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MOLES

MOLES. A collection of objects. Mole:. A collection of Avogadro’s number of objects. Avogadro’s number. 6.022 * 10 23. Atoms Particles Ions Cations Anions Molecules Formula Units. Mole Ratio.

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MOLES

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  1. MOLES A collection of objects Mole: A collection of Avogadro’s number of objects Avogadro’s number 6.022 * 1023 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 H2(g) 2 moles of Al reacts w/ 6 moles HCl to produce 2 moles AlCl3 and 3 moles H2 gas This states:

  3. 1 mole Argon atoms 1 mole Helium atoms MOLE CALCULATIONS One mole of argon atoms …. ….. is a number One mole of helium atoms …. ….. is a number 6.022 * 1023 Argon atoms 6.022 * 1023 Helium atoms ….. has a mass ….. has a mass 39.9 grams 4.00 grams

  4. RECAP: 1 mol of “anything” contains 6.022*1023 “parts” Elements on p.table = I mol Molar 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 = 1 mole Divide by Multiply by MASS

  6. Given mass, find moles How many moles are in 345.6 g NaNO3 ? Step 1: find formula wt. of NaNO3 1 Na = 23.0 g 1 N = 14.0 g 3 O = 3 * 16.0 = 48.0 g 1 mole of NaNO3 = 85.0 g Step 2 : Find # of moles Use factor-label method 4.07 moles NaNO3

  7. Given moles, find mass How many grams are in 0.6 moles N2O ? Step 1: find molecular wt. of N2O 2 N = 2 * 14.0 = 28.0 g 1 O = 16.0 g 1 mole of N2O = 44.0 g Step 2 : Find mass Use factor-label method 26.4 g N2O

  8. Multiply by Divide by MOLES 6.02*1023 Avogadro’s Number NUMBER of PARTICLES

  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 # molecules Use factor-label method 9.63 * 1023 molecules O2

  10. Given # atoms, find moles How many moles are in 3.01 *1012 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. Multiply by Divide by MOLES 6.02*1023 mass = 1 mole Avogadro’s Number Divide by Multiply by MASS NUMBER of PARTICLES There is not a direct relationship between MASS and PARTICLES

  12. 2 Step Conversion Problem Will need to use MASS at some point in the problem Given # formula units, find grams How many grams are in 1.208 * 1024 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 Step 3 : Find mass Use factor-label method 288 g AgCl Converts moles to MASS Converts formula units to MOLES

  13. 2 Step Conversion Problem Will need to use MASS at some point in the problem Given grams, find # atoms 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 Step 3 : Find atoms Use factor-label method 3.84*1023 atoms Hg Converts mass to MOLES Converts moles to ATOMS

  14. + PROBLEM What is the formula weight of sodium carbonate, Na2CO3. This is an industrial chemical used in making glass. Also, equivalent to 1mole of a substance SOLUTION 2 * 23.0 = 46.0 2 Na 1 C 3 O 1 * 12.0 = 12.0 3 * 16.0 = 48.0 106.0 g

  15. BALANCE EQUATIONS “STOICHIOMETRY” Sodium Phosphate (aq) + Barium Nitrate (aq) ------ Barium Phosphate (s) + Sodium Nitrate (aq) 2 Na3PO4(aq) + Ba(NO3)2(aq) ----- Ba3(PO4)2(s) + NaNO3 (aq) 3 3 6 Aluminum (s) + Hydrochloric Acid (aq) ----------> Aluminum Chloride (aq) + Hydrogen (g) 3 Al + HCl ---- AlCl3 + H2 2 6 2 3

  16. MASS - MASS CALCULATIONS & LIMITING REAGENT (amts of 2 reactants given: limiting reactants) Solid calcium metal burns in air (oxygen) to form a calcium oxide cmpd. Using 4.20 g Ca & 2.80 g O2 how much pdt is made? Find 1) grams of pdt made from g of Ca 2) grams of pdt made from g of O2 3) limiting reagent 4) how much pdt can be made 5) % yield if 3.85 g produced PLAN: Need ------> balanced chem. rxn STEPS: g Ca -----> mols Ca -----> mols pdt ------> g pdt M g Ca coeff X pdt/Y react M g pdt (same steps converting oxygen) 2 2 Ca(s) + O2(g) -------> CaO(s) M g Ca M g O2 coeff X pdt/Y react M g pdt 40.1 g Ca:CaO 2:2 56.1 g 32.0 g O2:CaO 1:2

  17. 4.20 g 2.80 g X g 5.89 g 9.82 g 2 Ca(s) + O2(g) -------> 2 CaO(s) M g : 40.1g 56.1 g Coeff : 2 1 2 mols : 0.105 0.175 0.105 0.0875 3) limiting reagent: Which produced the least amt-- Ca or O2? Ca ---> CaO = 5.89 g O2 ---> CaO = 9.82 g Ca, limiting reagent 5.89 g CaO 4) how much pdt can be made: 5) % yield=

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

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