1. What do I know about moles? What do I want to know about moles? What have I learned about moles today?

2.  This unit test contains 6 types of problems: 1. Molar mass and % composition—must be able to write a chemical formula 2. Grams to moles (using molar mass) 3. Moles to particles/atoms/molecules (using Avogadro’s #) 4. Grams to moles to particles (using molar mass and Avogadro’s #) 5. Empirical formulas 6. Molecular formulas

3.  The SI (metric) unit used to measure the amount of a substance  1 mole is always equal to: --Molar mass (g/mole) --Avogadro’s number of particles (6.02 x 10 23 ) --22. 4 Liters of a GAS (AKA molar volume) These may be used as conversion factors when working mole problems.

4.  Define molar mass AND Avogadro’s number.

5. 1 mole of this substance # of atoms/ molecules/ particles Mass of 1 mole (molar mass) g/mole Volume of 1 mole table sugar (sucrose)-- C 12 H 22 O 11 342.0 Sodium chloride-- NaCl 58.5 Cupric sulfate-- CuSO 4 159.6 Sulfur--S 32.1 Iron--Fe55.8 Water—H 2 O18.0

6.  Ex: Titanium 47.867 = 47.9 g/mole (this sample contains Avogadro’s number of atoms)  Ex: oxygen 15.999 = 16.0 g/mole (this sample contains Avogadro’s number of atoms)

7.  Multiply the # of atoms for each element by the atomic mass from periodic table Ex:Magnesium hydroxide Mg (OH) 2 Mg 1(24.3) = 24.3 O 2(16.0) = 32.0 H 2(1.0) = 2.0 58.3 g/mole (this mass also contains Avogadro’s number of molecules)

8.  Find the molar mass of aluminum sulfate Al 2 (SO 4 ) 3 Al 2(27.0) = 54.0 S 3 (32.1)= 96.3 O 12(16.0) = 192.0 342.3 g/mole (this mass also contains Avogadro’s number of molecules)

9.  Calculate the molar mass of diarsenic trioxide.

10.  Stannic carbonate  Diarsenic pentasulfide  Hydrofluoric acid

11.  Sucrose

12.  Shows the % of each element that makes up a compound  Must be calculate molar mass first. Ex: magnesium hydroxide Mg (OH) 2 Mg 1 x 24.3 =24.3 24.3/58.3 x 100 = 41.7% O 2 x 16= 32.0 32.0/58.3 x 100 = 54.9% H 2x1.0 = 2.0 58. 3 g/mole 2.0/58.3 x 100 = 3.4%

14.  Calculate the % composition of sulfurous acid

15.  SAVE YOUR WRAPPER FOR ENTIRE LAB!!  DO NOT START CHEWING UNTIL YOU SIT DOWN BACK AT YOUR DESK.  CHECK BALANCE TO MAKE SURE IT’S OK BEFORE YOU START!!

16.  READ PROBLEM:  MAKE A HYOTHESIS:  PROCEDURE 1—4  DATA TABLE 1—5

17.  After chewing: (KEEP YOUR SAME BALANCE)  Procedure 5—8  Data table 6—8  Conclusion  Questions 1—2

18.  Calculate the % composition for a sugar substitute called SUCRALOSE

19.  C 12( 12.0) = 144.0  H 19 (1.0) = 19.0  Cl 3 (35.5) = 106.5  O 8 (16.0) = 128.0 397. 5 g/mole % C= 144.0 / 397.5 x 100 = 36.2% % H= 19.0 / 397.5 x 100 = 4.8% % Cl= 106.5 / 397.5 x 100 = 26.8% % O= 128.0/ 397.5 x 100 = 32.2%

20.  1. Calculate the % composition of carbonic acid.  2. Calculate the % composition of diantimony trioxide.

21.  1. H 2 CO 3 H-2 (1.0) = 2.0 3.2% C- 1 (12.0) = 12.019.4% O – 3(16.0) = 48.077.4% 62.0 g/mole 2. Sb 2 O 3 Sb- 2(121.8) =243.683.5% O – 3(16.0) = 48.016.5% 291.6 g/mole

22.  Calculate the molar mass AND % composition of: 1. C 12 H 22 O 11 2. Cupric sulfate

23.  C: 12 (12.0) = 144.0  H: 22 (1.0) = 22.0  O: 11(16.0) = 176.0 CuSO 4 Cu: 1(63.5)= 63.5 S: 1 (32.1) = 32.1 O: 4(16.0) = 64.0

24. Will need to use unit conversion(cancellation) and molar mass will be used for the conversion factor. Ex: 2.50 grams of hydrochloric acid = ____moles H Cl 2.50 grams x 1 mole = 0.0685 moles(3sigfigs) 36.5 grams

25. Ex: 2.50 moles of HCl = __________grams 2.5 moles x 36.5 grams = 91 grams 1 mole (2 sig figs)

26.  Particles, atoms, molecules (synonyms)  Will have to use Avogadro’s number as a conversion factor  Ex: 5.25 x 10 25 atoms of Mg = _____moles 5. 25 x 10 25 atoms x 1 mole = 87.2 moles 6.02 x 10 23 (3 sig figs)

27. 2.50 moles MgO = _________molecules 2.50 moles x 6.02 x 10 23 molecules 1 mole = 1.50 x 10 24 molecules (3 sig figs)

28.  Will need to use both molar mass AND Avogadro’s number as conversion factors  Will be 2 steps instead of 1 step unit cancellation  Ex: 4.5 grams nitrous acid = __________molecules  HNO 2 4.5 g x 1 mole x 6.02 x 10 23 molecules = 47 g1 mole 5.8 x 10 22 molecules (2 sig figs)

29.  Ex: 9.35 x 10 21 particles of carbon tetrabromide = _____grams C Br 4 9.35 x 10 21 p x 1 mole x 154 grams = 6.02x10 23 p 1 mole 2.39 grams (3 sig figs)

30. 1. 0.14 mole (gram to moles) 2. 150 g (moles to grams) 3. 1.1 x 10 23 molecules (g to molecules) 4 5.30 x 10 25 molecules (moles to molecules) 5. 0.074 mole( gram to moles) 6. 0.619 g (particles to grams) 7. 0.49 mole (grams to moles) 8. 0.00083 mole (particles to moles)

31. 1. 22 g 2. 1.53 x 10 24 molecules 3. 0.014 g 4. 7.2 x 10 21 molecules 5. 7.2 x 10 23 molecules 6. 2.08 x 10 6 g 7. 56 g 8. 2.5 g 9. 31 g 10. 0.0029 mole 11. 167 g BONUS: 3.37 x 10 26 atoms

32. If grams are converted to moles, use _______________ to convert. If moles are converted to molecules, then use ______________to convert. ****Have calculator, periodic table, and best friend chart****

33. All members of your group must show their work on separate sheet of paper. When you calculate the answer, flip the card over to find a word. All of your words will make a sentence. First group to show all work and finish first, wins bonus!

34. Grams-----moles-----particles(atoms or molecules)

35. 1. 55.33 grams of sodium oxide = ____moles 2. 5.00 x 10 22 particles of sodium= _______moles 3. 2.49 x 10 26 atoms of acetic acid = _______________grams

36. Video sheet Problems #1—3, 5, 6

37. Fill in the blanks with multiply/divide OR molar mass/Avogadro’s number: ***When going from moles to grams, _______________ by _____________. ***When going from moles to particles, ____________by _________________. I AM CHECKING 5 HOMEWORK PROBLEMS!!!!

38. Briefly describe the steps for calculating an empirical formula AND molecular formula.

39. Define empirical formula ***both labs due today***

40. 1. Convert 5.03 x 10 24 molecules of phosphoric acid to grams. 2. Convert 35.75 grams of dinitrogen monoxide to moles. 3. Convert 5.0 moles of water to molecules.

41. 1. 819 grams of H 3 PO 4 2. 0.8125 moles of N 2 0 3. 3.0 x 10 24 molecules of water

42. Tell how to solve for each: 1. G to moles 2. Moles to G 3. Particles to moles 4. Moles to particles 5. G to particles 6. Particles to G

43.  HYPOTHESIS, DATA 1—8, CONCLUSION, QUESTIONS 1—4  3. MASS OF SUGAR (in grams—data #8)--------MOLES (SUGAR = C 12 H 22 O 11 )  4. MOLES----------PARTICLES

44.  Data Table: mass of empty vial AND substances mass (make sure you’ve subtracted empty vial each time!!)  SHOW WORK TO GET CREDIT  CONVERT GRAMS----------MOLES  CONVERT MOLES---------PARTICLES  ANSWER QUESTIONS 1---5, 6 (BONUS)

45. Convert 25.0 moles of water to grams.

46. A molecular formula is a whole number____________of the empirical formula.

47.  Shows the SIMPLEST, WHOLE NUMBER ratio of elements in a compound  Will give you % composition of compound and ask you to find the formulas

48. 1. Change % sign to grams (some problems may already give you grams instead of %) 2. Convert grams to moles (using molar mass) **round to 4 decimals*** 3. Simplify the mole ratio by dividing each one by the smallest 4. Round to the nearest whole number and assign these numbers to the appropriate element

49. A compound is 78.1% Boron and 21.9% H. Calculate the empirical formula. 78.1 grams B x 1 mole = 7.2315 moles B 10.8 g 21.9 grams H x 1mole =21.9 moles H 1.0 g 7.2315 : 21.9 7.2315 1: 3 = BH 3

50. Is a WHOLE NUMBER MULTIPLE of the empirical formula You must then first know the empirical formula

51. 1. You must first have empirical formula (if not, you will have to calculate it first!!) 2. Find the molar mass of the empirical formula. 3. Take the molar mass of the molecular formula that is given in the problem divided by the empirical formula’s molar mass. Round this to a whole number. 4. Distribute this number to the numbers within the empirical formula to get the new molecular formula.

52. Given the empirical formula of BH 3 and the molecular formula’s molar mass of 27.67 g/mole, find the molecular formula. Molar mass of BH 3 is 13.8 g/mole 27.67 divided by 13.8 = 2 So molecular formula is B 2 H 6

53. A compound is 4.04 grams of nitrogen and 11.46 grams of oxygen. The molecular molar mass is 108.0 g/mole. Find the molecular formula. 4.04 g N x 1 mole= 0.2886 mole N 14.0 g 11.46 g O x 1 mole =0.7163 mole O 16.0 g 0.2886 : 0.71630.2886 1: 2 = NO 2 empirical molar mass = 46.0 g/mole 108.0 divided by 46.0 = 2 Molecular formula = N 2 O 4