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Chapter #3 : Stoichiometry - Mole - Mass Relationships in Chemical Systems

Chapter #3 : Stoichiometry - 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

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Chapter #3 : Stoichiometry - Mole - Mass Relationships in Chemical Systems

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  1. Chapter #3 : Stoichiometry - 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

  2. Mole • The Mole is based upon the definition: • The amount of substance that contains as many elementary parts (atoms, molecules, or other?) as there are atoms in exactly • 12 grams of carbon -12. • 1 Mole = 6.022045 x 1023 particles

  3. Counting Objects of Fixed Relative Mass 12 red marbles @ 7g each = 84g 12 yellow marbles @4e each=48g 55.85g Fe = 6.022 x 1023 atoms Fe 32.07g S = 6.022 x 1023 atoms S Fig 3.1

  4. Mole - Mass Relationships of Elements Element Atom/Molecule Mass Mole Mass Number of Atoms 1 atom of H = 1.008 amu 1 mole of H = 1.008 g = 6.022 x 1023 atoms 1 atom of Fe = 55.85 amu 1 mole of Fe = 55.85 g = 6.022 x 1023 atoms 1 atom of S = 32.07 amu 1 mole of S = 32.07 g = 6.022 x 1023 atoms 1 atom of O = 16.00 amu 1 mole of O = 16.00 g = 6.022 x 1023 atoms 1 molecule of O2 = 32.00 amu 1 mole of O2 = 32.00 g = 6.022 x 1023 molecule 1 molecule of S8 = 256.52 amu 1 mole of S8 = 256.52 g = 6.022 x 1023 molecules

  5. Molecular Mass - Molar Mass ( M ) The Molecular mass of a compound expressed in amu is numerically the same as the mass of one mole of the compound expressed in grams. For water: H2O Molecular mass = (2 x atomic mass of H ) + atomic mass of O = 2 ( 1.008 amu) + 16.00 amu = 18.02 amu Mass of one molecules of water = 18.02 amu Molar mass = ( 2 x atomic mass of H ) + atomic mass of O = 2 ( 1.008 g ) + 16.00 g = 18.02 g 18.02 g H2O = 6.022 x 1023 molecules of water = 1 mole H2O

  6. One Mole of Common Substances CaCO3 100.09 g Oxygen 32.00 g Copper 63.55 g Water 18.02 g Fig. 3.2

  7. Information Contained in the Chemical Formula of Glucose C6H12O6 ( M = 180.16 g/mol) Carbon (C) Hydrogen (H) Oxygen (O) Atoms/molecule of compound Moles of atoms/ mole of compound Atoms/mole of compound Mass/moleculeof compound Mass/mole of compound 6 atoms 12 atoms 6 atoms 6 moles of 12 moles of 6 moles of atoms atoms atoms 6(6.022 x 1023) 12(6.022 x 1023) 6(6.022 x 1023) atoms atoms atoms 6(12.01 amu) 12(1.008 amu) 6(16.00 amu) =72.06 amu =12.10 amu =96.00 amu 72.06 g 12.10 g 96.00 g Table 3.2

  8. Mass - Mole Relationships of a Compound For an Element For a Compound Mass (g) of Element Mass (g) of compound Moles of Element Amount (mol) of compound Amount (mol) of compound Molecules (or formula units of compound) Atoms of Element

  9. Calculating the Number of Moles and Atoms in a Given Mass of Element Problem: Tungsten (W) is the element used as the filament in light bulbs, and has the highest melting point of any element 3680oC. How many moles of tungsten, and atoms of the element are contained in a 35.0 mg sample of the metal? Plan: Convert mass into moles by dividing the mass by the atomic weight of the metal, then calculate the number of atoms by multiplying by Avogadro’s number! Solution: Converting from mass of W to moles: Moles of W =

  10. Calculating the Number of Moles and Atoms in a Given Mass of Element Problem: Tungsten (W) is the element used as the filament in light bulbs, and has the highest melting point of any element 3680oC. How many moles of tungsten, and atoms of the element are contained in a 35.0 mg sample of the metal? Plan: Convert mass into moles by dividing the mass by the atomic weight of the metal, then calculate the number of atoms by multiplying by Avogadro’s number! Solution: Converting from mass of W to moles: Moles of W = 35.0 mg W x = 1 mol W 183.9 g W

  11. Calculating the Number of Moles and Atoms in a Given Mass of Element Problem: Tungsten (W) is the element used as the filament in light bulbs, and has the highest melting point of any element 3680oC. How many moles of tungsten, and atoms of the element are contained in a 35.0 mg sample of the metal? Plan: Convert mass into moles by dividing the mass by the atomic weight of the metal, then calculate the number of atoms by multiplying by Avogadro’s number! Solution: Converting from mass of W to moles: Moles of W = 35.0 mg W x = 0.00019032 mol 1.90 x 10 - 4 mol NO. of W atoms = = 1 mol W 183.9 g W

  12. Calculating the Number of Moles and Atoms in a Given Mass of Element Problem: Tungsten (W) is the element used as the filament in light bulbs, and has the highest melting point of any element 3680oC. How many moles of tungsten, and atoms of the element are contained in a 35.0 mg sample of the metal? Plan: Convert mass into moles by dividing the mass by the atomic weight of the metal, then calculate the number of atoms by multiplying by Avogadro’s number! Solution: Converting from mass of W to moles: Moles of W = 35.0 mg W x = 0.00019032 mol 1.90 x 10 - 4 mol NO. of W atoms = 1.90 x 10 - 4 mol W x = = 1 mol W 183.9 g W 6.022 x 1023 atoms 1 mole of W

  13. Calculating the Number of Moles and Atoms in a Given Mass of Element Problem: Tungsten (W) is the element used as the filament in light bulbs, and has the highest melting point of any element 3680oC. How many moles of tungsten, and atoms of the element are contained in a 35.0 mg sample of the metal? Plan: Convert mass into moles by dividing the mass by the atomic weight of the metal, then calculate the number of atoms by multiplying by Avogadro’s number! Solution: Converting from mass of W to moles: Moles of W = 35.0 mg W x = 0.00019032 mol 1.90 x 10 - 4 mol NO. of W atoms = 1.90 x 10 - 4 mol W x = = 1.15 x 1020 atoms of Tungsten 1 mol W 183.9 g W 6.022 x 1023 atoms 1 mole of W

  14. Calculating the Moles and Number of Formula Units in a Given Mass of Cpd. Problem: Trisodium Phosphate is a component of some detergents. How many moles and formula units are in a 38.6 g sample? Plan: We need to determine the formula, and the molecular mass from the atomic masses of each element multiplied by the coefficients. Solution: The formula is

  15. Calculating the Moles and Number of Formula Units in a Given Mass of Cpd. Problem: Trisodium Phosphate is a component of some detergents. How many moles and formula units are in a 38.6 g sample? Plan: We need to determine the formula, and the molecular mass from the atomic masses of each element multiplied by the coefficients. Solution: The formula is Na3PO4. Calculating the molar mass: M = 3x Sodium + 1 x Phosphorous = 4 x Oxygen = = 3 x 22.99 g/mol + 1 x 30.97 g/mol + 4 x 16.00 g/mol = 68.97 g/mol + 30.97 g/mol + 64.00 g/mol = 163.94 g/mol Converting mass to moles:

  16. Calculating the Moles and Number of Formula Units in a Given Mass of Cpd. Problem: Trisodium Phosphate is a component of some detergents. How many moles and formula units are in a 38.6 g sample? Plan: We need to determine the formula, and the molecular mass from the atomic masses of each element multiplied by the coefficients. Solution: The formula is Na3PO4. Calculating the molar mass: M = 3x Sodium + 1 x Phosphorous = 4 x Oxygen = = 3 x 22.99 g/mol + 1 x 30.97 g/mol + 4 x 16.00 g/mol = 68.97 g/mol + 30.97 g/mol + 64.00 g/mol = 163.94 g/mol Converting mass to moles: Moles Na3PO4 = 38.6 g Na3PO4 x (1 mol Na3PO4) 163.94 g Na3PO4

  17. Calculating the Moles and Number of Formula Units in a Given Mass of Cpd. Problem: Trisodium Phosphate is a component of some detergents. How many moles and formula units are in a 38.6 g sample? Plan: We need to determine the formula, and the molecular mass from the atomic masses of each element multiplied by the coefficients. Solution: The formula is Na3PO4. Calculating the molar mass: M = 3x Sodium + 1 x Phosphorous = 4 x Oxygen = = 3 x 22.99 g/mol + 1 x 30.97 g/mol + 4 x 16.00 g/mol = 68.97 g/mol + 30.97 g/mol + 64.00 g/mol = 163.94 g/mol Converting mass to moles: Moles Na3PO4 = 38.6 g Na3PO4 x (1 mol Na3PO4) 163.94 g Na3PO4 = 0.23545 mol Na3PO4 Formula units =

  18. Calculating the Moles and Number of Formula Units in a Given Mass of Cpd. Problem: Trisodium Phosphate is a component of some detergents. How many moles and formula units are in a 38.6 g sample? Plan: We need to determine the formula, and the molecular mass from the atomic masses of each element multiplied by the coefficients. Solution: The formula is Na3PO4. Calculating the molar mass: M = 3x Sodium + 1 x Phosphorous = 4 x Oxygen = = 3 x 22.99 g/mol + 1 x 30.97 g/mol + 4 x 16.00 g/mol = 68.97 g/mol + 30.97 g/mol + 64.00 g/mol = 163.94 g/mol Converting mass to moles: Moles Na3PO4 = 38.6 g Na3PO4 x (1 mol Na3PO4) 163.94 g Na3PO4 = 0.235 mol Na3PO4 Formula units = 0.235 mol Na3PO4 x 6.022 x 1023 formula units 1 mol Na3PO4

  19. Calculating the Moles and Number of Formula Units in a Given Mass of Cpd. Problem: Trisodium Phosphate is a component of some detergents. How many moles and formula units are in a 38.6 g sample? Plan: We need to determine the formula, and the molecular mass from the atomic masses of each element multiplied by the coefficients. Solution: The formula is Na3PO4. Calculating the molar mass: M = 3x Sodium + 1 x Phosphorous = 4 x Oxygen = = 3 x 22.99 g/mol + 1 x 30.97 g/mol + 4 x 16.00 g/mol = 68.97 g/mol + 30.97 g/mol + 64.00 g/mol = 163.94 g/mol Converting mass to moles: Moles Na3PO4 = 38.6 g Na3PO4 x (1 mol Na3PO4) 163.94 g Na3PO4 = 0.235 mol Na3PO4 Formula units = 0.23545 mol Na3PO4 x 6.022 x 1023 formula units 1 mol Na3PO4 = 1.42 x 1023 formula units

  20. Flow Chart of Mass Percentage Calculation Moles of X in one mole of Compound M (g / mol) of X Mass (g) of X in one mole of compound Divide by mass (g) of one mole of compound Mass fraction of X Multiply by 100 Mass % of X

  21. Mol wt and % Composition of NH4NO3 • 2 mol N x • 4 mol H x • 3 mol O x %N %H %O

  22. Mol wt and % Composition of NH4NO3 • 2 mol N x 14.01 g/mol = 28.02 g N • 4 mol H x 1.008 g/mol = 4.032 g H • 3 mol O x 15.999 g/mol = 48.00 g O 80.05 g/mol %N %H %O

  23. Mol wt and % Composition of NH4NO3 • 2 mol N x 14.01 g/mol = 28.02 g N • 4 mol H x 1.008 g/mol = 4.032 g H • 3 mol O x 15.999 g/mol = 48.00 g O 80.05 g/mol 28.02g N2 80.05g %N = x 100% = 35.00% 4.032g H2 80.05g %H = x 100% = 5.037% 48.00g O2 80.05g %O = x 100% = 59.96% 99.997%

  24. Calculating Mass Percentage and Masses of Elements in a Sample of a Compound - I Problem: Sucrose (C12H22O11) is common table sugar. (a) What is the mass percent of each element in sucrose? ( b) How many grams of carbon are in 24.35 g of sucrose? (a) Determining the mass percent of each element: mass of C = 12 x 12.01 g C/mol = mass of H = 22 x 1.008 g H/mol = mass of O = 11 x 16.00 g O/mol = Mass Fraction of C = =

  25. Calculating Mass Percentage and Masses of Elements in a Sample of a Compound - I Problem: Sucrose (C12H22O11) is common table sugar. (a) What is the mass percent of each element in sucrose? ( b) How many grams of carbon are in 24.35 g of sucrose? (a) Determining the mass percent of each element: mass of C = 12 x 12.01 g C/mol = 144.12 g C/mol mass of H = 22 x 1.008 g H/mol = 22.176 g H/mol mass of O = 11 x 16.00 g O/mol = 176.00 g O/mol 342.296 g/mol Finding the mass fraction of C in Sucrose& % C : Total mass of C mass of 1 mole of sucrose Mass Fraction of C = =

  26. Calculating Mass Percentage and Masses of Elements in a Sample of a Compound - I Problem: Sucrose (C12H22O11) is common table sugar. (a) What is the mass percent of each element in sucrose? ( b) How many grams of carbon are in 24.35 g of sucrose? (a) Determining the mass percent of each element: mass of C = 12 x 12.01 g C/mol = 144.12 g C/mol mass of H = 22 x 1.008 g H/mol = 22.176 g H/mol mass of O = 11 x 16.00 g O/mol = 176.00 g O/mol 342.296 g/mol Finding the mass fraction of C in Sucrose& % C : Total mass of C 144.12 g C mass of 1 mole of sucrose 342.30 g Cpd = 0.421046 To find mass % of C = 0.421046 x 100% = 42.11% Mass Fraction of C = =

  27. Empirical and Molecular Formulas Empirical Formula - The simplest formula for a compound that agrees with the elemental analysis! The smallest set of whole numbers of atoms. Molecular Formula - The formula of the compound as it exists, it may be a multiple of the Empirical formula.

  28. Steps to Determine Empirical Formulas Mass (g) of Element M (g/mol ) Moles of Element use no. of moles as subscripts Preliminary Formula change to integer subscripts Empirical Formula

  29. Some Examples of Compounds with the Same Elemental Ratio’s Empirical Formula Molecular Formula CH2(unsaturated Hydrocarbons) C2H4 , C3H6 , C4H8 OH or HO H2O2 S S8 P P4 Cl Cl2 CH2O (carbohydrates) C6H12O6

  30. Determining Empirical Formulas from Masses of Elements - I Problem: The elemental analysis of a sample compound gave the following results: 5.677g Na, 6.420 g Cr, and 7.902 g O. What is the empirical formula and name of the compound? Plan: First we have to convert mass of the elements to moles of the elements using the molar masses. Then we construct a preliminary formula and name of the compound. Solution: Finding the moles of the elements: Moles of Na = 5.678 g Na x = 0.2469 mol Na Moles of Cr = 6.420 g Cr x = 0.12347 mol Cr Moles of O = 7.902 g O x = 0.4939 mol O 1 mol Na 22.99 g Na 1 mol Cr 52.00 g Cr 1 mol O 16.00 g O

  31. Determining Empirical Formulas from Masses of Elements - I Problem: The elemental analysis of a sample compound gave the following results: 5.677g Na, 6.420 g Cr, and 7.902 g O. What is the empirical formula and name of the compound? Plan: First we have to convert mass of the elements to moles of the elements using the molar masses. Then we construct a preliminary formula and name of the compound. Solution: Finding the moles of the elements: Moles of Na = 5.678 g Na x = Moles of Cr = 6.420 g Cr x = Moles of O = 7.902 g O x = 1 mol Na 22.99 g Na 1 mol Cr 52.00 g Cr 1 mol O 16.00 g O

  32. Determining Empirical Formulas from Masses of Elements - I Problem: The elemental analysis of a sample compound gave the following results: 5.677g Na, 6.420 g Cr, and 7.902 g O. What is the empirical formula and name of the compound? Plan: First we have to convert mass of the elements to moles of the elements using the molar masses. Then we construct a preliminary formula and name of the compound. Solution: Finding the moles of the elements: Moles of Na = 5.678 g Na x = 0.2469 mol Na Moles of Cr = 6.420 g Cr x = 0.12347 mol Cr Moles of O = 7.902 g O x = 0.4939 mol O 1 mol Na 22.99 g Na 1 mol Cr 52.00 g Cr 1 mol O 16.00 g O

  33. Determining Empirical Formulas from Masses of Elements - II Constructing the preliminary formula: Na0.2469 Cr0.1235 O0.4939 Converting to integer subscripts (dividing all by smallest subscript): Na1.99 Cr1.00 O4.02 Rounding off to whole numbers: Na2CrO4 Sodium Chromate

  34. Determining the Molecular Formula from Elemental Composition and Molar Mass - I Problem: The sugar burned for energy in cells of the body is Glucose (M = 180.16 g/mol), elemental analysis shows that it contains 40.00 mass % C, 6.719 mass % H, and 53.27 mass % O. (a) Determine the empirical formula of glucose. (b) Determine the molecular formula. Plan: We are only given mass %, and no weight of the compound so we will assume 100g of the compound, and % becomes grams, and we can do as done previously with masses of the elements. Solution: Mass Carbon = 40.00% x 100g/100% = 40.00 g C Mass Hydrogen = 6.719% x 100g/100% = 6.719g H Mass Oxygen = 53.27% x 100g/100% = 53.27 g O 99.989 g Cpd

  35. Determining the Molecular Formula from Elemental Composition and Molar Mass - II Converting from Grams of Elements to moles: Moles of C = Mass of C x = 3.3306 moles C Moles of H = Mass of H x = 6.6657 moles H Moles of O = Mass of O x = 3.3294 moles O Constructing the preliminary formula C 3.33 H 6.67 O 3.33 Converting to integer subscripts, divide all subscripts by the smallest: 1 mole C 12.01 g C 1 mol H 1.008 g H 1 mol O 16.00 g O

  36. Determining the Molecular Formula from Elemental Composition and Molar Mass - II Converting from Grams of Elements to moles: Moles of C = Mass of C x = 3.3306 moles C Moles of H = Mass of H x = 6.6657 moles H Moles of O = Mass of O x = 3.3294 moles O Constructing the preliminary formula C 3.33 H 6.67 O 3.33 Converting to integer subscripts, divide all subscripts by the smallest: C 3.33/3.33 H 6.667 / 3.33 O3.33 / 3.33 = CH2O 1 mole C 12.01 g C 1 mol H 1.008 g H 1 mol O 16.00 g O

  37. Combustion Train for the Determination of the Chemical Composition of Organic Compounds. m 2 m 2 CnHm + (n+ ) O2 = n CO2(g) + H2O(g) Fig. 3.4

  38. Ascorbic Acid ( Vitamin C ) - I Contains C , H , and O • Upon combustion in excess oxygen, a 6.49 mg sample yielded 9.74 mg CO2 and 2.64 mg H2O • Calculate its Empirical formula! • C: 9.74 x10-3g CO2 x(12.01 g C/44.01 g CO2) = 2.65 x 10-3 g C • H: 2.64 x10-3g H2O x (2.016 g H2/18.02 gH2O) = 2.92 x 10-4 g H • Mass Oxygen = 6.49 mg - 2.65 mg - 0.292 mg = 3.54 mg O

  39. Vitamin C Combustion - II • C = 2.65 x 10-3 g C / ( 12.01 g C / mol C ) = = 2.21 x 10-4 mol C • H = 0.295 x 10-3 g H / ( 1.008 g H / mol H ) = = 2.92 x 10-4 mol H • O = 3.54 x 10-3 g O / ( 16.00 g O / mol O ) = = 2.21 x 10-4 mol O • Divide each by 2.21 x 10-4 • C = 1.00 Multiply each by 3 = 3.00 = 3.0 • H = 1.32 = 3.96 = 4.0 • O = 1.00= 3.00 = 3.0 C3H4O3

  40. Determining a Chemical Formula from Combustion Analysis - I Problem: Erythrose (M = 120 g/mol) is an important chemical compound as a starting material in chemical synthesis, and contains Carbon Hydrogen, and Oxygen. Combustion analysis of a 700.0 mg sample yielded 1.027 g CO2 and 0.4194 g H2O. Plan: We find the masses of Hydrogen and Carbon using the mass fractions of H in H2O, and C in CO2. The mass of Carbon and Hydrogen are subtracted from the sample mass to get the mass of Oxygen. We then calculate moles, and construct the empirical formula, and from the given molar mass we can calculate the molecular formula.

  41. Determining a Chemical Formula from Combustion Analysis - II Calculating the mass fractions of the elements: Mass fraction of C in CO2 = = = = 0.2729 g C / 1 g CO2 Mass fraction of H in H2O = = = = 0.1119 g H / 1 g H2O Calculating masses of C and H Mass of Element = mass of compound x mass fraction of element mol C xM of C mass of 1 mol CO2 1 mol C x 12.01 g C/ 1 mol C 44.01 g CO2 mol H xM of H mass of 1 mol H2O 2 mol H x 1.008 g H / 1 mol H 18.02 g H2O

  42. Determining a Chemical Formula from Combustion Analysis - III 0.2729 g C 1 g CO2 Mass (g) of C = 1.027 g CO2 x = 0.2803 g C Mass (g) of H = 0.4194 g H2O x = 0.04693 g H Calculating the mass of O: Mass (g) of O = Sample mass -( mass of C + mass of H ) = 0.700 g - 0.2803 g C - 0.04693 g H = 0.37277 g O Calculating moles of each element: C = 0.2803 g C / 12.01 g C/ mol C = 0.02334 mol C H = 0.04693 g H / 1.008 g H / mol H = 0.04656 mol H O = 0.37277 g O / 16.00 g O / mol O = 0.02330 mol O C0.02334H0.04656O0.02330 = CH2O formula weight = 30 g / formula 120 g /mol / 30 g / formula = 4 formula units / cpd = C4H8O4 0.1119 g H 1 g H2O

  43. Some Compounds with Empirical Formula CH2O (Composition by Mass 40.0% C, 6.71% H, 53.3%O) Molecular M Formula (g/mol) Name Use or Function CH2O 30.03 Formaldehyde Disinfectant; Biological preservative C2H4O2 60.05 Acetic acid Acetate polymers; vinegar ( 5% solution) C3H6O3 90.08 Lactic acid Causes milk to sour; forms in muscle during exercise C4H8O4 120.10 Erythrose Forms during sugar metabolism C5H10O5 150.13 Ribose Component of many nucleic acids and vitamin B2 C6H12O6 180.16 Glucose Major nutrient for energy in cells

  44. Two Compounds with Molecular Formula C2H6O Property Ethanol Dimethyl Ether M (g/mol) 46.07 46.07 Color Colorless Colorless Melting point - 117oC - 138.5oC Boiling point 78.5oC - 25oC Density (at 20oC) 0.789 g/mL 0.00195 g/mL Use Intoxicant in In refrigeration alcoholic beverages H H H H H C C O H H C O C H H H H H Table 3.4

  45. Fig. 3.7

  46. Molecular Formula Atoms Molecules Avogadro’s Number 6.022 x 1023 Moles Moles

  47. Chemical Equations Qualitative Information: Reactants Products States of Matter: (s) solid (l) liquid (g) gaseous (aq) aqueous 2 H2 (g) + O2 (g) 2 H2O (g)

  48. Balanced Equations • mass balance (atom balance)- same number of each element(1) start with simplest element(2) progress to other elements (3) make all whole numbers (4) re-check atom balance • charge balance (no “spectator” ions) 1 CH4 (g) + O2 (g) 1 CO2 (g) + H2O (g) 1 CH4 (g) + O2 (g) 1 CO2 (g) + 2 H2O (g) 1 CH4 (g) + 2O2 (g) 1 CO2 (g) + 2 H2O (g) Ca2+ (aq) + 2 OH- (aq) Ca(OH)2 (s) + Na+ + Na+

  49. C6H14 (l) + O2 (g) CO2 (g) + H2O(g) + Energy Balancing Chemical Equations - I Problem: The hydrocarbon hexane is a component of Gasoline that burns in an automobile engine to produce carbon dioxide and water as well as energy. Write the balanced chemical equation for the combustion of hexane (C6H14). Plan: Write the skeleton equation from the words into chemical compoundswith blanks before each compound. begin the balance with the most complex compound first, and save oxygen until last! Solution: C6H14 (l) + O2 (g) CO2 (g) + H2O(g) + Energy Begin with one Hexane molecule which says that we will get 6 CO2’s! 1 6

  50. C6H14 (l) + O2 (g) CO2 (g) + H2O(g) + Energy 2 C6H14 (l) + O2 (g) CO2 (g) + H2O(g) + Energy 12 14 2 C6H14 (l) + O2 (g) CO2 (g) + H2O(g) + Energy 12 14 Balancing Chemical Equations - II The H atoms in the hexane will end up as H2O, and we have 14 H atoms, and since each water molecule has two H atoms, we will get a total of 7 water molecules. 1 6 7 Since oxygen atoms only come as diatomic molecules (two O atoms, O2),we must have even numbers of oxygen atoms on the product side. We do not since we have 7 water molecules! Therefore multiply the hexane by 2, giving a total of 12 CO2 molecules, and 14 H2O molecules. This now gives 12 O2 from the carbon dioxide, and 14 O atoms from the water, which will be another 7 O2 molecules for a total of 19 O2 ! 19

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