Presentation on theme: "Chapter 11 Stoichiometry. Chocolate Chip Cookie Recipe Makes 42 cookies ½ c butter ½ c brown sugar ½ c granulated sugar 1 egg ½ tsp vanilla 1 cup flour."— Presentation transcript:
Chocolate Chip Cookie Recipe Makes 42 cookies ½ c butter ½ c brown sugar ½ c granulated sugar 1 egg ½ tsp vanilla 1 cup flour ½ tsp salt ½ tsp baking soda ½ c chopped walnuts ½ c chocolate chips What is the product of this recipe? How would you adjust the ingredients in the recipe to make 7 dozen cookies? What would you do if you only had ¼ c butter?
Stoichiometry Mass-mass relationships Quantitative study of chemical reactions
Balanced Equations give us: the relationship between the reactants and the products the ratios in which they combine The coefficients represent the number of moles 2 H 2 + 1 O 2 2 H 2 0 2 moles of hydrogen react w/ 1 mole of oxygen to form 2 moles of water
Law of Conservation of Mass Basis behind stoichiometry Mass of reactants= mass of product Ex. 4Fe + 3O 2 2 Fe 2 O 3 –4 moles of iron react w/ 3 moles of oxygen to form 2 moles of rust –Convert moles to grams 4 moles Fe 1mole Fe 55.8 g Fe = 223.4 g Fe 3 moles O 2 1 mole O 2 32.0 g O 2 = 96.0 g O 2 Total mass of reactants: 319.4 g
2 moles Fe 2 O 3 1 mole Fe 2 O 3 158.7 g Fe 2 O 3 = 319.4 g Fe 2 O 3 (mass of products) Mass of reactants= mass of products
Trail Mix In order to make trail mix, you need raisins, peanuts, and M&M's. What makes it good trail mix is having the right ratios! In the bags there are 1 dozen raisins (R), 2 dozen peanuts (P), and 3 dozen M&M's (MM). This is a perfect combination to make 1 package of trail mix (TM). If we were to write an equation to represent this recipe it would look like this: 1 dozen raisins + 2 dozen peanuts + 3 dozen M&M's = 1 trail mix or 1 R + 2 P + 3 MM 1 TM
1 R + 2 P + 3 MM 1 TM With this in mind, what if you only had ½ the contents of your bags… How many dozen peanuts would you need to mix them with 1/2 dozen raisins? By looking at the equation, you can come up with a conversion of 2 dozen peanuts for every 1 dozen raisins. SO… ½ dozen raisins x 2 dozen peanuts = 1 dozen peanuts needed 1 dozen raisins (or ½ your bag!) (Just like halving a recipe!)
1 R + 2 P + 3 MM 1 TM How many dozen M&M's would you need to combine them with 6 dozen peanuts? 6 dozen peanut 2 dozen peanuts 3 dozen MM 9 dozen m &m’s
1 R + 2 P + 3 MM 1 TM How many dozen raisins would you need to mix with 4 dozen peanuts? 4 dozen peanut 2 dozen peanut 1 dozen raisin = 2 dozen raisins
1 R + 2 P + 3 MM 1 TM How many dozen peanuts would you need to mix them with 18 M&M's?(careful!) 18 mm’s 12 mm’s 1 dozen mm 3 dozen mm’s 2 dozen peanuts 1 dozen peanuts
Chemists use ratios such as these ALL the time in the laboratory. Instead of dozens, they use moles – something we've already looked at. Just as in the equation above, the coefficients in front of the compounds provide the "mole ratios" the chemists need.
Mole Ratios Ratio between the numbers of moles of any 2 substances in a balanced equation You can have many different ratio scenarios for each equation
2 Al + 3 Cl 2 2 AlCl 3 Ratio of Al to Cl2 moles Al 3 moles Cl Ratio of Cl to Al3 moles Cl 2 moles Al Ratio of Cl to product3 moles Cl 2 moles AlCl 3 Ratio of product to Cl2 moles AlCl 3 3 moles Cl Ratio of Al to product2 moles Al 2 moles AlCl 3 Ratio of product to Al2 moles AlCl 3 2 moles Al
Stoichiometry Answers these ?’s: How much of 1 reactant is needed to combine w/ a given amount of another reactant How much product will be produced w/ a given amount of reactant How much reactant is needed to produce a given amount of product
Steps in Stoichiometry 1. Write the balanced equation 2. Write the given information under the equation (molar mass, # of moles, grams of known) 3. Convert grams of known substance to moles using molar mass
4. Determine the mole ratio using the coefficients from balanced equation moles of unknown moles of known 5. Convert the moles of unknown to grams using molar mass
Known mass g (from problem) 1 mole known g known (molar mass known) unknown moles known moles ( coef. from eqn.) (molar mass unknown) g unknown 1 mole unknown
How many grams of AlCl 3 are produced if 3.0g of Cl 2 react with excess aluminum? 1.) Balanced Equation: 2 Al + 3 Cl 2 2 AlCl 3 3.0 g 3 mole Cl 2 71.0 g/mole ? g 2 moles AlCl 3 133.5 g/mole 2.) Write in known info
3. Convert known grams to moles, put in mole ratio (from coefficients) and convert unknown moles to grams. 3.0g Cl 2 71.0 g Cl 2 1 mole Cl 2 3 moles Cl 2 2 moles AlCl 3 1 mole AlCl 3 133.5 g AlCl 3 (Molar mass known) (Mass known) (Mole Ratio) (molar mass unknown)
3. Multiply the top line and divide by the bottom (making sure to use parentheses correctly) (3.0 x 2 x 133.5) (71.0 x 3) = 3.8 g of AlCl 3 formed
Stoichiometry of Shuttle Launch NASA TV (Discovery Last Launch) NASA TV (Discovery Last Launch) You Tube (STS 119)
Percent Yield Quantities found in stoichiometry are theoretical or predicted amounts (the amount you are supposed to get if all the conditions are perfect) During actual experimentation, you will come up w/ less than expected % Yield = actual amt (experiment) x 100 Theoretical amt (calculated)
You calculated that you should get 82.2g of NaCl in a reaction. When you performed the experiment you produced only 30.7 g NaCl. What is your % yield? % Yield = 30.7 x 100 82.2 37.3%
Limiting Reactant (reagent) Reactant that is totally consumed during a reaction Limits the extent of a reaction Determines the amount of product that can be formed Once it is gone, the reaction stops
Excess Reactant (reagent) Reactant that remains after the reaction stops
Steps: 1. Balance equation 2. Using known amounts of each reactant, solve for mass of product (do 2 separate stoichiometry problems) 3. The one that produces the least amount of product is the limiting reactant.
75 g of CaO react with 30.0 g HCl to produce calcium chloride and water. What is the limiting reactant? How much water is formed? Balance equation and write in known info CaO + 2HCl CaCl 2 + H 2 0 75 g 1 mole 56.1 g/mole 30.0 g 2 mole 36.5 g/mole ? g 1 mole 18.0 g/mole
75 g CaO 56.1 g CaO 1 mole CaO 1 mole H 2 O 18.0 g H 2 O Do 2 stoichiometry problems, w/ each reactant as a known and the unknown being water =24 g H 2 O
30.0 g HCl 36.5 g HCl 1 mole HCl 2 mole HCl 1 mole H 2 O 18.0 g H 2 O Do 2 stoichiometry problems, w/ each reactant as a known and the unknown being water 7.40 g H 2 O
HCl is the limiting reactant, producing 7.40 g of water CaO is in excess (meaning some will be left over after the reaction)
Finding mass of excess reactant that remains: One can find the mass of excess reactants that will remain by doing another stoichiometry problem using your limiting reactant as your known and the excess reactant as your unknown Mass excess starting- mass excess used (stoich. Calc)= mass of left over excess react.
Using HCl as the known and CaO as the unknown, calc. the amt. Of CaO used. 30.0 g HCl 1 mole HCl 1 mole CaO 56.1 g CaO 36.5 g HCl2 mole HCl 1 mole CaO =23.1 CaO used 75 g CaO – 23.1g used = 51.9 g leftover
N 2 + 3H 2 2 NH 3 If you have 100.0 g of N 2 and 25.0 g of H 2, which is your limiting reactant in the reaction? How many grams of NH 3 can be produced from that limiting reactant? Calculate the mass of excess reactant that remains after the reaction is complete.
Ch. 11 Test Review (stoich song)stoich song Vocabulary –stoichiometry –Limiting reactant –Excess reactant –Mole ratio –Actual yield (experimental) –Theoretical yield (calculated) –% yield
Calculations –mole ratios –Law of conservation of mass –Stoichiometry –Limiting reactant problem –% yield Essay Practical application of stoichiometry
4HCl + O 2 2 H 2 O + 2 Cl 2 Interpret it in terms of moles What is the mole ratio between water and oxygen? Show that mass is conserved
_ Cu + _ AgNO 3 _ Cu(NO 3 ) 2 + _ Ag Balance the equation How many grams of copper are needed to react with 12.0 g of AgNO 3 ?
_ Al + _ Cl 2 __AlCl 3 Balance If you begin with 3.2 g of aluminum and 5.4g of chlorine what is the limiting reactant?
__Mg+ _ HCl _MgCl 2 + _ H 2 Balance If you reacted 10.0g of magnesium with excess hydrochloric acid. How many grams of MgCl 2 will be formed? After actually reacting this you formed 29.5 g of MgCl 2, what is the percent yield?