Stoichiometric Calculations

1. Stoichiometric Calculations

2. Chocolate Chip Cookies • 2.25 cups flour • 8 Tbsp butter • 0.5 cups shortening • 0.75 cups sugar • 0.75 cups brown sugar • 1 tsp salt • 1 tsp baking soda • 1 tsp vanilla • 0.5 cups Egg Beaters • 2 cups chocolate chips

3. How much? Of what? What units? Chocolate Chip Cookies • 2.25cupsflour • 8Tbspbutter • 0.5cupsshortening • 0.75cupssugar • 0.75cupsbrown sugar • 1tspsalt • 1tspbaking soda • 1tspvanilla • 0.5cupsEgg Beaters • 2cups Chocolate chips

4. How much? Of what? Chocolate Chip Cookies • 2.25flour • 8butter • 0.5shortening • 0.75sugar • 0.75brown sugar • 1salt • 1baking soda • 1vanilla • 0.5Egg Beaters • 2 Chocolate chips

5. How much? Of what? What units? Chocolate Chip Cookies • 2.25cupsflour • 8Tbspbutter • 0.5cupsshortening • 0.75cupssugar • 0.75cupsbrown sugar • 1tspsalt • 1tspbaking soda • 1tspvanilla • 0.5cupsEgg Beaters • 2cups Chocolate chips

6. Get on with it! What does this have to do with CHEMISTRY?

7. (177ºC) 1 batch of chocolate chip cookies! 2.25 cups flour + 8 Tbsp butter + 0.5 cups shortening + 0.75 cups sugar + 0.75 cups brown sugar + 1 tsp salt + 1 tsp baking soda + 1 tsp vanilla + 0.5 cups Egg Beaters + 2 cups Chocolate chips unit substance coefficient (a synthesis reaction)

8. Welcome to STOICHIOMETRY

9. The coefficients in chemical equations represent numbers of molecules or formula units, not masses of molecules or formula units. • When a reaction is to be run in a laboratory or chemicalplant, the amount of substances needed cannot be determined by counting molecules directly. • Counting is always done by weighing. • We will see how chemical equations can be used to determine the masses of reacting chemicals.

10. 3NaOH (aq) + H3PO4 (aq) Na3PO4 (aq) + 3 H2O • In the above reaction, the coefficients give us the ratiosby moles of the reactants and products. • The coefficients tell us that to make 1 mol of Na3PO4 from 1 mol of H3PO4, we must also use 3 mol of NaOH. • We don’t however, have to carry out the reaction with these actual numbers of moles. • Whatever quantities we choose must be in the proportionsset by the coefficients. • Regardless of the scale of the reaction, the coefficients of a chemical equation give the ratio in which the moles of one substance react with or produce molesof another.

11. 3NaOH (aq) + H3PO4 (aq) Na3PO4 (aq) + 3 H2O • We can look at this equation as a calculating tool, because its coefficients give us stoichiometric equivalencies between the substances involved. • For example, from the above equation we can generate the following equivalencies: • 3 mol NaOH ↔ 1 mol H3PO4 • 3 mol NaOH ↔ 1 mol Na3PO4 • 3 mol NaOH ↔ 3 mol H2O • 1 mol H3PO4↔ 1 mol Na3PO4 • 1 mol H3PO4↔ 3 mol H2O • Any of these can be used to construct conversion factors called mole ratios for stoichiometric calculations.

12. Example: How many moles of sodium phosphate, Na3PO4, can be made from 0.240 mol of NaOH by the following reaction? 3NaOH (aq) + H3PO4 (aq) Na3PO4 (aq) + 3H2O

13. In practical work, a chemist is often confronted by a question such as the following. • “If I start with so many grams of reactant A, how many grams of reactant B ought I use, and how many grams of a particular product should be produced?” • Notice that the question concerns grams, not moles, for the practical reason that masses in grams are delivered by laboratory balances. • The coefficients of the desired reaction, however, know nothing about grams, only about relative numbers of moles. • If we know two facts, namely, the balanced equation and the mass of any substance in it, we can calculate the required or expected mass of any other substance in the equation.

14. Example: Portland cement is a mixture of the oxides of calcium, aluminum, and silicon. The raw material for its calcium oxide is calcium carbonate, which occurs as the chief component of a natural rock, limestone. When calcium carbonate is strongly heated it decomposes by the following reaction. One product CO2, is driven off to leave the desired CaO as the only other product. • CaCO3 (s) CaO (s) + CO2 (g) • A chemistry student is to prepare 1.50 x 102 g of CaO in order to test a particular “recipe” for portland cement. How many grams of CaCO3 should be used, assuming that all will be converted? heat

15. Calculating Masses of Reactants and Products in Chemical Reactions Balance the equation for the reaction. Convert the known mass of the reactant or product to moles of that substance. Use the balanced equation to set up the appropriate mole ratios. Use the appropriate mole ratios to calculate the number of moles of the desired reactant or product. Convert from moles back to grams if required by the problem.

16. Example: One of the most spectacular reactions of aluminum, the thermite reaction, is with iron oxide, Fe2O3 , by which metallic iron is made. So much heat is generated that the iron forms in the liquid state. The equation is • Al (s) + Fe2O3 (s) Al2O3 (s) + Fe (l) • A certain welding operation, used over and over, requires that each time at least 86.0 g of Fe be produced. What is the minimum mass in grams of Fe2O3 that must be used for each operation? Calculate also how many grams of aluminum are needed.

17. The Wisdom of Gallagher Why are there Interstate Highways in Hawaii? Why are there floatation devices under plane seats instead of parachutes? Why do we drive on parkways and park on driveways? Why do hot dogs come ten to a package and hot dog buns only eight?

18. Hot Dogs in the News Takeru Kobayashi of Japan downed 44½ hot dogs in 12 minutes. One hot dog = one hot dog + one bun. WHAT IF… Mr. Kobayashi didn’t do his math correctly. He bought 5 packs of hot dogs (10 per package) and 5 packs of hot dog buns (8 per package). How many hot dogs (according to the official formula) could he have eaten? Source: CNN.com

19. Hot Dogs in the News One hot dog = one hot dog + one bun. WHAT IF… Mr. Kobayashi didn’t do his math correctly. He bought 5 packs of hot dogs (10 per package) and 5 packs of hot dog buns (8 per package). How many hot dogs (according to the official formula) could he have eaten? Excess 5 hot dog packs 10 hot dogs 50 hot dogs x = 1 hot dog pack Limiting 40 possible hot dogs 5 bun packs 8 buns 40 buns x = 1 bun pack Source: CNN.com

20. Let’s Revisit the Cookies (again)… For 1 batch: • In my pantry, I have: • 5.5 cups of flour • 16 Tbsp of butter • lots of everything else • 2.25 cups flour • 8 Tbsp butter • 0.5 cups shortening • 0.75 cups sugar • 0.75 cups brown sugar • 1 tsp salt • 1 tsp baking soda • 1 tsp vanilla • 0.5 cups Egg Beaters • 12 oz. Chocolate chips How many batches of cookies can I make?

21. LIMITING Let’s Revisit the Cookies (again)… For 1 batch: How many batches of cookies can I make? 5.5 cups • 2.25 cups flour • 8 Tbsp butter • 0.5 cups shortening • 0.75 cups sugar • 0.75 cups brown sugar • 1 tsp salt • 1 tsp baking soda • 1 tsp vanilla • 0.5 cups Egg Beaters • 12 oz. Chocolate chips 16 Tbsp EXCESS 1 batch cookies 5.5 c flour x = 2.25 c flour 2.4 batches 1 batch cookies 16 Tbsp butter x = 8 Tbsp butter 2.0 batches

22. 1 batch cookies 5.5 c flour x = 2.4 batches of cookies 2.25 c flour 1 batch 16 Tbsp butter x = 2.0 batches of cookies 8 Tbsp butter Now I Want to Bake a Cake! But do I have all the ingredients I need? How much flour do I have left after baking all those cookies? SOME FLOUR LEFT OVER… GONE! 2.25 cups flour 4.5 cups flour used 2.0 batches x = 1 batch cookies 5.5 cups – 4.5 cups = 1.0 cups left

23. Limiting Reactant Calculations • Many times a chemist will mix reactants in a mole ratio that does not agree with the coefficients of the equation. • Some reactions proceed better when one reactant is in stoichiometric excess, for example. • One such reaction is the preparation of ammonia, NH3, from its elements. • N2 (g) + 3H2 (g) 2NH3 (g) • Suppose a chemist mixed 1.00 mol of N2 with 5.00 mol of H2. What is the maximum number of moles of product that could form?

24. N2 (g) + 3H2 (g) 2NH3 (g) • Note the coefficients tell us that 1 mol of N2 consumes 3 mol of H2. • 1 mol N2 ↔ 3 mol H2 • But 5 mol of H2 was used, not 3, so there will be 2 mol of H2left over. • Once the 1 mol of N2 taken is consumed, no additional NH3 can form. • Therefore, the reactant that is completely consumed limits the amount of product that forms, so it is called the limitingreactant. • In this reaction, N2 is the limiting reactant, it limited the amount of NH3 that was formed.

25. Example: In an industrial process for making nitric acid, the first step is the reaction of ammonia with oxygen at high temperature in the presence of a platinum gauze. Nitrogen monoxide forms as follows: • 4NH3 + 5O2 4NO + 6H2O • How many grams of nitrogen monoxide can form if a mixture of 30.00 g of NH3 and 40.00 g of O2 is taken initially? How many grams of the excess reactant is left over?

26. Theoretical Yield and Percentage Yield

27. In most experiments designed for chemical synthesis, the amount of a product obtained falls short of the calculated maximum amount. • Losses occur for several reasons. • Some are mechanical, such as materials sticking to glassware. • But one of the most common causes of obtaining less than the stoichiometric amount of a product is the occurrence of a competing reaction. • It produces a by product, a substance made by a reaction that competes with the main reaction.

28. The actual yield of desired product is simply how much is obtained or isolated, stated in mass units. • The theoretical yield is the amount of product formed when the limiting reactant is completely consumed and no losses occur. • When less than the theoretical yield of product is obtained, chemists generally calculate the percentage yield or percent yield to describe how well the reaction went. • The percent yield is the actual yield calculated as a percentage of the theoretical yield.

29. Example: A chemist set up a synthesis of phosphorus trichloride by mixing 12.0 g P with 35.0 g Cl2 and obtained 42.4 g of PCl3. Calculate the percent yield of this compound. The equation for the main reaction is: • 2P (s) + 3Cl2 (g) 2PCl3 (l)