1. 1 © 2006 Brooks/Cole - Thomson Chapter 3 Stoichiometry of Formulas and Equations Chemical Reactions

2. 2 © 2006 Brooks/Cole - Thomson 2 3.1 The Mole An atom of carbon weighs 12 amu = 1.99 x 10 -23 g 1 mol of 12 C = 6.022 x 10 23 atoms of C = 12.00 g of C A mole is the amount of substance that contains as many elementary entities (atoms, molecules, or other particles) as there are atoms in exactly 12 g of the carbon-12 isotope. Just like a dozen = 12 items a pair = 2 items a mole = 6.022 x 10 23 items (also known as Avogadro’s number). The point—it takes a lot of atoms to have something we can weigh.

3. 3 © 2006 Brooks/Cole - Thomson 3 Avogadro’s Number used as conversion factors Avogadro’s number 6.02 x 10 23 can be written as an equality and two conversion factors. Equality: 1 mole = 6.02 x 10 23 particles Conversion Factors: 6.02 x 10 23 particles and 1 mole______ 1 mole6.02 x 10 23 particles

4. 4 © 2006 Brooks/Cole - Thomson 4 Molar Mass Molar Mass 1 mol of 12 C = 6.022 x 10 23 atoms of C = 12.00 g 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 Carbon is 12.011 g/mol

5. 5 © 2006 Brooks/Cole - Thomson 5 Where do we find molar mass of an element? It’s on your periodic table - It’s on your periodic table! The number that represents the weighted average atomic mass is the same amount a mole of those atoms would weigh in grams!

6. 6 © 2006 Brooks/Cole - Thomson 6 One-mole Amounts

7. 7 © 2006 Brooks/Cole - Thomson 7 PROBLEM: How many moles of Mg are in 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. 8 © 2006 Brooks/Cole - Thomson 8 Converting between moles, mass, & number of chemical entities Avogadro’s number is used to convert moles of a substance to # of entities of that substance How many Cu atoms are in 0.50 mole Cu? 0.50 mole Cu x 6.02 x 10 23 Cu atoms 1 mole Cu = 3.0 x 10 23 Cu atoms How many moles of CO 2 are in 2.50 x 10 24 molecules CO 2 ? 2.50 x 10 24 molecules CO 2 x 1 mole CO 2 6.02 x 10 23 molecules CO 2 = 4.15 moles CO 2 Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

9. 9 © 2006 Brooks/Cole - Thomson 9 MolecuLes, Compounds, & MOLAR MASS How many atoms of C are there in a can of beer if it contains 21.3 g of C 2 H 6 O?

10. 10 © 2006 Brooks/Cole - Thomson 10 MolecuLes, Compounds, & MOLAR MASS What is the molar mass of ethanol, C 2 H 6 O? 1 mol C 2 H 6 O contains 2 mol C (12.01 g C/1 mol) = 24.02 g C 6 mol H (1.01 g H/1 mol) = 6.06 g H 1 mol O (16.00 g O/1 mol) = 16.00 g O TOTAL = molar mass = 46.08 g/mol

11. 11 © 2006 Brooks/Cole - Thomson 11 How many moles of ethanol are there in a can of beer if it contains 21.3 g of C 2 H 6 O? (a) Molar mass of C 2 H 6 O = 46.08 g/mol (b) Calculate moles of alcohol

12. 12 © 2006 Brooks/Cole - Thomson 12 How many molecules of alcohol are there in a can of beer if it contains 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.

13. 13 © 2006 Brooks/Cole - Thomson 13 How many atoms of C are there in a can of beer if it contains 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

14. 14 © 2006 Brooks/Cole - Thomson 14 Mass Percent A pure compound always consists of the same elements combined in the same proportions by weight. Therefore, we can express molecular composition as PERCENT COMPOSITION BY WEIGHT Ethanol, C 2 H 6 O 52.13% C 13.15% H 34.72% O

15. 15 © 2006 Brooks/Cole - Thomson 15 Percent Composition Consider NO 2, molar mass = 46.0 g, what is the weight percent of N and of O? What are the weight percentages of N and O in NO?

16. 16 © 2006 Brooks/Cole - Thomson Percent Composition of a Hydrate Hydrate : A solid compound that includes water molecules in its crystal structure Anhydrous Compound : An ionic compound that is without water or from which water has been removed CuSO 4 5H 2 O (s)  CuSO 4 (s) + 5H 2 O (g) copper(II) sulfate pentahydrate copper(II) sulfate Cu 39.82% S 20.09% O 40.09% H 2 O 36.05%

17. 17 © 2006 Brooks/Cole - Thomson 3.3 Writing and Balancing Chemical Equations Chemical equations depict the kind of reactants and products and their relative amounts in a reaction. The numbers in the front are called stoichiometric coefficients The letters (s), (g), and (l) are the physical states of compounds.

18. 18 © 2006 Brooks/Cole - Thomson Balancing Chemical Equations ReactantsProducts 1 C atom =1 C atom 4 H atoms =4 H atoms 4 O atoms = 4 O atoms - start with the most complex substance - end with the least complex substance

19. 19 © 2006 Brooks/Cole - Thomson Balancing Polyatomic Ion as a Unit Na 3 PO 4 (aq) + MgCl 2 (aq)  NaCl(aq) + Mg 3 (PO 4 ) 2 (s) Balance PO 4 3- as a unit 2 Na 3 PO 4 (aq) Mg 3 (PO 4 ) 2 (s) 2 Na 3 PO 4 (aq) Mg 3 (PO 4 ) 2 (s) 2 PO 4 3- = 2 PO 4 3- 2 PO 4 3- = 2 PO 4 3- Balance Mg 3 MgCl 2 (aq) Mg 3 (PO 4 ) 2 (s) 3 MgCl 2 (aq) Mg 3 (PO 4 ) 2 (s) 3 Mg 2+ = 3 Mg 2+ Balance Na and Cl to complete balanced equation 3 MgCl 2 (aq) + 2 Na 3 PO 4 (aq)  6 NaCl(aq) + Mg 3 (PO 4 ) 2 (s) 6 Na + = 6 Na + 6 Na + = 6 Na + 6 Cl - = 6 Cl - 6 Cl - = 6 Cl -

20. 20 © 2006 Brooks/Cole - Thomson Balancing Equations __C 3 H 8 (g) + __ O 2 (g) ----> __CO 2 (g) + __ H 2 O(g) __B 4 H 10 (g) + __ O 2 (g) ----> __ B 2 O 3 (g) + __H 2 O(g) __Al(s) + __H 2 SO 4 (aq) ---- > __Al 2 (SO 4 ) 3 (aq) + __H 2 (g)

21. 21 © 2006 Brooks/Cole - Thomson 3.4 Calculating Quantites of Reactant and Product -the study of mass relationships in chemical reactions -It rests on the principle of the conservation of matter.

22. 22 © 2006 Brooks/Cole - Thomson 22 Molar Ratios from a Balanced Equation A molar ratio is used to relate the number of moles of one reactant or product to another. 4 Fe(s) + 3 O 2 (g) -------- > 2 Fe 2 O 3 (s) 4 Fe(s) + 3 O 2 (g) -------- > 2 Fe 2 O 3 (s) 4 moles Fe + 3 moles O 2 ------- > 2 moles Fe 2 O 3 Molar ratios: Fe and O 2 4 moles Fe and 3 moles O 2 3 moles O 2 4 moles Fe 3 moles O 2 4 moles Fe Fe and Fe 2 O 3 4 moles Fe and 2 moles Fe 2 O 3 2 moles Fe 2 O 3 4 moles Fe 2 moles Fe 2 O 3 4 moles Fe O 2 and Fe 2 O 3 3 moles O 2 and 2 moles Fe 2 O 3 2 moles Fe 2 O 3 3 moles O 2 2 moles Fe 2 O 3 3 moles O 2

23. 23 © 2006 Brooks/Cole - Thomson 23 Learning Check How many moles of Fe are needed for the reaction of 12.0 moles O 2 ? 4 Fe(s) + 3 O 2 (g) ------- > 2 Fe 2 O 3 (s)

24. 24 © 2006 Brooks/Cole - Thomson GENERAL PLAN FOR STOICHIOMETRY CALCULATIONS Mass of A Molar ratio (molesB/molesA) Moles of A Moles of B Mass Of B Molar mass of A Molar mass of B A is known; B is unknown

25. 25 © 2006 Brooks/Cole - Thomson PROBLEM: If 454 g of NH 4 NO 3 decomposes, how many grams of N 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 NH 4 NO 3 ---> N 2 O + 2 H 2 O STEP 2: Convert mass A to moles A STEP 3: Convert moles A to moles B STEP 4: Convert moles B to mass B ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY PROBLEMS!

26. 26 © 2006 Brooks/Cole - Thomson 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O STEP 4 Convert moles B (5.70 mol) to mass B: 250. g N 2 O This is called the THEORETICAL YIELD

27. 27 © 2006 Brooks/Cole - Thomson 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O In the above reaction, if only 131 g of N 2 O is collected, what is the percent yield? Reaction Yield Reaction Yield

28. 28 © 2006 Brooks/Cole - Thomson LIMITING REACTANTS eactants R eactantsProducts 2 NO(g) + O 2 (g) 2 NO 2 (g) Limiting reactant = ___________ Excess reactant = ____________

29. 29 © 2006 Brooks/Cole - Thomson PROBLEM: Mix 5.40 g of Al with 8.10 g of Cl 2. What mass of AlCl 3 can be formed?

30. 30 © 2006 Brooks/Cole - Thomson PROBLEM: Mix 5.40 g of Al with 8.10 g of Cl 2. What mass of AlCl 3 can be formed? 2 Al (s) + 3 Cl 2 (g)  2 AlCl 3 (s ) Solution: 5.40 g Al produces 0.200 mol AlCl 3 8.10 g Cl 2 produces 0.072 mol AlCl 3 Cl 2 is limiting reactant

31. 31 © 2006 Brooks/Cole - Thomson Cl 2 was the limiting reactant. Therefore, Al was present in excess. But how much?Cl 2 was the limiting reactant. Therefore, Al was present in excess. But how much? First, based on LR: Find the moles (or mass) of Al was required. First, based on LR: Find the moles (or mass) of Al was required. Then find the moles (or mass) of Al leftover (in excess).Then find the moles (or mass) of Al leftover (in excess). How much of which reactant will remain when reaction is complete?

32. 32 © 2006 Brooks/Cole - Thomson 2 Al + 3 Cl 2 products 0.200 mol 0.114 mol = LR Calculating Excess Al Excess Al = Al available - Al required = 0.200 mol - 0.0760 mol = 0.124 mol Al in excess

33. 33 © 2006 Brooks/Cole - Thomson In a solution: SOLVENT the component whose physical state is preserved when solution forms Usually in much larger quantity than the solute SOLUTESOLUTE the other solution component 3.5 Fundamentals of Solution Stoichiometry

34. 34 © 2006 Brooks/Cole - Thomson Molarity (M) Is a concentration term for solutions.Is a concentration term for solutions. Gives the moles of solute in 1 L solution.Gives the moles of solute in 1 L solution. Molar (M) = moles of solute liter of solution liter of solution

35. 35 © 2006 Brooks/Cole - Thomson Step 1: Calculate moles of acid required. (0.0500 mol/L)(0.250 L) = 0.0125 mol Step 2: Calculate mass of acid required. (0.0125 mol )(90.00 g/mol) = 1.13 g Amounts – Mass – Number Conversions for Solutions moles = MV What mass of oxalic acid, H 2 C 2 O 4, is required to make 250. mL of a 0.0500 M solution?

36. 36 © 2006 Brooks/Cole - Thomson Less than 1.0 L of water was used to make 1.0 L of solution. Notice the water left over. Concentration of Solute = The amount of solute in a solution Prepare Solution of Known Concentration

37. 37 © 2006 Brooks/Cole - Thomson PROBLEM: How many grams of NiCl 2 should be dissolved in enough water to make 250. mL of 0.154 M solution? Step 1: Calculate moles of NiCl 2 Step 2: Calculate mass of NiCl 2

38. 38 © 2006 Brooks/Cole - Thomson A shortcut A shortcut C initial V initial = C final V final Preparing Solutions by Dilution

39. 39 © 2006 Brooks/Cole - Thomson PROBLEM: You have 50.0 mL of 3.00 M NaOH and you want 0.500 M NaOH. What do you do? Add water to the 3.00 M solution to lower its concentration to 0.500 M Dilute the solution! But how much water do we add? Add 250. mL of water to 50.0 mL of 3.00 M NaOH to make 300. mL of 0.500 M NaOH. 3.00 M NaOH concentrated 0.500 M NaOH diluted 50.0 mL 250. mL water

40. 40 © 2006 Brooks/Cole - Thomson 40 Stoichiometry of Reactions in Solution 1.Write and balance the chemical equation 2.Find the moles of the known substance A 3.Find the moles of the unknown substance B, using the molar ratio 4.Convert moles B to the desired units

41. 41 © 2006 Brooks/Cole - Thomson 41 Problems 1) How many mL of a 0.150 M Na 2 S solution are needed to completely react 18.5 mL of 0.225 M NiCl 2 solution? NiCl 2 (aq) + Na 2 S(aq)  NiS(s) + 2NaCl(aq) 2) If 22.8 mL of 0.100 M MgCl 2 is needed to completely react 15.0 mL of AgNO 3 solution, what is the molarity of the AgNO 3 solution? MgCl 2 (aq) + 2AgNO 3 (aq)  2AgCl(s) + Mg(NO 3 ) 2 (aq)

42. 42 © 2006 Brooks/Cole - Thomson 42 3.2 Determining the Formula of an Unknown Compound 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. Then MOLECULAR formula is determined from empirical formula and molar mass. Molecular Formula Molar mass Assume 100 g compound

43. 43 © 2006 Brooks/Cole - Thomson 43 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.Solution: -the number of moles of each constituent (7.502 mol B and 18.75 mol H) -The mole ratio of these 2 elements -(2.500 mol H / 1 mol B) NEVER ROUND OFF!! -The empirical formula of the compound: B 2 H 5 A compound of B and H is 81.10% B. What is its empirical formula?

44. 44 © 2006 Brooks/Cole - Thomson 44 A compound of B and H is 81.10% B. Its empirical formula is B 2 H 5, and its molar mass is 53.3 g/mol. What is its molecular formula ? The molecular formula of a compound is found by determination of the number of empirical formula units in the molecule Molecular formula = B 4 H 10

45. 45 © 2006 Brooks/Cole - Thomson Determine the Formula of a compound by Combustion Determine the Formula of a compound by Combustion Burn 0.115 g of a hydrocarbon, C x H y, and produce 0.379 g of CO 2 and 0.1035 g of H 2 O. C x H y + some oxygen ---> 0.379 g CO 2 + 0.1035 g H 2 O What is the empirical formula of C x H y ?

46. 46 © 2006 Brooks/Cole - Thomson Determine the Formula of a Compound by Combustion Recognize that all C in CO 2 and all H in H 2 O is from C x H y. C x H y + some oxygen ---> 0.379 g CO 2 + 0.1035 g H 2 O

47. 47 © 2006 Brooks/Cole - Thomson Determine the Formula of a Compound by Combustion Solution: 8.61 mol CO 2 and 0.00575 mole H 2 O 8.61mol C and 0.01149 mol H Mole ratio: 1.149 x 10 -2 mol H/ 8.61 x 10 -3 mol C = 1.33 mol H / 1.00 mol C = 4 mol H / 3 mol C Empirical formula: C 3 H 4 C x H y + oxygen ---> 0.379 g CO 2 + 0.1035 g H 2 O

48. 48 © 2006 Brooks/Cole - Thomson Determine a Formula If this hydrocarbon has molar mass 80.0 g/mol. What is its molecular formula? Empirical formula: C 3 H 4 Molecular formula: (C 3 H 4 ) nMolecular formula: (C 3 H 4 ) n Mass of C 3 H 4 group: 40.0 g/C 3 H 4 groupMass of C 3 H 4 group: 40.0 g/C 3 H 4 group n = (80.0 g/mol)(1 group/40.0 g) = 2n = (80.0 g/mol)(1 group/40.0 g) = 2 Molecular formula: C 6 H 8Molecular formula: C 6 H 8 C x H y + oxygen ---> 0.379 g CO 2 + 0.1035 g H 2 O

49. 49 © 2006 Brooks/Cole - Thomson Determine Molar Mass 1.44 x 10 -3 mol If 1.44 x 10 -3 mol of this hydrocarbon was used in the experiment. What is its molar mass? 0.1035 g H 2 O = 1.149 x 10 -2 mol H 0.379 g CO 2 = 8.61 x 10 -3 mol C Total mass of the 1.44 x 10 -3 mol of compound: (1.149 x 10 -2 mol H x 1.008 g/mol H) + (8.61 x10 -3 mol C x 12.00 g/mol C) = 0.115 g compound Molar mass of hydrocarbon = 0.115 g / 1.44 x 10 -3 mol Molar mass of hydrocarbon = 0.115 g / 1.44 x 10 -3 mol Molar mass: 79.9 g/mol Molar mass: 79.9 g/mol 1.44 x 10 -3 mol C x H y + O 2 ---> 0.379 g CO 2 + 0.1035 g H 2 O