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Chapter 4 Chemical Equations and Stoichiometry

Chapter 4 Chemical Equations and Stoichiometry. Read/Study: Chapter 4. Memorization: Work on 3 x 5 Cards OWL Assignments: See OWL Website !. Chemical Equations - The sentences of the chemical language. Reactants: The original or starting substances in a

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Chapter 4 Chemical Equations and Stoichiometry

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  1. Chapter 4Chemical Equationsand Stoichiometry Read/Study:Chapter 4 Memorization:Work on 3 x 5 Cards OWL Assignments: See OWL Website!

  2. Chemical Equations -The sentences of the chemical language. Reactants:The original or starting substances in a chemical reaction; normally written on the left side of the chemical equation. Products: The new or formed substances in a chemical reaction; normally written on the right side of the chem- ical equation. Cu (s) + S (s) CuS (s) Copper + Sulfur Copper(II) sulfide 2 Na (s) + H2O (l) NaOH (aq) + H2 (g) 2 2 Sodium + Water Sodium Hydroxide Hydrogen

  3. The “Meaning” of a Chemical Equation: 4 CO (g) + 3 O2 (g)  4 CO2 (g) + O2 4 molecules 3 molecules 4 molecules 1 molecule 4 moles 3 moles 4 moles 1 mole

  4. Rules for Writing Chemical Equations - • 1. Identify reactants and products and write a “word • equation. Aqueous potassium + Aqueous lead(II) iodide nitrate Solid lead(II) + Aqueous potassium Iodide nitrate • 2. Write symbols and formulas for the elements and • compounds KI + Pb(NO3)2 PbI2 + KNO3

  5. 3. Balance by changing the coefficients in front of the • symbols and formulas. Do NOT change subscripts • in the formulas or add or remove substances. KI + Pb(NO3)2 PbI2 + KNO3 2 2 • 4. Check to see that the same number of each kind of • atom is shown on both sides. 2 KI + Pb(NO3)2 PbI2 + 2 KNO3 • 5. Add symbols showing the physical state of each sub- • stance; solid (s), liquid (l), gas (g) or aqueous solu- • tion (aq). 2 KI (aq) + Pb(NO3)2(aq) PbI2(s) + 2 KNO3 (aq)

  6. Practice Problem: Write the balanced chemical equation for the reaction of hot copper metal with oxygen to form solid copper(II) oxide. Heat 1. Hot copper metal + oxygen gas Solid copper(II) oxide Heat 2. Cu + O2 CuO Heat 3. Cu + O2 CuO 2 2 4. 2 copper atoms and 2 oxygen atoms on each side!! Heat 5. 2 Cu (s) + O2 (g)2CuO (s)

  7. Practice Problem: Burn liquid heptane, C7H16 , in oxygen gas to form carbon dioxide gas and liquid water. 1. C7H16 + O2 CO2 + H2O 2. C7H16 + O2 CO2 + H2O 11 7 8 3. 7 Carbons, 16 hydrogens, and 22 oxygens on each side of the equation. 4. C7H16 (l) + 11 O2 (g) 7 CO2 (g) + 8 H2O (l)

  8. Introduction to the Classification of Reactions: The last reaction was a “combustion reaction”. This type of reaction normally involves a rapid reaction of some substance (an element or compound) with oxygen. Reactions Combustion Formation Decomposition Other Types Two or more elements chemically combine to form one compound. A + B C N2 (g) + H2 (g) NH3 (g) 3 2

  9. Introduction to the Classification of Reactions: Reactions Combustion Formation Decomposition A single compound breaks apart to form two or more simpler substances; either simpler compounds or elements. C A + B CaCO3 (s) CaO (s) + CO2 (g) 2 H2O (l) 2 H2 (g) + O2 (g)

  10. Important Laws that Govern Chemical Reactions The Law of Conservation of Mass - There is no measurable change in total mass during a chemical reaction. “Mass is neither created nor destroyed” electrolysis 2 H2O (l) 2 H2 (g) + O2 (g) 88.8 g 11.2 g 100.0 g Cu (s) + S (s) CuS (s) D 63.55 g 32.07 g 95.62 g

  11. The Law of Constant Composition - The elemental composition by mass of a given compound is the same for ALL samples of that compound. The Law of Definite Proportions - When elements combine to form compounds, they do so in definite proportions by mass. Na (s) + Cl2 (g) NaCl (s) 39.3 g 60.7 g 100.0 g 23.0 g 35.5 g 58.5 g 39.3 g 60.7 g 23.0g 35.5 g 0.647 = =

  12. The Law of Multiple Proportions - In different com- pounds containing the same elements, the masses of one element combined with a fixed mass of the other element are in the ratio of small whole numbers. H O O/H 11.2 g 178 g 15.9 11.2 g 88.8 g 7.93 H2O2 H2O = 2 C O O/C 42.9 g 114 g 2.66 42.9 g 57.1 g 1.33 CO2 CO = 2

  13. Summary: The Law of Conservation of Mass The Law of Constant Composition The Law of Definite Proportions The Law of Multiple Proportions These laws that govern chemical reactions and com- pound composition allow us to do stoichiometric calculations. They were all well known by 1803 when John Dalton put forth his atomic theory of matter. This theory was able to “explain” why these laws that govern chemical reactions were true.

  14. Interpreting Chemical Equations: 2 Al (s) + Fe2O3 (s) Al2O3 (l) + 2 Fe (l) D 2 atoms 1 formula 1 formula 2 atoms unit unit 2 moles 1 mole 1 mole 2 moles 53.964 g 159.69 g 101.96 g 111.69 g 26.982 g 79.845 g 50.980 g 55.845 g Chemical equations express the quantitative relation- ships among the reactants and products in a chemical reaction.

  15. The “Meaning” of a Chemical Equation: 4 CO (g) + 3 O2 (g)  4 CO2 (g) + O2 4 molecules 3 molecules 4 molecules 1 molecule 4 moles 3 moles 4 moles 1 mole

  16. 2 H2 (g) + O2 (g) 2 H2O (l) 2 molecules 1 molecule 2 molecules 2 moles 1 mole 2 moles 4.0318 g 31.9988 g 36.0306 g Mass Relationships in Chemical Reactions: Calculate the number of grams of water produced when 58.63 grams of propane are burned. • 1. Write the balanced chemical equation and identify • the known and unknown quantities: 58.63 g ? g Problem: Equation: C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

  17. 2. Convert the known quantities to moles: (“When in doubt, calculate moles”) Problem: Equation: Formula Masses, u: 58.63 g ? g C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l) 44.097 18.0153 3 C @ 12.011 u = 36.033 u 8 H @ 1.00794 u = 8.063 52 u 1 C3H8 = 44.097 u 2 H @ 1.00794 u = 2.015 88 u 1 O @ 15.9994 u = 15.9994 u 1 H2O = 18.0153 u (58.63 g C3H8) (1 mol C3H8) (44.097 g C3H8) = 1.3296 mol C3H8

  18. 3. Use the mole ratio in the equation to find the moles • of the unknown material: Problem: Equation: Formula Masses, u: 58.63 g ? g C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l) 44.097 18.0153 moles of known moles of unknown (4 mol H2O) (1 mol C3H8) = 5.3184 mol H2O (1.3296 mol C3H8) • 4. Convert moles of the unknown material to the units • called for in the problem: (18.0153 g H2O) (1 mol H2O) (5.3184 mol H2O) = 95.81 g H2O

  19. Summary: Problem: Equation: Formula Masses, u: 58.63 g ? g C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l) 44.097 18.0153 moles of unknown moles of known (1 mol C3H8) (44.097 g C3H8) (4 mol H2O) (1 mol C3H8) (18.0153 g H2O) (1 mol H2O) (58.63 g C3H8) = 95.81 g H2O

  20. How many grams of Li3N (s) are formed when 73.6 g of Li (s) react with an excess of N2 (g)? Problem: Equation: Formula Masses, u: 73.6 g ? g 6 Li (s) + N2 (g) 2 Li3N (s) 34.830 6.941 moles of known moles of unknown (34.830 g Li3N) (mol Li3N) (1 mol Li) (6.941 g Li) (2 mol Li3N) (6 mol Li) (73.6 g Li) = 123 g Li3N

  21. Limiting Reactant (Limiting Reagent): The reactant in a chemical reaction that is completely used up. Excess Reactant (Excess Reagent): The reactant that is left over following a chemical reaction. 2 H2 (g) + O2 (g) 2 H2O (l) 11.2 g 110.0 g 100.0 g Excess Reactant Limiting Reactant When the hydrogen is all used up, the reaction HAS TO stop since no more of that reactant is available to react. 21.2 g left over 0.00 g left over

  22. Due to your enormous personal popularity, you have just been elected Social Chairperson of the class and are required to prepare hot dog meals for the class. You find, however, upon going to the grocery store that you have just enough money to buy 5 packages of hot dogs and 4 packages of buns. Each package of hot dogs contains 12 hot dogs but each package of buns contains 18 buns. If 62 people are in the class, will you have enough meals for everyone? Let 1 meal = 1 hot dog (Hd) + 1 Bun (Bu) 5 pkg 4 pkg ? meal Problem: Equation: Items/pkg: Hd (s) + Bu (s) Meal 12 18 1. Find the reactant that is limiting: (5 pkg Hd)(12 Hd/pkg Hd) = 60 Hd

  23. (4 pkg Bu)(18 Bu/pkg Bu) = 72 Bu Assume that Bu is the limiting reactant - (72 Bu)(1 Hd/1 Bu) = 72 Hd REQUIRED!! But Wait!! Only 60 Hd are AVAILABLE!! Thus, Hd, is, in fact, the limiting reactant. • 2. Use the limiting reactant to find the amount of • product that will be produced: (60 Hd)(1 meal/1 Hd) = 60 meals Since 60 < 62, there will NOT be enough meals to go around and your popularity will fall off.

  24. 3. If asked, calculate the amount of excess reactant left • over: 60 Bu REQUIRED! (60 Hd)(1 Bu/1 Hd) = However, 72 buns are actually available - 72 Bu available - 60 Bu required = 12 Bu in excess! Thus, 12 will have to be wasted and the Faculty Advisor will be upset with you for wasting money. Perhaps your days as the Social Chairperson are limited!!

  25. I2 (g) + 3 Cl2 (g)  2 ICl3 (g) Question 80 Pg. 172

  26. Question 80 Pg. 172

  27. Practice Exercise: A 100.00 g sample of Al (s) is heated with 100.00 g of O2 (g). How many grams of Al2O3 (s) will form? How many grams of the excess reactant will be left over? 1. Set up the information table: 100.00 g 100.00 g ? g Problem: Equation: Formula Masses, u: 4 Al (s) + O2 (g) Al2O3 (s) 2 3 26.9815 31.9988 101.9612 2. Find the limiting and excess reactants: (“When in doubt, calculate moles”) (100.00 g Al)(1 mol Al/26.9815 g Al) = 3.706 24 mol Al (100.00 g O2)(1 mol O2/31.9988 g O2) = 3.125 12mol O2

  28. Assume that O2 is the limiting reactant: 4.166 83 mol Al NEEDED (3.125 12mol O2)(4 mol Al/3 mol O2) = This is more than the Al (s) AVAILABLE! Therefore, Al is the limiting reactant and O2 is in excess. • 3. Use the limiting reactant (Al) to calculate the grams • of Al2O3 (s) that will be formed: (3.706 24 mol Al)(2 mol Al2O3/4 mol Al)(101.9612 g Al2O3/mol Al2O3) = 188.94 g Al2O3

  29. 4. Calculate the grams of excess reactant left over: (3.706 24 mol Al)(3 mol O2/4 mol Al)(31.9988 g O2/mol O2) = 88.9464 g O2 consumed 100.00 g O2 available - 88.9464 g O2 consumed = 11.05 g O2 excess Theoretical Yield - The amount of a product of a reaction calculated from the balanced chemical equation. Actual Yield - The amount of a product actually pro- duced in a chemical reaction.

  30. Percentage Yield - Actual Yield Theoretical Yield x 100% Practice Exercise: Copper metal is heated with an excess of sulfur to produce copper(I) sulfide. If the %-yield is 40.0%, how many grams of copper are required to produce an actual yield of 50.0 g of Cu2S? ? g xs 50.0 g @ 40.0%-yield Problem: Equation: Formula Masses, u: Cu (s) + S (s) Cu2S (s) 2 Theoretical Yield! 63.546 159.158 moles of known moles of unknown

  31. “When in doubt, calculate moles” AY = 50.0 g % - Y = 40.0% TY = ? AY x 100% %-Y AY TY x 100% = %-Y TY = TY = (50.0 g)(100%) (40.0 %) = 125.0 g Cu2S 0.7854 mol Cu2S (125.0 g Cu2S)(1 mol Cu2S/159.158 g Cu2S) = (0.7854 mol Cu2S)(2 mol Cu/1 mol Cu2S)(63.546 g Cu/mol Cu) = 99.8 g Cu

  32. The percent composition of a compound can also be determined by experiment. This type of experiment is known as quantitative analysis. An oxide of mercury (Hg) is decomposed to oxygen gas and mercury. If a 0.8349 g sample of the oxide gives 0.7732 g of Hg, find the percent by mass of Hg and oxygen in the compound. 0.7732 g Hg 0.8349 g sample 100 % = 92.61 % Hg 100 % total - 92.61 % Hg = 7.39 % O or (0.8349 g total - 0.7732 g Hg) (0.8349 g total) 100 % = 7.39 % O

  33. How Combustion Reactions are Done for Hydrocarbon Analyses: All of the Carbon atoms end up in CO2 and are, therefore, captured by the CO2 absorber. All of the hydrogen atoms end up in water. Therefore, they are captured by the water absorber.

  34. Images from your E-Textbook:

  35. Practice Problem: A hydrocarbon compound is analyzed by burning it with an excess of oxygen gas. In this analysis, 4.80 g of H2O were produced and 8.79 g CO2. What was the empirical formula of the hydrocarbon? • 1. First recognize that all of the carbon from the hydro- • carbon ends up in the CO2 and all of the hydrogen • ends up in the water. • 2. Calculate the moles of hydrogen in water and the • moles of carbon in CO2: (4.80 g H2O)(1 mol H2O/18.0153 g H2O)(2 mol H/mol H2O) = 0.5329 mol H

  36. (8.79 g CO2)(1 mol CO2/44.010 g CO2)(1 mol C/mol CO2) = 0.1997 mol C 3. Calculate the empirical formula of the hydrocarbon: C0.1997H0.5329 = (CH2.67) x 3 = C3H8 0.1997 0.1997 The initial result was a fraction. Since a molecule cannot have a partial atom, the formula must be multiplied by a factor that will make the subscripts whole numbers.

  37. A 0.018 69 g sample of an organic compound containing only C, H, and O gave 0.044 38 g of CO2 and 0.022 72 g of H2O on combustion. What is the percent composi- tion of the compound? First, recognize that ALL of the carbon in the compound ends up in CO2 and ALL of the hydrogen ends up in the water. 1 C @ 12.011 u = 12.011 u 2 O @ 15.9994 u = 31.9988 u 1 CO2 = 44.010 u (0.044 38 g CO2)(1 mol CO2) (1 mol C) (12.011 g C) (44.010 g CO2)(mol CO2) (1 mol C) = 0.012 112 g C

  38. ( 0.012 112 g C) (100%) (0.018 69 g sample) = 64.80 % C (0.022 72 g H2O)(1 mol H2O) (2 mol H) (1.007 94 g H) (18.015 g H2O) (mol H2O) (mol H) = 0.002 5424 g H (0.002 5424 g H) (100%) (0.018 69 g sample) = 13.60 % H 100.00 % total - (64.80 % C + 13.60 % H) = 21.60 % O

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