1 / 45

Chapter 9

StoIcHIomEtRY. Chapter 9. U sing E veryday E quations. Stoichiometry - the calculations of quantities of chemical reactions Its a form of bookkeeping for chemist. Mole to Mole Calculations.

heinrich-hunter
Download Presentation

Chapter 9

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. StoIcHIomEtRY Chapter 9

  2. Using Everyday Equations • Stoichiometry- the calculations of quantities of chemical reactions Its a form of bookkeeping for chemist

  3. Mole to Mole Calculations • If you know the number of moles of one substance, the balanced chemical equation allows you to determine the number of moles of all other substances in the reaction. • How to solve Mole to Mole problems: • write a balanced equation • use a Mole-to-Mole proportion to find moles of unknown (put unknown on top of proportion) • Note: The coefficients in the balanced equation are equal to the number of moles for that substance

  4. Mole to Mole Proportion • Coefficients from equation Problem • Desired = Desired • Given Given

  5. Mole toMole Examples A. How many moles of ammonia are produced when .60 mol of nitrogen reacts with hydrogen? B. How many moles of Oxygen are needed to produce 2.5 moles of Water? C. How many moles of Hydrogen are produced when 4.5 moles of Mg reacts with HCl?

  6. Mass to Mass Calculations • When given the number of grams of a substance in a chemical reaction you can calculate the specific number of grams of any other substance in that reaction. • How to Solve Mass to Mass Problems: • Write a balanced equation • Convert grams given to moles • Use a mole-mole proportion to find moles for unknown • Convert moles of unknown to grams

  7. Mass to Mass Examples 1. Calculate the number of grams of NH3 produced when 15.5 grams of nitrogen reacts with hydrogen 2. How many grams of Oxygen are produced when 40.5 grams of water decomposes? 3. How many grams of CO2 are produced during the combustion of 200 grams of methane?

  8. Calculating the Percent Yield • Theoretical yield- the maximum amount of product that will formed from given amounts of reactants. • Actual yield- the amount of product that actually forms when the reaction is carried out. This is often less then the theoretical yield. • Percent Yield- the ratio of the actual yield to the theoretical yield as a percent. This measures the efficiency of the reaction. percent yield = actual yield theoretical yield X 100

  9. 1X 3X 2X

  10. MOLES SUMMARY • We have recently learned that Avogadro’s number of particles is the same as a mole of a substance. • On the basis of the particle interpretation we just discussed, the equation also tells you the number of moles of reactants and products.

  11. MOLES SUMMARY • 1 mole of N2 molecules reacts with 3 moles of H2 molecules to make 2 moles of NH3 molecules. • The coefficients of the balanced chemical equation indicate the numbers of moles of reactants and products in a chemical reaction.

  12. MOLES SUMMARY • This is the most important information that a reaction equation provides. • Using this information, you can calculate the amounts of reactants and products.

  13. MASS SUMMARY • A balanced chemical equation must also obey the law of conservation of mass. • Mass can be neither created nor destroyed in ordinary chemical or physical processes. • Remember that mass is related to the number of atoms in a compound through the mole.

  14. MASS SUMMARY • The mass of 1 mol of N2 molecules is 28 g; the mass of 3 mols of H2 molecules is 6 g for a total mass of reactants of 34 g. • The mass of 2 moles of NH3 molecules is 2 * 17g or 34 g. • As you can see the reactants mass is equal to the mass of the products.

  15. 22.4 L 22.4 L 22.4 L 22.4 L 22.4 L 22.4 L

  16. USING EQUATIONS • You can see how much information is stored in a simple balanced reaction eqn • We can combine this information with our knowledge ofmole conversionsto perform important common stoichiometric calculations.

  17. MOLE – MOLE CALCULATIONS • A balanced rxn eqn is essential for all calculations involving amounts of reactants and products. • If you know the number of moles of 1 substance, the balanced eqn allows you to calc. the number of moles of all other substances in a rxn eqn.

  18. MOLE – MOLE CALCULATIONS • Let’s go back to our synthesis of ammonia rxn. N2(g) + 3H2(g)  2NH3(g) • The MOST important interpretation of this rxn is that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.

  19. MOLE – MOLE CALCULATIONS • These connections of the coefficients allows us to set up conversion factors called mole ratios. • The mole ratios are used to calculate the connections in moles of compounds in our reaction equation. • We can start calculating…

  20. MOLE – MOLE CALCULATIONS How many moles of ammonia are produced when .60 moles of N2 are reacted with H2? • Sample Mole – Mole problem: N2(g) + 3H2(g)  2NH3(g) Given: .60 moles of N2 Uknown: ____ moles of NH3

  21. MOLE – MOLE CALCULATIONS How many moles of ammonia are produced when .60 moles of N2 are reacted with H2? • Sample Mole – Mole problem: N2(g)+ 3H2(g)  2NH3(g) Given: .60 moles of N2 Uknnown: ____ moles of NH3

  22. MOLE – MOLE CALCULATIONS How many moles of ammonia are produced when .60 moles of N2 are reacted with H2? • Sample Mole – Mole problem: N2(g)+ 3H2(g) 2NH3(g) Given: .60 moles of N2 Uknown: ____ moles of NH3

  23. MOLE – MOLE CALCULATIONS • According to the reaction equation, for every 1 mole of N2 reacted we form2 mols of NH3. • To determine the number of moles of NH3, the given quantity of N2 is multiplied by the mole ratio from the rxn eqn in such a way that the units of “mol N2” cancel

  24. = 1.2 mol NH3 MOLE – MOLE CALCULATIONS N2(g) + 3H2(g) 2NH3(g) • Solve for the unknown: 2 mol NH3 .6 mol N2 1 mol N2

  25. MOLE – MOLE EXAMPLE 2 • This equation shows the formation of aluminum oxide. 4Al(s) + 3O2(g)  2Al2O3(s) • How many moles of aluminum are needed to form 3.7 mol Al2O3? Given: 3.7 moles of Al2O3 Uknown: ____ moles of Al

  26. MOLE – MOLE EXAMPLE 2 • This equation shows the formation of aluminum oxide. 4Al(s) + 3O2(g)  2Al2O3(s) • How many moles of aluminum are needed to form 3.7 mol Al2O3? Given: 3.7 moles of Al2O3 Uknown: ____ moles of Al

  27. Coefficients in the balanced equation Coefficients in the balanced equation = 7.4 mol Al MOLE – MOLE CALCULATIONS 4Al(s) + 3O2(g)  2Al2O3(s) • Solve for the unknown: 4 mol Al 3.7 mol Al2O3 2 mol Al2O3

  28. MASS – MASS CALCULATIONS • No lab balance measures moles directly, instead the mass of a substance is usually measured in grams. • From the mass of a reactant or product, the mass of any other reactant or product in a given chemical equation can be calculated.

  29. MASS – MASS CALCULATIONS • The mole – mole connection is still vital to do these calcs. • Then the mole ratio from the balanced equation can be used to calculate the number of moles of the unknown. • If the given sample is measured in grams, the mass can be converted to moles by using the molar mass.

  30. MASS – MASS CALCULATIONS • If it is mass of the unknown that needs to be determined, the number of moles of the unknown can be multiplied by the molar mass of the desired compound. • As in mole-mole calcs, the unknow can be either a reactant or a product.

  31. MASS – MASS CALCULATIONS • Again back to our synthesis of ammonia rxn. N2(g) + 3H2(g)  2NH3(g) N2(g) + 3H2(g) 2NH3(g) • Calculate the number of grams of NH3 produced by the reaction of 5.4 g of H2 with an excess of N2.

  32. MASS – MASS CALCULATIONS • What do we know? • Mass of H2 = 5.4 g H2 • 3 mol H2 = 2 mol NH3 (from balanced equation - AKA mole ratio) • Molar mass of H2 = 2.0 g H2 • Molar mass of NH3=17.0g NH3 • What are we asked for? • Mass of ammonia produced

  33. Coefficients in the balanced equation Molar mass of H2 = 2.7 mol H2 MASS – MASS CALCULATIONS Step 1: convert mass of given to moles of given using MM of G 1 mol H2 5.4 g H2 2.0 g H2

  34. Coefficients in the balanced equation Coefficients from the balanced equation = 1.8 mol NH3 MASS – MASS CALCULATIONS Step 2: convert mols of G tomols of U using the mole ratio N2(g) + 3H2(g)  2NH3(g) 2 mol NH3 2.7 mol H2 3 mol H2

  35. Coefficients in the balanced equation Molar mass of NH3 = 31.0 g NH3 MASS – MASS CALCULATIONS Step 3: convert moles of desired compound to mass using MM of U 17.0 g NH3 1.8 mol NH3 1 mol NH3

  36. MASS – MASS CALCULATIONS • Mass to mass calculations always follow those same three steps. • It uses the mole math that we have had lots of practice with (mass to moles and moles to mass) • The only difference is the new middle step where we use our newly acquired mole ratio

  37. MASS – MASS CALCULATIONS Acetylene gas (C2H2) is produced by adding water to calcium carbide (CaC2). • Let’s do another one: CaC2 + 2H2O  C2H2 + Ca(OH)2 CaC2 + 2H2O  C2H2 + Ca(OH)2 How many grams of acetylene are produced by adding water to 5.00 g CaC2?

  38. MASS – MASS CALCULATIONS • What do we know? • Mass of CaC2 = 5.0 g CaC2 • 1 mol CaC2 = 1 mol C2H2 (from balanced equation) • MM of CaC2 = 64.0 g CaC2 • MM of C2H2 = 26.0g C2H2 • What are we asked for? • Mass of C2H2 produced

  39. = .078mol CaC2 MASS – MASS CALCULATIONS Step 1: convert mass of given to molesof given using MM of G 1 mol CaC2 5.0 g CaC2 64.0 g CaC2

  40. = .078mol C2H2 MASS – MASS CALCULATIONS Step 2: convert mols of G to molsof U using the mole ratio CaC2 + 2H2O  C2H2 + Ca(OH)2 1 mol C2H2 .078mol CaC2 1 mol CaC2

  41. = 2.03 g C2H2 MASS – MASS CALCULATIONS Step 3: convert moles of desired compound to mass using MM of U 26.0 g C2H2 .078 mol C2H2 1 mol C2H2

  42. MORE MOLE CALCULATIONS • A balanced reaction equation indicates the relative number of moles of reactants and products. • We can expand our stoichiometric calculations to include any unit of measure-ment that is related to the mole.

  43. MORE MOLE CALCULATIONS • The given quantity can be expressed in numbers of particles, units of mass, or volumes of gases at STP. • The problems can include mass-volume, volume-volume, and particle-mass calculations.

  44. MORE MOLE CALCULATIONS • In any of these problems, the given quantity is first converted to moles. • Then the mole ratio from the balanced equationn is used to convert from the moles of given to the number of moles of the unknown

  45. MORE MOLE CALCULATIONS • Then the moles of the unknown are converted to the units that the problem requests. • The next slide summarizes these steps for all typical stoichiometric problems

More Related