# Stoichiometry It’s All Greek to Me TEKS 8.E and 9.B.

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Stoichiometry It’s All Greek to Me TEKS 8.E and 9.B

Stoichiometry u Greek for “measuring elements” u The calculations of quantities in chemical reactions based on a balanced equation. u We can interpret balanced chemical equations several ways.

How is it Used in Everyday Life? u Recipes (well, they make a good analogy) u Chemists and engineers take a balanced chemical equation for a product discovered in a research lab and scale it up so thousands, millions, even billions of pounds of good stuff can be produced for consumers For example ~ manufacturing of everything from medicines, to agriculture products, to gasoline, to plastic for your iPhone

More on Why I Have to Learn This u It is also important to know how efficient a manufacturing or other chemical process is (better the efficiency, the lower the cost, less the waste, etc.) u Another reason is to be sure you have enough – but not too much of a reactant (raw material or ingredient) - to be sure you can make the amount desired. This too ties to money, waste, etc.

The Arithmetic of Equations u TEKS OBJECTIVE 8.E: TLW perform stoichiometric calculations, including determination of mass relationships between reactants and products, calculation of limiting reagents, and percent yield

Lesson of the Cookies u When baking cookies, a recipe is usually used u Thus, a recipe is much like a balanced chemical equation. HOW? u HOW are recipes and balanced chemical equations different?

More Chemistry Lessons from Cookies u If you want to make more cookies you simply determine the ratio you need to apply to original ingredient amounts (If 2 eggs make 5 dozen, how many eggs do you need to make 10 dozen?) Proportional Relationships u What if your recipe says it will make 5 dozen cookies and you only get 4.5 dozen? Percent Yield = actual amount x 100% theoretical amount u If you don’t have enough of one ingredient how much can you make? (What if we only had 1 egg?) Limiting Reagent (& Excess Reagent)

Proportional Relationships u I have 5 eggs. How many cookies can I make? 3/4 c. brown sugar 1 tsp vanilla extract 2 eggs 2 c. chocolate chips Makes 5 dozen cookies. 2 1/4 c. flour 1 tsp. baking soda 1 tsp. salt 1 c. butter 3/4 c. sugar 5 eggs 5 doz. 2 eggs = 12.5 dozen cookies Ratio of eggs to cookies 2. 5 : 1 or 2. 5 1

Learning More from Cookies u In the example we need 2 eggs to make 5 dozen cookies u Soooooo – if we have 5 eggs we can increase our recipe by 5 =2.5 times 2 u HOWEVER ~ to truly duplicate this family favorite you have to multiple the amounts of ALL ingredients to keep the recipe in BALANCE and to make 12.5 it would look like….

New Proportional Relationships – everything is multiplied by 2.5 2 1/4 c. flour (6 5/8 c) 1 tsp. baking soda (2.5 t) 1 tsp. salt (2.5 t) 1 c. butter (2.5 c) 3/4 c. sugar (1 7/8 c) 3/4 c. brown sugar (1 7/8 c) 1 tsp vanilla extract (2.5 t) 2 eggs (5 eggs) 2 c. chocolate chips (5 c)

Let’s Try it with Snack Mix Ingredients (Reactants)Product Ingredients (Reactants) Product 1½ c. Oat Cereal- 4 c. Yummy and 1 c. Pretzels Healthy Snack ½ c. Nuts ½ c. M & Ms ½ c. Raisins Multiplier = What was our percent yield? Did we have a limiting reagent (ingredient)? If so, Which one? What would be excess reagents? What multiplier do we need to use to make ½ cup of mix for each person here?

Let’s Try it with Trail Mix Ingredients (Reactants)Product Ingredients (Reactants) Product 1 c. Mixed Dried Fruit- 6 c. Yummy and ½ c. Raisins and/or Healthy Snack Dried Cranberries 1 c. Unsalted Nuts peanuts, almonds, or walnuts 1 c. Shelled Seeds sunflower, or pumpkin ½ c. M & Ms 1 c. Cheerios 1 c. Pretzels What multiplier do we need to use to make ½ cup of mix for each person here?

YOU Try it – The Hamburger Analogy u Complete the top part of the worksheet titled “Hints for Solving Stoichiometry Problems”

Applying the Hamburger Recipe Concept to Chemical Recipes u Complete the bottom and back of the worksheet titled “Hints for Solving Stoichiometry Problems”

Stoichiometry u Greek for “measuring elements” u The calculations of quantities in chemical reactions based on a balanced equation. u We can interpret balanced chemical equations several ways.

1. In terms of Particles u Element- made of atoms u Molecular compound (made of only non- metals) = molecules u Ionic Compounds (made of a metal and non-metal parts) = formula units (ions)

2H 2 + O 2   2H 2 O u Two molecules of hydrogen and one molecule of oxygen form two molecules of water.  2 Al 2 O 3  Al + 3O 2 2formula unitsAl 2 O 3 form4 atoms Al and3diatomic molecules O2O2 2Na + 2H 2 O  2NaOH + H 2

Look at it differently  2H 2 + O 2   2H 2 O u 2 dozen molecules of hydrogen and 1 dozen molecules of oxygen form 2 dozen molecules of water. u 2 x (6.022 x 10 23 ) molecules of hydrogen and 1 x (6.022 x 10 23 ) molecules of oxygen form 2 x (6.022 x 10 23 ) molecules of water. u 2 moles of hydrogen and 1 mole of oxygen form 2 moles of water.

2. In terms of Moles  2 Al 2 O 3  Al + 3O 2  2Na + 2H 2 O  2NaOH + H 2 u The coefficients tell us how many moles of each substance

Mole to Mole conversions  2 Al 2 O 3  Al + 3O 2 u each time we use 2 moles of Al 2 O 3 we will also make 3 moles of O 2 2 moles Al 2 O 3 3 mole O 2 or 2 moles Al 2 O 3 3 mole O 2 These are possible conversion factors

Mole to Mole conversions u How many moles of O 2 are produced when 3.34 moles of Al 2 O 3 decompose?  2 Al 2 O 3  Al + 3O 2 3.34 mol Al 2 O 3 2 mol Al 2 O 3 3 mol O 2 =5.01 mol O 2

Practice: 2C 2 H 2 + 5 O 2  4CO 2 + 2 H 2 O If 3.84 moles of C 2 H 2 are burned, how many moles of O 2 are needed? (9.6 mol) How many moles of C 2 H 2 are needed to produce 8.95 mole of H 2 O? (8.95 mol) If 2.47 moles of C 2 H 2 are burned, how many moles of CO 2 are formed? (4.94 mol)

Mass-Mass Calculations u We do not measure moles directly, so what can we do? u We can convert grams to moles Use the Periodic Table for mass values u Then do the math with the mole ratio Balanced equation gives mole ratio! u Then turn the moles back to grams Use Periodic table values

3. In terms of Mass u The Law of Conservation of Mass applies u We can check using moles  2H 2 + O 2   2H 2 O 2 moles H 2 2.02 g H 2 1 mole H 2 =4.04 g H 2 1 mole O 2 32.00 g O 2 1 mole O 2 =32.00 g O 2 Total 36.04 g H 2 +O 2

In terms of Mass  2H 2 + O 2   2H 2 O 2 moles H 2 O 18.02 g H 2 O 1 mole H 2 O = 36.04 g H 2 O 2H 2 + O 2   2H 2 O 36.04 g H 2 + O 2 = 36.04 g H 2 O

For example... u If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how much solid copper would form?  2Fe + 3CuSO 4  Fe 2 (SO 4 ) 3 + 3Cu Answer = 17.2 g Cu

Chemical Calculations u TEKS OBJECTIVE 9.B: TLW perform stoichiometric calculations, including determination of mass and volume relationships between reactants and products for reactions involving gases.

4. In terms of Volume  2H 2 + O 2   2H 2 O u At STP, 1 mol of any gas = 22.4 L  (2 x 22.4 L H 2 ) + (1 x 22.4 L O 2 )  (2 x 22.4 L H 2 O) u NOTE: mass and atoms are always conserved- however, molecules, formula units, moles, and volumes will not necessarily be conserved!

Volume-Volume Calculations u How many liters of CH 4 at STP are required to completely react with 17.5 L of O 2 ?  CH 4 + 2O 2  CO 2 + 2H 2 O 17.5 L O2O2 22.4 L O2O2 1 mol O2O2 2 O2O2 1 CH 4 1 mol CH 4 22.4 L CH 4 = 8.75 L CH 4 22.4 L O 2 1 mol O 2 1 mol CH 4 22.4 L CH 4

More practice... u How many liters of CO 2 at STP will be produced from the complete combustion of 23.2 g C 4 H 10 ? Answer = 35.8 L CO 2 Answer = 58.2 L O 2 What volume of Oxygen would be required?

Avogadro told us: u Equal volumes of gas, at the same temperature and pressure contain the same number of particles. u Moles are numbers of particles u You can treat reactions as if they happen liters at a time, as long as you keep the temperature and pressure the same.

Shortcut for Volume-Volume: u How many liters of H 2 O at STP are produced by completely burning 17.5 L of CH 4 ?  CH 4 + 2O 2  CO 2 + 2H 2 O 17.5 L CH 4 1 L CH 4 2 L H 2 O = 35.0 L H 2 O Note: This only works for Volume- Volume problems.

Limiting Reagent & Percent Yield u TEKS OBJECTIVE 8.E: TLW identify and use the limiting reagent in a reaction to calculate the maximum amount of product(s) produced, and the amount of excess reagent.

Limiting Reagent & Percent Yield u TEKS OBJECTIVE 8.E: TLW calculate theoretical yield, actual yield, or percent yield, given appropriate information.

“Limiting” Reagent u If you are given one dozen loaves of bread, a gallon of mustard, and three pieces of salami, how many salami sandwiches can you make? u The limiting reagent is the reactant you run out of first. u The excess reagent is the reactant(s) you have left over. u The limiting reagent determines how much product you can make

How do you find limiting reagent? u Do two stoichiometry problems. u The one that makes the least product is the limiting reagent. u For example u Copper reacts with sulfur to form copper ( I ) sulfide. If 10.6 g of copper reacts with 3.83 g S how much product will be formed?

u If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed?  2Cu + S  Cu 2 S 10.6 g Cu 63.55g Cu 1 mol Cu 2 mol Cu 1 mol Cu 2 S 1 mol Cu 2 S 159.16 g Cu 2 S = 13.3 g Cu 2 S 3.83 g S 32.06g S 1 mol S 1 S 1 Cu 2 S 1 mol Cu 2 S 159.16 g Cu 2 S = 19.0 g Cu 2 S = 13.3 g Cu 2 S

Cu is Limiting Reagent

Another example u If 10.1 g of magnesium and 2.87 L of HCl gas are reacted, how many liters of gas will be produced? u How many grams of solid? u How much excess reagent remains?

Still another example u If 10.3 g of aluminum are reacted with 51.7 g of CuSO 4 how much copper will be produced? u How much excess reagent will remain?

Percent Yield u The amount of product made in a chemical reaction. u There are three types: 1. Actual yield- what you get in the lab when the chemicals are mixed 2. Theoretical yield- what the balanced equation tells should be made 3. Percent yield 3. Percent yield = Actual Theoretical X 100

Example u 6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate.  2Al + 3 CuSO 4  Al 2 (SO 4 ) 3 + 3Cu u What is the actual yield? u What is the theoretical yield? u What is the percent yield?

Details u Percent yield tells us how “efficient” a reaction is. u Percent yield can not be bigger than 100 %.

Labs u Soda Lab linklink u Balloon Races (Flinn Scientific Vol. 7, pages 72 – 74) In Periodic Groups, read entire procedure Determine potential hazards, precautions, PPE (if needed)

How do you get good at this? Group work on Guided Reading Book page 87

Independent Practice u Practice sets Mole to Mole conversions Mass to Mass conversions Mole to Mass conversions Mass to Mole conversions Volume conversions Limiting Reagents Percent Yield

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