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The Mole

The Mole. Stoichiometry: Cookbook Chemistry. 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. Counting atoms by counting moles.

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The Mole

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  1. The Mole Stoichiometry: Cookbook Chemistry

  2. The Mole • A mole is a number • Avogadro’s number = 6.02x1023 • 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 + Cl2 2NaCl • “Two moles sodium and one mole chlorine gas react to give two moles sodium chloride” • 4 moles sodium require 2 moles Cl2 • 5.2 moles sodium require 2.6 moles Cl2 • 3.1 moles Cl2 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  2FeCl3 + 3H2 • How many moles iron (III) chloride can be made using 4.3 moles HCl? • Set up a proportion • Coeff. 2mol FeCl3 = x mol FeCl3 prob. • side 6mol HCl 4.3mol HCl side • x=2(4.3)/6=1.43 mol FeCl3

  6. Limiting reagents • If mole quantities are not exact, one of the reactants will run out first – this reactant is the limiting reagent • 2H2 + O2 2H2O • If 3 moles H2 are reacted with 1 mole O2, what is the limiting reagent?

  7. Limiting reagents • Divide each mole quantity by the coefficient to find equivalents. • H2 3/2=1.5eq • O2 1/1=1eq  limiting reagent • The reactant with the fewest equivalents (O2) is the limiting reagent. The other (H2) is “in excess”.

  8. Limiting reagents • 2Fe + 6HCl  2FeCl3 + 3H2 • 0.0037 mol Fe is reacted with 0.017 mol HCl. What is the limiting reagent? • Fe  0.0037/2 = 0.00185eq Fe • 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 • 2H2 + O2 2H2O • 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. • H2: 4.5/2=2.25eq • O2: 1.9/1=1.9eq limiting reagent • Then set up a proportion between the limiting reagent and the desired product. • O2 1 = 1.9 • H2O 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 • H2O – (H) 2x1 = 2 • (O) 1x16 = 16 total = 18g/mol • NH3 – (N) 1x14 = 14 • (H) 3x1 = 3 total = 17g/mol • glucose (C6H12O6) • (C) – 6x12 = 72 • (H) – 12x1 = 12 • (O) – 6x16 = 96 total = 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) 35.5 total = 58.5g/mol • Mg(NO3)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 How many grams hydrogen peroxide (H2O2) are needed to make 2.43 grams iron (III) nitrate (Fe(NO3)3) according to the reaction below? 2HNO3 + H2O2 + 2Fe(NO3)2 2Fe(NO3)3 + 2H2O moles H2O2 moles iron (III) nitrate x by molar mass  by molar mass mole ratio (2:1) x g H2O2 2.43 g iron (III) nitrate

  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: H2SO4 (sulfuric acid) • total molar mass • H: 1x2=2 • S: 32x1=32 • O: 16x4=64 sum=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(NO3)2 • Applies to any type of compound • Molecular formula: mole ratio of elements in an actual molecule (all nonmetals), like H2O or NH3 • 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 – N2H4 • empirical formula – NH2 • Hydrogen peroxide • molecular formula – H2O2 • empirical formula – HO • Glucose, a sugar • molecular formula – C6H12O6 • empirical formula – CH2O

  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 NaC2H3O2 (sodium acetate).

  26. Ideal gas law • Boyle’s Law: PV = C (P1V1 = P2V2) • 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

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