1. Stoichiometry Dr. Ron Rusay

2. Chemical Stoichiometry ð Stoichiometry is the study of mass in chemical reactions. It deals with both reactants and products. ð It quantitatively and empirically relates the behavior of atoms and molecules in a balanced chemical equation to observable chemical change and measurable mass effects. ð It accounts for mass and the conservation of mass, just as the conservation of atoms in a balanced chemical equation.

3. Chemical Reactions Atoms, Mass & Balance: eg. Zn(s) + S(s)

4. Chemical Equation _ C 2 H 5 OH + _ O 2  _ CO 2 + _ H 2 O Reactants Products Reactants Products C=2; H =5+1=6; O=2+1C=1; H=2; O=2+1C=2; H =5+1=6; O=2+1C=1; H=2; O=2+1 1 C 2 H 5 OH + 3 O 2  2 CO 2 + 3 H 2 O1 C 2 H 5 OH + 3 O 2  2 CO 2 + 3 H 2 O Representation of a chemical reaction:

5. Chemical Equation The balanced equation can be completely stated as: ð 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water. C 2 H 5 OH + 3 O 2  2 CO 2 + 3 H 2 O

6. All Balanced Equations relate on a molar mass basis. For the combustion of octane: 2 C 8 H 18(l) + 25 O 2(g) 16 CO 2(g) +18 H 2 O (l) 2moles of octane react with 25 moles of oxygen to produce 16 moles of carbon dioxide and 18 moles of water. Chemical Equation

7. QUESTION The fuel in small portable lighters is butane (C 4 H 10 ). After using a lighter for a few minutes, 1.0 gram of fuel was used. How many moles of carbon dioxide would it produce? A.58 moles B.0.017 moles C.1.7  10 –24 moles D.0.068 moles

8. ANSWER The fuel in small portable lighters is butane (C 4 H 10 ). After using a lighter for a few minutes, 1.0 gram of fuel was used. How many moles of carbon dioxide would it produce? A.58 moles B.0.017 moles C.1.7  10 –24 moles D.0.068 moles The molar mass of butane: 4  12g/mol = 48 for carbon + 10  1g/mol = 10.0 for hydrogen. Total = 48.0 + 10.0 = 58.0 g/mol. Next; 1.0 gram of butane  1 mol/58.0 g = 0.017 mol

9. The Chemical Equation: Mole & Masses ð 46g (1 mole) of ethanol reacts with 3 moles of oxygen (96g) to produce 2 moles of carbon dioxide and 3 moles of water. ð How many grams of carbon dioxide and water are respectively produced from 46g (1 mole) of ethanol ? 2 mol x 44 g/mol = 88g 3 mol x 18 g/mol = 54 g C 2 H 5 OH + 3 O 2  2 CO 2 + 3 H 2 O

10. The Chemical Equation: Moles & Masses C 2 H 5 OH + 3 O 2  2 CO 2 + 3 H 2 OC 2 H 5 OH + 3 O 2  2 CO 2 + 3 H 2 O ð How many grams of oxygen are needed to react with 15.3g of ethanol in a 12oz. beer ? 15.3g ethanol x mol ethanol /46.0g ethanol = 0.333mol ethanol 0.333mol ethanol x 3mol oxygen / 1mol ethanol = 0.999mol oxygen 32.0g oxygen /mol oxygen x 0.999mol oxygen = 32.0g oxygen NOTE: It takes approximately 1 hour for the biologically equivalent amount of oxygen available from cytochrome p450 to consume the alcohol in a human in 1 beer to a level below the legal limit of 0.08%.

11. Chemical Stoichiometry ð Epsom salt (magnesium sulfate heptahydrate) is one of five possible hydrates: mono-, di-, tri-, hexa-, or hepta- hydrate. ð How can stoichiometry be used to determine, which hydrate is present in a pure unknown sample, by heating the sample in a kitchen oven at 400 o C for 45 minutes? MgSO 4. x H 2 O (s) MgSO 4 (s) + x H 2 O (g)

12. Mass Calculations All Balanced Equations relate on a molar and mass basis. For the combustion of octane: 2 C 8 H 18(l) + 25 O 2(g) 16 CO 2(g) +18 H 2 O (l) 228 g of octane (2 moles)* will react with 800 g of oxygen (25 moles) to produce (16 moles) 704 g of carbon dioxide and (18 moles) 324 g of water. *(2 moles octane x 114 g/mol = 228 g )

13. Mass Calculations: Reactants Products

14. 1.Balance the chemical equation. 2.Convert mass of reactant or product to moles. 3.Identify mole ratios in balanced equation: They serve as the “Gatekeeper”. 4.Calculate moles of desired product or reactant. 5.Convert moles to grams. Mass Calculations: Reactants Products

16. How many grams of salicylic acid are needed to produce 1.80 kg of aspirin?How many grams of salicylic acid are needed to produce 1.80 kg of aspirin? Balanced Equation:Balanced Equation: Mass Calculations: Reactants Products

17. grams (Aspirin) grams (Salicylic Acid) 1800 grams (A) grams (A) (Molecular Weight A) 1 mol (A) grams (SA) ? (SA) 1 mol (SA) (Molecular Weight SA) Avogadro's Number Atoms Molecules Stoichiometry 1 mol SA 1 mol A "Gatekeeper" Mass Calculations: Reactants Products

18. Mass Calculations: How many grams of salicylic acid are needed to produce 1.80 kg of aspirin? Balanced Equation:Balanced Equation: ?g C7 = 1.80 x 10 3 g C9 x [mol C9 /180.15g C9 ] x [ 1 mol C7 / 1 mol C9 ] x 138.12 g C7 /mol C7 1 1 ?g C7 = 1380 grams

19. Mass Calculations: Reactants Products

20. QUESTION The fuel in small portable lighters is butane (C 4 H 10 ). After using a lighter for a few minutes, 1.0 gram (0.017 moles) of fuel was used. How many grams of carbon dioxide would it produce? How many grams of carbon dioxide would this produce? A.) 750 mgB.) 6.0 g C) 1.5 gD.) 3.0 g

21. ANSWER D.) 3.0 g 1.0 g C 4 H 10 ?g CO 2 C 4 H 10 mol C 4 H 10 1.0 g 58 g C 4 H 10 mol CO 2 44 g CO 2 ?g CO 2 2 mol C 4 H 10 8 mol CO 2 3.0 g CO 2

22. Combustion Analysis SEE: COMPARISON to wt % CALCULATIONS COMPARISON to wt % CALCULATIONS Molecules with oxygen in their formula are more difficult to solve for O z knowing only the respective masses of C x H y O z sample, CO 2 and H 2 O. C x H y O z + (x + y/4 - z) O 2  x CO 2 + y/2 H 2 O C x H y + (x + y/4) O 2  x CO 2 + y/2 H 2 O

23. Combustion Analysis Calculation Ascorbic Acid ( Vitamin C ) Combustion of a 6.49 mg sample in excess oxygen, yielded 9.74 mg CO 2 and 2.64 mg H 2 OCombustion of a 6.49 mg sample in excess oxygen, yielded 9.74 mg CO 2 and 2.64 mg H 2 O Calculate it’s Empirical formula!Calculate it’s Empirical formula! C: 9.74 x10 -3 g CO 2 x(12.01 g C/44.01 g CO 2 )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 = 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)H: 2.64 x10 -3 g H 2 O x (2.016 g H 2 /18.02 gH 2 O) = 2.92 x 10 -4 g H = 2.92 x 10 -4 g H Mass Oxygen = 6.49 mg - 2.65 mg - 0.30 mgMass Oxygen = 6.49 mg - 2.65 mg - 0.30 mg = 3.54 mg O = 3.54 mg O

24. Vitamin C: Calculation (continued) C = 2.65 x 10 -3 g C / ( 12.01 g C / mol C ) =C = 2.65 x 10 -3 g C / ( 12.01 g C / mol C ) = = 2.21 x 10 -4 mol C = 2.21 x 10 -4 mol C H = 0.295 x 10 -3 g H / ( 1.008 g H / mol H ) =H = 0.295 x 10 -3 g H / ( 1.008 g H / mol H ) = = 2.92 x 10 -4 mol H = 2.92 x 10 -4 mol H O = 3.54 x 10 -3 g O / ( 16.00 g O / mol O ) =O = 3.54 x 10 -3 g O / ( 16.00 g O / mol O ) = = 2.21 x 10 -4 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 = 3.0C = 1.00 Multiply each by 3 = 3.00 = 3.0 H = 1.32 = 3.96 = 4.0H = 1.32 = 3.96 = 4.0 O = 1.00 = 3.00 = 3.0O = 1.00 = 3.00 = 3.0 C3H4O3C3H4O3

25. QUESTION Erythrose contains carbon, hydrogen and oxygen (MM = 120.0 g/mol). It is an important sugar that is used in many chemical syntheses. Combustion analysis of a 700.0 mg sample yielded 1.027 g CO 2 and 0.4194 g H 2 O. Mass Spectrometry produced a molecular ion @ 120 mass units (m/z). What is the molecular formula of erythrose? A)CH 2 O B)C 6 H 12 O 6 C)C 3 H 6 O 3 D)C 4 H 8 O 4

26. Alternative Solution Method Mass ratio of C in CO 2 = = = = 0.2729 g C / 1 g CO 2 Mass ratio 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 ratio 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 OANSWER D) C 4 H 8 O 4

27. Alternative Method 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 / cmpd = C 4 H 8 O 4 0.2729 g C 1 g CO 2 0.1119 g H 1 g H 2 O