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Chapter 3 Stoichiometry: Mole-Mass- Number Relationships in Chemical Systems

Chapter 3 Stoichiometry: Mole-Mass- Number Relationships in Chemical Systems

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Chapter 3 Stoichiometry: Mole-Mass- Number Relationships in Chemical Systems

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  1. Chapter 3Stoichiometry: Mole-Mass-Number Relationships inChemical Systems Read/Study:Chapter 3 Memorization:Work on 3 x 5 Cards ChemSkill Builder:Chapters 4, 5, & 6

  2. Atomic Masses: Definition - The relative mass of an atom usually expressed in relative atomic mass units ( amu or u). Atomic Mass Unit - A unit of mass that is equivalent to 1/12 of the mass of a single carbon-12 (12C) atom. A single carbon-12 atom has a mass of EXACTLY 12 u. A single oxygen-16 atom has a mass that is 1.332 9096 times that of a carbon-12 atom. 6 (12 u)(1.332 9096) = 15.994 915 u

  3. The Atomic Masses of Elements: The atomic masses shown on the Periodic Table are average masses of naturally occurring samples of the elements. These samples consist of mixtures of isotopes with different atomic masses. Isotope Mass (u) Fractional Atomic Mass Abundance (u) 16O 15.994 915 x 0.997 587 = 15.956 31 17O 16.999 133 x 0.000 374 = 0.006 358 18O 17.999 16 x 0.002 039 = 0.036 700 Weighted Average = 15.9994 u

  4. Mass Spectrum for Lead (Pb) 100 75 % of Most Abundant Isotope 50 25 0 203 205 207 209 Mass Number

  5. Practice Problem: Given the following data from the mass spectrometer, calculate the average atomic mass for the element, silicon (Si). 28Si - 92.23 % (Atomic mass - 27.976 93 u) 29Si - 4.67 % (Atomic mass - 28.976 49 u) 30Si - 3.10 % (Atomic mass - 29.973 76 u) (27.976 93 u)(92.23 u) + (28.976 49 u)(4.67 u) (100 u) (100 u) + (29.973 76 u)(3.10 u) = (100 u) 28.09 u

  6. Formula Masses: The relative mass of a compound is equal to the sum of the atomic masses of the elements forming the compound. Calculate the formula masses for NaCl, H2O, and H3PO4 - NaCl H2O 1 Na @ 22.989 77 u = 22.989 77 u 1 Cl @ 35.453 u = 35.453 u 1 NaCl = 58.443 u 2 H @ 1.007 94 u = 2.015 88 u 1 O @ 15.9994 u = 15.9994 u 1 H2O = 18.0153 u H3PO4 3 H @ 1.007 94 u = 3.023 82 u 1 P @ 30.973 76 u = 30.973 76 u 4 O @ 15.9994 u = 63.9976 u 1 H3PO4 = 97.9952 u

  7. Pair: A group containing 2 items or objects. Dozen: A group containing 12 items or objects. Gross: A group containing 144 items or objects. Mole: A group containing 6.0221 x 1023 items or objects. One Mole = the amount of substance (element or compound) that contains the same number of particles as there are atoms in EXACTLY 12 g of 12C. 6.0221 x 1023 = Avogadro’s Number

  8. “When in doubt, calculate moles!!” # of Particles Moles Avogadro’s Number How many atoms of carbon are in 2.6 moles of C? (2.6 mol C) (6.0221 x 1023 atom C) mol C = 1.6 x 1024 atom C How many moles of water are present when 2.673 x 1025 molecules of water are present? (2.673 x 1025 molecules H2O) (1 mol H2O) (6.0221 x 1023 molecules H2O) = 44.39 mol H2O

  9. Molar Mass: The mass in grams of one mole of any substance. The molar mass is numerically equal to the formula mass. Substance Formula Molar Mass Mass 2.015 88 u 2.015 88 g 55.847 u 55. 847 g 97.9952u 97.9952 g 80.06 u 80.06 g 58.443 u 58.443 g H2 Fe H3PO4 SO3 NaCl

  10. “When in doubt, calculate moles!!” # of Particles Moles Mass Avogadro’s Number Molar Mass How many moles of carbon are in 26.78 g of C? (26.78 g C) (1 mol C) (12.011 g C) = 2.230 mol C How many grams of water are in 32.6 moles of water? (32.6 mol H2O) (18.015 g H2O) (1 mol H2O) = 587 g H2O

  11. How many atoms of hydrogen are present in 36.5 g of water? “When in doubt, calculate moles!!” (36.5 g H2O)(1 mol H2O) (2 mol H) (6.0221 x 1023 atom H) (18.015 g H2O) (mol H2O) (mol H) = 2.44 x 1024 atom H How many grams of Au atoms are present in 1.45 x 1020 Au atoms? (1.45 x 1020 atom Au)(1 mol Au)(196.967 g Au) (6.0221 x 1023 atom Au) (mol Au) = 0.0474 g Au

  12. How many moles of uranium are there in 1.23 kg U? Molar Mass of U = 238.019 g/mol (1.23 kg U) (103 g U) (1 kg U) (1 mol U) (238.019 g U) = 5.17 mol U How many atoms of sulfur are there in 7.62 g S? (7.62 g S)(1 mol S/32.066 g S)(6.0221 x 1023 atom S/mol S) = 1.43 x 1023 atom S What is the mass of a single Pt atom? (1 Pt atom)(1 mol Pt/6.0221 x 1023 atom Pt) (195.08 g Pt) (mol Pt) = 3.239 x 10-22 g Pt

  13. “When in doubt, calculate moles!!” # of Particles Moles Mass Avogadro’s Number Molar Mass This allows you to do all types of calculations involving the conversion of particles to mass, mass to particles, mass to moles, moles to mass, particles to moles, or moles to particles. If the world contains 5.5 x 109 people, how many moles of people are on earth? (5.5 x 109 people)(1 mol people) (6.0221 x 1023 people) = 9.1 x 10 -15 mol people

  14. If the earth has a radius of 4 x 103 miles what is its surface area in ft2? If each person was allowed one ft2 of space on earth, how many moles of people would fit on the earth’s surface (counting the oceans)? Assume that the earth is a perfect sphere. Area of Sphere = 4pr2 1 mile = 5280 ft 4p(4000 mi)(4000 mi)(5280ft)(5280 ft) mi mi = 5.6 x 1015 ft2 (5.6 x 1015ft2) (1 person) ft2 (1 mol people) (6.0221 x 1023 persons) = 9 x 10-9 mol people = 0.000 000 0009 mol people

  15. Compound Formulas Empirical Formula - The formula of a compound that gives the simplest whole number ratio of the atoms in the compound. Molecular Formula -The formula that gives the actual number of atoms of each element in a molecule of a molecular compound. Compound Empirical Molecular Formula Formula Benzene CH C6H6 Water H2O H2O n-Hexane C3H7 C6H14 Hydrogen Peroxide HO H2O2 Sodium Chloride NaCl ----------

  16. Structural Formula - The formula of a molecular com- pound that shows the way the atoms in the molecules are joined together. O H H H H - C - H water H methane H2O CH4 H H H - C - C - O - H H H ethyl alcohol (ethanol) H H H - C - O -C - H H H dimethyl ether (CH3CH2OH) (CH3OCH3) C2H6O C2H6O “Condensed Structural Formulas”

  17. 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

  18. Rules for Writing Chemical Equations - • 1. Identify reactants and products and write a “word • equation. Aqueous potassium + Aqueous lead(II) chloride nitrate Solid lead(II) + Aqueous potassium chloride nitrate • 2. Write symbols and formulas for the elements and • compounds KCl + Pb(NO3)2 PbCl2 + KNO3

  19. 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. KCl + Pb(NO3)2 PbCl2 + KNO3 2 2 • 4. Check to see that the same number of each kind of • atom is shown on both sides. 2 KCl + Pb(NO3)2 PbCl2 + 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 KCl (aq) + Pb(NO3)2(aq) PbCl2(s) + 2 KNO3 (aq)

  20. 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)

  21. 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)

  22. 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 Combination Decomposition Other Types Two or more elements chemically combine to form one compound. A + B C N2 (g) + H2 (g) NH3 (g) 3 2

  23. Introduction to the Classification of Reactions: Reactions Combustion Combination 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)

  24. 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

  25. 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 = =

  26. 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

  27. 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.

  28. Composition Stoichiometry Percent Composition: The percent by mass of each element in a compound. It can be determined from the compound’s formula. Calculate the percent composition of ethyl alcohol (ethanol), C2H6O or CH3CH2OH. 1. Calculate the molar mass of C2H6O: 2 C @ 12.011 u = 24.022 u 6 H @ 1.007 94 u = 6.047 64 u 1 O @ 15.9994 u = 15.9994 u 1 C2H6O = 46.069 u

  29. 2. Calculate the percent by mass of each element • in the compound: (24.022 g C) (46.069 g ethanol) (100%) = 52.144 % C (6.047 64 g H) (46.069 g ethanol) (100%) = 13.127 % H (15.9994 g) (46.069 g ethanol) (100%) = 34.729 % O Total = 100.000 %

  30. Empirical Formulas from Percent Composition: What is the empirical formula for a compound with the following percent composition? C - 64.80 % H - 13.60 % O - 21.60 % Assume a 100.00 g sample! “When in doubt, calculate moles” (64.80 g C)(1 mol C/12.011 g C) = 5.395 mol C (13.60 g H)(1 mol H/1.007 94 g H) = 13.49 mol H (21.60 g O)(1 mol O/15.9994 g O) = 1.350 mol O C5.395H13.49O1.350 Find the smallest whole number ratio of subscripts!

  31. C5.395H13.49O1.350 1.350 1.350 1.350 = C4H10O Molecular Formulas from Empirical Formulas: The molecular mass of C4H10O was found by experiment to be 148.25 u. What is its molecular formula? The Formula Mass of the compound is: 4 C @ 12.011 u = 48.044 u 10 H @ 1.007 94 u = 10.0794 u 1 O @ 15.9994 u = 15.9994 u 1 C4H10O = 74.1228 u n(74.1228 u) = 148.25 u  n = 2 2 (C4H10O)= C8H20O2

  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. Copper(II) oxide is a catalyst.

  34. 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

  35. (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.

  36. 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

  37. ( 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

  38. 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.

  39. 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)

  40. 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

  41. 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

  42. 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

  43. 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

  44. 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

  45. 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

  46. (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.

  47. 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!!

  48. 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

  49. 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

  50. 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.