1. Afra Khanani Period 6 Honors Chemistry March 31 st

2. PROBLEM: A solution containing 6720 mg of H 2 0 is added to another containing 10.67 Liters of CO 2 at STP. Determine which reactant was in excess, as well as the number of grams over the amount required by the limiting species. Also, find the number of molecules of glucose that precipitated as well, as the theoretical and percent yield of glucose if 10.22 g C 6 H 12 O 6 was obtained.

3. The chemical reaction of water and carbon dioxide will produce oxygen gas, and an unknown element. This unknown element has an empirical formula of CH 2 O and a molecular mass of 180.15. Its molecular formula is the unknown product needed for your equation. Products

4. STEP 1 Find the unknown product Chapter 6

5. Givens: Empirical formula: CH 2 O Molecular mass: 180.15 STEP 1 Find the unknown product Chapter 6

6. Givens: Empirical formula: CH 2 O Molecular mass: 180.15 1. Find the mass of your empirical formula: CH 2 O = 30g STEP 1 Find the unknown product Chapter 6

7. Givens: Empirical formula: CH 2 O Molecular mass: 180.15 1. Find the mass of your empirical formula: CH 2 O = 30g 2. Divide the molecular mass by the empirical mass: 180/30 = 6 STEP 1 Find the unknown product Chapter 6

8. Givens: Empirical formula: CH 2 O Molecular mass: 180.15 1. Find the mass of your empirical formula: CH 2 O = 30g 2. Divide the molecular mass by the empirical mass: 180/30 = 6 3. Multiply that number (6) to your empirical formula (CH 2 O) : C 6 H 12 O 6 STEP 1 Find the unknown product Chapter 6

9. Givens: Empirical formula: CH 2 O Molecular mass: 180.15 1. Find the mass of your empirical formula: CH 2 O = 30g 2. Divide the molecular mass by the empirical mass: 180/30 = 6 3. Multiply that number (6) to your empirical formula (CH 2 O) : C 6 H 12 O 6 Molecular Formula and Unknown Product= C 6 H 12 O 6 STEP 1 Find the unknown product Chapter 6

10. STEP 2 Write and balance the equation Chapter 7

11. __ H 2 0 + __ CO 2 __ C 6 H 12 O 6 + __ 0 2 REACTANTS PRODUCTS STEP 2 Write and balance the equation Chapter 7

12. __ H 2 0 + __ CO 2 __ C 6 H 12 O 6 + __ 0 2 REACTANTS PRODUCTS STEP 2 Write and balance the equation Chapter 7

13. Water + Carbon Dioxide (+ energy) = Glucose + Oxygen

14. STEP 3 Start with one of the knowns (convert mg to g) Chapter 1 6720 mg of H 2 0 (Given)

15. 6720 mg H 2 0 1 gram H 2 0 1000 milligrams H 2 0 1 gram H 2 0 = 1000 mg H 2 0 STEP 3 Start with one of the knowns (convert mg to g) Chapter 1

16. STEP 4 Convert grams to moles Chapter 6

17. 6.72 g H 2 0 1 mole H 2 0 18 grams H 2 0 1 mole H 2 0 = 18 g H 2 0 STEP 4 Convert grams to moles Chapter 6

18. STEP 5 Convert mole to moles Chapter 6

19. 6 H 2 0 + 6 CO 2 C 6 H 12 O 6 + 6 0 2 STEP 5 Convert mole to moles Chapter 6

20. 6 H 2 0 + 6 CO 2 C 6 H 12 O 6 + 6 0 2.373 mole H 2 0 1 mole C 6 H 12 O 6 6 mole H 2 0 6 mole H 2 0 = 1 mole C 6 H 12 0 6 STEP 5 Convert mole to moles Chapter 6

21. STEP 6 Convert moles to grams Chapter 6

22. .062 mole C 6 H 12 0 6 180 grams C 6 H 12 0 6 1 mole C 6 H 12 0 6 1 mole C 6 H 12 0 6 = 180 g C 6 H 12 0 6 STEP 6 Convert moles to grams Chapter 6

23. Dimensional Analysis 6720 mg H 2 O1 g H 2 O1 mol H 2 O1 mol C 6 H 12 O 6 180 g C 6 H 12 O 6 1000 mg H 2 O18 g H 2 O6 mol H 2 O1 mol C 6 H 12 O 6 = 11.16 g C 6 H 12 O 6

24. You have figured out that: 6720 mg of H 2 0 = 11.16 grams C 6 H 12 O 6 Now figure out: 10.67 L CO 2 = ? grams C 6 H 12 O 6

25. STEP 1 Convert L at STP to moles Chapter 6 10.67 L of CO 2 (Given)

26. 10.67 L CO 2 1 mole CO 2 22.4 Liters CO 2 22.4 L CO 2 =1 mole CO 2 STEP 1 Convert L at STP to moles Chapter 6

27. STEP 2 Convert moles to moles Chapter 6 6 H 2 0 + 6 CO 2 C 6 H 12 O 6 + 6 0 2

28. .476 mole CO 2 1 mole C 6 H 12 O 6 6 mole CO 2 6 mole CO 2 = 1 mole C 6 H 12 0 6 STEP 2 Convert moles to moles Chapter 6

29. STEP 3 Convert moles to grams Chapter 6

30. .079 mole C 6 H 12 O 6 180 grams C 6 H 12 O 6 1 mole C 6 H 12 O 6 1 mole C 6 H 12 0 6 = 180 g C 6 H 12 0 6 STEP 3 Convert moles to grams Chapter 6

31. Dimensional Analysis 10.67 L CO 2 1 mol CO 2 1 mol C 6 H 12 O 6 180 g C 6 H 12 O 6 22.4 L CO 2 6 mol CO 2 1 mol CO 2 = 14.22 g C 6 H 12 O 6

32. You have figured out that: 6720 mg of H 2 0 = 11.16 grams C 6 H 12 O 6 10.67 L of CO 2 = 14.22 grams C 6 H 12 O 6 CO 2 is the excess reactant

33. You have figured out that: 6720 mg of H 2 0 = 11.16 grams C 6 H 12 O 6 10.67 L of CO 2 = 14.22 grams C 6 H 12 O 6 CO 2 is the excess reactant How much excess CO 2 ? (In grams)

34. CO 2 is the excess reactant How much excess CO 2 ? (In grams) 14.22 grams – 11.16 grams = You have figured out that: 6720 mg of H 2 0 = 11.16 grams C 6 H 12 O 6 10.67 L of CO 2 = 14.22 grams C 6 H 12 O 6 3.06 grams CO 2 in excess

35. Find the number of molecules of glucose that precipitated. What’s Next?

36. STEP 1 Convert moles to molecules Chapter 6 0.62 mole of C 6 H 12 O 6 (Found)

37. 0.62 mole of C 6 H 12 O 6 (Found) 0.62 moles C 6 H 12 O 6 6.02 x 10 23 C 6 H 12 O 6 1 mole C 6 H 12 O 6 1 mole C 6 H 12 0 6 = 6.02 x 10 23 molecules C 6 H 12 0 6 STEP 1 Convert moles to molecules Chapter 6

38. Find the number of molecules of glucose that precipitated. RESTATING THE QUESTION:

39. Find the number of molecules of glucose that precipitated. 3.73 E 22 molecules C 6 H 12 O 6 RESTATING THE QUESTION: 1.Scientific Notation 2.First Name 3.Last Name 4.Boxed In

40. THEORETICAL & PERCENT YIELD Chapter 8 Find theoretical percent yield of C 6 H 12 O 6 (Actual amount of Glucose obtained was 10.22 as stated before)

41. Theoretical Yield: (Already Found) 11.16 grams Find theoretical percent yield of C 6 H 12 O 6 (Actual amount of Glucose obtained was 10.22 as stated before) THEORETICAL & PERCENT YIELD Chapter 8

42. Theoretical Yield: (Already Found) 11.16 grams Percent Yield: ACTUAL YIELD/THEORETICAL x 100 Find theoretical percent yield of C 6 H 12 O 6 (Actual amount of Glucose obtained was 10.22 as stated before) THEORETICAL & PERCENT YIELD Chapter 8

43. Find theoretical percent yield of C 6 H 12 O 6 (Actual amount of Glucose obtained was 10.22 as stated before) Theoretical Yield: (Already Found) 11.16 grams Percent Yield: ACTUAL YIELD/THEORETICAL x 100 PERCENT YIELD : 10.22/11.16 x 100 = 92% THEORETICAL & PERCENT YIELD Chapter 8

44. Percent Error Chapter 2 %Error = (|Your Result - Accepted Value| / Accepted Value) x 100

45. %Error = (|Your Result - Accepted Value| / Accepted Value) x 100 How much should have been made = 11.16 g Glucose How much was made: 10.22 g Glucose Percent Error Chapter 2

46. %Error = (|Your Result - Accepted Value| / Accepted Value) x 100 How much should have been made = 11.16 g Glucose How much was made: 10.22 g Glucose (|10.22 – 11.16| / 11.16) x 100 = 8% Percent Error Chapter 2