1 / 29

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.

bianca-alford
Download Presentation

The Mole

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related