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Chapter 3

Chapter 3. Stoichiometry of Formulas and Equations. Stoichiometry is derived from the Greek words στοιχεῖον ( stoicheion , meaning element) and μέτρον ( metron , meaning measure.) . 화학양론. 몰. 2. 미지 화합물의 화학식 구하기 실험식 , 분자식. 3. 화학반응식과 균형 맞추기. 4. 반응물과 생성물의 양 한계반응물 , 수득량 , 수득 백분율.

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Chapter 3

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  1. Chapter 3 Stoichiometry of Formulas and Equations Stoichiometry is derived from the Greek words στοιχεῖον (stoicheion, meaning element) and μέτρον (metron, meaning measure.)

  2. 화학양론 • 몰 • 2. 미지 화합물의 화학식 구하기 • 실험식, 분자식 • 3. 화학반응식과 균형 맞추기 • 4. 반응물과 생성물의 양 • 한계반응물, 수득량, 수득 백분율 • 5. 용액에서의 화학양론 • 몰농도 • 용액 상태의 화학반응에서 화학량론

  3. Mole - Mass Relationships in Chemical Systems 3.1 The Mole 3.2 Determining the Formula of an Unknown Compound 3.3 Writing and Balancing Chemical Equations 3.4 Calculating the Amounts of Reactant and Product 3.5 Fundamentals of Solution Stoichiometry

  4. 화학전공 교수-학생 대화의 시간 3월 17일 (오늘) 오후 6:00 – 원천관 535호실 구체적인 진로지도 및 선배, 대학원생, 화학과 교수님과의 대화의 시간 김밥, 다과 준비

  5. The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of 12C 1 mol = NA = 6.02214179(30)×1023 Avogadro’s number (NA) The Mole(chemical quantity)

  6. 겁(劫)과 찰라(刹那) 겁[劫][<범어 kalpa] <명사> ≪불교≫ (‘천지가 개벽한 때부터 다음 개벽할 때까지의 동안’이라는 뜻으로) ‘지극히 길고 오랜 시간’을 일컫는 말. <반의어> 찰나. 힌두교에서는 1겁을 43억 2000만 년 (1.36x1017초)이라고 한다. =1.02×1019 찰나[刹那<범어 k¡a¡a] <명사> ① ≪불교≫ (75분의 1초에 해당하는) 지극히 짧은 시간. <반의어> 겁(劫). ② 어떤 사물 현상이 이루어지는 바로 그때. <참고> 순간(瞬間).

  7. The Mole(chemical quantity) • Avogadro’s number =6.0221023 만약 남북한 땅 위에 10원짜리 동전을 쌓는다면 그 높이는 얼마나 될까? 직경 23 mm 두께 1.5 mm 10 • 녹여 쌓으면 1,700 km • 사방으로 펼쳐 쌓으면 2,200 km • 삼각으로 펼쳐 쌓으면 1,900 km

  8. The Mole(chemical quantity) • Avogadro’s number =6.0221023 직경 23 mm 두께 1.5 mm 만약 10원 동전을 지구상(땅과 바다 위)에 쌓는다면 그 높이는 얼마나 될까? • 녹여 쌓으면 730 m • 사방으로 펼쳐 쌓으면 930 m • 삼각으로 펼쳐 쌓으면 800 m

  9. 1 mole 12C atoms = 6.022 ×1023 atoms = 12.00 g 1 12C atom = 12.00 amu 1 mole 12C atoms = 12.00 g 12C 1 mole lithium atoms = 6.941 g of Li atoms Molar mass is the mass of 1 mole of in grams watermelons For any element atomic mass (amu) = molar mass (grams)

  10. The Mole(chemical quantity) 10 kg 수박 1몰 수박을 쌓아보자!

  11. 6.02 × 1024kg ! • 지구질량 5.97 × 1024 kg

  12. One Mole of: S C Hg Cu Fe

  13. Information Contained in the Chemical Formula of Glucose C6H12O6 ( M = 180.16 g/mol) Oxygen (O) Carbon (C) Hydrogen (H) Atoms/molecule of compound 6 atoms 12 atoms 6 atoms Moles of atoms/ mole of compound 6 moles of atoms 12 moles of atoms 6 moles of atoms Atoms/mole of compound 6(6.022 × 1023) atoms 12(6.022 × 1023) atoms 6(6.022 × 1023) atoms Mass/moleculeof compound 6(12.01 amu) =72.06 amu 12(1.008 amu) =12.10 amu 6(16.00 amu) =96.00 amu Mass/mole of compound 72.06 g 12.10 g 96.00 g

  14. MASS(g) of atoms mass-mole-number of atoms. atomic mass M (g/mol) 1 amu×NA= 1 g(exact) AMOUNT(mol) of atoms Avogadro’s number (atoms/mol) 6.02214179×1023 NUMBER of atoms

  15. 107.9 g Ag mol Ag mol Fe 55.85g Fe 6.022x1023 mol Calculating the Mass and the Number of Atoms in a Given Number of Moles of an Element (a) Silver (Ag) is used in jewelry and tableware. How many grams of Ag are in 0.0342 mol of Ag? (b) Iron (Fe), the main component of steel, is the most important metal in industrial society. How many Fe atoms are in 95.8 g of Fe? (atomic mass(amu): Ag=107.9, Fe=55.85) SOLUTION : 0.0342 mol Ag × = 3.69g Ag 95.8g Fe× = 1.04x1024 atoms Fe ×

  16. MASS(g) of compound mass-mole-number of compounds. molecular (or formula) mass M (g/mol) chemical formula AMOUNT(mol) of compound AMOUNT(mol) of elements in compound Avogadro’s number (molecules/mol) MOLECULES (or formula units) of compound

  17. mol (NH4)2CO3 6.022x1023 96.09 g (NH4)2CO3 mol Calculating the Moles and Number of Formula Units in a Given Mass of a Compound Ammonium carbonate is white solid that decomposes with warming. Among its many uses, it is a component of baking powder, first extinguishers, and smelling salts. How many formula unit are in 41.6 g of ammonium carbonate? SOLUTION: Ammonium carbonate (NH4)2CO3. M = (2 ×14.01 g/mol N) + (8 × 1.008 g/mol H) + (12.01 g/mol C) + (3 × 16.00 g/mol O) = 96.09 g/mol × 41.6 g (NH4)2CO3 × = 2.61x1023 (NH4)2CO3

  18. Calculating the Mass Percents and Masses of Elements in a Sample of Compound Glucose (C6H12O6) is the most important nutrient in the living cell for generating chemical potential energy. (a) What is the mass percent of each element in glucose? (b) How many grams of carbon are in 16.55g of glucose? SOLUTION: • M(C6H12O6)= 6×12.01 + 12×1.008 + 6×16.00 = 180.16 g/mol Glucose mass percent of C = 6×12.01g C/180.16 g × 100 = 39.99 mass %C mass percent of H =12×1.008g H/180.16 g ×100 = 6.714 mass %H mass percent of O =6×16.00g O/180.16 g ×100 = 53.29 mass %O (b) 16.55g Glucose ×39.99 mass %C/100 % Glucose = 6.618 g C

  19. Empirical and Molecular Formulas Empirical Formula - The simplest formula for a compound that agrees with the elemental analysis and gives rise to the smallest set of whole numbers of atoms. Molecular Formula - The formula of the compound as it exists. Molecular Formula = (Empirical Formula )n

  20. Determining the Empirical Formula from Masses of Elements Elemental analysis of a sample of an ionic compound showed 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What are the empirical formula and name of the compound? SOLUTION: 0.123/0.123 = 1 Na 0.123/0.123 = 1 Cl 0.489/0.123 = 3.98 O 2.82 g Na/(22.99 g/mol) = 0.123 mol Na 4.35 g Cl/(35.45 g/mol) = 0.123 mol Cl 7.83 g O/(16.00 g/mol) = 0.489 mol O Na1 Cl1 O3.98→ NaClO4 NaClO4 is sodium perchlorate.

  21. 90.08 g molar mass of lactate 30.03 g mass of CH2O Determining a Molecular Formula from Elemental Analysis and Molar Mass During physical activity, lactic acid (M=90.08 g/mol) forms in muscle tissue and is responsible for muscle soreness. Elemental analysis shows that this compound contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O. Determine the empirical formula of lactic acid. (b) Determine the molecular formula. SOLUTION:Assuming there are 100. g of lactic acid, the constituents are: 3.33/3.33 = 1 C 6.66 /3.33 = 2 H 3.33 /3.33 = 1 O 40.0 g C/(12.01 g/mol) = 3.33 mol C 6.66 g H/(1.008 g/mol) = 6.66 mol H 53.3 g O/(16.00 g/mol) = 3.33 mol O CH2O : empirical formula C3H6O3 is the molecular formula =3 =

  22. Combustion analyzer for CH(O) of organic compounds. concept. Combustion Other gas absorber CO2(g) absorber H2O(g) absorber O2 CnHmOx+ (n + m/2 ) O2 = nCO2(g)+ (m/2) H2O(g)

  23. Combustion analyzer for CHNS of organic compounds. Modern version. CnHmNpSr+ (n + m/2 + p+ r) O2 →nCO2(g)+ (m/2) H2O(g)+ p NO2(g) + rSO2(g) CHNS (O) Analyzer modern instruments are automated for analysis. 3 mg of sample is sufficient to give satisfactory CHNS analysis.

  24. Determining a Molecular Formula from Combustion Analysis Vitamin C (M=176.12g/mol) is a compound of C,H, and O found in many natural sources especially citrus fruits. When a 1.000-g sample of vitamin C is placed in a combustion chamber and burned, the following data are obtained: mass of CO2after combustion =1.50 g mass of H2O after combustion =0.41 g What is the molecular formula of vitamin C? SOLUTION: • CO2 = 1.50 g → 1.50 g CO2 /(44.01 g CO2 /mol)× 12.01 g C/mol=0.409 g C • H2O = 0.41 g → 0.41 g H2O/(18.02 g H2O/mol)× 2.016 g H/mol=0.046 g H • O must be the difference: 1.000 g - (0.409 + 0.046) = 0.545 g O

  25. Determining a Molecular Formula from Combustion Analysis – cont. • 0.409 g C → 0.409/12.01=0.0341 mol C • 0.046 g H → 0.046/1.008=0.0456 mol H • 0.545 g O → 0.545/16.00=0.0341 mol O • 0.0341/0.0341 = 1 mol C • 0.0456/0.0341 = 1.33 mol H • 0.0341/0.0341 = 1 mol O C1H1.33O1 → C3H4O3→88.06 g/formula mass Molecular Formula = (Empirical Formula )n Molecular Formula mass = (Empirical Formula mass)n n = 176.12 g/88.06 g= 2 (C3H4O3)2=C6H8O6

  26. Constitutional Isomers of C2H6O

  27. Balancing Chemical Equations • Adjust the coefficient to the most simple whole numbers. • Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. • Check to make sure that you have the same number of each type of atom on both sides of the equation. • Set the equation for the coefficients (a, b, c & d) to make the number of atoms of each element the same on both sides of the equation. • Solve the equations. Usually start by setting the coefficient of the most complex compound be 1. 6. Specify the states of matter.

  28. X 2 Reactants Products 4 C 4 C 12 H 12 H 14 O 14 O Balancing Chemical Equations - example Ethane reacts with oxygen to form carbon dioxide and water a = 1 b = 7/2 c = 2 d = 3 a = 2 b = 7 c = 4 d = 6 aC2H6 + bO2 → c CO2 + d H2O C: 2 a = c, O: 2 b = 2 c +d, H: 6 a = 2 d. 2 C2H6 + 7 O2 → 4 CO2 + 6 H2O Let a = 1 2 = c, 2 b = 2 c +d, 6 = 2 d. => d = 3 6 = 2 d. => d = 3 2b = 2x2 + 3 = 7. b = 7/2 2 C2H6(g)+ 7 O2(g)→ 4 CO2(g) + 6 H2O(l)

  29. mass-mole-number relationships in a chemical reaction. MASS(g) of compound A MASS(g) of compound B M (g/mol) of compound A M (g/mol) of compound B molar ratio from balanced equation AMOUNT(mol) of compound A AMOUNT(mol) of compound B Avogadro’s number (molecules/mol) Avogadro’s number (molecules/mol) MOLECULES (or formula units) of compound A MOLECULES (or formula units) of compound B

  30. Calculating Amounts of Reactants and Products In a lifetime, the average American uses 1750 lb(794 g) of copper in coins, plumbing, and wiring. Copper is obtained from sulfide ores, such as chalcocite, or copper(I) sulfide, by a multistep process. After an initial grinding, the first step is to “roast” the ore (heat it strongly with oxygen gas) to form powdered copper(I) oxide and gaseous sulfur dioxide. (a) How many moles of oxygen are required to roast 10.0 mol of copper(I) sulfide? (b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide is roasted? (c) How many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide? • Write balanced chemical equation • Convert quantities of known substances into moles • Use coefficients in balanced equation to calculate the number of moles of the sought quantity • Convert moles of sought quantity into desired units

  31. 3mol O2 (a) 10.0mol Cu2S 2mol Cu2S 2mol SO2 64.07g SO2 2mol Cu2S mol SO2 10.0mol Cu2S (b) 103g Cu2O mol Cu2O kg Cu2O 143.10g Cu2O 3mol O2 32.00g O2 kg O2 (c) 2.86kg Cu2O 2mol Cu2O mol O2 103g O2 Calculating Amounts of Reactants and Products SOLUTION: 2Cu2S(s) + 3O2(g) → 2Cu2O(s) + 2SO2(g) (a) How many moles of oxygen are required to roast 10.0 mol of copper(I) sulfide? = 15.0mol O2 (b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide is roasted? = 641g SO2 (c) How many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide? = 20.0mol Cu2O 20.0mol Cu2O = 0.959kg O2

  32. 1 molecule C3H8 + 5 molecules O2 → 3 molecules CO2 + 4 molecules H2O molecules amount (mol) 1 mol C3H8 + 5 mol O2 → 3 mol CO2 + 4 mol H2O mass (amu) 44.09 amu C3H8 + 160.00 amuO2 → 132.03 amu CO2 + 72.06 amu H2O mass (g) 44.09 g C3H8 + 160.00 g O2 → 132.03 g CO2 + 72.06 g H2O total mass (g) 204.09 g → 204.09 g Balanced Equation Viewed in Terms of Reactants → Products C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

  33. Cl2 F2 Using Molecular Depictions to Solve a Limiting-Reactant Problem Nuclear engineers use chlorine trifluoride in the processing of uranium fuel for power plants. This extremely reactive substance is formed as a gas in special metal containers by the reaction of elemental chlorine and fluorine. (a) Suppose the reaction is run with 3.00 mol of Cl2 and 6.00 mol of F2, name the limiting reactant, and draw the container contents after the reaction is complete. (b)What will be the mass of remaining reactants and product? 1Cl2(g) + 3F2(g) → 2ClF3(g) M(g/mol)= 70.9 38.0 92.5

  34. Calculating Amounts of Reactant and Product in a Limiting-Reactant Problem A fuel mixture used in the early days of rocketry is composed of two liquids, hydrazine(N2H4) and dinitrogentetraoxide(N2O4), which ignite on contact to form nitrogen gas and water vapor. How many grams of nitrogen gas form when 1.00×102 g of N2H4 and 2.00×102 g of N2O4 are mixed? SOLUTION: 2 N2H4(l) + N2O4(l) → 3 N2(g) + 4 H2O(l) 1.00×102g N2H4 /(32.05g/mol) = 3.12mol N2H4 2.00×102g N2O4/(92.02g/mol)= 2.17mol N2O4 N2H4 is the limiting reactant. 3.12mol N2H4 /2= 1.56 < 2.17mol N2O4/1 1.56×3 = 4.48 mol N2 will be produced. 4.68mol N2×28.02 g/mol = 131 g N2

  35. % Yield = x 100 D Actual Yield (side products) Theoretical Yield The effect of side reactions on yield. Reaction Yield Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. A + B C (reactants) (main product) Actual Yield is the amount of product actually obtained from a reaction.

  36. 51.4 kg 66.73 kg Calculating Percent Yield Silicon carbide (SiC) is an important ceramic material that is made by allowing sand(silicon dioxide, SiO2) to react with powdered carbon at high temperature. Carbon monoxide is also formed. When 100.0 kg of sand is processed, 51.4 kg of SiC is recovered. What is the percent yield of SiC from this process? SOLUTION: SiO2(s) + 3C(s) → SiC(s) + 2CO(g) 100.0 kg SiO2 × (103 g /kg) ×(mol/60.09 g) = 1664 mol SiO2 nSiO2 = nSiC = 1664 1664 mol SiC ×(40.10 g/mol) × (kg/103 g) = 66.73 kg SiC × 100= 77.0%

  37. 0.715 mol glycine 1000mL 495 mL soln 1 L Calculating the Molarity of a Solution Glycine (H2NCH2COOH) is the simplest amino acid. What is the molarity of an aqueous solution that contains 0.715 mol of glycine in 495 mL? Molarity (M, mol/L) = nsolute/Vsolution SOLUTION: = 1.44 M glycine

  38. MASS (g) of compound in solution AMOUNT (mol) of compound in solution MOLECULES (or formula units) of compound in solution VOLUME (L) of solution mass-mole-number-volume relationships in solution. M (g/mol) Avogadro’s number (molecules/mol) Molarity (mol/L)

  39. 0.460 moles 1 L 141.96 g Na2HPO4 mol Na2HPO4 Calculating Mass of Solute in a Given Volumeof Solution A “buffered” solution maintains acidity as a reaction occurs. In living cells phosphate ions play a key buffering role, so biochemistry often study reactions in such solutions. How many grams of solute are in 1.75 L of 0.460 M sodium monohydrogen phosphate? SOLUTION: = 0.805 mol Na2HPO4 1.75 L× = 114 g Na2HPO4 0.805 mol Na2HPO4 ×

  40. Converting a concentrated solution to a dilute solution. The total number of solutes in the solution remains the same after dilution, But the number of solutes per unit volume changes...

  41. Preparing a Dilute Solution from a Concentrated Solution “Isotonic saline” is a 0.15 M aqueous solution of NaCl that simulates the total concentration of ions found in many cellular fluids. Its uses range from a cleaning rinse for contact lenses to a washing medium for red blood cells. How would you prepare 0.80 L of isotonic saline from a 6.0 M stock solution? the number of moles of solute does not change during the dilution but the volume does Mdil × Vdil= nsolute= Mconc × Vconc SOLUTION: nsolute required: 0.80 L soln ×0.15 M NaCl = 0.12 mol NaCl volume of conc. solution = 0.12 mol NaCl/6M = 0.020 L soln

  42. Calculating Amounts of Reactants and Products for a Reaction in Solution Specialized cells in the stomach release HCl to aid digestion. If they release too much, the excess can be neutralized with antacids. A common antacid contains magnesium hydroxide, which reacts with the acid to form water and magnesium chloride solution. As a government chemist testing commercial antacids, you use 0.10MHCl to simulate the acid concentration in the stomach. How many liters of “stomach acid” react with a tablet containing 0.10g of magnesium hydroxide? SOLUTION: Mg(OH)2(s) + 2HCl(aq) → MgCl2(aq) + 2H2O(l) n Mg(OH)2/1 = nHCl/2 0.10g / (58.33g/mol) = 1.7 × 10-3 mol Mg(OH)2 = n HCl/2 nHCl = 3.4 × 10-3 mol HCl = 0.10 M × VHCl VHCl= 3.4 × 10-2L HCl

  43. Solving Limiting-Reactant Problems for Reactions in Solution Mercury and its compounds have many uses, from fillings for teeth (as an alloy with silver, copper, and tin) to the industrial production of chlorine. Because of their toxicity, however, soluble mercury compounds, such mercury(II) nitrate, must be removed from industrial wastewater. One removal method reacts the wastewater with sodium sulfide solution to produce solid mercury(II) sulfide and sodium nitrate solution. In a laboratory simulation, 0.050L of 0.010M mercury(II) nitrate reacts with 0.020L of 0.10M sodium sulfide. How many grams of mercury(II) sulfide form? SOLUTION: Hg(NO3)2(aq) + Na2S(aq) → HgS(s) + 2NaNO3(aq) 0.050L×0.010M = 5.0×10-4mol HgS 0.020L×0.10M = 2.0x10-3 mol Na2S Since (5.0×10-4mol Hg(NO3)2/1) < (2.0x10-3 mol Na2S /1), Hg(NO3)2 is the limiting reagent. (5.0×10-4mol ) ×1 HgS will form. 5.0×10-4 mol × 232.7g/mol = 0.12 g HgS

  44. Limiting Reagent and theoretical Yield Products Reactants aA + bB → cC + dD

  45. 1. 다음 중 원소기호와 그 이름이 바르게 짝지어지지 않은 것을 고르고 바른 이름을 쓰시오. (a) Zn - Iron • (b) K - Potassium (c) Sn - Antimony • (d) Au - Gold (e) Ce - Cesium (f) Hg - Hydrogen (g) Fe - Steel (h) Ar -Arsenic

  46. 2. 다음의 순물질 중에서 원소가 아닌 것을 고르시오. 원소가 아닌 것의 이유를 쓰시오. (a) 수정 (b) 다이아몬드 (c) 흑연 (d) 풀러렌(C60) (e) 니켈 (f) 오존 (g) 사파이어 (i) 황동 (j) 청동 (h) 테플론 (j) 순금 (k) 브로민 (l) 주석 (m) 백금 (n) 지르코늄

  47. 메테인(CH4) 18.5 g이 염소 43.0 g과 반응하여 80% 수득률로염화메틸(CH3Cl)이 생성되었다. 생성된 반응물은 몇 g인가? 이 때 염화 수소도 같이 생성되었다. 원자질량(amu): C: 12.01, H:1.008, Cl: 35.45

  48. 뷰테인(부탄) 기체는 압축시켜서 일회용 라이터나 휴대용 가스버너의 액체 연료로 사용한다. 라이터에 뷰테인6.50mL가 들어있다(d＝0.579 g/mL). (a) 뷰테인을 전부 연소시키기 위해 필요한 최소량의산소는 몇 g인가? (b) 뷰테인이 전부 연소하면 CO2는 몇 mol이 생성되겠는가? (c) 뷰테인이 전부 연소하면, 전체 기체 분자는 얼마나 되겠는가? 원자질량(amu): C: 12.01, H:1.008, O: 16.00

  49. 진한 황산(18.3 M)은 밀도가 1.84g/mL이다. (a) 1 L용액 안에 황산 몇 mol이 들어있는가? (b) 용액에서 H2SO4의 질량 백분율은 얼마인가? 원자질량(amu): C: 12.01, H:1.008, O:16.00, S: 32.07

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