-12) – This is our “counting number” for atoms, molecules and ions much like a dozen is our counting number for cookies or doughnuts)
- 19. Copyright McGraw-Hill 2009 19 The Mole 2 H2(g) + O2(g) 2 H2O(l) 2 molecules H2(g) + 1 molecule O2(g) 2 molecules H2O(l) 2 moles H2(g) + 1 mole O2(g) 2 moles H2O(l) This relationship can be made because of Avogadro’s number (NA)
- 20. Copyright McGraw-Hill 2009 20 Moles and Atoms • Calculate the number of atoms found in 4.50 moles of silicon. • How many moles of silicon are in 2.45 x 1045 atoms?
- 21. Copyright McGraw-Hill 2009 21 Molar Mass • Molar mass - the mass of one mole of a substance in grams • Carbon = 12.0 grams/mole • Sodium = 22.9 grams/mole • What is the relationship between molar mass and atomic mass?
- 22. Copyright McGraw-Hill 2009 22 Molar Mass • What is molar mass for each of the following? Copper metal = Helium gas = Calcium metal =
- 23. Copyright McGraw-Hill 2009 23 Molar Mass for Compounds Calculate the molar mass for each of the following: H2O H 2 x 1.01 g/mol = 2.02 O 1 x 16.00 g/mol = 16.00 Molar mass = 18.02 g/mol
- 24. Copyright McGraw-Hill 2009 24 Your Turn! Calculate the molar mass for each of the following: Carbon dioxide Ammonia Oxygen gas (Don’t forget the diatomics!)
- 25. Copyright McGraw-Hill 2009 25 Conversions between grams, moles and atoms
- 26. Copyright McGraw-Hill 2009 26 Interconverting mass, moles and number of particles Determine the number of moles in 85.00 grams of sodium chlorate, NaClO3 1 mole NaClO 3 85.00 g NaClO 0.7986 mol NaClO 3 3 106.44 g NaClO 3
- 27. Copyright McGraw-Hill 2009 27 Another Determine the number of molecules in 4.6 moles of ethanol, C2H5OH. (1 mole = 6.022 x 1023) . 23 6.02 10 molecules C H OH 24 2 5 4.6 mol C H OH 2 8 10 molecules 2 5 1mol C H OH 2 5
- 28. Copyright McGraw-Hill 2009 28 Another • Determine how many H atoms are in 4.6 moles of ethanol. – Begin with the answer to the last problem 2.8 1024 molecules C 2 H 5 OH 6 H atoms 1 molecule C 2 H 5 OH 1.7 1025 atoms H
- 29. Copyright McGraw-Hill 2009 29 Your Turn Solve the following conversions How many atoms of silver are in 3.50 moles of silver? Determine the number of moles of carbon disulfide in 34.75 grams of CS2. Determine the number of sulfur atoms in 34.75 grams of CS2.
- 30. Copyright McGraw-Hill 2009 30 Another • How many grams of oxygen are present in 5.75 moles of aluminum oxide, Al2O3? Strategy:
- 31. Copyright McGraw-Hill 2009 31 Challenge Determine the number of fluorine atoms in 24.24 grams of sulfur hexafluoride. (hint: make a plan first!)
- 32. Copyright McGraw-Hill 2009 32 Empirical and Molecular Formulas • Empirical- simplest whole-number ratio of atoms in a formula • Molecular - the “true” ratio of atoms in a formula; often a whole-number multiple of the empirical formula • We can determine empirical formulas from % composition data; a good analysis tool.
- 33. Copyright McGraw-Hill 2009 33 Empirical Formulas • Steps for success – Convert given amounts to moles – Mole ratio (divide all moles by the smallest number of moles) – The numbers represent subscripts. • If the numbers are not whole numbers, multiply by some factor to make them whole.
- 34. Copyright McGraw-Hill 2009 34 Empirical Formula • Determine the empirical formula for a substance that is determined to be 85.63% carbon and 14.37% hydrogen by mass.
- 35. Copyright McGraw-Hill 2009 35 3.5 Combustion Analysis • Analysis of organic compounds (C,H and sometimes O) are carried using an apparatus like the one below
- 36. Copyright McGraw-Hill 2009 36 Combustion Analysis • The data given allows an empirical formula determination with just a few more steps. • The mass of products (carbon dioxide and water) will be known so we work our way back.
- 37. Copyright McGraw-Hill 2009 37 Combustion Analysis Suppose that 18.8 grams of glucose was burned in a combustion train resulting in 27.6 grams of carbon dioxide and 11.3 grams of water. Calculate the empirical and molecular formula of glucose. Molar mass = 180 g/mol (Assumptions: all C in CO2 originates from glucose; all H in H2O originates from glucose; O is found by difference)
- 38. Copyright McGraw-Hill 2009 38 Combustion Analysis Steps: Convert 27.6 g CO2 into g of C Convert 11.3 g H2O into g H Calculate g O = g sample - (g C + g H) Find empirical formula as before (g to moles, mole ratio)
- 39. Copyright McGraw-Hill 2009 39 Combustion Analysis Molecular formula = molecular mass/empirical mass = 180/30 = 6 Multiply through empirical formula to obtain new subscripts Molecular formula = C6H12O6
- 40. Copyright McGraw-Hill 2009 40 3.6 Calculations with Balanced Chemical Equations • Balanced equations allow chemists and chemistry students to calculate various amounts of reactants and products. • The coefficients in the equation are used as mole ratios.
- 41. Copyright McGraw-Hill 2009 41 Stoichiometry • Stoichiometry- using balanced equations to find amounts • How do the amounts compare in the reaction below?
- 42. Copyright McGraw-Hill 2009 42 Mole Ratios • Many mole ratios can be written from the equation for the synthesis of urea • Mole ratios are used as conversion factors 2molNH 3 1mol CO 2 or 1 molCO 2 2 mol NH 3
- 43. Copyright McGraw-Hill 2009 43 Calculations with Balanced Equations • How many moles of urea could be formed from 3.5 moles of ammonia? 2NH3(g)+ CO2(g) (NH2)2CO(aq)+ H2O(l) 3.5 mol NH3 1 mol (NH2 )2CO 2 mol NH3 = 1.8 mol (NH2 )2CO
- 44. Copyright McGraw-Hill 2009 44 Mass to Mass A chemist needs 58.75 grams of urea, how many grams of ammonia are needed to produce this amount? Strategy: Grams moles mole ratio grams 58.75 g (NH 2 ) 2 CO 1 mol(NH 2 ) 2 CO 58.06 g(NH 2 ) 2 CO 2 mol NH 3 1 mol (NH 2 ) 2 CO 17.04 g NH 3 1mol NH 3 = 34.49 g NH 3
- 45. Copyright McGraw-Hill 2009 45 Correction to the publisher’s prevous slide: MW urea 60.06 g/mol So mass is 33.34g.
- 46. Copyright McGraw-Hill 2009 46 You Try! How many grams of carbon dioxide are needed to produce 125 grams of urea?
- 47. Copyright McGraw-Hill 2009 47 3.7 Limiting Reactants • Limiting reactant - the reactant that is used up first in a reaction (limits the amount of product produced) • Excess reactant - the one that is left over – Industry often makes the more expensive reactant the limiting one to ensure its complete conversion into products
- 48. Copyright McGraw-Hill 2009 48 Limiting Reactant • If one loaf of bread contains 16 slices of bread and a package of lunchmeat contains 10 slices of turkey, how many sandwiches can be made with 2 pieces of bread and one slice of meat? • Which is the limiting reactant? How much excess reactant is left?
- 49. Copyright McGraw-Hill 2009 49 Limiting Reactant • How do you identify a limiting reactant problem? Example: If 5.0 moles of hydrogen react with 5.0 moles of oxygen, how many moles of water can be produced? Notice: both reactant amounts are given and a product amount is requested
- 50. Copyright McGraw-Hill 2009 50 Steps for Success • Step 1: write a balanced equation • Step 2: identify the limiting reactant – Must compare in terms of moles • Step 3: use a mole ratio to desired substance • Step 4: convert to desired units
- 51. Copyright McGraw-Hill 2009 51 Limiting Reactant How many molecules of water are formed when 7.50 grams of hydrogen gas react with 5.00 grams of oxygen gas? Step 1: 2H2(g) + O2(g) 2 H2O(l) Step 2: 7.50 g H2 /2.02 g/mol = 3.712 mol 5.00 g O2 / 32.00 g/mol = 0.1562 mol
- 52. Copyright McGraw-Hill 2009 52 Limiting Reactant Step 2 continued: Decide which is limiting - look at the mole ratio of reactants--it takes twice as much H2 as O2 so O2 limits in this case. Step 3 and step 4: 0.1562 molO 2 2 molH 2 O 1 molO 2 6.02 1023 molecules H 2 O 1 molH 2 O 1.88 1023 molecules H 2 O
- 53. Copyright McGraw-Hill 2009 53 Limiting Reactant • In the previous example, how many grams of hydrogen were left in excess? Step 1: how much H2 is used 0.1562 molO 2 2 molH 2 1 molO 2 2.02 g H 2 1 molH 2 0.63104 g H 2 used
- 54. Copyright McGraw-Hill 2009 54 Limiting Reactant • In the previous example, how many grams of hydrogen were left in excess? Step 2: initial H2 - used H2 7.50 g - 0.63 g = 6.87 g excess
- 55. Copyright McGraw-Hill 2009 55 Your Turn! • When 35.50 grams of nitrogen react with 25.75 grams of hydrogen, how many grams of ammonia are produced? • How many grams of excess reagent remain in the reaction vessel?
- 56. Copyright McGraw-Hill 2009 56 Reaction Yield • Theoretical yield: the maximum amount of product predicted by stoichiometry • Actual yield: the amount produced in a laboratory setting • Percent yield: a ratio of actual to theoretical (tells efficiency of reaction)
- 57. Copyright McGraw-Hill 2009 57 Percent Yield When a student reacted 3.75 grams of zinc with excess hydrochloric acid, 1.58 grams of zinc chloride were collected. What is the percent yield for this reaction? % yield = actual theoretical 100
- 58. Copyright McGraw-Hill 2009 58 Percent Yield • Step 1: Balanced equation • Step 2: Calculate theoretical yield • Step 3: Substitute into formula and solve
- 59. Copyright McGraw-Hill 2009 59 Percent Yield Zn(s) + 2 HCl(aq) ZnCl2(aq) + H2(g) Theoretical yield = 7.82 g ZnCl2 Actual yield = 1.58 g ZnCl2 Calculate % yield: (1.58g/7.82g)x100% = 20.2%
- 60. Copyright McGraw-Hill 2009 60 A Few Reaction Types • Combination: one product is formed • Decomposition: one reactant produces more than one product • Combustion: a hydrocarbon reacts with oxygen to produce carbon dioxide and water
- 61. Copyright McGraw-Hill 2009 61 Combination Reaction General formula: A + B AB Sodium + chlorine sodium chloride 2Na + Cl2 2 NaCl Sulfur dioxide + water sulfurous acid SO2 + H2O H2SO3
- 62. Copyright McGraw-Hill 2009 62 Decomposition Reaction General formula: AB A + B Copper (II) carbonate decomposes with heat into copper (II) oxide and carbon dioxide CuCO3 CuO + CO2 Potassium bromide decomposes into its elements 2KBr 2K + Br2
- 63. Copyright McGraw-Hill 2009 63 Combustion (hydrocarbons) General formula: CxHy + O2 CO2 + H2O Methane gas burns completely CH4 + 2O2 CO2 + 2H2O Butane liquid in a lighter ignites 2C4H10 + 13O2 8CO2 + 10H2O
- 64. Copyright McGraw-Hill 2009 64 Review • Molecular mass • Percent composition • Chemical equations – Reactants – Products – State symbols – Balancing
- 65. Copyright McGraw-Hill 2009 65 Review continued • Mole concept and conversions • Empirical and molecular formulas – Combustion analysis • Stoichiometry • Limiting reactant • % yield • Types of reactions