1. CMH 121 Luca Preziati Chapter 4: Chemical Reactions + + 4 Wood Boards6 Nails 1 Fence Panel + 4 Dozen Wood Boards 6 Dozen Nails 1 Dozen Fence Panels For a 12 panels fence…

2. CMH 121 Luca Preziati Chapter 4: Chemical Reactions + + 4C3H 2 C4H6C4H6 + 4 moles of C 3 moles of H 2 1 mole of C 4 H 6 For a mole of 1,3-Butadiene …

3. CMH 121 Luca Preziati Chapter 4: Chemical Reactions 1 Dozen = 12 items The mole is defined (since 1960) as the amount of substance of a system that contains as many entities as there are atoms in 12 g of carbon-12. Symbol: mol. Coined by Wilhelm Ostwald in 1893 1 mol = 12 g of carbon-12 1 mole = 6.0221415×10 23 items # of Molecules = # of moles X 6.022×10 23 A mole of carbon contains 6.0221415×10 23 atoms of carbon, but the same is true for any other element or molecule; in general: 6.0221415×10 23 is the Avogadro’s Number

4. CMH 121 Luca Preziati Chapter 4: Chemical Reactions m = number of nails X mass of 1 nail Example: 500g = 100 nails X 5g m = number of moles X mass of 1 mole of the substance m = n X M.M. Where: m = mass (g) n = number of moles (mol) M.M. = Molecular Mass (g/mol) = mass of 1 mole of the substance Sometime the Molecular Mass is measured in Daltons (1Da = 1g/mol)

5. CMH 121 Luca Preziati Chapter 4: Chemical Reactions The molar mass of an element is numerically the same as the atomic weight of the element. They are not however the same; they have different units: The atomic weight is defined as one twelfth of the mass of an isolated atom of carbon-12 and is therefore dimensionless The molar mass is measured in g/mol. The molar mass of a compound is given by the sum of the atomic weights of the atoms which form the compound. Example: molar mass of Ca(NO 3 ) 2

6. Write a Solution Map for converting the units : Information Given:1.1 x 10 22 Ag atoms Find:? moles Conv. Fact.: 1 mole = 6.022 x 10 23 Example: A silver ring contains 1.1 x 10 22 silver atoms. How many moles of silver are in the ring? atoms Ag moles Ag

7. Check the Solution: 1.1 x 10 22 Ag atoms = 1.8 x 10 -2 moles Ag The units of the answer, moles, are correct. The magnitude of the answer makes sense since 1.1 x 10 22 is less than 1 mole. Information Given:1.1 x 10 22 Ag atoms Find:? moles Conv. Fact.: 1 mole = 6.022 x 10 23 Sol’n Map: atoms  mole Example: A silver ring contains 1.1 x 10 22 silver atoms. How many moles of silver are in the ring?

8. Write a Solution Map for converting the units : Information Given:1.75 mol H 2 O Find:? g H 2 O C F: 1 mole H 2 O = 18.02 g H 2 O mol H 2 O g H 2 O Example: Calculate the mass (in grams) of 1.75 mol of water

9. Check the Solution: 1.75 mol H 2 O = 31.5 g H 2 O The units of the answer, g, are correct. The magnitude of the answer makes sense since 31.5 g is more than 1 mole. Information Given:1.75 mol H 2 O Find:? g H 2 O C F: 1 mole H 2 O = 18.02 g H 2 O Sol’n Map:mol  g Example: Calculate the mass (in grams) of 1.75 mol of water

10. Chemical Equations CH 4 and O 2 are the reactants, and CO 2 and H 2 O are the products the (g) after the formulas tells us the state of the chemical the number in front of each substance tells us the numbers of those molecules in the reaction called the coefficients CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g) ReactantsProducts State g=gas l=liquid s=solid Coefficients

11. Combustion of Methane Balanced to show the reaction obeys the Law of Conservation of Mass it must be balanced CH 4 (g) + 2 O 2 (g) ® CO 2 (g) + 2 H 2 O(g) H H C H H + O O C + OO OO + O HH O HH + 1 C + 4 H + 4 O

12. Tro's Introductory Chemistry, Chapter 712 Examples when magnesium metal burns in air it produces a white, powdery compound magnesium oxide Mg(s) + O 2 (g)  MgO(s) 1)count the number of atoms of on each side – count polyatomic groups as one “element” if on both sides Mg(s) + O 2 (g)  MgO(s) 1  Mg  1 2  O  1

13. Tro's Introductory Chemistry, Chapter 713 Examples Mg(s) + O 2 (g)  MgO(s) 2)pick an element to balance – avoid element in multiple compounds – do free elements last – since Mg already balanced, pick O 3)find least common multiple of both sides & multiply each side by factor so it equals LCM Mg(s) + O 2 (g)  MgO(s) 1  Mg  1 1 x 2  O  1 x 2

14. Tro's Introductory Chemistry, Chapter 714 Examples 4)use factors as coefficients in front of compound containing the element Mg(s) + O 2 (g)  2 MgO(s) 1  Mg  1 1 x 2  O  1 x 2

15. Tro's Introductory Chemistry, Chapter 715 Examples Mg(s) + O 2 (g)  MgO(s) 5)Recount – Mg not balanced now – That’s OK!! Mg(s) + O 2 (g)  2 MgO(s) 1  Mg  2 2  O  2 6)Repeat – attacking unbalanced element 2 Mg(s) + O 2 (g)  2 MgO(s) 2 x 1  Mg  2 2  O  2

16. CMH 121 Luca Preziati Chapter 4: Chemical Reactions The study of the numerical relationship between chemical quantities in a chemical reaction is called reaction stoichiometry stoichiometry Mole-to-Mole Conversions the balanced equation is the “recipe” for a chemical reaction the equation 3 H 2 (g) + N 2 (g)  2 NH 3 (g) tells us that 3 molecules of H 2 react with exactly 1 molecule of N 2 and make exactly 2 molecules of NH 3 or 3 molecules H 2  1 molecule N 2  2 molecules NH 3 (in this reaction)

17. Example: Sodium chloride, NaCl, forms by the following reaction between sodium and chlorine. How many moles of NaCl result from the complete reaction of 3.4 mol of Cl 2 ? Assume there is more than enough Na. 2 Na(s) + Cl 2 (g)  2 NaCl(s)

18. Check the Solution: 3.4 mol Cl 2  6.8 mol NaCl The units of the answer, moles NaCl, are correct. The magnitude of the answer makes sense since the equation tells us you make twice as many moles of NaCl as the moles of Cl 2. Information Given:3.4 mol Cl 2 Find:? moles NaCl CF: 1 mol Cl 2  2 mol NaCl SM: mol Cl 2  mol NaCl Example: How many moles of NaCl result from the complete reaction of 3.4 mol of Cl 2 in the reaction below? 2 Na(s) + Cl 2 (g)  2 NaCl(s)

19. 19 Mass-to-Mass Conversions we know there is a relationship between the mass and number of moles of a chemical 1 mole = Molar Mass in grams the molar mass of the chemicals in the reaction and the balanced chemical equation allow us to convert from the amount of any chemical in the reaction to the amount of any other

20. Example: In photosynthesis, plants convert carbon dioxide and water into glucose, (C 6 H 12 O 6 ), according to the following reaction. How many grams of glucose can be synthesized from 58.5 g of CO 2 ? Assume there is more than enough water to react with all the CO 2.

21. Write a Solution Map: g CO 2 Information Given:58.5 g CO 2 Find: g C 6 H 12 O 6 CF: 1 mol C 6 H 12 O 6 = 180.2 g 1 mol CO 2 = 44.01 g 1 mol C 6 H 12 O 6  6 mol CO 2 mol CO 2 mol C 6 H 12 O 6 g C 6 H 12 O 6 Example: How many grams of glucose can be synthesized from 58.5 g of CO 2 in the reaction? 6 CO 2 (g) + 6 H 2 O(l)  6 O 2 (g) + C 6 H 12 O 6 (aq)

22. Check the Solution: 58.5 g CO 2 = 39.9 g C 6 H 12 O 6 The units of the answer, g C 6 H 12 O 6, are correct. It is hard to judge the magnitude. Information Given:58.5 g CO 2 Find: g C 6 H 12 O 6 CF: 1 mol C 6 H 12 O 6 = 180.2 g 1 mol CO 2 = 44.01 g 1 mol C 6 H 12 O 6  6 mol CO 2 SM: g CO 2  mol CO 2  mol C 6 H 12 O 6  g C 6 H 12 O 6 Example: How many grams of glucose can be synthesized from 58.5 g of CO 2 in the reaction? 6 CO 2 (g) + 6 H 2 O(l)  6 O 2 (g) + C 6 H 12 O 6 (aq)

23. Limiting reactant and Yield Limiting Reagent (or reactant): The reactant that is completely consumed in a chemical reaction. Theoretical yield: The amount of product that can be made in a chemical reaction based on the amount of limiting reactant. Actual yield: The amount of product actually produced by a chemical reaction. Because of both controllable and uncontrollable factors, the actual yield of product will always be less than the theoretical yield.

24. Tro's Introductory Chemistry, Chapter 824 Example: When 11.5 g of C are allowed to react with 114.5 g of Cu 2 O in the reaction below, 87.4 g of Cu are obtained. Find the Limiting Reactant, Theoretical Yield and Percent Yield.

25. Tro's Introductory Chemistry, Chapter 825 Check the Solutions: Limiting Reactant = Cu 2 O Theoretical Yield = 101.7 g Percent Yield = 85.9% The Percent Yield makes sense as it is less than 100%. Information Given:11.5 g C, 114.5 g Cu 2 O 87.4 g Cu produced Find: Lim. Rct., Theor. Yld., % Yld. CF: 1 mol C = 12.01 g; 1 mol Cu = 63.54 g; 1 mol Cu 2 O = 143.08 g; 1 mol Cu 2 O  2 mol Cu; 1 mol C  2 mol Cu Example: When 11.5 g of C reacts with 114.5 g of Cu 2 O, 87.4 g of Cu are obtained. Find the Limiting Reactant, Theoretical Yield and Percent Yield. Cu 2 O(s) + C(s)  2 Cu(s) + CO(g)

26. Tro's Introductory Chemistry, Chapter 726 Dissociation when ionic compounds dissolve in water, the anions and cations are separated from each other - this is called dissociationdissociation – however not all ionic compounds are soluble in water! when compounds containing polyatomic ions dissociate, the polyatomic group stays together as one ion polyatomic ions dissociate

27. Tro's Introductory Chemistry, Chapter 727 Precipitation Reactions Pb(NO 3 ) 2 (aq) + 2 KI(aq)  2 KNO 3 (aq) + PbI 2 (s)

28. Tro's Introductory Chemistry, Chapter 728 Ionic Equations equations which describe the chemicals put into the water and the product molecules are called molecular equations 2 KOH(aq) + Mg(NO 3 ) 2 (aq)  2 KNO 3 (aq) + Mg(OH) 2 (s) equations which describe the actual dissolved species are called ionic equations – aqueous electrolytes are written as ions soluble salts, strong acids, strong bases – insoluble substances and nonelectrolytes written in molecule form solids, liquids and gases are not dissolved, therefore molecule form 2K +1 (aq) + 2OH -1 (aq) + Mg +2 (aq) + 2NO 3 -1 (aq)  K +1 (aq) + 2NO 3 -1 (aq) + Mg(OH) 2(s)

29. Tro's Introductory Chemistry, Chapter 729 Oxidation-Reduction Reactions We say that the element that loses electrons in the reaction is oxidized and the substance that gains electrons in the reaction is reduced you cannot have one without the other

30. Tro's Introductory Chemistry, Chapter 730 Reactions of Metals with Nonmetals Reactions of Metals with Nonmetals (Oxidation-Reduction) metals react with nonmetals to form ionic compounds – ionic compounds are solids at room temperature the metal loses electrons and becomes a cation – the metal undergoes oxidation the nonmetal gains electrons and becomes an anion – the nonmetal undergoes reduction In the reaction, electrons are transferred from the metal to the nonmetal 2 Na(s) + Cl 2 (g)  NaCl(s)

31. Tro's Introductory Chemistry, Chapter 731 Oxidation-Reduction Reactions any reaction that has an element that is uncombined on one side and combined on the other is a redox reaction – uncombined = free element – 2 CO + O 2  2 CO 2 – 2 N 2 O 5  4 NO 2 + O 2 – 3 C + Fe 2 O 3  3 CO + 2 Fe – Mg + Cl 2  MgCl 2 any reaction where a cation changes charge is redox – CuCl + FeCl 3  FeCl 2 + CuCl 2 – SnCl 2 + F 2  SnCl 2 F 2

32. Tro's Introductory Chemistry, Chapter 732 Combustion Reactions Reactions in which O 2 (g) is a reactant are called Combustion Reactions Combustion reactions release lots of energy Combustion reactions are a subclass of Oxidation- Reduction reactions 2 C 8 H 18 (g) + 25 O 2 (g)  16 CO 2 (g) + 18 H 2 O(g)