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CHAPTER THREE

CHAPTER THREE. Formulas, Equations, and Moles. Chemical Equations. The chemical equation is an attempt to show what is happening in the reaction at a molecular level. with heating. reactants or. starting materials. yields or produces. Chemical Equations. Consider a typical reaction:.

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CHAPTER THREE

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  1. CHAPTER THREE • Formulas, Equations, and Moles

  2. Chemical Equations • The chemical equation is an attempt to show what is happening in the reaction at a molecular level.

  3. with heating reactants or starting materials yields or produces Chemical Equations • Consider a typical reaction: products The coefficients represent the number of moles (or particles) needed for the complete reaction. No coefficient is understood to be “1”. Consider these numbers to be ratios.

  4. Chemical Equations • Remember the coefficients represent moles not mass 1 mole 3 moles 2 moles 3 moles 1 mol 3 mol 2 mol 3 mol 3 molecules 2 atoms 3 molecules 1 formula unit 159.7 g 84 g 111.7 g 132 g Equations are balanced for moles. Massmust be converted tomolesbefore you use the equation.

  5. The Mole • The term mole is a shorthand abbreviation of “gram formula molecular weight” • gram formula molecular weight  gram molecular weight  molecular weight  mol. weight mol.wt.  mol • mol. or mol was spelled mole (like it sounds) • Avogadro’s number = 6.022  1023 You need to make moles your friends (or at least allies) or you will not survive the course.

  6. The Mole • What is one mole? • It is the molecular or formula weight of the compound in grams. • Common formulas

  7. The Mole • Example 1: How many moles of Mg atoms are present in 73.4 g of Mg? You must know how to do these problems!

  8. Molar Masses of Compounds • Example 2: Calculate the molecular weight of propane, C3H8. • molecular weight and formula weight are virtually the same thing

  9. Example 3: Calculate the formula weight of calcium nitrate. First determine the chemical formula for calcium nitrate. calcium has a charge of +2 (it is a IIA element) nitrate has a formula and charge of NO3– Calcium nitrate must have the formula Ca(NO3)2 Molar Masses of Compounds

  10. Millimoles • Quantities are often given in millimoles or mmol. • 1000 mmol = 1 mol • Oxalic acid has a molecular weight of 90.04 • 1 mol oxalic acid = 90.04 g • 1 mmol oxalic acid = 0.09004 g or 90.04 mg

  11. Millimoles • Example 4: Calculate the number of mmol in 0.234 g of oxalic acid, (COOH)2. (mol.wt. oxalic acid = 90.04) • There are several different approaches to this problem An alternate approach is

  12. Law of Conservation of Matter • There is no detectable change in quantity of matter in an ordinary chemical reaction. • Balance chemical reactions to obey this law. • For example, propane,C3H8, burns in oxygen to give carbon dioxide and water. note that there are equal numbers of atoms of each element on both sides of the reaction (arrow) This is your goal for balancing an equation.

  13. Law of Conservation of Matter • Example 5: Ammonia, NH3, burns in oxygen to form NO and water. Write the balanced equation for this reaction. • First write as much of the equation as you can. Now pick an element and begin balancing This gives the balanced equation

  14. Law of Conservation of Matter • Example 6: Pentane, C5H12, burns in oxygen to form carbon dioxide & water. Write the balanced chemical equation for this reaction. • First write as much of the equation as you can. Now pick an element and begin balancing This requires lots of practice!

  15. Calculations Based on Chemical Equations • You will frequently be given problems to work involving chemical reactions. • If problem is presented in moles, molecules, ions, or particles, you may begin work directly. • Frequently a problem involves mass (grams, kg, pounds, or tons). Before you can solve the problem, you must convert this mass (weight) into relative proportion by converting this mass into “moles.” • You do not need to convert all units to grams, you may use any unit of measure as long as you convert these masses into molar units by dividing by the formula weight of the compound or element.

  16. Calculations Based on Chemical Equations • Example 7: What mass of CO is required to react with 146 g of iron (III) oxide? (f.wt. Fe2O3 = 159.6; mol.wt. CO = 28.0) Remember moles are your friends! mol Fe2O3

  17. Calculations Based on Chemical Equations • Example 8: What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide? (f.wt. Fe2O3 = 159.6; mol.wt. CO2 = 44.0)

  18. Calculations Based on Chemical Equations • Example 9: What mass of iron (III) oxide reacted with excess carbon monoxide if the carbon dioxide produced by the reaction had a mass of 8.65 grams? (f.wt. Fe2O3 = 159.6; mol.wt. CO2 = 44.0) mol. CO2

  19. Percent Yields from Reactions • The amount of material you calculated to be produced by the reaction is the theoretical yield. • The actual yield is the amount of product that is “actually” isolated from the reaction. • The percent yieldis based upon these two values.

  20. Percent Yields from Reactions • Example 10: A 10.0 g sample of ethanol, C2H5OH, was boiled with excess acetic acid, CH3CO2H, to produce 14.8 g of ethyl acetate, CH3CO2C2H5 . What is the percent yield? (mol.wt. C2H5OH = 46.0; mol.wt. CH3CO2C2H5 = 88.0) [1 mol ethanol makes 1 mol ethyl acetate]

  21. Sequential Reactions • Sequential reactions are multi-step reactions. • The amount of product formed in each step is considered to be the starting material for the next reaction. • Percent yield for the overall reaction is calculated by multiplying each of the % yields for the individual steps together.

  22. Limiting Reactant Concept • Example 11: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110 g of oxygen? (mol.wt. SO2 = 64.1; mol.wt. CS2 = 76.2; mol.wt. O2 = 32.0) By examination, there is not enough O2 to run the reaction. Therefore O2 is the limiting reagent.

  23. Limiting Reactant Concept • Example 11: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110 g of oxygen? (mol.wt. SO2 = 64.1; mol.wt. CS2 = 76.2; mol.wt. O2 = 32.0) Oxygen is the limiting reagent! The alternate method for solving this type of question is to solve the problem for each reactant listed. The one producing the least product is the limiting reagent.

  24. Limiting Reactant Concept • Example 12: How much carbon disulfide will remain unreacted after 95.6 g of carbon disulfide reacts with 110 g of oxygen? (mol.wt. SO2 = 64.1; mol.wt. CS2 = 76.2; mol.wt. O2 = 32.0) Oxygen is the limiting reagent! Depending upon the answer required:

  25. Concentration of Solutions • Nomenclature • solution • a mixture of two or more substances, usually one is a liquid • solvent • the material doing the dissolving (usually the most abundant material or the liquid) • solute • the material that is dissolved (usually the least abundant material) • concentration • the amount of solute dissolved in a solvent, can be expressed in percent (%), molarity (M), normality (N), or molality (m)

  26. Percent Solution • The percent solution is based on the mass of the solution. You may have to use density to determine the mass of the solution.

  27. Percent Solutions • Example 13: What mass of NaOH is required to prepare 250.0 g of solution that is 8.00% NaOH? • Convert the % to the decimal equivalent before beginning the calculation.

  28. Percent Solutions • Example 14: Calculate the mass of 8.00% NaOH solution that contains 32.0 g of NaOH. Remember to use the decimal fraction not %.

  29. Molarity • Molarity, M, is the most useful concentration. • This gives the concentration in moles which means that the values can be used (with little change) for chemical calculations. • [Ag+] is shorthand for moles of Ag+ per liter This can also be represented as

  30. Molarity • Example 15: Calculate the molarity of a solution that contains 12.5 g of sulfuric acid in 1.75 L of solution. (mol.wt. H2SO4 = 98.1) An alternative solution is

  31. Molarity • Example 3-13: Determine the mass of calcium nitrate required to prepare 3.50 L of 0.800 M Ca(NO3)2 . (f.wt. Ca(NO3)2 = 164 g)

  32. Molarity • Example17: The specific gravity of concentrated HCl is 1.185 and it is 36.31% w/w HCl. What is its molarity? • First calculate the weight of 1 L Then calculate the mass of HCl in this volume The number of moles of HCl equals molarity since this was calculated for a 1 L solution.

  33. Dilution of Solutions • Molarity dilution problems can be easy to solve. • Since the number of moles remains constant M1 = molarity solution 1 V1 = volume solution 1 M2 = molarity solution 2 V2 = volume solution 2 You may use either L or mL (these must be the same units) to solve the problem.

  34. Dilution of Solutions • Example 18: If 10 mL of 12 M HCl is added to enough water to give 100 mL of solution, what is concentration of solution?

  35. Dilution of Solutions • Example 19: What volume of 18.0 M sulfuric acid is required to make 2.50 L of a 2.40 M sulfuric acid solution?

  36. Stoichiometry with Solutions • Example 20: What volume of 0.500 M BaCl2 is required to completely react with 4.32 g of Na2SO4? (f.wt. Na2SO4= 142) • First write the chemical equation for the reaction 1 mol of Na2SO4 reacts with 1 mol of BaCl2

  37. Stoichiometry with Solutions • Example 21: What volume of 0.200 M NaOH will react with 50.0 mL of 0.200 M aluminum nitrate? What mass of aluminum hydroxide precipitates? (f.wt. Al(OH)3 = 78.0) 10 mmol 30 mmol 10 mmol 1 mol Al(NO3)3 react with 3 mol NaOH to make 1 mol Al(OH)3

  38. Titrations Titration is a method to exactly the determine the concentration of a solution. Titrations require standards – materials whose concentrations can be exactly (accurately) determined special glassware – burets and pipets, these have been calibrated to be accurate to  0.01 mL or less (0.001 mL) indicators – to “indicate” when the titration is finished This allows concentrations to be determined to 4 decimal places (min. 4 significant figures)

  39. Titrations • Example 22: What is the molarity of a KOH solution if 38.7 mL of KOH solution is required to react with 43.2 mL of 0.223 M HCl? 9.63 mmol 9.63 mmol 1 mol KOH reacts with 1 mol HCl

  40. Titrations • Example 23: What is the molarity of a barium hydroxide solution if 44.1 mL of 0.103 M HCl is required to react with 38.3 mL of the Ba(OH)2 solution? 2.27 mmol 4.54 mmol 1 mol Ba(OH)2 reacts with 2 mol HCl

  41. Chemical Formulas From Elemental Composition • empirical formula– simplest molecular formula • this shows the smallest whole number ratios of the elements present in the compound but notnecessarily the molecular formula • molecular formula– the actual formula • this is molecular formula of the compound showing all the elements present • The emperical and molecular formula of a compound can be determine from percent composition. • Percent composition is determined experimentally.

  42. Percent Composition and Chemical Formulas • Example 24: A compound contains 24.74% K; 34.76% Mn; and 40.50% O by mass. What is its empirical formula? • Assume that you have 100 units of the compound • Find the smallest whole number ratio of elements Formula: KMnO4

  43. Percent Composition and Chemical Formulas • Example 25: A sample of a compound contains 6.541g of Co and 2.368g of O. What is empirical formula for this compound? The empirical formula of the compound is Co3O4

  44. Percent Composition From Combustion Data • Example 26: Glucose, a sugar found in our blood, contains C, H, and O. On combustion analysis, a 1.457 g sample gives 2.1309 g CO2 and 0.8743 g H2O. What is the empirical formula of glucose from this data? The amount of O cannot be determined directly from this data. Additional calculations are needed.

  45. Percent Composition From Combustion Data • Example 26 continued: • sample wt. before combustion = 1.457 g Now it is possible to calculate the empirical formula.

  46. Percent Composition From Combustion Data • Example 26 continued: The empirical formula is CH2O

  47. Percent Composition From Combustion Data • Example 27: If glucose, empirical formula CH2O, has a formula weight of 180.18, what is the molecular formula of glucose? • From the previous problem, the empirical formula was found to be CH2O (f.wt. = 30.03) Actual molecular formula = (CH2O)6= C6H12O6

  48. Percent Composition • This is the mass of the element present in the sample divided by the mass of the sample ( 100%). • Example 28: What is the percent composition of C and H in C3H8. (mol. wt. C3H8 = 44.11) If you trusted you calculation for %C then… 100% – 81.68 % C = 18.32 % H

  49. Percent Composition • Example 29: Calculate the percent composition of the elements in Fe2(SO4)3 to 3 sig. fig. (f.wt. Fe2(SO4)3 = 399.9)

  50. End of Chapter 3

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