1. The Mole Stoichiometry: Cookbook Chemistry

2. The Mole  A mole is a number  Avogadro’s number = 6.02x10 23  Named after Amadeo Avogadro  Loschmidt determined the number of particles in one cubic centimeter of a gas at ordinary temperature and pressure

3. Counting atoms by counting moles  By counting moles, atoms or molecules are counted  Counting atoms by using moles eliminates waste in chemical reactions  Coefficients in chemical equations represent mole quantities

4. Counting atoms by counting moles 2Na + Cl 2  2NaCl  “Two moles sodium and one mole chlorine gas react to give two moles sodium chloride”  4 moles sodium require 2 moles Cl 2  5.2 moles sodium require 2.6 moles Cl 2  3.1 moles Cl 2 require 6.2 moles sodium  2:1 is the sodium/chlorine mole ratio

5. Counting atoms by counting moles  Counting atoms allows prediction of product quantities 2Fe + 6HCl  2FeCl 3 + 3H 2  How many moles iron (III) chloride can be made using 4.3 moles HCl?  Set up a proportion  Coeff. 2mol FeCl 3 = x mol FeCl 3 prob.  side 6mol HCl 4.3mol HCl side  x=2(4.3)/6=1.43 mol FeCl 3

6. Limiting reagents  If mole quantities are not exact, one of the reactants will run out first – this reactant is the limiting reagent  2H 2 + O 2  2H 2 O  If 3 moles H 2 are reacted with 1 mole O 2, what is the limiting reagent?

7. Limiting reagents  Divide each mole quantity by the coefficient to find equivalents.  H 2  3/2=1.5eq  O 2  1/1=1eq  limiting reagent  The reactant with the fewest equivalents (O 2 ) is the limiting reagent. The other (H 2 ) is “in excess”.

8. Limiting reagents 22Fe + 6HCl  2FeCl 3 + 3H 2 00.0037 mol Fe is reacted with 0.017 mol HCl. What is the limiting reagent? FFe  0.0037/2 = 0.00185eq Fe HHCl  0.017/6 = 0.00283eq HCl

9. Using limiting reagents  The quantity of product obtained is limited by the amount of the limiting reagent  2H 2 + O 2  2H 2 O  If 4.5 moles hydrogen gas and 1.9 moles oxygen are reacted, how many moles water will be formed?

10. Using limiting reagents  Solution: First determine the limiting reagent.  H 2 : 4.5/2=2.25eq  O 2 : 1.9/1=1.9eq limiting reagent  Then set up a proportion between the limiting reagent and the desired product.  O 2 1 = 1.9  H 2 O 2 x x=3.8 moles

11. Molar Mass  Molar mass is the mass of one mole of particles  Atomic mass – found in the bottom of each square of the periodic table – units are grams/mole  Atomic mass is the weighted average of the mass numbers of all the isotopes of an element.  Molecular mass – the mass of one mole of molecules  It is equal to the sum of the atomic masses of all the atoms in the molecule.

12. Molar mass  H 2 O – (H) 2x1 = 2  (O) 1x16 = 16total = 18g/mol  NH 3 – (N) 1x14 = 14  (H) 3x1 = 3total = 17g/mol  glucose (C 6 H 12 O 6 )  (C) – 6x12 = 72  (H) – 12x1 = 12  (O) – 6x16 = 96total = 180g/mol

13. Molar mass  Formula mass is the sum of all the atomic masses in a formula unit (for salts)  NaCl – (Na) 23  (Cl) 35.5total = 58.5g/mol  Mg(NO 3 ) 2  (Mg) 1x24.3 = 24.3  (N) 2x14 = 28  (O) 6x16 = 96  total = 148.3g/mol

14. Using molar mass  Mass to moles conversions  mass/(molar mass) = moles g  g/mol = g x mol/g = moles  Example: How many moles are represented by 2.5 grams of water?  Solution: 2.5g/(18g/mol) = 0.14mol

15. Using molar mass  Moles to mass conversions molesx(molar mass) = mass mol x g/mol = g  Example: What is the mass of 0.094 moles sodium chloride?  Solution: 0.094mol x 58.5g/mol = 5.5g

16. Mass-mass stoichiometry 2HNO 3 + H 2 O 2 + 2Fe(NO 3 ) 2  2Fe(NO 3 ) 3 + 2H 2 O How many grams hydrogen peroxide (H 2 O 2 ) are needed to make 2.43 grams iron (III) nitrate (Fe(NO 3 ) 3 ) according to the reaction below? x g H 2 O 2 2.43 g iron (III) nitrate moles H 2 O 2 moles iron (III) nitrate mole ratio (2:1)  by molar mass x by molar mass

17. Per cent yield  Mass obtained from calculations is “theoretical yield” – never obtained in practice Per cent yield = actual yield x 100% theoretical yield

18. Per cent yield  Jorma makes drugs for a hobby (aspirin, that is) and expects to obtain 2.13g aspirin from his synthesis reaction. In reality he only gets 1.89g. What is his % yield? (1.89/2.13)x100% = 88.7%

19. Per cent composition by mass  % composition by mass is a tool for compound identification  To calculate: divide the molar mass contribution of each element by the total molar mass and multiply by 100%

20. Per cent composition by mass  Example: H 2 SO 4 (sulfuric acid)  total molar mass  H: 1x2=2  S: 32x1=32  O: 16x4=64sum=98g/mol  %H=2(100%)/98=2.04%  %S=32(100%)/98=32.65%  %O=remainder=65.31%=64(100%)/98

21. Determining formulas from % composition  Formulas are a mole ratio of elements  Empirical formula: simplest mole ratio of elements, like NaCl or Ca(NO 3 ) 2  Applies to any type of compound  Molecular formula: mole ratio of elements in an actual molecule (all nonmetals), like H 2 O or NH 3  Often the molecular formula and the empirical formula are the same, but not always

22. Determining formulas from % composition  Hydrazine, a rocket fuel molecular formula – N 2 H 4 empirical formula – NH 2  Hydrogen peroxide molecular formula – H 2 O 2 empirical formula – HO  Glucose, a sugar molecular formula – C 6 H 12 O 6 empirical formula – CH 2 O

23. Determining formulas from % composition  % composition is a mass ratio – so by converting mass to moles, the empirical formula can be determined.  Example: Laboratory analysis finds a compound to consist of 28.05% Na, 29.27% C, 3.67% H, and 39.02% O. What is the empirical formula?

24. Determining formulas from % composition  Treat the % like grams  Convert grams to moles  Na: 28.05g/(23g/mol) = 1.22 mol  C: 29.27g/(12g/mol) = 2.44 mol  H: 3.67g/(1g/mol) = 3.67mol  O: 39.02g/(16g/mol) = 2.44 mol

25. Determining formulas from % composition  Convert to simplest whole number ratio – divide all mol quantities by the smallest one. These results become the subscripts in the formula.  Na: 1.22/1.22 = 1  C: 2.44/1.22 = 2  H: 3.67/1.22 = 3  O: 2.44/1.22 = 2  So the empirical formula is NaC 2 H 3 O 2 (sodium acetate).

26. Ideal gas law  Boyle’s Law: PV = C (P 1 V 1 = P 2 V 2 )  Factors that affect pressure/volume: Temperature (T) Amount (moles) of gas (Avogadro’s Principle) (n)  Ideal Gas Law: PV  nT  Constant of proportionality = R (gas constant)

27. Ideal Gas Law  Ideal gas Law: PV = nRT  V must be liters, T is Kelvins, n is moles  Values for gas constant (depends on pressure units) P in atm: R = 0.08206Latm/molK P in kPa: R = 8.314LkPa/molK P in torr: R = 62.4Ltorr/molK

28. Ideal Gas Law  Example: Find the moles of oxygen in a balloon of 2.3L volume and 1.3atm pressure if the temperature is 45ºC.  Solution: PV = nRT  1.3(2.3) = n(0.08206)(45+273)  n = 1.3(2.3)/(0.08206)(45+273)  n = 0.115 mol

29. Ideal Gas Law  Example 2: Find the molar volume of a gas at STP.  Solution: STP = standard temperature and pressure (273K and 1 atm)  PV = nRT  1V = 1(0.08206)(273) = 22.4L/mol