1. Chapter Three: Stoichiometry: Calculations with Chemical Formulas and Equations Chapter Three: Stoichiometry: Calculations with Chemical Formulas and Equations

2. Overview  Chemical Equations  Patterns/Reactions  Atomic/Molecular Weights  Moles/Molar Mass  Empirical/Molecular Formulas  Quantitative Relationships  Limiting Reactants/Theoretical Yields

3. Chemical Equations  chemical ‘sentences’ –reactants and products described by formulas or symbols combined with “punctuation” 2 H 2(g) + O 2(g) 2 H 2 O (l) reactant formulas product formula coefficients physical state ‘react to form’

4. “atoms can be neither created nor destroyed” –all equations must be ‘balanced’ with the same number of atoms on both sides of the reaction arrow H 2 O + O 2 Ù H 2 O 2 unbalanced 2 H & 3 O  2 H & 2 O 2H 2 O + O 2 Ù 2H 2 O 2 balanced 4 H & 4 O=4 H & 4 O

5. unbalanced balanced two formula units one formula unit H2OO2H2O2H2OO2H2O2

6. Examples  CH 3 OH (l) + O 2(g) Ù CO 2(g) + H 2 O (l)  Na (s) + H 2 O (l) Ù NaOH (aq) + H 2(g)  HBr (aq) + Ba(OH) 2(aq) Ù H 2 O (l) + BaBr 2(aq) 2243 222 2 2

7. Patterns of Chemical Reactivity  Because elements are grouped by chemical properties, their reactions can also be grouped: –alkali metals and water 2K (s) + 2H 2 O (l) Ù 2KOH (aq) + H 2(g) specific 2M (s) + 2H 2 O (l) Ù 2MOH (aq) + H 2(g) general

8. –Combustion in air C 3 H 8(g) + O 2(g) Ù CO 2(g) + H 2 O (l) 345 C x H y + O 2(g) Ù CO 2(g) + H 2 O (l) specific general hydrocarbon

9. –Combination Reactions 2Mg (s) + O 2(g) Ù 2MgO (s) X + Y Ù XY specific general

10. –Decomposition Reactions CaCO 3(s) Ù CaO (s) + CO 2(g) XY Ù X + Y specific general

11. Name the Reaction  PbCO 3(s) Ù PbO (s) + CO 2(g) decomposition  C (s) + O 2(g) Ù CO 2(g) combination  2NaN 3(s) Ù 2Na (s) + 3N 2(g) decomposition  2C 2 H 6(g) + 7O 2(g) Ù 4CO 2(g) + 6H 2 O (l) combustion

12. Atomic and Molecular Masses  Amu scale – defined by assigning the mass of 12 C as 12 amu exactly – 1 amu = 1.66054 x 10 -24 g – 1 g = 6.02214 x 10 23 amu  Average Atomic Masses – 12 C 98.892% abundant 13 C 1.1108% abundant (0.98892)(12 amu) + (0.01108)(13.00335 amu) = 12.011 amu atomic mass

13. Formula and Molecular Masses –sum of all atomic masses in the formula of an ionic or molecular compound vitamin C C 6 H 8 O 6 6 x 12.0=72.0 amu 8 x 1.0= 8.0 amu 6 x 16.0 =96.0 amu 176.0 amu formula mass of vitamin C (often called molecular mass)

14. Percentage Composition  Calculate the percent mass that each type of atom contributes to a molecule –% X = (no. X atoms)(X amu) x 100 formula mass cmpd –C6H8O6–C6H8O6 % C = (6)(12.01amu) x 100 = 40.94% C 176.0 amu % H = (8)(1.01amu) x 100 = 4.59% H 176.0 amu % O = (6)(16.00 amu) x 100 = 54.55% O 176.0 amu

15. The Mole  We can measure masses in amu but how do we relate that to mass in grams? We define a quantity of atoms – a mole – which has the same mass in grams as the mass of the element in amu.  So how many atoms does it take to make, say, 1.00 g of H? 1.0 g H x 1 atom H  6.0 x 10 23 atoms of H 1.7 x 10 -24 g H 12.0 g C x 1 atom C  6.0 x 10 23 atoms of C 2.0 x 10 -23 g C a mole

16. Avogadro’s Number  6.02214 x 10 23 units/mole –No. of atoms per mole of an element –No. of molecules per mole of molecular cmpd. –No. of formula units per mole of ionic cmpd. –No. of cows per mole of cows Memorize this number & what it means!

17.  1 C atom = 12 amu 1 mole C atoms = 12 g  1 Mg atom = 24 amu1 mole Mg atoms = 24 g  1 CO molecule = 28 amu1 mole CO molecules = 28 g  1 NaCl fm. unit = 58 amu1 mole NaCl fm.units = 58g

18. Molar Mass  From this information we can define something called the molar mass (MM) of an atom (or molecule or formula): from the equality: 1 mole C = 12.0 g C we define the molar mass of a substance 12.0 g C = MM orMolar Mass 1 mole C(Atomic Mass) (Molecular Mass) (Formula Mass) conversion factor

19. Problems  Practice Ex. 3.9: –How many mole in 508 g of NaHCO 3 ? Given: MM = 84.02 g/mol NaHCO 3 508 g NaHCO 3 508 g NaHCO 3 x 1 mole = 6.05 mole NaHCO 3 84.02 g NaHCO 3 – How many formula units of NaHCO 3 ? Given: 6.02 x 10 23 form. units/mole NaHCO 3 6.05 mole NaHCO 3 x 6.02 x 10 23 fm. units = 3.6 x 10 24 fm. 1 moleunitsNaHCO 3

20.  Molar Mass converts between moles and grams of a substance  Avogadro’s number converts between moles of a substance and atoms (or molecules or formula units) of that substance These are very important conversion factors, know & understand them!

21. Problems  How many moles of vitamin C are contained in 5.00 g of vitamin C? C 6 H 8 O 6 176.0 g/mol  17.5 mg of cocaine (C 17 H 21 NO 4 ) per kg of body weight is a lethal dose. How many moles is that? How many molecules?  In 25 g of C 12 H 30 O 2 THC (tetrahydrocannibinol) how many moles are there? How many molecules are there? How many C atoms are there?

22.  How many moles of O are contained in 1.50 moles of C 6 H 5 NO 3 ?  How many grams of nitrogen are contained in 70.0 g of C 6 H 5 NO 3 ? How many atoms?

23. Calculate the number of H atoms in 50.0 mg of acetominophen, C 8 H 9 O 2 N.

24. Determination Empirical Formulas  simplest ratio of atoms –change g of each element to moles or –assume 100 grams of substance & change the % of each element to moles –change the mole ratio of atoms to the simplest ratio by dividing by the smallest number of moles

25.  Practice Ex. 3.12: –5.325 g methyl benzoate contains 3.758 g C, 0.316 g H, 1.251 g O. Determine empirical formula. 3.758 g C x 1 mole = 0.313 mol C 12.01 g 0.316 g H x 1 mole = 0.313 mol H 1.01 g 1.251 g O x 1 mole = 0.0782 mol O 16.00 g C 0.313 H 0.313 O 0.0782 C4H4OC4H4O

26. Determination of Molecular Formulas  actual ratio of atoms –determine the empirical formula –divide the actual molar mass by the empirical formula mass to get ‘n’ –multiply the mole ratio in the empirical formula by ‘n’

27.  Practice Ex. 3.13: –Ethylene glycol is composed of 38.7% C, 9.7% H & 51.6% O by mass. Its true molar mass is 62.1 g/mol. What are the empirical and molecular formulas? 38.7 g C x 1 mole = 3.23 mole C 12.0 g 9.7 g H x 1 mole = 9.60 mole H 1.01 g 51.6 g O x 1 mole = 3.22 mole O 16.0 g C 3.23 H 9.60 O 3.22 CH 3 O empirical formula n = 2 molecular formula C 2 H 6 O 2

28. Formulas from Combustion Data  Formulas determined from products of combustion products –Menthol is composed of C, H, and O. A 0.1005 g sample of menthol is combusted, producing 0.2829 g of CO 2 and 0.1159 g H 2 O. What is the empirical formula? C x H y O z + O 2 Ù CO 2 + H 2 O 0.1005 g 0.2829 g 0.1159 g – Calculate moles CO 2 & C; moles H 2 O & H

29. 0.2829 g CO 2 x 1 mol x 1 mol C = 0.00643 mol C 44.0 g 1 mol CO 2 0.1159 g H 2 O x 1 mol x 2 mol H = 0.0129 mol H 18.0 g 1 mol H 2 O total mass of C + H = 0.0902 g mass of O = 0.1005 g - 0.0902 g = 0.0103 g O x 1 mol = 16.0 g 6.44 x 10 -4 mol O total mass of all C, H & O

30. C 0.00643 H 0.0129 O 0.000644 C 10 H 20 O 0.00643 mol C 0.0129 mol H 6.44 x 10 -4 mol O If the MM is 156 g/mol, what is the molecular formula? n=1 therefore molecular formula is C 10 H 20 O (empirical formula mass 156 g/mol)

31. Quantitative Stoichiometry  Determination of quantities from balanced chemical reaction equations –mole ratios from balanced chemical equation convert between species –if quantities are given for more than one reactant, the limiting reactant must be determined

32. –Given the following balanced equation: 1Mg(OH) 2 + 2HCl Ù 1MgCl 2 + 2H 2 O –Calculate the number of moles of HCl required to react completely with 0.42 mol of Mg(OH) 2 0.42 mol Mg(OH) 2 x 2 mol HCl = 0.84 mol HCl 1 mol Mg(OH) 2 The mole ratio comes from the balanced chemical equation

33. –How many grams of MgCl 2 can be produced? 0.42 mol Mg(OH) 2 x 1 mol MgCl 2 x 95.3 g MgCl 2 1 mol Mg(OH) 2 1 mol = 40.0 g MgCl 2 Theoretical Yield -- maximum amount that can be produced

34. General Sequence of Conversion: grams of reactant moles of reactant MM reactant moles of product mole ratio grams of product MM product

35. Practice Ex. 3.14:  How many grams of O 2 can be prepared from 4.50 g of KClO 3 ? 2KClO 3 Ù 2KCl + 3O 2 4.50 g KClO 3 x 1 mol x 3 mol O 2 x 32.0 g O 2 = 1.76 g O 2 122.6 g 2 mol KClO 3 1 mol

36. Limiting Reactant  given a non-stoichiometric amount of both reactants, you will have to determine which is the limiting reagent or reactant  example: you have 10 bicycle frames and 16 bicycle wheels and you need to put them together to produce as many bicycles as possible, how many bicycles can be produced, what is the limiting “reagent”, and how much excess “reagent” do you have left over?

37. — Balanced ‘Equation’ 1 (mole) frame + 2 (moles) wheels Ù 1 (mole) bicycles [10 (moles) frames] [16 (moles) wheels] [8(moles) bicycles] – Limiting Reactant -- will produce the least amount of product 10 mol frames x 1 mol bicycles = 10 bicycles 1 mol frames 16 mol wheels x 1 mol bicycles = 8 bicycles 2 mol wheels limiting reactant

38. Practice Ex. 3.16:  A mixture of 1.5 mol of Al and 3.0 mol of Cl 2 react. What is limiting & how many moles of AlCl 3 are formed? 2Al (s) + 3Cl 2(g) Ù 2AlCl 3(s) 1.5 mol 3.0 mol 1.5 mol Al x 2 mol AlCl 3 = 1.5 mol AlCl 3 2 mol Al 3.0 mol Cl 2 x 2 mol AlCl 3 = 2.0 mol AlCl 3 3 mol Cl 2 1.5 mol