1 / 50

Topic 1:Stoichiometry (12.5 hours)

Topic 1:Stoichiometry (12.5 hours). 1.1 The mole concept and Avogardo’s number 1.2 Formulas 1.3 Chemical equations 1.4 Mass and gas vol. relationships in chemical reactions. 1.1 The mole concept and Avogadro’s constant 1.2 Formulas. 1.1.1 Apply the mole concept to substances

dougal
Download Presentation

Topic 1:Stoichiometry (12.5 hours)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Topic 1:Stoichiometry(12.5 hours) 1.1 The mole concept and Avogardo’s number 1.2 Formulas 1.3 Chemical equations 1.4 Mass and gas vol. relationships in chemical reactions

  2. 1.1 The mole concept and Avogadro’s constant1.2 Formulas 1.1.1 Apply the mole concept to substances 1.1.2 Determine the number of particles and the amount of substance (in moles) 1.2.1 Define the terms relative atomic mass (Ar) and relative molecular mass (Mr) 1.2.2 Calculate the mass of one of mole of species from its formula 1.2.3 Solve problems involving the relationships between the amount of substance in moles, mass, and molar mass

  3. Quick Review: • Atoms are the smallest unit • The same atoms make up elements (Na or Cl2) • Different whole combination of atoms make up compounds (NaCl) or molecules (CH4)

  4. Atomic mass • Atoms contain different numbers of protons and neutrons, so they have different masses (isotopes) • Carbon-12 (6 p+, 6 no)has a mass 12 times greater than hydrogen-1(1 p+, 0 no) • It’s impossible to measure the mass of individual atoms on a balance, because it is so small • Chemists developed their own unit, called the mole

  5. The Mole • Similar to a dozen, except instead of 12, it’s 602,000,000,000,000,000,000,000 • 6.02 X 1023 mol-1(in scientific notation) • It applies to all kinds of particles: atoms, particles, molecules, ions, electrons, formula units depending the way the question is asked, so be careful. • This number is named in honor of Amedeo Avogadro (1776 – 1856)

  6. Just how big is Avogadro's constant? • The earth is estimated to be 4.54 billion years old, that is approximately 1.43 x 1017 s. Still much smaller than a mole. • If you were to count every grain of sand in the Sahara desert, assuming its 4000 km by 1500 km, and its depth is 10 m, and assuming that 1 cm3 contains 1000 grains of sand. It would contain 6 x 1022 grains of sand. Only one tenth the size of a mole!!!

  7. Relative atomic mass (Ar) • As the weighted mean mass of all the naturally occurring isotopes of an element relative to one twelfth of the mass of a carbon-12 atom. • Not whole numbers • No mass unit • Also applies to relative molecular mass (Mr)

  8. Molar Mass (or atomic mass) • The Mass of 1 mole (in grams) • Equal to the numerical value of the average atomic mass (get from periodic table), or add the atoms together for a molecule 1 mole of C atoms = 12.0 g 1 mole of Mg atoms = 24.3 g 1 mole of O2 molecules = 32.0 g

  9. Molar Mass of Compounds • The molar mass (MM) of a compound is determined the same way, except now you add up all the atomic masses for the molecule (or compound) • Ex. Molar mass of CaCl2 • Avg. Atomic mass of Calcium = 40.08g • Avg. Atomic mass of Chlorine = 35.45g • Molar Mass of calcium chloride = 40.08 g/mol Ca + (2 X 35.45) g/mol Cl 110.98 g/mol CaCl2 20 Ca40.08 17Cl 35.45

  10. Atoms or Molecules Divide by 6.02 X 1023 Multiply by 6.02 X 1023 Moles Multiply by atomic/molar mass from periodic table Divide by atomic/molar mass from periodic table Mass (grams)

  11. Practice • Calculate the Molar Mass of calcium phosphate • Formula = • Masses elements: • Ca: 3 Ca’s X 40.1 = • P: 2 P’s X 31.0 = • O: 8 O’s X 16.0 = • Molar Mass = Ca3(PO4)2 120.3 g 62.0 g 128.0 g 120.3g + 62.0g +128.0g 310.3 g/mol

  12. Calculations molar mass Avogadro’s numberGrams Moles particles Everything must go through Moles!!!

  13. Atoms/Molecules and Grams How many moles of Cu are present in 35.4 g of Cu? How many atoms? 35.4 g x 1 mol = 0.557 mol Cu 63.55 g 0.557 mol Cux 6.02x1023 atoms = 3.35x1023atoms 1 mol

  14. On your own! How many moles of Fe are present in 102.4 g of Fe? How many atoms? 102.4 g x 1 mol = 1.83 mol Fe 55.85 g 1.83 mol Fe x 6.02x1023 atoms = 1.10x1024 atoms 1 mol

  15. Work backwards! What is the mass (in grams) of 1.20x1024 molecules of glucose (C6H12O6)? 1.20x1024 molec. X 1 mol = 1.99 mol 6.02x1023 molec. 1.99 mol x 180.12 g = 358.4 g 1 mol From periodic table

  16. This is as tricky as it gets! • How many atoms of carbon are found in 2.6g of glucose (C6H12O6)? 2.6 g x 1 mole= 0.014 mol C6H12O6 180.12 g 0.014 mol x 6.02 x 1023 molecules = 8.4 x 1021 molecules mol 8.4 x 1021 molecules x 6 atoms of C = 5.06 x 1022 atoms 1 molecule C6H12O6

  17. 1.3 Chemical Equations 1.3.1 Deduce chemical equations when all reactants and products are given 1.3.2 Identify the mole ratio of any two species in a chemical equation 1.3.3 Apply the state symbols (s), (l), (g) and (aq)

  18. Chemical equations (review) • To deduce the products of an equation we must know the type of reaction occuring. • Single displacement • Double displacement • Combustion • Synthesis • Decomposition

  19. Single displacement: • either the metal changes position with the metal ion… AgNO3 (aq) + Cu(s) CuNO3(aq) + Ag(s) • or the non-metal changes position with the non-metal ion. F2(g) + NaCl(aq)  NaF(aq)+ Cl2(g)

  20. Double displacement • The cations (positively charged ions) change position with each other. NaOH + HCl  HOH + NaCl Fe(OH)2 + 2KBr  2KOH + FeBr2 Balancing an equation means that both sides must have the same number and type of atoms present. Coefficients are red (indicating total number of compounds), subscripts are blue (indicating number of atoms).

  21. Organic Combustion • A hydrocarbon reacting with oxygen to produce carbon dioxide and water CH4(g) + 2O2(g) 2H2O(g) + CO2(g) 2C2H6(g) + 7O2(g) 6H2O (g) + 4CO2(g) The coefficient is also related the mole concept. For the 1st reaction, we would say that one mole of methane reacts with two moles of oxygen to produce two moles of water and one mole of carbon dioxide gas.

  22. Synthesis • The combination of two or more particles to produce one compound. N2(g) + H2(g) NH3(l) Use (g) for gas, (l) for liquid, (s) for solids and (aq) if it dissolves in water/aqueous.

  23. Decomposition • A single large compound will be broken down into more simple parts. H2O (l) H2(g) + O2(g)

  24. 1.4 Mass relationships in chemical equations 1.4.1 Calculate theoretical yields from chemical equations 1.4.2 Determine the limiting reactant the reactant in excess when quantities of reacting substances are given. 1.4.3 Solve problems involving theoretical, experimental and percentage yield.

  25. Chemistry Recipes • Looking at a reaction tells us how much of something you need to react with something else to get a product • Be sure you have a balanced reaction before you start! • Example: 2 Na + Cl2  2 NaCl • This reaction tells us that by mixing 2 moles of sodium with 1 mole of chlorine we will get 2 moles of sodium chloride • What if we wanted 4 moles of NaCl? 10 moles? 50 moles?

  26. Practice • Write the balanced reaction for hydrogen gas reacting with oxygen gas. 2 H2 + O2 2 H2O • How many moles of reactants are needed? • What if we wanted 4 moles of water? • What if we had 3 moles of oxygen, how much hydrogen would we need to react and how much water would we get? • What if we had 50 moles of hydrogen, how much oxygen would we need and how much water produced? 2 mol H21 mol O2 4 mol H22 mol O2 6 mol H2, 6 mol H2O 25 mol O2, 50 mol H2O

  27. Mole Ratios • These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical • Example: How many moles of chlorine is needed to react with 5 moles of sodium (without any sodium left over)? 2 Na + Cl2 2 NaCl 5 moles Na 1 mol Cl2 2 mol Na = 2.5 moles Cl2

  28. Mole-Mole Conversions • How many moles of sodium chloride will be produced if you react 2.6 moles of chlorine gas with an excess (more than you need) of sodium metal? 2 Na + Cl2 2 NaCl 2.6 moles Cl2 2 mol NaCl 1 mol Cl2 = 5.2 moles NaCl

  29. You practice • Aluminum reacts with oxygen to produce aluminum oxide. If I have 2.6 mol of Al, how much Al2O3 would I produce? 4Al + 3O2 2Al2O3 2.6 mol Al x 2 mol Al2O3 = 1.3 mol Al2O3 4 mol Al

  30. Mole-Mass Conversions • Most of the time in chemistry, the amounts are given in grams instead of moles • We still go through moles and use the mole ratio, but now we also use molar mass to get to grams • Example: How many grams of chlorine are required to react completely with 5.00 moles of sodium to produce sodium chloride? 2 Na + Cl2 2 NaCl 5.00 moles Na 1 mol Cl2 70.90g Cl2 2 mol Na 1 mol Cl2 = 177g Cl2

  31. You Practice • Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum. 2 Al + 3 I2  2 AlI3

  32. Mass-Mole • We can also start with mass and convert to moles of product or another reactant • We use molar mass and the mole ratio to get to moles of the compound of interest • Calculate the number of moles of ethane (C2H6) needed to produce 10.0 g of water • 2 C2H6 + 7 O2 4 CO2 + 6 H20 10.0 g H2O 1 mol H2O 2 mol C2H6 18.0 g H2O 6 mol H20 = 0.185 mol C2H6

  33. Mass-Mass Conversions • Most often we are given a starting mass and want to find out the mass of a product we will get (called theoretical yield) or how much of another reactant we need to completely react with it (no leftover ingredients!) • Now we must go from grams to moles, mole ratio, and back to grams of compound we are interested in

  34. Mass-Mass Conversion • Ex. Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen. • N2 + 3 H2  2 NH3 2.00g N2 1 mol N2 2 mol NH3 17.06g NH3 28.02g N2 1 mol N2 1 mol NH3 = 2.4 g NH3

  35. Practice • How many grams of calcium nitride are produced when 2.00 g of calcium reacts with an excess of nitrogen? • __Ca + __N2 __Ca3N2

  36. Grilled Cheese Sandwich Bread + Cheese  ‘Cheese Melt’ 2 B + C  B2C 100 bread 30 slices ? sandwiches

  37. Limiting Reactants aluminum + chlorine gas  aluminum chloride Al(s) + Cl2(g)  AlCl3 2 Al(s) + 3 Cl2(g) 2 AlCl3 100 g 100 g ? g A. 200 g B. 125 g C. 667 g D. ???

  38. Limiting Reactant • Most of the time in chemistry we have more of one reactant than we need to completely use up other reactant. • That reactant is said to be in excess (there is too much). • The other reactant limits how much product we get. Once it runs out, the reaction s. This is called the limiting reactant.

  39. Limiting Reactant • To find the correct answer, we have to try all of the reactants. We have to calculate how much of a product we can get from each of the reactants to determine which reactant is the limiting one. • The lower amount of a product is the correct answer. • The reactant that makes the least amount of product is the limiting reactant. Once you determine the limiting reactant, you should ALWAYS start with it! • Be sure to pick a product! You can’t compare to see which is greater and which is lower unless the product is the same!

  40. LimitingReactant Limiting Reactant: Example • 10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced? 2 Al + 3 Cl2 2 AlCl3 • Start with Al: • Now Cl2: 10.0 g Al 1 mol Al 2 mol AlCl3 133.5 g AlCl3 27.0 g Al 2 mol Al 1 mol AlCl3 = 49.4g AlCl3 35.0g Cl2 1 mol Cl2 2 mol AlCl3 133.5 g AlCl3 71.0 g Cl2 3 mol Cl2 1 mol AlCl3 = 43.9g AlCl3

  41. LR Example Continued • We get 49.4g of aluminum chloride from the given amount of aluminum, but only 43.9g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 35.0g of chlorine is used up, the reaction comes to a complete .

  42. Limiting Reactant Practice • 15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made.

  43. Finding the Amount of Excess • By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess. • Can we find the amount of excess potassium in the previous problem?

  44. Finding Excess Practice • 15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I2 2 KI • We found that Iodine is the limiting reactant, and 19.6 g of potassium iodide are produced. 15.0 g I2 1 mol I2 2 mol K 39.1 g K 254 g I2 1 mol I2 1 mol K = 4.62 g K USED! 15.0 g K – 4.62 g K = 10.38 g K EXCESS Given amount of excess reactant Amount of excess reactant actually used Note that we started with the limiting reactant! Once you determine the LR, you should only start with it!

  45. Limiting Reactant: Recap • You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT. • Convert ALL of the reactants to the SAME product (pick any product you choose.) • The lowest answer is the correct answer. • The reactant that gave you the lowest answer is the LIMITING REACTANT. • The other reactant(s) are in EXCESS. • To find the amount of excess, subtract the amount used from the given amount. • If you have to find more than one product, be sure to start with the limiting reactant. You don’t have to determine which is the LR over and over again!

  46. Percent Yields • Theoretical yield: max amount of a product that is formed in a reaction. • Actual yield: amount of product that is actually obtained in a reaction • Usually less than theoretical. Why?

  47. Why? • Theoretical has assumed that all of limiting reagent has completely reacted. • Many reactions do not go to completion • Unexpected competing side reactions limit the formation of products. • Some reactants are lost during the separation process (remember in the lab, pouring off water, leaving silver behind!) • Impure reactants • Faulty measuring • Poor experimental design or technique

  48. How to calculate? Percent Yield = Actual Yield x 100% Theoretical yield

  49. Practice together: 20g of HBrO3 is reacted with excess HBr. • What is the theoretical yield of Br2? • What is the percent yield, if 47.3g is produced? HBrO3 + 5HBr  3H2O + 3 Br2

  50. Practice alone: When 35g of Ba(NO3)2 is reacted with excess Na2SO4, 29.8g of BaSO4 is recovered. Ba(NO3)2 +Na2SO4 BaSO4 + 2NaNO3 • Calculate the theoretical yield of BaSO4 • Calculate the percent yield of BaSO4

More Related