1. Stoichiometry Part 1: Calculating Quantities in Reactions

2. Objectives  Use proportional reasoning to determine mole ratios from a balanced chemical equation  Explain why mole ratios are central to solving stoichiometry problems  Solve stoichiometry problems involving mass by using molar mass  Solve stoichiometry problems involving the volume of a substance by using density  Solve stoichiometry problems involving the number of particles of a substance by using Avogadro’s number

3. Balanced Equations  Are similar to a recipe in that the coefficients in the balanced equation show the proportions of the reactants and products involved in the reaction  On a very small scale, the coefficients in a balanced equation represent the number of particles for each substance in the reaction  For example:  For the equation above, the coefficients show that two molecules of hydrogen react with one molecule of oxygen to form two molecules of water

4. Balanced Equations & Calculations  Calculations that involve chemical reactions use the proportions from balanced chemical equations to find the quantity of each reactant and product  For the calculations in this lesson you will assume that all reactions go into completion  You will also need to assume for this section that every reaction happens perfectly, with no loss of product

5. What is Stoichiometry?  It is the branch of chemistry that deals with the quantities of substances in chemical reactions  It is based on the understanding that the coefficients in a balanced equation also represent the number of moles of each substance  The mole ratio (a conversion factor) is the key to solving stoichiometry problems  For example:  This equation shows that 2 mol of C 8 H 18 react with 25 mol of O 2 to form 16 mol of CO 2 and 18 mol of H 2 O

6. Mole Ratio  Is a conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction.  2Al + 3H 2 SO 4 → Al 2 (SO 4 ) 3 + 3H 2  2 mol Al2 mol Al2 mol Al  3 mol H 2 SO 4 1 mol Al 2 (SO 4 ) 3 3 mol H 2  Then they can be flipped over  You can repeat this for the 2 nd, 3 rd and 4 th compound.

7.  Calculate the moles formed of each of the products in the following reaction if 3.30 mol of Fe 2 O 3 is used:  Fe 2 O 3 + 2Al → 2Fe + Al 2 O 3  Step 1: Gather the information  3.30 mol of Fe 2 O 3  Amt. of Fe = ?  Amt. of Al 2 O 3 = ? Using Mole Ratios Example

8.  Step 2: Plan your work  The mole ratio must cancel out the units of mol of Fe 2 O 3  2 mol Fe1 mol Al 2 O 3 1 mol Fe 2 O 3

9. Using Mole Ratios Example  Step 3: Calculate  3.30 mol of Fe 2 O 3 x 2 mol Fe = 6.6 mol Fe 1 mol Fe 2 O 3  3.30 mol of Fe 2 O 3 x 1 mol Al 2 O 3 = 3.30 mol Al 2 O 3 1 mol Fe 2 O 3  Step 4: Verify your result

10. Getting into Moles and Getting out of Moles  Substances are usually measured by mass or volume  So, normally you will have to convert your units to moles before using the mole ratio  We use the molar mass of elements and compounds to help us with this conversion.  Where is the molar mass found?  IN THE PERIODIC TABLE OF COURSE!

11. Problems Involving Mass, Volume, or Particles  For mass, we use molar mass to convert to moles first  For volume, we use density and molar mass to convert to moles  For the number of particles we use Avogadro’s number

12. Solving Mass to Mass Problems  Step 1: convert mass to moles using molar mass  Step 2: take known amount of mole and multiple by the mole ratio to find the unknown mole amount  Step 3: take the mole amount found and convert back to mass using the molar mass

13. Mass to Mass Example  2KClO 3  2KCl + 3O 2  How many grams of potassium chloride are produced if 25 g of potassium chlorate decompose?  Molar Mass of KClO 3 = 122.55 g/mol  K = 39.098 g/mol  Cl = 35. 453 g/mol  O = 15.999 g/mol x 3 1.25 g KClO 3 x 1 mol KClO 3 = 0.20 mol KClO 3 122.55 g KClO 3

14. Mass to Mass Example  2KClO 3  2KCl + 3O 2  How many grams of potassium chloride are produced if 25 g of potassium chlorate decompose? 2.0.20 mol KClO 3 x 2 mol KCl = 0.20 mol KCl 2 mol KClO 3 3.0.20 mol KCl x 74.55 g KCl = 14.91 g KCl 1 mol KCl

15. Solving Problems with Volume  When reactants are liquids, they are almost always measured by volume  Thus, we have to add two more steps to the process and five conversion factors in all!  To convert from volume we can use density  To convert from volume we can use the molar volume of a gas  The molar volume of any gas at STP (standard temperature & pressure ) is 22.41 L/mol or 22,410 mL/mol

16. Example Problem with Volume  Use the densities and balanced equation provided to answer the following: C 5 H 12 (l) → C 5 H 8 (l) + 2 H 2 (g)  How many milliliters of C 5 H 8 can be made from 366 mL C 5 H 12 ? Density of C 5 H 12 = 0.620 g/mL Density of C 5 H 8 = 0.681 g/mL

17. Example Problem with Volume  Step 1: Gather the information  Density of C 5 H 12 = 0.620 g/mL  Density of C 5 H 8 = 0.681 g/mL  366 mL C 5 H 12  # of mL of C 5 H 8 = ?

18. Example Problem with Volume  Step 2: Calculate using conversion factors A.Convert volume to mass using density B.Convert mass to moles using molar mass C.Find the unknown # of moles by multiplying the known number of moles by the mole ratio D.Convert the answer back to mass using the molar mass E.Convert the answer back to volume using density

19. Example Problem with Volume  Step 2: Calculate using conversion factors A.366 mL C 5 H 12 x 0.620 g C 5 H 12 = 226.92 g C 5 H 12 1 mL C 5 H 12 B.226. 92 g C 5 H 12 x 1 mol C 5 H 12 = 3.15 mol C 5 H 12 72.01 g C 5 H 12 C.3.15 mol C 5 H 12 x 1 mol C 5 H 8 = 3.15 mol C 5 H 8 1 mol C 5 H 12 D.3.15 mol C 5 H 8 x 68.01 g C 5 H 8 = 214.23 g C 5 H 8 1 mol C 5 H 8 E.214.23 g C 5 H 8 x 1 mL C 5 H 8 = 314.58 mL C 5 H 8 0.681 g C 5 H 8

20. Solving Problems with Particles  Sometimes in stoichiometry problems you will be asked to find the mass of a compound given a number of molecules, ions, atoms or particles  In these cases we make use of Avogadro's number.  You should know it to three significant figures: N A = 6.02 × 10 23  Amadeo Avogadro (1766-1856) never knew his own number!  Avogadro only originated the concept of this number, whose actual value was first estimated by Josef Loschmidt, an Austrian chemistry teacher, in 1895.

21. Example Problem with Particles  Given the following equation below, find how many molecules of BrF 5 form when 384 g Br 2 react with excess F 2 ? Br 2 (l) + 5F 2 (g) → 2 BrF 5 (l) Step 1: Gather the information 384 g Br 2 # of molecules of BrF 5 ?

22. Example Problem with Particles Br 2 (l) + 5F 2 (g) → 2 BrF 5 (l) Step 2: Plan your work A.Convert grams to moles B.Convert number of moles to moles of unknown using the mole ratio C.Take the moles of your answer and convert to molecules using Avogadro’s number

23. Example Problem with Particles Step 3: Calculate A.384 g Br 2 x 1 mol Br 2 = 2.40 mol Br 2 159.80 g Br 2 B.2. 40 mol Br 2 x 2 mol BrF 5 = 4.80 mol BrF 5 1 mol Br 2 C.4.80 mol BrF 5 x 6.022 x 10 23 molecules BrF 5 1 mol BrF 5 = 2.89 x 10 24 molecules of BrF 5

24. Important Things to Remember 1.Always write down the given information and what you are asked to find 2.Take the time to plan out how you are going to solve the problem In other words, what steps (calculations) do I have to take to get from point A to point B? 3.LABEL ! LABEL! LABEL!