1. MOLES Mole: A collection of objects A collection of Avogadro’s number of objects Avogadro’s number6.022 * 10 23 Atoms Particles Ions Cations Anions Molecules Formula Units

2. Mole Ratio The coefficient in front of each component in a balanced equation 2 Al (s) + 6 HCl (aq) -----  2 AlCl 3 (aq) + 3 H 2 (g) This states: 2 moles of Al reacts w/ 6 moles HCl to produce 2 moles AlCl 3 and 3 moles H 2 gas

3. MOLE CALCULATIONS 1 mole Helium atoms 1 mole Argon atoms One mole of helium atoms …. ….. is a number 6.022 * 10 23 Helium atoms One mole of argon atoms …. ….. is a number 6.022 * 10 23 Argon atoms ….. has a mass 4.00 grams 39.9 grams

4. RECAP: 1 mol of “anything” contains 6.022*10 23 “parts” Elements on p.table = I mol M olar mass (mass/1 mol) 1. Elements - decimal # on p.table Ti 47.87 amu or g 2. Molecules/Cmpd - sum of mass of all elements

5. MOLES MASS Multiply by Divide by mass = 1 mole

6. Given mass, find moles How many moles are in 345.6 g NaNO 3 ? Step 1: find formula wt. of NaNO 3 1 Na = 23.0 g 1 N = 14.0 g 3 O = 3 * 16.0 = 48.0 g 1 mole of NaNO 3 = 85.0 g Step 2 : Find # of moles Use factor-label method 4.07 moles NaNO 3

7. Given moles, find mass How many grams are in 0.6 moles N 2 O ? Step 1: find molecular wt. of N 2 O 2 N = 2 * 14.0 = 28.0 g 1 O = 16.0 g 1 mole of N 2 O = 44.0 g Step 2 : Find mass Use factor-label method 26.4 g N 2 O

8. MOLES NUMBER of PARTICLES 6.02*10 23 Multiply by Divide by Avogadro’s Number

9. Given moles, find # molecules How many molecules are in 1.6 moles oxygen? Step 1: Recognize that mass does not apply here. But Avogadro’s # is used Step 2 : Find # moleculesUse factor-label method 9.63 * 10 23 molecules O 2

10. Given # atoms, find moles How many moles are in 3.01 *10 12 atoms of Chromium? Step 1: Recognize that mass does not apply here. But Avogadro’s # is used Step 2 : Find # moles Use factor-label method 5.00 * 10 -12 mols Cr

11. MOLES MASS NUMBER of PARTICLES Multiply by Divide by mass = 1 mole 6.02*10 23 Multiply by Divide by There is not a direct relationship between MASS and PARTICLES Avogadro’s Number

12. Given # formula units, find grams 2 Step Conversion Problem Will need to use MASS at some point in the problem How many grams are in 1.208 * 10 24 formula units AgCl ? Step 1: find formula wt. of AgCl 1 Ag = 107.9 g 1 Cl = 35.5 g 1 mole of AgCl = 143.4 g Step 2 : Find moles Use factor-label method 288 g AgCl Converts formula units to MOLES Converts moles to MASS Step 3 : Find mass

13. Given grams, find # atoms Will need to use MASS at some point in the problem 2 Step Conversion Problem How many atoms are in 128.0 grams of Mercury ? Step 1: find molar mass of Hg 1 mole of Hg = 200.6 g Step 2 : Find moles Use factor-label method 3.84*10 23 atoms Hg Converts mass to MOLES Converts moles to ATOMS Step 3 : Find atoms

14. PROBLEMWhat is the formula weight of sodium carbonate, Na 2 CO 3. This is an industrial chemical used in making glass. SOLUTION 2 Na 1 C 3 O 2 * 23.0 = 46.0 1 * 12.0 = 12.0 3 * 16.0 = 48.0 + 106.0 g Also, equivalent to 1mole of a substance

15. Sodium Phosphate (aq) + Barium Nitrate (aq) ------  Barium Phosphate (s) + Sodium Nitrate (aq) Na 3 PO 4 (aq) + Ba(NO 3 ) 2 (aq) -----  Ba 3 (PO 4 ) 2 (s) + NaNO 3 (aq) 3632 BALANCE EQUATIONS “STOICHIOMETRY” Al + HCl ----  AlCl 3 + H 2 3 3622 Aluminum (s) + Hydrochloric Acid (aq) ----------> Aluminum Chloride (aq) + Hydrogen (g)

16. MASS - MASS CALCULATIONS & LIMITING REAGENT Solid calcium metal burns in air (oxygen) to form a calcium oxide cmpd. Using 4.20 g Ca & 2.80 g O 2 how much pdt is made? Find 1) grams of pdt made from g of Ca 2) grams of pdt made from g of O 2 3) limiting reagent 4) how much pdt can be made 5) % yield if 3.85 g produced PLAN: Need ------> balanced chem. rxn g Ca -----> mols Ca -----> mols pdt ------> g pdt M g Ca coeff X pdt/Y react M g pdt STEPS: Ca (s) + O 2 (g) -------> CaO (s) 2 M g Ca M g O 2 coeff X pdt/Y react M g pdt 40.1 g Ca:CaO 2:2 56.1 g 32.0 g O 2 :CaO 1:2 (same steps converting oxygen) (amts of 2 reactants given: limiting reactants)

17. 2 Ca (s) + O 2 (g) -------> 2 CaO (s) 4.20 g 2.80 g X g M g : 40.1g 56.1 g Coeff : 2 1 2 mols : 0.105 0.0875 0.175 5.89 g9.82 g 3) limiting reagent:Which produced the least amt-- Ca or O 2 ? Ca ---> CaO = 5.89 g O 2 ---> CaO = 9.82 g Ca, limiting reagent 4) how much pdt can be made: 5.89 g CaO 5) % yield=

18. RECAP: -- Balanced chem. equation -- then, GIVEN USE FIND grams M olar Mass mols mols coeff mols mols M olar Mass grams limiting reagent : the reactant that produces the least amt of moles or mass of a specific pdt