2. Unit 5: Stoichiometry http://www.unit5.org/chemistry/Stoichiometry.html Textbook: Chapter 10 & 11

3. Stoich…what?? Stoichiometry = part of chemistry that is concerned with the QUANTITATIVE relationship between reactants and products in a chemical reaction.

4. For this unit you should understand:  Moles, mass, representative particles (atoms, molecules, formula units), molar mass, and Avogadro’s number.  The percent composition of an element in a compound.  Empirical and Molecular Equations  Balanced chemical equations: for example, for a given mass of a reactant, calculate the amount of produced.  Identifying limiting reactants  The percent yield of a reaction.

5. Some common collections and the numbers of items in them. Timberlake, Chemistry 7 th Edition, page 179

6. Counting Atoms Chemistry is a quantitative science - we need a "counting unit." MOLEThe MOLE 1 mole is the amount of substance that contains as many particles (atoms or molecules) as there are in 12.0 g of C-12. 1mole = molar mass & 1 mole = 6.02 x 10 23 atoms or molecules

7. ? Visualizing a Chemical Reaction-The reason a “counting unit” is needed is atoms combine on a piece by piece basis! Na + Cl 2 NaCl ___ mole Cl 2 ___ mole NaCl___ mole Na 2 105 2 5

8. Amadeo Avogadro (1776 – 1856) 1 mole = 602213673600000000000000 or 6.02 x 10 23 particles thousands millions billionstrillions quadrillions ? This is Avogadro's number!! Number of particles (atoms or molecules or compounds) in 1 mole of any substance? Amedeo AvogadroAmedeo Avogadro (1766-1856) never knew his own number; it was named in his honor by a French scientist in 1909. its value was first estimated by Josef Loschmidt, an Austrian chemistry teacher, in 1895.

9. How Big is a Mole? One mole of marbles would cover the entire Earth (oceans included) for a depth of three miles. One mole of $100 bills stacked one on top of another would reach from the Sun to Pluto and back 7.5 million times. It would take light 9500 years to travel from the bottom to the top of a stack of 1 mole of $1 bills.

10. 1 Mole of Particles

11. How much is 1 mole of carbon? We use something called molar mass = the mass in grams of one mole of any pure substance. The masses of all atoms are established relative to the mass of carbon – carbon is our standard. The molar mass of any element is equal to its atomic mass, so C has a molar mass of? 12 grams

12. The Molar Mass and Number of Particles in One-Mole Quantities SubstanceMolar MassNumber of Particles in One Mole Carbon (C) 12.0 g 6.02 x 10 23 C atoms Sodium (Na) 23.0 g 6.02 x 10 23 Na atoms Iron (Fe) 55.9 g 6.02 x 10 23 Fe atoms NaF 42.0 g 6.02 x 10 23 NaF formula units CaCO 3 (antacid) 100.1 g 6.02 x 10 23 CaCO 3 formula units C 6 H 12 O 6 (glucose) 180.0 g 6.02 x 10 23 glucose molecules C 8 H 10 N 4 O 2 (caffeine) 194.0 g 6.02 x 10 23 caffeine molecules 1 mol of a substance = molar mass of a substance 1 mol of a substance = 6.02 x 10 23 particles of a substance

13. Welcome to Mole Island 1 mole = 22.4 L @ STP 1 mol = molar mass 1 mol = 6.02 x 10 23 particles

14. Mole Island Diagram

15. Percent Composition of Compounds % of an Element in a=mass of the element present in 1 mol of compound x 100 compound mass of 1 mol of compound Example: What is the mass percent of carbon, hydrogen and oxygen in a sample of ethanol, C 2 H 5 OH? Now You Try It! Carvone is a substance that occurs in two forms, both of which have the same molecular formula (C 10 H 14 O) and molar mass. One type of carvone gives caraway seeds their characteristic smell; the other is responsible for the smell of spearmint oil. Compute the mass percent of each element in carvone.

16. Molecular and Empirical Formulas Empirical formula is a compound with the smallest whole- number mole ratio of the elements. Simplest Ratio! Molecular formula- specifies the actual number of atoms of each element in one molecule or formula unit of the substance. Examples: For the following examples, the molecular formula for a compound is given; determine the empirical formula for each compound. C 6 H 6 H 2 O 2 C 12 H 4 Cl 4 O 2­ C 4 H 10 C 6 H 12 N 2 CCl 4

17. Steps for Determining the Empirical Formula 1. Write the symbol of each element present, next to it as a subscript write the given mass in grams. Obtain the mass of each element present (in grams). 2. Divide each mass in grams by the molar mass of the element. 3. Divide each number by the smallest number from step #2 to convert the smallest number to 1. If all of the numbers are integers (whole numbers), these are the subscripts in the empirical formula. If one or more of these numbers are not integers, go to step 4. 4. Multiply the numbers you derived in step 3 by the smallest integer that will convert all of them to whole numbers. This set of whole numbers represents the subscripts in the empirical formula

18. Calculating Empirical Formulas A 1.500 g a sample of a compound containing only carbon and hydrogen is found to contain 1.198 g of carbon. Determine the empirical formula for this compound. Examples: *Write the symbol of each element present, next to it as a subscript write the mass in grams. *Divide each number by the smallest number from step #2. Follow remaining directions.

19. Calculating Empirical Formulas Cisplatin, the common name for a platinum compound that is used to treat cancerous tumors, has the composition (mass percent) 65.02% platinum,9.34% nitrogen, 2.02% hydrogen, and 23.63% chlorine. Calculate the empirical formula for cisplatin. Examples:

20. Calculating Molecular Formula from the Empirical Formula A white powder is analyzed and found to have an empirical formula of P 2 O 5. The compound has a molar mass of 283.88 g. What is the compound’s molecular formula? Example: Molecular formula = n x empirical formula Where n is a small whole number Molar mass = n x empirical formula mass

21. Calculating Molecular Formula from the Empirical Formula A compound containing carbon, hydrogen, and oxygen is found to be 40.0% carbon and 6.700% hydrogen by mass. The molar mass of this compound is n 120.00 g/mol. Determine the molecular formula for this compound. Example: Molar mass = n x empirical formula mass

22. Stoichiometry Island Diagram Mass Particles Volume Mole Mass Volume Particles Known Unknown Substance A Substance B Stoichiometry Island Diagram 1 mole = molar mass (g) Use coefficients from balanced chemical equation 1 mole = 22.4 L @ STP 1 mole = 6.02 x 10 23 particles (atoms or molecules) 1 mole = 22.4 L @ STP 1 mole = 6.02 x 10 23 particles (atoms or molecules) 1 mole = molar mass (g) (gases)

23. Grilled Cheese Sandwich Bread + Cheese  ‘Cheese Melt’ 2 B + C  B 2 C 100 bread 30 slices? sandwiches Multiple Guess: 130 sandwiches 100 sandwiches 90 sandwiches 60 sandwiches 30 sandwiches Not enough information given 30 sandwiches

24. Limiting Reactants Limiting ReactantLimiting Reactant –used up in a reaction –determines the amount of product Excess ReactantExcess Reactant –added to ensure that the other reactant is completely used up –cheaper & easier to recycle Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

25. Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 270

26. Before and After Reaction 2 Before the reactionAfter the reaction Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 270 N 2 + H 2 NH 3 3 2 excess limiting LIMITING REACTANT DETERMINES AMOUNT OF PRODUCT

27. Limiting Reactants aluminum + chlorine gas  aluminum chloride Al(s) + Cl 2 (g)  AlCl 3 2 Al(s) + 3 Cl 2 (g)  2 AlCl 3 100 g 100 g ? g

28. How to Determine the Limiting Reactant 1.Begin by writing a correctly balanced chemical equation 2.Write down all quantitative values under equation (include units) 3.Convert ALL reactants to units of moles 4.Divide by the coefficient in front of each reactant 5.The smallest value is the limiting reactant!

29. Limiting Reactants aluminum + chlorine gas  aluminum chloride Al(s) + Cl 2 (g)  AlCl 3 2 Al(s) + 3 Cl 2 (g)  2 AlCl 3 100 g 100 g ? g ________ 27 g _______ 70.9 g 3.7037 1.41044 ______ 2 ______ 3 1.85185 0.47015 Limiting Reactant is Cl 2 Use Chlorine to determine the amount of product Cl 2 AlCl 3 ? g AlCl 3 : 100 g Cl 2 71 g Cl 2 = 125 g AlCl 3 1 mol Cl 2 2 mol AlCl 3 3 mol Cl 2 133.5 g AlCl 3 1 mol AlCl 3 AlAlCl 3 ? g AlCl 3 : 100 g Al 27 g Al = 494 g AlCl 3 1 mol Al 2 mol AlCl 3 2 mol Al 133.5 g AlCl 3 1 mol AlCl 3

31. Percent Yield calculated on paper measured in lab % yield = actual yield theoretical yield x 100

32. % yield = actual yield theoretical yield x 100 So using our previous problem, we calculated the theoretical yield as follows: Cl 2 AlCl 3 x g AlCl 3 = 100 g Cl 2 71 g Cl 2 1 mol Cl 2 2 mol AlCl 3 3 mol Cl 2 133.5 g AlCl 3 1 mol AlCl 3 = 125 g AlCl 3 If this was performed in the lab and the chemist measured the amount of aluminum chloride produced to be 102 g, what is the percent yield? % yield = x 100 102 g 125 g =82%

33. Stoichiometry Island Diagram Mass Particles VolumeMole Mass Known Unknown Substance A Substance B Stoichiometry Island Diagram Volume Particles M V P Mass Mountain Liter Lagoon Particle Place