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C n H m + (n+ )O 2(g) n CO 2(g) + H 2 O (g)

Fig. 3.4. m 2. m 2. C n H m + (n+ )O 2(g) n CO 2(g) + H 2 O (g). Ascorbic acid ( Vitamin C ) - I contains only C , H , and O. Upon combustion in excess oxygen, a 6.49 mg sample yielded 9.74 mg CO 2 and 2.64 mg H 2 O. Calculate its Empirical formula! C: H:

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C n H m + (n+ )O 2(g) n CO 2(g) + H 2 O (g)

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  1. Fig. 3.4 m 2 m 2 CnHm + (n+ )O2(g) n CO2(g) + H2O(g)

  2. Ascorbic acid ( Vitamin C ) - I contains only 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: • H: • Mass Oxygen =

  3. Ascorbic acid ( Vitamin C ) - I contains only 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) = 2.95 x 10-4 g H • Mass Oxygen = 6.49 mg - 2.65 mg - 0.30 mg = 3.54 mg O = 3.54x10-3 g O

  4. Vitamin C combustion - II • C = • H = • O = • Divide each by smallest: • C = • H = • O =

  5. 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 smallest (2.21 x 10-4 ): • C = 1.00 Multiply each by 3: C = 3.00 = 3.0 • H = 1.32 H = 3.96 = 4.0 • O = 1.00 O = 3.00 = 3.0 C3H4O3

  6. Determining a Chemical Formula from Combustion Analysis - I Problem: Erthrose (M = 120 g/mol) is an important chemical compound used often 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.

  7. Determining a Chemical Formula from Combustion Analysis - II Calculating the mass fractions of the elements: Mass fraction of C in CO2 = Mass fraction of H in H2O = Calculating masses of C and H: Mass of Element = mass of compound x mass fraction of element

  8. 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

  9. Determining a Chemical Formula from Combustion Analysis - III Mass (g) of C = Mass (g) of H = Calculating the mass of O: Calculating moles of each element: C = H = O =

  10. 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

  11. Fig. 3.5

  12. Molecular Formula Atoms Molecules Avogadro’s Number x6.022 x 1023 Moles Moles

  13. 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) But also Quantitative Information!

  14. Fig. 3.6

  15. Fig. 3.7

  16. Balanced Equations • Mass Balance (or Atom Balance)- same number of each element on each side of the equation:(1) start with simplest element(2) progress to other elements (3) make all whole numbers (4) re-check atom balance • Make charges balance. (Remove “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+

  17. Information Contained in a Balanced Equation Viewed in Reactants Products terms of: 2 C2H6 (g) + 7 O2 (g) = 4 CO2 (g) + 6 H2O(g) + Energy Molecules Amount (mol) Mass (amu) Mass (g) Total Mass (g)

  18. Information Contained in a Balanced Equation Viewed in Reactants Products terms of: 2 C2H6 (g) + 7 O2 (g) = 4 CO2 (g) + 6 H2O(g) + Energy Molecules 2 molecules of C2H6 + 7 molecules of O2 = 4 molecules of CO2 + 6 molecules of H2O Amount (mol) 2 mol C2H6 + 7 mol O2 = 4 mol CO2 + 6 mol H2O Mass (amu) 60.14 amu C2H6 + 224.00 amu O2 = 176.04 amu CO2 + 108.10 amu H2O Mass (g) 60.14 g C2H6 + 224.00 g O2 = 176.04 g CO2 + 108.10 g H2O Total Mass (g) 284.14g = 284.14g

  19. 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: Begin with one unit of most complex compound: Balance any elements that this forces:

  20. 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 C6H14 molecule which says that we will get 6 CO2’s! 1 6

  21. C6H14 (l) + O2 (g) CO2 (g) + H2O(g) + Energy C6H14 (l) + O2 (g) CO2 (g) + H2O(g) + Energy C6H14 (l) + O2 (g) CO2 (g) + H2O(g) + Energy Balancing Chemical Equations - II Next balance H atoms: 1 6 Balance O atoms last:

  22. 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 12+7 =19 O2 ! 19

  23. Fig. 3.8

  24. Chemical Equation Calc - I Atoms (Molecules) Avogadro’s Number 6.02 x 1023 Molecules Reactants Products

  25. Chemical Equation Calc - II Mass Atoms (Molecules) Molecular Weight Avogadro’s Number g/mol 6.02 x 1023 Molecules Reactants Products Moles

  26. Sample Problem: Calculating Reactants and Products in a Chemical Reaction - I Problem: Given the following chemical reaction between Aluminum Sulfide and water, if we are given 65.80 g of Al2S3: a) How many moles of water are required for the reaction? b) What mass of H2S & Al(OH)3 would be formed? Al2S3 (s) + 6 H2O(l) 2 Al(OH)3 (s) + 3 H2S(g) Plan: Calculate moles of Aluminum Sulfide using its molar mass, then from the equation, calculate the moles of Water, and then the moles of Hydrogen Sulfide, and finally the mass of Hydrogen Sulfide using it’s molecular weight. Solution: a) molar mass of Aluminum Sulfide = moles Al2S3 =

  27. Sample Problem: Calculating Reactants and Products in a Chemical Reaction - I Problem: Given the following chemical reaction between Aluminum Sulfide and water, if we are given 65.80 g of Al2S3: a) How many moles of water are required for the reaction? b) What mass of H2S & Al(OH)3 would be formed? Al2S3 (s) + 6 H2O(l) 2 Al(OH)3 (s) + 3 H2S(g) Plan: Calculate moles of Aluminum Sulfide using its molar mass, then from the equation, calculate the moles of Water, and then the moles of Hydrogen Sulfide, and finally the mass of Hydrogen Sulfide using it’s molecular weight. Solution: a) molar mass of Aluminum Sulfide = 150.17 g / mol moles Al2S3 = = 0.4382 moles Al2S3 65.80 g Al2S3 150.17 g Al2S3/ mol Al2S3

  28. Calculating Reactants and Products in a Chemical Reaction - II a) cont. H2O: 0.4382 moles Al2S3 x b)H2S: 0.4382 moles Al2S3 x ___ moles H2O 1 mole Al2S3 ___ moles H2S 1 mole Al2S3 molar mass of H2S = mass H2S = Al(OH)3: 0.4382 moles Al2S3 x molar mass of Al(OH)3 = mass Al(OH)3 =

  29. Calculating Reactants and Products in a Chemical Reaction - II a) cont. 0.4382 moles Al2S3 x = 2.629 moles H2O b) 0.4382 moles Al2S3 x = 1.314 moles H2S molar mass of H2S = 34.09 g / mol mass H2S = 1.314 moles H2S x = 44.81 g H2S 0.4382 moles Al2S3 x = 0.8764 moles Al(OH)3 molar mass of Al(OH)3 = 78.00 g / mol mass Al(OH)3 = 0.8764 moles Al(OH)3 x = = 68.36 g Al(OH)3 6 moles H2O 1 mole Al2S3 3 moles H2S 1 mole Al2S3 34.09 g H2S 1 mole H2S 2 moles Al(OH)3 1 mole Al2S3 78.00 g Al(OH)3 1 mole Al(OH)3

  30. Calculating the Amounts of Reactants and Products in a Reaction Sequence - I Problem: Calcium Phosphate could be prepared in the following reactionsequence: 4 P4 (s) + 10 KClO3 (s) 4 P4O10 (s) + 10 KCl (s) P4O10 (s) + 6 H2O (l) 4 H3PO4 (aq) 2 H3PO4 (aq) + 3 Ca(OH)2 (aq) 6 H2O(aq) + Ca3(PO4)2 (s) Given: 15.5 g P4 and sufficient KClO3 , H2O and Ca(OH)2. What mass of Calcium Phosphate could be formed? Plan: (1) Calculate moles of P4. (2) Use molar ratios to get moles of Ca3(PO4)2. (3) Convert the moles of product back into mass by using the molar mass of Calcium Phosphate.

  31. Calculating the Amounts of Reactants and Products in a Reaction Sequence - II Solution: moles of P4 = For Reaction #1 [ 4 P4 (s) + 10 KClO4 (s)4 P4O10 (s) + 10 KCl (s) ] For Reaction #2 [ 1 P4O10 (s) + 6 H2O (l)4 H3PO4 (aq) ] For Reaction #3 [ 2 H3PO4 + 3 Ca(OH)2 1 Ca3(PO4)2 + 6 H2O] moles Ca3(PO4)2 = moles P4 x

  32. Calculating the Amounts of Reactants and Products in a Reaction Sequence - II Solution: moles of Phosphorous = 15.50 g P4 x = 0.1251 mol P4 For Reaction #1 [ 4 P4 (s) + 10 KClO4 (s)4 P4O10 (s) + 10 KCl (s) ] For Reaction #2 [ 1 P4O10 (s) + 6 H2O (l)4 H3PO4 (aq) ] For Reaction #3 [ 2 H3PO4 + 3 Ca(OH)2 1 Ca3(PO4)2 + 6 H2O] 0.1251 moles P4 x x x = 0.2502 moles Ca3(PO4)2 1 mole P4 123.88 g P4 4 moles P4O10 4 moles P4 4 moles H3PO4 1 mole P4O10 1 mole Ca3(PO4)2 2 moles H3PO4

  33. Calculating the Amounts of Reactants and Products in a Reaction Sequence - III Molar mass of Ca3(PO4)2 = 310.18 g mole Mass of Ca3(PO4)2 product =

  34. Calculating the Amounts of Reactants and Products in a Reaction Sequence - III Molar mass of Ca3(PO4)2 = 310.18 g mole mass of product = 0.2502 moles Ca3(PO4)2 x = = 77.61 g Ca3(PO4)2 310.18 g Ca3(PO4)2 1 mole Ca3(PO4)2

  35. Fig. 3.9 Balanced reaction! Defines stoichiometric ratios! Unbalanced (i.e., non-stoichiometric) mixture! Limited by syrup!

  36. Limiting Reactant Problems a A + b B + c Cd D + e E +f F Steps to solve 1) Identify it as a limiting reactant problem - Information on the: mass, number of moles, number of molecules, volume and molarity of a solution is given for more than one reactant! 2) Calculate moles of each reactant! 3) Divide the moles of each reactant by stoic. coefficient (a,b,c etc...)! 4) Which ever is smallest, that reactant is the limiting reactant! 5) Use the limiting reactant to calculate the moles of product desired then convert to the units needed (moles, mass, volume, number of atoms etc....)!

  37. Acid - Metal Limiting Reactant - I • 2Al(s) + 6HCl(g) 2AlCl3(s) + 3H2(g) Consider the reaction above. If we react 30.0 g Al and 20.0 g HCl, how many moles of aluminum chloride will be formed? • 30.0 g Al • 20.0g HCl • Limiting reactant = one w/ fewer “equivalents” =

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