BICOL UNIVERSITY. College of Science Department of Chemistry. CHEM 1 GENERAL CHEMISTRY LECTURE HANDOUT Ver. 1.1 α

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1 BICOL UNIVERSITY College of Science Department of Chemistry CHEM 1 GENERAL CHEMISTRY LECTURE HANDOUT Ver. 1.1 α

2 1 mole = x of molecules (Avogadro's number) A mole is similar to a term like a dozen. If you have a dozen carrots, you have twelve of them. Similarly, if you have a mole of carrots, you have x carrots. 2 michaelvmontealegre

3 Molar mass (g/mol) Amount (mol) = mass (g) /MM (g/mol) 1 mol = x units 1. How many grams of lithium are in 3.50 moles of lithium? 2. How many moles of lithium are in 18.2 grams of lithium? 3. How many atoms of lithium are in 3.50 moles of lithium? 4. How many atoms of lithium are in 18.2 g of lithium? MOLE 3

4 5. Ammonium carbonate is a white solid that decomposes with warming. Among its many uses, it is a component of baking powder, fire extinguishers, and smelling salts. How many formula units are in 41.6 g of ammonium carbonate? Empirical formula: the lowest whole number ratio of atoms in a compound. Molecular formula: the true number of atoms of each element in the formula of a compound. molecular formula = (empirical formula)n molecular formula = C6H6 = (CH)6 empirical formula = CH Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio). Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio). 4 michaelvmontealegre

5 Calculate the percentage composition of magnesium carbonate, MgCO3. Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid? 1.Treat % as mass, and convert grams to moles 2. Divide each value of moles by the smallest of the values. MOLE 5

6 3. Multiply each number by an integer to obtain all whole numbers. The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 1. Find the formula mass of C3H5O2 3(12.01 g) + 5(1.01) + 2(16.00) = g 2. Divide the molecular mass by the mass given by the emipirical formula = 2 3. Multiply the empirical formula by this number to get the molecular formula. 6 michaelvmontealegre

7 1. A compound is 75.46% carbon, 4.43% hydrogen, and 20.10% oxygen by mass. It has a molecular weight of g/mol. What is the molecular formula for this compound? A hydrate containing copper, sulfur, oxygen, and water lost 9g upon heating. Originally the hydrate had weighed 25g. Analysis of the anhydrous substance revealed that the 6.4g of Cu, 3.2g of S, and 6.4g of O were present. Find the formula of the hydrate. MOLE 7

8 Two or more substances combine to form a new compound. A + X AX Reaction of elements with oxygen and sulfur Reactions of metals with Halogens Synthesis Reactions with Oxides There are others not covered here! A single compound undergoes a reaction that produces two or more simpler substances AX A + X Decomposition of: Binary compounds H2O(l ) 2H2(g) + O2(g) Metal carbonates CaCO3(s) CaO(s) + CO2(g) Metal hydroxides Ca(OH)2(s) CaO(s) + H2O(g) Metal chlorates 2KClO3(s) 2KCl(s) + 3O2(g) Oxyacids H2CO3(aq) CO2(g) + H2O(l ) A + BX AX + B BX + Y BY + X 2NaCl(s) + F2(g) 2NaF(s) + Cl2(g) Replacement of: Metals by another metal Hydrogen in water by a metal Hydrogen in an acid by a metal Halogens by more active halogens The ions of two compounds exchange places in an aqueous solution to form two new compounds. AX + BY AY + BX 8 michaelvmontealegre

9 One of the compounds formed is usually a precipitate, an insoluble gas that bubbles out of solution, or a molecular compound, usually water. A substance combines with oxygen, releasing a large amount of energy in the form of light and heat. Reactive elements combine with oxygen P4(s) + 5O2(g) P4O10(s) (This is also a synthesis reaction) The burning of natural gas, wood, gasoline C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) MOLE 9

10 The number of atoms of each element on the left hand side of the equation must be the same as the number of atoms of each element on the right hand side of the equation Reactants are written on the left hand side of the chemical equation. Products are written on right hand side of the chemical equation. You can't change the formula of reactant or product molecules in the chemical equation. You can only change the numbers of reactant or product molecules in the chemical equation. Al + Fe3O4---> Al2O3+ Fe 1. Count the number of each atom on the reactant and on the product side. 2. Determine a term to balance first. When looking at this problem, it appears that the oxygen will be the most difficult to balance so we'll try to balance the oxygen first. The simplest way to balance the oxygen terms is: Al + 3 Fe3O4---> 4 Al2O3+ Fe Be sure to notice that the subscript times the coefficient will give the number of atoms of that element. On the reactant side, we have a coefficient of three (3) multiplied by a subscript of four (4), giving 12 oxygen atoms. On the product side, we have a coefficient of four (4) multiplied by a subscript of three (3), giving 12 oxygen atoms. Now, the oxygens are balanced. 3. Choose another term to balance. We'll choose iron, Fe. Since there are nine (9) iron atoms in the term in which the oxygen is balanced we add a nine (9) coefficient in front of the Fe. We now have: Al +3 Fe3O4---> 4Al2O3+ 9Fe 4. Balance the last term. In this case, since we had eight (8) aluminum atoms on the product side we need to have eight (8) on the reactant side so we add an eight (8) in front of the Al term on the reactant side. Now, we're done, and the balanced equation is: 8Al + 3Fe3O4 ---> 4Al2O3 + 9 Fe 10 michaelvmontealegre

11 Synthesis reactions 1. CaO + H2O Ca(OH)2 2. P4 + O2 P2O5 3. Ca + O2 CaO 4. Cu + S8 CuS 5. CaO + H2O Ca(OH)2 6. S8 + O2 SO2 7. H2 + N2 NH3 8. H2 + Cl2 HCl 9. Ag + S8 Ag2S 10. Cr + O2 Cr2O3 11. Al + Br2 AlBr3 12. Na + I2 NaI 13. H2 + O2 H2O 14. Al + O2 Al2O3 Decomposition Reactions 15. BaCO3 BaO + CO2 16. MgCO3 MgO + CO2 17. K2CO3 K2O + CO2 18. Zn(OH)2 ZnO + H2O MOLE 11

12 19. Fe(OH)2 FeO + H2O 20. Ni(ClO3)2 NiCl2 + O2 21. NaClO3 NaCl + O2 22. KClO3 KCl + O2 23. H2SO4 H2O + SO3 24. H2CO3 H2O + CO2 25. Al2O3 Al + O2 26. Ag2O Ag + O2 Single Replacement Reactions 27. AgNO3 + Ni Ni(NO3)2 + Ag 28. AlBr3 + Cl2 AlCl3 + Br2 29. NaI + Br2 NaBr + I2 30. Ca + HCl CaCl2 + H2 31. Mg + HNO3 Mg(NO3)2 + H2 32. Zn + H2SO4 ZnSO4 + H2 33. K + H2O KOH + H2 34. Na + H2O NaOH + H2 Double Displacement Reactions 35. AlI3 + HgCl2 AlCl3 + HgI2(ppt) 36. HCl + NaOH NaCl H2O 37. BaCl2 + H2SO4 BaSO4 + HCl 12 michaelvmontealegre

13 38. Al2(SO4)3 + Ca(OH)2 Al(OH)3 + CaSO4 39. AgNO3 + K3PO4 Ag3PO4 + KNO3 40. CuBr2 + AlCl3 CuCl2 + AlBr3 41. Ca(C2H3O2)2 + Na2CO3 CaCO3 + NaC2H3O2 42. NH4Cl + Hg2(C2H3O2)2 NH4 C2H3O2 + Hg2Cl2 43. Ca(NO3)2 + HCl CaCl2 + HNO3 44. FeS + HCl FeCl2 + H2S 45. Cu(OH)2 + HC2H3O2 Cu(C2H3O2)2 + H2O 46. Ca(OH)2 + H3PO4 Ca3(PO4)2 + H2O 47. CaBr2 + KOH Ca(OH)2 + KBr Combustion Reactions 48. CH4 + O2 CO2 + H2O 49. C2H6 + O2 CO2 + H2O 50. C3H8 + O2 CO2 + H2O 51. C4H10 + O2 CO2 + H2O 52. C5H12 + O2 CO2 + H2O 53. C6H14 + O2 CO2 + H2O 54. C2H4 + O2 CO2 + H2O 55. C2H2 + O2 CO2 + H2O 56. C6H6 + O2 CO2 + H2O MOLE 13

14 The study of quantities of materials consumed and produced in chemical reactions. A balanced chemical equation can tell us: The ratio of the number of molecules of each type reacting and produced. The ratio of the moles of each reactant and product. The ratio of moles of each reactant and product in a reaction is known as the mole ratio (or stoichiometric ratio) The mole ratio can be used to calculate the mass of reactants and products. The mole ratio is the stoichiometric ratio of reactants and products and is the ratio of the coefficients for reactants and products found in the balanced chemical equation. For the reaction aa + bb cc + dd mol ratio for A : B : C : D is a : b : c : d 1. in the reaction 2Mg(s) + O2(g) 2MgO(s) the mole ratio of Mg : O2 : MgO is 2 : 1 : 2 That is, the complete reaction requires twice as many moles of magnesium as there are moles of oxygen. 2. in the reaction 2Al(OH)3 + 3H2SO4 Al2(SO4)3 + 6H2O the mole ratio of Al(OH)3 : H2SO4 : Al2(SO4)3 : H2O is 2 : 3 : 1 : 6 For each mole of Al2(SO4)3 produced, twice as many moles of Al(OH)3 are required to react with three times as many moles of H2SO michaelvmontealegre

15 For the balanced chemical reaction: 2Mg(s) + O2(g) 2MgO(s) the mole ratio of Mg : O2 : MgO is 2:1:2 That is, it requires 2 moles of magnesium and 1 mole of oxygen to produce 2 moles of magnesium oxide. If only 1 mole of magnesium was present, it would require 1 2 = ½ mole of oxygen gas to produce 2 2 = 1 mole magnesium oxide. If 10 moles of magnesium were present, it would require 1 2 x 10 = 5 moles of oxygen gas to produce 2 2 x 10 = 10 moles of magnesium oxide. For n moles of magnesium : 1 2 x n = ½n moles of oxygen gas are required 2 2 x n = n moles of magnesium oxide are produced Mg 2 : O2 1 : MgO 2 example mol 0.25 mol 0.50 mol example mol 0.50 mol 1.00 mol example mol 0.75 mol 1.50 mol example mol 1.00 mol 2.00 mol It is possible to calculate the mass of each reactant and product using the mole ratio from the balanced chemical equation and the equation moles = mass molecular mass For the balanced chemical equation: Given a mass of m grams of magnesium: 2Mg(s) + O2(g) 2MgO(s) mass O2 = moles(o2) x molecular mass(o2) Calculate moles Mg = mass(mg) MM(Mg) = m Use the balanced chemical equation to determine the mole ratio O2:Mg 1:2 Use the mole ratio to calculate moles O2 = 1 2 x moles(mg) Calculate moles of O2 = ½ x m mass O2 = moles(o2) x molecular mass(o2) = ½ x m x (2 x 16.00) = ½ x m x MOLE 15

16 mass MgO = moles(mgo) x molecular mass(mgo) Calculate moles Mg = mass(mg) MM(Mg) = m Use the balanced equation to determine the mole ratio MgO:Mg 2:2 = 1:1 Use the mole ratio to calculate moles MgO = 1 x moles(mg) Calculate moles of MgO = 1 x m mass MgO = moles(mgo) x molecular mass(mgo) = 1x m x ( ) = 1 x m x g of magnesium reacts completely with oxygen gas. Calculate the mass of oxygen consumed during the reaction and the mass of magnesium oxide produced. 1. Wite the balanced chemical equation: 2Mg(s) + O2(g) 2MgO(s) 2. Determine the mole ratio from the equation, Mg : O2 : MgO is 2:1:2 3. Use the mole ratios to calculate the mass of each reactant and product as shown below: mass O2 = moles(o2) x molecular mass(o2) Calculate moles Mg = mass(mg) MM(Mg) = = 0.50mol Use the balanced chemical equation to determine the mole ratio O2:Mg 1:2 Use the mole ratio to calculate moles O2 = 1 2 x moles(mg) Calculate moles of O2 = ½ x mol(mg) = ½ x 0.50 = 0.25mol mass O2 = moles(o2) x molecular mass(o2) = 0.25 x (2 x 16.00) = 0.25 x = 8.03g mass MgO = moles(mgo) x molecular mass(mgo) Calculate moles Mg = mass(mg) MM(Mg) = = 0.50mol Use the balanced equation to determine the mole ratio MgO:Mg 2:2 = 1:1 Use the mole ratio to calculate moles MgO = 1 x moles(mg) Calculate moles of MgO = 1 x mol(mg) = 0.50mol mass MgO = moles(mgo) x molecular mass(mgo) = 1x 0.50 x ( ) = 1 x 0.50 x = 20.16g 16 michaelvmontealegre

17 1. Balance the equation. 2. Convert mass or volume to moles, if necessary. 3. Set up mole ratios. 4. Use mole ratios to calculate moles of desired substituent. 5. Convert moles to mass or volume, if necessary grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? Balanced Equation: 1. Disposable lighters burn butane (C4H1O) and produce CO2 and H20. Balance the chemical equation for combustion of butane, and determine how many grams of CO2 are produced by burning 1.00 g of C4H1O. MOLE 17

18 2. Ammonia is used to make fertilizers for lawns and gardens by reacting nitrogen gas with hydrogen gas. a. Write a balanced equation with smallest whole-number coefficients for this reaction. b. How many moles of ammonia are formed when 1.34 mol of nitrogen react? c. How many grams of hydrogen are required to produce 2.75x10 3 g of ammonia? 18 michaelvmontealegre

19 d. How many molecules of ammonia are formed when 2.92 g of hydrogen react? Book References: Chang, Raymond. Chemistry. 9 th edn. Digital Content Manager Manning, Phillip. Essential Chemistry: Chemical Bonds. Infobase Publishing New York NY, USA Masterton, William L. and Hurley, Cecile N. Chemistry Principles and Reactions 6 th edn. Brooks/Cole Cengage Learning. CA USA Silberberg, Martin S. Chemistry: The molecular nature of matter and change. 5 th edn. The McGraw-Hill Companies, Inc. New York, NY USA Internet Resources: MOLE 19

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