2. STOICHIOMETRY Chapter 3 - the study of the quantitative aspects of chemical reactions.

3. Counting Atoms Chemistry is a quantitative science—we need a “counting unit.” 1 mole is the amount of substance that contains as many particles (atoms, molecules) as C atoms in 12.0 g of 12 C. MOLE

4. Particles in a Mole 6.02214199 x 10 23 Avogadro’s Number There is Avogadro’s number of particles in a mole of any substance. Amedeo Avogadro 1776-1856

5. Molar Mass 1 mol of 12 C = 12.00 g of C = 6.022 x 10 23 atoms of C 12.00 g of 12 C is its MOLAR MASS Taking into account all of the isotopes of C, the molar mass of C is 12.011 g/mol

6. One-mole Amounts

7. PROBLEM: What amount of Mg is represented by 0.200 g? How many atoms? Mg has a molar mass of 24.3050 g/mol. = 4.95 x 10 21 atoms Mg How many atoms in this piece of Mg?

8. MOLECULAR WEIGHT AND MOLAR MASS Molecular weight = sum of the atomic weights of all atoms in the molecule. Molar mass = molecular weight in grams per mol.

9. What is the molar mass of ethanol, C 2 H 6 O? 1 mol contains 2 moles of C (12.01 g C/1 mol) = 24.02 g C 6 moles of H (1.01 g H/1 mol) = 6.06 g H 1 mol of O (16.00 g O/1 mol) = 16.00 g O TOTAL = molar mass = 46.08 g/mol

10. How many moles of alcohol (C 2 H 6 O) are there in a “standard” can of beer if there are 21.3 g of C 2 H 6 O? (a) Molar mass of C 2 H 6 O = 46.08 g/mol (b) Calc. moles of alcohol

11. How many molecules of alcohol are there in a “standard” can of beer if there are 21.3 g of C 2 H 6 O? = 2.78 x 10 23 molecules We know there are 0.462 mol of C 2 H 6 O.

12. How many atoms of C are there in a “standard” can of beer if there are 21.3 g of C 2 H 6 O? = 5.57 x 10 23 C atoms There are 2.78 x 10 23 molecules. Each molecule contains 2 C atoms. Therefore, the number of C atoms is

13. Empirical & Molecular Formulas A pure compound always consists of the same elements combined in the same proportions by weight. Therefore, we can express the molecular composition as PERCENT BY WEIGHT Ethanol, C 2 H 6 O 52.13% C 13.15% H 34.72% O

14. Percent Composition Consider NO 2, Molar mass = ? What is the weight percent of N and of O? What are the weight percentages of N and O in NO?

15. Determining Formulas In chemical analysis we determine the % by weight of each element in a given amount of pure compound and derive the EMPIRICAL or SIMPLEST formula. PROBLEM : A compound of B and H is 81.10% B. What is its empirical formula?

16. Because it contains only B and H, it must contain 18.90% H.Because it contains only B and H, it must contain 18.90% H. In 100.0 g of the compound there are 81.10 g of B and 18.90 g of H.In 100.0 g of the compound there are 81.10 g of B and 18.90 g of H. Calculate the number of moles of each constitutent.Calculate the number of moles of each constitutent. A compound of B and H is 81.10% B. What is its empirical formula?

17. Calculate the number of moles of each element in 100.0 g of sample. A compound of B and H is 81.10% B. What is its empirical formula?

18. Now, recognize that atoms combine in the ratio of small whole numbers Find the ratio of moles of elements in the compound. A compound of B and H is 81.10% B. What is its empirical formula?

19. But we need a whole number ratio. 2.5 mol H/1.0 mol B = 5 mol H to 2 mol B EMPIRICAL FORMULA = B 2 H 5 Take the ratio of moles of B and H. Always divide by the smaller number. A compound of B and H is 81.10% B. What is its empirical formula?

20. A compound of B and H is 81.10% B. Its empirical formula is B 2 H 5. What is its molecular formula ? Is the molecular formula B 2 H 5, B 4 H 10, B 6 H 15, B 8 H 20, etc.? B 2 H 6 is one example of this class of compounds. B2H6B2H6

21. How to Determine the molar mass? Mass spectrometer

22. A compound of B and H is 81.10% B. Its empirical formula is B 2 H 5. What is its molecular formula ? We need to do an EXPERIMENT to find the MOLAR MASS. Here experiment gives 53.3 g/mol Compare with the mass of B 2 H 5 = 26.66 g/unit = 26.66 g/unit Find the ratio of these masses. Molecular formula = B 4 H 10

23. ELEMENTS are composed of identical particles, atoms CHEMICAL COMPOUNDS are formed when atoms of different elements combine with each other: A given compound always has the same relative numbers and types of atoms Law of definite proportion [Joseph Proust, 1754-1826] A given compound always contains the same proportion of elements by mass Law of multiple proportions [John Dalton, 1766-1844] When 2 elements form multiple compounds, the mass of the second element per gram of the first one can always be reduced to small whole numbers Law of conservation of mass [Lavoisier, 1743-1794] Mass is neither created nor destroyed in a chemical reaction

24. Chemical Equations Because the same atoms are present in a reaction at the beginning and at the end, the amount of matter in a system does not change.Because the same atoms are present in a reaction at the beginning and at the end, the amount of matter in a system does not change. The Law of the Conservation of MatterThe Law of the Conservation of Matter Demo of conservation of matter, See Screen 4.3.

25. Because of the principle of the conservation of matter, an equation must be balanced. It must have the same number of atoms of the same kind on both sides. Chemical Equations Lavoisier, 1788

26. Balancing Equations ___ Al(s) + ___ Br 2 (liq) ---> ___ Al 2 Br 6 (s)

27. Chemical Equations Depict the kind of reactants and products and their relative amounts in a reaction. 2 Al(s) + 3 Br 2 (g) ---> Al 2 Br 6 (s) The numbers in the front are called stoichiometric coefficients The letters (s), (g), and (l) are the physical states of compounds.

28. Chemical Equations 4 Al(s) + 3 O 2 (g) ---> 2 Al 2 O 3 (s) This equation means 4 Al atoms + 3 O 2 molecules ---give---> 2 molecules of Al 2 O 3 2 molecules of Al 2 O 3 4 moles of Al + 3 moles of O 2 ---give---> 2 moles of Al 2 O 3 2 moles of Al 2 O 3

29. STOICHIOMETRYSTOICHIOMETRY It rests on the principle of the conservation of matter. 2 Al(s) + 3 Br 2 (liq) ------> Al 2 Br 6 (s)

30. PROBLEM: If 454 g of NH 4 NO 3 decomposes, how much N 2 O and H 2 O are formed? What is the theoretical yield of products? STEP 1 Write the balanced chemical equation NH 4 NO 3 ---> N 2 O + 2 H 2 O

31. 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O STEP 2 Convert mass reactant (454 g) --> moles STEP 3 Convert moles reactant (5.68 mol) --> moles product

32. 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O STEP 3 Convert moles reactant --> moles product Relate moles NH 4 NO 3 to moles product expected. 1 mol NH 4 NO 3 --> 2 mol H 2 O Express this relation as the STOICHIOMETRIC FACTOR.

33. 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O = 11.4 mol H 2 O produced STEP 3 Convert moles reactant (5.68 mol) --> moles product

34. 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O STEP 4 Convert moles product (11.4 mol) --> mass product Called the THEORETICAL YIELD ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY PROBLEMS!

35. GENERAL PLAN FOR STOICHIOMETRY CALCULATIONS Mass reactant Stoichiometric factor Moles reactant Moles product Mass product

36. 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O STEP 5 How much N 2 O is formed? Total mass of reactants = total mass of products 454 g NH 4 NO 3 = ___ g N 2 O + 204 g H 2 O mass of N 2 O = 250. g

37. 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O STEP 6 Calculate the percent yield If you isolated only 131 g of N 2 O, what is the percent yield? This compares the theoretical (250. g) and actual (131 g) yields.

38. 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O STEP 6 Calculate the percent yield

39. PROBLEM: Using 5.00 g of H 2 O 2, what mass of O 2 and of H 2 O can be obtained? 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Reaction is catalyzed by MnO 2 Step 1: moles of H 2 O 2 Step 2: use STOICHIOMETRIC FACTOR to calculate moles of O 2 Step 3: mass of O 2

40. Reactions Involving a LIMITING REACTANT In a given reaction, there is not enough of one reagent to use up the other reagent completely.In a given reaction, there is not enough of one reagent to use up the other reagent completely. The reagent in short supply LIMITS the quantity of product that can be formed.The reagent in short supply LIMITS the quantity of product that can be formed.

41. LIMITING REACTANTS eactants R eactantsProducts 2 NO(g) + O 2 (g) 2 NO 2 (g) Limiting reactant = ___________ Excess reactant = ____________

42. LIMITING REACTANTS Demo of limiting reactants on Screen 4.7

43. 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 LIMITING REACTANTS React solid Zn with 0.100 mol HCl (aq) Zn + 2 HCl ---> ZnCl 2 + H 2 123 (See CD Screen 4.8)

44. Rxn 1Rxn 2Rxn 3 Rxn 1Rxn 2Rxn 3 mass Zn (g)7.003.271.31 mol Zn0.1070.0500.020 mol HCl0.1000.1000.100 mol HCl/mol Zn0.93/12.00/15.00/1 Lim ReactantLR = HClno LRLR = Zn LIMITING REACTANTS React solid Zn with 0.100 mol HCl (aq) Zn + 2 HCl ---> ZnCl 2 + H 2

45. Reaction to be Studied 2 Al + 3 Cl 2 ---> Al 2 Cl 6

46. PROBLEM: Mix 5.40 g of Al with 8.10 g of Cl 2. What mass of Al 2 Cl 6 can form? Mass reactant Stoichiometric factor Moles reactant Moles product Mass product

47. Determine the formula of a compound of Sn and I using the following data. Reaction of Sn and I 2 is done using excess Sn.Reaction of Sn and I 2 is done using excess Sn. Mass of Sn in the beginning = 1.056 gMass of Sn in the beginning = 1.056 g Mass of iodine (I 2 ) used = 1.947 gMass of iodine (I 2 ) used = 1.947 g Mass of Sn remaining = 0.601 gMass of Sn remaining = 0.601 g

48. Find the mass of Sn that combined with 1.947 g I 2. Mass of Sn initially = 1.056 g Mass of Sn recovered = 0.601 g Mass of Sn used = 0.455 g Find moles of Sn used: Tin and Iodine Compound

49. Now find the number of moles of I 2 that combined with 3.83 x 10 -3 mol Sn. Mass of I 2 used was 1.947 g. How many moles of iodine atoms ? = 1.534 x 10 -2 mol I atoms

50. Tin and Iodine Compound Now find the ratio of number of moles of moles of I and Sn that combined. Empirical formula is SnI 4