1. Stoichiometry © 2009, Prentice-Hall, Inc. Chapter 3 Stoichiometry Definition: Mathematical calculations for chemical formulas & equations. Mrs. Deborah Amuso Camden High School Camden, NY 13316 Chemistry, The Central Science, 11th edition Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten

2. Stoichiometry © 2009, Prentice-Hall, Inc. Chemical Equations Chemical equations are concise representations of chemical reactions. (example) CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

3. Stoichiometry © 2009, Prentice-Hall, Inc. Anatomy of a Chemical Equation CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g) Reactants: Products: appear on the appear on the left side of the right side of the equation equation The states are written in parentheses. Coefficients are inserted to balance the equation.

4. Stoichiometry © 2009, Prentice-Hall, Inc. Subscripts vs Coefficients Subscripts tell the number of atoms of each element in a molecule. H 2 O (The 2 means 2 atoms of H) Coefficients tell the number of molecules. 3 H 2 O (The 3 means 3 molecules of water)

5. Stoichiometry Counting Atoms Total # atoms = Coefficient x Subscript Practice: 3 H 2 SO 4 6-H 3-S 12-O or 3 – SO 4 2 Ca 3 (PO 4 ) 2 6-Ca 4-P 16-O or 4 – PO 4 © 2009, Prentice-Hall, Inc.

6. Stoichiometry Balancing Equations When balancing equations use coefficients to get the # of each type of atom on both sides of the equation the same. Consult your teacher for balancing tips. © 2009, Prentice-Hall, Inc.

7. Stoichiometry ReactionTypes 1.Combination 2.Decomposition 3.Combustion 4.Single Replacement 5.Double Replacement © 2009, Prentice-Hall, Inc.

8. Stoichiometry © 2009, Prentice-Hall, Inc. Combination Reactions Examples: 2 Mg (s) + O 2 (g)  2 MgO (s) N 2 (g) + 3 H 2 (g)  2 NH 3 (g) C 3 H 6 (g) + Br 2 (l)  C 3 H 6 Br 2 (l) In a combination reaction two or more substances react to form one product.

9. Stoichiometry © 2009, Prentice-Hall, Inc. In a decomposition reaction one reactant breaks down into two or more substances. Decomposition Reactions Examples: CaCO 3 (s)  CaO (s) + CO 2 (g) 2 KClO 3 (s)  2 KCl (s) + O 2 (g) 2 NaN 3 (s)  2 Na (s) + 3 N 2 (g)

10. Stoichiometry © 2009, Prentice-Hall, Inc. Combustion Reactions Examples: CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (g) C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (g) In a combustion reaction a carbon compound reacts with oxygen in the air to produce carbon dioxide & water. These are generally rapid reactions that produce a flame.

11. Stoichiometry Single Replacement Reactions In a single replacement reaction, an element & compound react to form a different element & compound. © 2009, Prentice-Hall, Inc. Examples: Zn (s) + CuSO 4 (aq)  Cu (s) + H 2 SO 4 (aq) 2 HCl (aq) + Mg (s)  MgCl 2 (aq) + H 2(g)

12. Stoichiometry Double Replacement Reactions In a double replacement reaction, two compounds in solution react by exchanging ions to form two different compounds. © 2009, Prentice-Hall, Inc. Examples: ZnCl 2 (aq) + 2 AgNO 3 (aq)  2AgCl (s) + Zn(NO 3 ) 2(aq) Ba(NO 3 ) 2(aq) + Na 2 SO 4(aq)  BaSO 4(s) + 2 NaNO 3 (aq)

13. Stoichiometry © 2009, Prentice-Hall, Inc. Formula Mass (FM) A formula mass is the sum of the atomic masses for the atoms in a chemical formula. So, the formula mass of calcium chloride, CaCl 2, would be Ca: 1(40.1 amu) = 40.1 + Cl: 2(35.5 amu) = 71.0 111.1 amu Formula masses are generally reported for ionic compounds.

14. Stoichiometry © 2009, Prentice-Hall, Inc. Molecular Mass (MM) A molecular mass is the sum of the atomic masses for the atoms in a molecule. For the molecule ethane, C 2 H 6, the molecular mass would be FM & MM are found the same way! C: 2(12.0 amu) = 24.0 30.0 amu + H: 6(1.0 amu) = 6.0

15. Stoichiometry © 2009, Prentice-Hall, Inc. Mass Percent or Percent Composition The mass percent of each element in a compound is found by using this equation: % element = (number of atoms)(atomic mass) (FM of the compound) x 100 Or simply: % = Part/Whole x 100

16. Stoichiometry © 2009, Prentice-Hall, Inc. Percent Composition So the percentage of carbon in ethane, C 2 H 6, is… %C = (2)(12.0 amu) (30.0 amu) 24.0 amu 30.0 amu = x 100 = 80.0%

17. Stoichiometry © 2009, Prentice-Hall, Inc. Moles & Avogadro’s Number 1 mole = 6.02 x 10 23 The mass of 1 molecule of water is 18.0 amu. The mass of 1 mole of water is 18.0 grams. Why? 1 gram = 6.02 x 10 23 amu.

18. Stoichiometry © 2009, Prentice-Hall, Inc. Molar Mass or GFM Molar mass is the mass of 1 mole of a substance in g/mol. –The molar mass of an element is the mass number that we find on the periodic table. example: Cu = 63.5 O 2 = 2(16.0) = 32 –The molar mass of a compound is the same number as its formula mass or molecular mass. example: C 2 H 6 = 30.0

19. Stoichiometry © 2009, Prentice-Hall, Inc. Using Moles Moles provide a bridge from the molecular scale to the real-world scale. # moles = given mass GFM GFM is the same as Molar Mass (It’s the mass in grams of 1 mole of the substance)

20. Stoichiometry Three Helpful Equations Use these 3 equations to convert between masses, moles, molecules (or units), & atoms (or ions): #moles = mass/GFM #molecules = (#moles) (6.02 x 10 23 ) (or # units) #atoms = (#molecules) (#atoms/molecule) (or # ions) © 2009, Prentice-Hall, Inc.

21. Stoichiometry Mole Relationships: Check Your Understanding How many atoms of oxygen are in 10.0 grams of CH 3 COOH? Think: Mass  Moles  Molecules  Atoms #moles = mass/GFM = 10.0g / 60.0 g/mol = 0.167 moles #molecules = (#moles)((6.02 x 10 23 ) = (0.167 moles) (6.02 x 10 23 ) = 1.00 x 10 23 molecules #atoms = (#molecules) (#atoms/molecule) = (1.00 x 10 23 molecules)(2 oxygen atoms/molecule) = 2.00 x 10 23 oxygen atoms © 2009, Prentice-Hall, Inc.

22. Stoichiometry Mole Relationships: Check Your Understanding How many hydroxide ions are in 10.0 grams of Ca(OH) 2 Think: Mass  Moles  Units  Ions #moles = mass/GFM = 10.0g / 74.1g/mol = 0.135 moles Ca(OH) 2 #units = (#moles) (6.02 x 10 23 ) = (0.135 mol) (6.02 x 10 23) = 8.12 x 10 22 #ions = (#units) (#ions/unit) = (8.12 x 10 22 ) (2 OH - /unit) = 1.62 x 10 23 OH - ions © 2009, Prentice-Hall, Inc.

23. Stoichiometry © 2009, Prentice-Hall, Inc. Calculating Empirical Formulas One can calculate the empirical formula from the percent composition of each element in the formula. Steps: 1)Assume percent of each element = its mass 2)Divide mass of each element by that element’s atomic mass to get # moles 3)Calculate the mole ratio by dividing by the smallest # of moles 4)These are the #’s in the empirical formula

24. Stoichiometry © 2009, Prentice-Hall, Inc. Calculating Empirical Formulas The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA. C:61.31 g / 12.00 g/mol = 5.105 mol C H: 5.14 g / 1.01 g/mol = 5.09 mol H N:10.21 g / 14.01 g/mol = 0.7288 mol N O:23.33 g / 16.00 g/mol = 1.456 mol O

25. Stoichiometry © 2009, Prentice-Hall, Inc. Calculating Empirical Formulas Calculate the mole ratio by dividing by the smallest number of moles: C:= 7.005  7 H:= 6.984  7 N:= 1.000 O:= 2.001  2 5.105 mol 0.7288 mol 5.09 mol 0.7288 mol 1.458 mol 0.7288 mol

26. Stoichiometry © 2009, Prentice-Hall, Inc. Calculating Empirical Formulas These are the subscripts for the empirical formula: C 7 H 7 NO 2

27. Stoichiometry © 2009, Prentice-Hall, Inc. Calculating Empirical Formulas Find the empirical formula of a compound whose mass percentage is 50.0% sulfur and 50.0% oxygen. Step 1 S:50.0 g / 32.1 g/mol = 1.56 mol S O:50.0 g / 16.0 g/mol = 3.13 mol O Step 2 S: 1.56 / 1.56 = 1 O: 3.13 / 1.56 = 2 Step 3 Empirical Formula is SO 2

28. Stoichiometry Calculating Molecular Formulas From Empirical Formulas & Molar Mass What is the MF if the EF is C 4 H 5 N 2 O and the MM is 194 g/mol? Step 1: Determine molar mass of empirical formula C 4 H 5 N 2 O = 97 g/mol Step 2: Divide: Molar Mass given/Molar Mass of empirical formula 194 / 97 = 2 Step 3: Multiply: Answer from Step 2 * Empirical Formula (2)(C 4 H 5 N 2 O) = C 8 H 10 N 4 O 2 © 2009, Prentice-Hall, Inc.

29. Stoichiometry Calculating Molecular Formulas From Empirical Formulas & Molar Mass What is the MF if the EF is CH 2 and the molecular mass is 84.16 amu? Step 1: Determine molecular mass of empirical formula CH 2 = 14.03 amu Step 2: Divide: Molecular mass given/molecular mass of emp formula 84.16 amu /14.03 amu = 6 Step 3: Multiply: Answer from Step 2 * Empirical Formula (6)(CH 2 ) = C 6 H 12 © 2009, Prentice-Hall, Inc.

30. Stoichiometry © 2009, Prentice-Hall, Inc. Combustion Analysis Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this. –C is determined from the mass of CO 2 produced. –H is determined from the mass of H 2 O produced. –O is determined by difference after the C and H have been determined. –Then you can determine the Empirical Formula of the compound

31. Stoichiometry Combustion Analysis When 0.255 g of caproic acid is burned, you get 0.512 g CO 2 and 0.209 g H 2 O. What is the emp formula of the acid? #mole CO 2 = mass/GFM = 0.512g / 44.0g/mol = 0.0116 mol 0.0116 mol CO 2 * (1 mole C/1 mol CO 2 ) = 0.0116 mol C 0.0116 mol C * (12.0 g/mol) = 0.140 g C #mole H 2 O = mass/GFM = 0.209g / 18.0g/mol = 0.0116 mol 0.0116 mol H 2 O * (2 mole H/1 mol H 2 O)= 0.0232 mol H 0.0232 mol H * (1.01 g/mol) = 0.0235 g H Mass O = Mass caproic acid – (mass C + mass H) Mass O = 0.225 g – (0.140 g + 0.0235 g) =.0615 g © 2009, Prentice-Hall, Inc.

32. Stoichiometry 2 nd Part of Problem For carbon: 0.140 g C / 12.0 g/mol = 0.0117 mol For hydrogen: 0.0235 g H / 1.01 g/mol = 0.0233 mol For oxygen: 0.0615 g O / 16.0 g/mol = 0.00384 mol For carbon: 0.0117 / 0.00384 = 3.05 For hydrogen: 0.0233 / 0.00384 = 6.07 For oxygen: 0 0.00384 / 0.00384 = 1 Empirical Formula = C 3 H 6 O © 2009, Prentice-Hall, Inc.

33. Stoichiometry © 2009, Prentice-Hall, Inc. Stoichiometric Calculations The coefficients in the balanced equation represent the # of molecules or # of moles of reactants and products. The coefficients tell the ‘recipe’ for the rxn.

34. Stoichiometry Stoichiometric Calculations © 2009, Prentice-Hall, Inc.

35. Stoichiometry © 2009, Prentice-Hall, Inc. Stoichiometric Calculations Starting with 1.00 g of C 6 H 12 O 6 … we calculate the moles of C 6 H 12 O 6 … use the coefficients to find the moles of H 2 O… and then turn the moles of water to grams. C 6 H 12 O 6 + 6 O 2  6 CO 2 + 6 H 2 O

36. Stoichiometry Stoichiometric Calculation Steps 1)Complete & balance equation using coefficients. 2) Identify Given & Unknown. 3) Convert Given mass into moles: #moles given = mass given /GFM given 4) Use mole ratio to find moles of Unknown. #moles given = #moles unk Coeff given Coeff unk 5) Convert moles of ‘unknown’ to answer. mass unknown = (#moles unk ) (GFM unk ) © 2009, Prentice-Hall, Inc.

37. Stoichiometry CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g) How many grams of water are produced when 32.0 grams of methane completely react? 1)Given = 32.0 g CH 4 Unk = ? g H 2 O 2)Find # moles CH 4 #moles CH 4 = mass/GFM = 32.0 g / 16.0 g/mol = 2.00 moles CH 4 3) Find # moles H 2 O 2.00 moles CH 4 = ? moles H 2 O 1 mole 2 mole Ans. = 4.00 moles H 2 O 4) Find mass of H 2 O mass H 2 O = (#moles)(GFM) = (4.00 mol)(18.0 g/mol) = Ans. = 72.0 g H 2 O © 2009, Prentice-Hall, Inc.

38. Stoichiometry 2 H 2 (g) + O 2 (g) 2 H 2 O (g) How many grams of water are produced when 6.0 grams of oxygen gas completely react? 1)Given = 6.0 g O 2 Unk = ? g H 2 O 2)Find # moles O 2 #moles O 2 = mass/GFM = 6.0 g / 32.0 g/mol = 0.19 moles O 2 3) Find # moles H 2 O 0.19 moles O 2 = ? moles H 2 O 1 mole 2 mole Ans. = 0.38 moles H 2 O 4) Find mass of H 2 O mass H 2 O = (#moles)(GFM) = (0.38 mol)(18.0 g/mol) = Ans. = 6.8 g H 2 O © 2009, Prentice-Hall, Inc.

39. Stoichiometry © 2009, Prentice-Hall, Inc. Limiting Reactants: How Many Cookies Can I Make? You can make cookies until you run out of one of the ingredients. Once you run out of sugar, you can’t make any more cookies. In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make.

40. Stoichiometry © 2009, Prentice-Hall, Inc. Limiting Reactants The limiting reactant (or limiting reagent) is the reactant you’ll run out of first What is the LR in this reaction? H 2

41. Stoichiometry How to Determine Which Reactant is the Limiting Reactant Step 1: Convert the masses given to moles. Step 2: Use these equations: # moles reactant A / coefficient A = x # moles reactant B / coefficent B = y Step 3: The smaller # (x or y) identifies the Limiting Reagent. Step 4: Use the # of moles of the Limiting Reagent to solve the rest of the problem. © 2009, Prentice-Hall, Inc.

42. Stoichiometry CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g) Given 50.0 g of CH 4 and 60.0 g of O 2 Which is the Limiting Reagent? Step 1: #moles CH 4 = mass/GFM = 50.0 g/16.0 g/mol = 3.13 moles #moles O 2 = mass/GFM = 60.0 g/32.0 g/mol = 1.88 moles Step 2: Use these equations: # moles reactant CH 4 / coefficient CH 4 = 3.13 / 1 = 3.12 # moles reactant O 2 / coefficent O 2 = 1.88 / 2 = 0.94 Step 3: The smaller # (x or y) identifies the Limiting Reagent 0.94 is smaller, so O 2 is the L.R. © 2009, Prentice-Hall, Inc.

43. Stoichiometry 2 CHCl 3 (g) + 2 Cl 2 (g) 2 CClO 4 (g) + 2 HCl (g) Given 15.9 g of CHCl 3 and 12.6 g of Cl 2 Which is the Limiting Reagent? Step 1: #moles CHCl 3 = mass/GFM = 15.9 g/119.5 g/mol = 0.133 mol #moles Cl 2 = mass/GFM = 12.6 g/71.0 g/mol = 0.177 mol Step 2: Use these equations: # mol reactant CHCl 3 / coefficient CHCl 3 = 0.133/ 2 = 0.0665 # mol reactant Cl 2 / coefficent Cl 2 = 0.177/ 2 = 0.0885 Step 3: The smaller # (x or y) identifies the Limiting Reagent 0.0665 is smaller, so CHCl 3 is the L.R. © 2009, Prentice-Hall, Inc.

44. Stoichiometry © 2009, Prentice-Hall, Inc. Theoretical Yield The theoretical yield is the maximum amount of product that can be made. –In other words it’s the amount of product possible as calculated through the stoichiometry problem. This is different from the actual yield, which is the amount one actually produces and measures.

45. Stoichiometry © 2009, Prentice-Hall, Inc. Percent Yield One finds the percent yield by comparing the amount actually obtained (actual yield) to the amount it was possible to make (theoretical yield). Actual Yield Theoretical Yield Percent Yield =x 100

46. Stoichiometry Percent Yield Problem CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g) Previously, you determined 72.0 g of water should be produced from 32.0 g of methane. Assume only 68.0 g of water are produced when this experiment is actually conducted. What is the percent yield? % yield = [actual yield/theoretical yield] x 100 % yield = [68.0g / 72.0 g] x 100 % yield = 94.4% © 2009, Prentice-Hall, Inc.