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Chemical Formulas and Composition Stoichiometry 化學計量法

2. Chemical Formulas and Composition Stoichiometry 化學計量法. Chapter Goals.  Chemical Formulas 化學式  Ions and Ionic Compounds 離子及離子化合物  Names and Formulas of Some Ionic Compounds  Atomic Weights 原子重  The Mole 莫耳數  Formula Weights, Molecular Weights, and Moles

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Chemical Formulas and Composition Stoichiometry 化學計量法

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  1. 2 Chemical Formulas and Composition Stoichiometry 化學計量法

  2. Chapter Goals  Chemical Formulas 化學式  Ions and Ionic Compounds 離子及離子化合物  Names and Formulas of Some Ionic Compounds  Atomic Weights 原子重  The Mole 莫耳數  Formula Weights, Molecular Weights, and Moles  Percent Composition and Formulas of Compounds  Derivation of Formulas from Elemental Composition  Determination of Molecular Formulas 分子式  Some Other Interpretations of Chemical Formulas  Purity of Samples

  3. Chemical Formulas 化學式 • Chemical formula shows the chemical composition of the substance. • ratio of the elements present in the molecule or compound • He, Ne, Na – monatomic elements • O2, H2, Cl2 – diatomic elements • O3, S4, P8 - more complex elements • H2O, C12H22O11 – compounds • Substance consists of two or more elements

  4. Chemical Formulas Allotropic modifications (allotropes)同素異形物: different forms of the same element in the same physical state(chapter 13)

  5. Chemical Formulas • Compound contain two or more elements in chemical combination in fixed proportions -Law of Definite Proportions 定組成定律 2 H atoms 1 O atom 1 H atom 1 Cl atom 4 C atoms 10 H atoms 1 O atom 1 N atom 3 H atoms 3 C atoms 8 H atoms 丙烷 乙醚 Organic compounds:contain C―C or C―H bonds or both, often in combination with nitrogen, oxygen, sulfur and other elements Inorganic compounds:do notcontain C―H bonds

  6. Chemical Formulas (比例模型) (球棍模型) Chemical formula the number of atoms of each type in the molecule • Structural Formula the order in which atoms are connected (chemical bonds between atoms)  Ball-and-Stick Model three-dimensional shape of molecules  Space-Filling Model Relative size of atoms and the shapes of molecules 四氯化碳 Fig 2-1 Formula and models for some molecules

  7. Example 2-1 Chemical Formula Look at each of following molecular models. For each one, write the structural formula and the chemical formula. (color code: black=carbon; white=hydrogen; red=oxygen; blue=nitrogen; light green=fluorine; dark green=chloride.) a)1-butanol丁醇 b)Freon-12二氯二氟代甲烷 (氟氯烷) c)Nitrogen mustard HN1 H H H H H C O H C C C H H H H Cl Cl F C F H H H H H Cl C O Cl C N C H H H H CHONFCl Occurs in some fruits, dried beans, cheese, and nuts; used as an additive in certain plastics, detergents, and some medicinal formulations C4H10O A highly toxic substance, used as a chemotherapy drug in the treatment of Hodgkin’s disease and of some forms of chronic leukemia Formerly used as a refrigerant; implicated in atmospheric ozone depletion CF2Cl2 C4H9NCl2

  8. Ions and Ionic Compounds • Ions are atoms or groups of atoms that possess an electric charge • Two basic types of ions: Positive ions or cations 正離子 one or more electrons less than neutral Na+, Ca2+, Al3+ NH4+-polyatomic cation Negative ions or anions 負離子 one or more electrons more than neutral F-, O2-, N3- SO42-, PO43--polyatomic anions Formula Unit: the small repeat of a substance- for non-ionic substances, the molecule NaCl, CaCl2 Fig 2-2 The arrangement of ions in NaCl

  9. Ions and Ionic Compounds Polyatomic ion (NH4)2SO4 Polyatomic compound Metals form more than one kind of ion

  10. Names and Formulas of Some Ionic Compounds • Formulas of ionic compounds are determined by the charges of the ions Charge on the cations = the charge on the anions (add to zero) • The compound must be neutral NaCl sodium chloride (Na1+ & Cl1-) KOH potassium hydroxide (K1+ & OH1-) CaSO4 calcium sulfate (Ca2+ & SO42-) Al(OH)3 aluminum hydroxide (Al3+ & 3 OH1-)

  11. H+ 1+ hydrogen Names and Formulas of Some Ionic Compounds You must know all of the molecular compounds from Table 2-2. Some examples are: H2SO4 - sulfuric acid FeBr2 - iron(II) bromide C2H5OH - ethanol

  12. Example 2-2 Formulas for Ionic Compounds Write the formula of the following ionic compounds: (a) sodium fluoride, (b) calcium fluoride, (c) iron(II) sulfate, (d) zinc phosphate (a) Na+ F- NaF (c) Fe2+ SO42- FeSO4 (d) Zn2+ PO43- Zn3(PO4)2 (b) Ca2+ F- CaF2 Example 2-3 Name for Ionic Compounds Name the following ionic compounds: (a) (NH4)2S, (b) Cu(NO3)2, (c) ZnCl2, (d) Fe2(CO3)3 (a) NH4+ S2- ammonium sulfide (c) Zn2+ Cl- Znic Chloride (b) Cu2+ NO3- copper(II) nitrate (d) Fe3+ CO32- iron(III) carbonate

  13. Names and Formulas of Some Ionic Compounds You do it! • What is the formula of nitric acid? HNO3 • What is the name of FeBr3? iron(III) bromide • What is the name of K2SO3? potassium sulfite • What is charge on sulfite ion? SO32- is sulfite ion

  14. Names and Formulas of Some Ionic Compounds You do it! • What is the formula of ammonium sulfide? (NH4)2S • What is charge on ammonium ion? NH41+ • What is the formula of aluminum sulfate? Al2(SO4)3 • What is charge on both ions? Al3+ and SO42-

  15. Atomic Weights 原子量 Atomic weights (AW) • an early observation was that carbon and hydrogen have relative atomic masses, of approximately 12 and 1 Atomic mass unit (amu) • exactly 1/12 of the mass of a particular kind of carbon atom, called carbon-12 H Hydrogen 1.00794 amu Na Sodium 22.989768 amu Mg Magnesium 24.305 amu

  16. The Mole莫耳數 1 mole = 6.022 x 1023 particles Avogadro’s number (NA) = 6.022 x 1023 亞佛加厥常數 Helium exists as discrete He atom  1 mole of He consist of 6.022x1023 He atoms Hydrogen commonly exists as diatomic (two-atom)H2  1 mole of Hydrogen is 6.022x1023 H2 molecule (2x(6.022x1023) H atoms) • Molar mass (g/mol) 莫耳質量 • the mass of 1 mole of atoms of a pure element in grams = the atomic weight of the element in atomic mass unit • H has an atomic weight of 1.00794 g • 1.00794 g of H atoms = 6.022 x 1023 H atoms • Mg has an atomic weight of 24.3050 g • 24.3050 g of Mg atoms = 6.022 x 1023 Mg atoms

  17. The Mole C 12 Ti 47.9 Au 197 H 1 S 32 Table 2-3, p. 57

  18. The Mole One mole of atoms Mercury (200.6g) 水銀 Bromine (79.9g) 溴 Aluminum (27.0g) 鋁 Copper (63.5g) 銅 Znic (65.4g) 鋅 iron (55.8g) 鐵 Sulfur (32.1g) 硫 Fig. 2-4, p. 57

  19. 55.85g Fe 1 mole atoms 1mole Fe atoms 55.85g Fe 1 mole atoms 55.85g Fe 6.022x1023 atoms 1 mole atoms 6.022x1023 atoms 1 mole atoms 1 mole atoms 6.022x1023 atoms Example 2-4; 2-5 Moles of atoms; Numbers of atoms How many moles of atoms does 136.9g of iron metal contain? And How many atoms? 136.9g Fe x ? Mole Fe atoms= =2.451 mol Fe atoms and ? Fe atoms= 2.451 mole atoms x =1.476x1024 Fe atoms Exercise 32, 40 and 42

  20. 55.85g Fe ? g Fe 1 mole Fe atoms = x 1 mole atoms Fe atom 6.022x1023 Fe atoms 6.022x1023 atoms 1 mole sheets 1.9 in. x 500 sheets 1 mi. 1ft x 12 in. 5,280ft Example 2-6 Masses of Atoms Calculate the average mass of one iron atom in grams. =9.274x10-23 g (the average mass of 1 Fe atom) Example 2-7 Avogadro’s Number A stack of 500 sheets of typing paper is 1.9 inches thick. Calculate the thickness, in inches and in miles, of a stack of typing paper that contains one mole of sheets. ? In.= 1 mole atoms sheets x =2.3x1021 in. ? mi.= 2.3x1021 in. x =3.6x1016 mi.

  21. ? g Mg = The Mole Example 2-1Calculate the mass of a single Mg atom in grams to 3 significant figures. 24.3g Mg 1 mole Mg atoms 1 Mg atom x x 1mole Mg atoms 6.022x1023 Mg atoms = 4.04 x10-23 g Mg

  22. ? Mg atoms = 1mole Mg 6.022x1023 Mg x 1 mole Mg 24.3g Mg The Mole Example 2-2Calculate the number of atoms in one-millionth of a gram of Mg to 3 significant figures. 1.00x10-6g Mg x = 2.48 x10-16 Mg atoms

  23. 6.022x1023 Mg atoms 1 mol Mg ? Mg atoms = The Mole Example 2-3How many atoms are contained in 1.67 moles of Mg? 1.67 mol Mg x = 1.0 x1024 Mg atoms

  24. 1mole Mg ? mol Mg = 24.3g Mg The Mole Example 2-4How many moles of Mg atoms are present in 73.4 g of Mg? 73.4g Mg x = 3.02 mol Mg IT IS IMPERATIVE THAT YOU KNOW HOW TO DO THESE PROBLEMS

  25. Formula Weights, Molecular Weights, and Moles 式量,分子量及莫耳數 • How do we calculate the molar mass of a compound? • add atomic weights of each atom Formula Weight (FW) • The molar mass of propane (丙烷), C3H8, is: 3 x C = 3 x 12.01 amu = 36.03 amu 8 x H = 8 x 1.01 amu = 8.08 amu Molar Mass = 44.11 amu

  26. Formula Weights, Molecular Weights, and Moles • The molar mass of calcium nitrate, Ca(NO3)2 , is: You do it! 1 x Ca = 1 x 40.08 amu = 40.08 amu 2 x N = 2 x 14.01 amu = 28.02 amu 6 x O = 6 x 16.00 amu = 96.00 amu Molar Mass = 164.10 amu

  27. Formula Weights, Molecular Weights, and Moles Molar mass of the substance = the formula weight of the substance

  28. Formula Weights, Molecular Weights, and Moles 28

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  30. Formula Weights, Molecular Weights, and Moles One Mole of Contains Cl2 or 70.90g 6.022 x 1023 Cl2 molecules 2(6.022 x 1023 ) Cl atoms C3H8 You do it! • 44.11 g • 6.022 x 1023 C3H8 molecules • 3 (6.022 x 1023 ) C atoms • 8 (6.022 x 1023 ) H atoms 30

  31. Formula Weights, Molecular Weights, and Moles 64.1g SO2 6.022x1023 SO2 molecule Example 2-9 Masses of Molecules What is the mass in grams of 10.0 million SO2 molecules? ? g SO2 = 1.0x106 SO2 molecules x =1.06x10-15 g SO2 Exercise 44 31

  32. Formula Weights, Molecular Weights, and Moles 1 mol O2 32.0g O2 6.02x1023 O2 molecules 1mol O2 6.02x1023 O2 molecules 32.0g O2 2 Oatoms x 1 O2 molecule Example 2-10 Moles How many (a) moles of O2 (b) O2 molecules, and (c) O atoms are contained in 40g of oxygen gas (dioxygen) at 25oC? One mole of O2 contains 6x1023 O2 molecules, and its mass is 32.0g (a) ? Mol O2 = 40.0g O2 x = 1.25 mol O2 (b) ? O2 molecules = 1.25 mol O2 x = 7.52x1023 O2 molecules ? O atoms = (c) 40.0g O2 x =1.5x1024 O atoms Exercise 36 32

  33. Formula Weights, Molecular Weights, and Moles 1 mol (NH4)2SO4 132.1g (NH4)2SO4 6.02x1023 formula units(NH4)2SO4 8 Hatoms x x 1 formula units (NH4)2SO4 1mol (NH4)2SO4 Example 2-11 Numbers of atoms Calculate the number of hydrogen atoms in 39.6g of ammonium sulfate, (NH4)2SO4 One mole of (NH4)2SO4 is 6x1023 formula units and has a mass of 132.1g g of (NH4)2SO4 mol of (NH4)2SO4 Formula units of (NH4)2SO4 H atoms ? H atoms = 39.6g (NH4)2SO4 x =1.44x1024 H atoms Exercise 34 33

  34. 1mole C3H8 ? C3H8 molecules = 44.11g C3H8 6.022x1023 C3H8 molecules x 1 mole C3H8 Formula Weights, Molecular Weights, and Moles Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane 74.6g C3H8 x = 1.02 x1024 molecules

  35. 1 mole C3H8 ?g C3H8 molecules = 6.022x1023 C3H8 molecules 44.11g C3H8 x 1mole C3H8 Formula Weights, Molecular Weights, and Moles Example 2-6. What is the mass of 10.0 billion propane molecules? 1.00x1010 molecules x = 7.32 x10-13 g of C3H8

  36. Formula Weights, Molecular Weights, and Moles Example 2-7. How many (a) moles, (b) molecules, and (c) oxygen atoms are contained in 60.0 g of ozone, O3? The layer of ozone in the stratosphere is very beneficial to life on earth. (a) 60.0g O3 x = 1.25 moles ? moles O3 ? molecules O3 ? O atoms = = = 3O atoms 1mole 6.022x1023 molecules (b) 1.25 moles x 1 O3 molecule 1 mole 48.0 gO3 = 7.53 x1023 molecules O3 (c) 7.53x1023 molecules O3 x = 2.26 x1024 atoms O

  37. Formula Weights, Molecular Weights, and Moles Example 2-8. Calculate the number of O atoms in 26.5 g of Li2CO3. 26.5g Li2CO3 x ? O atoms = x x 1 mol Li2CO3 3 O atoms 6.022x1023 form. units Li2CO3 73.8g Li2CO3 1 form. unit Li2CO3 1 mole Li2CO3 = 6.49 x1023 O atoms

  38. Formula Weights, Molecular Weights, and Moles • Occasionally, we will use millimoles • Symbol - mmol • 1000 mmol = 1 mol1 mmol = 10-3 mole • For example: oxalic acid 草酸 (COOH)2 • 1 mol = 90.04 g • 1 mmol = 0.09004 g or 90.04 mg milli10-3

  39. 1mmole (COOH)2 0.09004g (COOH)2 ? Mmol (COOH)2 = Formula Weights, Molecular Weights, and Moles Example 2-9Calculate the number of mmol in 0.234 g of oxalic acid, (COOH)2. 0.234g (COOH)2 x = 2.6mmol (COOH)2

  40. Mass C 3x12.01g = = % C 3x12.01+8x1.01g Mass C3H8 Percent Composition and Formulas of Compounds • % composition = mass of an individual element in a compound divided by the total mass of the compound x 100% • Determine the percent composition of C in C3H8. x 100% x 100% = 81.68%

  41. 8x1.01g Mass H = = % H Mass C3H8 3x12.01+8x1.01g Percent Composition and Formulas of Compounds • What is the percent composition of H in C3H8? x 100% x 100% = 18.32%

  42. Percent Composition and Formulas of Compounds Mass C Mass N Mass O 14.0g 48.0g 1.0g x 100% x 100% x 100% x 100% x 100% x 100% = = = Mass HNO3 Mass HNO3 Mass HNO3 63.0g 63.0g 63.0g Example 2-12 Percent Composition Calculate the percent composition by mass of HNO3 1xH=1x1.0g=1.0g 1xN=1x14.0g=14.0g3xO=3x16g=48g Mass of 1 mol of HNO3= 1+14.0+48.0 = 63.0g % H = = 1.6% H % N = = 22.2% N % O = = 76.2% O Total =100.0% Exercise 62 42

  43. 2x55.8g 3S 12O 2Fe = = % Fe 2x55.8+3x32.1+12x16.0 g Fe2(SO4)3 Fe2(SO4)3 Fe2(SO4)3 12x16g 2x32.1g = = = = % S % O 399.9g 399.9g Percent Composition and Formulas of Compounds Example 2-10Calculate the percent composition of Fe2(SO4)3 to 3 significant figures. x 100% x 100% = 27.9% x 100% = 24.1% x 100% x 100% x 100% = 48.0%

  44. Derivation of Formulas from Elemental Composition • Empirical Formula (實驗式) - smallest whole-number ratio of atoms present in a compound • CH2 is the empirical formula for alkenes • No alkene exists that has 1 C and 2 H’s • Molecular Formula - actual numbers of atoms of each element present in a molecule of the compound • Ethene – C2H4 (CH2)2 • Pentene – C5H10 (CH2)5 • We determine the empirical and molecular formulas of a compound from the percent composition of the compound. • percent composition is determined experimentally

  45. Molecular weight Simplest formula weight Determination of Molecular Formulas • Butane C4H10 Empirical formula C2H5 2x (C2H5) = C4H10 Molecular Formula • Benzene C6H6  Empirical formula CH 6x (CH) = C6H6 Molecular formula = n x simplest formula Molecular weight = n x simplest formula weight n=

  46. Derivation of Formulas from Elemental Composition 1 mol S atoms 1 mol O atoms 32.1g S 16.0g O Example 2-13 Simplest Formulas Compounds of containing sulfur and oxygen are serious air pollutants; they represent the major cause of acid rain. Analysis of a simple of a pure compound reveals that it contains 50.1% sulfur and 49.9% oxygen by mass. What is the simplest formula of the compound? ? mol S atoms = 50.1g S x = 1.56 mol S atoms ? mol O atoms = 49.9g O x = 3.12 mol O atoms S mol : O mol = 1.56 mol S atoms : 3.12 mol O atoms =1:2 SO2 Exercise 54

  47. Derivation of Formulas from Elemental Composition 1 mol O 1 mol Na 1 mol S 16.0g O 23.0g Na 32.1g S Example 2-14 Simplest Formulas A 20.882g sample of an ionic compound is found to contain 6.072g of Na, 8.474g of S, and 6.336g of O. What is its simplest formula? ? mol Na = 6.072g Na x = 0.264 mol Na ? mol S =8.474g S x = 0.264 mol S ? mol O =6.336g O x = 0.396 mol O Na mol: S mol : O mol = 0.264 : 0.264 : 0.396 =2 : 2 : 3 Na2S2O3 Exercise 56

  48. 1 mol K 1 mol Mn 1 mol O 39.10g K 32.1g Mn 16.0g O Derivation of Formulas from Elemental Composition Example 2-11A compound contains 24.74% K, 34.76% Mn, and 40.50% O by mass. What is its empirical formula? Let the compound weight 100g = 0.6327 mol K ? mol K = 24.74g K x = 0.6327 mol Mn ? mol Mn = 34.76g Mn x ? mol O = 40.50g O x = 2.531 mol O K mol: Mn mol : O mol = 0.6327 : 0.6327 : 2.531 =1 : 1 : 4 KMnO4

  49. 1 mol Co 1 mol O 58.93g Co 16.0g O Derivation of Formulas from Elemental Composition Example 2-12: A sample of a compound contains 6.541g of Co and 2.368g of O. What is the empirical formula for this compound? = 0.111 mol Co ? mol Co = 6.541g Co x ? mol O = 2.638g O x = 0.148 mol O Co mol: O mol = 0.111 : 0.148 = 3 : 4 Co3O4

  50. Derivation of Formulas from Elemental Composition Magnesium perchlorate 過氯酸鎂 Mg(ClO4)2 H Sodium hydroxid  C O?? A combustion train used for carbon-hydrogen analysis 燃燒反應器 50

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