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Chapter 9

Chapter 9. Quantitative Relationships and Stoichiometry. Quantitative Relationships. The Mole . The mole is the SI unit for “amount of substance.”. Atoms are so small, it is impossible to count them by the dozens, thousands, or even millions.

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Chapter 9

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  1. Chapter 9 Quantitative Relationships and Stoichiometry

  2. Quantitative Relationships The Mole The mole is the SI unit for “amount of substance.” Atoms are so small, it is impossible to count them by the dozens, thousands, or even millions. To count atoms, we use the concept of the mole. 1 mole of atoms = 602,000,000,000,000,000,000,000 atoms 6.02 x 1023 That is, 1 mole of atoms = _________ atoms.

  3. Amadeo Avogadro(1776 – 1856) Particles in a Mole 1 mole = 602213673600000000000000 or 6.022 x 1023 Amedeo Avogadro (1766-1856) never knew his own number; it was named in his honor by a French scientist in 1909. its value was first estimated by Josef Loschmidt, an Austrian chemistry teacher, in 1895. ? quadrillions thousands trillions billions millions There is Avogadro's number of particles in a mole of any substance.

  4. Careers in Chemistry - Philosopher Q: How much is a mole? A: A mole is a quantity used by chemists to count atoms and molecules. A mole of something is equal to 6.02 x 1023 “somethings.” 1 mole = 602 200 000 000 000 000 000 000 Q: Can you give me an example to put that number in perspective? A: A computer that can count 10,000,000 atoms per second would take 2,000,000,000 years to count 1 mole of a substance.

  5. Counting to 1 Mole Is that right? A computer counting 10 million atoms every second would need to count for 2 billion years to count just a single mole. Lets look at the mathematics. 365 days 24 hours 60 min 60 sec x sec = 1 year = 31,536,000 sec 1 year 1 day 1 hour 1 min Therefore 1 year has 31,536,000 seconds or 3.1536 x 107 sec. A computer counting 10,000,000 atoms every second could count 3.153 x 1014 atoms every year. Finally, 6.02 x 1023 atoms divided by 3.1536 x 1014 atoms every year equals 1,908,929,477 years or approximately 2 billion years!

  6. How Big is a Mole? …about the size of a chipmunk, weighing about 5 oz. (140 g), and having a length of about 7 inches (18 cm). BIG. I Meant, “How Big is 6.02 x 1023?” 6.02 x 1023 marbles would cover the entire Earth (including the oceans) …to a depth of 2 miles. 6.02 x 1023 $1 bills stacked face-to-face would stretch from the Sun to Pluto …7.5 million times. …and back It takes light 9,500 years to travel that far.

  7. 2 He 4.003 10 Ne 20.180 18 Ar 39.948 36 Kr 83.80 54 Xe 131.29 86 Rn (222) For any element on the Periodic Table, one mole of that element (i.e., 6.02 x 1023 atoms of that element) has a mass in grams equal to the mass number on the Table for that element. 1 mole of (i.e., 6.02 x 1023) helium atoms has a mass of 4.0 grams. 1 mol Ne = 20.2 g 1 mol Ar = 39.9 g 1 mol Kr = 83.8 g 1 mol Xe = 131.3 g 1 mol Rn = 222 g

  8. Mass (g) 1 mol = molar mass (in g) MOLE (mol) Particle (atoms) 1 mol = 6.02 x 1023 particles Island Diagram

  9. ( ( ) ) • Island Diagram Problems 1. How many moles is 3.79 x 1025 atoms of zinc? 1 mol 3.79 x 1025 at. = 63.0 mol Zn 6.02 x 1023 at. 2. How many atoms is 0.68 moles of zinc? 6.02 x 1023 at. 0.68 mol. = 4.1 x 1023 at. Zn 1 mol

  10. ( ) ( ( ) ) ( ( ) ) 3. How many grams is 5.69 moles of uranium? 238.0 g 5.69 mol = 1,354 g U 1 mol = 1.35 x 103 g U 4. How many grams is 2.65 x 1023 atoms of neon? 20.2 g 1 mol 2.65 x 1023 at. 1 mol 6.02 x 1023 at. = 8.9 g Ne 5. How many atoms is 421 g of promethium? 1 mol 6.02 x 1023 at. 421 g 1 mol 145 g = 1.75 x 1024 at. Pm

  11. 4 Na(s) + O2(g) 2 Na2O(s) Quantitative Relationships in Chemical Equations Coefficients of a balanced equation represent # of particles OR # of moles, but NOT # of grams. **

  12. Worksheet time • Then quiz time!

  13. Stoichiometry -- involves finding amts. of reactants & products in a reaction

  14. For generic equation: RA + RBP1+ P2 …one can find the… Given the… What can we do with stoichiometry? amount of RB (or RA) that is needed to react with it amount of RA (or RB) amount of P1 or P2 that will be produced amount of RA or RB amount of P1 or P2 you need to produce amount of RA and/or RB you must use

  15. Governing Equation: 2 patties + 3 bread 1 Big Mac® excess + 18 bread ? ? + ? 25 Big Macs® 4 patties + ? 6 bread 6 Big Macs® 50 patties 75 bread

  16. SUBSTANCE “A” SUBSTANCE “B” 1 mol = molar mass (in g) 1 mol = molar mass (in g) Use coefficients from balanced equation Mass (g) Mass (g) 1 mol = 22.4 L 1 mol = 22.4 L MOLE (mol) MOLE (mol) Volume (L or dm3) Volume (L or dm3) 1 mol = 22.4 dm3 1 mol = 22.4 dm3 Particle (at. or m’c) Particle (at. or m’c) 1 mol = 6.02 x 1023 particles 1 mol = 6.02 x 1023 particles Stoichiometry Island Diagram

  17. Molecular Weight and Molar Mass Molecular weight is the sum of atomic weights of all atoms in the molecule. example: NaCl has a molecular weight of 58.5 a.m.u. this is composed of a single molecule of NaCl Molar mass = molecular weight in grams. example: NaCl has a molar mass of 58.5 grams this is composed of a 6.02 x1023 molecules of NaCl

  18. The Molar Mass and Number of Particles in One-Mole Quantities Substance Molar Mass Number of Particles in One Mole Carbon (C) 12.0 g 6.02 x 1023 C atoms Sodium (Na) 23.0 g 6.02 x 1023 Na atoms Iron (Fe) 55.9 g 6.02 x 1023 Fe atoms NaF (preventative 42.0 g 6.02 x 1023 NaF formula units for dental cavities) CaCO3 (antacid) 100.1 g 6.02 x 1023 CaCO3 formula units C6H12O6 (glucose) 180.0 g 6.02 x 1023 glucose molecules C8H10N4O2 (caffeine) 194.0 g 6.02 x 1023 caffeine molecules

  19. mass mass vol. vol. MOL MOL part. part. 4 Na(s) + 1 O2(g) 2Na2O(s) When going from moles of one substance to moles of another, use coefficients from balanced equation. Use coeff. from balanced equation in crossing this bridge SUBSTANCE “B” SUBSTANCE “A” (unknown) (known)

  20. 4 Na(s) + O2(g) 2 Na2O(s) ( ) ( ( ) ) Na2O Na Na2O Na O2 How many moles oxygen will react with 16.8 moles sodium? 1 mol O2 16.8 mol Na = 4.2 mol O2 4 mol Na How many moles sodium oxide are produced from 87.2 moles sodium? 2 mol Na2O 87.2 mol Na = 43.6 mol Na2O 4 mol Na How many moles sodium are required to produce 0.736 moles sodium oxide? 4 mol Na 0.736 mol Na2O = 1.472 mol Na 2 mol Na2O

  21. ( ) ( ) ( ) 2 4 3 2 1 2 __TiO2 + __Cl2 + __C __TiCl4 + __CO2 + __CO How many mol chlorine will react with 4.55 mol carbon? 4 mol Cl2 4.55 mol C = 6.07 mol Cl2 3 mol C C Cl2 What mass titanium (IV) oxide will react with 4.55 mol carbon? 79.9 g TiO2 2 mol TiO2 4.55 mol C 1 mol TiO2 3 mol C C TiO2 = 242 g TiO2

  22. ( ) ( ( ) ) How many molecules titanium (IV) chloride can be made from 115 g titanium (IV) oxide? 1 mol 1 mol coeff. 1 mol 1 mol 1 mol 1 mol TiO2 TiCl4 2 mol TiCl4 1 mol TiO2 6.02 x 1023 m’c TiCl4 115 g TiO2 79.9 g TiO2 2 mol TiO2 1 mol TiCl4 = 8.7 x 1023 m’c TiCl4 Island Diagram helpful reminders: 2. The middle bridge conversion factor is the only one that has two different substances in it. The conversion factors for the other six bridges have the same substance in both the numerator and denominator. 3. The units on the islands at each end of the bridge being crossed appear in the conversion factor for that bridge. 1. Use coefficients from the equation only when crossing the middle bridge. The other six bridges always have “1 mol” before a substance’s formula.

  23. 2 Ir + Ni3P23 Ni + 2 IrP ( ) ( ( ) ) ( ) ( ) ( ) If 5.33 x 1028 m’cules nickel (II) phosphide react w/excess iridium, what mass iridium (III) phosphide is produced? Ni3P2 IrP 1 mol Ni3P2 2 mol IrP 223.2 g IrP 5.33 x 1028 m’c Ni3P2 1 mol IrP 1 mol Ni3P2 6.02 x 1023 m’c Ni3P2 = 3.95 x 107 g IrP How many grams iridium will react with 465 grams nickel (II) phosphide? Ni3P2 Ir 1 mol Ni3P2 2 mol Ir 465 g Ni3P2 192.2 g Ir 1 mol Ir 1 mol Ni3P2 238.1 g Ni3P2 = 751 g Ir

  24. 2 Ir + Ni3P23 Ni + 2 IrP ( ) ( ) iridium (Ir) nickel (Ni) How many moles of nickel are produced if 8.7 x 1025 atoms of iridium are consumed? Ir Ni 8.7 x 1025 at. Ir 1 mol Ir 3 mol Ni 2 mol Ir 6.02 x 1023 at. Ir = 217 mol Ni

  25. ( ( ( ) ) ) What volume hydrogen gas is liberated (at STP) if 50 g zinc react w/excess hydrochloric acid (HCl)? Zn H2 1 2 1 1 __ Zn + __ HCl __ H2 + __ ZnCl2 50 g excess X L 1 mol Zn 1 mol H2 22.4 L H2 50 g Zn 1 mol H2 1 mol Zn 65.4 g Zn = 17.1 L H2

  26. ( ) ( ) ( ) ( ) 2 atoms O 1 m’c O2 At STP, how many m’cules oxygen react with 632 dm3 butane (C4H10)? C4H10 O2 1 4 5 2 13 8 10 __ CO2 + __ H2O __ C4H10 + __ O2 1 mol C4H10 13 mol O2 632 dm3 C4H10 6.02 x 1023 m’c O2 22.4 dm3 C4H10 1 mol O2 2 mol C4H10 = 1.10 x 1026 m’c O2 Suppose the question had been “how many ATOMS of O2…” 1.10 x 1026 m’c O2 = 2.20 x 1026 at. O

  27. CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) + 891 kJ ( ) ( ) E E Energy and Stoichiometry A balanced eq. gives the ratios of moles-to-moles AND moles-to-energy. CH4 How many kJ of energy are released when 54 g methane are burned? 1 mol CH4 891 kJ 54 g CH4 = 3007 kJ 16 g CH4 1 mol CH4

  28. CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) + 891 kJ ( ) ( ( ) ) E E E E H2O O2 ( ) At STP, what volume oxygen is consumed in producing 5430 kJ of energy? 2 mol O2 22.4 L O2 5430 kJ = 273 L O2 891 kJ 1 mol O2 What mass of water is made if 10,540 kJ are released? 2 mol H2O 18 g H2O 10,540 kJ = 426 g H2O 891 kJ 1 mol H2O

  29. 3 B + 2 M + EE B3M2EE The Limiting Reactant A balanced equation for making a Big Mac® might be: With… …and… …one can make… 30 M excess B and excess EE 15 B3M2EE 30 B excess M and excess EE 10 B3M2EE 30 M 30 B and excess EE 10 B3M2EE

  30. 3 W + 2 P + S + H + F W3P2SHF A balanced equation for making a tricycle might be: …and… With… …one can make… 50 P excess of all other reactants 25 W3P2SHF 50 S excess of all other reactants 50 W3P2SHF 50 P 50 S + excess of all other reactants 25 W3P2SHF

  31. 2 Al(s) + 3 Cl2(g) 2 AlCl3(s) ( ( ) ) ( ( ) ) ( ( ) ) Solid aluminum reacts w/chlorine gas to yield solid aluminum chloride. If 125 g aluminum react w/excess chlorine, how many g aluminum chloride are made? Al AlCl3 1 mol Al 2 mol AlCl3 133.5 g AlCl3 125 g Al 27 g Al 1 mol AlCl3 2 mol Al = 618 g AlCl3 If 125 g chlorine react w/excess aluminum, how many g aluminum chloride are made? Cl2 AlCl3 1 mol Cl2 2 mol AlCl3 133.5 g AlCl3 125 g Cl2 71 g Cl2 1 mol AlCl3 3 mol Cl2 = 157 g AlCl3

  32. 2 Al(s) + 3 Cl2(g) 2 AlCl3(s) If 125 g aluminum react w/125 g chlorine, how many g aluminum chloride are made? (We’re out of Cl2.) 157 g AlCl3 limiting reactant (LR): the reactant that runs out first -- amount of product is based on LR Any reactant you don’t run out of is an excess reactant (ER). In a root beer float, the LR is usually the ice cream. (What’s the product?) Your enjoyment!

  33. From Examples Above… Limiting Reactant Excess Reactant(s) Big Macs tricycles Al / Cl2 / AlCl3 With… …and… …one can make… 30 M 30 B and excess EE 10 B3M2EE 50 P 50 S + excess of all other reactants 25 W3P2SHF 125 g Al 125 g Cl2 157 g AlCl3 B M, EE P W, S, H, F Cl2 Al

  34. For the generic reaction RA + RBP, assume that the amounts of RA and RB are given. Should you use RA or RB in your calculations? How to Find the Limiting Reactant 1. Calc. # of mol of RA and RB you have. 2. Divide by the respective coefficients in balanced equation. 3. Reactant having the smaller result is the LR.

  35. Step #2 Step #3 . . Step #1 _ _ . . ( ( ) ) For the Al / Cl2 / AlCl3 example: 2 Al(s) + 3 Cl2(g) 2 AlCl3(s) 1 mol Al = 2.31 125 g Al = 4.63 mol Al (HAVE) 2 27 g Al 1 mol Cl2 LR 125 g Cl2 3 = 0.58 = 1.76 mol Cl2 (HAVE) 71 g Cl2 “Oh, bee- HAVE !” (And start every calc. with the LR.)

  36. 2 Fe(s) + 3 Cl2(g) 2 FeCl3(s) ( ) . . _ _ . . ( ) ( ( ) ) 223 g Fe 179 L Cl2 Which is the limiting reactant: Fe or Cl2? LR 1 mol Fe = 2.0 223 g Fe = 4.0 mol Fe (HAVE) 2 55.8 g Fe = Fe 1 mol Cl2 179 L Cl2 3 = 2.66 = 8.0 mol Cl2 (HAVE) 22.4 L Cl2 How many g FeCl3 are produced? (“Oh, bee-HAVE!”) Fe FeCl3 2 mol FeCl3 162.3 g FeCl3 4.0 mol Fe = 649 g FeCl3 1 mol FeCl3 2 mol Fe

  37. 2 H2(g) + O2(g) 2 H2O(g) ( ) . . _ _ . . ( ( ) ) ( ) 13 g H2 80 g O2 Which is LR: H2 or O2? 1 mol H2 = 3.25 13 g H2 = 6.5 mol H2 (HAVE) 2 2 g H2 1 mol O2 LR 80 g O2 1 = 2.50 = 2.5 mol O2 (HAVE) 32 g O2 = O2 How many g H2O are formed? (“Oh, bee-HAVE!”) O2 H2O 2 mol H2O 18 g H2O 2.5 mol O2 = 90 g H2O 1 mol H2O 1 mol O2

  38. ( ) ( ) 3 g H2 left over Started with 13 g, used up 10 g… How many g O2 are left over? zero; O2 is the LR and therefore is all used up How many g H2 are left over? We know how much H2 we started with (i.e., 13 g). To find how much is left over, we first need to figure out how much was USED UP in the reaction. O2 H2 2 mol H2 2 g H2 2.5 mol O2 = 10 g H2 used up 1 mol H2 1 mol O2

  39. 2 Fe(s) + 3 Br2(g) 2 FeBr3(s) ( ) . . _ _ . . ( ) ( ( ) ) 181 g Fe 96.5 L Br2 Find LR. 1 mol Fe = 1.62 181 g Fe = 3.24 mol Fe (HAVE) 2 55.8 g Fe LR 1 mol Br2 96.5 L Br2 3 = 1.44 = 4.31 mol Br2 (HAVE) = Br2 22.4 L Br2 How many g FeBr3 are formed? (“Oh, bee-HAVE!”) Br2 FeBr3 2 mol FeBr3 295.5 g FeBr3 4.31 mol Br2 = 849 g FeBr3 1 mol FeBr3 3 mol Br2

  40. 2 Fe(s) + 3 Br2(g) 2 FeBr3(s) ( ) ( ) 181 g Fe 96.5 L Br2 How many g of the ER are left over? Br2 Fe 2 mol Fe 55.8 g Fe 4.31 mol Br2 = 160 g Fe used up 1 mol Fe 3 mol Br2 21 g Fe left over Started with 181 g, used up 160 g…

  41. solid aluminum oxide solid sodium oxide molten sodium solid aluminum . . _ _ . . ( ) ( ( ) ) ( ) Al3+ O2– Percent Yield Na1+ O2– 6 + Al2O3(s) 1 2 + Na2O(s) 3 Na(l) Al(l) Find mass of aluminum produced if you start w/575 g sodium and 357 g aluminum oxide. = 4.17 575 g Na = 25 mol Na (HAVE) 1 mol 6 23 g LR 1 mol 1 = 3.5 = 3.5 mol Al2O3 (HAVE) 357 g Al2O3 102 g Al2O3 Al 2 mol Al 27 g Al 3.5 mol Al2O3 = 189 g Al 1 mol Al 1 mol Al2O3

  42. 189 g This amt. of product (______) is the theoretical yield. -- -- amt. we get if reaction is perfect found by calculation Now suppose that we perform this reaction and get only 172 grams of aluminum. Why? -- -- -- couldn’t collect all Al not all Na and Al2O3 reacted some reactant or product spilled and was lost

  43. % yield can never be > 100%. -- Find % yield for previous problem. = 91.0%

  44. 2 H2(g) + O2(g) 2 H2O(g) + 572 kJ ( ) E E . . _ _ . . ( ( ) ) Reaction that powers space shuttle is: From 100 g hydrogen and 640 g oxygen, what amount of energy is possible? 1 mol H2 2 = 25 100 g H2 = 50 mol H2 (HAVE) 2 g H2 1 mol O2 LR 640 g O2 1 = 20 = 20 mol O2 (HAVE) 32 g O2 O2 572 kJ 20 mol O2 = 11,440 kJ 1 mol O2

  45. 2 H2(g) + O2(g) 2 H2O(g) + 572 kJ ( ) ( ) What mass of excess reactant is left over? O2 H2 2 mol H2 2 g H2 20 mol O2 = 80 g H2 used up 1 mol H2 1 mol O2 20 g H2 left over Started with 100 g, used up 80 g…

  46. CO2(g) + 2 LiOH(s) Li2CO3(s) + H2O(l) ( ) . _ . ( ( ) ) ( ) ( 0.75) On NASA spacecraft, lithium hydroxide “scrubbers” remove toxic CO2 from cabin. For a seven-day mission, each of four individuals exhales 880 g CO2 daily. If reaction is 75% efficient, how many g LiOH should be brought along? 880 g CO2 x (4 p) x (7 d) = 24,640 g CO2 person-day CO2 LiOH 1 mol CO2 2 mol LiOH 23.9 g LiOH 24,640 g CO2 1 mol LiOH 1 mol CO2 44 g CO2 (Need more than this!) (if reaction is perfect) = 26,768 g LiOH = 35,700 g LiOH REALITY: TAKE 100,000 g

  47. 2 NaN3(s) 3 N2(g) + 2 Na(s) ( ( ) ) E E . _ . __C3H8 + __O2 __CO2 + __H2O + __kJ ( ) ( 0.85) Automobile air bags inflate with nitrogen via the decomposition of sodium azide: At STP and a % yield of 85%, what mass sodium azide is needed to yield 74 L nitrogen? Shoot for… 74 L = 87.1 L N2 N2 NaN3 1 mol N2 2 mol NaN3 65 g NaN3 87.1 L N2 = 169 g NaN3 1 mol NaN3 3 mol N2 22.4 L N2 Conceptual question (How would you do it?) X g Y g ? kJ Find LR. Strategy: 1. 2. Calc. ? kJ. LR

  48. B2H6 + 3 O2B2O3 + 3 H2O 10 g 30 g X g ( ) . . _ _ . . ( ) ( ( ) ) 1 mol = 0.362 10 g B2H6 = 0.362 mol B2H6 (HAVE) 1 27.6 g LR 1 mol 30 g O2 3 = 0.313 = 0.938 mol O2 (HAVE) 32 g O2 B2O3 1 mol B2O3 69.6 g B2O3 0.938 mol O2 = 21.8 g B2O3 1 mol B2O3 3 mol O2

  49. ( ) . . _ _ . . ( ) ( ( ) ) 2 3 2 2 ___ZnS + ___O2___ZnO + ___SO2 100 g 100 g X g (assuming 81% yield) Balance and find LR. Strategy: 1. 2. 3. Use LR to calc. X g ZnO (theo. yield) Actual yield is 81% of theo. yield. LR 1 mol = 0.513 100 g ZnS = 1.026 mol ZnS (HAVE) 2 97.5 g 1 mol 100 g O2 3 = 1.042 = 3.125 mol O2 (HAVE) 32 g ZnS ZnO 2 mol ZnO 81.4 g ZnO 1.026 mol ZnS = 83.5 g ZnO 1 mol ZnO 2 mol ZnS Actual g ZnO = 83.5 (0.81) = 68 g ZnO

  50. ( ( ( ) ) ) . _ . ( ) ( ( ) ) ( 0.80) 2 1 2 1 ___Al + ___Fe2O3___Fe + ___Al2O3 X g X g 800 g needed **Rxn. has an 80% yield. Shoot for… 800 g Fe = 1000 g Fe Fe Al 1 mol Fe 2 mol Al 27 g Al 1000 g Fe = 484 g Al 1 mol Al 2 mol Fe 55.8 g Fe Fe Fe2O3 1 mol Fe 1 mol Fe2O3 159.6 g Fe2O3 1000 g Fe = 1430 g Fe2O3 1 mol Fe2O3 2 mol Fe 55.8 g Fe

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