1. C n H m + (n+ )O 2(g) n CO 2(g) + H 2 O (g) m 2 m 2 Fig. 3.4

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 CO 2 and 2.64 mg H 2 O. 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 CO 2 and 2.64 mg H 2 O. Calculate its Empirical formula! C: 9.74 x10 -3 g CO 2 x(12.01 g C/44.01 g CO 2 ) = 2.65 x 10 -3 g C H: 2.64 x10 -3 g H 2 O x (2.016 g H 2 /18.02 gH 2 O) = 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 C3H4O3C3H4O3

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 CO 2 and 0.4194 g H 2 O. Plan: We find the masses of Hydrogen and Carbon using the mass fractions of H in H 2 O, and C in CO 2. 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 CO 2 = Mass fraction of H in H 2 O = 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 CO 2 = = = = 0.2729 g C / 1 g CO 2 Mass fraction of H in H 2 O = = = = 0.1119 g H / 1 g H 2 O Calculating masses of C and H Mass of Element = mass of compound x mass fraction of element mol C x M of C mass of 1 mol CO 2 1 mol C x 12.01 g C/ 1 mol C 44.01 g CO 2 mol H x M of H mass of 1 mol H 2 O 2 mol H x 1.008 g H / 1 mol H 18.02 g H 2 O

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 Mass (g) of C = 1.027 g CO 2 x = 0.2803 g C Mass (g) of H = 0.4194 g H 2 O 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 C 0.02334 H 0.04656 O 0.02330 = CH 2 O formula weight = 30 g / formula 120 g /mol / 30 g / formula = 4 formula units / cpd = C 4 H 8 O 4 0.2729 g C 1 g CO 2 0.1119 g H 1 g H 2 O

13. Fig. 3.5

14. Moles Molecules Avogadro’s Number Molecular Formula Atoms x6.022 x 10 23

15. Chemical Equations ReactantsProducts Qualitative Information: States of Matter: (s) solid (l) liquid (g) gaseous (aq) aqueous 2 H 2 (g) + O 2 (g)2 H 2 O (g) But also Quantitative Information!

16. Fig. 3.6

17. Fig. 3.7

18. Balanced Equations 1 CH 4 (g) + O 2 (g)1 CO 2 (g) + H 2 O (g) 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 CH 4 (g) + 2 O 2 (g) 1 CO 2 (g) + 2 H 2 O (g) 1 CH 4 (g) + O 2 (g)1 CO 2 (g) + 2 H 2 O (g) Ca 2+ (aq) + 2 OH - (aq) Ca(OH) 2 (s)+ Na +

20. Information Contained in a Balanced Equation Viewed in Reactants Products terms of: 2 C 2 H 6 (g) + 7 O 2 (g) = 4 CO 2 (g) + 6 H 2 O (g) + Energy Molecules Amount (mol) Mass (amu) Mass (g) Total Mass (g)

21. Information Contained in a Balanced Equation Viewed in Reactants Products terms of: 2 C 2 H 6 (g) + 7 O 2 (g) = 4 CO 2 (g) + 6 H 2 O (g) + Energy Molecules 2 molecules of C 2 H 6 + 7 molecules of O 2 = 4 molecules of CO 2 + 6 molecules of H 2 O Amount (mol) 2 mol C 2 H 6 + 7 mol O 2 = 4 mol CO 2 + 6 mol H 2 O Mass (amu) 60.14 amu C 2 H 6 + 224.00 amu O 2 = 176.04 amu CO 2 + 108.10 amu H 2 O Mass (g) 60.14 g C 2 H 6 + 224.00 g O 2 = 176.04 g CO 2 + 108.10 g H 2 O Total Mass (g) 284.14g = 284.14g

22. 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 (C 6 H 14 ). Plan: Write the skeleton equation from the words into chemical compounds with 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:

23. 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 (C 6 H 14 ). Plan: Write the skeleton equation from the words into chemical compounds with blanks before each compound. Begin the balance with the most complex compound first, and save oxygen until last! Solution: C 6 H 14 (l) + O 2 (g) CO 2 (g) + H 2 O (g) + Energy Begin with one C 6 H 14 molecule which says that we will get 6 CO 2 ’s! 16

24. C 6 H 14 (l) + O 2 (g) CO 2 (g) + H 2 O (g) + Energy Next balance H atoms: Balancing Chemical Equations - II 16 Balance O atoms last: C 6 H 14 (l) + O 2 (g) CO 2 (g) + H 2 O (g) + Energy

25. The H atoms in the hexane will end up as H 2 O, and we have 14 H atoms, and since each water molecule has two H atoms, we will get a total of 7 water molecules. Balancing Chemical Equations - II 167 Since oxygen atoms only come as diatomic molecules (two O atoms, O 2 ),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 CO 2 molecules, and 14 H 2 O molecules. C 6 H 14 (l) + O 2 (g) CO 2 (g) + H 2 O (g) + Energy21214 This now gives 12 O 2 from the carbon dioxide, and 14 O atoms from the water, which will be another 7 O 2 molecules for a total of 12+7 =19 O 2 ! C 6 H 14 (l) + O 2 (g) CO 2 (g) + H 2 O (g) + Energy2121419

26. Fig. 3.8

27. Chemical Equation Calc - I ReactantsProducts Molecules Atoms (Molecules) Avogadro’s Number 6.02 x 10 23

28. Chemical Equation Calc - II ReactantsProducts Molecules Moles Mass Molecular Weight g/mol Atoms (Molecules) Avogadro’s Number 6.02 x 10 23

29. 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 Al 2 S 3 : a) How many moles of water are required for the reaction? b) What mass of H 2 S & Al(OH) 3 would be formed? Al 2 S 3 (s) + 6 H 2 O (l) 2 Al(OH) 3 (s) + 3 H 2 S (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 Al 2 S 3 =

30. 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 Al 2 S 3 : a) How many moles of water are required for the reaction? b) What mass of H 2 S & Al(OH) 3 would be formed? Al 2 S 3 (s) + 6 H 2 O (l) 2 Al(OH) 3 (s) + 3 H 2 S (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 Al 2 S 3 = = 0.4382 moles Al 2 S 3 65.80 g Al 2 S 3 150.17 g Al 2 S 3 / mol Al 2 S 3

31. Calculating Reactants and Products in a Chemical Reaction - II a) cont. H 2 O: 0.4382 moles Al 2 S 3 x b)H 2 S: 0.4382 moles Al 2 S 3 x ___ moles H 2 O 1 mole Al 2 S 3 ___ moles H 2 S 1 mole Al 2 S 3 molar mass of H 2 S = mass H 2 S = Al(OH) 3 : 0.4382 moles Al 2 S 3 x molar mass of Al(OH) 3 = mass Al(OH) 3 =

32. Calculating Reactants and Products in a Chemical Reaction - II a) cont. 0.4382 moles Al 2 S 3 x = 2.629 moles H 2 O b) 0.4382 moles Al 2 S 3 x = 1.314 moles H 2 S molar mass of H 2 S = 34.09 g / mol mass H 2 S = 1.314 moles H 2 S x = 44.81 g H 2 S 0.4382 moles Al 2 S 3 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 H 2 O 1 mole Al 2 S 3 3 moles H 2 S 1 mole Al 2 S 3 34.09 g H 2 S 1 mole H 2 S 2 moles Al(OH) 3 1 mole Al 2 S 3 78.00 g Al(OH) 3 1 mole Al(OH) 3

33. Calculating the Amounts of Reactants and Products in a Reaction Sequence - I Problem: Calcium Phosphate could be prepared in the following reaction sequence: 4 P 4 (s) + 10 KClO 3 (s) 4 P 4 O 10 (s) + 10 KCl (s) P 4 O 10 (s) + 6 H 2 O (l) 4 H 3 PO 4 (aq) 2 H 3 PO 4 (aq) + 3 Ca(OH) 2 (aq) 6 H 2 O (aq) + Ca 3 (PO 4 ) 2 (s) Given: 15.5 g P 4 and sufficient KClO 3, H 2 O and Ca(OH) 2. What mass of Calcium Phosphate could be formed? Plan: (1) Calculate moles of P 4. (2) Use molar ratios to get moles of Ca 3 (PO 4 ) 2. (3) Convert the moles of product back into mass by using the molar mass of Calcium Phosphate.

34. Calculating the Amounts of Reactants and Products in a Reaction Sequence - II Solution: moles of P 4 = For Reaction #1 [ 4 P 4 (s) + 10 KClO 4 (s) 4 P 4 O 10 (s) + 10 KCl (s) ] For Reaction #2 [ 1 P 4 O 10 (s) + 6 H 2 O (l) 4 H 3 PO 4 (aq) ] For Reaction #3 [ 2 H 3 PO 4 + 3 Ca(OH) 2 1 Ca 3 (PO 4 ) 2 + 6 H 2 O] moles Ca 3 (PO 4 ) 2 = moles P 4 x

35. Calculating the Amounts of Reactants and Products in a Reaction Sequence - II Solution: moles of Phosphorous = 15.50 g P 4 x = 0.1251 mol P 4 For Reaction #1 [ 4 P 4 (s) + 10 KClO 4 (s) 4 P 4 O 10 (s) + 10 KCl (s) ] For Reaction #2 [ 1 P 4 O 10 (s) + 6 H 2 O (l) 4 H 3 PO 4 (aq) ] For Reaction #3 [ 2 H 3 PO 4 + 3 Ca(OH) 2 1 Ca 3 (PO 4 ) 2 + 6 H 2 O] 0.1251 moles P 4 x x x = 0.2502 moles Ca 3 (PO 4 ) 2 1 mole P 4 123.88 g P 4 4 moles H 3 PO 4 1 mole P 4 O 10 4 moles P 4 O 10 4 moles P 4 1 mole Ca 3 (PO 4 ) 2 2 moles H 3 PO 4

36. Molar mass of Ca 3 (PO 4 ) 2 = 310.18 g mole Mass of Ca 3 (PO 4 ) 2 product = Calculating the Amounts of Reactants and Products in a Reaction Sequence - III

37. Molar mass of Ca 3 (PO 4 ) 2 = 310.18 g mole mass of product = 0.2502 moles Ca 3 (PO 4 ) 2 x = = 77.61 g Ca 3 (PO 4 ) 2 Calculating the Amounts of Reactants and Products in a Reaction Sequence - III 310.18 g Ca 3 (PO 4 ) 2 1 mole Ca 3 (PO 4 ) 2

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

39. Limiting Reactant Problems a A + b B + c C d 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....)!

40. 2Al (s) + 6HCl (g) 2AlCl 3(s) + 3H 2(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” = Acid - Metal Limiting Reactant - I