1. Stoichiometry

2. Information Given by the Chemical Equation  The coefficients in the balanced chemical equation show the molecules and mole ratio of the reactants and products  Since moles can be converted to masses, we can determine the mass ratio of the reactants and products as well.  The mass of the reactants must be equal to the mass of the products (Law of Conservation of Mass)

3. Information Given by the Chemical Equation 2 CO + O 2  2 CO 2 2 moles CO = 1 mole O 2 = 2 moles CO 2 Since 1 mole of CO = 28.01 g, 1 mole O 2 = 32.00 g, and 1 mole CO 2 = 44.01 g 2(28.01) g CO + 1(32.00) g O 2 = 2(44.01) g CO 2

4. Mole - Mole Conversions  Examines the ratio of the number of moles of chemicals in a reaction.  Every balanced equation tells you directly the mole ratio of any 2 reactants, any 2 products, or any reactant with product.  SO - you may directly convert moles of any one material to moles of any other material in the balanced equation.

5. Moles reactant or product looking for Moles reactant or product given Mole ratio =

6. Mass - Mass Calculations  Relates the mass of reactants and products in a chemical reaction.  For many lab problems, you will need to know the gram amounts of both substances used in your reaction: the given substance and the new substance.

7. Moles reactant Moles product Grams reactant Grams product x molar mass reactant / molar mass reactant x molar mass product / molar mass product Mole ratio

8.  Solving stoichiometry problems can be done in one step or in separate steps using dimensional analysis.

9. Example ± Use the molar mass of the given and the desired quantity to convert the moles to mass Determine the number of grams of carbon dioxide produced when 48.0 g oxygen is reacted with carbon monoxide.

10. Separate step method  1. 48.0 g / 32.00 g/mol = 1.5 mol O 2  2. 1.5 mol x 2 mol CO 2 /1 mol O 2 = 3.0 mol CO 2  3. 3.0 mol CO 2 x 44.01 g/mol = 132 g CO 2

11. Practice  Calculate the number of grams of ammonia that are produced when 6.5 grams of hydrogen react with nitrogen.  Calculate the number of grams of aluminum reacting with oxygen that are needed to produce 24.5 grams of aluminum oxide.

12. Limiting and Excess Reactants  A reactant which is completely consumed when a reaction is run to completion is called a limiting reactant (all reactant is gone)  A reactant which is not completely consumed in a reaction is called an excess reactant (some reactant is left)

13. Limiting and Excess Reactants  Steps for determining the limiting reactant: 1. Write and balance the chemical equation. 2. Convert from mass (or moles) of each reactant to moles of product. 3. Identify the reactant that produces the least amount of product.

14. Example Step 1: Balance the equation Cu(s) + 2 AgNO 3 (aq) --> Cu(NO 3 ) 2 (aq) + 2Ag(s) Determine the limiting reactant when 3.5 grams of copper is added to a solution containing 6.0 grams of silver nitrate.

15. Example Determine the limiting reactant when 3.5 grams of copper is added to a solution containing 6.0 grams of silver nitrate. Step 2: Calculate the mass of product formed by each reactant. Choose either Cu(NO 3 ) 2 or Ag. Convert 3.5 grams Cu to moles (grams) Ag Convert 6.0 grams AgNO 3 to moles (grams) Ag

16. Example Step 3: Identify which reactant produces the least amount of product. 3.5 g Cu -- 0.11 mol or 12 g Ag 6.0 g AgNO 3 – 0.035 mol or 3.8 g Ag AgNO 3 is the limiting reactant.

17. Practice  6.5 grams of methane gas (CH 4 ) is burned in the presence of 45 grams of oxygen gas. Determine the limiting reactant.

18. Percent Yield  Most reactions do not go to completion  The amount of product made in an experiment is called the actual yield  The maximum amount of a product that can be made when the limiting reactant is completely consumed is called the theoretical yield  The percentage of the theoretical yield that is actually made is called the percent yield

19. “ ” Percent Yield = x 100 % Actual Yield Theoretical Yield

20. Example ³ Divide the actual yield by the theoretical yield, then multiply by 100% The actual yield of MgO is 6.51 g The theoretical yield of MgO is 8.12 g Percent Yield is 6.51 g/8.12 g x 100 % = 80.2 % 4.92 grams of magnesium is burned in oxygen. The actual yield of magnesium oxide is 6.51 grams. The stoichiometric calculations give a theoretical yield of 8.12 grams of magnesium oxide. What is the percent yield?

21. Practice  5.15 g of potassium is heated in excess chlorine gas. The actual yield of 7.92 g of potassium chloride. What is the percent yield?  Hint: calculate the theoretical yield using the balanced equation.

22. Calculating excess reactant  Subtract the theoretical yield (in moles) of the limiting reagent from the theoretical yield of the excess reactant.  Then work backward from there to get the mass of the excess reactant.

23. Example  Determine the excess reactant when 3.2 moles oxygen reacts with 4.0 moles of carbon monoxide.  Theoretical Yields:

24. Example  Theoretical Yield O 2 (ER) - Theoretical Yield CO (LR) = 6.4 moles – 4.0 moles = 2.4 moles CO 2  Then use 2.4 moles CO 2 to calculate the amount of O 2 left over. 1 mole O 2 2 moles CO 2 = 1.2 mole O 2 2.4 moles CO 2 x

25. Practice  Determine the amount of excess reactant when 3.5 grams of copper is added to a solution containing 6.0 grams of silver nitrate.