1 / 109

Chapter 4 Stoichiometry : Quantitative Information about Chemical Reactions

Chapter 4 Stoichiometry : Quantitative Information about Chemical Reactions. Stoichiometry. The study of mass relationships in chemical reactions is called Stoichiometry . Stoichiometry provides quantitative information about chemical reactions.

oni
Télécharger la présentation

Chapter 4 Stoichiometry : Quantitative Information about Chemical Reactions

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. Chapter 4Stoichiometry: Quantitative Information about Chemical Reactions

  2. Stoichiometry The study of mass relationships in chemical reactions is called Stoichiometry. Stoichiometry provides quantitative information about chemical reactions. Mass must be conserved in a chemical reaction. Total mass of reactants Total mass of products = Chemical equations must therefore be balanced for mass. Numbers of atoms on the reactant side Numbers of atoms on the product side =

  3. Stoichiometry • The stoichiometric balancing coefficients are the numbers in front of the chemical formulas. • They give the ratio of reactants and products. • The balancing coefficients allow us to convert between numbers of reactants and products. Conversion factors ratio

  4. Stoichiometry Stoichiometry: The branch of chemistry that deals with the mole proportions of chemical reactions. The ratio of any two species (reactants or products) in a balanced chemical reaction. Stoichiometric ratio: 2A + 3B A2B3 2 A’s combine with 3B’s CONVERSION FACTORS!!! 2A 3B or 3B 2A

  5. Stoichiometry Given the following reaction 2C2H6(l) + 7 O2(g)  4CO2(g) + 6H2O(l) How many moles of water are produced when 3.0 mols of oxygen react?

  6. Stoichiometry Given the following reaction 2C2H6(l) + 7 O2(g)  4CO2(g) + 6H2O(l) How many moles of water are produced when 3.0 mols of oxygen react? mol O2  mol H2O

  7. Stoichiometry Given the following reaction 2C2H6(l) + 7 O2(g)  4CO2(g) + 6H2O(l) How many moles of water are produced when 3.0 mols of oxygen react? mol O2  mol H2O

  8. Stoichiometry Given the following reaction 2C2H6(l) + 7 O2(g)  4CO2(g) + 6H2O(l) How many moles of water are produced when 3.0 mols of oxygen react? mol O2  mol H2O 2 sig. figs. exact conversion factor 2 sig. figs.

  9. Stoichiometry Given the following reaction 2C2H6(l) + 7 O2(g)  4CO2(g) + 6H2O(l) How many moles of C2H6 must react to produce 1.75 mols of CO2?

  10. Stoichiometry Given the following reaction 2C2H6(l) + 7 O2(g)  4CO2(g) + 6H2O(l) How many moles of C2H6 must react to produce 1.75 mols of CO2? 1.75 mol CO2 = 0.875 mol C2H6

  11. Stoichiometry Given the following reaction 2C2H6(l) + 7 O2(g)  4CO2(g) + 6H2O(l) How many moles of C2H6 must react to produce 1.75 mols of CO2? 1.75 mol CO2 = 0.875 mol C2H6

  12. Stoichiometry • Mole/Mass relationships must be looked up first on the PT • It is not possible to relate masses in reactions without going through moles. Use the coefficients in the balanced equation

  13. Consider the following reaction: If 454 g of solid ammonium nitrate decomposes to form dinitrogen monoxide gas and liquid water, how many grams of dinitrogen monoxide and water are formed?

  14. Consider the following reaction: If 454 g of solid ammonium nitrate decomposes to form dinitrogen monoxide gas and liquid water, how many grams of dinitrogen monoxide and water are formed? Strategy Map for the solution to the problem:

  15. Consider the following reaction: If 454 g of solid ammonium nitrate decomposes to form dinitrogen monoxide gas and liquid water, how many grams of dinitrogen monoxide and water are formed? Data Information: Mass of reactant amount of reactant in moles Write the balanced chemical equation Convert moles reactant to moles product Step 1: Step 3: Eq. gives mole ratios (stoichiometry) amount of products in moles Convert mass to moles Convert moles of products to mass Step 2: Step 4: mass of products in grams

  16. Consider the following reaction: If 454 g of solid ammonium nitrate decomposes to form dinitrogen monoxide gas and liquid water, how many grams of dinitrogen monoxide and water are formed?

  17. Consider the following reaction: If 454 g of solid ammonium nitrate decomposes to form dinitrogen monoxide gas and liquid water, how many grams of dinitrogen monoxide and water are formed? Step 1: Write the balanced chemical equation

  18. Consider the following reaction: If 454 g of solid ammonium nitrate decomposes to form dinitrogen monoxide gas and liquid water, how many grams of dinitrogen monoxide and water are formed? Step 1: Write the balanced chemical equation NH4NO3(s)  N2O(g) + 2 H2O(l)

  19. Consider the following reaction: If 454 g of solid ammonium nitrate decomposes to form dinitrogen monoxide gas and liquid water, how many grams of dinitrogen monoxide and water are formed? Step 1: Write the balanced chemical equation NH4NO3(s)  N2O(g) + 2 H2O(l) Step 2:Convert mass of reactant (454 g NH4NO3) to moles.

  20. Consider the following reaction: If 454 g of solid ammonium nitrate decomposes to form dinitrogen monoxide gas and liquid water, how many grams of dinitrogen monoxide and water are formed? Step 1: Write the balanced chemical equation NH4NO3(s)  N2O(g) + 2 H2O(l) Step 2:Convert mass of reactant (454 g NH4NO3) to moles. = 5.67 mol NH4NO3

  21. NH4NO3(s)  N2O(g) + 2 H2O(g) Consider the following reaction: If 454 g of solid ammonium nitrate decomposes to form dinitrogen monoxide gas and liquid water, how many grams of dinitrogen monoxide and water are formed? Step 3: Using the balanced chemical equation, convert moles of reactant (NH4NO3) to moles of products.

  22. NH4NO3(s)  N2O(g) + 2 H2O(g) Consider the following reaction: If 454 g of solid ammonium nitrate decomposes to form dinitrogen monoxide gas and liquid water, how many grams of dinitrogen monoxide and water are formed? Step 3: Using the balanced chemical equation, convert moles of reactant (NH4NO3) to moles of products. 5.67 mol NH4NO3 = 5.67 mol N2O

  23. NH4NO3(s)  N2O(g) + 2 H2O(g) Consider the following reaction: If 454 g of solid ammonium nitrate decomposes to form dinitrogen monoxide gas and liquid water, how many grams of dinitrogen monoxide and water are formed? Step 3: Using the balanced chemical equation, convert moles of reactant (NH4NO3) to moles of products. 5.67 mol NH4NO3 = 5.67 mol N2O 5.67 mol NH4NO3 = 11.3 mol H2O

  24. NH4NO3(s)  N2O(g) + 2 H2O(g) Consider the following reaction: If 454 g of solid ammonium nitrate decomposes to form dinitrogen monoxide gas and liquid water, how many grams of dinitrogen monoxide and water are formed? Step 3: Using the balanced chemical equation, convert moles of reactant (NH4NO3) to moles of products. 5.67 mol NH4NO3 = 5.67 mol N2O 5.67 mol NH4NO3 = 11.3 mol H2O Recall that the coefficients represent the molar ratios of reactants to products and vice versa…

  25. Consider the following reaction: If 454 g of solid ammonium nitrate decomposes to form dinitrogen monoxide gas and liquid water, how many grams of dinitrogen monoxide and water are formed? Step 4: Now convert the moles of products (N2O and H2O) to grams using the molar mass of each.

  26. Consider the following reaction: If 454 g of solid ammonium nitrate decomposes to form dinitrogen monoxide gas and liquid water, how many grams of dinitrogen monoxide and water are formed? Step 4: Now convert the moles of products (N2O and H2O) to grams using the molar mass of each. 5.67 mol N2O = 250. g N2O

  27. Consider the following reaction: If 454 g of solid ammonium nitrate decomposes to form dinitrogen monoxide gas and liquid water, how many grams of dinitrogen monoxide and water are formed? Step 4: Now convert the moles of products (N2O and H2O) to grams using the molar mass of each. 5.67 mol N2O = 250. g N2O 11.3 mol H2O = 204 g H2O

  28. Consider the following reaction: If 454 g of solid ammonium nitrate decomposes to form dinitrogen monoxide gas and liquid water, how many grams of dinitrogen monoxide and water are formed? Step 4: Now convert the moles of products (N2O and H2O) to grams using the molar mass of each. 5.67 mol N2O = 250. g N2O 11.3 mol H2O = 204 g H2O = 204 g H2O …or 454g NH4NO3  250. g N2O

  29. Consider the following reaction: If 454 g of solid ammonium nitrate decomposes to form dinitrogen monoxide gas and liquid water, how many grams of dinitrogen monoxide and water are formed? Look what happens! Compound NH4NO3 N2O H2O Initial (g) Initial (mol) Change (mol) Final (mol) Final (g) 454 g 0 g 0 g 5.67 mol 0 mol 0 mol  5.67 mol + 5.67 mol +11.3 mol 0 mol + 5.67 mol +11.3 mol 0 g 250. g 204 g Mass is conserved but Moles are not!

  30. Stoichiometry Problems • In order to solve stoichiometry problems, one must go through moles using molar masses and mole ratios as conversion factors. • One cannot do this without writing a balanced chemical equation first. grams  moles  moles  grams  Using molar mass  Using mole ratios  Using molar mass

  31. Reactions Involving a Limiting Reactant • In some cases involving two or more reactants, there is insufficient amount of one reactant to consume the other reactants completely. • The reactant that is in short supply therefore LIMITSthe quantity of product that can be formed. • The theoretical yield of products is limited by this “Limiting Reactant”.

  32. Reaction Yields • The theoretical yieldis the maximum product yield that can be expected based on the masses of the reactants and the reaction stoichiometry. • The actual yieldis the experimentally measured amount of products that results upon completion of the reaction. • The percent yieldis a measure of the extent of the reaction in terms of the actual vs. the theoretical yield.

  33. Consider the reaction of aluminum and oxygen: • Which is the limiting reactant if we start with 50.0 g Al and 50.0 g O2? • What is the Theoretical Yield for the reaction?

  34. Consider the reaction of aluminum and oxygen: • Which is the limiting reactant if we start with 50.0 g • Al and 50.0 g O2? • What is the Theoretical Yield for the reaction? Al + O2Al?O? Figure out oxid #s and formula Al(s) + O2(g)  2Al2O3(s) Now balance 4Al(s) + 3O2(g)  2Al2O3(s)

  35. Consider the reaction of aluminum and oxygen: • Which is the limiting reactant if we start with 50.0 g Al and 50.0 g O2? • What is the Theoretical Yield for the reaction? 4Al(s) + 3O2(g)  2Al2O3(s) g Al(initial) mols Al(initial)  g O2(initial) mols O2(initial) 

  36. Consider the reaction of aluminum and oxygen: • Which is the limiting reactant if we start with 50.0 g Al and 50.0 g O2? • What is the Theoretical Yield for the reaction? 4Al(s) + 3O2(g)  2Al2O3(s) Compare moles of Al to moles of O2 What you have What you need Since 1.19 < 1.33, Aluminum is the Limiting Reactant. You don’t have enough Al to consume all that oxygen

  37. Consider the reaction of aluminum and oxygen: • Which is the limiting reactant if we start with 50.0 g Al and 50.0 g O2? • What is the Theoretical Yield for the reaction? 4Al(s) + 3O2(g)  2Al2O3(s) Calculate the moles and grams of product based on the limiting reactant, Al.

  38. Consider the reaction of Al and O2 previously discussed: From the calculations it was determined that 94.3g of Al2O3 could be formed from the reaction of 50.0 g of Al and 50.0 g of O2. This is the theoreticalyield.

  39. Consider the reaction of Al and O2 previously discussed: From the calculations it was determined that 94.3g of Al2O3 could be formed from the reaction of 50.0 g of Al and 50.0 g of O2. This is the theoreticalyield. If 48.5 g of Al2O3 was collected at the completion of the reaction (actualyield), what is the % yield for the reaction?

  40. Consider the reaction of Al and O2 previously discussed: From the calculations it was determined that 94.3g of Al2O3 could be formed from the reaction of 50.0 g of Al and 50.0 g of O2. This is the theoreticalyield. If 48.5 g of Al2O3 was collected at the completion of the reaction (actualyield), what is the % yield for the reaction? = 51.4 %

  41. Consider the reaction of Al and O2 previously discussed: From the calculations it was determined that 94.3g of Al2O3 could be formed from the reaction of 50.0 g of Al and 50.0 g of O2. This is the theoreticalyield. How many grams of oxygen remain?

  42. Consider the reaction of Al and O2 previously discussed: From the calculations it was determined that 94.3g of Al2O3 could be formed from the reaction of 50.0 g of Al and 50.0 g of O2. This is the theoreticalyield. How many grams of oxygen remain? By conservation of mass: mass of reactants = mass of products 4Al(s) + 3O2(g)  2Al2O3(s) 5.7g O2remains Only 44.3 gms of the 50 gms of O2 got consumed

  43. Chemical Analysis Find and weigh the product of reacting Na2SO4 with BaCl2

  44. Determining the Formula of a Hydrocarbon by Combustion Traps • Combustion involves the addition oxygen to another element. • When hydrocarbon molecules burn completely, the products are always carbon dioxide gas and water.

  45. Combustion Analysis A hydrocarbon is a compound that contains predominately carbon and hydrogen and possibly smaller amount of elements such as O, N and S. • A streaming flow of pure O2 is passed over a weighed mass of the unknown sample in an apparatus like the one above. • The heat from the furnace converts the carbon in the compound to CO2 and the hydrogen into H2O. • The CO2 and H2O are collected in pre-weighed traps.

  46. Combustion Analysis mass trap + H2O  mass trap = mass H2O mass trap + CO2  mass trap = mass CO2 When combustion is completed, the traps are weighed and the masses of CO2 and H2O are determined.

  47. Combustion Analysis • All of the carbon in the sample ends up in the CO2. • All of the hydrogen in the sample ends up in the water. • Using stoichiometry, one can relate the carbon in CO2 and the hydrogen in H2O back to the carbon and hydrogen in the original sample. • Any oxygen in the sample can be deduced via conservation of mass.

  48. Combustion Analysis Problem: • 1.516 g of a compound containing carbon, hydrogen and oxygen (CXHYOZ) is subjected to combustion analysis. • The results show that 2.082 g of CO2 and 1.705 g of H2O were produced. • What is the empirical formula for the compound? • If the molecular weight of the compound is 160.2 g/mol, what is the molecular formula of the compound? 1. Write out a tentative reaction CxHyOz+ O2(g)  CO2(g) + H2O (l)

  49. 1.516 g of a compound containing carbon, hydrogen and oxygen (CXHYOZ) is subjected to combustion analysis. • The results show that 2.082 g of CO2 and 1.705 g of H2O were produced. Figure out the moles involved g CO2 mol CO2 mol C

  50. 1.516 g of a compound containing carbon, hydrogen and oxygen (CXHYOZ) is subjected to combustion analysis. • The results show that 2.082 g of CO2 and 1.705 g of H2O were produced. Figure out the moles involved g CO2 mol CO2 mol C g H2O mol H2O mol H

More Related