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Cpt 4. CHEMICAL STOICHIOMETRY

Cpt 4. CHEMICAL STOICHIOMETRY. Definition : Study of proportions of * atoms inside substances ; * substances in solutions or chemical reactions Objectives: Calculate: * the composition & formulas of chemical compounds * the concentrations of solutions

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Cpt 4. CHEMICAL STOICHIOMETRY

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  1. Cpt 4. CHEMICAL STOICHIOMETRY • Definition:Study of proportions of * atoms inside substances; * substances in solutions or chemical reactions • Objectives: • Calculate: * the composition & formulas of chemical compounds * the concentrations of solutions * the quantitative outcome of chemical reactions

  2. 4.1. Mole and Molar mass • Thought teaser Weight Species Number 1 g H 6.022E23 atoms 16 g O , , , , 18 g H2O , , molecules 23 g Na(+) , , cations 35.5 g Cl(-) , , anions What trend is uncovered by the information from the table? See RQ2-19

  3. RQ2-19 What trend is uncovered by the information from the table on the previous slide? • Any amount of substance with a density equal to its atomic or molecular density contains the same number of components: 6.022E23 • Any amount of substance with a weight equal to its atomic or molecular weight contains the same number of components: 6.022E23 • Substances with the same weight contain the same number of components: 6.022E23

  4. The Mole • mole = 6.022E23 = Avogadro's number of components (atoms in an element, molecules in a compound) of a substance. • Number of components of a substance in a weight in grams equal to the atomic or the molecular weight of the substance. • Molar mass (MW) : mass of a mole : same magnitude but different units as the atomic or molecular mass. • unit of molar mass: g/mol • Mass - Molar mass relation: • # of moles = (mass/molar mass)

  5. RQ2-19B: Previous Material Review • On the chemical standpoint a mole is a: a, rodentthat lives in underground burrows; b, measure of amount of substance; c, measure of weight of substance • The number of moles of a substance shows how much of a substance there is by showing the: a, weight of substance; b, volume of substance; c, the number of components (atoms or molecules) of the substance.

  6. WORD OF WISDOMLearning Steps Resolve, Dare to make the first step in the learning process: REVIEW Reviewing leads to UNDERSTANDING Don’t understand? Ask quickly for help to keep the momentum

  7. WORD OF WISDOMBefore the Gain Unconfortable when studying? Expect the unconfort. Welcome the unconfort. It is beneficial to your learning You’re changing your thoughts From UNFOCUSED TO FOCUSED Say in your mind: “I’m happy to do it”. You feel a smile forming in your mind

  8. WORD OF WISDOM Not enoughtime? MAKE the time! Sacrifice something. Your better future: worth the sacrifice Tolerate Hardship on the way to Prosperity

  9. WORD OF WISDOMHard Work Work is too hard? Keep trying. It’ll geteasier and easier Ask for help When the going gets tough, the tough gets going

  10. Concept Illustration A 60 kg person has in average 36 kg of water. How many moles of water does that represent? #mol = wt / MW; MW(H2O) = 18.02 g/mol #mol = ?

  11. Illustration 2 • Assume you have 0.123 mol of C14H18O4, benzoyl peroxide. What mass does that correspond to? • Wt = #mol x MW MW (C14H18O4) = (14*12.01 + 18 + 4*16) g/mol = ? • Extra exercise: #81, 82, pg 80

  12. RQ2-20 • The adult human body has about 1014 cells. Does the body of an adult human have more or less than a mole of DNA? a. More, because the average adult has about 2000 moles of water and DNA is larger than water. b. Less, because the human body can have no more than 1014 DNA molecules, which is less than a mole c. Less, because the average adult has about 2000 moles of water and DNA is smaller than water.

  13. 4.2. Percent Composition • Restriction: wt/wt • Definition:% by weight = * (wt of atom / wt of molecule) x 100 * (wt of component / wt of whole item)x100 • Law of Definite Composition: “the proportion of each element in a sample of pure compound is constant.”

  14. Illustration • Calculate the weight % of iron in Fe2O3. What mass of iron is in 25.0 g of Fe2O3? • Extra Exercise: Find the % of C, H and O in glucose: C6H12O6. • Extra exercise: #69,pg 126

  15. Illustration (Continued) • Information provided • Formula: Fe2O3 • Information requested: % of Fe in Fe2O3. • Way to answer: use MW’s • % of Fe = 100 x (2 x MW(Fe) / MW(Fe2O3)) = ? • Extra example: #65, pg 126

  16. Illustration (Continued 2) • Information provided b: • Weight of Fe2O3: 25. 0 g • Information requested: mass of Fe in the Fe2O3 sample • Way to answer: • Wt(Fe) = % x Wt(Fe2O3) / 100 = ? • Extra exercise: # 77, pg 127

  17. RQ2-20B: Previous Material Review • The number of moles of a substance indicates the: a, weight; b, number of components; c, density of the substance • The percentage by weight of salt in a water solution indicates for example the number of grams of salt in: a, 100g; b, 100 mL; c,10 g of solution

  18. RQ2-20B2: Previous Material Review • If you are given the #mol of a substance for which you know the molar weight, you can find the weight of the substance by: a, multiplying; b, dividing: c, adding #mol and MW. • If you are given the % composition of solute in a solution for which you know the weight, you can find the weight of solute by using this expression: a, wt(solute) = % / wt(solution) / 100; b, wt(solute) = % x wt(solution) x 100; c, wt(solute) = % x wt(solution) / 100

  19. SUCCESS UNDERSTAND REMEMBER REVIEW PRACTICE WANT WISH

  20. WORD OF WISDOMHard Work Work is too hard? Keep trying. It’ll geteasier and easier Ask for help When the going gets tough, the tough gets going

  21. WORD OF WISDOMBefore the Gain Unconfortable when studying? Expect the unconfort. Welcome the unconfort. It is beneficial to your learning You’re changing your thoughts From UNFOCUSED TO FOCUSED Say in your mind: “I’m happy to do it”. You feel a smile forming in your mind

  22. WORD OF WISDOM Not enoughtime? MAKE the time! Sacrifice something. Your better future: worth the sacrifice Tolerate Hardship on the way to Prosperity

  23. 4.3. Derivation of Formulas • Types of formulas: • Empirical formula: shows the simplest ratios of atoms (groups of atoms) in a molecule • Molecular Formula: shows the actual numbers of atoms in a molecule a. Derivation of Empirical Formulas • General procedure: Weights of components Moles of components Raw Formula Intermediate formula Empirical formula

  24. Empirical Formula Determination Procedure Raw Formula

  25. Detailed Procedure • Raw formula: obtained using (calculated) # of moles for subscripts • Intermediate formula: obtained dividing all subscripts by the smallest subscript • Empirical formula: obtained after operation to round all fractional subscripts into the smallest whole numbers.

  26. Special Fractional Subscripts • Unrounded Subscripts that end with .33, .66 .25, .5 and .75 • Procedure: • Multiply the fractional subscript in the intermediate formula by the smallest number needed to produce the smallest whole number (or a number close to it).

  27. Illustration • Information provided: % composition of mandelic acid (an alpha hydroxy acid): C(63.15%), H(5.30%), and O(31.55%) • Information requested: EF of mandelic acid. • How do you proceed to solve the problem? See RQ2-20b

  28. RQ2-20b • How do you proceed to solve the problem in the previous question using the KNU method. Justify your answer. • a. Find RF from IF, itself from EF, itself found using the #mol of components of mandelic acid. • b. Find EF from IF, itself from RF, itself found using the #mol of components of mandelic acid. • c. Find IF from EF, itself from RF, itself found using the #mol of components of mandelic acid.

  29. RQ2-20C: Previous Material Review • The empirical formula of a compound is determined first by using as subscripts the: a, weight; b, volume; c, number of moles of the atoms that make up the compound • If one of the subscripts in the intermediate formula ends with .67, the subscript should be: a, rounded to the next whole number; b, multiplied by 3; c, divided by 3 to get the empirical formula

  30. Illustration Solution • Empirical Formula <- Intermediate Formula <- Raw Formula: CxHyOz x, y, z = #mol of C, H, and O • #mol = wt / MW • Wt = unknown. Use % values * Wt of C = ? MW = 12.0 g/mol * Wt of H = ? MW = 1.01 g/ mol * Wt of O = ? MW = 16.0 g/mol

  31. Illustration Solution (Continued) • Plug info of Wt and MW in the expression of #mol • Plug info of #mol in the expression of raw formula • Use the raw formula to find the intermediate formula • Use the intermediate formula to find the empirical formula • Extra exercise: #81, pg 127

  32. Molecular Formulas • Empirical Formula (EF)-> Empirical weight (EW) • Molecular Formula (MF)-> Molecular weight (MW) • MW/EW = Molecular/Empirical Ratio (MER) • MF/EF = MER -> MF = EF x MER

  33. Molecular Formulas (Example) • Information provided: * MW of cacodyl = 210 g/mol * % C: 22.88 % * % H: 5.76 % * As: 71.36 % • Information requested : Empirical and molecular formulas of cacodyl

  34. Molecular Formulas (Example Continued) • EF Intermediate formula <- Raw formula: CxHyAsz • x, y, and z = #mol of C, H, and As #mol = wt / MW; Find wt using %’s. What do you use the EF information for? See RQ2-20C

  35. EF and MF of Cacodyl (Continued) • EF -> EW = ? • MW / EW = MER = ? • MF = EF x MER = ?; • Extra exercise: #117, pg 128

  36. Molecular Formulas (Example 2) • Info provided: * wt of S: 1.256 g * wt of SFx: 5.722 g • Info requested: Value of x in SFx.

  37. Molecular Formulas (Example 2 Solution) • EF <- Raw formula: SmFn; n, m = #mol of S and F #mol = wt / MW; #mol of S = ? Find wt of F using wt of SFx – wt of S #mol of F =? • Extra example: #121, pg 128

  38. EF/MF Example 3: Using Results of reactions#120, pg 128 • Information provided * Weight of estrone (made of C, H, and O): 1.893 g * Weight of CO2 from combustion of estrone: 5.545 g * Weight of H2O from combustion of estrone : 1.388 g * MW of the estrone : 270.36 g/mol • Information requested: MF of estrone

  39. EF/MF Example 3 (Solution) • MF = EF x MER • EF <- IF <- Raw formula, CxHyOz. x, y, and z = #mol of C, H and O. • #mol of C = #mol of CO2. • #mol of H = 2 x #mol of H2O • #mol of O = wt/MW • Wt of O = total wt (sample) – wt of C and H • Continued on next slide

  40. EF/MF Example 3 (Solution, Continued) • Plug resulting weight into the expessions of #mol of O • Use #mol of C, H, and O to get the RF. • Use the RF to get the IF • Use the IF to get the EF • Use the EF to calculate the EW • Use the EW and the MW to get the MER • Use the MER and the EF to get the MF • Extra exercise: #119, pg 128.

  41. RQ2-21 • What type of information about moles does the molecular formula provide? • a. The actual number of moles ofcomponent atoms per mole of compound • b. The simplest ratio of moles ofcomponent atoms per mole of compound • c.It does not give any information about moles.Only the actual numbers ofcomponent atoms per molecule of compound

  42. 4.4. Solutions • Solution: homogeneous mixture of two or more substances. • Solvent: most abundant component. • Solute: least abundant • Concentration: amount of solute in a given amount of solvent • Molarity (M) = Measure of the concentration of a solution = moles of solute / volume (liters) of solution. M = #mol/V • Units: mol/L • Commercial unit : molar (M)

  43. Molarity (Illustration) Info provided: • Wt of K2Cr2O7: 2.335 g • Volume of final solution: 500 mL = 0.500 L Info requested: • Molar concentration of K2Cr2O7 • Molar concentrations of K(+) and Cr2O7(2-) • Extra ex: #53, pg 178

  44. Dilution/Concentration of Solutions • Dilution -> decreased final molarity • Concentration -> increased final molarity. • Total #mol of solute = constant = same before and after dilution/concentration Initial #mol(solute) = Final #mol (solute) Mi x Vi = Mf x Vf I = initial; f = final

  45. Dilution/Concentration of Solutions (Illustration) Info provided: • Volume of original CuSO4 solution: 4.00 mL • Molarity of original CuSO4 solution: 0.0250 M • Final volume of CuSO4 solution: 10.0 mL Info requested: final molarity of CuSO4 solution. Extra ex: #55, 178

  46. RQ2-21B: Previous Material Review • If you divide the #mol of solute by the volume (in L’s) of solution, you are determining the: a; density; b, percent composition; c, molarity of the solution • If you want to determine the #mol of solute in a solution, all you have to do is: a, multiply; b, divide; c, add the molarity by the volume of the solution

  47. RQ2-22 • Compare the stability of % composition by weight and molarityin function of temperature • a. % by weight is stableandmolarity is not. Unlike volumes, weights changewith temperature • b. molarity is stableand% by weightis not. Unlike weights, volumes do not changewith temperature • c. % by weight is stableandmolarity is not. Unlike volumes, weights do not changewith temperature

  48. 4.5. Calculations based on Equations and Reactions • Study case: C3H8 + 5O2 -> 3CO2 + 4H2O • Info fromEquations -> Identity & theoretical proportionsbetweenreactants & products of a reaction. Proportions -> mole ratios in equations • Info from Actual reactions: amounts of reactants actually used / products actually formed Weights -> # of moles -> mole ratios in actual reactions. • General Principle: mole ratios from actual reaction= mole ratios from equation

  49. a. Target # 1: amounts of reactants or products • Condition: Reactants proportions are as shown by the mole ratios from equations • General Principle: mole ratios from equation = mole ratios from actual reaction • Mol ratio = #mol of target substance / #mol of known substance • #mol of target substance = mol ratio of (Target/Known) x #mol of known substance • How do you determine the weight of target substance? See RQ2-23

  50. RQ2-23 • Use the expression of #mol of target substance to determine the weight of target substance • a. wt of target substance=mol ratio of (Target/Known)x #mol of knownx MW of target substance • b. wt of target substance=mol ratio of (Known/Target)x #mol of knownx MW of known substance • c. wt of target substance=mol ratio of (Target/Known)x #mol of known/ MW of target substance

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