2. THE MOLE Chapter 11

3. 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 Writing and balancing chemical equations

4. 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 Writing and balancing chemical equations Samples

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

6. 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 10 23 particles

7. mole - amount of substance that contains same # of entities as atoms in 12g of carbon-12. 1 mol contains 6.022x10 23 entities Avogadro’s number (N) The mole 1 mole H 2 O contains 6.022 x 10 23 H 2 O molecules 1 mole KNO 3 contains 6.022x10 23 KNO 3 formula units 1 mole Hg contains 6.022x10 23 Hg atoms Mole represents large quantity of microscopic particles.

8. mass of 1 mass of 1 Substance atom (molecule) mole of atoms (molecules) CaCO 3 100.09 amu 100.09 g O 2 32.00 amu 32.00 g H 2 O18.02 amu 18.02 g Copper63.55 amu 63.55 g Can weigh out grams using scale Use mass to ‘count’ entities. 6.022 x 10 23 entities Molar mass ( M ) (gmw)- mass of 1 mole of entities –M (g/mol) numerically equal to formula weight (amu) CH 4 = 1(12.10 g/mol) + 4(1.008 g/mol) = 16.04 g/mol The mole

9. Relating moles to chemical formulas Glucose C 6 H 12 O 6 ( M = 180.16 g/mol) Oxygen (O) Mass/mole of compound 6 atoms 96.00 g Table 3.2 Carbon (C)Hydrogen (H) Atoms/molecule of compound Moles of atoms/ mole of compound Atoms/mole of compound Mass/molecule of compound 6 atoms12 atoms 6 moles of atoms 12 moles of atoms 6 moles of atoms 6(6.022 x 10 23 ) atoms 12(6.022 x 10 23 ) atoms 6(6.022 x 10 23 ) atoms 6(12.01 amu) = 72.06 amu 12(1.008 amu) = 12.10 amu 6(16.00 amu) = 96.00 amu 72.06 g12.10 g 180.16 g/mole

10. Interconverting Moles, Mass, and # of Chemical Entities Mass (g) = no. of moles x gmw 1 mol No. of entities = no. of moles x 6.022x10 23 entities 1 mol

11. Mass % of element X = Mass Percent from Chemical Formula moles X in formula 1 mol compound (molar mass of X) molar mass of compound x 100% i.e. Mass % of H in H 2 O = 2 mol H 1.008 g H 1 mol H 2 O 1 mol H 18.02 g H 2 O 1 mol H 2 O x 100% = 11.19% H by mass

12. Particles Atoms –Single elements Formula units –Ionically bonded compounds Molecules –Covalent bonded compounds

13. Figure 3.6 macro micro Avogadro’s # takes us to/from macroscopic/microscopic Law of Conservation of Mass (6.02 x 10 23 molecules) (6.02 x 10 23 molecules) (1.20 x 10 24 molecules)

14. Particle  Mole Problems REMEMBER –The number of particles in 1 mole of ANY substance is ALWAYS the same (6.02 x 10 23 ) 1 mol = 6.02 x 10 23 particles –How many molecules are in 2.2 moles of water? 1 mol H 2 O = 6.02 x 10 23 particles Use this as your conversion factor!!! 1 mol H 2 O 6.02 x 10 23 molecules 1 mol H 2 OOR

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

16. Particle  Mole Practice Continued How many moles of sodium carbonate contain 7.9 x 10 24 formula units? 2.2 mol H 2 O 1 Take your given value and put it over ONE. Multiply by the conversion factor that allows you to cancel out the top unit. Leaving you with the unit you WANTED. 6.02 x 10 23 molecules 1 mol H 2 O X = 1.3244 x 10 24 ANSWER: 1.3 x 10 24 molecules of H 2 O

17. 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

18. Calculate moles of O 2 consumed when 10 mol of H 2 O are produced (using balanced equation from Table 3.5)? –C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(g) Calculate mass of CO 2 produced burning 1.00 g butane (C 4 H 10 ). –2C 4 H 10 (l) + 13O 2 (g)  8CO 2 (g) + 10H 2 O(l) Practice, practice, practice!!! Calculating amounts of reactant and product

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

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

21. 11.3 Moles of Compounds Formula mass – the sum of the atomic masses of all the atoms in a compound H2OH2O H: 2 x 1.01amu = 2.02 amu O: 1 x 16.00amu = 16.00 amu 18.02 amu

22. 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 Chemical rxts. that involve limiting reagents

23. What mass of NH 3 is produced from the rxt. of 1.00 g H 2 (g) w/ 1.00 g N 2 (g)? (Info on both reactants given) 3H 2 (g) + N 2 (g)  2NH 3 (g)

24. Is H 2 or O 2 the limiting reagent? –H2–H2 Chemical rxts. that involve limiting reagents Microscopic Picture

25. Moles of Compounds Continued Molar mass –The mass in grams of 1 mole of a substance –Formula mass of H 2 O = 18.02 amu –Molar mass of H 2 O = 18.02g/mol 1 mol of H 2 O = 18.02 g = 6.02 x 10 23 molecules

26. Mass  Mole Problems Changing the mass to moles or vice versa using the molar mass How many moles are in 11.2g of NaCl? 1.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

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

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

29. Multi-Step Conversions Mass-Particle g  mol  particles Particle-Mass Particles  mol  g What is the mass of 8.2 x 10 22 atoms of calcium? 1 mol Ca = 6.02 x 10 23 atoms Ca 1 mol Ca = 40.08g Ca

30. 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: 0 o C and 1 atm –1 mol = 22.4 liters

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

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

33. Percent Composition Problem Calculate the percent composition of hydrogen in water H2OH2O H: 2 x 1.01amu = 2.02 amu O: 1 x 16.00amu = 16.00 amu 18.02 amu