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IV Stoichiometry

IV Stoichiometry. Stoichiometry The relationship (mole ratio) between elements in a compound and between elements and compounds in a reaction e.g. H 2 O means: 1 molecule H 2 O contains 2 atom H and 1 atom O and 1 mole H 2 O contains 2 moles H atoms and 1 mole O atoms.

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IV Stoichiometry

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  1. IV Stoichiometry

  2. Stoichiometry • The relationship (mole ratio) between elements in a compound and between elements and compounds in a reaction e.g. H2O means: 1 molecule H2O contains 2 atom H and 1 atom O and 1 mole H2O contains 2 moles H atoms and 1 mole O atoms

  3. Chemical Compounds • Combination of elements • A compound is made up of specific elements in a specific ratio - Law of Constant Composition • Chemical Formula • Written representation of a chemical compound. So, a specific compound has a specific formula

  4. Terms • Formula Unit • Involves the lowest subscript which describes the ratio. • e.g. H2O; CH; CH2; CH4 • May or may not actually exist • Formula Weight • The mass of the formula unit • Empirical Formula • Written representation of the formula unit

  5. Terms continued • Molecule • Integral multiple of the formula unit (integer may be 1) that actually exists • e.g. C2H2; C2H4; C2H6 • Molecular Weight (Molar Mass) • Mass of the molecule • Molecular Formula • Written representation of the molecule

  6. Mass to mole conversions • Stoichiometry is in mole ratios • Most measurements are made in grams • So, you need to be able to get from grams to moles and moles to grams • The atomic weight listed on the periodic table is listed without units. Why?

  7. Units depend on what you want. • If you are looking on the atomic scale, atoms or molecules, units are amu. • If you are working on the macroscopic scale, moles of material, units are in grams.

  8. Convert 34.0 grams NH3 to moles. • Determine M of NH3. M = AW N + 3(AW H) = 14.0 + 3(1.01) = 17.0 2) Determine the moles of NH3

  9. How many molecules in 32.0 g of oxygen gas. • 2.0 • 1.0 • 0.5 • 6.02 x 1023 • 1.20 x 1024

  10. How many molecules in 32.0 g of oxygen gas. Oxygen occurs as O2 gas. AW of O = 16.00, so O2 = 32.00

  11. % Composition (% weight/ weight; %w/w)

  12. 1. Calculate the % composition form the Molecular Formula (or the Empirical Formula) What is the % composition by weight of C2H4O2? C = 12.01; H = 1.01; O = 16.00 Step 1: Find the molecular weight of the compound. MW = 2(AW C) + 4(AW H) + 2(AW O) = 2(12.01) + 4(1.01) + 2(16.00) = 60.06 g/ mol

  13. Step 2: Find the % of each element.

  14. Find the Chemical Formula from the % composition • Which Chemical formula will you get? • Empirical Formula, to get the molecular formula you would need more information than just the % composition

  15. A compound containing only carbon, hydrogen, and oxygen was found to contain 62.02 % C and 10.42% H. What is the formula of the compound? C = 12.01 H = 1.01 O = 16.00 • Find the amount of O %O = 100 - %C - %H = 100 - 62.02 - 10.42 = 27.57 % O

  16. 2) Determine the moles of each element (assume a 100 g sample)

  17. / 1.723 = 2.996 = 3 Mol C = 5.163 mol Mol H = 10.3168 mol Mol O = 1.723 mol 3) Divide by the smallest number to get whole number ratio / 1.723 = 5.9877 = 6 / 1.723 = 1 So, the compound has the ratio 3C:6H:1 O And the Empirical formula: C3H6O

  18. A further analysis of the compound found that the molar mass was 115.99 g/mol. What is the molecular formula of the compound? Molecular formula = (empirical formula)(# of formula units) You can find the # of formula units from MW/ FW

  19. Formula weight = weight of empirical formula: C3H6O FW = 3(12.01) + 6(1.01) + 16.00 = 58.048 # formula units = So the molecular formula contains 2 formula units. Multiply the subscripts by 2 2(C3H6O) = C6H12O2

  20. A compound is 84.80% uranium and the balance oxygen. What is the Empirical Formula of the compound? U = 238.0 O = 15.9994 • U2O5 • U3O8 • UO3 • UO2 • U3O

  21. A compound is 84.80% uranium and the balance oxygen. What is the Empirical Formula of the compound? U = 238.0 O = 15.9994 • Determine the amount of O 100 - 84.80 = 15.20 • Determine moles and mole ratio / 0.3563 = 1 / 0.3563 = 2.666 Clear denominator, multiply by 3 U1O8/3 U3O8 0.666 = 2/3 so O is 2 2/3 = 8/3

  22. 3) Determine an unknown element (X) from the % composition A compound XO2 is 78.8% X. What element is X? O = 16.0 • Ni • Co • P • Sn

  23. 3) Determine and unknown element (X) from the % composition A compound XO2 is 78.8% X. What element is X? O = 16.0 % O = 100 - 78.8 = 21.2 Since Compound is 78.8% X, 0.6625 mol X = 78.8 g X Go to Periodic Table 118.9 = Sn

  24. Reactions • Chemical Reaction is represented by a Chemical Equation • General Form: aA + bB cC + dD A/B are ? Reactants C/D are ? Products a,b,c,d are ? Stoichiometric coefficients

  25. Law of Conservation of Mass says? • No mass lost or gained • Total mass of reactants = total mass of products • Elements are re - arranged not changed • This allows us to “Balance” Equations • The number and kinds of atoms in the reactants have to show up as the same number and kind of atoms in the product

  26. Ca + H2O Ca(OH)2 + H2 1 Ca + H2O 1 Ca(OH)2 + H2 1Ca + 2H2O 1 Ca(OH)2 + H2 1 Ca + 2 H2O 1Ca(OH)2 + 1 H2 Ca + 2 H2O Ca(OH)2 + H2

  27. C3H7OH + O2 CO2 + H2O C3H7OH + O2 3 CO2 + H2O C3H7OH + O2 3 CO2 + 4 H2O C3H7OH + 9/2 O2 3 CO2 + 4 H2O 2 C3H7OH + 9 O2 6 CO2 + 8 H2O

  28. SiF4 + H2O HF + SiO2 SiF4 + H2O 4 HF + SiO2 SiF4 + 2 H2O 4 HF + SiO2

  29. When the reaction: C2H8N2 + N2O4 N2 + H2O + CO2 Is balanced using the smallest whole numbers, what is the coefficient of N2? • 1 • 2 • 3 • 4 • 5

  30. When the reaction: C2H8N2 + N2O4 N2 + H2O + CO2 Is balanced using the smallest whole numbers, what is the coefficient of N2? C2H8N2 + N2O4 N2 + H2O + CO2 3 2 2 4

  31. Hydrates • A compound (solid) that contains intact water molecules as part of the compound. • The water can be removed by heating to leave an anhydrous residue (solid). • Water can then be re - added to the anhydrate to yield the original hydrate

  32. CaSO4 • 2H2O Calcium sulfate dihydrate Each mole of compound contains: 1 mole calcium sulfate and 2 mole water or 1 mole Ca 1 mole S 6 mol O 4 mol H

  33. heat CaSO4 • 2H2O(s) CaSO4(s) + 2 H2O(g) 172 136 + 2(18) CaSO4(s) CaSO4 • 2H2O(s) Water vapor

  34. 17.0 g of CaSO4 • 2H2O(s) was heated to remove the water. What mass of residue (CaSO4) is left? • 13.4 • 13.44 • 3.56 • 15.0 • 15.03

  35. 17.0 g of CaSO4 • 2H2O(s) was heated to remove the water. What mass of residue (CaSO4) is left? What happened to the difference (17.0 - 13.4) = 3.6 grams of material? Lost as water vapor

  36. 15.00 grams of the hydrate Na2SO4 • XH2O(s) was heated to remove the water. After heating, 7.95 grams of material remained. What is the formula of the hydrate: (find the value of X) • Na2SO4 • H2O • Na2SO4 • 2H2O • Na2SO4 • 3H2O • Na2SO4 • 5H2O • Na2SO4 • 7H2O

  37. 15.00 7.95 15.00-7.95 7.05 Na2SO4 • XH2O(s) Na2SO4(s) + XH2O Find moles of both products and compare / 0.0559 = 1 / 0.0559 = 6.99 = 7 So, X = 7, formula = Na2SO4 • 7H2O

  38. Limiting Reactant • Limiting Reactant is that element or compound that determines the amount of product that you get. • It is the reactant that is used up

  39. You are the owner of a bike shop. A shipment came in with 183 frames, 150 seats, 252 pedals, 131 brake assemblies. How many bikes can you sell? (enter the number)

  40. Each bike needs, 1 frame, 1 seat, 2 pedals, 1 brake. Given: 183 frames 150 seats 252 pedals 131 brake assemblies Pedals will allow you to make only 126 bikes so that is the limiting reactant

  41. 3/2 3 2 NH3 + O2 N2 + H2O Balance NH3 + O2 N2 + H2O NH3 + O2 N2 + H2O 2 4 3 6

  42. 2 4 3 NH3 + O2 N2 + H2O 6 Remember: Stoichiometry is mole ratios How many moles of N2 can be formed from 4 mol NH3 and 4 mol O2? 1. Determine the LR. Mol O2 given > mol O2 required. So, NH3 is LR 2. Determine amount of product

  43. 2 4 3 NH3 + O2 N2 + H2O 6 How many moles of N2 can be formed from 6 mol NH3 and 4 mol O2? 1. Determine the LR. Mol O2 given < mol O2 required. So, O2 is LR 2. Determine amount of product

  44. 2 4 3 NH3 + O2 N2 + H2O 6 How many moles of N2 can be formed from 5 mol NH3 and 4 mol O2? 5 4 3.75 2.67 2.5

  45. 2 4 3 NH3 + O2 N2 + H2O 6 How many moles of N2 can be formed from 5 mol NH3 and 4 mol O2? 1. Determine the LR. Mol O2 given > mol O2 required. So, NH3 is LR 2. Determine amount of product

  46. 2 4 3 NH3 + O2 N2 + H2O 6 If 10.0 grams each of NH3 and O2 are reacted, how many grams of water and N2 are formed? • Find moles of each reactant. • Determine the Limiting Reactant Determine mole of O2 needed Compare to what was given: 0.442 mol required > 0.312 mol given So, O2 = LR

  47. 3. Determine the amount of product based on the LR total mass of products = 17.0 g What happened to conservation of mass? 3.0 g un-reacted NH3

  48. 10.0 grams Cr and 10.0 grams S are reacted to give chromic sulfide. How many grams of chromic sulfide are formed? Cr = 52.0 S = 32.0 • 20.9 g • 20.0 g • 19.2 g • 3.12 g • 0.80 g

  49. 10.0 grams Cr and 10.0 grams S are reacted to give chromic sulfide. How many grams of chromic sulfide are formed? Cr = 52.0 S = 32.0 • Write and balance the equation: 2Cr + 3S --> Cr2S3 2. Determine moles of each

  50. 3. Determine LR 4. Determine the amount of product based on the LR

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