Stoichiometry World of Chemistry: Chapter 9

Transcription

1 Stoichiometry World of Chemistry: Chapter 9

2 Chocolate Chip Cookies!! 1 cup butter 1/2 cup white sugar 1 cup packed brown sugar 1 teaspoon vanilla extract 2 eggs 2 1/2 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon salt 2 cups semisweet chocolate chips Makes 3 dozen How many eggs are needed to make 3 dozen cookies? How much butter is needed for the amount of chocolate chips used? How many eggs would we need to make 9 dozen cookies? How much brown sugar would I need if I had 1 ½ cups white sugar?

3 Cookies and Chemistry Huh!?!? O Just like chocolate chip cookies have recipes, chemists have recipes as well. O Instead of calling them recipes, we call them reaction equations. O Furthermore, instead of using cups and teaspoons, we use moles. O Lastly, instead of eggs, butter, sugar, etc. we use chemical compounds as ingredients.

4 Chemistry Recipes O Looking at a reaction tells us how much of something you need to react with something else to get a product (like the cookie recipe). O Be sure you have a balanced reaction before you start! O Example: 2 Na + Cl 2 2 NaCl O This reaction tells us that by mixing 2 moles of sodium with 1 mole of chlorine we will get 2 moles of sodium chloride O What if we wanted 4 moles of NaCl? 10 moles? 50 moles?

5 Practice O Write the balanced reaction for hydrogen gas reacting with oxygen gas. 2 H 2 + O 2 2 H 2 O O How many moles of reactants are needed? O What if we wanted 4 moles of water? O What if we had 3 moles of oxygen, how much hydrogen would we need to react and how much water would we get? O What if we had 50 moles of hydrogen, how much oxygen would we need and how much water produced?

6 Mole Ratios O These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical O Example: How many moles of chlorine is needed to react with 5 moles of sodium (without any sodium left over)? 2 Na + Cl 2 2 NaCl 5 moles Na 1 mol Cl 2 2 mol Na = 3 moles Cl 2

7 Mole-Mole Conversions O How many moles of sodium chloride will be produced if you react 2.6 moles of chlorine gas with an excess (more than you need) of sodium metal? 2 Na + Cl 2 2 NaCl 2.6 moles Cl 2 2 mol NaCl = 5.2 moles NaCl 1 mol Cl 2

8 Mole-Mass Conversions O Most of the time in chemistry, the amounts are given in grams instead of moles. O We still go through moles and use the mole ratio, but now we also use molar mass to get to grams O Example: How many grams of chlorine are required to react completely with 5.00 moles of sodium to produce sodium chloride? 2 Na + Cl 2 2 NaCl 5.00 mol Na 1 mol Cl g Cl 2 2 mol Na 1 mol Cl 2 = 177g Cl 2

9 Practice O Calculate the mass in grams of iodine required to react completely with 0.50 moles of aluminum. O Write balanced equation: 2 Al + 3 I 2 2 AlI mol Al 3 mol I g I 2 2 mol Al 1 mol I 2 = g I g I 2

10 Mass-Particle O We can also start with mass and convert to number of particles (atoms or molecules) of product or another reactant! O Calculate the number of molecules of ethane (C 2 H 6 ) needed to produce 10.0 g of water 2 C 2 H O 2 4 CO H g H 2 O 1 mol H 2 O 2 mol C 2 H x molecules C 2 H g H 2 O 6 mol H mol C 2 H 6 = 1.11 x molecules of C 2 H 6

11 Practice O Calculate how many molecules of oxygen are required to make 10.0 g of aluminum oxide. 4 Al + 3 O 2 2 Al 2 O g Al 2 O 3 1 mol Al 2 O 3 3 mol O x molecules O g Al 2 O 3 2 mol Al 2 O 3 1 mol O 2 = 8.86 x molecules O 2

12 Mass-Mass Conversions O However, we most often are given a starting mass and solve for the mass of a product we will get (called theoretical yield) or how much of another reactant we need to completely react with it (no leftover ingredients!)

13 Mass-Mass Conversion O Ex. Calculate how many grams of ammonia are produced when you react 2.00 g of nitrogen with excess hydrogen. N H 2 2 NH g N 2 1 mol N 2 2 mol NH g NH g N 2 1 mol N 2 1 mol NH 3 = 2.43 g NH 3

14 Practice How many grams of calcium nitride are produced when 2.00 g of calcium reacts with an excess of nitrogen? 3 Ca + N 2 Ca 3 N g Ca 1 mol Ca 1 mol Ca 3 N g Ca 3 N g Ca 3 mol Ca 1 mol Ca 3 N 2 = 2.47 g Ca 3 N 2

15 Percent Yield O The calculated amount of product is called the theoretical yield. It is the amounts of products predicted to form from the amounts of reactants used.

16 Percent Yield O Actually, the amount of product predicted is seldom obtained. O O One reason for this is side reactions (other reactions that consume one or more of the reactants or products). The amount of product actually obtained is called the actual yield. It must come from doing an experiment.

17 Percent Yield The actual yield is often compared to the theoretical yield. This comparison expressed as a percent is called percent yield. (Actual yield should never be greater than theoretical yield.) percent yield = actual yield theoretical yield 100%

18 Percent Yield O Example: If a reaction actually produced 6.63 grams of nitrogen instead of the predicted 10.6 grams, what is the percent yield? % yield = actual = 6.63 g x 100% = 62.5% theo g O Example: From starting reactant amounts, the theoretical yield was calculated to be 55.5 g of a product. What is the actual yield if the percent yield was 89.0%? 89.0% = x x 100% 55.5 g 0.89 = x 55.5 g x = 49.4 g