EXTRA CREDIT REMINDER

Transcription

1 EXTRA CREDIT REMINDER Due Tonight at Midnight (January 21 at 11:59 pm) via *** Kinesthetic: If you do not know how to use Prezi you may do a power point otherwise your Prezi link. This will count as a bonus in the test category approximately 10 points averaged with your other grades in the category. It can only HELP! Not harm your grade.

2 Important things to Remember

3 The Mole A counting unit Similar to a dozen, except instead of 12, it s 602,000,000,000,000,000,000, X (in scientific notation) This number is named in honor of Amedeo Avogadro ( )

4 Molar Mass The Mass of 1 mole (in grams) Equal to the numerical value of the average atomic mass (get from periodic table), or add the atoms together for a molecule 1 mole of C atoms = 12.0 g 1 mole of Mg atoms = 24.3 g 1 mole of O 2 molecules = 32.0 g Diatomic elements are: H, N, O, F, Cl, Br, I

5 Molar Mass of Compounds The molar mass (MM) of a compound is determined the same way, except now you add up all the atomic masses for the molecule (or compound) Ex. Molar mass of CaCl 2 Avg. Atomic mass of Calcium = 40.08g Avg. Atomic mass of Chlorine = 35.45g Molar Mass of calcium chloride = g/mol Ca + (2 X 35.45) g/mol Cl g/mol CaCl 2 20 Ca Cl 35.45

6 Atoms or Molecules Flowchart Divide by 6.02 X Multiply by 6.02 X Moles Divide by atomic/molar mass from periodic table Multiply by atomic/molar mass from periodic table Mass (grams)

7 Practice Calculate the Molar Mass of calcium phosphate Formula = Ca 3 (PO 4 ) 2 Masses elements: Ca: 3 Ca s X 40.1 = P: 2 P s X 31.0 = O: 8 O s X 16.0 = g 62.0 g g Molar Mass = 120.3g g g/mol g

8 Calculations molar mass Avogadro s number Grams Moles particles Everything must go through Moles!!!

9 Atoms/Molecules and Grams How many atoms of Cu are present in 35.4 g of Cu? 35.4 g Cu 1 mol Cu 6.02 X atoms Cu 63.5 g Cu 1 mol Cu = 3.4 X atoms Cu

10 Chocolate Chip Cookies!! 1 cup butter 1/2 cup white sugar 1 cup packed brown sugar 1 teaspoon vanilla extract 2 eggs 2 1/2 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon salt 2 cups semisweet chocolate chips Makes 3 dozen How many eggs are needed to make 3 dozen cookies? How much butter is needed for the amount of chocolate chips used? How many eggs would we need to make 9 dozen cookies? How much brown sugar would I need if I had 1 ½ cups white sugar?

11 Cookies and Chemistry Huh!?!? Just like chocolate chip cookies have recipes, chemists have recipes as well Instead of calling them recipes, we call them reaction equations Furthermore, instead of using cups and teaspoons, we use moles Lastly, instead of eggs, butter, sugar, etc. we use chemical compounds as ingredients

12 Chemistry Recipes Looking at a reaction tells us how much of something you need to react with something else to get a product (like the cookie recipe) Be sure you have a balanced reaction before you start! Example: 2 Na + Cl 2 2 NaCl This reaction tells us that by mixing 2 moles of sodium with 1 mole of chlorine we will get 2 moles of sodium chloride What if we wanted 4 moles of NaCl? 10 moles? 50 moles?

13 Mole Ratios These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical Example: How many moles of chlorine is needed to react with 5 moles of sodium (without any sodium left over)? 2 Na + Cl 2 2 NaCl 5 moles Na 1 mol Cl 2 2 mol Na = 2.5 moles Cl 2

14 Mole-Mole Conversions How many moles of sodium chloride will be produced if you react 2.6 moles of chlorine gas with an excess (more than you need) of sodium metal? 2 Na + Cl 2 2 NaCl

15 Mole-Mass Conversions Most of the time in chemistry, the amounts are given in grams instead of moles We still go through moles and use the mole ratio, but now we also use molar mass to get to grams Example: How many grams of chlorine are required to react completely with 5.00 moles of sodium to produce sodium chloride? 2 Na + Cl 2 2 NaCl 5.00 moles Na 1 mol Cl g Cl 2 2 mol Na 1 mol Cl 2 = 177g Cl 2

16 Practice Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum. 2 Al + 3 I 2 2 AlI 3

17 Mass-Mole We can also start with mass and convert to moles of product or another reactant We use molar mass and the mole ratio to get to moles of the compound of interest Calculate the number of moles of ethane (C 2 H 6 ) needed to produce 10.0 g of water 2 C 2 H O 2 4 CO H g H 2 O 1 mol H 2 O 2 mol C 2 H g H 2 O 6 mol H 2 0 = mol C 2 H 6

18 Practice Calculate how many moles of oxygen are required to make 10.0 g of aluminum oxide 4 Al + 3 O 2 2 Al 2 O 3

19 Practice Write the balanced reaction for hydrogen gas reacting with oxygen gas. H 2 + O 2 H 2 O How many moles of each reactants are needed? What if we wanted 4 moles of water how many moles of each reactant would you need? What if we had 3 moles of oxygen, how much hydrogen would we need to react and how much water would we get? What if we had 50 moles of hydrogen, how much oxygen would we need and how much water produced?

20 Practice Write the balanced reaction for hydrogen gas reacting with oxygen gas. 2 H 2 + O 2 2 H 2 O How many moles of reactants are needed? What if we wanted 4 moles of water? 4 mol H 2 2 mol O 2 2 mol H 2 1 mol O 2 What if we had 3 moles of oxygen, how much hydrogen would we need to react and how much water would we get? 6 mol H 2, 6 mol H 2 O What if we had 50 moles of hydrogen, how much oxygen would we need and how much water produced? 25 mol O 2, 50 mol H 2 O

21 Practice A 2 + 2B 2AB What is the mole ratio of substance A to substance AB? What is the mole ratio of substance A to substance B? What is the mole ratio of substance B to substance AB? Show the work for the problems below: If you have 4 moles of substance A, how many moles of substance AB can you produce? If you have 10 moles of substance B and an excess of substance A, how many moles of substance AB can you produce?

22 Mass-Mass Conversions Most often we are given a starting mass and want to find out the mass of a product we will get (called theoretical yield) Or how much of another reactant we need to completely react with it (no leftover ingredients!) Now we must go from grams to moles, mole ratio, and back to grams of compound we are interested in ga mol A mol B gb

23 Mass-Mass Conversion Ex. Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen. N 2 + H 2 NH 3 1. Make sure you have a balanced equation. 2. Convert grams of nitrogen to moles of nitrogen 3. Convert moles of nitrogen to moles of ammonia 4. Then, convert moles of ammonia to grams of ammonia With a partner discuss which values you will need for each step above. Then, work the problem. Hint: ga mol A mol B gb

24 Mass-Mass Conversion Ex. Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen. N H 2 2 NH 3 ga mol A mol B gb

25 Practice How many grams of calcium nitride are produced when 2.00 g of calcium reacts with an excess of nitrogen? Ca + N 2 Ca 3 N 2

26 Theoretical, Actual, and Percent Yield Theoretical yield: the maximum amount of product, which is calculated using the balanced equation. Actual yield: the amount of product obtained when the reaction takes place Percent yield: the ratio of actual yield to theoretical yield Percent yield = actual yield (g) x 100 theoretical yield (g) 26

27 Percent Yield 27

28 Calculating Percent Yield Suppose you have prepared cookie dough to make 5 dozen cookies. The phone rings and you answer. While you talk, a sheet of 12 cookies burns, and you have to throw them out. The rest of the cookies you make are okay. What is the percent yield of edible cookies? Theoretical yield: 60 cookies possible Actual yield: Percent yield: 48 cookies to eat 48 cookies x 100% = 80.% yield 60 cookies 28

29 Learning Check With a limited amount of oxygen, the reaction of carbon and oxygen produces carbon monoxide. 2C(g) + O 2 (g) 2CO(g) What is the percent yield if 40.0 g of CO are produced when 30.0 g of O 2 are used? 1) 25.0% 2) 75.0% 3) 76.2% 29

30 Solution STEP 1 Given: 40.0 g of CO produced (actual) 30.0 g of O 2 used Need: percent yield of CO STEP 2 Write a plan to calculate % yield of CO: g of O 2 moles of moles of g of CO O 2 CO (theoretical) Percent yield of CO = g of CO (actual) x 100% g of CO (theoretical) 30

31 Solution (continued) STEP 3 Write conversion factors: 1 mole of O 2 = 32.0 g of O 2 1 mole O 2 and 32.0 g O g O 2 1 mole O 2 1 mole of O 2 = 2 moles of CO 1 mole O 2 and 2 moles CO 2 moles CO 1 mole O 2 1 mole of CO = 28.0 g of CO 1 mole CO and 28.0 g CO 28.0 g CO 1 mole CO 31

32 Solution (continued) STEP 4 Setup to calculate theoretical yield in g of O 2 : 30.0 g O 2 x 1 mole O 2 x 2 moles CO x 28.0 g CO 32.0 g O 2 1 mole O 2 1 mole CO = 52.5 g of CO (theoretical) Setup to calculate percent yield: 40.0 g CO (actual) x 100 = 76.2% yield (3) 52.5 g CO (theoretical) 32

33 Learning Check When N 2 and 5.00 g of H 2 are mixed, the reaction produces 16.0 g of NH 3. What is the percent yield for the reaction? N 2 (g) + 3H 2 (g) 2NH 3 (g) 1) 31.3% of NH 3 2) 56.9% of NH 3 3) 80.0% of NH 3 33

34 Solution 2) 56.9% STEP 1 Given: 16.0 g of NH 3 produced (actual) 5.00 g of H 2 used Need: percent yield of NH 3 STEP 2 Write a plan to calculate % yield of NH 3 : g of H 2 moles of moles of g of NH 3 H 2 NH 3 (theoretical) Percent yield of NH 3 = g of NH 3 (actual) x 100% g of NH 3 (theoretical) 34

35 Solution (continued) STEP 3 Write conversion factors: 1 mole of H 2 = 2.02 g of H 2 1 mole H 2 and 2.02 g H g H 2 1 mole H 2 1 mole of H 2 = 2 moles of NH 3 1 mole H 2 and 2 moles NH 3 2 moles NH 3 1 mole H 2 1 mole of NH 3 = 17.0 g of NH 3 1 mole NH 3 and 17.0 g NH g NH 3 1 mole NH 3

36 Solution (continued) STEP 4 Setup to calculate theoretical yield of g of NH 3 : 5.00 g H 2 x 1 mole H 2 x 2 moles NH 3 x 17.0 g NH g H 2 3 moles H 2 1 mole NH 3 = 28.1 g of NH 3 (theoretical) Setup to calculate percent yield: Percent yield = 16.0 g NH 3 x 100 = 56.9% yield (2) 28.1 g NH 3 36

37 Guide to Calculations for Percent Yield 37

38 Limiting Reactant A limiting reactant in a chemical reaction is the substance that is used up limits the amount of product that can form and stops the reaction 38

39 Reacting Amounts In a table setting, there is 1plate, 1 fork, 1 knife, and 1 spoon. How many table settings are possible from 5 plates, 6 forks, 4 spoons, and 7 knives? What is the limiting item? 39

40 Reacting Amounts (continued) Only 4 place settings are possible. Initially Used Left over Plates Forks Spoons Knives The limiting item is the spoon. 40

41 Example 1 of an Everyday Limiting Reactant How many peanut butter sandwiches could be made from 8 slices of bread and 1 jar of peanut butter? With 8 slices of bread, only 4 sandwiches could be made. The bread is the limiting item. 41

42 Example 2 of an Everyday Limiting Reactant How many peanut butter sandwiches could be made from 8 slices bread and 1 tablespoon of peanut butter? With 1 tablespoon of peanut butter, only 1 sandwich could be made. The peanut butter is the limiting item. 42

43 Limiting Reactant When 4.00 moles of H 2 is mixed with 2.00 moles of Cl 2, how many moles of HCl can form? H 2 (g) + Cl 2 (g) 2HCl(g) 4.00 moles 2.00 moles??? Moles Calculate the moles of product that each reactant, H 2 and Cl 2, could produce. The limiting reactant is the one that produces the smaller number of moles of product. 43

44 Limiting Reactant (continued) HCl from H moles H 2 x 2 moles HCl = 8.00 moles of HCl 1 moles H 2 HCl from Cl moles Cl 2 x 2 moles HCl = 4.00 moles of HCl 1 mole Cl moles of HCl is the smaller number of moles produced. Thus, Cl 2 will be used up. The limiting reactant is Cl 2. 44

45 Limiting Reactants Using Mass If 4.80 moles Ca are mixed with 2.00 moles N 2, which is the limiting reactant? 3Ca(s) + N 2 (g) Ca 3 N 2 (s) Moles of Ca 3 N 2 from Ca 4.80 moles Ca x 1 mole Ca 3 N 2 = 1.60 moles of Ca 3 N 2 3 moles Ca (Ca is used up) Moles of Ca 3 N 2 from N moles N 2 x 1 mole Ca 3 N 2 = 2.00 moles of Ca 3 N 2 1 mole N 2 Ca is used up. Thus, Ca is the limiting reactant. 45

46 Learning Check What is the mass of water that can be produced when 8.00 g of H 2 and 24.0 g of O 2 react? 2H 2 (g) + O 2 (g) 2H 2 O(l) 1) 8.0 g of H 2 O 2) 27.0 g of H 2 O 3) 72 g of H 2 O 46

47 Solution 3) 72 g of H 2 O Moles of H 2 O from H 2 : 8.00 g H 2 x 1 mole H 2 x 2 moles H 2 O = 4.0 moles of H 2 O Moles of H 2 O from O 2 : 2.0 g H 2 2 moles H g O 2 x 1 mole O 2 x 2 moles H 2 O = 1.50 moles of H 2 O 32.0 g O 2 1 mole O 2 Smaller number of moles of H 2 O 1.50 moles H 2 O x 18.0 g H 2 O = 27.0 g of H 2 O 1 mole H 2 O 47

48 Guide to Calculating Product from a Limiting Reactant 48

49 Check Calculations Equation H 2 Cl 2 2HCl Initially 4.00 moles 2.00 moles 0 mole Reacted/ Formed Left after reaction 2.00 moles 2.00 moles moles 2.00 moles ( ) Excess 0 moles ( ) Limiting 4.00 moles ( ) Product possible 49

50 Limiting Reactant A limiting reactant in a chemical reaction is the substance that is used up limits the amount of product that can form and stops the reaction 50

51 Reacting Amounts In a table setting, there is 1plate, 1 fork, 1 knife, and 1 spoon. How many table settings are possible from 5 plates, 6 forks, 4 spoons, and 7 knives? What is the limiting item? 51

52 Reacting Amounts (continued) Only 4 place settings are possible. Initially Used Left over Plates Forks Spoons Knives The limiting item is the spoon. 52

53 Example 1 of an Everyday Limiting Reactant How many peanut butter sandwiches could be made from 8 slices of bread and 1 jar of peanut butter? With 8 slices of bread, only 4 sandwiches could be made. The bread is the limiting item. 53

54 Example 2 of an Everyday Limiting Reactant How many peanut butter sandwiches could be made from 8 slices bread and 1 tablespoon of peanut butter? With 1 tablespoon of peanut butter, only 1 sandwich could be made. The peanut butter is the limiting item. 54

55 Limiting Reactant When 4.00 moles of H 2 is mixed with 2.00 moles of Cl 2, how many moles of HCl can form? H 2 (g) + Cl 2 (g) 2HCl(g) 4.00 moles 2.00 moles??? Moles Calculate the moles of product that each reactant, H 2 and Cl 2, could produce. The limiting reactant is the one that produces the smaller number of moles of product. 55

56 Limiting Reactant (continued) HCl from H moles H 2 x 2 moles HCl = 8.00 moles of HCl 1 moles H 2 HCl from Cl moles Cl 2 x 2 moles HCl = 4.00 moles of HCl 1 mole Cl moles of HCl is the smaller number of moles produced. Thus, Cl 2 will be used up. The limiting reactant is Cl 2. 56

57 Limiting Reactants Using Mass If 4.80 moles Ca are mixed with 2.00 moles N 2, which is the limiting reactant? 3Ca(s) + N 2 (g) Ca 3 N 2 (s) Moles of Ca 3 N 2 from Ca 4.80 moles Ca x 1 mole Ca 3 N 2 = 1.60 moles of Ca 3 N 2 3 moles Ca (Ca is used up) Moles of Ca 3 N 2 from N moles N 2 x 1 mole Ca 3 N 2 = 2.00 moles of Ca 3 N 2 1 mole N 2 Ca is used up. Thus, Ca is the limiting reactant. 57

58 Learning Check What is the mass of water that can be produced when 8.00 g of H 2 and 24.0 g of O 2 react? 2H 2 (g) + O 2 (g) 2H 2 O(l) 1) 8.0 g of H 2 O 2) 27.0 g of H 2 O 3) 72 g of H 2 O 58

59 Solution 3) 72 g of H 2 O Moles of H 2 O from H 2 : 8.00 g H 2 x 1 mole H 2 x 2 moles H 2 O = 4.0 moles of H 2 O Moles of H 2 O from O 2 : 2.0 g H 2 2 moles H g O 2 x 1 mole O 2 x 2 moles H 2 O = 1.50 moles of H 2 O 32.0 g O 2 1 mole O 2 Smaller number of moles of H 2 O 1.50 moles H 2 O x 18.0 g H 2 O = 27.0 g of H 2 O 1 mole H 2 O 59

60 Guide to Calculating Product from a Limiting Reactant 60

61 Limiting Limiting Reactant: Example Reactant 10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced? 2 Al + 3 Cl 2 2 AlCl 3 Start with Al: 10.0 g Al 1 mol Al 2 mol AlCl g AlCl 3 Now Cl 2 : 27.0 g Al 2 mol Al 1 mol AlCl g Cl 2 1 mol Cl 2 2 mol AlCl g AlCl g Cl 2 3 mol Cl 2 1 mol AlCl 3 = 49.4g AlCl 3 = 43.9g AlCl 3

62 LR Example Continued We get 49.4g of aluminum chloride from the given amount of aluminum, but only 43.9g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 35.0g of chlorine is used up, the reaction comes to a complete.

63 Limiting Reactant Practice 15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made.

64 Finding the Amount of Excess By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess. Can we find the amount of excess potassium in the previous problem?

65 Finding Excess Practice 15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I 2 2 KI We found that Iodine is the limiting reactant, and 19.6 g of potassium iodide are produced g I 2 1 mol I 2 2 mol K 39.1 g K 254 g I 2 1 mol I 2 1 mol K = 4.62 g K USED! 15.0 g K 4.62 g K = g K EXCESS Given amount of excess reactant Amount of excess reactant actually used Note that we started with the limiting reactant! Once you determine the LR, you should only start with it!

66 Limiting Reactant: Recap 1. You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT. 2. Convert ALL of the reactants to the SAME product (pick any product you choose.) 3. The lowest answer is the correct answer. 4. The reactant that gave you the lowest answer is the LIMITING REACTANT. 5. The other reactant(s) are in EXCESS. 6. To find the amount of excess, subtract the amount used from the given amount. 7. If you have to find more than one product, be sure to start with the limiting reactant. You don t have to determine which is the LR over and over again!

67 1 gram of Chalk What is the formula for Chalk?