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Stoichiometry

Stoichiometry : study of the quantitative relations between amounts of reactants and products. Stoichiometry. Goals: Perform stoichiometry calculations . Understand the meaning of limiting reactant . Calculate theoretical and percent yields of a chemical reaction.

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Stoichiometry

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  1. Stoichiometry: study of the quantitative relations between amounts of reactants and products. Stoichiometry Goals: Perform stoichiometry calculations. Understand the meaning of limiting reactant. Calculate theoretical and percent yields of a chemical reaction. Use stoichiometry to analyze a mixture of compounds or to determine the formula of a compound. Define and use molarity in solution stoichiometry. Perform calculations for pH and titration problems.

  2. What is STOICHIOMETRY? The study of the quantitative aspects of chemical reactions. It rests on the principle of the _________ ________________. 2 Al(s) + 3 Br2(liq) ------> Al2Br6(s) You must always begin with a balanced equationbefore carrying out a stoichiometry calculation.

  3. Write a Chemical Equation • Phosphine, PH3 (g), combusts in oxygen gas to form gaseous water and solid tetraphosphorus decoxide. Always check (and REcheck) the balancing

  4. Information from a Balanced Equation Equation: 2 H2 (g) + O2 (g) 2 H2O (l) Molecules: 2 molecules H2 + 1 molecule O2 2 moleculesH2O Mass (amu): 4.0 amu H2 + 32 amu O2 36.0 amu H2O Amount (mol): Mass (g):

  5. General Plan for Stoichiometry Calculations Students should become familiar with stoichiometry calculations.

  6. Mole relationships in Chemical Equations • Stoichiometric factor – relates the amounts of any two substances involved in a chemical reaction, on a mole basis. C3H8 + O2 CO2 + H2O 3 4 5 propane 5 mol of O2 are required to burn ____ mol of C3H8 ____ mol of H2O are produced for every 1 mol of C3H8 burned ____ mol of CO2 are produced for every 1 mol of C3H8 burned

  7. If 454 g of NH4NO3 decomposes, how much N2O and H2O are formed? What is the theoretical yield of products? STEP 1 Write the balanced chemical equation. STEP 2Convert mass of reactant to moles of reactant.(454 g) --> moles STEP 3Convert moles reactant to moles product. A. Relate moles using coefficients, write a STOICHIOMETRIC FACTOR.

  8. STEP 3 B) Multiply moles of reactant by the stoichiometric factor. STEP 4Convert moles product to mass product = ___________________ YIELD

  9. STEP 5How much N2O is formed? Total mass of reactants = total mass of products

  10. STEP 6Calculate the percent yield. If you isolated only 131 g of N2O, what is the percent yield? Actual yield % yield = * 100 Theoretical yield Students should become familiar with % yield calculations.

  11. What is a Limiting Reactant? • In a given reaction, there is not enough of one reagent to use up the other reagent completely. • The reagent in short supply _______the quantity of product that can be formed.

  12. Which is the Limiting Reactant? 2 NO (g) + O2 (g) - - -> 2 NO2 (g) • Limiting reactant is • Excess reactant is Reactants Products

  13. C3H8 + O2 CO2 + H2O • How many grams of CO2 are produced from 50 g of propane? (theoretical yield) • How much O2 is required to completely burn the 50 g of propane? • How much H2O is formed? • If the actual yield of CO2 was 145 g. What is the % yield of the reaction? 50 g C3H8 (1 mol/ 44 g) = 1.14 mol C3H8 * (4 mol H2O / 1 mol C3H8)

  14. C3H8 + O2 CO2 + H2O • If you begin with 99.5 g of C3H8 and 211 g of O2, which one is the limiting reactant? • Assuming all the limiting reactant has reacted, how much CO2 will be formed?

  15. 1 2 3 Limiting Reactant React solid Zn with 0.100 mol HCl (aq) Zn + 2 HCl ---> ZnCl2 + H2 (See CD Screen 4.8) Rxn 1: Balloon inflates fully, some Zn left * More than enough Zn to use up the 0.100 mol HCl Rxn 2: Balloon inflates fully, no Zn left * Right amount of each (HCl and Zn) Rxn 3: Balloon does not inflate fully, no Zn left. * Not enough Zn to use up 0.100 mol HCl

  16. 1 2 3 Zn + 2 HCl ---> ZnCl2 + H2 0.10 mol of HCl, need ? mol Zn. 0.10 mol HCl 1 mol Zn 2 mol HCl = 0.050 mol Zn Rxn 1Rxn 2 Rxn 3 mass Zn (g) 7.00 3.27 1.31 mol Zn 0.107 0.050 0.020 mol HCl 0.100 0.100 0.100 mol HCl/mol Zn Lim Reactant

  17. Mass product Mass reactant Moles reactant Moles product Mix 5.40 g of Al with 8.10 g of Cl2. What mass of Al2Cl6 can form? ___Al + ____Cl2---> Al2Cl6 Stoichiometric factor

  18. Mix 5.40 g of Al with 8.10 g of Cl2. What mass of Al2Cl6 can form? STEP 1 FIND THE LIMITING REAGENT. Compare actual mole ratio of reactants to theoretical mole ratio. ___ Al + ___Cl2---> Al2Cl6 Students should become familiar with calculations using the concept of limiting reagent.

  19. Mix 5.40 g of Al with 8.10 g of Cl2. What mass of Al2Cl6 can form? Calculate moles of each reactant. Find the mole ratio of reactants:

  20. mass ____ mass Al2Cl6 moles ____ moles Al2Cl6 Mix 5.40 g of Al with 8.10 g of Cl2. What mass of Al2Cl6 can form? __ Al + ___ Cl2 ---> Al2Cl6 Limiting reactant = _____ BASE ALL CALCULATIONS on LR ____ Write conversion factor:

  21. Mix 5.40 g of Al with 8.10 g of Cl2. What mass of Al2Cl6 can form? STEP 2CALCULATE THE MASS OF THE PRODUCT. Calculate moles of Al2Cl6 expected based on LR. __ Al + __ Cl2 ---> Al2Cl6

  22. Mix 5.40 g of Al with 8.10 g of Cl2.How much of each reactant will remainwhen reaction is complete? __ Al + __ Cl2 ---> Al2Cl6

  23. Chemical Analysis • An impure sample of the mineral thenardite contains Na2SO4. A mass of mineral sample weights 0.123 g. The Na2SO4 in the sample is converted to insoluble BaSO4 by adding BaCl2. The recovered mass of BaSO4 is 0.177 g. What is the mass percentof Na2SO4 in the mineral?

  24. Chemical Analysis

  25. General Plan for Stoichiometry Calculations

  26. Chemical Analysis Balanced equation: Grams to moles Moles to moles Moles to grams Mass %

  27. Combustion Analysis of Hydrocarbons Active Figure 4.9

  28. Procedure for Calculating Empirical Formula Use Molar mass Moles of each element Grams of each element Calculate mole ratio Empirical Formula The central part of the calculation is determining the number of moles of each element in the compound. Remember: in the mole ratio, divide by smaller number, then multiply ‘til whole.

  29. What is the empirical formula of a hydrocarbon, CxHy, if 0.115 g burn and produce 0.379 g of CO2 and 0.1035 g of H2O. CxHy + some oxygen ---> 0.379 g CO2 + 0.1035 g H2O First, recognize that all C in CO2 and all H in H2O is from CxHy. 1. Calculate amount (in moles) of C in CO2 2. Calculate amount (in moles) of H in H2O 3. Find ratio of mol H/mol C to find values of x and y in CxHy.

  30. What is the empirical formula of a hydrocarbon, CxHy, if 0.115 g burn and produce 0.379 g of CO2 and 0.1035 g of H2O. 1. Calculate amount (moles) of C in CO2 2. Calculate amount (moles) of H in H2O 3. Ratio of mol H/mol C to find values of x and y in CxHy.

  31. Practice 4.67 Titanium (IV) oxide, TiO2, is heated in hydrogen gas to give water and a new titanium oxide, TixOy. If 1.598 g of TiO2 produces 1.438 g of TixOy, what is the formula of the new oxide? H2O + TixOy TiO2 + H2 • Calculate amount of Ti:

  32. Practice… 2. Calculate amount of O: 3. Calculate the molar ratio of O to T

  33. How are Reactions in Solution Quantified? In solution we need to define the - • SOLVENT the component whose physical state is ____________ when solution forms • SOLUTE the other solution component

  34. What is Molarity? The amount of solute in a solution is given by its concentration. Molarity = Students should become familiar with calculations using MOLARITY.

  35. Calculate molarity of a solution of 5.00 g of NiCl2•6 H2O dissolved in enough water to make 250 mL of solution. STEP 1 Calculate the number of moles of solute STEP 2 Calculate the molarity of the solution

  36. How many IONS are in the Solution?

  37. What mass of oxalic acid, H2C2O4, is required to make 250 mL of a 0.0500 M solution? STEP 1Calculate moles of acid required. STEP 2Calculate mass of acid required.

  38. Preparing Solutions • Weigh out a solid solute and dissolve in a given quantity of solvent. • Dilute a concentrated solution to give one that is __________ concentrated.

  39. Preparing Solutions

  40. You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Add water to the 3.0 M solution to lower its concentration to 0.50 M Dilute the solution! M = moles/L Notice that the amount of NaOH (moles of NaOH) present did not change.

  41. You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? But how much water do we add? Amount of NaOH in original sol. = Amount of NaOH in final sol. must also = Volume of final solution = moles of NaOH in ORIGINAL solution = moles of NaOH in FINAL solution

  42. You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Conclusion: add ______ of water to the 50.0 mL of 3.0 M NaOH to make ________ of 0.50 M NaOH. A shortcut: Cinitial • Vinitial = Cfinal • Vfinal • Principle of dilution: addition of solvent does not change the amount of solute in a solution but does change its concentration.

  43. Solution Stoichiometry • Zinc reacts with acids to produce H2 gas. • Have 10.0 g of Zn • What volume of 2.50 M HCl is needed to convert the Zn completely?

  44. General Plan for Solution Stoichiometry M = moles / volume Moles = M * volume

  45. Stoichiometric factor Zinc reacts with acids to produce H2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely?

  46. Zinc reacts with acids to produce H2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely? Step 1: Write the balanced equation Step 2: Calculate amount of Zn in moles Step 3: Use the stoichiometric factor

  47. Zinc reacts with acids to produce H2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely? Step 4: Calculate volume of HCl req’d * Use molarity (M) as conversion factor

  48. What is pH? • It is a concentration scale. • pH: a way to express acidity – the concentration of H+in solution. High pH: ____ [H+] Low pH: ______ [H+] Acidic solution pH < 7 Neutral pH = 7 Basic solution pH > 7

  49. The pH Scale pH = log (1/ [H+]) = - log [H+] In a neutral solution, [H+] = [OH-] = 1.00 x 10-7 M at 25 oC pH = - log [H+] = -log (1.00 x 10-7) = - [0 + (-7)] = 7 See CD Screen 5.17 for a tutorial See book Appendix A.3 for more on logs

  50. pH and [H+] If the [H+] of soda is 1.6 x 10-3 M, the pH is ____?

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