Mass Relationships in Chemical Reactions

1. Mass Relationships in Chemical Reactions Chapter 3

2. 화학양론 • 원자질량 • Avogadro 수와 몰질량 • 분자질량 • 질량분석기 • 화합물의 조성백분율 • 실험식 구하기 • 화학반응과 반응식 • 반응물과 생성물의 양 • 한계반응물 • 반응의 수득률

3. Micro World atoms & molecules Macro World grams Atomic mass is the mass of an atom in atomic mass units (amu) By definition: 1 atom 12C “weighs” 12 amu On this scale 1H = 1.008 amu 16O = 16.00 amu

4. 7.42×6.015 + 92.58×7.016 100 Natural lithium is: 7.42% 6Li (6.015 amu) 92.58% 7Li (7.016 amu) Average atomic mass of lithium: = 6.941 amu

5. Average atomic mass (6.941)

6. 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 × 1023 Avogadro’s number (NA) The Mole(chemical quantity)

7. 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(1.36×1017초)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).

8. 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

9. 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

10. 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)

11. The Mole(chemical quantity) 10 kg 수박 • 6.02 x 1024 kg • 지구질량 5.97×1024 kg 1몰 을 쌓자!

12. One Mole of: S C Hg Cu Fe

13. 1 12C atom 12.00 g 1.66×10-24 g = 12.00 amu 6.022×102312C atoms 1 amu = molar mass in g/mol NA= Avogadro’s number M × 1 amu = 1.66×10-24 g or 1 g = 6.022×1023 amu

14. 1 mol K = 39.10 g K 1 mol K = 6.022×1023 atoms K 1 mol K 6.022×1023 atoms K 0.551 g K × × = 1 mol K 39.10 g K Q. Do You Understand Molar Mass? How many atoms are in 0.551 g of potassium (K) ? 8.49×1021 atoms K

15. 1S 32.07 amu 2O + 2×16.00 amu SO2 SO2 64.07 amu Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule. For any molecule molecular mass (amu) = molar mass (grams) 1 molecule SO2 = 64.07 amu 1 mole SO2 = 64.07 g SO2

16. 1 mol C3H8O = (3 × 12) + (8 × 1) + 16 = 60 g C3H8O 1 mol C3H8O molecules = 8 mol H atoms 1 mol H = 6.022×1023 atoms H 8 mol H atoms 6.022×1023 H atoms 1 mol C3H8O × 72.5 g C3H8O × × = 1 mol C3H8O 1 mol H atoms 60 g C3H8O Q. Do You Understand Molecular Mass? How many H atoms are in 72.5 g of C3H8O ? 5.82 x 1024 atoms H

17. 1Na 22.99 amu NaCl 1Cl + 35.45 amu NaCl 58.44 amu Formula mass is the sum of the atomic masses (in amu) in a formula unit of an ionic compound. For any ionic compound formula mass (amu) = molar mass (grams) 1 formula unit NaCl = 58.44 amu 1 mole NaCl = 58.44 g NaCl

18. 1 formula unit of Ca3(PO4)2 3×40.08 3 Ca 2 P 2×30.97 8 O 310.18 amu + 8×16.00 Q. Do You Understand Formula Mass? What is the formula mass of Ca3(PO4)2 ?

19. Heavy Heavy Light Light KE = ½ ×m×v2 = qV v= (2 ×qV/m)1/2 F = q×v×B = mv2/R R =(mV/qB2)1/2

20. 2×(12.01 g) 6× (1.008 g) 1× (16.00 g) n× molar mass of element %C = %H = %O = × 100% = 34.73% × 100% = 13.13% × 100% = 52.14% × 100% 46.07 g 46.07 g 46.07 g molar mass of compound C2H6O Percent composition of an element in a compound = n is the number of moles of the element in 1 mole of the compound 52.14% + 13.13% + 34.73% = 100.0%

21. nK = 24.75 g K × = 0.6330 mol K 1 mol Mn nMn = 34.77 g Mn × = 0.6329 mol Mn 54.94 g Mn 1 mol O nO = 40.51 g O × = 2.532 mol O 16.00 g O 1 mol K 39.10 g K Percent Composition and Empirical Formulas Determine the empirical formula of a compound that has the following percent composition by mass: K 24.75, Mn 34.77, O 40.51 percent.

22. = 1.0 K : Mn : O : 1.0 4.0 0.6330 0.6329 2.532 ~ ~ ~ ~ 0.6329 0.6329 0.6329 Percent Composition and Empirical Formulas nK = 0.6330, nMn = 0.6329, nO = 2.532 KMnO4

23. g CO2 g H2O mol CO2 mol H2O mol C mol H g C g H Combust 11.5 g ethanol Collect 22.0 g CO2 and 13.5 g H2O 6.0 g C = 0.5 mol C 1.5 g H = 1.5 mol H g of O = g of sample – (g of C + g of H) 4.0 g O = 0.25 mol O Empirical formula C0.5H1.5O0.25 Divide by smallest subscript (0.25) Empirical formula C2H6O

24. 3 ways of representing the reaction of H2 with O2 to form H2O reactants products A process in which one or more substances is changed into one or more new substances is a chemical reaction A chemical equation uses chemical symbols to show what happens during a chemical reaction

25. How to “Read” Chemical Equations 2 Mg + O2 2 MgO 2 atoms Mg + 1 molecule O2 makes 2 formula units MgO 2 moles Mg + 1 mole O2 makes 2 moles MgO 48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO IS NOT 2 grams Mg + 1 gram O2 makes 2 g MgO

26. Ethane reacts with oxygen to form carbon dioxide and water a C2H6 + b O2 c CO2 + d H2O Balancing Chemical Equations • 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. • 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. C: 2 a = c, O: 2 b = 2 c +d, H: 6 a = 2 d.

27. Balancing Chemical Equations • Solve the equations. Usually start by setting the coefficient of the most complex compound be 1. C: 2 a = c, O: 2 b = 2 c +d, H: 6 a = 2 d. Let a = 1 a =1 b = 7/2 c = 2 d = 3 2 = c, 2 b = 2 c +d, 6 = 2 d. => d = 3 2b = 2×2 + 3 = 7. b = 7/2

28. × 2 2 C2H6 + 7 O2 4 CO2 + 6 H2O Balancing Chemical Equations • Adjust the coefficient to the most simple whole numbers. a =1 b = 7/2 c = 2 d = 3 a =2 b = 7 c = 4 d = 6

29. 4 C (2× 2) 4 C Reactants Products 12 H (2× 6) 12 H (6× 2) 14 O (7× 2) 14 O (4× 2 + 6) 4 C 4 C 12 H 12 H 14 O 14 O 2C2H6 + 7O2 4CO2 + 6H2O Balancing Chemical Equations • Check to make sure that you have the same number of each type of atom on both sides of the equation.

30. Amounts of Reactants and Products • 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. 2CH3OH + 3O2 2CO2 + 4H2O grams CH3OH moles CH3OH moles H2O grams H2O 4 mol H2O 18.0 g H2O 1 mol CH3OH × = × × 2 mol CH3OH 32.0 g CH3OH 1 mol H2O Q. Methanol burns in air according to the equation If 209 g of methanol are used up in the combustion, what mass of water is produced? molar mass CH3OH molar mass H2O coefficients chemical equation 209 g CH3OH 235 g H2O

32. 2NO + O2 2NO2 2NO + 1O2 2NO2 Limiting Reagents (한계반응물) NO: 7/2=3.5 O2 : 7/1 =7 NO is the limiting reagent O2 is the excess reagent NO2 : 3.5×2 =7

33. 2Al + Fe2O3 Al2O3 + 2Fe 124 g Al 601 g Fe2O3 27.0 g/mol Al 160.0 g/mol Fe2O3 = 3.756 = 2.296 2 1 Q. Limiting Reagents? In one process, 124 g of Al are reacted with 601 g of Fe2O3 Calculate the mass of Al2O3 formed. < Al is limiting reagent

34. 2Al + Fe2O3 Al2O3 + 2Fe Use limiting reagent (Al) to calculate amount of product that can be formed. 2.296 × 102. g Al2O3 = 234 g Al2O3

35. % Yield = ×100 Actual Yield Theoretical Yield Reaction Yield Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Actual Yield is the amount of product actually obtained from a reaction.

36. 3H2 (g) + N2 (g) 2NH3 (g) NH3 (aq) + HNO3 (aq) NH4NO3 (aq) 2Ca5(PO4)3F (s) + 7H2SO4 (aq) fluorapatite 3Ca(H2PO4)2 (aq) + 7CaSO4 (aq) + 2HF (g) Chemistry In Action: Chemical Fertilizers Plants need: N, P, K, Ca, S, & Mg

37. 2H2 + 1O2 2H2O 1 mol O2 2 mol H2 2 mol H2O 2 분량 2.0g H2 1 분량 32.0 g O2 2 분량 18.0g H2O