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

5. Givens: Empirical formula: CH 2 O Molecular mass: 180.15

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

7. STEP 1 Find the unknown product 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

8. STEP 1 Find the unknown product 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

9. STEP 1 Find the unknown product 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

10. STEP 1 Write and balance the equation

11. __ H 2 0 + __ CO 2 __ C 6 H 12 O 6 + __ 0 2 REACTANTS PRODUCTS

12. STEP 1 Write and balance the equation __ H 2 0 + __ CO 2 __ C 6 H 12 O 6 + __ 0 2 REACTANTS PRODUCTS

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

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

15. STEP 2 Start with one of the knowns (convert mg to g) 6720 mg of H 2 0 (Given) 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

16. STEP 3 Convert grams to moles

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

18. STEP 4 Convert mole to moles

19. 6 H 2 0 + 6 CO 2 C 6 H 12 O 6 + 6 0 2 STEP 4 Convert mole to moles

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

21. STEP 5 Convert moles to grams

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 STEP 5 Convert moles to grams 1 mole C 6 H 12 0 6 = 180 g C 6 H 12 0 6

23. 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

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

25. STEP 1 Convert L at STP to moles 10.67 L of CO 2 (Given) 10.67 L CO 2 1 mole CO 2 22.4 Liters CO 2 22.4 L CO 2 =1 mole CO 2

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

27. STEP 2 Convert moles to moles 6 H 2 0 + 6 CO 2 C 6 H 12 O 6 + 6 0 2.476 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

28. STEP 3 Convert moles to grams

29. .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

30. 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 mol C 6 H 12 O 6 CO 2 is the excess reactant

31. 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 mol C 6 H 12 O 6 CO 2 is the excess reactant How much excess CO 2 ? (In grams)

32. 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 mol C 6 H 12 O 6 3.06 grams CO 2 in excess

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

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

35. STEP 1 Convert moles to molecules 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

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

37. Find the number of molecules of glucose that precipitated. 3.73 E 22 molecules C 6 H 12 O 6 RESTATING THE QUESTION:

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

39. THEORETICAL & PERCENT YIELD 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)

40. THEORETICAL & PERCENT YIELD 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)

41. THEORETICAL & PERCENT YIELD 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%

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

43. Percent Error %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

44. Percent Error %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%