1. Stoichiometry Chapter 9 ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

2. Do you remember? What is a Mole? A mole (mol) is an amount of substance that contains the same number of particles as there are atoms in 12 g of carbon - 12. To four significant figures, there are 6.022 x 10 23 atoms in 12 g of carbon- 12. Thus a mole of natural carbon is the amount of carbon that contains 6.022 x10 23 carbon atoms. The number 6.022 x 10 23 is often called Avogadro’s number. ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

3. Stoichiometry Greek stoicheion ( to measure the elements) The proportional relationship between two or more substances during a chemical reaction Branch of chemistry that deals with quantities of substances in chemical reactions. ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

4. History of Stoichiometry Joseph Black (1728-1799) and his student Daniel Rutherford (1749-1819) quantitatively studied the following reaction in both the forward and reverse direction to measure the redistribution of mass during the reaction: CaCO 3 (s) → CaO(s) + CO 2 (g) Their findings laid the groundwork for modern stoichiometry. Daniel Rutherford ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

5. Balanced Equations Show Proportions A recipe for Kool-aid A recipe for Kool-aid 1 cup sugar 1 package of Kool-aid 2 quarts of water How do you make 4 quarts? A recipe for Water A recipe for Water 2H 2 + O 2 → 2H 2 O The coefficients show the relative amounts (ratio) in equations. These can be expressed in moles, molecules, or ions. The coefficients show the relative amounts (ratio) in equations. These can be expressed in moles, molecules, or ions. ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

6. 4 Questions There are 4 questions you need to ask yourself when solving problems, whether in real life or in class. 1. What do you know? 2. Where do you want to go? 3. How do you get there? 4. Does it make sense? ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

7. Mole Ratio is the Key to bridge the gap when converting from one substance to another. The mole to mole ratio is represented by The mole to mole ratio is represented by the coefficients in a balanced equation. the coefficients in a balanced equation. N 2 + 3H 2 → 2NH 3 ___mol known x mol unknown = ___mol unknown ___mol known x mol unknown = ___mol unknown mol known mol known Always start with what you know and make sure the units cancel !!! Always start with what you know and make sure the units cancel !!! ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

8. Sample Problem #1 Mole to Mole N 2 + 3H 2 → 2NH 3 What do we know? Where do we want to go? 312 moles NH 3 (known) _?_moles of H 2 (unknown) What do we know? Where do we want to go? 312 moles NH 3 (known) _?_moles of H 2 (unknown) How do we get there? 312 mol NH 3 x 3mol H 2 = _____mol H 2 2 mol NH 3 312 mol NH 3 x 3mol H 2 = _____mol H 2 2 mol NH 3 Does it make sense? yes 468 ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

9. Getting into Moles and getting out of Moles Substances are usually measured by mass (or volume). Substances are usually measured by mass (or volume). You may need to convert between units for mass, volume, and mole. You may need to convert between units for mass, volume, and mole. Steps to convert units Steps to convert units 1. Change units given into moles 2. Use mole ratio to determine moles of desired substance. 3. Change moles to whatever unit you need. ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

10. Molar Mass for Elements Use Atomic Mass from the Periodic Table for a conversion factor. Use Atomic Mass from the Periodic Table for a conversion factor. g elements g elements 1mol element 1mol element Molar Mass Calculation for Carbon Molar Mass Calculation for Carbon What do we know? Where do we want to go? What do we know? Where do we want to go? 0.55 g C ? mol C 0.55 g C ? mol C How do we get there? How do we get there? 0.55 g C x 1 mol C__ = 0.05 mol 0.55 g C x 1 mol C__ = 0.05 mol 12.01 g C 12.01 g C Does it make sense? yes ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

11. Sample Problem #2 Mass to Mass N 2 + 3H 2 → 2NH 3 What do we know? Where do we want to go? What do we know? Where do we want to go? 1221 g H 2 ? mass of NH 3 1221 g H 2 ? mass of NH 3 How do we get there? We use a Mole to Mole Ratio ( from the balanced equation) and Molar Mass Conversions (from the periodic table) and Molar Mass Conversions (from the periodic table) 1221 g H 2 x 1 mol H 2 x 2 mol NH 3 x 17.04 g NH 3 = __ g 2.02 g H 2 3 mol H 2 1 mol NH 3 1221 g H 2 x 1 mol H 2 x 2 mol NH 3 x 17.04 g NH 3 = __ g NH 3 2.02 g H 2 3 mol H 2 1 mol NH 3 Does it make sense? Yes Stoich Video Guest mass-mass Stoich Video Guest mass-mass 6867 ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

12. Volume-Volume Problems Conversions of volume to mass or mass to volume use Conversions of volume to mass or mass to volume use 1. Density as conversion factor for liquids 2. Molar volume of a gas, for gases at STP 3. Concentration of a solution, for an aqueous solution Don’t forget the units you want to cancel should be on the bottom of your conversion factor. Don’t forget the units you want to cancel should be on the bottom of your conversion factor. ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

13. Sample Problem #3 Volume to Volume POCl 3 (l) + 3 H 2 O → H 3 PO 4 (l) + 3 HCl(g) What do we know? Where do we want to go? 56 mL POCl 3 ___mL H 3 PO 4 1.67g/mL = density of POCl 3 153.32g/mol = molar mass of POCl 3 What do we know? Where do we want to go? 56 mL POCl 3 ___mL H 3 PO 4 1.67g/mL = density of POCl 3 153.32g/mol = molar mass of POCl 3 1.83 g/mL = density H 3 PO 4 98.00 g/mol = molar mass H 3 PO 4 1.83 g/mL = density H 3 PO 4 98.00 g/mol = molar mass H 3 PO 4 How do we get there? 56 mL POCl 3 x 1.67 g POCl 3 x 1 mol POCl 3 x 1 mol H 3 PO 4 x 98.00 g H 3 PO 4 x 1 mL H 3 PO 4 = _____ mL H 3 PO 4 1 mL POCl 3 153.32 g POCl 3 1 mol POCl 3 1 mol H 3 PO 4 1.83 g H 3 PO 4 Does it make sense? yes 33 ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

14. Particle Problems Use Avogadro’s number as conversion factor. Use Avogadro’s number as conversion factor. 6.022 x 10 23 particles 6.022 x 10 23 particles 1mole 1mole ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

15. Sample Problem #4 C 5 H 12 (l) → C 5 H 8 (l) + 2H 2 (g) What do we know? Where do we want to go? 1.89 x 10 24 molecules C 5 H 12 ? grams C 5 H 8 6.022 x 10 23 molecules/mol 68.13 g/mol molar mass C 5 H 8 1 mol C 5 H 12 = 1 mol C 5 H 8 How do we get there? 1.89 x 10 24 molecules C 5 H 12 x 1mol C 5 H 12 ____ x 1 mol C 5 H 8 x 68.13g C 5 H 8 = _214 g C 5 H 8 6.022 x 10 23 molecules 1 mol C 5 H 12 1 mol C 5 H 8 Does it Make sense? Does it Make sense? Yes Yes ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

16. Study Guide General Steps Convert the given unit to moles of the first substance. Convert the given unit to moles of the first substance. Convert moles of the first substance to mole of the second substance using the molar ratio derived from the formula for the compound. Convert moles of the first substance to mole of the second substance using the molar ratio derived from the formula for the compound. Convert moles of the second substance to the desired units of the second substance. Convert moles of the second substance to the desired units of the second substance. ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

17. Limiting Reactants and Percentage Yield 9-2 Chem Teachers Having Fun with Stoich Chem Teachers Having Fun with Stoich ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

18. Reactants Limiting Reactant Limiting Reactant ingredient that limits the amount of a product that can form. ingredient that limits the amount of a product that can form. The reactant that is used up first The reactant that is used up first Often the most expensive ingredient Often the most expensive ingredient Excess Reactant Excess Reactant Ingredient that there is more than enough of Ingredient that there is more than enough of Reactant that is left over after the reaction Reactant that is left over after the reaction ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

20. Yields Theoretical Yield Theoretical Yield Maximum quantity of a product that a reaction could theoretically make Maximum quantity of a product that a reaction could theoretically make Calculated based on limiting reactant Calculated based on limiting reactant The amount that should be made if it was perfect conditions, and a perfect reaction. Stoichiometry is what determines this amount. The amount that should be made if it was perfect conditions, and a perfect reaction. Stoichiometry is what determines this amount. ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

22. Actual Yield Actual Yield Determined experimentally Determined experimentally Mass of products actually formed usually less than expected (theoretical yield) Mass of products actually formed usually less than expected (theoretical yield) Why? Why? Many reactions don’t completely use limiting reactant Many reactions don’t completely use limiting reactant Product often needs to be purified and some is lost in this process. Product often needs to be purified and some is lost in this process. Other reaction (side reactions) are going on and using reactants Other reaction (side reactions) are going on and using reactants ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

23. Percentage Yield Percentage Yield Ratio of actual yield to theoretical yield. Ratio of actual yield to theoretical yield. ___Actual yield___ x 100 = percent yield Theoretical yield Theoretical yield Need more help check out this website. bhs.smuhsd.org/.../briolgastoich.htm l Need more help check out this website. bhs.smuhsd.org/.../briolgastoich.htm l bhs.smuhsd.org/.../briolgastoich.htm l bhs.smuhsd.org/.../briolgastoich.htm l ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright