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

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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.

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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

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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.

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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

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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?

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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)

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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

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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….

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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)

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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?

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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?

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

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

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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.

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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)

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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

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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.

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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

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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

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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

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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)

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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

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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

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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

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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

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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.

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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!

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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

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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?

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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.

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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.

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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.

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

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“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

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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?

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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

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Cu is Limiting Reagent

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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?

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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?

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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

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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?

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

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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)

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

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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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Chapter 12 Stoichiometry

Chapter 12 Stoichiometry

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