Stoichiometry The word stoichiometry derives from two Greek

Stoichiometry ¡ The word stoichiometry derives from two Greek words: stoicheion (meaning "element") and metron (meaning "measure").

Stoichiometry ¡ Stoichiometry deals with calculations about the masses (sometimes volumes) of reactants and products involved in a chemical reaction. Bring your calculator along for the ride! ¡

¡ Stoichiometry is the branch of chemistry and chemical engineering that deals with the quantities of substances that enter into, and are produced by, chemical reactions. ¡ Example: l 16 g of methane(CH 4) reacts with 64 g of oxygen(O 2) l 44 g of carbon dioxide (CO 2), and 36 g of water (H 2 O) are formed as reaction products.

Things to remember: • Moles & particles – everything can be compared in moles on a macroscopic scale and particles on a microscopic scale. • Stoichiometry uses moles to make conversions in a balanced equation. • Four basic problem types: page 275

Why again do we use moles? ü ü Using moles is a method to count a large number of microscopic particles. Also allows us to measure substances in a practical manner (using a balance to measure grams) in the lab or factory.

Things to remember: ¡ Particles – 6. 02 x 1023 particles (atoms, formula units, ions, molecules, etc. ) ¡ Mass/Grams – these can be converted to moles by dividing by the molar mass (found on the periodic table) ¡ Volume/liters – one mole of a gas contains 22. 4 liters.

Things to remember: Ø Equations must be balanced with the correct formulas for reactants and products. Example: Methane reacts with oxygen to form carbon dioxide and water. ___CH 4 +___O 2 ___CO 2 + ___H 2 O

Moles to Moles: Ratio ¡ In a balanced equation, the mole ratios can be converted to any other equal ratio value. ¡ N 2 + 3 H 2 2 NH 3 ¡ In the reactants, the ratio of nitrogen to hydrogen is: l 1: 3 or 2: 6 or 1. 5: 4. 5 or 3. 3: 9. 9 ¡ Any value can be used and the ratio will be a constant for an equation.

Mole to Mole problem ¡ The reaction of 8 moles aluminum with oxygen produces aluminum oxide. l l l (1) How many moles of O 2 are needed? (2) How many moles of aluminum oxide are formed? Write a balanced equation. from the balanced equation, what is the mole ratio for aluminum to oxygen and aluminum to aluminum oxide? What is the mole ratio of aluminum to oxygen and aluminum to aluminum oxide when you start with 8 moles of aluminum?

Mole to Mole problem ¡ 4 Al(s) + 3 O 2(g) 2 Al 2 O 3(s) ¡ Ratio of Al to oxygen is 4: 3 l l ¡ Do the “molar math” Set up a ratio: Ratio of Al to aluminum oxide is 4: 2 l l Do the “molar math” Set up a ratio:

¡ Work Problems: Page 295; 1, 2, 9, 10, 11

Mass to Mass Conversions 1. Convert the grams of the given substance to moles. ¡ 2. Compare the mole ratios. ¡ 3. Convert moles of unknown to grams. ¡

Mass to Mass Problem ¡ ¡ Calculate the number of grams of ammonia produced by the reaction of 6. 85 grams of nitrogen with excess hydrogen. Balanced equation: ¡ N 2 + 3 H 2 2 NH 3

N 2 + 3 H 2 2 NH 3 ¡ Knowns l l ¡ Mass of nitrogen = ¡ 6. 85 g of N 2 Mole ratio of N 2 to NH 3 = ¡ 1 : 2 Unknowns l l l Moles of N 2 = ¡ 6. 85 g / 28. 02 g/mol = 0. 244 mol Mole ratio of N 2 to NH 3 based on unknown = ¡ 0. 244 mol N 2 x 2 = 0. 489 mol NH 3 Mass of ammonia = ¡ 0. 489 mol NH 3 x 17. 03 g/mol = 8. 33 grams NH 3

Summary of steps 1. Balance the equation if needed. ¡ 2. Convert the grams, particles or volume of a substance into moles by dividing by your conversion factor. ¡ 3. Compare the mole ratios for the conversion substances using the balanced equation. ¡ 4. Convert the moles into what is required; grams, particles or volume by multiplying by your conversion factor. ¡

Your Path to Success! Convert From… Convert to Moles “Molar Math” Set up particles Mole ratio 6. 02 X 1023 mass Molar mass Gas volume 22. 4 Liters Convert to …. 6. 02 X 1023 particles Molar mass Moles mass Cross multiply to find moles 22. 4 Liters Gas volume

Let’s do one more!!! Nitrogen monoxide and oxygen gas combine to form the brown gas nitrogen dioxide that is found in smog. ¡ How many grams of nitrogen dioxide are produced when 8. 42 liters of nitrogen monoxide reacts with the oxygen in the atmosphere at STP. ¡

Smog Problem ¡ 1. Write a balanced equation ¡ 2. Convert 8. 42 liters of NO at STP to moles of NO ¡ 3. Compare the mole ratios ¡ 4. Convert moles of NO 2 to grams of NO 2

Smog Problem ¡ ¡ 2 NO(g) + O 2(g) 2 NO 2(g) Since the mole ratio is 1: 1, you will form 0. 376 moles of nitrogen dioxide.

Stoichiometry Hints 1. Balance the equation first and if you can’t balance the equation, check to make sure that your formulas are correct. 2. Anything that will allow you to calculate moles can be used as the basis for solving your problem. 3. Don’t forget to use the proportions given in the balanced equation!!!

Plan of Attack 1. Convert from the known g, L, or particles to moles of known by dividing by the conversion factor needed. 2. Compare the mole ratios of known and unknown to find the moles of unknown. 3. Convert from the moles of unknown by multiplying by the conversion factor to find g, L or particles of unknown.

¡ Problems: page 295; 3, 13 -15

Limiting Reagents Limiting reagents deter- mine or limit the amount of product that can be formed in a reaction. ¡ The reaction occurs until the limiting reagent is used up! ¡ Whatever reactant(s) is not used up will be the excess reagent. ¡

Limiting Reagents ¡ It is important again to convert everything into moles and compare mole ratios to determine which substance is used up and what substance is in excess.

Na. Cl - limiting reagent problem ¡ ¡ ¡ When sodium reacts with chlorine gas, sodium chloride is produced. Suppose that you react 4. 5 moles of sodium with 3. 1 moles of chlorine. 1. What is the limiting reagent? 2. How many moles of salt are produced?

Sodium Chloride Problem ¡ ¡ 1. Balanced equation: l 2 Na + Cl 2 2 Na. Cl 2. Mole ratio is 2 Na to 1 Cl 2. l To use all 4. 5 moles of sodium it 2. 25 would take ____ moles of chlorine. (We have 3. 1 moles of Cl 2) l To use all 3. 1 moles of chlorine it 6. 2 would take ____ moles of sodium. (We have 4. 5 moles of Na) l What is the limiting reagent?

Sodium Chloride Problem Since the sodium limits the reaction, the number of moles of sodium chloride produced will also be 4. 5 moles. ¡ How much of the chlorine will be in excess? ¡ l Since you will use 2. 25 moles of chlorine the excess will be 0. 85 moles (the difference between what you use and what you have left over)

Limiting Reagent Problem ¡ 3 NH 4 NO 3 + Na 3 PO 4 (NH 4)3 PO 4 + 3 Na. NO 3 ¡ Assuming we started with 30 grams of ammonium nitrate and 50 grams of sodium phosphate, determine: l a)Which of the reagents is the limiting reagent? l b)What is the maximum amount of each product that can be formed? l c)How much of the other reagent is left over after the reaction is complete?

Percent Yield Ø On paper we can predict amounts of reactants and products. This is theoretical yield of a reaction. Ø In the lab, reality strikes! Labs do not give 100% theoretical yield. The actual yield is usually something less. Ø What would cause a lab to not have 100% yield? Keep in mind some of the lab processes that we have done involving measurements.

Percent Yield ¡ It a lot like a batting average, free throw percentage or your grade on a test.

Percent Yield – A Problem Ø Ø Ø Methane combusts with oxygen to produce water and carbon dioxide. What is theoretical yield of water in this reaction if 23. 4 grams of methane reacts in excess oxygen? What is the percent yield if you produce 50. 0 grams of water?

Percent Yield – A Problem ¡ CH 4(g) + 2 O 2(g) 2 H 2 O(g) + CO 2(g) ¡ 23. 4 g 1. 46 moles methane ¡ ¡ 1 methane to 2 waters means 2. 92 moles of water 52. 6 grams of water

Percent Yield – A Problem What is the percent error then? ¡ You are right - it’s the left over amount. ¡ 4. 9% ¡

Percent Error

¡ Homework problems p. 296; 22 -25, 27, 29, 30

The Hamburger

First the hamburger analogy ¡ ¡ ¡ My recipe for a bacon double cheeseburger is: 1 hamburger bun 2 hamburger patties 2 slices of cheese 4 strips of bacon Ketchup & mustard

Based on this recipe: 1. a. b. c. d. If I have five bacon double cheeseburgers: How many hamburger buns do I have? How many hamburger patties do I have? How many slices of cheese do I have? How many strips of bacon do I have?

How many bacon double cheeseburgers can you make if you start with: 1 bun, 2 patties, 2 slices of cheese, 4 strips of bacon a. Basic burger equation (1 : 2 : 4) 2 buns, 4 patties, 4 slices of cheese, 8 strips of bacon Double quantities (2 : 4 : 8) 2 buns, 3 patties, 4 slices of cheese, 8 strips of bacon Limiting Reagent (2 : 3 : 4 : 8) 2 buns, 4 patties, 4 slices of cheese, 10 strips of bacon Excess Reagent (2 : 4 : 10)

1. If you had fixings for 100 bacon double cheeseburgers, but when you were cooking you ruined 10 of them. What percentage of the bacon double cheeseburgers do you actually make?