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THE MOLE

THE MOLE. Chapter 11. Writing and balancing chemical equations. Chemical equation : identities and quantities of substances involved in chemical/physical change Balance using Law of Conservation of Mass and Law of Definite Composition same # of atoms on each side

iliana-house
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THE MOLE

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  1. THE MOLE Chapter 11

  2. Writing and balancing chemical equations • Chemical equation: identities and quantities of substances involved in chemical/physical change • Balance using Law of Conservation of Mass and Law of Definite Composition • same # of atoms on each side • fixed ratio of elements in compound

  3. Writing and balancing chemical equations 1. translate statement: reactants  products 2. balance atoms using stoichiometric coefficients 3. Adjust stoichiometric coefficients (if necessary) • Smallest whole # preferred 4. Check 5. Specify states of matter Samples

  4. Stoichiometry: study of quantitative aspects of chemical formulas/reactions • Mole: unit chemists use to count chemical entities by weighing them

  5. 11.1 Measuring Matter • The Mole • SI Unit for the amount of a substance • The number of particles equal to the number of atoms in exactly 12.0 grams of carbon-12 • Also called Avogadro’s number • 1 mol = 6.02 x 1023 particles

  6. The mole mole - amount of substance that contains same # of entities as atoms in 12g of carbon-12. 1 mol contains 6.022x1023 entities Avogadro’s number (N) 1 mole H2O contains 6.022 x 1023 H2O molecules 1 mole KNO3 contains 6.022x1023 KNO3formula units 1 mole Hg contains 6.022x1023 Hg atoms Mole represents large quantity of microscopic particles.

  7. 6.022 x 1023 entities Can weigh out grams using scale Use mass to ‘count’ entities. The mole • Molar mass (M) (gmw)- mass of 1 mole of entities • M (g/mol) numerically equal to formula weight (amu) CH4 = 1(12.10 g/mol) + 4(1.008 g/mol) = 16.04 g/mol mass of 1 mass of 1 Substance atom (molecule) mole of atoms (molecules) CaCO3 100.09 amu 100.09 g O2 32.00 amu 32.00 g H2O 18.02 amu 18.02 g Copper 63.55 amu 63.55 g

  8. Relating moles to chemical formulas Glucose C6H12O6 ( M = 180.16 g/mol) Table 3.2 Oxygen (O) Carbon (C) Hydrogen (H) Atoms/molecule of compound 6 atoms 12 atoms 6 atoms Moles of atoms/ mole of compound 6 moles of atoms 12 moles of atoms 6 moles of atoms Atoms/mole of compound 6(6.022 x 1023) atoms 12(6.022 x 1023) atoms 6(6.022 x 1023) atoms 6(16.00 amu) = 96.00 amu Mass/moleculeof compound 6(12.01 amu) = 72.06amu 12(1.008 amu) = 12.10 amu Mass/mole of compound 72.06 g 12.10 g 96.00 g 180.16 g/mole

  9. gmw Mass (g) = no. of moles x 1 mol 6.022x1023 entities No. of entities = no. of moles x 1 mol Interconverting Moles, Mass, and # of Chemical Entities

  10. i.e. Mass % of H in H2O = 2 mol H1.008 g H 1 mol H2O 1 mol H 18.02 g H2O 1 mol H2O x 100% = 11.19% H by mass Mass Percent from Chemical Formula Mass % of element X = moles X in formula 1 mol compound (molar mass of X) x 100% molar mass of compound

  11. Particles • Atoms • Single elements • Formula units • Ionically bonded compounds • Molecules • Covalent bonded compounds

  12. (6.02 x 1023 molecules) (1.20 x 1024 molecules) (6.02 x 1023 molecules) Avogadro’s # takes us to/from macroscopic/microscopic Law of Conservation of Mass macro micro Figure 3.6

  13. Particle  Mole Problems 1 mol H2O 6.02 x 1023 molecules 6.02 x 1023 molecules 1 mol H2O • REMEMBER • The number of particles in 1 mole of ANY substance is ALWAYS the same (6.02 x 1023) 1 mol = 6.02 x 1023 particles • How many molecules are in 2.2 moles of water? 1 mol H2O = 6.02 x 1023 particles Use this as your conversion factor!!! OR

  14. Calculating amounts of reactant and product • Balanced equation needed for stoichiometric calculations • Ratios of reactants/products to calculate amounts of reactants/products

  15. Particle  Mole Practice Continued 2.2 mol H2O 1 6.02 x 1023 molecules 1 mol H2O Multiply by the conversion factor that allows you to cancel out the top unit. Leaving you with the unit you WANTED. Take your given value and put it over ONE. How many moles of sodium carbonate contain 7.9 x 1024 formula units? = 1.3244 x 1024 X ANSWER: 1.3 x 1024 molecules of H2O

  16. 11.2 Mass and the Mole • Atomic mass • The mass of an atom relative to the mass assigned to carbon-12 • Molar mass • The mass in grams of one mole of any pure substance • Use the average atomic mass off the periodic table • Molar mass of an element • Atomic mass in grams per mole (g/mol) • Molar mass of oxygen = 16.00 g/moL • Molar mass of helium = 4.00 g/moL

  17. Calculating amounts of reactant and product • Calculate moles of O2 consumed when 10 mol of H2O are produced (using balanced equation from Table 3.5)? • C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g) • Calculate mass of CO2 produced burning 1.00 g butane (C4H10). • 2C4H10(l) + 13O2(g)  8CO2(g) + 10H2O(l) • Practice, practice, practice!!!

  18. Mass & Mole Problems • How many moles are in 82.2g of aluminum? 1 mole of Al = 26.98g Al

  19. Mass & Mole Problems • How many grams are in 3.5 mol of neon? 1 mole of Ne = 20.18g Ne

  20. 11.3 Moles of Compounds • Formula mass – the sum of the atomic masses of all the atoms in a compound H2O H: 2 x 1.01amu = 2.02 amu O: 1 x 16.00amu = 16.00 amu 18.02 amu

  21. Chemical rxts. that involve limiting reagents • Limiting reagent- reactant that forms fewer moles of product • Limits amount of product formed (chair analogy) • To decide which reagent is limiting reagent: 1. If information given in g, convert to moles 2. Use ratio from balanced equation to find moles of final product possibly produced 3. Reagent that produces least possible moles of product is limiting reagent

  22. What mass of NH3 is produced from the rxt. of 1.00 g H2(g) w/ 1.00 g N2(g)? (Info on both reactants given) 3H2(g) + N2(g)  2NH3(g)

  23. Chemical rxts. that involve limiting reagents Microscopic Picture • Is H2 or O2 the limiting reagent? • H2

  24. Moles of Compounds Continued • Molar mass • The mass in grams of 1 mole of a substance • Formula mass of H2O = 18.02 amu • Molar mass of H2O = 18.02g/mol 1 mol of H2O = 18.02 g = 6.02 x 1023 molecules

  25. Mass  Mole Problems • Changing the mass to moles or vice versa using the molar mass • How many moles are in 11.2g of NaCl? • Determine the molar mass. - Na: 1 x 23.00 = 23.00g - Cl: 1 x 35.45 = 35.45g = 58.45g/mol Therefore, 1 mol NaCl = 58.45g

  26. Mass  Mole Problems 1 mol of NaCl 58.45g NaCl 11.2g NaCl 1 58.45g NaCl 1 mol of NaCl 1 mol NaCl 58.45g NaCl 2. Convert between the molar mass and the moles. 1 mol of NaCl = 58.45g OR Multiply by the conversion factor that allows you to cancel out the top unit. Leaving you with the unit you WANTED. Take your given value and put it over ONE. =0.1916167665 mol NaCl X ANSWER: 0.192 mol NaCl

  27. More Practice 1 mol of NaCl 58.45g NaCl 2.50 mol NaCl 1 58.45g NaCl 1 mol of NaCl 58.45g NaCl 1 mol NaCl • What is the mass of 2.50 mol of NaCl? • Find the molar mass… 1 mol of NaCl = 58.45g OR Multiply by the conversion factor that allows you to cancel out the top unit. Leaving you with the unit you WANTED. Take your given value and put it over ONE. = 146.125g NaCl X ANSWER: 146.13g NaCl

  28. Multi-Step Conversions • Mass-Particle g  mol  particles • Particle-Mass Particles mol  g • What is the mass of 8.2 x 1022 atoms of calcium? 1 mol Ca = 6.02 x 1023 atoms Ca 1 mol Ca = 40.08g Ca

  29. Mole Volume Problems • Equal volumes of gases at the same temperature and pressure contain the same number of particles. • Molar volume • The volume of 1 mol of gas at standard conditions (STP) • STP • standard temperature and pressure: 0oC and 1 atm • 1 mol = 22.4 liters

  30. Mole  Volume Practice Problem 0.35 mol He 1 22.4 L 1 mol He • What is the volume of 0.35 moles of helium gas at STP? Multiply by the conversion factor that allows you to cancel out the top unit. Leaving you with the unit you WANTED. Take your given value and put it over ONE. = 7.84L He X ANSWER: 7.84L He

  31. Percent Composition mass of element mass of compound • Percent Composition • The percent by mass of each element in a compound = % composition X 100

  32. Percent Composition Problem • Calculate the percent composition of hydrogen in water H2O H: 2 x 1.01amu = 2.02 amu O: 1 x 16.00amu = 16.00 amu 18.02 amu

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